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[ "# Tag Info\n\nThe question in its current form has a simple answer: This is an instance of the Assignment Problem, or equivalently of finding a minimum-weight matching in a bipartite graph. Decide on a cost function $f(x, y)$ to measure the dissimilarity between two given RGB tuples $x$ and $y$ -- that is, this function should assign identical RGB tuples a cost of zero, ..." ]

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[ "PICCA Frederic-Emmanuel frederic-emmanuel.picca at synchrotron-soleil.fr\nThu Feb 18 14:57:50 UTC 2016\n\n```Hello\n\nI try to mix my C library and the Pipe module\n\nhere the code I am using\n\nsolveTraj :: Factory -> Geometry -> Detector -> Sample -> Pipe Engine Geometry IO ()\nsolveTraj f g d s = do\ne <- await\nlet name = engineName e\nwithSample s \\$ \\sample ->\nwithDetector d \\$ \\detector ->\nwithGeometry f g \\$ \\geometry ->\nwithEngineList f \\$ \\engines ->\nwithCString name \\$ \\cname -> do\nc_hkl_engine_list_init engines geometry detector sample\nengine <- c_hkl_engine_list_engine_get_by_name engines cname nullPtr\nn <- c_hkl_engine_pseudo_axis_names_get engine >>= darrayStringLen\nyield \\$ solve' engine n e >>= getSolution0\n\nwhere\n\ngetSolution0 :: ForeignPtr HklGeometryList -> IO Geometry\n\nAnd I am using this like this\n\nrunEffect \\$ for (each engines) >-> solveTraj factory geometry detector sample >-> P.print\n\nwhere\n\n[Engine]\nengines\n\nBut When I compile the code I get this error.\n\nsrc/Hkl/C.hsc:83:3:\nCouldn't match type `IO' with `Proxy () Engine () Geometry IO'\nExpected type: Proxy () Engine () Geometry IO ()\nActual type: IO ()\nIn a stmt of a 'do' block:\nwithSample s\n\\$ \\ sample\n-> withDetector d\n\\$ \\ detector -> withGeometry f g \\$ \\ geometry -> ...\nIn the expression:\ndo { e <- await;\nlet name = engineName e;\nwithSample s \\$ \\ sample -> withDetector d \\$ \\ detector -> ... }\nIn an equation for `solveTraj':\nsolveTraj f g d s\n= do { e <- await;\nlet name = ...;\nwithSample s \\$ \\ sample -> withDetector d \\$ ... }\n\nsrc/Hkl/C.hsc:91:19:\nCouldn't match type `Proxy x'0 x0 () (IO Geometry) m0' with `IO'\nExpected type: IO ()\nActual type: Proxy x'0 x0 () (IO Geometry) m0 ()\nIn a stmt of a 'do' block:\nyield \\$ solve' engine n e >>= getSolution0\nIn the expression:\ndo { c_hkl_engine_list_init engines geometry detector sample;\nengine <- c_hkl_engine_list_engine_get_by_name\nengines cname nullPtr;\nn <- c_hkl_engine_pseudo_axis_names_get engine >>= darrayStringLen;\nyield \\$ solve' engine n e >>= getSolution0 }\nIn the second argument of `(\\$)', namely\n`\\ cname\n-> do { c_hkl_engine_list_init engines geometry detector sample;\nengine <- c_hkl_engine_list_engine_get_by_name\nengines cname nullPtr;\n.... }'\n\nI do not understand why the yield does not produce the right type as output.\n\nI think I missed something big :), but..." ]

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[ "## G io ii i2\n\nFigure 3.1. Mass fractions of different constituents of the outer envelope of a newly born neutron star versus matter density in beta equilibrium at different temperatures Tg = T/ (109 K) (after Haensel et al. 1996). Calculations are performed for the Lattimer & Swesty (1991) model of nucleon matter with a specific choice (K0 = 220 MeV) of the incompressibility of cold symmetric nuclear matter at the saturation density.\n\nWe will summarize the results using a more recent version of the compressible liquid-drop model, formulated and developed by Lattimer & Swesty (1991), selecting a specific value of the incompressibility of symmetric nuclear matter at saturation (equilibrium) density, K0 = 220 MeV (for the definition of K0, see § 5.4). We assume nuclear equilibrium as well as beta equilibrium of the matter. The assumption of nuclear equilibrium is justified by high temperature. Beta equilibrium is adopted for simplicity; a very rapid cooling of matter at highest temperatures can produce deviations from beta equilibrium.\n\nIn Fig. 3.1 we show the composition of the hot matter of the neutron star envelope for T = 5 x 109 K, 8 x 109 K, and 1.2 x 1010 K. We restrict ourselves to p < 1013 g cm-3, because at higher densities the thermal effects on matter composition are negligible. At T > 5 x 109 K, the shell and pairing effects, so visible in the T = 0 (ground state) approximation, particularly through jumps in the density dependence of various quantities (see § 3.2), are washed out by the thermal effects.\n\nAt T = 1.2 x 1010K, the nuclei evaporate completely for p < 109 g cm-3. This can be understood within the compressible liquid-drop model; the nuclei are then considered as droplets of nuclear matter. At p < 1011 g cm-3, these droplets have to coexist with a vapor of neutrons, protons and a-particles. However, the coexistence of two different nucleon phases (denser - nuclear liquid, less dense - vapor of nucleons and a-particles) is possible only at T lower than some critical temperature at given a density, Tcrit(p). For p < 109 g cm-3, one has Tcrit(p) < 1.2 x 1010 K.\n\nWith decreasing temperature, the mass fraction of evaporated nucleons and a-particles decreases. For T = 8 x 109 K, a-particles are present at p <\n\n1010 g cm-3, while free protons appear at even lower p. Free neutrons are present at all densities, but their fraction does not exceed one percent for p <\n\nAt T = 5 x 109 K the thermal effects are weak and imply mainly the appearance of a small fraction of free neutrons (\"neutron vapor\") below zero-temperature neutron drip density pND; this fraction falls below 10-5 at p = 1010 g cm-3. Further decrease of T leads to the disappearance of neutrons at p < pND, and to switching-on of shell effects. Another important effect will be the onset of superfluidity of neutrons (both inside and outside the nuclei) and protons. The composition freezes and does not change with further decrease of the temperature. An initially fluid element of the matter solidifies if its temperature falls below the melting temperature Tm that depends on local density and composition (§ 2.3.3)." ]

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[ "## INDEX function\n\nFunction Type: Lookup and Reference Description: INDEX function returns the value of a given location in a table range or array. This function is commonly used with MATCH function which give us the location of the lookup_value. Function Structure:  =INDEX(array, row_number, [column_number]) Argument Breakdown: array – This is the group of cells where your excelContinue reading “INDEX function”\n\n## VLOOKUP\n\nFunction Type: Lookup and Reference Description: VLOOKUP is one of the most popular functions in Excel. It is used to lookup a data and retrieve the corresponding data from a specific column. You may either lookup for the exact or approximate match. The “V” in Vlookup means Vertical. Function Structure:  =VLOOKUP(lookup_value, table_array, column_index_number, [range_lookup]) ArgumentContinue reading “VLOOKUP”" ]

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[ "# Solang Solidity Examples¶\n\nHere are two examples of Solidity contracts.\n\n## Flipper¶\n\nThis is the ink! flipper example written in Solidity:\n\n```contract flipper {\nbool private value;\n\n/// Constructor that initializes the `bool` value to the given `init_value`.\nconstructor(bool initvalue) {\nvalue = initvalue;\n}\n\n/// A message that can be called on instantiated contracts.\n/// This one flips the value of the stored `bool` from `true`\n/// to `false` and vice versa.\nfunction flip() public {\nvalue = !value;\n}\n\n/// Simply returns the current value of our `bool`.\nfunction get() public view returns (bool) {\nreturn value;\n}\n}\n```\n\n## Full Example¶\n\nThis example exists to show the language features that Solang supports.\n\n```// full_example.sol\n\n/*\nThis is an example contract to show all the features that the\nSolang Solidity Compiler supports.\n*/\n\ncontract full_example {\n// Process state\nenum State {\nRunning,\nSleeping,\nWaiting,\nStopped,\nZombie,\nStateCount\n}\n\n// Variables in contract storage\nState state;\nint32 pid;\nuint32 reaped = 3;\n\n// Constants\nint32 constant first_pid = 1;\n\n// Our constructors\nconstructor(int32 _pid) {\n// Set contract storage\npid = _pid;\n}\n\n// Reading but not writing contract storage means function\n// can be declared view\nfunction is_zombie_reaper() public view returns (bool) {\n/* must be pid 1 and not zombie ourselves */\nreturn (pid == first_pid && state != State.Zombie);\n}\n\n// Returning a constant does not access storage at all, so\n// function can be declared pure\nfunction systemd_pid() public pure returns (uint32) {\n// Note that cast is required to change sign from\n// int32 to uint32\nreturn uint32(first_pid);\n}\n\n/// Convert celcius to fahrenheit\nfunction celcius2fahrenheit(int32 celcius) pure public returns (int32) {\nint32 fahrenheit = celcius * 9 / 5 + 32;\n\nreturn fahrenheit;\n}\n\n/// Convert fahrenheit to celcius\nfunction fahrenheit2celcius(int32 fahrenheit) pure public returns (int32) {\nreturn (fahrenheit - 32) * 5 / 9;\n}\n\n/// is this number a power-of-two\nfunction is_power_of_2(uint n) pure public returns (bool) {\nreturn n != 0 && (n & (n - 1)) == 0;\n}\n\n/// calculate the population count (number of set bits) using Brian Kerningham's way\nfunction population_count(uint n) pure public returns (uint count) {\nfor (count = 0; n != 0; count++) {\nn &= (n - 1);\n}\n}\n\n/// calculate the power of base to exp\nfunction power(uint base, uint exp) pure public returns (uint) {\nreturn base ** exp;\n}\n\n/// returns true if the address is 0\n}\n\n/// reverse the bytes in an array of 8 (endian swap)\nfunction byte8reverse(bytes8 input) public pure returns (bytes8 out) {\nout = ((input << 56) & hex\"ff00_0000_0000_0000\") |\n((input << 40) & hex\"00ff_0000_0000_0000\") |\n((input << 24) & hex\"0000_ff00_0000_0000\") |\n((input << 8) & hex\"0000_00ff_0000_0000\") |\n((input >> 8) & hex\"0000_0000_ff00_0000\") |\n((input >> 24) & hex\"0000_0000_00ff_0000\") |\n((input >> 40) & hex\"0000_0000_0000_ff00\") |\n((input >> 56) & hex\"0000_0000_0000_00ff\");\n}\n\n/// This mocks a pid state\nfunction get_pid_state(int64 _pid) pure private returns (State) {\nint64 n = 8;\nfor (int16 i = 1; i < 10; ++i) {\nif ((i % 3) == 0) {\nn *= _pid / int64(i);\n} else {\nn /= 3;\n}\n}\n\nreturn State(n % int64(State.StateCount));\n}\n\n/// Overloaded function with different return value!\nfunction get_pid_state() view private returns (uint32) {\nreturn reaped;\n}\n\nfunction reap_processes() public {\nint32 n = 0;\n\nwhile (n < 100) {\nif (get_pid_state(n) == State.Zombie) {\n// reap!\nreaped += 1;\n}\nn++;\n}\n}\n\nfunction run_queue() public pure returns (uint16) {\nuint16 count = 0;\n// no initializer means its 0.\nint32 n=0;\n\ndo {\nif (get_pid_state(n) == State.Waiting) {\ncount++;\n}\n}\nwhile (++n < 1000);\n\nreturn count;\n}\n\n// cards\nenum suit { club, diamonds, hearts, spades }\nenum value { two, three, four, five, six, seven, eight, nine, ten, jack, queen, king, ace }\nstruct card {\nvalue v;\nsuit s;\n}\n\ncard card1 = card(value.two, suit.club);\ncard card2 = card({s: suit.club, v: value.two});\n\n// This function does a lot of copying\nfunction set_card1(card c) public returns (card previous) {\nprevious = card1;\ncard1 = c;\n}\n\n/// return the ace of spades\nfunction ace_of_spaces() public pure returns (card) {\nreturn card({s: suit.spades, v: value.ace });\n}\n\n/// score card\nfunction score_card(card c) public pure returns (uint32 score) {\nif (c.s == suit.hearts) {\nif (c.v == value.ace) {\nscore = 14;\n}\nif (c.v == value.king) {\nscore = 13;\n}\nif (c.v == value.queen) {\nscore = 12;\n}\nif (c.v == value.jack) {\nscore = 11;\n}\n}\n// all others score 0\n}\n}\n```" ]

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[ "# Servable Types¶\n\nDLHub can serve many different kinds of functions and machine learning models. Each type of servable has a different tool (a “Model Class”) that will aid you in collecting the data needed to run the servable. Here, we detail the types of servables available in DLHub and how to describe them.\n\n## Python Functions¶\n\nIt is possible to publish any Python function as a servable in DLHub. DLHub currently supports two types of Python functions: static functions and class methods. Static functions call members of Python modules and class methods involve calling a function of a specific Python object. Using numpy as an example, numpy.sum(x) involves calling a static function of the numpy module and x.sum() calls the sum class method of the ndarray x.\n\n### Python Static Functions¶\n\nModel Class: PythonStaticMethodModel\n\nServing a Python function requires specifying the name of the module defining the function, the name of the function, and the inputs/outputs of the function. As an example, documenting the max function from numpy would start with:\n\nmodel = PythonStaticMethodModel.create_model('numpy', 'max', autobatch=False,\nfunction_kwargs={'axis': 0})\n\n\nThe first arguments define the module and function name, and are followed by how the command is executed. autobatch=True would tell DLHub to run the function on each member of a list. function_kwargs defines the default keyword arguments for the function (in our case, axis=0)\n\nThe next step is to define the arguments to the function:\n\nmodel.set_inputs('ndarray', 'Matrix', shape=[None, None])\nmodel.set_outputs('ndarray', 'Max of a certain axis', shape=[None])\n\n\nEach of these functions takes the type of input and a short description for that input. Certain types of inputs require further information (e.g., ndarrays require the shape of the array). See Argument Types for a complete listing of argument types.\n\nFunctions that take more than one argument (e.g., f(x, y)) require you to tell DLHub to unpack the inputs before running the function. As an example:\n\nfrom dlhub_sdk.utils.types import compose_argument_block\nmodel = PythonStaticMethodServable.from_function_pointer(f)\nmodel.set_inputs('tuple', 'Two numbers', element_types=[\ncompose_argument_block('float', 'A number'),\ncompose_argument_block('float', 'A second number')\n])\nmodel.set_unpack_inputs(True)\n\n\nNote that we used an input type of “tuple” to indicate that the function takes a fixed number of arguments. You can also use a type of “list” for functions that take a variable number of inputs.\n\n#### Using Static Functions to Create Special Interfaces¶\n\nSome servables required a specialized interface to make the software servable via DLHub. For example, some preprocessing of the input may need to occur before execution.\n\nFor these cases, we define a static function in file named app.py and create a Python servable for that interface function.\n\nSee our interface to SchNet as an example: link.\n\n### Python Class Method¶\n\nModel Class: PythonClassMethodModel\n\nPython class methods are functions associated with a specific Python object. DLHub needs both the file containing the object itself and documentation for the function. As an example, consider a Python object using the following code:\n\nclass ExampleClass:\ndef __init__(self, a):\nself.a = a\ndef f(self, x, b=1):\nreturn self.a * x + b\nx = ExampleClass(1)\nwith open('pickle.pkl', 'wb') as fp:\npkl.dump(x, fp)\n\n\nThe code to serve function f would be:\n\nmodel = PythonClassMethodModel.create_model('pickle.pkl', 'f')\nmodel.set_inputs('float', 'Input value')\nmodel.set_outputs('float', 'Output value')\n\n\nThis code defines the file containing the serialized object (pickle.pkl), the name of the function to be run, and the types of the inputs and outputs. Note that the syntax for defining inputs and outputs is the same as the static functions.\n\nFor this example, it is necessary to include a module defining ExampleClass in the required libraries:\n\nmodel.add_requirement('fake_module_with_exampleclass_on_pypi')\n\n\nor adding the code that defines the class to a seperate file (e.g., example.py) and adding that to the list of files required by DLHub:\n\nmodel.add_file('example.py')\n\n\nAs with the Python static methods, you can specify the functions with multiple arguments using set_unpack_inputs.\n\n## Keras Models¶\n\nModel Class: KerasModel\n\nDLHub serves Keras models using the HDF5 file saved using the Model.save function (see Keras FAQs). The methods described here also work with tf.keras though you should use the Tensorflow loader if you saved the model into Tensorflow’s SavedModel format. As an example, the description for a Keras model created using:\n\nmodel = Sequential()\nmodel.compile(optimizer='rmsprop', loss='mse')\nmodel.fit(X, y)\nmodel.save('model.h5')\n\n\ncan be generated from only the h5 model:\n\nmodel_info = KerasModel.create_model('model.h5')\n\n\nModels with weights and architecture as separate files can be described using:\n\nmodel_info = KerasModel.create_model('model.h5', arch_path='arch.json')\n\n\nKeras also allows users to add their own custom layers to their models for any custom operation that has trainable weights. Use this when the Keras Lambda layer does not apply. In Keras, these layers can be added when loading the model:\n\nmodel = load_model('model.h5', custom_objects={'CustomLayer': CustomLayer})\n\n\nAdding custom layers to a DLHub description can be achived with the add_custom_object method, which takes the name and class of the custom layer:\n\nmodel_info.add_custom_object('CustomLayer', CustomLayer)\n\n\nThe DLHub SDK reads the architecture in the HDF5 file and determines the inputs and outputs automatically:\n\n{\n\"methods\": {\n\"run\": {\n\"input\": {\n\"type\": \"ndarray\", \"description\": \"Tensor\", \"shape\": [null, 1]\n},\n\"output\": {\n\"type\": \"ndarray\", \"description\": \"Tensor\", \"shape\": [null, 1]\n},\n\"parameters\": {},\n\"method_details\": {\n\"method_name\": \"predict\"\n}\n}\n}\n}\n\n\nWe recommended changing the descriptions for the inputs and outputs from their default values:\n\nmodel_info['servable']['methods']['run']['output']['description'] = 'Response'\n\n\nbut the model is ready to be served without any modifications.\n\nThe SDK also determines the version of Keras on your system, and saves that in the requirements.\n\n## PyTorch Models¶\n\nModel Class: TorchModel\n\nDLHub serves PyTorch models using the .pt file saved using the torch.save function (see PyTorch FAQs). As an example, the description for a PyTorch model created using:\n\nclass Net(nn.Module):\ndef __init__(self):\nsuper(Net, self).__init__()\nself.conv1 = nn.Conv2d(1, 20, 5, 1)\nself.conv2 = nn.Conv2d(20, 50, 5, 1)\nself.fc1 = nn.Linear(4*4*50, 500)\nself.fc2 = nn.Linear(500, 10)\n\ndef forward(self, x):\nx = F.relu(self.conv1(x))\nx = F.max_pool2d(x, 2, 2)\nx = F.relu(self.conv2(x))\nx = F.max_pool2d(x, 2, 2)\nx = x.view(-1, 4*4*50)\nx = F.relu(self.fc1(x))\nx = self.fc2(x)\nreturn F.log_softmax(x, dim=1)\n\nmodel = Net()\ntorch.save(model, 'model.pt')\n\n\ncan be generated from the .pt file and the shapes of the input and output arrays.\n\nmodel_info = TorchModel.create_model('model.pt', (None, 1, 28, 28), (None, 10))\n\n\nDLHub will need the definition for the Net module in order to load and run it. You must add the Python libraries containing the module definitions as requirements, or add the files defining the modules to the servable definition.\n\nmodel_info.add_file('Net.py')\n\n\nAs with Keras, we recommended changing the descriptions for the inputs and outputs from their default values:\n\nmodel_info['servable']['methods']['run']['output']['description'] = 'Response'\n\n\nbut the model is ready to be served without any modifications.\n\nIn some cases, you may need to specify the data types of your input array(s) using the keyword arguments of create_model. The type specifications are needed because PyTorch does not do type casting automatically. If in doubt, the data type is float and you can use the default settings.\n\nThe SDK also determines the version of Torch on your system, and saves that in the requirements.\n\n## TensorFlow Graphs¶\n\nModel Class: TensorFlowModel\n\nDLHub uses the same information as TensorFlow Serving for serving a TensorFlow model.\n\nDLHub supports multiple functions to be defined for the same SavedModel servable, but requires one function is marked with DEFAULT_SERVING_SIGNATURE_DEF_KEY.\n\nThe SDK also determines the version of TensorFlow installed on your system, and lists it as a requirement.\n\nHow these models are created is very different between TF1 and TF2.\n\n### TF1¶\n\nSave your model using the SavedModelBuilder as described in the TensorFlow v1.0. As an example, consider a graph expressing $$y = x + 1$$:\n\n# Create the graph\nwith tf.Session() as sess:\nx = tf.placeholder('float', shape=(None, 3), name='Input')\ny = x + 1\n\n# Prepare to save the function\nbuilder = tf.saved_model.builder.SavedModelBuilder('./export')\n\n# Make descriptions for the inputs and outputs\nx_desc = tf.saved_model.utils.build_tensor_info(x)\ny_desc = tf.saved_model.utils.build_tensor_info(y)\n\n# Create a function signature\nfunc_sig = tf.saved_model.signature_def_utils.build_signature_def(\ninputs={'x': x_desc},\noutputs={'y': y_desc},\nmethod_name='run'\n)\n\n# Add the session, graph, and function signature to the saved model\nsess, [tf.saved_model.tag_constants.SERVING],\nsignature_def_map={\ntf.saved_model.signature_constants.DEFAULT_SERVING_SIGNATURE_DEF_KEY: func_sig\n}\n)\n\n# Write the files\nbuilder.save()\n\n\nThe DLHub SDK reads the ./export directory written by this code:\n\nmetadata = TensorFlowModel.create_model(\"./export\")\n\n\nto generate metadata describing which functions were saved:\n\n{\n\"methods\": {\n\"run\": {\n\"input\": {\n\"type\": \"ndarray\", \"description\": \"x\", \"shape\": [null, 3]\n},\n\"output\": {\n\"type\": \"ndarray\", \"description\": \"y\", \"shape\": [null, 3]\n},\n\"parameters\": {},\n\"method_details\": {\n\"input_nodes\": [\"Input:0\"],\n}\n}\n}\n}\n\n\n### TF2¶\n\nFollow the instructions in Tensorflow’s documentation to save your model into the SavedModel format. DLHub requires you to specify the signatures for each of your function you wish to serve, which means you must either specify the input signature when defining the tf.function or create a concrete version of the function (see documentation).\n\nThe following example shows how to save a tf.Module with one function without a signature and a second with a signature.\n\nclass CustomModule(tf.Module):\n\ndef __init__(self):\nsuper().__init__()\nself.m = tf.Variable([1.0, 1.0, 1.0], name='slope')\n\[email protected]\ndef __call__(self, x):\ny = self.m * x + 1\nreturn y\n\[email protected](input_signature=[tf.TensorSpec([], tf.float32),\ntf.TensorSpec((None, 3), tf.float32)])\ndef scalar_multiply(self, z, x):\nreturn tf.multiply(z, x, name='scale_mult')\n\nmodule = CustomModule()\n\n# Make a concrete version of __call__\ncall = module.__call__.get_concrete_function(tf.TensorSpec((None, 3)))\n\ntf.saved_model.save(\nmodule, \"./export\", signatures={\ntf.saved_model.DEFAULT_SERVING_SIGNATURE_DEF_KEY: call,\n'scalar_multiply': module.scalar_multiply\n}\n)\n\n\nThe DLHub SDK will automatically recognize the function signatures and use them to construct a servable accordingly:\n\nmetadata = TensorFlowModel.create_model(\"./export\")\n\n\nwill generate metadata describing which functions were saved:\n\n{\n\"run\": {\n\"input\": {\n\"type\": \"ndarray\", \"description\": \"x:0\", \"shape\": [null, 3],\n\"item_type\": {\"type\": \"float\"}\n},\n\"output\": {\n\"type\": \"ndarray\", \"description\": \"Identity:0\", \"shape\": [null, 3],\n\"item_type\": {\"type\": \"float\"}\n},\n}, \"scalar_multiply\": {\n\"input\": {\n\"type\": \"tuple\", \"description\": \"Several tensors\",\n\"element_types\": [{\n\"type\": \"ndarray\", \"description\": \"x:0\", \"shape\": [null, 3], \"item_type\": {\"type\": \"float\"}\n}, {\n\"type\": \"ndarray\", \"description\": \"z:0\", \"shape\": [], \"item_type\": {\"type\": \"float\"}\n}\n]\n},\n\"output\": {\n\"type\": \"ndarray\", \"description\": \"Identity:0\", \"shape\": [null, 3],\n\"item_type\": {\"type\": \"float\"}\n},\n}\n}\n\n\n## Scikit-Learn Models¶\n\nModel Class: ScikitLearnModel\n\nDLHub supports scikit-learn models saved using either pickle or joblib. The saved models files do not always contain the number of input features for the model, so they need to provided along with the serialization method and, for classifiers, the class names:\n\n# Loading SVC trained on the iris dataset\nmodel_info = ScikitLearnModel.create_model('model.pkl', n_input_columns=4, classes=3)\n\n\nGiven this information, the SDK generates documentation for how to invoke the model:\n\n{\n\"methods\": {\n\"run\": {\n\"input\": {\n\"type\": \"ndarray\",\n\"shape\": [null, 4],\n\"description\": \"List of records to evaluate with model. Each record is a list of 4 variables.\",\n\"item_type\": {\"type\": \"float\"}\n},\n\"output\": {\n\"type\": \"ndarray\",\n\"shape\": [null, 3],\n\"description\": \"Probabilities for membership in each of 3 classes\",\n\"item_type\": {\"type\": \"float\"}\n},\n\"parameters\": {},\n\"method_details\": {\n\"method_name\": \"_predict_proba\"\n}\n}\n}\n}\n\n\nThe SDK will automatically document the type of model and extract the scikit-learn version used to save the model, which it includes in the requirements." ]

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[ "# #5dffed Color Information\n\nIn a RGB color space, hex #5dffed is composed of 36.5% red, 100% green and 92.9% blue. Whereas in a CMYK color space, it is composed of 63.5% cyan, 0% magenta, 7.1% yellow and 0% black. It has a hue angle of 173.3 degrees, a saturation of 100% and a lightness of 68.2%. #5dffed color hex could be obtained by blending #baffff with #00ffdb. Closest websafe color is: #66ffff.\n\n• R 36\n• G 100\n• B 93\nRGB color chart\n• C 64\n• M 0\n• Y 7\n• K 0\nCMYK color chart\n\n#5dffed color description : Light cyan.\n\n# #5dffed Color Conversion\n\nThe hexadecimal color #5dffed has RGB values of R:93, G:255, B:237 and CMYK values of C:0.64, M:0, Y:0.07, K:0. Its decimal value is 6160365.\n\nHex triplet RGB Decimal 5dffed `#5dffed` 93, 255, 237 `rgb(93,255,237)` 36.5, 100, 92.9 `rgb(36.5%,100%,92.9%)` 64, 0, 7, 0 173.3°, 100, 68.2 `hsl(173.3,100%,68.2%)` 173.3°, 63.5, 100 66ffff `#66ffff`\nCIE-LAB 91.665, -46.021, -3.872 55.555, 79.957, 92.621 0.244, 0.35, 79.957 91.665, 46.184, 184.809 91.665, -62.99, 1.367 89.419, -45.581, 1.179 01011101, 11111111, 11101101\n\n# Color Schemes with #5dffed\n\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #ff5d6f\n``#ff5d6f` `rgb(255,93,111)``\nComplementary Color\n• #5dff9c\n``#5dff9c` `rgb(93,255,156)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #5dc0ff\n``#5dc0ff` `rgb(93,192,255)``\nAnalogous Color\n• #ff9c5d\n``#ff9c5d` `rgb(255,156,93)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #ff5dc0\n``#ff5dc0` `rgb(255,93,192)``\nSplit Complementary Color\n• #ffed5d\n``#ffed5d` `rgb(255,237,93)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #ed5dff\n``#ed5dff` `rgb(237,93,255)``\n• #6fff5d\n``#6fff5d` `rgb(111,255,93)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #ed5dff\n``#ed5dff` `rgb(237,93,255)``\n• #ff5d6f\n``#ff5d6f` `rgb(255,93,111)``\n• #11ffe5\n``#11ffe5` `rgb(17,255,229)``\n• #2affe7\n``#2affe7` `rgb(42,255,231)``\n• #44ffea\n``#44ffea` `rgb(68,255,234)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #77fff0\n``#77fff0` `rgb(119,255,240)``\n• #90fff3\n``#90fff3` `rgb(144,255,243)``\n• #aafff6\n``#aafff6` `rgb(170,255,246)``\nMonochromatic Color\n\n# Alternatives to #5dffed\n\nBelow, you can see some colors close to #5dffed. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #5dffc5\n``#5dffc5` `rgb(93,255,197)``\n• #5dffd2\n``#5dffd2` `rgb(93,255,210)``\n• #5dffe0\n``#5dffe0` `rgb(93,255,224)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #5dfffb\n``#5dfffb` `rgb(93,255,251)``\n• #5df6ff\n``#5df6ff` `rgb(93,246,255)``\n• #5de9ff\n``#5de9ff` `rgb(93,233,255)``\nSimilar Colors\n\n# #5dffed Preview\n\nThis text has a font color of #5dffed.\n\n``<span style=\"color:#5dffed;\">Text here</span>``\n#5dffed background color\n\nThis paragraph has a background color of #5dffed.\n\n``<p style=\"background-color:#5dffed;\">Content here</p>``\n#5dffed border color\n\nThis element has a border color of #5dffed.\n\n``<div style=\"border:1px solid #5dffed;\">Content here</div>``\nCSS codes\n``.text {color:#5dffed;}``\n``.background {background-color:#5dffed;}``\n``.border {border:1px solid #5dffed;}``\n\n# Shades and Tints of #5dffed\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000f0d is the darkest color, while #fafffe is the lightest one.\n\n• #000f0d\n``#000f0d` `rgb(0,15,13)``\n• #00221e\n``#00221e` `rgb(0,34,30)``\n• #003630\n``#003630` `rgb(0,54,48)``\n• #004941\n``#004941` `rgb(0,73,65)``\n• #005d53\n``#005d53` `rgb(0,93,83)``\n• #007164\n``#007164` `rgb(0,113,100)``\n• #008476\n``#008476` `rgb(0,132,118)``\n• #009887\n``#009887` `rgb(0,152,135)``\n• #00ab98\n``#00ab98` `rgb(0,171,152)``\n• #00bfaa\n``#00bfaa` `rgb(0,191,170)``\n• #00d3bb\n``#00d3bb` `rgb(0,211,187)``\n• #00e6cd\n``#00e6cd` `rgb(0,230,205)``\n``#00fade` `rgb(0,250,222)``\n• #0fffe4\n``#0fffe4` `rgb(15,255,228)``\n• #22ffe6\n``#22ffe6` `rgb(34,255,230)``\n• #36ffe9\n``#36ffe9` `rgb(54,255,233)``\n• #49ffeb\n``#49ffeb` `rgb(73,255,235)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\n• #71ffef\n``#71ffef` `rgb(113,255,239)``\n• #84fff1\n``#84fff1` `rgb(132,255,241)``\n• #98fff4\n``#98fff4` `rgb(152,255,244)``\n• #abfff6\n``#abfff6` `rgb(171,255,246)``\n• #bffff8\n``#bffff8` `rgb(191,255,248)``\n• #d3fffa\n``#d3fffa` `rgb(211,255,250)``\n• #e6fffc\n``#e6fffc` `rgb(230,255,252)``\n• #fafffe\n``#fafffe` `rgb(250,255,254)``\nTint Color Variation\n\n# Tones of #5dffed\n\nA tone is produced by adding gray to any pure hue. In this case, #a8b4b3 is the less saturated color, while #5dffed is the most saturated one.\n\n• #a8b4b3\n``#a8b4b3` `rgb(168,180,179)``\n• #a2bab8\n``#a2bab8` `rgb(162,186,184)``\n• #9bc1bd\n``#9bc1bd` `rgb(155,193,189)``\n• #95c7c1\n``#95c7c1` `rgb(149,199,193)``\n• #8fcdc6\n``#8fcdc6` `rgb(143,205,198)``\n• #89d3cb\n``#89d3cb` `rgb(137,211,203)``\n``#82dad0` `rgb(130,218,208)``\n• #7ce0d5\n``#7ce0d5` `rgb(124,224,213)``\n• #76e6da\n``#76e6da` `rgb(118,230,218)``\n• #70ecde\n``#70ecde` `rgb(112,236,222)``\n• #69f3e3\n``#69f3e3` `rgb(105,243,227)``\n• #63f9e8\n``#63f9e8` `rgb(99,249,232)``\n• #5dffed\n``#5dffed` `rgb(93,255,237)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #5dffed is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Ortho and Para hydrogen\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nA molecule of dihydrogen contains two atoms, in which the nuclei of both the atoms are spinning. Depending upon the direction of the spin of the nuclei, the hydrogens are of two types:\n\nOrtho hydrogen molecules are those in which the spins of both the nuclei are in the same direction. Molecules of hydrogen in which the spins of both the nuclei are in the opposite direction are called para hydrogen.", null, "Figure 1: Ortho and para hydrogen\n\nOrdinary dihydrogen is an equilibrium mixture of ortho and para hydrogen.\n\n$\\text{ortho hydrogen} \\ce{<=>} \\text{para hydrogen} \\nonumber$\n\nThe amount of ortho and para hydrogen varies with temperature as:\n\n• At 0°K, hydrogen contains mainly para hydrogen which is more stable.\n• At the temperature of liquefaction of air, the ratio of ortho and para hydrogen is 1 : 1.\n• At the room temperature, the ratio of ortho to para hydrogen is 3 : 1.\n• Even at very high temperatures, the ratio of ortho to para hydrogen can never be more than 3 : 1.\n\nThus, it has been possible to get pure para hydrogen by cooling ordinary hydrogen gas to a very low temperature (close to 20 K) but it is never possible to get a sample of hydrogen containing more than 75% of ortho hydrogen.\n\n## Contributors and Attributions\n\nOrtho and Para hydrogen is shared under a not declared license and was authored, remixed, and/or curated by Binod Shrestha." ]

[ null, "https://chem.libretexts.org/@api/deki/files/28753/ortho-para-hydrogen.jpg", null ]

[ "", null, ", 12.11.2019 08:31, strikeboystorm\n\n# If you know the equation of a proportional relationship how can you draw the graph of the equation", null, "", null, "", null, "### Other questions on the subject: Mathematics", null, "Carmen begins her next painting on a rectangular canvas that is 82.7 cm long and has a area of 8,137.68 cm2. will the painting fit in a frame with an opening that is 82.7 cm long and 95 cm wide? explain", null, "Mathematics, 21.06.2019 18:30, jb141553\nIf 3x+8=3x+8 is it one solution or no solution", null, "Mathematics, 21.06.2019 19:10, chancler\nFind the roots of the polynomial function f(x) = x^3 + 2x^2 + x", null, "Mathematics, 21.06.2019 21:00, XxXisaiahXxX\nTwo electrical lines are parallel to each other. one of the lines is represented by the equation –4x + y = 8. what is the slope of the other electrical line?\nDo you know the correct answer?\nIf you know the equation of a proportional relationship how can you draw the graph of the equation...\n\n### Questions in other subjects:", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Mathematics, 21.02.2020 00:56\nTotal solved problems on the site: 8122374" ]

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[ "News\n1. Home\n2. -\n3. ball-mill\n4. -\n5. ball load calculation in ball mill\n\n# ball load calculation in ball mill", null, "Pulp samples collected around the ball mill or rod mill and hydrocyclones screen or classifier classification system are screened and the cumulative weight percentage retained is calculated for several mesh sizes to obtain the Ball Mill Circulating Load\n\nGet a Quote Send Message\n\n## Hot Products\n\nOur products sell well all over the world, and have advanced technology in the field of crushing sand grinding powder.", null, "For example your ball mill is in closed circuit with a set of cyclones The grinding mill receives crushed ore feed The pulp densities around your cyclone are sampled and known over an 8hour shift allowing to calculate corresponding to circulating load ratios and circulating load tonnage on tonsday or tonshour\n\nMore Details", null, "calculation of ball mill residence time\n\nRTD Holdup Ball mill Ball load Slurry Scientific Academic estimated and the calculated values compared reasonably with the measured ones within the limits of and Ball Loading on Mill Residence Time Distribution\n\nMore Details", null, "Ball Mill Loading Dry Milling Ball Mill Loading dry milling When charging a ball mill ceramic lined mill pebble mill jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product\n\nMore Details", null, "CALCULATION OF BALL MILL GRINDING EFFICIENCY Page\n\nMar 08 2013 · calculation of ball mill grinding efficiency dear experts please tell me how to calculate the grinding efficiency of a closed ckt open ckt ball mill in literatures it is written that the grinding efficiency of ball mill is very less less than 10 please expalin in a n excel sheet to calcualte the same thanks sidhant reply\n\nMore Details", null, "definition circulating load in a ball mill\n\nCirculating Load Calculation Formula Mineral Processing Jul 9 2016 Here is a formula that allows you to calculate the circulating load ratio around a ball mill\n\nMore Details", null, "Media and Product Ball Mill Loading Guide Percentages are based on total volume of cylinder NOTE With media load at 50 voids are created equal to 20 of cylinder volume These voids are filled when product is loaded into the mill Mills can be loaded by volume or\n\nMore Details", null, "ball size calculation for ball mill\n\nBall mill Wikipedia the free encyclopedia A ball mill is a type of grinder used to grind and blend materials for use in mineral dressing Principleedit A ball mill works on the principle of impact and attrition size reduction is done by impact as the balls drop from near the top of the shell Chat Now\n\nMore Details", null, "MODELING THE SPECIFIC GRINDING ENERGY AND BALL\n\n2 MODELLING THE SPECIFIC GRINDING ENERGY AND BALLMILL SCALEUP Ballmill scale up Bond’s LawData zBond work index w i zFeed D f and product d size both 80 cumulative passing Result The specific grinding energy w Mill power draw P wT where T the mill capacity Mill dimensions from Tables or charts\n\nMore Details", null, "PDF Circulating load calculation in grinding circuits\n\nAn industrial ball mill operating in closed–circuit with hydrocyclones was studied by five sampling surveys algorithm for circulating load calculation in mineral processing closed circuits\n\nMore Details", null, "A Method to Determine the Ball Filling in Miduk Copper\n\nload percentage calculation that belongs to mill diameter\n\nMore Details", null, "TECHNICAL NOTES 8 GRINDING R P King Mineral Tech\n\nthe mill is used primarily to lift the load medium and charge Additional power is required to keep the mill rotating 813 Power drawn by ball semiautogenous and autogenous mills A simplified picture of the mill load is shown in Figure 83 Ad this can be used to establish the essential features of a model for mill\n\nMore Details", null, "Optimum choice of the makeup ball sizes for maximum\n\nThe calculation of d ij includes the aforementioned S and B functions which can be determined in a laboratory batch mill However these parameters are sensitive to milling conditions such as the mill rotational speed ball filling powder filling mill diameter and ball diameter\n\nMore Details", null, "volume calculation for ball mill SicuPlus\n\nDesign Method of Ball Mill by Sumitomo Chemical The physical constants used in these calculations are the mill was stopped and small volume Effects of Fins on Ball Motion in the Mill To predict ball mill Chat Now Volume Calculation For Ball Mill iescoachingdelhiin\n\nMore Details", null, "ball mill circulating load formula Popular Education\n\nBall Mill Instruction Manual PDF BICO Inc The FC Bond Ball Mill is a small universal laboratory mill used in calculating the dry in clos€d circuit witb l0O per cent circulating load iD a l2in dia by 24in\n\nMore Details", null, "calculation of ball mill charge volume\n\nBall Mill Instruction Manual PDF BICO Inc The FC Bond Ball Mill is a small universal laboratory mill used in calculating the made in the Bico Mill rurming at 70 revolutions per minute with a charge of285 iron balls lhe work requircd is proportioDal to the rcdultioo in volume Read More\n\nMore Details", null, "Ball Mill an overview ScienceDirect Topics\n\nThe ball mill is a tumbling mill that uses steel balls as the grinding media The length of the cylindrical shell is usually 1–15 times the shell diameter Figure 811The feed can be dry with less than 3 moisture to minimize ball coating or slurry containing 20–40 water by weight\n\nMore Details", null, "Nov 18 2014 · Slide 2 Content Introduction Circulating load calculate Most factories the purpose of control Effective factors for circulating load Test design Conclusion References 3 Slide 3 circulating load ratio As the ratio of the amount of solids going through the ball mill divided by the amount of solids going through the circuit\n\nMore Details", null, "Ball Mill SlideShare\n\nNov 18 2008 · Summary The Ball Mill is designed to grind materials by turning the cylindrical shell with grinding medium eg steel balls put in the shell and has a simple structure and ease of handling Furthermore The Ball Mill of a large capacity has been available to a very extensive range of applications in both dry and wet 14 References 1\n\nMore Details", null, "Home\n\nis home to a collection of both free and subscriptionbased calculation tools to aid metallurgical process engineers perform comminution calculations Grinding circuit design tools including for SAG millball mill circuits and geometallurgy energy models are available to subscribers\n\nMore Details", null, "Ball mill calculations tube mill calculations separator\n\nAll Ball mill or tube mill calculation Critical speed Ball Size calculations Separator efficiency Mill power cnsumption calculation production at blain All Ball mill or tube mill calculation Critical speed Ball Size calculations Separator efficiency Mill power cnsumption calculation production at blain Circulation load CL Go To\n\nMore Details", null, "Calculation of circulating load of a grinding mill\n\nCalculation of circulating load of a grinding mill grinding mill circulating load calculation is a ratio of the mill drive power to the milling circuit output Ball Mill Circulating Load Pdf Get Price And Support Online ball load calculation in ball mills Circulating load calculation in grinding circuits SciELO 2008 working with predictive control models applied to ball mills\n\nMore Details", null, "Ball Mill Load feed Page 1 of 1\n\nDec 03 2010 · Re Ball Mill Load feed Dear xxxx When the mill is chocked it only rotates like a feed material balls thrown cetrifugally outward toward mill liners and remain stick over is no grinding process going on but a choked process\n\nMore Details", null, "Calculation Of Circulating Load Of A Grinding Mill Pdf\n\nMore About calculate ball mill circulating load pdf 2012 ball mill dynamic load calculation Crushing and grinding machine supplier in jaw crusher Get Price And Support Online calculation of recirculating load in ball mill circuit ball mill circulating load calculation Ball Mill Recirculating Load for Grinding Fundamentals to give\n\nMore Details", null, "FC BOND BICO BALL MILL\n\nFC BOND BICO BALL MILL The FC Bond Ball Mill is a small universal laboratory mill used in calculating the grindability of all ores GRINDABILITY IS THE NUMBER OF NET GRAMS OF SCREEN UNDERSIZE PRODUCED PER REVOLUTION This Ball Mill can be used continuously or it can be used for any number of revolutions according to the type of grind desired\n\nMore Details", null, "Design Construction and Performance Analysis of a 5\n\nlaboratory ball mill This method is based on two power calculation approaches used in ball and rod mill design processes due to its simplicity and workability The first approach which is specific power calculation determines the power required to grind an ore from a given feed size to a specific product size The second\n\nMore Details", null, "circulating load calculation in closed circuit ball mill\n\nCirculating Load Calculation Formula Here is a formula that allows you to calculate the circulating load ratio around a ball mill and hydrocylone as part of a grinding circuit For example your ball mill is in closed circuit with a set of cyclones The grinding mill receives crushed ore feed\n\nMore Details", null, "Ball Mill Circulating Load Mineral Processing Metallurgy The circulating loads generated in a typical ball mill cyclone circuit contain a small fraction of bypassed fines The concept that high circulating loads will result in overgrinding can be refuted by regarding increases in circulating load in the same vein as multistage grinding\n\nMore Details", null, "Calculation Of Filling Volume In Ball Mill\n\nball mill volume calculation wet ball mill calculations for fill volume ball mill volume calculationFull Text RS Publicatio Issue 4 volume 1 JanuaryFebruary 2014 AvailableDesign and Optimization of Ball Mill for Clinker Grinding in Cement Plant Sanjeevequations offerquick parameters estimation useful for practicing engineers A Ball Mill\n\nMore Details", null, "vrm and ball mill circulating load page 1 of 1 sep 07 2011· re vrm and ball mill circulating load mainly in usa the term circulating load is more often used than the circulation ating load is percentage of coarse return in relation to fines it can be calculated by coarse return tph x 100 mill output range of cirulating load in a conventional close circuit\n\nMore Details", null, "PDF Grinding in Ball Mills Modeling and Process Control\n\nGrinding in Ball Mills Modeling and Process Control equipment used to load the starting Distribusi partikel hasil penggerusan dengan ball mill replika ini mirip dengan yang dari standard\n\nMore Details", null, "AMIT 135 Lesson 7 Ball Mills Circuits – Mining Mill\n\nNumber size and mass of each ball size depends on mill load and whether or not the media is being added as the initial charge For the initial chargin of a mill Coghill and DeVaney 1937 defined the ball size as a function of the top size of the feed ie d↓V 040 K√F\n\nMore Details", null, "Calculation of circulating load of a grinding mill Henan\n\nBall mill circulating load Circulating Load Calculation Formula 2019815 Here is a formula that allows you to calculate the circulating load ratio around a ball mill and hydrocylone as part of a grinding circuit For example your ball mill is in closed circuit with a set of cyclones The grinding mill\n\nMore Details", null, "Technical Calculation Selection Method Ball Screws\n\nBall Screw Length L 720 62 60 72 914mm Extra Margin is for overrun countermeasure and is normally 152 times the screw lead Lead 20 x 15 x 2 both ends 60 2 Evaluating Allowable Axial Load Since the Load Applicable Span Length 1 is 820mm Allowable Axial Load P according to the formula on P1895 “4\n\nMore Details", null, "power calculation formula of ball mill\n\nBall Mill DesignPower Calculation Diese Seite übersetzen The ball mill motor power requirement calculated above as 1400 HP is the power that must be applied at the mill drive in order to grind the tonnage of feed from one size distribution\n\nMore Details", null, "Optimization of mill performance by using\n\nto the volumetric mill filling which influences grinding media wear rates throughput power draw and product grind size from the circuit Each of these performance parameters peaks at different filling values In order to continuously optimize mill operation it is vital to obtain regular measurements of the ball load and pulp position\n\nMore Details", null, "ball mill capacity calculations\n\nBall Mill DesignPower Calculation LinkedIn 20161212 The basic parameters used in ball mill design power calculations rod mill or any tumbling mill sizing are material to be ground characteristics Bond Work Index bulk density specific density desired mill tonnage capacity DTPH operating Mill sizing method\n\nMore Details", null, "Page 1 Ball Milling Theory\n\ninvolve grinding With Lloyds ball milling book having sold over 2000 copies there are probably over 1000 home built ball mills operating in just America alone This article borrows from Lloyds research which was obtained from the commercial ball milling industry and explains some of the key design criteria for making your own ball mill\n\nMore Details", null, "For example your ball mill is in closed circuit with a set of cyclones The grinding mill receives crushed ore feed The pulp densities around your cyclone are sampled and known over an 8hour shift allowing to calculate corresponding to circulating load ratios and circulating load tonnage on tonsday or tonshour\n\nMore Details", null, "Ball load calculation in ball mills Ball load calculation in ball mills Ball mill wikipedia a ball mill a type of grinder is a cylindrical device used in grinding or mixing materials like ores chemicals ceramic raw materials and paintsball mills rotate around a horizontal axis partially filled with the material to be g\n\nMore Details", null, "Calculate and Select Ball Mill Ball Size for Optimum Grinding\n\nIn Grinding selecting calculate the correct or optimum ball size that allows for the best and optimumideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do Often the ball used in ball mills is oversize “just in case”\n\nMore Details", null, "Ball Mill Loading dry milling When charging a ball mill ceramic lined mill pebble mill jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product Charging a Dry Mill – The general operation of a grinding mill is to have the product impacted between the balls as they tumble\n\nMore Details", null, "size in the rod mill discharge ball mill discharge and cyclone feed is 27 percent 5 percent and 14 percent has metal grade 02 percent calculate mass of concentrate and tailing Seven a floatation circulating load is 025\n\nMore Details", null, "Free Calculation Ball Mill Ball Load\n\nfree calculation ball mill ball load Wet Ball Mill Calculations For Fill Volumeball mill dynamic load charge calculation for cement mill circulating load ratio PDF 1044 K Journal of Mining and Environment power index SPI and a formula for the SAG mill necessary to calculate the proportion of each mills mills\n\nMore Details", null, "Ball Mill DesignPower Calculation\n\nBall Mill Power Calculation Example 1 A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80 passing ¼ inch 6350 microns The required product size distribution is to be 80 passing 100 mesh 149 microns\n\nMore Details", null, "CALCULATION OF BALL MILL GRINDING EFFICIENCY Page\n\nMar 08 2013 · calculation of ball mill grinding efficiency dear experts please tell me how to calculate the grinding efficiency of a closed ckt open ckt ball mill in literatures it is written that the grinding efficiency of ball mill is very less less than 10 please expalin in a n excel sheet to calcualte the same thanks sidhant reply\n\nMore Details", null, "Mill Steel Charge Volume Calculation\n\nThe loading or change volume can then be read off the graph below or approximated from the equation and calculation Steel Charge Loading in Mill 113 – 126 H D Ball Mill Charge Volume Calculation Charge Volume of a Grinding Mill Method 1 Charge Volume of a Grinding Mill\n\nMore Details", null, "Ball Mill Parameter Selection Calculation Power\n\nThe ball loading capacity of the mill can be calculated according to the formula 114 Gra Quantity of Grinding Medium T Rho s loose density of grinding medium tm3\n\nMore Details", null, "Calculation Of A Ball Mill Load\n\nCalculation Of A Ball Mill Load 20191027load percentage calculation that belongs to mill diameter and distance between top point in the mill and load surface h 57his equation is profitable for determining of whole mill load volume which is included stony and metallic ballsill load 113\n\nMore Details", null, "calculation of ball mill charge volume\n\nIn wet ball mill measurement accuracy of mill load ML is very important load to mill volume ratio replaces the medium charge ratio The relations between Calculate the input of\n\nMore Details", null, "ball mill grinding sediment ball load calculation in ball\n\n20041013 Calculation of the power draw of dry multicompartment ball mills 225 The mill load that is the volume of charge in the mill is the principal determinant of power draw Estimation of the ball load that is mixed with the cement charge is difficult and can be highly erroneous\n\nMore Details", null, "definition circulating load in a ball mill\n\nA Bond Ball Mill Index Test BBMWI is a standard test for determining the Ball Mill Work Index of a sample of ore produce a 250 circulating load Chat Now PDF 660 K Iranian Journal of Science and Technology Transactions\n\nMore Details", null, "ball size calculation for ball mill\n\nBall mill Wikipedia the free encyclopedia A ball mill is a type of grinder used to grind and blend materials for use in mineral dressing Principleedit A ball mill works on the principle of impact and attrition size reduction is done by impact as the balls drop from near the top of the shell Chat Now\n\nMore Details", null, "ball mill recirculating load calculation pdf\n\nGrinding Ball Mill Load Calculation Formula ball mill recirculating load calculation pdf Calculating a grinding circuits circulating loads based on Screen A\n\nMore Details", null, "Ball charges calculators\n\nBall top size bond formula calculation of the top size grinding media balls or cylpebsModification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency\n\nMore Details", null, "calculate ball mill circulating load pdf\n\nball mill circulating load circulating load calculation in ball mill Circulating Load Calculation FormulaHere is a formula that allows you to calculate the circulating load ratio around a ball mill and hydrocylone as part of a grinding circuit For example your ball mill is in closed circuit with a set of cyclones circulating load calculation in ball millgrinding mill\n\nMore Details", null, "calculation of ball mill design\n\nCirculating Load Calculation In Closed Circuit Ball Mill calculating re rod mill design calculation of rod mill grinding capacity ball mill capacity Get price 3 easy steps to calculate ball mill capacity bulkonline Forums Let us present an easy and convenient service for calculating ball mill capacity Ball mills are one of the\n\nMore Details", null, "capacity calculation of ball mill\n\nball mill volume calculation calculation of grinding media charging ball mill of ZMEIt is the main ball mill volume calculation Mill grinding Wikipedia the volume of ball vs volume of ball The ball mill is key equipment in grinding industry\n\nMore Details", null, "ball mill circulating load formula Popular Education\n\nBall Mill Instruction Manual PDF BICO Inc The FC Bond Ball Mill is a small universal laboratory mill used in calculating the dry in clos€d circuit witb l0O per cent circulating load iD a l2in dia by 24in\n\nMore Details", null, "Ball Load Calculation In Ball Mill greenvboutiquein load calculation in ball mill antriksharaliasorgin Bearing Load Calculation 43 Mean load The load on bearings used in machines under normal circumstances will\n\nMore Details", null, "End Mill Speed and Feed Calculator Martin Chick Associates\n\nSpeed And Feed Calculators Ball Mill Finish Calculator Part Spacing Calculator G And M Code Characters Standard End Mill Sizes Standard Drill Sizes Drill And Chip Load In I am creating a new calculator based on your feedback Please fill out the form below with feeds and speeds that work for you and I will place them into a new\n\nMore Details", null, "Ball Mill Parameter Selection Calculation Power\n\n1 Calculation of ball mill capacity The production capacity of the ball mill is determined by the amount of material required to be ground and it must have a certain margin when designing and selecting There are many factors affecting the production capacity of the ball mill in addition to the nature of the material grain size hardness density temperature and humidity the degree of\n\nMore Details", null, "Ball Mill Recirculating Load Calculation Pdf\n\nCirculating load calculation in ball millirculating load calculation formulahere is a formula that allows you to calculate the circulating load ratio around a ball mill and hydrocylone as part of a grinding circuit for example your ball mill is in closed circuit with a set of cyclones circulating load calculation in ball millgrinding mill\n\nMore Details", null, "Circulating Load Ball Mill Inneneinrichtungen Kreienbühl\n\nball mill circulating load circulating load calculation in ball mill Circulating Load Calculation FormulaHere is a formula that allows you to calculate the circulating load ratio around a ball mill and hydrocylone as part of a grinding circuit For example your ball mill is in closed circuit with a set of cyclones\n\nMore Details", null, "ball mill grinding sediment ball load calculation in ball\n\n20041013 Calculation of the power draw of dry multicompartment ball mills 225 The mill load that is the volume of charge in the mill is the principal determinant of power draw Estimation of the ball load that is mixed with the cement charge is difficult and can be highly erroneous\n\nMore Details", null, "Simple calculation in ball calculation in the ball mill simple calculation in the ball mill calculate and select ball mill ball size for in grinding selecting calculate the correct or optimum ball size that allows for the best and optimumideal or target grind size to be\n\nMore Details", null, "ball mill recirculating load calculation pdf\n\nQuality ball mill design calculation pdf is the best formula calculation recirculating load in a mill to the choice and calculation of ball media load calculation Circulating load SlideShare Slide 2 Content Introduction Circulating load calculate Most factories As the ratio of the amount of solids going through the ball mill divided\n\nMore Details", null, "how to ball mill feed capacity calculate for example\n\nCalculation Feed capacity → Mill dimensions Filling ratio 3045 Mill dimensions → Bulk volume of the balls ball size D2 k Dp where k ranges from 09\n\nMore Details", null, "A Method to Determine the Ball Filling in Miduk Copper\n\nload percentage calculation that belongs to mill diameter\n\nMore Details", null, "ball mill filling degree calculation\n\nCalculate and Select Ball Mill Ball Size for Aug 4 2016 ball mill filling degree formula Description ball mill filling calculation formula to calculate filling of ball millStone Crusher Machine From\n\nMore Details", null, "calculation of torque for ball mill calculation pdf\n\ntorque in rolling mills calculation formula with example critical speed of ball mill calculation india ball mill circulating load Get Price And Support Online Rolling mill speed calculation formula pdf mill ExampleCalculate the rolling load if steel sheet is hot rolled 30 from Read More\n\nMore Details", null, "Calculation Of A Ball Mill Load\n\nCalculation Of A Ball Mill Load 20191027load percentage calculation that belongs to mill diameter and distance between top point in the mill and load surface h 57his equation is profitable for determining of whole mill load volume which is included stony and metallic ballsill load 113 A Method To Determine The Ball Filling In Miduk Copper\n\nMore Details", null, "calculate ball mill circulating load pdf\n\nBall Mill Instruction Manual PDF BICO Inc The FC Bond Ball Mill is a small universal laboratory mill used in calculating the ofminus 6 mesh stage crushed dry feed was used and the circulation load\n\nMore Details", null, "Ball Mill Calculation Crusher Mills Cone Crusher Jaw\n\nball load calculation in ball mills – Gold Ore Crusher CALCULATION OF THE POWER DRAW OF DRY MULTI–COMPARTMENT BALL calculation of mill load A direct measurement of the load entails the crash stopping of the ball\n\nMore Details", null, "", null, "", null, "" ]

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[ "# Need Help with Brain Boot Camp problem\n\n#### Kristin H 0309\n\n##### New member\n\" A car and a motorcycle set off from the same point to travel the same journey. The car sets off two minutes before the motorcycle. If the car travels at 60km/hr and the motorcycle travels at 80km/hr how many kilometers from the starting point will they draw level? \"\nThank you so much for your help! It has been many, many years since I learned Algebra in school!\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\n\" A car and a motorcycle set off from the same point to travel the same journey. The car sets off two minutes before the motorcycle. If the car travels at 60km/hr and the motorcycle travels at 80km/hr how many kilometers from the starting point will they draw level? \"\nThank you so much for your help! It has been many, many years since I learned Algebra in school!\n\nPlease share your work/thoughts and context of the problem (what is the subject topic?) - so that we know where to begin to help you.\n\nHint: The \"find\" of the problem is:\n\n\"how many kilometers from the starting point will they draw level?\"\n\nAssume:\n\nThe distance to draw level = d km\n\nHow long the car will take to travel this distance = d/60 hr\n\nHow long the MC will take to travel this distance = d/80 hr\n\nContinue.....\n\n#### Kristin H 0309\n\n##### New member\nI'm still very frustrated by this problem but will make a stab at the solution:\nd/.75 + 2/1 = d/1\n1.5= 2d\n3=d\n\n#### Romsek\n\n##### Full Member\n$$\\displaystyle \\text{Supposing the car starts at time t=0~hr, it's distance from start will be d_c(t) = 60t~km}$$\n\n$$\\displaystyle \\text{Similarly the motorcycle distance is given by d_m(t) = 80\\left(t-\\dfrac{2}{60}\\right)~km}$$\n\n$$\\displaystyle \\text{We want to find out the time these two distances are equal}\\\\ 60t=80\\left(t - \\dfrac{1}{30}\\right) = 80t - \\dfrac 8 3\\\\ \\dfrac 8 3 = 20 t\\\\ t = \\dfrac{8}{60} = \\dfrac{2}{15}~hr\\\\ d_c(t) = d_m(t) = 60 \\dfrac{2}{15} = 8~km$$\n\n#### Subhotosh Khan\n\n##### Super Moderator\nStaff member\nI'm still very frustrated by this problem but will make a stab at the solution:\nd/.75 + 2/1 = d/1\n1.5= 2d\n3=d\nHow did you get those equations?\n\nThose are so misguided that those are NOT EVEN WRONG!\n\nI'll do this problem for you but I am afraid you are not ready for these types of problems.\n\n\" A car and a motorcycle set off from the same point to travel the same journey. The car sets off two minutes before the motorcycle. If the car travels at 60km/hr and the motorcycle travels at 80km/hr how many kilometers from the starting point will they draw level? \"\n\nOne important thing to note in this problem is to note that car spends more time on road than the motorcycle.\n\nHow much longer (in time) did the car spend on the road? = 2 minutes = 1/30 hours\n\nThen\n\nd/60 = d/80 + 1/30\n\nd/60 - d/80 = 1/30\n\n$$\\displaystyle \\dfrac{4*d - 3*d}{240} \\ = \\dfrac{1}{30}$$\n\nd = 240/30 = 8 km" ]

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[ "# Bounds on the distribution of the number of gaps when circles and lines are covered by fragments: Theory and practical application to genomic and metagenomic projects\n\n## Abstract\n\n### Background\n\nThe question of how a circle or line segment becomes covered when random arcs are marked off has arisen repeatedly in bioinformatics. The number of uncovered gaps is of particular interest. Approximate distributions for the number of gaps have been given in the literature, one motivation being ease of computation. Error bounds for these approximate distributions have not been given.\n\n### Results\n\nWe give bounds on the probability distribution of the number of gaps when a circle is covered by fragments of fixed size. The absolute error in the approximation is typically on the order of 0.1% at 10× coverage depth. The method can be applied to coverage problems on the interval, including edge effects, and applications are given to metagenomic libraries and shotgun sequencing.\n\n## Background\n\nThe question of how a circle becomes covered when random arcs are marked off has arisen repeatedly in bioinformatics. As an example, a prokaryotic chromosome is typically circular and the clones extracted from it for genomic libraries or shotgun sequencing projects are randomly positioned arcs. The number of uncovered gaps is of particular interest: a genomic library ideally has no gaps, while one might seek to stop the undirected part of a shotgun sequencing project when a small number of gaps remain (we call this the 'stopping problem'). Coverage problems also arise in the culture-independent methods of metagenomics, since the number of clones coming from each genome in a mixed community is random. Accordingly, the question of the number of gaps has been treated by many authors, in both mathematical and biological contexts.\n\nWe refer the reader to for a review of the mathematical literature on circle covering problems and the exact distribution of the number of gaps when all arcs are of equal length. Driven by practical considerations, approximate distributions for the number of gaps have been given in the genomics literature: see for example . These approximate distributions are easier to compute than the exact distributions. Some address modified coverage problems with particular biological relevance, such as the 'edge effects' which arise when certain arc positions cannot occur. Bounds for the probability of completely covering the circle are given in , but to the knowledge of the authors no bounds have been given for the distribution of the number of gaps. In this paper we give bounds for the probability distribution of the number of gaps in circle covering problems. The method can be applied to coverage problems on the interval, including edge effects. Applications are given to metagenomic libraries and the stopping problem in shotgun sequencing.\n\n## Results\n\nProposition 1 Suppose that n arcs, each of length s, are placed uniformly and independently at random on a circle of circumference 1. Then the number of gaps has approximately the Poisson distribution with parameter m = n(1 - s)n-1. The error in the approximation is given in Proposition 2.\n\nProposition 2 In the setting of Proposition 1, let W denote the number of gaps and let Y denote a Poisson random variable with parameter m. Then for any nonnegative integer w,\n\n|P(Ww) - P(Yw)| ≤ ε\n\nwhere\n\n$ε = n 2 ( 1 − s ) 2 ( n − 1 ) − n ( n − 1 ) ( 1 − 2 s ) + n − 1 max ( 1 , m ) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWF1oqzcqGH9aqpdaWcaaqaaiabd6gaUnaaCaaaleqabaGaeGOmaidaaOGaeiikaGIaeGymaeJaeyOeI0Iaem4CamNaeiykaKYaaWbaaSqabeaacqaIYaGmcqGGOaakcqWGUbGBcqGHsislcqaIXaqmcqGGPaqkaaGccqGHsislcqWGUbGBcqGGOaakcqWGUbGBcqGHsislcqaIXaqmcqGGPaqkcqGGOaakcqaIXaqmcqGHsislcqaIYaGmcqWGZbWCcqGGPaqkdaqhaaWcbaGaey4kaScabaGaemOBa4MaeyOeI0IaeGymaedaaaGcbaGagiyBa0MaeiyyaeMaeiiEaGNaeiikaGIaeGymaeJaeiilaWIaemyBa0MaeiykaKcaaaaa@57B5@$\n\nand x+ = max(x, 0).\n\nCorollary 1 In the setting of Propositions 1 and 2, the probability that the circle is completely covered is approximately e-m and the absolute error in this approximation is at most ε.\n\nProposition 3 Suppose that N arcs, each of length S, are placed uniformly and independently at random on an interval of length 1 (so that the whole of each arc lies on the interval). Then the number of gaps (excluding the gap at each end) has approximately the Poisson distribution with parameter m = n(1 - s)n-1, where s = $S 1 − S MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaiabdofatbqaaiabigdaXiabgkHiTiabdofatbaaaaa@30F7@$and n = N - 1. The error in the approximation is just as in Proposition 2, again taking s = $S 1 − S MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaiabdofatbqaaiabigdaXiabgkHiTiabdofatbaaaaa@30F7@$and n = N - 1.\n\nCorollary 2 In the setting of Proposition 3, the probability that no gaps exist except end gaps of length at most d is approximately e -M , where\n\n$M = ( N − 1 ) ( 1 − 2 S 1 − S ) N + 2 ( 1 − S − d 1 − S ) N MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWGnbqtcqGH9aqpcqGGOaakcqWGobGtcqGHsislcqaIXaqmcqGGPaqkdaqadaqaamaalaaabaGaeGymaeJaeyOeI0IaeGOmaiJaem4uamfabaGaeGymaeJaeyOeI0Iaem4uamfaaaGaayjkaiaawMcaamaaCaaaleqabaGaemOta4eaaOGaey4kaSIaeGOmaiZaaeWaaeaadaWcaaqaaiabigdaXiabgkHiTiabdofatjabgkHiTiabdsgaKbqaaiabigdaXiabgkHiTiabdofatbaaaiaawIcacaGLPaaadaahaaWcbeqaaiabd6eaobaaaaa@4A9D@$\n\nand the absolute error in this approximation is at most\n\n$M 2 − 2 ( 1 − S − 2 d 1 − S ) + N − 4 ( N − 1 ) ( 1 − 2 S − d 1 − S ) + N − ( N − 1 ) ( N − 2 ) ( 1 − 3 S 1 − S ) + N MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@66DC@$\n\n## Discussion\n\nWe have given bounds on the probability distribution of the number of uncovered gaps when arcs of fixed length are placed randomly on a circle or interval. As discussed in , one motivation for such approximations is the issue of computational overflow arising when the exact solution is applied. Typically they involve simple, well-known probability distributions and this aids both computation and further mathematical analysis. Our own motivation in beginning this work was seeing certain quite poor approximations used in practice, both locally and in the literature. For a cautionary example we take s = 10-2 and n = 750, values which arise when 0.3% of the clones in a metagenomic library consisting of 2.5 × 105 40-kilobase fosmid inserts are from the genome of interest, and the genome of interest has length 4 megabases. Here ad-hoc approximations using elementary probability theory can indicate a 95% probability that the library completely covers that genome, while the true probability is 66% (neglecting biologically related bias; calculations are to the nearest integer and are not given).\n\nWe stress the similarity between our approximations and those already given in the literature. In [2, 3, 7] simplifying assumptions are made which give a binomial distribution for the number of gaps. Our approximation is a Poisson distribution, and it is well known in probability theory that the binomial distribution converges to the Poisson distribution in certain limits; one such convergence is proved in . Indeed, these different approximations are generally numerically close. The contribution of the present paper is therefore to provide an approximation with error bounds.\n\nWe refer the reader to for a simple modification to coverage problems when a certain minimum overlap is required between arcs, and give a worked example in Additional file 2. Note that in the expected number of gaps is calculated approximately; as shown in Additional file 1, the exact value for the expected number of gaps is m = n(1 - s)n-1. This is the parameter of the Poisson distribution in Proposition 1. Our approximating Poisson distribution therefore has the same expectation as the exact distribution, although its variance differs (the exact variance of the number of gaps is given in Additional file 1).\n\nOur results may be applied to the stopping problem. Suppose we desire p% probability that no more than w gaps remain at the end of the undirected part of a shotgun sequencing project. By inverting Proposition 1 (see Additional file 2 for details) we obtain an approximate value for the number of clones, and hence coverage depth, required. Here again, our contribution is not the estimate but rather the lower bound given by Proposition 2 for the probability that no more than w gaps remain (neglecting biologically related bias). Other solutions to the stopping problem have been proposed: see for example and for alternative strategies and further discussion of the stopping problem.\n\nThe practical relevance of our approximations clearly depends on the size of the error bound. Figure 1 plots the error bound against coverage depth for arc lengths s = 10-1, 10-2, 10-3 and 10-7 (the curves for s = 10-4, 10-5 and 10-6 are almost indistinguishable from the s = 10-7 curve at this scale). We emphasise that s is the relative arc length – so for genomic applications, s is the actual arc length divided by the length of the genome. These arc lengths are intended to represent the full range of typical genomic projects: for example the smallest, s = 10-7, would correspond to covering the largest known eukaryote genome, the amoeba Chaos chaos , with 400 kilobase bacterial artificial chromosome (BAC) inserts. A lookup table of relative arc lengths for recent shotgun sequencing projects is given in . It can be seen that for these relative arc lengths, error bounds on the order of 0.1% are achieved at 10× coverage. Further discussion of Figure 1 is given in Additional file 1; it should also be remarked that the error bound at 5× coverage is considerably larger. For the particular experimental parameters relevant to the user, the spreadsheet in Additional file 3 may be used to obtain values for the approximation and error bounds.\n\nThe validity of coverage problems in general for genomic applications depends of course on the extent to which they capture the actual problem. For example, a given genome might contain one gene which is toxic to the E. coli host in a BAC library. Since library fragments will not contain this gene, the corresponding coverage problem is on the interval rather than the circle. With a priori knowledge of an unclonable region we may therefore apply Proposition 3 rather than Proposition 1; without such information we may choose either to neglect this effect or to model it (for example using Propositions 1 and 3 and conditional probability). Pathological cases certainly exist, for example the highly repetitive maize genome for which as many as 80% of arc positions may not be cloned . The approach should therefore be chosen using the best available information. Certain other biases are not so dependent on the particular target genome, for example the inevitable small variations in clone length and position bias. These have been discussed in [3, 4, 10] using empirical data and simulations and the consensus is that they may be neglected. The interested mathematical reader may check that by first conditioning on the fragment lengths, our method gives a lower bound for P(Ww) in Proposition 2 when the arc lengths are random, although we do not pursue this.\n\nAnother modelling issue arises in metagenomics, which is the culture-independent study of a mixed community of genomes. In a metagenomic library the number of clones n from a genome in the community is random, having a binomial distribution (in the absence of bias). If the composition of the community is known then, from the central limit theorem of probability, n is well approximated by its average and this value may be used in Proposition 1 (see Additional file 2). Further, since the distribution of n is concentrated at a few values around its mean, it is typically computationally inexpensive to obtain satisfactory bounds for the distribution of the number of gaps by an application of conditional probability.\n\n## References\n\n1. 1.\n\nSolomon H: Geometric probability. Philadelphia, Pa.: Society for Industrial and Applied Mathematics; 1978.\n\n2. 2.\n\nLander ES, Waterman MS: Genomic mapping by fingerprinting random clones: a mathematical analysis. Genomics 1988, 2(3):231–239. 10.1016/0888-7543(88)90007-9\n\n3. 3.\n\nRoach JC: Random subcloning. Genome Research 1995, 5: 464–473.\n\n4. 4.\n\nWendl MC, Waterston RH: Generalized gap model for bacterial artificial chromosome clone fingerprint mapping and shotgun sequencing. Genome Research 2002, 12: 1943–1949. 10.1101/gr.655102\n\n5. 5.\n\nShepp LA: Covering the circle with random arcs. Israel J Math 1972, 11: 328–345.\n\n6. 6.\n\nWendl MC: Occupancy modeling of coverage distribution for whole genome shotgun sequencing. Bulletin of Mathematical Biology 2006, 68: 179–196. 10.1007/s11538-005-9021-4\n\n7. 7.\n\nWendl MC, Barbazuk WB: Extension of Lander-Waterman theory for sequencing filtered DNA libraries. BMC Bioinformatics 2005, 6: 245. 10.1186/1471-2105-6-245\n\n8. 8.\n\nSiegel AF, Holst L: Covering the circle with random arcs of random sizes. J Appl Probab 1982, 19(2):373–381. 10.2307/3213488\n\n9. 9.\n\nFriz CT: The biochemical composition of the free-living amoebae Chaos chaos, Amoeba dubia, and Amoeba proteus. Comparative Biochemistry and Physiology 1968, 26: 81–90. 10.1016/0010-406X(68)90314-9\n\n10. 10.\n\nKupfer K, Smith MW, Quackenbush J, Evans GA: Physical mapping of complex genomes by sampled sequencing: a theoretical analysis. Genomics 1995, 27: 90–100. 10.1006/geno.1995.1010\n\n## Acknowledgements\n\nThe research of all the authors except JRM was supported by Science Foundation Ireland under Grant Number SFI 04/RP1/I512. JRM's work was supported by the Irish Government under the National Development Plan (2000–2006) and Science Foundation Ireland. The authors would like to thank Andrew Haugh for providing numerical simulations during this project and Professor Armand Makowski for the seminar which was its inspiration, and the reviewers for their helpful comments.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to John Moriarty.\n\n### Authors' contributions\n\nJRM brought the problem to the attention of the other authors and posed the worked examples [see Additional file 2]. AM wrote the mathematical proofs [see Additional file 1]. JM supervised the study and wrote the remainder. All authors read and approved the final manuscript.\n\n## Electronic supplementary material\n\n### Methods\n\nAdditional File 1: . Mathematical proofs. (PDF 126 KB)\n\n### Practical application to genomics and metagenomics\n\nAdditional File 2: . Worked example applications in genomics and metagenomics. (PDF 68 KB)\n\n### Calculator\n\nAdditional File 3: . A calculator implementing the formulas in Propositions 1 and 2. (XLS 22 KB)\n\n## Authors’ original submitted files for images\n\nBelow are the links to the authors’ original submitted files for images.\n\n## Rights and permissions\n\nReprints and Permissions\n\nMoriarty, J., Marchesi, J.R. & Metcalfe, A. Bounds on the distribution of the number of gaps when circles and lines are covered by fragments: Theory and practical application to genomic and metagenomic projects. BMC Bioinformatics 8, 70 (2007). https://doi.org/10.1186/1471-2105-8-70", null, "" ]

