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https://tenminutetutor.com/computer-science/gcse/data-representation/numbers/bitwise-operations/	[ "# Bitwise logical operations\n\nMartin McBride, 2017-02-24\nTags binary bit bit field and bitwise and or bitwise or bit mask not bitwise not xor bitwise xor\nCategories data representation numbers\n\nYou may be familiar with logic gates which use AND, OR, NOT and XOR logic.\n\nYou may also have used and, or, and not combinations in programming (for example if statements and while loops).\n\nBitwise logical operators follow the same rules but they act on the individual bits in a byte. They are useful for bit manipulation.\n\n## Bit fields\n\nOften, a byte or word is used to contain a number value. But sometimes we are more interested in the individual bits within the byte.\n\nFor example, if you were making a computer game using a Raspberry Pi, Arduino or similar, you might connect push-buttons to the computers I/O (input/output) pins for left, right and fire functions.\n\nWhen you check to see which buttons are pressed, you might get back a byte value, where:\n\n• Bit 0 is set if the left button is pressed.\n• Bit 1 is set if the right button is pressed.\n• Bit 2 is set if the fire button is pressed.\n\nSo if the right button is pressed, you would get a value 00000010 binary (2 decimal).\n\nAlternatively, if the left button and fire button were both pressed at the same time, you would get a value 00000101 binary (5 decimal).\n\nIf a byte (or a word of any size) is used as a collection of bits, we call it a bit field value. Bitwise logical operators are very useful for dealing with bit fields.\n\n## Bitwise AND\n\nThe bitwise AND operator often uses the symbol &.\n\nIt combines values by ANDing together each pair of bits in the two input bytes:", null, "So in this example:\n\n• Value A bit 0 is 0, Value B bit 0 is 0, so the result bit 0 is 0.\n• Value A bit 1 is 0, Value B bit 1 is 1, so the result bit 1 is 0.\n• Value A bit 2 is 1, Value B bit 2 is 0, so the result bit 2 is 0.\n• Value A bit 3 is 1, Value B bit 3 is 1, so the result bit 3 is 1.\n• etc\n\nA particular result bit will only be 1 if both input bits are 1.\n\nWe can do this in pseudocode like this:\n\n```a = 01111100b\nb = 01101010b\nresult = a & b\nPRINT(result)\n```\n\nWhich will print the value 01101000 binary (104 in denary).\n\n## Using bitwise AND as a bit mask\n\nA bit mask is used to mask out unwanted bits in a byte, and just leave the ones you need. For example, suppose you have a byte value, and you want to find the value of the lowest hexadecimal digit. To do this you need to take the value of the lower 4 bits, ignoring the higher 4 bits.\n\nFor the AND function, we know that:\n\n• ANDing anything with 0 is always 0, ie A & 0 = 0.\n• ANDing anything with 1 leaves the value unchanged, ie A & 1 = A.\n\nWe can think of bitwise AND as a masking function. Here, we are using input B as a mask:", null, "In this case:\n\n• For the low 4 bits, where the mask bits are 1, the bit values in A are passed through to the result.\n• For the high 4 bits, where the mask bits are 0, the bits in the result are forced to 0.\n\n## Bitwise OR\n\nThe bitwise OR operator often uses the symbol \|.\n\nIt combines values by ORing together each pair of bits in the two input bytes:", null, "So in this example:\n\n• Value A bit 0 is 0, Value B bit 0 is 0, so the result bit 0 is 0.\n• Value A bit 1 is 0, Value B bit 1 is 1, so the result bit 1 is 1.\n• Value A bit 2 is 1, Value B bit 2 is 0, so the result bit 2 is 1.\n• Value A bit 3 is 1, Value B bit 3 is 1, so the result bit 3 is 1.\n• etc\n\nA particular result bit will be 1 if either or both input bits are 1.\n\nWe can do this in pseudocode like this:\n\n```a = 01111100b\nb = 01101010b\nresult = a \| b\nPRINT(result)\n```\n\nWhich will print the value 01111110 binary (126 in denary).\n\n## Using bitwise OR to combine bit fields\n\nSometimes you might have two bit fields which need to be combined. For example:\n\n• Value A has 4 bits of data in the lower 4 bits, and the upper 4 bits are set to 0.\n• Value B has 4 bits of data in the upper 4 bits, and the lower 4 bits are set to 0.\n\nWe can think of bitwise AND as a masking function. Here, we are using input B as a mask:", null, "In this case:\n\n• For the lower 4 bits of data in A are ORed with lower 4 bits of zeros in B, so the bit values in A are passed through to the result.\n• For the upper 4 bits of data in B are ORed with upper 4 bits of zeros in A, so the bit values in B are passed through to the result.\n\n## Bitwise NOT\n\nThe bitwise NOT operation often uses the symbol ~. It takes one value, and returns the result of inverting each bit. For example:\n\n```a = 10101110b\nresult = ~a\nPRINT(result)\n```\n\nprints 01010001 binary (each bit is the opposite of the input bit).\n\n## Bitwise XOR\n\nThe bitwise XOR operation often uses the symbol ^.\n\nIt behaves in a similar way to bitwise OR. The difference is that the output bit will be 1 if either (but not both) input bits are 1. The output bit will be zero if both inputs are 0, or if both inputs are 1." ]	[ null, "https://tenminutetutor.com/img/computer-science/gcse/data-representation/numbers/and-bits.png", null, "https://tenminutetutor.com/img/computer-science/gcse/data-representation/numbers/and-mask-bits.png", null, "https://tenminutetutor.com/img/computer-science/gcse/data-representation/numbers/or-bits.png", null, "https://tenminutetutor.com/img/computer-science/gcse/data-representation/numbers/or-combine-bits.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8658709,"math_prob":0.98580354,"size":4750,"snap":"2021-43-2021-49","text_gpt3_token_len":1294,"char_repetition_ratio":0.15233882,"word_repetition_ratio":0.32798395,"special_character_ratio":0.28694737,"punctuation_ratio":0.105166055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995128,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T03:26:18Z\",\"WARC-Record-ID\":\"<urn:uuid:80befccb-2d00-4162-9f19-7126b0138518>\",\"Content-Length\":\"13553\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b7af5b0-9fe3-44b1-ba87-976bff200d3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d37f19a-9767-44a3-bac0-d6be6bf277bb>\",\"WARC-IP-Address\":\"192.236.209.132\",\"WARC-Target-URI\":\"https://tenminutetutor.com/computer-science/gcse/data-representation/numbers/bitwise-operations/\",\"WARC-Payload-Digest\":\"sha1:QZBM5TZJ3A6BGN4HXCCRFAW55DEKMVAF\",\"WARC-Block-Digest\":\"sha1:GLGHCAVW4N3EFUY6XIY4G2SGL4VAUHYN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585231.62_warc_CC-MAIN-20211019012407-20211019042407-00664.warc.gz\"}"}
https://www.get-digital-help.com/how-to-use-the-bitrshift-function/	[ "Author: Oscar Cronquist Article last updated on May 11, 2022", null, "The BITRSHIFT function calculates the number where the binary equivalent is shifted right by a specified number of bits and then converted back to a number.\n\nFormula in cell D3:\n\n=BITRSHIFT(B3, C3)\n\nNumber 5 is 00000101 binary. Adding one zero to the left and removing the last digit you get 00000010 binary which is number 2.\n\n### Excel Function Syntax\n\nBITRSHIFT(numbershift_amount)\n\n### Arguments\n\n number Required. The number you want to shift. shift_amount Required. How many digits you want to shift the binary representation of the number." ]	[ null, "https://www.get-digital-help.com/wp-content/uploads/2018/03/How-to-use-the-BITRSHIFT-function.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.768588,"math_prob":0.97270817,"size":2163,"snap":"2022-05-2022-21","text_gpt3_token_len":498,"char_repetition_ratio":0.16442798,"word_repetition_ratio":0.1388102,"special_character_ratio":0.2196024,"punctuation_ratio":0.0906801,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9893829,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T19:45:14Z\",\"WARC-Record-ID\":\"<urn:uuid:1e6d61b9-c129-4254-bc58-c0402a732d6d>\",\"Content-Length\":\"67965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:175dc933-20f7-4220-9719-edea8e46c52b>\",\"WARC-Concurrent-To\":\"<urn:uuid:48796deb-f29b-40cf-9a74-939844eff7f2>\",\"WARC-IP-Address\":\"104.26.9.201\",\"WARC-Target-URI\":\"https://www.get-digital-help.com/how-to-use-the-bitrshift-function/\",\"WARC-Payload-Digest\":\"sha1:F2NYZCSQEXKMLRJTTA25ORNUYWWM7K6G\",\"WARC-Block-Digest\":\"sha1:L4WLFCCRIEGKK4JG5KRS6VH3KDMNH55F\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522309.14_warc_CC-MAIN-20220518183254-20220518213254-00312.warc.gz\"}"}
https://neuron.yale.edu/phpBB/viewtopic.php?p=9698	[ "## NetStim.interval variable\n\nModerator: wwlytton\n\npatoorio\nPosts: 81\nJoined: Wed Jan 30, 2008 12:46 pm\n\n### NetStim.interval variable\n\nHi,\n\nI want to have a NetStim that delivers random pulses but with a variable mean interval. Specifically, I want to have sinusoid form with max=0 and min=20 (or anything resembling that).\nI have tried by playing a sine vector on Net1.interval, i.e.\n\nCode: Select all\n\n``sinvec.play(&Net1.interval,1,1) ``\nBut it doesn't work; I get a segmentation violation. (sinevec is a vector previously loaded from a text file)\n(BTW I also tried with sinvec.play(Net1.interval,1,1), without the & and it doesn't work either but without the error).\n\nWhat is the proper way of doing what I want? I have thought of modifying the .mod file but I want to leave that as the last resource.\n\nThanks!\nted\nPosts: 5797\nJoined: Wed May 18, 2005 4:50 pm\nLocation: Yale University School of Medicine\nContact:\n\n### Re: NetStim.interval variable\n\npatoorio wrote:I want to have a NetStim that delivers random pulses but with a variable mean interval.\nYou need to find or create an algorithm that does what you want.\n\nConsider a very simple problem: you have a simulation with fixed time step dt, and you want pulses that occur at an average interval of 2dt. One way to do this is:\nat every new time step:\npick a number from the uniform distribution over [0,1]\nif the number is <=0.5, generate a pulse\n\nTo generalize, suppose you want an average interval of Ndt, where N>1. You could do this by:\nat every new time step\npick a number from the uniform distribution over [0,1]\nif the number is <=1/N, generate a pulse.\n\nTo generalize even further, suppose you want the average interval to vary with time. This means that N must be a function of time\nN = f(t) where f is always >=1\nSo at every new time step,\npick a number from the uniform distribution over [0,1]\nif the number is <= 1/f(t), generate a pulse.\n\nThis will generate pulses whose instantaneous average interval (discoverable by averaging results of an ensemble of simulations) is f(t)dt\nAnd it will have to be implemented with a mod file.\npatoorio\nPosts: 81\nJoined: Wed Jan 30, 2008 12:46 pm\n\n### Re: NetStim.interval variable\n\nThanks a lot for telling me to do what I didn't want to do ;)\nBut I mean it, because it turned out to be easier than I thought... until I got into trouble.\n\nI have something like this:\n\nCode: Select all\n\n``````INITIAL {\nr_interval = exprand(1)\nlast_event = 0\n}\n\nBREAKPOINT {\nif (t>start && t<start+dur) {\ninterval = (1000)1/(minRate + (maxRate-minRate)cos(23.14159freq(t-phase)/(1000)))\nnext_ev = last_event + invl(interval)\nif (t>=next_ev) {event()}\n}\n}\n\nFUNCTION invl(mean (ms)) (ms) {\ninvl = (1. - noise)mean + noisemeanr_interval\n}\n\nPROCEDURE event() {\nnet_event(t)\nlast_event = t\nr_interval = exprand(1)\n}``````\n(for better readability I have deleted some lines that check that some things are greater than 0)\nThe idea is to keep the (exponentially distributed) random number r_interval unchanged until there is a new event; what changes from step to step is the mean of the distribution.