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https://www.crazy-numbers.com/en/5129	[ "Discover a lot of information on the number 5129: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!\n\n## Mathematical properties of 5129\n\nIs 5129 a prime number? No\nIs 5129 a perfect number? No\nNumber of divisors 4\nList of dividers 1, 23, 223, 5129\nSum of divisors 5376\nPrime factorization 23 x 223\nPrime factors 23, 223\n\n## How to write / spell 5129 in letters?\n\nIn letters, the number 5129 is written as: Five thousand hundred and twenty-nine. And in other languages? how does it spell?\n\n5129 in other languages\nWrite 5129 in english Five thousand hundred and twenty-nine\nWrite 5129 in french Cinq mille cent vingt-neuf\nWrite 5129 in spanish Cinco mil ciento veintinueve\nWrite 5129 in portuguese Cinco mil cento vinte e nove\n\n## Decomposition of the number 5129\n\nThe number 5129 is composed of:\n\n1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5\n\n1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1\n\n1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2\n\n1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9\n\nOther ways to write 5129\nIn letter Five thousand hundred and twenty-nine\nIn roman numeral MMMMMCXXIX\nIn binary 1010000001001\nIn octal 12011\nIn US dollars USD 5,129.00 (\\$)\nIn euros 5 129,00 EUR (€)\nSome related numbers\nPrevious number 5128\nNext number 5130\nNext prime number 5147\n\n## Mathematical operations\n\nOperations and solutions\n51292 = 10258 The double of 5129 is 10258\n51293 = 15387 The triple of 5129 is 15387\n5129/2 = 2564.5 The half of 5129 is 2564.500000\n5129/3 = 1709.6666666667 The third of 5129 is 1709.666667\n51292 = 26306641 The square of 5129 is 26306641.000000\n51293 = 134926761689 The cube of 5129 is 134926761689.000000\n√5129 = 71.617037079176 The square root of 5129 is 71.617037\nlog(5129) = 8.5426659873893 The natural (Neperian) logarithm of 5129 is 8.542666\nlog10(5129) = 3.7100326990658 The decimal logarithm (base 10) of 5129 is 3.710033\nsin(5129) = 0.93937510808136 The sine of 5129 is 0.939375\ncos(5129) = -0.34289124561168 The cosine of 5129 is -0.342891\ntan(5129) = -2.739571570005 The tangent of 5129 is -2.739572" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6732245,"math_prob":0.98496866,"size":2272,"snap":"2023-40-2023-50","text_gpt3_token_len":765,"char_repetition_ratio":0.170194,"word_repetition_ratio":0.03954802,"special_character_ratio":0.4322183,"punctuation_ratio":0.15384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9923804,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T17:52:28Z\",\"WARC-Record-ID\":\"<urn:uuid:77f50a54-38e9-4e18-a44d-24bf0eeb9403>\",\"Content-Length\":\"33719\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f8eb1a3-655c-45dc-b08f-f444b4b4040e>\",\"WARC-Concurrent-To\":\"<urn:uuid:46aa23c3-305e-4f26-a006-611d7bbefb29>\",\"WARC-IP-Address\":\"128.65.195.174\",\"WARC-Target-URI\":\"https://www.crazy-numbers.com/en/5129\",\"WARC-Payload-Digest\":\"sha1:PKWG7MZ2AZMDF452MZMF4WPW37TCFV7E\",\"WARC-Block-Digest\":\"sha1:ALLWJOKOJ4XMVP6LMMTHEJKS2WMJTUTA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510924.74_warc_CC-MAIN-20231001173415-20231001203415-00448.warc.gz\"}"}
https://proofwiki.org/wiki/Exclusive_Or_is_Negation_of_Biconditional	[ "# Exclusive Or is Negation of Biconditional\n\n## Theorem\n\nExclusive or is equivalent to the negation of the biconditional:\n\n$p \\oplus q \\dashv \\vdash \\neg \\paren {p \\iff q}$\n\n## Proof\n\n $\\displaystyle p \\oplus q$ $\\dashv \\vdash$ $\\displaystyle \\paren {p \\lor q} \\land \\neg \\paren {p \\land q}$ Definition of Exclusive Or $\\displaystyle$ $\\dashv \\vdash$ $\\displaystyle \\neg \\paren {p \\iff q}$ Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction\n\n$\\blacksquare$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7543202,"math_prob":0.9989884,"size":717,"snap":"2019-51-2020-05","text_gpt3_token_len":229,"char_repetition_ratio":0.13464236,"word_repetition_ratio":0.0,"special_character_ratio":0.31241283,"punctuation_ratio":0.171875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9896806,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T10:01:33Z\",\"WARC-Record-ID\":\"<urn:uuid:bf18d138-cf24-49e4-bc45-1d3568ff6c09>\",\"Content-Length\":\"33517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b4b6f484-9ba0-46a2-a148-03a93a4fc15f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed33aa28-9e9f-4a02-a115-cf9651d686fa>\",\"WARC-IP-Address\":\"104.27.168.113\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Exclusive_Or_is_Negation_of_Biconditional\",\"WARC-Payload-Digest\":\"sha1:YGK4PRWIVHHDF32STFCJENVJQ735TQKD\",\"WARC-Block-Digest\":\"sha1:MXQBVEXLFQPQSRIJ55BPOINF4VM3OMGS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251687958.71_warc_CC-MAIN-20200126074227-20200126104227-00445.warc.gz\"}"}
https://www.mechanics-industry.org/articles/meca/full_html/2020/06/mi180087/mi180087.html	[ "Open Access\n Issue Mechanics & Industry Volume 21, Number 6, 2020 608 14 https://doi.org/10.1051/meca/2020083 30 October 2020", null, "This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## 1 Introduction\n\nIn rotating machinery such as motors and mechanical transmission, ball bearings are extensively used. This is often justified by their good performance, associated with the significance of their maximum load. Incidents caused by undetected defects in the operation of this component can have very serious consequences on both safety and maintenance. A number of studies, , have been devoted to the detection of bearing failure and some methods based on acoustic and vibration analyses have been developed to predict the life cycle of a bearing. Abnormal levels of temperature rise remain a very effective way to detect failure in a bearing. Generally, when two surfaces are moving, it appears inevitably a heat generation at the contact. In the literature on bearing problems, two types of heat transfer exchanges are reported: (i) heat conduction between different rolling elements such as the inner ring, the rolling elements, the outer ring and (ii) thermal effects mainly due to different heat fluxes generated and which affect the bearing lifespan.\n\n– Conduction heat transfer between the different rolling elements is based on the principle of thermal contact on the rolling elements. The fundamental aspects of thermal rolling contact were highlighted by Bejan . The rolling elements contacts situated on rings forces the heat flux lines to concentrate on the contact area. This concept of macro constriction was studied by Holm and the one of static constriction was established by Chow and Yovanovich who studied the heat transfer in a given geometry of contacts. Further this theory was performed by Lemczyk and Yovanovich where they introduces the use of the Biot number to characterize the heart exchange. This constriction thermal resistance was generalized to a dynamic resistance constriction model that takes into account the rotation of the elements on each other. Their model is based on the Hertz theory in case of statics and smooth contacts which tends to increase thermal conductance with movement of the elements in contacts.\n\n– Heat generated by the normal operation of a structure has been the object of a number of studies aimed at evaluating the constraints associated with heat flux and as a result predicting fatigue and bearings lifespan. Sliding effects (friction effect) was originally analyzed by Jones and further investigated by Harris . The latter uses the Hertz theory for calculating contact areas between bearing elements, and determines a relation of heat flux generated by friction in the presence of a lubricant. He then uses a nodal model to determine temperature distribution in the rolling elements and subsequently, in the structure. Liao and Lin investigated the case of the overheating of ball bearings at high speed. They determined contact angles and surfaces when the bearing is overheated, i.e. when elements dilate and thus, the performance of ball bearing is evaluated. Böhmer et al. demonstrated the influence of the generated heat flux by friction on the fatigue behavior of a ball bearing. Muzychka and Yovanovitch built their research on Blok's findings to estimate the contact resistance associated with maximum temperature in case of an elliptical contact. They set another thermal resistance formula; nevertheless, they noted that this approach is only valid for a relatively high Peclet number. Baïri et al. investigated the stationary thermal behaviour of a ball bearing and they established a thermal map of the ball bearing, the flash and the average temperature of the heated region. In Reference , Pouly et al. developed a thermal network model to evaluated power loss attributed to rolling friction and sliding friction between balls and cage during the movement. In their model, the resistance between balls and oil, the ball and cage drag as well as hydrodynamic forces are considered. They analysed too the influence of an air-oil mixture in rolling bearing on the thermal behaviour of rolling bearing elements operating at high speed. Yan et al. showed that the presence of the cage affects significantly air-oil flow and heat dissipation within the bearing cavity. Takabi and Khonsari developed a theoretical model to study the temperature of the ball bearing, considering several parameters like heat transfer mechanism, frictional heat generation, preload and thermal response of the system. Neurouth et al. developed two models of grease lubricated thrust ball bearing using thermal network model to predict the power loss and heat transfer with lubricant taking into account different assumptions. The obtained numerical results showed that grease and air-oil mixture do not have any significant role on thermal transfer in the bearing. A third thermal model, which takes into the account rings, balls and contact areas, was proposed. Takabi and Khonsari [20,21] developed a dynamic simulation of rolling bearing thermal failure that considers the thermal expansion. In this case two types of failure were investigated for a deep groove ball bearing rotating at high speed with a large radial load. The results show that the unstable motion of the cage induces a seizure in the bearing. For cylindrical roller bearing used in high speed machine tools, thermal failure is due to thermal preload, which affect lubricant film thickness and causes a thermal seizure. Neurouth et al. proposed a thermo-mechanical model of the FZG test rig based on the thermal network method. The results show that the inner and outer rings of a rolling bearing are at different temperatures. In this study, two approaches were proposed to investigate the thermal behaviour of rolling bearing elements. Yan et al. developed a transient thermal network model of a spindle bearing system, which consider the thermal behaviour and the structure interactions. The temperature rise of the outer ring as a function of rotational speed using, is obtained in steady and transient regime. The obtained results was discussed and compared to the experimental ones. Zheng and Chen proposed a multi-node thermal network model for an angular ball bearing to study the heat generation and the thermal transfer in bearings, while taking into consideration, the presence of the coolant lubricant, and the radial and axial structural constraints.\n\nHowever, it has to be mentioned that no model in the literature accounts for the case of heat flow related to the presence of defect on bearing tracks. Indeed, this impact results in a tearing off of the material attributed to a plastic transformation, which amounts to heat dissipation at this point. In this context, the number of parameters relevant to heat transfer is important, and among these parameters, we can list rotational speed, load, lubrication, various frictions (rolling elements, rings, cages) and bearing failure.\n\nThus, the objective of this work is to determine the temperature field generated not only by the heating of the bearing by friction alone but also by the effect of the presence of a defect on the bearing (in our case the defect is located on the outer ring). The proposed model allows obtaining a direct correlation between the severity of the defect and temperature variation of the structure studied via a comprehensive equation that links all the parameters of the system. The model permits to determine the severity of the damage via thermal measurements.\n\nIn this work, the thermal model allows the determination of the temperature field in the presence of a defect. The final resolution is based on the nodal method.\n\n## 2 Thermal modeling\n\nIn this study, we present a heat transfer model to simulate, in a steady state regime, the temperature distribution in the ball bearings, with and without defect, operating at moderate speed. The system as represented in Figure 1 is divided into isothermal elements designated as nodes, which are interconnected by their thermal resistances. These resistances are generally dependent on a heat transfer mechanisms in the bearing: conduction, free convection, forced convection and radiation. However, the contribution of radiation heat transfer is assumed very small with respect to fluid convection and conduction between elements. Such as that sketched in Figures 1 and 2, the modeled system is made up of a house, inner ring, outer ring, balls, cage and shaft. A small amount of grease is used as a lubricant between the balls and the cage and the system is surrounded by air (Fig. 2). We have used 21 nodes in our system. The nodes T11, T12 and T13 are symmetrical to the nodes T14, T15 and T16.", null, "Fig. 1Node distribution in the ball bearing.", null, "Fig. 2Thermal network elements.\n\n### 2.1 Nodal distribution\n\nThe discretization of the system components, ball bearing, shaft and housing is shown in Figures 1 and 2. On the shaft are placed three volumetric nodes: (i) at the free end of the shaft, (ii) in middle, (iii) at the other end of the shaft. This modelling, coupled with that of the bearing obtained in a previous section, results in a 21 × 21 square system, whose resolution gives the temperature field in the bearing.\n\nThe housing, shaft, grease and air temperatures are considered as input parameters and determined from experimental data. Load direction is tested to investigate the effect of the loaded part on temperature rise of the system. Different loads are then used in a series of experiments to investigate the effect of rotation speed on temperature rise.\n\nThe formulae used to calculate thermal resistances are given below.\n\n### 2.2 Thermal resistance of conduction\n\nThe outer and inner rings are considered as two cylinders and the thermal resistance is given by:", null, "(1)\n\nwith K is the thermal conductivity", null, "### 2.3 Thermal resistance of constriction heat transfer\n\nAs explained in the introduction, because of the constriction phenomenon, the source of heat generated by friction is concentrated on a very small contact area. As a result, and following this model, one node is placed on the surface whereas another is in the bulk , and then connected by a constriction resistance as detailed by Muzychka and Yovanovitch . Thermal constriction resistance is:", null, "(2)\n\n### 2.4 Heat exchange coefficient by convection heat transfer\n\nConvection heat transfer between the ball bearing and the surrounding air or between the ball bearing and the lubricant is difficult to quantify. For each forced convection process, thermal resistance is estimated after the average Nusselt number was determined. The expression of convection thermal resistance is given by the following expression:", null, "(3)\n\nWhere s: exchange heat surface [m2]\n\nNu: Nusselt number\n\nhv: Convection heat transfer coefficient [w/m2.K]\n\n### 2.5 Natural convection heat transfer between the outer ring and the ambient atmosphere\n\nIn the case where the heat flux is generated by convection, we use Newton law given by:", null, "In stable air conditions, the convection heat transfer coefficient between the ball bearing surface and the atmosphere as estimated by Jeng and Huang is given by:", null, "(4)\n\nTa: the air temperature.\n\nPalmgren proposed for the outer ring surface as follows:", null, "(5)\n\nDh: diameter of the outer ring of the ball bearing\n\nWh: width of the outer ring of ball bearing\n\nFrom Equations (4) and (5) it follows that the heat flux is given by:", null, "### 2.6 Forced convection heat transfer between the inner ring and compressed air flow\n\nThe rotation of the inner ring in contact with the outside atmosphere induces a forced convection heat transfer with the surrounding air. For a disc in rotation, Wagner proposed the average Nusselt number as follows:", null, "(6)with Pr = 0.71 for the air.\n\n### 2.7 Convection heat transfer in presence of grease\n\nFor use of a small amount of grease as a lubricant Using a small amount of grease as a lubricant, Neurouth et al. evaluated the convection heat transfer coefficient with Nusselt number expressions as follows:", null, "(7)With", null, "(8)", null, "(9)", null, "", null, "", null, "## 3 Heat Generation\n\nThermal and mechanical processes in a motion system, especially in bearings systems are coupled and all generated heat must be taken into account. This part deals with the mechanical effects that can generate heat.\n\nIn our case, two sources of heat generation are considered and the first, which is the most common, corresponds to the heat generated by frictional effects: inner ring/ball friction (Q1) and outer ring/ball friction (Q2). The second is due to impact of balls on the defect seen on the bearing, which has not been studied in other works: heat generated by the impact of ball on a defect seen on the bearing (Q3); hence, the originality of this presented work.\n\nTo calculate power losses due to frictions at the different points of contact: inner ring/ball and outer ring/ball, it is necessary to carry out kinetic calculations, which allows to obtain the velocity and forces to be integrated in the model. The grease is used to avoid dry friction between the elements, in our case, we assume that the grease has no influence on friction power loss and the slip between ball bearings and the raceway is negligible.\n\n### 3.1 Bearing frictional power loss\n\nThe bearing frictional power loss is expressed by the product of speed by the friction torque as detailed by Harris .", null, "(10)\n\nThe friction torque includes, torque due to applied load M1, viscous friction torque Mν and torque due to the flange Mf, Palamgren proposed a formula to calculate the viscous friction torque as follows:", null, "(11)", null, "(12)", null, "(13)\n\nIn this study, we have Mf = 0.\n\n### 3.2 Heat source generated by defect\n\nA second large family of heat sources can be considered in bearings with defect, attributable to the impacts between the rolling elements and the singularity generated on the raceway. This type of distribution necessitates first of all a modeling of the defect, which is represented by a hole in the outer ring material (Fig. 3). The singularities are the source of impact during the passage of the rolling elements, and the repeated impacts exert a pull on the material, thereby resulting in a plastic deformation.\n\nThe force exerted by a rolling element Fm on the singularity can be expressed in different ways, depending on the assumptions. The most common form is the one proposed by Tandon and Choudhury : the shape of the excitation signal is considered to be a combination of three signals having triangular, rectangular and semi sinusoidal forms. Bogard et al. have shown that the proposed three forms of the signal separately give similar results. Thereby, we consider the form of the signal to be triangular (Fig. 4).\n\nIn this case, the defect model is based on these assumptions:\n\n• When a ball passes the location of a defect in outer ring, the exciting forces expressed by the amplitude A, is precisely equal to the forces acting on the ball.\n\n• The rotational speed of the center of the ball is constant.\n\n• The sliding phenomenon is ignored.\n\n• The rebound phenomenon due to the shock is ignored.\n\nFor the static problem, the rotational speed of the center of the ball is expressed as:", null, "(14)\n\na: the half radius of the defect in the rolling direction,\n\nΔTs: The shock period\n\nD1: the distance between the middle of two balls\n\nTd0: the periodicity of the passage of rolling elements upon the outer ring defect.\n\nThe shock period is given by:", null, "(15)", null, "(16)\n\nTherefore", null, "(17)\n\nThe defect situated on the outer ring of the ball bearing is revealed by its characteristic frequency, which is equal to the inverse of the characteristic period given as:", null, "(18)\n\nEquation (17) is substituted in Equation (18) and the shock period can thus be written as:", null, "(19)\n\nTherefore, the power of the shock is expressed as follows:", null, "(20)\n\nFrom Equations (19) and (20), the power of the shock is stated as:", null, "(21)\n\nAccording to the previous assumptions, the external excitation force or the pulse amplitude generated each time the rolling element strikes the defect on the raceway of the outer ring, is equal to the forces acting on the ball.\n\nReplacing Fm by Qmax in Equation (21) we obtained:", null, "(22)\n\nUsing Steinbeck's formula for ball bearings under pure radial load and nominal diametrical clearance, defined as:", null, "(23)", null, "(24)\n\nThe incident heat flux density is expressed as:", null, "(25)\n\nWith", null, "(26)\n\nThen from Equations (24)(26), we obtained:", null, "(27)", null, "Fig. 3Representation of the impact region in ball bearing.", null, "Fig. 4Form of the excitation signal.\n\n## 4 Numerical resolution\n\nThe first law of thermodynamics for a closed system is expressed by this equation:", null, "(28)\n\nwhere Q represents the heat transfer rate expressed in (W) taken as positive when it enters in the system, W is the work transfer rate expressed in (J/s) and u is the internal energy variation with time, t, in (J/s).\n\nChangenet et al. rely on the following analysis: the work rate is substituted by the transferred heat from the system to the other elements and the rate of internal energy is substituted by the rate of absorbs heat (thermal inertia). Then, the equation (28) can be written as in referense :", null, "(29)", null, "(30)", null, "(31)", null, "(32)\n\nEquation (30) is the basis used to solve the studied problem. It was applied for each node of the thermal network, i, whereas j represents each node connected to i. In the steady-state case we have:", null, "(33)\n\nHence, the equation can be stated as:", null, "(34)\n\nFor all nodes we obtain a system of nonlinear equations, which is difficult to solve. Each node, i, generates a balance sheet of heat flux φij giving us the system of equations which is non-linear. In this case we solve this system using the Newton-Raphson method :", null, "(35)\n\nFor a series of nonlinear functions qi, the Newton-Raphson method permits to have a system of simultaneous linear Equation (36). These equations can be solved for an error ϵj.", null, "(36)", null, "(37)\n\nThis non-linear equation can be solved by iteration. After initializing the parameters, each iteration is translated into a linear system equation (36) whose unknowns are the components of the error vector ϵj between the corresponding temperatures of two successive iterations. The resolution is stopped when errors ϵj are reasonably low.\n\nIn our case the system is composed by 21 nodes and we want to simulate the temperature evolutions in the ball bearing in steady state regime depending on the radial load, and the rotational velocity in different configurations. All thermal resistance values are calculated by given formulae developed before.\n\nAll heat dissipation sources are Q1 corresponding to the node T3 (Inner ring/ball contacts), Q2 corresponding to the node T5 (outer ring/ball contacts) and Q3 defect temperature. This considered values are considered as inputs and calculated by the use of formulae developed in last section.\n\nA computer program on Matlab software language was written to solve Equation (36) based on the theoretical model in steady state regime described in the last section. The temperature rise in the nodes and the heat generations under different operating parameters is calculated. Figure 5 illustrates the flow chart of the program to calculate the temperature distributions and heat generation in the ball bearing.", null, "Fig. 5Flow-chart for the algorithm of the resolution procedure.\n\n## 5 Nodal distribution\n\nFor the resolution we use the nodal method. The nodal model of the ball bearing consists of 21 nodes as shown in Table 1 and Figure 2. Two nodes are used to illustrate heat distribution in the loaded part of the outer ring (T6) as well as in the unloaded part (T9).\n\nThe discretization of the system components, ball bearing, shaft and housing is shown in Figure 8. On the shaft, are placed three volumetric nodes: (i) at the free end of the shaft, (ii) in middle, (iii) at the other end of the shaft. This modeling, coupled with that of the bearing obtained in a previous section, results in a 21 × 21 square system, whose resolution gives the temperature field in the bearing.\n\nTable 1\n\nDescription of the nodes.\n\n## 6 Nodal distribution of generated heat flux due to the defect\n\nEnergy, which evolves over time attributed to impacts of the rolling elements on the defect, and takes the form of heat flux, must be considered in the balance of heat energy in nodal modelling. Therefore, these connection nodes are added to a portion of the heat flux generated by successive impacts, as shown in Figure 6.\n\nThese energies are distributed over connections of nodes between rolling elements and the defective ring. This distribution is proportional to the load distribution in the ball bearing. Consequently, the energy injected into each node depends not only on the number of nodes but also on their selected location during nodal discretization (see Fig. 7).", null, "Fig. 6Energetic contribution of the impact on defect.", null, "Fig. 7Schematic of energy intake of the impact.\n\n## 7 Experimental investigations\n\nThe study is focused on a rotating machine consisting on the use of ball bearings, type SKF 6206 to quantify energy intake linked to bearing defect localized on the outer rice. The test was carried out in two stages with healthy bearing, and then with a defect bearing of 17 mm2 of surface (Fig. 8). Lubrication is achieved by greasing of the two bearings, the bearings and grease properties are defined in Table 2.", null, "Fig. 8Image of a defect on the ball bearing.\nTable 2\n\nFeatures of ball bearing and its grease.\n\n### 7.1 Experimental setup\n\nA schematic illustration of the experimental setup is presented in Figure 9. An electric motor of 10 kW at variable speeds rotates the assembly system. The load carried by the hydraulic cylinder (Fig. 10) is considered to be radial.\n\nFor thermal measurements we need to know the value of emissivity, for that, the visible faces of the bearing are painted in black. The defective bearing is mounted in such a way that the defect is on the loaded portion to facilitate the visualization of the temperature rise. For thermal measurements, an infrared camera (CEDIP Titanium, 640 × 512 InSb detectors, with a pitch of 15 μm) is used to realize thermal maps of the outer ring, simultaneously submitted to rotation and radial loading. The used of 50 mm lens allows to obtain a spatial resolution compatible with the size of the region of interest, but also to be able to register simultaneously the temperature variations far from the system. This external ring is covered with black paint (emissivity equal to 0.92) in order to standardize and increase its uniformity, the stainless steel is well known to have a very low emissivity. Then, the adherence of the paint has to be regularly checked, because a local abruption would immediately change the measured signal in significant proportions. Due to the short distance between the object and the camera, the atmosphere is assumed to be transparent. However, a particular attention is paid to both uniformity and stability of the environment: choice of the device orientation relatively to the thermal sources, screening of the thermal parasites, and simultaneous recording of thermal reference points taken in the field of the camera. First of all, images are recorded at very low speed, until the thermal equilibrium is reached. This first step shows that the time constant of such system is of several tenths of minutes, due to the thermal inertia induced by the high mass of the metallic mechanical device. Once the thermal equilibrium is reached, a thermal sequence of 300 images is registered at 10 Hz, and a mean image is computed in order to reduce the noise. Then, a small area is placed on the infrared image, at the aplomb of the defect. The thermal experimental values given in the present paper correspond to the average of this area. An example of thermogram obtained by infrared thermography is presented on Figure 11.", null, "Fig. 9Schematic illustration of the experimental setup.", null, "Fig. 10Experimental setup image.", null, "Fig. 11Experimental Infrared thermogram obtained by infrared camera CEDIP.\n\n## 8 Results and discussions\n\n### 8.1 Results of bearing without defect\n\nThis study examines the influence of rotational speed variations on the temperature distributions of a ball bearing. The obtained results from the simulation of increased temperatures in case of a ball bearing without defect, where the radial load is set at 500 N are shown in Figure 12. The temperature of different nodes of bearing increases when the rotational speed varies from 300 rpm to 700 rpm and the maximum value is measured at contact point located between the ball and the inner ring (node T3). The direction of the applied radial load affects strongly the temperature rise materialized by nodes T2, T4 and then T5. This increase of heat flux in the contact zone is due, first, to the stress applied by the load and, second, to the phenomenon of constriction of the heat flux. In Figure 13, the temperature of the loaded part increases greatly whereas the unloaded part exhibits no such behavior; this shows the effect of the applied load direction on the temperature of the bearing parts. Figure 14 compares the generated heat (at node T5) obtained from both the numerical and experimental results. Both results show that the heat generated by node T5 increases with the increase of the rotation speed and exhibit the same trend.\n\nFigure 15 presents the predicted results of the increase of the temperature as a function of the variation of the radial load. The increase of temperature is found to be proportional to the increase of the applied radial load in a non-linear relationship with a maximum value measured at contact point between the ball and the inner ring (node T3). The direction of the applied radial load affects strongly the increase in temperature of the bearing as shown in Figure 16. In Figure 17 we present a comparison between the numerical and the experimental results of heat generated at (node T5) of ball bearing without a defect as it depends on the variation of the load. As we can see, the result show that the two curves have a similar pattern when the radial load increases, although a very small difference caused by the parasite temperature of the surrounding system can be observed.", null, "Fig. 12Rotational speed effect on temperature rise of the bearing.", null, "Fig. 13Temperature evolution with rotational speed of the loaded and unloaded parts.", null, "Fig. 14Numerical and experimental comparison of temperature of node T5 with variation of the rotational speed.", null, "Fig. 15Radial load effect on temperature rise.", null, "Fig. 16Influence of the radial load variations on the temperature rises of the loaded and unloaded parts.", null, "Fig. 17Comparison between numerical and experimental results of temperature rise of node T5 as a function of radial load.\n\n### 8.2 Ball bearing with defect\n\nDamaged ball bearing refers to a ball bearing with a defect of 17 mm2 located on its outer ring. As illustrated in Figure 18, we present the evolution of the temperature of node T5 obtained by the numerical and the experimental measurements realized by the infrared camera. The results are given under radial load equal to 500 N. Figure 18 shows that both curves have the same pattern; the temperature increases with the increase of rotational speed. With a varying of the applied radial load where the force maintained in a fixed direction, both numerical and experimental with a defect are compared. In Figure 19 we show that as the radial load increases the temperature in the bearing increases as well. Both curves have the same evolution and both results demonstrate the validation of the model to predict temperature increases with a localized defect on the bearing.\n\nIn Figure 20, we present a comparison of the temperature evolutions in between the loaded and unloaded parts for a ball bearing with a defect. We notices that the increase in rotation speed affects considerably the temperature rise in the loaded part. It can be noted that the presence of a defect in the bearing contributes considerably to this increase. The same results are observed in Figure 21, where the application of a radial load on the defective bearing results in a drastic temperature rise compared to that of the bearing without defect.\n\nIn Figure 22, we compare the results obtained from experimental tests: two ball bearings with and without defect. Both temperature curves increase with increasing rotational speed and exhibit the same pattern. Moreover, in a defective ball bearing, the temperature rise is considerably higher than that in a bearing without defect, which shows that our theory is founded. Another comparison of temperature evolution obtained from experimental tests on the two ball bearings, with and without defect, depends on the radial load as shown in Figure 23. Indeed, it can be seen that with increasing radial load, the dissipated heat increases as well. At high load value, the effect of the thermal response on the dynamic system is significant.", null, "Fig. 18Comparison between numerical and experimental results of temperature rise of node T5 as a function of rotational speed for the damaged ball bearing.", null, "Fig. 19Comparison between numerical and experimental results of temperature rise of node T5 as a function of radial load for the damaged ball bearing.", null, "Fig. 20Comparison of temperature evolution as a function of rotational speed between the loaded and unloaded parts for the damaged ball bearing.", null, "Fig. 21Temperature evolution as a function of radial load between the loaded and unloaded parts for the damaged ball bearing.", null, "Fig. 22Experimental temperature rise of node T5 results as a function of rotational speed of a damaged and healthy ball bearing.", null, "Fig. 23Comparison of the temperature rise of node T5 as a function of radial load between the damaged and healthy ball bearings.\n\n## 9 Conclusion\n\nA well-maintained production equipment is necessary for both safety and production concerns and requires accurate and effective damage detection techniques and procedures. This study reported on the thermal analysis of defected or not defected bearings. A model validated by an experiment was developed to predict heat distribution in the bearings and quantify the severity of the defect or damage through temperature increases. The model is based on the nodal theory and incorporates a defect on the outer ring of the bearing. The experiment was performed on a housing made of two ball bearings and resulted in a temperature of the damaged bearing higher than that of the good one. In addition, the temperature at the loaded part of the ball bearing was higher than that of the unloaded part. These experimental results concur with the model's ones. The model quantified the severity of defects appearing on the bearings with a good efficiency depending on the conditions of use of the system. As we have seen, the developed model also allows to analyses the influence of operating conditions on a ball bearing without defect.\n\n## Nomenclature\n\na : Half-elliptical radius in the rolling direction (m)\n\nA : Exchange surface (m2)\n\nb : Half-elliptical radius in the direction transverse to the rolling (m)\n\nCp : Specific heat (J/kg.K)\n\nd : Diameter of the rolling element (m)\n\nD : Diameter of the bearing ball (m)\n\nDh : Diameter of the outer ring or housing of the ball bearing (m)\n\nf0 : The factor depending upon type of bearing and lubrication system (dimensionless)\n\nFm : The force exerted by a rolling element on the singularity (N)\n\nhv : Convective heat transfer coefficient (W/m2.°C)\n\nK : Thermal conductivity (W/m.°C)\n\nL : Bearing length (m)\n\ndext : Outer ring diameter (m)\n\ndint : Inner ring diameter (m)\n\nMfrot : Torque acting on the ball bearing (N.m)\n\nMl : Friction torque due to applied load (N.m)\n\nMv : Viscous friction torque (N.m)\n\nMf : Friction torque due to flange load (N.m)\n\nFβ : Dynamical equivalent load (N)\n\nQroul : Bearing frictional power loss (W)\n\nωroul : Rotational speed (rev/min)\n\nν : Kinematic viscosity of lubricant (m2/s)\n\nα : Thermal diffusivity (m2/s)\n\nPe: Peclet number (dimensionless)\n\nNu : Nusselt number (dimensionless)\n\nRthk: Thermal resistance of conduction\n\nRth : Thermal resistance of constriction (°C/W)\n\nRthV: Thermal resistance of convection\n\nT : Temperature (°C)\n\nTa : Ambient Temperature (°C)\n\nS : Area of outer ring (m2)\n\nWh : Width of the outer ring of the ball bearing (m)\n\nVc : Relative velocity of the cage (m/s)\n\nV : Relative velocity of contact (m/s)\n\nγair : Kinematic viscosity of air (m2/s)\n\nχ : Thermal effusivity ‎(J.K−1.m−2.s−1/2)\n\nRe: Reynolds number (dimensionless)\n\nPr: Prandtl number (dimensionless)\n\nfl : Factor depending upon bearing design (dimensionless)\n\ndm : Pitch diameter (m)\n\nVcb : Normal velocity of the ball (m/s)\n\nΔTs : Shock period (s)\n\nTdo : Periodicity of the passage of rolling elements upon the outer ring defect (s)\n\nD1 : Distance between the middle of two balls (m)\n\nrm : Mean radius of the bearing (m)\n\nZ : Number of rolling elements (dimensionless)\n\nfab : The passing frequency of rolling element on outer ring defect (s−1)\n\nfr : The rotational frequency of the inner ring (s−1)\n\nPs : Power of shock (W)\n\nSd : Surface defect (m2)\n\nt : Time (s)\n\nQ : Generated heat rate (W)\n\nU : Internal energy (J)\n\nZ : Number of rolling elements (balls)\n\nRij : Thermal resistance between nodes i and j (°C/W)\n\nSij : Thermal conductance between nodes i and j (W/°C)\n\nmi : Mass at node i (kg)\n\nCp,i : Specific heat of node i (J/kg.°C)\n\nQi : Heat transfer rate by the node i (W)\n\nφij : Energy flows exchanged between two nodes i and j (W)\n\nqi : Heat flux (W/m2)\n\nεj : Error vector (°C)\n\nn : Total number of nodes (dimensionless)\n\nΨdo : Incident surface flux (W/m2)\n\nα : Contact angle\n\nΨj : The energy of node j (W)\n\n## References\n\n1. 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In the text", null, "Fig. 5Flow-chart for the algorithm of the resolution procedure. In the text", null, "Fig. 6Energetic contribution of the impact on defect. In the text", null, "Fig. 7Schematic of energy intake of the impact. In the text", null, "Fig. 8Image of a defect on the ball bearing. In the text", null, "Fig. 9Schematic illustration of the experimental setup. In the text", null, "Fig. 10Experimental setup image. In the text", null, "Fig. 11Experimental Infrared thermogram obtained by infrared camera CEDIP. In the text", null, "Fig. 12Rotational speed effect on temperature rise of the bearing. In the text", null, "Fig. 13Temperature evolution with rotational speed of the loaded and unloaded parts. In the text", null, "Fig. 14Numerical and experimental comparison of temperature of node T5 with variation of the rotational speed. In the text", null, "Fig. 15Radial load effect on temperature rise. In the text", null, "Fig. 16Influence of the radial load variations on the temperature rises of the loaded and unloaded parts. In the text", null, "Fig. 17Comparison between numerical and experimental results of temperature rise of node T5 as a function of radial load. In the text", null, "Fig. 18Comparison between numerical and experimental results of temperature rise of node T5 as a function of rotational speed for the damaged ball bearing. In the text", null, "Fig. 19Comparison between numerical and experimental results of temperature rise of node T5 as a function of radial load for the damaged ball bearing. In the text", null, "Fig. 20Comparison of temperature evolution as a function of rotational speed between the loaded and unloaded parts for the damaged ball bearing. In the text", null, "Fig. 21Temperature evolution as a function of radial load between the loaded and unloaded parts for the damaged ball bearing. In the text", null, "Fig. 22Experimental temperature rise of node T5 results as a function of rotational speed of a damaged and healthy ball bearing. In the text", null, "Fig. 23Comparison of the temperature rise of node T5 as a function of radial load between the damaged and healthy ball bearings. In the text\n\nCurrent usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.\n\nData correspond to usage on the plateform after 2015. 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https://economics.stackexchange.com/questions/33730/notation-inverse-of-demand-function	[ "# Notation? Inverse of demand function\n\nI'm looking through The Investment Decisions of Firms by S.J. Nickell, and I've come across some notation that I don't quite understand. Any clarification would be very welcome.\n\nNickell first assumes that firms face a downward sloping demand curve:\n\n$$z\\{p(t)\\}\\beta(t) \\textrm{ where } z_p<0.$$ He says:\n\nThe demand curve itself thus has a constant shape defined by $$z\\{p(t)\\}$$ and is shifted up and down by the time function $$\\beta(t)$$.\n\nSo $$z$$ is some kind of functional form whose first derivative in $$p$$ is negative.\n\nNow comes the part that confuses me. He assumes that firms adjust output according to demand.\n\nAs a result we have $$z\\{p(t)\\}\\beta(t) = F\\{K(t),L(t)\\}.$$ Solving for the output price we obtain $$p(t) = p[F\\{K(t), L(t)\\}/\\beta(t)]$$\n\nHow did he get this? $$F$$ equals quantity demanded. Fine, it's the assumption. But why is $$p$$ the inverse of $$z$$?\n\nThen he derives these partial derivatives, which are also confusing me because he brings $$z$$ back in:\n\n$$\\frac{\\partial p}{\\partial K} = \\frac{F_K}{z_p \\beta(t)} \\quad\\textrm{and}\\quad \\frac{\\partial p}{\\partial L} = \\frac{F_L}{z_p \\beta(t)}.$$\n\nLike I said, any help is most welcome. I think what's mostly confusing me is the inverse function bit. I can see how he gets the partial derivatives otherwise.\n\nThanks.\n\n• Is it possible that it is just a typo? – Walrasian Auctioneer Jan 26 at 20:34\n\nAs a result we have $$z\\{p(t)\\}\\beta(t) = F\\{K(t),L(t)\\}.$$ Solving for the output price we obtain $$p(t) = p[F\\{K(t), L(t)\\}/\\beta(t)]$$\nI am pretty sure that what is meant here is that there is a $$p(t)$$ trajectory for which $$z\\{p(t)\\} = F\\{K(t),L(t)\\} / \\beta(t)$$ (see first equation)\nGiven the function $$z$$, this depends only on the right hand side of the equation, hence let us denote the trajectory $$p(t)$$ as a function of this, i.e. $$p(t) = p[F\\{K(t), L(t)\\}/\\beta(t)]$$ A different notation would be $$p(t) = z^{-1}\\left( F\\{K(t), L(t)\\}/\\beta(t) \\right).$$ This second notation is useful because all the partial derivatives follow from this notation and the chain rule." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9316643,"math_prob":0.9999758,"size":1281,"snap":"2020-45-2020-50","text_gpt3_token_len":359,"char_repetition_ratio":0.10963195,"word_repetition_ratio":0.0,"special_character_ratio":0.28883684,"punctuation_ratio":0.10769231,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000087,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T02:15:42Z\",\"WARC-Record-ID\":\"<urn:uuid:229c0fbf-dbc5-4499-b705-d1dbbb5d0b14>\",\"Content-Length\":\"148320\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1376055-f7a6-4c75-9e62-7d54e2ddedb4>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f194a7e-6a04-4aba-a2a5-df86ec7b11f4>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://economics.stackexchange.com/questions/33730/notation-inverse-of-demand-function\",\"WARC-Payload-Digest\":\"sha1:D32VE36IWZ4HKZWSZH3A4JE5QL4AIPMO\",\"WARC-Block-Digest\":\"sha1:YFG6WXYFUKQOORBHWKHP7HTAFHLPLZZZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141180636.17_warc_CC-MAIN-20201125012933-20201125042933-00574.warc.gz\"}"}