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[ "# Old seminars, 1990–2014\n\nSeminars 2014:\n\nDecember 4, 2014, 10:00\nGerard Freixas (Jussieu, Paris) : Reciprocity laws on Riemann surfaces, connections on Deligne pairings and holomorphic torsio.\nAbstract:\nIn this talk I will recall classical reciprocity laws on Riemann surfaces and explain how they translate into the language of connections on Deligne pairings. I will give a couple of applications. One explains a construction of Hitchin on hyperkähler varieties of lambda-connections, the other explains a result of Fay on holomorphic extensions of analytic torsion on spaces of characters of the fundamental group of a Riemann surface. The contents of this talk are based on joint work with Richard Wentworth (University of Maryland).\n\nDecember 3, 2014, 10:00\n\nEleonora Di Nezza (Imperial College London):Regularizing properties and uniqueness of the Kaehler-Ricci flow\n\nAbstract: Let X be a compact Kaehler manifold. I will show that the Kaehler-Ricci flow, as well as its twisted version, can be run from an arbitrary positive closed current, and that it is immediately smooth in a Zariski open subset of X. Moreover, if the initialdata has positive Lelong number we indeed have propagation of singularities for short time.Finally, I will prove a uniqueness result in the case of zero Lelong numbers.\n(This is a joint work with Chinh Lu)\n\nNovember 26, 2014, 10:00\nChinh H. Lu (Göteborg): The completion of the space of Kähler metrics.\nAbstract: We will talk about recent results of T. Darvas concerning the completion of the space of Kähler metrics\n\nJune 18, 2014, 10:00\nIvan Cheltsov (Edinburgh): Cylinders in del Pezzo surfaces.\n\nAbstract: For an ample divisor H on a variety V, an H-polar cylinder\nin V is an open ruled affine subset whose complement is a support of an\neffective Q-divisor that is Q-rationally equivalent to H.\nIn the case when V is a Fano variety and H is its anticanonical divisor,\nthis notion links together affine, birational and Kahler geometries.\nI prove existence and non-existence of H-polar cylinders in smooth and\nmildly singular (with at most du Val singularities) del Pezzo surfaces\nfor different ample divisors H. In particular, I will answer an old\nquestion of Zaidenberg and Flenner about additive group actions on the cubic Fermat affine\nthreefold cone. This is a joint work with Park and Won.\n\nMay 21, 2014, 10:00\nBo Berndtsson: Yet another proof of the Ohsawa-Takegoshi theorem.\nAbstract: A very simple proof of  optimal lower bounds for the Bergman kernel was given recently by Blocki and Lempert. I will show how their method generalizes to at least one version of the Ohsawa-Takegoshi theorem.\n\nMay 15, 2014, 10:00\nRobert Berman (Göteborg) : A primer on toric manifolds – from the analytic point of view\nAbstract: In this talk I will give an elementary and “hands-on” introduction to toric manifolds, from an analytic point of view. More precisely, I will explain how to build a polarized compact complex manifold X from a given polytope P (satisfying the Delzant conditions). Topologically, the manifold X is fibered over the polytope P in real tori, in much the same way as the Riemann sphere may be fibered over the unit-interval by embedding it in R^3 and projecting onto a coordinate axis.\n\nApril 30, 2014, 10:00\nChinh H. Lu (Göteborg): Complex Hessian equations on compact Kähler manifolds.\nAbstract:  The complex Hessian equation is an interpolation between the Laplace and complex Monge-Ampère equation. An analogous version of the Calabi-Yau equation has been recently solved by Dinew and Kolodziej.\nIn this talk we first survey some known results about existence and regularity of solutions in the non-degenerate case. We then discuss how to solve the degenerate equation by a variational approach due to Berman-Boucksom-Guedj-Zeriahi. The key (and new) point is a regularization result for m-subharmonic functions. This is joint work with Van-Dong Nguyen (Ho Chi Minh city University of Pedagogy).\n\nApril 23, 2014, 10:00\nYanir Rubinstein (University of Maryland): Logzooet - beyond elephants\nAbstract: This is the second talk in the series. It should be of interest to differential/algebraic/complex geometers\n\nThe compact four-manifolds that admit a Kähler metric with positive Ricci curvature have been classified in the 19th century: they come in 10 families. In analogy with conical Riemann surfaces (e.g., football, teardrop) and hyperbolic 3-folds with a cone singularity along a link appearing in Thurston's program, one may consider 4-folds with a Kähler metric having \"edge singularities\", namely admitting a 2-dimensional cone singularity transverse to an immersed minimal surface, a complex edge'. What are all the pairs (4-fold, immersed surface) that admit a Kähler metric with positive Ricci curvature away from the edge? In joint work with I. Cheltsov (Edinburgh) we classify all such pairs under some assumptions. These now come in infinitely-many families and we then pose the \"Calabi problem\" for these pairs: when do they admit Kähler-Einstein edge metrics? This problem is far from being solved, even in this low dimension, but we report on some initial progress: some understanding of the non-existence part of the conjecture, as well as several existence results.​\n\nApril 9, 2014, 10:00\nYanir Rubinstein (University of Maryland): Logzooet\nAbstract: This is the first talk out of two and will require little prerequisites.\nThe compact four-manifolds that admit a Kähler metric with positive Ricci curvature have been classified in the 19th century: they come in 10 families. In analogy with conical Riemann surfaces (e.g., football, teardrop) and hyperbolic 3-folds with a cone singularity along a link appearing in Thurston's program, one may consider 4-folds with a Kähler metric having \"edge singularities\", namely admitting a 2-dimensional cone singularity transverse to an immersed minimal surface, a complex edge'. What are all the pairs (4-fold, immersed surface) that admit a Kähler metric with positive Ricci curvature away from the edge? In joint work with I. Cheltsov (Edinburgh) we classify all such pairs under some assumptions. These now come in infinitely-many families and we then pose the \"Calabi problem\" for these pairs: when do they admit Kähler-Einstein edge metrics? This problem is far from being solved, even in this low dimension, but we report on some initial progress: some understanding of the non-existence part of the conjecture, as well as several existence results.\n\nApril 2, 2014, 10:00\nChoi Young-Jun (KIAS): Variations of Kähler-Einstein metrics on strongly pseudoconvex domains\nAbstract: By a celebrated theorem of Cheng and Yau, every bounded strongly pseudo-convex domain with smooth boundary admits a unique complete Kähler-Einstein metric. In this talk, we discuss the plurisubharmonicity of variations of the Kähler- Einstein metrics of strongly pseudoconvex domains.\n\nMarch 19, 2014, 10.00\nNikolay Shcherbina (Wuppertal University): Bounded plurisubharmonic functions, cores and Liouville type property.\nAbstract. For a domain G in a complex manifold M of dimension n let F be the family of all bounded above plurisubharmonic functions. Define a notion of the core c(G) of G as:\n\nc(G) = {z \\in G: rank L(z,f) < n for all functions f \\in F},\n\nwhere L(z,f) is the Levi form of f at z. The main purpose of the talk is to discuss a Liouville type property of c(G), namely, to investigate if it is true or not that every function g from F has to be a constant on each connected component of c(G).​\n\nMarch 12, 2014, 10.00\nRobert Berman (Göteborg): The convexity of the K-energy on the space of all Kähler metrics and applications\nAbstract: The K-energy is a functional on the space H of all Kähler metrics in a given cohomology class and was  introduced by Mabuchi in the 80's. Its critical points are Kähler metrics with constant scalar curvature. As shown by Mabuchi the K-energy is convex along smooth geodesics in the space H. This was result was later put into a the framework of geometric invariant theory in infinite dimensions, by Donaldson.   However, when studying the geometry of H one is, in general, forced to work with a weaker notion of geodesics introduced by Chen (due to lack of a higher order regularity theory for the PDE describing the corresponding geodesic equation). In this talk, which is based on a joint work with Bo Berndtsson, it will be shown that the K-energy remains convex along weak geodesics, thus confirming a conjecture by Chen. Some applications in Kähler geometry will also be briefly discussed.\n\nFebruary 26, 2014 9.30\nBo Berndtsson (Göteborg): Complex interpolation of real norms\nAbstract: The method of complex interpolation is a way to, given a family of complex Banach norms, find intermediate  or averaged norms. I will describe an extension of this to real norms and also mention the relation to the boundary value problem for geodesics in the space of metrics on a line bundle. This is joint work with Bo'az Klartag, Dario Cordero and Yanir Rubinstein.​\n\nFebruary 12, 2014, 10.00\nXu Wang (Tongji University, Shanghai): Variation of the Bergman kernels of pseudoconvex domains.\nAbstract: We shall give a variational formula of the full Bergman kernels associated to smoothly bounded strongly pseudoconvex domains. An equivalent criterion for the triviality of holomorphic motions of planar domains in terms of the Bergman kernel is given as an application.\n\nJanuary 27, 2014, 10.00\n\nHåkan Samuelsson Kalm (Göteborg): On analytic structure in maximal ideal spaces\nAbstract: John Wermer's classical maximality theorem says the following: Let $f$ be a continuous function on the unit circle $b\\Delta$, where $\\Delta \\subset \\mathbb{C}$ is the unit disk. Then, either $f$ is the boundary values of a holomorphic function on $\\Delta$ or the uniform algebra generated by $z$ and $f$ on the unit circle equals the algebra of all continuous functions on the unit circle. I will discuss to what extent Wermer's maximality theorem extends to the setting of several complex variables. In particular, we will answer a question posed by Lee Stout concerning the presence of analytic structure for a uniform algebra whose maximal ideal space is a manifold. This is joint work with A. Izzo (Bowling Green) and E. F. Wold (Oslo).\n\nPublished: Fri 26 Apr 2019." ]

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[ "# Conductor is cut into four equal parts. What will be the new resistivity\n\n(i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity ‘ρ’. Now this conductor is cut into four equal parts. What will be the new resistivity of each of these parts? Why?\n\n(ii) Find the resistance if all of these parts are connected in:\n\n1. Parallel\n2. Series\n\n(iii) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which combination will consume more power and why?\n\n(i) Resistivity will not change as it depends on the nature of the material of the conductor.\n\n(ii) The length of each part becomes L/4. ρ, A constant.\n\nR = ρL/A\n\nResistance of each part = Rpart = (ρL/4)/A = R/4\n\n(a) In parallel, Reqv = R/16 Ω\n\n(b) In series, Reqv = R Ω\n\n(iii) P = V2/R\n\nIf Reqv is less, power consumed will be more.\n\nIn the given case, Reqv is lesser in the parallel and power consumed will be more." ]

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[ "# balancing chemical equations calculator\n\nInstructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. knop. The balancer is case sensitive. Examples of Equations you can enter: … Balancing chemical equations calculator works in sensible manners as there is artificial intelligence is doing work. This chemical equation balancer can help you to balance an unbalanced equation. 2H2 + O2 = 2H2O 3. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. Balance Chemical Equation - Online Balancer. Balancing any chemical equations is made simple with this chemical formula balancer alias calculator. The balanced equation will appear above. There is an algorithm used named Gauss-Jordan elimination but slightly modified. Enter a chemical equation to balance [Chemical Equations Examples: H 2 + O 2 = H 2 O Na 2 + … Enter the equation directly into the Balancing Chemical Equations Calculator to balance the given chemical equations. To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. Use uppercase for the first character in the element and lowercase for the second character. Use this Calculator to balance Chemistry Equations. This balancer can also help you check whether the equation is balanced or not, thus you may edit the equation and check it's balance. The answer will appear below; Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide This online chemistry calculator balances equations of chemical reactions person_outline Timur schedule 2017-04-22 16:48:34 Articles that describe this calculator Voer een reactievergelijking in om te balanceren: Instructies over het balanceren van chemische vergelijkingen: Voer een vergelijking van een chemische reactie in en druk op de 'Balance!' You can also enter the equations by clicking the elements in the table given in the chemical equation balancer. H2 + O2 = H2O 2. Following are some equation input format examples: 1. Het antwoord verschijnt eronder. balancing equations calculator performs as to balance the given equation, it calculates the coefficients." ]

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[ "# 1: Introduction to Statistics\n\n•", null, "• Contributed by David Lane\n• Associate Professor (Psychology, Statistics, and Management) at Rice University\n\nThis first chapter begins by discussing what statistics are and why the study of statistics is important. Subsequent sections cover a variety of topics all basic to the study of statistics. One theme common to all of these sections is that they cover concepts and ideas important for other chapters in the book.\n\n• 1.1: What are Statistics?\nStatistics include numerical facts and figures, but also involves math and relies upon calculations of numbers. It also relies heavily on how the numbers are chosen and how the statistics are interpreted.\n• 1.2: Importance of Statistics\nIt is important to properly evaluate data and claims that bombard us every day. If you cannot distinguish good from faulty reasoning, then you are vulnerable to manipulation and to decisions that are not in your best interest. Statistics provides tools that you need in order to react intelligently to information you hear or read. In this sense, statistics is one of the most important things that you can study.\n• 1.3: Descriptive Statistics\nDescriptive statistics are numbers that are used to summarize and describe data. The word \"data\" refers to the information that has been collected from an experiment, a survey, a historical record, etc. Descriptive statistics are just descriptive. They do not involve generalizing beyond the data at hand. Generalizing from our data to another set of cases is the business of inferential statistics.\n• 1.4: Inferential Statistics\nIn statistics, we often rely on a sample --- that is, a small subset of a larger set of data --- to draw inferences about the larger set. The larger set is known as the population from which the sample is drawn.\n• 1.5: Sampling Demonstration\nThis demonstration is used to teach students how to distinguish between simple random sampling and stratified sampling and how often random and stratified sampling give exactly the same result.\n• 1.6: Variables\nVariables are properties or characteristics of some event, object, or person that can take on different values or amounts (as opposed to constants such as π that do not vary). When conducting research, experimenters often manipulate variables. When a variable is manipulated by an experimenter, it is called an independent variable. The experiment seeks to determine the effect of the independent variable on a dependent variable.\n• 1.7: Percentiles\nA test score in and of itself is usually difficult to interpret. For example, if you learned that your score on a measure of shyness was 35 out of a possible 50, you would have little idea how shy you are compared to other people. More relevant is the percentage of people with lower shyness scores than yours. This percentage is called a percentile.\n• 1.8: Levels of Measurement\nBefore we can conduct a statistical analysis, we need to measure our dependent variable. Exactly how the measurement is carried out depends on the type of variable involved in the analysis. Different types are measured differently. To measure the time taken to respond to a stimulus, you might use a stop watch. Stop watches are of no use, of course, when it comes to measuring someone's attitude towards a political candidate.\n• 1.9: Measurements\nThis is a demonstration of a very complex issue. Experts in the field disagree on how to interpret differences on an ordinal scale, so do not be discouraged if it takes you a while to catch on. In this demonstration you will explore the relationship between interval and ordinal scales. The demonstration is based on two brands of baked goods.\n• 1.10: Distributions\nDefine \"distribution\" Interpret a frequency distribution Distinguish between a frequency distribution and a probability distribution Construct a grouped frequency distribution for a continuous variable\n• 1.11: Summation Notation\nMany statistical formulas involve summing numbers. Fortunately there is a convenient notation for expressing summation. This section covers the basics of this summation notation.\n• 1.12: Linear Transformations\nOften it is necessary to transform data from one measurement scale to another. For example, you might want to convert height measured in feet to height measured in inches.\n• 1.13: Logarithms\nThe log transformation reduces positive skew. This can be valuable both for making the data more interpretable and for helping to meet the assumptions of inferential statistics.\n• 1.14: Statistical Literacy\n• 1.E: Introduction to Statistics (Exercises)" ]

[ null, "https://stats.libretexts.org/@api/deki/files/1247/Lane.jpg", null ]

[ "", null, "Скачать презентацию Approximation Algorithms for Non-Uniform Buy-at-Bulk Network Design Problems\n\n51a75716159298372f28a1c9df64335a.ppt\n\n• Количество слайдов: 30", null, "Approximation Algorithms for Non-Uniform Buy-at-Bulk Network Design Problems Mohammad. Taghi Hajiaghayi Carnegie Mellon University Joint work with Chandra Chekuri (UIUC) Guy Kortsarz (Rutgers, Camden) Mohammad R. Salavatipour (University of Alberta)", null, "Motivation Suppose we are given a network and some nodes have to be connected by cables Each cable has a cost (installation or cost of usage) Question: Install cables satisfying demands at minimum cost 5 3 14 8 21 9 10 11 16 21 7 27 12 This is the well-studied Steiner forest problem and is NP-hard 2", null, "Motivation (cont’d) Consider buying bandwidth to meet demands between pairs of nodes. The cost of buying bandwidth satisfy economies of scale The capacity on a link can be purchased at discrete units: Costs will be: Where 3", null, "Motivation (cont’d) So if you buy at bulk you save More generally, we have a non-decreasing monotone concave (or even sub-additive) function where f (b) is the minimum cost of cables with bandwidth b. Question: Given a set of bandwidth demands between nodes, install sufficient capacities at minimum total cost bandwidth 4", null, "Motivation (cont’d) The previous problem is equivalent to the following problem: There a set of pairs to be connected For each possible cable connection e we can: Buy it at b(e): and have unlimited use Rent it at r(e): and pay for each unit of flow A feasible solution: buy and/or rent some edges to connect every si to ti. Goal: minimize the total cost 5", null, "Motivation (cont’d) If this edge is bought its contribution to total cost is 14. 14 If this edge is rented, its contribution to total cost is 2 x 3=6 3 10 Total cost is: where f(e) is the number of paths going over e. 6", null, "Cost-Distance These problems are also known as cost-distance problems: cost function length function Also a set of pairs of nodes each with a demand for every i Feasible solution: a set s. t. all pairs are connected in 7", null, "Cost-Distance (cont’d) The cost of the solution is: where is the shortest path in The cost is the start-up cost and is the per-use cost (length). Goal: minimize total cost. 8", null, "Multicommodity Buy At Bulk The problem is called Multi-Commodity Buy-at-Bulk (MC-BB) Note that the solution may have cycles 5 11 8 12 21 9", null, "Special Cases If all si (sources) are equal we have the single-source case (SS-BB) If the cost and length functions on the edges are all the same, i. e. each edge e has cost c + l f(e) for constants c, l : Uniform-case Single-source 21 8 12 5 11 10", null, "Previous Work Formally introduced by F. S. Salman, J. Cheriyan, R. Ravi and S. Subramanian, 1997 O(log n) approximation for the uniform case, i. e. each edge e has cost c+l f(e) for some fixed constants c, l (B. Awerbuch and Y. Azar, 1997; Y. Bartal, 1998) O(log n) randomized approximation for the single-sink case: A. Meyerson, K. Munagala and S. Plotkin, 2000 O(log n) deterministic approximation for the single-sink case: C. Chekuri, S. Khanna and S. Naor, 2001 11", null, "Hardness Results for Buy-at-Bulk Problems Hardness of Ω(log n) for the singlesink case J. Chuzhoy, A. Gupta, J. Naor and A. Sinha, 2005 Ω(log 1/2 - n) in general M. Andrews 2004, unless NP ZPTIME(npolylog(n)) 12", null, "Algorithms for Special Cases Steiner Forest A. Agrawal, P. Klein and R. Ravi, 1991 M. X. Goemans and D. P. Williamson, 1995 Single source S. Guha, A. Meyerson and K. Munagala , 2001 K. Talwar, 2002 A. Gupta, A. Kumar and T. Roughgarden, 2002 A. Goel and D. Estrin, 2003 13", null, "Multicommodity Buy at Bulk Multicommodity Uniform Case: Y. Azar and B. Awerbuch, 1997 Y. Bartal, 1998 A. Gupta, A. Kumar, M. Pal and T. Roughgarden, 2003 The only known approximation for the general case M. Charikar, A. Karagiozova, 2005. The ratio is: exp( O(( log D )1/2 )) 14", null, "Our Main Result Theorem: If h is the number of pairs of si, ti then there is a polytime algorithm with approximation ratio O(log 4 h). For simplicity we focus on the unit-demand case (i. e. di=1 for all i’s) and we present O(log 5 n loglog n). 15", null, "Overview of the Algorithm The algorithm iteratively finds a partial solution connecting some of the residual pairs The new pairs are then removed from the set; repeat until all pairs are connected (routed) Density of a partial solution = cost of the partial solution # of new pairs routed The algorithm tries to find low density partial solution at each iteration 16", null, "Overview of the Algorithm (cont’d) The density of each partial solution is at most Õ(log 4 n) (OPT / h') where OPT is the cost of optimum solution and h' is the number of unrouted pairs A simple analysis (like for set cover) shows: Total Cost Õ(log 4 n) OPT (1/n 2 + 1/(n 2 - 1) +…+ 1) Õ(log 5 n) OPT 17", null, "Structure of the Optimum How to compute a low-density partial solution? Prove the existence of low-density one with a very specific structure: junction-tree Junction-tree: given a set P of pairs, tree T rooted at r is a junction tree if It contains all pairs of P r For every pair si, ti P the path connecting them in T goes through r 18", null, "Structure of the Optimum (cont’d) So the pairs in a junction tree connect via the root We show there is always a partial solution with low density that is a junction tree Observation: If we know the pairs participating in a junction-tree it reduces to the single-source BB problem r Then we could use the O(log n) approximation of [MMP’ 00] 19", null, "Summary of the Algorithm So there are two main ingredients in the proof Theorem 2: There is always a partial solution that is a junction tree with density Õ (log 2 n) (OPT / h') Theorem 3: There is an O (log 2 n) approximation for the problem of finding lowest density junction tree (this is low density SS-BB). Corollary: We can find a partial solution with density Õ (log 4 n) (OPT / h') This implies an approximation Õ (log 5 n) for MC-BB. 20", null, "More Details of the Proof of Theorem 2: We want to show there is a partial solution that is a junction tree with density Õ (log 2 n) (OPT / h') Consider an optimum solution OPT. Let E* be the edge set of OPT, OPTc be its cost and OPTl be its length. By the result of Elkin, Emek, Spielman and Tang 2005 on probabilistic distribution on spanning trees and by loosing a factor Õ (log 2 n) on length, we can assume that E* is a forest T (WLOG we assume T is connected). 21", null, "More Details of the Proof of Theorem 2: From T we obtain a collection of rooted subtrees T 1, …, Ta such that any edge e of T is in at most O(log h) of the subtrees For every pair there is exactly one index i such that both vertices are in Ti; further the root of Ti is their least common ancestor The total cost of the junction trees is at most Õ (log 2 h) OPT (O (log h) OPTc + Õ (log 2 h) OPTl) Thus at least one of junction trees of T 1, …, Ta has the desired density of Õ (log 2 h) (OPT / h') 22", null, "More Details of the Proof of Theorem 2: Given T, we pick a centeroid r 1 (i. e. , largest remaining component has at most 2/3 |V(T)| vertices). Add tree T rooted at r 1 to the collection Remove r 1 from T and apply the procedure recursively to each of the resulting component Each pair is on exactly one subtree in the collection The depth of the recursion is in O (log h) 23", null, "Some Details of the Proof of Theorem 3: There is an approximation for finding lowest density junction tree. This is very similar to SS-BB except that we have to find a lowest density solution. Here we have to connect a subset of the pairs to the root r with lowest density (= cost of solution / # of pairs in sol). Let denote the set of paths from r to i. We formulate the problem as an IP and then consider the LP relaxation of the problem 24", null, "Some Details of the Proof of Theorem 3: We solve the LP by setting ys=yt for each pair (s, t), and then find a subset of nodes to solve the SS-BB We find a class of y among O (log n) classes of almost equal yi with maximum sum (loose a factor O (log n)) We use the O (log n) approx of [MMP, CKN] for SS-BB 25 (indeed it is the upper bound on integrality gap of the LP)", null, "Some Remarks: For the polynomially bounded demand case we can find low density junction-trees using a more refined region growing technique and also using a greedy algorithm (within O (log 4 n)) Hajiaghayi, Kortsarz and Salavatipour, ECCC 2006 The greedy algorithm is based on an algorithm for the k-shallow-light tree problem Hajiaghayi, Kortsarz and Salavatipour, APPROX 2006 There is a conjectured upper bound of O (log n) for distortion in embedding a graph metric into a probability distribution over its spanning tree (Alon, Karp, Peleg and West, 1991) If true, that would improve our approximation factor for arbitrary demands to O (log 4 n) 26", null, "Some Remarks (cont’d): Indeed, as suggested by Racke, our current approach can be applied via Bartal’s trees (and interestingly not FRT) to obtain an O(log h) factor instead of Õ (log 2 h) factor For a constant fraction of the pairs, we use strong diameter property which is true in Bartal’s construction It is more technical, but we can obtain factor O (log 4 h) for general demands (solving an open problem) 27", null, "Recent Extensions The result O(log 4 n) can be extended to the vertex-weighted case but requires some new ideas and some extra work [CHKS’ 07]. Especially we obtain the tight result O(log n) for the single-sink vertex-weighted case via LP rounding Also it needs some subtle change of vertex weights to edge weights in the junction tree lemma Also our results can be extended to the stochastic versions with non-uniform inflation (by loosing an extra factor O(log n)) [Gupta, Hajiaghayi, Kumar’ 06]. Some technique has been used in the Dial-a-Ride problem [Gupta, Hajiaghayi, Ravi, Nagarajan’ 06]. 28", null, "Open Problems There are still quite large gaps between upper bounds (approx alg) and lower bounds (hardness) For MC-BB: vs For SS-BB: vs It would be nice to upper bound the integrality gap for MC-BB. Emphasize on the conjecture of Alon, Karp, Peleg and West, 1991 29", null, "Thanks for your attention… 30", null, "" ]

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[ "#126be9 – hex color\n\n# #126be9 hex color information\n\nhex: #126BE9\nrgb(18, 107, 233)\nhsl(215, 86%, 49%)\n\nIn the additive RGB color model, color #126be9 (hexadecimal – hex triplet) has values of 18 (7% red), 107 (42% green) and 233 (91% blue). In the CMYK (subtractive color model), color #126be9 has values 92% cyan, 54% magenta, 0% yellow and 9% black. In the HSL model, it is represented by 215° hue, 86% saturation and 49% lightness. Win32 representation: DWORD COLORREF C=0x009eb621. Decimal value is 1207273.\n\n#126be9 is not web-safe (Netscape). Nearest web-safe color is #0066ff. #126be9 isn't X11 color, closest X11 color is RoyalBlue (#4169e1).\n\nRGB\nred = 18 (7%)\ngreen = 107 (42%)\nblue = 233 (91%)\nHSL\nhue = 0.598 (215°)\nsaturation = 0.857 (86%)\nlightness = 0.492 (49%)\nCMYK\ncyan = 0.923\nmagenta = 0.541\nyellow = 0.000\nblack = 0.086\n\nRelated colors (alternatives): Cerulean Blue, Azul, Bright Navy Blue, Blue (Crayola), Brandeis Blue, Clear Blue, True Blue, Gradus Blue, Azure, Electric Blue.\n\n## #126be9 color spaces, conversions\n\nThis table contains information about #126be9 color values in the most popular color spaces: RGB (additive color model), CMY, CMYK (subtractive color model), HSL, HSI, CIE XYZ. This data can be downloaded in JSON (126be9.json) and CSV (126be9.csv) formats.\n\nRGB is opposite of CMY (CMYK). CMY colors are complementary to RGB colors.\nR = 18 G = 107 B = 233\n7.06% 41.96% 91.37%\nC = 92.94% M = 58.04% Y = 8.63%\n0.92941176470588 0.58039215686275 0.086274509803922\nHSL (HSI) – hue, saturation, lightness/intensity\nH = 215° S = 85.66% L = 49.22%\n0.59767441860465 0.85657370517928 0.4921568627451\nHSV (HSB) – hue, saturation, brightness/value\nH = 215° S = 92.27% V = 91.37%\n0.59767441860465 0.92274678111588 0.91372549019608\nCMYK – cyan, magenta, yellow, key/black\nC = 92.27 M = 54.08 Y = 0.00 K = 8.63\n0.92274678111588 0.54077253218884 0 0.086274509803922\nCIE – International Commission on Illumination\nCIE XYZ X = 20.215 Y = 16.527 Z = 79.215\nCIE 1931 XYZ color space 20.215129552317 16.527180541374 79.215405953526\nYxy Y = 16.527 x = 0.174 y = 0.143\nCIE (x, y, Y) 16.527180541374 0.17433190512381 0.14252764804926\nHunter-Lab L = 40.65 a = 17.62 b = -87.07\nHunter 1948 color space 40.653635189703 17.615744004353 -87.071642291535\nCIELAB L = 47.66 a = 24.07 b = -70.12\nCIE 1976 (L*, a*, b*) 47.658672764246 24.066815219679 -70.122414621473\nCIELUV L = 47.66 u = -23.52 v = -107.95\nCIE 1976 (L*, u*, v*) 47.658672764246 -23.520745312819 -107.95283885979\nCIELCH L = 47.66 C = 74.14 H = 288.94\nCylindrical representation 47.658672764246 74.137471140874 288.94280715407\n\n## #126be9 in HTML and CSS\n\nThe following declarations are allowed (CSS 3.0):\n\n``` color: #126be9; color: rgb(18, 107, 233); color: rgb(7%, 42%, 91%); color: rgba(18, 107, 233, 1); color: hsl(215, 86%, 49%); color: hsla(215, 86%, 49%, 1); ```\n\nUsing #126be9 as text color and element's border color:\n\n``` div {   border: 4px solid #126be9;   color: #126be9; } ```\nfont color & element border color\n\n...as background color:\n\n``` div { background-color: #126be9; } ```\nbackground-color\n\n``` div {   box-shadow: 7px 7px 10px 0px rgba(18, 107, 233, 1);   text-shadow: #126be9 4px 4px 3px; }```\n``` div.linear-gradient { background: linear-gradient(45deg, #126be9, #126be9, #fff); } div.radial-gradient { background: radial-gradient(ellipse farthest-corner, #126be9 0%, #fff 90%); } ```" ]

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[ "Getting Image", null, "", null, "", null, "", null, "Register now for special offers", null, "+91\n\nHome\n\n>\n\nEnglish\n\n>\n\nClass 11\n\n>\n\nMaths\n\n>\n\nChapter\n\n>\n\nTrigonometric Functions\n\n>\n\nIf 2sec^2A-sec^4A-2cos e c^2A+...\n\n# If 2sec^2A-sec^4A-2cos e c^2A+cos e c^4A=(15)/4,t h e n tan A is equal 1//sqrt(2) (b) 1/2 (c) 1/2sqrt(2) (d) -1/(sqrt(2))", null, "Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke!", null, "Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.", null, "Transcript\n\nHindi question if to sec square A minus x power 4 - 2 cos square A + cos power 4 equal to 15 by 4 and 10 equals to ok to sex per a minus power 4A - 2 cos square A + cos power 4 a Kuchh to 15 by 45 value to sec squared minus cosec Square x square minus cosec square A + cosec square minus x square minus Cos power 4 minus x square x power 4 + cosec square A minus sin square A into cos square A + sec square which is equal to 15\n\n4 yahan se kya Banaye Ja sakte hain kaun sak square A minus sin square a common Lene theek hai hamare pass cosec square A + sin square A minus 2 minus b square minus x square is equals to 15 by 4 so cosec square A Jaega Tujhko mein le sakte hain isko photomultiplier Karenge se 2-4 inter main ho sex mere ko kaise Khel Sakte Court square A minus 1 minus x square is 10 square cos square x square + 1 into cosec Square x cos square A + 1 + x square x square + 1 - 2 = 215\n\nX 17 cancel out 1 + 1 is to 2 is equal to four times of court square A minus 10 square into cot square A + 10 square equal to 15 15 hamare pass kya karte a square minus a minus b into a + b 4 times of court power 4 minus 10 power 4A equals to equal to 15154 of a sakte hain 1 minus 10 power 8 14:45 10 power 4A power 444\n\n8k + 15 10 power 4 into 10 power minus 4 equal to zero equation ban gai hai following Mil Gaye 8444 10 power 4 a minus 1 into 10 Mai ko karna isko time we cannot Ke Humne To ki Tanki value chahie the Civil chahie Hamen time ok hatane ke liye Humne Kote Kote niche kya hai theek hai so isliye hamare pass Ganesh ko 10:00 Baje uski hi ho gaya tha 10 power 4A equals to 1 by 4 aur 10 power for equals to minus 4 Ke Khiladi 10 square kya Gaya se 10 square equals to plus minus 1 by 2 + 1 by 2 and 10\n\nsquare A + value OK good night image compressor to 10kb 10 plus minus 1 by root 2 to 10 51 bus stand to plus minus 1 by root 2 ok so sweet Anarchy value for GT 210 to be plus minus 1 by root ok given that to sec squared minus sec 44 44 A equal to 55 41010 equal to 1 plus minus 1 by root 2 minutes of India the correct option Mysore Management by root 2 and 1 by root ok shops in A and B are the correct option ok so this is the answer for this question\n\n## Related Videos\n\n22048\n\n36\n\n4.4 K\n\n4:48\nIf 2sec^2A-sec^4A-2cos e c^2A+cos e c^4A=(15)/4,t h e n tan A is equal 1//sqrt(2) (b) 1/2 (c) 1/2sqrt(2) (d) -1/(sqrt(2))\n642528930\n\n26\n\n6.0 K\n\n6:21\nIf 2sec^2A-sec^4A-2cos e c^2A+cos e c^4A=(15)/4,t h e n tan A is equal 1//sqrt(2) (b) 1/2 (c) 1/2sqrt(2) (d) -1/(sqrt(2))\n646267433\n\n12\n\n8.3 K\n\n5:56\nIf 2sec^2A-sec^4A-2cos e c^2A+cos e c^4A=(15)/4,then tan A is equal\n642566607\n\n24\n\n4.1 K\n\n2:27\nProve the following identities: cot^4A-1=cos e c^4A-2cos e c^2A\n642570232\n\n100\n\n4.9 K\n\n2:17\ncos^4A-sin^4A is equal to <br>(a)2cos^2A+1 <br>(b) 2cos^2A-1 <br>(c) 2sin^2A-1 <br>(d) 2sin^2A+1\n642578428\n\n9\n\n4.1 K\n\n5:21\nFor the principal values, evaluate each of the following: tan^(-1)sqrt(3)-sec^(-1)(-2)+cos e c^(-1)2/(sqrt(3)) 2sec^(-1)(2)-2cos e c^(-1)(-2)" ]

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[ "# 181 23.2 Faraday’s Law of Induction: Lenz’s Law\n\n### Summary\n\n• Calculate emf, current, and magnetic fields using Faraday’s Law.\n• Explain the physical results of Lenz’s Law\n\nFaraday’s experiments showed that the emf induced by a change in magnetic flux depends on only a few factors. First, emf is directly proportional to the change in flux $\\boldsymbol{\\Delta \\phi}$. Second, emf is greatest when the change in time $\\boldsymbol{\\Delta t}$ is smallest—that is, emf is inversely proportional to $\\boldsymbol{\\Delta t}$. Finally, if a coil has $\\boldsymbol{N}$ turns, an emf will be produced that is $\\boldsymbol{N}$ times greater than for a single coil, so that emf is directly proportional to $\\boldsymbol{N}$. The equation for the emf induced by a change in magnetic flux is\n\n$\\boldsymbol{\\textbf{emf} = -N}$ $\\boldsymbol{\\frac{\\Delta \\phi}{\\Delta t}}$\n\nThis relationship is known as Faraday’s law of induction. The units for emf are volts, as is usual.\n\nThe minus sign in Faraday’s law of induction is very important. The minus means that the emf creates a current I and magnetic field B that oppose the change in flux $\\boldsymbol{\\Delta \\phi}$ —this is known as Lenz’s law. The direction (given by the minus sign) of the emf is so important that it is called Lenz’s law after the Russian Heinrich Lenz (1804–1865), who, like Faraday and Henry,independently investigated aspects of induction. Faraday was aware of the direction, but Lenz stated it so clearly that he is credited for its discovery. (See Figure 1.)", null, "Figure 1. (a) When this bar magnet is thrust into the coil, the strength of the magnetic field increases in the coil. The current induced in the coil creates another field, in the opposite direction of the bar magnet’s to oppose the increase. This is one aspect of Lenz’s law—induction opposes any change in flux. (b) and (c) are two other situations. Verify for yourself that the direction of the induced Bcoil shown indeed opposes the change in flux and that the current direction shown is consistent with RHR-2.\n\n### Problem-Solving Strategy for Lenz’s Law\n\nTo use Lenz’s law to determine the directions of the induced magnetic fields, currents, and emfs:\n\n1. Make a sketch of the situation for use in visualizing and recording directions.\n2. Determine the direction of the magnetic field B.\n3. Determine whether the flux is increasing or decreasing.\n4. Now determine the direction of the induced magnetic field B. It opposes the change in flux by adding or subtracting from the original field.\n5. Use RHR-2 to determine the direction of the induced current I that is responsible for the induced magnetic field B.\n6. The direction (or polarity) of the induced emf will now drive a current in this direction and can be represented as current emerging from the positive terminal of the emf and returning to its negative terminal.\n\nFor practice, apply these steps to the situations shown in Figure 1 and to others that are part of the following text material.\n\n# Applications of Electromagnetic Induction\n\nThere are many applications of Faraday’s Law of induction, as we will explore in this chapter and others. At this juncture, let us mention several that have to do with data storage and magnetic fields. A very important application has to do with audio and video recording tapes. A plastic tape, coated with iron oxide, moves past a recording head. This recording head is basically a round iron ring about which is wrapped a coil of wire—an electromagnet (Figure 2). A signal in the form of a varying input current from a microphone or camera goes to the recording head. These signals (which are a function of the signal amplitude and frequency) produce varying magnetic fields at the recording head. As the tape moves past the recording head, the magnetic field orientations of the iron oxide molecules on the tape are changed thus recording the signal. In the playback mode, the magnetized tape is run past another head, similar in structure to the recording head. The different magnetic field orientations of the iron oxide molecules on the tape induces an emf in the coil of wire in the playback head. This signal then is sent to a loudspeaker or video player.\n\nSimilar principles apply to computer hard drives, except at a much faster rate. Here recordings are on a coated, spinning disk. Read heads historically were made to work on the principle of induction. However, the input information is carried in digital rather than analog form – a series of 0’s or 1’s are written upon the spinning hard drive. Today, most hard drive readout devices do not work on the principle of induction, but use a technique known as giant magnetoresistance. (The discovery that weak changes in a magnetic field in a thin film of iron and chromium could bring about much larger changes in electrical resistance was one of the first large successes of nanotechnology.) Another application of induction is found on the magnetic stripe on the back of your personal credit card as used at the grocery store or the ATM machine. This works on the same principle as the audio or video tape mentioned in the last paragraph in which a head reads personal information from your card.\n\nAnother application of electromagnetic induction is when electrical signals need to be transmitted across a barrier. Consider the cochlear implant shown below. Sound is picked up by a microphone on the outside of the skull and is used to set up a varying magnetic field. A current is induced in a receiver secured in the bone beneath the skin and transmitted to electrodes in the inner ear. Electromagnetic induction can be used in other instances where electric signals need to be conveyed across various media.", null, "Figure 3. Electromagnetic induction used in transmitting electric currents across mediums. The device on the baby’s head induces an electrical current in a receiver secured in the bone beneath the skin. (credit: Bjorn Knetsch)\n\nAnother contemporary area of research in which electromagnetic induction is being successfully implemented (and with substantial potential) is transcranial magnetic simulation. A host of disorders, including depression and hallucinations can be traced to irregular localized electrical activity in the brain. In transcranial magnetic stimulation, a rapidly varying and very localized magnetic field is placed close to certain sites identified in the brain. Weak electric currents are induced in the identified sites and can result in recovery of electrical functioning in the brain tissue.\n\nSleep apnea (“the cessation of breath”) affects both adults and infants (especially premature babies and it may be a cause of sudden infant deaths [SID]). In such individuals, breath can stop repeatedly during their sleep. A cessation of more than 20 seconds can be very dangerous. Stroke, heart failure, and tiredness are just some of the possible consequences for a person having sleep apnea. The concern in infants is the stopping of breath for these longer times. One type of monitor to alert parents when a child is not breathing uses electromagnetic induction. A wire wrapped around the infant’s chest has an alternating current running through it. The expansion and contraction of the infant’s chest as the infant breathes changes the area through the coil. A pickup coil located nearby has an alternating current induced in it due to the changing magnetic field of the initial wire. If the child stops breathing, there will be a change in the induced current, and so a parent can be alerted.\n\n### Making Connections: Conservation of Energy\n\nLenz’s law is a manifestation of the conservation of energy. The induced emf produces a current that opposes the change in flux, because a change in flux means a change in energy. Energy can enter or leave, but not instantaneously. Lenz’s law is a consequence. As the change begins, the law says induction opposes and, thus, slows the change. In fact, if the induced emf were in the same direction as the change in flux, there would be a positive feedback that would give us free energy from no apparent source—conservation of energy would be violated.\n\n### Example 1: Calculating Emf: How Great Is the Induced Emf?\n\nCalculate the magnitude of the induced emf when the magnet in Figure 1(a) is thrust into the coil, given the following information: the single loop coil has a radius of 6.00 cm and the average value of $\\boldsymbol{B \\;\\textbf{cos} \\;\\theta}$ (this is given, since the bar magnet’s field is complex) increases from 0.0500 T to 0.250 T in 0.100 s.\n\nStrategy\n\nTo find the magnitude of emf, we use Faraday’s law of induction as stated by $\\boldsymbol{\\textbf{emf} = -N \\frac{\\Delta \\phi}{\\Delta t}}$, but without the minus sign that indicates direction:\n\n$\\boldsymbol{\\textbf{emf} = N}$ $\\boldsymbol{\\frac{\\Delta \\phi}{\\Delta t}}$\n\nSolution\n\nWe are given that $\\boldsymbol{N = 1}$ and $\\boldsymbol{\\Delta t=0.100 \\;\\textbf{s}}$, but we must determine the change in flux $\\boldsymbol{\\Delta \\phi}$ before we can find emf. Since the area of the loop is fixed, we see that\n\n$\\boldsymbol{\\Delta \\phi = \\Delta (BA \\;\\textbf{cos} \\theta) = A \\Delta(B \\;\\textbf{cos} \\;\\theta)}$\n\nNow $\\boldsymbol{\\Delta (B \\;\\textbf{cos} \\;\\theta) = 0.200 \\;\\textbf{T}}$, since it was given that $\\boldsymbol{B \\;\\textbf{cos} \\;\\theta}$ changes from 0.0500 to 0.250 T. The area of the loop is $\\boldsymbol{A = \\pi r^2 = (3.14 \\dots)(0.060 \\;\\textbf{m})^2 = 1.13 \\times 10^{-2} \\;\\textbf{m}^2}$. Thus,\n\n$\\boldsymbol{\\Delta \\phi = (1.13 \\times 10^{-2} \\;\\textbf{m}^2)(0.200 \\;\\textbf{T})}.$\n\nEntering the determined values into the expression for emf gives\n\n$\\boldsymbol{\\textbf{Emf} =N}$ $\\boldsymbol{\\frac{\\Delta \\phi}{\\Delta t}}$ $\\boldsymbol{=}$ $\\boldsymbol{\\frac{(1.13 \\times 10^{-2} \\;\\textbf{m}^2)(0.200 \\;\\textbf{T})}{0.100 \\;\\textbf{s}}}$ $\\boldsymbol{= 22.6 \\; \\textbf{mV}}$\n\nDiscussion\n\nWhile this is an easily measured voltage, it is certainly not large enough for most practical applications. More loops in the coil, a stronger magnet, and faster movement make induction the practical source of voltages that it is.\n\n### PhET Explorations: Faraday’s Electromagnetic Lab\n\nPlay with a bar magnet and coils to learn about Faraday’s law. Move a bar magnet near one or two coils to make a light bulb glow. View the magnetic field lines. A meter shows the direction and magnitude of the current. View the magnetic field lines or use a meter to show the direction and magnitude of the current. You can also play with electromagnets, generators and transformers!\n\n# Section Summary\n\n• Faraday’s law of induction states that the emfinduced by a change in magnetic flux is\n\n$\\boldsymbol{\\textbf{emf = -N}}$ $\\boldsymbol{\\frac{\\Delta \\phi}{\\Delta t}}$\n\n• when flux changes by $\\boldsymbol{\\Delta \\phi}$ in a time $\\boldsymbol{\\Delta t}$.\n• If emf is induced in a coil, $\\boldsymbol{N}$ is its number of turns.\n• The minus sign means that the emf creates a current $\\boldsymbol{I}$ and magnetic field $\\boldsymbol{B}$ that oppose the change in flux $\\boldsymbol{\\Delta \\phi}$ —this opposition is known as Lenz’s law.\n\n### Conceptual Questions\n\n1: A person who works with large magnets sometimes places her head inside a strong field. She reports feeling dizzy as she quickly turns her head. How might this be associated with induction?\n\n2: A particle accelerator sends high-velocity charged particles down an evacuated pipe. Explain how a coil of wire wrapped around the pipe could detect the passage of individual particles. Sketch a graph of the voltage output of the coil as a single particle passes through it.\n\n### Problems & Exercises\n\n1: Referring to Figure 5(a), what is the direction of the current induced in coil 2: (a) If the current in coil 1 increases? (b) If the current in coil 1 decreases? (c) If the current in coil 1 is constant? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenz’s Law.", null, "Figure 5. (a) The coils lie in the same plane. (b) The wire is in the plane of the coil\n\n2: Referring to Figure 5(b), what is the direction of the current induced in the coil: (a) If the current in the wire increases? (b) If the current in the wire decreases? (c) If the current in the wire suddenly changes direction? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenz’s Law.\n\n3: Referring to Figure 6, what are the directions of the currents in coils 1, 2, and 3 (assume that the coils are lying in the plane of the circuit): (a) When the switch is first closed? (b) When the switch has been closed for a long time? (c) Just after the switch is opened?\n\n4: Repeat the previous problem with the battery reversed.\n\n5: Verify that the units of $\\boldsymbol{\\Delta \\phi / \\Delta t}$ are volts. That is, show that $\\boldsymbol{1 \\;\\textbf{T} \\cdot \\;\\textbf{m}^2 \\textbf{/s} = 1 \\;\\textbf{V}}$.\n\n6: Suppose a 50-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of\n$\\boldsymbol{0.250 \\;\\textbf{m}^2}$. It is stretched to have no area in 0.100 s. What is the direction and magnitude of the induced emf if the uniform magnetic field has a strength of 1.50 T?\n\n7: (a) An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner’s 2.00 T field with his fingers pointing in the direction of the field. Find the average emf induced in his wedding ring, given its diameter is 2.20 cm and assuming it takes 0.250 s to move it into the field. (b) Discuss whether this current would significantly change the temperature of the ring.\n\n8: Integrated Concepts\n\nReferring to the situation in the previous problem: (a) What current is induced in the ring if its resistance is $\\boldsymbol{0.0100 \\;\\Omega}$? (b) What average power is dissipated? (c) What magnetic field is induced at the center of the ring? (d) What is the direction of the induced magnetic field relative to the MRI’s field?\n\n9: An emf is induced by rotating a 1000-turn, 20.0 cm diameter coil in the Earth’s $\\boldsymbol{5.00 \\times 10^{-5} \\;\\textbf{T}}$ magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 10.0 ms?\n\n10: A 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.\n\n11: Integrated Concepts\n\nApproximately how does the emf induced in the loop in Figure 5(b) depend on the distance of the center of the loop from the wire?\n\n12: Integrated Concepts\n\n(a) A lightning bolt produces a rapidly varying magnetic field. If the bolt strikes the earth vertically and acts like a current in a long straight wire, it will induce a voltage in a loop aligned like that in Figure 5(b). What voltage is induced in a 1.00 m diameter loop 50.0 m from a $\\boldsymbol{2.00 \\times 10^6 \\;\\textbf{A}}$ lightning strike, if the current falls to zero in $\\boldsymbol{25.0 \\;\\mu \\textbf{s}}$? (b) Discuss circumstances under which such a voltage would produce noticeable consequences.\n\n## Glossary\n\nthe means of calculating the emf in a coil due to changing magnetic flux, given by $\\boldsymbol{\\textbf{emf} =-N \\frac{\\Delta \\phi}{\\Delta t}}$\nLenz’s law\nthe minus sign in Faraday’s law, signifying that the emf induced in a coil opposes the change in magnetic flux\n\n### Solutions\n\nProblems & Exercises\n\n1: (a) CCW\n\n(b) CW\n\n(c) No current induced\n\n3: (a) 1 CCW, 2 CCW, 3 CW\n\n(b) 1, 2, and 3 no current induced\n\n(c) 1 CW, 2 CW, 3 CCW\n\n7: (a) 3.04 mV\n\n(b) As a lower limit on the ring, estimate R = 1.00 mΩ. The heat transferred will be 2.31 mJ. This is not a significant amount of heat.\n\n9: 0.157 V\n\n11: proportional to $\\boldsymbol{\\frac{1}{r}}$", null, "" ]

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[ "# tests/testthat/test-one-chrom.R In junctions: The Breakdown of Genomic Ancestry Blocks in Hybrid Lineages\n\n```context(\"one_chromosome\")\ntest_that(\"one chrom, use\", {\ntestthat::skip_on_os(\"solaris\")\npopulation_size <- 10000\nrun_time <- 10\n\nvx <- sim_phased_unphased(pop_size = population_size,\ntotal_runtime = run_time,\nmarkers = 1000,\ntime_points = run_time)\n\nfound <- c()\n\nfor (i in unique(vx\\$individual)) {\nfocal_data <- subset(vx, vx\\$time == run_time & vx\\$individual == i)\ntime1 <- estimate_time_one_chrom(J = sum(abs(diff(focal_data\\$anc_chrom_1))),\nN = population_size,\nH_0 = 0.5,\nmarker_distribution = focal_data\\$location)\ntime2 <- estimate_time_one_chrom(J = sum(abs(diff(focal_data\\$anc_chrom_2))),\nN = population_size,\nH_0 = 0.5,\nmarker_distribution = focal_data\\$location)\n\nfound <- c(found, c(time1, time2))\n}\n\ntestthat::expect_equal(mean(found), run_time, tolerance = 3)\n\n# induce marker error\ntestthat::expect_error(\nestimate_time_one_chrom(J = sum(abs(diff(focal_data\\$anc_chrom_1))),\nN = 1000,\nH_0 = 0.5)\n)\n\n# induce warning:\npopulation_size <- 100\nrun_time <- 100\n\nvx <- sim_phased_unphased(pop_size = population_size,\ntotal_runtime = run_time,\nmarkers = 1000,\ntime_points = run_time)\nfocal_data <- subset(vx, vx\\$time == run_time & vx\\$individual == 0)\ntestthat::expect_warning(\nestimate_time_one_chrom(J = sum(abs(diff(focal_data\\$anc_chrom_2))),\nN = population_size,\nH_0 = 0.5,\nmarker_distribution = focal_data\\$location,\nupper_lim = 10)\n)\n})\n```\n\n## Try the junctions package in your browser\n\nAny scripts or data that you put into this service are public.\n\njunctions documentation built on March 18, 2022, 6:28 p.m." ]

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[ "Search a number\nBaseRepresentation\nbin10110010111100000\n311122200012\n4112113200\n510412431\n61544052\n7531050\noct262740\n9148605\n1091616\n1162918\n1245028\n1332915\n1425560\n151c22b\nhex165e0\n\n91616 has 24 divisors (see below), whose sum is σ = 206640. Its totient is φ = 39168.\n\nThe previous prime is 91591. The next prime is 91621. The reversal of 91616 is 61619.\n\nIt is a junction number, because it is equal to n+sod(n) for n = 91591 and 91600.\n\nIt is a congruent number.\n\nIt is an unprimeable number.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 20 + ... + 428.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (8610).\n\n291616 is an apocalyptic number.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 91616, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (103320).\n\n91616 is an abundant number, since it is smaller than the sum of its proper divisors (115024).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n91616 is a wasteful number, since it uses less digits than its factorization.\n\n91616 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 426 (or 418 counting only the distinct ones).\n\nThe product of its digits is 324, while the sum is 23.\n\nThe square root of 91616 is about 302.6813505983. The cubic root of 91616 is about 45.0806783146.\n\nIt can be divided in two parts, 91 and 616, that added together give a palindrome (707).\n\nThe spelling of 91616 in words is \"ninety-one thousand, six hundred sixteen\"." ]

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[ "Sunteți pe pagina 1din 7\n\n# MA 105 Spoiltgenius Week 8 Assignment\n\nQUESTION 1\n\nSelect the correct description of right-hand and left-hand behavior of the graph of the\npolynomial function.\nRises to the left, falls to the right\nRises to the right, rises to the left\nFalls to the left, rises to the right\nFalls to the right\nQUESTION 2\n\nSelect the correct description of right-hand and left-hand behavior of the graph of the\npolynomial function.\nf(x) = 42 5x + 4\nFalls to the left, rises to the right\nFalls to the left, falls to the right\nRises to the left, rises to the right\nRises to the left, falls to the right\nFalls to the left\nQUESTION 3\n\nf(x) = x2 25\n-25\n5\n-5\n25\n5\nQUESTION 4\n\n## Use synthetic division to divide.\n\n(43 + x2 11x + 6) (x + 2)\n42 5x 6\n42 7x + 3\n42 2x 2\n42 + 5x 12\n42 + 7x 4\nQUESTION 5\nUse the Remainder Theorem and synthetic division to find the function value. Verify\nh(x) = x3 62 5x + 7\nh(-8)\n-849\n-847\n-851\n-848\n-845\nQUESTION 6\nFind all the rational zeroes of the function.\nx3 122 + 41x 42\n-2, -3, -7\n2, 3, 7\n2, -3, 7\n-2, 3, 7\n-2, 3, -7\nQUESTION 7\nThe total revenue R earned (in thousands of dollars) from manufacturing handheld video\ngames is given by\nR(p) = -25p2 + 1700p\nwhere p is the price per unit (in dollars).\nFind the unit price that will yield a maximum revenue.\n\\$38\n\\$35\n\\$36\n\n\\$37\n\\$34\nQUESTION 8\nFind the domain of the function\nDomain: all real numbers x except x = 7\nDomain: all real numbers x except x = 49\nDomain: all real numbers x except x = 8\nDomain: all real numbers x except x = -7\nDomain: all real numbers x except x = 7\nQUESTION 9\nFind the domain of the function and identify any vertical and horizontal asymptotes.\nDomain: all real numbers x\nVertical asymptote: x = 0\nHorizontal asymptote: y = 0\nDomain: all real numbers x except x = 2\nVertical asymptote: x = 0\nHorizontal asymptote: y = 0\nDomain: all real numbers x except x = 5\nVertical asymptote: x = 0\nHorizontal asymptote: y = 2\nDomain: all real numbers x\nVertical asymptote: x = 0\nHorizontal asymptote: y = 2\nDomain: all real numbers x except x = 5\nVertical asymptote: x = 5\nHorizontal asymptote: y = 0\nQUESTION 10\nSimplify f and find any vertical asymptotes of f.\nx+3; vertical asymptote: x = -3\nx; vertical asymptote: none\nx; vertical asymptote: x = -3\nx-3; vertical asymptote: none\nx2; vertical asymptote: none\nQUESTION 11\n\n## Determine the equations of any horizontal and vertical asymptotes of\n\nhorizontal: y = 5; vertical: x = 0\nhorizontal: y = 1; vertical: x = -5\nhorizontal: y = 1; vertical: x = 1 and x = -5\nhorizontal: y = -1; vertical: x = -5\nhorizontal: y = 0; vertical: none\nQUESTION 12\nIdentify all intercepts of the following function.\nx-intercepts: (3, 0)\nno intercepts\nx-intercepts: (-3,0)\nx-intercepts: (0,0)\nx-intercepts: (3,0)\nQUESTION 13\nSelect the correct graph of the function.\n\nQUESTION 14\nThe game commission introduces 100 deer into newly acquired state game lands. The\npopulation N of the herd is modeled by\nwhere t is the time in years. Find the populations when t=40. (Round your answer to\nthe nearest whole number.)\n1,442 deer\n1,632 deer\n1,594 deer\n1,550 deer\n1,500 deer\nQUESTION 15\nEvaluate the function at the indicated value of x. Round your result to three decimal\nplaces.\nFunction: f(x) = 6000(6x)\nValue: x = -1.3\n\n584.191\n784.191\n-584.191\n684.191\n-784.191\nQUESTION 16\nSelect the graph of the function.\n\nQUESTION 17\nUse the One-to-One Property to solve the equation for x.\nex2-6 = e5x\nx = -6\nx=5\nx = 6, -1\nx = -6, -1\nx = -6,1\nQUESTION 18\nlog366 = 1/2\n36 = -6\n36 = 6\n6 = 36\n36 = -1/6\n36 = 1/6\nQUESTION 19\nWrite the exponential equation in logarithmic form.\n272 = 729\nlog27729 = 2\nlog27729 = 1/2\nlog72927 = 2\nlog27729 = -2\nlog272 = 729\nQUESTION 20\n\nFind the exact value of the logarithmic expression without using a calculator.\n4 ln e7\n7\n28\n4\ne\n1\nQUESTION 21\nCondense the expression to the logarithm of a single quantity.\nln310 + ln3x\nln3(10 x)\nln310/x\nln3(10 + x)\nln310x\nln310x\n4 points\nQUESTION 22\nSolve for x.\n6x = 1,296\n6\n10\n4\n-6\n-4\nQUESTION 23\nSolve the exponential equation algebraically. Approximate the result to three decimal\nplaces.\nex 8 = 12\nln20 2.485\nln20 2.996\nln20 -2.485\nln20 2.079\nln20 -2.996\nQUESTION 24\n\n## An initial investment of \\$9000 grows at an annual interest rate of 5% compounded\n\ncontinuously. How long will it take to double the investment?\n1 year\n14.40 years\n13.86 years\n14.86 years\n13.40 years\nQUESTION 25\nThe populations (in thousands) of Pittsburgh, Pennsylvania from 2000 through 2007 can\nbe modele by\nwhere t represents the year, with t = 0\ncorresponding to 2000. Use the model to find the numbers of cell sites in the year\n2001.\n2,418,774\n2,419,774\n2,421,774\n2,420,774\n2,422,774" ]

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[ "# Present Value of $18 in 9 Years When you have a single payment that will be made to you, in this case$18, and you know that it will be paid in a certain number of years, in this case 9 years, you can use the present value formula to calculate what that $18 is worth today. Below is the present value formula we'll use to calculate the present value of$18 in 9 years.\n\n$$Present\\: Value = \\dfrac{FV}{(1 + r)^{n}}$$\n\nWe already have two of the three required variables to calculate this:\n\n• Future Value (FV): This is the $18 • n: This is the number of periods, which is 9 years So what we need to know now is r, which is the discount rate (or rate of return) to apply. It's worth noting that there is no correct discount rate to use here. It's a very personal number than can vary depending on the risk of your investments. For example, if you invest in the market and you earn on average 8% per year, you can use that number for the discount rate. You can also use a lower discount rate, based on the US Treasury ten year rate, or some average of the two. The table below shows the present value (PV) of$18 paid in 9 years for interest rates from 2% to 30%.\n\nAs you will see, the present value of $18 paid in 9 years can range from$1.70 to $15.06. Discount Rate Future Value Present Value 2%$18 $15.06 3%$18 $13.80 4%$18 $12.65 5%$18 $11.60 6%$18 $10.65 7%$18 $9.79 8%$18 $9.00 9%$18 $8.29 10%$18 $7.63 11%$18 $7.04 12%$18 $6.49 13%$18 $5.99 14%$18 $5.54 15%$18 $5.12 16%$18 $4.73 17%$18 $4.38 18%$18 $4.06 19%$18 $3.76 20%$18 $3.49 21%$18 $3.24 22%$18 $3.01 23%$18 $2.79 24%$18 $2.60 25%$18 $2.42 26%$18 $2.25 27%$18 $2.09 28%$18 $1.95 29%$18 $1.82 30%$18 $1.70 As mentioned above, the discount rate is highly subjective and will have a big impact on the actual present value of$18. A 2% discount rate gives a present value of $15.06 while a 30% discount rate would mean a$1.70 present value.\n\nThe rate you choose should be somewhat equivalent to the expected rate of return you'd get if you invested \\$18 over the next 9 years. Since this is hard to calculate, especially over longer periods of time, it is often useful to look at a range of present values (from 5% discount rate to 10% discount rate, for example) when making decisions.\n\nHopefully this article has helped you to understand how to make present value calculations yourself. You can also use our quick present value calculator for specific numbers." ]

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[ "Solutions by everydaycalculation.com\n\n## Divide 3/9 with 8/10\n\n3/9 ÷ 8/10 is 5/12.\n\n#### Steps for dividing fractions\n\n1. Find the reciprocal of the divisor\nReciprocal of 8/10: 10/8\n2. Now, multiply it with the dividend\nSo, 3/9 ÷ 8/10 = 3/9 × 10/8\n3. = 3 × 10/9 × 8 = 30/72\n4. After reducing the fraction, the answer is 5/12\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# How big can a set of integers be if all pairs have small gcd?\n\nSuppose $A\\subset[1,N]$ is a set of integers. If for any distinct $a,b\\in A$ we have $(a,b)\\leq M$ then how big can $|A|$ be?\n\nIf $M=1$ then $|A|$ is at most $\\pi(N)$ since the map $a\\mapsto P_+(a)$ (which sends $a$ to its largest prime factor) is an injection to the primes, and so the primes themselves are the most efficient. If $M=N$ then we can take $|A|=N$. What about $M$ in between? Can you beat the primes? It would be safe to assume that $A$ consists solely of $M$-smooth numbers since we can decompose $A=A_1\\cup A_2$ where each $a\\in A_1$ is $M$-smooth and each $a\\in A_2$ has a prime factor which at least $M+1$. The large prime factors appearing as divisors in $A_2$ cannot be repeated, so $|A_2|\\leq \\pi(N)$.\n\n• Even more efficiently, you may beat the primes by adding all pairwise products of primes not exceeding $M$ (including their squares). – Ilya Bogdanov Oct 21 '15 at 16:31\n• Actually, for $M= 1$ you have $|A| = \\pi(N) + 1$: you can have $1$ as well as the primes. @IlyaBogdanov: there's no reason to restrict to pairwise products, you can throw in products of arbitrarily many (not necessarily distinct) primes $x = p_1 \\ldots p_n$, $p_1 \\le \\ldots \\le p_n$, as long as $x/p_1 \\le M$. – Robert Israel Oct 21 '15 at 18:21\n• Thus for $M=5$, in addition to the primes $\\le N$ you have $\\{1,4,6,8,9,10,15,25\\}$. – Robert Israel Oct 21 '15 at 18:28\n\nWe'll prove that the maximal cardinality of such a set for $M^2\\leq N$ has size equal to $$\\pi(N)+\\sum_{1<n\\leq M} \\pi(p(n))$$ where $p(n)$ is the smallest prime factor of $n$. Since $$\\sum_{1<n\\leq M} \\pi(p(n))\\sim \\frac{M^2}{2\\log^2 M},$$ this proves that Ilya Bogdanov's example of including all pairwise products of primes not exceeding $M$ is nearly optimal in terms of asymptotics.\n\nAs you suggest in the question, this problem is equivalent to constructing the largest subset $A\\subset S_M(N)$ such that $\\gcd(a,b)\\leq M$ for every $a,b\\in A$ where $S_M(N)$ is the set of $n\\leq N$ whose largest prime factor is at most $M$.\n\nTo see why, suppose that $A$ satisfies $\\gcd(a,b)\\leq M$ for every $a,b\\in A$. Then every prime that is greater than $M$ can divide at most one element of $A$. If $p>M$ divides $a\\in A$, then making $a=p$ only helps create a larger set $A$. Since these primes do not interact with the $M$-smooth numbers, the proof is complete.\n\nLet $T(N,M)$ denote the maximum size of such a set $A\\subset S_M(N)$. Then the size of the largest subset of $[1,N]$ with pairwise $\\gcd$'s bounded by $M$ is $$\\pi(N)-\\pi(M)+T(N,M).$$\n\nAs mentioned by Fedja in the comments s, the reasoning above for the primes extends to all integers. By considering those primes $p,q\\leq M$ such that $pq>M$, we see that there can only be exactly one such number divisible by $pq$. Similarly for any $p,q,r$ with $pq,qr,rp\\leq M$ and $pqr>M$ there can only be one number in our set divisible by $pqr$. Thus we find that the maximal size is the sum over those integers whose largest proper divisor is less than $M$. Since the largest proper divisor of $n$ equals $n/p(n)$ where $p(n)$ is the least prime factor, we can group things based on this, and we have that $$T(N,M)=\\sum_{1<p\\leq M}\\sum_{n\\leq M:\\ p(n)\\geq p}1.$$ Rearranging this equals $$\\sum_{1<n\\leq M}\\sum_{p\\leq p(n)}1=\\sum_{n\\leq M}\\pi\\left(p(n)\\right),$$ and for composite $n$ $p(n)\\leq\\sqrt{n}$, so the primes dominate this sum. Thus asymptotically we have $$T(N,M)\\sim\\sum_{p\\leq M}\\pi(p)\\sim\\frac{M^{2}}{2\\log^{2}M}.$$\n\n• You can continue in the same spirit: consider pairs of primes $p,q\\le M$ with $pq>M$. Any such pair can enter just one number, so we can as well have all pairs there. Next consider all triples such that the product of any 2 is less than or equal to $M$ but the triple product is $>M$. Again, any triple can enter at most one number, so take them all, etc. After that, of course, just add $[1,M]$. So to describe a maximal cardinality set it easy. As to the size, it looks like we just need to play with the prime number theorem carefully to figure the asymptotics but I have to teach a class now... – fedja Oct 21 '15 at 18:08\n• This set can be also described as the set of all numbers whose largest proper divisor is not greater than $M$, which allows to estimate the size of the complement way faster than with my prime count idea. Now it is time to run on the stairs :-) – fedja Oct 21 '15 at 18:13\n• @Fedja: Great, that works nicely! – Eric Naslund Oct 21 '15 at 18:22\n• There's still something off if $M>\\sqrt{N}$ which you should account for. Fedja's description of numbers with largest proper divisor not larger than $M$ is very nice. For example, this includes all numbers with no prime factors below $N/M$ which has interesting asymptotics when $M$ is larger than $\\sqrt{N}$ (eventually getting to $N$ when $M=N$). So you could either restrict attention to small ranges of $M$, or perhaps flesh out the argument a little bit more ... – Lucia Oct 22 '15 at 3:19\n• @Lucia: You're right, I have edited the question for now, but I'll add in a nice form for the sum when $M>\\sqrt{N}$ – Eric Naslund Oct 22 '15 at 11:09" ]