\nThe error I get when trying to compile is that net_event (as well as net_send) has to be within the INITIAL or a NET_RECEIVE block. I cannot see how to accomplish that since I need interval to change every time, and every time check if last_event + invl(interval) is lower than t. In other words, I have to be able to generate (in any way) an event from within the BREAKPOINT block.\nHow can I do that?\n\nThanks!\npatoorio\nPosts: 81\nJoined: Wed Jan 30, 2008 12:46 pm\n\n### Re: NetStim.interval variable\n\nI did some research and found that the WATCH command may be useful here.\nNow I have this:\n\nCode: Select all\n\n``````INITIAL {\nif (noise < 0) {noise = 0}\nif (noise > 1) {noise = 1}\nif (minRate <=0) {minRate = 0.0001}\nr_interval = exprand(1)\nlast_event = 0\ninterval = (1000)1/(minRate + (maxRate-minRate)cos(23.14159freq(t-phase)/(1000)))\nnext_ev = last_event + invl(interval)\nnet_send(0,2)\n}\n\nBREAKPOINT {\nif (t>start && t<start+dur) {\ninterval = (1000)1/(minRate + (maxRate-minRate)cos(23.14159freq(t-phase)/(1000)))\nnext_ev = last_event + invl(interval)\n}\n}\n\nif (flag==1) {\nnet_event(t)\nlast_event = t\nr_interval = exprand(1)\nnext_ev = last_event + invl(interval)\n}\nif (flag==2) {}\nWATCH (t>next_ev) 1\n}\n\nFUNCTION invl(mean (ms)) (ms) {\nif (mean <= 0.) {\nmean = .01 (ms)\n}\nif (noise == 0) {\ninvl = mean\n}else{\ninvl = (1. - noise)mean + noisemeanr_interval\n}\n}``````\nIt compiles successfuly but just trying to initialize the model gives a segmentation violation error.\n\nBTW, I'm using this mechanism in a very simple way:\n\nCode: Select all\n\n``````objref Net1,Con1,IF1\n\nNet1 = new SinStim()\nNet1.start = 0\nNet1.noise = 0\n\nIF1 = new IntFire1(0.5)\nCon1 = new NetCon(Net1,IF1)\nCon1.weight = 0.7``````\npatoorio\nPosts: 81\nJoined: Wed Jan 30, 2008 12:46 pm\n\n### Re: NetStim.interval variable\n\nWell, I found the solution... and a possible bug?\n\nIt turns out that WATCH doesn't work in an ARTIFICIAL_CELL!! I just changed the process to a POINT_PROCESS, created a mock section to insert it and now it works like a charm!\nIs this normal? BTW I haven't tried with the last alpha version, I'm working with NEURON 7.1\n\nBest wishes\nted\nPosts: 5797\nJoined: Wed May 18, 2005 4:50 pm\nLocation: Yale University School of Medicine\nContact:\n\n### Re: NetStim.interval variable\n\nI wonder if that would work as an ARTIFICIAL_CELL with version 7.2. By the way, 7.2 has now (finally) been released officially.\n\nI really don't like this algorithm for execution during a simulation--requires code execution at every fadvance() and forces fixed dt integration. It may be better to precalculate the spike times and save them to one or more files, then read them into one or more Vectors prior to calling run() and use VecStim to play the events into NetCons--for source code and an example of the use of VecStim see\nnrn/share/nrn/examples/nrniv/netcon/vecevent\n(c:/nrnxx/examples/nrniv/netcon/vecevent for MSWin users). This would allow the use of adaptive integration--even though the event times themselves were calculated in a way that forces them to lie at integer multiples of some sample rate.\npatoorio\nPosts: 81\nJoined: Wed Jan 30, 2008 12:46 pm\n\n### Re: NetStim.interval variable\n\nted wrote:I wonder if that would work as an ARTIFICIAL_CELL with version 7.2. By the way, 7.2 has now (finally) been released officially. .\nNo, it doesn't. And now there isn't any segmentation violation message, neuron just quits.\nted wrote:I really don't like this algorithm for execution during a simulation--requires code execution at every fadvance() and forces fixed dt integration. It may be better to precalculate the spike times and save them to one or more files, then read them into one or more Vectors prior to calling run() and use VecStim to play the events into NetCons--for source code and an example of the use of VecStim see\nnrn/share/nrn/examples/nrniv/netcon/vecevent\n(c:/nrnxx/examples/nrniv/netcon/vecevent for MSWin users). This would allow the use of adaptive integration--even though the event times themselves were calculated in a way that forces them to lie at integer multiples of some sample rate.\nSounds interesting. I will give it a try and see how much the simulation is sped up.\nThanks!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9243999,"math_prob":0.92478836,"size":1147,"snap":"2020-45-2020-50","text_gpt3_token_len":273,"char_repetition_ratio":0.12423447,"word_repetition_ratio":0.13170731,"special_character_ratio":0.23975588,"punctuation_ratio":0.101626016,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97521025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T23:31:05Z\",\"WARC-Record-ID\":\"<urn:uuid:c40cb112-a311-40cf-9a83-944ff6d3bc83>\",\"Content-Length\":\"48595\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f73d6a7f-3e93-4ca2-b519-dc88efa7961f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7fd87bce-2ec2-4673-b0d2-1bef3b1d934d>\",\"WARC-IP-Address\":\"128.36.111.212\",\"WARC-Target-URI\":\"https://neuron.yale.edu/phpBB/viewtopic.php?p=9698\",\"WARC-Payload-Digest\":\"sha1:TJSAVLX7K42VGRB3IANCXOXNJH5QH36J\",\"WARC-Block-Digest\":\"sha1:LZDETG754W3CMJ5HWKOE5KLNXZKBGGL2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107911792.65_warc_CC-MAIN-20201030212708-20201031002708-00502.warc.gz\"}"}
https://www.studytonight.com/java-programs/java-program-to-print-the-left-arrow-star-pattern	[ "# Java Program to Print the Left Arrow Star Pattern\n\nIn this tutorial, we will see how to print the left arrow star pattern in Java. First, we will ask the user to initialize the number of rows. Then, we will use loops to print the pattern. But before moving further, if you are not familiar with the concept of the loops in java, then do check the article on Loops in Java.\n\nInput: Enter the number of rows: 6\n\nOutput: The pattern is:\n\n****\n\n*\n\n\n\n\n\n\n\n\n\n\n\n\n\n*\n\n*\n\n***\n\n## Program 1: Print the Left Arrow Star Pattern\n\nIn this program, we will see how to print the left arrow star pattern in Java using a for loop.\n\n### Algorithm:\n\n1. Start\n2. Create an instance of the Scanner class.\n3. Declare variables to store the number of rows and the pattern symbol.\n4. Ask the user to initialize these variables.\n5. Use two for loops to print the pattern.\n6. The first for loop displays the upper pattern of “left arrow” and the second for loop displays the lower pattern.\n7. First, check the condition i<n at for loop, if it is true, then it executes the first inner for the loop until the condition is false, after that, it executes the second inner for the loop until the condition is false.\n8. The first inner for loop display space and the second inner for loop displays character which we have given to display.\n9. After the execution of the first outer for loop, the second outer for loop will be executed.\n10. Check the condition at for loop, if it is true, then execute the inner loops until the condition i<n.\n11. Display the result.\n12. Stop\n\nThe below example illustrates the implementation of the above algorithm.\n\n``````//Java Program to Print the Left Arrow Star Pattern\nimport java.util.Scanner;\npublic class Main\n{\npublic static void main(String[] args)\n{\n//Take input from the user\nScanner sc=new Scanner(System.in);\nSystem.out.println(\"Enter the number of rows: \");\nint n=sc.nextInt();\nfor(int i=1;i<=n;i++)\n{\nfor(int j=1;j<=n-i;j++)\n{\nSystem.out.print(\" \");\n}\nfor(int j=i;j<=n;j++)\n{\nSystem.out.print(\"\");\n}\nSystem.out.println();\n}\nfor(int i=1;i<n;i++)\n{\nfor(int j=0;j<i;j++)\n{\nSystem.out.print(\" \");\n}\nfor(int j=0;j<=i;j++)\n{\nSystem.out.print(\"\");\n}\nSystem.out.println();\n}\n}\n}``````\n\nEnter the number of rows: 6\n***\n*\n\n\n\n\n\n\n*\n*\n***\n\n## Program 2: Print the Left Arrow Star Pattern\n\nIn this program, we will see how to print the left arrow star pattern in Java using a while loop.\n\n### Algorithm:\n\n1. Start\n2. Create an instance of the Scanner class.\n3. Declare variables to store the number of rows and the pattern symbol.\n4. Ask the user to initialize these variables.\n5. Use two while loops to print the pattern.\n6. First, check the condition i<=n at while, if it is true then it executes the code in the while loop.\n7. The first while will execute until i<=n is false.\n8. After the execution of the first while loop, the second while loop will be executed.\n9. Display the result.\n10. Stop\n\nThe below example illustrates the implementation of the above algorithm.\n\n``````//Java Program to Print the Left Arrow Star Pattern\nimport java.util.Scanner;\npublic class Main\n{\npublic static void main(String[] args)\n{\n//Take input from the user\nScanner sc=new Scanner(System.in);\nSystem.out.println(\"Enter the number of rows: \");\nint n=sc.nextInt();\nint i=1;\nint j;\nwhile(i<=n)\n{\nj=1;\nwhile(j<=n-i)\n{\nSystem.out.print(\" \");\nj++;\n}\nj=i;\nwhile(j<=n)\n{\nSystem.out.print(\"\");\nj++;\n}\nSystem.out.println();\ni++;\n}\ni=1;\nwhile(i<n)\n{\nj=0;\nwhile(j<i)\n{\nSystem.out.print(\" \");\nj++;\n}\nj=0;\nwhile(j<=i)\n{\nSystem.out.print(\"\");\nj++;\n}\nSystem.out.println();\ni++;\n}\n}\n}``````\n\nEnter the number of rows: 6\n***\n*\n\n\n\n\n\n\n*\n*\n***\n\n## Program 3: Print the Left Arrow Star Pattern\n\nIn this program, we will see how to print the left arrow star pattern in Java using a do-while loop.\n\n### Algorithm:\n\n1. Start\n2. Create an instance of the Scanner class.\n3. Declare variables to store the number of rows and the pattern symbol.\n4. Ask the user to initialize these variables.\n5. Use two do-while loops to print the pattern.\n6. At first, the do-while loop will be executed until the condition is false i<=n. Inner do-while loops will be executed until the condition is false.\n7. After the execution of the first do-while loop, the second do-while loop will be executed until the condition i<n is false. Inner do-while loops will be executed until the condition is false.\n8. Display the result.\n9. Stop\n\nThe below example illustrates the implementation of the above algorithm.\n\n``````//Java Program to Print the Left Arrow Star Pattern\nimport java.util.Scanner;\npublic class Main\n{\npublic static void main(String[] args)\n{\n//Take input from the user\nScanner sc=new Scanner(System.in);\nSystem.out.println(\"Enter the number of rows: \");\nint n=sc.nextInt();\nint i=1;\nint j;\ndo\n{\nj=1;\ndo\n{\nSystem.out.print(\" \");\n}while(j++<=n-i);\nj=i;\ndo\n{\nSystem.out.print(\"\");\nj++;\n}while(j<=n);\nSystem.out.println();\ni++;\n} while(i<=n);\ni=1;\ndo\n{\nj=0;\ndo\n{\nSystem.out.print(\" \");\n}while(j++<i);\nj=0;\ndo\n{\nSystem.out.print(\"\");\nj++;\n} while(j<=i);\nSystem.out.println();\ni++;\n}while(i<n);\n}\n}``````\n\nEnter the number of rows: 6\n***\n*\n\n\n\n\n\n\n*\n*\n****" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61893785,"math_prob":0.810059,"size":5112,"snap":"2021-31-2021-39","text_gpt3_token_len":1254,"char_repetition_ratio":0.15935788,"word_repetition_ratio":0.5432243,"special_character_ratio":0.31005478,"punctuation_ratio":0.1846722,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9915204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T10:57:16Z\",\"WARC-Record-ID\":\"<urn:uuid:c0cd4876-9333-414a-b028-280f6b9f2a59>\",\"Content-Length\":\"159840\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06e24422-4995-40eb-bbf0-7fe023d1d20b>\",\"WARC-Concurrent-To\":\"<urn:uuid:d317f1fa-88e3-4655-a0ac-1c0fcb494a0f>\",\"WARC-IP-Address\":\"65.2.112.116\",\"WARC-Target-URI\":\"https://www.studytonight.com/java-programs/java-program-to-print-the-left-arrow-star-pattern\",\"WARC-Payload-Digest\":\"sha1:7YVESJ4J3NTAV3FFVFFF5RUY7FRWH7QN\",\"WARC-Block-Digest\":\"sha1:IXVA73OSZD6JY6P3CM24JRW26OY4V4WM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057421.82_warc_CC-MAIN-20210923104706-20210923134706-00591.warc.gz\"}"}