https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/statug_nlmixed_details30.htm	[ "### Covariance Matrix\n\nThe estimated covariance matrix of the parameter estimates is computed as the inverse Hessian matrix, and for unconstrained problems it should be positive definite. If the final parameter estimates are subjected to", null, "active linear inequality constraints, the formulas of the covariance matrices are modified similar to Gallant (1987) and Cramer (1986, p. 38) and additionally generalized for applications with singular matrices.\n\nThere are several steps available that enable you to tune the rank calculations of the covariance matrix.\n\n1. You can use the ASINGULAR=, MSINGULAR=, and VSINGULAR= options to set three singularity criteria for the inversion of the Hessian matrix", null, ". The singularity criterion used for the inversion is", null, "where", null, "is the diagonal pivot of the matrix", null, ", and ASING, VSING, and MSING are the specified values of the ASINGULAR=, VSINGULAR=, and MSINGULAR= options, respectively. The default values are as follows:\n\n• ASING: the square root of the smallest positive double-precision value\n\n• MSING: 1E–12 if you do not specify the SINGHESS= option and", null, "otherwise, where", null, "is the machine precision\n\n• VSING: 1E–8 if you do not specify the SINGHESS= option and the value of SINGHESS otherwise\n\nNote that, in many cases, a normalized matrix", null, "is decomposed, and the singularity criteria are modified correspondingly.\n\n2. If the matrix", null, "is found to be singular in the first step, a generalized inverse is computed. Depending on the G4= option, either a generalized inverse satisfying all four Moore-Penrose conditions is computed (a", null, "-inverse) or a generalized inverse satisfying only two Moore-Penrose conditions is computed (a", null, "-inverse, Pringle and Rayner 1971). If the number of parameters n of the application is less than or equal to G4=i, a", null, "-inverse is computed; otherwise, only a", null, "-inverse is computed. The", null, "-inverse is computed by the (computationally very expensive but numerically stable) eigenvalue decomposition, and the", null, "-inverse is computed by Gauss transformation. The", null, "-inverse is computed using the eigenvalue decomposition", null, ", where", null, "is the orthogonal matrix of eigenvectors and", null, "is the diagonal matrix of eigenvalues,", null, ". The", null, "-inverse of", null, "is set to", null, "where the diagonal matrix", null, "is defined using the COVSING= option:", null, "If you do not specify the COVSING= option, the nr smallest eigenvalues are set to zero, where nr is the number of rank deficiencies found in the first step.\n\nFor optimization techniques that do not use second-order derivatives, the covariance matrix is computed using finite-difference approximations of the derivatives." ]	[ null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0352.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0048.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0353.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0354.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0048.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0355.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0044.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0356.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0048.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0357.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0358.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0357.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0358.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0357.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0358.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0357.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0359.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0313.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0360.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0361.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0357.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0048.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0362.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0363.png", null, "https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/images/statug_nlmixed0364.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8503861,"math_prob":0.9755809,"size":2261,"snap":"2021-43-2021-49","text_gpt3_token_len":470,"char_repetition_ratio":0.15551618,"word_repetition_ratio":0.0058479533,"special_character_ratio":0.19239275,"punctuation_ratio":0.10796915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99885136,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,1,null,5,null,1,null,1,null,5,null,1,null,3,null,1,null,5,null,5,null,3,null,5,null,3,null,5,null,3,null,5,null,1,null,1,null,1,null,1,null,5,null,5,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T21:41:20Z\",\"WARC-Record-ID\":\"<urn:uuid:4b982f15-e8aa-4a08-bd76-99a1e4d8272a>\",\"Content-Length\":\"23457\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f05495ff-3a5a-4aac-b432-1db2228453b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:15cab5df-49f4-44fb-a751-c1ed1ec435e9>\",\"WARC-IP-Address\":\"104.68.247.134\",\"WARC-Target-URI\":\"https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/statug_nlmixed_details30.htm\",\"WARC-Payload-Digest\":\"sha1:QSZ6SHHM5Y7IWAXFUN375EFWHXECHDXA\",\"WARC-Block-Digest\":\"sha1:3SN5SSIRARNQLEIREZH74POHNVMB2BOY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585768.3_warc_CC-MAIN-20211023193319-20211023223319-00204.warc.gz\"}"}
https://www.numbers.education/450.html	[ "Is 450 a prime number? What are the divisors of 450?\n\n## Parity of 450\n\n450 is an even number, because it is evenly divisible by 2: 450 / 2 = 225.\n\nFind out more:\n\n## Is 450 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 450 is about 21.213.\n\nThus, the square root of 450 is not an integer, and therefore 450 is not a square number.\n\n## What is the square number of 450?\n\nThe square of a number (here 450) is the result of the product of this number (450) by itself (i.e., 450 × 450); the square of 450 is sometimes called \"raising 450 to the power 2\", or \"450 squared\".\n\nThe square of 450 is 202 500 because 450 × 450 = 4502 = 202 500.\n\nAs a consequence, 450 is the square root of 202 500.\n\n## Number of digits of 450\n\n450 is a number with 3 digits.\n\n## What are the multiples of 450?\n\nThe multiples of 450 are all integers evenly divisible by 450, that is all numbers such that the remainder of the division by 450 is zero. There are infinitely many multiples of 450. The smallest multiples of 450 are:\n\n## How to determine whether an integer is a prime number?\n\nTo determine the primality of a number, several algorithms can be used. The most naive technique is to test all divisors strictly smaller to the number of which we want to determine the primality (here 450). First, we can eliminate all even numbers greater than 2 (and hence 4, 6, 8…). Then, we can stop this check when we reach the square root of the number of which we want to determine the primality (here the square root is about 21.213). Historically, the sieve of Eratosthenes (dating from the Greek mathematics) implements this technique in a relatively efficient manner.\n\nMore modern techniques include the sieve of Atkin, probabilistic algorithms, and the cyclotomic AKS test.\n\n## Numbers near 450\n\n• Preceding numbers: …448, 449\n• Following numbers: 451, 452\n\n### Nearest numbers from 450\n\n• Preceding prime number: 449\n• Following prime number: 457\nFind out whether some integer is a prime number" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92728156,"math_prob":0.9982569,"size":1817,"snap":"2020-24-2020-29","text_gpt3_token_len":612,"char_repetition_ratio":0.25261998,"word_repetition_ratio":0.33333334,"special_character_ratio":0.43423226,"punctuation_ratio":0.1402715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991121,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-11T05:18:19Z\",\"WARC-Record-ID\":\"<urn:uuid:d54d4fa8-ec02-443a-a9ef-fb096d3adcab>\",\"Content-Length\":\"20221\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b43377dc-cdab-42b4-949a-660844d63f21>\",\"WARC-Concurrent-To\":\"<urn:uuid:009da1f0-2fca-4b87-8e0e-452305700844>\",\"WARC-IP-Address\":\"213.186.33.19\",\"WARC-Target-URI\":\"https://www.numbers.education/450.html\",\"WARC-Payload-Digest\":\"sha1:73ME2Y6IUNIRMJKZA4YF4DICOFIMFMBJ\",\"WARC-Block-Digest\":\"sha1:EJIWPF5JQ4BO6QTUWDHULXAF4DEM6UZF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655921988.66_warc_CC-MAIN-20200711032932-20200711062932-00452.warc.gz\"}"}
http://www.toontricks.com/2019/06/tutorial-is-there-more-elegant.html	[ "# Tutorial :Is there a more elegant replacement for this MATLAB loop?", null, "### Question:\n\nI want to get better at vectorizing my loops in MATLAB. At the moment, I'm trying to count the occurrences of values in a list of ints. My code is similar to this:\n\n``list = [1 2 2 3 1 3 2 2 2 1 5]; occurrence_list = zeros(1,max(list)); for x=list occurrence_list(x) = occurrence_list(x) + 1; end ``\n\nIs there a simple vectorized replacement for that for loop? (Or is there a built in MATLAB function that I'm missing?) I'm doing this on pretty small data sets, so time isn't an issue. I just want to improve my MATLAB coding style.\n\n### Solution:1\n\nIn addition to the HIST/HISTC functions, you can use the ACCUMARRAY to count occurrence (as well as a number of other aggregation operations)\n\n``counts = accumarray(list(:), 1) %# same as: accumarray(list(:), ones(size(list(:))), [], @sum) ``\n\nAnother way is to use TABULATE from the Statistics Toolbox (returns value,count,frequency):\n\n``t = tabulate(list) t = 1 3 27.273 2 5 45.455 3 2 18.182 4 0 0 5 1 9.0909 ``\n\nNote that in cases where the values don't start at 1m or in case there's large gaps between the min and the max, you will get a lot of zeros in-between the counts. Instead use:\n\n``list = [3 11 12 12 13 11 13 12 12 12 11 15]; v = unique(list); table = [v ; histc(list,v)]' table = 3 1 11 3 12 5 13 2 15 1 ``\n\nrepresenting the unique values and their counts (this will only list values with at least one occurrence)\n\n### Solution:2\n\nYou can use the hist function for that. Specify an output, and force the bins to be integers over the range of your array.\n\n``list = [1 2 2 3 1 3 2 2 2 1 5]; bins = min(list):max(list); counts = hist(list,bins); ``\n\n### Solution:3\n\nSo, this is basically a histogram. Off my memory - look for the HIST function.\n\nNote:If u also have question or solution just comment us below or mail us on [email protected]\nPrevious\nNext Post »" ]	[ null, "https://4.bp.blogspot.com/-te3nq62xJRQ/WIzbeIW_hiI/AAAAAAAAAaU/T9jgUxascMELRPlYee11mkSEA3KFhImRQCLcB/s1600/lucky%2Brathore%2B%2Buser%2Basked%2BQuestion_logo.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84170187,"math_prob":0.9488338,"size":1784,"snap":"2022-27-2022-33","text_gpt3_token_len":531,"char_repetition_ratio":0.11011236,"word_repetition_ratio":0.055045873,"special_character_ratio":0.31838566,"punctuation_ratio":0.1325,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9909735,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T16:14:29Z\",\"WARC-Record-ID\":\"<urn:uuid:32e5ea4b-c487-4cec-b1fa-2b4a8165f7a2>\",\"Content-Length\":\"175291\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e7427a8-751e-423e-bb25-fdcef123aa6a>\",\"WARC-Concurrent-To\":\"<urn:uuid:384b00cd-cb18-4249-aecc-888dcddb6504>\",\"WARC-IP-Address\":\"142.251.111.121\",\"WARC-Target-URI\":\"http://www.toontricks.com/2019/06/tutorial-is-there-more-elegant.html\",\"WARC-Payload-Digest\":\"sha1:YEEQUVH6AUG4MWNSYKVGBN5BZND4MQUK\",\"WARC-Block-Digest\":\"sha1:XWHWAN65PK3NF47YAO4TCHYDQCRVQ2NG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572192.79_warc_CC-MAIN-20220815145459-20220815175459-00764.warc.gz\"}"}
https://www.alverno.edu/assessment/initialassessments/cplinitialassessmentforql156.php	[ "# Mathematical Connections CPL Assessment\n\n###### Mathematical Connections (QL 156) CPL Assessment\n\nThe Assessment and Outreach Center will contact students who are eligible to complete the QL 156 CPL assessment. The assessment consists of three sections: representing data, predicting data, and measurement concepts. Each part consists of word problem applications.\n\n###### Representing Data\n• Using descriptive statistics (mean, median, mode) to represent a set of data\n• Using graphs (bar graph, line graph, circle graph, histogram, frequency polygon) to represent a set of data\n###### Predicting Data\n• Reading and interpreting the normal curve to make predictions\n• Using definition of probability and odds to make predictions\n• Using compound probability to make predictions\n• Using counting concepts (counting principal, permutation, combination) to make predictions\n###### Measurement Concepts\n• Converting measures within and between the American and Metric measurement systems (approximate conversion of common units of length, volume, mass and temperature)\n• Using Pythagorean Theorem to determine unknown lengths\n• Determining geometric quantities of perimeter, area, surface area and volume (2D and 3D figures based on rectangles, triangles and circles)\n• Connecting concepts of volume, capacity and mass\n###### Sample CPL Assessment\n\nBelow is a link to a sample assessment. The sample is designed to reflect the length and level of expectation of the assessment. Geometric shapes, measurement units, and data presentations will vary on the actual assessment.\n\nQL 156 Practice Assessment\n\nQL 156 Practice Assessment w/solutions" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9234926,"math_prob":0.9031521,"size":1110,"snap":"2019-51-2020-05","text_gpt3_token_len":234,"char_repetition_ratio":0.17359856,"word_repetition_ratio":0.0,"special_character_ratio":0.21081081,"punctuation_ratio":0.11219512,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9888892,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T18:44:44Z\",\"WARC-Record-ID\":\"<urn:uuid:745d6a75-d4fa-4fdc-8578-ce3a773e80f1>\",\"Content-Length\":\"24652\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4da7509e-8965-4d5c-87c5-edee7264c688>\",\"WARC-Concurrent-To\":\"<urn:uuid:1044cabc-2458-4ed5-86de-04a2053b132a>\",\"WARC-IP-Address\":\"159.135.38.231\",\"WARC-Target-URI\":\"https://www.alverno.edu/assessment/initialassessments/cplinitialassessmentforql156.php\",\"WARC-Payload-Digest\":\"sha1:U45RITHH6NQSPIOHBB4PAVKIACEUMHZO\",\"WARC-Block-Digest\":\"sha1:IXIRVH3RKSO7U2IMHSQCIBT45SM2CTMZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541309137.92_warc_CC-MAIN-20191215173718-20191215201718-00233.warc.gz\"}"}
https://www.shaalaa.com/search-important-solutions/university-of-mumbai-be-mechanical-engineering-semester-2-fe-first-year_29?subjects=applied-physics-2_72	[ "# BE Mechanical Engineering Semester 2 (FE First Year) - University of Mumbai Important Questions for Applied Physics 2\n\nSubjects\nTopics\nSubjects\nPopular subjects\nTopics\nApplied Physics 2\n< prev 1 to 20 of 80 next >\n\nWhy the Newton’s rings are circular and centre of interference pattern (reflected) is dark?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Newton’S Rings\n\nWhat is Rayleigh’s criterion of resolution? Define resolving power of grating?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Resolving Power of a Grating\n\nWhat will be the fringe pattern if wedge shaped air film is illuminated with white light?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Wedge Shaped Film(Angle of Wedge and Thickness Measurement)\n\nObtain the condition for maxima and minima of the light reflected from a thin transparent film of uniform thickness. Why is the visibility of the fringe much higher in the reflected system than in the transmitted system.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Interference in Thin Film of Constant Thickness Due to Reflected and Transmitted Light\n\nExplain the experimental method to determine the wavelength of spectral lines using grating. A diffraction grating has 5000 lines / cm and the total ruled with is 5cm. Calculate dispersion for a wavelength of 5000 A0 in the second order.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Application of Diffraction > Determination of Wavelength of Light with a Plane Transmission Grating\n\nA wedge shaped air film having angle 40 seconds is illuminated by monochromatic light. Fringes are observed vertically through a microscope. The distance between 10 consecutive dark fringes is 1.2 cm. find the wavelength of monochromatic light used.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Wedge Shaped Film(Angle of Wedge and Thickness Measurement)\n\nDescribe in detail the concept of anti-reflecting film with a proper ray diagram.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Anti-reflecting Films and Highly Reflecting Film\n\nWhy the Newton’s rings are circular and fringes in wedge shaped film are straight?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Newton’S Rings\n\nWhat is grating and grating element?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Diffraction Grating\n\nWhat is meant by thin film? Comment on the colours in thin film in sunlight.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Interference in Thin Film of Constant Thickness Due to Reflected and Transmitted Light\n\nDerive the conditions of maxima and minima due to interference of light transmitted from thin film of uniform thickness.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Interference in Thin Film of Constant Thickness Due to Reflected and Transmitted Light\n\nExplain the experimental method to determine the wavelength of spectral line using diffraction grating.\nWhat is the highest order spectrum which can be seen with monochromatic light of wavelength 6000 Ao by means of a diffraction grating with 5000 lines/cm?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Diffraction Grating\n\nTwo optically plane glass strips of length 10 cm are placed one over the other.A thin foil of thickness 0.01 mm is introduced between them at one end to form an air film.\nIf the light used has wavelength 5900 Ao,find the separation between consecutive bright fringes.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Diffraction of Light\n\nWith Newton’s ring experiment explain how to determine the refractive index of liquid.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Newton’S Rings\n\nExplain how interference in wedge shaped film is used to test optical flatness of given glass plate.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Wedge Shaped Film(Angle of Wedge and Thickness Measurement)\n\nWhat is diffraction grating? What is the advantage of increasing the number of lines in the grating?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Diffraction Grating\n\nIn Newtons’s ring experiment,what will be the order of dark ring which will have double the diameter of 40th dark ring?\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Newton’S Rings\n\nDerive the conditions of maxima and minima due to interference of light reflected from thin film of uniform thickness.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Interference in Thin Film of Constant Thickness Due to Reflected and Transmitted Light\n\nDiscuss the Fraunhoffer diffraction at single slit and obtain the condition for minima.\nIn plane transmission grating the angle of diffraction for second order principal maxima for wavelength 5 x 10-5 is 35o. Calculate number of lines/cm for this diffraction grating.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Diffraction Grating\n\nA wedged shaped air film having an angle of 40 seconds is illuminated by monochromatic light and fringes are observed vertically through a microscope. The distance measured between consecutive fringes is 0.12 cm. Calculate wavelength of light used.\n\nAppears in 1 question paper\nChapter: Interference and Diffraction of Light\nConcept: Wedge Shaped Film(Angle of Wedge and Thickness Measurement)\n< prev 1 to 20 of 80 next >" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87851655,"math_prob":0.49180865,"size":4047,"snap":"2023-14-2023-23","text_gpt3_token_len":861,"char_repetition_ratio":0.14123176,"word_repetition_ratio":0.21290323,"special_character_ratio":0.18680504,"punctuation_ratio":0.086280055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789318,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T18:59:26Z\",\"WARC-Record-ID\":\"<urn:uuid:f725ad36-5879-496b-8a89-96effd3164a7>\",\"Content-Length\":\"92296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5650599-68b2-4c48-8ce2-59d53580aebf>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c213b9e-96b8-48d7-a1d0-162afd7ae667>\",\"WARC-IP-Address\":\"172.67.75.172\",\"WARC-Target-URI\":\"https://www.shaalaa.com/search-important-solutions/university-of-mumbai-be-mechanical-engineering-semester-2-fe-first-year_29?subjects=applied-physics-2_72\",\"WARC-Payload-Digest\":\"sha1:G2KIFPJWG3RFCF4VQQMB7UFS5UQKRR25\",\"WARC-Block-Digest\":\"sha1:APVRNJQY24A5X6JLR7QNQO6S6LARU5TR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649302.35_warc_CC-MAIN-20230603165228-20230603195228-00317.warc.gz\"}"}
https://cs.stackexchange.com/questions/51362/are-there-any-problems-that-get-easier-as-they-increase-in-size/51386	[ "# Are there any problems that get easier as they increase in size?\n\nThis may be a ridiculous question, but is it possible to have a problem that actually gets easier as the inputs grow in size? I doubt any practical problems are like this, but maybe we can invent a degenerate problem that has this property. For instance, perhaps it begins to \"solve itself\" as it gets larger, or behaves in some other bizarre way.\n\n• One real problem with this property that comes to mind is unsalted password hash cracking when it is formulated like “given n hashes, crack at least one hash”. Since cracking speed would scale linearly with n, running time would be proportional to 1/n – except that we can't actually assign a definitive time since cracking is stochastic and does not have a constant upper bound on time. – amon Jan 2 '16 at 9:09\n• @amon The running time doesn't scale like $1/n$. It takes time $n$ just to read the $n$ hashes you've been given as input! – David Richerby Jan 2 '16 at 9:33\n• Do you mean easier in absolute or relative terms? Which cost measures do you permit? Do you require strictly decreasing cost, or is non-increasing (from some point on) enough? – Raphael Jan 2 '16 at 10:25\n• @DavidRicherby In this example, it is legitimate to ignore the cost of reading the input as long as I don't make any statement about the absolute cost. Instead, the speed increases linearly with the input. Therefore, n•T(1)>T(n) even when considering the cost of reading the input. I.e. for this problem it is easier to solve a large input at once rather than splitting up the input, even though the problem is divisible. I am not saying that T(n)>T(n+1) for all n. – amon Jan 2 '16 at 10:48\n• To everybody who wants to post yet another answer of the form, \"Some problem where the input is a question plus a big bunch of hints about the answer\": This does not work. The hardest inputs of length $n$ are the ones where you use all $n$ bits to ask the question and give no hints. The fact that it's easy to deal with short questions with a lot of hints does not mean the worst-case running time is good. – David Richerby Jan 2 '16 at 15:51\n\nNo, it's not possible: at least, not in an asymptotic sense, where you require the problem to keep getting strictly easier, forever, as $n \\to \\infty$.\n\nLet $T(n)$ be the best possible running time for solving such a problem, where $n$ is the size of the input. Note that the running time is a count of the number of instructions executed by the algorithm, so it has to be a non-negative integer. In other words, $T(n) \\in \\mathbb{N}$ for all $n$. Now if we consider a function $T: \\mathbb{N} \\to \\mathbb{N}$, we see there is no such function that is strictly monotonically decreasing. (Whatever $T(0)$ is, it has to be finite, say $T(0)=c$; but then since $T$ is monotonically strictly decreasing, $T(c) \\le 0$ and $T(c+1) \\le -1$, which is impossible.) For similar reasons, there is no function that is asymptotically strictly decreasing: we can similarly prove that there's no running time function $T(n)$ where there exists $n_0$ such that for all $n \\ge n_0$, $T(n)$ is monotonically strictly decreasing (any such function would have to become eventually negative).\n\nSo, such a problem cannot exist, for the simple reason that running times have to be non-negative integers.\n\nNote that this answer covers only deterministic algorithms (i.e., worst-case running time). It doesn't rule out the possibility of randomized algorithms whose expected running time is strictly monotonically decreasing, forever. I don't know whether it's possible for such an algorithm to exist. I thank Beni Cherniavsky-Paskin for this observation.\n\n• This is a nice proof, but I disagree with the premise. Instead of asking for a strictly monotonically decreasing runtime, the question might more reasonably be requiring a function where there exists an a, b with a<b so that T(a)>T(b), i.e. its non-strictly monotonically decreasing. Then, it's of course possible to find suitable integer functions. But why integers? I was under the impression that running time denoted a time, not an instruction count (except of course for Turing machines), and that the T expression could use the non-integer operations like log() or non-integer exponents. – amon Jan 2 '16 at 8:56\n• @amon \"running time denoted a time, not an instruction count\" Absolutely not. Running time is always an instruction count. Anything else would be impossible to reason about as it would depend on infeasibly many implementation details. – David Richerby Jan 2 '16 at 9:29\n• As vague as the question is, I don't see how it excludes a cost function of, say, $T(n) = n^2 \\cdot (1+\\epsilon)^{-n} + n$. Now, $T(n) \\sim n$ but $T(n) \\approx n^2$ for \"small\" $n$, so the problem \"gets easier\", relatively speaking. (Absolute costs grow asymptotically, of course). – Raphael Jan 2 '16 at 10:25\n• @Raphael, $T(n) \\sim n$ is not a problem getting easier: $T(n)$ gets larger as $n$ gets larger, so the problem gets harder as $n$ gets larger, once $n$ is large enough. I did state in the first sentence of my answer that no problem can keep getting easier forever. Of course, a problem can get easier for a little while ($T(n)$ can be decreasing for $n \\le c$, say), but but it can't keep getting easier forever. – D.W. Jan 2 '16 at 22:56\n• Even with integer times, for a randomized algorithm the expected time (or any other measure of the distribution) can be fractional, and could gradually approach some constant from above. [This doesn't mean such problems actually exist, only that the \"no such function $T$ exists\" argument is insufficient.] – Beni Cherniavsky-Paskin Jan 5 '16 at 14:23\n\nAlthough it's not quite an answer to your question, the Boyer-Moore string search algorithm comes close. As Robert Moore says on his web page about the algorithm,\n\nOur algorithm has the peculiar property that, roughly speaking, the longer the pattern is, the faster the algorithm goes.\n\nIn other words, generally speaking the algorithm searches for an instance of a target string in a source string and for a fixed source string, the longer the target string is, the faster the algorithm runs.\n\n• Arguably, the pattern isn't the size of the problem but instead the length of the string being searched is. As in David Richerby's comment above, I would argue that the length of the pattern is more of a hint about how to solve the problem (got about searching the string) than the problem itself (seeing if a pattern matches a string of a particular length.) – Kevin Jan 3 '16 at 2:08\n• @Kevin The statement suggests that searching a pattern of length $\\sqrt{n}$ in a length $n$ text is faster than searching a pattern of length $\\log n$. Looking at such fixed-relation inputs (i.e. pairs of strings), I think Rick has given a suitable answer to the question (if not in the classical, asymptotic sense). – Raphael Jan 4 '16 at 9:50\n\nClearly, from a pure mathematical, purely CS algorithm viewpoint this is impossible. But in fact there are several real-world examples of when scaling up your project makes it easier, many which are not intuitive to end-users.\n\nDirections: the longer your directions get, they can sometimes get easier. For example, if I want Google Maps to give me directions for going west 3000 miles, I could drive to the West Coast -- and would get cross-country driving instructions. But if I wanted to go 6000 miles west, I would end up with significantly simpler instructions: get on a plane from NYC to Hokkaido. Giving me a cross-country route that incorporates traffic, roads, weather, etc. is rather difficult algorithmically, but telling me to get on a plane and looking up flights in a database is comparatively significantly simpler. ASCII graph of difficulty vs distance:\n\n \| /\n\| /\nDifficulty \| / ____-------\n\| / ____----\n\| / ____----\n---------------------------------\nDistance\n\n\nRendering: say I want a render of one face and a rendering of 1000 faces; this is for a billboard ad so both final images must be 10000px by 5000px. Rendering one face realistically would be hard -- at the resolution of several thousand pixels across you have to use really powerful machines -- but for the crowd of 1000 faces each face need only be ten pixels across, and can easily be cloned! I could probably render 1000 faces on my laptop, but rendering a realistic face 10000px across would take a very long time and powerful machines. ASCII graph of difficulty vs. objects rendered, showing how difficulty of rendering n objects to an image of a set size drops off quickly but then returns slowly:\n\n \| -\n\|- - _________\nDifficulty \| -- ______-------\n\| ------\n\|\n---------------------------------\nObjects\n\n\nHardware control: many things with hardware get much easier. \"Move motor X 1 degree\" is hard and/or impossible, and you have to deal with all kinds of things that you wouldn't have to deal with for \"move motor X 322 degrees\".\n\nShort duration tasks: Say you want item X to be on for (very small amount of time) every second. By increasing the amount of time that X runs, you will need less complex software as well as hardware.\n\n• In your \"directions\" example, please state exactly what is the computational problem and what is the instance. It's not at all clear to me that your 6k miles example is a larger instance or just an example of an easy part of something (e.g., if I give you a large graph connected graph plus one isolated vertex, asking for shortest paths in general is \"difficult\" but asking for a shortest path from the isolated vertex to anywhere is trivial). Again, for your rendering example, what is the actual computational problem? What is the instance against which you're measuring complexity? – David Richerby Jan 3 '16 at 9:24\n• The rendering example doesn't seem to be instances of the same problem: the first one is rendering a single image; the second one is rendering many small images and then pasting multiple copies of those images into some area. – David Richerby Jan 3 '16 at 9:26\n• I think wrt to travel the parameters would be the name of the 2 cities and n would be the number of characters to encode them. – emory Jan 5 '16 at 4:40\n\nThere are cases. They are the cases where the success criteria is a function of the data, rather than trying to find a single answer. For example, statistical processes whose results are phrased with confidence intervals can become easier.\n\nOne particular case I'm thinking of is problems which have a transition from discrete behaviors to continuous behaviors, like fluid flows. Solving the small problem to within a degree of error can involve modeling all of the discrete interactions, which may call for a supercomputer. The continuous behaviors often permit simplifications without yielding results outside of a related error bound.\n\nThe question is interesting and USEFUL, because our philosophy in informatics is to solve problems the more we read the more dificult is. But, in fact, the MOST of the problems that are presented in the typical way (difficult) can be easily represented in the \"easy\" way; even knowing the response of D.W (which is in a wrong considering that easy does not mean faster, means \"less slow\"; so you do not have to find negative times, you hace to find asymptotic time).\n\nThe trick to find one is putting the part of the solution like hints as an entry, and considering the entry of the problem like a constant parameter.\n\nExample: What is the longest way in car between London and Paris avoiding to visit twice a French and a British town and not visiting other country? considerin, you have to go to Birmingham before Ashford, Orleans before Versailles, La Rochelle before Limoge, etc...\n\nIt is clear that this problem with long entries will be easier that with short ones.\n\nExample of use: Imagine a playgame managed by the machine, and the IA of the computer have to determine if he has to explore more in the play to find more hints or else, if now is time to deduce what is the best decision to assume.\n\n• Your example doesn't work. Instances that are large because they have so many hints that the hints determine a linear order of the graph's vertices are indeed easy. However, instances that are large because they give a large graph with almost no hints are just as hard as the ordinary Hamiltonian path problem. Thus, the worst-case running time of any algorithm that solves this problem is going to be at least as bad as the worst-case running time of the best algorithm for Hamiltonian path, which doesn't seem to be \"super easy\". – David Richerby Jan 2 '16 at 9:26\n• @David, your response is completely incorrect: 1. The entry is not a graph: the large graph is a PARAMETER. So the hamiltonian problem is converted to a constant (very large, but a constant). 