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[ "", null, "# WORD RECOGNITION METHOD AND WORD RECOGNITION PROGRAM\n\nImported: 10 Mar '17 | Published: 27 Nov '08\n\nTomoyuki Hamamura\n\nUSPTO - Utility Patents\n\n## Abstract\n\nA word recognition method of performing recognition processing with respect to each word candidate obtained by reading characters in character information written in a reading material is provided. This word recognition method includes a matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word, a character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character, and a correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step.\n\n## Description\n\n### CROSS-REFERENCE TO RELATED APPLICATIONS\n\nThis is a Continuation Application of PCT Application No. PCT/JP2008/053433, filed Feb. 27, 2008, which was published under PCT Article 21(2) in Japanese.\n\nThis application is based upon and claims the benefit of priority from prior Japanese Patent Application No. 2007-065522, filed Mar. 14, 2007, the entire contents of which are incorporated herein by reference.\n\n### BACKGROUND OF THE INVENTION\n\n1. Field of the Invention\n\nThe present invention relates to a word recognition method of recognizing words and a word recognition program that is used to execute word recognition processing in, e.g., an optical character reading apparatus that optically reads a word including a plurality of characters written in, e.g., a reading material.\n\n2. Description of the Related Art\n\nFor example, in an optical character reading apparatus, when reading a character written in a reading material, accurate reading can be generally performed by using knowledge of words even though an accuracy for recognizing each character is low. Various kinds of methods have been conventionally proposed as implementation methods.\n\nAmong others, there is a method disclosed in Jpn. Pat. Appln. KOKAI Publication No. 2001-283157 as a method of using a posteriori probability as an evaluation value for each word to enable accurate word recognition even if the number of characters is not fixed.\n\nHowever, the method disclosed in the above-explained document is provided on the assumption that a position where a word is written is already known, and it cannot be said that word recognition can be performed with a sufficient accuracy when the position where the word is written is unknown. For example, when a correct word is disorderly written or an incorrect word is fairly written, an evaluation value (a matching score) of a word in a dictionary similar to the incorrect word rises, and erroneous recognition is thereby apt to be carried out.\n\n### BRIEF SUMMARY OF THE INVENTION\n\nTherefore, it is an object of the present invention to provide a word recognition method and a word recognition program that can accurately perform word recognition even if a position where a word is written is unknown.\n\nAccording to the present invention, there is provided a word recognition method of performing recognition processing with respect to each word candidate obtained by reading characters in character information written in a reading material, comprising: a matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word; a character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character; and a correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step.\n\nAccording to the present invention, there is provided a word recognition program that allows a computer to perform recognition processing with respect to a word candidate obtained by reading characters in character information written in a reading material, comprising: a matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word; a character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character; and a correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step.\n\n### DETAILED DESCRIPTION OF THE INVENTION\n\nAn embodiment according to the present invention will now be explained hereinafter with reference to the accompanying drawings.\n\nFIG. 1 schematically shows a structure of a word recognition system that realizes a word recognition method according to an embodiment of the present invention.\n\nIn FIG. 1, this word recognition system is formed of a CPU (a central processing unit) 1, an input device 2, a scanner 3 as image inputting means, a display device 4, a first memory 5 as storing means, a second memory 6 as storing means, a reading device 7, and others.\n\nThe CPU 1 executes an operating system program stored in the second memory 6 and an application program (e.g., a word recognition program) stored in the second memory 6 to perform, e.g., word recognition processing that will be explained later in detail.\n\nThe input device 2 is formed of, e.g., a keyboard, a mouse, and others, and utilized by a user to perform various kinds or operations or input various kinds of data.\n\nThe scanner 3 reads each character in a word written in a reading material based on optical scanning, and inputs the read character.\n\nThe display device 4 is formed of, e.g., a display device or a printer, and outputs various kinds of data.\n\nThe first memory 5 is formed of, e.g., an RAM (a random access memory), and utilized as a working memory for the CPU 1 to temporarily store various kinds of data that are being processed. For example, the first memory 5 temporarily stores, a character dictionary 9, a word dictionary 10, a probability table 11, and others which will be explained later.\n\nThe second memory 6 is formed of, e.g., a hard disk device, and stores various kinds of programs and others to operate the CPU 1. The second memory 6 stores an operating system program that is used to operate the input device 2, the scanner 3, the display device 4, the first memory 5, the second memory 6, the reading device 7, and others, a word recognition program, the character dictionary 9 for recognition of characters constituting each word, the word dictionary 10 for word recognition, the probability table 11 storing appearance probabilities of characters constituting each word, and others. The word dictionary 10 stores a plurality of candidates of words that should be recognized in advance, and it is a city name dictionary in which areas where the word recognition systems are installed, e.g., city names in states are registered.\n\nThe reading device 7 is formed of, e.g., a CD-ROM drive device and reads the word recognition program or the word dictionary 10 for word recognition stored (saved) in a CD-ROM 8 as a storage medium. The word recognition program, the character dictionary 9, the word dictionary 10, and the probability table 11 read by the reading device 7 are stored (saved) in the second memory 6.\n\nAn outline of the word recognition method will now be explained hereinafter with reference to a flowchart in FIG. 2.\n\nFirst, image fetching processing of fetching (reading) an image in a postal matter P by the scanner 3 is carried out (step ST1). Region detection processing of detecting a region where an address is written based on the image fetched by this image fetching processing is performed (step ST2). Cutout processing of using vertical projection or horizontal projection to cut out a character pattern in a rectangular region in accordance with each character in a word corresponding to a city name from the address written region detected by this region detection processing is performed (step ST3). Character recognition processing of obtaining character recognition candidates based on a degree of similarity obtained by comparing the character pattern of each character in the word cut out by this cutout processing with a character pattern stored in the character dictionary 9 is effected (step ST4). Word recognition processing of calculating posteriori probabilities of respective city names in the word dictionary 10 and recognizing a city name having the highest posteriori probability as a word by using a recognition result of each word obtained by the character recognition processing, each character in the city names stored in the word dictionary 10, and the probability table 11 is carried out (step ST5). Each processing is controlled by the CPU 1.\n\nA description will now be given as to an example of reading a city name from an address written in, e.g., a form in alphabet-using countries as a specific example.\n\nFIG. 3 is a view showing an example of an image of, e.g., a form in alphabet-using countries fetched by, e.g., a scanner. FIG. 4 is a view showing an example of word candidates detected from the image fetched by, e.g., the scanner. FIG. 5 is a view showing an example of a word dictionary having city name words registered therein. In this case, such word candidates as shown in the image in FIG. 4 are detected from the image in FIG. 3, these word candidates are searched for words registered in the word dictionary (e.g., city names registered in the word dictionary in FIG. 5), and words written in the form or the like (e.g., city names) are specified. At this time, predetermined correction processing is performed with respect to a result of matching processing for the word candidates and the words in the word dictionary in this embodiment in particular.\n\nHere, the word recognition method including the correction processing according to this embodiment will now be explained with reference to a flowchart in FIG. 6.\n\nWhen such an image of, e.g., a form as depicted in FIG. 3 is fetched through, e.g., a scanner, each word candidate included in the image is detected as shown in FIG. 4 (step 11). A serial number is given to each detected word candidate.\n\nWord matching processing for each word candidate is carried out (step 12). The word matching processing is processing of collating each word candidate with a plurality of words in the word dictionary to calculate, every word, a matching score indicative of a degree that the word candidate matches with a word in the dictionary.\n\nA known technology called DP (Dynamic Programming) matching (or dynamic planning method) can be applied to this word matching processing. This DP matching is well known as an algorithm that minimizes computational processing of calculating, e.g., a degree of similarity when collating objects (e.g., words) having different element numbers (e.g., character numbers), and it performs association between elements (alignment) while considering, e.g., a deviance of the elements to realize optimum collating processing. Availing this technology enables realizing, e.g., an improvement in a speed of the matching processing.\n\nMatching scores calculated by the word matching processing are stored in a matching score table. FIG. 7 is a view showing an example of matching scores stored in the matching score table. This matching score table stores matching scores, every word, as matching processing results of respective word candidates having serial numbers 1 to 7 and city name words STOCKHOLM, TOCHICA, MOHEDA, . . . in the word dictionary.\n\nTOSHIBA indicative of a town name is written in an actual form as the word candidate having the serial number 3, and it can be expected that its matching score with respect to each city name word in the word dictionary becomes low. However, the word dictionary has a city name word TOCHICA in which five characters, match with TOSHIBA and two characters are different from the same. Since each character constituting TOSHIBA is fairly written in the form, a matching score (reference character a) with respect to the city name word TOCHICA in the word dictionary becomes high because of the matching five characters.\n\nOn the other hand, StockHolm indicative of a city name is written in the actual form as the word candidate having the serial number 6, and it can be expected that its matching score with respect to a city name word STOCKHOLM in the word dictionary becomes high. However, STOCKHOLM is disorderly written in the form, its matching score (reference numeral b) is not that high as expected.\n\nAt this time, the matching score denoted by reference numeral a is higher than the matching score designated by reference numeral b, and hence TOSHIBA written in the form may be possibly erroneously recognized as the city name word TOCHICA. This embodiment avoids this erroneous recognition based on the following respective steps.\n\nAfter the word matching processing, character quality score calculation processing for each word candidate is carried out (step 13). This character quality score calculation processing is processing of calculating a character quality score indicative of a degree that each character constituting each word candidate matches with an arbitrary character (any character in alphabets in the character dictionary). For example, when calculating a quality score of a given character candidate in a given word candidate, a probability (or a degree of similarity) that this character candidate matches with any one of alphabets in the character dictionary is calculated. A result of addition of individual quality scores acquired by performing such calculation processing with respect to each character candidate is adopted as a character quality score of this word candidate.\n\nIn this character quality score calculation processing, the above-explained DP matching can be likewise applied when performing the character candidate matching processing.\n\nThe character quality scores calculated by the character quality score calculation processing are stored in a character quality score table. FIG. 8 is a view showing an example of character quality scores stored in the character quality score table. This character quality score table stores character quality scores with respect to the word candidates having the serial numbers 1 to 7.\n\nFor example, in the word candidate having the serial number 3 or the word candidate having the serial number 7 written in the form, each character is fairly written. In particular, since a boundary between characters is clear, each character can be specified without fail. Since a shape of each character to be specified substantially matches with an arbitrary alphabet in the word dictionary, a character quality score of such a candidate is higher than those of other candidates. For example, in case of the word candidate TOSHIBA having the serial number 3, hand-written characters T, O, S, H, I, B, and A are fair so that they substantially match with alphabets T, O, S, H, I, B, and A in the character dictionary (a degree that these characters match with or similar to these alphabets is high), and hence a high score is given. In case of the word candidate having the serial number 7, a high score is likewise given.\n\nOn the other hand, in the word candidate having the serial number 4 or the word candidate having the serial number 6 written in the form, each character is disorderly written in a running hand. In particular, since a boundary between characters is unclear, specifying each character is difficult, and each character is apt to be erroneously specified. Even if each character can be correctly specified, it is often the case that a shape of each character does not substantially match with an arbitrary alphabet in the character dictionary. Therefore, such a word candidate has a lower character quality score than those of the other candidates. For example, the disorderly written word candidate StockHolm having the serial number 6 corresponds to this case, and it has a low score. The word candidate having the serial number 4 likewise has a low score.\n\nAfter the character quality score calculation processing, corrected score calculation processing with respect to each word candidate is executed (step 14). The corrected score calculation processing is processing of correcting the matching scores obtained by the word matching processing based on the character quality scores acquired by the character quality score calculation processing. For example, processing of subtracting a character quality score obtained by the character quality score calculation processing from a matching score acquired by the word matching processing. As a result, a matching score of a word candidate having fairly written characters is greatly reduced by the correction processing and, on the other hand, a matching score of a word candidate having disorderly written characters is slightly reduced by the correction processing.\n\nCorrected scores calculated by the corrected score calculation processing are stored in a corrected score table. FIG. 9 is a view showing an example of corrected scores stored in the corrected score table. This corrected score table stores corrected scores of respective words as a result of correcting matching scores which are matching processing results of the respective word candidates having the serial numbers 1 to 7 and the city name words STOCKHOLM, TOCHICA, MOHEDA, . . . in the word dictionary.\n\nAt a point in time where the word matching processing is executed, the matching score (reference character a) of the word candidate having the serial number 3 and the word TOCHICA in the dictionary is higher than the matching score (reference character b) of the word candidate having the serial number 6 and the word STOCKHOLM in the dictionary, but levels of the scores are reversed after the corrected score calculation processing. That is, after the corrected score calculation processing, the matching score (the corrected score) (reference character b) of the word candidate having the serial number 6 and the word STOCKHOLM in the dictionary is higher than the matching score (the corrected score) (reference character a) of the word candidate having the serial number 3 and the word TOCHICA in the dictionary. As a result, obtaining a correct recognition result can be expected.\n\nAfter the corrected score calculation processing, a city name word having the highest corrected score in the corrected score table is selected from the word dictionary, and the selected city name word is output as a recognition result (step 15).\n\nDetails of the word matching processing depicted in FIG. 6 will now be explained with reference to a flowchart in FIG. 10.\n\nFirst, 1 is set to a number i of word candidates (step 21). Then, an ith word candidate is selected (step 22).\n\nSubsequently, 1 is set to a number j of words in the word dictionary (step 23). Further, a jth word in the word dictionary is selected (step 24).\n\nThen, matching processing with respect to the selected ith word candidate and jth word in the dictionary is executed to calculate a matching score (step 25). Furthermore, the matching score is written at a position (i, j) in the matching score table (step 26).\n\nHere, j is compared with the number of all words in the dictionary (step 27). If j is smaller than the number of all words in the dictionary, 1 is added to j (step 28), and the processing from the step 24 is repeated. On the other hand, if j is not smaller, i is compared with the number of all word candidates (step 29). If i is smaller than the number of all word candidates, 1 is added to i (step 30), and the processing from the step 22 is repeated. On the other hand, if i is not smaller, the word matching processing is terminated.\n\nDetails of the character quality score calculation processing in FIG. 6 will now be explained with reference to a flowchart in FIG. 11.\n\nFirst, 1 is set to a number i of word candidates (step 41). Moreover, an ith word candidate is selected (step 42).\n\nThen, a character quality score of character candidates constituting the word candidate is calculated (step 43). Additionally, the character quality score is written at a position i in the character quality score table (step 44).\n\nSubsequently, i is compared with the number of all word candidates (step 45). If i is smaller than the number of all word candidates, 1 is added to i (step 46), and the processing from the step 42 is repeated. On the other hand, if i is not smaller, the character quality score calculation processing is terminated.\n\nDetails of the corrected score calculation processing in FIG. 6 will now be explained with reference to a flowchart in FIG. 12.\n\nFirst, 1 is set to a number i of word candidates (step 51). Further, a value at the position i in the character quality score table is read as a correction value h (step 52).\n\nThen, 1 is set to a number j of words in the word dictionary (step 53). Furthermore, a value at the position (i, j) in the matching score table is read as a score s (step 54). Moreover, the correction value h is subtracted from the score s, and a calculation result is written at a position (i, j) in the corrected score table (step 55).\n\nSubsequently, j is compared with the number of all words in the dictionary (step 56). If j is smaller than the number of all words in the dictionary, 1 is added to j (step 57), and the processing from the step 54 is repeated. On the other hand, if j is not smaller, i is compared with the number of all word candidates (step 58). If i is smaller than the number of all word candidates, 1 is added to i (step 59), and the processing from the step 52 is repeated. On the other hand, if i is not smaller, the corrected score calculation processing is terminated.\n\nThen, specific examples of computational expressions that are used to calculate the various kinds of scores will now be explained.\n\nHere, a word in the word dictionary, a character recognition result of all character candidates, a universal set of positions of a character candidate, and a universal set of paths from a left end to a right end of a given position are defined as follows, respectively.\n\nWord in the dictionary: wi=ci1, ci2, . . . ), cijC (C is a set of alphabets)\n\nCharacter recognition result of all character candidates: r=(r1, r2, . . . )\n\nUniversal set of positions: LL\n\nUniversal set of paths from a left end to a right end of a position L: nN\n\nTo calculate a matching score of a given word candidate and a given word in the dictionary, a calculation using a posteriori probability ratio is performed. In this case, as a basic computational expression, the following Expression (1) is adopted, for example.\n\n$log P ( w i | r ) P ( w i ) i log P ( r i | c ij ) P ( r i ) Expression ( 1 )$\n\nA numerator of a fractional part on a left-hand side in Expression 1 corresponds to a posteriori probability, and a denominator of the same corresponds to a priori probability.\n\nTo calculate a matching score (corresponding to the score table in FIG. 7) in the word matching processing based on Expression (1), the following Expression (2) is used.\n\n$match ( L , w i ) max n [ j { P ( r n j | c ij ) / P ( r n j ) } ] Expression ( 2 )$\n\nFurther, to calculate a character quality score (corresponding to the score table in FIG. 8) in the character quality score calculation processing, the following Expression (3) is used.\n\n$match ( L , C * ) max n [ j { P ( r n j | C ) / P ( r n j ) } ] Expression ( 3 )$\n\nwhere, C* in Expression (3) is indicative of an arbitrary alphabet.\n\nAt last, to calculate a maximum value of a corrected score (corresponding to the score table in FIG. 9) in the corrected score calculation processing, the following Expression (4) is used as an evaluation function.\n\n$value ( w i ) P ( w i ) max L { match ( L , w i ) / match ( L , C * ) } Expression ( 4 )$\n\nAccording to the foregoing embodiment, even if a position where a word is written is unknown, word recognition can be accurately performed. For example, even if a correct word is disorderly written and an incorrect word is fairly written, a matching score is corrected, and an appropriate evaluation value can be obtained, thereby avoiding occurrence of erroneous recognition.\n\nIt is to be noted that the present invention is not restricted to the foregoing embodiment as it is, and constituent elements can be modified and embodied on an embodying stage without departing from the scope of the invention. Furthermore, various inventions can be formed based on appropriate combinations of a plurality of constituent elements disclosed in the foregoing embodiment. For example, some constituent elements can be eliminated from all constituent elements disclosed in the embodiment. Moreover, constituent elements in different embodiments can be appropriately combined.\n\nAccording to the present invention, when performing word recognition in, e.g., an optical character reading apparatus that optically reads a word formed of a plurality of characters written in a reading material, word recognition can be accurately performed even if a position where the word is written is unknown.\n\n## Claims\n\n1. A word recognition method of performing recognition processing with respect to each word candidate obtained by reading characters in character information written in a reading material, comprising:\na matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word;\na character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character; and\na correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step.\na matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word;\na character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character; and\na correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step.\n2. The method according to claim 1, wherein the correcting step includes processing of subtracting the character quality score obtained at the character quality score calculating step from the matching score acquired at the matching processing step.\n3. The method according to claim 1, wherein the arbitrary character is one of alphabetical letters.\n4. The method according to claim 1, further comprising a step of outputting a word having the highest matching score in respective matching scores corrected at the correcting step as a recognition result.\n5. A word recognition program that allows a computer to perform recognition processing with respect to a word candidate obtained by reading characters in character information written in a reading material, comprising:\na matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word;\na character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character; and\na correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step.\na matching processing step of collating each word candidate with a plurality of words in a word dictionary and calculating, every word, a matching score indicative of a degree that each word candidate matches with a word;\na character quality score calculating step of calculating a character quality score indicative of a degree that a character candidate constituting each word candidate matches with an arbitrary character; and\na correcting step of correcting a matching score obtained at the matching processing step based on a character quality score acquired at the character quality score calculating step." ]

[ null, "https://pixel.quantserve.com/pixel/p-S5j449sRLqmpu.gif", null ]

[ "# Homomorphism\n\nA homomorphism, (or sometimes simply morphism) from one mathematical object to another of the same kind, is a mapping that is compatible with all relevant structure. The notion of morphism is studied abstractly in category theory.\n\nFor example, if one object consists of a set X with an ordering <= and the other object consists of a set Y with an ordering [=, then it must hold for the function f : X -> Y that\n\nif    u <= v    then    f(u) [= f(v).\n\nOr, if on these sets the binary operations * and @ are defined, respectively, then it must hold that\n\nf(u) @ f(v) = f(u * v).\n\nAn example of homomorphism is given by group homomorphism.\n\nA homomorphism which is also a bijection is called an isomorphism; two isomorphic objects are completely indistinguishable as far as the structure in question is concerned. A homomorphism from a set to itself is called an endomorphism, and if it is also an isomorphism is called an automorphism.\n\nAny homomorphism f : X -> Y defines an equivalence relation on X by a ~ b iff f(a) = f(b). The quotient set X / ~ can then be given an object-structure in a natural way, e.g. [x] * [y] =[x * y]. In that case the image of X is necessarily isomorphic to X / ~. Note in some cases (e.g. groups) a single equivalence class U suffices to specify the structure of the quotient, so we write it X / U." ]

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[ "You are on page 1of 5\n\n## HSPA+/LTE-A Turbo Decoder on GPU and Multicore CPU\n\nMichael Wu, Guohui Wang, Bei Yin, Christoph Studer, and Joseph R. Cavallaro Dept. of Electrical and Computer Engineering, Rice University, Houston, TX e-mail: {mbw2, gw2, by2, studer, cavallar}@rice.edu\n\nAbstract—This paper compares two implementations of re- configurable and high-throughput turbo decoders. The first implementation is optimized for an NVIDIA Kepler graphics processing unit (GPU), whereas the second implementation is for an Intel Ivy Bridge processor. Both implementations support max-log-MAP and log-MAP turbo decoding algorithms, various code rates, different interleaver types, and all block-lengths, as specified by HSPA+ and LTE-Advanced. In order to ensure a fair comparison between both implementations, we perform device-specific optimizations to improve the decoding throughput and error-rate performance. Our results show that the Intel Ivy Bridge processor implementation achieves up to 2× higher decoding throughput than our GPU implementation. In addition our CPU implementation requires roughly 4× fewer codewords to be processed in parallel to achieve its peak throughput.\n\nI. INTRODUCTION\n\nTurbo codes are capacity-approaching channel codes that can be decoded at high throughput and low power using dedicated hardware accelerators. Hence, turbo codes are used in a large number of cellular wireless standards, such as 3GPP HSPA+ and LTE-Advanced . Recently, a number of software-based wireless testbeds have been developed to demonstrate the feasibility of software-based real-time com- munication systems on general purpose processors –. Im- plementations on such architectures are attractive for multiple reasons. First, they are inherently flexible (with respect to code rates, block lengths, etc.) and capable of supporting multiple standards. Second, they use off-the-shelf components, without the need for dedicated hardware accelerator blocks. Although turbo codes offer superior error-rate performance over convolutional codes, turbo decoding requires higher computational complexity . To meet the throughput re- quirements of existing wireless standards, turbo decoding is typically carried out with specialized hardware accelerators, such as ASIC designs – or FPGA implementations . As a consequence, SDR systems such as the ones in – rely on convolution codes instead of turbo codes to avoid the high complexity of channel decoding. The use of con- volutional codes, however, results in inferior error-correction performance (compared to turbo codes). In addition, LTE- Advanced, specifies the use of turbo codes for both uplink and downlink. Hence, corresponding SDR receiver designs necessitate the development of software-based turbo decoder solutions that are capable of achieving high throughput.\n\nThis work was supported in part by Broadcom and by the US National Science Foundation under grants CNS-1265332, ECCS-1232274, EECS- 0925942 and CNS-0923479. We would also like to thank Nvidia and Intel for their generous hardware donations.", null, "Figure 1.\n\nHigh-level structure of a rate- 1 /3 3GPP turbo decoder.\n\nContributions: In this paper, we evaluate the performance of turbo decoder implementations on two different high per- formance programmable processors, namely on a quad-core Intel i7-3770K (Ivy Bridge) and a Nvidia GeForce GTX 680 (Kepler GK104). We design two parallel turbo decoder implementations with a similar feature set. Both proposed implementations support HSPA+ and LTE-Advanced and take advantage of unique features of both platforms. In particular, we perform a variety of device specific optimizations, such as the use of linear-MAP approximation on the CPU and the use of shuffle instructions on the GPU, to maximize the throughput and/or to improve the error-rate performance. We conclude by comparing the throughput of both implementations.\n\nII. OVERVIEW OF TURBO DECODING\n\nThe high-level structure of a rate- 1 /3 3GPP turbo decoder is shown in Figure 1. The turbo decoder consists of two concatenated component decoders exchanging soft informa- tion in terms of the log-likelihood ratio (LLR) for each transmitted information bit through an interleaver (denoted by ) and a deinterleaver (denote by 1 ). HSPA+ uses intra-row and inter-row permutations to generate the inter- leaver addresses , whereas LTE-Advanced (LTE-A) uses a quadratic permutation polynomial (QPP) interleaver . The component decoders are the same for HSPA+ and LTE-A.\n\nA. Algorithm outline\n\nTurbo decoding is carried out in multiple iterations (de- noted by I) where each iteration consists of two component decoding phases. In each phase, a component decoder per- forms maximum a-posteriori (MAP) decoding using the BCJR algorithm , which generates so-called extrinsic LLRs given the LLRs obtained by the detector and a-priori LLRs obtained from the other component decoder. The BCJR algorithm consists of one forward and one backward traversal on a trellis, which is defined by the underlying code. Specifically, to decode a codeword of N information bits, the BCJR algorithm performs the following steps: (i) In the forward traversal step,", null, "Figure 2. Structure of the 3GPP trellis. There are 8 states per trellis step and one step per transmitted information bit. The vector s k consists of all state metrics at trellis step k. The values u k and p k , are the k th information bit and the parity bit (both ±1) respectively.\n\nit iteratively computes N sets of forward state metrics for each transmitted information bit. (ii) In the backward traversal step, it iteratively computes N sets of backward state metrics for each transmitted information bit. To compute the extrinsic LLRs, the BCJR algorithm then combines the forward and backward state metrics.\n\nB.\n\nBranch-metric computation\n\nHSPA+ and LTE-Advanced both operate on a 8-state trellis,\n\nwhich is illustrated in Figure 2. Let s\n\nk+1\n\nj\n\nbe the j th state\n\nassociated with information bit k +1. There are two incoming\n\nbranches into state s\n\nj\n\nk+1\n\n. Each incoming branch is associated\n\nwith values u k and p k , the k th information bit and the parity bit (both ±1), respectively. The branch metrics associated with\n\nstates s\n\nk\n\ni\n\nand s k+1 are computed as follows:\n\nγ s\n\nk\n\ni\n\n, s\n\nk+1\n\nj\n\n= 0.5(L", null, "k\nsys + L\n\nk\n\na )u k + 0.5(L\n\nk\n\np\n\np k ).\n\nHere, L\n\nk\n\nsys\n\nk\n\nand L a are the systematic channel LLR and the\n\na-priori LLR for k th trellis step, respectively. In addition, the\n\nparity LLRs for the k th trellis step are L\n\nk\n\np\n\nk\n\n= L p0 for MAP\n\ndecoder 0 and L\n\nk\n\np\n\nk\n\n= L p1 for MAP decoder 1. Note that we\n\ndo not need to evaluate the branch metric γ (s k , s k+1 ) for all\n\n16 possible branches (see Figure 2), as there are only four\n\ndifferent branch metrics: γ\n\n0\n\nk\n\n= 0.5(L\n\n0.5(L\n\nk sys + L\n\nk\n\na L\n\nk\n\np\n\n), γ\n\n0\n\nk\n\n1\n\n, and -γ k .", null, "k\nsys + L\n\nk\n\na\n\n+ L\n\nk\n\np\n\n), γ\n\n1\n\nk\n\n=\n\nC.\n\nForward and backward state metric computation\n\nThe forward state metrics can be computed iteratively from trellis step to trellis step. The forward state metrics of step k+1\n\ncorrespond to the vector α k+1 = [α k+1 ,\n\n1\n\nj th forward state metric α\n\nk+1\n\nj\n\nk+1\n\n8\n\n], where the\n\n. . .\n\n, α\n\nonly depends on two forward\n\nstate metrics of stage k. These state metrics are computed by\n\nα\n\nk+1\n\nj\n\n= max\n\nk iF α i\n\n+ γ(s\n\nk\n\ni\n\n, s\n\nk+1\n\nj\n\n)\n\n(1)\n\nwhere the set F contains the two indices of the states in step k\n\nk+1\n\nconnected to state s operator is defined as\n\nj\n\n(as defined by the trellis). The max {·}\n\nmax {a, b} = max{a, b} + log (1 + exp(−|a b|)) , (2)\n\n where log (1 + exp(−|a − b|)) is a correction term. For the max-log approximation, we approximate max ∗ by max ∗ (a, b) ≈ max(a, b). In this case, one can scale the\n\nextrinsic LLRs by a factor of 0.7 to to partially recover the error-rate performance loss induced by the approximation (see, e.g., , for additional details). Computation of the backward state metrics is similar to that of the forward trellis traversal in (1). The vector of backward\n\nstate metrics, denoted by β k = [β\n\nk\n\nβ\n\nj\n\n= max\n\nk+1 iB β i\n\n. , β , s\n\n1\n\n,\n\n.\n\n.\n\n+ γ(s\n\nj\n\nk\n\n8\n\nk\n\nk\n\n], is computed as\n\nk+1\n\ni\n\n) .\n\n(3)\n\nHere, the set B contains the indices of states in step k + 1\n\nconnected to state s\n\nk\n\nj\n\nas defined by the trellis.\n\n• D. LLR computation\n\nAfter the forward and backward iterations have been carried\n\nout, the extrinsic LLRs for the k th bit are computed as\n\nL\n\nk\n\ne\n\n= max\n\n{s k ,s k+1 }∈U 1 α i\n\nk\n\n+ β\n\nj\n\nk+1\n\n+ γ s\n\nk\n\ni\n\n, s\n\nk+1\n\nj\n\nmax\n\nk {s k ,s k+1 }∈U 1 α i\n\nL\n\nk sys L\n\nk\n\np ,\n\n+ β\n\nj\n\nk+1\n\n+ γ s\n\nk\n\ni\n\n, s\n\nk+1\n\nj\n\nwhere\n\nthe\n\nsets\n\nU 1\n\nand\n\nU 1\n\nconnected by paths where u k\n\ndesignate the set\n\n=\n\n1\n\nand\n\nthe\n\nset\n\nof states\n\nof states\n\nconnected by paths where u k = 1, respectively.\n\nIII. TURBO DECODER IMPLEMENTATIONS\n\nAt a high level, both the Ivy-Bridge and Nvidia Kepler architectures can be viewed as multi-core SIMD processors. For the Intel CPU, we explicitly deploy SIMD instructions via Intel intrinsics to vectorize the MAP decoding algorithm. For the NVIDIA GPU, we used CUDA to parallelize\n\nthe workload. The CUDA compiler can easily vectorize the\n\nGPU computations, since in the CUDA programming model, threads execute the same set of computations, just on different input data. To achieve high decoding throughput, the BCJR algorithm outlined in Section II needs to be vectorized. We deploy the vectorization scheme put forward in , which vectorizes the BCJR algorithm into SIMD operations on vectors with\n\neight 8 bit elements to accelerate a UMTS turbo decoder on\n\nan Analog Devices DSP. In the following subsections, we\n\ncompare and contrast our turbo decoder implementations for the Intel CPU and the NVIDIA GPU.\n\n• A. SIMD data types\n\nFor the quad-core Intel Ivy-Bridge processor, each core can\n\nexecute SIMD instructions, supporting operations on various vector data types. Most hardware implementations of turbo decoders carry out fixed point calculations and use 10 bit- to-12 bit precision to achieve an error-rate performance close\n\nto a floating point implementation –. To achieve high throughput on the CPU, while maintaining good error-rate per- formance, we used vectors with eight 16 bit integer elements. The targeted Nvidia GTX 680 consists of 8 Next Generation SM (SMX) units, where each SMX unit is roughly equivalent to an Intel Ivy-Bridge core. An SMX unit can issue multiple 1024 bit SIMD instructions in parallel, where each instruction operates on vectors with 32 elements each having 32 bit. The architecture is optimized for single-precision floating-point\n\noperations (integer operations can be up to 6× slower). As a result, we used single-precision floating point operations in our GPU implementation. Since the computations for turbo- decoding consists of operations on vectors with 8 elements, we also decode at least 4 codewords in parallel to ensure a full utilization of the 1024 bit SIMD instruction.\n\n• B. Memory allocation\n\nAmong all possible codes in HSPA+ and LTE-A, the longest code is the LTE codeword with K = 6144 information bits. The length of the rate- 1 /3 encoded data is 18444 bit, which is the largest amount of memory required among all codewords. For our CPU implementation, we store all data as 16 bit values. The implementation requires 48 KB for input/output LLRs, and 96 KB for forward state metrics. Since the amount of L2 cache per core is 256 KB, all data fits into the cache. On the GPU, shared memory, a small amount of mem- ory (48KB per SMX) managed using explicit load and store instructions, can be used to cache data locally. Unfortunately, we cannot store data in the shared memory. This is because we decode at least 4 codewords in parallel to ensure the full utilization of the 1024 bit SIMD instruction and requires at least 4× the amount of storage, which outstrip the amount of available shared memory. Therefore, we store the input/output LLRs and forward state metrics in the device’s memory, which has high access latency, reducing the throughput of the design.\n\n• C. Multi-mode interleaver lookup table\n\nTo support HSPA+ and LTE-A, we need to support both interleaver types. Generating the HSPA interleaver addresses is rather complicated . To achieve high throughput, we decided to generate lookup tables which contain all possible interleaved and deinterleaved memory addresses instead of computing the addresses on-the-fly. For the Intel architecture, we store the lookup table in memory and rely on the fact that the entries in the lookup table will be cached. For the Nvidia architecture, we explicitly copy the correct entries of the lookup table at the start of the decoding iteration into constant memory, a small amount of read-only cache available on the GPU; this enables efficient lookup table accesses.\n\n• D. Max operation\n\nFor the max-log approximation, we simply omit the correc- tion term of max operator, as defined in (2), and approximate max as a simple max operation followed by scaling the extrinsic LLRs by a factor of 0.7. For both the CPU and GPU implementations, the max-log approximation corresponds to a vertical (element-wise) max instruction. Since the GPU supports element-wise logarithm and expo- nential functions, we can implement the log-MAP algorithm directly. Overall, the log-MAP algorithm requires one vertical max instruction, one vector subtraction, one vector absolute value, one call to the element-wise log function, one call to element-wise exponential function, and one vector addition. The Intel architecture does not support scalar or vec- tor fixed-point logarithm or exponential functions. We therefore approximate the correction term c(a, b) =\n\nlog (1 + exp(−|a b|)) with a piece-wise linear function as c(a, b) = max{0.25(2.77 − |a b|), 0} . We then add this correction term to max{a, b}. This approximation requires 6 additional instructions compared to the max-log approxima- tion: one vector subtraction, one vector absolute value, one vector shift, one vector maximum, and one vector addition.\n\nE. Branch-metric computation\n\nIn order to compute all branch metrics for every state k,\n\nwe need to compute four branch metrics, γ\n\n• 0 and γ k and the\n\nk\n\n1\n\nnegated versions, γ\n\n• 0 and γ k\n\nk\n\n1\n\n(see Section II-B), which\n\nrequires scalar operations only. To parallelize the workload on the CPU and GPU, we fetch 8 consecutive systematic channel LLRs, 8 consecutive\n\nparity LLRs and 8 consecutive a priori LLRs at a time.\n\nWe then compute the branch metrics, {γ k+1 ,\n\n{γ\n\n0\n\nk+1 ,\n\n. . .\n\n0\n\n1\n\n1\n\n, γ k+8 } and\n\n. . .\n\n, γ k+8 }, in parallel. Finally, we compute the negated\n\nversions. In total, the branch metric computation requires two\n\nvector additions and three vector subtractions.\n\nF. Forward and backward traversal\n\nFigures 3(a) and 3(b) depict the vectorized implementation of the forward and backward state-metric computation units in (1) and (3), respectively. Compared to the implementa- tion in , we rearranged the placement of shuffles (data exchange between SIMD lanes) to increase the instruction- level parallelism (ILP). This approach does not increase the number of required shuffle operations and is beneficial for the Intel architecture, since multiple shuffles can be executed in parallel. Figure 3(c) depicts the computations used to generate the extrinsic LLRs, where β + and β are intermediate values computed while computing β k (see Figure 3(b)). On the Intel architecture, the α k and β k computations can be implemented very efficiently using intrinsics. The vector γ k is constructed using one shuffle instruction. The rest of the α k and β k computation consists of two vector additions, two 128 bit shuffles and two element-wise max operations. Since we use 16 bit for the forward state metrics, these metrics can overflow. Hence, we re-normalize the metrics by subtracting α k (1) from α k . To reduce the number of instructions required by this re-normalization step, we normalize the state metric only every 8 trellis steps. As a result, the overhead of renormalization is low, requiring three additional instructions (extract the first element, broadcast, and vector subtract) every 8 trellis steps. The same normalization scheme is used during the backward state metric computation phase. In our previous work , we emulated shuffle instructions with shared memory load and store instructions on the Nvidia architecture. One new feature of Kepler is that it explicitly supports shuffle instructions. We therefore replaced the shared memory operations with shuffles. As a result, the number of instructions for α k and β k computation is similar to that of the CPU implementation. Since all computations are carried out in floating point arithmetic, the forward and backward state metrics do not need to be normalized.", null, "Figure 3.\n\n(a) Vectorized computation of α k+1 for the 3GPP turbo code. The block vmax implements the vectorized element-wise max operator. (b)\n\nVectorized computation of β k1 for the 3GPP turbo code. (c) Vectorized LLR computation for the 3GPP turbo code. The block hmax reduces 8 elements in the input vector to one element using the max operator.\n\n• G. LLR computation\n\nAs shown in Figure 3(c), the LLR computation consists of two vector addition instructions, two hmax functions, and a few scalar operations. The function hmax reduces the elements to one element using the max operator; this can be accomplished using a tree reduction, which requires 3 shuffle instructions and 3 max operations on the CPU. Unfortunately, a tree reduction does not use all lanes of the SIMD instruction. To increase computational efficiency, we buffer the inputs to hmax for 8 consecutive stages in shared memory column wise, instead of evaluating hmax one stage at a time. We then apply the element-wise max operation row-wise to find the minimum in each column.\n\n• H. Multi-codeword decoding\n\nTransmitted frame for both LTE-A and HSPA+ consists of multiple codewords. Since both CPU and GPU are multi-core processors, we parallelize the workload across all available cores to maximize the peak decoding throughput. For the CPU, such a parallelization is straightforward. We assigned at least one codeword per core using OpenMP to maximize the core utilization and, hence, the throughput. The GPU heavily relies on multi-threading to hide pipeline stalls and memory-access stalls. A corresponding implemen- tation requires a large number of independent sets of in- structions . As a result, we assigned a large number of codewords to each SMX using CUDA. To reduce the number of codewords needed to achieve high throughput on the GPU, we employed a windowed decoding approach, which divides a codeword into P sections (or windows) and processes these sections in parallel . Since the forward and backward state metrics are computed from the first trellis stage to the last stage (in a continuous fashion), we exchange the forward and backward state metrics among different windows between the iterations. This approach reduces the performance loss associated with parallel processing. Nevertheless, there is a tradeoff between the number of windows and the decoding latency, which we will discuss in the following section.\n\nIV. IMPLEMENTATION RESULTS\n\nWe now show our implementation results on an Intel Core i7-3770K, a quad core Ivy Bridge processor, and a Nvidia", null, "Figure 4. FER performance of (a) log-MAP decoding and (b) max-log-MAP decoding for K = 6144 and 6 decoding iterations.\n\nGTX 680, a Kepler GK104 GPU. In our experiments, we used the Intel C++ Compiler 14.0 and the CUDA 5.5 toolkit. In the following, we first compare the error-rate performance in terms of the frame error rate (FER) with a floating point reference decoder , and then, we compare the throughput of our optimized turbo decoder implementations.\n\nA. FER performance\n\nWe evaluate the FER in an additive white Gaussian noise (AWGN) channel. For the following experiments, we used the longest LTE code-word, which consists of K = 6144 information bits, and a code rate of 1 /3. Since the GPU turbo decoder supports windowed decoding, we consider a different number of windows (all of equal size), where the number of windows is P ∈ {1, 32, 64, 96, 128}. Figure 4(a) compares the FER of the CPU decoder using the linear approximation detailed in Section II-B and the GPU decoder using log-MAP decoding. Figure 4(b) compares the FER of all turbo decoders using the max-log-MAP algorithm. As expected, the use of 16 bit precision for the CPU imple- mentation leads to a small loss (0.14 dB for the max-log- MAP case) in terms of FER. The linear approximation results in only a very small FER degradation (0.12 dB). For the GPU implementation, the FER performance with P = 1 matches the\n\nTable I\n\nPEAK THROUGHPUT FOR THE K = 6144, RATE- 1 /3 LTE TURBO CODE.\n\nCPU (Mbps)\n\nGPU (Mbps)\n\nI\n\nmax-log-MAP\n\nlinear log-MAP\n\nmax-log-MAP\n\nlog-MAP\n\n4\n\n6\n\n8\n\n10\n\n122.6\n\n76.2\n\n55.6\n\n46.9\n\n62.3\n\n40.0\n\n30.8\n\n25.1\n\n54.2\n\n37.0\n\n30.0\n\n24.9\n\n55.2\n\n35.5\n\n27.8\n\n23.7\n\nreference (floating-point) implementation; for larger window sizes P, the FER performance degrades only slightly. We note that the same behavior applies to the HSPA+ codes, which are not shown here for the sake of brevity.\n\nB. Peak decoding throughput\n\nTable I shows the peak throughput of our CPU and GPU implementations. The results for HSPA+ are similar; for log-MAP algorithm and 6 decoding iterations, we achieved 41.6 Mbps and 36.7 Mbps on CPU and GPU respectively. As expected, the throughput for both implementation is inversely proportional to the number of decoding iterations I. The CPU implementation appears to be instruction bounded as additional instructions increase the runtime proportionally. The number of instructions required to compute a forward state metric is 7 using the max-log approximation, whereas the use of the linear log-MAP approximation requires 6 additional instructions. As a result, the number of instructions required for linear log-MAP is 2× of max-log-MAP. For 6 decoding iterations, the throughput of the linear log-MAP approximation is 40 Mbps. The throughput of the linear log-MAP approxima- tion is approximately 2× lower than the throughput of max- log-MAP implementation, which is 76.2 Mbps. For the GPU implementation, we found the instructions per cycle (IPC) is low, typically 1. Using the Nvidia profiler, we found that instructions operands are often unavailable (due to memory latency), leading to a low execution unit utilization. As a result, additional instructions (that do not require ad- ditional memory access) can execute on the idling execution units and thus do not significantly degrade the throughput. Hence, the throughput of log-MAP is not significantly slower than that of the max-log-MAP algorithm on the GPU. For the CPU implementation, a workload of 8 parallel codewords is sufficient to reach the peak throughput. For the GPU implementation, significantly more codewords need to be processed in parallel. Given a codeword, increasing the number of windows, P, does not increase peak throughput as the number of computation and memory transactions required to process the codeword stays the same. Increasing P, how- ever, is an effective way of reducing the number of codewords required to reach peak throughput. This trend is similar to our previous implementation . On the Kepler architecture, we require 512 codewords for P = 1, 16 codewords for P = 32 and above, to reach the peak performance. In summary, our GPU implementation is up to 2× slower than that of our CPU implementation for the max-log ap- proximation, and only 1.2× slower for the optimal log-MAP\n\nalgorithm. For the Nvidia Kepler architecture, the maximum IPC is 6-to-7, while the achieved IPC is 1. The achieved IPC is low as operands of the instructions are often not ready due to memory access latency, leading to low execution unit utilization and low throughput. Coupled by the fact that CPU is clocked much faster than the GPU, the CPU implementation was able to outperform the GPU implementation. We emphasize that it is possible to further improve the decoding throughput on the GPU. For example, we can reduce the number of memory accesses via data compression, which reduces the time for which execution units wait for operands. Nevertheless, the CPU implementation seems to be better suited for SDR applications since we achieve higher throughput with 2× fewer number of parallel codewords, which significantly reduces the latency of the entire system.\n\nREFERENCES\n\n A. Vosoughi, G. Wang, H. Shen, J. R. Cavallaro, and Y. Guo, “Highly scalable on-the-fly interleaved address generation for UMTS/HSPA+ parallel turbo decoder,” in IEEE ASAP, June 2013, pp. 356–362. The 3rd Generation Partnership Project (3GPP), “Evolved universal terrestrial radio access (E-UTRA); multiplexing and channel coding, Tech. Spec. 36.212 Release-11,” 2012.\n\n\n\nK. Tan, J. Zhang, J. Fang, H. Liu, Y. Ye, S. Wang, Y. Zhang, H. Wu,\n\n• W. Wang, and G. M. Voelker, “Sora: High performance software radio\n\nusing general purpose multi-core processors,” in USENIX NSDI, Apr. 2009, pp. 75–90. C. R. Berger, V. Arbatov, Y. Voronenko, F. Franchetti, and M. Puschel, “Real-time software implementation of an IEEE 802.11 a baseband receiver on Intel multicore,” in IEEE ICASSP, May 2011, pp. 1693–\n\n\n\n1696.\n\nK. Tan, J. Zhang, J. Fang, H. Liu, Y. Ye, S. Wang, Y. Zhang, H. Wu,\n\n• W. Wang, and G. M. Voelker, “Soft-LTE: A software radio implementa-\n\ntion of 3GPP Long Term Evolution based on Sora platform,” in Demo in ACM MobiCom 2009, Sept. 2009. C. Berrou and A. Glavieux, “Near optimum error correcting coding and decoding: Turbo-codes,” IEEE Trans. on Commun., vol. 44, pp. 1261 – 1271, Oct. 1996. C. Benkeser, A. Burg, T. Cupaiuolo, and Q. Huang, “Design and optimization of an HSDPA turbo decoder ASIC,” IEEE JSSC, vol. 44, no. 1, pp. 98–106, Jan. 2009. C. Studer, C. Benkeser, S. Belfanti, and Q. Huang, “Design and implementation of a parallel turbo-decoder ASIC for 3GPP-LTE,” IEEE JSSC, vol. 46, no. 1, pp. 8–17, Jan. 2011. Y. Sun and J. R. Cavallaro, “Efficient hardware implementation of a highly-parallel 3GPP LTE/LTE-advance turbo decoder,” INTEGRATION, the VLSI journal, vol. 44, no. 4, pp. 305–315, Sept. 2011. Xilinx Corporation, 3GPP LTE turbo decoder v2.0, 2008.\n\n\n\n[Online]. Available: http://www.xilinx.com/products/ipcenter/DO-DI- TCCDEC-LTE.htm A. Nimbalker and Y. W. Blankenship and B. K. Classon and K. T. Blankenship, “ARP and QPP interleavers for LTE turbo coding,” in IEEE\n\nWCNC, Apr. 2008, pp. 1032–1037. L. Bahl, J. Cocke, F. Jelinek, and J. Raviv, “Optimal decoding of linear codes for minimizing symbol error rate,” IEEE Trans. on Inf. Theory, vol. IT-20, pp. 284–287, Mar. 1974. J. Vogt and A. Finger, “Improving the max-log-MAP turbo decoder,” IET Electron. Letters, vol. 36, no. 23, pp. 1937–1939, 2000. NVIDIA Corporation, CUDA Compute Unified Device Ar- chitecture Programming Guide, 2008. [Online]. Available:\n\nhttp://www.nvidia.com/object/cuda_develop.html K. Loo, T. Alukaidey, and S. Jimaa, “High Performance Parallelised 3GPP Turbo Decoder,” in IEEE Personal Mobile Commun. Conf., Apr. 2003, pp. 337–342. J.-F. Cheng and T. Ottosson, “Linearly approximated log-MAP algo- rithms for turbo decoding,” in IEEE VTC Spring, vol. 3, May 2000, pp.\n\n\n\n2252–2256.\n\nM. Wu, Y. Sun, G. Wang, and J. R. Cavallaro, “Implementation of a high throughput 3GPP turbo decoder on GPU,” J. of Signal Process. Syst., vol. 65, no. 2, pp. 171–183, Nov. 2011." ]

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[ "## Why aren't these equivalent?\n\nDiscussion of Common Lisp\njordan\nPosts: 4\nJoined: Wed Oct 22, 2008 1:47 pm\n\n### Why aren't these equivalent?\n\nHello, noob question here. I'm coding up some DO loop examples to help me understand it. I eventually got my example working, but I'm curious why a previous version did NOT work.\n\nThe working one:\n\nCode: Select all\n\n``````(defun add-inputs ()\n\"Add a series of numbers from input\"\n(do ((input (get-integer-from-input) (get-integer-from-input))\n(sum 0 (+ sum input)))\n((= input 0) sum)))``````\nThe non-working one, that I thought would be equivalent:\n\nCode: Select all\n\n``````(defun add-inputs-broken ()\n\"Add a series of numbers from input\"\n(do ((input (get-integer-from-input))\n(sum 0 (+ sum input)))\n((= input 0) sum)\n(setf input (get-integer-from-input))))``````\n\nJasper\nPosts: 209\nJoined: Fri Oct 10, 2008 8:22 am\nLocation: Eindhoven, The Netherlands\nContact:\n\n### Re: Why aren't these equivalent?\n\nCode: Select all\n\n``````(do ((i 0 (+ i 1)) (list nil (cons list i)) ((= i 10) list))\n(do ((i 0) (list nil (cons list i)) ((= i 10) list) (setf i (+ i 1))``````\nCompare those. the 'do body is done before the 'increment' operations.\n\njordan\nPosts: 4\nJoined: Wed Oct 22, 2008 1:47 pm\n\n### Re: Why aren't these equivalent?\n\nAh, I see your point, I was clobbering my initial value before making use of it with the other variable's increment statement.\n\nThanks!" ]

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[ "Directory: Mathimatics-Numerical algorithms\nPlat: matlab\nSize: 45KB\nDescription:   Matlab or C language is adopted to realize the programming of the marginal price calculation example of distribution network node, and various schemes are solved respectively to obtain the planning results. This paper analyzes the calculation results of marginal price of distribution network nodes under different conditions, and draws the research conclusion of marginal price of distribution network nodes based on game theory\n\nFile list:" ]

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[ "# `unreal.SoundWaveSpectralData`¶\n\nclass `unreal.``SoundWaveSpectralData`(frequency_hz=0.0, magnitude=0.0, normalized_magnitude=0.0)\n\nSound Wave Spectral Data\n\nC++ Source:\n\n• Module: Engine\n\n• File: SoundWave.h\n\nEditor Properties: (see get_editor_property/set_editor_property)\n\n• `frequency_hz` (float): [Read-Write] The frequency (in Hz) of the spectrum value\n\n• `magnitude` (float): [Read-Write] The magnitude of the spectrum at this frequency\n\n• `normalized_magnitude` (float): [Read-Write] The normalized magnitude of the spectrum at this frequency\n\nproperty `frequency_hz`\n\n[Read-Write] The frequency (in Hz) of the spectrum value\n\nType\n\n(float)\n\nproperty `magnitude`\n\n[Read-Write] The magnitude of the spectrum at this frequency\n\nType\n\n(float)\n\nproperty `normalized_magnitude`\n\n[Read-Write] The normalized magnitude of the spectrum at this frequency\n\nType\n\n(float)" ]

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[ "# This Is\n\n0 Comment\n\nThe diagram below shows an angle, 0, graphed in the xy-coordinate plane.\nSegment RT is the initial side of the angle, and segment RM is the terminal side.\nSegments RT and RM are radii of the unit circle centered at the origin R(0, 0).\nT\n\nD\nSE\nThe x-coordinate of point M is 1. What is the measure of 0 to the nearest" ]

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[ "# do engineers need linear algebra\n\nYou are viewing the article: do engineers need linear algebra at audreysalutes.com\n\n## do engineers need linear algebra\n\nLinear algebra is important to engineers because it enables an easier way of problem solving. Using matrices to solve a large system of equations makes the process much easier.", null, "## Do engineers need linear algebra?\n\nLinear algebra is important to engineers because it enables an easier way of problem solving. Using matrices to solve a large system of equations makes the process much easier.\n\n## Do engineers use linear equations?\n\nEngineering is one of the most well-known fields for using linear equations.24-Apr-2018\n\n## Is Linear Algebra for engineers hard?\n\nLinear algebra is hard. Linear algebra is one of the most difficult courses that most STEM majors will study in university. Linear algebra is not an easy class because it is a very abstract course and it requires strong analytical and logical skills.29-Oct-2021\n\n## Do engineers use linear algebra?\n\nLinear algebra is a branch of mathematics that is used by engineers and applied scientists to design and analyze complex systems. Civil engineers use linear algebra to design and analyze load-bearing structures such as bridges.12-Apr-2009\n\n## Why is algebra important in engineering?\n\nAlgebra is used throughout engineering, but it is most commonly used in mechanical, electrical, and civil branches due to the variety of obstacles they face. Engineers need to find dimensions, slopes, and ways to efficiently create any structure or object.\n\n## Why is linear algebra so useful?\n\nIn simpler words, linear algebra helps you understand geometric concepts such as planes, in higher dimensions, and perform mathematical operations on them. It can be thought of as an extension of algebra into an arbitrary number of dimensions. Rather than working with scalars, it works with matrices and vectors.24-Jul-2020\n\n## How do engineers use linear equations?\n\nLinear equations are used to calculate measurements for both solids and liquids. An electrical engineer, for example, uses linear equations to solve problems involving voltage, current and resistance.24-Apr-2018\n\n## Why is Linear Algebra important in electrical engineering?\n\nLinear algebra is essential to nearly every sub-discipline of electrical engineer. Linear algebra is commonly associated with vector spaces but is more simply a means to solving systems of linear equations. Using Kirchoff's Voltage/Current Laws, a system of equations can be formed for any electrical circuit.\n\n## Why is Linear Algebra important for engineering?\n\nLinear algebra is important to engineers because it enables an easier way of problem solving. Using matrices to solve a large system of equations makes the process much easier.\n\n## What kind of equations do engineers use?\n\nAnalytic geometry uses the principles of calculus and trigonometry to determine limits, vectors, integrals, mean values and derivatives. One of the more advanced math functions engineers must understand is differential equations.\n\n## Who uses linear equations in real-life?\n\nIn real-life situations where there is an unknown quantity or identity, the use of linear equations comes into play, for example, figuring out income over time, calculating mileage rates, or predicting profit. Most of the time mental calculations are used in some real-life situations without drawing a line graph.\n\n## Do engineers use linear equations?\n\nEngineering is one of the most well-known fields for using linear equations. Engineers include architects, surveyors and a variety of engineers in fields such as: biomedical.24-Apr-2018\n\n## Is Linear Algebra good for engineering?\n\nLinear algebra is important to engineers because it enables an easier way of problem solving. Using matrices to solve a large system of equations makes the process much easier.\n\n## Is Linear Algebra hard or easy?\n\nThe pure mechanics of Linear algebra are very basic, being far easier than anything of substance in Calculus. The difficulty is that linear algebra is mostly about understanding terms and definitions and determining the type of calculation and analysis needed to get the required result.23-Oct-2020\n\n## What is the hardest part of linear algebra?\n\nThe hardest part in linear algebra is defining a mathematical structure via a set of axioms and understanding baffling concepts such as linear independence, eigenvectors, and abstract vector space. Students generally struggle with understanding abstract concepts in linear algebra.29-Oct-2021\n\n## Is Linear Algebra for engineers hard?\n\nLinear algebra is hard. Linear algebra is one of the most difficult courses that most STEM majors will study in university. Linear algebra is not an easy class because it is a very abstract course and it requires strong analytical and logical skills.29-Oct-2021\n\napplications of linear algebra in engineering\n\napplications of linear algebra in engineering pdf\n\napplications of linear algebra in software engineering\n\napplications of linear algebra in civil engineering\n\nexplain any five applications of linear algebra to design and analyze the engineering systems\n\napplications of linear algebra in electrical engineering\n\napplications of linear algebra in engineering ppt\n\nhow is algebra used in civil engineering\n\nSee more articles in the category: Engine" ]

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", 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[ "# Equations and Inequalities Questions and Answers\n\n#### Overview:\n\n Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Equations and Inequalities Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.\n\nDirections: Two equations (I) and (II) are given in each question. On the basis of these equations, you have to decide the relation between $$x$$ and $$y$$.\n\n1. (I). $$\\frac{6}{\\sqrt{x}} + \\frac{4}{\\sqrt{x}} = \\sqrt{x}$$, (II). $$y^2 - \\frac{10^\\frac{5}{2}}{\\sqrt{y}} = 0$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (e) $$x = y$$\n\nSolution: $$\\frac{6}{\\sqrt{x}} + \\frac{4}{\\sqrt{x}} = \\sqrt{x}....(1)$$ $$\\frac{6 + 4}{\\sqrt{x}} = \\sqrt{x}$$ $$6 + 4 = \\sqrt{x} \\times \\sqrt{x}$$ $$x = 10$$ $$y^2 - \\frac{10^\\frac{5}{2}}{\\sqrt{y}} = 0....(2)$$ $$\\frac{y^{2 + \\frac{1}{2}} - 10^\\frac{5}{2}}{\\sqrt{y}} = 0$$ $$y^{\\frac{5}{2}} - 10^\\frac{5}{2} = 0$$ $$y^{\\frac{5}{2}} = 10^\\frac{5}{2}$$ $$y = 10$$ Hence, $$x = y$$.\n\n1. (I). $$\\sqrt{49} x + \\sqrt{441} = 0$$, (II). $$8y - \\sqrt{256} = 0$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (b) $$x \\lt y$$\n\nSolution: $$\\sqrt{49} x + \\sqrt{441} = 0......(1)$$ $$7x + 21 = 0$$ $$7x = -21$$ $$x = - \\frac{21}{7}$$ $$x = -3$$ $$8y - \\sqrt{256} = 0......(2)$$ $$8y - 16 = 0$$ $$8y = 16$$ $$y = 2$$ Hence, $$x \\lt y$$.\n\n1. (I). $$\\frac{12}{\\sqrt{x}} - \\frac{2}{\\sqrt{x}} = 5 \\sqrt{x}$$, (II). $$\\frac{\\sqrt{y}}{3} + \\frac{3 \\sqrt{y}}{18} = \\frac{1}{\\sqrt{y}}$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (e) $$x = y$$\n\nSolution: $$\\frac{12}{\\sqrt{x}} - \\frac{2}{\\sqrt{x}} = 5 \\sqrt{x}...(1)$$ $$\\frac{12 - 2}{\\sqrt{x}} = 5 \\sqrt{x}$$ $$10 = 5x$$ $$x = 2$$ $$\\frac{\\sqrt{y}}{3} + \\frac{3 \\sqrt{y}}{18} = \\frac{1}{\\sqrt{y}}...(2)$$ $$\\frac{6 \\sqrt{y} + 3 \\sqrt{y}}{18} = \\frac{1}{\\sqrt{y}}$$ $$\\frac{9 \\sqrt{y}}{18} = \\frac{1}{\\sqrt{y}}$$ $$9y = 18$$ $$y = 2$$ Hence, $$x = y$$.\n\n1. (I). $$x^2 - 256 = 144$$, (II). $$y^2 - 35y + 306 = 0$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (e) No relation can be established between $$x$$ and $$y$$\n\nSolution: $$x^2 - 256 = 144......(1)$$ $$x^2 = 144 + 256$$ $$x^2 = 400$$ $$x = \\sqrt{400}$$ $$x = \\pm 20$$ $$y^2 - 35y + 306 = 0......(2)$$ $$y^2 - 17y - 18y + 306 = 0$$ $$y \\ (y - 17) - 18 \\ (y - 17) = 0$$ $$(y - 17) \\ (y - 18) = 0$$ $$y = 17 \\ and \\ y = 18$$ Hence, No relation can be established.\n\n1. (I). $$2 \\sqrt{x} - \\frac{1}{3 \\sqrt{x}} = \\sqrt{x}$$, (II). $$\\frac{1}{\\sqrt{y}} + \\frac{2}{\\sqrt{y}} = 12 \\sqrt{y}$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (a) $$x \\gt y$$\n\nSolution: $$2 \\sqrt{x} - \\frac{1}{3 \\sqrt{x}} = \\sqrt{x}...(1)$$ $$\\frac{6x - 1}{3 \\sqrt{x}} = \\sqrt{x}$$ $$6x - 1 = 3x$$ $$3x = 1$$ $$x = \\frac{1}{3}$$ $$\\frac{1}{\\sqrt{y}} + \\frac{2}{\\sqrt{y}} = 12 \\sqrt{y}...(2)$$ $$\\frac{1 + 2}{\\sqrt{y}} = 12 \\sqrt{y}$$ $$12y = 3$$ $$y = \\frac{1}{4}$$ Hence, $$x \\gt y$$.\n\n1. (I). $$\\frac{12}{x^2} + \\frac{5}{x^2} - \\frac{7}{x^2} = \\frac{20}{x}$$, (II). $$5.25 + 2.50 = y + 7$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (b) $$x \\lt y$$\n\nSolution: $$\\frac{12}{x^2} + \\frac{5}{x^2} - \\frac{7}{x^2} = \\frac{20}{x}...(1)$$ $$\\frac{12 + 5 + 7}{x^2} = \\frac{20}{x}$$ $$\\frac{10}{x^2} = \\frac{20}{x}$$ $$20x^2 = 10x$$ $$x = \\frac{1}{2} = 0.5$$ $$5.25 + 2.50 = y + 7.....(2)$$ $$7.75 = y + 7$$ $$7.75 - 7 = y$$ $$y = 0.75$$ $$Hence, \\ x \\lt y$$\n\n1. (I). $$\\frac{x^{\\frac{1}{2}}}{64} = \\frac{81}{x^{\\frac{3}{2}}}$$, (II). $$y^{\\frac{1}{3}} \\times y^{\\frac{2}{3}} \\times 576 = 8 \\times y^2$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (d) $$x \\le y$$\n\nSolution: $$\\frac{x^{\\frac{1}{2}}}{64} = \\frac{81}{x^{\\frac{3}{2}}}...(1)$$ $$x^{\\frac{1}{2}} \\times x^{\\frac{3}{2}} = 81 \\times 64$$ $$x^2 = 81 \\times 64$$ $$x = \\sqrt{81 \\times 64}$$ $$x = \\pm \\ 9 \\times 8$$ $$x = \\pm \\ 72$$ $$y^{\\frac{1}{3}} \\times y^{\\frac{2}{3}} \\times 576 = 8 \\times y^2...(2)$$ $$y \\times 576 = 8 \\times y^2$$ $$576 = 8y$$ $$y = 72$$ $$Hence, \\ x \\le y$$\n\n1. (I). $$\\frac{2}{\\sqrt{x}} + \\frac{3}{\\sqrt{x}} = \\sqrt{x}$$, (II). $$y^2 - \\frac{5^\\frac{5}{2}}{\\sqrt{y}} = 0$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (e) $$x = y$$\n\nSolution: $$\\frac{2}{\\sqrt{x}} + \\frac{3}{\\sqrt{x}} = \\sqrt{x}....(1)$$ $$\\frac{2 + 3}{\\sqrt{x}} = \\sqrt{x}$$ $$2 + 3 = \\sqrt{x} \\times \\sqrt{x}$$ $$x = 5$$ $$y^2 - \\frac{5^\\frac{5}{2}}{\\sqrt{y}} = 0....(2)$$ $$\\frac{y^{2 + \\frac{1}{2}} - 5^\\frac{5}{2}}{\\sqrt{y}} = 0$$ $$y^{\\frac{5}{2}} - 5^\\frac{5}{2} = 0$$ $$y^{\\frac{5}{2}} = 5^\\frac{5}{2}$$ $$y = 5$$ Hence, $$x = y$$.\n\n1. (I). $$5x - 2x = 22.50 + 4.50$$, (II). $$\\sqrt{y + 40} = 4 + 3$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (e) $$x = y$$\n\nSolution: $$5x - 2x = 22.50 + 4.50....(1)$$ $$3x = 27$$ $$x = 9$$ $$\\sqrt{y + 40} = 4 + 3....(2)$$ $$\\sqrt{y + 40} = 7$$ $$y + 40 = 7^2$$ $$y + 40 = 49$$ $$y = 9$$ $$Hence, \\ x = y$$\n\n1. (I). $$\\sqrt{x + 12} = \\sqrt{100} - \\sqrt{36}$$, (II). $$y^2 = 235 - 210$$\n\n1. $$x \\gt y$$\n2. $$x \\lt y$$\n3. $$x \\ge y$$\n4. $$x \\le y$$\n5. $$x = y$$ or No relation can be established between $$x$$ and $$y$$\n\nAnswer: (e) No relation can be established between $$x$$ and $$y$$\n\nSolution: $$\\sqrt{x + 12} = \\sqrt{100} - \\sqrt{36}...(1)$$ $$\\sqrt{x + 12} = \\pm \\ (10 - 6)$$ $$\\sqrt{x + 12} = \\pm \\ 4$$ $$x + 12 = (\\pm \\ 4)^2$$ $$x + 12 = 16$$ $$x = 4$$ $$y^2 = 235 - 210....(2)$$ $$y^2 = 25$$ $$y = \\sqrt{25}$$ $$y = \\pm \\ 5$$ Hence, No relation can be established." ]

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[ "# Faraday’s Law of Electromagnetic Induction\n\nFaraday’s Law states that, whenever the flux of magnetic field through the area bounded by a closed conducting loop changes, an emf is produced in the loop. The produced emf is given by\n\nξ = -dØ/dt\nwhere Ø = ∫B.ds is the flux of the magnetic field through the area. Ø is called the magnetic flux.\n\nIt is the law of electromagnetic induction.The SI unit of magnetic flux is Weber which is equivalent to Tesla meter2Faraday’s Law is basically an experimental result. Faraday performed a sequence of experiments to arrive at the result mentioned above. We will discuss the experiment to better understand how emf is produced due to changing magnetic flux.\n\nMaterial Required: Conducting Loop, Bar Magnet and Galvanometer\n\nA conduction loop is connected to the Galvanometer and the Bar Magnet is slowly brought toward the loop along the axis of the loop. Now the bar magnet is taken away from the loop but again along the axis of the loop. Please perform the following steps from the simulator below. Be sure to tick the the field lines checkbox in the simulator below.\n\n• Take the bar magnet closer to the coil.\n• Take the bar magnet away from the coil.\n• Hold the bar magnet stationary.\n\n#### Observations:\n\n• As the bar magnet is brought nearer to the loop, there is a deflection in the Galvanometer needle. This means that a current is flowing in the loop in a particular direction as shown by the deflection of Galvanometer. But current can only flow if there is some emf. This means that there must have been some induced emf due to movement of bar magnet.\n• As the bar magnet is taken away from the loop, again there is a deflection in the Galvanometer needle but this time in opposite direction. What does this mean? This means that current flowing in the loop is in opposite direction. Thus the direction or polarity of emf changed which caused the flow of current in opposite direction.\n• When the bar magnet is held stationary, there is no deflection in the needle of Galvanometer.\n\n#### Conclusion and formulation of Faraday’s Law\n\nThus we have these three observations and we need to formulate how Faraday came to a conclusion.  See, as we bring the bar magnet toward the conducting loop, the strength of magnetic field increases. But does this mean that induced emf depend on strength of magnetic field alone? No, because we saw in experiment that when the bar magnet is held stationary near the loop, no current flows through the loop. Thus, the induced emf doesn’t only depend on the strength of magnetic field.\n\nNext, when we bring the bar magnet faster toward or away from the loop, the magnitude of deflection of Galvanometer increases. Well, this means induced emf depends on the rate of change of magnetic field passing through the conducting loop. Mind the word magnetic field passing through the loop. Thus more the number of magnetic field lines passing through the loop, more will be the magnitude of induced emf. Actually this is the magnetic flux. Magnetic flux is the number of magnetic field lines passing through a closed curve.\n\nTherefore, we can now say that induced emf depends on the rate of change of magnetic flux linking the conducting loop. What about the negative sign?\n\nFrom observation (1) and (2), we can at least say that the induced emf is directional in nature and depends on whether the flux linkage through the loop is increasing or decreasing. The significance of this negative sign is explained by Lenz in his famous Lenz’s Law.\n\n#### Summery\n\nThus to summarize, as per Faraday’s Law the induced emf is dependent on rate of change of flux linking through a conducting coil or loop. Therefore, to have an induced emf the flux through the coil must change.\n\nAs Magnetic Flux Ø = B.A where bold sign means vector form. Thus flux can be changed by varying the following:\n\n• Cross sectional area of the coil.\n• Magnitude or direction of magnetic field.\n• Both the cross sectional area and magnetic field.\n\nHope you fully understood the Faraday’s Law. If still you have any doubt, please write in comment box. Thank you!" ]

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[ "", null, "An Ecosystem of δ-Potentials - I An ethereal energy source in Blip 19th July 2020 HOME \"Why, for example, do we still not have an international center for climate predictions, which by current estimates would cost “only” \\$1 billion spread over 10 years? That’s peanuts compared to what particle physics sucks up, yet vastly more important. Or why, you may have wondered recently, do we not have a center for epidemic modeling? It’s because too much science funding is handed out on the basis of inertia. In the past century, particle physics has grown into a large, very influential and well-connected community. They will keep on building bigger particle colliders as long as they can, simply because that’s what particle physicists do, whether that makes sense or not. It’s about time society takes a more enlightened approach to funding large science projects than continuing to give money to those they have previously given money to. We have bigger problems than measuring the next digit on the mass of the Higgs boson.\" - Sabine Hossenfelder  in The World Doesn’t Need a New Gigantic Particle Collider, Scientific American, June 19, 2020. *** So far we have developed the idea of a discrete measurement space or j-space, which is solely based on the physical measurements of an infinite source.  Since physical measurements do not allow the precise determination of origin, j-space has to have lower bounds for the measurements of various physical quantities.  Similarly upper bounds must also exist.      In j-space the resource management is very important and different observers have different resources and thus different capacities to perform precise measurements.  The observer Obsc, with measurement capacity (v/c ~1), determines the lower and the upper bounds, based on which the guidelines for the measurements in Aku's universe Blip, are established.      The fabric of j-space consists of j-pixels,  which we have discussed earlier.  In next few blogs we will be discussing, how to introduce physical structures, cosmic and elementary both, in j-space fabric.  We will be using basic quantum mechanics concepts, such as Levinson's theorem,  bound states and  δ-potentials, along with Kruskal-Szekeres coordinates, in j-space. δ-potentials and bound states:      The presence of δ-potentials and subsequent bound states for a particle, is a standard problem in Quantum Mechanics.  The case of a single delta potential and the corresponding bound state for a particle of mass m, is shown below:", null, "The bound state Eb is same as the ground state if the potential is symmetric.   If there are more than one delta functions, there can be more than one bound states depending upon the magnitude of αδ(x).  Our objective in next few blogs, is to discuss the basic conjecture behind the discrete measurement space or j-space, that every object in j-space is in a state of measurement, thus performing a measurement as well as being measured itself. Why do elementary particles exist in j-space, to begin with?     Let us review the expression for the bound state energy in the presence of a delta function:", null, "The important quantities here are m, α, and ħ.  But first we need to understand what do we mean when we make the statement that \"a particle of mass m exists\"?     In a measurement space, a measurement is completed when the corresponding circuit", null, "is completed i.e. the observer has arrived back at its point of origin.      In the discrete measurement space the objective is to measure the Least-Energy Surface (LES) with the absolute precision, but clearly not all the observers can perform this measurement.  We also note that the physical world exists in region-I of Kruskal-Szekeres coordinates, shown below:", null, "The position of the measurement circuit in Kruskal-Szekeres coordinates is around the Event-Horizon Point (EHP), Rsch = 2MG.  In the following diagram various measurement circuits around EHP are shown.  We note that the conventional space and time exist in the external Region-1 only, or in other words a macroscopic observer ObsM, exists in the external Region-I only.", null, "The measurement circuits are shown in blue.  The macroscopic observer's capability to make measurements is decreasing as we move from the left to right in the external Region-1, and it is represented by the hyperbolic paths in black.  Sharper the curvature at the vertex, more difficult the measurement, i.e. more and more resources are needed to make a measurement as we move from the right to the left in the external Region-1.      In j-space, the measurement circuit is around LES and the intersection of an hyperbola and an arc of measurement circuit (shown in red) in the external Region-1, represents the measurement of an elementary particle, made by ObsM1.  Thus we have elementary particles corresponding to each red-arc (or equivalently the part of a blue-measurement circuit in Region-I) which potentially can be measured by a macroscopic observer ObsM, given ObsM has sufficient resources to measure it2.  In the following diagram, an observer can measure m2 and m3, but not m1.", null, "To measure the particle m1, the observer making measurement needs to be provided additional resources in the external Region-1.  This situation is shown by the dashed path in the following diagram.", null, "This is what particle-colliders are trying to do in a rather rudimentary way.  However it must be clearly understood that no matter how much resources are provided to an observer making measurements in the external region-1, LES can never be measured with an absolute precision.  At best we will have an enormous catalog of particles, afflicted with la maladie exotique, but no new fundamental principle is likely to be discovered.  For that we will have to analyze the results obtained by Astronomy and Cosmology. Nature of bound states in j-space     Let us consider an observer ObsM in j-space, with resources  in the range (Ei, Ef).  Providing a precise value of the energy to an observer in j-space is not possible, as it implies ΔE = 0.  According to uncertainty principle,  ΔEΔt > ħ,  ΔE = 0 means a state with infinite life-time is measured by ObsM.  An infinite life-time measured by ObsM, is finite (∞j) per the measurements of Obsc or Obsi, and hence Δt is always finite and subsequently ΔE is always greater than 0.      The observer ObsM will be measuring the potentials ranging from a shallow-well to a deep-well as shown below:", null, "We note that as the depth of the potential well increases the observer ObsM has less and less freedom in the free space.          Next we consider an observer Obsc who has infinite, ∞j, resources in j-space.  This observer Obsc will have values for energy ranging from 0j to ∞j, in j-space corresponding to the current information space (q = 3).   We want Obsc to measure a shallow- potential well in a higher information space (q = 2).3 In other words we want Obsc, if possible, to perform a zero entropy measurement in q = 2 information space.     For Obsc since it has ∞j resources, all the potential wells in q = 3 information space, are equivalent to shallow-wells and represent zero-entropy measurements.  However in the case of q = 2 information space, the situation becomes rather interesting.  A shallow-well in q = 2 space is equivalent to δ-potential in q = 3 space even for Obsc.  And δ-potential is, what Obsc is measuring even though Obsc has infinite, ∞j, resources  in q = 3 space, as shown below:", null, "In this case, all the states are inside the δ-potential and they form in essence a single bound state or a \"coherent\" state of the width ΔE ~ 0.  Thus the nature of the bound state in q = 3 measurement space, while measuring a shallow well from q = 2 space, is that of a coherent state Ecoh.  The measurement of δ-potential in Blip     Our interest is in understanding how the measurements of the coherent state Ecoh, manifest themselves into physical parameters measured in Blip.  Blip is represented by the measurements made by the macroscopic observer ObsM,  in the external region-I in K-S coordinates.      Since the energy of the coherent state Ecoh in q = 3 information space, has a minimum limit ∞j, the corresponding rest mass mo = E × (c2)-1, will be infinite or extremely heavy, even though the physical dimensions may be in Planck's domain, in the q = 3 space or equivalently in the external region-I of KS Coordinates.     The coherent state will have a minimum spread ΔEcoh ~ 0j, per ObsM measurements.  The uncertainty principle correlates ΔEcoh with the coherent state life-time tcoh in external region-I, as ΔEcoh × tcoh ~ ħ.  Since ΔEcoh is very small, the coherent state life-time tcoh will be exceeding large (~∞j).       Next, in external Region-I, an infinitesimal change in entropy of the coherent state dScoh and the change in the energy to bring about the infinitesimal change dScoh in the entropy dEcoh,  are correlated to the temperature Tcoh of the coherent state as:", null, "It will take infinite amount of resources (dEcoh ~ ∞), to bring about infinitesimal change in the entropy (dScoh ~ 0), of the coherent state under discussion.  Hence the temperature of the coherent state as measured by the macroscopic observer ObsM, will be exceedingly high. Finally an infinite energy-source in Blip     Therefore in the discrete measurement space, the shallow-well from a higher-information-space|q=2 is equivalent to a δ-potential in the current-information-space|q=3.  The information corresponding to the shallow-well|q=2, will be measured as a coherent-state|q=3 by a macroscopic observer ObsM, in Blip (equivalently q=3 information state or the external region-I of KS coordinates).", null, "This coherent state will result in an extremely massive entity in Blip, with an infinite life-time and infinite temperature.  The infiniteness of the coherent state Ecoh, disappears as Ecoh slowly radiates away, in the external Region-I (in a few billion years probably).    Lastly, the \"measured-mass\" and the temperature of the coherent state, as they are determined in Blip, are statistical quantities inversely related to the spread in the energy ΔEcoh, of the coherent state|q=3, similar to the life-time tcoh. ____________________ 1. The physical world of Blip, exists at Λ-plane. 2.  Action in j-space is defined as the resources provided to an observer traveling the path PQ, at the point P, such that the observer can arrive at the point Q unassisted.  In external region-I of KS coordinates, action translates into the physical variables, energy E or the momentum k.  3. Different information spaces do not agree upon the definitions of Lorentz and Möbius invariances. *** To be continued.. Previous Blogs: EPR Paradox-I  Nutshell-2015", null, "Information on www.ijspace.org is licensed under a Creative Commons Attribution 4.0 International License. Attribution — You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. No additional restrictions — You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits. This is a human-readable summary of (and not a substitute for) the license.", null, "" ]