https://research.chalmers.se/publication/208010	[ "# A Solution to the Pole Problem for the Shallow Water Equations on a Sphere Artikel i vetenskaplig tidskrift, 2014\n\nWe consider a reduced gridding technique for the shallow water equations on a sphere, based on spherical coordinates. In a small vicinity of the poles, a longitudinal derivative is discretized at a grid-point on a parallel, by using points on the great circle through the grid-point and tangent to the parallel. Centered one-dimensional interpolation formulas are used in this process and also in connecting adjacent segments in the reduced grid. The remaining spatial discretization is obtained by simply replacing derivatives by centered equidistant finite difference approximations. Numerical experiments for scalar advection equations and for the well-known Rossby-Haurwitz test example indicate that the methods developed work surprisingly well. Some advantages are that (i) a fairly uniform grid, with many reductions or segments, can be used, (ii) order of approximation \\$2p\\$ in the spatial discretizations requires only \\$4p+1\\$ points and (iii) the local and simple structure of the schemes will make efficient implementation on massively parallel computer systems possible. The paper is an attempt towards global numerical weather prediction models, by first analyzing the pole problem for reduced latitude-longitude grids.\n\nsphere\n\nnumerical weather prediction.\n\nreduced grid\n\nshallow water equations\n\nsegment\n\npole problem\n\n## Författare\n\n### Göran Christer Starius\n\nChalmers, Matematiska vetenskaper\n\nGöteborgs universitet\n\n2076-2585 (ISSN)\n\nVol. 5 2 152-170\n\nMatematik\n\n### Fundament\n\nGrundläggande vetenskaper\n\n2019-02-18" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85659033,"math_prob":0.92656744,"size":1440,"snap":"2020-45-2020-50","text_gpt3_token_len":298,"char_repetition_ratio":0.10027855,"word_repetition_ratio":0.0,"special_character_ratio":0.17222223,"punctuation_ratio":0.06896552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97327083,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T01:05:38Z\",\"WARC-Record-ID\":\"<urn:uuid:6cd3a030-bd52-48b1-a0e8-0f3ce87e1050>\",\"Content-Length\":\"32611\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49e6152b-b2d2-45e0-a324-b1e9768a0439>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc461659-2fd3-40da-879d-edfdad26b6dc>\",\"WARC-IP-Address\":\"40.113.65.9\",\"WARC-Target-URI\":\"https://research.chalmers.se/publication/208010\",\"WARC-Payload-Digest\":\"sha1:4L6POD3EN7PTHAQQNS2AAHD3DYG24ZAK\",\"WARC-Block-Digest\":\"sha1:ROQNC3OH3HB6KWTTBZDMS7BE7XDMBAJI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141685797.79_warc_CC-MAIN-20201201231155-20201202021155-00563.warc.gz\"}"}
https://slideplayer.com/slide/234197/	[ "", null, "# Chapter 11 Trees Graphs III (Trees, MSTs) Reading: Epp Chp 11.5, 11.6.\n\n## Presentation on theme: \"Chapter 11 Trees Graphs III (Trees, MSTs) Reading: Epp Chp 11.5, 11.6.\"— Presentation transcript:\n\nChapter 11 Trees Graphs III (Trees, MSTs) Reading: Epp Chp 11.5, 11.6\n\nOutline 1.Trees 1.1 Definition of a tree. 1.2 Examples of trees. 1.3 Theorem #1: Tree Characterization 2.Rooted Trees 2.1 Definitions 2.2 Definition: n-ary trees and binary trees. 2.3 Definition: Full n-ary trees 2.4 Theorem #2: Full Tree 2.5 Theorem #3: Leaves-Height 2.6 Examples 3.Spanning Trees 3.1 Motivation 3.2 Definition 3.3 Theorem #4: Spanning Tree 3.4 Minimum Spanning Tree (MST) 3.5 Kruskals Algorithm 3.6 Prims Algorithm\n\n1. Trees 1.1 Definition: Let G=(V,E). G is a tree IFF (a) G is connected; and (b) G does not have any circuits (acyclic). Comment: –The textbook distinguishes between trivial and non-trivial circuits. They define a trivial circuit as a circuit of length 0. –As far as we are concerned, unless otherwise stated, when we say circuit we mean NON- TRIVIAL circuit. i.e. the default meaning of circuit is a non-trivial circuit.\n\n1. Trees 1.1 Definition: Let G=(V,E). G is a tree IFF (a) G is connected; and (b) G does not have any circuits (acyclic). 1.2.1 Example of a tree\n\n1. Trees 1.2.2 Examples of trees in real life usage –Family Tree –Tournaments –Directory Tree –Syntax Tree –Execution Tree –Decision Tree –Search Tree –B-Tree (Databases)\n\n1. Trees 1.3 Theorem (Tree Characterization Theorem): Let G=(V,E). G is a tree IFF G is connected and \|E\|=\|V\|-1 n Proof Strategy: –( ) Assume G is a tree Prove that G is connected(Trivial) Prove that \|E\| = \|V\| - 1(Prove by induction on \|V\|) –Lemma 1: A tree with more than 1 vertex has at least 1 vertex of degree 1 –( ) Assume G is connected and \|E\| = \|V\| -1 (Prove that G is a tree. How? Show that G fits the definition of a tree.) Prove that G is connected(Trivial) Prove that G has no circuits(Prove by contradiction) –Lemma 2: Deletion of edge from a circuit of a connected graph does not violate connectedness.\n\nOutline 1.Trees 1.1 Definition of a tree. 1.2 Examples of trees. 1.3 Theorem #1: Tree Characterization 2.Rooted Trees 2.1 Definitions 2.2 Definition: n-ary trees and binary trees. 2.3 Definition: Full n-ary trees 2.4 Theorem #2: Full Tree 2.5 Theorem #3: Leaves-Height 2.6 Examples 3.Spanning Trees 3.1 Motivation 3.2 Definition 3.3 Theorem #4: Spanning Tree 3.4 Minimum Spanning Tree (MST) 3.5 Kruskals Algorithm 3.6 Prims Algorithm\n\n2.1 Rooted Trees (Definition) n A rooted tree is a tree in which one vertex is distinguished from the others and is called the root. n The level of a vertex v is the path length from the root to v. The height of the tree is the maximum level to any vertex of the tree. root Level 0 Level 1 Level 2 Level 3 Level 4\n\n2.1 Rooted Trees (Definition) n Given any vertex v in a rooted tree: –The children of v are the vertices adjacent to v, 1 level further away from the root. –The parent of v is the vertex adjacent to v, 1 level nearer to the root. –The siblings of v are the vertices which have the same parent as v. v Parent of v Children of v Siblings of v\n\n2.1 Rooted Trees (Definition) n Given any vertex v in a rooted tree: –The ancestor of v are the vertices which lie in the path from v to the root. –If u is the ancestor of v, then v is the descendant of u. v Ancestors of v Descendants of v\n\n2.1 Rooted Trees (Definition) n Given any rooted tree: –The internal vertices of the tree are the vertices which have at least 1 child. –The external vertices of the tree are the vertices which have no children. External vertices are also known as the leaves of the tree, or terminal vertices. External Vertices (Leaves) The rest are Internal Vertices\n\n2.2 m-ary trees and binary trees (Defn) n A m-ary tree is a rooted tree in which every vertex has at most m children. Example of a 4-ary Tree Example of a 3-ary Tree\n\n2.2 m-ary trees and binary trees (Defn) n A m-ary tree is a rooted tree in which every vertex has at most m children. –A binary tree is a m-ary tree with n=2. Each child of the binary tree is designated either the left child or the right child. Given a vertex v of a binary tree, the left subtree of v (right subtree of v) is the binary tree whose root is the left child of v (right child of v). v Left subtree of v Right subtree of v Left child of v Right child of v Example of a Binary Tree\n\n2.3 Full m-ary Trees (Definition) n A m-ary tree is FULL iff every vertex has either 0 or m children. (OR every internal vertex has m children). n Examples of full binary trees. Full Tree?YesNoYesNo Yes\n\nProof: 2.4 Full Tree Theorem n Full Tree theorem: A full m-ary tree with k internal vertices has mk + 1 vertices. Let T=(V,E) be a full m-ary tree, with k internal vertices. Total number of vertices in T Number of vertices that HAVE a parent Number of vertices that DO NOT HAVE a parent = + Q: How many vertices HAVE a parent? 1.Observe for a 2-ary tree with 7 internal vertices 2. Each internal vertex has 2 children\n\n2.5 Leaves-Height Theorem. n Leaves-Height Theorem for Binary Trees: Let T=(V,E) be a binary tree that has t leaves, and height h. Then t 2 h. Proof: (by using induction on the height of the tree) Base Case: h = 0 T has 1 vertex, which is a leaf. t = 1 1 = 2 0 = 2 h Base case is true.\n\n2.5 Leaves-Height Theorem. n Leaves-Height Theorem for Binary Trees: Let T=(V,E) be a binary tree that has t leaves, and height h. Then t 2 h. Proof: (by using induction on the height of the tree) Inductive Case: Assume that t 2 h for h = 0,1,2,…,k (STRONG!) Let T be any binary tree of height k+1. (Need to show t 2 k+1 ) k 1 TLTL TRTR With respect to the root vertex, let the left and right subtrees be T L and T R respectively. Let the number of leaves in T L and T R be t L and t R respectively. (t = t L + t R ) Height of T L and T R are both < k+1. By inductive hypothesis, t L 2 (T L Height) 2 k and t R 2 (T R Height) 2 k. t = t L + t R 2 k + 2 k = 2 k+1.\n\n2.5 Leaves-Height Theorem. n Corollary to the Leaves-Height Theorem: Let T=(V,E) be a binary tree that has t leaves, and height h. Then log 2 t h. Proof: Using leaves-height theorem, we have t 2 h. Taking logarithms on both sides will yield log 2 t log 2 2 h log 2 t h\n\n2.5 Leaves-Height Theorem. IN GENERAL: n Leaves-Height Theorem for m-ary Trees: Let T=(V,E) be a m-ary tree that has t leaves, and height h. Then t m h n Corollary to the Leaves-Height Theorem: Let T=(V,E) be a m-ary tree that has t leaves, and height h. Then log m t h n Proof left as exercise (follows very closely to the proofs shown before)\n\n2.6 Examples n Q: Is there a binary tree that has height 5 and 38 external vertices? n A: No, since 38 > 2 5 which violates the leaves-height theorem.\n\n2.6 Examples n Q: Is there a full binary tree with 10 internal vertices and 13 external vertices? n A: No. Using the full-tree theorem, a binary tree with 10 internal vertices has 21 vertices in total. Therefore there should be 21-10 = 11 external vertices.\n\nOutline 1.Trees 1.1 Definition of a tree. 1.2 Examples of trees. 1.3 Theorem #1: Tree Characterization 2.Rooted Trees 2.1 Definitions 2.2 Definition: n-ary trees and binary trees. 2.3 Definition: Full n-ary trees 2.4 Theorem #2: Full Tree 2.5 Theorem #3: Leaves-Height 2.6 Examples 3.Spanning Trees 3.1 Motivation 3.2 Definition 3.3 Theorem #4: Spanning Tree 3.4 Minimum Spanning Tree (MST) 3.5 Kruskals Algorithm 3.6 Prims Algorithm\n\n3.1 Spanning Trees. n Example of use is in IP multicasting (studied in networking). n Do a web-search on the keywords IP multicasting spanning tree and more info will be available.\n\n3.1 Spanning Trees. n Layer 2 routing of packets through network switches. n Multiple connections from one switch to the rest of network to increase fault tolerance. n When all links are operational, redundacy in connection occurs. n Network forms a spanning tree so that packets will not be redundantly routed. R n Network will elect a root. n Root will broadcast packets to all other switches. n Each switch will select the best link to use.\n\n3.1 Spanning Trees. n Layer 2 routing of packets through network switches. n Multiple connections from one switch to the rest of network to increase fault tolerance. n When all links are operational, redundacy in connection occurs. n Network forms a spanning tree so that packets will not be redundantly routed. R n Network will elect a root. n Root will broadcast packets to all other switches. n Each switch will select the best link to use.\n\n3.1 Spanning Trees. n Layer 2 routing of packets through network switches. n Multiple connections from one switch to the rest of network to increase fault tolerance. n When all links are operational, redundacy in connection occurs. n Network forms a spanning tree so that packets will not be redundantly routed. R n Network will elect a root. n Root will broadcast packets to all other switches. n Each switch will select the best link to use.