2. The entry is the solution of the problem, so: if greater, you are offering a combinatorial explossion of hints. An entry of one hint gives a help, two hints the double, three hints will be close to 4 twices..., because you are eliminating possible solutions. So, this was not a Hamiltonian, this is a solution from an especific graph and the problem is WHAT to do with parts of the solutions. – Juan Manuel Dato Jan 2 '16 at 12:26\n• I think your argument is interesting as larger instances are \"easier\" in some sense, but I think the answer to the original question is ultimately \"no\". Since the graph is finite, there are thus only finitely many possible hints. Therefore every instance can be solved in constant time (for instance, using a lookup table). Even though larger instances are (intuitively) easier in the (asymptotic) computer science view all instances are equally hard (solvable in constant time). – Tom van der Zanden Jan 2 '16 at 13:01\n• @Tom, I agree your consideration about the complexity will be constant, but the problem is how we are accepting the new hints: if with our philosophy of calculating the long entry is not better than a short entry, then we have to change our philosophy - because that is a fact: long entries implies easier problems. So we cannot work in that way... I would recommend my book, but I have no reputation... – Juan Manuel Dato Jan 2 '16 at 13:26\n• @TomvanderZanden Actually, it can't be solved in constant time because the list of \"hints\" might contain duplicates. It requires something like time $n\\log n$ to deduplicate before you can do the table look-up. So, guess what? Solving bigger instances is still harder. – David Richerby Jan 2 '16 at 13:37\n\nConsider a program that takes as input what you know about a password and then tries to crack it. I think this does what you want. For example:\n\n• No input-> Brute force crack over all symbols and a word of any length\n• Length of password -> Brute force all symbols in a word of that length\n• Contained symbols -> Shrinks list of symbols to check\n• ...\n• Contained Symbols including multiple occurrences and length -> Only compute permutations\n• All symbols in correct order -> basically solved itself\n\nI should add that this is a trick, since the problem stated like this is inverse to to input size. You could leave out one layer of abstraction and say that the input size is large for no input (check all symbols and lengths of words) and small if you enter the correct password in the beginning.\n\nSo it all comes down on how much abstraction you allow.\n\n• This doesn't work. First, it's not a well-defined computational problem: suppose your input is just \"The password has 8 characters\" -- that doesn't uniquely determine any password. Second, suppose you make it well-defined by including a hash and asking for any password that matches the hash. In this case, the hardest instances are the ones of the form \"The hash is: [$b$-bit hash]\" with no extra information. These instances don't get easier to solve as the input length increases: they get exponentially harder. – David Richerby Jan 2 '16 at 15:21\n\nAs a matter of fact, I do have a problem that gets smaller as the data increases. One of my application records attributes of a particular product, say cheese. Attributes are for instance CheeseType, Brand, Country, Area, MilkType, etc. Every month or so, I get a list of new cheeses that came into the market during that time, along with their attributes. Now, these attributes are typed by hand by a group of humans. Some make typos, or just don't know the value for all attributes.\n\nWhen you make a search in my database, I try to predict from statistics what the cheese tastes like, based on these attributes. What happens, is that for each attribute, I end up with a range of values ; some are valid some are invalid. Eliminating or correcting these invalid ones is only possible if I have enough data. It's about making the difference between real values and noise, without eliminating rare but valid values.\n\nAs you can imagine, with low volume, the noise is too important to fix things properly. If you have 5 instances of Cheddar, 1 of Brie , 1 of Bri, and 1 of Chedar, how do I tell which is correct and which is a typo? With more volume, the typos tend to keep very low, but the rare values get a few crucial increments, making them escape from the noise (backed by experience). In this case, I could imagine 50000 Cheddar, 3000 Brie, 5 Bri, 15 Chedar, for instance.\n\nSo yes, some problems solve themselves eventually, when you have enough data.\n\n• This fails for the usual reason. A large input might be one in which people tell you about many different types of cheese, rather than one in which they tell you about a few kinds of cheese but some of them mis-spell it. Also, it's not clear that \"easier\" is supposed to be interpreted as \"allow higher confidence in the result\". – David Richerby Jan 3 '16 at 23:38\n• This is a real-life problem (I have had it twice already), that cannot be solved with low amount of data. It can, and it even gets easier to tell apart, good values from wrong ones, as volume gets high. It has the merit of answering the question \"Are there any problems that get easier as they increase in size?\" It doesn't matter how many types of cheeses come up, eventually, with enough volume, they will have more \"hits\" than the typos. This is cs.stackexchange, not maths, so problems are different, and solving them sometimes is simply about having higher confidence in the results. – chris Jan 4 '16 at 18:56\n• Isn't this also kind of the premise of the TV show Numbers? Or at least some episodes - I know I specifically recall one scene where the maths guy remarks that the algorithm he's using to solve the problem at hand gets more effective with a larger dataset. – Dan Henderson Jan 4 '16 at 19:02\n• \"Gets more effective\" != \"gets easier\". – David Richerby Jan 4 '16 at 19:06\n\nConsider the NP-complete problem 3-SAT. If you keep augmenting the problem by providing inputs of the form x_i = true/false, you either end up converting the individual disjunctions into two-variable clauses, thereby creating a 2-SAT problem which is decidedly P, or you simply end up getting a true/false answer.\n\nFor the case where there is redundancy in the x_i = true/false inputs (same input provided a lot of times, or contradictory inputs) you can easily sort the inputs and either ignore the redundant values, or report an error if the values contradict.\n\nIn any case, I think this represents a 'realistic' problem that gets easier to solve as the number of inputs grows. The 'easier' aspect is in converting an NP-complete problem to a P problem. You can still game the system by providing ridiculous inputs such that just the sorting would take longer than brute forcing the problem.\n\nNow, a really cool scenario would be if we are willing to accept T(0) (utilizing D.W.'s notation in the answer above) can be infinite. For example, T(0) could be equivalent to solving Turing's Halting Problem. If we could devise a problem such that adding more inputs converts it into a solvable problem, we have struck gold. Note that it is not enough to convert it into a solvable problem asymptotically - because that is just as bad as brute forcing the problem.\n\n• These particular inputs get easier. However, when you consider all possible inputs 3SAT in general gets significantly harder as you add more clauses: the hard inputs are the ones without these \"hint\" clauses. If you're disallowing general inputs, you need to state exactly what inputs you're allowing. – David Richerby Jan 3 '16 at 9:18\n• First off: we are in agreement that adding more inputs can increase the running time. I say essentially the same thing above. Second, I clearly say we are taking an existing 3-SAT and adding only inputs of the form x_i = true/false. I think this is clear enough and I don't need to make any further clarifications. I think you are taking the trouble to form the most misconstrued interpretation of what I have written. Please don't trouble yourself. – v vv cvvcv Jan 3 '16 at 10:51\n• No, seriously. What computational problem are you solving? A computational problem is deciding the membership of a set of strings (let's say a set of formulas to avoid annoyances about coding). What is the set of formulas for which you're claiming that deciding whether a long formula is in the set is easier than deciding that a short formula is in the set? As soon as you try to make this precise, I'm pretty sure your claim will fall apart. – David Richerby Jan 3 '16 at 11:26\n• Can you please make explicit your understanding of 'my claim'? As soon as you try to make this precise, I'm pretty sure you will stop wasting internet bandwidth. – v vv cvvcv Jan 4 '16 at 12:05\n• I'm a computer scientist, not a mind-reader. Making your claim precise is your job, not mine. – David Richerby Jan 4 '16 at 12:20\n\nThe question asks: \"is it possible to have a problem that actually gets easier as the inputs grow in size?\" What if the inputs are resources to be used by the algorithm to work on a job. It is common knowledge that the more the resources the better. Below is an example, in which the more there are employees the better.\n\n1) You are given two inputs:\ni) The number of employees in an industry. This is a natural number and is the main input $n$.\nii) Information about the industry. There are $t$ tasks to be done by the workers, labeled A, B, C, D... There are $p$ places that connect the tasks enabling the employees to switch between tasks. They are labeled 0, 1, 2, 3... The information given is a simple directed graph made up of routes, for example: A-1, 1-2, 2-B, C-3... For simplicity each route has a cost of 1.\n\n2) Problem:\nStarting from the first tasks A, and with $n$ employees, you are to find an optimal strategy for the employees to use in order to visit all the tasks. Tasks can be visited more than once. Recall that you cannot switch between tasks unless you find a path between them. The path from task A to B is independent of that from B to A.\n\n3) Output:\nThe output is the paths between tasks to be taken by the employees. Each path is associated with the number of employees taking it. For example:\n\nA to B with $n1$ employees (a forward path)\nA to C with $n2$ employees (a forward path)\nB to D with $n3$ employees (a forward path)\nD to B with $n4$ employees (a reversed path)\nB to E with $n5$ employees (a forward path)\n\n4) Possible solution:\nOne possible solution is to first compute the shortest path to the closest nodes from A. This will be a forward path. Then recursively compute the forward path for each visited tasks. The result is a tree. For example:\n\n A\nB C\nD E\n\n\nNow is the time to determine how the employees are going to traverse the tree so to visit the tasks. Starting from task A with $n$ employees, $n1$ are sent to the left subtree and $n2$ are sent to the right subtree. If $n2 \\neq 0$ then none from the left subtree will ever need to go to the right subtree.\n\nFor $n=\\infty$ the employees will all just move forward. For $n = 1$ the employee will have to use reverse paths in order to visit other tasks. For D to B for example the algorithm will compute that shortest path. This is extra computation. Why not directly compute a shortest path from D to E?! Fine, hopefully that is still extra computation.\n\nBut of course computation time will not decrease infinitely (by the way for $n$ too large it will lead to poor resource management and stuff).\n\n• Thank you for sharing your thoughts. Normally in computer science an algorithm is understood to accept a sequence of bits as its input, and output another sequence of bits. With that standard understanding, I don't see how this answer can make sense. If you have a different notion of algorithm in mind, I think it would help if you edited the question to describe what you mean by algorithm (since it sounds like you're not using the term in a way that corresponds to standard usage of the term, as I understand it). – D.W. Jan 2 '16 at 7:22\n• The input can simply be a number (the number of resources). This will affect the number of extra computation the algorithm will have to go through. I will edit the answer to provide a more concrete example. – yemelitc Jan 2 '16 at 7:52\n• Thanks for your edit -- that makes it much clearer. I now see that you're not confusing the cost of computing the solution with the cost of executing it as I originally thought. But now we're in the usual situation. First, it takes at least linear time to read the input. Second, the hard instances aren't the ones where you give a small tree and a gazillion people but where you give a large tree and relatively few people. (E.g., if you allow me a million bits, I'll choose a tree with about a thousand vertices and give you five people, not a tree with five vertices and a thousand people.) – David Richerby Jan 3 '16 at 20:53\n• I agree. It seems we all ended up quite critical about it, unlike what the original question tempted us to! But hopefully you get my idea of 'input as a resource': no matter how large the work, the more people the better. Still in an asymptotic sense you are definitely right, I should just blame it on non-negative integers. – yemelitc Jan 4 '16 at 4:08" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9309945,"math_prob":0.9275376,"size":26854,"snap":"2021-04-2021-17","text_gpt3_token_len":6327,"char_repetition_ratio":0.13493483,"word_repetition_ratio":0.034710392,"special_character_ratio":0.2435019,"punctuation_ratio":0.10501419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97826624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T14:41:31Z\",\"WARC-Record-ID\":\"<urn:uuid:847e6a9c-0b34-4b11-98aa-fc1a2cd8f473>\",\"Content-Length\":\"254866\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b52d27a3-e3cd-44d3-9298-85a3851ecbf8>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2c19db5-cc85-4a6a-9d1d-689c2e6124f8>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/51362/are-there-any-problems-that-get-easier-as-they-increase-in-size/51386\",\"WARC-Payload-Digest\":\"sha1:YS2CU7ATZXFPHDZWLB5V7OOSUZCE5CEH\",\"WARC-Block-Digest\":\"sha1:RF77OMSED32AFZTSJXTMZG3WK3SK5NE4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703538082.57_warc_CC-MAIN-20210123125715-20210123155715-00056.warc.gz\"}"}
http://mochan.info/reinforcement-learning/machine-learning/2020/06/23/PGT-for-One-Step-MDPs.html	[ "Let us consider the simplified case of one step MDP to see the policy gradient theorem in use. Every episode is just a single step. The episode starts, takes an action and then ends with a reward for that step. We present a simplified version of the policy gradient theorem for this one step episode setting with simplified proofs. We then present a simplified example of rock, paper and scissors which only lasts a single episode. OneStepMDP_PGT\n\n### Definitions\n\n#### Markov Decision Process (1)\n\nA Markov decision process $M$ is a tuple where\n$M = (S, A, R_a)$\nand\n\n• $S$ is the finite set of states\n• $A$ is the finite set of actions_ and $A_s$ is the set of actions for state $s \\in S$\n• $R_a(s)$ is the reward from state $s$ taking action $a$\n\nSuppose that $M$ is a one-step MDP. The episode starts at state $s \\in S$. The episode ends after one step and with reward $r$.\n\n#### Policy\n\nGiven a state $s$, the policy $\\pi(s,a) \\in [0,1]$ gives us the probability that we will take action $a$ from state $s$.\n\nLet $\\theta \\in \\mathbb{R}^n$ be some set of parameters that determine our policy. We denote such a policy by $\\pi_\\theta$.\n\n### Theorem\n\nLet $r$ be the random variable of the reward we get from taking an action. Let $J(\\theta)$ be the expected reward from following policy $\\pi_\\theta$ with parameters $\\theta$.\n$J(\\theta) = \\mathbb{E}_{\\pi_\\theta} [r]$\n\nSuppose we start at state $s$. Then,\n$\\nabla_\\theta J(\\theta) = \\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right]$\n\n#### Notation Clarifications\n\n• The notation $\\mathbb{E}_{\\pi_\\theta} [r]$ is different than the standard method of writing expectations since we have a subscript $\\pi_\\theta$ on the expectation. Usually, we have a random variable $X$ and then $\\mathbb{E}[X] = \\sum_{i=1}^k x_i p_i$. In our case, $r$ denotes the random variable of the reward for each action taken. We could have used $r_{\\pi_\\theta}$ as the random variable where the probability of each action taken is given by $\\pi_\\theta$ and the expectation would be written as $\\mathbb{E}[r_{\\pi_\\theta}]$ for the reward for taking actions according to $\\pi_\\theta$.\n• The notation $\\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right]$ has inside it term ($\\nabla_\\theta \\log \\pi_\\theta(s,a)$) that are not random variables. $\\mathbb{E}_{\\pi_\\theta}$ is the expectation over all actions that are possible to take. The $\\nabla_\\theta \\log \\pi_\\theta(s,a)$ in this context should be interpreted as the random variable with given state $s$ and over all possible actions $a$. We are slightly abusing the notation to use the same symbols with the function as the random variable.\n\n## Proof\n\nBy our definition of expectation of the reward, we have\n\\begin{aligned} J(\\theta) &= \\mathbb{E}_{\\pi_\\theta} [r] \\\\ &= \\sum_{a \\in A} \\pi_\\theta(a,s) R_a(s) \\end{aligned}\n\n\\begin{aligned} \\nabla_\\theta J(\\theta) &= \\nabla_\\theta \\left[ \\sum_{a \\in A} \\pi_\\theta(a,s) R_a(s) \\right] \\\\ &= \\sum_{a \\in A} \\nabla_\\theta \\pi_\\theta(a,s) \\; R_a(s) \\end{aligned}\nThe reward function does not depend on the $\\theta$ parameter and thus we don't take its gradient. The reward we get is determined by the action which is determined by $\\theta$ but the reward function itself is independent of $\\theta$.\n\nThe gradient of the policy can be written in terms of log using the identity $\\frac{\\nabla f}{f} = \\nabla \\log f$ as follows\n\\begin{aligned} \\nabla_\\theta \\pi_\\theta(s, a) &= \\pi_\\theta(s,a) \\frac{\\nabla_\\theta \\pi_\\theta(s, a)}{\\pi_\\theta(s,a)} \\\\ &= \\pi_\\theta(s,a) \\nabla_\\theta \\log \\pi_\\theta(s, a) \\end{aligned}\n\nUsing the log gradient substitution, we have\n\\begin{aligned} \\nabla_\\theta J(\\theta) &= \\sum_{a \\in A} \\nabla_\\theta \\pi_\\theta(s,a) \\; R_a(s) \\\\ &= \\sum_{a \\in A} \\pi_\\theta(s,a) \\nabla_\\theta \\log \\pi_\\theta(s, a) \\; R_a(s) \\\\ &= \\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right] \\end{aligned}\n\n## Rock Paper Scissors MDP\n\nAs an example of one step MDP, we will use rock, paper and scissors game. We play the game for one step and we either win, lose or draw.\n\nFor the first example of the game we will assume\n\n• we know what the opponent is going to play.\n• opponent will play rock (since we specify a starting state in our theorem - we will generalize it to any starting state below)\n\n### States\n\nWe have 3 states for what we know the opponent will play.\n\n• $s_{r}$: opponent plays rock\n• $s_{p}$: opponent plays paper\n• $s_{s}$: opponent plays scissors\n\n### Actions\n\nWe either rock, paper or scissors.\n\n• $a_r$ : play rock\n• $a_p$ : play paper\n• $a_s$ : play scissors\n\n### Rewards\n\nWe get +1 points if we win the game, 0 if the we draw the game and -1 if we lose the game.\n\n### Policy\n\nLet $\\theta \\in \\mathbb{R}$ be the parameter for our policy. We define $\\pi_\\theta$ as the following:\n\\begin{aligned} \\pi_\\theta(s_r, a_p) &= \\sigma(\\theta) \\\\ \\pi_\\theta(s_r, a_r) &= \\frac{1 - \\sigma(\\theta)}{2} \\\\ \\pi_\\theta(s_r, a_s) &= \\frac{1 - \\sigma(\\theta)}{2} \\\\ \\end{aligned}\nwhere $\\sigma(x)$ is the sigmoid function given by\n$\\sigma(x) = \\frac{1}{1+e^{-x}}$\n\nThe sigmoid function maps $(-\\infty, \\infty)$ to $(0, 1)$. Then, the above policy is a valid probability as the sum over all actions sums up to 1.\n\nWhen $\\theta \\rightarrow -\\infty$, $\\pi_\\theta(s_r, a_p) \\rightarrow 0$, and $\\theta \\rightarrow \\infty$, $\\pi_\\theta(s_r, a_p) \\rightarrow 1$. Thus, the smaller the value of $\\theta$, the smaller the probability of us choose the action $a_p$ for paper.\n\nThen, we have reward 1 for playing paper and -1 for playing scissors,\n\\begin{aligned} \\nabla_\\theta J(\\theta) &= \\sum_{a \\in A} \\nabla_\\theta \\pi_\\theta(s,a) \\; R_a(s) \\\\ &= 1 \\cdot \\nabla_\\theta \\pi_\\theta(s_r, a_p) + (-1) \\cdot \\nabla_\\theta \\pi_\\theta(s_r, a_s) \\\\ &= \\sigma^\\prime(\\theta) + \\frac{\\sigma^\\prime(\\theta)}{2} \\\\ &= \\frac{3}{2} \\sigma^\\prime(\\theta) \\end{aligned}\n\nWe do the parameter update using the rule\n$\\theta = \\theta + \\alpha \\nabla_\\theta J(\\theta)$\nwhere $\\alpha \\in \\mathbb{R}$ is the learning rate.\n\nSince $\\sigma^\\prime > 0$, $\\theta$ is always increasing. Thus, $\\pi_\\theta(s_r, a_p)$ goes towards 1 and the probability for other actions towards 0. Thus, the policy gradient will choose to play paper for the rock that the opponent plays.\n\n## Generalization (1) - With Transition Probability\n\nWe can generalize the MDP to have a transition probability.\n\nIn the first definition of MDP, if we are in state $s$ and we take action $a$, we would transition to a state $s^\\prime$ and obtain single reward for the action.\n\nHere the generalization results is that when taking an action $a$, it now gives us a probability of transitioning to each of the states in $S$. Before an action from state $s$ would only take you another specific state $s^\\prime$. Now, there is a probability for transition to every state of $S$. If the probability of transitioning to state $s^\\prime$ is 1 and others 0, it is effectively the same as without the generalization.\n\nFor reward, if we are in state $s$ and we transition to state $s^\\prime$, the reward is given by $R_a(s, s^\\prime)$.\n\nA Markov decision process (2) $M$ is a tuple where\n$M = (S, A, P_a, R_a)$\nbe\n\n• $S$ is the finite set of states\n• $A$ is the finite set of actions_ and $A_s$ is the set of actions for state $s \\in S$\n• $P_a(s, s^\\prime)$ is the transition probability from state $s$ to another state $s^\\prime$ by taking action $a$\n• $R_a(s, s^\\prime)$ is the reward from state $s$ to $s^\\prime$\n\nLet $M$ me an MDP(2) as defined above. Let $r$ be the random variable of the reward we get from taking an action. Let $J(\\theta)$ be the expected reward from following policy $\\pi_\\theta$ with parameters $\\theta$.\n$J(\\theta) = \\mathbb{E}_{\\pi_\\theta} [r]$\n\nSuppose we start at state $s$. Then,\n$\\nabla_\\theta J(\\theta) = \\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right]$\n\n### Proof\n\nThe definition of MDP has changed and thus, when we take an action $a$, we have a transition probability of going from state $s$ to state $s^\\prime$ with probability $P_a(s, s^\\prime)$.\n\nThus, we have to sum over all subsequent states that we might go to after an action is taken.\n\\begin{aligned} J(\\theta) &= \\mathbb{E}_{\\pi_\\theta} [r] \\\\ &= \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\end{aligned}\n\nNow, taking the gradient and noting that the transition probability and reward does not depend on $\\theta$, and so we have,\n\\begin{aligned} \\nabla_\\theta J(\\theta) &= \\nabla_\\theta \\left( \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\right)\\\\ &= \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\nabla_\\theta \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\\\ &= \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\left[ \\nabla_\\theta \\log \\pi_\\theta(s, a) \\right] \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\\\ &= \\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right] \\end{aligned}\n\n### Rock Paper Scissors\n\nIn the previous example of rock, paper, scissors, we knew what the opponent was going to play. With the new formulation of the MDP, we now divide the state into what we believe the opponent will play and what the opponent plays after we perform our action.\n\n#### States\n\nWe have the 6 states, 3 for before we play and 3 after we play. The 3 before we play will stand for what we think the opponent will play. The 3 after we play will be what the opponent played.\n\n• $s_{r_1}$: we think opponent plays rock\n• $s_{p_1}$: we think opponent plays paper\n• $s_{s_1}$: we think opponent plays scissors\n• $s_{r_2}$: opponent played rock\n• $s_{p_2}$: opponent played paper\n• $s_{s_2}$: opponent played scissors\n\n#### Actions\n\nThe actions are the same as before. We have $a_r$, $a_p$ and $a_s$ for playing rock, paper and scissors respectively.\n\n#### Transition Probabilities\n\nThe transition probability is non-zero from $s_{r_1}, s_{p_1}, s_{s_1}$ to $s_{r_2}, s_{p_2}, s_{s_2}$. If our guess was random, then all the transition probabilities would be $\\frac{1}{3}$.\n\nFor example, a random transition probability could be\n\\begin{aligned} P(s_{r_1}, s_{r_2}) &= \\frac{1}{3} \\\\ P(s_{r_1}, s_{p_2}) &= \\frac{1}{3} \\\\ P(s_{r_1}, s_{s_2}) &= \\frac{1}{3} \\\\ \\end{aligned}\n\n#### Rewards\n\nThe reward is similar. We still get +1 for winning, 0 for drawing and -1 for losing the game. However, in this case, the reward depends on the starting state and the ending state and the action taken.\n\nWe can illustrate the rewards in term of a matrix. Suppose our action was $a_r$. Then,\n$\\begin{matrix} & s_{r_2} & s_{p_2} & s_{s_2} \\\\ s_{r_1} & 0 & -1 & 1 \\\\ s_{p_1} & 0 & -1 & 1 \\\\ s_{s_1} & 0 & -1 & 1 \\\\ \\end{matrix}$\nNote the ending state and the action only determines the reward. Whatever, we believe the opponent was going to play does not matter in the reward if the transition probability is equally random.\n\n#### Policy\n\nLet $\\theta \\in \\mathbb{R}^3$. We will denote $\\theta = [\\theta_1, \\theta_2, \\theta_2]$.\n\nWe assume that we start at state $s_{r_1}$. Then, the policy is given by\n\\begin{aligned} \\pi_\\theta(s_{r_1}, a_r) &= \\frac{e^{\\theta_1}}{\\sum_{i=1}^3 e^{\\theta_i}} \\\\ \\pi_\\theta(s_{r_1}, a_p) &= \\frac{e^{\\theta_2}}{\\sum_{i=1}^3 e^{\\theta_i}} \\\\ \\pi_\\theta(s_{r_1}, a_s) &= \\frac{e^{\\theta_3}}{\\sum_{i=1}^3 e^{\\theta_i}} \\\\ \\end{aligned}\n\nThus, the higher the probability of $\\theta_i$, the more likely that we will take that action.\n\nSince we have 3 parameters in $\\theta$, the gradient is a $3 \\times 1$ matrix.\n\n$\\nabla_\\theta \\pi_\\theta(s_{r_1}, a_r) = \\begin{bmatrix} \\frac{\\partial \\pi_\\theta(s_{r_1}, a_r)}{d \\theta_1} \\\\ \\frac{\\partial \\pi_\\theta(s_{r_1}, a_r)}{d \\theta_2} \\\\ \\frac{\\partial \\pi_\\theta(s_{r_1}, a_r)}{d \\theta_3} \\\\ \\end{bmatrix} = \\begin{bmatrix} \\frac{e^{\\theta_1} (e^{\\theta_2} + e^{\\theta_3})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ -\\frac{e^{\\theta_1+\\theta_2}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ -\\frac{e^{\\theta_1+\\theta_3}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ \\end{bmatrix}$\n\n$\\nabla_\\theta \\pi_\\theta(s_{r_1}, a_p) = \\begin{bmatrix} \\frac{\\partial \\pi_\\theta(s_{r_1}, a_p)}{d \\theta_1} \\\\ \\frac{\\partial \\pi_\\theta(s_{r_1}, a_p)}{d \\theta_2} \\\\ \\frac{\\partial \\pi_\\theta(s_{r_1}, a_p)}{d \\theta_3} \\\\ \\end{bmatrix} = \\begin{bmatrix} -\\frac{e^{\\theta_1+\\theta_2}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ \\frac{e^{\\theta_2} (e^{\\theta_1} + e^{\\theta_3})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ -\\frac{e^{\\theta_2+\\theta_3}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ \\end{bmatrix}$\n\n$\\nabla_\\theta \\pi_\\theta(s_{r_1}, a_s) = \\begin{bmatrix} \\frac{\\partial \\pi_\\theta(s_{r_1}, a_s)}{d \\theta_1} \\\\ \\frac{\\partial \\pi_\\theta(s_{r_1}, a_s)}{d \\theta_2} \\\\ \\frac{\\partial \\pi_\\theta(s_{r_1}, a_s)}{d \\theta_3} \\\\ \\end{bmatrix} = \\begin{bmatrix} -\\frac{e^{\\theta_1+\\theta_3}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ -\\frac{e^{\\theta_2+\\theta_3}}{(e^{\\theta_2} + e^{\\theta_3} + e^{\\theta_3})^2} \\\\ \\frac{e^{\\theta_3} (e^{\\theta_1} + e^{\\theta_2})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ \\end{bmatrix}$\n\n### Optimality Under Random Choice\n\nSuppose we do not know how the opponent plays and thus all transition probabilities are $\\frac{1}{3}$.\n\nLet us first look at the partial derivative of $J(\\theta)$ with respect to $\\theta_1$ instead of the whole $\\nabla_\\theta J(\\theta)$.\n\n\\begin{aligned} \\frac{\\partial J(\\theta)}{\\partial \\theta_1} &= \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\frac{\\partial \\pi_\\theta(s, a)}{\\partial \\theta_1} P_a(s, s^\\prime) R_a(s, s^\\prime) \\\\ &= \\frac{1}{3} \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\frac{\\partial \\pi_\\theta(s, a)}{\\partial \\theta_1} R_a(s, s^\\prime) \\\\ &= \\frac{1}{3} \\sum_{a \\in A} \\frac{\\partial \\pi_\\theta(s, a)}{\\partial \\theta_1} \\sum_{s^\\prime \\in S} R_a(s, s^\\prime) \\\\ &= 0 \\end{aligned}\n\nThe policy only depends on the starting state and not the state we transition to. Here, the reward is dependent on the state we start with and the state we end up with. Since for each action, we lose one and win one, the total is 0. Thus, we can choose any value of $\\theta$ and we will have the same expected reward since the gradient is 0 everywhere.\n\n### Policy Improvement Under Scissors Only\n\nSuppose that the opponent only plays scissors. Then, we should always play rock and thus, $\\theta_1$ should be as large as possible and $\\theta_2, \\theta_3$ should be as small as possible.\n\n\\begin{aligned} \\frac{\\partial J(\\theta)}{\\partial \\theta_1} &= \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\frac{\\partial \\pi_\\theta(s, a)}{\\partial \\theta_1} P_a(s, s^\\prime) R_a(s, s^\\prime)\\\\ &= \\sum_{a \\in A} \\frac{\\partial \\pi_\\theta(s, a)}{\\partial \\theta_1} P_a(s, s_{s_2}) R_a(s, s_{s_2})\\\\ &= \\frac{\\partial \\pi_\\theta(s, a_r)}{\\partial \\theta_1} R_{a_r}(s, s_{s_2}) + \\frac{\\partial \\pi_\\theta(s, a_p)}{\\partial \\theta_1} R_{a_p}(s, s_{s_2})\\\\ &= \\frac{\\partial \\pi_\\theta(s, a_r)}{\\partial \\theta_1} - \\frac{\\partial \\pi_\\theta(s, a_p)}{\\partial \\theta_1}\\\\ \\end{aligned}\n\nThe first step follows from definition. The second step follows that the opponent is only playing scissors and all other transitions that do not end in scissors is 0. The third step follows from the fact that action of rock or paper will result a positive or negative reward. If both the opponent and the agent draw scissors, it is a draw and thus has reward 0. The fourth step is just reward of 1 for winning and -1 for losing.\n\nThus, the derivative for $\\theta_1$\n\\begin{aligned} \\frac{\\partial J(\\theta)}{\\partial \\theta_1} &= \\frac{\\partial \\pi_\\theta(s, a_r)}{\\partial \\theta_1} - \\frac{\\partial \\pi_\\theta(s, a_p)}{\\partial \\theta_1}\\\\ &= \\frac{e^{\\theta_1} (e^{\\theta_2} + e^{\\theta_3})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} - \\left( -\\frac{e^{\\theta_1+\\theta_2}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\right) \\\\ &= \\frac{e^{\\theta_1} (2e^{\\theta_2} + e^{\\theta_3})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ &> 0 \\end{aligned}\n\nNext, for $\\theta_2$, we have\n\\begin{aligned} \\frac{\\partial J(\\theta)}{\\partial \\theta_2} &= \\frac{\\partial \\pi_\\theta(s, a_r)}{\\partial \\theta_2} - \\frac{\\partial \\pi_\\theta(s, a_p)}{\\partial \\theta_2}\\\\ &= -\\frac{e^{\\theta_1+\\theta_2}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} - \\frac{e^{\\theta_2} (e^{\\theta_1} + e^{\\theta_3})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ &< 0 \\end{aligned}\n\nFinally, for $\\theta_3$, we have\n\\begin{aligned} \\frac{\\partial J(\\theta)}{\\partial \\theta_3} &= \\frac{\\partial \\pi_\\theta(s, a_r)}{\\partial \\theta_3} - \\frac{\\partial \\pi_\\theta(s, a_p)}{\\partial \\theta_3}\\\\ &= -\\frac{e^{\\theta_1+\\theta_3}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} + \\frac{e^{\\theta_2+\\theta_3}}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\\\ &= \\frac{e^{\\theta_3}(e^{\\theta_2}-e^{\\theta_1})}{(e^{\\theta_1} + e^{\\theta_2} + e^{\\theta_3})^2} \\end{aligned}\n\nThis follows from $\\theta_1 > \\theta_2$, and hence $\\frac{\\partial J(\\theta)}{\\partial \\theta_3} < 0$.\n\nThus, $\\theta_1$ increases and $\\theta_2, \\theta_3$ decreases when updating $\\theta$ along the policy gradient. Thus, the policy favors playing rock over the other two alternatives.\n\n## Generalization (2) - No specified start state\n\nIn the above theorem statements, we have required to specify a start state.\n\nIn order to not have to specify a start state, we have to define a distribution $d(s)$ that represents the probability of starting at state $s$.\n\nLet $M$ me an MDP(2). Let $r$ be the random variable of the reward we get from taking an action. Let $J(\\theta)$ be the expected reward from following policy $\\pi_\\theta$ with parameters $\\theta$.\n$J(\\theta) = \\mathbb{E}_{\\pi_\\theta} [r]$\n\nThen,\n$\\nabla_\\theta J(\\theta) = \\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right]$\n\n### Proof\n\nThe random variable $r$ covers the reward starting from any state $s \\in S$ and the initial starting state is not specified.\n\n\\begin{aligned} J(\\theta) &= \\mathbb{E}_{\\pi_\\theta} [r] \\\\ &= \\sum_{s \\in S} d(s) \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\end{aligned}\n\nNow, taking the gradient and noting that the initial distribution, transition probability and reward does not depend on $\\theta$, we have\n\\begin{aligned} \\nabla_\\theta J(\\theta) &= \\nabla_\\theta \\left( \\sum_{s \\in S} d(s) \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\right)\\\\ &= \\sum_{s \\in S} d(s) \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\nabla_\\theta \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\\\ &= \\sum_{s \\in S} d(s) \\sum_{a \\in A} \\sum_{s^\\prime \\in S} \\left[ \\nabla_\\theta \\log \\pi_\\theta(s, a) \\right] \\pi_\\theta(s, a) P_a(s, s^\\prime) R_a(s, s^\\prime) \\\\ &= \\mathbb{E}_{\\pi_\\theta} \\left[ \\left( \\nabla_\\theta \\log \\pi_\\theta(s,a) \\right) r \\right] \\end{aligned}\n\n## Python Code Example\n\nUsing $\\theta \\in \\mathbb{R}^9$, we have probability for every action and initial state, the given example can start from any state (which only signifies the belief of what the opponent will play). It transitions to the state that the opponent's play after the action is taken. We show the improvement of the $\\theta$ parameters using policy gradients and policy improvements." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86489606,"math_prob":0.99994266,"size":12214,"snap":"2022-40-2023-06","text_gpt3_token_len":4350,"char_repetition_ratio":0.15986896,"word_repetition_ratio":0.09994712,"special_character_ratio":0.27337483,"punctuation_ratio":0.112869196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000009,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T23:43:19Z\",\"WARC-Record-ID\":\"<urn:uuid:3cef65a6-c853-49ac-a6b7-ea876c68a6e6>\",\"Content-Length\":\"749175\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9765c34-1dd8-48e6-86da-08654d12c9a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:445f4516-73b8-4e29-9635-4d0471222e74>\",\"WARC-IP-Address\":\"192.30.252.153\",\"WARC-Target-URI\":\"http://mochan.info/reinforcement-learning/machine-learning/2020/06/23/PGT-for-One-Step-MDPs.html\",\"WARC-Payload-Digest\":\"sha1:PE7B3EL4TDAORX774PVQNALX3AQARYBP\",\"WARC-Block-Digest\":\"sha1:SB5U73IX3OVBWJJQ5D23BORWEB4JKDVR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500983.76_warc_CC-MAIN-20230208222635-20230209012635-00195.warc.gz\"}"}
https://studysoup.com/tsg/18374/discrete-mathematics-and-its-applications-7-edition-chapter-4-4-problem-60e	[ "×\n×\n\n# Show that if p is an odd prime, then there are exactly (p", null, "ISBN: 9780073383095 37\n\n## Solution for problem 60E Chapter 4.4\n\nDiscrete Mathematics and Its Applications \| 7th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Discrete Mathematics and Its Applications \| 7th Edition\n\n4 5 0 323 Reviews\n18\n3\nProblem 60E\n\nShow that if p is an odd prime, then there are exactly (p ‒ l)/2 quadratic residues of p among the integers 1,2,…, p‒ 1.\n\nIf p is an odd prime and a is an integer not divisible by p. the Legendre symbol", null, "is defined to be 1 if a is a quadratic residue of p and ‒ 1 otherwise.\n\nStep-by-Step Solution:\nStep 1 of 3\n\nChemistry Lecture 7 Chapter 15: Chemical Equilibrium Dynamics of Chemical Reactions and Microscopic Reversibility  Reactions can be reversible and dynamic  when rate of forward and reverse reaction are equal  reaction arrows in each direction can denote this Equilibrium  does not mean same amount of reactant and product...\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780073383095\n\nSince the solution to 60E from 4.4 chapter was answered, more than 241 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 60E from chapter: 4.4 was answered by , our top Math solution expert on 06/21/17, 07:45AM. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7. Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. This full solution covers the following key subjects: odd, quadratic, prime, among, Integer. This expansive textbook survival guide covers 101 chapters, and 4221 solutions. The answer to “Show that if p is an odd prime, then there are exactly (p ? l)/2 quadratic residues of p among the integers 1,2,…, p? 1.If p is an odd prime and a is an integer not divisible by p. the Legendre symbol is defined to be 1 if a is a quadratic residue of p and ? 1 otherwise.” is broken down into a number of easy to follow steps, and 59 words.\n\nUnlock Textbook Solution" ]	[ null, "https://studysoup.com/cdn/54cover_2450429", null, "https://studysoup.com/cdn/54cover_2450429", null, "https://lh6.googleusercontent.com/3235mdlL6NuCGFByC4Cn1YFBALXqXFjZp4Qa-NuOMVKmgxILSTvTLNabnx6SWaM8hExKIDqIQYu6EfL51wXPbXFFX0Z_hXa0Z9MSN21f-PhuvaV3zF6xFPmQu32ArvZdSz0nawQ2", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83193374,"math_prob":0.8146741,"size":593,"snap":"2020-10-2020-16","text_gpt3_token_len":140,"char_repetition_ratio":0.108658746,"word_repetition_ratio":0.0,"special_character_ratio":0.21922429,"punctuation_ratio":0.09322034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9926405,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-29T12:01:36Z\",\"WARC-Record-ID\":\"<urn:uuid:ebf55bcf-9881-4fab-932e-f3be8440e56a>\",\"Content-Length\":\"79210\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81319842-f41b-452c-b3be-a91862cbba81>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9ce1809-294a-45ce-b83c-7cd6f16dcdf5>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/18374/discrete-mathematics-and-its-applications-7-edition-chapter-4-4-problem-60e\",\"WARC-Payload-Digest\":\"sha1:4OZRCSL4D34M2JKEL4KEM3SPBJXZRJDS\",\"WARC-Block-Digest\":\"sha1:YXFFJWYCEXGZPOOWBVUJKQ54CJ22KQE6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875149238.97_warc_CC-MAIN-20200229114448-20200229144448-00265.warc.gz\"}"}