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[ "Gyroscope\n\n## Homework Statement\n\n3.14\n(0.5) How is the fractional change in temperature related to $$\\frac{dT}{T}$$ the fractional change in pressure $$\\frac{dP}{P}$$ ?\n(IPhO 2006)\n\n## Homework Equations\n\n$$PV^{\\gamma}=\\rm constant$$\n\n## The Attempt at a Solution\n\nnC_vdT=-PdV (1), dV/dP=-nRT/(P^2)\n\nIf I substitute dV in the equation (1) it does not give the right result.\n\nCan someone help me, please? :tongue2:\n\nThanks my dear friends.\n\nmarcusl\nGold Member\nIt appears you are talking about an ideal gas? Then in\n$$P=nRT/V$$\nwe see right away that P changes the same way as T. This is shown mathematically as\n\n$$\\frac{dP}{dT}=\\frac{nR}{V},$$\n\n$$\\frac{dP}{P}=\\frac{nRdT}{PV}=\\frac{dT}{T}.$$\n\nGyroscope\nThanks marcusl for your help. But you are wrong. :( I already solved it. Thanks anyway my dear friend.\n\nmarcusl" ]

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[ "# zbMATH — the first resource for mathematics\n\nNumerical solution of fuzzy differential equations under generalized differentiability. (English) Zbl 1181.34005\nSummary: We interpret a fuzzy differential equation by using the strongly generalized differentiability concept. Utilizing the Generalized Characterization Theorem, we investigate the problem of finding a numerical approximation of solutions. Then we show that any suitable numerical method for ODEs can be applied to solve numerically fuzzy differential equations under generalized differentiability. The generalized Euler approximation method is implemented and its error analysis, which guarantees pointwise convergence, is given. The method’s applicability is illustrated by solving a linear first-order fuzzy differential equation.\n\n##### MSC:\n 34A07 Fuzzy ordinary differential equations 65L99 Numerical methods for ordinary differential equations\nFull Text:\n##### References:\n Buckley, J.J.; Feuring, T., Fuzzy differential equations, Fuzzy sets and systems, 110, 43-54, (2000) · Zbl 0947.34049 Chalco-Cano, Y.; Román-Flores, H., Comparison between some approaches to solve fuzzy differential equations, Fuzzy sets and systems, 160, 1517-1527, (2009) · Zbl 1198.34005 Nieto, J.J.; Rodríguez-López, R., Euler polygonal method for metric dynamical systems, Information sciences, 177, 587-600, (2007) Prakash, P.; Sudha Priya, G.; Kim, J.H., Third-order three-point fuzzy boundary value problems, Nonlinear analysis: hybrid systems, (2009) · Zbl 1196.34006 Kaleva, O., Fuzzy differential equations, Fuzzy sets and systems, 24, 301-317, (1987) · Zbl 0646.34019 Nieto, J.J.; Rodríguez-López, R.; Franco, D., Linear first-order fuzzy differential equation, International journal of uncertainty fuzziness knowledge-based systems, 14, 687-709, (2006) · Zbl 1116.34005 Song, S.; Wu, C., Existence and uniqueness of solutions to the Cauchy problem of fuzzy differential equations, Fuzzy sets and systems, 110, 55-67, (2000) · Zbl 0946.34054 Hüllermeier, E., An approach to modelling and simulation of uncertain systems, International journal of uncertainty fuzziness knowledge-based system, 5, 117-137, (1997) · Zbl 1232.68131 Bede, B.; Gal, S.G., Almost periodic fuzzy-number-valued functions, Fuzzy sets and systems, 147, 385-403, (2004) · Zbl 1053.42015 Bede, B.; Gal, S.G., Generalizations of the differentiability of fuzzy number value functions with applications to fuzzy differential equations, Fuzzy sets and systems, 151, 581-599, (2005) · Zbl 1061.26024 Bede, B.; Rudas, I.J.; Bencsik, A.L., First order linear fuzzy differential equations under generalized differentiability, Information sciences, 177, 1648-1662, (2007) · Zbl 1119.34003 Chalco-Cano, Y.; Román-Flores, H., On new solutions of fuzzy differential equations, Chaos, solitons & fractals, 38, 112-119, (2008) · Zbl 1142.34309 Hüllermeier, E., Numerical methods for fuzzy initial value problems, International journal of uncertainty fuzziness knowledge-based systems, 7, 439-461, (1999) · Zbl 1113.65311 Abbasbandy, S.; Allahviranloo, T., Numerical solutions of fuzzy differential equations by Taylor method, Journal of computational methods in applied mathematics, 2, 113-124, (2002) · Zbl 1019.34061 Abbasbandy, S.; Allahviranloo, T.; Lopez-Pouso, O.; Nieto, J.J., Numerical methods for fuzzy differential inclusions, Journal of computer and mathematics with applications, 48, 1633-1641, (2004) · Zbl 1074.65072 Allahviranloo, T.; Ahmadi, N.; Ahmadi, E., Numerical solution of fuzzy differential equations by predictor – corrector method, Information sciences, 177, 1633-1647, (2007) · Zbl 1183.65090 Friedman, M.; Ma, M.; Kandel, A., Numerical solution of fuzzy differential and integral equations, Fuzzy sets and systems, 106, 35-48, (1999) · Zbl 0931.65076 Khastan, A.; Ivaz, K., Numerical solution of fuzzy differential equations by Nyström method, Chaos, solitons & fractals, 41, 859-868, (2009) · Zbl 1198.65113 Bede, B., Note on “numerical solutions of fuzzy differential equations by predictor corrector method”, Information sciences, 178, 1917-1922, (2008) · Zbl 1183.65092 Diamond, P.; Kloeden, P., Metric spaces of fuzzy sets, (1994), World Scientific Singapore · Zbl 0843.54041 Puri, M.L.; Ralescu, D.A., Differentials of fuzzy functions, Journal of mathematical analysis and applications, 91, 552-558, (1983) · Zbl 0528.54009 Kaleva, O., A note on fuzzy differential equations, Nonlinear analysis, 64, 895-900, (2006) · Zbl 1100.34500\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

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[ "Solutions by everydaycalculation.com\n\n1st number: 2 0/2, 2nd number: 15/45\n\n4/2 + 15/45 is 7/3.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 2 and 45 is 90\n2. For the 1st fraction, since 2 × 45 = 90,\n4/2 = 4 × 45/2 × 45 = 180/90\n3. Likewise, for the 2nd fraction, since 45 × 2 = 90,\n15/45 = 15 × 2/45 × 2 = 30/90\n180/90 + 30/90 = 180 + 30/90 = 210/90\n5. After reducing the fraction, the answer is 7/3\n6. In mixed form: 21/3\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "#### CBSE CLASS 7 Mathematics NCERT Videos\n\n##### Tags: CBSE NCERT Class 7 Congruence of Triangles Textbook pdf, CBSE NCERT Class 7 Congruence of Triangles Solutions pdf, NCERT Solutions Class 7 Mathematics Congruence of Triangles pdf, NCERT Solutions Class 7 Mathematics pdf, NCERT Solutions Class 7 Congruence of Triangles pdf, NCERT Textbook Class 7 Congruence of Triangles pdf, NCERT Textbook Class 7 Mathematics Congruence of Triangles pdf, CBSE Class 7 Mathematics Congruence of Triangles videos, CBSE Class 7 Mathematics videos, CBSE NCERT Class 7 Mathematics Congruence of Triangles videos\n\nBuy CBSE CLASS7 Mathematics books from Amazon", null, "" ]

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[ " \"Conjecture\" related terms, short phrases and links", null, "Web keywen.com\nConjecture       Article     History   Tree Map\n Encyclopedia of Keywords > Proof > Conjecture Michael Charnine\n\n Keywords and Sections\n Review of Short Phrases and Links\n\nThis Review contains major \"Conjecture\"- related terms, short phrases and links grouped together in the form of Encyclopedia article.\n\n### DefinitionsUpDw('Definitions','-Abz-');\n\n1. The conjecture is fundamental to topology, the branch of math that deals with shapes, sometimes described as geometry without the details.\n2. Conjecture is to form an opinion from incomplete Information; to guess.\n3. The conjecture is true if the Hilbert space H is not separable (i.e.\n4. The conjecture is known to hold for curves, surfaces and abelian varieties.\n5. This conjecture was proved, as the Prime Number Theorem, in 1896, simultaneously by Jacques Hadamard and by Charles-Jean de la Vallée Poussin. (Web site)\n\n### FavorUpDw('FAVOR','-Abz-');\n\n1. We present various positive and negative results in favor of the conjecture.\n\n### Important Variables RelatedUpDw('IMPORTANT_VARIABLES_RELATED','-Abz-');\n\n1. The plan must identify important variables related to the conjecture, and specify how they are to be measured.\n\n### Gerd FaltingsUpDw('GERD_FALTINGS','-Abz-');\n\n1. It was eventually proved by Gerd Faltings in 1983, about six decades after the conjecture was made; it is now known as Faltings' theorem.\n\n### YearsUpDw('YEARS','-Abz-');\n\n1. This conjecture rests on the number of portraits of the same young woman painted in the early years of his stay in Amsterdam and before he met his bride. (Web site)\n2. The source of the money for his projects is a matter of some conjecture, with his supporters saying that it came from his years as a businessman in Moscow.\n\n### Original ConjectureUpDw('ORIGINAL_CONJECTURE','-Abz-');\n\n1. Freedman proved the conjecture for n = 4 in 1982 but the original conjecture remains open.\n\n### Original EnumerationUpDw('ORIGINAL_ENUMERATION','-Abz-');\n\n1. Contains the original enumeration of the 92 solids and the conjecture that there are no others. (Web site)\n\n1. In mathematics, the gradient conjecture, due to René Thom, was proved in 2000 by K. Kurdyka, T. Mostowski and A. Parusinski.\n\n### Equitable LiensUpDw('EQUITABLE_LIENS','-Abz-');\n\n1. It is a matter of conjecture how far equitable liens extend outside of the unpaid vendor's lien. (Web site)\n\n### Arnold ConjectureUpDw('ARNOLD_CONJECTURE','-Abz-');\n\n1. Another application is a new proof of the Arnold conjecture for area preserving diffeomorphisms closed oriented surfaces.\n\n### Infinite Dimensional Complex Hilbert SpaceUpDw('INFINITE_DIMENSIONAL_COMPLEX_HILBERT_SPACE','-Abz-');\n\n1. Conjecture 1 (Invariant Subspace Problem, ISP0) Let be an infinite dimensional complex Hilbert space, and let be a bounded linear operator.\n\n### Abelian VarietiesUpDw('ABELIAN_VARIETIES','-Abz-');\n\n1. For most abelian varieties, the algebra is generated in degree one, so the Hodge conjecture holds. (Web site)\n\n### Break BlowfishUpDw('BREAK_BLOWFISH','-Abz-');\n\n1. CONCLUSIONS I conjecture that the most efficient way to break Blowfish is through exhaustive search of the keyspace. (Web site)\n\n### Volsung CycleUpDw('VOLSUNG_CYCLE','-Abz-');\n\n1. I conjecture that it belonged originally to the Volsung cycle, and to the wer-wolf Sinfjoetli. (Web site)\n\n### Weil ConjectureUpDw('WEIL_CONJECTURE','-Abz-');\n\n1. The Weil conjecture on Tamagawa numbers proved resistant for many years. (Web site)\n2. Freitag and Kiehl Étale cohomology and the Weil conjecture. (Web site)\n\n### Taniyama-Shimura ConjectureUpDw('TANIYAMA-SHIMURA_CONJECTURE','-Abz-');\n\n1. The proof of the Taniyama-Shimura conjecture is a huge achivement. (Web site)\n\n### Inverse ConjectureUpDw('INVERSE_CONJECTURE','-Abz-');\n\n1. This result is sometimes referred to as the inverse conjecture for the Gowers norm (in high, but bounded, characteristic).\n2. The inverse conjecture has a number of consequences.\n3. Exercise 20 (Inverse conjecture for the Gowers norm, case) Let be such that, and let.\n\n### Hoop ConjectureUpDw('HOOP_CONJECTURE','-Abz-');\n\n1. The hoop conjecture is well confirmed in momentarily static spaces, but it has not been investigated systematically for the system with relativistic motion.\n\n### Goldbach ConjectureUpDw('GOLDBACH_CONJECTURE','-Abz-');\n\n1. High school pupils might know that the problem of Goldbach Conjecture is named after the Prussian born number theorist Christian Goldbach (1690-1764).\n\n### Birch-Swinnerton-Dyer ConjectureUpDw('BIRCH-SWINNERTON-DYER_CONJECTURE','-Abz-');\n\n1. Besides the research papers, there is a letter of Parshin and a paper of Zagier with is interpretations of the Birch-Swinnerton-Dyer Conjecture. (Web site)\n\n### Swinnerton-Dyer ConjectureUpDw('SWINNERTON-DYER_CONJECTURE','-Abz-');\n\n1. An investigation of the Birch and Swinnerton-Dyer conjecture, which ties together the constellation of invariants attached to an abelian variety.\n\n### DeligneUpDw('DELIGNE','-Abz-');\n\n1. Today I'll talk about a conjecture by Deligne on Hochschild cohomology and the little 2-cubes operad. (Web site)\n2. Cattani, Deligne, and Kaplan proved that this is always true, without assuming the Hodge conjecture. (Web site)\n3. The strongest evidence in favor of the Hodge conjecture is the algebraicity result of Cattani, Deligne and Kaplan. (Web site)\n\n### Collatz ConjectureUpDw('COLLATZ_CONJECTURE','-Abz-');\n\n1. If the Collatz conjecture is true, the program will always halt no matter what positive starting integer is given to it.\n2. The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.\n\n### Baum-Connes ConjectureUpDw('BAUM-CONNES_CONJECTURE','-Abz-');\n\n1. The Baum-Connes conjecture, now a long-standing problem, has been joined by others in a group known as the isomorphism conjectures in K-theory.\n\n### Poincare ConjectureUpDw('POINCARE_CONJECTURE','-Abz-');\n\n1. He plays topology and algebra off against each other to get a deceptively simple-looking reformulation of the Poincare Conjecture. (Web site)\n\n### Geometrization ConjectureUpDw('GEOMETRIZATION_CONJECTURE','-Abz-');\n\n1. In dimension three, the conjecture had an uncertain reputation until the geometrization conjecture put it into a framework governing all 3-manifolds. (Web site)\n2. The Fields Medal was awarded to Thurston in 1982 partially for his proof of the geometrization conjecture for Haken manifolds. (Web site)\n3. The geometrization conjecture is an analogue for 3-manifolds of the uniformization theorem for surfaces. (Web site)\n\n### Main ConjectureUpDw('MAIN_CONJECTURE','-Abz-');\n\n1. The Hauptvermutung of geometric topology (German for main conjecture), is the conjecture that topological manifolds and piecewise linear manifolds coincide.\n\n### Smooth Poincaré ConjectureUpDw('SMOOTH_POINCARÉ_CONJECTURE','-Abz-');\n\n1. The statement that they do not exist is known as the \"smooth Poincaré conjecture\". (Web site)\n2. Milnor 's exotic spheres show that the smooth Poincaré conjecture is false in dimension seven, for example. (Web site)\n\n### Novikov ConjectureUpDw('NOVIKOV_CONJECTURE','-Abz-');\n\n1. Lie algebras and algebraic K-theory and an introduction to Connes'work and recent results on the Novikov conjecture. (Web site)\n2. Novikov also carried out important research in geometric topology, and posed the Novikov conjecture. (Web site)\n3. Using the solution of the Novikov conjecture for special groups some applications to the classification of low dimensional manifolds are given. (Web site)\n\n### Epsilon ConjectureUpDw('EPSILON_CONJECTURE','-Abz-');\n\n1. The proof of epsilon conjecture was a significant step towards the proof of Fermat's Last Theorem. (Web site)\n\n### Calabi ConjectureUpDw('CALABI_CONJECTURE','-Abz-');\n\n1. His work on the Calabi conjecture for Kähler metrics led to the development of Calabi-Yau manifolds.\n2. When the Chern class is zero it was proved by Yau as an easy consequence of the Calabi conjecture.\n\n### Kepler ConjectureUpDw('KEPLER_CONJECTURE','-Abz-');\n\n1. In 1998 Thomas Hales, following the approach suggested by László Fejes Tóth in 1953, announced the proof of the Kepler conjecture.\n2. For the problem of finding the densest packing of spheres in three-dimensional Euclidean space, see Kepler conjecture.\n3. The Kepler conjecture, named after Johannes Kepler, is a mathematical conjecture about sphere packing in three-dimensional Euclidean space.\n\n### PresenceUpDw('PRESENCE','-Abz-');\n\n1. The presence of the eggs does not, in fact, produce more itching; this conjecture is only myth.\n\n### ParticularUpDw('PARTICULAR','-Abz-');\n\n1. Theorem (Lefschetz theorem on (1,1)-classes) Any element of is the cohomology class of a divisor on X. In particular, the Hodge conjecture is true for H 2. (Web site)\n2. In particular, we will discuss a conjecture of Hurewicz which is analogous to the Borel conjecture about strong measure zero sets of reals.\n3. Some familiarity with global analysis and partial differential equations is assumed, in particular in the part on the Calabi conjecture.\n\n### RelationshipUpDw('RELATIONSHIP','-Abz-');\n\n1. This book is full of conjecture about Maxfield Parrish's private life and his relationship with his model, Sue Lewin. (Web site)\n\n### GeneralityUpDw('GENERALITY','-Abz-');\n\n1. This conjecture stood for decades before being proved in generality.\n\n### CombinationUpDw('COMBINATION','-Abz-');\n\n1. We prove the Weinstein Conjecture on the solid torus via a combination of results due to Hofer et al.\n\n### UniversityUpDw('UNIVERSITY','-Abz-');\n\n1. The conjecture looks as though it has been solved by Wolfgang Haken and Kenneth Appel, both of the University of Illinois.\n2. It was not until 1976 that the four-color conjecture was finally proven by Kenneth Appel and Wolfgang Haken at the University of Illinois. (Web site)\n3. In 1998 Thomas Hales, currently Andrew Mellon Professor at the University of Pittsburgh, announced that he had a proof of the Kepler conjecture.\n\n### IndiaUpDw('INDIA','-Abz-');\n\n1. How long he remained there and how long he traversed the entire length and breadth of India, Nepal, Burma and Sri Lanka could only be a matter of conjecture. (Web site)\n2. Thus it is difficult to conjecture the extent to which ancient scientists of India were influenced by the scientific knowledge of Greeks. (Web site)\n\n### ProblemsUpDw('PROBLEMS','-Abz-');\n\n1. Finite fields are an active area of research, including recent results on the Kakeya conjecture and open problems on the size of the smallest primitive root. (Web site)\n2. Number theory also holds two problems widely considered to be unsolved: the twin prime conjecture and Goldbach's conjecture. (Web site)\n\n### ShowUpDw('SHOW','-Abz-');\n\n1. On the other hand, we show that the conjecture is valid if the closure of the set of points of uncountable character is scattered. (Web site)\n\n### LoveUpDw('LOVE','-Abz-');\n\n1. Fuller, another Anglican historian, describes it as the mere conjecture \"of those who love to feign what they cannot find\". (Web site)\n\n### Poincaré ConjectureUpDw('POINCARÉ_CONJECTURE','-Abz-');\n\n1. Institute for his proof of the Poincaré conjecture, which asserts that “Every simply connected, closed 3-manifold is homeomorphic to the 3 - sphere.\n2. He was responsible for formulating the Poincaré conjecture, one of the most famous problems in mathematics. (Web site)\n3. Henri Poincaré studied three-dimensional manifolds and raised a question, today known as the Poincaré conjecture. (Web site)\n\n### DimensionsUpDw('DIMENSIONS','-Abz-');\n\n1. The Generalized Poincaré conjecture is true topologically, but false smoothly in some dimensions.\n\n### RemarksUpDw('REMARKS','-Abz-');\n\n1. Cleomedes is accurate in some of his remarks on lunar eclipses, especially his conjecture that the shadow on the Moon suggests a spherical Earth. (Web site)\n2. By Michael Brooks; also includes a brief history of time travel in relativity, including remarks on wormholes and the chronology protection conjecture.\n\n### AnalyticUpDw('ANALYTIC','-Abz-');\n\n1. The conjecture relies on analytic and arithmetic objects defined by the elliptic curve in question. (Web site)\n2. The proof (the Solution of the Poincaré conjecture) is analytic, not topological. (Web site)\n\n### GeometryUpDw('GEOMETRY','-Abz-');\n\n1. Geometry is being used here in a precise manner akin to Klein 's notion of geometry (see Geometrization conjecture for further details). (Web site)\n\n### CategoriesUpDw('Categories','-Abz-');\n\n1. Proof\n2. Science > Mathematics > Topology > Algebraic Topology\n3. Information > Science > Mathematics > Mathematicians\n4. Eacute\n5. Encyclopedia of Keywords > Nature > Matter\n\n### Related Keywords\n\n* Algebraic Topology * Andrew Wiles * Calabi-Yau Manifolds * Cantor * Case * Claim * Cohomology * Complex * Correct * Curves * Dimension * Eacute * Elliptic Curves * Existence * False * Fields * Flawed * Formula * Gauss * Generalization * General Relativity * Graph * Groups * History * Hodge Conjecture * Homotopy Groups * Isomorphism * Manifold * Manifolds * Martyr * Mathematician * Mathematicians * Mathematics * Matter * Null Hypothesis * One-Way Functions * Origin * Outline * Poincar * Problem * Prof * Proof * Proofs * Proving * Result * Rumor * Space * Spacetime * Special Case * Special Cases * Speculation * Subject * Sufficient * Support * Terms * Theorem * Theory * Topology * True * Truth * Way * Whether * Wiles * Word Games * Work * Yau\n1. Books about \"Conjecture\" in Amazon.com", null, "", null, "Short phrases about \"Conjecture\"   Originally created: August 01, 2010.   Links checked: January 31, 2013.   Please send us comments and questions by this Online Form   Please click on", null, "to move good phrases up.\n0.0228 sec. a=1.." ]

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[ "", null, "Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °\n\nYou are not logged in.\n\n## #1 2020-08-24 08:32:59\n\nOtsdarva\nMember", null, "Registered: 2010-12-21\nPosts: 10\n\nFaced with this quadratic I was tasked with finding the vertex and y-intercept: -3/2x^2-6x\n\nI could just use the formula -b/2a but because I'm learning to fully understand rather than just memorize shortcuts, I take the long way.  Get rid of -3/2 by multiplying everything by -2/3 and left with: x^2+4x\n\nI complete the square to get (x+2)^2-4 (half of 4 is 2, 2 squared is 4, subtract a 4 to maintain equality).. but this is wrong and I still don't know why it's wrong.  This gives me the vertex (-2,-4) but the correct vertex is (-2,6).  Where am I messing up?  The only way I can reach (-2,6) as an answer is if I had -2/3 on the other side of the equation in the first step, to end up with x^2+4x=-2/3.  Then multiply the lone -4 at the end by -3/2.. but isn't that not possible?  If the equation is set to equal zero and I multiply everything by -2/3 in the beginning, wouldn't I just be left with zero on the other side of the equation?\n\nLast edited by Otsdarva (2020-08-24 08:34:55)\n\nOffline\n\n## #2 2020-08-24 19:23:01\n\nBob", null, "Registered: 2010-06-20\nPosts: 8,914\n\nhi Otsdarva\n\nMy method is\n\nFrom this you get roots of -4 and 0, and hence the x coordinate of the vertex is -2, leading to a y coordinate of +6.\n\nTaking out the -3/2 is OK but you need to maintain the factor as you complete the square, otherwise you are changing the quadratic.\n\nBob\n\nChildren are not defined by school ...........The Fonz\nYou cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei\nSometimes I deliberately make mistakes, just to test you!  …………….Bob", null, "Offline\n\n## #3 2020-08-25 04:15:16\n\nOtsdarva\nMember", null, "Registered: 2010-12-21\nPosts: 10" ]

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[ "# What's the difference between the automorphism and isomorphism of graph?\n\nWhat's the difference between the automorphism and isomorphism of graph?\n\n1. In graph theory, an isomorphism of graphs $G$ and $H$ is a bijection between the vertex sets of $G$ and $H$, $f \\colon V(G) \\to V(H) \\,\\!$ such that any two vertices $u$ and $v$ of $G$ are adjacent in $G$ if and only if $ƒ(u)$ and $ƒ(v)$ are adjacent in $H$.\n\n2. In the mathematical field of graph theory, an automorphism of a graph is a form of symmetry in which the graph is mapped onto itself while preserving the edge–vertex connectivity.\n\nCould you give a example to explain the difference of the automorphism and isomorphism from the graph $G$ to $G$ itself? Since not all the isomorphism from the graph $G$ to $G$ itself is automorphism.\n\nI have this question when I read this post, please find the key word An isomorphism is a bijective structure-preserving map...and there is a paragraph says that\n\nThe graph representation also bring convenience to counting the number of isomorphisms (the pre-factor). For example, for a graph $g$ with $V$ vertices, $E$ edges from some scalar theory, permutation of vertices is equivalent to relabeling of vertices, hence does not change the graph structure. The same goes to permutation of edges. Therefore, the number of graphs isomorphic to $g$ is $V!⋅E!$.\n\nMaybe the isomorphism here does not demand to preserve the vertex-edge realtion, only demand to preserve the structure.\n\nBy definition, an automorphism is an isomorphism from $G$ to $G$, while an isomorphism can have different target and domain.\n\nIn general (in any category), an automorphism is defined as an isomorphism $f:G \\to G$.\n\n• Not all the isomorphism from the graph $G$ to $G$ itself is automorphism. Since automorphism preserving the edge–vertex connectivity. – Eden Harder May 11 '14 at 11:11\n• @EdenHarder I don't understand what you mean. An isomorphism is structure-preserving as well, so it preservers the edge-vertex connectivity. – Fredrik Meyer May 11 '14 at 11:26\n• Thanks! I update the question. – Eden Harder May 11 '14 at 12:34\n• @EdenHarder They key word in the page you're linking to is the word \"structure preserving\". That means exactly that all the vertex-edge incidences are preserved. – Fredrik Meyer May 11 '14 at 12:48\n• Thanks! I update the question. – Eden Harder May 11 '14 at 12:55\n\nAs an example, consider the graphs $G$ and $G'$ on 4 vertices, labelled 1, 2, 3 and 4, where $G$ has edge set $\\{\\{1,2\\},\\{1,3\\},\\{2,3\\},\\{3,4\\}\\}$ and $G'$ has edge set $\\{\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{3,4\\}\\}$. Sketch both of these graphs !\n\nThen the permutation $\\alpha = (1, 2, 3, 4)$ (in disjoint cycle notation) is an isomorphism from $G$ to $G'$. Why ? Because, applying $\\alpha$ to the vertex labels in the edge set of $G$ we obtain the edge set of $G'$. Check this ! Since an isomorphism from $G$ to $G'$ exists, $G$ and $G'$ are isomorphic. However, since their edge sets are different, $G$ and $G'$ are not equal.\n\nThe permutation $\\beta = (1, 2)$ is an isomorphism from $G$ to $G$ itself, that is, an automorphism of $G$, also known as a symmetry of $G$. Check this, as above !\n\n• If you are reading this answer and didn't understand that parantheses notation have a look here: youtube.com/watch?v=X4_4Bqj6EdA It basically means map 1 to 2, 2 to 3, 3 to 4 and 4 to 1. – VenkiPhy6 Feb 25 '19 at 5:19\n\nAn automorphism of a graph $\\Gamma$ is an isomorphism from $\\Gamma$ to itself.\n\n• Not all the isomorphism from the graph G to G itself is automorphism. Since automorphism preserving the edge–vertex connectivity. – Eden Harder May 11 '14 at 11:12\n• @EdenHarder, an isomorphism takes and edge of $G$ to an edge of $H$ (read the definition that YOU wrote in the question!). Thus every isomorphism from $G$ to $G$ is an automorphism of $G$. BTW, your 2. is an excerpt from Wikipedia entry Graph Automorphism, if you'd have bothered to read the next sentence you would see: \"...That is, it is a graph isomorphism from G to itself.\" – DKal May 11 '14 at 11:27\n• Thanks! I update the question. – Eden Harder May 11 '14 at 12:34\n\nIf you want to visualize , think about the adjacency matrix of the graph. After applying the automorphism, it will look same as previous.\n\nSay, $A$ is the adjacency matrix of graph $G$ and $P$ is an automorphism matrix (permutation matrix). Then $P^{-1}AP = A$; Here $P$ is an automorphism (informally)." ]

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[ "### Annual report pursuant to Section 13 and 15(d)\n\n#### Derivative Financial Instruments (Details) - Schedule of conversion feature derivative valuations\n\nv3.23.1\nDerivative Financial Instruments (Details) - Schedule of conversion feature derivative valuations\n12 Months Ended\nDec. 31, 2022\n\\$ / shares\nInception [Member]\nFair Value Measurement Inputs and Valuation Techniques [Line Items]\nCommon stock price \\$ 25.5\nExpected volatility 100.00%\nExpected dividend yield 0.00%\nDiscount rate 37.70%\nInception [Member] | Minimum [Member]\nFair Value Measurement Inputs and Valuation Techniques [Line Items]\nExercise price \\$ 28.05\nRisk free rate 2.30%\nExpected term (years) 6 months 18 days\nPV factor \\$ 0.67\nInception [Member] | Maximum [Member]\nFair Value Measurement Inputs and Valuation Techniques [Line Items]\nExercise price \\$ 30.6\nRisk free rate 3.00%\nExpected term (years) 1 year 3 months\nPV factor \\$ 0.84\nConversion Derivative Valuations [Member]\nFair Value Measurement Inputs and Valuation Techniques [Line Items]\nCommon stock price \\$ 8.5\nExpected volatility 100.00%\nExpected dividend yield 0.00%\nDiscount rate 37.70%\nConversion Derivative Valuations [Member] | Minimum [Member]\nFair Value Measurement Inputs and Valuation Techniques [Line Items]\nExercise price \\$ 28.05\nRisk free rate 4.50%\nExpected term (years) 3 months 14 days\nPV factor \\$ 0.8\nConversion Derivative Valuations [Member] | Maximum [Member]\nFair Value Measurement Inputs and Valuation Techniques [Line Items]\nExercise price \\$ 30.6\nRisk free rate 4.70%\nExpected term (years) 8 months 12 days\nPV factor \\$ 0.91" ]

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[ "### Bobcat crz 52 price\n\nMar 27, 2014 · Inverse functions are used every day in real life. For example, when a computer reads a number you type in, it converts the number to binary for internal storage, then it prints the number out again onto the screen that you see – it’s utilizing an inverse function. A basic example involves converting temperature from Fahrenheit to Celsius. Here is a set of practice problems to accompany the Inverse Functions section of the Graphing and Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University.\n\n### Bowtech realm sr6\n\nA set of rules or Laws of Boolean Algebra expressions have been invented to help reduce the number of logic gates needed to perform a particular logic operation resulting in a list of functions or theorems known commonly as the Laws of 1 = 0 The Inverse (Complement) of a 1 is always equal to 0.\nfc_unit_4.2_test_prep_key.pdf: File Size: 548 kb: File Type: pdf Inverse Laplace Calculator\n\n### Fitbit versa keeps turning off\n\nSee more ideas about inverse functions, algebra, function. This is a foldable I use when first introducing inverse functions to my Algebra 1 students. Presented symbolically, graphically using the horizontal and vertical line test, and algebraically.\nPROPERTIES OF INVERSE FUNCTIONS • The graph of the inverse function is the reflection of the graph of the original relation. The line of reflection is y = x. • Functions f and g are inverses of each other provided: f(g(x)) = xandg(f(x)) = x • The inverse of a function is written as f ‑1(x), which is read If you're given a function and must find its inverse, first remind yourself that domain and range swap places in the functions. When you make that change, you call the new f ( x ) by its true name — f -1 ( x ) — and solve for this function. For example, follow the steps to find the inverse of this function\n\n### Multi step equations answers\n\nThe math functions available for structured text programs are listed in Figure 270. It is worth noting that these functions match the structure of those available for ladder logic. Other, more advanced, functions are also available - a general rule of thumb is if a function is available in one lan-guage, it is often available for others ...\n• If a function has an inverse then the graphs of y = f(x) and y = f−1(x) are symmetric about the line y = x. • The horizontal line test: a function f has an inverse, if and only if no horizontal line intersects its graph more than once. • If the domain of f is an interval if f is either an increasing or decreasing function on that ... Inverse of a function 'f ' exists, if the function is one-one and onto, i.e, bijective. Since trigonometric functions are many-one over their domains, we restrict their. (iii) Whenever no branch of an inverse trigonometric function is mentioned, we mean the principal value branch.\n\n### John deere 625 moco\n\nKuta Software - Infinite Algebra 2. Function Inverses. State if the given functions are inverses.\n1. Form composite functions. 2. Verify inverse functions. 3. Find the inverse of a function. 4. Use horizontal line test to determine if a function has an inverse function. 5. Use the graph of a one to one function to graph its inverse function. Forming Composite Functions STEPS: 1. Place one function inside the other and simplify. Inverse functions and their derivatives can be tricky. One of the trickiest topics on the AP Calculus AB/BC exam is the concept of inverse functions and their derivatives. In this review article, we'll see how a powerful theorem can be used to find the derivatives of inverse functions.\n\n### Syair hk sniperangka hari ini\n\nCBSE Test Papers class 12 Mathematics Inverse Trigonometric Functions. CBSE chapter wise practice papers with solution for class 12 Mathematics chapter 2 Inverse Trigonometric Functions for free download in PDF format. 12th Mathematics chapter 2 Inverse Trigonometric Functions have many topics.\nFor Parents. ASPEN Portal; Bullying Prevention & Intervention; Financial Assistance; Equity at Lexington Public Schools; Food Services; LPS Print Center; Online Payments 1. Relations and Functions 7 2. Inverse Trigonometric Functions 15 3. Matrices 20 4. Determinants 20 5. Continuity and Differentiation 33 6. Applications of Derivatives 39 7. Integrals 50 8. Applications of Integrals 69 9. Differential Equation 73 10. Vectors 82 11. Three-Dimensional Geometry 89 12. Linear Programming 99 13. Probability 102 ...\n\n### Bcm chevy colorado\n\n• Osprey ultralight padded organizer\n• #### Vddk error 30007 failed to connect to host or proxy\n\n• What is leah ashes roblox password\n\n• Nac hormones\n\n• #### Splunk case\n\nIndex of mkv 21 bridges\n\n### Juliantina full episodes eng sub\n\nNov 18, 2019 · Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 2 Inverse Trigonometric Functions. Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Maths Inverse Trigonometric Functions MCQs Pdf with Answers to know their preparation […]\nInverse functions make solving algebraic equations possible, and this quiz/worksheet combination will help you test your understanding of this vital process. The quiz questions will ensure that ..." ]

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[ "# Moments and Linear Motion\n\n## Homework Statement\n\nJust in case my interpretation of the diagram does not give a clear representation, here is a link to all of the questions. http://imgur.com/a/jsv9e\n\nThe question begins by asking me to state the necessary conditions for a body to be in a state of equilibrium.\n\nIt is a 1.5m wall on the left hand side to which a cable is connected from. This cable extends 2.5m to the right and is run over or connected to a beam (the diagram is very unclear to me but could you kindly help with both possibilities?), which is diagonally connected to a hinge at the bottom of the 1.85m wall, to where it is connected to a mass. The angle between the beam and cable, in the area where the cable runs over or is connected to the beam, is unknown and represented by θ. The beam has a weight of 900N in its centre and the mass has a weight of 4500N. The mass of the cable and beam are negligible.\n\nDraw a free body diagram showing the forces acting on the system. (3 marks)\n\nCalculate the tension T in the cable. (2 marks)\n\nCalculate the net force on the beam from the hinge. (3 marks)\n\nThe other question states that during the initial phase of take off, a BWIA jet has an acceleration of 4.0 ms^-2 lasting 5 seconds. The burner engines are then turned up to full power for an acceleration of 10 ms^-2. The speed needed for take off is 300 ms^-1.\n\nCalculate the:\n\n(i) length of the runway\n(ii) total time of take off\n\nHence sketch a velocity-time graph for the motion of the jet\n\n## Homework Equations\n\nI am not sure if this is regarding equations that the question gave me to use or equations that I personally used. The question did not give me any equations to use as I had to use my own brain to figure which ones were needed. However, we do not use calculus at our level so I used:\n\nMoments: F1 * d1 = F2 * d2\nLinear Motion: v^2 = u^2 + 2as and s = ut + 1/2 at^2\n\n## The Attempt at a Solution\n\nFor moments, I figured that the cable provides an anti-clockwise force to balance the clockwise forces of the beam and the mass (I included them both as I thought they were both connected to each other and the cable pulled them up). Thus T * 2.5 = (900cos35.6 * 2.5) + (4500 * 2.5)\n\nI got 35.6° from using tan θ = opposite/adjacent > tan θ = 1.85/2.5 > θ = tan inverse 1.8/2.5 = 35.6°\n\nI used cos as the vertical component is adjacent to the angle.\n\nSo T = 5231.79\nT = 5230N (3 s.f.)\n\nThen I figured that another force would be needed to balance the system as the cable would be pulling the entire system down to the left.\n\nSo to find the upward force:\ntotal upward force - total downward force = 0\nupward force - (5231.79 + 900cos35.6 + 4500) = 0\nupward force = 10463.58\nupward force = 10500N (3 s.f.)\n\nThen for linear motion:\n\n(i) u = 4 * 5 = 20ms^-1, v = 300ms^-1, a = 10ms^-2\n\nv^2 = u^2 + 2as\n300^2 = 20^2 + 2(10)s\n(300^2 - 20^2)/20 = s\n4480m = s\n\n(ii) s = ut + 1/2 at^2\n4480 = 20t + 1/2 (10)t^2\n4480 = 20t + 5t^2\n5t^2 + 20t - 4480 = 0\n\nUsing quadratic formula [-b +/- sqrt (b^2 - 4ac)]/2a\n\na = 5\nb = 20\nc = -4480\n\nI worked out t = 28s and t = -32s, discarded the negative answer and used 28s as the time for the second part of the motion. The first part took 5s as it accelerated from 0 ms^-1 to 20ms^-1 in 5s. So 28s + 5s = 33s (the total time of take off).\n\nRelated Introductory Physics Homework Help News on Phys.org\nSteamKing\nStaff Emeritus\nHomework Helper\nFor Problem 1, you should draw a free body diagram of the boom with all loads and reactions indicated.\n\nApply the equations of equilibrium to the boom in order to find the tensions and reaction at the attachment point of the boom.\n\nBTW, your moment equations are incorrect. If you take moments about the attachment point of the boom with the wall, you will see a couple of mistakes in your moment equation. Why did you multiply the weight of the boom by the cosine of the angle between the boom and the cable? Why didn't you do the same for the load?\n\nSteamKing\nStaff Emeritus\nHomework Helper\nFor Problem 2, you are not applying the equations of motion correctly.\n\nFor the first part of the takeoff, the jet accelerates at a constant velocity for a fixed amount of time.\n\nIt's not stated explicitly, but you should assume that when t = 0, the position of the jet s = 0 and the velocity of the jet v = 0. Applying the constant acceleration of 4 m/s^2 over the time interval of 5 sec. should give you the velocity directly at the end of this interval. You can also determine how far the jet has rolled in this time. This distance is part of the total distance required for takeoff.\n\nOnce the jet is rolling and the engines are throttled to max. thrust, then you have a situation where there is an initial velocity and a known acceleration and you have to find the time required to reach takeoff v = 300 m/s and find the distance it takes to reach this velocity.\n\nAhh, so for problem 2 I have to factor in the distance travelled from rest to 20 ms^-1 which is 50m. So the total length would be 50 + 4480 = 4530m.\n\nHowever for problem 1 I am still lost. I also do not understand what you mean by \"boom.\" Is that the beam? Are you saying that the cosine of the angle is irrelevant? This particular question is causing me serious headache." ]

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[ "## How many moles of CO2 are in 132 g of CO2?\n\nThe answer is 44.0095. We assume you are converting between grams CO2 and mole. You can view more details on each measurement unit: molecular weight of CO2 or mol This compound is also known as Carbon Dioxide.\n\nHow many moles of O are in 88g of CO2?\n\nHENCE – UNDER 44 GMS — NO OF MOLES OF O ITEM – 2 ( I.E O2) THEREFORE – UNDER 88 GMS – 2 x 2 i.e. 4 moles .\n\n### How many moles are in CO2?\n\nIf 1 mole of CO2 weighs 12+2X16 = 44g, then there’s 12/44 = 0.272 moles of CO2 in 12g. In 1 mole of CO2 there are 2 moles of oxygen atoms, and 1 mole of carbon atoms – so two thirds of the total moles.\n\nREAD ALSO:   What is prevention diabetes?\n\nHow do you calculate moles of CO2?\n\nCalculate the number of moles of CO2 by the formula n=PV/RT, where P is the pressure from Step 3, V is the volume from Step 2, T is the temperature from Step 1 and R is a proportionality constant equal to 0.0821 L atm / K mol.\n\n## How many atoms are in 88g of CO2?\n\n88g of CO2 is equal to 2 moles of CO2. one mole cotains 6.023 × 10²³ atoms.\n\nHow many o2 atoms will be present in 88g of CO2?\n\n44 gms of CO2 has 2 oxygen atoms as the molecular weight of CO2 is 44gms then 88gms means 2CO2 which has 4 oxygen atoms.\n\n### How many atoms are in a mole of CO2?\n\nLet’s consider the carbon dioxide molecule. We know it has the formula CO2, and this tells us that: 1 mole carbon dioxide contains 6.02 x 1023 molecules. 1 mole of carbon atoms per mole of carbon dioxide = 6.02 x 1023 carbon atoms.\n\nHow do you find moles of C in CO2?\n\nREAD ALSO:   How do you say love in Japanese?\n\nBecause 1 mol of carbon is contained in 1 mol of carbon dioxide, the moles of carbon are equal to the moles of carbon dioxide. To find the mass of carbon present in the CO2 in grams, we need only to multiply the moles of carbon by the molar mass of carbon.\n\n## How many oxygen atoms are present in 3 moles of CO2?\n\n1 molecule of CO2 = 1 C atom + 2 O atom. So, 3 moles of CO2 contains (3 x 2 x 6.02 x 10*23) O atoms = 3.61 x 10*24 O atoms. How many moles of Oxygen atoms are present in 5 moles of Na2SO4?\n\nWhat is the molar mass of 1g CO2?\n\nmolecular weight of CO2 or mol This compound is also known as Carbon Dioxide. The SI base unit for amount of substance is the mole. 1 grams CO2 is equal to 0.022722366761722 mole.\n\n### What is the difference between grams CO2 and moles?\n\nWe assume you are converting between grams CO2 and mole. You can view more details on each measurement unit: molecular weight of CO2 or mol This compound is also known as Carbon Dioxide. The SI base unit for amount of substance is the mole. 1 grams CO2 is equal to 0.022722366761722 mole." ]

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[ "# What does it mean that Ouroboros is formally verified?\n\nI was talking with some friends about Cardano the other day and I brought up the fact that Ouroboros is formally verified. One of my friends asked me to explain what that means and why it's significant. I did my best but I realized I wasn't confident with my grasp of the concept. Here's how I explained it:\n\nFormal verification has 2 aspects:\n\n1. the algorithm is described in a mathematical way explaining assumptions and outcomes (like the threat model). Proofs and lemmas are provided which provide a theoretical baseline for the algorithm's properties. Then this mathematical description is transcribed into a computer language specifically designed for proof evaluation to make sure the humans who wrote the proof didn't forget/miss anything.\n2. Once the theoretical model has been validated it needs to be implemented in code and this implementation leaves a lot of room for error. After the developers are finished implementing the protocol in their language of choice (in this case Haskell?) the implementation is run through another piece of software (something like static analysis) which somehow evaluates all possible code paths (in a general heuristic sense, maybe not literally) and compares those code paths with the computer model created in step 1.\n\nIf the implementation matches the theoretical model, then we can say the code is formally verified.\n\nIs this explanation correct? How accurate is my explanation? What's missing? How can I improve it?" ]

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[ "# How do you graph, find the zeros, intercepts, domain and range of f(x)=abs(x+4)?\n\nJul 26, 2018\n\nBelow\n\n#### Explanation:\n\nFor zeroes, let $y = 0$\n$0 = \\left\\mid x + 4 \\right\\mid$\n$0 = x + 4$ or $- \\left(x + 4\\right) = 0$\nSince both equations will get me the same value, I'll use $0 = x + 4$\n\n$x + 4 = 0$\n$x = - 4$\n\nFor intercepts, let $x = 0$\n$f \\left(0\\right) = \\left\\mid 0 + 4 \\right\\mid$\n$f \\left(0\\right) = 4$\n\nDomain: all reals $x$\nRange: $y \\ge 0$\n\ngraph{abs(x+4) [-10, 10, -5, 5]}" ]

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[ "# T-SQL implementation of FizzBuzz\n\n• Eric M Russell\n\nSSC Guru\n\nPoints: 125088\n\nComments posted to this topic are about the item T-SQL implementation of FizzBuzz\n\n\"Do not seek to follow in the footsteps of the wise. Instead, seek what they sought.\" - Matsuo Basho\n\n• gsc_dba\n\nSSCertifiable\n\nPoints: 5407\n\nNice solution.\n\nOne less line of code: 🙂\n\n`SELECT`\n\n``` CASE WHEN [x].[n] % 15 = 0 THEN 'FizzBuzz' WHEN [x].[n] % 3 = 0 THEN 'Fizz' WHEN [x].[n] % 5 = 0 THEN 'Buzz' ELSE CAST([x].[n] AS VARCHAR(9)) END AS [p] FROM ( SELECT ROW_NUMBER() OVER ( ORDER BY [id] ) AS [n] FROM [sys].[sysobjects] ) [x] WHERE [x].[n] <= 100 ORDER BY ```\n\n` [x].[n];`\n\ngsc_dba\n\n• akljfhnlaflkj\n\nSSC Guru\n\nPoints: 76202\n\nNever heard of that interview question before.\n\n• Velveeta22\n\nTen Centuries\n\nPoints: 1289\n\nIf you are going for fewest number of characters in the exercise you can save one more with:\n\nWHERE n < 101\n\n• Eric M Russell\n\nSSC Guru\n\nPoints: 125088\n\nNever heard of that interview question before.\n\nI've never heard it or asked it during an actual interview. Someone else on the team had suggested using it once for interviews, but when I saw their baseline solution was a cursor with 30 lines of code, I wanted something better, so I came up with this.\n\n\"Do not seek to follow in the footsteps of the wise. Instead, seek what they sought.\" - Matsuo Basho\n\n• WILLIAM MITCHELL\n\nSSChampion\n\nPoints: 13692\n\nOne more version, using a \"single\" select statement 😉\n\n``` SELECT TOP 100 CASE WHEN ROW_NUMBER() OVER (ORDER BY object_id) % 15 = 0 THEN 'FizzBuzz' WHEN ROW_NUMBER() OVER (ORDER BY object_id) % 3 = 0 THEN 'Fizz' WHEN ROW_NUMBER() OVER (ORDER BY object_id) % 5 = 0 THEN 'Buzz' ELSE CAST( ROW_NUMBER() OVER (ORDER BY object_id) AS varchar) END FROM sys.objects ```\n\nSSChampion\n\nPoints: 11956\n\nIn an empty database I only got 97 numbers from sys.objects.\n\n``` WITH x(y) AS (SELECT top (100) ROW_NUMBER() OVER (ORDER BY id) FROM sys.syscolumns s) SELECT CASE WHEN y % 15 = 0 THEN 'FizzBuzz' WHEN y % 3 = 0 THEN 'Fizz' WHEN y % 5 = 0 THEN 'Buzz' ELSE CAST(y AS varchar) END FROM x ```\n\n• austin.harrison\n\nSSC Journeyman\n\nPoints: 81\n\nNon set-based solution.\n\nDoesn't depend on selecting from a system table that has the desired number of rows. (What if it was for numbers 1-1000000?)\n\nPreserves the integer datatype of the numbers. (If that is needed.)\n\n``` DECLARE @Count INT = 0 WHILE @Count < 100 BEGIN SET @Count = @Count + 1 IF (@Count%15 = 0) PRINT 'FizzBuzz' ELSE IF (@Count%3 = 0) PRINT 'Fizz' ELSE IF (@Count%5=0) PRINT 'Buzz' ELSE PRINT @Count END ```\n\nSSChampion\n\nPoints: 11956\n\nSimple. Just add a few cross joins for however many you need.\n\n``` WITH x(y) AS (SELECT top (1000) ROW_NUMBER() OVER (ORDER BY s.object_id) FROM sys.objects s CROSS JOIN sys.objects t) SELECT CASE WHEN y % 15 = 0 THEN 'FizzBuzz' WHEN y % 3 = 0 THEN 'Fizz' WHEN y % 5 = 0 THEN 'Buzz' ELSE CAST(y AS varchar) END FROM x ```\n\n• Eric M Russell\n\nSSC Guru\n\nPoints: 125088\n\nSimilar in concept to Oracle [DUAL] table, it would be useful if SQL Server had a system tally table, perhaps containing only a single INT column with rows from -2,147,483,648 to 2,147,483,647. Currently, a lot of debate is centered around how best to implement such a thing on the fly.\n\n\"Do not seek to follow in the footsteps of the wise. Instead, seek what they sought.\" - Matsuo Basho\n\nSSChampion\n\nPoints: 11956\n\nI currently use a table function to generate a range of numbers for joining to queries. I find it is easy for others to understand and is fast enough for everything we do. It includes an offset parameter to generate any range you may need. The code is below.\n\nSince sql server 2012 introduced Sequence objects which are brilliant by the way, I've been wondering if a Sequence may be the way to go.\n\n``` -- return range of rows for ss2012 or higher IF OBJECT_ID('dbo.GetOneMillionNumbers', 'IF') IS NOT NULL DROP FUNCTION dbo.GetOneMillionNumbers; GO CREATE FUNCTION dbo.GetOneMillionNumbers(@count AS int, @offset int) RETURNS TABLE AS RETURN ( WITH x (y) AS ( SELECT n1.na FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n1 (na) CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n2 (na) CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n3 (na) CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n4 (na) CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n5 (na) CROSS JOIN (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) AS n6 (na) ) SELECT top (@count) ROW_NUMBER() OVER(ORDER by (SELECT NULL)) + @offset AS MySequence from x ); GO -- SELECT * FROM dbo.GetOneMillionNumbers(775, 1000); -- SELECT max(MySequence) FROM dbo.GetOneMillionNumbers(1000000, -1000001); ```\n\nViewing 11 posts - 1 through 11 (of 11 total)" ]

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[ "Name:\n\n# Energy Lab\n\n## Background\n\n1. Work refers to an activity involving a force and movement in the directon of the force. A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work. 1\n2. Energy is the capacity for doing work. You must have energy to accomplish work - it is like the \"currency\" for performing work. To do 100 joules of work, you must expend 100 joules of energy. 1\n3. Power is the rate of doing work or the rate of using energy, which are numerically the same. If you do 100 joules of work in one second (using 100 joules of energy), the power is 100 watts. 1\n4. An open system is a system which continuously interacts with its environment.\n5. An isolated system (closed) is a physical system that does not interact with its surroundings. An isolated system obeys conservation laws: its total energy and mass stay constant.\n6. I can describe energy being transferred from one place to another in a system.\n7. I can describe energy being transformed from one form to another in a system and explain how energy is conserved in a closed system.\n\n1. http://hyperphysics.phy-astr.gsu.edu\n\n## Directions\n\n1. Lab station work: At each station draw and label a simple diagram showing:\n\n1. The device and its major components.\n2. The forms of energy present. There may be several forms at each station.\n3. How energy is being transformed. There may be transformations at each station.\n4. How energy is being transferred. There may be transferences at each station.\n5. Explain how this is an open or closed system.\n2. Completed Transfer Table\n3. Completed Transform table\n\nNote: A device may need your interaction for it to operate. You do not need to include the system that is you beyond the initial interaction.\n\nExample: the windup flashlight needs the handle to be rotated. Limit discussion of you to just that initial turning of the handle. This is about the device.\n\n## Stations\n\n1. Battery flashlight\n2. Windup flashlight\n3. Butane lighter\n4. Pendulum\n5. Ball and ramp\n6. Glow stick", null, "7. Spooldozer", null, "8. Mechanical", null, "timer", null, "9. Magnetic Coin detector\n10. Mechanical clicker\n11. Solar cell plant\n12. Real plant" ]

[ null, "https://ricksci.com/images/external.png", null, "https://ricksci.com/images/external.png", null, "https://ricksci.com/images/external.png", null, "https://ricksci.com/images/external.png", null ]

[ "### Description of Installment Loan Calendar Systems\n\n In the calculation of Installment Loans (including Home Equity loans, but not Mortgages), there are three common methods in use to measure the number of days between payments when computing the interest charge and/or Annual Percentage Rate.", null, "Our LoanMaker 23a App for the iPhone/iPad has all three of these calendar systems. Download on the App Store.\n\n#### Actual-over-365 (US Rule)\n\nThe actual-over-365 method counts the number of days in the first period (between the loan date and the first payment due date),  and then computes the first period interest charge by applying 1/365th of the annual rate for each day in the first period.    Interest is accrued using the U.S. Rule.\n\nFor each subsequent month, the actual number of days is counted between each payment due date and the next, and the interest charge for the period is computed by applying 1/365th of the annual rate for each day within the period (28, 29, 30 or 31 days).\n\nThis method is considerably harder to calculate.  It provides a slightly more precise payment than the Actual-to-first or Federal Calendar methods, but the gain in precision does not materially improve the accuracy of the payment and is usually of little value for these reasons:\n\n• Payments often cannot be made on their exact due date because of weekends, holidays, etc.,\n\n• Borrowers are often early or late in making the payments, and\n\n• The actual interest charge is assessed on a day-by-day basis based on the days the actual payments are made.  This means that the final payment will nearly always be different than the regular payment due to these variations in the payment pattern.\n\nThus, even when an exact payment is computed by this method, the borrower would still end up with an odd, final payment.\n\nTo check loans that were written with this method, we suggest you use the Actual-to-first method.\n\n#### Actual-to-First-Payment (US Rule)\n\nThe actual-to-first method counts the number of days between the loan date and the first payment due date,  and then computes the first period interest charge by applying 1/365th of the annual rate for each day in the first period.  Interest is accrued using the U.S. Rule.\n\nFor each subsequent month, interest is charged by applying 1/12th of the annual rate to the balance due during the month.  (This results in an average interest charge for 30.4166667 days per month.)\n\nThis method can be used to compute the payments for both simple-interest and pre-computed loans.\n\nThis method produces a payment and interest charge very close to the theoretical payment computed by the Actual-over-365 method but without the complexity of that method.  The APR can be disclosed either by the U.S. Escrow Rule by the Federal Calendar.\n\n#### \"Federal\" Calendar (US Rule / Actuarial)\n\nThe \"Federal\" calendar is the method described in U.S. Federal Reserve Regulation Z to measure the first period for the actuarial method of computing the Annual Percentage Rate.  This method is commonly used by recent software vendors, mostly because it is published.  It is well suited for pre-computed loans, however for simple-interest loans, the Actual-to-first or Actual-over-365 method would be preferred. Interest is accrued using the U.S. Rule or a simple/actuarial method.\n\nHowever the Federal calendar's intended purpose was to be used to calculate the A.P.R. by the Actuarial method, and not as a method to charge interest.  Under Regulation Z, lenders have the choice of using the actuarial method or the U. S. Escrow Rule to disclose the A.P.R.  Regulation Z clearly states that is has no authority over the charging of interest -- that is a matter for the states to regulate.  Because a majority of states require the use of the U. S. Escrow Rule to compute and charge interest on consumer loans (that is, the charging of interest on unpaid interest is not allowed), it makes more sense to use either the Actual-to-first or Actual-over-365 calendar systems to compute and disclose consumer loans.\n\n(The U.S. Escrow Rule prevents the charging of interest on unpaid interest, which can occur when a loan has a first period that is longer than one month.  Unpaid interest is held aside in an escrow account, and no additional interest is charged upon it.  In contrast, the actuarial method simply adds unpaid interest to the principal amount, allowing it to negatively amortize.  This results in additional interest charges accruing on the unpaid interest.)\n\nOne interesting feature of the Federal calendar is that it always computes 1 month to the first payment if the loan and first payment due date fall on the same day of the month.  E.g., February 15th to March 15th is considered 1 month, as is March 15th to April 15th.  For a simple-interest loan, these first periods would be 28 (in a non-leap year) days and 31 days, resulting is a significantly different interest charge for the first period (see note 2).\n\n#### Other Calendar Systems\n\nIn commercial lending (which is not subject to Regulation Z), it is common to see an Actual over 360 calendar system.  This system uses a daily rate that is 1/360th of the annual rate, and then assesses that rate for all 365 days in the year.\n\nThis results in an extra 5 days' interest for the lender. effectively increasing the interest rate.  E.g., a 12.5% loan on this calendar system will net the lender an actual rate of 12.674%.\n\nAlso, the old 360-day year calendar is occasionally found, mostly in commercial lending.  This system assumes every month has 30 days, and charges 1/360th of the annual rate per day.\n\nIf a loan has no odd-days in the first period, i.e., the loan date and first payment due date are 1 month apart, this calendar system will give the same result as the Federal calendar.  However, if there is a short or long first period, the 360-day year calendar is not acceptable for assessing an interest charge or disclosing the A.P.R.\n\n#### Which Calendar Method Should Be Used?\n\nIt depends on whether you are calculating loans or verifying loans that  have already been calculated, and also on what the type of loan is (simple-interest or pre-computed).\n\nTo calculate the payment for simple-interest loans, the most accurate payment will be calculated by the Actual-over-365 or Actual-to-first methods, although as noted above the payment will not be materially more accurate.  The actual-to-first method is sometimes preferred because the results are consistent from one month to another when the number of days to the first payment is the same.  The method is also far simpler to compute.  The results are very close to the Actual-over-365 method.  (Furthermore, Regulation Z states that one can ignore the fact that months have a different number of days when making disclosures.)\n\nTo calculate the payment for pre-computed loans, the most common method in use is the Federal calendar (mostly because the formulae are published in Regulation Z).  For indirect loans, you will find the Federal calendar is much more commonly used, however the actual-to-first method will provide a more accurate count of the days in the first period (resulting in a more accurate interest charge).  Also, pre-computed loans are much more common in indirect lending.\n\nWe note that in early 2005, GMAC switched to a simple-interest method for their auto loans abandoning the pre-computed method they had been using for years.\n\nIf you are verifying loans from an indirect (or direct) source, you should probably use the same method that was used to compute the loan.  If you use another method, you will see small differences in the payment and A.P.R.  These differences are not really differences in the A.P.R., but are caused by the variations in the way the lengths of the periods are measured and the effect upon the calculated payment.  The A.P.R. is a time-value measure of the cost of borrowing, and it is affected by both the Finance Charge and the amount of time the borrower has the use of the money.\n\n#### Pre-Computed vs. Simple-Interest Loans\n\nWith a pre-computed loan, if the borrower makes all the scheduled payments on time, he has fulfilled the loan.  If the borrower is late past the end of a grace period (typically 10 days) the lender can assess a late charge, however, no additional interest is charged.  Thus the interest charge has been \"pre-computed\" and will not change.\n\nSimple interest loans charge the borrower one day's interest for each day the loan is outstanding.  The interest is computed for each day between one payment and the next.\n\nPre-computed loans were commonly called Rule-of-78 loans by older loan processing systems because the pre-computed interest was accrued a month at a time using the Rule of 78's.  (Using the Rule of 78's is no longer allowed in nearly all, if not all, states.)\n\nHowever, one can still have pre-computed loans and use the Actuarial method to accrue the interest charges.  Today, it is common to have such pre-computed loans. The pre-computed interest is commonly apportioned a month at a time using an actuarial method (1/12th of the annual rate is applied to the balance to compute the interest for the month).  If a borrower elects to pay off the loan in the middle of a payment period, the lender is required to compute a payoff amount that only charges the borrower for the actual number of days that have elapsed in the period.\n\n Actuarial Method:  When used to compute an A.P.R., the actuarial method will result in either the same or slightly lower percentage than the U.S. Escrow Rule (thus making it slightly less stringent on lenders).  The actuarial method was probably adopted for this reason, as well as because it is easier to calculate than the U.S. Rule method.  Only the formulae for the actuarial method are published in Regulation Z. However, when the actuarial method formulae are \"turned-around\", that is, used to compute the interest charge rather than the A.P.R., the resulting charge can be higher than that developed with the U.S. Rule method.\n Note 2:  States with simple-interest laws typically specify that only 1 day's interest can be charged for each day (24-hour period).  Use of the Federal calendar to charge interest can result in 2, 3 or more days of interest being charged for one day. For pre-computed loans with one month to the first period, use of the Federal calendar is usually accepted." ]

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[ "Solutions by everydaycalculation.com\n\n## 8 is what percent of 30?\n\n8 of 30 is 26.67%\n\n#### Steps to solve \"what percent is 8 of 30?\"\n\n1. 8 of 30 can be written as:\n8/30\n2. To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100\n\n8/30 × 100/100\n3. = (8 × 100/30) × 1/100 = 26.67/100\n4. Therefore, the answer is 26.67%\n\nIf you are using a calculator, simply enter 8÷30×100 which will give you 26.67 as the answer.\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to work with percentages in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# Physics of the Car Accident. Building a Safe Campus by Solving Physics\n\n20 Slides915.50 KB", null, "Physics of the Car Accident. Building a Safe Campus by Solving Physics Problem Service-Learning Component of General Physics Course Elena Flitsiyan Department of Physics", null, "The Problem: Students have difficulty seeing connections between physics class and the “real world” Opportunities to help students these connections are not being fully realized in the current course design Implementing service learning elements in introductory physics course will result in improving the interactive learning component and also educate student community how prevent the car accidents", null, "Forces on an inclined road Often when solving problems involving Newton’s laws we will need to deal with resolving acceleration due to gravity on an inclined surface", null, "Forces on an inclined road What normal force does the surface exert? mgsin mgcos W mg y x", null, "Forces on an inclined road F F x mg sin y n mg cos", null, "Forces on an inclineed road F F x mg sin ma y n mg cos 0 Equilibrium", null, "Forces on an inclined road If the car is just stationary on the incline what is the (max) coefficient of static friction? Fx mg sin s n ma 0 F y n mg cos 0 mg sin s n s mg cos sin s tan cos", null, "Horizontal (Flat) Curve The force of static friction supplies the centripetal force v2 f S m r 2 v f S n S mg m r Solving for the maximum speed at which the car can negotiate the curve gives: v gr Fr Note, this does not depend on the mass of the car", null, "Banked Curve These are designed with friction equaling zero There is a component of the normal force that supplies the centripetal force (1), and component of the normal force that supplies the gravitational force (2). Dividing (1) by (2) gives: 2 v tan rg mv 2 n sin r (1) n cos mg ( 2)", null, "Suppose that a 1 800-kg car passes over a bump in a roadway that follows the arc of a circle of radius 20.4 m as in Figure. (a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 30.0 km/h? (b) What is the maximum speed the car can have as it passes this highest point without losing contact with the road?", null, "n mg mv 2 Fy may n mg r 2 2 8.33 m s v 2 n m g 1 800 kg 9.8 m s r 20.4 m 1 h 1 000 m v 30 km h 8.33 m s 3 600 s 1 km 1.15 10 4 N up n 0 mv 2 mg r v gr 9.8 m s 20.4 m 14.1 m s 2 50.9 km h", null, "“Centrifugal” Force From the frame of the passenger (b), a force appears to push her toward the door From the frame of the Earth, the car applies a leftward force on the passenger The outward force is often called a centrifugal force It is a fictitious force due to the acceleration associated with the car’s change in direction", null, "If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μs 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard?", null, "If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μs 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard? We adopt the view of an inertial observer. If it is on the verge of sliding, the cup is moving on a circle with its centripetal acceleration caused by friction. Fy may : n mg 0 Fx max : mv 2 f s n s mg r v s gr 0.8 9.8 m s 2 30 m 15.3 m s If you go too fast the cup will begin sliding straight across the dashboard to the left.", null, "Impulse Approximation In many cases, one force acting on a particle acts for a short time, but is much greater than any other force present When using the Impulse Approximation, we will assume this is true Especially useful in analyzing collisions The force will be called the impulsive force The particle is assumed to move very little during the collision p i and p f represent the momenta immediately before and after the collision", null, "Impulse-Momentum: Crash Test Example Categorize Assume force exerted by wall is large compared with other forces Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum Can apply impulse approximation", null, "Crash Test Example Analyze The momenta before and after the collision between the car and the wall can be determined Find Initial momentum Final momentum Impulse Average force Check signs on velocities to be sure they are reasonable", null, "Two-Dimensional Collision Example Conceptualize See picture Choose East to be the positive x-direction and North to be the positive y-direction Categorize Ignore friction Model the cars as particles The collision is perfectly inelastic The cars stick together", null, "Two dimensional collision m1 1500.0kg m2 2500.0 kg Find vf .", null, "Two dimensional collision m1 800.0kg m2 1400.0 kg Find vf . m1 v1 m2 v2 (m1 m2) vf (800kg) (25m/s) 0 (2200kg) vf cosθ – x-component (1400kg) (20m/s) 0 (2200kg) vf sinθ - y-component 1400 0.8 tan 800 54.50 (1400)(20) (2200)v f sin 54.50 v f (9.07iˆ)m / s (12.72 ˆj )m / s v f 15.63m / s 12.75 0 tan 89.6 9.07 1" ]

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[ "Anda di halaman 1dari 17\n\n# Questions on Thevenin Equivalent Circuits\n\nFall 2004\n2. Thevenin Circuits (25 points)\n\nLet V1=12V, R1=50 ohms, R2=10K ohms, R3=2K ohms, and R4=500 ohms. RL\nrepresents the load placed on the circuit between points Aand B.\n\n## b) Find the Thevenin resistance between A and B (7 points)\n\nc) Draw the Thevenin equivalent to the circuit shown on the previous page. Include the\nload resistor RL. (3 points)\n\n## e) What would be the voltage from A to B if RL were 10Meg ohms? (2 points)\n\nf) You use the DMM to measure the voltage across the load resistor given in part c (2K\nohms) and part d (10Meg ohms). In which circuit would the measured voltage be closer\nto the voltage you calculated? Why? (4 points)\nFall 2004 solution\n(none available)\nSpring 2004\n2) Thevenin circuits (20 points)\n\nIn the circuit above, V1=6 volts. R1= 50Ω, R2= 500Ω, R3= 800Ω, R4= 3000Ω\n\n## b) Find the Thevenin Resistance (8 points)\n\nc) If you place a load resistor of 2K between A and B, what would be the voltage at point A? ( 4 points)\nSpring 2004 solution\n2) Thevenin circuits (20 points)\n\nA: In the circuit above, V1=6 volts. R1= 50Ω, R2= 500Ω, R3= 800Ω, R4= 3000Ω\n\n## Vth = VA-VB = 5.78V\n\nb) Find the Thevenin Resistance (8 points)\n\nA ---R4--+------R1-----+-------B\n| |\n+----R2----R3--+\n\n## Rth = R4 + [(R1*R23)/(R1+R23)] R23=500+800=1300\n\nRth=3000+[(50*1300)/(50+1300)]=3048.15 ohms Rth=3048 ohms\nc) If you place a load resistor of 2K between A and B, what would be the voltage at point A? ( 4 points)\n\n## VA = [RL/(RL+Rth)]Vth = [2K/(2K+3048)]5.78 = 2.29V VA = 2.29V\n\nB: In the circuit above, V1=6 volts. R1= 50Ω, R2= 1000Ω, R3= 500Ω, R4= 2000Ω\na) Find the Thevenin Voltage (Voc) of the Circuit (8 points)\n\n## Vth = VA-VB = 5.806V\n\nb) Find the Thevenin Resistance (8 points) [see pictures for A]\n\n## Rth = R4 + [(R1*R23)/(R1+R23)] R23=1000+500=1500\n\nRth=2000+[(50*1500)/(50+1500)]=2048.38 ohms Rth=2048 ohms\nc) If you place a load resistor of 2K between A and B, what would be the voltage at point A? ( 4 points)\n\n## VA = [RL/(RL+Rth)]Vth = [3K/(3K+2048)]5.81 = 3.45V VA = 3.45V\n\nFall 2003\n2. Thevenin Circuits (20 points)\n\nR1 R4\n\nA\nV2\nR2 R3\n\nB\nLet V2=12V, R1=50 ohms, R2=1K ohms, R3=2K ohms, and R4=500 ohms.\n\n## e) Draw the Thevenin equivalent circuit with a load of 4K ohms. (4 points)\n\nFall 2003 solution\n2. Thevenin Circuits (20 points)\n\nR1 R4\n\nA\nV2\nR2 R3\n\nB\nLet V2=12V, R1=50 ohms, R2=1K ohms, R3=2K ohms, and R4=500 ohms.\n\n## Test A: Vth = VR23 R23=R2//R3=1K*2K/(1K+2K)=0.66667K\n\nVAB = 0.66667K/(0.66667K+50)(12V)=11.16V\nVAB = 11.16V\n\n## Test B: Vth = VR23 R23=R2//R3=2K*0.5K/(2K+0.5K)=0.4K\n\nVAB = 0.4K/(0.4K+50)(12V)=10.67V\nVAB = 10.67V\n\n## Test A: Rth = R4 + R123\n\nR123 = R1//R2//R3 Æ 1/R123 = 1/50 + 1/1K + 1/2K Æ R123 = 46.5 ohms\nRth = 500+46.5 = 546.5 ohms\nRth = 546.5 ohms\n\n## Test B: Rth = R4 + R123\n\nR123 = R1//R2//R3 Æ 1/R123 = 1/50 + 1/2K + 1/0.5K Æ R123 = 44.4 ohms\nRth = 3000+44.4 = 3044.4 ohms\nRth = 3044.4 ohms\nDraw the Thevenin equivalent circuit with a load of 4K ohms. (4 points)\n\nTEST A TEST B\nSpring 2003\n2. Thevenin circuits (20 points)\n\na) (6 points) Find the Thevenin voltage (Voc) of the circuit assuming the load will be\nconnected between A and B.\n\n## c) (4 points) Draw the Thevinen equivalent circuit with a load of 1K ohms.\n\nd) (4 points) Find the voltage between A and B for this circuit with a load of 1K ohms.\nSpring 2003 solution\n2. Thevenin circuits (20 points)\n\na) (6 points) Find the Thevenin voltage (Vth) of the circuit assuming the load will be\nconnected between A and B.\n\n## R245 = R2+R4+R5 = 1k+5k+3k = 9k\n\nR2345 = R3 // R245 = (9kx4k)/(9k+4k) = 2.77k\nV2345 = V1(R2345)/(R1+R2345) = 15x2.77/(2+2.77) = 8.71 volts\nV4 = V2345xR4/(R2+R4+R5) = (8.71x5k)/9k = 4.84 volts\n\n## R13 = R1//R3=(R1xR3)/(R1+R3) = (2kx4k)/(2k+4k) = 1.33k\n\nR1235 = R13 + R2 + R5 = 1.33K + 1K + 3k = 5.33 K\nR12345 = R1235//R4 = (5.33x5)/(5.33+5) = 2.58K ohms\n\n## c) (4 points) Draw the Thevinen equivalent circuit with a load of 1K ohms.\n\nd) (4 points) Find the voltage between A and B for this circuit with a load of 1K ohms.\n\nFall 2002\n\nR1 R3\n\nV1\nA B\n\nR2 R4\n\n## Given: V1=12 V; R1=2 kΩ; R2=4 kΩ; R3= 8 kΩ; R4= 4 kΩ\n\na) Find the Thevenin Voltage (Voc) of the circuit assuming the load will be connected\nbetween A and B. (6 points)\n\n## c) Find the current going from A to B if a 4 kΩ resistor is connected between A and B.\n\n(8 points)\nFall 2002 solution\n\n## 2. Thevenin circuits (20 points)\n\nR1 R3\n\nV1\nA B\n\nR2 R4\n\n0\nGiven: V1=12 V; R1=2 kΩ; R2=4 kΩ; R3= 8 kΩ; R4= 4 kΩ\n\na) Find the Thevenin Voltage (Voc) of the circuit assuming the load will be connected\nbetween A and B. (6 points)\n\n## Vth = VA – VB = V1*R2/(R1+R2) – V1*R4/(R3+R4)\n\nVth = 12*4K/6K – 12*4K/(8K+4K) = 8-4 = 4 volts\n\nRedraw circuit:\n\n## Rth = R1//R2 + R3//R4 = R1*R2/(R1=R2) + R3*R4/(R3+R4)\n\nRth = 2K*4K(2K+4K) + (8K*4K)/(8K+4K) = 4/3 + 8/3 = 12/3 = 4 Kohms\n\n(8 points)\n\n## V=IR Vth = I* (Rth+4K) 4 = I * (4K+4K)\n\nI = 0.5 mamps\nSpring 2002\n2. Thevenin circuits (20 points)\n\nR1 R2\nA\n2k 4k\n\nV1\n10V\nR3 R4\n1k 5k\n\n## c) Draw the Standard Thevenin Circuit, including a load Resistor ( 4 points)\n\nSpring 2002 solution\nFall 2001 solution" ]