\n\n3.1 Spanning Trees. n Layer 2 routing of packets through network switches. n Multiple connections from one switch to the rest of network to increase fault tolerance. n When all links are operational, redundacy in connection occurs. n Network forms a spanning tree so that packets will not be redundantly routed. R n Network will elect a root. n Root will broadcast packets to all other switches. n Each switch will select the best link to use. n When a link goes down, the network reconfigures again.\n\n3.2 Spanning Trees (Definition). n Definition: Let G=(V,E). A spanning tree for G is a subgraph T=(V,E) of G, such that T is a tree and T contains every vertex of G. R\n\n3.3 Spanning Tree theorem. n Spanning Tree theorem: A graph is connected IFF it has a spanning tree. Proof: ( ) Assume that G=(V,E) is connected. We will show that G has a spanning tree. Step 1: Let H = G Step 2: while (H has a circuit C) { Step 2a: Remove an edge from C to form new graph H. Step 2b: Let H = H } Step 3: Output H. 1. Algorithm will terminate because G is finite and there is a finite number of edges to delete\n\n3.4 Minimum Spanning Tree A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 Let the following graph depict the scenario where the vertices are cities and the weighted edges are distances (km) between the cities. Lets say that the country wants to connect up the cities by building roads between them. The longer the road, the more money it has to spend. How do we connect up the cities and spend the LEAST AMOUNT OF MONEY? Ans: Find the MINIMUM spanning tree.\n\n3.4 Minimum Spanning Tree. n Definition: A weighted graph is a graph where each edge has a number associated with it. G = (V,E), E Z x { {x,y} \| x,y V} n The total weight of the graph is the sum of all the weights of the edges in the graph. n A minimum spanning tree (MST) for a weighted graph is a spanning tree that has the least possible total weight compared to all other spanning trees for the graph\n\n3.4 Minimum Spanning Tree. n How to find the minimum spanning tree? –Kruskals Algorithm –Prims Algorithm\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. Idea: To add edges of the smallest weights which do not cause a circuit.\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 m=0 \|V\|-1 = 12\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 m=1 \|V\|-1 = 12\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 m=2 \|V\|-1 = 12\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 m=3 \|V\|-1 = 12\n\n3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 m=4 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 8 10 6 12 m=7 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 8 10 6 12 m=8 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 8 10 6 12 m=9 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 8 10 6 12 m=10 \|V\|-1 = 12 CIRCUIT!!!\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 2 6 6 8 10 6 12 m=10 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 2 6 6 8 10 6 12 m=11 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 2 6 6 8 10 6 12 m=12 \|V\|-1 = 12\n\n3 3.5 MST: Kruskals Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), E={}, m=0 n 3. while (m < \|V\| - 1) { –a. Find edge e in E of least weight. –b. E = E - {e} –c. If E {e} does not produce circuit E = E {e} m = m + 1 } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 m=12 \|V\|-1 = 12 Algorithm Halts. Cost = 39\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. Idea: To grow a spanning tree.\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 \|V\| = 12 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 \|V\| = 12 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 \|V\| = 11 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 \|V\| = 11 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 12 \|V\| = 10 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 \|V\| = 9 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 \|V\| = 8 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 \|V\| = 7 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 4 2 6 6 3 8 10 6 \|V\| = 6 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 6 6 3 8 10 6 \|V\| = 5 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 6 6 3 8 10 \|V\| = 4 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 6 6 3 8 10 \|V\| = 3 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 6 6 3 10 \|V\| = 2 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 6 6 3 \|V\| = 1 > 0\n\n3.6 MST: Prims Algorithm n 1. Input: G=(V,E) n 2. Let T=(V,E), V={v}, E={}, m=0, V=V-{v} n 3. while (\|V\| > 0) { –a. Find an edge e such that e = {x,y}, x in V, y in V. e connects T to some vertex in V. e has least weight of all edges connecting T to a vertex in V. –b. V=V {y}, E = E {e}, V = V - {y} } n 4. Output T. A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 3 \|V\| = 0 Algorithm Halts. Cost = 39\n\nDifferent output possible 3 A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 A GH BCD I J K F L E M 5 5 2 2 2 3 3 4 4 4 2 3 Kruskals Algorithm Prims Algorithm Cost of Tree = 39" ]	[ null, "https://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82745594,"math_prob":0.99731326,"size":22262,"snap":"2021-31-2021-39","text_gpt3_token_len":8761,"char_repetition_ratio":0.14367868,"word_repetition_ratio":0.69610244,"special_character_ratio":0.41114905,"punctuation_ratio":0.15091108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995647,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T06:50:29Z\",\"WARC-Record-ID\":\"<urn:uuid:b0128af5-005f-4043-9061-857f1406d9ce>\",\"Content-Length\":\"253656\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67aea20f-6acb-4694-bfae-74d7e015f0d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:d142f4c7-3554-422d-a0d9-21a27dbc7b3e>\",\"WARC-IP-Address\":\"144.76.166.55\",\"WARC-Target-URI\":\"https://slideplayer.com/slide/234197/\",\"WARC-Payload-Digest\":\"sha1:RRG2WJVOXA4ONWCQDZZJXQCK6JP2C6H7\",\"WARC-Block-Digest\":\"sha1:Y4VRBNXMJXPEDLVSNAAPASPDTH47WAZ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055601.25_warc_CC-MAIN-20210917055515-20210917085515-00603.warc.gz\"}"}
https://www.colorhexa.com/770147	[ "# #770147 Color Information\n\nIn a RGB color space, hex #770147 is composed of 46.7% red, 0.4% green and 27.8% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 99.2% magenta, 40.3% yellow and 53.3% black. It has a hue angle of 324.4 degrees, a saturation of 98.3% and a lightness of 23.5%. #770147 color hex could be obtained by blending #ee028e with #000000. Closest websafe color is: #660033.\n\n• R 47\n• G 0\n• B 28\nRGB color chart\n• C 0\n• M 99\n• Y 40\n• K 53\nCMYK color chart\n\n#770147 color description : Dark pink.\n\n# #770147 Color Conversion\n\nThe hexadecimal color #770147 has RGB values of R:119, G:1, B:71 and CMYK values of C:0, M:0.99, Y:0.4, K:0.53. Its decimal value is 7799111.\n\nHex triplet RGB Decimal 770147 `#770147` 119, 1, 71 `rgb(119,1,71)` 46.7, 0.4, 27.8 `rgb(46.7%,0.4%,27.8%)` 0, 99, 40, 53 324.4°, 98.3, 23.5 `hsl(324.4,98.3%,23.5%)` 324.4°, 99.2, 46.7 660033 `#660033`\nCIE-LAB 24.951, 49.309, -6.951 8.756, 4.4, 6.349 0.449, 0.226, 4.4 24.951, 49.796, 351.976 24.951, 56.948, -14.982 20.975, 37.81, -3.264 01110111, 00000001, 01000111\n\n# Color Schemes with #770147\n\n• #770147\n``#770147` `rgb(119,1,71)``\n• #017731\n``#017731` `rgb(1,119,49)``\nComplementary Color\n• #6c0177\n``#6c0177` `rgb(108,1,119)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #77010c\n``#77010c` `rgb(119,1,12)``\nAnalogous Color\n• #01776c\n``#01776c` `rgb(1,119,108)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #0c7701\n``#0c7701` `rgb(12,119,1)``\nSplit Complementary Color\n• #014777\n``#014777` `rgb(1,71,119)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #477701\n``#477701` `rgb(71,119,1)``\n• #310177\n``#310177` `rgb(49,1,119)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #477701\n``#477701` `rgb(71,119,1)``\n• #017731\n``#017731` `rgb(1,119,49)``\n• #2b001a\n``#2b001a` `rgb(43,0,26)``\n• #440129\n``#440129` `rgb(68,1,41)``\n• #5e0138\n``#5e0138` `rgb(94,1,56)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #900156\n``#900156` `rgb(144,1,86)``\n• #aa0165\n``#aa0165` `rgb(170,1,101)``\n• #c30274\n``#c30274` `rgb(195,2,116)``\nMonochromatic Color\n\n# Alternatives to #770147\n\nBelow, you can see some colors close to #770147. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #770165\n``#770165` `rgb(119,1,101)``\n• #77015b\n``#77015b` `rgb(119,1,91)``\n• #770151\n``#770151` `rgb(119,1,81)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #77013d\n``#77013d` `rgb(119,1,61)``\n• #770133\n``#770133` `rgb(119,1,51)``\n• #770129\n``#770129` `rgb(119,1,41)``\nSimilar Colors\n\n# #770147 Preview\n\nThis text has a font color of #770147.\n\n``<span style=\"color:#770147;\">Text here</span>``\n#770147 background color\n\nThis paragraph has a background color of #770147.\n\n``<p style=\"background-color:#770147;\">Content here</p>``\n#770147 border color\n\nThis element has a border color of #770147.\n\n``<div style=\"border:1px solid #770147;\">Content here</div>``\nCSS codes\n``.text {color:#770147;}``\n``.background {background-color:#770147;}``\n``.border {border:1px solid #770147;}``\n\n# Shades and Tints of #770147\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, #020001 is the darkest color, while #ffeef8 is the lightest one.\n\n• #020001\n``#020001` `rgb(2,0,1)``\n• #16000d\n``#16000d` `rgb(22,0,13)``\n• #290019\n``#290019` `rgb(41,0,25)``\n• #3d0124\n``#3d0124` `rgb(61,1,36)``\n• #500130\n``#500130` `rgb(80,1,48)``\n• #64013b\n``#64013b` `rgb(100,1,59)``\n• #770147\n``#770147` `rgb(119,1,71)``\n• #8a0153\n``#8a0153` `rgb(138,1,83)``\n• #9e015e\n``#9e015e` `rgb(158,1,94)``\n• #b1016a\n``#b1016a` `rgb(177,1,106)``\n• #c50275\n``#c50275` `rgb(197,2,117)``\n• #d80281\n``#d80281` `rgb(216,2,129)``\n• #ec028d\n``#ec028d` `rgb(236,2,141)``\n• #fd0498\n``#fd0498` `rgb(253,4,152)``\n• #fd18a0\n``#fd18a0` `rgb(253,24,160)``\n• #fd2ba8\n``#fd2ba8` `rgb(253,43,168)``\n• #fd3fb0\n``#fd3fb0` `rgb(253,63,176)``\n• #fe52b8\n``#fe52b8` `rgb(254,82,184)``\n• #fe66c0\n``#fe66c0` `rgb(254,102,192)``\n• #fe79c8\n``#fe79c8` `rgb(254,121,200)``\n• #fe8dd0\n``#fe8dd0` `rgb(254,141,208)``\n• #fea0d8\n``#fea0d8` `rgb(254,160,216)``\n• #feb3e0\n``#feb3e0` `rgb(254,179,224)``\n• #ffc7e8\n``#ffc7e8` `rgb(255,199,232)``\n• #ffdaf0\n``#ffdaf0` `rgb(255,218,240)``\n• #ffeef8\n``#ffeef8` `rgb(255,238,248)``\nTint Color Variation\n\n# Tones of #770147\n\nA tone is produced by adding gray to any pure hue. In this case, #40383d is the less saturated color, while #770147 is the most saturated one.\n\n• #40383d\n``#40383d` `rgb(64,56,61)``\n• #44343e\n``#44343e` `rgb(68,52,62)``\n• #492f3e\n``#492f3e` `rgb(73,47,62)``\n• #4d2b3f\n``#4d2b3f` `rgb(77,43,63)``\n• #522640\n``#522640` `rgb(82,38,64)``\n• #572141\n``#572141` `rgb(87,33,65)``\n• #5b1d42\n``#5b1d42` `rgb(91,29,66)``\n• #601843\n``#601843` `rgb(96,24,67)``\n• #651344\n``#651344` `rgb(101,19,68)``\n• #690f44\n``#690f44` `rgb(105,15,68)``\n• #6e0a45\n``#6e0a45` `rgb(110,10,69)``\n• #720646\n``#720646` `rgb(114,6,70)``\n• #770147\n``#770147` `rgb(119,1,71)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #770147 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":0.50422996,"math_prob":0.46225977,"size":3654,"snap":"2021-31-2021-39","text_gpt3_token_len":1588,"char_repetition_ratio":0.13315068,"word_repetition_ratio":0.011111111,"special_character_ratio":0.57498634,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9928855,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T14:21:06Z\",\"WARC-Record-ID\":\"<urn:uuid:1150c774-294f-45df-bc59-f875d41ef404>\",\"Content-Length\":\"36088\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce913684-74b1-4adb-9ec6-b7e2f3bb0d34>\",\"WARC-Concurrent-To\":\"<urn:uuid:b391aea8-a4a7-4de9-aaac-387c96f092a0>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/770147\",\"WARC-Payload-Digest\":\"sha1:TC4FM4JLWDSJXSQ76POGNY2PT5X2B7UZ\",\"WARC-Block-Digest\":\"sha1:SKQBZY7GAUPJFYEIE2PT7UC3PFK4GMTS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058450.44_warc_CC-MAIN-20210927120736-20210927150736-00638.warc.gz\"}"}