https://forum.arduino.cc/t/furniture-with-extras-solenoid-motors-led-strip/699250	[ "# Furniture with extras (solenoid, motors, LED strip)\n\nI'm helping my brother with creating a furniture piece that has doors that open and colourful LED strips.\n(The problem is that I only know how to program but have to read up all electronics related things.)\n\nAt the moment, I'm stuck on the following decisions:\nQ1. Which power source should I use for the solenoid.\n(Q2. Can I power the Arduino and the LED strip both with the same 5V@10A power supply.)\n\nI have two power sources:\nA) 5V for Arduino (because of the LED strip: 5V@10A)\nB) 24V@1A (we have a sponsor how gives us specialised motors for opening cupboards, therefore the constraint of using a 24V power supply.)\n\nNow, both power supplies have more than enough power for a small solenoid, but I'm worried about how to reduce the power in the circuit to avoid too much power at the solenoid. (I am aiming at 8W power at the solenoid.)\nA) I would buy a 6V@1A (HS-0530B; DC 6V 1A) solenoid. I guess that running it with only 5V will only reduce the power of the solenoid?\nIf I understand it correctly, the solenoid will pull more current if available. Therefore, I would add a resistor to keep the current low enough?\n(see schematics attached)\n\nB) If I use the 24V power supply, I match the specifications of a 24V solenoid (JF-0530B 24V 300mA), but again, I think I need to add a resistor to limit the power at the solenoid. On the other hand, I read that at 24V the solenoids are less powerful.\n\nI want to use 3m of WS2812B with 60LEDs/m. According to the seller that needs 18W/m, hence 54W. I will reduce the brightness a bit (or shorten it) to stay below 45W. The remaining 5V@1A should be sufficient to run the other devices and the Arduino.\n\nsteffenpl:\n...\nIf I understand it correctly, the solenoid will pull more current if available. ...\n\nSorry, but i think this is the point where you fall ... current is totally depending from voltage AND resistance ... if a load have a resistance of (example) 10 ohm, and you power it with (example) 5V, there is no way that it can pull more than 500mA (Ohm law, I = V / R), also if your power supply can give 100A, the load still draw its own 500mA and no more (until you don't increase the voltage ofcourse)\n\npowering it with 5V instead of 6, simply DECREASE the nominal current (and you have to check if at 5V it still have enough force, cause ofcourse also that is decreased a bit)\n\nThe only thing you need to take in consideration is the fact that being an inductive load, it store energy that release at turn-off ... so, usually is always used a diode in antiparallel (K on +, A on -) on every inductive load, especially if used with logics (included your relay)\n\nAbout solenoid forces, they depend from mechanic construction, better or worse use of magnetic flux in them, used materials ... nothing say that a 24V solenoid have to be less strong than a 5V one ...\n\nThank you very much! Yes, I definitely lack basics, thanks for your explanation, that helps me a lot!\n\nSorry, but i think this is the point where you fall ... current is totally depending from voltage AND resistance ... if a load have a resistance of (example) 10 ohm, and you power it with (example) 5V, there is no way that it can pull more than 500mA (Ohm law, I = V / R), also if your power supply can give 100A, the load still draw its own 500mA and no more (until you don't increase the voltage ofcourse)\n\nI get how I was wrong. What I don't understand now how solenoids are regulated?\n\nThe datasheets often show plots of force/displacement for different levels of power (i.e. 4W, 8W, 16W etc).\nThey state that a higher power level is only allowed if the duty cycle is low enough.\n\nI have a solenoid with 6V and rated current 300mA => 1.8W (and R ~ 20Ω).\nIf want to use it with 3.6W, then the way to archive this would be to use a \"6*sqrt(2) V\" power source? Since then P = U²/R = 12 / 20 W = 3.6W. Is that correct?\nSomehow, I imagined that it would be easier if solenoids have less resistance but instead one would limit the current to control how strong they push.\n\nsteffenpl:\nI get how I was wrong. What I don't understand now how solenoids are regulated?\n\nThe datasheets often show plots of force/displacement for different levels of power (i.e. 4W, 8W, 16W etc).\nThey state that a higher power level is only allowed if the duty cycle is low enough.\n\nYou can vary the power level by varying the voltage supplied to the solenoid. The solenoid has a fixed (DC) resistance, so more voltage will translate to more current and thus greater power. The caveat is that greater power will also cause greater heat rise within the coil, and if this is excessive it will destroy the coil. This is where duty cycle comes into the equation. Also, you cannot supply just any arbitrarily chosen higher voltage, because at some point you will cause breakdown of the wire insulation... with unfortunate results.\nS.\n\nSounds good! I think I know what to do now. Thanks for your help!\n\nThis topic was automatically closed 120 days after the last reply. New replies are no longer allowed." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9325823,"math_prob":0.83538693,"size":4730,"snap":"2023-14-2023-23","text_gpt3_token_len":1233,"char_repetition_ratio":0.12759204,"word_repetition_ratio":0.3020595,"special_character_ratio":0.2551797,"punctuation_ratio":0.12649402,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9557305,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T11:13:06Z\",\"WARC-Record-ID\":\"<urn:uuid:6ba973df-2bfe-4a32-a3f2-142174c0a824>\",\"Content-Length\":\"43292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:361bf2c1-f8ca-42b5-9766-2a9909d8be8d>\",\"WARC-Concurrent-To\":\"<urn:uuid:00a4f839-afb5-4cee-a173-81cae9b7351b>\",\"WARC-IP-Address\":\"184.104.178.75\",\"WARC-Target-URI\":\"https://forum.arduino.cc/t/furniture-with-extras-solenoid-motors-led-strip/699250\",\"WARC-Payload-Digest\":\"sha1:3255KJL76I2OQHHCQVFTV5OVJWECVKDG\",\"WARC-Block-Digest\":\"sha1:LIF6FXYYBOIBFFYQL46RMSGCVE76NLOA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949181.44_warc_CC-MAIN-20230330101355-20230330131355-00360.warc.gz\"}"}
https://studentvip.com.au/notes/24232/math1002-math1002-linear-algebra-revision	[ "USYD\n\nSemester 2, 2016\n\nMath1002 Linear Algebra Revision\n\n40 pages\n\n4,000 words\n\n\\$29.00\n\n3\n\nDescription\n\nNote: These are handwritten notes.\n\nThis is a revision summarised from 21 lectures\nTopics Covered:\n VECTORS:\n- Draw vectors\n- Unit vectors\n- Parallel vectors\n- Cartesian form\n- Length of a vector\n- Linear independence of two vectors\n\n DOT PRODUCTS:\n- Algebraic definition\n- Geometric definition\n- Cauchy Schwarz Inequality\n- Dot product rules\n\n CROSS PRODUCTS:\n- Algebraic definition\n- Geometric definition\n- Cross product rules\n\n LINES L:\n- Parametric vector equation of L\n- Parametric scalar equations of L\n- Cartesian equation\n- From point to lines\n\n PLANES P:\n- Vector equation of a plane P\n- Cartesian equation of a plane P\n- From point to plane\n\n INTERSECTIONS:\n- Point of intersection of a line and a plane\n- Intersections of planes\n\n SYSTEMS OF LINEAR EQUATIONS:\n- Row Echelon Form examples\n- Reduced Row Echelon Form examples\n\n MATRICES:\n- Addition, subtraction, scalar multiplication\n- Powers of matrix\n- Matrices and systems of operations\n- Matrix inverses\n- Manipulating matrix equations\n- Determinants\n- Eigenvalues and Eigenvectors\n- Eigenspace\n- Diagonalisation\n\nAuthor\n\nJenny\n\nCampus\n\nUSYD, Camperdown/Darlington\n\nMember since\n\nMarch 2015" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74092543,"math_prob":0.9290276,"size":1318,"snap":"2019-43-2019-47","text_gpt3_token_len":348,"char_repetition_ratio":0.1217656,"word_repetition_ratio":0.057142857,"special_character_ratio":0.23899849,"punctuation_ratio":0.0880829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T04:38:31Z\",\"WARC-Record-ID\":\"<urn:uuid:5cc2df93-48ee-4e33-a90f-535797e2948f>\",\"Content-Length\":\"26633\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f23cd963-1d80-44e6-85be-269a3656c6ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d64f7fb-7f00-4aa6-afe2-e2f124094c88>\",\"WARC-IP-Address\":\"3.104.5.24\",\"WARC-Target-URI\":\"https://studentvip.com.au/notes/24232/math1002-math1002-linear-algebra-revision\",\"WARC-Payload-Digest\":\"sha1:3AWFONFBL6RF4JPPC7XVT2ZRM3H7S4SE\",\"WARC-Block-Digest\":\"sha1:VCOFEGR54FHHXK7OASHYEQFA44LPBKFT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987798619.84_warc_CC-MAIN-20191022030805-20191022054305-00184.warc.gz\"}"}
https://javascript.tutorialink.com/how-to-set-up-a-function-dynamically-without-calling-it-in-javascript/	[ "# How to set up a function dynamically without calling it in javascript?\n\nI have a function which calls multiple functions at the same time and waits until all of them are finished. It’s working as intended. It’s labeled as `parallel` below\n\nI’ve got an array which I want to map into a function with the param for each value in the array. However as you might imagine, the function is being called right when it’s being mapped so the results are not being returned to the parallel results. Below is an example of what I’m trying to do.\n\nAny ideas on how I can do this & return the results to the parallel function?\n\nCurrently doing\n\n```let array = [1,2,3,4,5];\nr = await parallel(\ndoStuff(array),\ndoStuff(array),\ndoStuff(array),\ndoStuff(array),\ndoStuff(array),\n);\nif(r.err.code) return r;\n```\n\nWould like to do\n\n```let array = [1,2,3,4,5];\n\nr = await parallel(array.map(v => doStuff(v)))\nif(r.err.code) return r;\n\n```\n\nParallel\n\n```async function waitForAll(...functions) {\nreturn await Promise.all(functions)\n}\n\nconst parallel = async (...functions) => {\nconst d = {err: {code:0,message:\"\"},res:{}}; let r,sql,vars;\n\nr = await waitForAll(...functions)\nfor(let i in r){ if(r[i].err.code) return r[i]; }\n\nd.res = r;\n\nreturn d;\n}\n\nmodule.exports = {\nparallel\n}\n```\n\nEdit: `doStuff()` is async\n\nSince `parallel` accepts any number of separate arguments, it looks like all you need to do is spread the result of mapping into the call. Right now you’re only passing a single array as the first argument.\n\n```const array = [1,2,3,4,5];\nconst r = await parallel(...array.map(v => doStuff(v)))\nif(r.err.code) return r;\n```\n\nOr change `parallel`‘s implementation to accept an array as a single argument\n\n```const parallel = async (functions) => {\n```\n\nand do\n\n```const r = await parallel(array.map(v => doStuff(v)))\n```\n\nUnless `doStuff` has different behavior depending on whether its second or more arguments are defined, you might be able to simplify to\n\n```const r = await parallel(array.map(doStuff))\n```\n\nremoving the inline call." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80801886,"math_prob":0.97645754,"size":1892,"snap":"2022-27-2022-33","text_gpt3_token_len":478,"char_repetition_ratio":0.16896187,"word_repetition_ratio":0.040677965,"special_character_ratio":0.27536997,"punctuation_ratio":0.18465228,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99098825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T00:53:20Z\",\"WARC-Record-ID\":\"<urn:uuid:a3a708cd-0192-4d94-8322-1e6596dc4a5e>\",\"Content-Length\":\"39174\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:069cb172-8ba7-4b76-a385-318ba8edcd24>\",\"WARC-Concurrent-To\":\"<urn:uuid:84339928-3161-4ed6-b3c2-0f3262c9fe3c>\",\"WARC-IP-Address\":\"172.67.191.217\",\"WARC-Target-URI\":\"https://javascript.tutorialink.com/how-to-set-up-a-function-dynamically-without-calling-it-in-javascript/\",\"WARC-Payload-Digest\":\"sha1:OS7XPCC56TKPGT4UZ6YTKJJE7YXLW7KL\",\"WARC-Block-Digest\":\"sha1:RKNJ2O3MPNPWTKFU2U7DG4X3E66HM7D4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103619185.32_warc_CC-MAIN-20220628233925-20220629023925-00422.warc.gz\"}"}
https://unix.stackexchange.com/questions/498680/date-arithmetic-and-ternary-operator-in-one-line	[ "# Date, arithmetic, and ternary operator in one line\n\nI have a simple code to ensure a script takes at least x seconds (500 here) on Ubuntu\n\n``````t1=\\$(date +%s)\n# script is here\nt2=\\$(date +%s)\nlet \"t = 500 - \\$t2 + \\$t1\"\n(( t = t>0 ? t : 1 ))\nsleep \\$t\n``````\n\nThe code works perfectly, but I believe my coding is not efficient, and these three lines\n\n``````t2=\\$(date +%s)\nlet \"t = 500 - \\$t2 + \\$t1\"\n(( t = t>0 ? t : 1 ))\n``````\n\nshould be expressed in one single line. My question is how to improve the code.\n\n• Is this a Linux platform, or some UNIX? (More specifically, do you have GNU `date`?) – roaima Feb 4 '19 at 21:48\n• There is much to be said for not sacrificing readability for perceived efficiency. If you compress that all into an inscrutable one-liner running as a subshell expansion as a parameter for `sleep`, how are you going to unravel that when someone calls you at 4 in the morning the day after a most raucous party to debug it when it goes sideways? – DopeGhoti Feb 4 '19 at 21:53\n• You could lose the `t` and `t2` entirely, but make sure your phone is switched off at 4am. `sleep \\$( printf \"%d\\n\" \\$(( 500 - (\\$(date +%s) - t1) )) \| sed 's/^-.*/1/' )` – roaima Feb 4 '19 at 22:01\n• @roaima Sorry for being vague. I updated the question. I haven't got stuck anywhere. I just want to improve my code, as I believe it is not the efficient way. – Googlebot Feb 4 '19 at 22:01\n• @DopeGhoti very good point indeed. – Googlebot Feb 4 '19 at 22:02\n\n## 2 Answers\n\nHere is how I would do it using `bash` special parameter `SECONDS`:\n\n``````#!/bin/bash\n\nSECONDS=0\n# script is here\nsleep \"\\$(( 500 > SECONDS ? 500 - SECONDS : 1 ))\"\n``````\n\nNormally `SECONDS` returns time (in seconds) since the script has started, but one can assign any value to (re)set the timer.\n\n• Zeroing `SECONDS` doesn't hurt, and at least makes the initial value explicit, but since it tells the time since the script started you don't have to zero it. Unless the script does something else first, of course. – ilkkachu Feb 4 '19 at 22:25\n\nWhat your script is, is non-portable (since you're using `let` and `(( .. ))`), confusing (since you're using both `let` and `(( .. ))`), lacking in documentation (there are no comments, and the variable names are non-descriptive), and marginally unsafe (since you haven't quoted the expansion of `\\$t`).\n\nIf you want a rewrite, here's mine:\n\n``````#!/bin/sh\n\nmin_duration=500\nt_start=\\$(date +%s)\n# script is here\nt_end=\\$(date +%s)\nelapsed=\\$(( t_end - t_start ))\n\n# sleep long enough to make sure 'min_duration' seconds has elapsed,\n# but at least 1 second\nsleep \"\\$(( elapsed < min_duration ? min_duration - elapsed : 1 ))\"\n``````\n• Is there any practical difference between your `: \"\\$(( elapsed = t_end - t_start ))\"` and simply `(( elapsed = t_end - t_start ))`? – roaima Feb 4 '19 at 22:11\n• @roaima The exit status when the calculation results in a zero. It matters when running under `set -e`. Personally, I think I would have used `elapsed=\\$(( ... ))`. Also notice that he's writing for `/bin/sh`, not `bash`. – Kusalananda Feb 4 '19 at 22:16\n• @roaima, I was just thinking about the portability point. At least `dash` and `busybox` don't support `(( .. ))`. With Bash, I'd use `(( .. ))` since it looks less ugly to me. The exit status shouldn't matter (well, unless someone uses `set -e`, but let's not go there...) – ilkkachu Feb 4 '19 at 22:18\n• @Kusalananda, argh, of course with `elapsed=\\$((..))`... Thanks. – ilkkachu Feb 4 '19 at 22:19\n• @Kusalananda ah yes. Thankyou. – roaima Feb 5 '19 at 0:18" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84589356,"math_prob":0.8310765,"size":811,"snap":"2021-21-2021-25","text_gpt3_token_len":279,"char_repetition_ratio":0.10161091,"word_repetition_ratio":0.19767442,"special_character_ratio":0.40197286,"punctuation_ratio":0.09659091,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95296216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T11:00:23Z\",\"WARC-Record-ID\":\"<urn:uuid:05b83485-2622-4c38-a7bb-f394d72fbd9f>\",\"Content-Length\":\"188818\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d014b46c-0874-43ca-a02d-1aa93d51973b>\",\"WARC-Concurrent-To\":\"<urn:uuid:1282cbb4-d6bf-42b0-8137-745ce2886050>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://unix.stackexchange.com/questions/498680/date-arithmetic-and-ternary-operator-in-one-line\",\"WARC-Payload-Digest\":\"sha1:D53AZ63AB3DLP2XMI737RM2D65X5EJIT\",\"WARC-Block-Digest\":\"sha1:CGC5UGA6VJXYIDEJCL7IJ2AF3NAG2IP3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487620971.25_warc_CC-MAIN-20210615084235-20210615114235-00177.warc.gz\"}"}
http://slideplayer.com/slide/4586606/	[ "", null, "# Multiply complex numbers\n\n## Presentation on theme: \"Multiply complex numbers\"— Presentation transcript:\n\nMultiply complex numbers\nEXAMPLE 4 Multiply complex numbers Write the expression as a complex number in standard form. a. 4i(–6 + i) b. (9 – 2i)(–4 + 7i) SOLUTION a. 4i(–6 + i) = –24i + 4i2 Distributive property = –24i + 4(–1) Use i2 = –1. = –24i – 4 Simplify. = –4 – 24i Write in standard form.\n\nMultiply complex numbers\nEXAMPLE 4 Multiply complex numbers b. (9 – 2i)(–4 + 7i) = – i + 8i – 14i2 Multiply using FOIL. = – i – 14(–1) Simplify and use i2 = – 1 . = – i + 14 Simplify. = – i Write in standard form.\n\nDivide complex numbers\nEXAMPLE 5 Divide complex numbers 7 + 5i Write the quotient in standard form. 1  4i 7 + 5i 1 – 4i = 1 + 4i Multiply numerator and denominator by 1 + 4i, the complex conjugate of 1 – 4i. 7 + 28i + 5i + 20i2 1 + 4i – 4i – 16i2 = Multiply using FOIL. 7 + 33i + 20(–1) 1 – 16(–1) = Simplify and use i2 = -1. – i 17 = Simplify.\n\nDivide complex numbers\nEXAMPLE 5 Divide complex numbers 13 17 = + 33 i Write in standard form.\n\nGUIDED PRACTICE for Examples 3, 4 and 5 Write the expression as a complex number in standard form. 5 1 + i 11. i(9 – i) 13. 5 2 i ANSWER 1 + 9i ANSWER 14. 5 + 2i 12. (3 + i)(5 – i) 3 – 2i 11 13 + 16 i ANSWER 16 + 2i ANSWER" ]	[ null, "http://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67986655,"math_prob":0.9999294,"size":1316,"snap":"2021-43-2021-49","text_gpt3_token_len":510,"char_repetition_ratio":0.16615854,"word_repetition_ratio":0.12543555,"special_character_ratio":0.4224924,"punctuation_ratio":0.11235955,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99974734,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T02:00:41Z\",\"WARC-Record-ID\":\"<urn:uuid:2cc60c06-3e0a-46f3-a1d1-39c5ad33baa2>\",\"Content-Length\":\"142488\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5383c50d-24c9-4468-9bcb-77440689f587>\",\"WARC-Concurrent-To\":\"<urn:uuid:c4cb3575-a796-4e0c-a66f-c2dbeb08ada8>\",\"WARC-IP-Address\":\"138.201.58.10\",\"WARC-Target-URI\":\"http://slideplayer.com/slide/4586606/\",\"WARC-Payload-Digest\":\"sha1:3LUF4FIUOSK3TCV2GA5I5KEQZFKIAIWL\",\"WARC-Block-Digest\":\"sha1:WKNW6WX7PEFFQOZF7F6FFOXGPLSWPBHG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363420.81_warc_CC-MAIN-20211207232140-20211208022140-00491.warc.gz\"}"}
https://www.asvabtestbank.com/math-knowledge/flash-cards/967804/10	[ "## Math Knowledge Flash Card Set 967804\n\n Cards 10 Topics Angles Around Lines & Points, Calculations, Classifications, Coordinate Grid, Dimensions, Parallel Lines, Pythagorean Theorem\n\n#### Study Guide\n\n###### Angles Around Lines & Points\n\nAngles around a line add up to 180°. Angles around a point add up to 360°. When two lines intersect, adjacent angles are supplementary (they add up to 180°) and angles across from either other are vertical (they're equal).\n\n###### Calculations\n\nThe circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2.\n\n###### Classifications\n\nA monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms. Linear expressions have no exponents. A quadratic expression contains variables that are squared (raised to the exponent of 2).\n\n###### Coordinate Grid\n\nThe coordinate grid is composed of a horizontal x-axis and a vertical y-axis. The center of the grid, where the x-axis and y-axis meet, is called the origin.\n\n###### Dimensions\n\nA circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter (AC, CB, CD). A chord is a line segment that connects any two points along its perimeter (AB, AD, BD). The diameter of a circle is the length of a chord that passes through the center of the circle (AB) and equals twice the circle's radius (2r).\n\n###### Parallel Lines\n\nParallel lines are lines that share the same slope (steepness) and therefore never intersect. A transversal occurs when a set of parallel lines are crossed by another line. All of the angles formed by a transversal are called interior angles and angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°) and are called corresponding angles. Alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°) and all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other. Same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°).\n\n###### Pythagorean Theorem\n\nThe Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, $$c = \\sqrt{a + b}$$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90863246,"math_prob":0.998944,"size":2297,"snap":"2020-24-2020-29","text_gpt3_token_len":589,"char_repetition_ratio":0.120366335,"word_repetition_ratio":0.011961723,"special_character_ratio":0.2716587,"punctuation_ratio":0.11208791,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995072,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-29T14:12:32Z\",\"WARC-Record-ID\":\"<urn:uuid:e320250b-b51f-4c8c-8560-6e991ca16b3d>\",\"Content-Length\":\"12610\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d2c5581-76b0-48a4-b130-1757ece75b04>\",\"WARC-Concurrent-To\":\"<urn:uuid:f710e3c2-365d-4d5a-928d-430dbce2f3be>\",\"WARC-IP-Address\":\"68.183.116.70\",\"WARC-Target-URI\":\"https://www.asvabtestbank.com/math-knowledge/flash-cards/967804/10\",\"WARC-Payload-Digest\":\"sha1:FWOFCUMGAKNGQDA6VMNMC4VFRGSS7EVK\",\"WARC-Block-Digest\":\"sha1:XVJEVDHI7WMRQNEDAGIGEELWEKJR52XG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347404857.23_warc_CC-MAIN-20200529121120-20200529151120-00045.warc.gz\"}"}
https://www.mmhub.cn/blog/2016/10/21/mathematical-model/	[ "# Mathematical model(数学模型)\n\nA mathematical model is an abstract model that uses mathematical language to describe the behavior of a system.\n\nMathematical models are used particularly in the natural sciences and engineering disciplines (such as physics, biology, and electrical engineering) but also in the social sciences (such as economics, sociology and political science); physicists, engineers, computer scientists, and economists use mathematical models most extensively.\n\nEykhoff (1974) defined a mathematical model as a representation of the essential aspects of an existing system (or a system to be constructed) which presents knowledge of that system in usable form.\n\nEykhoff (1974)把数学模型定义为一个现存（或准备建立的）系统的本质特征以实用的信息形式表达出来的一种描述。\n\nMathematical models can take many forms, including but not limited to dynamical systems, statistical models, differential equations, or game theoretic models.\n\nThese and other types of models can overlap, with a given model involving a variety of abstract structures.\n\nThere are six basic groups of variables: decision variables, input variables, state variables, exogenous variables, random variables, and output variables.\n\nSince there can be many variables of each type, the variables are generally represented by vectors.\n\nMathematical modelling problems are often classified into black box or white box models, according to how much a prior information is available of the system.\n\nA black-box model is a system of which there is no a prior information available.\n\nA white-box model (also called glass box or clear box) is a system where all necessary information is available.\n\nPractically all systems are somewhere between the black-box and white-box models, so this concept only works as an intuitive guide for approach.\n\nUsually it is preferable to use as much a prior information as possible to make the model more accurate.\n\nvia:https://www.sciencedaily.com/terms/mathematical_model.htm" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6491779,"math_prob":0.9405529,"size":2347,"snap":"2022-40-2023-06","text_gpt3_token_len":838,"char_repetition_ratio":0.13828425,"word_repetition_ratio":0.0,"special_character_ratio":0.17000426,"punctuation_ratio":0.1046832,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99475086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T18:13:04Z\",\"WARC-Record-ID\":\"<urn:uuid:6712483d-ef5c-4203-8ec2-05c9918b0e14>\",\"Content-Length\":\"66945\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb493d05-8e12-4f31-a240-6ea426288cb6>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2104dcf-a538-4088-92c7-f537d4590b5d>\",\"WARC-IP-Address\":\"123.56.64.175\",\"WARC-Target-URI\":\"https://www.mmhub.cn/blog/2016/10/21/mathematical-model/\",\"WARC-Payload-Digest\":\"sha1:N6WEDVEBETDC3S6B7UEFB7YHOOTU2YOU\",\"WARC-Block-Digest\":\"sha1:V675UZJKD57XHTOQPUJQBQAIQSIWG4RX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337428.0_warc_CC-MAIN-20221003164901-20221003194901-00552.warc.gz\"}"}
http://www.codewod.com/2012/05/fibonacci-sequence.html	[ "## How to\n\nWhen solving these problems, learning by doing is much better than learning by reading. I encourage to you read only as far in the solution as you need, then trying to solve the problem. If you get stuck, try reading a little further. And of course, let me know if you find a better solution!\n\n## Tuesday, May 15, 2012\n\n### The Fibonacci Sequence\n\nProblem: Write a recursive function to prints the xth Fibonacci number.\n\nint finNum(int num) {...}\n\nSolution:\n\nThis is a canonical recursive problem, but can be encountered. Let's start this problem with a review of the Fibonacci sequence. This is a sequence of numbers where each number is the sum of the previous two numbers, and the first two numbers in the sequence are 0 and 1.\n\nA mathematician expresses this as:", null, "$F_n = F_{n-1} + F_{n-2},\\!\\,$", null, "$F_0 = 0,\\; F_1 = 1.$\n\nNow, let's do an example of the first say, 7, Fibonacci numbers to make sure that we get the pattern:\n\n0\n1\n1\n2\n3\n5\n8\n\nAnd we could go more.\n\nWell, with this, it seems set up for recursion. We have our two base cases, which is if the num is 1 or 2. And, otherwise, we have the sum of the two previous numbers.\n\nWith this, we can code it up, and we need to check for base cases:\n\nint fibNum(int num) {\n\n/* Check for the error case of negative number or 0 /\nif (num <= 0) return ERROR;\n\n/ Now, enter the two base cases /\nif (num == 1) return 0;\nelse if (num == 2) return 1;\n\n/ and now, recursive call to figure out the number /\nelse return (fibNum(num-1) + fibNum(num-2));\n}\n\nAnd, that's it. Let's do an example, say 7 above, and walk through it to make sure that it works.\n\nIt does!\n\nSo, this solution works recursively. What is it's runtime?\n\nWell, it calls fibNum two times more as num grows. This means that the function is O(2^n). This is horrible for even small sizes of n. Wow. Just awful!\n\nThink about how to make this better, while still recurring. That's tomorrow's problem.\n\n/ Please let me know if you find any bugs here. */" ]	[ null, "http://upload.wikimedia.org/wikipedia/en/math/0/c/e/0cebc512d9a3ac497eda6f10203f792e.png ", null, "http://upload.wikimedia.org/wikipedia/en/math/a/9/2/a92c5f0981136ba333124cdfe6d3c3ce.png ", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92228496,"math_prob":0.98590493,"size":2253,"snap":"2021-43-2021-49","text_gpt3_token_len":570,"char_repetition_ratio":0.09915518,"word_repetition_ratio":0.009569378,"special_character_ratio":0.26098534,"punctuation_ratio":0.15040651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986598,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T21:27:17Z\",\"WARC-Record-ID\":\"<urn:uuid:858f4ad6-c555-44c8-8fb4-f66d8b8f043d>\",\"Content-Length\":\"48211\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9afe6871-5c09-4e69-83f8-05ee3c90ce7e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e74ddec3-17b6-4d63-8ff9-f3c9a5eccd72>\",\"WARC-IP-Address\":\"142.250.65.83\",\"WARC-Target-URI\":\"http://www.codewod.com/2012/05/fibonacci-sequence.html\",\"WARC-Payload-Digest\":\"sha1:5VDVYXLHLRLOMG25U4FGI2H7SEYJXGJ7\",\"WARC-Block-Digest\":\"sha1:Z2FZNV3UJLVC22KNU6TADUGMFEXJXG6Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585281.35_warc_CC-MAIN-20211019202148-20211019232148-00082.warc.gz\"}"}
http://quantum.ustc.edu.cn/web/index.php/publications?page=6	[ "# 论文发表\n\n1. Chen, L. -K., Da Li, Z. -, Yao, X. -C., Huang, M., Li, W., Lu, H., Yuan, X., Zhang, Y. -B., Jiang, X., Peng, C. -Z., Li, L., Liu, N. -L., Ma, X., Lu, C. -Y., Chen, Y. -A. & Pan, J. -W. Observation of ten-photon entanglement using thin BiB\\_3O\\_6 crystals. Optica 4, 77 (2017).\n2. Zou, W. -J., Li, Y. -H., Wang, S. -C., Cao, Y., Ren, J. -G., Yin, J., Peng, C. -Z., Wang, X. -B. & Pan, J. -W. Protecting entanglement from finite-temperature thermal noise via weak measurement and quantum measurement reversal. Physical Review A 95, 042342 (2017).\n3. Yang, B., Chen, Y. -Y., Zheng, Y. -G., Sun, H., Dai, H. -N., Guan, X. -W., Yuan, Z. -S. & Pan, J. -W. Quantum criticality and the Tomonaga-Luttinger liquid in one-dimensional Bose gases. Physical Review Letters 119, 165701 (2017).\n4. He, Y., He, Y. -M., Wei, Y. -J., Jiang, X., Chen, K., Lu, C. -Y., Pan, J. -W., Schneider, C., Kamp, M. & Hofling, S. Quantum State Transfer from a Single Photon to a Distant Quantum-Dot Electron Spin. Physical Review Letters 119, 060501 (2017).\n5. Wu, C., Bai, B., Liu, Y., Zhang, X., Yang, M., Cao, Y., Wang, J., Zhang, S., Zhou, H., Shi, X., Ma, X., Ren, J. -G., Zhang, J., Peng, C. -Z., Fan, J. -Y., Zhang, Q. & Pan, J. -W. Random Number Generation with Cosmic Photons. Physical Review Letters 118, 140402 (2017).\n6. Yin, J., Cao, Y., Li, Y. -H., Liao, S. -K., Zhang, L., Ren, J. -G., Cai, W. -Q., Liu, W. -Y., Li, B., Dai, H., Li, G. -B., Lu, Q. -M., Gong, Y. -H., Xu, Y., Li, S. -L., Li, F. -Z., Yin, Y. -Y., Jiang, Z. -Q., Li, M., Jia, J. -J., Ren, G., He, D., Zhou, Y. -L., Zhang, X. -X., Wang, N., Chang, X., Zhu, Z. -C., Liu, N. -L., Chen, Y. -A., Lu, C. -Y., Shu, R., Peng, C. -Z., Wang, J. -Y. & Pan, J. -W. Satellite-based entanglement distribution over 1200 kilometers. Science 356, 1140-1144 (2017).\n7. Yin, J., Cao, Y., Li, Y. -H., Ren, J. -G., Liao, S. -K., Zhang, L., Cai, W. -Q., Liu, W. -Y., Li, B., Dai, H., Li, M., Huang, Y. -M., Deng, L., Li, L., Zhang, Q., Liu, N. -L., Chen, Y. -A., Lu, C. -Y., Peng, C., Wang, J. -Y., Shu, R. & Pan, J. -W. Satellite-to-Ground Entanglement-Based Quantum Key Distribution. Physical Review Letters 119, 200501 (2017).\n8. Liao, S. -K., Cai, W. -Q., Liu, W. -Y., Zhang, L., Li, Y., Ren, J. -G., Yin, J., Shen, Q., Cao, Y., Li, Z. -P., Li, F. -Z., Chen, X. -W., Sun, L. -H., Jia, J. -J., Wu, J. -C., Jiang, X. -J., Wang, J. -F., Huang, Y. -M., Wang, Q., Zhou, Y. -L., Deng, L., Xi, T., Ma, L., Hu, T., Zhang, Q., Chen, Y. -A., Liu, N. -L., Wang, X. -B., Zhu, Z. -C., Lu, C. -Y., Shu, R., Peng, C. -Z., Wang, J. -Y. & Pan, J. -W. Satellite-to-ground quantum key distribution. Nature 549, 43–47 (2017).\n9. Zhou, N., Jiang, W. -H., Chen, L. -K., Fang, Y. -Q., Da Li, Z. -, Liang, H., Chen, Y. -A., Zhang, J. & Pan, J. -W. Sine wave gating silicon single-photon detectors for multiphoton entanglement experiments. Review of Scientific Instruments 88, 083102 (2017).\n10. Zheng, Y., Song, C., Chen, M. -C., Xia, B., Liu, W., Guo, Q., Zhang, L., Xu, D., Deng, H., Huang, K., Wu, Y., Yan, Z., Zheng, D., Lu, L., Pan, J. -W., Wang, H., Lu, C. -Y. & Zhu, X. Solving Systems of Linear Equations with a Superconducting Quantum Processor. Physical Review Letters 118, 210504 (2017).\n11. Liao, S. -K., Lin, J., Wang, J. -Y., Liu, W. -Y., Qiang, J., Yin, J., Li, Y., Shen, Q., Zhang, L., Liang, X. -F., Yong, H. -L., Li, F. -Z., Yin, Y. -Y., Cao, Y., Cai, W. -Q., Zhang, W. -Z., Jia, J. -J., Wu, J. -C., chen, xiao-wen & Pan, J. -W. Space-to-Ground Quantum Key Distribution Using a Small-Sized Payload on Tiangong-2 Space Lab. Chinese Physics Letters 34, 090302 (2017).\n12. Yang, B., Dai, H. -N., Sun, H., Reingruber, A., Yuan, Z. -S. & Pan, J. -W. Spin-dependent optical superlattice. Physical Review A 96, 011602 (2017).\n13. He, Y., Ding, X., Su, Z. -E., Huang, H. -L., Qin, J., Wang, C., Unsleber, S., Chen, C., Wang, H., He, Y. -M., Wang, X. -L., Zhang, W., Chen, S. -J., Schneider, C., Kamp, M., You, L., Wang, Z., Hofling, S., Lu, C. -Y. & Pan, J. -W. Time-Bin-Encoded Boson Sampling with a Single-Photon Device. Physical Review Letters 118, 190501 (2017).\n14. Xu, P., Yong, H. -L., Chen, L. -K., Liu, C., Xiang, T., Yao, X. -C., Lu, H., Da Li, Z. -, Liu, N. -L., Li, L., Yang, T., Peng, C. -Z., Zhao, B., Chen, Y. -A. & Pan, J. -W. Two-Hierarchy Entanglement Swapping for a Linear Optical Quantum Repeater. Physical Review Letters 119, 170502 (2017).\n15. Wang, X., Liang, F. -T., Miao, P., Qian, Y. & Jin, G. 10-Gbps true random number generator accomplished in ASIC. 1-4 (2016). doi:10.1109/RTC.2016.7543082\n16. Navarro, C. A., Huang, W. & Deng, Y. Adaptive multi-GPU Exchange Monte Carlo for the 3D Random Field Ising Model. Computer Physics Communications 205, 48-60 (2016).\n17. Shangguan, M., Xia, H., Wang, C., Qiu, J., Shentu, G., Zhang, Q., Dou, X. & Pan, J. -W. All-fiber upconversion high spectral resolution wind lidar using a Fabry-Perot interferometer. Optics Express 24, 19322-19336 (2016).\n18. Yang, S. -J., Wang, X. -J., Bao, X. -H. & Pan, J. -W. An efficient quantum light–matter interface with sub-second lifetime. Nature Photonics 10, 381-384 (2016).\n19. Sanders, B. Asked to speak in a developing country? Say yes!. Physics Today 69, 10-12 (2016).\n20. Xia, H., Shangguan, M., Shentu, G., Wang, C., Qiu, J., Zheng, M., Xie, X., Dou, X., Zhang, Q. & Pan, J. -W. Brillouin optical time-domain reflectometry using up-conversion single-photon detector. Optics Communications 381, 37-42 (2016)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59174675,"math_prob":0.8422235,"size":2293,"snap":"2020-24-2020-29","text_gpt3_token_len":604,"char_repetition_ratio":0.17169069,"word_repetition_ratio":0.0,"special_character_ratio":0.31966856,"punctuation_ratio":0.16624041,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96814424,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T16:31:44Z\",\"WARC-Record-ID\":\"<urn:uuid:a33e7434-5931-4331-aa45-3be154fb892d>\",\"Content-Length\":\"37366\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8db26ca6-02d5-4794-8672-22757cca164a>\",\"WARC-Concurrent-To\":\"<urn:uuid:85de12c0-5809-4485-b0c9-345ff89dc011>\",\"WARC-IP-Address\":\"218.104.71.173\",\"WARC-Target-URI\":\"http://quantum.ustc.edu.cn/web/index.php/publications?page=6\",\"WARC-Payload-Digest\":\"sha1:VA3VPQEDL3COYR5Z5SFXKM6BA74BLPDU\",\"WARC-Block-Digest\":\"sha1:UR6UUJZY4W2OYTCYB3C3X66C5JM66MXD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655911092.63_warc_CC-MAIN-20200710144305-20200710174305-00275.warc.gz\"}"}
https://blogs.msdn.microsoft.com/calvin_hsia/2007/09/05/querying-northwind-with-group-by-bug-fixed/	[ "# Querying Northwind with Group By bug fixed\n\nIn this post:\n\nI mentioned how a bug with Group By and composite keys made more local calculations. Using the AsEnumerable operator caused the query to execute locally.\n\nI now have a build of Visual Studio that has the bug fixed and I modified the query to remove the AsEnumerable().\n\nThe performance is very quick, as it’s all executed on the back end, as seen by this log:\n\nSELECT [t7].[CustomerID], [t7].[CompanyName], [t7].[value] AS [CustTotal], [t7].[value2] AS [PctTotal]\n\nFROM (\n\nSELECT SUM([t6].[value2]) AS [value], SUM([t6].[value]) AS [value2], [t6].[CustomerID], [t6].[CompanyName]\n\nFROM (\n\nSELECT (@p1 * (CONVERT(Real,(CONVERT(Decimal(29,4),CONVERT(Int,[t3].[Quantity]))) * [t3].[UnitPrice])) * (@p2 - [t3].[Discount])) / ((\n\nSELECT SUM([t5].[value])\n\nFROM (\n\nSELECT (CONVERT(Real,(CONVERT(Decimal(29,4),CONVERT(Int,[t4].[Quantity]))) * [t4].[UnitPrice])) * (@p3 - [t4].[Discount]) AS [value]\n\nFROM [dbo].[Order Details] AS [t4]\n\n) AS [t5]\n\n)) AS [value], [t3].[CustomerID], [t3].[CompanyName], [t3].[value] AS [value2]\n\nFROM (\n\nSELECT (CONVERT(Real,(CONVERT(Decimal(29,4),CONVERT(Int,[t2].[Quantity]))) * [t2].[UnitPrice])) * (@p0 - [t2].[Discount]) AS [value], [t2].[Quantity], [t2].[UnitPrice], [t2].[Discount], [t0].[CustomerID], [t0].[CompanyName]\n\nFROM [dbo].[Customers] AS [t0]\n\nINNER JOIN [dbo].[Orders] AS [t1] ON [t0].[CustomerID] = [t1].[CustomerID]\n\nINNER JOIN [dbo].[Order Details] AS [t2] ON [t1].[OrderID] = [t2].[OrderID]\n\n) AS [t3]\n\n) AS [t6]\n\nGROUP BY [t6].[CustomerID], [t6].[CompanyName]\n\n) AS [t7]\n\nORDER BY [t7].[value2] DESC, [t7].[CustomerID]\n\n-- @p0: Input Real (Size = 0; Prec = 0; Scale = 0) \n\n-- @p1: Input Real (Size = 0; Prec = 0; Scale = 0) \n\n-- @p2: Input Real (Size = 0; Prec = 0; Scale = 0) \n\n-- @p3: Input Real (Size = 0; Prec = 0; Scale = 0) \n\n-- Context: SqlProvider(Sql2005) Model: AttributedMetaModel Build: 3.5.20828.0\n\nThe entire code is posted here:\n\nImports System.IO\n\nPublic Class Form1\n\nPrivate Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load\n\nMe.Width = 1024\n\nMe.Height = 768\n\nDim NWind As New DataClasses1DataContext\n\nNWind.Log = Console.Out\n\nDim q = From cust In NWind.Customers _\n\nJoin ord In NWind.Orders On cust.CustomerID Equals ord.CustomerID _\n\nJoin det In NWind.Order_Details On ord.OrderID Equals det.OrderID _\n\nGroup By cust.CustomerID, cust.CompanyName Into _\n\nCustTotal = Sum(det.Quantity * det.UnitPrice * (1 - det.Discount)), _\n\nPctTotal = Sum(100 * (det.Quantity * det.UnitPrice * (1 - det.Discount)) / _\n\nAggregate d2 In NWind.Order_Details _\n\nInto Sum(d2.Quantity * d2.UnitPrice * (1 - d2.Discount))) _\n\nOrder By CustomerID _\n\nSelect CustomerID, CompanyName, CustTotal, PctTotal _\n\nOrder By PctTotal Descending\n\nBrowse(q)\n\nEnd Sub\n\nSub Browse(Of t)(ByVal seq As IEnumerable(Of t))\n\nDim GridView As New DataGridView\n\nGridView.Width = Me.Width\n\nGridView.Height = Me.Height\n\nDim pl = New List(Of t)(seq)\n\nGridView.DataSource = pl\n\nMe.Text = pl.Count.ToString\n\nGridView.Dock = DockStyle.Fill\n\nGridView.AutoResizeColumns()\n\nEnd Sub\n\nEnd Class\n\nTags\n\n1.", null, "Calvin Hsia s WebLog Querying Northwind with Group By bug fixed \| fix my credit says:" ]	[ null, "https://secure.gravatar.com/avatar/3651e719491c63d4424134d66fe53db9", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5172929,"math_prob":0.9684279,"size":3261,"snap":"2020-10-2020-16","text_gpt3_token_len":1017,"char_repetition_ratio":0.12496162,"word_repetition_ratio":0.086124405,"special_character_ratio":0.33701319,"punctuation_ratio":0.22408536,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99537635,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-28T22:17:36Z\",\"WARC-Record-ID\":\"<urn:uuid:5b780eb2-75d1-454c-9d11-57af2e89118c>\",\"Content-Length\":\"69228\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fbc29e55-441a-4b76-bce6-1d2aee53b307>\",\"WARC-Concurrent-To\":\"<urn:uuid:939621c7-4dcd-49c1-b02e-48b166ed908d>\",\"WARC-IP-Address\":\"23.47.22.65\",\"WARC-Target-URI\":\"https://blogs.msdn.microsoft.com/calvin_hsia/2007/09/05/querying-northwind-with-group-by-bug-fixed/\",\"WARC-Payload-Digest\":\"sha1:I2EGIQJJVMBBES4IHISSN6KPCZXHXIE7\",\"WARC-Block-Digest\":\"sha1:VN4IRBFHP3L43JA2LHWA7HHK2E2DFQ5V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370493120.15_warc_CC-MAIN-20200328194743-20200328224743-00003.warc.gz\"}"}