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[ "1 2 Previous Next 29 Replies Latest reply on Jul 17, 2018 5:26 AM by Karthikeyan Masialamani Go to original post\n• ###### 15. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nHere's the polygon approach. The data set is quite simplistic. It's an Excel file with one two tabs as shown below. I've also attached the file.\n\nSections Tab", null, "Model Tab", null, "We'll use Model in Tableau to create the data densification required to draw the curves. Once in Tableau, we're using bins (on the Range field) and some calculated fields, plus a bit of trigonometry (for a refresher on trig, please read my blog post: Beyond \"Show Me\" Part 2: Trigonometry - Ken Flerlage: Analytics Architecture, Strategy, & Visualization), to draw the polygons. I'm attaching the workbook. All the calculations are fairly well commented in the calcs themselves, so I won't detail them here. If you have any questions, let me know.", null, "• ###### 18. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nGreat. Let me know if you have any questions.\n\n• ###### 19. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nwhat is the use of Range(Bin) here? I can see the 'Size of Bins' value as 1, if I change it to 0.5 I get the below one. If I change I to 2, it is totally different. can you explain what is going on behind the scene?", null, "", null, "2)  Angle Spacing :\n\nWINDOW_MAX(MAX(60/99))\n\n// Angle spacing between each point. Each slice is 60 degrees and we have 100 points.\n\n// But we want the first and last points of adjacent slices to appear on the same point...\n\n// ...so we adjust it as if there were only 99 points.", null, "Getting 0.6061 - is this value same for all the rows? it seems like that.\n\nHow do you test the value of all the calculated field values step by step to ensure the numbers are correct?\n\nI'm also not clear on 99 instead of 100. I can understand the bolded statement, but the second part is not clear i.e .100 point..\n\n60*3 ( slices) = 180 ???  points stands for something here?\n\n3) How to view/get X & Y Values outside tableau? If I click on 'ViewData', I am just getting below the columns only, but usually all the calculated field value should be displayed. Right? or If I go back and see the 'DataSource' part I can see at least the value of calculated field values. But here I can see 'Undefined'.", null, "", null, "How to test the output of these function for row by row?  I mean\n\nStep1: some number in Angel\n\nHow do you basically draw circle like this ? I mean do you test your numbers outside tableau ? or Verifying the output in tableau and then based on it modifying the calculated field numbers accordingly.\n\n• ###### 20. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nThere's a lot here. I will try to answer these properly some time this evening, if I am able to find time.\n\n• ###### 22. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nLet me try to answer a couple of these questions.\n\n1) Why Bins? I'm using bins to artificially create missing values. Essentially, I want 101 individual points. Point 101 will be at the center (0,0), and the rest of the points will be used to draw the curve. When you create bins on the Range field, which has 2 values--1 and 101, then use an \"Index\" table calculation, telling it to compute using the bins, it will essentially create those 101 points. This is kind of an advanced topic, but you can take a look at the following blog I wrote for more details. It doesn't specifically detail bins, but you'll see how I use them there and that may help with understanding: Beyond \"Show Me\" Part 3: Parametric Equations - Ken Flerlage: Analytics Architecture, Strategy, & Visualization . You can also check out this blog: https://www.thedataschool.co.uk/david-sanchez/densification-bins-index-tableau-normal-distribution/\n\nSince I want 101 points, I'm using a bin size of 1. If you used a bin size of 0.5 then you'd get 202 points, but the angle calculations are designed for 101 points, so that causes your polygons to get out of whack.\n\nInstead of using bins, you could just create a data set that has 101 records with values 1, 2, 3, 4...101. Then that would allow you to do something similar.\n\n2) Angle Spacing. The value will always be the same. I'm trying to find the spacing of the angle between each point. We want them equidistant so we get a smooth curve, so the angle spacing is always the same. The chart we're creating is half of a circle. A circle is 360 degrees, so half is 180 degrees. Since each slice is one third of the 180 degrees, they are each 60 degrees. We'll use 100 points to approximate each curve, but we want the first point to be at the beginning and the last to be and the end. Take for example the following line with 3 points plotted on it:", null, "Let's assume that the line is length of 1. If we wanted to plot 3 equidistant points, then we'd assume that you just divide the length by the number of dots. In our case, 1/3. But that would give you spacing of 1/3 for each dot, which would look like this when plotted:", null, "What we really want is the spacing to be 1/2. So instead of dividing 1 by the number of dots (let's call that \"n\"), we want to divide 1 by the n-1. In this case that would be 1/(3-1) or 1/2. Same with our example of 100 dots. We'll divide the total degrees of each slice (60) by n-1 or 99.\n\n3) I don't think there is a way to export the data with the x and y coordinates. This is because the table calculations used in this can be set up to operate different ways. For instance, Index is set up to be computed using Range (bins). I could change this and the results would be different (much like when you changed the bin size). This does not actually impact the underlying data. It only changes the view itself.\n\n4) When I first started doing this kind of thing in Tableau, I couldn't quite grasp the concept of using bins for data densification, so I did all of the math in Excel. Over time, I began to understand how to do it in Tableau and I now try to do as much of it in Tableau as possible, so people can download my work and understand what's being done. If I need to verify the values, I tend to just drop those fields on tooltip and hover over each one to make sure I'm getting what I expect. I also take small steps to get to the end result. For example, when I'm doing curves, I tend to just get everything laid out like the following first. Then I do the tooltip thing and make sure things are right before actually plotting the x and y coordinates.", null, "As I said, this is all pretty complex stuff, so it takes some time to figure out. I'd recommend checking out my series of blogs called \"Beyond Show Me\" which go into these in small steps. It starts by introducing the concept of creating custom visualizations by plotting x & y coordinates, then talks about how to leverage trigonometry to draw curves, then discusses some complex curves which are defined by parametric equations. You can find them here:\n\nBeyond “Show Me” Part 1: It’s All About the X & Y - Ken Flerlage: Analytics Architecture, Strategy, & Visualization\n\nBeyond \"Show Me\" Part 2: Trigonometry - Ken Flerlage: Analytics Architecture, Strategy, & Visualization\n\nBeyond \"Show Me\" Part 3: Parametric Equations - Ken Flerlage: Analytics Architecture, Strategy, & Visualization\n\nAlso, as noted earlier, you can do this without bins. So I'm attaching an Excel document and Tableau workbook which do it that way.\n\nI hope this helps. If this has answered your questions, please mark this as helpful or the \"correct answer\". This will help others with similar questions in the future. Thanks!!\n\n• ###### 23. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nKen,\n\n1) I can understand about BIN now, but i still need to play with it in tableau to get more knowledge. Out of curiosity, I am just asking the BIN will use Histogram kind of algorithm behind the scene ?\n\nTo generate the range of values we have to use BIN along with INDEX(), what about RANK() in BIN ? Does BIN support only INDEX()?\n\nCan I get few use cases where we think BIN while new requirement comes ? shall we use BIN to identify and fill the missing years , say for example a security's performance data was received in a file, there was no record for 2015 and from 2010 to 2018 data was filled in that file, but 2015 is missing, in this scenario, we can use BIN to identify the missing YEAR? Correct?\n\n2) We'll use 100 points to approximate each curve....\n\nI Can understand the line and 3 points example, but when i come to our case..it is bit unclear for me...can we also use 50, 75 or any number here ?...how you defined 100 here ? i mean based on some equation? please let me know.\n\n• ###### 24. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\n1) If you create a histogram in Tableau, it will actually create Bins, just like we've done here. It's pretty much the same thing--we're just using it for a slightly different use case here. You can use lots of different functions in conjunction with Bins. INDEX() is a simple one as it allows us to create numbers 1, 2, 3, 4, etc. You can use lots of other functions though, depending on your use case. For the missing year example, it may not be necessary to use Bins as use of a continuous year/date field may sort of take care of that. All depends on your data though.\n\n2) When we draw curves in Tableau, we're really just approximating a curve. We're just plotting points then connecting them with straight lines. We could draw this curve with only 3 points if we like, but it would not look like a smooth curve. By using 100 points, it looks like a smooth curve and our eyes can't really tell the difference. You could use 50 or 75 or 1000 or whatever you like. In fact, 50 would probably give you a quite a smooth curve as well. I just chose 100 because I knew I'd get that smoothness from it, but you can play around with different numbers.\n\n1 of 1 people found this helpful\n• ###### 25. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nThanks Ken! I am on vacation leave today so I will be back to work tomorrow.\n\nI don’t see correct answer or useful kind of option from my phone so as soon as I back I will mark it accordingly\n\n• ###### 26. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nYeah, the mobile interface can be a bit problematic. I typically change it to the desktop view, which gives the full interface.\n\n• ###### 27. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nKen\n\nIf you have some free time can you please check my another question topic “which chart will consume more memory” and provide your thoughts\n\n• ###### 28. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n\nWhich question is that?\n\n• ###### 29. Re: Full Round - Gauge/Speedometer Kind of Circular Chart\n1 2 Previous Next" ]

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[ "VICES Q6 Q7 Plot the following poin...\nQuestion", null, "", null, "# VICES Q6 Q7 Plot the following points on a graph paper. A(4,-5); B(-5,-4); C(-9, 8); D(7,6); E(0,-5); F(-8, 0); G(4,0); H(O, 3). Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the larger sides lie on the x-axis and one of the vertices lies in the third quadrant.\n\n11th - 12th Class\nMaths\nSolution", null, "146", null, "4.0 (1 ratings)", null, "", null, "", null, "Quick and Stepwise Solutions Just click and Send", null, "OVER 20 LAKH QUESTIONS ANSWERED Download App for Free" ]

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[ "# BEP-7\n\nViews:\n\nCategory: Education\n\n## Presentation Description\n\nNo description available.\n\n## Presentation Transcript\n\n### BREAK EVEN POINT & ANALYSIS:\n\nBREAK EVEN POINT & ANALYSIS\n\n### DEFINITION:\n\nDEFINITION The break even point is the point where the gains equal the losses. The point defines when an investment will generate a positive return. The point where sales or revenues equal expenses. The point where total costs equal total revenues. There is no profit made or loss incurred at the break even point. It is the lower limit of profit when prices are set and margins are determined.\n\n### DEFINITION:\n\nDEFINITION At this point the income of the business exactly equals its expenditure. If production is enhanced beyond this level, profit shall accrue to the business and if it is decreased from this level, loss shall be suffered by the business.\n\n### FORMULA:\n\nFORMULA Break even point (of output) = (fixed cost) / (contribution per unit) Where, Contribution=selling cost-variable cost Fixed cost= Contribution- profit\n\n### FORMULA:\n\nFORMULA Break even point of Sales= 1. Fixed price x SP per unit Contribution per unit Fixed Cost x Total Sales Total Contribution\n\n### CALCULATION OF THE BREAK EVEN POINT:\n\nCALCULATION OF THE BREAK EVEN POINT VARIABLE COST- They are directly related to the volume of sales: that is these cost increase in proportion to the increase in sales and vice versa. FIXED COST - Fixed costs continue regardless of how much you can sell or not sell, and can be made up of such expenses as rent, wages, telephone account and insurance. These cost can be estimated by using last years figure as a basis, because they typically do not change. Formula= FIXED COST VARIABLE COSTS\n\n### PowerPoint Presentation:\n\nAt break even point, the desired profit is zero. In case the volume of output or sales is to be computed for a desired profit, the amount of desired profit should be added to fixed cost is the formula given above. Units for a desired profit= Fixed cost+ desired profit Contribution per unit\n\n### PowerPoint Presentation:\n\nSales for a desired profit= (Fixed cost + Desired profit) (P/V Ratio)or profit volume ratio Where as, P/V ratio= Contribution per unit Selling price per unit = Total contribution Total sales\n\n### CASH BREAK EVEN POINT:\n\nCASH BREAK EVEN POINT It is the point where cash breaks even i.e. the volume of sales where cash realization on account of sales will be sufficient to meet the immediate cash liabilities. The label of activities where the total costs under two alternatives are same While calculating this point cash fixed costs (i.e. excluding fixed share of depreciation and deferred expenses) and cash contribution (i.e. after making adjustments for variable share of depreciation etc.) are considered.\n\n### FORMULA FOR CASH BREAK EVEN POINT:\n\nFORMULA FOR CASH BREAK EVEN POINT Cash break even point (in units) = (Cash fixed cost) / (cash contribution per unit) Cash break even point (in sales Rs.) = (Cash fixed cost) / (cash contribution per unit) x selling price per unit\n\n### BREAK EVEN ANALYSIS:\n\nBREAK EVEN ANALYSIS It refers to the ascertainment of level of operations where total revenue equals to total costs. Analytical tool to determine probable level of operation. Method of studying the relationship among sales, revenue, variable cost, fixed cost to determine the level of operation at which all the costs are equal to the sales revenue and there is no profit and no loss situation. Important techniques is profit planning and managerial decision making.\n\n### DEFINATIONS USED IN BREAK EVEN POINT-:\n\nDEFINATIONS USED IN BREAK EVEN POINT- Fixed Cost: The sum of all costs required to produce the first unit of a product. This amount does not vary as production increases or decreases, until new capital expenditures are needed. Variable Unit Cost: Costs that vary directly with the production of one additional unit. Expected Unit Sales: Number of units of the product projected to be sold over a specific period of time. Unit Price: The amount of money charged to the customer for each unit of a product or service.\n\n### ADVANTAGE :\n\nADVANTAGE It is cheap to carry out and it can show the profits/losses at varying levels of output. It provides a simple picture of a business - a new business will often have to present a break-even analysis to its bank in order to get a loan.\n\n### LIMITATIONS:\n\nLIMITATIONS Break-even analysis is only a supply side ( i.e. costs only) analysis, as it tells you nothing about what sales are actually likely to be for the product at these various prices. It assumes that fixed costs (FC) are constant It assumes average variable costs are constant per unit of output, at least in the range of likely quantities of sales. ( i.e. linearity) It assumes that the quantity of goods produced is equal to the quantity of goods sold (i.e., there is no change in the quantity of goods held in inventory at the beginning of the period and the quantity of goods held in inventory at the end of the period). In multi-product companies, it assumes that the relative proportions of each product sold and produced are constant ( i.e. , the sales mix is constant).\n\n### Profit Forecasting:\n\nProfit Forecasting Profit forecasting can not be done without proper profit forecasting, profit forecasting means projection of future earnings taking into consideration all the factors affecting the size of business profits, e.g., Firm’s pricing policies, costing policies, depriciation policy, and so on. Joel Dean has pointed out three approaches to profit forecasting.\n\n### 1. Spot Projection:\n\n1. Spot Projection Projecting the entire profits and loss statement for a specified future period by forecasting each element separately; forecasts are made about sales volume and prices and costs of producing the anticipated sales. Since profits are residual resulting from the forces that shape demand for the company’s products and govern the behavior of is costs.\n\n### 2. Break – even Analysis:\n\n2. Break – even Analysis Analysis identifying functional relations of both revenues and costs to output rate, with profits related to output as a residual; or alternatively, relating profits to output directly by the usual data used in break-even analysis.\n\n### 3. Environmental Analysis:\n\n3. Environmental Analysis Analysis relating the company’s profits to key variables in the economic environment, such as the general business activity and the general price level. These variables are general to the company.\n\n### THANKS !:\n\nTHANKS !", null, "" ]

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[ "# An object with a mass of 21 g is dropped into 400 mL of water at 0^@C. If the object cools by 160 ^@C and the water warms by 12 ^@C, what is the specific heat of the material that the object is made of?\n\nNov 27, 2017\n\nThe specific heat of the object is $= 5.98 k J k {g}^{-} 1 {K}^{-} 1$\n\n#### Explanation:\n\nThe heat is transferred from the hot object to the cold water.\n\nFor the cold water, Delta T_w=12ºC\n\nFor the object DeltaT_o=160ºC\n\n${m}_{o} {C}_{o} \\left(\\Delta {T}_{o}\\right) = {m}_{w} {C}_{w} \\left(\\Delta {T}_{w}\\right)$\n\nthe specific heat of water ${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$\n\nLet, ${C}_{o}$ be the specific heat of the object\n\nThe mass of the object is ${m}_{o} = 0.021 k g$\n\nThe mass of the water is ${m}_{w} = 0.4 k g$\n\n$0.021 \\cdot {C}_{o} \\cdot 160 = 0.40 \\cdot 4.186 \\cdot 12$\n\n${C}_{o} = \\frac{0.40 \\cdot 4.186 \\cdot 12}{0.021 \\cdot 160}$\n\n$= 5.98 k J k {g}^{-} 1 {K}^{-} 1$" ]

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[ "Find the next number in each sequence\n W I N I O N O W\nNumber Series\nThe Contest Center\nWappingers Falls, NY 12590\n W I N I O N O W\n\nNumber Series 1\nWhat are the next two numbers in this series? (Each number determines the next number. You do not need any outside knowledge, like subway stops, the periodic table, sizes of planets, etc.)\n\n1, 4, 8, 13, 21, 30, 36, 45, 54, 63, 73, ?, ?\n\nNumber Series 2\nWhat are the next two numbers in this series? (This is a purely arithmetic series, where each number determines the next number.)\n\n30, 33, 42, 50, 55, 80, 88, 152, 162, 174, ?, ?\n\nNumber Series 3\nWhat are the 12th, 21st, and 23rd numbers in this series? (Each number is uniquely determined by its position on the list. You do not need any outside knowledge, like the words to a song, the order of states entering the union, etc.)\n\n1, 4, 3, 11, 15, 13, 17, 24, 23, 73, 101, ?, ...\n\nNumber Series 4\nWhat are the next two numbers in this series? (Each number is uniquely determined by its position on the list. You will need some knowledge of mathematics for this puzzle.)\n\n3, 5, 11, 17, 29, 41, 59, 71, 101, ?, ?, ...\n\nNumber Series 5\nWhat are the next two numbers in this series? (These are successive numbers sharing a common property. No math or outside knowledge is needed.)\n\n3, 39, 41, 43, 45, 49, 51, 53, 55, 64, ?, ?, ...\n\nNumber Series 6\nWhat are the next six numbers in this series? (Each number is determined from the number before by a simple mathematical relationship. You will need some math.)\n\n2, 3, 4, 6, 8, 10, 13, 16, 20, 24, ?, ?, ?, ...\n\nNumber Series 7\nWhat are the next two numbers in this series? (This series is not mathematical. Some patriotism is required.)\n\n1, 6, 10, 2, 3, 4, 2, 3, 6, 6, 2, 7, ?, ?, ...\n\nNumber Series 8\nWhat are the next three numbers in this series? (Each number is determined from the number before by a simple mathematical relationship. You will need some math.)\n\n1, 2, 3, 5, 7, 10, 12, 14, 17, 19, 22, 25, ?, ?, ?, ...", null, "Send us an email to submit answers to these puzzles, or to submit new number series puzzles. We welcome new puzzles from our website visitors. Be sure to change the \\$ to an @ in our email address." ]

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[ "# sklearn.neighbors.DistanceMetric¶\n\nclass sklearn.neighbors.DistanceMetric\n\nDistanceMetric class\n\nThis class provides a uniform interface to fast distance metric functions. The various metrics can be accessed via the get_metric class method and the metric string identifier (see below).\n\nExamples\n\n>>> from sklearn.neighbors import DistanceMetric\n>>> dist = DistanceMetric.get_metric('euclidean')\n>>> X = [[0, 1, 2],\n[3, 4, 5]]\n>>> dist.pairwise(X)\narray([[ 0. , 5.19615242],\n[ 5.19615242, 0. ]])\n\n\nAvailable Metrics\n\nThe following lists the string metric identifiers and the associated distance metric classes:\n\nMetrics intended for real-valued vector spaces:\n\n identifier class name args distance function “euclidean” EuclideanDistance sqrt(sum((x - y)^2)) “manhattan” ManhattanDistance sum(|x - y|) “chebyshev” ChebyshevDistance max(|x - y|) “minkowski” MinkowskiDistance p sum(|x - y|^p)^(1/p) “wminkowski” WMinkowskiDistance p, w sum(|w * (x - y)|^p)^(1/p) “seuclidean” SEuclideanDistance V sqrt(sum((x - y)^2 / V)) “mahalanobis” MahalanobisDistance V or VI sqrt((x - y)' V^-1 (x - y))\n\nMetrics intended for two-dimensional vector spaces: Note that the haversine distance metric requires data in the form of [latitude, longitude] and both inputs and outputs are in units of radians.\n\n identifier class name distance function “haversine” HaversineDistance 2 arcsin(sqrt(sin^2(0.5*dx) + cos(x1)cos(x2)sin^2(0.5*dy)))\n\nMetrics intended for integer-valued vector spaces: Though intended for integer-valued vectors, these are also valid metrics in the case of real-valued vectors.\n\n identifier class name distance function “hamming” HammingDistance N_unequal(x, y) / N_tot “canberra” CanberraDistance sum(|x - y| / (|x| + |y|)) “braycurtis” BrayCurtisDistance sum(|x - y|) / (sum(|x|) + sum(|y|))\n\nMetrics intended for boolean-valued vector spaces: Any nonzero entry is evaluated to “True”. In the listings below, the following abbreviations are used:\n\n• N : number of dimensions\n\n• NTT : number of dims in which both values are True\n\n• NTF : number of dims in which the first value is True, second is False\n\n• NFT : number of dims in which the first value is False, second is True\n\n• NFF : number of dims in which both values are False\n\n• NNEQ : number of non-equal dimensions, NNEQ = NTF + NFT\n\n• NNZ : number of nonzero dimensions, NNZ = NTF + NFT + NTT\n\n identifier class name distance function “jaccard” JaccardDistance NNEQ / NNZ “matching” MatchingDistance NNEQ / N “dice” DiceDistance NNEQ / (NTT + NNZ) “kulsinski” KulsinskiDistance (NNEQ + N - NTT) / (NNEQ + N) “rogerstanimoto” RogersTanimotoDistance 2 * NNEQ / (N + NNEQ) “russellrao” RussellRaoDistance NNZ / N “sokalmichener” SokalMichenerDistance 2 * NNEQ / (N + NNEQ) “sokalsneath” SokalSneathDistance NNEQ / (NNEQ + 0.5 * NTT)\n\nUser-defined distance:\n\n identifier class name args “pyfunc” PyFuncDistance func\n\nHere func is a function which takes two one-dimensional numpy arrays, and returns a distance. Note that in order to be used within the BallTree, the distance must be a true metric: i.e. it must satisfy the following properties\n\n1. Non-negativity: d(x, y) >= 0\n\n2. Identity: d(x, y) = 0 if and only if x == y\n\n3. Symmetry: d(x, y) = d(y, x)\n\n4. Triangle Inequality: d(x, y) + d(y, z) >= d(x, z)\n\nBecause of the Python object overhead involved in calling the python function, this will be fairly slow, but it will have the same scaling as other distances.\n\nMethods\n\n Convert the true distance to the reduced distance. Get the given distance metric from the string identifier. Compute the pairwise distances between X and Y Convert the Reduced distance to the true distance.\n__init__(self, /, *args, **kwargs)\n\nInitialize self. See help(type(self)) for accurate signature.\n\ndist_to_rdist()\n\nConvert the true distance to the reduced distance.\n\nThe reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance.\n\nget_metric()\n\nGet the given distance metric from the string identifier.\n\nSee the docstring of DistanceMetric for a list of available metrics.\n\nParameters\nmetricstring or class name\n\nThe distance metric to use\n\n**kwargs\n\nadditional arguments will be passed to the requested metric\n\npairwise()\n\nCompute the pairwise distances between X and Y\n\nThis is a convenience routine for the sake of testing. For many metrics, the utilities in scipy.spatial.distance.cdist and scipy.spatial.distance.pdist will be faster.\n\nParameters\nXarray_like\n\nArray of shape (Nx, D), representing Nx points in D dimensions.\n\nYarray_like (optional)\n\nArray of shape (Ny, D), representing Ny points in D dimensions. If not specified, then Y=X.\n\nReturns\n——-\ndistndarray\n\nThe shape (Nx, Ny) array of pairwise distances between points in X and Y.\n\nrdist_to_dist()\n\nConvert the Reduced distance to the true distance.\n\nThe reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance." ]

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[ "Computes the discrete Fourier Transform sample frequencies for a signal of size n.\n\n## Usage\n\ntorch_fft_fftfreq(\nn,\nd = 1,\ndtype = torch_get_default_dtype(),\nlayout = torch_strided(),\ndevice = NULL,\n)\n\n## Arguments\n\nn\n\n(integer) – the FFT length\n\nd\n\n(float, optional) – the sampling length scale. The spacing between individual samples of the FFT input. The default assumes unit spacing, dividing that result by the actual spacing gives the result in physical frequency units.\n\ndtype\n\n(default: torch_get_default_dtype()) the desired data type of returned tensor.\n\nlayout\n\n(default: torch_strided()) the desired layout of returned tensor.\n\ndevice\n\n(default: NULL) the desired device of returned tensor. Default: If NULL, uses the current device for the default tensor type.\n\n(default: FALSE) If autograd should record operations on the returned tensor.\n\n## Note\n\nBy convention, torch_fft_fft() returns positive frequency terms first, followed by the negative frequencies in reverse order, so that f[-i] for all 0 < i <= n/2 gives the negative frequency terms. For an FFT of length n and with inputs spaced in length unit d, the frequencies are: f = [0, 1, ..., (n - 1) // 2, -(n // 2), ..., -1] / (d * n)\n\nFor even lengths, the Nyquist frequency at f[n/2] can be thought of as either negative or positive. fftfreq() follows NumPy’s convention of taking it to be negative.\n\n## Examples\n\nif (torch_is_installed()) {\ntorch_fft_fftfreq(5) # Nyquist frequency at f is positive\ntorch_fft_fftfreq(4) # Nyquist frequency at f is given as negative\n\n}\n#> torch_tensor\n#> 0.0000\n#> 0.2500\n#> -0.5000\n#> -0.2500\n#> [ CPUFloatType{4} ]" ]

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[ "Message-ID: <1170998030.16363.1635021151342.JavaMail.tomcat@d7e90b4c3fee> Subject: Exported From Confluence MIME-Version: 1.0 Content-Type: multipart/related; boundary=\"----=_Part_16362_1948227177.1635021151341\" ------=_Part_16362_1948227177.1635021151341 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Content-Location: file:///C:/exported.html PhpStorm 193.5662.16 Release Notes\n\n# PhpStorm 193.5662.16 Release Notes\n\n=20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 = =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20 =20\n PHP Feature WI-49636 = Remove \"php\" from const order in PHP = class when formatting code PHP Profiler Feature WI-40687 = Show memory column for Xdebug > 2.6 Bug WI-49773 Memory view: remove percents + add bytes to the = column PHPDoc Feature WI-48105 = Allow to configure PHPDoc reference/type and variable/params highlighting=20 Usability WI-50091 Missing example of PHPDoc|Variable coloring Code Analysis. Inspection Feature IDEA-227586 = = Warn on returning Nullable from Flux/Mono transformation should support met= hod reference Bug IDEA-227141 Quickfixes in the Inspections view are reordered= Bug IDEA-227840 Intention does nothing when rethrow exception=20 Code Analysis. Structural Searc= h Bug IDEA-226388 Existing template \"\n• not contained= in\nor\n\" doesn't work Core. Installation Task IDEA-227756 Bundle Kotlin 1.3.61 with IntelliJ IDEA Core. Platform API Bug IDEA-226561 Disposer.register should be documented to throw = IncorrectOperationException when the parent disposable is disposed Core. Plugin Management Bug IDEA-225789 Keymap is not added if Apply button is pressed a= fter keymap plugin installation Bug IDEA-224855 Plugin Manager: a lot of plugins are falsely mar= ked as uncompatible Bug IDEA-225468 \"Notification group Profiling is already re= gistered\" on installing first theme plugin Core. Run. Configurations Bug IDEA-224132 Run Configurations: the tasks added as beforeRun= tasks seem to be executed in parallel Cosmetics IDEA-227160 Add text disclaimer to run/debug configuration t= emplate page Editor. Editing Text Feature IDEA-225455 = = \\$SELECTION\\$ does not work in Live Template when part of another variable=20 Bug IDEA-227376 The \"File Structure\" popup no longer c= orrectly locates the function you select when dealing with collapsed areas<= /td>=20 Bug IDEA-226412 column mode is not revealed on status bar (regre= ssion) Bug IDEA-226837 Caret jumps to another line when moving using le= ft/right arrows Bug IDEA-227271 InlayHints not properly disabled Bug IDEA-227963 \"Enter/Smart indent\" option sets \"= ;AUTOCOMPLETE_ON_SMART_TYPE_COMPLETION\" instead of \"SMART_INDENT_= ON_ENTER\" Editor. Intention Actions Performance IDEA-227453 Freeze on showing available intentions Editor. TextMate Bundles Bug IDEA-226563 Navigating to php/bin/set_error_handler after fi= nd-in-path locks up Lang. XML Bug IDEA-226310 XSD validation ignores \"anyAttribute\"<= /td>=20 Lang. YAML Bug IDEA-226689 Helm. java.lang.AssertionError: Should come here= only for YAMLKeyValue with no value and a following indent at adding a mis= sing key from values.schema.json Tools. Docker Feature IDEA-200863 = = Ability to inspect the image of the running Docker container. Bug IDEA-227236 Dockerfile, image: the deployed container doesn'= t have log if chosen in another project Bug IDEA-226467 Docker. Absent errors in case of running a conta= iner with already allocated ports Bug IDEA-227975 Doing an docker-compose down using the Docker se= rvices panel causes named volumes to be removed/deleted (new in 2019.3)=20 Bug IDEA-227705 Add *.dockerfile file type pattern for dockerfil= e. Bug IDEA-203230 Docker. 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Git Bug IDEA-227242 Deleting directory with unversioned files (by ac= tion from Project View) does not remove corresponding unversioned files fro= m Local Changes Bug IDEA-226627 Git Clone Dialog: Trim the URL Version Control. Log Bug IDEA-227019 Git log tab bug with a branch named with parenth= esis Version Control. Subversion= Bug IDEA-224516 Intellij IDEA never ends (or as best, is very sl= ow) \"performing VCS refresh\". Se attached image. Problem appears = in all 2019.3 versions. Not a problem in version 2019.2.3. DB Connectivity Bug DBE-2684 call trim() on connection dialog fields DB Introspection Feature DBE-5036 = Create \"fake\" foreign key reference Data Views Bug DBE-9593 MongoDB collection view: ISODate truncated to ju= st YYYY-MM-DD format No subsystem Bug WEB-42670 Invalid destructuring type evaluation for an arr= ow function parameter with explicit annotation CSS Performance WEB-42613 CSS: CssColorGutterRenderer can cause freezes=20 Dart Feature WEB-41912 =20 = Support Dart web apps debugging using 'webdev daemon'-based workflow and VM= Service protocol (like Dart DevTools) Bug WEB-42526 NPE from DartServerRootsHandler.getEnclosingDart= PackageDirectory() Bug WEB-41978 NPE on autocompletion for Dart (SDK 2.2 or 2.3)<= /td>=20 Debugger Bug WEB-42393 Support debugger frames for dependencies install= ed with Yarn PnP File Watchers Bug WEB-42549 \"Enable File Watcher\" is suggested for= files inside node_modules JavaScript Feature WEB-21053 =20 = ES6 Template Strings Don't Collapse Bug WEB-42662 Color methods in optional chains as normal metho= ds Bug WEB-42469\n, , and do not work in JavaScript quick documentation Bug WEB-42593 Type of Map entries iterated with for..of inferr= ed as any Bug WEB-42466 Unwanted autocomplete on rest operator for args = collecting Bug WEB-42431 JSdoc: Better coverage of @implements with exter= nal files Bug WEB-42672 Global definitions are not resolved in strict mo= de Performance WEB-42708 outOfMemory on tuple types analyzing Cosmetics WEB-28832 Structure diagram shows right arrow buttons when= there are no sub members to that item JavaScript. 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[ "# Differential Analytic Turing Automata\n\nAuthor: Jon Awbrey\n\nThe task ahead is to chart a course from general ideas about transformational equivalence classes of graphs to a notion of differential analytic turing automata (DATA). It may be a while before we get within sight of that goal, but it will provide a better measure of motivation to name the thread after the envisioned end rather than the more homely starting place.\n\nThe basic idea is as follows. One has a set $$\\mathcal{G}$$ of graphs and a set $$\\mathcal{T}$$ of transformation rules, and each rule $$\\mathrm{t} \\in \\mathcal{T}$$ has the effect of transforming graphs into graphs, $$\\mathrm{t} : \\mathcal{G} \\to \\mathcal{G}.$$ In the cases that we shall be studying, this set of transformation rules partitions the set of graphs into transformational equivalence classes (TECs).\n\nThere are many interesting excursions to be had here, but I will focus mainly on logical applications, and and so the TECs I talk about will almost always have the character of logical equivalence classes (LECs).\n\nAn example that will figure heavily in the sequel is given by rooted trees as the species of graphs and a pair of equational transformation rules that derive from the graphical calculi of C.S. Peirce, as revived and extended by George Spencer Brown.\n\nHere are the fundamental transformation rules, also referred to as the arithmetic axioms, more precisely, the arithmetic initials.\n\nThat should be enough to get started.\n\n## Cactus Language\n\nI will be making use of the cactus language extension of Peirce's Alpha Graphs, so called because it uses a species of graphs that are usually called \"cacti\" in graph theory. The last exposition of the cactus syntax that I've written can be found here:\n\nThe representational and computational efficiency of the cactus language for the tasks that are usually associated with boolean algebra and propositional calculus makes it possible to entertain a further extension, to what we may call differential logic, because it develops this basic level of logic in the same way that differential calculus augments analytic geometry to handle change and diversity. There are several different introductions to differential logic that I have written and distributed across the Internet. You might start with the following couple of treatments:\n\nI will draw on those previously advertised resources of notation and theory as needed, but right now I sense the need for some concrete examples.\n\n## Example 1\n\nLet's say we have a system that is known by the name of its state space $$X\\!$$ and we have a boolean state variable $$x : X \\to \\mathbb{B},\\!$$ where $$\\mathbb{B} = \\{ 0, 1 \\}.\\!$$\n\nWe observe $$X\\!$$ for a while, relative to a discrete time frame, and we write down the following sequence of values for $$x.\\!$$\n\n\\begin{array}{c|c} t & x \\'\"UNIQ-MathJax1-QINU\"' in other words'\"UNIQ-MathJax2-QINU\"' =='\"UNIQ--h-10--QINU\"'Notions of Approximation== {| cellpadding=\"2\" cellspacing=\"2\" width=\"100%\" | width=\"60%\" | | width=\"40%\" | for equalities are so weighed<br> that curiosity in neither can<br> make choice of either's moiety. |- | height=\"50px\" | | valign=\"top\" | — ''King Lear'', Sc.1.5–7 (Quarto) |- | | for qualities are so weighed<br> that curiosity in neither can<br> make choice of either's moiety.<br> |- | height=\"50px\" | | valign=\"top\" | — ''King Lear'', 1.1.5–6 (Folio) |} Justifying a notion of approximation is a little more involved in general, and especially in these discrete logical spaces, than it would be expedient for people in a hurry to tangle with right now. I will just say that there are ''naive'' or ''obvious'' notions and there are ''sophisticated'' or ''subtle'' notions that we might choose among. The later would engage us in trying to construct proper logical analogues of Lie derivatives, and so let's save that for when we have become subtle or sophisticated or both. Against or toward that day, as you wish, let's begin with an option in plain view. Figure 1.4 illustrates one way of ranging over the cells of the underlying universe \\(U^\\bullet = [u, v]\\! and selecting at each cell the linear proposition in $$\\mathrm{d}U^\\bullet = [\\mathrm{d}u, \\mathrm{d}v]\\!$$ that best approximates the patch of the difference map $${\\mathrm{D}f}\\!$$ that is located there, yielding the following formula for the differential $$\\mathrm{d}f.\\!$$\n\n $$\\begin{array}{*{11}{c}} \\mathrm{d}f & = & \\mathrm{d}\\texttt{((} u \\texttt{)(} v \\texttt{))} & = & uv \\cdot 0 & + & u \\texttt{(} v \\texttt{)} \\cdot \\mathrm{d}u & + & \\texttt{(} u \\texttt{)} v \\cdot \\mathrm{d}v & + & \\texttt{(} u \\texttt{)(} v \\texttt{)} \\cdot \\texttt{(} \\mathrm{d}u \\texttt{,~} \\mathrm{d}v \\texttt{)} \\end{array}$$\n o---------------------------------------o | | | o | | / \\ | | / \\ | | / \\ | | o o | | / \\ / \\ | | / \\ / \\ | | / \\ / \\ | | o o o | | / \\ / \\ / \\ | | / \\ / \\ / \\ | | / \\ / \\ / \\ | | o o o o | | / \\ /%\\ /%\\ / \\ | | / \\ /%%%\\ /%%%\\ / \\ | | / \\ /%%%%%\\ /%%%%%\\ / \\ | | o o%%%%%%%o%%%%%%%o o | | |\\ /%\\%%%%%/ \\%%%%%/%\\ /| | | | \\ /%%%\\%%%/ \\%%%/%%%\\ / | | | | \\ /%%%%%\\%/ \\%/%%%%%\\ / | | | | o%%%%%%%o o%%%%%%%o | | | | |\\%%%%%/%\\ /%\\%%%%%/| | | | | | \\%%%/%%%\\ /%%%\\%%%/ | | | | | u | \\%/%%%%%\\ /%%%%%\\%/ | v | | | o---+---o%%%%%%%o%%%%%%%o---+---o | | | \\%%%%%/ \\%%%%%/ | | | | \\%%%/ \\%%%/ | | | | du \\%/ \\%/ dv | | | o-------o o-------o | | \\ / | | \\ / | | \\ / | | o | | | o---------------------------------------o Figure 1.4. df = linear approx to Df \n\nFigure 2.4 illustrates one way of ranging over the cells of the underlying universe $$U^\\bullet = [u, v]\\!$$ and selecting at each cell the linear proposition in $$\\mathrm{d}U^\\bullet = [du, dv]\\!$$ that best approximates the patch of the difference map $$\\mathrm{D}g\\!$$ that is located there, yielding the following formula for the differential $$\\mathrm{d}g.\\!$$\n\n $$\\begin{array}{*{11}{c}} \\mathrm{d}g & = & \\mathrm{d}\\texttt{((} u \\texttt{,} v \\texttt{))} & = & uv \\cdot \\texttt{(} \\mathrm{d}u \\texttt{,} \\mathrm{d}v \\texttt{)} & + & u \\texttt{(} v \\texttt{)} \\cdot \\texttt{(} \\mathrm{d}u \\texttt{,} \\mathrm{d}v \\texttt{)} & + & \\texttt{(} u \\texttt{)} v \\cdot \\texttt{(} \\mathrm{d}u \\texttt{,} \\mathrm{d}v \\texttt{)} & + & \\texttt{(} u \\texttt{)(} v \\texttt{)} \\cdot \\texttt{(} \\mathrm{d}u \\texttt{,} \\mathrm{d}v \\texttt{)} \\end{array}$$\n o---------------------------------------o | | | o | | / \\ | | / \\ | | / \\ | | o o | | /%\\ /%\\ | | /%%%\\ /%%%\\ | | /%%%%%\\ /%%%%%\\ | | o%%%%%%%o%%%%%%%o | | /%\\%%%%%/ \\%%%%%/%\\ | | /%%%\\%%%/ \\%%%/%%%\\ | | /%%%%%\\%/ \\%/%%%%%\\ | | o%%%%%%%o o%%%%%%%o | | / \\%%%%%/ \\ / \\%%%%%/ \\ | | / \\%%%/ \\ / \\%%%/ \\ | | / \\%/ \\ / \\%/ \\ | | o o o o o | | |\\ /%\\ / \\ /%\\ /| | | | \\ /%%%\\ / \\ /%%%\\ / | | | | \\ /%%%%%\\ / \\ /%%%%%\\ / | | | | o%%%%%%%o o%%%%%%%o | | | | |\\%%%%%/%\\ /%\\%%%%%/| | | | | | \\%%%/%%%\\ /%%%\\%%%/ | | | | | u | \\%/%%%%%\\ /%%%%%\\%/ | v | | | o---+---o%%%%%%%o%%%%%%%o---+---o | | | \\%%%%%/ \\%%%%%/ | | | | \\%%%/ \\%%%/ | | | | du \\%/ \\%/ dv | | | o-------o o-------o | | \\ / | | \\ / | | \\ / | | o | | | o---------------------------------------o Figure 2.4. dg = linear approx to Dg \n\nWell, $$g,\\!$$ that was easy, seeing as how $$\\mathrm{D}g\\!$$ is already linear at each locus, $$\\mathrm{d}g = \\mathrm{D}g.\\!$$\n\n## Analytic Series\n\nWe have been conducting the differential analysis of the logical transformation $$F : [u, v] \\mapsto [u, v]\\!$$ defined as $$F : (u, v) \\mapsto ( ~ \\texttt{((} u \\texttt{)(} v \\texttt{))} ~,~ \\texttt{((} u \\texttt{,~} v \\texttt{))} ~ ),\\!$$ and this means starting with the extended transformation $$\\mathrm{E}F : [u, v, \\mathrm{d}u, \\mathrm{d}v] \\to [u, v, \\mathrm{d}u, \\mathrm{d}v]\\!$$ and breaking it into an analytic series, $$\\mathrm{E}F = F + \\mathrm{d}F + \\mathrm{d}^2 F + \\ldots,\\!$$ and so on until there is nothing left to analyze any further.\n\nAs a general rule, one proceeds by way of the following stages:\n\n $$\\begin{array}{*{6}{l}} 1. & \\mathrm{E}F & = & \\mathrm{d}^0 F & + & \\mathrm{r}^0 F \\\\ 2. & \\mathrm{r}^0 F & = & \\mathrm{d}^1 F & + & \\mathrm{r}^1 F \\\\ 3. & \\mathrm{r}^1 F & = & \\mathrm{d}^2 F & + & \\mathrm{r}^2 F \\\\ 4. & \\ldots \\end{array}$$\n\nIn our analysis of the transformation $$F,\\!$$ we carried out Step 1 in the more familiar form $$\\mathrm{E}F = F + \\mathrm{D}F\\!$$ and we have just reached Step 2 in the form $$\\mathrm{D}F = \\mathrm{d}F + \\mathrm{r}F,\\!$$ where $$\\mathrm{r}F\\!$$ is the residual term that remains for us to examine next.\n\nNote. I'm am trying to give quick overview here, and this forces me to omit many picky details. The picky reader may wish to consult the more detailed presentation of this material at the following locations:\n\nLet's push on with the analysis of the transformation:\n\n $$\\begin{matrix} F & : & (u, v) & \\mapsto & (f(u, v), g(u, v)) & = & ( ~ \\texttt{((} u \\texttt{)(} v \\texttt{))} ~,~ \\texttt{((} u \\texttt{,~} v \\texttt{))} ~) \\end{matrix}$$\n\nFor ease of comparison and computation, I will collect the Figures that we need for the remainder of the work together on one page.\n\n### Computation Summary for Logical Disjunction\n\nFigure 1.1 shows the expansion of $$f = \\texttt{((} u \\texttt{)(} v \\texttt{))}\\!$$ over $$[u, v]\\!$$ to produce the expression:\n\n $$\\begin{matrix} uv & + & u \\texttt{(} v \\texttt{)} & + & \\texttt{(} u \\texttt{)} v \\end{matrix}$$\n\nFigure 1.2 shows the expansion of $$\\mathrm{E}f = \\texttt{((} u + \\mathrm{d}u \\texttt{)(} v + \\mathrm{d}v \\texttt{))}\\!$$ over $$[u, v]\\!$$ to produce the expression:\n\n $$\\begin{matrix} uv \\cdot \\texttt{(} \\mathrm{d}u ~ \\mathrm{d}v \\texttt{)} & + & u \\texttt{(} v \\texttt{)} \\cdot \\texttt{(} \\mathrm{d}u \\texttt{(} \\mathrm{d}v \\texttt{))} & + & \\texttt{(} u \\texttt{)} v \\cdot \\texttt{((} \\mathrm{d}u \\texttt{)} \\mathrm{d}v \\texttt{)} & + & \\texttt{(} u \\texttt{)(} v \\texttt{)} \\cdot \\texttt{((} \\mathrm{d}u \\texttt{)(} \\mathrm{d}v \\texttt{))} \\end{matrix}$$\n\nIn general, $$\\mathrm{E}f\\!$$ tells you what you would have to do, from wherever you are in the universe $$[u, v],\\!$$ if you want to end up in a place where $$f\\!$$ is true. In this case, where the prevailing proposition $$f\\!$$ is $$\\texttt{((} u \\texttt{)(} v \\texttt{))},\\!$$ the indication $$uv \\cdot \\texttt{(} \\mathrm{d}u ~ \\mathrm{d}v \\texttt{)}\\!$$ of $$\\mathrm{E}f\\!$$ tells you this: If $$u\\!$$ and $$v\\!$$ are both true where you are, then just don't change both $$u\\!$$ and $$v\\!$$ at once, and you will end up in a place where $$\\texttt{((} u \\texttt{)(} v \\texttt{))}\\!$$ is true.\n\nFigure 1.3 shows the expansion of $$\\mathrm{D}f$$ over $$[u, v]\\!$$ to produce the expression:\n\n $$\\begin{matrix} uv \\cdot \\mathrm{d}u ~ \\mathrm{d}v & + & u \\texttt{(} v \\texttt{)} \\cdot \\mathrm{d}u \\texttt{(} \\mathrm{d}v \\texttt{)} & + & \\texttt{(} u \\texttt{)} v \\cdot \\texttt{(} \\mathrm{d}u \\texttt{)} \\mathrm{d}v & + & \\texttt{(} u \\texttt{)(} v \\texttt{)} \\cdot \\texttt{((} \\mathrm{d}u \\texttt{)(} \\mathrm{d}v \\texttt{))} \\end{matrix}$$\n\nIn general, $${\\mathrm{D}f}\\!$$ tells you what you would have to do, from wherever you are in the universe $$[u, v],\\!$$ if you want to bring about a change in the value of $$f,\\!$$ that is, if you want to get to a place where the value of $$f\\!$$ is different from what it is where you are. In the present case, where the reigning proposition $$f\\!$$ is $$\\texttt{((} u \\texttt{)(} v \\texttt{))},\\!$$ the term $$uv \\cdot \\mathrm{d}u ~ \\mathrm{d}v\\!$$ of $${\\mathrm{D}f}\\!$$ tells you this: If $$u\\!$$ and $$v\\!$$ are both true where you are, then you would have to change both $$u\\!$$ and $$v\\!$$ in order to reach a place where the value of $$f\\!$$ is different from what it is where you are.\n\nFigure 1.4 approximates $${\\mathrm{D}f}\\!$$ by the linear form $$\\mathrm{d}f\\!$$ that expands over $$[u, v]\\!$$ as follows:\n\n \\begin{matrix} \\mathrm{d}f & = & uv \\cdot 0 & + & u \\texttt{(} v \\texttt{)} \\cdot \\mathrm{d}u & + & \\texttt{(} u \\texttt{)} v \\cdot \\mathrm{d}v & + & \\texttt{(} u \\texttt{)(} v \\texttt{)} \\cdot \\texttt{(} \\mathrm{d}u \\texttt{,~} \\mathrm{d}v \\texttt{)} \\'\"UNIQ-MathJax3-QINU\"' Another way to convey the same information is by means of the extended proposition'\"UNIQ-MathJax4-QINU\"' {| align=\"center\" cellpadding=\"8\" style=\"text-align:center\" | \\(\\text{Orbit 2}\\! $$\\begin{array}{c|cc|cc|cc|} t & u & v & \\mathrm{d}u & \\mathrm{d}v & \\mathrm{d}^2 u & \\mathrm{d}^2 v \\\\[8pt] 0 & 0 & 0 & 0 & 1 & 1 & 0 \\\\ 1 & 0 & 1 & 1 & 1 & 1 & 1 \\\\ 2 & 1 & 0 & 0 & 0 & 0 & 0 \\\\ 3 & {}^\\shortparallel & {}^\\shortparallel & {}^\\shortparallel & {}^\\shortparallel & {}^\\shortparallel & {}^\\shortparallel \\end{array}$$\n\nA more fine combing of the second Table brings to mind a rule that partly covers the remaining cases, that is, $$\\mathrm{d}u = v, ~ \\mathrm{d}v = \\texttt{(} u \\texttt{)}.\\!$$ This much information about Orbit 2 is also encapsulated by the extended proposition $$\\texttt{(} uv \\texttt{)((} \\mathrm{d}u \\texttt{,} v \\texttt{))(} \\mathrm{d}v, u \\texttt{)},\\!$$ which says that $$u\\!$$ and $$v\\!$$ are not both true at the same time, while $$\\mathrm{d}u\\!$$ is equal in value to $$v\\!$$ and $$\\mathrm{d}v\\!$$ is opposite in value to $$u.\\!$$\n\n## Turing Machine Example\n\nSee Theme One Program for documentation of the cactus graph syntax and the propositional modeling program used below.\n\nBy way of providing a simple illustration of Cook's Theorem, namely, that “Propositional Satisfiability is NP-Complete”, I will describe one way to translate finite approximations of turing machines into propositional expressions, using the cactus language syntax for propositional calculus that I will describe in more detail as we proceed.\n\n$$\\mathrm{Stilt}(k) =\\!$$\nSpace and time limited turing machine,\nwith $$k\\!$$ units of space and $$k\\!$$ units of time.\n$$\\mathrm{Stunt}(k) =\\!$$\nSpace and time limited turing machine,\nfor computing the parity of a bit string, with number of tape cells of input equal to $$k.\\!$$\n\nI will follow the pattern of discussion in Herbert Wilf (1986), Algorithms and Complexity, pp. 188–201, but translate his logical formalism into cactus language, which is more efficient in regard to the number of propositional clauses that are required.\n\nA turing machine for computing the parity of a bit string is described by means of the following Figure and Table.", null, "$$\\text{Figure 3.} ~~ \\text{Parity Machine}\\!$$\n\n Table 4. Parity Machine o-------o--------o-------------o---------o------------o | State | Symbol | Next Symbol | Ratchet | Next State | | Q | S | S' | dR | Q' | o-------o--------o-------------o---------o------------o | 0 | 0 | 0 | +1 | 0 | | 0 | 1 | 1 | +1 | 1 | | 0 | # | # | -1 | # | | 1 | 0 | 0 | +1 | 1 | | 1 | 1 | 1 | +1 | 0 | | 1 | # | # | -1 | * | o-------o--------o-------------o---------o------------o \n\nThe TM has a finite automaton (FA) as one component. Let us refer to this particular FA by the name of $$\\mathrm{M}.$$\n\nThe tape head (that is, the read unit) will be called $$\\mathrm{H}.$$ The registers are also called tape cells or tape squares.\n\n### Finite Approximations\n\nTo see how each finite approximation to a given turing machine can be given a purely propositional description, one fixes the parameter $$k\\!$$ and limits the rest of the discussion to describing $$\\mathrm{Stilt}(k),\\!$$ which is not really a full-fledged TM anymore but just a finite automaton in disguise.\n\nIn this example, for the sake of a minimal illustration, we choose $$k = 2,\\!$$ and discuss $$\\mathrm{Stunt}(2).$$ Since the zeroth tape cell and the last tape cell are both occupied by the character $$^{\\backprime\\backprime}\\texttt{\\#}^{\\prime\\prime}$$ that is used for both the beginning of file $$(\\mathrm{bof})$$ and the end of file $$(\\mathrm{eof})$$ markers, this allows for only one digit of significant computation.\n\nTo translate $$\\mathrm{Stunt}(2)$$ into propositional form we use the following collection of basic propositions, boolean variables, or logical features, depending on what one prefers to call them:\n\nThe basic propositions for describing the present state function $$\\mathrm{QF} : P \\to Q$$ are these:\n\n $$\\begin{matrix} \\texttt{p0\\_q\\#}, & \\texttt{p0\\_q*}, & \\texttt{p0\\_q0}, & \\texttt{p0\\_q1}, \\\\[6pt] \\texttt{p1\\_q\\#}, & \\texttt{p1\\_q*}, & \\texttt{p1\\_q0}, & \\texttt{p1\\_q1}, \\\\[6pt] \\texttt{p2\\_q\\#}, & \\texttt{p2\\_q*}, & \\texttt{p2\\_q0}, & \\texttt{p2\\_q1}, \\\\[6pt] \\texttt{p3\\_q\\#}, & \\texttt{p3\\_q*}, & \\texttt{p3\\_q0}, & \\texttt{p3\\_q1}. \\end{matrix}$$\n\nThe proposition of the form $$\\texttt{pi\\_qj}$$ says:\n\n At the point-in-time $$p_i,\\!$$ the finite state machine $$\\mathrm{M}$$ is in the state $$q_j.\\!$$\n\nThe basic propositions for describing the present register function $$\\mathrm{RF} : P \\to R$$ are these:\n\n $$\\begin{matrix} \\texttt{p0\\_r0}, & \\texttt{p0\\_r1}, & \\texttt{p0\\_r2}, & \\texttt{p0\\_r3}, \\\\[6pt] \\texttt{p1\\_r0}, & \\texttt{p1\\_r1}, & \\texttt{p1\\_r2}, & \\texttt{p1\\_r3}, \\\\[6pt] \\texttt{p2\\_r0}, & \\texttt{p2\\_r1}, & \\texttt{p2\\_r2}, & \\texttt{p2\\_r3}, \\\\[6pt] \\texttt{p3\\_r0}, & \\texttt{p3\\_r1}, & \\texttt{p3\\_r2}, & \\texttt{p3\\_r3}. \\end{matrix}$$\n\nThe proposition of the form $$\\texttt{pi\\_rj}$$ says:\n\n At the point-in-time $$p_i,\\!$$ the tape head $$\\mathrm{H}$$ is on the tape cell $$r_j.\\!$$\n\nThe basic propositions for describing the present symbol function $$\\mathrm{SF} : P \\to (R \\to S)$$ are these:\n\n $$\\begin{matrix} \\texttt{p0\\_r0\\_s\\#}, & \\texttt{p0\\_r0\\_s*}, & \\texttt{p0\\_r0\\_s0}, & \\texttt{p0\\_r0\\_s1}, \\\\[4pt] \\texttt{p0\\_r1\\_s\\#}, & \\texttt{p0\\_r1\\_s*}, & \\texttt{p0\\_r1\\_s0}, & \\texttt{p0\\_r1\\_s1}, \\\\[4pt] \\texttt{p0\\_r2\\_s\\#}, & \\texttt{p0\\_r2\\_s*}, & \\texttt{p0\\_r2\\_s0}, & \\texttt{p0\\_r2\\_s1}, \\\\[4pt] \\texttt{p0\\_r3\\_s\\#}, & \\texttt{p0\\_r3\\_s*}, & \\texttt{p0\\_r3\\_s0}, & \\texttt{p0\\_r3\\_s1}, \\\\[12pt] \\texttt{p1\\_r0\\_s\\#}, & \\texttt{p1\\_r0\\_s*}, & \\texttt{p1\\_r0\\_s0}, & \\texttt{p1\\_r0\\_s1}, \\\\[4pt] \\texttt{p1\\_r1\\_s\\#}, & \\texttt{p1\\_r1\\_s*}, & \\texttt{p1\\_r1\\_s0}, & \\texttt{p1\\_r1\\_s1}, \\\\[4pt] \\texttt{p1\\_r2\\_s\\#}, & \\texttt{p1\\_r2\\_s*}, & \\texttt{p1\\_r2\\_s0}, & \\texttt{p1\\_r2\\_s1}, \\\\[4pt] \\texttt{p1\\_r3\\_s\\#}, & \\texttt{p1\\_r3\\_s*}, & \\texttt{p1\\_r3\\_s0}, & \\texttt{p1\\_r3\\_s1}, \\\\[12pt] \\texttt{p2\\_r0\\_s\\#}, & \\texttt{p2\\_r0\\_s*}, & \\texttt{p2\\_r0\\_s0}, & \\texttt{p2\\_r0\\_s1}, \\\\[4pt] \\texttt{p2\\_r1\\_s\\#}, & \\texttt{p2\\_r1\\_s*}, & \\texttt{p2\\_r1\\_s0}, & \\texttt{p2\\_r1\\_s1}, \\\\[4pt] \\texttt{p2\\_r2\\_s\\#}, & \\texttt{p2\\_r2\\_s*}, & \\texttt{p2\\_r2\\_s0}, & \\texttt{p2\\_r2\\_s1}, \\\\[4pt] \\texttt{p2\\_r3\\_s\\#}, & \\texttt{p2\\_r3\\_s*}, & \\texttt{p2\\_r3\\_s0}, & \\texttt{p2\\_r3\\_s1}, \\\\[12pt] \\texttt{p3\\_r0\\_s\\#}, & \\texttt{p3\\_r0\\_s*}, & \\texttt{p3\\_r0\\_s0}, & \\texttt{p3\\_r0\\_s1}, \\\\[4pt] \\texttt{p3\\_r1\\_s\\#}, & \\texttt{p3\\_r1\\_s*}, & \\texttt{p3\\_r1\\_s0}, & \\texttt{p3\\_r1\\_s1}, \\\\[4pt] \\texttt{p3\\_r2\\_s\\#}, & \\texttt{p3\\_r2\\_s*}, & \\texttt{p3\\_r2\\_s0}, & \\texttt{p3\\_r2\\_s1}, \\\\[4pt] \\texttt{p3\\_r3\\_s\\#}, & \\texttt{p3\\_r3\\_s*}, & \\texttt{p3\\_r3\\_s0}, & \\texttt{p3\\_r3\\_s1}. \\end{matrix}$$\n\nThe proposition of the form $$\\texttt{pi\\_rj\\_sk}$$ says:\n\n At the point-in-time $$p_i,\\!$$ the tape cell $$r_j\\!$$ bears the mark $$s_k.\\!$$\n\n### Initial Conditions\n\nGiven but a single free square on the tape, there are just two different sets of initial conditions for $$\\mathrm{Stunt}(2),$$ the finite approximation to the parity turing machine that we are presently considering.\n\n#### Initial Conditions for Tape Input \"0\"\n\nThe following conjunction of 5 basic propositions describes the initial conditions when $$\\mathrm{Stunt}(2)$$ is started with an input of \"0\" in its free square:\n\n $$\\begin{array}{l} \\texttt{p0\\_q0} \\\\ \\\\ \\texttt{p0\\_r1} \\\\ \\\\ \\texttt{p0\\_r0\\_s\\#} \\\\ \\texttt{p0\\_r1\\_s0} \\\\ \\texttt{p0\\_r2\\_s\\#} \\end{array}$$\n\nThis conjunction of basic propositions may be read as follows:\n\n At time $$p_0,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_0,\\!$$ At time $$p_0,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ At time $$p_0,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ At time $$p_0,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{0},$$ At time $$p_0,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#}.$$\n\n#### Initial Conditions for Tape Input \"1\"\n\nThe following conjunction of 5 basic propositions describes the initial conditions when $$\\mathrm{Stunt}(2)$$ is started with an input of \"1\" in its free square:\n\n $$\\begin{array}{l} \\texttt{p0\\_q0} \\\\ \\\\ \\texttt{p0\\_r1} \\\\ \\\\ \\texttt{p0\\_r0\\_s\\#} \\\\ \\texttt{p0\\_r1\\_s1} \\\\ \\texttt{p0\\_r2\\_s\\#} \\end{array}$$\n\nThis conjunction of basic propositions may be read as follows:\n\n At time $$p_0,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_0,\\!$$ At time $$p_0,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ At time $$p_0,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ At time $$p_0,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{1},$$ At time $$p_0,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#}.$$\n\n### Propositional Program\n\nA complete description of $$\\mathrm{Stunt}(2)$$ in propositional form is obtained by conjoining one of the above choices for initial conditions with all of the following sets of propositions, that serve in effect as a simple type of declarative program, telling us all that we need to know about the anatomy and behavior of the truncated TM in question.\n\n#### Mediate Conditions\n\n $$\\begin{array}{l} \\texttt{(~p0\\_q\\#~(~p1\\_q\\#~))} \\\\ \\texttt{(~p0\\_q*~(~p1\\_q*~))} \\\\ \\\\ \\texttt{(~p1\\_q\\#~(~p2\\_q\\#~))} \\\\ \\texttt{(~p1\\_q*~(~p2\\_q*~))} \\end{array}$$\n\n#### Terminal Conditions\n\n $$\\begin{array}{l} \\texttt{((~p2\\_q\\#~)(~p2\\_q*~))} \\end{array}$$\n\n#### State Partition\n\n $$\\begin{array}{l} \\texttt{((~p0\\_q0~),(~p0\\_q1~),(~p0\\_q\\#~),(~p0\\_q*~))} \\\\ \\texttt{((~p1\\_q0~),(~p1\\_q1~),(~p1\\_q\\#~),(~p1\\_q*~))} \\\\ \\texttt{((~p2\\_q0~),(~p2\\_q1~),(~p2\\_q\\#~),(~p2\\_q*~))} \\end{array}$$\n\n#### Register Partition\n\n $$\\begin{array}{l} \\texttt{((~p0\\_r0~),(~p0\\_r1~),(~p0\\_r2~))} \\\\ \\texttt{((~p1\\_r0~),(~p1\\_r1~),(~p1\\_r2~))} \\\\ \\texttt{((~p2\\_r0~),(~p2\\_r1~),(~p2\\_r2~))} \\end{array}$$\n\n#### Symbol Partition\n\n $$\\begin{array}{l} \\texttt{((~p0\\_r0\\_s0~),(~p0\\_r0\\_s1~),(~p0\\_r0\\_s\\#~))} \\\\ \\texttt{((~p0\\_r1\\_s0~),(~p0\\_r1\\_s1~),(~p0\\_r1\\_s\\#~))} \\\\ \\texttt{((~p0\\_r2\\_s0~),(~p0\\_r2\\_s1~),(~p0\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r0\\_s0~),(~p1\\_r0\\_s1~),(~p1\\_r0\\_s\\#~))} \\\\ \\texttt{((~p1\\_r1\\_s0~),(~p1\\_r1\\_s1~),(~p1\\_r1\\_s\\#~))} \\\\ \\texttt{((~p1\\_r2\\_s0~),(~p1\\_r2\\_s1~),(~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{((~p2\\_r0\\_s0~),(~p2\\_r0\\_s1~),(~p2\\_r0\\_s\\#~))} \\\\ \\texttt{((~p2\\_r1\\_s0~),(~p2\\_r1\\_s1~),(~p2\\_r1\\_s\\#~))} \\\\ \\texttt{((~p2\\_r2\\_s0~),(~p2\\_r2\\_s1~),(~p2\\_r2\\_s\\#~))} \\end{array}$$\n\n#### Interaction Conditions\n\n $$\\begin{array}{l} \\texttt{((~p0\\_r0~) ~p0\\_r0\\_s0~ (~p1\\_r0\\_s0~))} \\\\ \\texttt{((~p0\\_r0~) ~p0\\_r0\\_s1~ (~p1\\_r0\\_s1~))} \\\\ \\texttt{((~p0\\_r0~) ~p0\\_r0\\_s\\#~ (~p1\\_r0\\_s\\#~))} \\\\ \\\\ \\texttt{((~p0\\_r1~) ~p0\\_r1\\_s0~ (~p1\\_r1\\_s0~))} \\\\ \\texttt{((~p0\\_r1~) ~p0\\_r1\\_s1~ (~p1\\_r1\\_s1~))} \\\\ \\texttt{((~p0\\_r1~) ~p0\\_r1\\_s\\#~ (~p1\\_r1\\_s\\#~))} \\\\ \\\\ \\texttt{((~p0\\_r2~) ~p0\\_r2\\_s0~ (~p1\\_r2\\_s0~))} \\\\ \\texttt{((~p0\\_r2~) ~p0\\_r2\\_s1~ (~p1\\_r2\\_s1~))} \\\\ \\texttt{((~p0\\_r2~) ~p0\\_r2\\_s\\#~ (~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r0~) ~p1\\_r0\\_s0~ (~p2\\_r0\\_s0~))} \\\\ \\texttt{((~p1\\_r0~) ~p1\\_r0\\_s1~ (~p2\\_r0\\_s1~))} \\\\ \\texttt{((~p1\\_r0~) ~p1\\_r0\\_s\\#~ (~p2\\_r0\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r1~) ~p1\\_r1\\_s0~ (~p2\\_r1\\_s0~))} \\\\ \\texttt{((~p1\\_r1~) ~p1\\_r1\\_s1~ (~p2\\_r1\\_s1~))} \\\\ \\texttt{((~p1\\_r1~) ~p1\\_r1\\_s\\#~ (~p2\\_r1\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r2~) ~p1\\_r2\\_s0~ (~p2\\_r2\\_s0~))} \\\\ \\texttt{((~p1\\_r2~) ~p1\\_r2\\_s1~ (~p2\\_r2\\_s1~))} \\\\ \\texttt{((~p1\\_r2~) ~p1\\_r2\\_s\\#~ (~p2\\_r2\\_s\\#~))} \\end{array}$$\n\n#### Transition Relations\n\n $$\\begin{array}{l} \\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s0~~(~p1\\_q0~~p1\\_r2~~p1\\_r1\\_s0~))} \\\\ \\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s1~~(~p1\\_q1~~p1\\_r2~~p1\\_r1\\_s1~))} \\\\ \\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s\\#~~(~p1\\_q\\#~~p1\\_r0~~p1\\_r1\\_s\\#~))} \\\\ \\texttt{(~p0\\_q0~~p0\\_r2~~p0\\_r2\\_s\\#~~(~p1\\_q\\#~~p1\\_r1~~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{(~p0\\_q1~~p0\\_r1~~p0\\_r1\\_s0~~(~p1\\_q1~~p1\\_r2~~p1\\_r1\\_s0~))} \\\\ \\texttt{(~p0\\_q1~~p0\\_r1~~p0\\_r1\\_s1~~(~p1\\_q0~~p1\\_r2~~p1\\_r1\\_s1~))} \\\\ \\texttt{(~p0\\_q1~~p0\\_r1~~p0\\_r1\\_s\\#~~(~p1\\_q*~~p1\\_r0~~p1\\_r1\\_s\\#~))} \\\\ \\texttt{(~p0\\_q1~~p0\\_r2~~p0\\_r2\\_s\\#~~(~p1\\_q*~~p1\\_r1~~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{(~p1\\_q0~~p1\\_r1~~p1\\_r1\\_s0~~(~p2\\_q0~~p2\\_r2~~p2\\_r1\\_s0~))} \\\\ \\texttt{(~p1\\_q0~~p1\\_r1~~p1\\_r1\\_s1~~(~p2\\_q1~~p2\\_r2~~p2\\_r1\\_s1~))} \\\\ \\texttt{(~p1\\_q0~~p1\\_r1~~p1\\_r1\\_s\\#~~(~p2\\_q\\#~~p2\\_r0~~p2\\_r1\\_s\\#~))} \\\\ \\texttt{(~p1\\_q0~~p1\\_r2~~p1\\_r2\\_s\\#~~(~p2\\_q\\#~~p2\\_r1~~p2\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{(~p1\\_q1~~p1\\_r1~~p1\\_r1\\_s0~~(~p2\\_q1~~p2\\_r2~~p2\\_r1\\_s0~))} \\\\ \\texttt{(~p1\\_q1~~p1\\_r1~~p1\\_r1\\_s1~~(~p2\\_q0~~p2\\_r2~~p2\\_r1\\_s1~))} \\\\ \\texttt{(~p1\\_q1~~p1\\_r1~~p1\\_r1\\_s\\#~~(~p2\\_q*~~p2\\_r0~~p2\\_r1\\_s\\#~))} \\\\ \\texttt{(~p1\\_q1~~p1\\_r2~~p1\\_r2\\_s\\#~~(~p2\\_q*~~p2\\_r1~~p2\\_r2\\_s\\#~))} \\end{array}$$\n\n### Interpretation of the Propositional Program\n\nLet us now run through the propositional specification of $$\\mathrm{Stunt}(2),$$ our truncated TM, and paraphrase what it says in ordinary language.\n\n#### Mediate Conditions\n\n $$\\begin{array}{l} \\texttt{(~p0\\_q\\#~(~p1\\_q\\#~))} \\\\ \\texttt{(~p0\\_q*~(~p1\\_q*~))} \\\\ \\\\ \\texttt{(~p1\\_q\\#~(~p2\\_q\\#~))} \\\\ \\texttt{(~p1\\_q*~(~p2\\_q*~))} \\end{array}$$\n\nIn the interpretation of the cactus language for propositional logic that we are using here, an expression of the form $$\\texttt{(p(q))}$$ expresses a conditional, an implication, or an if-then proposition, commonly read in one of the following ways:\n\n $$\\begin{array}{l} \\mathrm{not}~ p ~\\mathrm{without}~ q \\\\[4pt] p ~\\mathrm{implies}~ q \\\\[4pt] \\mathrm{if}~ p ~\\mathrm{then}~ q \\\\[4pt] p \\Rightarrow q \\end{array}$$\n\nA text string expression of the form $$\\texttt{(p(q))}$$ corresponds to a graph-theoretic data-structure of the following form:\n\n o---------------------------------------o | | | p q | | o---o | | | | | @ | | | o---------------------------------------o | ( p ( q )) | o---------------------------------------o \n\nTaken together, the Mediate Conditions state the following:\n\n If $$\\mathrm{M}$$ at $$p_0\\!$$ is in state $$q_\\#,\\!$$ then $$\\mathrm{M}$$ at $$p_1\\!$$ is in state $$q_\\#,\\!$$ and If $$\\mathrm{M}$$ at $$p_0\\!$$ is in state $$q_*,\\!$$ then $$\\mathrm{M}$$ at $$p_1\\!$$ is in state $$q_*,\\!$$ and If $$\\mathrm{M}$$ at $$p_1\\!$$ is in state $$q_\\#,\\!$$ then $$\\mathrm{M}$$ at $$p_2\\!$$ is in state $$q_\\#,\\!$$ and If $$\\mathrm{M}$$ at $$p_1\\!$$ is in state $$q_*,\\!$$ then $$\\mathrm{M}$$ at $$p_2\\!$$ is in state $$q_*.\\!$$\n\n#### Terminal Conditions\n\n $$\\begin{array}{l} \\texttt{((~p2\\_q\\#~)(~p2\\_q*~))} \\end{array}$$\n\nIn cactus syntax, an expression of the form $$\\texttt{((p)(q))}$$ expresses the disjunction $$p ~\\mathrm{or}~ q.$$ The corresponding cactus graph, here just a tree, has the following shape:\n\n o---------------------------------------o | | | p q | | o o | | \\ / | | o | | | | | @ | | | o---------------------------------------o | ((p) (q)) | o---------------------------------------o \n\nIn effect, the Terminal Conditions state the following:\n\n At time $$p_2\\!$$ machine $$\\mathrm{M}$$ is in state $$q_\\#,\\!$$ or At time $$p_2\\!$$ machine $$\\mathrm{M}$$ is in state $$q_*.\\!$$\n\n#### State Partition\n\n $$\\begin{array}{l} \\texttt{((~p0\\_q0~),(~p0\\_q1~),(~p0\\_q\\#~),(~p0\\_q*~))} \\\\ \\texttt{((~p1\\_q0~),(~p1\\_q1~),(~p1\\_q\\#~),(~p1\\_q*~))} \\\\ \\texttt{((~p2\\_q0~),(~p2\\_q1~),(~p2\\_q\\#~),(~p2\\_q*~))} \\end{array}$$\n\nIn cactus syntax, an expression of the form $$\\texttt{((} e_1 \\texttt{),(} e_2 \\texttt{),(} \\ldots \\texttt{),(} e_k \\texttt{))}\\!$$ expresses a statement to the effect that exactly one of the expressions $$e_j\\!$$ is true, for $$j = 1 ~\\mathit{to}~ k.$$ Expressions of this form are called universal partition expressions, and the corresponding painted and rooted cactus (PARC) has the following shape:\n\n o---------------------------------------o | | | e_1 e_2 ... e_k | | o o o | | | | | | | o-----o--- ... ---o | | \\ / | | \\ / | | \\ / | | \\ / | | \\ / | | \\ / | | \\ / | | \\ / | | @ | | | o---------------------------------------o | ((e_1),(e_2),(...),(e_k)) | o---------------------------------------o \n\nThe State Partition segment of the propositional program consists of three universal partition expressions, taken in conjunction expressing the condition that $$\\mathrm{M}$$ has to be in one and only one of its states at each point in time under consideration. In short, we have the constraint:\n\n At each of the points in time $$p_i,\\!$$ for $$i\\!$$ in the set $$\\{ 0, 1, 2 \\},\\!$$ $$\\mathrm{M}$$ can be in exactly one state $$q_j,\\!$$ for $$j\\!$$ in the set $$\\{ 0, 1, \\#, * \\}.\\!$$\n\n#### Register Partition\n\n $$\\begin{array}{l} \\texttt{((~p0\\_r0~),(~p0\\_r1~),(~p0\\_r2~))} \\\\ \\texttt{((~p1\\_r0~),(~p1\\_r1~),(~p1\\_r2~))} \\\\ \\texttt{((~p2\\_r0~),(~p2\\_r1~),(~p2\\_r2~))} \\end{array}$$\n\nThe Register Partition segment of the propositional program consists of three universal partition expressions, taken in conjunction saying that the read head $$\\mathrm{H}$$ must be reading one and only one of the registers or tape cells available to it at each of the points in time under consideration. In sum:\n\n At each of the points in time $$p_i,\\!$$ for $$i = 0, 1, 2,\\!$$ $$\\mathrm{H}$$ is reading exactly one cell $$r_j,\\!$$ for $$j = 0, 1, 2.\\!$$\n\n#### Symbol Partition\n\n $$\\begin{array}{l} \\texttt{((~p0\\_r0\\_s0~),(~p0\\_r0\\_s1~),(~p0\\_r0\\_s\\#~))} \\\\ \\texttt{((~p0\\_r1\\_s0~),(~p0\\_r1\\_s1~),(~p0\\_r1\\_s\\#~))} \\\\ \\texttt{((~p0\\_r2\\_s0~),(~p0\\_r2\\_s1~),(~p0\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r0\\_s0~),(~p1\\_r0\\_s1~),(~p1\\_r0\\_s\\#~))} \\\\ \\texttt{((~p1\\_r1\\_s0~),(~p1\\_r1\\_s1~),(~p1\\_r1\\_s\\#~))} \\\\ \\texttt{((~p1\\_r2\\_s0~),(~p1\\_r2\\_s1~),(~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{((~p2\\_r0\\_s0~),(~p2\\_r0\\_s1~),(~p2\\_r0\\_s\\#~))} \\\\ \\texttt{((~p2\\_r1\\_s0~),(~p2\\_r1\\_s1~),(~p2\\_r1\\_s\\#~))} \\\\ \\texttt{((~p2\\_r2\\_s0~),(~p2\\_r2\\_s1~),(~p2\\_r2\\_s\\#~))} \\end{array}$$\n\nThe Symbol Partition segment of the propositional program for $$\\mathrm{Stunt}(2)$$ consists of nine universal partition expressions, taken in conjunction stipulating that there has to be one and only one symbol in each of the registers at each point in time under consideration. In short, we have:\n\n At each of the points in time $$p_i,\\!$$ for $$i\\!$$ in $$\\{ 0, 1, 2 \\},\\!$$ in each of the tape registers $$r_j,\\!$$ for $$j\\!$$ in $$\\{ 0, 1, 2 \\},\\!$$ there can be exactly one sign $$s_k,\\!$$ for $$k\\!$$ in $$\\{ 0, 1, \\# \\}.\\!$$\n\n#### Interaction Conditions\n\n $$\\begin{array}{l} \\texttt{((~p0\\_r0~) ~p0\\_r0\\_s0~ (~p1\\_r0\\_s0~))} \\\\ \\texttt{((~p0\\_r0~) ~p0\\_r0\\_s1~ (~p1\\_r0\\_s1~))} \\\\ \\texttt{((~p0\\_r0~) ~p0\\_r0\\_s\\#~ (~p1\\_r0\\_s\\#~))} \\\\ \\\\ \\texttt{((~p0\\_r1~) ~p0\\_r1\\_s0~ (~p1\\_r1\\_s0~))} \\\\ \\texttt{((~p0\\_r1~) ~p0\\_r1\\_s1~ (~p1\\_r1\\_s1~))} \\\\ \\texttt{((~p0\\_r1~) ~p0\\_r1\\_s\\#~ (~p1\\_r1\\_s\\#~))} \\\\ \\\\ \\texttt{((~p0\\_r2~) ~p0\\_r2\\_s0~ (~p1\\_r2\\_s0~))} \\\\ \\texttt{((~p0\\_r2~) ~p0\\_r2\\_s1~ (~p1\\_r2\\_s1~))} \\\\ \\texttt{((~p0\\_r2~) ~p0\\_r2\\_s\\#~ (~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r0~) ~p1\\_r0\\_s0~ (~p2\\_r0\\_s0~))} \\\\ \\texttt{((~p1\\_r0~) ~p1\\_r0\\_s1~ (~p2\\_r0\\_s1~))} \\\\ \\texttt{((~p1\\_r0~) ~p1\\_r0\\_s\\#~ (~p2\\_r0\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r1~) ~p1\\_r1\\_s0~ (~p2\\_r1\\_s0~))} \\\\ \\texttt{((~p1\\_r1~) ~p1\\_r1\\_s1~ (~p2\\_r1\\_s1~))} \\\\ \\texttt{((~p1\\_r1~) ~p1\\_r1\\_s\\#~ (~p2\\_r1\\_s\\#~))} \\\\ \\\\ \\texttt{((~p1\\_r2~) ~p1\\_r2\\_s0~ (~p2\\_r2\\_s0~))} \\\\ \\texttt{((~p1\\_r2~) ~p1\\_r2\\_s1~ (~p2\\_r2\\_s1~))} \\\\ \\texttt{((~p1\\_r2~) ~p1\\_r2\\_s\\#~ (~p2\\_r2\\_s\\#~))} \\end{array}$$\n\nIn briefest terms, the Interaction Conditions simply express the circumstance that the mark on a tape cell cannot change between two points in time unless the tape head is over the cell in question at the initial one of those points in time. All that we have to do is to see how they manage to say this.\n\nConsider a cactus expression of the following form:\n\n $$\\begin{array}{l} \\texttt{((}~ p_i\\_r_j ~\\texttt{)}~ p_i\\_r_j\\_s_k ~\\texttt{(}~ p_{i+1}\\_r_j\\_s_k ~\\texttt{))} \\end{array}$$\n\nThis expression has the corresponding cactus graph:\n\n o---------------------------------------o | | | p_r p_r_s | | o o | | \\ / | | p_r_s o | | | | | @ | | | o---------------------------------------o \n\nA propositional expression of this form can be read as follows:\n\n $$\\mathrm{If}$$ At the time $$p_i,\\!$$ the tape cell $$r_j\\!$$ bears the mark $$s_k,\\!$$ $$\\mathrm{But}$$ it is not the case that: At the time $$p_i,\\!$$ the tape head is on the tape cell $$r_j,\\!$$ $$\\mathrm{Then}$$ At the time $$p_{i+1},\\!$$ the tape cell $$r_j\\!$$ bears the mark $$s_k.\\!$$\n\nThe eighteen clauses of the Interaction Conditions simply impose one such constraint on symbol changes for each combination of the times $$p_0, p_1,\\!$$ registers $$r_0, r_1, r_2,\\!$$ and symbols $$s_0, s_1, s_\\#.\\!$$\n\n#### Transition Relations\n\n $$\\begin{array}{l} \\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s0~~(~p1\\_q0~~p1\\_r2~~p1\\_r1\\_s0~))} \\\\ \\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s1~~(~p1\\_q1~~p1\\_r2~~p1\\_r1\\_s1~))} \\\\ \\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s\\#~~(~p1\\_q\\#~~p1\\_r0~~p1\\_r1\\_s\\#~))} \\\\ \\texttt{(~p0\\_q0~~p0\\_r2~~p0\\_r2\\_s\\#~~(~p1\\_q\\#~~p1\\_r1~~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{(~p0\\_q1~~p0\\_r1~~p0\\_r1\\_s0~~(~p1\\_q1~~p1\\_r2~~p1\\_r1\\_s0~))} \\\\ \\texttt{(~p0\\_q1~~p0\\_r1~~p0\\_r1\\_s1~~(~p1\\_q0~~p1\\_r2~~p1\\_r1\\_s1~))} \\\\ \\texttt{(~p0\\_q1~~p0\\_r1~~p0\\_r1\\_s\\#~~(~p1\\_q*~~p1\\_r0~~p1\\_r1\\_s\\#~))} \\\\ \\texttt{(~p0\\_q1~~p0\\_r2~~p0\\_r2\\_s\\#~~(~p1\\_q*~~p1\\_r1~~p1\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{(~p1\\_q0~~p1\\_r1~~p1\\_r1\\_s0~~(~p2\\_q0~~p2\\_r2~~p2\\_r1\\_s0~))} \\\\ \\texttt{(~p1\\_q0~~p1\\_r1~~p1\\_r1\\_s1~~(~p2\\_q1~~p2\\_r2~~p2\\_r1\\_s1~))} \\\\ \\texttt{(~p1\\_q0~~p1\\_r1~~p1\\_r1\\_s\\#~~(~p2\\_q\\#~~p2\\_r0~~p2\\_r1\\_s\\#~))} \\\\ \\texttt{(~p1\\_q0~~p1\\_r2~~p1\\_r2\\_s\\#~~(~p2\\_q\\#~~p2\\_r1~~p2\\_r2\\_s\\#~))} \\\\ \\\\ \\texttt{(~p1\\_q1~~p1\\_r1~~p1\\_r1\\_s0~~(~p2\\_q1~~p2\\_r2~~p2\\_r1\\_s0~))} \\\\ \\texttt{(~p1\\_q1~~p1\\_r1~~p1\\_r1\\_s1~~(~p2\\_q0~~p2\\_r2~~p2\\_r1\\_s1~))} \\\\ \\texttt{(~p1\\_q1~~p1\\_r1~~p1\\_r1\\_s\\#~~(~p2\\_q*~~p2\\_r0~~p2\\_r1\\_s\\#~))} \\\\ \\texttt{(~p1\\_q1~~p1\\_r2~~p1\\_r2\\_s\\#~~(~p2\\_q*~~p2\\_r1~~p2\\_r2\\_s\\#~))} \\end{array}$$\n\nThe Transition Relation segment of the propositional program for $$\\mathrm{Stunt}(2)$$ consists of sixteen implication statements with complex antecedents and consequents. Taken together, these give propositional expression to the TM Figure and Table that were given at the outset.\n\nJust by way of a single example, consider the clause:\n\n $$\\texttt{(~p0\\_q0~~p0\\_r1~~p0\\_r1\\_s1~~(~p1\\_q1~~p1\\_r2~~p1\\_r1\\_s1~))}$$\n\nThis complex implication statement can be read to say:\n\n $$\\mathrm{If}$$ At the time $$p_0,\\!$$ the machine $$\\mathrm{M}$$ is in the state $$q_0,\\!$$ and At the time $$p_0,\\!$$ the scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ and At the time $$p_0,\\!$$ the tape cell $$r_1\\!$$ contains a $$\\texttt{1},$$ $$\\mathrm{Then}$$ At the time $$p_1,\\!$$ the machine $$\\mathrm{M}$$ is in the state $$q_1,\\!$$ and At the time $$p_1,\\!$$ the scanner $$\\mathrm{H}$$ is reading cell $$r_2,\\!$$ and At the time $$p_1,\\!$$ the tape cell $$r_1\\!$$ contains a $$\\texttt{1}.$$\n\n### Computation\n\nThe propositional program for $$\\mathrm{Stunt}(2)$$ uses the following set of $$9 + 12 + 36 = 57\\!$$ basic propositions or boolean variables:\n\n $$\\begin{matrix} \\texttt{p0\\_r0}, & \\texttt{p0\\_r1}, & \\texttt{p0\\_r2}, \\\\[6pt] \\texttt{p1\\_r0}, & \\texttt{p1\\_r1}, & \\texttt{p1\\_r2}, \\\\[6pt] \\texttt{p2\\_r0}, & \\texttt{p2\\_r1}, & \\texttt{p2\\_r2}. \\end{matrix}$$\n $$\\begin{matrix} \\texttt{p0\\_q\\#}, & \\texttt{p0\\_q*}, & \\texttt{p0\\_q0}, & \\texttt{p0\\_q1}, \\\\[6pt] \\texttt{p1\\_q\\#}, & \\texttt{p1\\_q*}, & \\texttt{p1\\_q0}, & \\texttt{p1\\_q1}, \\\\[6pt] \\texttt{p2\\_q\\#}, & \\texttt{p2\\_q*}, & \\texttt{p2\\_q0}, & \\texttt{p2\\_q1}. \\end{matrix}$$\n $$\\begin{matrix} \\texttt{p0\\_r0\\_s\\#}, & \\texttt{p0\\_r0\\_s*}, & \\texttt{p0\\_r0\\_s0}, & \\texttt{p0\\_r0\\_s1}, \\\\[4pt] \\texttt{p0\\_r1\\_s\\#}, & \\texttt{p0\\_r1\\_s*}, & \\texttt{p0\\_r1\\_s0}, & \\texttt{p0\\_r1\\_s1}, \\\\[4pt] \\texttt{p0\\_r2\\_s\\#}, & \\texttt{p0\\_r2\\_s*}, & \\texttt{p0\\_r2\\_s0}, & \\texttt{p0\\_r2\\_s1}, \\\\[12pt] \\texttt{p1\\_r0\\_s\\#}, & \\texttt{p1\\_r0\\_s*}, & \\texttt{p1\\_r0\\_s0}, & \\texttt{p1\\_r0\\_s1}, \\\\[4pt] \\texttt{p1\\_r1\\_s\\#}, & \\texttt{p1\\_r1\\_s*}, & \\texttt{p1\\_r1\\_s0}, & \\texttt{p1\\_r1\\_s1}, \\\\[4pt] \\texttt{p1\\_r2\\_s\\#}, & \\texttt{p1\\_r2\\_s*}, & \\texttt{p1\\_r2\\_s0}, & \\texttt{p1\\_r2\\_s1}, \\\\[12pt] \\texttt{p2\\_r0\\_s\\#}, & \\texttt{p2\\_r0\\_s*}, & \\texttt{p2\\_r0\\_s0}, & \\texttt{p2\\_r0\\_s1}, \\\\[4pt] \\texttt{p2\\_r1\\_s\\#}, & \\texttt{p2\\_r1\\_s*}, & \\texttt{p2\\_r1\\_s0}, & \\texttt{p2\\_r1\\_s1}, \\\\[4pt] \\texttt{p2\\_r2\\_s\\#}, & \\texttt{p2\\_r2\\_s*}, & \\texttt{p2\\_r2\\_s0}, & \\texttt{p2\\_r2\\_s1}. \\end{matrix}$$\n\nThis means that the propositional program itself is nothing but a single proposition or boolean function of the form $$p : \\mathbb{B}^{57} \\to \\mathbb{B}.$$\n\nAn assignment of boolean values to the above set of boolean variables is called an interpretation of the proposition $$p,\\!$$ and any interpretation of $$p\\!$$ that makes the proposition $$p : \\mathbb{B}^{57} \\to \\mathbb{B}$$ evaluate to $$1\\!$$ is referred to as a satisfying interpretation of the proposition $$p.\\!$$ Another way to specify interpretations, instead of giving them as bit vectors in $$\\mathbb{B}^{57}$$ and trying to remember some arbitrary ordering of variables, is to give them in the form of singular propositions, that is, a conjunction of the form $$e_1 \\cdot \\ldots \\cdot e_{57}$$ where each $$e_j\\!$$ is either $$v_j\\!$$ or $$\\texttt{(} v_j \\texttt{)},$$ that is, either the assertion or the negation of the boolean variable $${v_j},\\!$$ as $$j\\!$$ runs from 1 to 57. Even more briefly, the same information can be communicated simply by giving the conjunction of the asserted variables, with the understanding that each of the others is negated.\n\nA satisfying interpretation of the proposition $$p\\!$$ supplies us with all the information of a complete execution history for the corresponding program, and so all we have to do in order to get the output of the program $$p\\!$$ is to read off the proper part of the data from the expression of this interpretation.\n\n### Output\n\nOne component of the $$\\begin{smallmatrix}\\mathrm{Theme~One}\\end{smallmatrix}$$ program that I wrote some years ago finds all the satisfying interpretations of propositions expressed in cactus syntax. It's not a polynomial time algorithm, as you may guess, but it was just barely efficient enough to do this example in the 500 K of spare memory that I had on an old 286 PC in about 1989, so I will give you the actual outputs from those trials.\n\n#### Output Conditions for Tape Input \"0\"\n\nLet $$p_0\\!$$ be the proposition that we get by conjoining the proposition that describes the initial conditions for tape input \"0\" with the proposition that describes the truncated turing machine $$\\mathrm{Stunt}(2).$$ As it turns out, $$p_0\\!$$ has a single satisfying interpretation. This interpretation is expressible in the form of a singular proposition, which can in turn be indicated by its positive logical features, as shown in the following display:\n\n o-------------------------------------------------o | | | p0_q0 | | p0_r1 | | p0_r0_s# | | p0_r1_s0 | | p0_r2_s# | | p1_q0 | | p1_r2 | | p1_r2_s# | | p1_r0_s# | | p1_r1_s0 | | p2_q# | | p2_r1 | | p2_r0_s# | | p2_r1_s0 | | p2_r2_s# | | | o-------------------------------------------------o \n\nThe Output Conditions for Tape Input \"0\" can be read as follows:\n\n At the time $$p_0,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_0,\\!$$ and At the time $$p_0,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ and At the time $$p_0,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_0,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{0},$$ and At the time $$p_0,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_1,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_0,\\!$$ and At the time $$p_1,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_2,\\!$$ and At the time $$p_1,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_1,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{0},$$ and At the time $$p_1,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_2,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_\\#,\\!$$ and At the time $$p_2,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ and At the time $$p_2,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_2,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{0},$$ and At the time $$p_2,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#}.$$\n\nThe output of $$\\mathrm{Stunt}(2)$$ being the symbol that rests under the tape head $$\\mathrm{H}$$ if and when the machine $$\\mathrm{M}$$ reaches one of its resting states, we get the result that $$\\mathrm{Parity}(0) = 0.$$\n\n#### Output Conditions for Tape Input \"1\"\n\nLet $$p_1\\!$$ be the proposition that we get by conjoining the proposition that describes the initial conditions for tape input \"1\" with the proposition that describes the truncated turing machine $$\\mathrm{Stunt}(2).$$ As it turns out, $$p_1\\!$$ has a single satisfying interpretation. This interpretation is expressible in the form of a singular proposition, which can in turn be indicated by its positive logical features, as shown in the following display:\n\n o-------------------------------------------------o | | | p0_q0 | | p0_r1 | | p0_r0_s# | | p0_r1_s1 | | p0_r2_s# | | p1_q1 | | p1_r2 | | p1_r2_s# | | p1_r0_s# | | p1_r1_s1 | | p2_q* | | p2_r1 | | p2_r0_s# | | p2_r1_s1 | | p2_r2_s# | | | o-------------------------------------------------o \n\nThe Output Conditions for Tape Input \"1\" can be read as follows:\n\n At the time $$p_0,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_0,\\!$$ and At the time $$p_0,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ and At the time $$p_0,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_0,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{1},$$ and At the time $$p_0,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_1,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_1,\\!$$ and At the time $$p_1,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_2,\\!$$ and At the time $$p_1,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_1,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{1},$$ and At the time $$p_1,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_2,\\!$$ machine $$\\mathrm{M}$$ is in the state $$q_*,\\!$$ and At the time $$p_2,\\!$$ scanner $$\\mathrm{H}$$ is reading cell $$r_1,\\!$$ and At the time $$p_2,\\!$$ cell $$r_0\\!$$ contains the symbol $$\\texttt{\\#},$$ and At the time $$p_2,\\!$$ cell $$r_1\\!$$ contains the symbol $$\\texttt{1},$$ and At the time $$p_2,\\!$$ cell $$r_2\\!$$ contains the symbol $$\\texttt{\\#}.$$\n\nThe output of $$\\mathrm{Stunt}(2)$$ being the symbol that rests under the tape head $$\\mathrm{H}$$ when and if the machine $$\\mathrm{M}$$ reaches one of its resting states, we get the result that $$\\mathrm{Parity}(1) = 1.$$" ]