http://mywcct.com/5th-grade-fraction-worksheets/	[ "", null, "Printables\n\n# 5th Grade Fraction Worksheets\n\nFractions worksheets printable for teachers worksheets. Fractions worksheets printable for teachers worksheets. Fractions worksheets printable for teachers worksheets. Grade 5 addition subtraction of fractions worksheets free adding worksheet. 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numbers free equivalent worksheet", null, "## Free fraction worksheets for 5th grade that and printables", null, "## Simplify fractions worksheet 5th grade 6th printable fraction worksheets subtracting fractions", null, "## 1000 images about worksheets on pinterest english for kids 5th grade math and printable multiplication worksheets", null, "## 5th grade fractions worksheets free printables education com math worksheet fraction fruit", null, "## Fraction worksheets 5th grade kids activities compare fractions", null, "## Adding fractions 1 5th grade fraction worksheets grace worksheets", null, "## How to divide fractions free printable fraction worksheets dividing 3", null, "## Worksheets for 5th grade scalien fractions scalien", null, "## Fractions worksheets printable for teachers the converting mixed to improper all math worksheet from page at", null, "## Fractions worksheets printable for teachers worksheets", null, "## Worksheets for 5th grade scalien fractions scalien", null, "## 1000 images about math fractions on pinterest 5th grade number worksheets and teaching", null, "## 5th grade addition worksheets davezan long division for math worksheets", null, "## Fractions worksheets printable for teachers worksheets", null, "## Fraction worksheets 5th grade kids activities reducing fractions click here free for grade", null, "## 5th grade fractions worksheets free printables education com fifth worksheet fraction review addition subtraction and inequalities", null, "Related Posts\n\n### Months Of The Year Worksheets", null, "" ]	[ null, "http://www.math-salamanders.com/image-files/printable-fraction-worksheets-subtracting-fractions-ld-3.gif", null, "http://www.math-aids.com/images/equivalent-fractions.png", null, "http://www.math-aids.com/images/fractions_visual.png", null, "http://www.math-aids.com/images/adding-two-fractions.png", null, "http://www.k5learning.com/sites/all/files/worksheets/math/grade-5-adding-fractions-worksheet.gif", null, 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https://ham.stackexchange.com/questions/2213/can-i-improve-reception-with-a-more-sensitive-antenna-preamplifier-better-fe?noredirect=1	[ "# Can I improve reception with a more sensitive antenna / preamplifier / better feedline?\n\nI am trying to receive a station, but I can only barely hear it through the noise. Will doing anything that increases the signal from my antenna help, such as:\n\n• using a more sensitive antenna\n• installing a shorter or higher quality feedline\n• improving SWR, thus reducing losses\n• AM, FM, SSB, CW, Digital? Can you control receiver band width? What frequency being received? What antenna type? What's the local time you listen? Do you have digital filtering? Any other info may help. – Optionparty Oct 8 '14 at 17:15\n• @Optionparty, I'm aware there are many things I could do to improve reception, in general. The question is if these particular things improve reception. – Phil Frost - W8II Oct 8 '14 at 20:27\n• All your suggestions are good. A directional antenna reduces noise reception from other directions. Preamp can increases S/N ratio. Low SWR really brings up a weak signal. Once signal & noise mix in the first stage of your receiver, they seem inseparable, like to much salt in soup. – Optionparty Oct 9 '14 at 12:20\n• Did you read the answer below, @Optionparty? These things only help under specific circumstances which aren't usually true. If you have comments regarding the answer, I'd love to hear them. – Phil Frost - W8II Oct 9 '14 at 15:18\n\nThese things will significantly help under one condition: the RF noise floor is below the receiver's noise floor. This is almost never the case on HF, where low-noise amplifiers are easy and atmospheric noise is high. On VHF and UHF the natural noise floor is lower, and low-noise receivers are more expensive, so it is possible but not guaranteed. It will depend on the quality of your receiver, and noise in that particular location and frequency. The 2.4 GHz ISM band, for example, is full of man-made noise.\n\nHere's a simple test: compare the noise floor with a dummy load versus an antenna connected. If the noise floor is higher with the antenna connected, then the RF noise floor is higher than the receiver noise floor, and a more sensitive antenna will not significantly help. In this case you need to investigate ways of increasing the signal to noise ratio in other ways, such as by increasing transmitter power or installing directional antennas.\n\nIf we think of the radio's output as the combination of three components:\n\n1. The desired signal, free of any noise\n2. RF noise received by the antenna. This could be atmospheric noise, interfering stations, or noise from electronics, powerlines, etc.\n3. Noise generated internally by the receiver. That is, thermal noise in its resistors, its amplifiers, etc.\n\nWe might think that the radio is contributing noise in all circumstances, and if we can boost the noise and signal from the antenna, then the radio's noise becomes relatively less significant, and SNR is improved. However, when the RF noise floor is above the receiver's noise floor, that improvement will be negligible.\n\nThe reason is that RF noise and the receiver's noise are uncorrelated, like random white noise. As I will demonstrate, even if the RF noise is just a little above the receiver noise, because of the way uncorrelated noise adds, the receiver's noise will make very little contribution to the total noise. As such, there is very little to be gained by reducing the receiver's noise, or equivalently, making the antenna \"louder\".\n\n## Orly? Show Me The Math.\n\n$$P_1 + P_2 = P_\\text{total}$$\n\nBut in this context our noise power is probably specified in dBm. To add these, we must first convert them to plain milliwatts (without the decibel), add, then convert back to dBm:\n\n$$10 \\log_{10}\\left( 10^{P_\\text{1(dBm)}/10} + 10^{P_\\text{2(dBm)}/10} \\right) = P_\\text{total(dBm)}$$\n\nSo for example, if the receiver noise floor is -96 dBm and the RF noise floor is 6 dB above that (-90 dBm), then the total noise floor is effectively:\n\n$$10 \\log_{10}\\left( 10^{-90\\:\\mathrm{dBm}/10} + 10^{-96\\:\\mathrm{dBm}/10} \\right) = -89.03\\:\\mathrm{dBm}$$\n\nSo, the receiver's noise, which was only 6 dB below the RF noise, increased the total noise floor from -90 to -89.03, or by 0.97 dB. Not a terribly big deal.\n\nAn equivalent, and perhaps more intuitive way to think about it is this: uncorrelated noise amplitudes (that is, RMS voltage or current) add like orthogonal vectors:", null, "Geometrically, we can see that as the RF noise becomes greater than the receiver noise, the triangle gets increasingly thin, and the total noise isn't much longer than the RF noise.\n\nMathematically, that's:\n\n$$E_\\text{total} = \\sqrt{E_1^2 + E_2^2}$$\n\nAnd we can convert between dBm ($P_\\text{dBm}$) and RMS volts ($E_\\text{RMS}$), assuming a 50Ω impedance, with:\n\n\\begin{align} E_\\text{RMS} &= \\sqrt{ 0.001\\:\\mathrm{W} \\cdot 50 \\:\\Omega \\cdot 10^{P_\\text{dBm}/10} }\\\\ &= \\sqrt{0.05\\:\\mathrm V^2} \\cdot 10^{P_\\text{dBm}/20} \\\\ &\\approx 0.2236\\:\\mathrm V \\cdot 10^{P_\\text{dBm}/20} \\end{align}\n\nand:\n\n\\begin{align} P_\\text{dBm} &= 10 \\log_{10}\\left( E_\\text{RMS}^2 / (0.001\\:\\mathrm W \\cdot 50 \\:\\Omega) \\right)\\\\ &= 20 \\log_{10}(E_\\text{RMS}) - 10 \\log_{10}(0.05\\:\\mathrm V^2) \\\\ &\\approx 20 \\log_{10}(E_\\text{RMS}) + 13.01 \\end{align}\n\nOne might still say that even this small improvement is an improvement, and is worth doing. However, there are two circumstances where that would be a bad idea:\n\nFirstly, if you are increasing your antenna sensitivity by adding a preamplifier, this preamplifier will introduce noise of its own. Unless the preamplifier's noise figure is lower than that of your receiver, or your receiver lacks sufficient gain to bring the RF noise floor above the receiver noise floor, then you are making things worse.\n\nSecondly, increasing the signal+noise from the antenna by any means in many situations decreases dynamic range. This is especially a concern with SDRs that are recently popular, because they have such a wide receive bandwidth. Consider that at some point, this signal in an SDR ends up at an analog to digital converter. Say it's a 16 bit converter. If the noise floor is high (because the antenna is so sensitive), then the least significant of these bits contain nothing but noise, and are essentially wasted. Furthermore, you approach overloading the converter and running into clipping." ]	[ null, "https://i.stack.imgur.com/XVrGc.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91226435,"math_prob":0.9432233,"size":4962,"snap":"2020-10-2020-16","text_gpt3_token_len":1298,"char_repetition_ratio":0.15631303,"word_repetition_ratio":0.007633588,"special_character_ratio":0.2682386,"punctuation_ratio":0.12487206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96205354,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-08T16:27:31Z\",\"WARC-Record-ID\":\"<urn:uuid:2794a615-1cca-4b32-8903-3817f7903ece>\",\"Content-Length\":\"150760\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8d7a885-6369-4fa8-86a5-034eb2189b66>\",\"WARC-Concurrent-To\":\"<urn:uuid:c287d254-a1ca-41d9-9e7d-1539b1391b93>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://ham.stackexchange.com/questions/2213/can-i-improve-reception-with-a-more-sensitive-antenna-preamplifier-better-fe?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:UUVIBKNCPUMH7UILVF7CLPCCLSSQ43RA\",\"WARC-Block-Digest\":\"sha1:3SCQBM47KAN4GFNLHHCTME7YXWAKU2C6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371818008.97_warc_CC-MAIN-20200408135412-20200408165912-00537.warc.gz\"}"}