https://appliedprobability.blog/2019/01/26/dynamic-programming-4/	[ "# Dynamic Programming\n\nWe briefly explain the principles behind dynamic programming and then give its definition.\n\n### An Introductory Example\n\nIn the figure below there is a tree consisting of a root node labelled", null, "$R$ and two leaf nodes colored grey. For each edge, there is a cost. Your task is to find the lowest cost path from the root node to a leaf.", null, "There are a number of ways to solve this, such as enumerating all paths. However, we are interested in one approach where the problem is solved backwards, through a sequence of smaller sub-problems. Specifically, once we reach the penultimate node on the left (in the dashed box) then it is clearly optimal to go left with a cost of", null, "$1$. This solves an easier sub problem and, after solving each sub problem, we can then attack a slightly bigger problem. If we solve for each leaf in this way we can solve the problem for the antepenultimate nodes (the node before the penultimate node).\n\nThus the problem of optimizing the cost of the original tree can be broken down to a sequence of much simpler optimizations given by the shaded boxed below.", null, "From this we see the optimal path has a cost of", null, "$5$ and consists of going right, then left, then right.\n\nLet’s consider the problem a little more generally in the next figure. The tree on the righthand-side has a lowest cost path of value", null, "$L_{rhs}$ and the lefthand-side tree has lowest cost", null, "$L_{lhs}$ and the edges leading to each, respective tree, have costs", null, "$l_{rhs}$ and", null, "$l_{lhs}$. Once the decision to go left or right is made (at cost", null, "$l_{rhs}$ or", null, "$l_{lhs}$) it is optimal to follow the lowest cost path (at cost", null, "$L_{rhs}$ or", null, "$L_{lhs}$). So", null, "$L$, the minimal cost path from the root to a leaf node satisfies", null, "Similarly, convince yourself that the same argument applies from any node", null, "$x$ in the tree network that is", null, "where", null, "$L_{x}$ is the minimum cost from", null, "$x$ to a leaf node and where for", null, "$a\\in\\{lhs,rhs \\}$", null, "$x(a)$ is the node to the lefthand-side or righthand-side of", null, "$x$. The equation above is an example of the Bellman equation for this problem, and in our example, we solved this equation recursively starting from leaf nodes and working our way back to the root node.\n\nThe idea of solving a problem from back to front and the idea of iterating on the above equation to solve an optimisation problem lies at the heart of dynamic programming.\n\n## Definition\n\nWe now give a general definition of a dynamic programming:\n\nTime is discrete", null, "$t=0,1,...,T$;", null, "$x_t\\in \\mathcal{X}$ is the state at time", null, "$t$;", null, "$a_t\\in \\mathcal{A}_t$ is the action at time", null, "$t$; The state evolves according to functions", null, "$f_t: \\mathcal{X}\\times \\mathcal{A}_t \\rightarrow \\mathcal{X}$. Here", null, "This is called the Plant Equation. A policy", null, "$\\pi$ choses an action", null, "$\\pi_t$ at each time", null, "$t$. The (instantaneous) reward for taking action", null, "$a$ in state", null, "$x$ at time", null, "$t$ is", null, "$r_t(a,x)$ and", null, "$r_T(x)$ is the reward for terminating in state", null, "$x$ at time", null, "$T$.\n\nDef.[Dynamic Program] Given initial state", null, "$x_0$, a dynamic program is the optimization", null, "Further, let", null, "$R_\\tau(x_{\\tau},{\\bf a})$ (Resp.", null, "$V_\\tau(x_\\tau)$) be the objective (Resp. optimal objective) for when the summation is started from", null, "$t=\\tau$, rather than", null, "$t=0$.\n\nThrm [Bellman’s Equation]", null, "$V_T(x) = r_T(x)$ and for", null, "$t=T-1,...,0$", null, "where", null, "$x_t\\in\\mathcal{X}$ and", null, "$x_{t+1}=f_t(x_t,a_t)$.\n\nProof. Let", null, "${\\bf a}_t=(a_t,...,a_{T-1})$. Note that", null, "$R_{t}(x_{t},{\\mathbf a}_t) = r_t(x_t,a_t) + R_{t+1}(x_{t+1},{\\mathbf a}_{t+1}).$", null, "", null, "$\\square$" ]	[ null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/tree_practice_b_box.jpg", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/tree_practice_c-1.jpg", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/tree_practice_3.jpg", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/7609143500d3608b1ea4f141e9a3da3e.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/a2c69936a0847683029129b35bbfc573.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/screenshot-2019-01-26-at-10.20.12.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/screenshot-2019-01-26-at-10.22.18.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://appliedprobability.files.wordpress.com/2019/01/2c8305be01cb578eebd26d2d321d8d0c.png", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92860353,"math_prob":0.9998565,"size":2867,"snap":"2021-31-2021-39","text_gpt3_token_len":613,"char_repetition_ratio":0.12818721,"word_repetition_ratio":0.0038461538,"special_character_ratio":0.20927799,"punctuation_ratio":0.088652484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000015,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108],"im_url_duplicate_count":[null,null,null,5,null,null,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,null,null,null,null,null,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T23:35:05Z\",\"WARC-Record-ID\":\"<urn:uuid:508d7365-7c4c-41fe-a43d-2ccc7d7634bd>\",\"Content-Length\":\"89328\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e31df30d-e92c-4f0b-b642-94f398792f87>\",\"WARC-Concurrent-To\":\"<urn:uuid:830f35d0-ec3e-436b-b099-2d2e34fc8329>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://appliedprobability.blog/2019/01/26/dynamic-programming-4/\",\"WARC-Payload-Digest\":\"sha1:KXI7L2K55SCFIYPV37VOY2ZH754EC6VF\",\"WARC-Block-Digest\":\"sha1:A5PAN776I7D5KZCEO2QK4EJCPK353QP3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055808.78_warc_CC-MAIN-20210917212307-20210918002307-00284.warc.gz\"}"}
http://qgcpkjw.com/Ofxt/joefy.iunm?dje=7	[ "# 新年新展望，长安汽车旗下轿车长安逸动PLUS即将登场！", null, "# 打消冬季行车疑虑昂希诺纯电动为“电车”吹来暖风", null, "1月18日，北京现代首款纯电动SUV昂希诺纯电动，给了我们大家一个深入体验冬季行车的机会，来一场说走就走的冬季自驾纯电动新能源汽车的新体验。是由北京现代胜鸿都特约经销店举办的昂希诺纯电动车友会自驾活动，...\n\n# 中国老炮与德系新品之间的对决新H6 Coupe更胜一筹", null, "# 除了速度，菲斯塔凭什么敢叫板思域\n\n“时尚、操控、智能”,时下年轻人购车的三大需求。思域历经十代演变,已经是本田造车史上最成熟的车型之一。而菲斯塔则更迎合年轻人的需求,可以说是专门为年轻消费者...\n\n# 奇瑞新能源S61冬季高寒测试曝光，预估2020年上半年上市\n\n020年1-2月份期间，S61测试团队前往素有“神州边陲”之称的黑龙江黑河，在-35°的极寒环境中进行首次高寒测试。高寒测试是奇瑞新能源对S61整车“三高测试”的项目之...\n\n\n• 快速找车\n• 选择品牌\n• 选择品牌\n• A 奥迪\n• A 阿斯顿·马丁\n• A 阿尔法·罗密欧\n• B 宝沃\n• B 布加迪\n• B 巴博斯\n• B 保时捷\n• B 宾利\n• B 奔驰\n• B 宝马\n• B 本田\n• B 别克\n• B 标致\n• B 比亚迪\n• B 宝骏\n• B 北汽制造\n• B 北汽新能源\n• B 北汽幻速\n• B 北汽威旺\n• B 北京汽车\n• B 奔腾\n• B 北汽绅宝\n• C 长安\n• C 长安商用\n• C 长城\n• C 昌河\n• D 大众\n• D 道奇\n• D DS\n• D 东南\n• D 东风风神\n• D 东风风行\n• D 东风小康\n• D 东风风度\n• D 东风\n• F 福特\n• F 丰田\n• F 菲亚特\n• F 法拉利\n• F 福田\n• F 福迪\n• F 福汽启腾\n• G 观致\n• G 广汽传祺\n• G 广汽吉奥\n• G GMC\n• H 红旗\n• H 汉腾汽车\n• H 哈弗\n• H 哈飞\n• H 海格\n• H 海马\n• H 华颂\n• H 黄海\n• H 华泰\n• H 恒天\n• J 吉利汽车\n• J 捷豹\n• J Jeep\n• J 江淮\n• J 江铃\n• J 金杯\n• J 九龙\n• J 金旅\n• K 凯翼\n• K 凯迪拉克\n• K 克莱斯勒\n• K 科尼塞克\n• K 卡威\n• K 开瑞\n• L 路虎\n• L 林肯\n• L 劳斯莱斯\n• L 兰博基尼\n• L 雷克萨斯\n• L 铃木\n• L 雷诺\n• L 理念\n• L 力帆\n• L 莲花汽车\n• L 猎豹\n• L 路特斯\n• L 陆风\n• M 马自达\n• M MG\n• M MINI\n• M 玛莎拉蒂\n• M 摩根\n• M 迈凯轮\n• N 纳智捷\n• O 欧宝\n• O 讴歌\n• O 欧朗\n• Q 奇瑞\n• Q 起亚\n• Q 启辰\n• R 日产\n• R 荣威\n• R 瑞麒\n• S 三菱\n• S 斯威汽车\n• S 萨博\n• S smart\n• S 斯柯达\n• S 斯巴鲁\n• S 思铭\n• S 双龙\n• S 上汽大通\n• S 双环\n• T 特斯拉\n• T 腾势\n• W 沃尔沃\n• W 五菱汽车\n• W 五十铃\n• W 威兹曼\n• W 威麟\n• X 现代\n• X 雪佛兰\n• X 雪铁龙\n• X 西雅特\n• Y 一汽\n• Y 英菲尼迪\n• Y 英致\n• Y 依维柯\n• Y 野马汽车\n• Y 永源\n• Z 众泰\n• Z 中华\n• Z 中兴\n• Z 知豆\n• 选择车系\n• 选择车系\n• 车型对比\n• 选择品牌\n• 选择品牌\n• A 奥迪\n• A 阿斯顿·马丁\n• A 阿尔法·罗密欧\n• B 宝沃\n• B 布加迪\n• B 巴博斯\n• B 保时捷\n• B 宾利\n• B 奔驰\n• B 宝马\n• B 本田\n• B 别克\n• B 标致\n• B 比亚迪\n• B 宝骏\n• B 北汽制造\n• B 北汽新能源\n• B 北汽幻速\n• B 北汽威旺\n• B 北京汽车\n• B 奔腾\n• B 北汽绅宝\n• C 长安\n• C 长安商用\n• C 长城\n• C 昌河\n• D 大众\n• D 道奇\n• D DS\n• D 东南\n• D 东风风神\n• D 东风风行\n• D 东风小康\n• D 东风风度\n• D 东风\n• F 福特\n• F 丰田\n• F 菲亚特\n• F 法拉利\n• F 福田\n• F 福迪\n• F 福汽启腾\n• G 观致\n• G 广汽传祺\n• G 广汽吉奥\n• G GMC\n• H 红旗\n• H 汉腾汽车\n• H 哈弗\n• H 哈飞\n• H 海格\n• H 海马\n• H 华颂\n• H 黄海\n• H 华泰\n• H 恒天\n• J 吉利汽车\n• J 捷豹\n• J Jeep\n• J 江淮\n• J 江铃\n• J 金杯\n• J 九龙\n• J 金旅\n• K 凯翼\n• K 凯迪拉克\n• K 克莱斯勒\n• K 科尼塞克\n• K 卡威\n• K 开瑞\n• L 路虎\n• L 林肯\n• L 劳斯莱斯\n• L 兰博基尼\n• L 雷克萨斯\n• L 铃木\n• L 雷诺\n• L 理念\n• L 力帆\n• L 莲花汽车\n• L 猎豹\n• L 路特斯\n• L 陆风\n• M 马自达\n• M MG\n• M MINI\n• M 玛莎拉蒂\n• M 摩根\n• M 迈凯轮\n• N 纳智捷\n• O 欧宝\n• O 讴歌\n• O 欧朗\n• Q 奇瑞\n• Q 起亚\n• Q 启辰\n• R 日产\n• R 荣威\n• R 瑞麒\n• S 三菱\n• S 斯威汽车\n• S 萨博\n• S smart\n• S 斯柯达\n• S 斯巴鲁\n• S 思铭\n• S 双龙\n• S 上汽大通\n• S 双环\n• T 特斯拉\n• T 腾势\n• W 沃尔沃\n• W 五菱汽车\n• W 五十铃\n• W 威兹曼\n• W 威麟\n• X 现代\n• X 雪佛兰\n• X 雪铁龙\n• X 西雅特\n• Y 一汽\n• Y 英菲尼迪\n• Y 英致\n• Y 依维柯\n• Y 野马汽车\n• Y 永源\n• Z 众泰\n• Z 中华\n• Z 中兴\n• Z 知豆\n• 选择车系\n• 选择车系\n• 选择车型\n• 选择车型\n• 意见反馈" ]	[ null, "http://www.chextx.com/Uploads/news/2020/01/5e26a581dcfad1003014166.jpg", null, "http://www.chextx.com/Uploads/news/2020/01/5e269eb7dd98f1313652373.jpg", null, "http://www.chextx.com/Uploads/news/2020/01/5e2577790549d772666926.jpg", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.81984174,"math_prob":0.68694323,"size":1615,"snap":"2019-51-2020-05","text_gpt3_token_len":1738,"char_repetition_ratio":0.12911235,"word_repetition_ratio":0.0,"special_character_ratio":0.26749226,"punctuation_ratio":0.160221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99977404,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T15:34:41Z\",\"WARC-Record-ID\":\"<urn:uuid:642c09ea-f0c0-4fd9-a774-f2641b87fa8e>\",\"Content-Length\":\"45064\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93417682-5729-4296-9a68-95ccf58d551d>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcd4c7a0-76fc-4e6a-bb02-80d1fe2e5aa2>\",\"WARC-IP-Address\":\"154.208.223.90\",\"WARC-Target-URI\":\"http://qgcpkjw.com/Ofxt/joefy.iunm?dje=7\",\"WARC-Payload-Digest\":\"sha1:7RVRLH62GGQWCLI77VAVTRQQ45G6TASU\",\"WARC-Block-Digest\":\"sha1:76LXMD6KWLSWKBRQPZISEWBQAGTJXGWL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250604397.40_warc_CC-MAIN-20200121132900-20200121161900-00004.warc.gz\"}"}
https://newpathworksheets.com/math/grade-4/pattern-blocks/ohio-common-core-standards	[ "## ◂Math Worksheets and Study Guides Fourth Grade. Pattern Blocks\n\n### The resources above correspond to the standards listed below:\n\n#### Ohio Learning Standards\n\nOH.4.NF. NUMBER AND OPERATIONS—FRACTIONS\nBuild fractions from unit fractions by applying and extending previous understandings of operations on whole numbers limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. (Fractions need not be simplified).\n4.NF.4. Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.\n4.NF.4.a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4) or 5/4 = (1/4) + (1/4) + (1/4) + (1/4) + (1/4)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84600246,"math_prob":0.9865597,"size":647,"snap":"2021-21-2021-25","text_gpt3_token_len":200,"char_repetition_ratio":0.16329704,"word_repetition_ratio":0.03846154,"special_character_ratio":0.32457495,"punctuation_ratio":0.1724138,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9965421,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-10T19:44:58Z\",\"WARC-Record-ID\":\"<urn:uuid:71b66d5f-f940-4934-b702-eeb4ee4d554a>\",\"Content-Length\":\"32267\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71db14e7-3150-4de0-b78c-d0e2405f61e9>\",\"WARC-Concurrent-To\":\"<urn:uuid:0125f549-da3c-4389-abb5-2d87250d1b64>\",\"WARC-IP-Address\":\"13.58.13.153\",\"WARC-Target-URI\":\"https://newpathworksheets.com/math/grade-4/pattern-blocks/ohio-common-core-standards\",\"WARC-Payload-Digest\":\"sha1:WERAIA27MA3OMMFNNRIGNU7AO7637PNX\",\"WARC-Block-Digest\":\"sha1:GDBOF3ESRQUWM4JG3FFG2EN3JGTAW7J7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991759.1_warc_CC-MAIN-20210510174005-20210510204005-00466.warc.gz\"}"}
https://www.codevscolor.com/c-plus-plus-smallest-second-smallest	[ "# C++ program to find the smallest and the second smallest numbers", null, "## Find the smallest and the second smallest numbers in an array in C++:\n\nWe will learn how to find the smallest and second smallest numbers in an array of numbers in C++. For example, if the array is [1, 3, 5, 7, 2], the smallest number will be 1 and the second smallest number will be 2.\n\nWe can solve it in different ways. One traditional approach is to use two variables to hold the smallest and second smallest numbers and update the values with a loop. Another way is to sort the array and pick the first and the second element from the sorted array. Let’s learn how to write the programs for both of these approaches.\n\n### Method 1: By using a loop:\n\nThe program will follow the below steps:\n\n• Initialize two variables to hold the smallest and second smallest numbers.\n• Assign INT_MAX to both of these variables. So, the smallest and the second smallest values will be the maximum integer value when the program starts.\n• Start a loop from the first element to the end of the array. On each iteration, compare the current value pointed by the loop with the smallest and second smallest variables.\n• If it is smaller than both of these variables, assign the smallest variable value to the second smallest variable and assign the current number to the smallest variable.\n• If it is smaller than the second smallest variable, assign it to the second smallest variable.\n• Once the loop ends, print the smallest and second smallest variables, which are the required values.\n``````#include <iostream>\n#include <climits>\nusing namespace std;\n\nint main()\n{\nint arr = {1, 4, 2, 7, 4, 12, 0, 9};\nint smallest = INT_MAX;\nint secondSmallest = INT_MAX;\n\nfor (int i = 0; i < size(arr); i++)\n{\nif (arr[i] < smallest && arr[i] < secondSmallest)\n{\nsecondSmallest = smallest;\nsmallest = arr[i];\n}\nelse if (arr[i] < secondSmallest)\n{\nsecondSmallest = arr[i];\n}\n}\n\ncout << \"Smallest value: \" << smallest << endl;\ncout << \"Second smallest value: \" << secondSmallest << endl;\n}``````\n\nThe arr is the given array of integers. It is following the same steps we have discussed above. It will print:\n\n### Method 2: By using loop and user input values:\n\nWe can take the numbers as inputs from the user. The following program takes the numbers as inputs and insert these numbers to the array before it finds the smallest and second smallest values:\n\n``````#include <iostream>\n#include <climits>\nusing namespace std;\n\nint main()\n{\nint size, arr;\nint smallest = INT_MAX;\nint secondSmallest = INT_MAX;\n\ncout << \"Enter the size of the array: \" << endl;\ncin >> size;\n\nfor (int i = 0; i < size; i++)\n{\ncout << \"Enter the number for index \" << i << \": \";\ncin >> arr[i];\n}\n\nfor (int i = 0; i < size; i++)\n{\nif (arr[i] < smallest && arr[i] < secondSmallest)\n{\nsecondSmallest = smallest;\nsmallest = arr[i];\n}\nelse if (arr[i] < secondSmallest)\n{\nsecondSmallest = arr[i];\n}\n}\n\ncout << \"Smallest value: \" << smallest << endl;\ncout << \"Second smallest value: \" << secondSmallest << endl;\n}``````\n\nIf you run this program, it will give results as below:\n\n### Method 3: By sorting the array elements:\n\nIf we sort the array elements in ascending order, the first element of the sorted array will be the smallest element and the second element will be the second smallest element of the array. We can use the inbuilt std::sort() function to sort the array.\n\n``````#include <iostream>\nusing namespace std;\n\nint main()\n{\nint size, arr;\ncout << \"Enter the size of the array: \" << endl;\ncin >> size;\n\nfor (int i = 0; i < size; i++)\n{\ncout << \"Enter the number for index \" << i << \": \";\ncin >> arr[i];\n}\n\nsort(arr, arr + size);\n\nint smallest = arr;\nint secondSmallest = arr;\n\ncout << \"Smallest value: \" << smallest << endl;\ncout << \"Second smallest value: \" << secondSmallest << endl;\n}``````\n\nIt will print similar output.\n\nYou can use any of the approaches we discussed above. The sorting method is a quick and concise way to find the smallest and second smallest values." ]	[ null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABQAAAALCAIAAADwazoUAAAACXBIWXMAAAsTAAALEwEAmpwYAAACb0lEQVQoz2P4UGDwpcTgV6XB/yaDY8mqmoK86gIgpCHIpyHIB2FDRQR4lXl55bh4BBm5Z8SI/l8hw/Cl0OBBls6aYOVtUSqZxuLyPDwQ/ap83Cq8XGr8PDCzeJX5eG3l+PPsRJLNhXYVSfxfIsPwp8xobaCKLDe3NBe3Ei+PBsQefh5DaQkzRXltUSFNYQENQT4tIV5Jdp4qV9H/C5X+LpL/u0Du5zxphg8FBu+L9N8W67Y7ysiBrdUQ5JPnYLFRV4l0d/W2NFPn51HkYod4QVOIV0eYb36C+P8lct/mSjN8Kza8mKzZ7SQboCqiCLKZR0tYwNPUyM/aMsDW2sfSLMzV2cfSXJ2fW4mX10tdoDNA7Fy99P+Fct/nSjP8LjXaGqpqKi5gLSWgJcSnxs+jIcRno6Ea5urkbWnmYWoU7GjvYqinzsclw8mTZyfysl/+Sa/sjzmyP+eDnf25RP9bhd6nKj1/FUENMbH00OBYH8+8+NikoIAYbw8fS/NgB1t/WytNIX4VPh5ZLh59Ub7HfdL/FskwvMs3+FVuuDFUxU9Z2EiUX5Wf105Lzcvc1NvSzNPM2N/G0tfS3N/G0kBKXFOQV56bJ9dO+HKTDMjmedIgzT/LDbeGqzrKCqoL8KnycpnKy/pamYc6OfhZW0R5uEW6u8T5eoGiXYBHnpsn3174YJnUnS7pvwtlGD4WGHws0v9eqX86VV1TkFdNgFeNn0eZm0OJmwNEcrErcrIpcbHDk4oKL484K8+USHAi+VtmtMhHyVpKwEpKQI0fkbY0hfghDDsNeQ8jNWcdRSdtRSNpke25Ug+6ZN9Pk/k9XxoAxgyoCciv5TUAAAAASUVORK5CYII=", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7160797,"math_prob":0.9863351,"size":4315,"snap":"2023-14-2023-23","text_gpt3_token_len":1059,"char_repetition_ratio":0.21967061,"word_repetition_ratio":0.31944445,"special_character_ratio":0.28088066,"punctuation_ratio":0.1360294,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986064,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T21:48:52Z\",\"WARC-Record-ID\":\"<urn:uuid:3e2e7d38-a534-4ff3-9a82-047ddf1fa5f5>\",\"Content-Length\":\"92639\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:103d8d72-23dc-4773-a2cd-ab6a0366aad6>\",\"WARC-Concurrent-To\":\"<urn:uuid:04d1b5e2-95cb-4f17-8abd-1cac36418d81>\",\"WARC-IP-Address\":\"3.210.81.252\",\"WARC-Target-URI\":\"https://www.codevscolor.com/c-plus-plus-smallest-second-smallest\",\"WARC-Payload-Digest\":\"sha1:ZQHWKOOY5ARQYL3TNTRXQLT2NVOI7STC\",\"WARC-Block-Digest\":\"sha1:42DEXXU54WLBNDRLZJSPZNR5Z7LCWBKF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648858.14_warc_CC-MAIN-20230602204755-20230602234755-00286.warc.gz\"}"}
https://studylib.net/doc/11889465/linear-algebra-2270-homework-11-problems-	[ "# Linear Algebra 2270 Homework 11 Problems:\n\nadvertisement", null, "```Linear Algebra 2270\nHomework 11\na preparation for the last quiz on 08/05/2015\nProblems:\n1. Perform Gram-Schmidt orthogonalization of the following vectors:\n⎡1⎤\n⎡−1⎤\n⎡−2⎤\n⎡2⎤\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\nv1 = ⎢ 0 ⎥ , v2 = ⎢ 1 ⎥ , v3 = ⎢ 1 ⎥ , v4 = ⎢0⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢ ⎥\n⎢−1⎥\n⎢3⎥\n⎢4⎥\n⎢4⎥\n⎣ ⎦\n⎣ ⎦\n⎣ ⎦\n⎣ ⎦\nAnd using it find matrices Q̃, R̃ such that matrix\n⎡∣\n∣\n∣\n∣ ⎤⎥\n⎢\n⎥\n⎢\nA = ⎢v1 v2 v3 v4 ⎥ = Q̃R̃\n⎥\n⎢\n⎢∣\n∣\n∣\n∣ ⎥⎦\n⎣\nUse this factorization of A to find the solution set of the linear system:\nAx = 0\n2. Perform Gram-Schmidt orthonormalization of the following vectors:\n⎡ √2 ⎤\n⎡ 1 √2⎤\n⎡−√2⎤\n⎡√2⎤\n⎢\n⎥\n⎢2 ⎥\n⎢\n⎥\n⎢ ⎥\n⎢\n⎥\n⎢\n⎥\n⎢\n⎥\n⎢ ⎥\n0 ⎥ , v2 = ⎢ √\n0 ⎥ , v3 = ⎢ √\n0 ⎥ , v4 = ⎢√1 ⎥\nv1 = ⎢ √\n⎢\n⎥\n⎢3 ⎥\n⎢\n⎥\n⎢ ⎥\n⎢− 2⎥\n⎢ 2⎥\n⎢2 2⎥\n⎢ 2⎥\n⎣\n⎦\n⎣2 ⎦\n⎣\n⎦\n⎣ ⎦\nAnd find QR decomposition of matrix\n⎡∣\n∣\n∣\n∣ ⎤⎥\n⎢\n⎢\n⎥\nA = ⎢v1 v2 v3 v4 ⎥\n⎢\n⎥\n⎢∣\n∣\n∣\n∣ ⎥⎦\n⎣\n1\n3. (a) Consider space R2 with the dot product and a set W = {[ ]}. Find W ⊥ and (W ⊥ )⊥\n−1\n1\n]). Find W ⊥ and (W ⊥ )⊥\n−1\n⎡ 1 ⎤ ⎡0⎤\n⎢ ⎥ ⎢ ⎥\n⎢ ⎥ ⎢ ⎥\n(c) Consider space R3 with the dot product and a set W = {⎢−1⎥ , ⎢0⎥}. Find W ⊥ and (W ⊥ )⊥\n⎢ ⎥ ⎢ ⎥\n⎢ 0 ⎥ ⎢1⎥\n⎣ ⎦ ⎣ ⎦\n(d) Let V be any inner product space. Let W be any subset of V , W ⊂ V . Prove that W ⊥ is a\nsubspace.\n(b) Consider space R2 with the dot product and a set W = span([\n(e) Let V be a finite dimensional inner product space. Let W be any subspace of V , W ≺ V .\nProve that W = (W ⊥ )⊥ .\nHint: Start by considering an orthonormal basis of W , enrich it to an orthonormal basis of V ,\nshow that the additional vectors form a basis of W ⊥ .\n4. (a) Find a determinant of A by performing row or column operations (or both). [In additional\nnotes posted on the website, we prove that one can perform row operations. The use of column\noperations have been justified in the lecture.]\n⎡2 2 2 ⎤\n⎢\n⎥\n⎢\n⎥\nA = ⎢1 0 1 ⎥\n⎢\n⎥\n⎢2 −2 1⎥\n⎣\n⎦\n1\n(b) Find a determinant of A by performing expansions according to rows or columns. For 2 × 2\nmatrices you may use a formula given in the lecture.\n⎡2\n⎢\n⎢1\n⎢\nA=⎢\n⎢2\n⎢\n⎢3\n⎣\n2 2\n0 1\n0 1\n0 −1\n1⎤⎥\n2⎥⎥\n⎥\n0⎥⎥\n1⎥⎦\nHint: It might be a good idea to start with an expansion according to the second column.\n(c) Using all available methods (expansion according to a row or a column or row and column\noperations), find the determinant of the matrix:\n⎡2\n⎢\n⎢2\n⎢\nA=⎢\n⎢2\n⎢\n⎢3\n⎣\n3 2 1⎤⎥\n3 4 1⎥⎥\n⎥\n1 −1 0⎥⎥\n0 −1 1⎥⎦\nHint: It might be a good idea to start with operations: r2 + = (−1)r1 , r2 = 21 r2 and then an\nexpansion according to one of the rows.\n2\n```" ]	[ null, "https://s2.studylib.net/store/data/011889465_1-568a658ef238e9f22eb9a197b8def500.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7634673,"math_prob":0.98884636,"size":2448,"snap":"2021-04-2021-17","text_gpt3_token_len":1386,"char_repetition_ratio":0.1333879,"word_repetition_ratio":0.2056962,"special_character_ratio":0.36723855,"punctuation_ratio":0.094736844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9761767,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-19T03:42:47Z\",\"WARC-Record-ID\":\"<urn:uuid:4b641f71-6ce7-4aaa-bb3d-c66de96e4ff6>\",\"Content-Length\":\"48388\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eaba9040-b49d-4864-8061-0d07e71fffef>\",\"WARC-Concurrent-To\":\"<urn:uuid:707e41ad-d02c-4f21-a230-7c4d6c1e63a1>\",\"WARC-IP-Address\":\"104.21.72.58\",\"WARC-Target-URI\":\"https://studylib.net/doc/11889465/linear-algebra-2270-homework-11-problems-\",\"WARC-Payload-Digest\":\"sha1:RA53AHHJBQFY7MLQHQDKH5V52FCJ5DT4\",\"WARC-Block-Digest\":\"sha1:5QUM2B2GNQYKGDC2WL3JGN4SJODLB6EG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038863420.65_warc_CC-MAIN-20210419015157-20210419045157-00152.warc.gz\"}"}
https://www.nuclear-power.net/nuclear-power/reactor-physics/atomic-nuclear-physics/fundamental-particles/beta-particle/spectrum-beta-particles/	[ "# Spectrum of Beta Particles\n\n## Spectrum of beta particles", null, "The shape of this energy curve depends on what fraction of the reaction energy (Q value-the amount of energy released by the reaction) is carried by the electron or neutrino.\n\nIn the process of beta decay, either an electron or a positron is emitted. This emission is accompanied by the emission of antineutrino (β- decay) or neutrino (β+ decay), which shares energy and momentum of the decay. The beta emission has a characteristic spectrum. This characteristic spectrum is caused by the fact that either a neutrino or an antineutrino is emitted with emission of beta particle. The shape of this energy curve depends on what fraction of the reaction energy (Q value-the amount of energy released by the reaction) is carried by the massive particle. Beta particles can therefore be emitted with any kinetic energy ranging from 0 to Q. By 1934, Enrico Fermi had developed a Fermi theory of beta decay, which predicted the shape of this energy curve.\n\n## See previous:\n\nNature of Interaction of Beta Particles\n\nBeta Particle\n\nBremsstrahlung" ]	[ null, "https://www.nuclear-power.net/wp-content/uploads/2015/03/beta-decay-spectrum.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92304,"math_prob":0.8284932,"size":799,"snap":"2020-45-2020-50","text_gpt3_token_len":170,"char_repetition_ratio":0.13584906,"word_repetition_ratio":0.0,"special_character_ratio":0.19774719,"punctuation_ratio":0.07534247,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9776623,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T13:56:31Z\",\"WARC-Record-ID\":\"<urn:uuid:06af2b5a-69a4-466c-aec2-517a5650eb12>\",\"Content-Length\":\"436776\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:760f3c79-65ae-4262-9835-ef791219b452>\",\"WARC-Concurrent-To\":\"<urn:uuid:af66de41-a575-4754-aff6-4ddf348ed0fe>\",\"WARC-IP-Address\":\"172.67.164.94\",\"WARC-Target-URI\":\"https://www.nuclear-power.net/nuclear-power/reactor-physics/atomic-nuclear-physics/fundamental-particles/beta-particle/spectrum-beta-particles/\",\"WARC-Payload-Digest\":\"sha1:PW7K4FVRVIS623F6AKCFHB5KZZ5XDTQC\",\"WARC-Block-Digest\":\"sha1:5IYGMWFZPIE2WTUPOAIODWTYULOVHZRO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141216175.53_warc_CC-MAIN-20201130130840-20201130160840-00669.warc.gz\"}"}
https://xbuzznet.com/qa/how-do-you-handle-missing-data.html	[ "", null, "# How Do You Handle Missing Data?\n\n## How do you find the missing data percentage?\n\nE.g.\n\nthe number of missing data elements for the read variable (cell G6) is 15, as calculated by the formula =COUNT(B4:B23).\n\nSince there are 20 rows in the data range the percentage of non-missing cells for read (cell G7) is 15/20 = 75%, which can be calculated by =G6/COUNTA(B4:B23)..\n\n## How do you handle missing or corrupted data in a dataset?\n\nhow do you handle missing or corrupted data in a dataset?Method 1 is deleting rows or columns. We usually use this method when it comes to empty cells. … Method 2 is replacing the missing data with aggregated values. … Method 3 is creating an unknown category. … Method 4 is predicting missing values.\n\n## How do you know if data is missing randomly?\n\nThe only true way to distinguish between MNAR and Missing at Random is to measure the missing data. In other words, you need to know the values of the missing data to determine if it is MNAR. It is common practice for a surveyor to follow up with phone calls to the non-respondents and get the key information.\n\n## How do you solve missing values in time series data?\n\nIntroduction. In time series data, if there are missing values, there are two ways to deal with the incomplete data: omit the entire record that contains information. Impute the missing information.\n\n## How do you explain missing data?\n\nMissing data (or missing values) is defined as the data value that is not stored for a variable in the observation of interest. The problem of missing data is relatively common in almost all research and can have a significant effect on the conclusions that can be drawn from the data .\n\n## How do you deal with missing categorical data?\n\nThere is various ways to handle missing values of categorical ways.Ignore observations of missing values if we are dealing with large data sets and less number of records has missing values.Ignore variable, if it is not significant.Develop model to predict missing values.Treat missing data as just another category.\n\n## What is missing at random?\n\n‘Missing at random’ means that there might be systematic differences between the missing and observed blood pressures, but these can be entirely explained by other observed variables.\n\n## When should you impute data?\n\nImputation works best when many variables are missing in small proportions such that a complete case analysis might render 60-30% completeness, but each variable is perhaps only missing 10% of its values.\n\n## How do you handle missing data values?\n\nIn statistical language, if the number of the cases is less than 5% of the sample, then the researcher can drop them. In the case of multivariate analysis, if there is a larger number of missing values, then it can be better to drop those cases (rather than do imputation) and replace them.\n\n## What percentage of missing data is acceptable?\n\n@shuvayan – Theoretically, 25 to 30% is the maximum missing values are allowed, beyond which we might want to drop the variable from analysis. Practically this varies.At times we get variables with ~50% of missing values but still the customer insist to have it for analyzing.\n\n## How do you fill missing values in a data set?\n\nFilling missing values using fillna() , replace() and interpolate() In order to fill null values in a datasets, we use fillna() , replace() and interpolate() function these function replace NaN values with some value of their own. All these function help in filling a null values in datasets of a DataFrame.\n\n## When should missing values be removed?\n\nIt’s most useful when the percentage of missing data is low. If the portion of missing data is too high, the results lack natural variation that could result in an effective model. The other option is to remove data. When dealing with data that is missing at random, related data can be deleted to reduce bias." ]	[ null, "https://mc.yandex.ru/watch/69930196", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86050975,"math_prob":0.9457398,"size":4320,"snap":"2021-43-2021-49","text_gpt3_token_len":915,"char_repetition_ratio":0.20227063,"word_repetition_ratio":0.13484646,"special_character_ratio":0.21597221,"punctuation_ratio":0.099520385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9701861,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T00:54:58Z\",\"WARC-Record-ID\":\"<urn:uuid:db9f2c09-55e4-4962-b935-46aa393b9e02>\",\"Content-Length\":\"30922\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:269a80b0-ee08-4941-aeb3-86238b02a8ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:2fc73220-3840-401f-bec9-740c95fc3359>\",\"WARC-IP-Address\":\"45.130.40.26\",\"WARC-Target-URI\":\"https://xbuzznet.com/qa/how-do-you-handle-missing-data.html\",\"WARC-Payload-Digest\":\"sha1:NTJAK4PE5NI2EJA7QTDENDA3KVCWFWC6\",\"WARC-Block-Digest\":\"sha1:IDUWOWRLYMX22KF46FWJ7GV2EHI5XQWP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585290.83_warc_CC-MAIN-20211019233130-20211020023130-00145.warc.gz\"}"}
https://proceedings.neurips.cc/paper/2018/file/801fd8c2a4e79c1d24a40dc735c051ae-Reviews.html	[ "NIPS 2018\nSun Dec 2nd through Sat the 8th, 2018 at Palais des Congrès de Montréal\nPaper ID: 2866 Deep Generative Models for Distribution-Preserving Lossy Compression\n\n### Reviewer 1\n\nThe paper proposes a novel problem formulation for lossy compression, namely distribution-preserving lossy compression (DPLC). For a rate constrained lossy compression scheme, for large enough rate of the compression scheme, it is possible to (almost) exactly reconstruct the original signal from its compressed version. However, as the rate gets smaller, the reconstructed signal necessarily has very high distortion. The DPLC formulation aims to alleviate this issue by enforcing an additional constraint during the design of the encoder and the decoder of the compression scheme. The constraint requires that irrespective of the rate of the compression the distribution of the reconstructed signal is the same as that of the original signal. This ensures that for the very small rate the decoder simply outputs a new sample from the source distribution and not a useless signal with very large distortion. In particular, a DPLC scheme needs to have a stochastic decoder which is hard to train using the existing techniques in the literature. The authors propose a way to deal with this and employ Wasserstein distance to measure the divergence between the original signal distribution and that of the output of the decoder. The authors build upon the recent line of work on utilizing neural networks to model the signal distributions. The authors rely on both generative adversarial networks and autoencoders to design a DPLC scheme, which gives a novel object called Wassserstein++. The authors theoretical quantify the performance of their proposed procedure for encoder and decoder design by relating its distortion performance to a suitable Wasserstein distance. They further, implement their procedure with two GAN datasets and show that it indeed allows them to map between the exact reconstruction and sampling from the generative model as the rate of the compression scheme approaches zero. Overall, I find the problem formulation unique with a reasonably comprehensive treatment of the problem presented in the paper. The paper is well written. Minor comment: It's not clear to me what $G$ refers to in the statement of Theorem 1. Should it be $G^{\\star}$? Line 148: \".....$F$ in (6).....\" should be \"......$F$ in (7).....\"?\n\n### Reviewer 2\n\nSummary ======= This paper proposes a formulation of lossy compression in which the goal is to minimize distortion subject to the constraint that the reconstructed data follows the data distribution (and subject to a fixed rate). As in Agustsson et al. (2018), a stochastic decoder is used to achieve this goal, but trained for a combination of WGAN/WAE instead of LS-GAN. The encoder is trained separately by minimizing MMD. The authors furthermore provide a theorem which provides bounds on the achievable distortion for a given bit rate. Good ==== The proposed formulation is a slightly different take on lossy compression, which has the potential to inspire novel solutions. Constraining the reconstruction distribution – although difficult in practice – would ensure that reconstructions are always of high perceptual quality. Bad === The empirical evaluation is quite limited. In particular, the comparison is limited to toy datasets (CelebA, LSUN). It would have been nice to see how the proposed approach performs compared to other published approaches or standardized codecs on more realistic datasets (e.g., Kodak). Furthermore, why were qualitative comparisons with Agustsson et al. (2018) omitted in Figure 1 when they were clearly available (Figure 2)? Despite being a more complicated approach than Agustsson et al. (2018), it does not seem to perform better (Figure 2). The only demonstrated difference is an increase in conditional variance, which in itself may not be a desirable property for a practical compression system nor does it provide strong evidence that the reconstruction follows the target distribution more closely. The paper seems to lack important details. For example, how is the rate limitation of the encoder achieved? How is the output of the encoder quantized? In Figure 2, how are the quantized outputs encoded into bits? I find it quite difficult to follow the train of thought on page 4. Please provide more detail on why joint optimization of G and F is worse than optimizing G first and subsequently optimizing F (L128-L137), in particular elaborate the claim that \"constraining F implies that one only optimizes over a subset of couplings.\" This constraint still seems in place in the two-step optimization (Equation 8), so why is it not an issue then? The role and implications of Theorem 1 could be explained more clearly. The authors claim that \"Theorem 1 states that the distortion incurred by the proposed procedure is equal to W(P_X, P_G) [...]\". However, this assumes that we solve Equation 8 exactly, which may be as difficult as solving the compression problem itself, so the bound isn't very practical. It seems to me the main purpose is therefore to show that the achievable distortion is limited by the Wasserstein distance of G, which justifies minimizing the Wasserstein distance of G* via WGAN/WAE. Correct me if I missed something. Minor ===== Constraining the perceptual quality to be perfect limits the achievable distortion (Blau & Michaeli, 2017). In some applications it may be desirable to trade off perceptual quality for lower distortion, which is not possible with the proposed formulation.\n\n### Reviewer 3\n\nThis works uses generative models for the task of compression. The premise behind this paper is that distribution preserving lossy compression should be possible at all bitrates. At low bitrates, the scheme should be able to recover data from the underlying distribution, while at higher bitrates it produces data (images) that reconstructs the input samples. That the scheme works as advertised can be seen rather convincingly in section 5 containing the evaluations. The paper is well written and constitutes a very interesting exercise. The compression setup consists of an encoder to compress the input image E, and a stochastic component B, and a generator G recovers the input. Together, they are trained to minimize a distortion measure. The objective function is formulated as a Lagrangian consisting of the distortion measure between input and reconstruction and a divergence term, implemented as a GAN which takes effect at low bitrates so that it recovers images from the underlying distribution. Several other generative compressive setups are compared against. But the closest comparison (in terms of approach) seems to be the work \"Generative Adversarial Networks for learned image compression\" (termed GC in the results). In this connection, they point out that if one were simply to minimize the distortion measure as done in GC, it results in the setup largely ignoring the noise injected except at very low bitrates where it is the only factor involved. I think I can intuitively understand this part, but a little clarification would be useful. The reasoning in section 3, on why one cannot simply replace the distortion metric using a Wasserstein distance seems to be that the encoder is rate constrained and therefore optimizing this metric would not result in optimizing over all couplings. They then go on to decompose the distortion measure (equation (4)) as the sum of two terms, one of which is the wasserstein distance and the other, a term containing the distortion rate which goes to zero at large values.. By doing so (the generator is learnt separately) they can replace the distortion measure with the said Wasserstein term and optimize it. The connection where they jointly train the setup to function essentially as a VAE-GAN by combining the two terms in the objective - the distortion as the primal, and the divergence as the dual is a most interesting trick for future use. One aspect that I could not piece together (it is likely that I am missing it) is how the Wasserstein++ algorithm manages to enforce the 1-Lipshitz constraint for the GAN part of the objective. The 1-Lipshitz requirement (as one would expect) is explicitly mentioned in equation 11 for the GAN term. As far as I can see, weight clipping/gradient penalty does not seem to be mentioned in Wasserstein++. As regards the evaluation studies produced in section 5, I am curious about use of the Frechet Inception Distance as a metric for use in GANs. On the whole, I am very impressed with this work. It presents a way to attain generative compression at both low and high bitrates, with reconstructions getting better at higher bitrates and the model recovering distribution images (in GAN like fashion) at low bitrates. The use of Wasserstein autoencoders is also quite interesting. Answers to specific review criteria below: 1) Quality: Excellent. 2) Clarity: Very good. The mathematical details could perhaps be worked out as an extended tutorial. 3) Originality: Very good. The material presents Wasserstein autoencoders and GANs (combined as Wasserstein++) as the approach to effect compression. This is rather unique. 4) Significance: Very good. Generative compression is a significant area of research." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9408288,"math_prob":0.90294254,"size":9229,"snap":"2021-04-2021-17","text_gpt3_token_len":1860,"char_repetition_ratio":0.13430895,"word_repetition_ratio":0.0013623978,"special_character_ratio":0.19709611,"punctuation_ratio":0.09669523,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96531975,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-12T13:16:21Z\",\"WARC-Record-ID\":\"<urn:uuid:816f9371-954f-4b72-b1c4-9260591a2375>\",\"Content-Length\":\"10716\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9b714e5f-0f3b-47d6-b6ff-596b9ea15cb6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9b27798-278b-4ef6-81ec-db5393d092c0>\",\"WARC-IP-Address\":\"198.202.70.94\",\"WARC-Target-URI\":\"https://proceedings.neurips.cc/paper/2018/file/801fd8c2a4e79c1d24a40dc735c051ae-Reviews.html\",\"WARC-Payload-Digest\":\"sha1:KDG7BHTCCPXQES2AWYKORAYL4FCWHGZN\",\"WARC-Block-Digest\":\"sha1:4FYQTHHFTREVAVBY2BC6PRPWIFRSBIY2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038067400.24_warc_CC-MAIN-20210412113508-20210412143508-00199.warc.gz\"}"}