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[ "# NAG Library Routine Document\n\n## 1Purpose\n\nc06rcf computes the discrete quarter-wave Fourier sine transforms of $m$ sequences of real data values.\n\n## 2Specification\n\nFortran Interface\n Subroutine c06rcf ( m, n, x, work,\n Integer, Intent (In) :: m, n Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (Inout) :: x(m*(n+2)), work(*) Character (1), Intent (In) :: direct\nC Header Interface\n#include <nagmk26.h>\n void c06rcf_ (const char *direct, const Integer *m, const Integer *n, double x[], double work[], Integer *ifail, const Charlen length_direct)\n\n## 3Description\n\nGiven $m$ sequences of $n$ real data values ${x}_{\\mathit{j}}^{\\mathit{p}}$, for $\\mathit{j}=1,2,\\dots ,n$ and $\\mathit{p}=1,2,\\dots ,m$, c06rcf simultaneously calculates the quarter-wave Fourier sine transforms of all the sequences defined by\n $x^kp = 1n ∑ j=1 n-1 xjp × sin j 2k-1 π2n + 12 -1 k-1 xnp , if ​ direct='F' ,$\nor its inverse\n $xkp = 2n ∑ j=1 n x^ j p × sin 2j- 1 k π2n , if ​ direct='B' ,$\nwhere $k=1,2,\\dots ,n$ and $p=1,2,\\dots ,m$.\n(Note the scale factor $\\frac{1}{\\sqrt{n}}$ in this definition.)\nA call of c06rcf with ${\\mathbf{direct}}=\\text{'F'}$ followed by a call with ${\\mathbf{direct}}=\\text{'B'}$ will restore the original data.\nThe transform calculated by this routine can be used to solve Poisson's equation when the solution is specified at the left boundary, and the derivative of the solution is specified at the right boundary (see Swarztrauber (1977)).\nThe routine uses a variant of the fast Fourier transform (FFT) algorithm (see Brigham (1974)) known as the Stockham self-sorting algorithm, described in Temperton (1983), together with pre- and post-processing stages described in Swarztrauber (1982). Special coding is provided for the factors $2$, $3$, $4$ and $5$.\nBrigham E O (1974) The Fast Fourier Transform Prentice–Hall\nSwarztrauber P N (1977) The methods of cyclic reduction, Fourier analysis and the FACR algorithm for the discrete solution of Poisson's equation on a rectangle SIAM Rev. 19(3) 490–501\nSwarztrauber P N (1982) Vectorizing the FFT's Parallel Computation (ed G Rodrique) 51–83 Academic Press\nTemperton C (1983) Fast mixed-radix real Fourier transforms J. Comput. Phys. 52 340–350\n\n## 5Arguments\n\n1:     $\\mathbf{direct}$ – Character(1)Input\nOn entry: if the forward transform as defined in Section 3 is to be computed, direct must be set equal to 'F'.\nIf the backward transform is to be computed, direct must be set equal to 'B'.\nConstraint: ${\\mathbf{direct}}=\\text{'F'}$ or $\\text{'B'}$.\n2:     $\\mathbf{m}$ – IntegerInput\nOn entry: $m$, the number of sequences to be transformed.\nConstraint: ${\\mathbf{m}}\\ge 1$.\n3:     $\\mathbf{n}$ – IntegerInput\nOn entry: $n$, the number of real values in each sequence.\nConstraint: ${\\mathbf{n}}\\ge 1$.\n4:     $\\mathbf{x}\\left({\\mathbf{m}}×\\left({\\mathbf{n}}+2\\right)\\right)$ – Real (Kind=nag_wp) arrayInput/Output\nOn entry: the data must be stored in x as if in a two-dimensional array of dimension $\\left(1:{\\mathbf{m}},1:{\\mathbf{n}}+2\\right)$; each of the $m$ sequences is stored in a row of the array. In other words, if the data values of the $\\mathit{p}$th sequence to be transformed are denoted by ${x}_{\\mathit{j}}^{\\mathit{p}}$, for $\\mathit{j}=1,2,\\dots ,n$ and $\\mathit{p}=1,2,\\dots ,m$, the first $mn$ elements of the array x must contain the values\n $x11 , x12 ,…, x1m , x21 , x22 ,…, x2m ,…, xn1 , xn2 ,…, xnm .$\nThe $\\left(n+1\\right)$th and $\\left(n+2\\right)$th elements of each row ${x}_{n+1}^{\\mathit{p}},{x}_{n+2}^{\\mathit{p}}$, for $\\mathit{p}=1,2,\\dots ,m$, are required as workspace. These $2m$ elements may contain arbitrary values as they are set to zero by the routine.\nOn exit: the $m$ quarter-wave sine transforms stored as if in a two-dimensional array of dimension $\\left(1:{\\mathbf{m}},1:{\\mathbf{n}}+2\\right)$. Each of the $m$ transforms is stored in a row of the array, overwriting the corresponding original sequence. If the $n$ components of the $\\mathit{p}$th quarter-wave sine transform are denoted by ${\\stackrel{^}{x}}_{\\mathit{k}}^{\\mathit{p}}$, for $\\mathit{k}=1,2,\\dots ,n$ and $\\mathit{p}=1,2,\\dots ,m$, the $m\\left(n+2\\right)$ elements of the array x contain the values\n5:     $\\mathbf{work}\\left(*\\right)$ – Real (Kind=nag_wp) arrayWorkspace\nNote: the dimension of the array work must be at least ${\\mathbf{m}}×{\\mathbf{n}}+2×{\\mathbf{n}}+2×{\\mathbf{m}}+15$.\nThe workspace requirements as documented for c06rcf may be an overestimate in some implementations.\nOn exit: ${\\mathbf{work}}\\left(1\\right)$ contains the minimum workspace required for the current values of m and n with this implementation.\n6:     $\\mathbf{ifail}$ – IntegerInput/Output\nOn entry: ifail must be set to $0$, . If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.\nFor environments where it might be inappropriate to halt program execution when an error is detected, the value  is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value  is used it is essential to test the value of ifail on exit.\nOn exit: ${\\mathbf{ifail}}={\\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).\n\n## 6Error Indicators and Warnings\n\nIf on entry ${\\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).\nErrors or warnings detected by the routine:\n${\\mathbf{ifail}}=1$\nOn entry, ${\\mathbf{m}}=〈\\mathit{\\text{value}}〉$.\nConstraint: ${\\mathbf{m}}\\ge 1$.\n${\\mathbf{ifail}}=2$\nOn entry, ${\\mathbf{n}}=〈\\mathit{\\text{value}}〉$.\nConstraint: ${\\mathbf{n}}\\ge 1$.\n${\\mathbf{ifail}}=3$\nOn entry, ${\\mathbf{direct}}=〈\\mathit{\\text{value}}〉$.\nConstraint: ${\\mathbf{direct}}=\\text{'F'}$ or $\\text{'B'}$.\n${\\mathbf{ifail}}=4$\nAn internal error has occurred in this routine. Check the routine call and any array sizes. If the call is correct then please contact NAG for assistance.\n${\\mathbf{ifail}}=-99$\nAn unexpected error has been triggered by this routine. Please contact NAG.\nSee Section 3.9 in How to Use the NAG Library and its Documentation for further information.\n${\\mathbf{ifail}}=-399$\nYour licence key may have expired or may not have been installed correctly.\nSee Section 3.8 in How to Use the NAG Library and its Documentation for further information.\n${\\mathbf{ifail}}=-999$\nDynamic memory allocation failed.\nSee Section 3.7 in How to Use the NAG Library and its Documentation for further information.\n\n## 7Accuracy\n\nSome indication of accuracy can be obtained by performing a subsequent inverse transform and comparing the results with the original sequence (in exact arithmetic they would be identical).\n\n## 8Parallelism and Performance\n\nc06rcf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.\nc06rcf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.\nPlease consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.\n\n## 9Further Comments\n\nThe time taken by c06rcf is approximately proportional to $nm\\mathrm{log}\\left(n\\right)$, but also depends on the factors of $n$. c06rcf is fastest if the only prime factors of $n$ are $2$, $3$ and $5$, and is particularly slow if $n$ is a large prime, or has large prime factors.\n\n## 10Example\n\nThis example reads in sequences of real data values and prints their quarter-wave sine transforms as computed by c06rcf with ${\\mathbf{direct}}=\\text{'F'}$. It then calls the routine again with ${\\mathbf{direct}}=\\text{'B'}$ and prints the results which may be compared with the original data.\n\n### 10.1Program Text\n\nProgram Text (c06rcfe.f90)\n\n### 10.2Program Data\n\nProgram Data (c06rcfe.d)\n\n### 10.3Program Results\n\nProgram Results (c06rcfe.r)" ]

[ null ]

[ "Translator Disclaimer\nJuly 1996 Random Fourier series and continuous additive functionals of Lévy processes on the torus\nMichael B. Marcus, Jay Rosen\nAnn. Probab. 24(3): 1178-1218 (July 1996). DOI: 10.1214/aop/1065725178\n\n## Abstract\n\nLet X be an exponentially killed Lévy process on $T^n$ , the $n$ -dimensional torus, that satisfies a sector condition. (This includes symmetric Lévy processes.) Let$\\mathscr{F}_e$ denote the extended Dirichlet space of X. Let $h \\subset \\mathscr{F}_e$ and let ${h_y, y \\ subset T^n}$ denote the set of translates of $h$. That is, $h_y(\\dot) = h(\\dot - y)$. We consider the family of zero-energy continuous additive functions ${N_t^{[h_y]}, (y,t) \\subset T^n \\times R^+}$ defined by Fukushima. For a very large class of random functions h we show that\n\n$$J_\\rho (T^n) = \\int (\\log N_\\rho (T^n,\\varepsilon))^{1/2} d\\varepsilon < \\infty$$\n\nis a necessary and sufficient condition for the family ${N_t^{[h_y]}, (y,t) \\subset T^n \\times R^+}$ to have a continuous version almost surely. Here $N_p(T^n, \\varepsilon)$ is the minimum number of balls of radius $\\varepsilon$ in the metric p that covers $T^n$, where the metric p is the energy metric. We argue that this condition is the natural extension of the necessary and sufficient condition for continuity of local times of Lévy processes of Barlow and Hawkes.\n\nResults on the bounded variation and p-variation (in t )of $N_t^{[h_y]}$, for y fixed, are also obtained for a large class of random functions h.\n\n## Citation\n\nMichael B. Marcus. Jay Rosen. \"Random Fourier series and continuous additive functionals of Lévy processes on the torus.\" Ann. Probab. 24 (3) 1178 - 1218, July 1996. https://doi.org/10.1214/aop/1065725178\n\n## Information\n\nPublished: July 1996\nFirst available in Project Euclid: 9 October 2003\n\nzbMATH: 0862.60066\nMathSciNet: MR1411491\nDigital Object Identifier: 10.1214/aop/1065725178\n\nSubjects:\nPrimary: 42A61 , 60G15 , 60J45 , 60J55\n\nKeywords: Continuous additive functionals , Dirichlet spaces , random Fourier series", null, "", null, "" ]

[ null, "https://projecteuclid.org/Content/themes/SPIEImages/Share_black_icon.png", null, "https://projecteuclid.org/images/journals/cover_aop.jpg", null ]

[ "JDK7和JDK8中HashMap的大致变化是（这其实也是一个常被问道的面试题~）：\n\n1.7中采用数组+链表，1.8采用的是数组+链表/红黑树，即在1.7中链表长度超过一定长度后就改成红黑树存储。\n\n1.7扩容时需要重新计算哈希值和索引位置，1.8并不重新计算哈希值，巧妙地采用和扩容后容量进行&操作来计算新的索引位置。\n\n1.7是采用表头插入法插入链表，1.8采用的是尾部插入法。\n\n## 目录\n\n• 什么是哈希表\n• HashMap实现原理\n• 为何HashMap的数组长度一定是2的次幂？\n• 重写equals方法需同时重写hashCode方法\n• 总结\n\n## 一、什么是哈希表", null, "## 二、HashMap实现原理\n\nHashMap的主干是一个Entry数组。Entry是HashMap的基本组成单元，每一个Entry包含一个key-value键值对。\n\n//HashMap的主干数组，可以看到就是一个Entry数组，初始值为空数组{}，主干数组的长度一定是2的次幂，至于为什么这么做，后面会有详细分析。\ntransient Entry<K,V>[] table = (Entry<K,V>[]) EMPTY_TABLE;\n\nEntry是HashMap中的一个静态内部类。代码如下\n\nstatic class Entry<K,V> implements Map.Entry<K,V> {\nfinal K key;\nV value;\nEntry<K,V> next;//存储指向下一个Entry的引用，单链表结构\nint hash;//对key的hashcode值进行hash运算后得到的值，存储在Entry，避免重复计算\n\n/**\n* Creates new entry.\n*/\nEntry(int h, K k, V v, Entry<K,V> n) {\nvalue = v;\nnext = n;\nkey = k;\nhash = h;\n}", null, "//实际存储的key-value键值对的个数\ntransient int size;\nint threshold;\n//负载因子，代表了table的填充度有多少，默认是0.75\n//用于快速失败，由于HashMap非线程安全，在对HashMap进行迭代时，如果期间其他线程的参与导致HashMap的结构发生变化了（比如put，remove等操作），需要抛出异常ConcurrentModificationException\ntransient int modCount;\n\npublic HashMap(int initialCapacity, float loadFactor) {\n//此处对传入的初始容量进行校验，最大不能超过MAXIMUM_CAPACITY = 1<<30(230)\nif (initialCapacity < 0)\nthrow new IllegalArgumentException(\"Illegal initial capacity: \" +\ninitialCapacity);\nif (initialCapacity > MAXIMUM_CAPACITY)\ninitialCapacity = MAXIMUM_CAPACITY;\nthrow new IllegalArgumentException(\"Illegal load factor: \" +\n\nthreshold = initialCapacity;\n\n}\n\nOK,接下来我们来看看put操作的实现吧\n\npublic V put(K key, V value) {\n//如果table数组为空数组{}，进行数组填充（为table分配实际内存空间），入参为threshold，此时threshold为initialCapacity 默认是1<<4(24=16)\nif (table == EMPTY_TABLE) {\ninflateTable(threshold);\n}\n//如果key为null，存储位置为table或table的冲突链上\nif (key == null)\nreturn putForNullKey(value);\nint hash = hash(key);//对key的hashcode进一步计算，确保散列均匀\nint i = indexFor(hash, table.length);//获取在table中的实际位置\nfor (Entry<K,V> e = table[i]; e != null; e = e.next) {\n//如果该对应数据已存在，执行覆盖操作。用新value替换旧value，并返回旧value\nObject k;\nif (e.hash == hash && ((k = e.key) == key || key.equals(k))) {\nV oldValue = e.value;\ne.value = value;\ne.recordAccess(this);\nreturn oldValue;\n}\n}\nmodCount++;//保证并发访问时，若HashMap内部结构发生变化，快速响应失败\nreturn null;\n}\n\nprivate void inflateTable(int toSize) {\nint capacity = roundUpToPowerOf2(toSize);//capacity一定是2的次幂\ntable = new Entry[capacity];\ninitHashSeedAsNeeded(capacity);\n}\n\ninflateTable这个方法用于为主干数组table在内存中分配存储空间，通过roundUpToPowerOf2(toSize)可以确保capacity为大于或等于toSize的最接近toSize的二次幂，比如toSize=13,则capacity=16;to_size=16,capacity=16;to_size=17,capacity=32.\n\nprivate static int roundUpToPowerOf2(int number) {\n// assert number >= 0 : \"number must be non-negative\";\nreturn number >= MAXIMUM_CAPACITY\n? MAXIMUM_CAPACITY\n: (number > 1) ? Integer.highestOneBit((number - 1) << 1) : 1;\n}\n\nroundUpToPowerOf2中的这段处理使得数组长度一定为2的次幂，Integer.highestOneBit是用来获取最左边的bit（其他bit位为0）所代表的数值.\n\nhash函数\n\n//这是一个神奇的函数，用了很多的异或，移位等运算，对key的hashcode进一步进行计算以及二进制位的调整等来保证最终获取的存储位置尽量分布均匀\nfinal int hash(Object k) {\nint h = hashSeed;\nif (0 != h && k instanceof String) {\nreturn sun.misc.Hashing.stringHash32((String) k);\n}\n\nh ^= k.hashCode();\n\nh ^= (h >>> 20) ^ (h >>> 12);\nreturn h ^ (h >>> 7) ^ (h >>> 4);\n}\n\n/**\n* 返回数组下标\n*/\nstatic int indexFor(int h, int length) {\nreturn h & (length-1);\n}\n\nh&（length-1）保证获取的index一定在数组范围内，举个例子，默认容量16，length-1=15，h=18,转换成二进制计算为\n\n 1 0 0 1 0\n& 0 1 1 1 1\n__________________\n0 0 0 1 0 = 2", null, "void addEntry(int hash, K key, V value, int bucketIndex) {\nif ((size >= threshold) && (null != table[bucketIndex])) {\nresize(2 * table.length);//当size超过临界阈值threshold，并且即将发生哈希冲突时进行扩容\nhash = (null != key) ? hash(key) : 0;\nbucketIndex = indexFor(hash, table.length);\n}\n\ncreateEntry(hash, key, value, bucketIndex);\n}\n\n## 三、为何HashMap的数组长度一定是2的次幂？\n\nvoid resize(int newCapacity) {\nEntry[] oldTable = table;\nint oldCapacity = oldTable.length;\nif (oldCapacity == MAXIMUM_CAPACITY) {\nthreshold = Integer.MAX_VALUE;\nreturn;\n}\n\nEntry[] newTable = new Entry[newCapacity];\ntransfer(newTable, initHashSeedAsNeeded(newCapacity));\ntable = newTable;\nthreshold = (int)Math.min(newCapacity * loadFactor, MAXIMUM_CAPACITY + 1);\n}\n\nvoid transfer(Entry[] newTable, boolean rehash) {\nint newCapacity = newTable.length;\n//for循环中的代码，逐个遍历链表，重新计算索引位置，将老数组数据复制到新数组中去（数组不存储实际数据，所以仅仅是拷贝引用而已）\nfor (Entry<K,V> e : table) {\nwhile(null != e) {\nEntry<K,V> next = e.next;\nif (rehash) {\ne.hash = null == e.key ? 0 : hash(e.key);\n}\nint i = indexFor(e.hash, newCapacity);\n//将当前entry的next链指向新的索引位置,newTable[i]有可能为空，有可能也是个entry链，如果是entry链，直接在链表头部插入。\ne.next = newTable[i];\nnewTable[i] = e;\ne = next;\n}\n}\n}\n\nhashMap的数组长度一定保持2的次幂，比如16的二进制表示为 10000，那么length-1就是15，二进制为01111，同理扩容后的数组长度为32，二进制表示为100000，length-1为31，二进制表示为011111。", null, "", null, "", null, "get方法\n\npublic V get(Object key) {\n//如果key为null,则直接去table处去检索即可。\nif (key == null)\nreturn getForNullKey();\nEntry<K,V> entry = getEntry(key);\nreturn null == entry ? null : entry.getValue();\n}\n\nget方法通过key值返回对应value，如果key为null，直接去table处检索。我们再看一下getEntry这个方法\n\nfinal Entry<K,V> getEntry(Object key) {\n\nif (size == 0) {\nreturn null;\n}\n//通过key的hashcode值计算hash值\nint hash = (key == null) ? 0 : hash(key);\n//indexFor (hash&length-1) 获取最终数组索引，然后遍历链表，通过equals方法比对找出对应记录\nfor (Entry<K,V> e = table[indexFor(hash, table.length)];\ne != null;\ne = e.next) {\nObject k;\nif (e.hash == hash &&\n((k = e.key) == key || (key != null && key.equals(k))))\nreturn e;\n}\nreturn null;\n}\n\n## 四、重写equals方法需同时重写hashCode方法\n\n/**\n* Created by chengxiao on 2016/11/15.\n*/\npublic class MyTest {\nprivate static class Person{\nint idCard;\nString name;\n\npublic Person(int idCard, String name) {\nthis.idCard = idCard;\nthis.name = name;\n}\n@Override\npublic boolean equals(Object o) {\nif (this == o) {\nreturn true;\n}\nif (o == null || getClass() != o.getClass()){\nreturn false;\n}\nPerson person = (Person) o;\n//两个对象是否等值，通过idCard来确定\nreturn this.idCard == person.idCard;\n}\n\n}\npublic static void main(String []args){\nHashMap<Person,String> map = new HashMap<Person, String>();\nPerson person = new Person(1234,\"乔峰\");\n//put到hashmap中去\nmap.put(person,\"天龙八部\");\n//get取出，从逻辑上讲应该能输出“天龙八部”\nSystem.out.println(\"结果:\"+map.get(new Person(1234,\"萧峰\")));\n}\n}" ]

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[ "English | Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. 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[ "Welcome Forums Gravitation Einstein Reply To: Einstein\n\n#524\nGyula Szász\nModerator\n\nEinstein’s relativity theories, SR and GR, are physically invalid theories. The quantum theories (QT) based on the light quantum hypothesis are also invalid.\n\nThe physics based on four kinds of stable elementary particles, e, p, P and E, which carry two kinds of conserved elementary charges and the interactions are non-quantized, non-conservative interactions and propagate with the constant velocity c.\n\nThe elementary masses, me and mP, of the elementary particles are not equivalent of energy; only for the stable elementary particles are the inertial and gravitational masses equal, mi(elementary particle) = mg(elementary particle). For the other entire particle system are the inertial and gravitation masses different. The inertial masses are\nmi = (N(P) + N(E))∙mP + (N(e) + N(p))∙me – E(bound)/c^2,\nand the gravitational masses\nmg = |(N(P) – N(E))∙mP + (N(p) – N(e))∙me |.\nMasses, mi and mg, are greater or equal zero and the bound energy E(bound) is radiated the bonding of the elementary particles, e, p, P and E." ]

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[ "Latest Articles\n\n# One less than the previous - For 7 the Multiplicand is 143\n\nInfo Post\nPART  2 : FOR FINDING DECIMALS OF FRACTIONS\n\nFor 7 the Multiplicand is 143\n\nIt is also called as ' Kevalaih Saptakam Gunyat ' which is the corollary to Paraavartya Yojayet, the fourth sutra of Vedic mathematics. This sutra is helpful for decimals for which the denominator is 7.\n\nExample 1\nWhat is the result of 1/7 ? Convert to a decimal\n\nWe multiply the multiplicand (143) by 999 using Ekanyunena Purvena or One less than the Previous sutra.We use the same number of digits for multiplying ,so '999'\n\nNow ,143 * 999 =?\n\nUsing One less than the Previous sutra,\n143 - 1=142\n999 - 142 =857\nWe get Result = 142857\n\nNow multiply the result by the numerator of (1/7) ,\n142857 * 1 =142857\n\nPlace a decimal point at the beginning as 0.142857\n\nAS SIMPLE AS THAT\n\nTherefore , 1/7 = 0.142857\n\nExample 2\nWhat is the result of 2/7 ? Convert to a decimal\n\nWe multiply the multiplicand (143) by 999 using Ekanyunena Purvena or One less than the Previous sutra.We use the same number of digits for multiplying ,so '999'\nNow ,143 * 999 =?\n\nUsing One less than the Previous sutra,\n143 - 1=142\n999 - 142 =857\nWe get Result = 142857\n\nNow multiply the result by the numerator of ( 2/7) ,\n142857 * 2 =285714\nPlace a decimal point at the beginning as 0.285714\n\nAS SIMPLE AS THAT\n\nTherefore , 1/7 = 0.285714\n\nSimilarly the result can be found out for any numerical value for fractions by '7'.Sounds fun!Is'nt it?...Work out and Enjoy!\n\nNext time we can see how to find decimals of any number when divided by 13 ,17 and 19.i.e; Fractions of 13 ,17 and 19.\n\nYou will love this : Decimals of any number for fractions of 9\n1/9 = .111... - (1 *.111)\n2/9 = .222... - (2 *.111)\n3/9 = .333... - (3 *.111)\n4/9 = .444... - (4 *.111)\n5/9 = .555... - (5 *.111)\n6/9 = .666... - (6 *.111)\n7/9 = .777... - (7 *.111)\n8/9 = .888...\n- (8 *.111)\n\nFrom the above we know that (1/9) =.111 , so any number divided by 9 say 21/9 gives you 21 * 0.111 =2.3333 and so on\nSimilarly for 10  and 11,the decimals are\n 1/10 = .1 2/10 = .2 3/10 = .3 4/10 = .4 5/10 = .5 6/10 = .6 7/10 = .7 8/10 = .8 9/10 = .9 1/11 = .090909... 2/11 = .181818... 3/11 = .272727... 4/11 = .363636... 5/11 = .454545... 6/11 = .545454... 7/11 = .636363... 8/11 = .727272... 9/11 = .818181... 10/11 = .909090...\n\nDecimals for fractions of 13 in a minute\n\nShare:" ]

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[ "Contents - Index\n\nTools: General - Quick Calc\n\nThis is used to create a lease or loan calculation.\n\nLayout / How To\nThis screen closely resembles the New Deal Calculate page, and acts much in the same way. Fill out the data on the screen and click the Calculate button. Upon a successful calculation, the schedules and quote on the tabs on the right will become available.\n\nBase Values\nFill out the data in the boxes in the left column of the screen (on the Calculations tab). In the Base Calculation Value box, provide only 4 of the 5 values and the 5th one will be calculated for you. To back calculate an interest rate, leave it zero and uncheck the box after it.\n\nIf the end value of the deal is \\$0.00, then you must check the box after it. The same applies to the interest rate.\n\nThe Term is the number of payments to be made and must correspond to the payment frquency selected.\n\nThere are two buttons available that will pull in customer data. Use the Import Current Calculations button to retrieve the original calculation data for this customer. Use the Import Current Balance button to start the Capitalized Cost at the customer's current balance.\n\nWhen using the Load Current Balance button, some assumptions are made. For a lease, the end value will be set to 40% of the capitalized cost, the term and interest rate will remain the same, the Contract Date will be set to today's date, the First Payment Due Date will be set to today's date, and the Second Payment Due Date will be set on payment period out. For a loan, the end value will be zero, the term and interest rate will remain the same, the Contract Date will be set to today's date, the First Payment Due Date will be set to 30 days out, and the Second Payment Due Date will be set one payment period out.\n\nCalculation Options\nIn the Calculation Options box, there are four items that need addressing. First, you need to select the Calculation Type. \"Spreadsheet\" is the most common. (\"Standard\" is not recommended; please use \"Standard with Interest\" instead.) Second, you need to tell select the Payment Frequency from the list. Monthly is the most common selection. Third, you will need to answer the Prorate question. \"No\" is the most common selection. If you select \"Yes\", then a prorated amount will automatically be calculated for you and added to the deal.\n\nDates\nThe Contract Date is the contract date of the deal. Interest will start as of this date.\n\nThe First Payment Due Date is the date on which the first payment is due. If prorating, interest will be calculated from the Deal Date to this date.\n\nBased on the payment frequency and the First Payment Due Date, the Second Payment Due Date will be calculated for you. You may manually change this.\n\nOther Information\nThis is used for the reports that are generated on the tabs at the top of the screen. If you import calculation data from the current deal, then these fields will be filled in from the current deal.\n\nCalculated Results\nThis is where the resulting calculations will display. If there is a periodic tax amount, then fill in the tax percentage where indicated. You must click the Calculate button to get the results. If the Net Capitalized Cost includes fees, then enter the total of those fees into the APR Calculation box and a more accurate APR will be calculated for you.\n\nPlanned Long Term Growth\nThese fields are provided on the PLTG tab that will allow you to perform the PLTG calculation. You must calculate the deal, first. Then provide the PLTG input fields and click the Calculate PLTG button. The results will display and the PLTG tab will contain printable results." ]

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[ "Item\nproduct name\namount\nUnit price (netto)\n\nAmount netto:0.00 zł\n\nAmount brutto:0.00 zł\n\n## Bearing Temperature Sensors\n\n### Applications\n\n• Temperature range: -40 .. +250°C (depends on sensor construction)\n• Temperature measurement of bearings in steam and gas turbines, power generators, etc.\n• Power industry\n\n### Features\n\n• Mineral insulated construction\n• Adapted to individual customer requirements\n• Small diameters (from Ø3 mm)\n• Short response time\n• Possibility of bending the sensor\n• Resistant to vibration\n\n### ATEX versions\n\nIntrinsically safe designs are available for applications in hazardous areas. These models are provided with a type-examination certificate for „intrinsically safe“ type of protection according to Directive 2014/34/UE (ATEX) for gases and dust.", null, "Miniature Bearing Sensors are designed to monitor temperature of embedded bearings critical to machine performance in industrial rotating equipment. The most reliable indicator of bearing condition is the temperature of the metal beneath the bearing shoe. Rising temperatures indicates the breakdown of the lubricating oil film. Quick identification of this condition allows machine shutdown and maintenance to occur before catastrophic failure of the bearing itself. Typical applications for bearing sensors include motors, turbines, pumps, gear boxes and other journal bearing pedestals.\n\n\nKalkulator\nUnit converter\n\nRTDs and TCs\ncharacteristic calculator\n\nResistance thermometers\n\nThermocouples\n\nCold Junction Compensation = °C\n\nCalculate value of resistance to temperature\n(values according to ITS-90 scale)\n\nR = Ω  T = 0 °C\n\nCalculate value of temperature to resistance\n(values according to ITS-90 scale)\n\nT = °C R = 0 Ω\n\nLimit values for the calculator\n\n-200 °C - 850 °C\n\n18,52 Ω - 390,48 Ω\n\nCold Junction Compensation = °C\n\nCalculate value of resistance to temperature\n(values according to ITS-90 scale)\n\nR = Ω  T = 0 °C\n\nCalculate value of temperature to resistance\n(values according to ITS-90 scale)\n\nT = °C R = 0 Ω\n\nLimit values for the calculator\n\n-60 °C - 250 °C\n\n69,52 Ω - 289,16 Ω\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-210 °C - 1200 °C\n\n-8,095 mV - 69,553 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-270 °C - 1372 °C\n\n-6,458 mV - 54,886 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-270 °C - 1300 °C\n\n-4,345 mV - 47,513 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-270 °C - 1000 °C\n\n-9,835 mV - 76,373 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-270 °C - 400 °C\n\n-6,258 mV - 20,872 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-50 °C - 1768.1 °C\n\n-0,226 mV - 21,103 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n-50 °C - 1768.1 °C\n\n-0,236 mV - 18,694 mV\n\nCold Junction Compensation = °C\n\nCalculate value of thermoelectric emf to temperature\n\nE = mV T = 0 °C\n\nCalculate value of temperature to thermoelectric emf\n\nT = °C  E = 0 mV\n\nLimit values for the calculator\n\n0°C - 1820 °C\n\n-0,003 mV - 13,820 mV\n\nPomoc. Instrukcja użytkownika.\n\nUnit\nconverter\nRTDs and TCs\ncharacteristic calculator" ]

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[ "### Stats\n\n7,451 visits, 18,538 views\n\n### Translations\n\nThis tutorial hasn't been translated.\n\n### Bit Logic\n\nTo check, if our gauntlet may fit on the hand, we'll do a bitwise AND with these two numbers 5 AND 1. And big suprise, the result will be 1 which is equal to true!!\n\nWhy does that work?\n\nThe bitwise AND works like this:\n\n1 AND 1 = 1\n\n0 AND 1 = 0\n\n1 AND 0 = 0\n\nSo if we have 101 (our slot) and 001 (the gauntlet) this will happen:\n\n1. Bit - 1 AND 0 = 0\n\n2. Bit - 0 AND 0 = 0\n\n3. Bit - 1 AND 1 = 1\n\nThe result is 1 and one is equal to true - just like 0 is the \"same\" as false\n\nNow, how do we set the bits? There are two ways, one is adding the decimal values of the \"position\" the bit is in, the other is to combine bit groups with the binary OR.\n\nTo make it easy on you, I have provided the decimal values you need as instance variables within the slot sprite - even thou this example does not use this variables.\n\nIf you want to create a hat, that you could store in the bag as well as hold it in the hand or wear it on the head, you'd have 49.\n\n1 (for the bag - or all), 16 (for the hand) and 32 for (for the head)\n\nWe don't use the bitwise OR in this example, but to be complete here's how it works:\n\n1 OR 1 = 1\n\n0 OR 1 = 1\n\n1 OR 0 = 1\n\n0 OR 0 = 0\n\nYou see, that the OR sets the bit, if it is not already set." ]

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[ "# Yambo input file for GW calculations explained", null, "Example of a yambo input file for GW calculation in a solid. In order to generate the following input file you can use the command “yambo -g n -p p -V par\" with yambo 4.0.2 (and with yambo 3.4.2  “yambo -g n -p p”).\n\n```gw0 # [R GW] GoWo Quasiparticle energy levels\nppa # [R Xp] Plasmon Pole Approximation\nrim_cut # [R RIM CUT] Coulomb potential\nHF_and_locXC # [R XX] Hartree-Fock Self-energy and Vxc\nem1d # [R Xd] Dynamical Inverse Dielectric Matrix\nNLogCPUs=0 # [PARALLEL] Live-timing CPU`s (0 for all)\n\nX_all_q_CPU= \"2 2 2 2\" # [PARALLEL] CPUs for each role\nX_all_q_ROLEs= \"q k c v\" # [PARALLEL] CPUs roles (q,k,c,v)\nParallelization of the dieletric constant the product of each role\nin X_all_q_CPU should be equal to the total number of processors.\nIn this example I use 16 processors 2x2x2x2. Notice that the parallelization in q\nis more efficient but use more memory.\n\nX_all_q_nCPU_invert=4 # [PARALLEL] CPUs for matrix inversion\nNumber of processors to invert the dielectric constant.\nIn general matrix inversion is very inefficiently parallelized, therefore\ndo not use many processor for this part, I advice about 4.\n\nIf your calculation uses to much memory you can run with openmp and increase\nhere the number of Threads for the dipoles, xhi and the self-energy.\nUse the same number in all of them.\n\nSE_CPU= \"2 4 2\" # [PARALLEL] CPUs for each role\nSE_ROLEs= \"q qp b\" # [PARALLEL] CPUs roles (q,qp,b)\nSimilar to the previous parallelization.\nThe total product should equal to the number of processors. The parallelization\nin qp is the more efficient while the one on b uses less memory.\n\nFFTGvecs= 20065 RL # [FFT] Plane-waves\nTotal number of G-vectors used in the calculation.\nDecrease this number to use less memory.\n\nRandQpts= 3000000 # [RIM] Number of random q-points in the BZ\nRandGvec= 1 RL # [RIM] Coulomb interaction RS components\nParameters for numerical integration\nof the Coulomb potentianl at q=0. Do not change them.\n\nEXXRLvcs= 20065 RL # [XX] Exchange RL components\nNumber of G-vectors in the exchange, Eq. 104 in Ref.This number should be less than\nthe total number of G-vectors FFTGvecs otherwise yambo automatically\ndecreases it to FFTGvecs.\n\nChimod= \"\" # [X] IP/Hartree/ALDA/LRC/BSfxc\n% BndsRnXp\n1 | 1600 | # [Xp] Polarization function bands\n%\nNumber of bands in the dielectric constant calculation.\nSee eq. 102 in Ref. \n\nNGsBlkXp= 3002 mHa # [Xp] Response block size\nNumber of G-vectors in the dielectric constant, see Eq. 26 and 27 in Ref. \n\n% LongDrXp\n1.000000 | 0.000000 | 0.000000 | # [Xp] [cc] Electric Field\n%\nDirection of the electric field for the calculation of the q=0 component\nof the dielectric constant e(q,w)\n\nPPAPntXp= 27.21138 eV # [Xp] PPA imaginary energy\nSecond frequency used to fit the Godby-Needs plasmon-pole model (PPM).\nIf results change by changing this frequency, the PPM is not adequate for your calculation.\n\n% GbndRnge\n1 | 1600 | # [GW] G[W] bands range\n%\nNumber of bands used to expand the Green's function, Eq. 110 in Ref. \nThis number is usually larger than the number of bands used to calculated the dielectric\nconstant. You can use the same number for both.\n\nGDamping= 0.10000 eV # [GW] G[W] damping\nDamping in the Green's function definition, the delta\nparameter in Eq. 49 in Ref. . Few people investigate the effect of this parameters\non the final results, usually everybody assume it equal to 0.1 eV.\n\ndScStep= 0.10000 eV # [GW] Energy step to evaluate Z factors\nDysSolver= \"n\" # [GW] Dyson Equation solver (\"n\",\"s\",\"g\")\nParameters related to the linearized solution\nof the Dyson equation, see Eq. 58 in Ref..\n\nGTermKind= \"BRS\" # [GW] GW terminator (\"none\",\"BG\" Bruneval-Gonze,\"BRS\" Berger-Reining-Sottile)\nTerminator for the self-energy and the dielectric constant.\nThis flag speeds-up the convergence with the number of conduction bands,\nsee Phys. Rev. B 82, 041103(R).\n\n%QPkrange # [GW] QP generalized Kpoint/Band indices\n1| 2|144|432|\n%\nK-points and band range where you want to calculate\nthe GW correction. The syntax is\nfirst kpoint | last kpoint | first band | last band\n```\n\nIf you are working with isolated molecules consider the following  tricks and suggestions in your input file.\n1) Add the Coulomb cut-off to your input by adding the -r (-c in yambo 3.4.1) flag in the yambo comand line\n\n```CUTGeo= \"box XYZ\" # [CUT] Coulomb Cutoff geometry: box/cylinder/sphere X/Y/Z/XY..\n% CUTBox\n10.00 | 10.00 | 20.00 | # [CUT] [au] Box sides\n%\nCUTCylLen= 0.000000 # [CUT] [au] Cylinder length\nThese parameters specify a cut-off for the Coulomb interaction\nin such a way to speed-up convergence with the supercell. My advise is to use\nthe box cut-off with sides slightly smaller than the cell. In this\nexample I suppose to have a supercell (12x12x22 a.u.) and I choose\na cutoff 10x10x20 a.u. (see PRB 73, 205119)```\n\n2) Parallelization for isolated systems:\n\n```when you a single-kpoint (for an isolated molecule for example) you cannot\nparallelize in q and k you have\nto set the number of processors for these roles equal to 1.\n```\n\nReferences:\n Quasiparticle calculations in solids\nAULBUR W. G., JÖNSSON L., WILKINS J. W.\nhttp://www.physics.ohio-state.edu/~wilkins/vita/gw_review.ps\n\n The GW method\nhttp://xxx.lanl.gov/pdf/cond-mat/9712013v1\n\n## 10 thoughts on “Yambo input file for GW calculations explained”\n\n1.", null, "Simil\n\nHi,\nI have a system with an organic molecule on top of Au surface. Is it possible to get Projected Density Of States (PDOS) of the molecule after GW calculations in Yambo?\n\n1.", null, "attacc Post author\n\nDear Simil\n\nas fas as I know, it is not possible. You can try to look if you can force\nquantum-espresso to read the quasi-particle correction before calculation the PDOS,\nbut I don’t know how much it can be complicated\n\nregards\nClaudio\n\n2.", null, "Fabio\n\nHi, i have a question regarding the GW convergence.\n\nBndsRnXp, NGsBlkXp, GbndRnge cannot be converged independently, right?\n\nFor what i have saw in my results, these quantities are inter-dependent.\nAm i doing something wrong? If not, is there a way to reduce the number of calculations we have to do\nfor reach convergence? Aren’t BndsRnXp and NGsBlkXp related somehow?\n\nFabio\n\n1.", null, "attacc Post author\n\nDear Fabio\n\nregarding BndsRnXp, NGsBlkXp you are right, they are inter-dependent,\nbut GbndRnge should be independent from the other two.\nTry to converge BndsRnXp, NGsBlkXp together and then converge GbndRnge.\nUsually the gap converges faster than the single band values.\n\nLet me know\nClaudio\n\n1.", null, "Fabio\n\nDear Claudio,\n\nI converged BndsRnXp, NGsBlkXp together with GbndRnge (100) fixed.\nI found that BndsRnXp=500 , and NGsBlkXp=8 Ry is enough to reach convergence.\n\nThen I converged GbndRnge, with BndsRnXp=500, NGsBlkXp=8, 10, 12, 14 Ry.\nI found that the convergence is achieved for GbndRnge=500, BndsRnXp=500, NGsBlkXp=12 Ry and not NGsBlkXp=8 Ry.\n\nI considered a convergence criterion within 0.05 eV.\n\nThis shows that there is an inter-dependence between\nGbndRnge and NGsBlkXp. Is that normal?\n\nThanks you,\n\nFabio\n\n1.", null, "attacc Post author\n\nDear Fabio\n\nyes this is normal, in the sense that usually the inter-dependence between GbndRnge and NGsBlkXp\nis small but not zero.\nNotice that in general number of bands to converge the Green’s functions\nis larger than the one for the dielectric constant GbndRnge >= BndsRnXp.\nDid you converge the absolute value of the bands or their energy difference (as the gap)?\nBecause in general the energy difference converge fast, while the absolute value is slower.\n\nClaudio\n\nClaudio\n\n1.", null, "Fabio\n\nDear Claudio,\n\nI converged the energy difference. But the inter-dependence\nbetween GbndRnge and NGsBlkXp is indeed small.\n\nAnd my convergence criterion is <= 0.05 eV. If i choose <= 0.1 eV a smaller\nGbndRnge and NGsBlkXp would be enough.\n\nThank you,\nFabio\n\n3.", null, "Shi\n\nHi Claudio,\nI have a general question regarding GW and BSE.\nSay I have finished those heavy calculations, which files/folder should I back up in order to reproduce those calculations in the future while reducing the computational time? Is that even possible?\nIt is hard to back up all the files (in many cases > 10Gb).\nI used yambo -J job_string_dir when running yambo. So some files are stored in job_string_dir as well besides SAVE directory.\n\nThanks!\n\nShi\n\n1.", null, "attacc Post author\n\nDear Shi\nit depends from your calculation. If you performed GW calculation, the imporat results\nare in the ndb.QP that contains all information about the quasi-particle corrections.\nIf you performed BSE, the files are the ndb.BS*, that contain the BSE matrix.\nAnother important file could be the dielectric constant ndb.pp*\n\nbest\nClaudio\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# 5036923\n\n## 5,036,923 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 5036923 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.\n\n5036923 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.\n\n## Prime factorization of 5036923:\n\n### 29 × 173687\n\nSee below for interesting mathematical facts about the number 5036923 from the Numbermatics database.\n\n### Names of 5036923\n\n• Cardinal: 5036923 can be written as Five million, thirty-six thousand, nine hundred twenty-three.\n\n### Scientific notation\n\n• Scientific notation: 5.036923 × 106\n\n### Factors of 5036923\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 2\n• Sum of prime factors: 173716\n\n### Divisors of 5036923\n\n• Number of divisors d(n): 4\n• Complete list of divisors:\n• Sum of all divisors σ(n): 5210640\n• Sum of proper divisors (its aliquot sum) s(n): 173717\n• 5036923 is a deficient number, because the sum of its proper divisors (173717) is less than itself. Its deficiency is 4863206\n\n### Bases of 5036923\n\n• Binary: 100110011011011011110112\n• Base-36: 2ZYIJ\n\n### Squares and roots of 5036923\n\n• 5036923 squared (50369232) is 25370593307929\n• 5036923 cubed (50369233) is 127789724956353662467\n• The square root of 5036923 is 2244.3090250677\n• The cube root of 5036923 is 171.4174790773\n\n### Scales and comparisons\n\nHow big is 5036923?\n• 5,036,923 seconds is equal to 8 weeks, 2 days, 7 hours, 8 minutes, 43 seconds.\n• To count from 1 to 5,036,923 would take you about twelve weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 5036923 cubic inches would be around 14.3 feet tall.\n\n### Recreational maths with 5036923\n\n• 5036923 backwards is 3296305\n• The number of decimal digits it has is: 7\n• The sum of 5036923's digits is 28\n• More coming soon!" ]

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[ "# AUC.Song_Zhou: AUC estimator proposed by Song and Zhou In survAUC: Estimators of prediction accuracy for time-to-event data.\n\n## Description\n\nSong and Zhou's estimators of AUC for right-censored time-to-event data\n\n## Usage\n\n ```1 2 3 4``` ```AUC.sh(Surv.rsp, Surv.rsp.new=NULL, lp, lpnew, times, type=\"incident\", savesensspec=FALSE) sens.sh(Surv.rsp, lp, lpnew, times, type=\"incident\") spec.sh(Surv.rsp, lp, lpnew, times) ```\n\n## Arguments\n\n `Surv.rsp` A `Surv(.,.)` object containing to the outcome of the training data. `Surv.rsp.new` A `Surv(.,.)` object containing the outcome of the test data. `lp` The vector of predictors estimated from the training data. `lpnew` The vector of predictors obtained from the test data. `times` A vector of time points at which to evaluate AUC. `type` A string defining the type of true positive rate (TPR): `\"incident\"` refers to incident TPR , `\"cumulative\"` refers to cumulative TPR. `savesensspec` A logical specifying whether sensitivities and specificities should be saved.\n\n## Details\n\nThe `sens.sh` and `spec.sh` functions implement the estimators of time-dependent true and false positive rates proposed by Song and Zhou (2008).\n\nThe `AUC.sh` function implements the estimators of cumulative/dynamic and incident/dynamic AUC proposed by Song and Zhou (2008). These estimators are given by the areas under the time-dependent ROC curves estimated by `sens.sh` and `spec.sh`. In case of cumulative/dynamic AUC, the `iauc` summary measure is given by the integral of AUC on [0, max(`times`)] (weighted by the estimated probability density of the time-to-event outcome). In case of incident/dynamic AUC, `iauc` is given by the integral of AUC on [0, max(`times`)] (weighted by 2 times the product of the estimated probability density and the estimated survival function of the time-to-event outcome).\n\nThe results obtained from `spec.sh`, `spec.sh` and `AUC.sh` are valid as long as `lp` and `lpnew` are the predictors of a correctly specified Cox proportional hazards model. In this case, the estimators remain valid even if the censoring times depend on the values of the predictors.\n\n## Value\n\n`AUC.sh` returns an object of class `survAUC`. Specifically, `AUC.sh` returns a list with the following components:\n\n `auc` The cumulative/dynamic or incident/dynamic AUC estimates (evaluated at `times`). `times` The vector of time points at which AUC is evaluated. `iauc` The summary measure of AUC.\n\n`sens.sh` and `spec.sh` return matrices of dimensions `times` x `lpnew + 1`. The elements of these matrices are the sensitivity and specificity estimates for each threshold of `lpnew` and for each time point specified in `times`.\n\n## References\n\nSong, X. and X.-H. Zhou (2008).\nA semiparametric approach for the covariate specific ROC curve with survival outcome.\nStatistica Sinica 18, 947–965.\n\n`AUC.uno`, `AUC.cd`, `AUC.hc`, `GHCI`, `IntAUC`\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14``` ```TR <- ovarian[1:16,] TE <- ovarian[17:26,] train.fit <- coxph(Surv(futime, fustat) ~ age, x=TRUE, y=TRUE, method=\"breslow\", data=TR) lp <- predict(train.fit) lpnew <- predict(train.fit, newdata=TE) Surv.rsp <- Surv(TR\\$futime, TR\\$fustat) Surv.rsp.new <- Surv(TE\\$futime, TE\\$fustat) times <- seq(10, 1000, 10) AUC_sh <- AUC.sh(Surv.rsp, Surv.rsp.new, lp, lpnew, times) names(AUC_sh) AUC_sh\\$iauc ```" ]

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[ "## Welcome to the Treehouse Community\n\nWant to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community!\n\n### Looking to learn something new?\n\nTreehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.", null, "", null, "# Not understanding functions and arguements\n\nI'm not sure if i just don't understand how this question is phrased or if I'm actually not understanding it? Can anyone give me some insight on what I'm doing wrong and mind clarifying exactly what I'm doing wrong?\n\nscript.js\n```function max(5,2) {\nif(5 > 2) {\nreturn 5\n}\n}\n```", null, "Hi, The function should work for any two numbers, not just 5 and 2. The two numbers that your function takes in are known as arguments. You can name these arguments whatever you like. The function should look something like the following:\n\n```function max(num1, num2){\nif(num1 > num2){\nreturn num1;\n}else{\nreturn num2;\n}\n}\n```\n\nNow, you can call the function using any two numbers, and the function will return the larger of the two numbers. e.g. max(5,2);", null, "```// function name \"max\"\nfunction max(num1, num2) { // arguments num1, num2( you can name them \"a\" and \"b\", or x, y, lemon, anything )\nif (num1 > num2) {\nreturn num1;\n} else {\nreturn num2;\n}\n};\n\n// you can use any numbers you want\nmax(6,10);\n// just changed the arguments\n```\n\nSo, basically a function is a set of instructions .In this case our function is set to give the max value every time an argument is passed. So now what are arguments? Well, arguments are nothing just you can say values which you want to use.\n\nYou just created a function \"max\" which takes two arguments and you can name it anything you want. What this function is doing basically it just returns the number which has the higher value than the other.\n\nAdded if else statements because what if the user types a higher value in the 1st argument and later on decides that may be for no reason the higher value looks nice in the second argument, then what? thats why an else statement.\n\nI hope it helps ;)\n\nYou are doing a wonderful job!", null, "There is still a couple problems, though. So first, you cannot have numbers as the first letter in a variable or argument name. Second, you need to add an else clause to the \"if\" condition. That's it for the first task. For the second task, you should alert the result of \"max(<num1>, <num2>)\". You can use any integer (number) to replace the <num1> and <num2>, but I choose 10 and 15. This is the full code:\n\n```function max(a, b) {\nif (a > b) {\nreturn a;\n} else {\nreturn b;\n}\n}", null, "Hope that helps!" ]

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[ "How do solve the following linear system?: -x - y = 8 , 8x+2y=2 ?\n\nMar 13, 2016\n\nThe solution for the system of equations is:\n color(green)(x =3\ncolor(green)(y = -11\n\nExplanation:\n\n$- x - y = 8$ , multiplying this equation by $8$\n$\\textcolor{g r e e n}{- 8 x} - 8 y = 64$..........equation $\\left(1\\right)$\n\n$\\textcolor{g r e e n}{8 x} + 2 y = 2$.................equation $\\left(2\\right)$\n\nSolving by elimination would result in elimination of the terms $\\textcolor{g r e e n}{8 x}$:\n\nAdding the two equations:\n\n$\\cancel{\\textcolor{g r e e n}{- 8 x}} - 8 y = 64$\n$\\cancel{\\textcolor{g r e e n}{8 x}} + 2 y = 2$\n\n$- 6 y = 66$\n\n$y = \\frac{66}{- 6}$\n\ncolor(green)(y = -11\n\nFinding the value of $x$ from equation $\\left(1\\right)$:\n$- x - y = 8$\n\n$x = - y - 8$\n\n$x = - \\textcolor{g r e e n}{\\left(- 11\\right)} - 8$\n\n$x = 11 - 8$\n\n color(green)(x =3" ]

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[ "", null, "# Why this boilerplate code?\n\nI have been preparing for competitive programming for a while now, and I have been looking at a lot of solutions that people have submitted to problems. One thing I have noticed is that most submissions which take low time to execute have weird looking boilerplate code at the start. This is a good example of what I am talking about.\n\nIf anyone feels too lazy to click on that link, then here you go:\n\n```\n#include\n#include\n#include\nusing namespace std;\n\ntemplate\nusing ordered_set = __gnu_pbds::tree<T, __gnu_pbds::null_type, less, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;\n#define ll long long int\n#define el \"\\n\"\n#define sqr(x) ((x) * (x))\n#define all(vec) (vec).begin(),(vec).end()\ntemplate inline void ckmax(T &x, T y) {if (y > x) x = y; }\ntemplate inline void ckmin(T &x, T y) {if (y < x) x = y; }\n#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator _it(_ss); err(_it, args); }\nvoid err(istream_iterator it) {}\ntemplate\nvoid err(istream_iterator it, T a, Args... args) {\ncerr << *it << \" =: \" << a << endl;\nerr(++it, args...);\n}\ntemplate\ninline std::ostream& operator << (std::ostream& os, const std::pair& p) {\nreturn os << \"(\" << p.first << \": \" << p.second << \")\";\n}\ntemplate\ninline std::ostream &operator << (std::ostream & os,const std::vector& v) {\nbool first = true;\nos << \"[\";\nfor(unsigned int i = 0; i < v.size(); i++) {\nif(!first) os << \", \";\nos << v[i];\nfirst = false;\n}\nreturn os << \"]\";\n}\ntemplate\ninline std::ostream &operator << (std::ostream & os,const std::set& v) {\nbool first = true;\nos << \"{\";\nfor (typename std::set::const_iterator ii = v.begin(); ii != v.end(); ++ii) {\nif(!first) os << \", \";\nos << *ii;\nfirst = false;\n}\nreturn os << \"}\";\n}\ntemplate\ninline std::ostream &operator << (std::ostream & os,const std::map& v) {\nbool first = true;\nos << \"{\";\nfor (typename std::map::const_iterator ii = v.begin(); ii != v.end(); ++ii) {\nif(!first) os << \", \";\nos << *ii ;\nfirst = false;\n}\nreturn os << \"}\";\n}\ntemplate\ninline std::ostream &operator << (std::ostream & os,const std::unordered_set& v) {\nreturn os << std::set(v.begin(), v.end());\n}\ntemplate\ninline std::ostream &operator << (std::ostream & os,const ordered_set& v) {\nreturn os << std::set(v.begin(), v.end());\n}\ntemplate\ninline std::ostream &operator << (std::ostream & os,const std::unordered_map& v) {\nreturn os << std::map(v.begin(), v.end());\n}\n\nconst int MOD = 1e9 + 7;\nconst long long INF = 1e18;\nconst double EPS = 1e-6;\n```\n\nTo me, it looks like a whole lot of mumbo-jumbo, probably because I am not too familiar with C++. Could anybody please explain, if not line-by-line, then at least on an overall how each statement or preprocessor directive benefits the execution of the program.\n\nThose `#include`'s are for importing policy based data structures you can read about here.\nThen, they are a bunch of `#define`'s to save some extra keystrokes.\nThe rest part is the boilerplate code, which helps (a lot!) in debugging.\nYou can print `vector`s, `set`s, `map`s, etc, using `cout` with these codes.\n\nIn short, if you just want to increase I/O speed of your C++ program, these 2 lines are enough for that.\n\n``````int main()\n{\nios_base::sync_with_stdio(false);\ncin.tie(NULL);\n\n// you code\n}\n``````\n\nAlso, I just realised that the solution I linked to was yours", null, "Sorry if I am going off-topic", null, "" ]