https://physics.stackexchange.com/questions/386966/how-many-particles-are-there-in-the-entire-universe	[ "# How many particles are there in the entire universe? [duplicate]\n\nHow many particles are in the entire (rather than just the observable) universe?\n\nI would guess that if the universe is open, or infinite, then there is a countable infinity of particles. But if the universe is closed, or finite, then there is only a finite number of particles.\n\nIs this correct? Or could a finite universe have a countable infinity of particles or an infinite universe an uncountable infinity of particles.\n\n• I've read that the number of photons depends on the observer, in particular on the observer's acceleration. So maybe the answer would depend on the observer. Feb 17 '18 at 12:51\n• I suppose that's right. But I presume that such effects would not alter whether the answer is finite, countably or uncountably infinite?\n– Ben\nFeb 17 '18 at 12:57\n• Related: physics.stackexchange.com/q/98595/2451 and links therein. Feb 17 '18 at 13:10\n• Possible duplicate of Dumbed-down explanation how scientists know the number of atoms in the universe? Feb 17 '18 at 17:42\n• There is no simultaneity across the whole universe: one can not talk about the entire universe at a given time, like \"now\". So, strictly speaking, there is no number of particles to be associated with such an ill-defined physical system. Feb 18 '18 at 13:41\n\nIf the universe is infinite and is even vaguely homogeneous, the answer is trivially infinite. If the universe is finite, the number is finite, but we have no idea how big. The best we can do is set a lower limit based on the number of particles in the observable universe and the minimum size of the universe based on cosmological observations.\n\n• Thanks Chris - in the case that the universe is infinite, does it follow that the number of particles is COUNTABLY infinite? Or could they be UNCOUNTABLY infinite?\n– Ben\nFeb 20 '18 at 7:06\n• That depends on the topology of the universe, I guess? (Or to put it another way, it depends how infinite the universe is.) If the universe is a Lindelöf space, there should be a countable number. Otherwise uncountable. I have no idea if there is any way to tell that about the universe, though.\n– Chris\nFeb 20 '18 at 7:26\n• I'm inclined to think there would be no way experimentally to tell the difference, though.\n– Chris\nFeb 20 '18 at 7:52\n\nIf we talk of atoms then we can say that it is estimated that the there are between $10^{78}$ to $10^{82}$ atoms as per\n\nhttps://www.universetoday.com/36302/atoms-in-the-universe/amp/\n\nElse we can also state that :-\n\nThe answer to the question depends on what is meant by the universe. The standard cosmological model is that the universe is infinite. The only way the universe could be finite if it has a constant positive curvature, but the current measurement of the curvature implies that the universe is flat and therefore infinite.\n\nHowever, the observable universe is finite. The observable universe is the part of the universe that we can see - and since the universe is only 13.7 billion years old, we can only see photons that reach us in less than 13.7 billion years. Therefore the observable universe is defined as only the parts of the universe that are within 13.7 billion light years of us.\n\nThe commonly accepted answer for the number of particles in the observable universe is $10^{80}$. This number would include the total of the number of protons, neutrons, neutrinos and electrons.\n\nNow most of the photons in our universe are the photons from the cosmic microwave background radiation and it is estimated that there are $10^{9}$ photons for every particle in the universe so that would make $10^{89}$ photons in the universe.\n\nUntil we know what the dark matter particle is, we cannot make an accurate estimate of the number of dark matter particles. We do know that the total mass of the dark matter is about 6 times the mass of the particles in the universe. Currently, the favored theoretical candidate for the dark matter particle is the WIMP - the weakly interacting massive particle. These particles are assumed to be much heavier (x100?) than a proton, so if this is the dark matter particle then it would not significantly increase the number of particles in the universe. On the other hand, if the dark matter particle is the axion, it may be 1/1000th the mass of a proton (or less) so it could push up the particle count by several powers of 10.\n\nWe know even less about the dark energy in the universe, but the leading estimate is that it is \"just\" a small constant vacuum energy density. If the dark energy is just vacuum energy, then that would not increase the particle count for the universe.\n\nSource:-\n\nhttps://www.quora.com/How-many-particles-are-there-in-the-universe" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.935448,"math_prob":0.8423204,"size":2472,"snap":"2021-31-2021-39","text_gpt3_token_len":541,"char_repetition_ratio":0.18679093,"word_repetition_ratio":0.01946472,"special_character_ratio":0.22734627,"punctuation_ratio":0.08865979,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99124634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-29T01:07:26Z\",\"WARC-Record-ID\":\"<urn:uuid:edd579c3-e39e-4db4-8638-a59f1061318f>\",\"Content-Length\":\"173827\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0e15ced-145c-49ad-bfaa-1258c805d0f9>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8080656-a8d3-485f-aea8-b120f89129de>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/386966/how-many-particles-are-there-in-the-entire-universe\",\"WARC-Payload-Digest\":\"sha1:NLJX3KWXEZVVAEHBOO2RIH5Y6R7X76N4\",\"WARC-Block-Digest\":\"sha1:M7QHU7B2AVGNNPSVPIYGXVGC6YMGYIU4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780061350.42_warc_CC-MAIN-20210929004757-20210929034757-00131.warc.gz\"}"}
https://www.percentagecal.com/answer/what-is-percentage-difference-from-10.11-to-13.61	[ "What is the percentage increase/decrease\n\n#### Solution for What is the percentage increase/decrease from 10.11 to 13.61:\n\n(13.61-10.11):10.11100 =\n\n(13.61:10.11-1)100 =\n\n134.61918892186-100 = 34.61918892186\n\nNow we have: What is the percentage increase/decrease from 10.11 to 13.61 = 34.61918892186\n\nWhat is the percentage increase/decrease\n\n#### Solution for What is the percentage increase/decrease from 13.61 to 10.11:\n\n(10.11-13.61):13.61100 =\n\n(10.11:13.61-1)100 =\n\n74.283614988979-100 = -25.716385011021\n\nNow we have: What is the percentage increase/decrease from 13.61 to 10.11 = -25.716385011021\n\nCalculation Samples" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.876896,"math_prob":0.97373366,"size":1012,"snap":"2023-40-2023-50","text_gpt3_token_len":362,"char_repetition_ratio":0.23313493,"word_repetition_ratio":0.25316456,"special_character_ratio":0.5118577,"punctuation_ratio":0.17948718,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99880224,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T00:50:48Z\",\"WARC-Record-ID\":\"<urn:uuid:f1094883-1cd4-4647-a441-88f402c0c8b8>\",\"Content-Length\":\"10339\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0fea95a3-1d82-490f-93b3-97857be8420e>\",\"WARC-Concurrent-To\":\"<urn:uuid:b12a2e66-71e2-4242-87c3-511d66aebf34>\",\"WARC-IP-Address\":\"217.23.5.136\",\"WARC-Target-URI\":\"https://www.percentagecal.com/answer/what-is-percentage-difference-from-10.11-to-13.61\",\"WARC-Payload-Digest\":\"sha1:25ZIGYPUQQB67PNHN4HD7WA24IVBDB7E\",\"WARC-Block-Digest\":\"sha1:JFIRXCZ7YHE7JAP63WC22QG2T4C6AJUL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510462.75_warc_CC-MAIN-20230928230810-20230929020810-00093.warc.gz\"}"}
https://id.scribd.com/document/82803617/A-Parallel-Ant-Colony-Optimization-Algorithm-for-the-Salesman-Problem	[ "Anda di halaman 1dari 74\n\n# A PARALLEL ANT COLONY OPTIMIZATION ALGORITHM FOR THE TRAVELLING SALESMAN PROBLEM: IMPROVING PERFORMANCE USING CUDA\n\nby Octavian Nitica\n\nA thesis submitted to the Faculty of the University of Delaware in partial fulfillment of the requirements for the degree of Honors Bachelor of Science in Computer and Information Science with Distinction.\n\nSpring 2011\n\nA PARALLEL ANT COLONY OPTIMIZATION AGLORITHM FOR TRAVELLING SALESMAN PROBLEM: IMPROVING PERFORMANCE USING CUDA\n\nby Octavian Nitica\n\nApproved:\n\n__________________________________________________________ John Cavazos, Ph.D. Professor in charge of thesis on behalf of the Advisory Committee\n\nApproved:\n\n__________________________________________________________ Louis Rossi, Ph.D. Committee member from the Department of Mathematical Sciences\n\nApproved:\n\n__________________________________________________________ Pak-Wing Fok, Ph.D. Committee member from the Board of Senior Thesis Readers\n\nApproved:\n\n## __________________________________________________________ Michael Arnold, Ph.D. Director of the University Honors Program\n\nACKNOWLEDGMENTS I would like to thank John Cavazos for helping me as an advisor and mentor here at Delaware. Your help has been invaluable to the development of me as a researcher. I would like to thank Scott-Grauer Gray for helping me with CUDA programming and with the development of the multi-GPU portion of the work. I would like to thank the UD PetaApps Cloud Physics group (especially Lou Rossi and Lian-Ping Wang) for funding me the past summer and helping me get started on this research. Finally, I would like to thank my friends and family for their constant support and for making college an unforgettable experience.\n\niii\n\nTABLE OF CONTENTS LIST OF TABLES ........................................................................................................ vii LIST OF FIGURES ..................................................................................................... viii ABSTRACT ................................................................................................................... x CHAPTERS 1 INTRODUCTION ....................................................................................................... 1 2 BACKGROUND AND MOTIVATION ..................................................................... 3 2.1 2.2 2.3 NP-Hard Problems..................................................................................... 3 Search Algorithms ..................................................................................... 5 Travelling Salesman Problem .................................................................... 7 2.3.1 2.4 Nearest Neighbor List .................................................................... 8\n\n## Heuristics ................................................................................................. 10 2.4.1 A Suboptimal Heuristic Solution ................................................ 11\n\n2.5 2.6\n\nMetaheuristics.......................................................................................... 13 Ant Colony Optimization ........................................................................ 14 2.6.1 2.6.2 2.6.3 2.6.4 Termination Condition ................................................................ 15 Construction Stage ...................................................................... 16 Pheromone Update ...................................................................... 17 Daemon Actions .......................................................................... 19\n\niv\n\n2.7\n\n## Parallel ACO ........................................................................................... 19\n\n3 GPU COMPUTING .................................................................................................. 22 3.1 3.2 SIMT Model ............................................................................................ 23 GPU Memory .......................................................................................... 25 3.2.1 3.3 3.4 Memory Transfer ......................................................................... 28\n\n## Synchronization Issues ............................................................................ 28 CUDA for Evolutionary Computing ....................................................... 29\n\n4 INITIAL IMPLEMENTATION ................................................................................ 30 4.1 4.2 4.3 Analyzing the Original Code ................................................................... 30 TSPLIB .................................................................................................... 31 Issues in GPU Programming ................................................................... 32 4.3.1 4.3.2 4.3.3 4.4 4.5 4.6 4.7 4.8 Memory Coalescing..................................................................... 32 Thread Divergence ...................................................................... 33 Random Number Generation ....................................................... 