https://socratic.org/questions/what-is-the-hybridization-for-c2h6-my-teacher-wrote-sp-s-sigma-bond-c-h-and-sp-s	[ "What is the hybridization for C2H6? my teacher wrote sp-s sigma bond [C-H] and sp-sp sigma bond[C-C]\n\nMar 1, 2016\n\n${\\text{C\"_2\"H}}_{6}$ has an $s {p}^{3}$ hybridization on each carbon because of the four electron groups surrounding each carbon. It made four identical bonds in a perfect tetrahedral geometry, which means it needed four identical orbitals to make those bonds.\n\nEach carbon has to hybridize one $2 s$ and three $2 p$ orbitals in order to generate four identical $s {p}^{3}$ orbitals that are compatible in symmetry with hydrogen's $1 s$ orbitals.", null, "Therefore, each $\\text{C\"-\"H}$ bond in ${\\text{C\"_2\"H}}_{6}$ is between an $s {p}^{3}$ of carbon (YELLOW) and a $1 s$ of hydrogen (BLUE sphere), i.e. an $s {p}^{3} - s$ connection (the YELLOW/BLUE overlap in (b)), and each $\\text{C\"-\"C}$ bond is an $s {p}^{3} - s {p}^{3}$ connection (the YELLOW overlap in (b)).\n\nIF YOUR TEACHER WAS CORRECT ON THE HYBRIDIZATIONS\n\nOn the other hand, ${\\text{C\"_2\"H}}_{2}$ would match your teacher's observations.\n\nWith two electron groups on a POLYatomic molecule, each carbon requires only two $s p$ lobes and hence only one $s p$ hybridized orbital to bond with the other carbon AND a single hydrogen.", null, "In this case we can see that:\n\n• The $s p$ of carbon 1 (YELLOW dumbbell) overlaps with one hydrogen's $1 s$ (BLUE sphere) to make a $\\setminus m a t h b f \\left(\\sigma\\right)$ bond.\n• The $s p$ of carbon 1 (YELLOW dumbbell) overlaps with the $s p$ of carbon 2 (YELLOW dumbbell) and vice versa to make a $\\setminus m a t h b f \\left(\\sigma\\right)$ bond.\n• The $s p$ of carbon 2 (YELLOW dumbbell) overlaps with the second hydrogen's $1 s$ (BLUE sphere) to make a $\\setminus m a t h b f \\left(\\sigma\\right)$ bond.\n\nEach triple bond incorporates an additional ${p}_{x} \\text{/} {p}_{x}$ and ${p}_{y} \\text{/} {p}_{y}$ overlap between carbons 1 and 2, accounting for two $\\pi$ bonds (i.e. $p - p$ connections).\n\nHence, when including those two $p - p$ $\\pi$ bonds with the $s p - s p$ $\\sigma$ bond between carbons 1 and 2, we have accounted for the the triple bond between carbons 1 and 2.\n\n(One triple bond = 1 $\\sigma$ + 2 $\\pi$ bonds)" ]	[ null, "https://useruploads.socratic.org/ulLoNm3NSjqMsoz2hQXy_CNX_Chem_08_02_ethane.jpg", null, "https://useruploads.socratic.org/cV07hL8yT2OCFfRggnuU_CNX_Chem_08_03_C2H2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8384027,"math_prob":0.9997867,"size":971,"snap":"2022-05-2022-21","text_gpt3_token_len":258,"char_repetition_ratio":0.11582213,"word_repetition_ratio":0.18831168,"special_character_ratio":0.23789908,"punctuation_ratio":0.056179777,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99979395,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T20:00:07Z\",\"WARC-Record-ID\":\"<urn:uuid:e75c3756-1672-4b3d-a0eb-86075ece6cf3>\",\"Content-Length\":\"38931\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ee3cb3b-b02b-4324-af24-31aa7369bb32>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e718bdb-e58c-4e38-ac0a-e34e8e81d690>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/what-is-the-hybridization-for-c2h6-my-teacher-wrote-sp-s-sigma-bond-c-h-and-sp-s\",\"WARC-Payload-Digest\":\"sha1:BXRMLF5QL2LOTKMNQNKZHX4MAJAM2BYP\",\"WARC-Block-Digest\":\"sha1:WNI6UUD776F22MLMDKCA7JXUNDKVRVCM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306335.77_warc_CC-MAIN-20220128182552-20220128212552-00406.warc.gz\"}"}
https://www.colorhexa.com/0a04b5	[ "# #0a04b5 Color Information\n\nIn a RGB color space, hex #0a04b5 is composed of 3.9% red, 1.6% green and 71% blue. Whereas in a CMYK color space, it is composed of 94.5% cyan, 97.8% magenta, 0% yellow and 29% black. It has a hue angle of 242 degrees, a saturation of 95.7% and a lightness of 36.3%. #0a04b5 color hex could be obtained by blending #1408ff with #00006b. Closest websafe color is: #0000cc.\n\n• R 4\n• G 2\n• B 71\nRGB color chart\n• C 94\n• M 98\n• Y 0\n• K 29\nCMYK color chart\n\n#0a04b5 color description : Strong blue.\n\n# #0a04b5 Color Conversion\n\nThe hexadecimal color #0a04b5 has RGB values of R:10, G:4, B:181 and CMYK values of C:0.94, M:0.98, Y:0, K:0.29. Its decimal value is 656565.\n\nHex triplet RGB Decimal 0a04b5 `#0a04b5` 10, 4, 181 `rgb(10,4,181)` 3.9, 1.6, 71 `rgb(3.9%,1.6%,71%)` 94, 98, 0, 29 242°, 95.7, 36.3 `hsl(242,95.7%,36.3%)` 242°, 97.8, 71 0000cc `#0000cc`\nCIE-LAB 21.897, 60.312, -82.455 8.507, 3.487, 43.938 0.152, 0.062, 3.487 21.897, 102.158, 306.184 21.897, -6.028, -86.941 18.673, 48.645, -126.438 00001010, 00000100, 10110101\n\n# Color Schemes with #0a04b5\n\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #afb504\n``#afb504` `rgb(175,181,4)``\nComplementary Color\n• #0457b5\n``#0457b5` `rgb(4,87,181)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #6204b5\n``#6204b5` `rgb(98,4,181)``\nAnalogous Color\n• #57b504\n``#57b504` `rgb(87,181,4)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #b56204\n``#b56204` `rgb(181,98,4)``\nSplit Complementary Color\n• #04b50a\n``#04b50a` `rgb(4,181,10)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #b50a04\n``#b50a04` `rgb(181,10,4)``\nTriadic Color\n• #04afb5\n``#04afb5` `rgb(4,175,181)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #b50a04\n``#b50a04` `rgb(181,10,4)``\n• #afb504\n``#afb504` `rgb(175,181,4)``\nTetradic Color\n• #06026a\n``#06026a` `rgb(6,2,106)``\n• #070383\n``#070383` `rgb(7,3,131)``\n• #09039c\n``#09039c` `rgb(9,3,156)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #0b05ce\n``#0b05ce` `rgb(11,5,206)``\n• #0d05e7\n``#0d05e7` `rgb(13,5,231)``\n• #140cfa\n``#140cfa` `rgb(20,12,250)``\nMonochromatic Color\n\n# Alternatives to #0a04b5\n\nBelow, you can see some colors close to #0a04b5. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #042ab5\n``#042ab5` `rgb(4,42,181)``\n• #041cb5\n``#041cb5` `rgb(4,28,181)``\n• #040db5\n``#040db5` `rgb(4,13,181)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #1904b5\n``#1904b5` `rgb(25,4,181)``\n• #2704b5\n``#2704b5` `rgb(39,4,181)``\n• #3604b5\n``#3604b5` `rgb(54,4,181)``\nSimilar Colors\n\n# #0a04b5 Preview\n\nText with hexadecimal color #0a04b5\n\nThis text has a font color of #0a04b5.\n\n``<span style=\"color:#0a04b5;\">Text here</span>``\n#0a04b5 background color\n\nThis paragraph has a background color of #0a04b5.\n\n``<p style=\"background-color:#0a04b5;\">Content here</p>``\n#0a04b5 border color\n\nThis element has a border color of #0a04b5.\n\n``<div style=\"border:1px solid #0a04b5;\">Content here</div>``\nCSS codes\n``.text {color:#0a04b5;}``\n``.background {background-color:#0a04b5;}``\n``.border {border:1px solid #0a04b5;}``\n\n# Shades and Tints of #0a04b5\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, #000008 is the darkest color, while #f4f4ff is the lightest one.\n\n• #000008\n``#000008` `rgb(0,0,8)``\n• #02011b\n``#02011b` `rgb(2,1,27)``\n• #03012f\n``#03012f` `rgb(3,1,47)``\n• #040142\n``#040142` `rgb(4,1,66)``\n• #050255\n``#050255` `rgb(5,2,85)``\n• #060268\n``#060268` `rgb(6,2,104)``\n• #07037b\n``#07037b` `rgb(7,3,123)``\n• #08038f\n``#08038f` `rgb(8,3,143)``\n• #0904a2\n``#0904a2` `rgb(9,4,162)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\n• #0b04c8\n``#0b04c8` `rgb(11,4,200)``\n• #0c05db\n``#0c05db` `rgb(12,5,219)``\n• #0d05ef\n``#0d05ef` `rgb(13,5,239)``\nShade Color Variation\n• #160efa\n``#160efa` `rgb(22,14,250)``\n• #2821fa\n``#2821fa` `rgb(40,33,250)``\n• #3b34fb\n``#3b34fb` `rgb(59,52,251)``\n• #4d47fb\n``#4d47fb` `rgb(77,71,251)``\n• #605bfb\n``#605bfb` `rgb(96,91,251)``\n• #736efc\n``#736efc` `rgb(115,110,252)``\n• #8581fc\n``#8581fc` `rgb(133,129,252)``\n• #9894fd\n``#9894fd` `rgb(152,148,253)``\n• #aaa7fd\n``#aaa7fd` `rgb(170,167,253)``\n• #bdbbfd\n``#bdbbfd` `rgb(189,187,253)``\n• #cfcefe\n``#cfcefe` `rgb(207,206,254)``\n• #e2e1fe\n``#e2e1fe` `rgb(226,225,254)``\n• #f4f4ff\n``#f4f4ff` `rgb(244,244,255)``\nTint Color Variation\n\n# Tones of #0a04b5\n\nA tone is produced by adding gray to any pure hue. In this case, #5a5960 is the less saturated color, while #0a04b5 is the most saturated one.\n\n• #5a5960\n``#5a5960` `rgb(90,89,96)``\n• #535267\n``#535267` `rgb(83,82,103)``\n• #4c4b6e\n``#4c4b6e` `rgb(76,75,110)``\n• #464475\n``#464475` `rgb(70,68,117)``\n• #3f3d7c\n``#3f3d7c` `rgb(63,61,124)``\n• #383683\n``#383683` `rgb(56,54,131)``\n• #322f8a\n``#322f8a` `rgb(50,47,138)``\n• #2b2891\n``#2b2891` `rgb(43,40,145)``\n• #252099\n``#252099` `rgb(37,32,153)``\n• #1e19a0\n``#1e19a0` `rgb(30,25,160)``\n• #1712a7\n``#1712a7` `rgb(23,18,167)``\n• #110bae\n``#110bae` `rgb(17,11,174)``\n• #0a04b5\n``#0a04b5` `rgb(10,4,181)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0a04b5 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" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5467324,"math_prob":0.7067361,"size":3636,"snap":"2019-35-2019-39","text_gpt3_token_len":1663,"char_repetition_ratio":0.12527533,"word_repetition_ratio":0.011111111,"special_character_ratio":0.55390537,"punctuation_ratio":0.23276836,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9888292,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T19:59:49Z\",\"WARC-Record-ID\":\"<urn:uuid:47fb8fab-2c06-4eac-a360-74374bb750f9>\",\"Content-Length\":\"36183\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14d6f035-ec4c-4b96-bb9a-8c9616da14d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:15dc0ff8-15b7-4160-8a9a-380a6ec70768>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0a04b5\",\"WARC-Payload-Digest\":\"sha1:N5F2HYNURVTTU2GM5Q4LKOYXBSRBWAJ2\",\"WARC-Block-Digest\":\"sha1:ZVVHGGWUVVQHYP4U4RQCBPVI4BUJG5OI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027318986.84_warc_CC-MAIN-20190823192831-20190823214831-00290.warc.gz\"}"}
https://tex.stackexchange.com/questions/182911/problem-aligning-multiple-pgfplots-due-to-different-axis-label-sizes/182921#182921	[ "# Problem aligning multiple pgfplots due to different axis label sizes\n\nI have two plots that comes back to back in my paper. One of these plots is generated from 1000 data samples while the other one from 320. The problem that I'm, facing right now, is that plots are aligned by right-most axis labels (1,000 and 320). This result in a situation where actual right side border of the plots are not aligned. To make it clear, please consider the following MWE:\n\n\\documentclass[11pt,twocolumn]{article}\n\\usepackage{lipsum}\n\\usepackage{pgfplots}\n\\usepackage{tikz}\n\n\\begin{document}\n\\lipsum[1-3]\n\\begin{figure}[t]\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}\n\\def\\UpperBound{2.112528}\n\\def\\LowerBound{-0.092346}\n\\begin{axis}[\nscale only axis,\nxmin = 0,\nxmax = 1000,\nymin = -0.292346,\nymax = 2.312528,\ny = 30,\nwidth = 0.5\\textwidth,\nheight = 0.25\\textwidth,\ngrid style = {dashed, ultra thin},\ngrid = major,\nxtick = {0,250,...,1000},\nytick = \\empty,\ntick label style = {font=\\small}\n]\n\\addplot[thick, blue] table [x expr=\\coordindex, y index=0]{values.csv};\n\n\\draw[thin, green, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\UpperBound) --\n(axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\UpperBound);\n\\draw[thin, red, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\LowerBound) --\n(axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\LowerBound);\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{A caption.}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}\n\\begin{axis}[\nscale only axis,\ngrid = major,\nxmin = 0,\nxmax = 320,\nwidth = 0.5\\textwidth,\nheight = 0.15\\textwidth,\ngrid style = {dashed, ultra thin},\nxtick = {0,80,...,320},\nytick = \\empty,\ntick label style = {font=\\small}\n]\n\\addplot[thick, blue] table [x expr=\\coordindex, y index=0]{error_file.csv};\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{Another caption.}\n\\end{figure}\n\\end{document}\n\n\nresulting in:", null, "", null, "Is there any option to bring the right-most value inside the canvas of the plot such that outer most of the plot becomes the right border? (and other labels as well) Or perhaps rotate the labels like gnuplots (although I prefer not to resort to the latter).", null, "Data for MWE File1, File2\n\n• Would it be an option to let the labels protrude over the right edge of the plot area, making the right edge of the plot area flush with the text area?\n– Jake\nJun 3, 2014 at 11:48\n• @Jake, not sure if I got what you mean but if it aligns my plots it would be fine! I actually don't like any of my own ideas, so please don't hesitate to come up with your solution to the problem. Jun 3, 2014 at 12:11\n\nBy adding trim axis right to the tikzpicture options, everything that protrudes past the right edge of the plot area will be ignored for calculating the size of the plot and for positioning the plot on the page. That way, you allow the rightmost labels to protrude into the margin of your document, which may or may not be acceptable for you:", null, "\\documentclass[11pt,twocolumn]{article}\n\\usepackage{lipsum}\n\\usepackage{pgfplots}\n\\usepackage{tikz}\n\n\\begin{document}\n\\lipsum[1-3]\n\\begin{figure}[t]\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}[trim axis right]\n\\def\\UpperBound{2.112528}\n\\def\\LowerBound{-0.092346}\n\\begin{axis}[\nscale only axis,\nxmin = 0,\nxmax = 1000,\nymin = -0.292346,\nymax = 2.312528,\ny = 30,\nwidth = 0.5\\textwidth,\nheight = 0.25\\textwidth,\ngrid style = {dashed, ultra thin},\ngrid = major,\nxtick = {0,250,...,1000},\nytick = \\empty,\ntick label style = {font=\\small}\n]\n\\addplot[thick, blue, domain=0:1000] {rnd};\n\n\\draw[thin, green, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\UpperBound) --\n(axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\UpperBound);\n\\draw[thin, red, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\LowerBound) --\n(axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\LowerBound);\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{A caption.}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}[trim axis right]\n\\begin{axis}[\nscale only axis,\ngrid = major,\nxmin = 0,\nxmax = 320,\nwidth = 0.5\\textwidth,\nheight = 0.15\\textwidth,\ngrid style = {dashed, ultra thin},\nxtick = {0,80,...,320},\nytick = \\empty,\ntick label style = {font=\\small}\n]\n\\addplot[thick, blue, domain=0:320] {rnd};\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{Another caption.}\n\\end{figure}\n\\end{document}\n\n\nYou can set the xticklabels explicitly using \\llap for the last xticklabel\n\nxtick = {0,250,...,1000},\nxticklabels ={0,250,...,750,\\llap{1000}},\n\n\nand\n\nxtick = {0,80,...,320},\nxticklabels = {0,80,...,240,\\llap{320}},", null, "Code\n\n\\documentclass[11pt,twocolumn]{article}\n\\usepackage{lipsum}\n\\usepackage{pgfplots}\n\n\\pgfplotsset{\nlast xtick/.style={\nextra x ticks = #1,\nextra x tick style = {tick label style={anchor=north east,inner xsep=0pt}}\n}\n}\n\n\\begin{document}\n\\lipsum[1-3]\n\\begin{figure}[t]\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}\n\\def\\UpperBound{2.112528}\n\\def\\LowerBound{-0.092346}\n\\begin{axis}[\nscale only axis,\nxmin = 0,\nxmax = 1000,\nymin = -0.292346,\nymax = 2.312528,\ny = 30,\nwidth = 0.5\\textwidth,\nheight = 0.25\\textwidth,\ngrid style = {dashed, ultra thin},\ngrid = major,\nxtick = {0,250,...,1000},\nxticklabels ={0,250,...,750,\\llap{1000}},\nytick = \\empty,\ntick label style = {font=\\small},\n]\n\\addplot[thick, blue] table [x expr=\\coordindex, y index=0]{values.csv};\n\n\\draw[thin, green, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\UpperBound) --\n(axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\UpperBound);\n\\draw[thin, red, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\LowerBound) --\n(axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\LowerBound);\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{A caption.}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}\n\\begin{axis}[\nscale only axis,\ngrid = major,\nxmin = 0,\nxmax = 320,\nwidth = 0.5\\textwidth,\nheight = 0.15\\textwidth,\ngrid style = {dashed, ultra thin},\nxtick = {0,80,...,320},\nxticklabels = {0,80,...,240,\\llap{320}},\nlast xtick = 320,\nytick = \\empty,\ntick label style = {font=\\small}\n]\n\\addplot[thick, blue] table [x expr=\\coordindex, y index=0]{error_file.csv};\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{Another caption.}\n\\end{figure}\n\\end{document}\n\n\nOr if there are no other extra x ticks you can use this option to define a different style for the last xtick:\n\n\\pgfplotsset{\nlast xtick/.style={\nextra x ticks = #1,\nextra x tick style = {tick label style={anchor=north east,inner xsep=0pt}}\n}\n}\n\n\nand then for the first plot\n\nxtick = {0,250,...,750},\nlast xtick = 1000,\n\n\nand for the second plot\n\nxtick = {0,80,...,300},\nlast xtick = 320,\n\n\nAn alternative is use of scaled x ticks=base 10:-4 and 10:-2, respectively in the axis options.", null, "Code\n\n\\documentclass[11pt,twocolumn]{article}\n\\usepackage{lipsum}\n\\usepackage{pgfplots}\n\\usepackage{tikz}\n\n\\begin{document}\n\\lipsum[1-3]\n\\begin{figure}[t]\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}\n\\def\\UpperBound{2.112528}\n\\def\\LowerBound{-0.092346}\n\\begin{axis}[\nscale only axis,\nxmin = 0,\nxmax = 1000,\nymin = -0.292346,\nymax = 2.312528,\ny = 30,\nwidth = 0.5\\textwidth,\nheight = 0.25\\textwidth,\ngrid style = {dashed, ultra thin},\ngrid = major,\nxtick = {0,250,...,1000},\nytick = \\empty,\ntick label style = {font=\\small},\nscaled x ticks=base 10:-4,\n]\n\\addplot[thick, blue] table [x expr=\\coordindex, y index=0]{values.csv};\n\\draw[thin, green, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\UpperBound) -- (axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\UpperBound);\n\\draw[thin, red, dashed] (axis cs:\\pgfkeysvalueof{/pgfplots/xmin},\\LowerBound) -- (axis cs:\\pgfkeysvalueof{/pgfplots/xmax},\\LowerBound);\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{A caption.}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\resizebox{\\linewidth}{!}{\n\\begin{tikzpicture}\n\\begin{axis}[\nscale only axis,\ngrid = major,\nxmin = 0,\nxmax = 320,\nwidth = 0.5\\textwidth,\nheight = 0.15\\textwidth,\ngrid style = {dashed, ultra thin},\nxtick = {0,80,...,320},\nytick = \\empty,\ntick label style = {font=\\small},\nscaled x ticks=base 10:-2,\n]\n\\addplot[thick, blue] table [x expr=\\coordindex, y index=0]{error_file.csv};\n\\end{axis}\n\\end{tikzpicture}\n}\n\\caption{Another caption.}\n\\end{figure}\n\\end{document}\n\n• But the alignment would still be slightly off in this case (3.2 protrudes further than 10)\n– Jake\nJun 3, 2014 at 12:42\n• Indeed, Jake, thanks. In such case, I would suggest 10:-4 for the first plot then two plots will have the same formats. Jun 3, 2014 at 12:50" ]	[ null, "https://i.stack.imgur.com/nnj8g.png", null, "https://i.stack.imgur.com/yE0KW.png", null, "https://i.stack.imgur.com/RKKmA.png", null, "https://i.stack.imgur.com/8cfnK.png", null, "https://i.stack.imgur.com/JbGEW.png", null, "https://i.stack.imgur.com/pAKce.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62424815,"math_prob":0.99705994,"size":2065,"snap":"2022-40-2023-06","text_gpt3_token_len":664,"char_repetition_ratio":0.09849588,"word_repetition_ratio":0.115830116,"special_character_ratio":0.3108959,"punctuation_ratio":0.20603015,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9975173,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T18:51:57Z\",\"WARC-Record-ID\":\"<urn:uuid:77efc5ab-ba25-437f-b9ef-4934d3b653e5>\",\"Content-Length\":\"256887\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b5daa57-27f5-4350-8b3c-1632bd01a9be>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bd2c1ca-e446-4818-af91-a1ba5d6693d1>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/182911/problem-aligning-multiple-pgfplots-due-to-different-axis-label-sizes/182921#182921\",\"WARC-Payload-Digest\":\"sha1:4AC5T5L3ZKRIMZWHJYKEGH7ZKFP2Z4IQ\",\"WARC-Block-Digest\":\"sha1:LLHPBVKJDJYTYP6CFZKXNMMXDB3HZSHU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337339.70_warc_CC-MAIN-20221002181356-20221002211356-00470.warc.gz\"}"}
https://dumpz.org/aBMw45B324Am	[ "# include iostream include math using namespace std double area double r\n\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18``` ```#include #include using namespace std; double area(double r) { return M_PI * pow(r, 2); } int main() { double r; cout << \"Enter the radius of the circle: \"; cin >> r; cout << \"The area of a circle with a radius of \" << r << \" is equal to: \" << area(r) << endl; return 0; } ```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7101068,"math_prob":0.6701464,"size":292,"snap":"2021-43-2021-49","text_gpt3_token_len":86,"char_repetition_ratio":0.121527776,"word_repetition_ratio":0.0,"special_character_ratio":0.38698632,"punctuation_ratio":0.18333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9882577,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T22:01:43Z\",\"WARC-Record-ID\":\"<urn:uuid:e030f94b-6bc9-4831-b3da-a853333a3c75>\",\"Content-Length\":\"9809\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2e433f4-0578-4a0d-b1f1-59ab86dbb023>\",\"WARC-Concurrent-To\":\"<urn:uuid:731af1e4-7a41-4326-9ee3-c06297be667b>\",\"WARC-IP-Address\":\"89.248.172.2\",\"WARC-Target-URI\":\"https://dumpz.org/aBMw45B324Am\",\"WARC-Payload-Digest\":\"sha1:S4J442IRDEMIDYRPA6SSEEBGSQ4YXN6T\",\"WARC-Block-Digest\":\"sha1:NQM3RMHLIPJ7OGR53INAURF2EMJAEDMB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587926.9_warc_CC-MAIN-20211026200738-20211026230738-00078.warc.gz\"}"}
https://www.daniweb.com/programming/software-development/threads/340738/string-class-operator-problem	[ "Here is the full code relating to my performance problem I’m having with my operator+= which I described in this adjacent thread….\n\nAs a quick review, I’m finding that the operator+= function in the Standard C++ Library’s String Class, i.e., <string>, is performing almost impossibly fast compared to the one in my own String Class, and I’m looking for help or explanations why this is so. In the thread referred to above I provided a stripped down version of my String Class with piles of console output statements in the various implicated String Class member functions to help folks see what is going on, and hopefully provide ideas/comment on which I could work. However, below are four full programs with included performance timings from the Windows Api GetTickCount() function. Included in the programs are output statements which clearly show what is happening rather quickly, and what is taking a lot of time. Before I post the code, however, here is the string algorithm being processed…\n\n'1) Create a 500000 character string of dashes or spaces;\n'2) Change every 7th dash or space to a \"P\";\n'3) SEARCH FOR AND Replace every \"P\" with a \"PU\" (hehehe);\n'4) Replace every dash or space with an \"8\";\n'5) Put in a carriage return & line feed every 90 characters;\n'6) Output last 4000 characters.\n\nHere is my first implementation which uses the C++ Standard Library String Class…\n\n``````#include \"windows.h\"\n#include <stdio.h>\n#include <string>\n#define NUMBER 500000\nusing namespace std;\n\nstring& ReplaceAll(string& context, const string& from, const string& to) //This is Bruce Eckel's\n{ //function from \"Thinking\nsize_t lookHere=0; //In C++\", Volume Two.\nsize_t foundHere;\n\nwhile((foundHere=context.find(from,lookHere)) != string::npos)\n{\ncontext.replace(foundHere, from.size(), to);\nlookHere=foundHere+to.size();\n}\n\nreturn context;\n}\n\nint main(void)\n{\nunsigned t1=0,t2=0;\nint iCount=0;\nstring s2;\n\nt1=GetTickCount(), t2=t1;\nputs(\"Starting....\");\nstring s1(NUMBER,'-');\nt2=GetTickCount()-t2;\nprintf(\"Finished Creating String With %u Of These - : milliseconds elapsed - %u\\n\",NUMBER,t2), t2=t1;\nfor(int i=0; i<NUMBER; i++)\n{\niCount++;\nif(iCount%7==0) s1[iCount-1]='P';\n}\nt2=GetTickCount()-t2;\nprintf(\"Finished Inserting 'P's In s1! : milliseconds elapsed - %u\\n\",t2), t2=t1;\nReplaceAll(s1,\"P\",\"PU\");\nt2=GetTickCount()-t2;\nprintf(\"Finished Replacing 'P's With PU! : milliseconds elapsed - %u\\n\",t2), t2=t1;\nReplaceAll(s1,\"-\",\"8\");\nt2=GetTickCount()-t2;\nprintf(\"Finished Replacing '-'s With 8! : milliseconds elapsed - %u\\n\",t2), t2=t1;\ns2.reserve(750000);\nputs(\"Now Going To Create Lines With CrLfs!\");\nfor(int i=0; i<NUMBER; i=i+90)\ns2+=s1.substr(i,90)+\"\\r\\n\";\nt2=GetTickCount()-t2;\nprintf(\"Finished Creating Lines! : milliseconds elapsed - %u\\n\",t2);\ns1=s2.substr(s2.length()-4000,4000);\nt1=GetTickCount()-t1;\nprintf(\"t1 = %u\\n\",(unsigned)t1);\ngetchar();\n\nreturn 0;\n}``````\n\nIt compiles to 46592 bytes statically linked to all necessary library code and optimized fully for speed. Here are results from a couple runs on my old Win 2000 Compaq laptop, which is rather slow…\n\n``````//Starting....\n//Finished Creating String With 500000 Of These - : milliseconds elapsed - 0\n//Finished Inserting 'P's In s1! : milliseconds elapsed - 20\n//Finished Replacing 'P's With PU! : milliseconds elapsed - 17365\n//Finished Replacing '-'s With 8! : milliseconds elapsed - 17435\n//Now Going To Create Lines With CrLfs!\n//Finished Creating Lines! : milliseconds elapsed - 17445\n//t1 = 17445\n\n//Starting....\n//Finished Creating String With 500000 Of These - : milliseconds elapsed - 0\n//Finished Inserting 'P's In s1! : milliseconds elapsed - 10\n//Finished Replacing 'P's With PU! : milliseconds elapsed - 17335\n//Finished Replacing '-'s With 8! : milliseconds elapsed - 17415\n//Now Going To Create Lines With CrLfs!\n//Finished Creating Lines! : milliseconds elapsed - 17425\n//t1 = 17425``````\n\nAs can be easily seen above, almost all the time is being burnt up replacing the Ps with PUs through this call…\n\nReplaceAll(s1,\"P\",\"PU\");\n\nI’ll do the arithmetic for you. 17345 of the total time of 17425 is going down that sink hole. Note that the big concatenation at the end is taking next to no time at all…\n\nfor(int i=0; i<NUMBER; i=i+90)\ns2+=s1.substr(i,90)+\"\\r\\n\";\n\nTen ticks is about as close to instantaneous as you are going to get. That’s all but unbelievable. But there are the real numbers I’m seeing.\n\nNow here is the same program using my String Class. My String Class is rather large in code it seems (Strings.cpp is 779 lines), but compiles to 20 K less (27136 bytes) than the C++ Standard String Class. Here is Main.cpp followed by Strings.h and Strings.cpp required by Main.cpp…\n\n``````#include <windows.h>\n#include <tchar.h>\n#include <stdio.h>\n#include \"Strings.h\"\n#define NUMBER 500000\n\nint main()\n{\nconst TCHAR* const CrLf=_T(\"\\r\\n\");\nDWORD t1,t2;\nString s2;\n\nputs(\"Starting....\");\nt1 = GetTickCount(), t2=t1;\nString s1(_T('-'),NUMBER);\nt2=GetTickCount()-t2;\nprintf(\"s1 has %u of these - milliseconds elapsed %u\\n\",NUMBER,(unsigned)t2), t2=t1;\nfor(int i=1; i<=NUMBER; i++)\nif(i%7==0) s1.SetChar(i,_T('P'));\nt2=GetTickCount()-t2;\nprintf(\"s1 has 'P' every 7th char milliseconds elapsed %u\\n\",(unsigned)t2), t2=t1;\ns2=s1.Replace((TCHAR)_T(\"P\"),(TCHAR)_T(\"PU\"));\nt2=GetTickCount()-t2;\nprintf(\"s2 created with PUs in place of Ps milliseconds elapsed %u\\n\",(unsigned)t2), t2=t1;\ns1=s2.Replace((TCHAR)_T(\"-\"),(TCHAR)_T(\"8\"));\nt2=GetTickCount()-t2;\nprintf(\"dashes replaced with 8s milliseconds elapsed %u\\n\",(unsigned)t2), t2=t1;\ns2.SetChar(1,'\\0');\nprintf(\"Now Do Concatenation...\\n\");\nfor(int i=1; i<NUMBER; i+=90)\ns2+=s1.Mid(i,90)+CrLf;\nt2 = GetTickCount() - t1;\nprintf(\"Finished Creating Lines milliseconds elapsed %u\\n\",(unsigned)t2), t2=t1;\ns1=s2.Right(4000);\nt1 = GetTickCount() - t1;\nprintf(\"t1 = %u\\n\",(unsigned)t1);\ngetchar();\n\nreturn 0;\n}``````\n``````//Strings.h //My String Class\n#if !defined(STRINGS_H)\n#define STRINGS_H\n#define EXPANSION_FACTOR 4\n#define MINIMUM_ALLOCATION 128\n\nclass String\n{\npublic:\nString(); //Uninitialized Constructor\nString(const TCHAR); //Constructor Initializes String With TCHAR\nString(const TCHAR); //Constructor Initializes String With TCHAR\nString(const String&); //Constructor Initializes String With Another String (Copy Constructor)\nString(int); //Constructor Initializes Buffer To Specific Size\nString(const TCHAR, const int); //Constructor initializes String with int # of chars\nString& operator=(const TCHAR); //Assigns TCHAR To String\nString& operator=(const TCHAR); //Assigns TCHAR To String\nString& operator=(const String&); //Assigns one String to another (this one)\nString& operator=(int); //Converts And Assigns An Integer to A String\nString& operator=(unsigned int); //Converts And Assigns An Unsigned Integer to A String\nString& operator=(long); //Converts And Assigns A Long to A String\nString& operator=(double); //Converts And Assigns A double to A String\nString& operator+(const TCHAR); //For adding TCHAR to String\nString& operator+(const TCHAR); //For adding null terminated TCHAR array to String\nString& operator+(const String&); //For adding one String to Another\nString& operator+=(const String&); //For Concatenating another String Onto this\nbool operator==(const String); //For comparing Strings\nString Left(unsigned int); //Returns String of iNum Left Most TCHARs of this\nString Right(unsigned int); //Returns String of iNum Right Most TCHARs of this\nString Mid(unsigned int, unsigned int); //Returns String consisting of number of TCHARs from some offset\nString& Make(const TCHAR, int); //Creates (Makes) a String with iCount TCHARs\nString Remove(const TCHAR, bool); //Returns A String With A Specified TCHAR* Removed\nString Remove(TCHAR* pStr); //Returns A String With All The TCHARs In A TCHAR* Removed (Individual char removal)\nString Retain(TCHAR* pStr); //Seems to return a String with some characters retained???\nString Replace(TCHAR, TCHAR); //Replace a match string found in this with a new string\nint InStr(const TCHAR); //Returns one based offset of a specific TCHAR in a String\nint InStr(const TCHAR, bool); //Returns one based offset of a particular TCHAR pStr in a String\nint InStr(const String&, bool); //Returns one based offset of where a particular String is in another String\nvoid LTrim(); //Returns String with leading spaces/tabs removed\nvoid RTrim(); //Returns String with spaces/tabs removed from end\nvoid Trim(); //Returns String with both leading and trailing whitespace removed\nunsigned int ParseCount(const TCHAR); //Returns count of Strings delimited by a TCHAR passed as a parameter\nvoid Parse(String, TCHAR); //Returns array of Strings in first parameter as delimited by 2nd TCHAR delimiter\nint iVal(); //Returns int value of a String\nint LenStr(void); //Returns length of string\nTCHAR* lpStr(); //Returns address of pStrBuffer member variable\nTCHAR GetChar(unsigned int); //Returns TCHAR at one based index\nvoid SetChar(unsigned int, TCHAR); //Sets TCHAR at one based index\nvoid Print(bool); //Outputs String to Console with or without CrLf\n~String(); //String Destructor\n\nprivate:\nTCHAR* pStrBuffer; //Holds String Buffer Allocated In Various Overloaded Constructors And Members\nint iAllowableCharacterCount; //Persists Number of characters which will fit in this->pStrBuffer\n};\n\nString operator+(TCHAR* lhs, String& rhs); //global function\n#endif //#if !defined(STRINGS_H)``````\n``````//Strings.cpp == My String Class\n#include <tchar.h>\n#include <stdlib.h>\n#include <stdio.h>\n#include <math.h>\n#include <string.h>\n#include \"Strings.h\"\n\nString operator+(TCHAR* lhs, String& rhs) //Global Function Which\n{ //Operates On String\nString sr=lhs; //Objects\nsr=sr+rhs;\nreturn sr;\n}\n\nString::String() //Uninitialized Constructor\n{\npStrBuffer=new TCHAR[MINIMUM_ALLOCATION];\npStrBuffer=_T('\\0');\nthis->iAllowableCharacterCount=MINIMUM_ALLOCATION-1;\n}\n\nString::String(const TCHAR ch) //Constructor: Initializes with TCHAR\n{\npStrBuffer=new TCHAR[MINIMUM_ALLOCATION];\npStrBuffer=ch;\npStrBuffer=_T('\\0');\niAllowableCharacterCount=MINIMUM_ALLOCATION-1;\n}\n\nString::String(const TCHAR* pStr) //Constructor: Initializes with TCHAR\n{\nint iLen,iNewSize;\n\niLen=_tcslen(pStr);\niNewSize=(iLen/16+1)16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n_tcscpy(pStrBuffer,pStr);\n}\n\nString::String(const String& s) //Constructor Initializes With Another String, i.e., Copy Constructor\n{\nint iLen,iNewSize;\n\niLen=_tcslen(s.pStrBuffer);\niNewSize=(iLen/16+1)16;\nthis->pStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n_tcscpy(this->pStrBuffer,s.pStrBuffer);\n}\n\nString::String(int iSize) //Constructor Creates String With Custom Sized\n{ //Buffer (rounded up to paragraph boundary)\nint iNewSize;\n\niNewSize=(iSize/16+1)16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\nthis->pStrBuffer=_T('\\0');\n}\n\nString::String(const TCHAR ch, int iCount)\n{\nint iNewSize;\n\niNewSize=(iCount/16+1)16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\nfor(int i=0; i<iCount; i++)\npStrBuffer[i]=ch;\npStrBuffer[iCount]=_T('\\0');\n}\n\nString& String::operator=(const TCHAR ch) //Overloaded operator = for assigning a TCHAR to a String\n{\nthis->pStrBuffer=ch;\nthis->pStrBuffer=_T('\\0');\n\nreturn this;\n}\n\nString& String::operator=(const TCHAR* pStr) //Constructor For If Pointer To Asciiz String Parameter\n{\nint iLen,iNewSize;\n\niLen=_tcslen(pStr);\nif(iLen<this->iAllowableCharacterCount)\n_tcscpy(pStrBuffer,pStr);\nelse\n{\ndelete [] pStrBuffer;\niNewSize=(iLen/16+1)16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n_tcscpy(pStrBuffer,pStr);\n}\n\nreturn this;\n}\n\nString& String::operator=(const String& strRight) //Overloaded operator = for\n{ //assigning another String to\nint iRightLen,iNewSize; //a String\n\nif(this==&strRight)\nreturn this;\niRightLen=_tcslen(strRight.pStrBuffer);\nif(iRightLen < this->iAllowableCharacterCount)\n_tcscpy(pStrBuffer,strRight.pStrBuffer);\nelse\n{\niNewSize=(iRightLen/16+1)16;\ndelete [] this->pStrBuffer;\nthis->pStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n_tcscpy(pStrBuffer,strRight.pStrBuffer);\n}\n\nreturn this;\n}\n\nbool String::operator==(const String strCompare)\n{\nif(_tcscmp(this->pStrBuffer,strCompare.pStrBuffer)==0) //_tcscmp\nreturn true;\nelse\nreturn false;\n}\n\nString& String::operator+(const TCHAR ch) //Overloaded operator + (Puts TCHAR in String)\n{\nint iLen,iNewSize;\nTCHAR pNew;\n\niLen=_tcslen(this->pStrBuffer);\nif(iLen<this->iAllowableCharacterCount)\n{\nthis->pStrBuffer[iLen]=ch;\nthis->pStrBuffer[iLen+1]='\\0';\n}\nelse\n{\niNewSize=((this->iAllowableCharacterCountEXPANSION_FACTOR)/16+1)16;\npNew=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n_tcscpy(pNew,this->pStrBuffer);\ndelete [] this->pStrBuffer;\nthis->pStrBuffer=pNew;\nthis->pStrBuffer[iLen]=ch;\nthis->pStrBuffer[iLen+1]='\\0';\n}\n\nreturn this;\n}\n\n{ //or pointers to Asciiz Strings)\nint iLen,iNewSize;\nTCHAR pNew;\n\niLen=_tcslen(this->pStrBuffer)+_tcslen(pChar);\nif(iLen<this->iAllowableCharacterCount)\n{\nif(this->pStrBuffer)\n_tcscat(this->pStrBuffer,pChar);\nelse\n_tcscpy(this->pStrBuffer, pChar);\n}\nelse\n{\niNewSize=(iLenEXPANSION_FACTOR/16+1)16;\npNew=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount = iNewSize-1;\nif(this->pStrBuffer)\n{\n_tcscpy(pNew,this->pStrBuffer);\ndelete [] pStrBuffer;\n_tcscat(pNew,pChar);\n}\nelse\n_tcscpy(pNew,pChar);\nthis->pStrBuffer=pNew;\n}\n\nreturn this;\n}\n\n{ //Another String to the left\nint iLen,iNewSize; //operand\nTCHAR pNew;\n\niLen=_tcslen(this->pStrBuffer) + _tcslen(strRight.pStrBuffer);\nif(iLen < this->iAllowableCharacterCount)\n{\nif(this->pStrBuffer)\n_tcscat(this->pStrBuffer,strRight.pStrBuffer);\nelse\n_tcscpy(this->pStrBuffer,strRight.pStrBuffer);\n}\nelse\n{\niNewSize=(iLenEXPANSION_FACTOR/16+1)16;\npNew=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\nif(this->pStrBuffer)\n{\n_tcscpy(pNew,this->pStrBuffer);\ndelete [] pStrBuffer;\n_tcscat(pNew,strRight.pStrBuffer);\n}\nelse\n_tcscpy(pNew,strRight.pStrBuffer);\nthis->pStrBuffer=pNew;\n}\n\nreturn this;\n}\n\nString& String::operator+=(const String& strRight)\n{\nint iLen,iNewSize;\nTCHAR pNew;\n\niLen=_tcslen(this->pStrBuffer) + _tcslen(strRight.pStrBuffer);\nif(iLen < this->iAllowableCharacterCount)\n{\nif(this->pStrBuffer)\n_tcscat(this->pStrBuffer,strRight.pStrBuffer);\n}\nelse\n{\niNewSize=(iLenEXPANSION_FACTOR/16+1)16;\npNew=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\nif(this->pStrBuffer)\n{\n_tcscpy(pNew,this->pStrBuffer);\ndelete [] pStrBuffer;\n_tcscat(pNew,strRight.pStrBuffer);\n}\nelse\n_tcscpy(pNew,strRight.pStrBuffer);\nthis->pStrBuffer=pNew;\n}\n\nreturn this;\n}\n\nString String::Left(unsigned int iNum)\n{\nunsigned int iLen,i,iNewSize;\n\niLen=_tcslen(this->pStrBuffer);\nif(iNum<iLen)\n{\niNewSize=(iNumEXPANSION_FACTOR/16+1)16;\nString sr((int)iNewSize);\nsr.iAllowableCharacterCount=iNewSize-1;\nsr.pStrBuffer=new TCHAR[iNewSize];\nfor(i=0;i<iNum;i++)\nsr.pStrBuffer[i]=this->pStrBuffer[i];\nsr.pStrBuffer[iNum]='\\0';\nreturn sr;\n}\nelse\n{\nString sr(this);\nreturn sr;\n}\n}\n\nString String::Right(unsigned int iNum) //Returns Right\\$(strMain,iNum)\n{\nunsigned int iLen,iNewSize;\n\niLen=_tcslen(this->pStrBuffer);\nif(iNum<iLen)\n{\niNewSize=(iNumEXPANSION_FACTOR/16+1)16;\nString sr((int)iNewSize);\nsr.iAllowableCharacterCount=iNewSize-1;\nsr.pStrBuffer=new TCHAR[iNewSize];\n_tcsncpy(sr.pStrBuffer,this->pStrBuffer+iLen-iNum,iNum);\nsr.pStrBuffer[iNum]='\\0';\nreturn sr;\n}\nelse\n{\nString sr(this);\nreturn sr;\n}\n}\n\nString String::Mid(unsigned int iStart, unsigned int iCount)\n{\nunsigned int iLen,iNewSize;\n\niLen=_tcslen(this->pStrBuffer);\nif(iStart && iStart<=iLen)\n{\nif(iCount && iStart+iCount-1<=iLen)\n{\niNewSize=(iCountEXPANSION_FACTOR/16+1)16;\nString sr((int)iNewSize);\nsr.iAllowableCharacterCount=iNewSize-1;\n_tcsncpy(sr.pStrBuffer,this->pStrBuffer+iStart-1,iCount);\nsr.pStrBuffer[iCount]='\\0';\nreturn sr;\n}\nelse\n{\nString sr(this);\nreturn sr;\n}\n}\nelse\n{\nString sr(this);\nreturn sr;\n}\n}\n\nString& String::Make(const TCHAR ch, int iCount) //Creates (Makes) a String with iCount TCHARs\n{\nif(iCount>this->iAllowableCharacterCount)\n{\ndelete [] pStrBuffer;\nint iNewSize=(iCountEXPANSION_FACTOR/16+1)16;\nthis->pStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n}\nfor(int i=0; i<iCount; i++)\npStrBuffer[i]=ch;\npStrBuffer[iCount]=_T('\\0');\n\nreturn this;\n}\n\nString String::Remove(const TCHAR* pToRemove, bool blnCaseSensitive)\n{\nint i,j,iParamLen,iReturn=0;\n\nif(pToRemove==0)\nreturn this;\nString strNew(this->LenStr());\niParamLen=_tcslen(pToRemove);\ni=0, j=0;\ndo\n{\nif(pStrBuffer[i]==0)\nbreak;\nif(blnCaseSensitive)\niReturn=_tcsncmp(pStrBuffer+i,pToRemove,iParamLen); //_tcsncmp\nelse\niReturn=_tcsnicmp(pStrBuffer+i,pToRemove,iParamLen); //__tcsnicmp\nif(iReturn==0)\ni=i+iParamLen;\nstrNew.pStrBuffer[j]=pStrBuffer[i];\nj++, i++;\nstrNew.pStrBuffer[j]=_T('\\0');\n}while(1);\n\nreturn strNew;\n}\n\nString String::Remove(TCHAR* pStr)\n{\nunsigned int i,j,iStrLen,iParamLen;\nTCHAR pThis, pThat, p;\n\niStrLen=this->LenStr(); //The length of this\nString sr((int)iStrLen); //Create new String big enough to contain original String (this)\niParamLen=_tcslen(pStr); //Get length of parameter (pStr) which contains chars to be removed\npThis=this->pStrBuffer;\np=sr.lpStr();\nfor(i=0; i<iStrLen; i++)\n{\npThat=pStr;\nfor(j=0; j<iParamLen; j++)\n{\nif(pThis==pThat)\n{\nbreak;\n}\npThat++;\n}\n{\np=pThis;\np++;\np=_T('\\0');\n}\npThis++;\n}\n\nreturn sr;\n}\n\nString String::Retain(TCHAR* pStr)\n{\nunsigned int i,j,iStrLen,iParamLen;\nTCHAR pThis, pThat, p;\nbool blnFoundGoodChar;\n\niStrLen=this->LenStr(); //The length of this\nString sr((int)iStrLen); //Create new String big enough to contain original String (this)\niParamLen=_tcslen(pStr); //Get length of parameter (pStr) which contains chars to be retained\npThis=this->pStrBuffer; //pThis will point to this String's buffer, and will increment through string.