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[ "# Square Root : Find square Root of any number in a 5 seconds\n\n## Square number\n\nA square number or perfect square is the product of some integer with itself. For example, 25 is a square number, since it can be written as 5 × 5.\n\n##", null, "To Find square of a number, just multiply it by itself.\n\nFor Example, square of 2 is 2 x 2 = 4\n\n## Square Root\n\nThe square root of a number is a value that, when multiplied by itself, gives the number.\n\n## Square Root Symbol\n\nThe symbol of Square root is √\n\n## Finding Square Roots\n\nThe easiest way to find a square root is to use a calculator, but you can do it without one. Here’s one way, using 12 as an example of the squared number:\n\n✓ Pick a number that when squared, comes close to (but is less than) the number whose square root you’re finding: 3 x 3 = 9. This is a better choice than 4: 4 × 4 = 16\n\n✓ Divide the number you’re finding the square root of (12) by the number you squared (3) in step 1: 12 ÷ 3 = 4\n✓ Average the closest square root (3) and the answer of step 2 (4): 3 + 4 = 7. 7 ÷ 2 = 3.5\n✓ Square the average to see how close the number is to 12:\n\n3.5 × 3.5 = 12.25—Close, but not close enough!\n\nRepeat steps 2 and 3 until the number squared is very close to 12:\n\nDivide: 12 ÷ 3.5 = 3.43\n\nAverage: 3.5 + 3.43 = 6.935\n\n6.935 ÷ 2 = 3.465\n\n3.465 × 3.465 = 12.006, close enough!\n\n## Problems based on Square Root\n\nHow to Calculate Square of a Number\n\nExample1: 6 × 6 = 36, so a square root of 36 is 6.\n\nExample2: √25 = 5 (because 5 x 5 = 25)\n\nExample3: √9 = 3 (because 3 x 3 = 9 )\n\nExample 4: √256 = 16 (because 16 x 16 = 256 )\n\nExample 5: √2500 = 50 (because 50 x 50 = 2500)\n\nExample 6: √3025 = 55 (because 55 x 55 = 3025)\n\nExample 7: √4489 = 67 (because 67 x 67 = 4489)\n\nExample 8: √7744 = 88 (because 88 x 88 = 7744)\n\nExample 9: √8649 = 93 (because 93 x 93 = 8649)\n\nExample 10: √10000 = 100 (because 100 x 100 = 10000)\n\nNote:\n1. We can also square negative numbers.\nFor Example (-3) X (-3)= 9\n\n2. Square Root is also applied for decimals\nFor Example √15.5 = 3.94\n\n## Perfect Square\n\nA number made by squaring a whole number (0,1,2,3,4_ _ _ _)\n\nFor Example 9 is a perfect square because it can be expressed as 3 x 3 (the product of two equal integers)\n\n16 is a perfect square because it can be expressed as 4 * 4 (the product of two equal integers)\n\n## Square roots of numbers 1-100\n\nGiven below are the square numbers up to 100\n\n Number Square Square root 1 1 1.000 2 4 1.414 3 9 1.732 4 16 2.000 5 25 2.236 6 36 2.449 7 49 2.646 8 64 2.828 9 81 3.000 10 100 3.162 11 121 3.317 12 144 3.464 13 169 3.606 14 196 3.742 15 225 3.873 16 256 4.000 17 289 4.123 18 324 4.243 19 361 4.359 20 400 4.472 21 441 4.583 22 484 4.690 23 529 4.796 24 576 4.899 25 625 5.000 26 676 5.099 27 729 5.196 28 784 5.292 29 841 5.385 30 900 5.477 31 961 5.568 32 1,024 5.657 33 1,089 5.745 34 1,156 5.831 35 1,225 5.916 36 1,296 6.000 37 1,369 6.083 38 1,444 6.164 39 1,521 6.245 40 1,600 6.325 41 1,681 6.403 42 1,764 6.481 43 1,849 6.557 44 1,936 6.633 45 2,025 6.708 46 2,116 6.782 47 2,209 6.856 48 2,304 6.928 49 2,401 7.000 50 2,500 7.071 51 2,601 7.141 52 2,704 7.211 53 2,809 7.280 54 2,916 7.348 55 3,025 7.416 56 3,136 7.483 57 3,249 7.550 58 3,364 7.616 59 3,481 7.681 60 3,600 7.746 61 3,721 7.810 62 3,844 7.874 63 3,969 7.937 64 4,096 8.000 65 4,225 8.062 66 4,356 8.124 67 4,489 8.185 68 4,624 8.246 69 4,761 8.307 70 4,900 8.367 71 5,041 8.426 72 5,184 8.485 73 5,329 8.544 74 5,476 8.602 75 5,625 8.660 76 5,776 8.718 77 5,929 8.775 78 6,084 8.832 79 6,241 8.888 80 6,400 8.944 81 6,561 9.000 82 6,724 9.055 83 6,889 9.110 84 7,056 9.165 85 7,225 9.220 86 7,396 9.274 87 7,569 9.327 88 7,744 9.381 89 7,921 9.434 90 8,100 9.487 91 8,281 9.539 92 8,464 9.592 93 8,649 9.644 94 8,836 9.695 95 9,025 9.747 96 9,216 9.798 97 9,409 9.849 98 9,604 9.899 99 9,801 9.950 100 10,000 10.000\n\n# Find Square Root Quickly\n\nWatch the video below to learn some quick  tricks for calculation of square root." ]

[ null, "http://teachmedaily.com/wp-content/uploads/2014/12/Square-Root-Find-square-Root-of-any-number-in-a-5-seconds-300x150.jpg", null ]

[ "## The inverse fallacy: an account of deviations from Bayes's theorem and the additivity principle\n\nVillejoubert, Gaelle and Mandel, David R. (2002) The inverse fallacy: an account of deviations from Bayes's theorem and the additivity principle. Memory & Cognition, 30(2), pp. 171-178. ISSN (print) 0090-502X\n\n## Abstract\n\nIn judging posterior probabilities, people often answer with the inverse conditional probability�a tendency named the inverse fallacy. Participants (N = 45) were given a series of probability problems that entailed estimating both p(H|D) and p(~H|D). The findings revealed that deviations of participants' estimates from Bayesian calculations and from the additivity principle could be predicted by the corresponding deviations of the inverse probabilities from these relevant normative benchmarks. Methodological and theoretical implications of the distinction between inverse fallacy and base-rate neglect and the generalization of the study of additivity to conditional probabilities are discussed.\n\n### Actions (Repository Editors)", null, "Item Control Page" ]

[ null, "https://eprints.kingston.ac.uk/style/images/action_view.png", null ]

[ " Multiplicative inverse : définition de Multiplicative inverse et synonymes de Multiplicative inverse (anglais)\n\nPublicité ▼\n\n# définition - Multiplicative inverse\n\nmultiplicative inverse (n.)\n\n1.(mathematics) one of a pair of numbers whose product is 1: the reciprocal of 2/3 is 3/2; the multiplicative inverse of 7 is 1/7\n\n## définition (complément)", null, "voir la définition de Wikipedia\n\nPublicité ▼\n\n# synonymes - Multiplicative inverse\n\nmultiplicative inverse (n.)\n\nreciprocal\n\n## dictionnaire analogique", null, "inverse; opposite[ClasseHyper.]", null, "mathematical sciences[Classe]", null, "(abstract)[termes liés]", null, "mathematics[Domaine]", null, "FieldOfStudy[Domaine]", null, "oppositeness, opposition[Hyper.]", null, "mathematician - mathematical - mathematic, mathematical[Dérivé]", null, "science, scientific discipline[Domaine]", null, "[ à l'inverse de ] (fr)[Syntagme]", null, "inverse, opposite[Hyper.]", null, "math, mathematics, maths[Domaine]", null, "multiplicative inverse (n.)\n\nPublicité ▼\n\nWikipedia\n\n# Multiplicative inverse", null, "The reciprocal function: y = 1/x. For every x except 0, y represents its multiplicative inverse.\n\nIn mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x−1, is a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fraction a/b is b/a. For the multiplicative inverse of a real number, divide 1 by the number. For example, the reciprocal of 5 is one fifth (1/5 or 0.2), and the reciprocal of 0.25 is 1 divided by 0.25, or 4. The reciprocal function, the function f(x) that maps x to 1/x, is one of the simplest examples of a function which is self-inverse.\n\nThe term reciprocal was in common use at least as far back as the third edition of Encyclopædia Britannica (1797) to describe two numbers whose product is 1; geometrical quantities in inverse proportion are described as reciprocall in a 1570 translation of Euclid's Elements.\n\nIn the phrase multiplicative inverse, the qualifier multiplicative is often omitted and then tacitly understood (in contrast to the additive inverse). Multiplicative inverses can be defined over many mathematical domains as well as numbers. In these cases it can happen that ab ≠ ba; then \"inverse\" typically implies that an element is both a left and right inverse.\n\n## Practical applications\n\nThe multiplicative inverse has innumerable applications in algorithms of computer science, particularly those related to number theory, since many such algorithms rely heavily on the theory of modular arithmetic. As a simple example, consider the exact division problem where you have a list of odd word-sized numbers each divisible by k and you wish to divide them all by k. One solution is as follows:\n\n1. Use the extended Euclidean algorithm to compute k−1, the modular multiplicative inverse of k mod 2w, where w is the number of bits in a word. This inverse will exist since the numbers are odd and the modulus has no odd factors.\n2. For each number in the list, multiply it by k−1 and take the least significant word of the result.\n\nOn many machines, particularly those without hardware support for division, division is a slower operation than multiplication, so this approach can yield a considerable speedup. The first step is relatively slow but only needs to be done once.\n\n## Examples and counterexamples\n\nIn the field of real numbers, Zero does not have a reciprocal because no real number multiplied by 0 produces 1. With the exception of zero, reciprocals of every complex number are complex, reciprocals of every real number are real, and reciprocals of every rational number are rational. The imaginary units, ±i, are the only complex numbers with additive inverse equal to multiplicative inverse. For example, additive and multiplicative inverses of i are −(i) = −i and 1/i = −i, respectively.\n\nTo approximate the reciprocal of x, using only multiplication and subtraction, one can guess a number y, and then repeatedly replace y with 2y − xy2. Once the change in y becomes (and stays) sufficiently small, y is an approximation of the reciprocal of x.\n\nIn constructive mathematics, for a real number x to have a reciprocal, it is not sufficient that x ≠ 0. There must instead be given a rational number r such that 0 < r < |x|. In terms of the approximation algorithm in the previous paragraph, this is needed to prove that the change in y will eventually become arbitrarily small.\n\nIn modular arithmetic, the modular multiplicative inverse of a is also defined: it is the number x such that ax ≡ 1 (mod n). This multiplicative inverse exists if and only if a and n are coprime. For example, the inverse of 3 modulo 11 is 4 because 4 · 3 ≡ 1 (mod 11). The extended Euclidean algorithm may be used to compute it.\n\nThe sedenions are an algebra in which every nonzero element has a multiplicative inverse, but which nonetheless has divisors of zero, i.e. nonzero elements x, y such that xy = 0.\n\nA square matrix has an inverse if and only if its determinant has an inverse in the coefficient ring. The linear map that has the matrix A−1 with respect to some base is then the reciprocal function of the map having A as matrix in the same base. Thus, the two distinct notions of the inverse of a function are strongly related in this case, while they must be carefully distinguished in the general case (see below).\n\nThe trigonometric functions are related by the reciprocal identity: the cotangent is the reciprocal of the tangent; the secant is the reciprocal of the cosine; the cosecant is the reciprocal of the sine.\n\nIt is important to distinguish the reciprocal of a function ƒ in the multiplicative sense, given by 1/ƒ, from the reciprocal or inverse function with respect to composition, denoted by ƒ−1 and defined by ƒ o ƒ−1 = id. Only for linear maps are they strongly related (see above), while they are completely different for all other cases. The terminology difference reciprocal versus inverse is not sufficient to make this distinction, since many authors prefer the opposite naming convention, probably for historical reasons (for example in French, the inverse function is preferably called application réciproque).\n\nA ring in which every nonzero element has a multiplicative inverse is a division ring; likewise an algebra in which this holds is a division algebra.\n\n## Pseudo-random number generation\n\nThe expansion of the reciprocal 1/q in any base can also act as a source of pseudo-random numbers, if q is a \"suitable\" safe prime, a prime of the form 2p + 1 where p is also a prime. A sequence of pseudo-random numbers of length q − 1 will be produced by the expansion.\n\n## Reciprocals of irrational numbers\n\nEvery number excluding zero has a reciprocal, and reciprocals of certain irrational numbers often can prove useful for reasons linked to the irrational number in question. Examples of this are the reciprocal of e which is special because no other positive number can produce a lower number when put to the power of itself, and the golden ratio's reciprocal which, being roughly 0.6180339887, is exactly one less than the golden ratio and in turn illustrates the uniqueness of the number.\n\nThere are an infinite number of irrational reciprocal pairs that differ by an integer (giving the curious effect that the pairs share their infinite mantissa). These pairs can be found by simplifying n+√(n2+1) for any integer n, and taking the reciprocal.\n\n## Further remarks\n\nIf the multiplication is associative, an element x with a multiplicative inverse cannot be a zero divisor (meaning for some y, xy = 0 with neither x nor y equal to zero). To see this, it is sufficient to multiply the equation xy = 0 by the inverse of x (on the left), and then simplify using associativity. In the absence of associativity, the sedenions provide a counterexample.\n\nThe converse does not hold: an element which is not a zero divisor is not guaranteed to have a multiplicative inverse. Within Z, all integers except −1, 0, 1 provide examples; they are not zero divisors nor do they have inverses in Z. If the ring or algebra is finite, however, then all elements a which are not zero divisors do have a (left and right) inverse. For, first observe that the map ƒ(x) = ax must be injective: ƒ(x) = ƒ(y) implies x = y:", null, "\\begin{align} ax &= ay &\\quad \\rArr & \\quad ax-ay = 0 \\\\ & &\\quad \\rArr &\\quad a(x-y) = 0 \\\\ & &\\quad \\rArr &\\quad x-y = 0 \\\\ & &\\quad \\rArr &\\quad x = y. \\end{align}\n\nDistinct elements map to distinct elements, so the image consists of the same finite number of elements, and the map is necessarily surjective. Specifically, ƒ (namely multiplication by a) must map some element x to 1, ax = 1, so that x is an inverse for a.\n\nThe multiplicative inverse of a fraction", null, "$\\tfrac{a}{b}$ is simply", null, "$\\tfrac{b}{a}.$", null, "Toutes les traductions de Multiplicative inverse\n\nContenu de sensagent\n\n• définitions\n• synonymes\n• antonymes\n• encyclopédie\n\n• definition\n• synonym\n\ndictionnaire et traducteur pour sites web\n\nAlexandria", null, "Une fenêtre (pop-into) d'information (contenu principal de Sensagent) est invoquée un double-clic sur n'importe quel mot de votre page web. LA fenêtre fournit des explications et des traductions contextuelles, c'est-à-dire sans obliger votre visiteur à quitter votre page web !\n\nEssayer ici, télécharger le code;\n\nSolution commerce électronique\n\nAugmenter le contenu de votre site\n\nAjouter de nouveaux contenus Add à votre site depuis Sensagent par XML.\n\nParcourir les produits et les annonces\n\nObtenir des informations en XML pour filtrer le meilleur contenu.\n\nIndexer des images et définir des méta-données\n\nFixer la signification de chaque méta-donnée (multilingue).\n\nRenseignements suite à un email de description de votre projet.\n\nJeux de lettres\n\nLes jeux de lettre français sont :\n○   Anagrammes\n○   jokers, mots-croisés\n○   Lettris\n○   Boggle.\n\nLettris\n\nLettris est un jeu de lettres gravitationnelles proche de Tetris. Chaque lettre qui apparaît descend ; il faut placer les lettres de telle manière que des mots se forment (gauche, droit, haut et bas) et que de la place soit libérée.\n\nboggle", null, "Il s'agit en 3 minutes de trouver le plus grand nombre de mots possibles de trois lettres et plus dans une grille de 16 lettres. Il est aussi possible de jouer avec la grille de 25 cases. Les lettres doivent être adjacentes et les mots les plus longs sont les meilleurs. Participer au concours et enregistrer votre nom dans la liste de meilleurs joueurs ! Jouer\n\nDictionnaire de la langue française\nPrincipales Références\n\nLa plupart des définitions du français sont proposées par SenseGates et comportent un approfondissement avec Littré et plusieurs auteurs techniques spécialisés.\nLe dictionnaire des synonymes est surtout dérivé du dictionnaire intégral (TID).\nL'encyclopédie française bénéficie de la licence Wikipedia (GNU).\n\nChanger la langue cible pour obtenir des traductions.\nAstuce: parcourir les champs sémantiques du dictionnaire analogique en plusieurs langues pour mieux apprendre avec sensagent.\n\n5547 visiteurs en ligne\n\ncalculé en 0,062s\n\nJe voudrais signaler :\nsection :\nune faute d'orthographe ou de grammaire\nun contenu abusif (raciste, pornographique, diffamatoire)\nune violation de copyright\nune erreur\nun manque\nautre\nmerci de préciser :\nallemand anglais arabe bulgare chinois coréen croate danois espagnol espéranto estonien finnois français grec hébreu hindi hongrois islandais indonésien italien japonais letton lituanien malgache néerlandais norvégien persan polonais portugais roumain russe serbe slovaque slovène suédois tchèque thai turc vietnamien\nallemand anglais arabe bulgare chinois coréen croate danois espagnol espéranto estonien finnois français grec hébreu hindi hongrois islandais indonésien italien japonais letton lituanien malgache néerlandais norvégien persan polonais portugais roumain russe serbe slovaque slovène suédois tchèque thai turc vietnamien" ]

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[ "# FCOS网络解析\n\n## 0 前言\n\n1. 检测器的性能和Anchor的size以及aspect ratio相关，比如在RetinaNet中改变Anchor（论文中说这是个超参数hyper-parameters）能够产生约4%的AP变化。换句话说，Anchor要设置的合适才行。\n2. 一般Anchor的size和aspect ratio都是固定的，所以很难处理那些形状变化很大的目标（比如一本书横着放w远大于h，竖着放h远大于w，斜着放w可能等于h，很难设计出合适的Anchor）。而且迁移到其他任务中时，如果新的数据集目标和预训练数据集中的目标形状差异很大，一般需要重新设计Anchor。\n3. 为了达到更高的召回率（查全率），一般需要在图片中生成非常密集的Anchor Boxes尽可能保证每个目标都会有Anchor Boxes和它相交。比如说在FPN（Feature Pyramid Network）中会生成超过18万个Anchor Boxes（以输入图片最小边长800为例），那么在训练时绝大部分的Anchor Boxes都会被分为负样本，这样会导致正负样本极度不均。下图是我随手画的样例，红色的矩形框都是负样本，黄色的矩形框是正样本。\n4. Anchor的引入使得网络在训练过程中更加的繁琐，因为匹配正负样本时需要计算每个Anchor Boxes和每个GT BBoxes之间的IoU。", null, "", null, "## 1 FCOS网络结构\n\nit has been shown that positioning it on the regression branch can obtain better performance.", null, "", null, "(\n\nc\n\nx\n\n,\n\nc\n\ny\n\n)\n\n(c_x, c_y)\n\n(cx,cy)，特征图相对原图的步距是s，那么网络预测该点对应的目标边界框坐标为：\n\nx\n\nm\n\ni\n\nn\n\n=\n\nc\n\nx\n\nl\n\ns\n\n,\n\ny\n\nm\n\ni\n\nn\n\n=\n\nc\n\ny\n\nt\n\ns\n\nx\n\nm\n\na\n\nx\n\n=\n\nc\n\nx\n\n+\n\nr\n\ns\n\n,\n\ny\n\nm\n\na\n\nx\n\n=\n\nc\n\ny\n\n+\n\nb\n\ns\n\nx_{min}=c_x - l cdot s , y_{min}=c_y - t cdot s \\ x_{max}=c_x + r cdot s , y_{max}=c_y + b cdot s\n\nxmin=cxls ,  ymin=cytsxmax=cx+rs ,  ymax=cy+bs\n\nc\n\ne\n\nn\n\nt\n\ne\n\nr\n\nn\n\ne\n\ns\n\ns\n\n=\n\nm\n\ni\n\nn\n\n(\n\nl\n\n,\n\nr\n\n)\n\nm\n\na\n\nx\n\n(\n\nl\n\n,\n\nr\n\n)\n\n×\n\nm\n\ni\n\nn\n\n(\n\nt\n\n,\n\nb\n\n)\n\nm\n\na\n\nx\n\n(\n\nt\n\n,\n\nb\n\n)\n\ncenterness^*=sqrt{frac{min(l^*,r^*)}{max(l^*,r^*)} times frac{min(t^*,b^*)}{max(t^*,b^*)}}\n\ncenterness=max(l,r)min(l,r)×max(t,b)min(t,b)", null, "## 2 正负样本的匹配\n\nSpecifically, location (x, y) is considered as a positive sample if it falls into the center area of any ground-truth box, by following . The center area of a box centered at (cx , cy ) is defined as the sub-box (cx − rs, cy − rs, cx + rs, cy + rs) , where s is the total stride until the current feature maps and r is a hyper-parameter being 1.5 on COCO. The sub-box is clipped so that it is not beyond the original box. Note that this is different from our original conference version, where we consider the locations positive as long as they are in a ground-truth box.\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n(x,y)，只要它落入GT box中心区域，那么它就被视为正样本（其实在2019年的文章中，最开始说的是只要落入GT内就算正样本）。对应的参考文献就是2019年发表的FCOS版本。但在2020年发表的FCOS版本中，新加了一条规则，在满足以上条件外，还需要满足点\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n(x,y)\n\n(\n\nc\n\nx\n\nr\n\ns\n\n,\n\nc\n\ny\n\nr\n\ns\n\n,\n\nc\n\nx\n\n+\n\nr\n\ns\n\n,\n\nc\n\ny\n\n+\n\nr\n\ns\n\n)\n\n(c_x - rs, c_y - rs, c_x + rs, c_y + rs)\n\n(cxrs,cyrs,cx+rs,cy+rs)这个sub-box范围内，其中\n\n(\n\nc\n\nx\n\n,\n\nc\n\ny\n\n)\n\n(c_x, c_y)\n\n(cx,cy)是GT的中心点，s是特征图相对原图的步距，r是一个超参数控制距离GT中心的远近，在COCO数据集中r设置为1.5，关于r的消融实验可以看2020版论文的表6。换句话说点\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n(x,y)不仅要在GT的范围内，还要离GT的中心点\n\n(\n\nc\n\nx\n\n,\n\nc\n\ny\n\n)\n\n(c_x, c_y)\n\n(cx,cy)足够近才能被视为正样本。\n\n(\n\nc\n\nx\n\nr\n\ns\n\n,\n\nc\n\ny\n\nr\n\ns\n\n,\n\nc\n\nx\n\n+\n\nr\n\ns\n\n,\n\nc\n\ny\n\n+\n\nr\n\ns\n\n)\n\n(c_x - rs, c_y - rs, c_x + rs, c_y + rs)\n\n(cxrs,cyrs,cx+rs,cy+rs)这个sub-box范围内，所以右侧的feature map中打勾的位置都被视为正样本。", null, "", null, "## 3 损失计算\n\nL\n\nc\n\nl\n\ns\n\nL_{cls}\n\nLcls、定位损失\n\nL\n\nr\n\ne\n\ng\n\nL_{reg}\n\nLreg以及center-ness损失\n\nL\n\nc\n\nt\n\nr\n\nn\n\ne\n\ns\n\ns\n\nL_{ctrness}\n\nLctrness三部分共同组成：\n\nL\n\n(\n\n{\n\np\n\nx\n\n,\n\ny\n\n}\n\n,\n\n{\n\nt\n\nx\n\n,\n\ny\n\n}\n\n,\n\n{\n\ns\n\nx\n\n,\n\ny\n\n}\n\n)\n\n=\n\n1\n\nN\n\np\n\no\n\ns\n\nx\n\n,\n\ny\n\nL\n\nc\n\nl\n\ns\n\n(\n\np\n\nx\n\n,\n\ny\n\n,\n\nc\n\nx\n\n,\n\ny\n\n)\n\n+\n\n1\n\nN\n\np\n\no\n\ns\n\nx\n\n,\n\ny\n\n1\n\n{\n\nc\n\nx\n\n,\n\ny\n\n>\n\n0\n\n}\n\nL\n\nr\n\ne\n\ng\n\n(\n\nt\n\nx\n\n,\n\ny\n\n,\n\nt\n\nx\n\n,\n\ny\n\n)\n\n+\n\n1\n\nN\n\np\n\no\n\ns\n\nx\n\n,\n\ny\n\n1\n\n{\n\nc\n\nx\n\n,\n\ny\n\n>\n\n0\n\n}\n\nL\n\nc\n\nt\n\nr\n\nn\n\ne\n\ns\n\ns\n\n(\n\ns\n\nx\n\n,\n\ny\n\n,\n\ns\n\nx\n\n,\n\ny\n\n)\n\nbegin{aligned} L({p_{x,y}},{t_{x,y}}, {s_{x,y}}) & = frac{1}{N_{pos}}sum_{x,y}L_{cls}(p_{x,y},c^*_{x,y}) \\ & + frac{1}{N_{pos}}sum_{x,y} 1_{{c^*_{x,y}>0}}L_{reg}(t_{x,y},t^*_{x,y}) \\ & + frac{1}{N_{pos}}sum_{x,y} 1_{{c^*_{x,y}>0}}L_{ctrness}(s_{x,y}, s^*_{x,y}) end{aligned}\n\nL({px,y},{tx,y},{sx,y})=Npos1x,yLcls(px,y,cx,y)+Npos1x,y1{cx,y>0}Lreg(tx,y,tx,y)+Npos1x,y1{cx,y>0}Lctrness(sx,y,sx,y)\n\n• p\n\nx\n\n,\n\ny\n\np_{x,y}\n\n表示在特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点处预测的每个类别的score\n\n• c\n\nx\n\n,\n\ny\n\nc^*_{x,y}\n\n表示在特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点对应的真实类别标签\n\n• 1\n\n{\n\nc\n\nx\n\n,\n\ny\n\n>\n\n0\n\n}\n\n1_{{c^*_{x,y}>0}}\n\n当特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点被匹配为正样本时为1，否则为0\n\n• t\n\nx\n\n,\n\ny\n\nt_{x,y}\n\n表示在特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点处预测的目标边界框信息\n\n• t\n\nx\n\n,\n\ny\n\nt^*_{x,y}\n\n表示在特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点对应的真实目标边界框信息\n\n• s\n\nx\n\n,\n\ny\n\ns_{x,y}\n\n表示在特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点处预测的center-ness\n\n• s\n\nx\n\n,\n\ny\n\ns^*_{x,y}\n\n表示在特征图\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n点对应的真实center-ness\n\nL\n\nc\n\nl\n\ns\n\nL_{cls}\n\nLcls采用bce_focal_loss，即二值交叉熵损失配合focal_loss，计算损失时所有样本都会参与计算（正样本和负样本）。定位损失\n\nL\n\nr\n\ne\n\ng\n\nL_{reg}\n\nLreg采用giou_loss（在2019版中采用iou_loss，但在2020版中说采用giou_loss会更好一点），计算损失时只有正样本参与计算。center-ness损失采用二值交叉熵损失，计算损失时只有正样本参与计算。\n\n(\n\nx\n\n,\n\ny\n\n)\n\n(x,y)\n\n(x,y)点处对应的GT信息\n\nc\n\nx\n\n,\n\ny\n\nc^*_{x,y}\n\ncx,y\n\nt\n\nx\n\n,\n\ny\n\nt^*_{x,y}\n\ntx,y比较好得到，只要匹配到某一GT目标则\n\nc\n\nx\n\n,\n\ny\n\nc^*_{x,y}\n\ncx,y对应GT的类别，\n\nt\n\nx\n\n,\n\ny\n\nt^*_{x,y}\n\ntx,y对应GT的bbox。而获取真实的center-ness\n\ns\n\nx\n\n,\n\ny\n\ns^*_{x,y}\n\nsx,y）要复杂一点，下面是\n\ns\n\nx\n\n,\n\ny\n\ns^*_{x,y}\n\nsx,y的计算公式，前面有提到过。\n\nc\n\ne\n\nn\n\nt\n\ne\n\nr\n\nn\n\ne\n\ns\n\ns\n\n=\n\nm\n\ni\n\nn\n\n(\n\nl\n\n,\n\nr\n\n)\n\nm\n\na\n\nx\n\n(\n\nl\n\n,\n\nr\n\n)\n\n×\n\nm\n\ni\n\nn\n\n(\n\nt\n\n,\n\nb\n\n)\n\nm\n\na\n\nx\n\n(\n\nt\n\n,\n\nb\n\n)\n\ncenterness^*=sqrt{frac{min(l^*,r^*)}{max(l^*,r^*)} times frac{min(t^*,b^*)}{max(t^*,b^*)}}\n\ncenterness=max(l,r)min(l,r)×max(t,b)min(t,b)\n\nl\n\n,\n\nt\n\n,\n\nr\n\n,\n\nb\n\nl^*,t^*,r^*,b^*\n\nl,t,r,b 再套用上面的公式就能得到\n\ns\n\nx\n\n,\n\ny\n\ns^*_{x,y}\n\nsx,y（这里的\n\nl\n\n,\n\nt\n\n,\n\nr\n\n,\n\nb\n\nl^*,t^*,r^*,b^*\n\nl,t,r,b无论是计算特征图尺度上的还是原图尺度上的都无所谓，因为\n\nc\n\ne\n\nn\n\nt\n\ne\n\nr\n\nn\n\ne\n\ns\n\ns\n\ncenterness^*\n\ncenterness对尺度不敏感）。", null, "## 4 其他\n\n### 4.1 Ambiguity问题\n\nAnother concern about the FCN-based detector is that it may have a large number of ambiguous samples due to the overlap in ground-truth boxes.", null, "", null, "(\n\nc\n\nx\n\nr\n\ns\n\n,\n\nc\n\ny\n\nr\n\ns\n\n,\n\nc\n\nx\n\n+\n\nr\n\ns\n\n,\n\nc\n\ny\n\n+\n\nr\n\ns\n\n)\n\n(c_x - rs, c_y - rs, c_x + rs, c_y + rs)\n\n(cxrs,cyrs,cx+rs,cy+rs)这个sub-box范围内）能够进一步降低ambiguous samples的比例（小于3%）。在论文表2中（2020版）有给出ambiguous samples比例的消融实验结果。", null, "### 4.2 Assigning objects to FPN\n\nk\n\n=\n\nk\n\n0\n\n+\n\nl\n\no\n\ng\n\n2\n\n(\n\nw\n\nh\n\n/\n\n224\n\n)\n\nk = left lfloor {k_0 + log_2(sqrt{wh} / 224)} right rfloor\n\nk=k0+log2(wh\n\n/224)\n\nAs shown in Table 7, this strategy results in degraded performance ( 37.7% AP). We conjecture that it may be because the strategy cannot make sure the complete object be within the receptive field of the target FPN level.", null, "m\n\na\n\nx\n\n(\n\nl\n\n,\n\nt\n\n,\n\nr\n\n,\n\nb\n\n)\n\nmax(l^*,t^*,r^*,b^*)\n\nmax(l,t,r,b)策略，其中\n\nl\n\n,\n\nt\n\n,\n\nr\n\n,\n\nb\n\nl^*,t^*,r^*,b^*\n\nl,t,r,b分别代表某点（特征图映射在原图上）相对GT Box左边界、上边界、右边界以及下边界的距离（在上面3 损失计算中有讲过）。关于这个策略在2020版论文的2.2章节中介绍的很清楚，对于不同的预测特征层只要满足以下公式即可，比如说对于P4特征图只要\n\nm\n\na\n\nx\n\n(\n\nl\n\n,\n\nt\n\n,\n\nr\n\n,\n\nb\n\n)\n\nmax(l^*,t^*,r^*,b^*)\n\nmax(l,t,r,b)\n\n(\n\n64\n\n,\n\n128\n\n)\n\n(64,128)\n\n(64,128)之间即为正样本：\n\nm\n\ni\n\n1\n\n<\n\nm\n\na\n\nx\n\n(\n\nl\n\n,\n\nt\n\n,\n\nr\n\n,\n\nb\n\n)\n\n<\n\nm\n\ni\n\nm_{i-1} < max(l^*,t^*,r^*,b^*) < m_i\n\nmi1<max(l,t,r,b)<mi", null, "THE END", null, "", null, "" ]

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[ "Content-type: text/html Man page of X-GEN\n\n# X-GEN\n\nSection: X-GEN Commands (1)\nUpdated: April 2005\n\n## NAME\n\nX-GEN - refine1\n\n## DESCRIPTION\n\n\"refine1\" provides for a line-oriented interactive refinement of crystal and detector parameters. The current definitions of the run parameter file, the calibration mapping, and the border definition are invoked upon mapping.\n\nWhen you invoke refine1 a brief help menu and a prompt appear. The user can specify a variety of sample properties, operational designations, and refinement commands. The user specifies the command or property by entering a one-letter command designator followed, in some cases, by one or more command arguments.\n\nCommands:\n\nThe command designators, together with their associated arguments (shown here in italics), are:\n\na\nNonlinear refinement with specified weight-factors for integerness and for RMS errors in X, Y, and omega.\nb [q]\nPerform a Bravais lattice determination. If an\nargument (e.g. q) is specified, then the program will prompt the user for acceptance of the Bravais lattice with the best figure of merit; otherwise, it simply informs the user of the choices.\nc\nChoose an auto-indexing solution.\nd param val ...\nSpecify one or more parameters by value.\nFor example,\nd a 46.5 beta 107.5 plated 12.65\nwould set a to 46.5 Angstroms, beta to 107.5 degrees, and the sample-to-detector distance to 12.65 cm.\ne\nWrite out the current parameter values and exit.\nf [q]\nPerform a Bravais determination and pick the solution\nspecified by the letter given as q. Thus if after the Bravais determination the user prefers the Bravais lattice that gives the third-best figure of merit, then the correct specification would be f c. If the user specifies \"f -50\", then the symmetry imposed will be that of the most recently specified spacegroup (e.g., from a \"d spacegroup 96\" command). If the user specifies \"f -51\", then the symmetry imposed will be that of a recently specified crystal system (e.g. \"d system 3\").\ng\nRead in the parameters from the file whose environment variable is UPARAMS.\nh\nThis does a detector remapping based on a polynomial fit of the errors in detector (X,Y) positions of the form\ndX = u0 + u1*X + u2*Y + u3*X*X + u4*X*Y + u5*Y*Y,\ndY = v0 + v1*X + v2*Y + v3*X*X + v4*X*Y + v5*Y*Y,\nThese twelve parameters (u0-u5 and v0-v5) are determined by least-squares fit to the dX and dY values.\ni [axerr angerr angle q]\nAuto-index. The user can specify\nthe fractional error in axis lengths and axis angles, and the minimum angle between the difference-vector basis vectors. Ordinarily the auto-indexing operation is followed by a two-dimensional optimization--a determination of the minimal residual as a function of the X and Y-center offsets. If the user specifies a fourth argument to the command, then the two-dimensional optimization is omitted.\nj param y/n ...\nTurn on or off the refinement of one or\nmore parameters. Thus \"j beta n xcen y\" would turn off the refinement of beta and turn on the refinement of the X-center value.\nk\nDo a linear least-squares refinement of the active\nparameters based on a minimization of the scanning-angle residual. This particular style of refinement is lightly tested and may be faulty.\nl [a]\nDo a linear least-squares refinement of the active\nparameters based on a minimization of the integerness residual. This is a carefully tested option. Note that linear refinements can prove to be unstable if the unit cell lengths and the detector parameters are both active.\nm\nRecreate the pixel-to-centimeter mapping. The technique is straightforward: the signed error in X and Y in several hundred \"neighborhoods\" across the detector face is determined, and the nominal position of the fiducial point at the center of each neighborhood is moved in (X,Y) by that amount.\nn\nAn internal command; don't use it.\no\nPrint out the reflections under refinement, including the observed and predicted values of X, Y, and omega and the non-integer value of (h,k,l).\np\nAlter the management of reflections in ice-rings. By default ice-ring reflections are treated the same as any other reflections. \"P\" with no arguments toggles that behavior, i.e. if ice-ring reflections are currently treated as normal, then invoking \"P\" will cause ice-ring reflections to be excluded. If ice-ring reflections are currently being excluded, then invoking \"P\" will cause them to be treated normally. If P is invoked with arguments, they change this from a toggle to a selector, viz. P ON unconditionally turns on exclusion of ice-rings, and P OFF unconditionally turns it off. The list of resolution ranges associated with ice-rings is read from the file with environment variable name ICERING. If that file is absent, the excluded ranges are set to be (3.97 to 3.60 Å), (3.48 to 3.40Å), (2.70 to 2.63Å), and (2.28 to 2.21Å). There are prominent ice-rings at even higher resolution than 2.21 Angstroms; if those prove to be a problem during refinement, we recommend setting up your own file.\nq\nQuit the program without writing the parameters out.\nr\nRocking-curve refinement: the only parameters under refinement are gamma0, gamma1, and gamma2.\ns\nPerform an indexing of a crystal based on an explicit indexing of three or more reflections.\nt [x]\nTransform the unit cell into a different orientation.\n\"t q\" prints out the specific transformations allowed; other values (e.g. \"t c\") perform specific transformations, e.g. permuting (a, b, c) into (b, c, a).\nu\nUpdate the parameters and print the (X,Y,omega,index)\nstatistics based on that update.\nv\nList the available commands.\nw\nWrite out the current parameters but do not exit.\nx\nLinear refinement based on a minimization of the \"X\" residual. This option has not been carefully tested.\ny\nLinear refinement based on a minimization of the \"Y\"\nresidual. This option has not been carefully tested.\nz\nLinear refinement based on a minimization of the \"Z\"\nresidual. This option has not been carefully tested.\n0\nNonlinear refinement based entirely on minimizing the scattering-angle residual.\n9\nNonlinear refinement based entirely on minimizing the integerness residual.\n1-8\nNonlinear refinement as a weighted mixture of indexing and scanning-angle refinement; 1 involves primarily scanning-angle refinement, 8 involves primarily indexing. The weight with which the index residual enters into the total residual is specified by the numerical value parameter; the weight associated with the omega error residual is 1 - (index weight). Thus if we use a command value u, 0 <= u <= 9, then\nwx = 0, wh = 0.11 * u, wp = 1 - wh .\n\nRefinable or adjustable parameters:\n\nThe specific parameters whose values can be adjusted with a \"d\" command, or whose refinement can be turned on or off with the \"j\" command, are:\n\n* a, b, c:\nThe unit cell lengths a, b, and c in Angstroms.\nIn space groups where cell lengths are tied to one another, the output will reflect those ties: thus in tetragonal spacegroups, even if the user enters values of a and b that are not equal, the program will force b to equal a.\n* alpha, beta, gamma:\nThe unit cell angles alpha, beta, and gamma\nin degrees. The software will force values of the cell angles appropriate to the crystal system: thus in hexagonal spacegroups it will force alpha = beta = 90, gamma = 120. In rhombohedral spacegroups indexed rhombohedrally, gamma is taken to be the independent angle, so the software sets beta to gamma and alpha to gamma.\n* omega, chi, phi: The pseudo-goniostat Euler angles omega, chi, and\nphi in degrees. These angles specify the rotation from the goniostat's true (omega = chi = phi = 0) position to one at which a will lie along X and b mostly along Y, where X is the direction pointing from the crystal toward the source, Z is the rotation axis, and Y is forms a right-handed (X,Y,Z) system with the others.\n* plateD: the sample-to-detector distance in cm.\n* Xcen: the X offset of the main-beam from the detector center at\n2theta = 0, in centimeters.\n* Ycen: the Y offset of the main-beam from the detector center at\n2theta = 0, in centimeters.\n* tilt: the angle between the detector's nominal vertical axis and\nthe crystal goniostat's omega direction, in degrees. This value is typically close to zero, but with a detector turned on its side it could be +/- 90.\n* swing: the angle between the direct beam direction and the\nnormal to the detector face. The sign of this angle is typically opposite to that of the 2theta angle defined in the data acquisition software of on most goniostats.\n* gamma0, gamma1, gamma2: Three parameters characterizing the\ndependence of the rocking width of the reflections on their (X,Y) position. The first two parameters are specified in frames; the third in degrees. The definitions of these parameters are given in Harrison et al, Methods in Enzymology 114: 226-230 (1985), except that what they call gamma(y) is called gamma0 here; their gamma(z) is gamma1; and their gamma(1) is gamma2 here.\n* xerror, yerror, zerror: These define the maximum errors\nallowed in reflection index, detector X, detector Y, and scanning angle. Any reflection violating these criteria will be excluded from refinement. These errors are ignored for rocking-curve refinements and during auto-indexing. For linear refinements on index, the index errors are examined individually. For nonlinear refinements and linear refinements on scanning angle, all the error limits must be satisfied in order for a reflection to be included in the residual calculations.\n* resmin, resmax: These define the minimum resolution (maximum D\nspacing) and maximum resolution (minimum D spacing) for reflections used in refinement. The default values of these parameters will be taken somewhat outside the lowest-resolution and largest-resolution reflections encountered in the CENTROIDS file, so that small changes in detector parameters will not cause any reflections to drop off the refinable list. Thus if the data extend from 24 to 1.8 Angstroms, the program will set the limits to about 29 and 1.6.\n* isys: This defines the crystal system as an integer between 1 and 7.\n1 is triclinic, 2 is monoclinic, 3 is orthorhombic, 4 is tetragonal, 5 is cubic, 6 is trigonal or hexagonal, and 7 is rhombohedral indexed rhombohedrally. Rhombohedral spacegroups may be specified as being in either a rhombohedral (7) or a hexagonal (6) system, depending on how you wish to index the crystal.\n* spacegroup: This defines the spacegroup as an integer between 1\nand 230. The number is the International Tables numerical designation for the spacegroup; thus P212121 is spacegroup 19, and P61 is spacegroup 169. Rhombohedral spacegroups may be specified either with rhombohedral or hexagonal crystal systems; the International Tables number given in this slot will be unaffected.\n* stepsize: This is the stepsize between images, in degrees. The\nsign convention is opposite to that found on most three- and four-axis goniostats, so if the data acquisition program is set up to step by +0.25 degrees per frame, the value specified here should often be -0.25.\n* lambda: This defines the X-radiation wavelength in Angstroms\n(0.1nm).\n* startomega, startchi, startphi: These define the position,\nin degrees, of the goniostat as of frame zero of the current data run. Thus if frame 1 was collected at (omega, chi, phi) = (40.0, 45.0, 180.0) and the stepsize is -0.2 degrees, the start values should be (40.2, 45.0, 180.0).\n\nExamples\n\nA refine1 script to auto-index and refine a data set is given in the refall documentation.\n\n## REPORTING BUGS\n\nReport bugs to Andy Howard at [email protected] or 312-567-5881.\n\nCopyright © 2002, Illinois Institute of Technology. See the file 'LICENSE' for information on usage and redistribution of this file, and for a DISCLAIMER OF ALL WARRANTIES\n\nNAME\nDESCRIPTION\nREPORTING BUGS" ]

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[ "# A Prufer-Sequence Based Algorithm for Calculating the Size of Ideal Randomly Branched Polymers\n\n### Citation:\n\nSingaram SW, Gopal A, Ben-Shaul A. A Prufer-Sequence Based Algorithm for Calculating the Size of Ideal Randomly Branched Polymers. JOURNAL OF PHYSICAL CHEMISTRY B. 2016;120 :6231-6237.", null, "2016_pruefer.pdf 1.34 MB\n\nJUL 7\n\n### Abstract:\n\nBranched polymers can be represented as tree graphs. A one-to-one correspondence exists between a tree graph comprised of N labeled vertices and a sequence of N 2 integers, known as the Prufer sequence. Permutations of this sequence yield sequences corresponding to tree graphs with the same vertex-degree distribution but (generally) different branching patterns. Repeatedly shuffling the Prufer sequence we have generated large ensembles of random tree graphs, all with the same degree distributions. We also present and apply an efficient algorithm to determine graph distances directly from their Prufer sequences. From the (Prufer sequence derived) graph distances, 3D size metrics, e.g., the polymer's radius of gyration, R-g, and average end-to-end distance, were then calculated using several different theoretical approaches. Applying our method to ideal randomly branched polymers of different vertex-degree distributions, all their 3D size measures are found to obey the usual N-1/4 scaling law. Among the branched polymers analyzed are RNA molecules comprised of equal proportions of the four-randomly distributed-nucleotides. Prior to Prufer shuffling, the vertices of their representative tree graphs, these ``random-sequence'' RNAs exhibit an R-g similar to N-1/3 scaling.\n\nLast updated on 12/18/2016" ]

[ null, "https://scholars.huji.ac.il/profiles/openscholar/modules/os/modules/os_files/icons/application-pdf.svg", null ]

[ "Home | Menu | Get Involved | Contact webmaster", null, "", null, "", null, "", null, "", null, "# Number 909023\n\nnine hundred nine thousand twenty three\n\n### Properties of the number 909023\n\n Factorization 909023 Divisors 1, 909023 Count of divisors 2 Sum of divisors 909024 Previous integer 909022 Next integer 909024 Is prime? YES (71946th prime) Previous prime 909019 Next prime 909031 909023rd prime 13982443 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 11011101111011011111 Octal 3357337 Duodecimal 37a07b Hexadecimal ddedf Square 826322814529 Square root 953.42697675281 Natural logarithm 13.72012567537 Decimal logarithm 5.9585748718322 Sine -0.024088809456234 Cosine -0.99970982252801 Tangent 0.024095801515003\nNumber 909023 is pronounced nine hundred nine thousand twenty three. Number 909023 is a prime number. The prime number before 909023 is 909019. The prime number after 909023 is 909031. Number 909023 has 2 divisors: 1, 909023. Sum of the divisors is 909024. Number 909023 is not a Fibonacci number. It is not a Bell number. Number 909023 is not a Catalan number. Number 909023 is not a regular number (Hamming number). It is a not factorial of any number. Number 909023 is a deficient number and therefore is not a perfect number. Binary numeral for number 909023 is 11011101111011011111. Octal numeral is 3357337. Duodecimal value is 37a07b. Hexadecimal representation is ddedf. Square of the number 909023 is 826322814529. Square root of the number 909023 is 953.42697675281. Natural logarithm of 909023 is 13.72012567537 Decimal logarithm of the number 909023 is 5.9585748718322 Sine of 909023 is -0.024088809456234. Cosine of the number 909023 is -0.99970982252801. Tangent of the number 909023 is 0.024095801515003\n\n### Number properties\n\nExamples: 3628800, 9876543211, 12586269025" ]

[ null, "https://www.numberempire.com/images/graystar.png", null, "https://www.numberempire.com/images/graystar.png", null, "https://www.numberempire.com/images/graystar.png", null, "https://www.numberempire.com/images/graystar.png", null, "https://www.numberempire.com/images/graystar.png", null ]

[ "### Make Gauss Proud\n\n##### Score: 3 Points\n\nA very long thread is oriented along the axis of a circle of radius $r=10\\space m$. The thread is charged, having a uniform charge density $\\lambda=2\\space C/m$. What is the electric flux across the circular area??\n\n$\\epsilon_0 = 1$.\n\nElectrostatics\n\n#### Statistics\n\nTried 60\n\nSolved 35\n\nFirst Solve @Nahian9696" ]

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[ "# Another Look at Large-Cap Stock Return Comovement: A Semi-Markov-Switching Approach\n\n## Abstract\n\nThe paper revisits the question of how stock return comovement varies with volatility and market returns. I propose an eigenvalue-based measure of comovement implied by the state-dependent correlation matrix estimated using a novel multivariate semi-Markov-switching approach. I show that compared to a basic Markov-switching structure the refined model performs very well in terms of capturing the well-known stylized facts of stock returns such as volatility clustering. With a focus on large-cap stocks, I illustrate the significance of comovement differential across states and document the different comovement patterns in different industries. Although the financial sector tends to conform to the conventional sentiment that comovement is highest when market is down and volatile, the conclusion should be tempered with caution when applied to other industries. In some cases, it is the high return state that registers the highest comovement.\n\nThis is a preview of subscription content, access via your institution.\n\n1. 1.\n\nEstimation of the degrees of freedom requires numerically solving a digamma function at each step, which adds another layer of complexity in terms of updating a weighted average for the conditional mean and covariance matrix. Analytical solutions do not exist; see the appendix in Bulla and Bulla (2006).\n\n## References\n\n1. Ang, A., & Chen, J. (2002). Asymmetric correlations of equity portfolios. Journal of Financial Economics, 63(3), 443–494.\n\n2. Billio, M., & Caporin, M. (2005). Multivariate Markov switching dynamic conditional correlation GARCH representations for contagion analysis. Statistical Methods and Applications, 14(2), 145–161.\n\n3. Bekaert, G., Hodrick, R.J., & Zhang, X. (2008). Stock return comovements. ECB working paper 931, European Central Bank.\n\n4. Bulla, J., & Bulla, I. (2006). Stylized facts of financial time series and hidden semi-Markov Models. Computational Statistics & Data Analysis, 51(4), 2192–2209.\n\n5. Cho, J. S., & White, H. (2007). Testing for regime switching. Econometrica, 75(6), 1671–1720.\n\n6. Croux, C., Forni, M., & Reichlin, L. (2001). Measure of comovemnet for economic variables: Theory and empirics. The Review of Economics and Statistics, 83(2), 232–241.\n\n7. Edwards, S., & Susmel, R. (2003). Interest-rate volatility in emerging markets. Review of Economics and Statistics, 85(2), 325–348.\n\n8. Ferguson, J. D. (1980). Variable duration models for speech. In J. D. Ferguson (Ed.) Proceedings of the symposium on the applications of hidden Markov models to text and speech (pp. 143–179). New York, NJ: Princeton.\n\n9. Gray, S. (1996). Modeling the conditional distribution of interest rates as a regime-switching process. Journal of Financial Economics, 42(1), 27–62.\n\n10. Guedon, Y. (2003). Estimating hidden semi-Markov chains from discrete sequences. Journal of Computational and Graphical Statistics, 12(3), 604–639.\n\n11. Guidolin, M., & Timmermann, A. (2007). Asset allocation under multivariate regime switching. Journal of Economic Dynamics and Control, 31(11), 3503–3544.\n\n12. Hamilton, J., & Sumsel, R. (1994). Autoregressive conditional heteroskedasticity and change in regime. Journal of Econometrics, 64(1), 307–333.\n\n13. Hong, Y., Tu, J., & Zhou, G. (2007). Asymmetries in stock returns: Statistical tests and economic evaluation. Review of Financial Studies, 20(5), 1547–1581.\n\n14. Laloux, L., Cizeau, P., Potters, M., & Bouchaud, J. P. (2000). Random matrix theory and financial correlations. International Journal of Theoretical and Applied Finance, 3(3), 391–397.\n\n15. MacDonald, I. L., & Zucchini, W. (1997). Hidden Markov and other models for discrete-valued time series. London: Chapman & Hall.\n\n16. Maheu, J. M., & McCurdy, T. H. (2000). Identifying bull and bear markets in stock returns. Journal of Business & Economic Statistics, 18(1), 100–112.\n\n17. Peel, D., & McLachlan, G. J. (2000). Robust mixture modelling using the t distribution. Statistics and Computing, 10(4), 339–348.\n\n18. Pelletier, D. (2006). Regime switching for dynamic correlations. Journal of Econometrics, 131(1), 445–473.\n\n19. Reigneron, P. A., Allez, R., & Bouchaud, J. P. (2011). Principal regression analysis and the index leverage effect. Physica A, 390(17), 3026–3035.\n\n20. Rua, A. (2010). Measuring comovement in the time-frequency space. Journal of Macroeconomics, 32(2), 685–691.\n\n21. Rydén, T., Teräsvirta, T., & Asbrink, S. (1998). Stylized facts of daily return series and the hidden Markov model. Journal of Applied Econometrics, 13(3), 217–244.\n\n22. Smith, A., Naik, P. A., & Tsai, C. L. (2005). Markov-switching model selection using Kullback–Leibler divergence. Journal of Econometrics, 134(2), 553–577.\n\n23. Syllignakis, M. N., & Kouretas, G. P. (2011). Dynamic correlation analysis of financial contagion: Evidence from the Central and Eastern European markets. International Review of Economics & Finance, 20(4), 717–732.\n\n24. Timmermann, A. (2000). Moments of Markov switching models. Journal of Econometrics, 96(1), 75–111.\n\n## Acknowledgments\n\nI thank the editors, Hans Amman and Shu-Heng Chen, and two anonymous referees for insightful comments. I also thank Christophe Deissenberg, Gary Anderson and seminar participants at the 21st International Conference on Computing in Economics and Finance for encouraging feedback. All errors are mine.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Kaihua Deng.\n\n## Appendices\n\n### Appendix 1\n\nThe goal of this appendix is to show that, in a multivariate context, the semi-Markov-switching framework is numerically stable, identifies states consistent with the market data, captures duration dependence (discussed in Sect. 4), and considerably reproduces the volatility clustering effect. I also consider the effect of prewhitening and outlier reduction. These findings together with the fact that a semi-Markov-switching model has been documented to account for other stock market stylized facts provide economic ground for its use in the full sample study.\n\n### Basic Results\n\nA preliminary study is carried out and the results are given in Table 7. In addition, I compare the EM-sMSw model with the plain EM-MSw model. The three measures of stock return comovement are arranged in a way so that they run from the low state to the high state based on the estimated state-dependent volatilities. Since individual stock returns are often noisy, I also match the ordering of comovement with market data by computing the state-dependent market returns and volatilities over the same sample period. The ordering of state-dependent mean returns for individual stocks (not reported) are mostly in line with that of the market mean returns but exceptions are not rare: When the market is down, not all stock returns are low at the same time.\n\nIn an unreported exercise, sMSw appears to be more robust to outliers compared to the basic MSw model. Estimation of EM-sMSw models, is not sensitive to starting values of the conditional mean or the transition probability matrix, but is somewhat influenced by starting values of the variance-covariance matrix and the choice of duration distributions. For the latter, I use the versatile gamma distribution because its scale and shape parameters can vary independently and the result corroborates an earlier finding in the literature on duration dependence of states. Of practical concern is the computational cost, at which direct ML scores very badly, whereas EM-MSw and EM-sMSw are much faster. In the case of two regimes, the latter converges very fast and starting values do not seem to matter. When the order of switching is greater than two, to guarantee proper convergence I use the EM-MSw estimates as starting values.\n\nI also use $$\\hbox {VAR}\\left( 1 \\right)$$ prewhitening/recoloring to check the potential influence of residual serial correlation and spurious ARCH effect. In the case of $${\\mathrm {m}}=3$$ and $${\\mathrm {m}}=2$$ (not reported), the results are stable across different model specifications and estimation techniques. It is found that $$\\hbox {VAR}\\left( 1 \\right)$$-prewhitening does not appear to yield noticeable differences and that, for a given model, conclusion is the same for all three measures of comovement. In terms of order selection, MSwC and BIC pick either $${\\mathrm {m}}=2$$ or $${\\mathrm {m}}=3$$, whereas AIC and AIC$$_{\\mathrm{c}}$$ always pick the highest order of switching.\n\n### Model Evaluation\n\nSince the four-regime model is not supported by BIC or MSwC, I only focus on two and three regimes in this section. In Table 7, I have conducted a Wald test of parameter equality on the average pairwise correlation measure of comovement ($$Cm_{avg}$$) in the Gaussian specification. The ordering of comovement is statistically insignificant for NYSE Fin Top 4, whereas for DJIA Top 4 and DAX 30 Top 4, significant but opposite conclusions are found.\n\nFor lack of a simple relationship between the model parameters and the top eigenvalue of the state-dependent correlation matrix, approximate confidence intervals for Cm and $$Cm_{diff}$$ cannot be directly imputed using, say, the second-order delta method. So I conduct a bootstrap experiment to see whether eigenvalue-based comovement differentials are statistically significant. I limit the case to two regimes selected by MSwC. Figure 1 shows the distribution of Cm based on 2000 parametric bootstrap samples generated from the two-state Gaussian NYSE Fin Top 4 model. As the paper is primarily concerned with comovement, I briefly report on the correlation parameters and transition probabilities in Table 8. Note that for DAX30 and DJIA the probability of staying in the low-return state is not high by conventional standard.\n\nThe small standard errors in Table 8 lead to the conjecture that the implied sampling distribution of Cm and $$Cm_{diff}$$ should also have small standard errors, as the latter is fully determined by the correlation matrix. To see this more clearly, two normal distributions centered around the bootstrap means are superimposed onto Fig. 1, both of which have a standard deviation of 0.02. This, together with the bootstrap distribution, suggests that a difference of 0.05–0.1 or more can be viewed as solid evidence for comovement differential across states. For example, in the case of $${\\mathrm {m}}=3$$, NYSE Financial Top 4 does not deliver a clear message, whereas DAX30 Top 4 yields much more significant result and does not accord with the conventional wisdom of low return-high comovement, at least in the 4-year sample.\n\nTo highlight the semi-Markov-switching model’s ability to capture the slow decay of volatility dynamics, Figs. 4 and 5 illustrates the model-implied autocorrelations for squared returns and the sample ACF for squared residuals. In the Gaussian Markov-switching case, exact autocorrelation functions for squared returns can be derived and is plotted in Fig. 5; see Appendix 2 for the expressions. In the semi-Markov case, I approximate the model-implied ACF for squared returns by simulating a sample of 40,000 data points. A polynomial is fitted and superimposed onto each graph. Take DJIA Top 4, Fig. 4 shows that the semi-Markov model can explain a substantial amount of volatility clustering as the filtered squared returns (squared residuals) are very much whitened.\n\nIn Fig. 5, it is clear that in a standard Markov-switching model, the model-implied ACF for squared returns decreases too fast due to the geometric declining nature of the duration probability. Bulla and Bulla (2006) find the same result in their univariate analysis.\n\nOne remaining question is how well the states are identified since the high and low volatility states may be poorly separated thus confounding the comovement measures across states. Three commonly used methods in this area are filtering, smoothing and Viterbi’s algorithm. I take the filtered (smoothed) state to be the one that has the highest filtered (smoothed) probability; in Viterbi’s algorithm, a single most probable state is produced at each point in time as a byproduct of the global maximum likelihood. In the following simulation study, I compare filtered states, smoothed states and global most probable path inference by computing the percentage of correct matches for the case of $${\\mathrm {m}}=3$$. Table 9 is based on 1000 simulated samples of size 1000 from the corresponding estimated models. A match is said to occur if the estimated state coincides with the true state.\n\nIt can be seen from Table 9 that the medium volatility state is the hardest to estimate. All three methods are good at estimating the low volatility state, while smoothed inference and Viterbi’s algorithm perform better for the high and medium volatility states. The pattern changes little regardless of the model specification used and using smoothed or Viterbi states does not make any noticeable difference. In the next section, results are based on global most probable states using Viterbi’s algorithm.\n\n### A Note on Student’s t Distribution\n\nAt times, a t-distribution appears to be accommodating too much to outliers, whereas outlier reduction fixes this problem but imperfectly by introducing a mild yet unnecessary clustering of high volatility states at the cutoff points. To check this possibility, I simulate data from MSw models estimated with a t-distribution and find that extreme returns occur much too often compared to the frequency seen in real data, the latter of which actually looks much more like a simulation done using Gaussian returns plus just a handful of extreme events. When outlier reduction is combined with t-distributed returns, the tendency to over-accommodate to extreme returns overwhelms the use of outlier reduction because the extra parameter, degree of freedom, changes accordingly and absorbs most of the impact of cutoff points.\n\nAnother undesirable side-effect of using Student’s t returns is that the estimated transition matrix is much more concentrated along the diagonal for all states, with the off-diagonal probabilities being very small and thus susceptible to creating spuriously persistent states. By contrast, Gaussian returns are much more robust to one-time big swings and the estimated transition matrix is well-behaved in the sense that the probability of staying in one state is never insanely high especially for the high volatility state. As each row of a transition matrix sums to one and the diagonal element measures the persistence of a certain state, its columns are also meaningful in an important sense. Suppose, there is one highly persistent state with the corresponding diagonal probability close to one and other elements in the same column extremely small. This means that the probability of entering this state is extremely low, then such persistent state might just be the result of influential outliers. In the extreme case if the true data are such that all diagonal elements are extremely high and all off-diagonal probabilities extremely low, the system is then quite “disconnected” and the researcher might as well model the subperiods independently instead of pooling all the data together.\n\n### Exact Moments of Markov-Switching Models\n\nLet $${\\Gamma }$$ be the transition matrix, $$p_{ij} \\left( k \\right)$$ the $$\\left( {i,j} \\right)$$ element of $${\\Gamma }^{k}$$. Denote the stationary distribution and conditional mean vector as $${\\varvec{\\pi }} =\\left( {\\pi _1 ,\\pi _2 ,\\ldots ,\\,\\pi _m } \\right) ^{\\prime }$$ and $${\\varvec{\\mu }} =\\left( {\\mu _1 ,\\mu _2 ,\\ldots ,\\,\\mu _m } \\right) ^{\\prime }$$.\n\n\\begin{aligned}&\\displaystyle \\mu =\\mathop {\\sum } \\limits _{i=1}^m \\mu _i \\pi _i ,\\\\&\\displaystyle Var\\left( {X_t } \\right) =\\mathop {\\sum } \\limits _{i=1}^m \\left( {\\sigma _i^2 +\\mu _i^2 } \\right) \\pi _i -\\left( {\\mathop {\\sum } \\limits _{i=1}^m \\mu _i \\pi _i } \\right) ^{2},\\\\&\\displaystyle Cov\\left( {X_t X_{t+k} } \\right) =E\\left( {X_t X_{t+k} } \\right) -\\mu ^{2}=\\mathop {\\sum } \\limits _{i,j=1}^m E\\left( {X_t | i} \\right) \\pi _i p_{ij} \\left( k \\right) E\\left( {X_{t+k} | j} \\right) -\\mu ^{2}\\\\&\\displaystyle ={{\\varvec{\\pi }} }'diag ({\\varvec{\\mu }})\\Gamma ^k{\\varvec{\\mu }}- ({\\varvec{\\pi }}'{\\varvec{\\mu }})^2,\\\\&\\displaystyle Kurtosis =\\frac{E\\left( {X_t -\\mu } \\right) ^{4}}{Var^{2}\\left( {X_t } \\right) }=\\frac{E\\left( {X_t^4 } \\right) -4\\mu E\\left( {X_t^3 } \\right) +6\\mu ^{2}E\\left( {X_t^2 } \\right) -3\\mu ^{4}}{\\left( {\\mathop {\\sum } \\nolimits _{i=1}^m \\left( {\\sigma _i^2 +\\mu _i^2 } \\right) \\pi _i -\\left( {\\mathop {\\sum } \\nolimits _{i=1}^m \\mu _i \\pi _i } \\right) ^{2}} \\right) ^{2}}\\\\&\\displaystyle =\\frac{\\mathop {\\sum } \\nolimits _{i=1}^m \\pi _i \\left( {\\mu _i^4 +6\\mu _i^2 \\pi _i^2 +3\\sigma _i^4 } \\right) -4\\mu \\mathop {\\sum } \\nolimits _{i=1}^m \\pi _i \\left( {\\mu _i^3 +3\\mu _i \\sigma _i^2 } \\right) +6\\mu ^{2}\\mathop {\\sum } \\nolimits _{i=1}^m \\pi _i \\left( {\\sigma _i^2 +\\mu _i^2 } \\right) -3\\mu ^{4}}{\\left( \\mathop {\\sum } \\nolimits _{\\mathrm {i}=1}^{\\mathrm {m}} \\left( {\\sigma _i^2 +\\mu _i^2 } \\right) \\pi _i - ({\\varvec{\\pi }}'{\\varvec{\\mu }})^2 \\right) ^2},\\\\&\\displaystyle \\rho _k =\\frac{Cov\\left( {X_t X_{t+k} } \\right) }{Var\\left( {X_t } \\right) }=\\frac{{{\\varvec{\\pi }} }'diag ({\\varvec{\\mu }}) \\Gamma ^k{\\varvec{\\mu }}- ({\\varvec{\\pi }}'{\\varvec{\\mu }})^2}{\\mathop {\\sum } \\nolimits _{{i}=1}^{{m}} \\left( {\\sigma _i^2 +\\mu _i^2 } \\right) \\pi _i - ({\\varvec{\\pi }}'{\\varvec{\\mu }})^2},\\\\&\\displaystyle \\rho _k^{sqr} =\\frac{Cov\\left( {X_t^2 X_{t+k}^2 } \\right) }{Var\\left( {X_t^2 } \\right) }=\\frac{E\\left( {X_t^2 X_{t+k}^2 } \\right) -E^{2}\\left( {X_t^2 } \\right) }{E\\left( {X_t^4 } \\right) -E^{2}\\left( {X_t^2 } \\right) },\\\\&\\displaystyle E\\left( {X_t^2 X_{t+k}^2 } \\right) =\\mathop {\\sum } \\limits _{i,j=1}^m E\\left( {X_t^2 | i} \\right) \\pi _i p_{ij} \\left( k \\right) E\\left( {X_{t+k}^2 | j} \\right) . \\end{aligned}\n\nThe last expression can be computed from the conditional second moment of the state-dependent distribution.\n\n### Estimating the Benchmark Model\n\nUsing notation of Sect. 2.1 and treating the unobserved states as missing data one gets,\n\n\\begin{aligned} L\\left( T \\right) =P\\left( {{\\varvec{X}}^{\\left( {\\varvec{T}} \\right) }={\\varvec{x}}^{\\left( T \\right) }} \\right) =\\mathop {\\sum } \\limits _{s_1 ,s_2 ,\\ldots ,\\,\\,s_T \\in \\left\\{ {1,2} \\right\\} } P\\left( {{\\varvec{X}}^{\\left( {\\varvec{T}} \\right) }={\\varvec{x}}^{\\left( T \\right) },{\\varvec{S}}^{\\left( T \\right) }={\\varvec{s}}^{\\left( T \\right) }} \\right) , \\end{aligned}\n\nand by the strong Markov property, a particular realization of states contributes to the likelihood by as much as\n\n\\begin{aligned}&P\\left( {{\\varvec{X}}^{\\left( {{\\varvec{T}}} \\right) } ={\\varvec{x}}^{\\left( T \\right) },{\\varvec{S}}^{\\left( T \\right) } ={\\varvec{s}}^{\\left( T \\right) }} \\right) \\\\&\\quad =\\pi \\left( {s_1 } \\right) P\\left( {x_1 | s_1 } \\right) p\\left( {s_2 | s_1 } \\right) P\\left( {x_2 | s_2 } \\right) \\ldots p\\left( {s_T | s_{T-1} } \\right) P\\left( {x_T | s_T } \\right) . \\end{aligned}\n\nA full permutation of the histories of states thus yields\n\n\\begin{aligned} L\\left( T \\right)= & {} \\mathop {\\sum } \\limits _{s_1 ,s_2 ,\\ldots ,\\,\\,s_T \\in \\left\\{ {1,2} \\right\\} } \\pi \\left( {s_1 } \\right) P\\left( {x_1 | s_1 } \\right) p\\left( {s_2 | s_1 } \\right) \\ldots p\\left( {s_T | s_{T-1} } \\right) P\\left( {x_T | s_T } \\right) \\\\= & {} {{\\varvec{\\pi }}}'P(x_1)\\Gamma P(x_2)\\ldots \\Gamma P(x_x)\\mathbf{1}, \\end{aligned}\n\nin which $${\\Gamma }$$ is the $$m\\times m$$ transition matrix, $$P\\left( \\cdot \\right)$$ is the diagonal conditional probability matrix of observation $$x_t$$ given each possible state $$s_t$$. For a general homogenous and irreducible m-state conditionally serially uncorrelated Markov-switching model, the likelihood function can be written conveniently in matrix form as\n\n\\begin{aligned} L\\left( {\\varvec{\\theta }} \\right) ={{\\varvec{\\pi }}}'P(x_1)\\Gamma P(x_2)\\ldots \\Gamma P(x_x)\\mathbf{1}. \\end{aligned}\n\nThus a model of n assets has in total $$m\\left( {m-1+n\\left( {n+3} \\right) /2} \\right)$$ parameters: $$m^{2}-m$$ to determine $${\\Gamma }$$, mn for the state-dependent means, and $$mn\\left( {n+1} \\right) /2$$ for the state-dependent variances and covariances. These parameters are collected in $${\\varvec{\\theta }}$$. The transition probability matrix dictates that $$p_{ii}^{u-1} \\left( {1-p_{ii} } \\right)$$ is the duration probability of staying in state i for u periods, $$u=1,2,\\ldots$$.\n\nAn alternative way to estimate the model is the EM algorithm based on the reknowned Baum-Welch forward-backward procedure. The EM algorithm can be used to check the influence of initial state on subsequent estimates. It does not evaluate the likelihood function directly and is fastest when the E-step can be broken down into a collection of terms involving subsets of parameters of interest, each of which is simple to maximize either numerically, or even better, analytically. This is true when stationarity is not imposed on the initial state but not otherwise. In the latter case, the E-step cannot be neatly seperated and no analytic solution exists. The E step and M step in the basic model are simplified versions of the semi-Markov-switching algorithm and can be found in MacDonald and Zucchini (1997). While it is common to assume that the process starts from its stationary distribution using ML-MSw, I do not impose this restriction on the EM algorithm.\n\nTo avoid boundary problems, I transform natural parameters into working ones before each run of estimation. Thus the algorithm returns a Hessian in terms of working parameters. I then recover the variance-covariance matrix with respect to natural parameters using the sandwich estimator:\n\n\\begin{aligned} \\widehat{\\hbox {Var}} \\left( \\widehat{{\\varvec{\\theta }}} \\right) =\\widehat{\\hbox {H}}^{-1} \\left( \\widehat{{\\varvec{\\theta }}} \\right) \\hat{I}_{BHHH} \\left( \\widehat{{\\varvec{\\theta }}} \\right) \\widehat{\\hbox {H}}^{-1} \\left( \\widehat{{\\varvec{\\theta }}} \\right) , \\end{aligned}\n\nLet $${\\varvec{\\phi }}$$ and $${\\varvec{\\theta }}$$ denote the vector of working parameters and natural parameters respectively, then\n\n\\begin{aligned} I= & {} E\\left( {G'G} \\right) = \\mathop {\\sum } \\limits _{t = 1}^n E\\left( {G_t G_t'} \\right) = E \\left( {\\varvec{g}}{\\varvec{g}}' \\right) ,\\\\ \\hat{I}_{BHHH}= & {} G'\\left( \\widehat{{\\varvec{\\theta }}} \\right) G \\left( \\widehat{{\\varvec{\\theta }}} \\right) = \\left( \\begin{array}{ccc} \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\theta _1}}}} &{} \\quad \\cdots \\quad &{} \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\theta _1}}}} \\\\ \\vdots \\quad &{} \\ddots \\quad &{} \\vdots \\\\ \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\theta _n}}}} &{} \\quad \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\theta _n}}}} \\\\ \\end{array} \\right) _{\\widehat{{\\varvec{\\theta }}}}' \\left( \\begin{array}{ccc} \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\theta _1}}}} \\quad &{} \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\theta _1}}}} \\\\ \\vdots \\quad &{} \\ddots &{} \\quad \\vdots \\\\ \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\theta _n}}}} &{} \\quad \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\theta _n}}}} \\\\ \\end{array} \\right) _{\\widehat{{\\varvec{\\theta }}}},\\\\ \\displaystyle \\frac{{\\partial {{\\varvec{L}}_{1:T}}}}{{\\partial {\\varvec{\\theta }} }}= & {} \\displaystyle \\frac{{\\partial {\\varvec{\\phi }} '}}{{\\partial {\\varvec{\\theta }} }}\\displaystyle \\frac{{\\partial {{\\varvec{L}}_{1:T}}}}{{\\partial {\\varvec{\\phi }} }} = \\left( \\begin{array}{ccc} \\displaystyle {\\frac{{\\partial {\\phi _1}}}{{\\partial {\\theta _1}}}} \\quad &{} \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {\\phi _n}}}{{\\partial {\\theta _1}}}} \\\\ \\vdots &{} \\ddots &{} \\vdots \\\\ \\displaystyle {\\frac{{\\partial {\\phi _1}}}{{\\partial {\\theta _n}}}} \\quad &{} \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {\\phi _n}}}{{\\partial {\\theta _n}}}} \\\\ \\end{array} \\right) \\left( \\begin{array}{ccc} \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\phi _1}}}} \\quad &{} \\cdots &{}\\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\phi _1}}}} \\\\ \\vdots &{} \\ddots &{} \\vdots \\\\ \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\phi _n}}}} &{} \\quad \\cdots &{}\\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\phi _n}}}} \\\\ \\end{array}\\right) \\\\&= \\left( \\begin{array}{ccc} \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\theta _1}}}} &{} \\quad \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\theta _1}}}} \\\\ \\vdots \\quad &{} \\ddots &{}\\quad \\vdots \\\\ \\displaystyle {\\frac{{\\partial {l_1}}}{{\\partial {\\theta _n}}}} &{} \\quad \\cdots &{} \\quad \\displaystyle {\\frac{{\\partial {l_T}}}{{\\partial {\\theta _n}}}} \\\\ \\end{array} \\right) ,\\quad \\widehat{\\hbox {H}} \\left( \\widehat{{\\varvec{\\theta }}} \\right) =\\left( \\frac{\\partial {{\\varvec{\\phi }} }'}{\\partial {\\varvec{\\theta }}} \\widehat{\\hbox {H}} \\left( \\widehat{{\\varvec{\\phi }}}\\right) \\frac{\\partial {\\varvec{\\phi }}}{\\partial {\\varvec{\\theta }}}\\right) _{\\widehat{{\\varvec{\\theta }}}}. \\end{aligned}\n\nAs an example, in the case of bivariate normal MSw model with two states, $$m=2$$, the natural parameters are\n\n\\begin{aligned} {\\varvec{\\theta }} =\\left( {\\mu _1^1 ,\\,\\,\\,\\mu _1^2 ,\\,\\,\\,\\mu _2^1 ,\\,\\,\\,\\mu _2^2 ,\\,\\,\\,\\sigma _1^1 ,\\,\\,\\,\\sigma _1^2 ,\\,\\,\\,\\sigma _2^1 ,\\,\\,\\,\\sigma _2^2 ,\\,\\,\\,\\rho ^{1},\\,\\,\\,\\rho ^{2},\\,\\,\\,p_{21} ,\\,\\,\\,p_{12} } \\right) ^{{{\\prime }}}, \\end{aligned}\n\nand the working parameters are\n\n\\begin{aligned} {\\varvec{\\phi }}= & {} \\Bigg ( \\ldots ,\\log \\sigma _1^1 ,\\log \\sigma _1^2 ,\\log \\sigma _2^1 ,\\log \\sigma _2^2 ,\\log \\frac{\\left( {1+\\rho ^{1}} \\right) }{\\left( {1-\\rho ^{1}} \\right) },\\log \\frac{\\left( {1+\\rho ^{2}} \\right) }{\\left( {1-\\rho ^{2}} \\right) },\\\\&\\log \\frac{p_{21} }{\\left( {1-p_{21} } \\right) },\\log \\frac{p_{12} }{\\left( {1-p_{12} } \\right) } \\Bigg )^{{\\prime }}. \\end{aligned}\n\n### Appendix 3\n\nIn this appendix I report the summary statistics for the four-year, the 15-year and the 30-year samples. Data are multiplied by 100.\n\n## Rights and permissions\n\nReprints and Permissions" ]