34\n\nAbstraction of Implementation ................................................................ 35 Platform ................................................................................................... 36 Experiments and Results ......................................................................... 36 Asymmetric TSP Results ......................................................................... 38 Analysis of Results .................................................................................. 40\n\n## Pheromone Updating ............................................................................... 43 Second Implementation Results .............................................................. 44 Solution Quality....................................................................................... 45\n\n6 MULTI-GPU IMPLEMENTATION ......................................................................... 48 6.1 6.2 6.3 Threading ................................................................................................. 49 Multi-GPU Results .................................................................................. 50 Improving the Multi-GPU Implementation ............................................. 52\n\n7 RELATED AND FUTURE WORK.......................................................................... 54 7.1 7.2 Related Work ........................................................................................... 54 Future Work............................................................................................. 55 7.2.1 ACO Instruction Scheduling ....................................................... 56\n\n## 8 CONCLUSION.......................................................................................................... 58 BIBLIOGRAPHY ......................................................................................................... 59\n\nvi\n\nLIST OF TABLES Table 1: A table showing the max and average RPD results for the second implementation on different problem sizes. ............................................ 46 Table 2: Shows the RPD for the multi-GPU implementation on a trial run of different problems.................................................................................... 52\n\nvii\n\nLIST OF FIGURES Figure 1: An example of a weighted graph. ................................................................. 11 Figure 2: A step-through of a greedy algorithm on the graph. ..................................... 12 Figure 3: An example of coalesced versus uncoalesced memory access. .................... 27 Figure 4: Shows how the memory from a structure may be mapped into a single array for use on a GPU. .......................................................................... 33 Figure 5: Pseudo-code of the algorithm when running the construction stage on the GPU. ................................................................................................. 35 Figure 6: Speedup gained from running the initial implementation over the sequential algorithm on different problem sizes. Each graph represents a different number of ants. .................................................... 38 Figure 7: Speedup gained from running the initial implementation over the sequential algorithm on asymmetric problems. Each graph represents a different number of ants. .................................................... 39 Figure 8: Speedup gained from the second implementation over the sequential implementation. Each graph represents a different number of ants. ...... 45\n\nviii\n\nFigure 9: A graph showing the number of iterations per a minute the multi-GPU version runs at for different number of GPUs. ....................................... 51\n\nix\n\nABSTRACT\n\nThe ant colony optimization (ACO) algorithm is a metaheuristic algorithm used for combinatorial optimization problems. It is a good choice for many hard combinatorial problems because it is more efficient that brute force methods and produces better solutions than greedy algorithms. However, ACO is computationally expensive, and it can still take a long time to compute a solution on large problem sets. Fortunately, the structure of the ACO algorithm is amenable to being parallelized. By using CUDA to implement an ACO algorithm, I achieved significant improvement in performance over a highly-tuned sequential CPU implementation. I propose several different ways to improve the performance of the ACO algorithm using GPUs, including a multi-GPU approach. I ran the CUDA-parallelized ACO algorithm on the travelling salesman problem (TSP) problem and compared our performance to a highly-tuned sequential CPU implementation.\n\nChapter 1 INTRODUCTION The main topic of this thesis is to improve the performance of the ACO algorithm using Graphics Processing Units (GPUs). I use the CUDA platform to program Nvidia GPUs to achieve this goal, and my work involves applying ACO to the Travelling Salesman Problem (TSP). The main contribution of this thesis is a faster ACO implementation for the TSP and a framework for future improvement on this problem. In another aspect of this thesis, I explore many issues with porting such an algorithm to a GPU. It is also my goal to show that ACO algorithms are amenable to being run in parallel on GPU architectures. The second chapter deals with background to the algorithm and the problem it applies to, and why such research might be relevant. Next, the third chapter deals with GPU computing and explains programming with CUDA. I use the well known ACOTSP package as a baseline to compare against, and I also port several stages of this package to run on the GPU. The fourth chapter explains my first implementation of the work, where I port the construction stage of the algorithm in the ACOTSP package. My second implementation is detailed in the fifth chapter, where I port all the stages of the ACO algorithm to the GPU. The sixth chapter covers my work in\n\napplying a multi-GPU approach, where the algorithm runs on several GPUs at once to further improve execution time. The seventh chapter analyzes some related work and outlines an option for future work by applying the ACO algorithm to instruction scheduling. Finally, the thesis ends with a chapter containing my conclusions.\n\nChapter 2 BACKGROUND AND MOTIVATION In this thesis a parallel ant colony optimization algorithm is applied to the Travelling Salesman Problem. This section highlights the reason behind why such an implementation is beneficial. It discusses the difficulty of solving NP-Hard problems such as the Travelling Salesman Problem. It also covers the reasoning behind search algorithms such as the ant colony algorithm and why such an algorithm might be applied. Finally, it brings up why such an algorithm might benefit from execution on parallel platforms such as GPUs. 2.1 NP-Hard Problems\n\nComplexity classes can be thought of as problems that are relatively the same difficulty to solve in terms of resources. In this thesis, the term is used exclusively in regards to computation (or clock cycles used). The ACO algorithm was developed to solve problems belonging to the complexity class known as NP-Hard. Problems belonging to this class are often not possible to solve with brute force calculation due to the scaling of computation in regards to input. A notation called Big O is commonly used to represent algorithm complexity. A simple way to look at Big O notation is it is a representation of the approximate running time of an algorithm with respect to its input. Big O notation does not deal with absolute notions, but a Big O running time\n\ndenotes the upper bound of computation ignoring constants. In other words, Big O denotes the worst-case running time with respect to some input. Since Big O is relative to the input, it uses the input size as part of the notation. Let n denote the size of input given to an algorithm. An algorithm of O(n) complexity has linear complexity. This means the algorithm runs approximately c computations for every input element where c is a constant. An example of this would be an algorithm to sum all the elements in an array. Every extra element x adds one more iteration the algorithm would have to run to add x to the total. In this case the algorithm has a c value of 1. Now an algorithm of complexity O(nm), where m is a constant and m > 1, has polynomial time complexity. These types of algorithms perform at worst-case cnm iterations to solve a problem for an input size of n, where c and m are constants. A problem belonging to the NP complexity class means that a solution to the problem can be verified to be correct in polynomial time. NP-Hard problems are at least as hard as problems belonging to the NP complexity class. While solutions in this complexity class may be checked fairly rapidly, they may have much larger computational requirements for generating all the solutions of a problem. If all the solutions to a problem are not generated, it may be impossible to guarantee the best solution has been found. A solution with the best candidate solution value to a problem is often referred to as the optimal solution. A problem may have more than one optimal solution. Take for example when finding the shortest path between two\n\nvertices in a graph: there may be two different routes that have the same lowest cost value. Many algorithms have been developed to compute good, yet not optimal, solutions to NP-Hard problems in a reasonable amount of time. For more information on algorithm complexity, please refer to the following book, Introduction to Algorithms (pgs. 41-61, 966-986).\n\n2.2\n\nSearch Algorithms\n\nThe type of algorithm used in this paper to find a good solution is called a search algorithm. A search algorithm tries to find an object with specified properties among a collection of objects. The set of all candidate solutions to a problem that a search algorithm tries to solve is called the search space. I give a more formal definition here. Let D = {d1, d2, ..., dn} be a set of data elements. Then let S = {s1, s2, ..., sm} define the search space, where every element sm is a subset of D related by some mathematical formula or procedure. A search algorithm constructs objects in the search space (also called candidate solutions) from the data and then finds the object sbest which has the best value. The method that solutions are better than others is problem specific. The basic idea is to enumerate a value for every candidate solution so that it is comparable to other solutions. Once solutions are comparable, it is a straightforward to choose the best one.\nAnother important thing to note about search algorithms is that the search space is often represented as a graph. Taking our definition of grouping you can think of each data element\n\nas a node in the graph. The mathematical relation between the nodes represents the edges in the graph. Then a valid path through the graph is a candidate solution to the problem. The properties of the graph are dependent on the problem the search algorithm is attempting to solve. Looking at search spaces this way also helps in developing search algorithms. Also, for many problems, it may not even be known how many data elements are in a solution. Therefore trying to generate solutions with a holistic approach is difficult. The graph visualization makes an iterative approach intuitive with the idea of expanding the graph. An algorithm starts at an arbitrary data element. It then finds all the possible nodes (called available nodes) that it can use as the next element in a candidate solution at that step. Then it selects one element from the set of available nodes and repeats the process until a solution is constructed. For a more detailed example of expanding graphs, please refer to Chapter 3 of Artificial Intelligence: A Modern Approach .