\np=sr.lpStr(); //p will start by pointing to new String's buffer and will increment through new string\nfor(i=0; i<iStrLen; i++)\n{\npThat=pStr;\nblnFoundGoodChar=false;\nfor(j=0; j<iParamLen; j++)\n{\nif(pThis==pThat)\n{\nblnFoundGoodChar=true;\nbreak;\n}\npThat++;\n}\nif(blnFoundGoodChar)\n{\np=pThis;\np++;\np=_T('\\0');\n}\npThis++;\n}\n\nreturn sr;\n}\n\nString String::Replace(TCHAR* pMatch, TCHAR* pNew)\n{\nint iLenMatch,iLenNew,iLenMainString,iCountMatches,iExtra,iExtraLengthNeeded,iAllocation,iCtr;\nString sr;\n\niCountMatches=0, iAllocation=0, iCtr=0;\niLenMatch=_tcslen(pMatch);\niLenNew=_tcslen(pNew);\niLenMainString=this->LenStr();\nif(iLenNew==0)\nsr=this->Remove(pMatch); //return\nelse\n{\niExtra=iLenNew-iLenMatch;\nfor(int i=0; i<iLenMainString; i++)\nif(_tcsncmp(pStrBuffer+i,pMatch,iLenMatch)==0) iCountMatches++; //Count how many\niExtraLengthNeeded=iCountMatchesiExtra; //match strings\niAllocation=iLenMainString+iExtraLengthNeeded;\nString strNew(iAllocation);\nfor(int i=0; i<iLenMainString; i++)\n{\nif(_tcsncmp(pStrBuffer+i,pMatch,iLenMatch)==0)\n{\n_tcscpy(strNew.pStrBuffer+iCtr,pNew);\niCtr=iCtr+iLenNew;\ni=i+iLenMatch-1;\n}\nelse\n{\nstrNew.pStrBuffer[iCtr]=this->pStrBuffer[i];\niCtr++;\n}\nstrNew.pStrBuffer[iCtr]=_T('\\0');\n}\nsr=strNew;\n}\n\nreturn sr;\n}\n\nint String::InStr(const TCHAR ch)\n{\nint iLen,i;\n\niLen=_tcslen(this->pStrBuffer);\nfor(i=0;i<iLen;i++)\n{\nif(this->pStrBuffer[i]==ch)\nreturn (i+1);\n}\n\nreturn 0;\n}\n\nint String::InStr(const TCHAR pStr, bool blnCaseSensitive)\n{\nint i,iParamLen,iRange;\n\nif(pStr==0)\nreturn 0;\niParamLen=_tcslen(pStr);\niRange=_tcslen(pStrBuffer)-iParamLen;\nif(iRange>=0)\n{\nfor(i=0;i<=iRange;i++)\n{\nif(blnCaseSensitive)\n{\nif(_tcsncmp(pStrBuffer+i,pStr,iParamLen)==0) //_tcsncmp\nreturn i+1;\n}\nelse\n{\nif(_tcsnicmp(pStrBuffer+i,pStr,iParamLen)==0) //__tcsnicmp\nreturn i+1;\n}\n}\n}\n\nreturn 0;\n}\n\nint String::InStr(const String& s, bool blnCaseSensitive)\n{\nint i,iParamLen,iRange,iLen;\n\niLen=_tcslen(s.pStrBuffer);\nif(iLen==0)\nreturn 0;\niParamLen=iLen;\niRange=_tcslen(pStrBuffer)-iParamLen;\nif(iRange>=0)\n{\nfor(i=0;i<=iRange;i++)\n{\nif(blnCaseSensitive)\n{\nif(_tcsncmp(pStrBuffer+i,s.pStrBuffer,iParamLen)==0) //_tcsncmp\nreturn i+1;\n}\nelse\n{\nif(_tcsnicmp(pStrBuffer+i,s.pStrBuffer,iParamLen)==0) //__tcsnicmp\nreturn i+1;\n}\n}\n}\n\nreturn 0;\n}\n\nvoid String::LTrim()\n{\nunsigned int i,iCt=0,iLenStr;\n\niLenStr=this->LenStr();\nfor(i=0;i<iLenStr;i++)\n{\nif(pStrBuffer[i]==32\|\|pStrBuffer[i]==9)\niCt++;\nelse\nbreak;\n}\nif(iCt)\n{\nfor(i=iCt;i<=iLenStr;i++)\npStrBuffer[i-iCt]=pStrBuffer[i];\n}\n}\n\nvoid String::RTrim()\n{\nunsigned int iCt=0, iLenStr;\n\niLenStr=this->LenStr()-1;\nfor(unsigned int i=iLenStr; i>0; i--)\n{\nif(this->pStrBuffer[i]==9\|\|this->pStrBuffer[i]==10\|\|this->pStrBuffer[i]==13\|\|this->pStrBuffer[i]==32)\niCt++;\nelse\nbreak;\n}\nthis->pStrBuffer[this->LenStr()-iCt]=0;\n}\n\nvoid String::Trim()\n{\nthis->LTrim();\nthis->RTrim();\n}\n\nunsigned int String::ParseCount(const TCHAR c) //returns one more than # of\n{ //delimiters so it accurately\nunsigned int iCtr=0; //reflects # of strings delimited\nTCHAR p; //by delimiter.\n\np=this->pStrBuffer;\nwhile(p)\n{\nif(p==c)\niCtr++;\np++;\n}\n\nreturn ++iCtr;\n}\n\nvoid String::Parse(String* pStr, TCHAR delimiter)\n{\nunsigned int i=0;\nTCHAR* pBuffer=0;\nTCHAR* c;\nTCHAR* p;\n\npBuffer=new TCHAR[this->LenStr()+1];\nif(pBuffer)\n{\np=pBuffer;\nc=this->pStrBuffer;\nwhile(c)\n{\nif(c==delimiter)\n{\npStr[i]=pBuffer;\np=pBuffer;\ni++;\n}\nelse\n{\np=c;\np++;\np=0;\n}\nc++;\n}\npStr[i]=pBuffer;\ndelete [] pBuffer;\n}\n}\n\nint String::iVal()\n{\nreturn _ttoi(this->pStrBuffer); //_ttoi\n}\n\nString& String::operator=(int iNum)\n{\nif(this->iAllowableCharacterCount<16)\n{\nint iNewSize;\ndelete [] this->pStrBuffer;\niNewSize=16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n}\n_stprintf(this->pStrBuffer,_T(\"%d\"),iNum);\n\nreturn this;\n}\n\nString& String::operator=(unsigned int iNum)\n{\nif(this->iAllowableCharacterCount<16)\n{\nint iNewSize;\ndelete [] this->pStrBuffer;\niNewSize=16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n}\n_stprintf(this->pStrBuffer,_T(\"%d\"),iNum);\n\nreturn this;\n}\n\nString& String::operator=(long iNum)\n{\nif(this->iAllowableCharacterCount<16)\n{\nint iNewSize;\ndelete [] this->pStrBuffer;\niNewSize=16;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n}\n_stprintf(this->pStrBuffer,_T(\"%ld\"),iNum);\n\nreturn this;\n}\n\nString& String::operator=(double dblNum)\n{\nif(this->iAllowableCharacterCount<16)\n{\nint iNewSize;\ndelete [] this->pStrBuffer;\niNewSize=32;\npStrBuffer=new TCHAR[iNewSize];\nthis->iAllowableCharacterCount=iNewSize-1;\n}\n_stprintf(this->pStrBuffer,_T(\"%10.14f\"),dblNum);\n\nreturn this;\n}\n\nint String::LenStr(void)\n{\nreturn _tcslen(this->pStrBuffer);\n}\n\nTCHAR String::lpStr()\n{\nreturn pStrBuffer;\n}\n\nTCHAR String::GetChar(unsigned int iOffset)\n{\nreturn this->pStrBuffer[iOffset-1];\n}\n\nvoid String::SetChar(unsigned int iOneBasedOffset, TCHAR tcChar)\n{\nif((int)iOneBasedOffset<=this->iAllowableCharacterCount)\nthis->pStrBuffer[iOneBasedOffset-1]=tcChar;\n}\n\nvoid String::Print(bool blnCrLf)\n{\n_tprintf(_T(\"%s\"),pStrBuffer);\nif(blnCrLf)\n_tprintf(_T(\"\\n\"));\n}\n\nString::~String() //String Destructor\n{\ndelete [] pStrBuffer;\npStrBuffer=0;\n}``````\n\nWhew! I know…..\n\nAnyway, here are the timings on that…\n\n``````//Starting....\n//s1 has 500000 of these - milliseconds elapsed 0\n//s1 has 'P' every 7th char milliseconds elapsed 10\n//s2 created with PUs in place of Ps milliseconds elapsed 80\n//dashes replaced with 8s milliseconds elapsed 150\n//t1 = completed concatenation 12538``````\n\nIts beating the Standard C++ Library by almost 5000 ticks or five seconds. But that’s where things really start to get weird! My string class is only taking 70 ticks to replace the Ps with PUs (see my String::Replace() member function), whereas the ReplaceAll() function from Bruce Eckel which takes Std. C++ Library String Class parameters is taking like 17000 ticks or 17 seconds to do it! However, my String Class is taking some 12,400 ticks or twelve seconds to do what the Std. Library String Class is doing in 10 ticks or 0.010 seconds!!!! Go figure! All in all I’m coming out way ahead with my String Class, but I’m completely vexed by the time difference and apparent lack of performance of my operator+= String Class member compared to what I’m seeing with the Std. Lib. String Class.\n\nIn an attempt to see what I could do to lessen the time I thought I’d do just simple strcpy(), strcat() calls from the Standard C String Library, which I suppose are sitting right on top of the asm string primitives that blast memory blocks about blindingly fast, so here is that iteration of the program which is exactly the same as the above one except for the for loop at bottom with strcpy(), strcat() calls instead of my String Class member function calls…\n\n``````#include <windows.h>\n#include <tchar.h>\n#include <stdio.h>\n#include \"Strings.h\"\n#define NUMBER 500000\n\nint main()\n{\nString s1(_T(' '),NUMBER);\nDWORD t1=0,t2=0;\nString s2;\n\nt1 = GetTickCount();\nfor(int i=1; i<=NUMBER; i++)\nif(i%7==0) s1.SetChar(i,_T('P'));\nt2=GetTickCount()-t1;\nprintf(\"t2=%u\\tAfter Setting Ps\\n\",(unsigned)t2), t2=t1;\ns2=s1.Replace((TCHAR)_T(\"P\"),(TCHAR)_T(\"PU\"));\nt2=GetTickCount()-t1;\nprintf(\"t2=%u\\tAfter Doing PUs\\n\",(unsigned)t2), t2=t1;\ns1=s2.Replace((TCHAR)_T(\" \"),(TCHAR)_T(\"8\"));\nt2=GetTickCount()-t1;\nprintf(\"t2=%u\\tAfter Doing 8s\\n\",(unsigned)t2), t2=t1;\ns2.Make(_T('\\0'),2200000);\nt2=GetTickCount()-t1;\nprintf(\"t2=%u\\tAfter Making s2\\n\",(unsigned)t2), t2=t1;\nTCHAR* pBuf1=s1.lpStr();\nTCHAR* pBuf2=s2.lpStr();\nfor(int i=0; i<NUMBER; i+=90)\n{\n_tcsncpy(pBuf2,pBuf1,90);\n_tcscat(pBuf2,(TCHAR)_T(\"\\r\\n\"));\npBuf2=pBuf2+92;\npBuf1=pBuf1+90;\n}\nt2=GetTickCount()-t1;\nprintf(\"t2=%u\\tAfter Concatenation\\n\",(unsigned)t2), t2=t1;\ns1=s2.Right(4000);\nt2 = GetTickCount()-t1, printf(\"t2=%u\\n\",(unsigned)t2);\ngetchar();\n\nreturn 0;\n}``````\n\nt2=10 After Setting Ps\nt2=80 After Doing PUs\nt2=150 After Doing 8s\nt2=160 After Making s2\nt2=160 After Concatenation\nt2=160\n\nWell, that did the trick, how about it!?! Who says C isn’t worth learning? But it still doesn’t answer my original question about why my String Class is doing so bad at the concatenation compared to the Standard C++ Library String Class. Just as a matter of interest, here is a PowerBASIC program implementing the same algorithm. It does this in 40 ticks and compiles to 9216 bytes! That’s 1/3 rd the size of my smallest C++ version and four times faster…\n\n``````#Compile Exe\n#Dim All\nDeclare Function GetTickCount Lib \"Kernel32.dll\" Alias \"GetTickCount\" () As Dword\nDeclare Function printf CDecl Lib \"msvcrt.dll\" Alias \"printf\" (szFmtStr As Asciiz, lpVar1 As Any) As Long\n\nDeclare Function MessageBox Lib \"User32.dll\" Alias \"MessageBoxA\" _\n(\nByval hWnd As Dword, _\nlpText As Asciiz, _\nlpCaption As Asciiz, _\nByval dwType As Dword _\n) As Long\n\nFunction PBMain() As Long\nRegister i As Long, iCnt As Long\nLocal currPos As Long\nLocal s, s1 As String\nLocal tick As Dword\n\ntick = getTickCount()\ns = String\\$(500000, \"-\") '1) Make String With 500000 dashes\nFor i = 1 To 500000\nIncr iCnt\nIf iCnt = 7 Then\niCnt = 0\nAsc(s, i) = &h50 '2) Put Ps Every 7th character\nEnd If\nNext\nReplace \"P\" With \"PU\" In s '3) Replace Ps with PUs\nReplace Any \"-\" With \"8\" In s '4) Change dashes to 8s\ns1 = String\\$(600000, \\$NUL) '5) 2nd string for CRLF's a little bigger\ncurrPos = 1 'for s2 position tracking\nFor i = 1 To 500000 Step 90\nMid\\$(s1, currPos) = Mid\\$(s, i, 90) & \\$CRLF\ncurrPos = currPos + 92\nNext ii\ns=Right\\$(s, 4000)\ntick = getTickCount() - tick\nprintf(\"tick = %u\",Byval tick)\nMessageBox(0,Byval Strptr(s),\"Last 4000 Characters Of Result\",%MB_OK)\n\nPBMain=0\nEnd Function``````\n\nStore the string length before hand, add the length of the two strings to find the new length, if the new length is greater than the current capacity then you call to increase the size, then call to have the the data of the new string concatenated to yours.\n\n``````\n``````\n\n## All 4 Replies\n\nStore the string length before hand, add the length of the two strings to find the new length, if the new length is greater than the current capacity then you call to increase the size, then call to have the the data of the new string concatenated to yours.\n\n``````class String\n{\nprivate:\nsize_t length;\nTCHAR data;\npublic:\nString size() { return(length); }\n\n&String::operator+=(const String& str) {\nsize_t new_length = this->size() + str.size();\n\nif(new_length > this.capacity())\nthis->reallocate_reserve(new_length);\n\nconcatenate(...);\n}\n};``````\n\nThat's pretty much what's in the g++(well technically +=() calls append(), but whatever)\n\nBTW, you should read this: http://www.agner.org/optimize/optimizing_cpp.pdf\n\nThanks for looking at my posts MosaicFuneral. That link you gave is interesting. I’m going to save and study it. In terms of this…\n\nStore the string length before hand, add the length of the two strings to find the new length, if\nthe new length is greater than the current capacity then you call to increase the size, then call\nto have the the data of the new string concatenated to yours.\n\nI do believe I’m already doing that. Here are the private data members of my String Class…\n\n``````private:\nTCHAR* pStrBuffer; //Holds String Buffer Allocated In Various Overloaded Constructors And Members\nint iAllowableCharacterCount; //Persists Number of characters which will fit in this->pStrBuffer``````\n\nAll my member functions which allocate memory (constructors, overloaded operator functions, etc.) allocate it on 16 byte paragraph boundaries and always round up to multiples of 16 bytes. Here is my operator+= member. You can see the very first line obtains the length of the string controlled by ‘this’ plus the length to be added, i.e., strRight. If the buffer indicated by my iAllowableCharacterCount member is big enough, the strRight is strcat()’ed to it. If not, a memory re-allocation occurs, and both strings are moved to the new and larger buffer. I might point out that in the real program there would be next to none of this time wasting re-allocation and string movement taking place. My swap/temp space is s2. Here it would have been made large enough to accommodate the larger string with the ‘PUs’ instead of just the ‘Ps’…\n\ns2=s1.Replace((TCHAR)_T(\"P\"),(TCHAR)_T(\"PU\"));\n\nAlso note my EXPANSION_FACTOR define in the string class header. It is set to ‘4’ for this example which is extremely, extremely aggressive. Also, the MINIMUM_ALLOCATION was set to 128 which is also ridiculously high. I did that just to protect myself against the 92 byte chunk copies. Still no dice. The Std Lib String class is killing me.\n\n``````String& String::operator+=(const String& strRight)\n{\nint iLen=_tcslen(this->pStrBuffer) + _tcslen(strRight.pStrBuffer); //get length of both\nif(iLen < this->iAllowableCharacterCount) //compare with present\n_tcscat(this->pStrBuffer,strRight.pStrBuffer); //allocation\nelse //present alloc isn’t\n{ //enough, so re-alloc\nint iNewSize=(iLenEXPANSION_FACTOR/16+1)16; //and copy both to new\nTCHAR* pNew=new TCHAR[iNewSize]; //memory\nthis->iAllowableCharacterCount=iNewSize-1;\nif(this->pStrBuffer)\n{\n_tcscpy(pNew,this->pStrBuffer);\ndelete [] pStrBuffer;\n_tcscat(pNew,strRight.pStrBuffer);\n}\nelse\n_tcscpy(pNew,strRight.pStrBuffer);\nthis->pStrBuffer=pNew;\n}\n\nreturn *this;\n}``````\n\nYou hsve given me an idea though that I want to check out further…\n\nYes, it's pretty much a similar concept but you're shoving it all in one oversized function. The example I showed you kept things small, clean, and minimal.\n\nI finally got your point MosaicFuneral. I haven't been saving the string length of 'this' in any of my member functions. That I haven't no doubt goes all the way back to when I first started to develop my own string class 3 or 4 years ago, I was only concerned with getting it to work at all; I wasn't concerned with speed or efficiency. Until now, that is. The way it is now my string lengths are being calculated in every member function call, and each concatenation in a loop is making about four calls or so. So I've added a private string length member to my class, and I'm working through the code now making the necessary changes. All the while I'm thinking, \"Wow! \"Should have done this before!\". When I get it done I come back with the performance improvements. I can't help thinking they'll be substantial.\n\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of 1.19 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.53175527,"math_prob":0.83032495,"size":27717,"snap":"2020-34-2020-40","text_gpt3_token_len":8387,"char_repetition_ratio":0.19128206,"word_repetition_ratio":0.14172228,"special_character_ratio":0.29083234,"punctuation_ratio":0.19770335,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95815194,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T03:31:05Z\",\"WARC-Record-ID\":\"<urn:uuid:34bc1936-3727-4c17-a2a6-5ed3b962afe7>\",\"Content-Length\":\"104138\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6f392a8-d5c2-4cb7-b21e-99ca8e561be1>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8bb99bc-1265-424e-b9e1-f39d94d7c016>\",\"WARC-IP-Address\":\"104.22.4.5\",\"WARC-Target-URI\":\"https://www.daniweb.com/programming/software-development/threads/340738/string-class-operator-problem\",\"WARC-Payload-Digest\":\"sha1:ECVJPW65JIATLQDZBKGFALD7SSNGQBQM\",\"WARC-Block-Digest\":\"sha1:FEWDDX5ZONLY3VOQSYFKVKEKZIFX2444\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198887.3_warc_CC-MAIN-20200921014923-20200921044923-00616.warc.gz\"}"}
http://stpaulschurchmumbai.com/milking-shorthorn-ggtojis/kyg8z.php?c254b0=excel-unique-function-multiple-columns	[ "# excel unique function multiple columns\n\nList3:=Sheet1!\\$E\\$1:\\$E\\$7 I was able to extract a unique distinct list from the column of the sheet using the following formula: If the value is in one column but not in another column, then marked as TRUE, otherwise, marked as FALSE. Very helpful article - many thanks for posting the formula and the explanation. Assuming your values in range A2: C9, please enter the following formula into cell E2: 2. Are there any quick tricks for you to extract unique values from multiple columns in Excel? 2. At first, please insert one new blank column at the left of your data, in this example, I will insert column A beside the original data. I am using a helper column on my \"summary\" sheet for each of the sheets I pull data from. Question: I have cell values spanning over several columns and I want to create a unique list from that range. Excel Array Formula to Extract a List of Unique Values from a Column Update : If you have Office 365 you can use the new UNIQUE Function to extract a distinct or unique list. Extract unique distinct values A to Z from a range and ignore blanks Extract a unique distinct list sorted from A-Z from range Extract unique distinct values from … Using either of your array formulas below for refering to a 4-column list, I'm having trouble with a \"0\" (zero) being placed when blank cells are in the referenced columns. Related Lessons. Supposing you have several columns with multiple values, some values are repeated in same column or different column. Ignores logical values and text, MIN(IF(COUNTIF(\\$B\\$12:B12, tbl_text)=0, ROW(tbl_text)-MIN(ROW(tbl_text))+1)), Step 4 - Part two, identify array values in current row, INDEX(array,row_num,[column_num]) returns a value or reference of the cell at the intersection of a particular row and column, in a given range, INDEX(tbl_text, MIN(IF(COUNTIF(\\$B\\$12:B12, tbl_text)=0, ROW(tbl_text)-MIN(ROW(tbl_text))+1)), , 1)), INDEX(tbl_text, MIN({1,1,1;2,2,2;3,3,3}), , 1)), INDEX(tbl_text, 1, , 1)) returns array {\"Apple\", \"Banana\", \"Lemon\"}. The FILTER function (as you will see) can test multiple criteria, and from any column. MIN(number1,[number2]) unique values. what a waste of time..... formula does NOT work. For example, I have a range of data which contains Name and Order columns, now, to sum only unique values in Order column based on the Name column as following screenshot shown. =LARGE(IF(COUNTIF(\\$D\\$12:D12, tbl_num)=0, tbl_num, \"\"), 1), Step 1 - Remove previously extracted values above current cell with an array with boolean values, Step 2 - Convert boolean values to numeric values, IF(logical_test;[value_if:true];[value_if_false]) checks whether a condition is met, and returns one value if TRUE, and another value if FALSE, IF(COUNTIF(\\$D\\$12:D12, tbl_num)=0, tbl_num, \"\"), IF({TRUE,TRUE,TRUE;TRUE,TRUE,TRUE;TRUE,TRUE,TRUE}, {1, 2, 1;2, 4, 3;1, 3, 1}, \"\"), Step 3 - Convert boolean values to numeric values. In this accelerated training, you'll learn how to use formulas to manipulate text, work with dates and times, lookup values with VLOOKUP and INDEX & MATCH, count and sum with criteria, … This tutorial describes multiple ways to extract a unique or distinct list from a column in Excel. at the end there is a #N/A in this file can you please suggest me how to get rid of it. To remove duplicate values, click Data > Data Tools > Remove Duplicates. However, not everybody has the subscription and will upgrade when it is sensible for their business, often combining it with a hardware refresh. To return unique values from in the range A1:A10, you can use a formula like this: = UNIQUE( A1:A10) To return unique values from the horizontal range A1:E1, set the by_col argument to TRUE or 1: = UNIQUE( A1:E1,1) // extract unique from horizontal array. But how to do it, if you need multiple columns to recognize the duplicate? (Excel 97-2003 Workbook .xls). `Insert your formula here.` Create a list of duplicates where adjacent cell value meets a condition, Extract unique distinct records from two data sets, Filter unique distinct records using criteria. Microsoft and the Office logo are trademarks or registered trademarks of Microsoft Corporation in the United States and/or other countries. In DAX is quite simple to get the number o unique values - just use the DISTINCTCOUNT function. Select the Copy To Another Location option (Figure B), select the Unique Records Only check box, and type B1 in the Copy To field. 1. In this first example we want to apply a Conditional Formatting to the data set below. Source. Perhaps, you'll find it more convenient to have the results in a single cell? I attach an image of an example of what I need. For example here I´d like to know, how many animals and colors are there. Perform previous steps again, but this time fill in Column Name MergeMetrics and Formula =Metrics.For both new columns: by the way, thanks so much for the info on your website. See screenshot: 4. 2. This is a little bit complex but it is fast. To highlight unique or duplicate values, use the Conditional Formatting command in the Style group on the Home tab. Is that possible? This number may or may not be repeated on any of the other sheets. I really appreciate your reply. (Note:Don't click a cell in the first row.). The value for MultiPC shown below the formula is the absolute references for MultiPC (comma separated between sheets, e.g. Hi Oscar, =INDEX(tbl_text, 1, 1) returns value \"Apple\" in cell B13. Put your VBA code here. Gets a value in a specific cell range based on a row and column number. 6. Thanks to Eero, who contributed the original array formula! Count Unique Values Cell C13 formula: Counts number of Unique values (each distinct value is counted only once whether it appears once or multiple times). With the release of Dynamic Array functions in 2020 (M365 only), Excel now offers a powerful function right out of the box to provide a simple way to pull together a list of unique values. Recommended Articles. The list has some duplicate values. The formulas create several errors. does anyone know, for the output, how to make it into several lines but not into one line ? And now you want to find the values which are present in either column only once. I need to enlarge that range for a bigger table. There are also a few functions, such as TRANSPOSE, that must always be an array formula to work correctly. More Than Two Columns Meet Criteria. IFERROR (IFERROR (formula1, formula2), formula3) I guess technically a blank is a unique value in the tbl but I'm trying to make sure only relevant numbers are returned. unique values. The UNIQUE function returns an array of unique values from a given array or range. Count unique values based on one criteria. The \"summary\" page is getting quite processor intensive with all the helper columns I am using. Count of unique values in multiple columns. The following detailed steps may help you. Back in March 2018, Gašper wrote a brilliant two part article on Merge Queries being an alternative to VLOOKUP . You can use a formula based on the IF function, the ISNA function and the VLOOKUP function. Step 1 - Find new unique distinct text values, =INDEX(tbl_text, MIN(IF(COUNTIF(\\$B\\$12:B12, tbl_text)=0, ROW(tbl_text)-MIN(ROW(tbl_text))+1)), MATCH(0, COUNTIF(\\$B\\$12:B12, INDEX(tbl_text, MIN(IF(COUNTIF(\\$B\\$12:B12, tbl_text)=0, ROW(tbl_text)-MIN(ROW(tbl_text))+1)), , 1)), 0), 1), COUNTIF(range,criteria) - Counts the number of cells within a range that meet the given condition, COUNTIF(\"Text\", {\"Apple\",\"Banana\",\"Lemon\";\"Orange\",\"Lemon\",\"Apple\";\"Lemon\",\"Banana\",\"Orange\"})=0, {TRUE,TRUE,TRUE;TRUE,TRUE,TRUE;TRUE,TRUE,TRUE}. =INDIRECT(TEXT(MIN(IF((\\$A\\$2:\\$C\\$9<>\"\")(COUNTIF(\\$E\\$1:E1,\\$A\\$2:\\$C\\$9)=0),ROW(\\$2:\\$9)100+COLUMN(\\$A:\\$C),7^8)),\"R0C00\"),)&\"\", Kutools for Excel Solves Most of Your Problems, and Increases Your Productivity by However, this array formula should be done the Ctrl+Shift+Enter in first cell and copy down. Using the SUM/SUMPRODUCT Function for Multiple Columns. Step 2 - Convert boolean array to row numbers, IF(logical_test,[value_if:true],[value_if_false]) checks whether a condition is met, and returns one value if TRUE, and another value if FALSE, IF(COUNTIF(\\$B\\$12:B12, tbl_text)=0, ROW(tbl_text)-MIN(ROW(tbl_text))+1), IF({TRUE,TRUE,TRUE;TRUE,TRUE,TRUE;TRUE,TRUE,TRUE}, {2;3;4}-MIN({2;3;4})+1), IF({TRUE,TRUE,TRUE;TRUE,TRUE,TRUE;TRUE,TRUE,TRUE}, {2;3;4}-2+1), IF({TRUE,TRUE,TRUE;TRUE,TRUE,TRUE;TRUE,TRUE,TRUE}, {1;2;3}) and returns {1,1,1;2,2,2;3,3,3}. =IFERROR(IFERROR(IFERROR(IFERROR(INDEX(List1, MATCH(0, COUNTIF(\\$E\\$1:E1, List1), 0)), INDEX(List2, MATCH(0, COUNTIF(\\$E\\$1:E1, List2), 0))), INDEX(List3, MATCH(0, COUNTIF(\\$E\\$1:E1, List3), 0))), INDEX(List4, MATCH(0, COUNTIF(\\$E\\$1:E1, List4), 0))), \"\"), =IFERROR(INDEX(\\$B\\$15:\\$D\\$64, MIN(IF((COUNTIF(\\$F\\$14:\\$F14, \\$B\\$15:\\$D\\$64)=0)(\\$B\\$15:\\$D\\$64\"\"), ROW(\\$B\\$15:\\$D\\$64)-MIN(ROW(\\$B\\$15:\\$D\\$64))+1)), MATCH(0, COUNTIF(\\$F\\$14:\\$F14, INDEX(\\$B\\$15:\\$D\\$64, MIN(IF((COUNTIF(\\$F\\$14:\\$F14, \\$B\\$15:\\$D\\$64)=0)*(\\$B\\$15:\\$D\\$64\"\"), ROW(\\$B\\$15:\\$D\\$64)-MIN(ROW(\\$B\\$15:\\$D\\$64))+1)), , 1)), 0), 1),\"\"), See this workbook: I'm guessing it's because the named range doesn't consitute an array (not rectangular? […], The following array formula in cell B11 extracts duplicates from cell range B3:E8, only one instance of each duplicate is […], Your formula now looks like this: {=array_formula}. When you will be doing some complex data analysis, you might be needed to analyze more than one conditions at a time. The formula in cell G3 is: =UNIQUE(tblExam[First]) As the UNIQUE function is referencing the entire column, when the column expands or retracts, so does the result of the formula. 2. Another option if you have the new LET function. Formulas are the key to getting things done in Excel. UNIQUE Function. Thank You for posting and keeping it here. #4 Count Cells with Multiple Criteria – Between Two Values. In this blog post though we take things further to see example of testing and formatting multiple columns. In the Advanced Filter dialog, choose Copy To A New Location. So yeah guys, this how you count unique values in a range on multiple conditions. Column C formulas Return COUNT of Unique or Duplicate Values in column A. List1:=Sheet1!\\$A\\$1:\\$A\\$7 If you have any doubts regarding this excel formula article then let me know in the comments section below. There are multiple ways to do this in Excel and in this tutorial I will show you some of these (such as comparing using VLOOKUP formula or IF formula or Conditional formatting). Open and create multiple documents in new tabs of the same window, rather than in new windows. Muchas gracias por la macro!!! I am then combining this into a unique distinct sorted list without blanks. I have a matrix of data in Excel and I want to write in a row the unique values in that data using only formulas. I would like to create the list of uniques in the column of the summary worksheet's table. Then check the field Value or drag the Value to the Rows label, now you will get the unique values from the multiple columns as follows: With the following VBA code, you can also extract the unique values from multiple columns. This array formula is CORRECT. Any ideas of how to eliminate this zero? I have several worksheets, each with a table inserted. This function is very helpful in eliminating items that occur multiple times. How to achieve it. 3 columns, each with the same 7 names. Assuming that you have a list of data in two different columns and you want to find all unique values in two given columns. COUNTIFS extends the COUNTIF function which only allows one criteria. Returns the smallest number in a set of values. Instead of using the AutoSum feature you can also use the SUM function directly to calculate the total sales for a month. See this example: Unique distinct values from a cell range [...], […] As the name also implies, the data in G2:J14 is expected to be text with length > 0. To create a list of unique values, you simply reference the range that contains duplicates in the array argument for UNIQUE. I realize this is an old thread but its the closest I have been able to get to a solution for my problem. I have tried the formula […], Create a unique distinct alphabetically sorted list, The array formula in cell D3 extracts unique distinct values sorted A to Z, from column B to column D. […], Extract a unique distinct list and sum amounts based on a condition, Anura asks: Is it possible to extend this by matching items that meet a criteria? 5 easy ways to extract Unique Distinct Values, First, let me explain the difference between unique values and unique distinct values, it is important you know the difference […], Extract a unique distinct list from two columns, Question: I have two ranges or lists (List1 and List2) from where I would like to extract a unique distinct […], Vlookup – Return multiple unique distinct values, Ahmed Ali asks: How to return multiple values using vlookup in excel and removing duplicates? In this final example, we would like to apply the Conditional Formatting rule if more than 2 of the columns have values of 100 or more. When I enter more entries in the array, the list updates with new entries in the order of first looking through the row, then going down the column. I am really amazed to see your examples in the thread \"Extract a unique distinct list from three columns with possible blanks\". Note:To apply this Extract cells with unique values (include the first duplicate), firstly, you should download the Kutools for Excel, and then apply the feature quickly and easily. In many cases when using Excel with large duplicate data tables, to filter the unique list with multiple columns we use the Advanced Filter function with the Unique select attribute in Excel. I have more than 20 columns with true and false values. I got through and had the 343 different combinations, but since the data in my 3 columns is the same, some of them are duplicates. So type say \"Unique items from the table\" in A1 and enter the following formula as an array into A2 and copy it down as far as necessary. Sheet1!\\$B\\$2:\\$B\\$31,Sheet2!\\$B\\$2:\\$B:23...). Sometimes, you need to extract the unique values from a single column, the above methods will not help you, here, I can recommend a useful tool-Kutools for Excel, with its Extract cells with unique values (include the first duplicate) utility, you can quickly extract the unique values. Step 1: Create a Helper Column, and type the formula =IF(OR(A2=B2,B2=C2,C2=D2), \"Duplicate\",\"No Duplicate\") in the first cell; Supposing, you have the following data range that you want to list only the unique names of column B based on a specific criterion of column A to get the result as below screenshot shown. < becomes < and > becomes > How to add VBA code to your comment Array Formula. Returns a specific value if TRUE and another specific value if FALSE. In Excel, there are several ways to filter for unique values—or remove duplicate values: To filter for unique values, click Data > Sort & Filter > Advanced. How to calculate the numbers of Apple , Banana , Lemon , Orange. With the help of Countif function first, we will get the cell with unique values and then with the help of Sum and If function we will get the sum of the count of unique cells. AF, AN, AV, BD, BL, BT, CB, CJ). Ask Question Asked 5 years, 4 months ago. So yeah guys, this how you can sum multiple columns with condition without using sumif function. Great Formula! What's your opinion? If we use the formula =TEXTJOIN(“, “, TRUE, B4:C12) then Excel will give the following result: Example 3. Any suggestions? Like, it works perfectly and then at a certain point it just returns me errors for every row.. which is the worst! Then click Next button, check Create a single page field for me option in wizard step2, see screenshot: 4. The MMULT function is … I've seen how to do this with the \"ctrl+shift+enter\" arrays, but I was hoping to achieve it with a more efficient formula. (it is supposed column A to be free), =INDEX(tbl,MIN(IF(COUNTIF(\\$A\\$1:A1,tbl)=0,ROW(tbl)-MIN(ROW(tbl)) This function can count how many cells contain a value of 100 or more. https://www.get-digital-help.com/wp-content/uploads/2014/10/Unique-distinct-values-from-four-ranges-with-blanks.xlsx, Named ranges This is again an array formula to count distinct values in a range. The COUNTIF function can be used to compare two lists and return the number of items within both lists.Let’s look at an example. Make sure you subscribe to my newsletter so you don't miss new blog articles. Count unique values based on two given dates. How could you deal with this task in Excel quickly and easily? Click one cell in your data, and press Alt+D keys, then press P key immediately to open the PivotTable and PivotChart Wizard, choose Multiple consolidation ranges in the wizard step1, see screenshot: 3. After that, you simply enclose the formula in the ROWS function to calculate the number of rows. Can you please submit the correct formula... the VBA function works just fine. These new formulas are being rolled out to Office 365 subscribers over the next few months. From B1, choose Data, Filter, Advanced. Hi, Then press Shift + Ctrl + Enter keys together, and then drag the fill handle to extract the unique values until blank cells appear. Unique values in any column are those value which has only one occurrence and that can be counted with the help of countif function is used along with SumIf function or Sum and If function. I believe they are blank as I ran the GoTo special function to move all populated cells to the left Hi, See the snip below. Copy the headings from B1 and D1 to a blank section of the worksheet. Is there a way to add a single criteria element to the formula? We will use the COUNTIF function for this. Click OK to close this dialog, and all the unique values have been extracted at once. VLOOKUP is one of the lookup and reference functions in Excel and Google Sheets used to find values in a specified range by “row”.It compares them row-wise until it finds a match. In general one removes duplicates in the data and then count the number of distinct values. Increases your productivity by The trick is to \"feed\" the entire range to UNIQUE so that it finds the unique combinations of values in multiple columns. Thank you for writing this, it works like a charm! Scenario Suppose you have a list of customer names. 3. Array formulas allows you to do advanced calculations not possible with regular formulas. In the Formulas Helper dialog box, please do the following operations: 4. i've adjusted to my sheet but am only returning the first value in the defined array... what am i missing? This article, I will take some examples for you to count unique values based on one or more criteria in a worksheet. Let’s take an example where we are given the name, surname, and date of anniversary in separate columns and we wish to join them. Multiple versions written by the same person do not count as unique, but articles with the same title written by different people do count as unique. Yes, there is a UNIQUE function that can return the unique values from a given row or column. UNIQUE is not restricted to a single column. Thank you!! And then we can test if the answer to that is more than two. I would prefer the list first list the unique values in the column going downward, then then next column downward etc. In this Excel tutorial, I will show you different methods to compare two columns in Excel and look for matches or differences. Trying to get a list of the top 10 values of R for each of the names in Column A. After that, you simply enclose the formula in the ROWS function to calculate the number of rows. Register To Reply. [/vb]. When the formula is entered, the results will automatically spill down into the cells below. Specify the headings from Step 1 as the Output range. For this type =SUM(B2:B9).. Now after pressing Enter, drag this formulated cell in cell C10 and D10 to calculate the total sales for the month of February and March.. Excel Vlookup to Return Multiple Values. Create an INDEX Function in Excel. I then added a name in the name manager that specifies only the data in the formula column. I have 4 and I'd rather understand the \"general\" approach than keep creating ever more convoluted formulas as columns increase. In Excel, we call this the lookup value. is this the case with all non-contiguous ranges?). To achieve this, instead of referencing the entire range, use the ampersand (&) to concatenate the columns and put the desired delimiter in between. So, I almost have it set. Here are two methods of doing this: METHOD 1: Using COUNTIFS function. Syntax: =UNIQUE(array, [by_col], [occurs_once]) array is the range or array you want the unique … You can also ask queries regarding Excel 2019, 2016, 2013 and older. Open a workbook in Excel and copy below data into the sheet. In this tutorial, we will look at how to use VLOOKUP on multiple columns with multiple criteria. For example, to count the unique rows in the range A2:C10, we use this formula: =ROWS(UNIQUE(A2:C10)) How to solve this task quickly and easily In Excel? TRANSPOSE(COLUMN(names) ^ 0)) The COLUMN function is used to create a numeric array with 3 columns and 1 row, and TRANSPOSE converts this array to 1 column and 3 rows. List2:=Sheet1!\\$C\\$1:\\$C\\$7 Related Article: How to Use SUMIF Function in Excel SUMIFS with dates in Excel ... My preference would be an Excel formula (or even a built in function) as opposed to VBA. Thank you! Upload picture to postimage.org or imgur Since we want all the columns of the original table except the Salary column, we will redefine our Array argument to go from Department (column A) to Position (column E). This array is 4 rows by 3 columns, matching the structure of “Names”. UNIQUE From Non-Adjacent Columns. Before the UNIQUE function was released, Excel users were left using more complex methods to compile a list of unique values from a range. The ARRAY formula in F2 is this. I am pulling unique distinct a-z sorted values from a column with a single criteria. If I reference 3 columns only, there's no problem. How? For this type =SUM(B2:B9).. Now after pressing Enter, drag this formulated cell in cell C10 and D10 to calculate the total sales for the month of February and March.. Re: Create unique list from multiple columns using UNIQUE. Unique-distinct-values-from-multiple-columns-using-array-formulas-without-blanks.xls. How to add a picture to your comment: If you’re not fortunate enough to have Office 365 and these handy dynamic array formulas, you can use one of the techniques described here. Instead of SUM function, you can also … , do you have a list of customer names it 's one of the other part returns row numbers the! Are duplicates because the sequencing is different from other articles and comments, but using... This Excel tutorial excel unique function multiple columns i will show you different methods to compare two columns in worksheet. The DISTINCTCOUNT function Basic Excel formulas so it is fast article then let know. Examples and a downloadable Excel template the info on your website ( not rectangular defined array... what am missing! To Eero, who contributed the original array formula the number of cells that meet a cell! Two columns in Excel counts the number of cells in a specific cell range based a... 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Conditional Formatting command in the name manager that specifies only the data set spread multiple... Values based on the Home tab ( or even a built in function ) as to. Have worked for my issue helper column on my `` summary '' is! Newsletter so you do n't miss new blog articles they appear automatically you... Just returns me errors for every row.. which is the absolute references for MultiPC shown below the formula the! Returns row numbers numbers in the formulas i have cell values spanning over several columns and i can ``. Tbl, the arrays for our unique formulas are range D2: respectively! Calculations not possible with regular formulas, which can create unique lists sort! Array formulas to return multiple values, use the SUM function directly to the! Soon as new items are added to the new dynamic array formulas allows you to count distinct values in columns!\n\nTOP", null, "" ]	[ null, "http://www.stpaulschurchmumbai.com/wp-content/plugins/wp-news-bulletin/lib/img/close.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82092685,"math_prob":0.8950364,"size":32008,"snap":"2021-31-2021-39","text_gpt3_token_len":7340,"char_repetition_ratio":0.16857268,"word_repetition_ratio":0.0910624,"special_character_ratio":0.23662834,"punctuation_ratio":0.14783654,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97793806,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T06:09:22Z\",\"WARC-Record-ID\":\"<urn:uuid:0e02a40b-af80-468e-ac36-caf5dd0fb20c>\",\"Content-Length\":\"119164\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2892adbd-ee93-4aa8-b432-e608971b6263>\",\"WARC-Concurrent-To\":\"<urn:uuid:81d1e9f9-7cf7-4097-89e3-aa5112765995>\",\"WARC-IP-Address\":\"206.183.111.202\",\"WARC-Target-URI\":\"http://stpaulschurchmumbai.com/milking-shorthorn-ggtojis/kyg8z.php?c254b0=excel-unique-function-multiple-columns\",\"WARC-Payload-Digest\":\"sha1:Q5UE2D2BIWT73WSNN5JQD3BUHPH3DMST\",\"WARC-Block-Digest\":\"sha1:EBUU6B6YTQUW7L2YMZRARWOLCYUHRIGX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154796.71_warc_CC-MAIN-20210804045226-20210804075226-00661.warc.gz\"}"}