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[ "Eletiofe What's a Semi-Log Plot and How Can You Use...\n\n# What’s a Semi-Log Plot and How Can You Use It for Covid Data?\n\n-\n\nIt’s quite clear that things aren’t going so well with this Covid-19 pandemic. I mean, it’s bad, and it seems to be getting worse. The number of infected humans is just getting stupid-large. As of today, there have been more than 10 million confirmed cases of Covid-19 in just the United States. But some other countries (like South Korea) have significantly fewer cases—under 30,000. Or maybe you want to look at cases in the US in March, when they numbered in the hundreds, and compare them to October, when they numbered in the millions. So, how do you display data for stuff with such a huge range of values? The answer is to use a semi-log plot. I’m going to explain exactly how this works.\n\nLet’s start with a simple plot of the number of confirmed cases as a function of day number. In this example, Day 1 is the first day that USA (and South Korea) had their first positive Covid case. Since South Korea has a smaller population than the USA (51 million vs. 328 million), it might be useful to also include some larger countries. I’m going to use Brazil (209 million) and India (1.4 billion). This is just a normal (not semi-log) plot for these four countries. Oh, you can get all this data from the Covid-19 Data Repository at Johns Hopkins University.\n\nWhat do you notice from this plot? Other than the US doing very poorly with its confirmed cases? The thing that should jump out at you is that you can’t even see the data for South Korea. It’s there, it’s just too small to see. If you want to examine both the huge USA data and the not-so-huge South Korea data, you need a semi-log plot. (In this case “log” is short for logarithm—not an actual wooden log.)\n\nWhat the heck is a semi-log plot? I guess the first thing is to explain logs. Let me start with a number—a big number. How about 1 million? I could write this in the most common way as a 1 followed by 6 zeros. Like this: 1,000,000. But I could also write that as a power of ten.\n\nJust to be clear, 106 means 10 x 10 x 10 x 10 x 10 x 10. But what if I want to do the inverse of 10 raised to some power? It’s much easier to write big numbers by raising them to some power—this is exactly what we do with numbers in scientific notation. Finding the power of 10 that a number is raised to is exactly what a logarithm does. If I take the log of 1,000,000, it gives the result of 6. Oh, here is an important note. If we are talking about 10 raised to some power, that means we are using a log base of 10. The two most common bases are 10 (because we write numbers in base-10) or e, the natural number where e is approximately 2.718 (it’s irrational). Here is a more detailed explanation of e.\n\nBut wait! You can also take the logarithm for numbers that aren’t integer powers of 10. Let’s just pick a number—I’m going with 1,234. If I take the logarithm of this number, I get:\n\nThis means that if you raise 10 to the power of 3.09132, you get 1,234. But why? Why would you do that? OK, let’s go back to our terrible Covid data. Suppose that instead of plotting the number of confirmed infections, I plot the log (base 10) of the number of infections. I can then plot the log of the number vs. the day number. Here’s what that looks like.\n\nJust to be clear—this is the same data as the first plot, but there is a big difference. You can actually see the data for South Korea even though that country’s numbers are so much lower than those in the USA. Why? Well, let’s look at the total number of confirmed cases as of November 17, 2020. For the USA, it’s 11,036,935 and for South Korea it’s 28,769. Now let’s take the log (base 10) of both of these numbers.\n\nUnless you want a very rough estimate of a log function, you are going to need to just punch this in your calculator (or use a log-table where you look up a value). Now instead of having numbers that are very far apart, we now have values in the same range (7.04 and 4.46) such that they will easily fit in the same scale on a graph. But you don’t have to actually take the log of the data. There is another option—use a non-linear vertical axis. Here’s what that looks like.\n\nNotice that the numbers on the vertical axis are not fixed step sizes. The markings on the vertical axis show an increase by a power of 10 instead of an additive increase of, say 1,000 more cases. Since the horizontal scale is still linear, this is called a semi-log plot. A log-log plot would have both axis in an exponential scale.\n\nThen when should you use a semi-log plot and when should you use a linear-axis plot? If you have data that spans a very wide range of values (different orders of magnitude) then you pretty much have to create a semi-log plot so that you can see all of it. If the range of data is in the same magnitude (even if it’s super large numbers) then you can just make a normal plot. But you need to be careful. If you look at a semi-log plot and assume it has a linear axis, it makes South Korea look like it’s pretty bad when it’s really not.\n\nSadly, for Covid confirmed cases you pretty much need a semi-log plot.\n\nMore Great WIRED Stories\n\n### If Covid-19 Did Start With a Lab Leak, Would We Ever Know?\n\n“We find ourselves ten months into one of the most catastrophic global health events of our lifetime,” wrote Stanford...\n\n### The FBI Has Made Over 100 Arrests Related to the Capitol Riot\n\nIn yet another week that felt like a month, the world continues to feel the reverberations of the ...\n\n### Can This Group Revive the Finicky Corpse Flower?\n\nThis story originally appeared on Undark and is part of the Climate Desk collaboration.The alien-like blooms and putrid...\n\n### The Race Is On to Identify and Stop Inauguration Rioters\n\nThey came prepared for war. Leroy Coffman parked his Red GMC Sierra pickup truck a stone's throw from the...\n\n### The Physics of Reddit’s Spinning Solar System Icon\n\nWhile waiting for Reddit to load on my phone, I wondered if I could do some physics with..." ]

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[ "# Propositions and Logical Operations\n\n## Propositions\n\nIn logic, a proposition is a statement that evaluates true or false, such as “all cars are blue” or “4 + 4 = 8.” A phrase such as “go outside” is a command and not a proposition because it does not have a true or false value. Propositional statements can be denoted by variables, such as `Q = \"all cars are blue,\"` or `P = \"4 + 4 = 8,\" `which evaluate to `Q = False` and `P = True`.\n\n## Compound Propositions\n\nCompound propositions are the result of combining multiple propositions. For example, `K = Q and P`. That compound proposition evaluates to false because, as we’ve seen, Q is false. We can make K evaluate to true by changing the compound proposition to `K = Q or P`. The condition here is that only one variable must evaluate true for K to be true because we are saying “or” instead of “and.”\n\n## Logic Operators\n\nlogical operators refer to operations that combine propositions, such as the “and” and “or” operators we’ve previously used. Below is a list of logical operators and their denotations in discrete mathematics.\n\nThe conjunction operation is denoted by ∧ and read as “and,” such as:\n\nK = Q ∧ P\nK = Q and P\n\nThe disjunction operation reads as “or” and has two distinctions, the “exclusive or” and the “inclusive or.”\n\nThe “inclusive or” denoted by ∧ and reads as “or” dictates that a compound proposition is true if either statement is true.\n\nInclusive-or example:\nK = True\nP = False\nN = True\nR = K ∧ P ∧ N\nR evaluates to True when either K, P, or N is true.\n\nThe “exclusive or” denoted by ⊕ and reads as “or” dictates that only one operation must evaluate true for the compound proposition to be true.\n\nExclusive-or example:\nK = True\nP = False\nN = True\nR = K ∧ P ∧ N\nR evaluates to False when more than variables are true.\n\nThe negation operation, denoted by ¬ and read as “not,” reverses the value of a proposition. ¬Q is read as “not Q.”\n\nNegation operation example:\nP = True\nT = False\nQ = P ∧ T (Q evaluates to true because P is true)\nQ = ¬P ∧ T (Q evaluates to false because both P and T are false)\nThe negation of ¬P reversed P’s value from True to False.\n\n## Order of Operation\n\nThe order of operation for compound propositions is (¬) negation first, (∧) conjunction second, and (∨) disjunction last.\n\n1. ¬ (not)\n2. ∧ (and)\n3. ∨ (or)\n\nSimilar to the order of operation in mathematics, everything in parenthesis is applied first except that the negation operation (¬) always takes precedence.\n\nOrder of operation examples\nExample #1\n\n`F ∨ T ∧ T`\n`F ∨ (T ∧ T)` (Conjunction has precedence over disjunction)\n`F ∨ (T)` (True and True evaluate to True)\n`T` (False or True evaluate to True)\nExample #2\n`¬F ∨ T ∧ T`\n`(¬F) ∨ T ∧ T` (Negation has precedence over conjunction and disjunction\n`(¬F) ∨ (T ∧ T)` (Conjunction has precedence over disjunction)\n`(¬F) ∨ T` (True and True evaluate to True)\n`T ∨ T ` (Negation of False evaluates to True)\n`T ` (True and True evaluate to True)" ]

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[ "# Convolutional neural networks backpropagation\n\nMy question is regarding the answer to this question: Training a convolution neural network\n\nIt seems like the answer is saying to change all the weights in a given filter by the same amount in the same direction?\n\nWon't the CNN (convolutional neural network) be unable to learn appropriate filters this way?\n\nSee the diagrams representing the weights here: http://jefkine.com/general/2016/09/05/backpropagation-in-convolutional-neural-networks/\n\nThe 2x2 delta is backpropagated onto a 2x2 filter which creates a 3x3 grid of the change in weights If you are supposed to add up those 9 values and then add them to the original 2x2 filter at all positions it will just change each value in the 2x2 filter by the same amount and in the same directions. Is that the correct way to update weights in a CNN? Or are you supposed to update each weight (each value in the 2x2 filter) individually by its own gradient like in a normal artificial neural network ANN?\n\n• could you briefly summarize the answer you are asking about, so as to make your question self-contained? (if the answer you are referring to is deleted in the future, for example). Thanks – Antoine Jun 16 '17 at 15:20\n\nA convolutional network layer is a just a fully-connected layer where two things are true:\n\n• certain connections are removed; means their weights are forced to be constant zero. So you can just ignore these connections/weights\n• each of the non-zero weights is shared across multiple connections, ie across multiple pairs of input/output neurons\n\nWhen a weight is shared across connections, the gradient update for that weight is the sum of the gradients across all the connections that share the same weight.\n\nSo then looking at your questions:\n\nOr are you supposed to update each weight (each value in the 2x2 filter) individually by its own gradient like in a normal artificial neural network ANN?\n\nEach weight is updated just like in a 'normal ann', by which case I interpret this to mean in a fully-connected, or 'dense', or 'linear' layer, right? However, as stated, the update applied to each of the weights will be the sum of the gradients from all the connections that share that weight.\n\nThe 2x2 delta is backpropagated onto a 2x2 filter which creates a 3x3 grid of the change in weights If you are supposed to add up those 9 values and then add them to the original 2x2 filter at all positions it will just change each value in the 2x2 filter by the same amount and in the same directions. Is that the correct way to update weights in a CNN?\n\nThe gradients 'flow back' to where they came. In the forward pass, each weight in the CNN 'kernel' will be used to calculate the output of multiple connections. In the backward pass, the gradients will 'flow backwards', along those exact same connections, onto the originating weights.\n\nIn the worst case, rather than thinking of the CNN as some magical thing, you can, as stated, consider a CNN to be a standard fully-connected/linear layer, with many connections/weights forced to be zero, and the remaining weights shared across multiple connections. Then, you can use your existing knowledge for fully-connected / linear layers, to handle the CNN layer.\n\nFor background, some conceptual aims/motivations of a CNN layer compared to a fully-connected layer are:\n\n• enforces a prior on adjacency. Since pixels in images tend to have more correlation and relationships with adjacent pixels, so we keep only connections between adjacent input/output pixels, and remove the other connections\n• partly to remove parameters, to avoid over-fitting, and partly because we want to enforce a prior on invariance to translation, we share the weights such that input pixels in one region will give identical outputs, no matter where in the image they are, but simply translated in the output image\n\nIn the answer you reference, the goal is to calculate $w_k$, one of the weights in one filter in a (convolutional) neural network. The update formulas for any parameter in a neural network, using gradient descent, is quite simple if you just express it as a derivative of the cost function $J$:\n\n$\\Delta w_k = -\\eta \\frac{\\partial J}{\\partial w_k}$\n\nBut, computing $\\frac{\\partial J}{\\partial w_k}$ is where things get tricky. In a convolutional neural network, this weight is shared, so you need to calculate\n\n$\\frac{\\partial J}{\\partial w_k} = \\sum \\frac{\\partial J}{\\partial w_l}$\n\nwhere the sum is taken over all the occurrences of that weight. So, if you have a 2x2 filter and an 8x8 image (stride 1), the filter is applied 7x7=49 times and in each application of this filter the top left weight is used. So, the sum is over these 49 values. For the top right weight, however, the sum is still over 49 values, but it's now a different set of values and so the update is different." ]

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[ "#### Would you like to use Vecteezy in English?\n\nView in different language\n\n# How are PPD Free Earnings Calculated?", null, "# PPD Free Earnings Overview\n\nUnlike PPD Pro earnings, PPD (Pay Per Download) Free earnings are calculated based on the total number of downloads a contributor receives across their portfolio.\n\nThis model is calculated based on a set rate: \\$5.00 paid per 1,000 free downloads.\n\n# Calculating PPD Free Earnings\n\nTo calculate PPD Free Earnings, Vecteezy takes the total number of downloads a contributor receives and multiplies this amount by \\$0.005.\n\n• Total downloads x \\$0.005 = amount earned\n\nWe use this formula for each contributor monthly to calculate total PPD Free contributor earnings.\n\n# PPD Free Earnings Examples\n\nHere are two examples that illustrate PPD Free payouts:\n\nContributor A:\n\nReceives 10,000 downloads in a month and earns \\$50.00 for that month\n\n• 10,000 downloads received x \\$0.005 = \\$50.00\n\nContributor B:\n\nReceives 250,000 downloads in a month and earns \\$1,250.00 for that month\n\n• 250,000 total downloads x \\$0.005 = \\$1,250.00", null, "Learn more about our PPD Free and PPD Pro Contributor programs here.\n\nInterested in becoming a contributor for Vecteezy? Apply Here!" ]

[ null, "https://static.vecteezy.com/system/assets/asset_files/000/001/423/original/PPD_Free_Header.jpg", null, "https://static.vecteezy.com/system/assets/asset_files/000/001/435/original/PPD_Free_AB.jpg", null ]

[ "# Mixed logit models\n\nStata fits discrete choice models. Stata 15 will fit them with random coefficients. Discrete choice is another way of saying multinomial or conditional logistic regression. The word \"mixed\" is used by statisticians whenever some coefficients are random and others are fixed. Therefore, Stata 15 fits mixed logit models. These models are fit with the new `asmixlogit` command.\n\nRandom coefficients arise for many reasons, but there is a special reason researchers analyzing discrete choices might be interested in them. Random coefficients are a way around the IIA assumption. If you have a choice among walking, public transportation, or a car and you choose walking, the other two alternatives are irrelevant. Take one of them away, and you would still choose walking. Human beings sometimes violate this assumption, at least judged by their behavior.\n\nMathematically speaking, IIA makes alternatives independent after conditioning on covariates. If IIA is violated, then the alternatives would be correlated. Random coefficients allow that." ]

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[ "irrigation pipe flow rate in liechtenstein Despriction", null, "How much water can flow through a pipe?How much water can flow through a pipe?How Much Water Can Flow Through A Pipe (GPM/GPH)? How Much Water Can Flow Through A Pipe (GPM/GPH)? We regularly get asked about the water flow capacity of different pipe sizes, and which is the best roof drain for a specific pipe size.How Much Water Can Flow Through A Pipe (GPM/GPH)? Water-Flow uniformity through irrigation gated-pipes.\n\ngated pipe. Results also showed great agreement between the theoretical gated pipe flow rate, based on newly derived equation and the corresponding fieldwork. Rate of discharge through the gate was found to be a power function of the head for gate sizes 2.84, 5.67 and 11.34 cm2 in the following form q =b h0.37 Where q is in m3/s , h irrigation pipe flow rate in liechtenstein What is a flow factor pipe?What is a flow factor pipe?FRICTION FACTOR PIPE SIZING This Friction Factoris used to determine the maximum flow in gallons per minute through any section of lateral line pipe while not exceeding a predetermined pressure loss (pressure variation).THE HANDBOOK OF TECHNICAL IRRIGATION INFORMATION THE irrigation pipe flow rate in liechtenstein\n\nWhat type of pipe is used for irrigation?What type of pipe is used for irrigation? Habit, since here in La La Land (Los Angeles, California) we use mostly PVC pipe for irrigation.   However some types of pipe are technically defined as tube.    The difference is the material they are constructed of.   Steel and PVC plastic are generally called pipe.   Polyethylene, PEX, and copper are usually called tube or tubing.Irrigation Pipe Sizing Chart for Laterals#1 FlexPVC&Water Flow Charts Based on Pipe Size\n\nWater Flow Chart #1 The chart below takes into consideration the potential damage from hydraulic hammer (shock) and noise considerations due to excessive fluid velocity. For more detailed information click here for our pipe selection based on pipe size and flow requirement Nomograph.You can flow more than what is shown in the chart (see Chart #2 below) however, you may run into problems if you\n\nliechtenstein\n\nMust include:\n\nliechtensteinExplore furtherFlow Rate Calculator - calculate the flow rate of a pipegigacalculatorPipe Flow Calculator HazenWilliams EquationomnicalculatorWater Pipe Flow Rate Table Chili PepperchilipepperappFLOW RATE CALCULATOR1728Pipe Volume Calculator - Inch CalculatorwwwchcalculatorRecommended to you based on what's popular Pipe and riser irrigation systems Irrigation Water irrigation pipe flow rate in liechtensteinWhat flow rate do you want or need? Generally the response is in the range of 1220ML per day. The answer to this helps to determine the diameter of pipe required and the likely pump specifications. You should seek advice from an irrigation surveyor and designer to ensure the flow rate meets the requirements of the farm's physical characteristics and/or the crop being grown.\n\nliechtenstein\n\nMust include:\n\nliechtensteinFlow Rate Calculator - calculate the flow rate of a pipe\n\nUsing The Flow Rate CalculatorFlow Rate FormulaCalculation ExamplesThis pipe flow rate calculator calculates the volumetric flow rate (discharge rate) a gas or liquid going through a round or rectangular pipe of known dimensions with a known velocity. If the substance is a liquid and its volumetric density is known the calculator will also output the mass flow rate (more information is required to calculate it for gases and it is currently not supported). The output is in either imperial or metric units, depending on your selection. Some of the output units include m3/h, m3/min, See more on gigacalculator\n\nliechtenstein\n\nMust include:\n\nliechtensteinProper Water Pressure Flow In Drip Irrigation SystemsAll drip irrigation systems are designed to operate in a specific flow and pressure range. The flow rate through each emitter is directly related to the feed pressure, and proper operating pressure facilitates accurate distribution to emitters. As feed pressure is increased, the water velocity within the pipes\n\nliechtenstein\n\nMust include:\n\nliechtensteinHydraulics and Pumping - Irrigation NZIf flow is fully turbulent, the Hazen-Williams equation is recommended instead h = K x (100 C) 1.852 x Q1.852 D4.866 x L 100 Where h = head loss, K = unit coefficient (2.38× 108 for SI unit), C = coefficient of retardation and depends on pipe material,Q = flow rate,D = inside diameter and L = pipe length.\n\nliechtenstein\n\nMust include:\n\nliechtensteinIrrigation Pipe Sizing Chart for LateralsUp to 76 GPM = 2 size pipe. Now go back and look at the flow for each section of pipe on your plan. Then based on the GPM flow, insert the pipe size from the schedule you made. So the section with a flow of 2.5 GPM will be 3/4 pipe. The section with 1.3 GPM will also be 3/4 pipe. The section with 3.8 GPM will be 3/4 pipe.\n\nliechtenstein\n\nMust include:\n\nliechtensteinPeople also askWhat is the flow rate of a rectangular pipe?What is the flow rate of a rectangular pipe?Example 2 A rectangular pipe has a height of 2cm and width of 4cm and a gas running through it at a speed of 15 m/s. What is the discharge rate of this pipe? First, we find the cross-section area which is simply 2 4 = 8 cm 2 or 0.0008 m 2. To find the flow rate Q, we multiply 0.0008 by 15 to get 0.012 cubic meters per second.Flow Rate Calculator - calculate the flow rate of a pipe\n\nliechtenstein\n\nMust include:\n\nliechtensteinTHE HANDBOOK OF TECHNICAL IRRIGATION\n\nHUNTER Handbook of Technical Information FORMULAS GENERAL 2 GENERAL SLOPE Slope, as used in irrigation, is a measure of the incline of an area. It can be described as (1) a percent, formula A, (2) a degree, formulas B and C, or (3) a ratio, formula D.\n\nliechtenstein\n\nMust include:\n\nliechtensteinUsing Flexible Pipe (Poly-pipe) with Surface IrrigationPoly-pipes also come in larger thicknesses (15 mil), allowing more pressure to be contained (up to 5 feet of water head or 2.15 psi). Pipe diameter should be selected based on irrigation flow-rate. Table 1 provides some approximate diameters and thicknesses needed for selected flow-rates.\n\nliechtenstein\n\nMust include:\n\nliechtensteinWater Measurement for Agricultural Irrigation and\n\nSince flow rate is volume per unit time, the volume of water applied during irrigation can be calculated if the flow rate and irrigation duration (time) are known. Example Find the volume of water applied and depth of application for a pump that discharges 800 gpm Calculating Horsepower Requirements and Sizing velocity of flow in the suction pipe of centrifugal pumps should be kept between 2 and 3 ft/s in order to prevent cavitation. Table 5 lists the maximum flow rates recommended for different ID (internal diameter) pipe sizes using the 5 ft/s rule. Many friction loss tables give both the friction loss and velocity for any given gpm and pipe size.Drip Irrigation System |Install Drip - Garden Drip irrigation pipe flow rate in liechtensteinThese issuers offer a fixed rate within a more or less wide pressure range. The usefulness of these emitters is the ability of homogenization irrigation along an irrigation line as the last emitter line typically have a lower pressure than the first because the pressure drop due to friction of water with pipe\n\nDrip Tubing - Hose - Drip Irrigation - Irrigation - Products\n\nFlow rates 0.4, 0.6 and 0.9 gph (1.5 l/h, 2.3 l/h and 3.5 l/h) Accepts Rain Bird Easy Fit Compression Fittings, XF Dripline Insert Fittings and 17mm insert fittings. Be the First to ReviewHandbook of Technical Irrigation Information UNIVERSITY109 100' Head Plastic Irrigation Pipe (PIP) 110 SDR-81 50 PSI Plastic Irrigation Pipe (PIP) 111 SDR-51 80 PSI Plastic Irrigation Pipe (PIP) irrigation pipe flow rate in liechtenstein A to point B with a flow rate of 40 GPM. According to the manufacturer, the control valve will lose 1.0 PSI at 40 GPM.Home - NaanDanJainNaanDanJain Irrigation Ltd. is the leading global producer and provider of tailor-made irrigation solutions. The company offers the widest range of cost-effective and customized technologies across more than 120 countries worldwide.\n\nHow Much Water Can Flow Through A Pipe (GPM/GPH)?\n\nThat said, we put together the following tables to serve as general guides for estimating a pipe's water flow capacity through a pipe or roof drain. If you have questions, please call our Drain Wizard at 800-635-0384. Water Flow (GPM/GPH) based on Pipe Size and Inside/Outside DiametersHydraulics and Pumping - Irrigation NZiBOK 10HYD Volume 1FwHmalHdend1ap i 5 If flow is fully turbulent, the Hazen-Williams equation is recommended instead h = K x ( 100 C) 1.852 x Q1.852 D4.866 x L 100 Where h = head loss, K = unit coefficient (2.38× 108 for SI unit), C = coefficient of retardation and depends on pipe material,Q = flow rate,D = inside diameter and L = pipe length. However for turbulent flow in smooth irrigation pipe flow rate in liechtensteinInstallation calculators - Pope IrrigationAn easy way to calculate your flow rate is by doing a bucket test. All you need is your garden tap, a standard 9 litre bucket and a stopwatch. Before you start, make sure you measure your water flow at times of peak water usage. If using a tap timer or pressure reducer, make sure you measure the flow\n\nInstallation calculators - Pope Irrigation\n\nAn easy way to calculate your flow rate is by doing a bucket test. All you need is your garden tap, a standard 9 litre bucket and a stopwatch. Before you start, make sure you measure your water flow at times of peak water usage. If using a tap timer or pressure reducer, make sure you measure the flow Irrigation Pipes For Sale in KenyaEunidrip Online ShopEunidrip irrigation HDPE pipe. Irrigation HDPE pipes for sale in Kenya are considered, be the best choice for irrigation on the various project due to its high quality and reliability. It made up of high density, polyethylene particles that are closely packed. Usually made up of high-pressure rate known as pressure normal (PN).Irrigation Supplies Horizon Distributors - Irrigation irrigation pipe flow rate in liechtensteinFlow rate .64 GPM to 5.3 GPM Radius 15 to 37 Pressure range 20 psi to 100 psi Arc setting 40 to 360 degrees Precipitation rates Approximately 0.60 per hour at 40 psi Nozzle trajectory Approximately 14° 1/2 female inlet NPT\n\nIrrigation Water Pumps Publications\n\nHowever, the type of irrigation system, distance from the water source and size of the piping system will determine the flow rate and total dynamic head. Basic Pump Operating Characteristics Head is a term commonly used with pumps.Landscape Irrigation Formulas - Irrigation TutorialsWater flow (in cubic meters per second) = 0.55 x pump power (watts) / pressure (pascals) Water flow (in liters per second) = 5.43 x pump power (kilowatts) / pressure (bars) Link to top of Page. Miscellaneous Irrigation Formulas, Conversions, Equations, and Calculations Sprinkler Precipitation RateLandscape Irrigation Formulas - Irrigation TutorialsWater flow (in cubic meters per second) = 0.55 x pump power (watts) / pressure (pascals) Water flow (in liters per second) = 5.43 x pump power (kilowatts) / pressure (bars) Link to top of Page. Miscellaneous Irrigation Formulas, Conversions, Equations, and Calculations Sprinkler Precipitation Rate\n\nMeasuring Irrigation Flow - uaex.edu\n\nto select, install and use irrigation flow meters, as well as estimate flow manually using simple hydraulic formulas. In almost all cases, for flow meters to be accurate, pipes must be full of water (full pipe flow). Propeller Flow Meters Propeller flow meters are the most common devices used for measuring water flow rate and totalization irrigation pipe flow rate in liechtensteinMeasuring the Flow of Water in Pipes Irrigation AgronomySeveral devices are used for measuring the flow of water in pipes. A few of them, which may find application on the farm, are as follows 1. Volume Measurement 2. Flow Rate Measurement 3. Measurement of Cumulative Flow 4. Water Level Recording. Method # 1. Volume Measurement A simple method for measuring small irrigation streams Minimum Required Pipe Size - Farm Irrigation SuppliesCalculate minimum required pipe size based on flow rate, pipe length, material and maximum allowable pressure loss. Use right size of pipe and save money sand energy.\n\nPipe Flow Calculator HazenWilliams Equation\n\nFeb 03, 2021The hydraulic radius, R, is the proportion between the area and the perimeter of your pipe. If the pipe is circular, you will find it according to the following equation R = A / P = r² / 2r = r / 2 = d / 4. where r is the pipe radius, and d is the pipe diameter. You can view and modify all these parameters (area, perimeter, hydraulic radius) in the advanced mode of this pipe flow calculator.Pipe and riser irrigation systems Irrigation Water irrigation pipe flow rate in liechtensteinIt is possible to reduce capital expenditure by using a smaller pipe and increasing the size of the pump/motor set required to deliver the desired flow rate. For example, to increase the pump/motor size on a pump may cost \\$1000 (subject to power requirements and pump specifications) while , to increase the pipe size from 355mm to 400mm may cost irrigation pipe flow rate in liechtensteinProper Water Pressure Flow In Drip Irrigation SystemsAll drip irrigation systems are designed to operate in a specific flow and pressure range. The flow rate through each emitter is directly related to the feed pressure, and proper operating pressure facilitates accurate distribution to emitters. As feed pressure is increased, the water velocity within the pipes\n\nRural Irrigation Pipe Systems - Vinidex\n\nRural Irrigation Pipe Systems P 1 2 50 L L T V P L C C T 1 T epr C 2 P irrigation pipe flow rate in liechtenstein 20190916 PIPE FLOW CALCULATOR Three factors should be known before an appropriate choice of PE pipe can be made 1. The length of the pipeline. 2. The quantity of water required. 3. Nett pressure - taking into account available head, differences in level overSome results are removed in response to a notice of local law requirement. For more information, please see here.Some results are removed in response to a notice of local law requirement. For more information, please see here.Chapter 6 Irrigation System DesignChapter 6 Irrigation System Design Part 652 Irrigation Guide (210-vi-NEH 652, IG Amend. NJ1, 06/2005) NJ6-1 NJ652.06 Irrigation System Design a) General A properly designed irrigation system addresses uniform irrigation application in a timely manner while minimizing losses and damage to soil, water, air, plant, and animal resources.\n\nUseful Irrigation Formulas Charts\n\nApplication Rate (inches/hour) GPH X 1.604 Area (square feet) = Useful Irrigation Formulas Charts Application Rates of Emitter-Line Grids This simple formula allows the precise calculation of application rates in inches per hour. Just enter the emitter rate (GPH), the spac-ing of the emitter-line that you are using in inches, and theWater Application Rate CalculatorWater Application Rate. Use this form to calculate the water application rate using the flow rate onto an area. This calculator assumes perfect irrigation efficiency and uniformity. Learn more about the units used on this page.Water flow rate in drip irrigation - SAB spaWhen it comes to irrigation, one of the factors that cannot be underestimated is certainly the water flow rate. To better understand this topic, it is necessary to start from the definition of flow rate In fluid dynamics, flow rate is the amount of substance that crosses a section of area A in a unit of time.\n\nWater-Flow uniformity through irrigation gated-pipes.\n\ngated pipe. Results also showed great agreement between the theoretical gated pipe flow rate, based on newly derived equation and the corresponding fieldwork. Rate of discharge through the gate was found to be a power function of the head for gate sizes 2.84, 5.67 and 11.34 cm2 in the following form q =b h0.37 Where q is in m3/s , h irrigation pipe flow rate in liechtensteinWhy does the size of your pipe matter? Southern's Water irrigation pipe flow rate in liechtensteinOct 16, 2016Flow rate. The flow rate (usually measured in L/min, m3/hr or L/sec) is the volume of water that is travelling past a fixed point in the pipe. You measure flow rate with a paddle or electronic water meter. As an equivalent in electricity is the Amps or current. Pressure (measured in kilopascals or pounds per square inch) is the amount of irrigation pipe flow rate in liechtenstein" ]

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[ "# Created Variables CC table\n\nThe Created_variables_CC table in CFE contains the most of the profile-level derived Indicators and duplicates some related stored data for convenience . It is populated and used by the statistician for reporting and data extracts. Primarily these are the APACHE score and its components, e.g. Glasgow Coma Scale\n\nfield type description source\nD_ID text link to L_Log table relational derived\nBirth date DOB from L Log table relational derived\nage integer Age calculated\nage_pts integer as per VBA Function ApScore_Age calculated\nApTotal_Age byte twin of age_pts calculated\nAp_Chronic text(30) Chronic Health APACHE relational derived\nchronic_pts long as per VBA Function Ap_Chronic calculated\nAp_Temp single Temperature from L Log table relational derived\ntemp_pts integer as per VBA Function ApScore_Temp calculated\nAp_APSysBP long Sys BP from L Log table relational derived\nAp_APDiasBP long Dias BP from L Log table relational derived\nMBP integer Mean BP via VBA function AP_MeanBP calculated\nBP_pts integer twin of MBP calculated\nAp_HeartRate integer HR from L Log table relational derived\nheart_pts integer as per VBA Function ApScore_HeartRate calculated\nAp_RespRate long RR from L Log table relational derived\nresp_pts integer as per VBA Function ApScore_RespRate calculated\nAp_Eye text(13) AP Eye field from L Log table relational derived\neye_pts integer as per VBA Function GCS_Eye_pts calculated\nAp_Motor text(19) AP Motor field from L Log table relational derived\nmotor_pts integer as per VBA Function GCS_Motor_pts calculated\nAp_Verbal text(26) AP Verbal field from L Log table relational derived\nverbal_pts integer as per VBA Function GCS_Verbal_pts calculated\nApTotal_Neuro byte 15-(eye_pts + motor_pts + verbal_pts]) calculated\nAp_FIO2 single FIO2 from L Log table relational derived\nAp_PO2 single PO2 from L Log table relational derived\nPO2_score integer as per VBA Function ApScore_PO2 calculated\nAp_CO2 single CO2 from L Log table relational derived\nAp_pH single PH from L Log table relational derived\nph_score integer as per VBA Function ApScore_pH calculated\nAp_SerCO2 single Serum CO2 from L Log table relational derived\nSerCO2_score integer as per VBA Function ApScore_SerCO2 calculated\nabg_score integer as per VBA Function ApScore_ABG calculated\nAp_Na single Na from L Log table relational derived\nna_score integer as per VBA Function ApScore_Na calculated\nAp_K single K from L Log table relational derived\nk_score integer as per VBA Function ApScore_K calculated\nAp_Hemat single HCT from L Log table relational derived\nhemat_score integer as per VBA Function ApScore_Hemat calculated\nAp_WBC integer WBC from L Log table relational derived\nwbc_score integer as per VBA Function ApScore_WBC calculated\nAp_Creat single Creatinine from L Log table relational derived\nAp_ARF text(5) ARF from L Log table relational derived\ncreat_score integer as per VBA Function ApScore_Creat calculated\nApTotal_Physiological byte as per VBA Function ApTotal_Physiological calculated\nCH text(255) IIf([Ap_Chronic]=\"9No CH\",\"N\",\"Y\") calculated\nApTotal_Chronic byte as per VBA Function AP_Chronic calculated\nApTotal byte as per VBA Function APACHE_Total calculated\nSite string from L Log table and s_dispo table relational derived\nWard string from L Log table and s_dispo table", null, "_dev_CFE remove as discussed 2019-09-29. added: 2019-09-29 action: 2019-09-29\n• Cargo\n\n• Categories\n|| relational derived\nProgram string from L Log table and s_dispo table relational derived\nactiveLocation yes/no from L Log table and s_dispo table", null, "_dev_CFE remove as discussed 2019-09-29. added: 2019-09-29 action: 2019-09-29\n• Cargo\n\n• Categories\n|| relational derived\nRecordStatus text(20) RecordStatus from L Log table relational derived\n\n## Implementation\n\nThis table is implemented to be filled by Created _Variables_CC maker query as needed.\n\n## Log\n\n2021-05? - made query Created_Variables_CC maker_2021 to test new derivations of fields" ]

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[ "", null, "# format core dump 5353, WT_BTREE.compressor == NULL\n\nXMLWordPrintable\n\n#### Details\n\n• Type:", null, "Task\n• Status: Closed\n• Resolution: Fixed\n• Affects Version/s: None\n• Fix Version/s:\n• Component/s: None\n• Labels:\n\n#### Description\n\nFormat core dump 5353, here's the CONFIG:\n\n `############################################` `# RUN PARAMETERS` `# stress run WT-5353` `############################################` `auto_throttle=1` `firstfit=0` `bitcnt=3` `bloom=1` `bloom_bit_count=45` `bloom_hash_count=20` `bloom_oldest=0` `cache=59` `checkpoints=1` `checksum=off` `chunk_size=1` `compaction=0` `compression=zlib` `data_extend=0` `data_source=file` `delete_pct=42` `dictionary=0` `evict_max=5` `file_type=variable-length column-store` `backups=0` `huffman_key=0` `huffman_value=0` `insert_pct=11` `internal_key_truncation=1` `internal_page_max=10` `isolation=read-uncommitted` `key_gap=16` `key_max=49` `key_min=13` `leak_memory=0` `leaf_page_max=17` `logging=1` `merge_max=17` `merge_threads=2` `mmap=1` `ops=100000` `prefix_compression=1` `prefix_compression_min=2` `repeat_data_pct=15` `reverse=0` `rows=100000` `runs=100` `split_pct=65` `statistics=0` `threads=8` `value_max=209` `value_min=0` `wiredtiger_config=` `write_pct=41` ```############################################ ```\n\nI've got thousands of runs of this without a failure, so it's not an obvious failure.\nHere's the stack:\n\n `(gdb) where` `#0 0x00000000004bc91d in __wt_bt_write (session=0x251e400, ` ` buf=0x7fd71819eb58, addr=0x7fd757ffe6f0 \"\\030\", addr_sizep=0x7fd757ffe7f8, ` ` checkpoint=0, compressed=1) at ../src/btree/bt_io.c:160` `WT-1 0x00000000004715f8 in __rec_split_write (session=0x251e400, ` ` r=0x7fd71819eb40, bnd=0x7fd7180b6ce0, buf=0x7fd71819eb58, last_block=1)` ` at ../src/btree/rec_write.c:2786` `WT-2 0x0000000000470ba9 in __rec_split_finish_std (session=0x251e400, ` ` r=0x7fd71819eb40) at ../src/btree/rec_write.c:2501` `WT-3 0x0000000000470c71 in __rec_split_finish (session=0x251e400, ` ` r=0x7fd71819eb40) at ../src/btree/rec_write.c:2532` `WT-4 0x000000000047527b in __rec_row_leaf (session=0x251e400, r=0x7fd71819eb40, ` ` page=0x2554cd0, salvage=0x0) at ../src/btree/rec_write.c:4493` `WT-5 0x000000000046d085 in __wt_rec_write (session=0x251e400, ref=0x2554c90, ` ` salvage=0x0, flags=0) at ../src/btree/rec_write.c:411` `WT-6 0x00000000004605b4 in __sync_file (session=0x251e400, syncop=8)` ` at ../src/btree/bt_sync.c:135` `WT-7 0x0000000000460aa7 in __wt_cache_op (session=0x251e400, ` ` ckptbase=0x7fd71804dd00, op=8) at ../src/btree/bt_sync.c:354` `WT-8 0x000000000044b4e4 in __checkpoint_worker (session=0x251e400, ` ` cfg=0x7fd757ffed70, is_checkpoint=1) at ../src/txn/txn_ckpt.c:841` `WT-9 0x000000000044b72f in __wt_checkpoint (session=0x251e400, ` ` cfg=0x7fd757ffed70) at ../src/txn/txn_ckpt.c:899` `WT-10 0x000000000044a620 in __wt_txn_checkpoint (session=0x251e400, ` ` cfg=0x7fd757ffed70) at ../src/txn/txn_ckpt.c:402` `WT-11 0x0000000000440b1d in __session_checkpoint (wt_session=0x251e400, ` ` config=0x0) at ../src/session/session_api.c:793` `WT-12 0x0000000000406cc5 in ops (arg=0x2544600) at ../../../test/format/ops.c:289` `WT-13 0x000000339e807c53 in start_thread (arg=0x7fd757fff700)` ` at pthread_create.c:308` `WT-14 0x000000339e0f5dbd in clone ()` ` at ../sysdeps/unix/sysv/linux/x86_64/clone.S:113` `(gdb) frame 0` `#0 0x00000000004bc91d in __wt_bt_write (session=0x251e400, ` ` buf=0x7fd71819eb58, addr=0x7fd757ffe6f0 \"\\030\", addr_sizep=0x7fd757ffe7f8, ` ` checkpoint=0, compressed=1) at ../src/btree/bt_io.c:160` `160 WT_ERR(btree->compressor->decompress(` `(gdb) p btree->compressor` `\\$12 = (WT_COMPRESSOR *) 0x0` `(gdb) p btree->dhandle->name` `\\$13 = 0x2548bf0 \"file:WiredTiger.wt\"` `(gdb) frame 1` `WT-1 0x00000000004715f8 in __rec_split_write (session=0x251e400, ` ` r=0x7fd71819eb40, bnd=0x7fd7180b6ce0, buf=0x7fd71819eb58, last_block=1)` ` at ../src/btree/rec_write.c:2786` `2786 WT_ERR(__wt_bt_write(session,` `(gdb) p *bnd` `\\$14 = {start = 0x7fd71819ee28 \"!file:wt\", recno = 0, entries = 2, addr = {` ` addr = 0x0, size = 0 '\\000', type = 3 '\\003', reuse = 0 '\\000'}, ` ` size = 1629, cksum = 0, dsk = 0x0, skip = 0x0, skip_next = 0, ` ` skip_allocated = 0, key = {data = 0x0, size = 0, flags = 0, mem = 0x0, ` ``` memsize = 0}, already_compressed = 1} ```\n\nThis is diagnostic code that decompresses and verifies compressed buffers before writing them:\n\n `#ifdef HAVE_DIAGNOSTIC` ` /*` ` * We're passed a table's disk image. Decompress if necessary and` ` * verify the image. Always check the in-memory length for accuracy.` ` */` ` dsk = buf->mem;` ` if (compressed) {` ` WT_ERR(__wt_scr_alloc(session, dsk->mem_size, &tmp));` ` ` ` memcpy(tmp->mem, buf->data, WT_BLOCK_COMPRESS_SKIP);` ``` WT_ERR(btree->compressor->decompress( ```\n\nWhat happened is we're writing the WiredTiger.wt file (which isn't compressed), but the WT_BOUNDARY we're using to write the block has the WT_BOUNDARY.already_compressed flag set, presumably from a previous use of the WT_BOUNDARY structure to write a compressed block. (It must be from a previous run: WT_BOUNDARY.already_compressed is only set in *rec_split_raw_worker(), and we're not going through the raw-compression code here, the call to *rec_split_finish_std() tells us this reconciliation isn't configured for raw-compression.\n\nI can make this bug happen by explicitly not resetting WT_BOUNDARY.already_compressed inside __rec_bnd_cleanup(), but I don't yet see any other path where we might end up with it not reset.\n\n#### People\n\nAssignee:", null, "Keith Bostic\nReporter:", null, "Keith Bostic" ]

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[ "Moscow University Mathematics Bulletin\n\n, Volume 74, Issue 3, pp 127–130\n\n# On Some Analytic Method for Approximate Solution of Systems of Second Order Ordinary Differential Equations\n\nArticle\n\n## Abstract\n\nAn approach to using Chebyshev series to solve canonical second-order ordinary differential equations is described. This approach is based on the approximation of the solution to the Cauchy problem and its first and second derivatives by partial sums of shifted Chebyshev series. The coefficients of the series are determined by an iterative process using the Markov quadrature formula. It is shown that the described approach allows one to propose an approximate analytical method of solving the Cauchy problem. A number of canonical second-order ordinary differential equations are considered to represent their approximate analytical solutions in the form of partial sums of shifted Chebyshev series.\n\n## Preview\n\nUnable to display preview. Download preview PDF.\n\n## References\n\n1. 1.\nS. F. Zaletkin, “Numerical Integration of Ordinary Differential Equations Using Orthogonal Expansions,” Matem. Model. 22 (1), 69 (2010).\n2. 2.\nO. B. Arushanyan, N. I. Volchenskova, and S. F. Zaletkin, “Application of Orthogonal Expansions for Approximate Integration of Ordinary Differential Equations,” Vestn. Mosk. Univ., Matem. Mekhan., No. 4, 40 (2010) [Moscow Univ. Math. Bull. 65 (4), 172 (2010)].Google Scholar\n3. 3.\nO. B. Arushanyan, N. I. Volchenskova and S. F. Zaletkin, “Calculation of Expansion Coefficients of Series in Chebyshev Polynomials for a Solution to a Cauchy Problem,” Vestn. Mosk. Univ. Matem. Mekhan., No. 5, 24 (2012).Google Scholar\n4. 4.\nO. B Arushanyan and S. F. Zaletkin, “Application of Markov’s Quadrature in Orthogonal Expansions,” Vestn. Mosk. Univ., Matem. Mekhan., No. 6, 18 (2009) [Moscow Univ. Math. Bull. 64 (6), 244 (2009)].Google Scholar\n5. 5.\nS. F. Zaletkin, “Markov’s Formula with Two Fixed Nodes for Numerical Integration and Its Application in Orthogonal Expansions,” Vychisl. Metody Programm. 6, 1 (2005).Google Scholar\n6. 6.\nO. B. Arushanyan and S. F. Zaletkin, “Justification of an Approach to Application of Orthogonal Expansions for Approximate Integration of Canonical Systems of Second Order Ordinary Differential Equations,” Vestn. Mosk. Univ., Matem. Mekhan., No. 3, 29 (2018)Google Scholar\n7. 7.\nO. B. Arushanyan and S. F. Zaletkin, “To the Theory of of Calculation of Orthogonal Expansion for Solution to Cauchy Problem for Second Order Ordinary Differential Equations,” Vychisl. Metody Programm. 19, 178 (2018).Google Scholar\n8. 8.\nO. B. Arushanyan, N. I. Volchenskova and S. F. Zaletkin, “Approximate Integration of Ordinary Differential Equations on the Base of Orthogonal Expansions,” Differ. Uravn. Processy Upravl. 14 (4), 59 (2009).\n9. 9.\nO. B. Arushanyan, N. I. Volchenskova and S. F. Zaletkin, “Approximate Solution of Ordinary Differential Equations Using Chebyshev Series,” Sib. Elektron. Matem. Izv., 7, 122 (2010).\n10. 10.\nO. B. Arushanyan, N. I. Volchenskova and S. F. Zaletkin, “Calculation of Coefficients of Chebyshev Series for Solution of Ordinary Differential Equations,” Sib. Elektron. Matem. Izv., 8, 273 (2011).", null, "" ]

[ null, "https://rd.springer.com/track/controlled/article/denied/10.3103/S0027132219030057", null ]

[ "Optimal control is one of the most popular decision-making tools recently in many researches and in many areas. The Lorenz-R &#246;ssler model is one of the interesting models because of the idea of consolidation of the two models : Lorenz and &#246;ssler. This paper discusses the Lorenz-R&#246;ssler model from the bifurcation phenomena and the optimal control problem (OCP). The bifurcation property at the system equilibrium", null, "is studied and it is found that saddle-node and Hopf bifurcations can be holed under some conditions on the parameters. Also, the problem of the optimal control of Lorenz-R&#246;ssler model is discussed and it u ses the Pontryagin’s Maximum Principle (PMP) to derive the optimal control inputs that achieve the optimal trajectory. Numerical examples and solutions for bifurcation cases and the optimal controlled system are carried out and shown graphically to show the effectiveness of the used procedure.\n\nLorenz-Rössler Model Bifurcation Pontryagin Principle Optimal Control Problem\n1. Introduction\n\nPrediction of any system’s development is a very important goal, especially in the case of chaotic systems which exist frequently in several real-life and various fields.\n\nThese systems are very important for the service of mankind. Those systems include psychology , secure communications , economics system modeling , medicine , etc. Due to the importance of these models, these deserved to be studied. Despite the recent trend of forecasting, this task is often not easy. The chaotic models usually need to study the optimization or the OCP. The Principle of Optimality means to take the best choice (procedure) to minimize the cost (maximize the profit) of the current time (stage) and all subsequent times (stages). The OCPs in general does not have a perfect solution, so solutions are often approximate, which is another aspect of difficulty and sensitivity.\n\nLorenz system is a reduced version of a larger system studied earlier by Barry Saltzman . This model is a system of three non-linear ordinary differential equations, is extremely sensitive to perturbation of the initial conditions, and this system has strange attractors; thus this system has a chaotic behavior for different values of parameters and different initial conditions as it is in the famous butterfly attractor which is produced for the special values of parameters of (10, 28, 8/3) respectively as these parameters shown sequentially in the model in . Rössler attractor is another famous attractor which is one of the products of the works of a German biochemist Otto Eberhard Rössler. There is a similarity between Lorenz and Rössler attractors, where the latter follows an outward spiral around two fixed points. Although each variable in the system is oscillating around specified values, the oscillations are chaotic . There is a merging of the two models of Lorenz and Rössler in a non-additive form, and because the new additive system losses the chaotic behavior property, some switching between variables was done and some transactions were manipulated . Although the chaotic systems are difficult to predict in a long-term, this paper discusses the optimal control problem of the Lorenz-Rössler through some external inputs. It is important to mention that the Lorenz-Rössler which should be studied is that which is presented in and . We use some procedures in the optimal control problem to determine some sets of numbers of parameters and initial conditions that achieve the optimal behavior to the goal state.\n\nThe qualitative changes in the trajectories in the phase space due to the change in one or more control parameters are called bifurcations. The bifurcating study is possible for a one-dimensional system with one parameter. But it is difficult in higher-dimensional cases, especially with several parameters. so there is little research in this area .\n\nIn the next section, we provide a necessary mathematical introduction. The mathematical system of the Lorenz-Rössler model is presented and a brief discussion of the stability of the system is in Section 3. In Section 4, an analytical investigation of some cases of bifurcation is discussed and many diagrams for those cases are presented. In Section 5, the optimal control problem is discussed followed by many digital examples that were made through simulations. The conclusion is presented in Section 6.\n\n2. Mathematical Introduction\n\nIt is known that the optimal control problem requires: A mathematical form for the system to be controlled, description of the constraints, determination of the goal to be accomplished, usually it is an additional boundary condition and determination for the performance measure . Firstly, we consider the simplest example to find the shortest length between two specified locations A = ( t 0 , x 1 ) and B = ( t f , x 2 ) in R2. Consider the case of no constraints on the variables, and for simplicity, we assume that the minimal curve is given as the graph of a smooth function x ( t ) , t ∈ R . The problem now can be illustrated as minimizing the integral which is given by\n\nJ = ∫ A B 1 + ( d x / d t ) 2 d t = ∫ A B 1 + x ˙ 2 d t = ∫ A B g ( t , x ( t ) , x ˙ ( t ) ) d t (2.1)\n\nwhere J is called a functional or the objective function, that must be minimized with respect to t. In a simple case of one dependent variable x ( t ) and no constraints, and conditionally that the optimal curve x ¯ ( t ) exist and unique, the general problem now is to find the optimal curve that minimizes the functional\n\nJ = ∫ t 0 t f g ( t , x ( t ) , x ˙ ( t ) ) d t (2.2)\n\nwhere g is a continuous function in all its variables and has a continuous first and second orders partial derivatives with respect to all its variables. Moreover, t 0 and t f are fixed. If we consider a small variation in the carve, then\n\nx ( t ) = x ¯ ( t ) + ϵ δ ( t )     ∀ t (2.3)\n\nTherefore\n\nx ˙ ( t ) = x ¯ ˙ ( t ) + ϵ δ ˙ ( t ) (2.4)\n\nwhere ϵ is a small parameter and δ ( t ) is an arbitrary real function of t s.t. δ ( t 0 ) = δ ( t f ) = 0 . It is clear that the optimal curve x ¯ ( t ) is a member of the family (2.3) at ϵ = 0 . See Figure 1.\n\nThus, the functional in (2.2) can be written as\n\nJ = ∫ t 0 t f g ( t , x ¯ ( t ) + ϵ δ ( t ) , x ¯ ˙ ( t ) + ϵ δ ˙ ( t ) ) d t (2.5)\n\nThe necessary condition in order to be extremum function is \n\nd J d ϵ | ϵ = 0 = 0 (2.6)\n\nUnder the assumption that x and all its derivatives are continuous, with some mathematical processes and considering that x ( t ) = x ¯ ( t ) and x ˙ ( t ) = x ¯ ˙ ( t ) at ϵ = 0 , the condition (2.6) can be\n\nδ ( t f ) ∂ g ∂ x ˙ | t f − δ ( t 0 ) ∂ g ∂ x ˙ | t 0 + ∫ t 0 t f δ ( t ) ( ∂ g ∂ x − d d t ( ∂ g ∂ x ˙ ) ) d t = 0 (2.7)\n\nbut at ϵ = 0 , δ ( t f ) = δ ( t 0 ) = 0 and δ ( t ) ≠ 0 ∀ t ∈ ] t 0 , t f [ , then\n\n∂ g ∂ x − d d t ( ∂ g ∂ x ˙ ) = 0 , ∀ t ∈ ] t 0 , t f [ , s.t. x ( t 0 ) = x 0 and x ( t f ) = x f (2.8)\n\nEquation (2.8) gives the necessary condition to minimize J and it is known as the Euler-Lagrange (E-L) equation, which associates with the vibrational problem (2.3).\n\nLet us now consider the n first ordinary differential equations (ODEs) as the following forms x ˙ ( t ) = ( x ˙ 1 , ⋯ , x ˙ n ) c , and suppose that the optimal state x ¯ ( t ) = ( x ¯ 1 , ⋯ , x ¯ n ) c exist and unique, that is the vector of n twice differentiable function. In this case, the functional (2.2) can take the following form\n\nJ ( x ) = ∫ t 0 t f g ( x ( t ) , x ˙ ( t ) , t ) d t ,     x ( t 0 ) = x 0 ,     x ( t f ) = x f (2.9)\n\nUnder the same conditions in the case of one dependent variable, the function g ( x ) which makes the integral (2.9) an extremum must satisfy the n simultaneous E-L equations, which are given by\n\n∂ g ∂ x i − d d t ( ∂ g ∂ x ˙ i ) = 0 ,     ∀ t ∈ ] t 0 , t f [ , (2.10)\n\nwith the boundary conditions\n\nx i ( t 0 ) = x i 0 and x i ( t f ) = x i f ,     i = 1 , 2 , ⋯ , n (2.11)\n\nSee and for more details. Now, in the case when there are constraints on the states and control variables. The functional that need to maximize (minimize) takes the following form\n\nJ * ( x , u ) = G ( x f , t f ) + ∫ t 0 t f [ g 0 ( x , u , t ) + λ c ( g − x ˙ ) ] d t (2.12)\n\nwhere G : ℝ n × ℝ → ℝ and g 0 : ℝ n × ℝ m × ℝ → ℝ are real valued functions that can be selected to weight the terminal and transient performance respectively. G ( x f ) can be called the terminal cost, and g 0 can be the instantaneous loss per unit of time. λ c = ( λ 1 , ⋯ , λ n ) is called Lagrange multipliers (L-m) vector, by integrating the term λ c x ˙ in (2.12) we get\n\nJ * = G ( X f , t f ) − λ c x | t 0 t f + ∫ t 0 t f [ H + λ ˙ c x ] d t (2.13)\n\nwhere\n\nH = g 0 + λ c g (2.14)\n\nis called the Hamiltonian function (H.f). Some-times H takes the form\n\nH ( x , u , λ , λ 0 , t ) = λ 0 g 0 ( x , u , t ) + λ c g ( x , u , t ) (2.15)\n\nwhere λ 0 ≥ 0 and one can get λ 0 = 1 for maximization .\n\nTheorem: Assume u * ( . ) is the optimal function that maximizes the objective function J * and x * ( . ) is the corresponding trajectory, then u * must satisfy the following conditions \n\nH ( x * , u * , λ * , λ 0 , t ) ≥ H ( x * , u , λ * , λ 0 , t ) (2.16.1)\n\nλ ˙ j ( t ) = − ∂ H ∂ x j ,     λ j ( t f ) = ∂ G ∂ x j | t = t f ,     ∂ H ∂ u j | u * = 0 ,   ∀ ( u ∈ U , t ∈ [ t 0 , t f ] , j = 1 , 2 , ⋯ , n ) (2.16.2)\n\nThis system consists of 2n nonlinear differential equations with n initial conditions x j ( t 0 ) and n terminal conditions λ j ( t f ) . For more details about this theorem and its proof see and . Notable: An additional equation is required if t f is indeterminate.\n\n3. Lorenz-Rössler Mathematical\n\nThe Lorenz-Rössler system is a three-dimensional system with five parameters. This system is described by the following equations as presented in and \n\nx ˙ 1 = a 1 ( x 2 − x 1 ) − x 2 − x 1 x ˙ 2 = a 2 x 1 − x 2 − 20 x 1 x 3 + x 1 + a 3 x 2 x ˙ 3 = 5 x 1 x 2 − b 1 x 3 + b 2 + x 1 ( x 3 − b 3 ) (3.1)\n\nwhere x 1 , x 2 and x 3 are the state variables of the system, a 1 , a 2 , a 3 , b 1 , b 2 and b 3 are the system parameters. Clearly, the zero-state is not a solution of the system (3.1) because the system is not homogeneous, and with a few mathematical calculations we can be sure that this system has the following possible equilibrium states\n\nE 1 = ( 0 , 0 , b 2 / b 1 ) (3.2)\n\nE 2 = ( 0 , x 2 , b 2 / b 1 ) ,   a 1 = a 3 = 1 (3.3)\n\nE 3 = ( x 1 , f ( x 1 ) , B ) (3.4)\n\nwhere\n\nf ( x 1 ) = A x 1 ,   A = a 1 + 1 a 1 − 1 ,   a 1 ≠ 1 ,   B = 1 20 ( a 2 + 1 + A ( a 3 − 1 ) ) , (3.5)\n\nx 1 = ( − ( B − b 3 ) ± ( ( B − b 3 ) 2 − 20 A ( b 2 − B b 1 ) ) 1 2 ) / 10 A (3.6)\n\nIt is easy to show that the system (3.1) under some conditions, is unstable at least at one of its steady-states, so be it E 1 . The Jacobian matrix W of the model (3.1) is given by\n\nW = ( − a 1 − 1 a 1 − 1 0 a 2 + 1 − 20 x 3 a 3 − 1 − 20 x 1 5 x 2 + x 3 − b 3 5 x 1 x 1 − b 1 ) ,     i , j = 1 , 2 , 3\n\nAnd W valued at stationary state E 1 is given by\n\nW 1 = ( − ( a 1 + 1 ) a 1 − 1 0 a 2 + 1 − 20 b 2 / b 1 a 3 − 1 0 − b 3 + b 2 / b 1 0 − b 1 ) (3.7)\n\nAccording the linear stability analysis and theory of linear differential equations, we strive to find the eigenvalues of W 1 . The determinant equation of W 1 is given by the following equation:\n\n| λ I − W 1 | = ( λ + b 1 ) [ λ 2 + θ 1 λ + θ 2 ] = 0 (3.8)\n\nwhere\n\nθ 1 = 2 + a 1 − a 3 (3.9)\n\nθ 2 = ( 1 + a 1 ) ( 1 − a 3 ) − ( a 1 − 1 ) ( a 2 + 1 − 20 b 2 / b 1 ) (3.10)\n\nIn general, the eigenvalues of W 1 are complex numbers. In this regard, we are not concerned with the values of the solutions of (3.8) but with their signs. Based on the linear stability theory, if there are at least one of the eigenvalues in (3.8) is positive, the equilibrium point E 1 is unstable. So, for the linear part in (3.8), the eigenvalue is λ 1 = − b 1 < 0 , while for the quadratic polynomial part, and according of the Descartes’ rule of the number of the positive real roots of a polynomial, the quadratic polynomial in (3.8) has at least one positive root if θ 1 < 0 , i.e. a 3 > a 1 + 2 . That proof that the Lorenz-Rössler for different values of parameters is unstable at least at E 1 .\n\n4. Bifurcation of Lorenz-Rössler System\n\nIn this section, we discuss the bifurcation phenomenon of the considered system. At the first equilibrium point, E 1 depends upon the following characteristic Equation (3.8), that can be rewritten as the following:\n\n( λ + b 1 ) [ λ 2 + ( 2 + a 1 − a 3 ) λ + 2 − a 1 a 3 − a 3 − a 1 a 2 + a 2 + 20 b 2 ( a 1 − 1 ) / b 1 ] = 0 (4.1)\n\nthen the values of λ are λ 1 = − b 1 , and\n\nλ 2 , 3 = − ( 2 + a 1 − a 3 ) / 2 ± [ ( 2 + a 1 − a 3 ) 2 − 4 { ( 2 − a 1 a 3 − a 3 − a 2 a 1 + a 2 ) + 20 b 2 ( a 1 − 1 ) / b 1 } ] 1 2 / 2 (4.2)\n\nThe bifurcation phenomena arise when one or more of the eigenvalues equal to zero, by analyzing the values of the last two eigenvalues many cases hold:\n\nCase 1: when ( 2 − a 1 a 3 − a 3 − a 2 a 1 + a 2 ) + 20 b 2 ( a 1 − 1 ) / b 1 = 0 , then λ 2 = − ( 2 + a 1 − a 3 ) and λ 3 = 0 , this case is a Saddle-Node bifurcation (SNB).\n\nWe chose the parameter a 3 as a bifurcation parameter, with giving fixed values of the other parameters the bifurcation diagrams can be drawn as in Figure 2.\n\nCase 2: when ( 2 − a 1 a 3 − a 3 − a 2 a 1 + a 2 ) + 20 b 2 ( a 1 − 1 ) / b 1 > 0 and ( 2 + a 1 − a 3 ) = 0 , then\n\nλ 2 , 3 = ∓ i ( − 4 { ( 2 − a 1 a 3 − a 3 − a 2 a 1 + a 2 ) − 20 b 2 ( a 1 − 1 ) / b 1 } ) 1 / 2 / 2\n\nwhere Hopf bifurcation (HB) holds. Choosing b 1 as a bifurcation parameter with fixed values of the other parameters give a picture of this case in the bifurcation diagram as in Figure 3.\n\nNext we chose the parameter a 3 as a bifurcation parameter, with giving fixed values of the other parameters the bifurcation diagrams can be drawn as in Figure 4.\n\n5. Optimal Control Problem in a Period of Time\n\nIn the case of constraints on the control variables, Pontryagin maximum principleis considered as a design tool to get the best possible trajectory for a dynamical system by providing a necessary condition that must hold for an optimum, but not (in general) sufficient conditions . So, the basic problem is to find the controllers that maximize/minimize the functional J in (2.12), which is called a co-state function and considered as a type of L-m. We use the PMP to find the best possible controllers with respect to a choice of specified measure.\n\nThe selected measure can be presented as the following forms:\n\nminmize   ∅ = 1 2 ∫ t 0 T ∑ i = 1 3 ( α i w i 2 + β i u i 2 ) d t (5.1)\n\nSubject to:\n\nThe controlled system of (3.1) that is given by\n\nx ˙ 1 = a 1 ( x 2 − x 1 ) − x 2 − x 1 + e 1 x ˙ 2 = a 2 x 1 − x 2 − 20 x 1 x 3 + x 1 + a 3 x 2 + e 2 x ˙ 3 = 5 x 1 x 2 − b 1 x 3 + b 2 + x 1 ( x 3 − b 3 ) + e 3 (5.2)\n\nAnd the initial and terminal conditions\n\nx i | t 0 = x i 0 ,     x i | T = x ¯ i ,     i = 1 , 2 , 3 (5.3)\n\nwhere:\n\nw i = ( x i − x ¯ i ) and u i = ( e i − e ¯ i ) (5.4)\n\n- α i , β i , i = 1 , 2 , 3 are positive control constants.\n\n- x ¯ is any steady-states of the system as E 1 , E 2 or E 3 that are defined in Equations (3.2)-(3.4).\n\n- e i are the controlling inputs that be determined by the PMP with respect to the optimality measure for the system (3.1) near its stead-states.\n\n- e ¯ i are the optimal control inputs.\n\nThe selected measure or the objective function (5.1) represents the sum of squares of the deviations of x i from their goal levels x ¯ i and deviations of the control inputs e i from their goal levels e ¯ i , ( i = 0 , 1 , 2 ).\n\nNow, our aim is to keep the system states x i , i = 1 , 2 , 3 to their goal levels x ¯ i and the control inputs e i to their goal levels (optimal controllers) e ¯ i over time as close as possible. Let us consider the following an additional variable as a replacement of the cost function (5.1)\n\nx ˙ * ( t ) = 1 2 ∑ i = 1 3 ( α i w i 2 + β i u i 2 ) (5.5)\n\nwith the initial condition x * | t 0 = 0 and the terminal condition x * | T = ∅ .\n\nThen, introduce the co-state variables γ = ( γ 1 , γ 2 , γ 3 , γ * ) c that are related to the state variables of the system (3.1) and the additional state variable (5.5) respectively. Then the H.f takes the following form\n\nH = γ * x ˙ * + ∑ i = 1 3 γ i x ˙ i = ∑ i = 1 3 γ * 2 ( α i w i 2 + β i u i 2 ) + γ 1 [ a 1 ( x 2 − x 1 ) − x 2 − x 1 + e 1 ]     + γ 2 [ a 2 x 1 − x 2 − 20 x 1 x 3 + x 1 + a 3 x 2 + e 2 ]     + γ 3 [ 5 x 1 x 2 − b 1 x 3 + b 2 + x 1 ( x 3 − b 3 ) ] (5.6)\n\nThe Hamiltonian equations are given by:\n\n∂ γ * ∂ t = − ∂ H ∂ x * = 0 (5.7)\n\n∂ γ i ∂ t = − ∂ H ∂ x i ,     i = 1 , 2 , 3\n\nFrom Equation (5.7), clearly γ * is a constant, so for minimization, we can choose γ * = − 1 . Using Equations (5.6) and (5.7) with m = 2 for simplicity we can get the co-state differential equations as\n\nγ ˙ 1 = α 1 w 1 + γ 1 ( a 1 + 1 ) − γ 2 ( 1 + a 2 − 20 x 3 ) − γ 3 ( 5 x 2 + x 3 − b 3 ) (5.8)\n\nγ ˙ 2 = α 2 w 2 − γ 1 ( a 1 − 1 ) − γ 2 ( a 3 − 1 ) − 5 γ 3 x 1 (5.9)\n\nγ ˙ 3 = α 3 w 3 + 20 γ 2 x 1 − γ 3 ( x 1 − b 1 ) (5.10)\n\nFor minimizing the H.f w.r.t e i , ∀ i through the conditions ∂ H / ∂ e i = 0 , we can get\n\ne i = e ¯ i + γ i β i ,     i = 1 , 2 , 3 (5.11)\n\nBy substituting (5.11) in the controlled system in (3.1) with Equations (5.8)-(5.10) we get the following system of seven nonlinear differential equations\n\nx ˙ 1 = a 1 ( x 2 − x 1 ) − x 2 − x 1 + e ¯ 1 + γ 1 β 1\n\nx ˙ 2 = a 2 x 1 − x 2 − 20 x 1 x 3 + x 1 + a 3 x 2 + e ¯ 2 + γ 2 β 2\n\nx ˙ 3 = 5 x 1 x 2 − b 1 x 3 + b 2 + x 1 ( x 3 − b 3 ) + e ¯ 3 + γ 3 β 3\n\nx ˙ * = 1 2 ∑ i = 1 3 ( α i ( x i − x ¯ i ) 2 + ( γ i 2 / β i ) ) (5.12)\n\nγ ˙ 1 = α 1 ( x 1 − x ¯ 1 ) + γ 1 ( a 1 + 1 ) − γ 2 ( 1 + a 2 − 20 x 3 )     − γ 3 ( 5 x 2 + x 3 − b 3 )\n\nγ ˙ 2 = α 2 ( x 2 − x ¯ 2 ) − γ 1 ( a 1 − 1 ) − γ 2 ( a 3 − 1 ) − 5 γ 3 x 1\n\nγ ˙ 3 = α 3 ( x 3 − x ¯ 3 ) + 20 γ 2 x 1 − γ 3 ( x 1 − b 1 )\n\nwith the following boundary conditions: x i | t 0 = x i 0 , x i | T = x ¯ i , γ i | T = 0 , i = 1 , 2 , 3 .\n\n6. Numerical Simulation\n\nIn the following, some numerical solutions of the system in Equations (5.12), that display how the system states converge to the goal state in different cases, and how the co-state variables disappear at the end of time T.\n\n- The optimal control to the stationary state E 1 = ( 0 , 0 , b 2 b 1 = 5 ) is shown in Figure 5.\n\n- The optimal control to the stationary state E 3 = ( x 1 , A x 1 , B ) is shown in Figure 6 where\n\nA = a 1 + 1 a 1 − 1 ,   a 1 ≠ 1 ,     B = ( a 2 + 1 + A ( a 3 − 1 ) ) 20 , and\n\nx 1 = ( − ( B − b 3 ) ± ( B − b 3 ) 2 − 20 A ( b 2 − B b 1 ) 1 2 ) / 10 A\n\nFigure 5 and Figure 6 indicate that, according to the assumed values of the parameters, the optimal controlled state x 1 , x 2 , x 3 converge with time to the assumed goal level 0, 0, 5, respectively in Figure 5 and to 0.450, 0.675, 0.325 respectively in Figure 6. Also, in each case the co-state variables γ 1 , γ 2 , γ 3 disappear with time. The assumed goal levels are represented by the dotted lines. All results indicate the possibility of the optimal control of the Lorenz-Rössler system, and the PMP has shown excellent results in achieving the optimal behavior of the system.\n\n7. Conclusion\n\nMany studies can be implemented on the Lorenz-Rössler model, but in this paper, we have focused on the issues of the bifurcations and the optimal control problem of the system. The bifurcation analysis of the system at the equilibrium state ( 0 , 0 , b 2 / b 1 ) was discussed and it was found that a saddle-node bifurcation and a Hopf bifurcation can be holed under some conditions. Many bifurcation diagrams have verified those cases using examples that are showing graphically for some chosen parameters. The procedure of the Pontryagin Maximum Principle is considered to solve the optimal control problem. The optimal control inputs were analytically derived and it is found that they are functions of the co-state variables which disappear when the system arrives at the ideal state. Analytical methods are used to solve the necessary conditions, while the non-linear differential equations of the optimal controlled system are solved numerically by the math software Maple, and then some illustrative solutions are shown graphically.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest regarding the publication of this paper.\n\nCite this paper\n\nAlwan, S.M., Al-Mahdi, A.M. and Odhah, O.H. (2020) Optimal Control and Bifurcation Issues for Lorenz-Rössler Model. Open Journal of Optimization, 9, 71-85. https://doi.org/10.4236/ojop.2020.93006\n\nNomenclatureReferencesRobertson, R. and Combs, A. (1995) Chaos Theory in Psychology and the Life Sciences. Lawrence Erlbaum Associates, Mahwah.Guan, X.P., Fan, Z.P. and Chen, C.L. 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