https://it.mathworks.com/matlabcentral/answers/480273-how-use-matlab-to-data-generation	[ "# how use MATLAB to data generation\n\n7 views (last 30 days)\ndodo shu on 13 Sep 2019\nAnswered: Cam Salzberger on 13 Sep 2019\nIs it possible to use MATLAB to generate a set of data with the corresponding relation of y=sin(x), and then import it into excel?\n\nCam Salzberger on 13 Sep 2019\nHello Dodo,\nYes, if you are looking to evaluate the sine function at particular points, then export that data to an Excel spreadsheet, you can do all that in MATLAB. For a very basic example:\nx = linspace(0, 2*pi, 100);\ny = sin(x);\nwritematrix(y, \"MyExcelFile.xlsx\")\nIf you have an older verison of MATLAB, you may want to use xlswrite instead of writematrix. You can modify the input datapoints to the sine function by changing how you define x, and even do degrees as units if you use \"sind\" instead of \"sin\". You can also modify where and how the data is written to the Excel file by adding options to writematrix.\n-Cam" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87764525,"math_prob":0.85578144,"size":653,"snap":"2019-51-2020-05","text_gpt3_token_len":166,"char_repetition_ratio":0.104776576,"word_repetition_ratio":0.0,"special_character_ratio":0.24042879,"punctuation_ratio":0.1294964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956737,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T18:27:41Z\",\"WARC-Record-ID\":\"<urn:uuid:39745be4-7eca-4f19-b1e5-05927f96d5ba>\",\"Content-Length\":\"85534\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8288222-35e6-4056-8c9f-201c45ba790c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0a382640-fa81-40b1-b56c-77073a22b131>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://it.mathworks.com/matlabcentral/answers/480273-how-use-matlab-to-data-generation\",\"WARC-Payload-Digest\":\"sha1:T76MKDHKXZSBPSAZD57ANOJXYH42KHZP\",\"WARC-Block-Digest\":\"sha1:ZQ2Z34KWGRYZY7AMK7TAWSRVXPEUANVB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251779833.86_warc_CC-MAIN-20200128153713-20200128183713-00161.warc.gz\"}"}
https://zr9558.com/2013/10/15/	[ "# Plane Hyperbolic Geometry\n\nAssume", null, "$\\mathbb{D}=\\{ z: \|z\|<1\\}$ is the unit disc on the complex plane", null, "$\\mathbb{C},$", null, "$\\mathbb{H}=\\{z: \\Im{z}>0\\}$ is the upper half plane on the complex plane,", null, "$\\mathbb{B}=\\{ z: \|\\Im{z}\|<\\pi/2\\}$ is the band between", null, "$y=-\\pi/2$ and", null, "$y=\\pi/2$.\n\nDefinition 1. Hyperbolic metric on the unit disc.\n\nThe hyperbolic metric on the unit disc", null, "$\\mathbb{D}$ is defined as", null, "$\\rho_{\\mathbb{D}}(z)=\\frac{2}{1-\|z\|^{2}} \|dz\|$ for all", null, "$z \\in \\mathbb{D} .$\n\nIf", null, "$\\phi : U \\rightarrow \\mathbb{D}$ is a conformal mapping, where", null, "$U \\subseteq \\mathbb{C}$, then we can also define the hyperbolic metric on the domain U,", null, "$\\rho_{U}(z)=\\frac{2 \|\\phi^{'}(z)\|}{1-\|\\phi(z)\|^{2}} \|dz\|$ for all", null, "$z\\in U.$\n\nFrom above and", null, "$\\phi(z)=(z-i)/(z+i)$ is a conformal mapping which maps the upper half plane", null, "$\\mathbb{H}$ onto the unit disc", null, "$\\mathbb{D}.$ From above formula, we can calculate the hyperbolic metric on", null, "$\\mathbb{H}$ is", null, "$\\rho_{\\mathbb{H}}(z)=\\frac{1}{\\Im{z}} \|dz\|$ for all", null, "$z\\in \\mathbb{H}.$\n\nThe hyperbolic metric on the band", null, "$\\mathbb{B}$ is", null, "$\\rho_{\\mathbb{B}}(z)=\\frac{1}{\\cos \\Im{z} } \|dz\|$ for all", null, "$z\\in \\mathbb{B}.$\n\nSimilarly, we can define the one dimensional hyperbolic metric. On the real line", null, "$\\mathbb{R}$, if the interval", null, "$I=(-1,1)$, then the restriction of the hyperbolic metric on the unit disc", null, "$\\mathbb{D}$ is", null, "$\\rho_{I}(x)= \\frac{2}{1-x^{2}} dx$ for all", null, "$x \\in (-1,1).$\n\nThis is called the hyperbolic metric of the interval I.\n\nUsing the same idea, we can extend the definition of hyperbolic metric on any real interval", null, "$I=(a,b)$. Since there exists a linear map", null, "$\\phi$ which maps a to -1 and b to 1, i.e.", null, "$\\phi(x)=(2x-b-a)/(b-a)$. Its derivative is", null, "$\\phi^{'}(x)= 2/(b-a)$. Therefore, the hyperbolic metric on the interval I is", null, "$\\rho_{(a,b)}(x)=\\frac{2\|\\phi^{'}(x)\|}{1-\|\\phi(x)\|^{2}} dx= \\frac{b-a}{(x-a)(b-x)} dx= (\\frac{1}{x-a}+ \\frac{1}{b-x}) dx$ for all", null, "$x\\in (a,b).$\n\nMoreover, assume", null, "$(c,d) \\subseteq (a,b)$, then the hyperbolic distance between c and d is", null, "$\\int_{c}^{d} \\rho_{(a,b)}(x) dx = \\int_{c}^{d} (\\frac{1}{x-a} + \\frac{1}{b-x}) dx = (\\ln\\frac{x-a}{b-x}) \|_{x=c}^{x=d} = \\ln \\frac{(d-a)(b-c)}{(b-d)(c-a)}.$\n\nIf we use the notation of cross ratio, then assume", null, "$l=(a,c), j=(c,d), r=(d,b),$", null, "$t=(a,b)$. Therefore, the hyperbolic distance between c and d in the interval (a,b) equals to", null, "$\\ln \\frac{(\|l\|+\|j\|)\\cdot (\|j\|+\|r\|)}{\|l\| \\cdot \|r\|} = \\ln (1+ \\frac{\|t\|\\cdot \|j\|}{\|l\| \\cdot \|r\|}) = \\ln (1+ Cr(t,j)),$\n\nwhere", null, "$Cr(t,j)= (\|t\|\\cdot \|j\|) / (\|l\| \\cdot \|r\|).$\n\nDefinition 2. (Curvature of conformal metric)\n\nLet", null, "$\\rho$ be a", null, "$C^{2}$ positive function on an open subset", null, "$U \\subseteq \\mathbb{C}$. Then the curvature of the metric", null, "$\\rho(z)\|dz\|$ is given by", null, "$K(z)=-\\frac{(\\Delta \\ln \\rho)(z)}{\\rho^{2}(z)},$\n\nwhere", null, "$\\Delta$ is the Laplacian operator", null, "$\\Delta= \\frac{\\partial^{2}}{\\partial x^{2}} + \\frac{\\partial^{2}}{\\partial y^{2}}.$\n\nRemark. Use the identities", null, "$\\frac{\\partial}{\\partial \\overline{z}} =\\frac{1}{2} (\\frac{\\partial}{\\partial x} + i \\frac{\\partial}{\\partial y}),$", null, "$\\frac{\\partial}{\\partial z} =\\frac{1}{2} (\\frac{\\partial}{\\partial x} - i \\frac{\\partial}{\\partial y}),$\n\nwe get", null, "$\\Delta=\\frac{\\partial^{2}}{\\partial x^{2}} + \\frac{\\partial^{2}}{\\partial y^{2}} = 4 \\frac{\\partial^{2}}{\\partial z \\partial \\overline{z}}.$\n\nTheorem 1.\n\nThe curvature of hyperbolic metric of the unit disc", null, "$\\mathbb{D}$, the upper half plane", null, "$\\mathbb{H}$ and the band", null, "$\\mathbb{B}$ is -1.\n\nTheorem 2.\n\nIf", null, "$\\phi: U\\rightarrow \\mathbb{D}$ is a conformal mapping, where", null, "$U \\subseteq \\mathbb{C}$ is an open subset of the complex plane", null, "$\\mathbb{C}$. From above, the hyperbolic metric on U is", null, "$\\rho_{U}(z)=\\frac{2 \|\\phi^{'}(z)\|}{1-\|\\phi(z)\|^{2}} \|dz\|$ for all", null, "$z \\in U\\subseteq \\mathbb{C}.$\n\nThen the curvature of the metric", null, "$\\rho_{U}(z)$ is", null, "$-1$.\n\nTheorem 3.\n\nOn the complex sphere", null, "$\\hat{\\mathbb{C}}$, the sphere metric on", null, "$\\hat{\\mathbb{C}}$ is defined as", null, "$\\rho(z)=\\frac{1}{1+\|z\|^{2}} \|dz\|$ for all", null, "$z\\in \\hat{\\mathbb{C}}.$\n\nThen the curvature of the sphere metric is 1." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79324144,"math_prob":1.000009,"size":1833,"snap":"2020-45-2020-50","text_gpt3_token_len":453,"char_repetition_ratio":0.20940405,"word_repetition_ratio":0.098265894,"special_character_ratio":0.22094926,"punctuation_ratio":0.10817942,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T19:36:34Z\",\"WARC-Record-ID\":\"<urn:uuid:b366c9c9-0ae7-4d96-96aa-c703c0ba39f5>\",\"Content-Length\":\"120880\",\"Content-Type\":\"application/http; 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https://codeforces.com/topic/96954/?mobile=true&locale=en	[ "", null, "", null, "Prefix sums k times\n\nRevision en1, by Skeef79, 2021-10-26 17:45:42\n\nHello codeforces!\n\nI am wondering is there a way to caculate prefix sums $k$ times modulo $p$ fast, especially when we can't use matrix multiplication?\n\nBy calculating prefix sums $k$ times I mean this piece of code:\n\nfor(int it = 0; it < k; it++) {\nfor(int i = 1; i < n; i++)\npref[i]+= pref[i-1];\n}\n\n\nSo if we start with array: $1, 0, 0, 0$ we will get:\n\n1. $1, 1, 1, 1$;\n\n2. $1, 2, 3, 4$;\n\n3. $1, 3, 6, 10$;\n\n4. $1, 4, 10, 20$; ...\n\n#### History\n\nRevisions", null, "Rev. Lang. By When Δ Comment\nen1", null, "Skeef79 2021-10-26 17:45:42 500 Initial revision for English translation\nru1", null, "Skeef79 2021-10-26 17:41:50 500 Первая редакция (опубликовано)" ]	[ null, "https://cdn.codeforces.com/s/61771/images/icons/icon-bars.png", null, "https://cdn.codeforces.com/s/61771/images/icons/icons-ellipsis-v.png", null, "https://cdn.codeforces.com/s/61771/images/icons/control.png", null, "https://cdn.codeforces.com/s/24838/images/flags-16/en.png", null, "https://cdn.codeforces.com/s/24838/images/flags-16/ru.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6268113,"math_prob":0.9988234,"size":444,"snap":"2022-27-2022-33","text_gpt3_token_len":162,"char_repetition_ratio":0.11590909,"word_repetition_ratio":0.0,"special_character_ratio":0.48423424,"punctuation_ratio":0.27966103,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9706845,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T15:18:42Z\",\"WARC-Record-ID\":\"<urn:uuid:aecc5436-a811-4e59-a8b1-c7f7971c99dc>\",\"Content-Length\":\"76615\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:601dc12e-6619-4f7c-bc16-731832be7f04>\",\"WARC-Concurrent-To\":\"<urn:uuid:97ac96f2-d756-475b-a323-e33714face87>\",\"WARC-IP-Address\":\"213.248.110.126\",\"WARC-Target-URI\":\"https://codeforces.com/topic/96954/?mobile=true&locale=en\",\"WARC-Payload-Digest\":\"sha1:FBPOJHSAAZXCLUOMZSUWKCPQYSG43BWC\",\"WARC-Block-Digest\":\"sha1:RXEJPSKTH75OJYIHPCRF75SFXYDLGBW5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103334753.21_warc_CC-MAIN-20220627134424-20220627164424-00539.warc.gz\"}"}
https://www.scribd.com/document/239989903/General-Solution-of-Equations-of-Motion-of-Axisymmetric-Problem-of-Micro-Isotropic-Micro-Elastic-Solid	[ "You are on page 1of 11\n\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.\n\ncom\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 80 \| P a g e\n\nGeneral Solution of Equations of Motion of Axisymmetric\nProblem of Micro-Isotropic, Micro-Elastic Solid\n\nR. Srinivas\n1,*\n, M. N. Rajashekar\n2\nand K. Sambaiah\n3\n\n1\nDepartment of Mathematics, BITS, Narsampet, Warangal-506 331, Telangana, India.\n2\nDepartment of Mathematics, JNTUHCEJ, Karimnagar-505 501, Telangana, India.\n3\nDepartment of Mathematics, Kakatiya University, Warangal-506 009, Telangana, India.\n\nABSTRACT\nIn this paper, we obtain the general solution of equations of motion of axisymmetric problem of micro-isotropic,\nmicro-elastic solid in static case. The equations of motion of axisymmetric problem are converted into vector\nmatrix differential equations using the Hankel transform. Applying the technique of solving the eigen value\nproblem, the general solution of the said problem is obtained. The results of the corresponding problem in linear\nmicropolar elasticity are obtained as a particular case of this paper.\nKEYWORDS: Micro-isotropic, micro-elastic solid, Eigen value, Hankel transform.\n\nI. INTRODUCTION\nThe classical theory of elasticity describes well the behavior of construction materials provided the stresses\ndo not exceed the elastic limit and no stress concentration occurs. The discrepancy between the results of the\nclassical theory of elasticity and the experiments appears in all the cases when the microstructure of the body\nis significant. The materials having microstructure are metals, polymers, rocks and concrete. The influence of\nmicrostructure is particularly evident in the case of elastic vibrations of high frequency and/or small wave\nlength. To remove the short comings of the classical theory of elasticity, Eringen introduced the theory of\nmicromorphic materials which includes the micromotion. This theory was simplified by Koh extending the\nconcept of coincidence of principal directions of stresses and strain of classical elasticity and named it as the\ntheory of micro-isotropic, micro-elastic materials. Nowacki has shown that the equations of motion of\naxisymmetric problem of micropolar solid can be decomposed into two mutually independent sets of three\nequations. Das et al. [5, 6] have obtained general solution of equations of motion in thermoelasticity and\nmagnetothermo-elasticity using eigen value approach to solve vector matrix differential equation.\nIn the present paper, we apply the technique of solving an eigen value problem to obtain the general\nsolution of the axisymmetric problem of micro-isotropic, micro-elastic solid. The results of the corresponding\nproblem in micropolar theory are obtained as a particular case of it.\n\nII. BASIC EQUATIONS\nThe equations of motion and the constitute equations of micro-isotropic, micro-elastic solid under the\nabsence of body forces and body couples are given by Parameshwaran and Koh \nThe displacement equations of motion are\n( ) ( ) u A u A A u A A A \| = V + V + + V V +\n3\n2\n3 2 3 2 1\n2 ) ( . ) ( (1)\n( ) ( ) \| \| \| \|\n\nj A u A B B B = V V + V V +\n3 3\n2\n3 5 4\n4 2 2 . ) ( 2 (2)\n) ( ) ( 5 4 ), ( 2 , 1\n2\n1\n2 2\nij ij ij pp kk ij ij kk pp\nj A A B B \| \| o \| \| o \|\n\n= + (3)\nThe stress, couple-stress and stress moment are as follows.\nkm km pp km\ne A e A t\n2 1 ) (\n2 + = o\n(4)\n) ( 2\n3 ] [ ] [ p p pkm km km\nr A t \| c o + = =\n(5)\n) ( 5 4 ) (\n2\nkm km pp km\nA A \| o \| o =\n(6)\nk mn mn k pp mn k\nB B t\n), ( 2 , 1 ) (\n2 \| o \| + = (7)\n) ( 2\n, 5 , 4 , 3 kl p p l k k l kl\nB B B m o \| \| \| + + = (8)\nRESEARCH ARTICLE OPEN ACCESS\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 81 \| P a g e\nwhere\n1 1\no + = A ,\n3 1\nt = B ,\n2 2\no + = A\n, , 2\n10 7 2\nt t + = B\n,\n5 3\no = A\n\n, 2 2\n10 7 9 4 3\nt t t t + + = B\n\n, 2\n4 4\nt = B\n\n,\n2 5\no = A\n\n9 5\n2t = B\n. (9)\nsubject to the conditions\n, 0 2 3\n2 1\n> + A A\n\n, 0\n2\n> A\n\n, 0\n3\n> A\n, 0 2 3\n5 4\n> + A A\n\n, 0\n5\n> A\n\n, 0 2 3\n2 1\n> + B B\n, 0\n2\n> B\n\n, 0\n3\n> B ,\n3 4 3\nB B B < < . 0\n5 4 3\n> + + B B B (10)\nwhere\n\nis the average mass density, j is the micro-inertia. The macro displacement in the micro elastic\ncontinuum is denoted by\nk\nu and the micro deformation by ,\n) ( mn\n\| for the linear theory we have the macro-strain\n) , ( m k km\ne e = , the macro rotation vector\nm n kmn k\nu r\n,\n2\n1\nc = , the micro-strain\n) (mn\n\| and micro-rotation\nkm pkm p\n\| c \|\n2\n1\n= . The stress measures are the asymmetric stress (macro-stress)\nmn\nt , the relative stress (micro-\nstress)\nkm\no and the stress moment\nkmn\nt and also the couple stress tensor\nkmn pnm kp\nt m c = . The symbol ( )\nappeared in suffix of a quantity indicate that the quantity is symmetric and [ ] shows the quantity is skew-\nsymmetric. ,, o\n1\n, o\n2,\no\n5,\nt\n3,\nt\n4,\nt\n7,\nt\n9\nand t\n10\nare the ten elastic moduli. Further,\npkm\nc is the permutation\nsymbol and\nkm\no is the Kronecker delta. The ( . ) denotes the derivatives with respect to time.\n\nIII. FORMULATION AND SOLUTION OF THE PROBLEM\nThe problem is to find the general solutions of axisymmetric equations of static micro-isotropic, micro-\nelastic material under the absence of body forces and body couples, we take 0 = =\|\n\nu and the cylindrical\ncoordinates u , r and z are introduced.\nThe equations of motion (1) and (2) for the static case are as follows.\n( ) ( ) 0\n1\n2\n2\n3 3 2 1\n2 2\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+\nc\nc\n+ +\n(\n\nc\nc\nV +\nz r\nA\nr\ne\nA A A\nu\nr r\nu\nu A A\nz r\nr\nu u\n\|\nu\n\|\nu\n(11)\n( ) ( ) 0 2\n1 2\n3 3 2 1\n2 2\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+\nc\nc\n+ +\n(\n\nc\nc\n+ V +\nr z\nA\ne\nr\nA A A\nu\nr r\nu\nu A A\nz r r\n\| \|\nu u\nu\nu\n(12)\n( )\| \| ( ) ( ) 0\n1\n2\n3 3 2 1\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+\nc\nc\n+ + V +\nu\n\|\n\|\nu\nr\nz\nr\nr r\nA\nz\ne\nA A A u A A (13)\n( ) 0\n1\n2 2 2 4\n2\n2\n3 5 4 3\n2 2\n2\n3\n=\n(\n\nc\nc\n\nc\nc\n\nc\n' c\n+ +\n(\n\nc\nc\nV\nz\nu u\nr\nA\nr\ne\nB B A\nr r\nB\nz\nr\nr\nr\nu u\nu\n\|\nu\n\| \|\n\| (14)\n( ) 0 2\n1\n2 2 4\n2\n2\n3 5 4 3\n2 2\n2\n3\n=\n(\n\nc\nc\n\nc\nc\n\nc\n' c\n+ +\n(\n\nc\nc\n+ V\nr\nu\nz\nu\nA\ne\nr\nB B A\nr r\nB\nz r r\nu\n\|\nu\n\| \|\n\|\nu\nu\nu\n(15)\n\| \| ( ) ( ) 0\n1\n2 2 2 4 2\n3 5 4 3\n2\n3\n=\n(\n\nc\nc\n\nc\nc\n\nc\n' c\n+ + V\nu\n\| \|\nu\nr\nz z\nu\nru\nr r\nA\nz\ne\nB B A B (16)\n,\n1 4\no = A\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 82 \| P a g e\nwhere\n( )\nz\nu u\nr\nru\nr r\ne\nz\nr\nc\nc\n+\nc\nc\n+\nc\nc\n=\nu\nu\n1 1\n\n( )\nz r\nr\nr r\ne\nz\nr\nc\nc\n+\nc\nc\n+\nc\nc\n= '\n\|\nu\n\|\n\|\nu\n1 1\n\n2\n2\n2\n2\n2 2\n2\n2\n1 1\nz r r r r c\nc\n+\nc\nc\n+\nc\nc\n+\nc\nc\n= V\nu\n\nIn case of the vectors of macro-displacement u\n\nand micro-rotation \| depend only on the coordinates r and z ,\nthe equations (11) to (16) reduce to\n( ) ( ) 0 2\n3 3 2 1\n2\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+ +\n(\n\nV +\nz\nA\nr\ne\nA A A\nr\nu\nu A A\nr\nr\nu\n\|\n(17)\n( ) 0 2\n3\n2\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+ +\n(\n\nV +\nr z\nA\nr\nu\nu A A\nz r\n\| \|\nu\nu\n(18)\n( )\| \| ( ) ( ) 0\n1\n2\n3 3 2 1\n2\n3 2\n=\n(\n\nc\nc\n+\nc\nc\n+ + V +\nu\n\| r\nr r\nA\nz\ne\nA A A u A A\nz\n(19)\n( ) 0 2 2 2 4 2\n3 5 4 3\n2\n2\n3\n=\n(\n\nc\nc\n+\nc\n' c\n+ +\n(\n\nV\nz\nu\nA\nr\ne\nB B A\nr\nB\nr\nr\nr\nu\n\|\n\|\n\| (20)\n0 2 4 2\n3 3\n2\n2\n3\n=\n(\n\nc\nc\n\nc\nc\n\n(\n\nV\nr\nu\nz\nu\nA A\nr\nB\nz r\nu\nu\nu\n\|\n\|\n\| (21)\n\| \| ( ) ( ) 0\n1\n2 2 2 4 2\n3 5 4 3\n2\n3\n=\n(\n\nc\nc\n\nc\n' c\n+ + V\nu\n\| \| ru\nr r\nA\nz\ne\nB B A B\nz z\n(22)\nwhere\n( )\nz\nu\nru\nr r\ne\nz\nr\nc\nc\n+\nc\nc\n=\n1\n\n( )\nz\nr\nr r\ne\nz\nr\nc\nc\n+\nc\nc\n= '\n\|\n\|\n1\n\n2\n2\n2\n2\n2\n1\nz r r r c\nc\n+\nc\nc\n+\nc\nc\n= V\nEquations given by (17) to (22) can be split into two sets of equations. One of these is coupled in\nu\n\| , ,\nz r\nu u\n\nand other set is coupled in . , ,\nu u\n\| \| u\nr\n\nThese two sets are given by the equations (23) to (25) and\n(26) to (28).\n( ) ( ) 0 2\n3 3 2 1\n2\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+ +\n(\n\nV +\nz\nA\nr\ne\nA A A\nr\nu\nu A A\nr\nr\nu\n\|\n(23)\n( )\| \| ( ) ( ) 0\n1\n2\n3 3 2 1\n2\n3 2\n=\n(\n\nc\nc\n+\nc\nc\n+ + V +\nu\n\| r\nr r\nA\nz\ne\nA A A u A A\nz\n(24)\n0 2 4 2\n3 3\n2\n2\n3\n=\n(\n\nc\nc\n\nc\nc\n\n(\n\nV\nr\nu\nz\nu\nA A\nr\nB\nz r\nu\nu\nu\n\|\n\|\n\| (25)\nand\n( ) 0 2 2 2 4 2\n3 5 4 3\n2\n2\n3\n=\n(\n\nc\nc\n+\nc\n' c\n+ +\n(\n\nV\nz\nu\nA\nr\ne\nB B A\nr\nB\nr\nr\nr\nu\n\|\n\|\n\| (26)\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 83 \| P a g e\n\| \| ( ) ( ) 0\n1\n2 2 2 4 2\n3 5 4 3\n2\n3\n=\n(\n\nc\nc\n\nc\n' c\n+ + V\nu\n\| \| ru\nr r\nA\nz\ne\nB B A B\nz z\n(27)\n( ) 0 2\n3\n2\n2\n3 2\n=\n(\n\nc\nc\n\nc\nc\n+ +\n(\n\nV +\nr z\nA\nr\nu\nu A A\nz r\n\| \|\nu\nu\n(28)\nThe equations (23), (24) and (25) are three mutually independent functions\nz r\nu u , and\nu\n\| involved. Multiplying\n(24) by ( ) r rJ\n0\nand (23), (25) by ( ) r rJ\n1\nand integrating between the limits 0 to we find that the system of\npartial differential equations (23) to (25) reduces to the following system of ordinary differential equations.\n( ) { } ( ) { } \| \| ( ) 0 2 2\n3 3 2 1\n2\n2 1\n2\n3 2\n= + + +\nu\n\| D A u D A A A u A A D A A\nz r\n(29)\n( ) ( ) { } ( ) { } \| \| 0 2 2\n3\n2\n3 2\n2\n2 1 3 2 1\n= + + + + +\nu\n\| A u A A D A A u D A A A\nz r\n(30)\n\n( ) \| \| 0 4 2 2 2\n3\n2 2\n3 3 3\n= + +\nu\n\| A D B u A u D A\nz r\n(31)\n\nwhere\nz r\nu u , and\nu\n\| are the Hankel transforms of the functions\nz r\nu u ,\n\nand\nu\n\|\n\nrespectively and are given\nby\n( )\n}\n\n=\n0\n1\n, dr r J ru u\nr r\n( )\n}\n\n=\n0\n0\n, dr r J ru u\nz z\n( )\n}\n\n=\n0\n1\ndr r J r \| \|\nu u\n.\nFurther,\ndz\nd\nD = and\n2\n2\n2\ndz\nd\nD = .\nWe represent the equations (29) to (31) as a matrix differential equation\n\n\| \| 0\n2\n= X R QD PD (32)\nwhere\n\n( )\n( )\n(\n(\n(\n\n+\n+\n=\n3\n2 1\n3 2\n2 0 0\n0 2 0\n0 0\nB\nA A\nA A\nP ,\n( )\n( )\n(\n(\n(\n\n+\n+\n=\n0 0 2\n0 0\n2 0\n3\n3 2 1\n3 3 2 1\nA\nA A A\nA A A A\nQ\n\n( )\n( )\n(\n(\n(\n\n+\n+\n+\n=\n3\n2\n3 3\n3\n2\n3 2\n2\n2 1\n4 2 2 0\n2 0\n0 0 2\nA B A\nA A A\nA A\nR\n\nand\n(\n(\n(\n\n=\nu\n\|\nz\nr\nu\nu\nX .\nWe suppose\n(\n(\n(\n\n= = '\nu\n\|\nz\nr\nu\nu\nD DX X ,\n(\n(\n(\n\n= = ' '\nu\n\|\nz\nr\nu\nu\nD X D X\n2 2\n(33)\nNow the equation (32) can be expressed as\n\n0 = ' ' ' RX X Q X P (34)\nMultiplying the equation (34) by\n1\nP we have\n\nX L X L X\n2 1\n+ ' = ' ' (35)\nwhere\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 84 \| P a g e\n( )\n( )\n( )\n(\n(\n(\n(\n(\n(\n(\n\n+\n+\n+\n\n+\n+\n= =\n\n0 0\n0 0\n2\n) (\n2\n) (\n0\n3\n3\n2 1\n3 2 1\n3 2\n3\n3 2\n3 2 1\n1\n1\nB\nA\nA A\nA A A\nA A\nA\nA A\nA A A\nQ P L\n\n( )\n( )\n( )\n( ) ( )\n( )\n(\n(\n(\n(\n(\n(\n(\n\n+\n+ +\n+\n+\n+\n= =\n\n3\n3\n2\n3\n3\n3\n2 1\n3\n2 1\n2\n3 2\n3 2\n2\n2 1\n1\n2\n2\n0\n2\n2\n2\n0\n0 0\n2\nB\nA B\nB\nA\nA A\nA\nA A\nA A\nA A\nA A\nR P L\n\nWe express equations (33) and (35) as a single matrix differential equation\n(\n\n'\n(\n\n=\n(\n\n'\nX\nX\nO I\nL L\nX\nX\ndz\nd\n2 1\n(36)\nwhere I and O\n\nare unit and zero matrices of order 3 respectively.\nSuppose\n(\n\n=\nO I\nL L\nE\n2 1\nand\n(\n\n'\n=\nX\nX\nY (37)\nIn view of equation (37) the equation (36) can be written as\nEY Y\ndz\nd\n= (38)\nwhere\n\n\| \|\n6 6x\nij\ne E =\nAssume\nsz\nWe Y = be a solution of equation (38) then we have sW EW = (39)\n( )\n( )\n3 2\n3 2 1\n12\nA A\nA A A\ne\n+\n+\n=\n\n,\n( )\n3 2\n3\n13\n2\nA A\nA\ne\n+\n\n= ,\n( )\n( )\n3 2\n2\n2 1\n14\n2\nA A\nA A\ne\n+\n+\n=\n\n( )\n( )\n2 1\n3 2 1\n21\n2A A\nA A A\ne\n+\n+\n=\n\n,\n( )\n( )\n2 1\n2\n3 2\n25\n2A A\nA A\ne\n+\n+\n=\n\n,\n( )\n2 1\n3\n26\n2\n2\nA A\nA\ne\n+\n=\n\n3\n3\n31\nB\nA\ne\n\n= , ,\n3\n3\n35\nB\nA\ne\n\n=\n\n3\n3\n2\n3\n36\n2\nB\nA B\ne\n+\n=\n\n.\n\nHence, s\n\nis an eigen value of the matrix E\n\nand W\n\nis the corresponding eigen vector.\n\nThe eigen values of the matrix E\n\nare the roots of the equation\n( ) 0 det = sI E (40)\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 85 \| P a g e\n0\n0 0 1 0 0\n0 0 0 1 0\n0 0 0 0 1\n0 0\n0 0\n0 0\n., .\n36 35 31\n26 25 21\n14 13 12\n=\n\ns\ns\ns\ne e s e\ne e s e\ne e e s\ne i (41)\nExpanding the determinant of equation (41) we get the characteristic equation as\n( ) ( ) 0 2 2\n2 4 2 2 2 4 4 2 2 6\n= + + + , , , s s s (42)\nwhere\n2 2 2\nl + = , and\n( )\n( )\n3 2 3\n3 2 3 2\n2 2\nA A B\nA A A\nl\n+\n+\n=\nThe eigen values of the matrix E\n\nare the roots of the equation (42) and they are , , ,\n.\n\nLet , = = =\n3 2 1\n, , s s s and , =\n4\ns\nThe four eigen vectors corresponding to four distinct eigen values\n4 3 2 1\n, , , s s s s of the matrix E are obtained by\nsolving the following homogeneous equations.\n0\n) (\n) (\n) (\n) (\n) (\n) (\n0 0 1 0 0\n0 0 0 1 0\n0 0 0 0 1\n0 0\n0 0\n0 0\n6\n5\n4\n3\n2\n1\n36 35 31\n26 25 21\n14 13 12\n=\n(\n(\n(\n(\n(\n(\n(\n(\n\n(\n(\n(\n(\n(\n(\n(\n(\n\ns W\ns W\ns W\ns W\ns W\ns W\ns\ns\ns\ne e s e\ne e s e\ne e e s\n(43)\nfor\n. 4 3 2 1\n, , , s s s s s =\nIf we denote the cofactors of the elements of the first row of the coefficient matrix of the equation (43) by\n) (s E\ni\n, for 6 , 5 , 4 , 3 , 2 , 1 = i\n\nthen\n\| \|\nT\ns E s E s E s E s E s E s W ) ( ) ( ) ( ) ( ) ( ) ( ) (\n6 5 4 3 2 1\n= (44)\nare the solutions of the equation (43) and hence they are eigen vectors corresponding to the eigen values\n4 3 2 1\n, , , s s s s of the matrix E . The elements of equation (44) are given by\n( )\n( )\n( )\n( )\n( )\n( )\n(\n\n+\n+\n+\n+\n+\n+\n(\n\n+\n+\n+ +\n+ =\n2\n2 1 3\n3 2 3 4\n2 1\n3 2\n3\n3 2\n2 1\n3 2 1 3 5\n1\n2\n2 2\n2\n2\n2\n3\n) (\nA A B\nA A A\nA A\nA A\ns\nB\nA\nA A\nA A A\ns s s E\n( )\n( )\n( )\n( )\n( )\n(\n\n+\n+\n\n+\n+\n=\n2 1 3\n3 2 1 3 2 2\n2 1\n3 2 1 2\n2\n2\n2 2\n2\n) (\nA A B\nA A A A\ns\nA A\nA A A\ns s E\n(\n\n=\n3\n2 2\n2\n3 3\n) (\nB\ns\ns A s E\n\n( )\n( )\n( )\n( )\n2 1\n3 2 2 2\n3\n3 2\n2 1\n3 2 1 2 4\n4\n2\n2\n2\n3\n) (\nA A\nA A\ns\nB\nA\nA A\nA A A\ns s s E\n+\n+\n\n+\n+\n+ +\n+ =\n( )\n( )\n( )\n( )\n( )\ns\nA A B\nA A A A\ns\nA A\nA A A\ns E\n(\n\n+\n+\n\n+\n+\n=\n2 1 3\n3 2 1 3 2 2\n2 1\n3 2 1\n5\n2\n3 2\n2\n) (\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 86 \| P a g e\n(\n\n=\n3\n2 2\n3 6\n) (\nB\ns\ns A s E\n\nThus, the solution of the differential equation (38) is given by Das et al. \n\| \| \| \|\n2 1\n) exp( ) ( ) exp( ) ( ) exp( ) ( ) exp( ) (\n4 2 2 3 2 1 1 1 s s s s\nsz s W\ndz\nd\nc z s z s W c sz s W\ndz\nd\nc z s s W c Y\n= =\n+ + + =\n) exp( ) ( ) exp( ) (\n4 4 6 3 3 5\nz s s W c z s s W c + + (45)\nwhere\n6 5 4 3 2 1\n, , , , , c c c c c c are arbitrary constants which are to be determined from the boundary conditions.\n\nThe equation (45) can be expressed as\n( ) ( ) ) exp( ) ( ) exp( ) ( ) exp( ) ( ) exp( ) (\n2 2 4 2 2 4 3 1 1 2 1 1 2 1\nz s s W c z s s W z c c z s s W c z s s W z c c Y ' + + + ' + + =\n) exp( ) ( ) exp( ) (\n4 4 6 3 3 5\nz s s W c z s s W c + + (46)\nwhere ( ) represents the differentiation with respect to z .\nFor the half space , 0 > z the equation (46) reduces to the form\n( ) ) exp( ) ( ) exp( ) ( ) exp( ) (\n4 4 6 2 2 4 2 2 4 3\nz s s W c z s s W c z s s W z c c Y + ' + + = (47)\nwhere the constants\n6 4 3\n, , c c c\n\nare to be determined from the boundary conditions.\nEquating the corresponding elements of matrices of equation (47), we obtain\n( ) ( ) \| \| ) exp( ) ( ) exp( ) ( ) (\n4 4 1 6 2 2 1 4 2 1 4 3\n'\nz s s E c z s s E c s E z c c z u\nr\n+ + + =\n( ) ( ) \| \| ) exp( ) ( ) exp( ) ( ) (\n4 4 21 6 2 2 2 4 2 2 4 3\n'\nz s s E c z s s E c s E z c c z u\nz\n+ + + =\n( ) ( ) \| \| ) exp( ) ( ) exp( ) ( ) (\n4 4 3 6 2 2 3 4 2 3 4 3\n'\nz s s E c z s s E c s E z c c z + + + =\nu\n\|\n( ) ( ) \| \| ) exp( ) ( ) exp( ) ( ) (\n4 4 4 6 2 2 4 4 2 4 4 3\nz s s E c z s s E c s E z c c z u\nr\n+ + + = (48)\n( ) ( ) \| \| ) exp( ) ( ) exp( ) ( ) (\n4 4 5 6 2 2 5 4 2 5 4 3\nz s s E c z s s E c s E z c c z u\nz\n+ + + = (49)\n( ) ( ) \| \| ) exp( ) ( ) exp( ) ( ) (\n4 4 6 6 2 2 6 4 2 6 4 3\nz s s E c z s s E c s E z c c z + + + =\nu\n\| (50)\nThe equations (26), (27) and (28) three mutually independent functions\nz r\nu u ,\n\nand\nu\n\|\n\nare involved.\nMultiplying (27) by ( ) r rJ\n0\n\nand (26), (28) by ( ) r rJ\n1\n\nand integrating between the limits 0\n\nto . We find\nthat the system of partial differential equations (26) to (28) reduces to the following system of ordinary\ndifferential equations.\n( ) \| \| ( ) 0 2 2 4 2 2\n3 5 4 3\n2\n5 4 3\n2\n3\n= + + +\nu\n\| \| u D A D B B A B B B D B\nz r\n(51)\n( ) ( ) \| \| 0 2 4 2 2 2\n3 3\n2\n3\n2\n5 4 3 5 4\n= + + + + +\nu\n\| \| u A A B D B B B D B B\nz r\n(52)\n( )( ) 0 2 2\n2 2\n3 2 3 3\n= + +\nu\n\| \| u D A A A D A\nz r\n(53)\nwhere\nz r\n\| \| , and\nu\nu are the Hankel transforms of the functions\nz r\n\| \| ,\n\nand\nu\nu\n\nrespectively and are given\nby\n( )\n}\n\n=\n0\n1\n, dr r J r\nr r\n\| \| ( )\n}\n\n=\n0\n0\n, dr r J r\nz z\n\| \| ( )\n}\n\n=\n0\n1\ndr r J ru u\nu u\n.\nWe repeat the method adopted for solving the first set of equations. The equations (51) to (53) can be expressed\nas vector matrix differential equation.\nFY Y\ndz\nd\n= (54)\nwhere\nT\nz r z r\nu u Y\n(\n\n' ' '\n\n=\nu u\n\| \| \| \| and \| \|\nij\nf F = (55)\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 87 \| P a g e\nThe elements of the matrix F\n\nare given by\n0\n16 15 11\n= = = f f f ,\n( )\n3\n5 4\n12\nB\nB B\nf\n+\n= ,\n3\n3\n13\nB\nA\nf = ,\n( )\n3\n3\n2\n5 4 3\n14\n2\nB\nA B B B\nf\n+ + +\n=\n\n.\n0\n24 23 22\n= = = f f f ,\n( )\n( )\n5 4 3\n5 4\n21\nB B B\nB B\nf\n+ +\n+\n=\n\n,\n( )\n5 4 3\n3\n2\n3\n25\n2\nB B B\nA A\nf\n+ +\n+\n=\n\n,\n( )\n5 4 3\n3\n26\nB B B\nA\nf\n+ +\n\n=\n\n.\n0\n34 33 32\n= = = f f f ,\n( )\n3 2\n3\n31\n2\nA A\nA\nf\n+\n= ,\n( )\n3 2\n3\n35\n2\nA A\nA\nf\n+\n=\n\n,\n2\n36\n= f .\n1\n41\n= f , . 0\n46 45 44 43 42\n= = = = = f f f f f\n1\n52\n= f , . 0\n56 55 54 53 51\n= = = = = f f f f f\n1\n63\n= f , . 0\n66 65 64 62 61\n= = = = = f f f f f (56)\nAgain, if g is an eigen value of the matrix F\n\nthen ) exp(gz U Y =\n\nis a solution of equation (54). Hence, U\n\nis the corresponding eigen vector.\nThe eigen values of F are the roots of the characteristic equation\n0 = gI F (57)\n0\n0 0 1 0 0\n0 0 0 1 0\n0 0 0 0 1\n0 0\n0 0\n0 0\n36 35 31\n26 25 21\n14 13 12\n=\n\ng\ng\ng\nf f g f\nf f g f\nf f f g\n(58)\nNow simplifying equation (58) and using equation (56) therein we obtain the characteristic equation as\n( ) ( ) 0\n2\n2\n2\n1\n2 2\n2\n2\n1\n2\n2\n2 2\n1\n2 2 2\n2\n2\n1\n2 4 6\n= + + + + + g g g (59)\nwhere\n2\n1\n2 2\n1\nk + = ,\n2\n2\n2 2\n2\nk + = ,\n( )\n5 4 3\n3 2\n1\nB B B\nA\nk\n+ +\n= ,\n( )\n( )\n.\n2 2\n3 3 2\n3 2 3 2\n2\nB A A\nA A A\nk\n+\n+\n=\nThe roots are\n2 1\n, ,\n\nwhich are the distinct eigen values of the matrix F .\nThe corresponding eigen vectors are obtained by solving the following homogeneous equation.\n0\n) (\n) (\n) (\n) (\n) (\n) (\n0 0 1 0 0\n0 0 0 1 0\n0 0 0 0 1\n0 0\n0 0\n0 0\n6\n5\n4\n3\n2\n1\n36 35 31\n26 25 21\n14 13 12\n=\n(\n(\n(\n(\n(\n(\n(\n(\n\n(\n(\n(\n(\n(\n(\n(\n(\n\ng U\ng U\ng U\ng U\ng U\ng U\ng\ng\ng\nf f g f\nf f g f\nf f f g\n(60)\nfor . 6 , 5 , 4 , 3 , 2 , 1 , = = i g g\ni\n\nhere . , , , , ,\n2 6 2 5 1 4 1 3 2 1\n= = = = = = g g g g g g\nWe denote the cofactors of the elements of the first row of the coefficient matrix in equation (60) by\n. 6 , 5 , 4 , 3 , 2 , 1 ), ( = i g F\ni\nthen\n\| \|\nT\ng F g F g F g F g F g F g U ) ( ) ( ) ( ) ( ) ( ) ( ) (\n6 5 4 3 2 1\n= are the solutions of the equation (60) and\nhence they are eigen vectors corresponding to the eigen values 6 , 5 , 4 , 3 , 2 , 1 , = i g\ni\nof the matrix . F\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 88 \| P a g e\nThe expressions of 6 , 5 , 4 , 3 , 2 , 1 ), ( = i g F\ni\nare\n( )\n( )\n( ) ( )\n( )( )\n(\n\n+ + +\n+ + +\n\n+ +\n+ + +\n+ =\n3 2 3 5 4\n3 2 3\n2\n3 3 2 2\n3 5 4\n3 3 5 4\n2\n3 5\n1\n2 2 2 2\n) (\nA A B B B\nA A A B A A\ng\nB B B\nA B B B\ng g g F\n\n( )\n( )\n( )( )\n( )( )\n(\n\n+ + +\n+ +\n\n+ +\n+\n=\n3 2 3 5 4\n2\n3\n2\n3 2 5 4 2\n3 5 4\n5 4 4\n2\n2\n) (\nA A B B B\nA A A B B\ng\nB B B\nB B\ng g F\n\n( )\n( )\n( )( )\n(\n\n+ + +\n+ +\n+\n+\n=\n3 2 3 5 4\n2\n3 4 3 5\n2\n3 2\n3 2\n3 4\n3\n2 2 2\n) (\nA A B B B\nA B B B A\ng\nA A\nA\ng g F\n\n( )\n( )\n( ) ( )\n( )( )\n(\n\n+ + +\n+ + +\n\n+ +\n+ + +\n+ =\n3 2 3 5 4\n3 2 3\n2\n3 2 3 2\n3 5 4\n3\n2\n4 3 5 2 4\n4\n2 2 2 2\n) (\nA A B B B\nA A A A A B\nB B B\nA B B B\ng g g F\n\n( )\n( )\n( )\n( ) ( )( )\n(\n\n+ + +\n\n+ +\n+\n\n+ +\n+\n=\n3 2 3 5 4\n2\n3\n3 5 4\n3\n5 4\n3 5 4\n5 4 3\n5\n2\n) (\nA A B B B\nA\nB B B\nB B\ng\nB B B\nB B\ng g F\n\n( )\n( )\n( )\n( )( )\n(\n\n+ + +\n+\n+\n+\n=\n3 2 3 5 4\n2\n3 4 3 5\n2\n3\n3 2\n3 3\n6\n2 2 2\n) (\nA A B B B\nA B B B A\ng\nA A\nA\ng g F\n\nSince g\n1\n, g\n2\n, g\n3\n, g\n4\n, g\n5\n, g\n6\nare the distinct roots of characteristic equation (59).\nThe general solution of the differential equation (55) is given by\n+ + + + = ) exp( ) ( ) exp( ) ( ) exp( ) ( ) exp( ) (\n4 4 4 3 3 3 2 2 2 1 1 1\nz g g U C z g g U C z g g U C z g g U C Y\n\n) exp( ) ( ) exp( ) (\n6 6 6 5 5 5\nz g g U C z g g U C + (61)\n\nwhere C\ni\n, i=1,2,3,4,5,6 are arbitrary constants.\nFrom equation (61), we have\n( )\n\n=\n=\n6\n1\n1\n'\n) exp( ) (\ni\ni i i r\nz g g F C z \|\n( )\n\n=\n=\n6\n1\n2\n'\n) exp( ) (\ni\ni i i z\nz g g F C z \|\n( )\n\n=\n=\n6\n1\n3\n'\n) exp( ) (\ni\ni i i\nz g g F C z u\nu\n\n( )\n\n=\n=\n6\n1\n4\n) exp( ) (\ni\ni i i r\nz g g F C z \| (62)\n( )\n\n=\n=\n6\n1\n5\n) exp( ) (\ni\ni i i z\nz g g F C z \| (63)\n( )\n\n=\n=\n6\n1\n6\n) exp( ) (\ni\ni i i\nz g g F C z u\nu\n(64)\nNow, we consider the equations of motion corresponding to micro-strains.\nThe equations of motion under the absence of body forces and couples the equation (3) involving micro-strains\ncan be expressed as\n0 2 2\n5 4\n2\n2\n2\n1\n= V + V\nrr pp rr pp\nA A B B \| \| \| \| (65)\n0 2 2\n5 4\n2\n2\n2\n1\n= V + V\nuu uu\n\| \| \| \| A A B B\npp pp\n(66)\n0 2 2\n5 4\n2\n2\n2\n1\n= V + V\nzz pp zz pp\nA A B B \| \| \| \| (67)\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 89 \| P a g e\n( ) ( )\n0 2 2\n5\n2\n2\n= V\nu u\n\| \|\nr r\nA B (68)\n( ) ( )\n0 2 2\n5\n2\n2\n= V\nrz rz\nA B \| \| (69)\n( ) ( )\n0 2 2\n5\n2\n2\n= V\nz z\nA B\nu u\n\| \| (70)\nwhere\n2\n2\n2\n2\n2\n1\nz r r r c\nc\n+\nc\nc\n+\nc\nc\n= V and\nzz rr pp\n\| \| \| \|\nuu\n+ + =\nAdding equations (65), (66) and (67) we have\n( ) ( ) 0 2 3 2 3\n5 4\n2\n2 1\n= + V +\npp pp\nA A B B \| \| (71)\nSubtract equation (66) from equation (65) and subtract equation (67) from equation (66) we get\n( ) ( ) 0 2 2\n5\n2\n2\n= V\nuu uu\n\| \| \| \|\nrr rr\nA B (72)\n( ) ( ) 0 2 2\n5\n2\n2\n= V\nzz zz\nA B \| \| \| \|\nuu uu\n(73)\nFrom equation (68) we have\n( ) ( )\n0\n2\n1\n2\n= V\nu u\n\| \|\nr r\nm (74)\nwhere\n2\n5 2\n1\nB\nA\nm =\nThe general solution of equation (74) is given by\n( )\n( ) \| \|\| \| ) sin( ) cos( ) ( ) ( ,\n2 1 1 0 2 1 0 1\nnz b nz b r m K a r m I a z r\nr\n+ + =\nu\n\| (75)\nAs the equations (69) and (70) are similar to equation (68). Hence the solutions of\n( ) rz\n\| and\n( ) z u\n\| are given by\n( )\n( ) \| \|\| \| ) sin( ) cos( ) ( ) ( ,\n4 3 1 0 4 1 0 3\nnz b nz b r m K a r m I a z r\nrz\n+ + = \| (76)\n( )\n( ) \| \|\| \| ) sin( ) cos( ) ( ) ( ,\n6 5 1 0 6 1 0 5\nnz b nz b r m K a r m I a z r\nz\n+ + =\nu\n\| (77)\nwhere ( ) r m I\n1 0\n\nand ( ) r m K\n1 0\n\nare modified Bessel function of the first and second kinds respectively.\nFrom equation (72) we have\n( ) ( ) 0\n2\n1\n2\n= V\nuu uu\n\| \| \| \|\nrr rr\nm (78)\nThe general solution of equation (78) is given by\n( ) \| \|\| \| ) sin( ) cos( ) ( ) ( ) , (\n8 7 1 0 8 1 0 7\nnz b nz b r m K a r m I a z r\nrr\n+ + =\nuu\n\| \| (79)\nSimilarly, the general solution of equation (73) is given by\n( ) \| \|\| \| ) sin( ) cos( ) ( ) ( ) , (\n10 9 1 0 10 1 0 9\nnz b nz b r m K a r m I a z r\nzz\n+ + = \| \|\nuu\n(80)\nFrom equation (71) we have\n0\n2\n2\n2\n= V\npp pp\nm \| \| (81)\nwhere\n\n( )\n( )\n2 1\n5 4 2\n2\n2 3\n2 3\nB B\nA A\nm\n+\n+\n=\n\nThe general solution of equation (81) is given by\n\| \|\| \| ) sin( ) cos( ) ( ) ( ) , (\n12 11 2 0 12 2 0 11\nnz b nz b r m K a r m I a z r\npp\n+ + = \| (82)\nNow, solving equations (79), (80) and (82) we get\n\| \|\| \| \| \| ) ( ) ( ) sin( ) cos( ) ( ) ( ) , (\n1 0 10 1 0 9 8 7 1 0 8 1 0 7\nr m K a r m I a nz b nz b r m K a r m I a z r\nrr\n+ + + + = \|\n\| \|\| \| ) sin( ) cos( ) ( ) ( )] sin( ) cos( [\n12 11 2 0 12 2 0 11 10 9\nnz b nz b r m K a r m I a nz b nz b + + + + (83)\n\| \|\| \| ) sin( ) cos( ) ( ) ( ) , (\n8 7 1 0 8 1 0 7\nnz b nz b r m K a r m I a z r + + =\nuu\n\| (84)\n\nR. Srinivas et al Int. Journal of Engineering Research and Applications www.ijera.com\nISSN : 2248-9622, Vol. 4, Issue 7( Version 1), July 2014, pp.80-90\n\nwww.ijera.com 90 \| P a g e\n\| \|\| \| \| \| ) ( ) ( ) sin( ) cos( ) ( ) ( ) , (\n1 0 10 1 0 9 8 7 1 0 8 1 0 7\nr m K a r m I a nz b nz b r m K a r m I a z r\nzz\n+ + + + = \|\n\| \| ) sin( ) cos(\n10 9\nnz b nz b + (85)\nhere\n0\na to\n12\na and\n0\nb to\n12\nb are arbitrary constants.\nThe arbitrary constants involved in equations (48), (49), (50), (62), (63), (64), (83), (84), (85), (75), (76)\nand (77) can be determined using specified boundary conditions of a particular problem. Once, these constants\nare found, it is possible to find displacements, micro-rotation, stress and mirco-streses. The results of micropolar\ntheory Mahalanabis and Manna are obtained as particular case of this paper when\n, 2 , 2 , 2\n3 5 4\n\| o = = = B B B k =\n3\n2A , , 2\n3 1\n= + A A =\n3 2\nA A and . 0 , , ,\n5 4 2 1\nA A B B\n\nREFERENCES\n Eringen, A.C.: Theory of micropolar elasticity in fracture-II, Academic Press, New York, pp.621-729\n(1968).\n Eringen, A.C.: Mechanics of micromorphic materials, In Proceedings of the 11\nth\nInternational\nCongress of Applied Mechanics, Munich, Berlin, pp.131-138 (1964b).\n Koh, S.L.: A special theory of microelasticity, International Journal of Engineering Science, 8, pp.583-\n593 (1970).\n Nowacki, W.: The axisymmetric Lambs problem in a semi-infinite micropolar elastic solid, Proc.\nVibr. Probl. Warsaw, 10 (2), p.97 (1969).\n Das, N.C., Das, S.N. and Das, B.: Eigen value approach to thermoelasticity, Journal of Thermal\nStresses, 6 (1), pp.35-43 (1983).\n Das, N.C., Mitra, A.K. and Mahalanabis, R.K.: Eigen value approach to magnetothermo-elasticity,\nJournal of Thermal Stresses, 6 (1), pp.73-92 (1983).\n Mahalanabis, R.K. and Manna, J.: Eigen value approach to linear micropolar elasticity, Indian Journal\nof Pure and Applied Mathematics, 20 (12), pp.1237-1250 (1989).\n Parameshwaran, S. and Koh, S.L.: Wave propagation in a micro-isotropic, micro-elastic solid,\nInternational Journal of Engineering and Science, 11, pp.95-107 (1973)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90898466,"math_prob":0.9994555,"size":665,"snap":"2019-43-2019-47","text_gpt3_token_len":124,"char_repetition_ratio":0.12405446,"word_repetition_ratio":0.043010753,"special_character_ratio":0.16541353,"punctuation_ratio":0.10810811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99813324,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T07:11:11Z\",\"WARC-Record-ID\":\"<urn:uuid:957ee9d2-f8cf-441f-936e-3be67c7298ec>\",\"Content-Length\":\"370196\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dfc9fbc6-a081-44a9-92d9-3ae879ad8a86>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e993407-c41d-4877-a91b-5d62d83e0d9e>\",\"WARC-IP-Address\":\"151.101.202.152\",\"WARC-Target-URI\":\"https://www.scribd.com/document/239989903/General-Solution-of-Equations-of-Motion-of-Axisymmetric-Problem-of-Micro-Isotropic-Micro-Elastic-Solid\",\"WARC-Payload-Digest\":\"sha1:DPEOWQIFOYEVWANBUMV2HQJ57FW7JP6I\",\"WARC-Block-Digest\":\"sha1:HGRMMOJCNTZN64E7XY76JUMWKABWBXNO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987803441.95_warc_CC-MAIN-20191022053647-20191022081147-00311.warc.gz\"}"}