\n\nA basic type of search algorithm is an uninformed search. This type of search finds a solution with little to no information. An exhaustive search, which is an uninformed search, simply enumerates every candidate solution iteratively and then selects the best one (also known as a brute force approach). Since an exhaustive search checks every candidate solution to a problem, it is guaranteed to find the best solution. However, a search space can be too large to compute all the solutions in a reasonable amount of time. This leads to the idea of reducing the search space with smarter algorithms. By reducing the search space in intelligent ways, someone can still get a good solution while being able to solve the problem within a feasible time frame. This is the benefit behind using heuristic and metaheuristic functions.\n\n2.3\n\n## Travelling Salesman Problem\n\nA classic example of a computer science problem is the Travelling Salesman Problem (TSP). The problem is to find the shortest possible cycle through a graph of cities that visits each city exactly once. Let G = (V,E) be a graph where V is a set of vertices and E is a set of edges. Then, the travelling salesman problem can be represented as G, where every element v1, v2, .., vn belonging to V represents a city in the problem and every element e1, e2, ..., en belonging to E represents an edge between two of the cities. The graph has several properties. For example, it is weighted. This means that every edge belonging to E has some value x, that makes it comparable to other edges (in this case, the value is an integer denoting physical distance between the cities). Because it is weighted, the total distance of a cycle can be computed by summing all of the edges in a candidate solution. Also, this means a solution can be checked if it is the shortest cycle in linear time by comparing the computed solution cycle value against the shortest cycle value. The graph is also strongly-connected. That means for every two vertices vx and vy belonging to V, there exists and edge exy belonging to E. Thus, you can get to any city from any other city in the problem. Another interesting property of the graph is the symmetry of the edges. A TSP can be symmetric, which means the edges are undirected. Let A (vA) and B (vB) be two cities in the problem. Then if the problem is symmetric, the distance to get from A to B (eAB) is the same as the distance from B to A (eBA). However, a TSP can also be asymmetric, in which the edges are directed. This\n\nmeans the value of the edge from city A to city B differs from the value of the edge from B to A. The TSP falls into the category of NP-Hard problems. As said before, a cycle can be checked if it is the shortest cycle in linear time given a solution path and the shortest path value. However, there are an extremely large number of possible cycles since the graph is strongly-connected. In fact, given the number of cities in the problem as n, there are exactly n! candidate solution cycles! Thus, every single cycle would have to be checked to guarantee finding the optimal solution. A basic brute force algorithm would have complexity O(n!), which means that for an input of just 100 cities you would need to check ~10158 cycles. Even with dynamic programming, where similar segments of cycles are stored in memory and are not recalculated for every path, you would have a complexity of O(cn). This makes the problem infeasible to solve with a brute force search. Therefore reducing the search space for TSPs is crucial.\n\n2.3.1\n\n## Nearest Neighbor List\n\nThe nearest neighbor list is a way to reduce computational and possibly memory overhead when computing a tour (or in other words, a cycle through the all the cities) in the TSP instance. The nearest neighbor list is obtained by sorting all the distances from one city to all the other cities in the problem. Then a subset of lowest cost cities, which we can denote by nn, is taken. This number is typically between 15\n\nand 40 . There are several tradeoffs for using the nearest neighbor list. For one, it can reduce the complexity of some of the steps from linear complexity (O(n)) to a constant complexity (O(1)). This is because only the cities in the nearest neighbor list are searched instead of all the cities in the graph. Only if all the nearest neighbors have been chosen must the rest of the graph be searched. This helps reduce computational overhead. The use of a nearest neighbor list can also affect memory overhead. By only storing the costs of the nearest neighbors, and then computing other distances only when necessary, a large amount of memory overhead can be avoided. However, my approach does not use this method because it is concerned with optimal performance. An important note when using a nearest neighbor list is that the solution quality may degrade. Logically, an optimal tour will uses as many lowest cost edges (or the lowest cost edge available) as possible. In fact, many heuristic functions use this assumption as the basis for generating a solution. By taking a subset of lowest cost edges to search, the whole optimal tour or at least a majority of it is usually included in the search space. However, if the nearest neighbor list is too small, it can be impossible to find the optimal tour because it ends up lying outside the search space. Using a neighbor list is a heuristic, and there is no guarantee of optimality if one is used.\n\n2.4\n\nHeuristics\n\nHeuristics are used for informed search, where informed guesses are made to reduce the search space. Let the search space be an arbitrary graph. As defined before, at each step of constructing a candidate solution there exist a set of available nodes. A heuristic function ranks the available nodes using problem specific information. So a heuristic function f(n) takes in an node x off the available node list and gives it an enumerable value. This information can then be used to reduce the search space to one candidate solution. While this solution is often not guaranteed to be optimal, heuristic functions often produce pretty good results. A big factor in how good a heuristic algorithm will be is the quality of the heuristic used and how good the rankings are. A poor heuristic may produce a solution that is far from optimal. One very popular heuristic algorithm is the greedy best-first choice. A greedy algorithm is one that always makes the locally optimal choice. Choosing the heuristically best ranked node from the list of available nodes is called making the locally optimal choice, but it may not lead to the best overall solution to the problem, or making the globally optimal choice. After ranking the available nodes a greedy algorithm always chooses the node with the best ranking. An example of a greedy algorithm would be using the distances between cities in the TSP as a heuristic, and then always choosing the closest city as the next one in the tour. The idea of best-first means that the next node is chosen according to a specific rule. An example rule involves always selecting the node predicted to have the least cost to a solution if\n\n10\n\nchosen. Some best-first searches can involve backtracking and keeping track of past available nodes. A greedy best-first choice combines the two ideas: rank the currently available nodes by a heuristic that predicts the lowest cost to the final solution, and then always make the greedy choice and choose the best ranked node.\n\n2.4.1\n\n## A Suboptimal Heuristic Solution\n\nWe now go through an example where a greedy search may not give the best global optimal solution because of local optima. Figures 1 and 2 on the next page depict an example problem.\n\n11\n\n## Figure 2: A step-through of a greedy algorithm on the graph.\n\nFigure 1 denotes a simple problem graph. The graph above could easily represent a TSP graph, where each node is a city. The best solution is the cycle with the lowest cost. The heuristic used here is unimportant, other than the fact that it gives an enumerable value to each edge. When using a greedy search, the available edge with the lowest possible value is always chosen. Figure 2 steps through the choices the greedy search will make to create a cycle starting from node A. The final cycle created using the greedy search is A->B->D->C->A, with a total value of 13. However, the best cycle starting at A is A->C->B->D->A, with a total value of 12. This highlights a fundamental problem of many heuristic functions, the best choice made at each local node, may not be the best with respect to the global problem.\n\n12\n\n2.5\n\nMetaheuristics Metaheuristics are algorithms that use an underlying heuristic function and try\n\nto improve the solution quality. They attempt to diversify the search space and avoid converging to a poor solution because of local optima. Metaheuristics also iteratively try to improve their solution quality, which means that they do not terminate after just generating one solution. They will generate a number of candidate solutions, evaluate the respective solution qualities, and then use the information gained to improve the quality when generating new candidate solutions. There are many different types of metaheuristics, including but not limited to simulated annealing, particle swarm optimization, genetic algorithms, and ant colony optimization . An important method that many metaheuristic searches use is the idea of stochastic optimization. This means that random variables are used when constructing solutions. This is a common tactic to diversify the search space. By adding randomness when choosing the elements of a solution, local optima can be avoided, and it is more likely a global optimum will be found. This is because the best value determined by the underlying heuristic is not always chosen. The Ant Colony Optimization algorithm is a metaheuristic that uses the idea of stochastic optimization to improve the solution quality.\n\n13\n\n2.6\n\n## Ant Colony Optimization\n\nThe Ant Colony Optimization (ACO) algorithm gets its name from real ants, because it models the way ants search for food. Ants in the real world begin by randomly searching for food. As ants find sources of food, they leave pheromone trails that allow other ants to find the food. Ants are influenced to travel along paths with pheromone trails. Over time, the closest food source will develop a stronger pheromone trail than other food sources. This is because ants can travel to the closest food location and back faster, depositing a stronger pheromone trail than other food sources. Since the trail is stronger, more ants will be influenced to travel along the path, further reinforcing the trail until the majority of ants converge to the trail. This is how ants find the closest food source. Pheromones also evaporate over time, which is important so ants do not continue to go to the same food source after it has disappeared. The ACO algorithm is a metaheuristic that models how real ants find food when generating a solution to a problem. The ACO algorithm works in a similar way. It has two main stages: the construction stage and the pheromone update stage. There is also a daemon actions stage, which allows for statistical observation or adding additional heuristics, but this is not crucial to the algorithm. In the construction stage, individual ants construct a candidate solution to the problem using a probability function. The function uses a heuristic function and the amount of pheromone on the edges to decide which city to choose next. Once all ants have constructed their respective solution, the ants enter the\n\n14\n\nnext stage. In the pheromone update stage, certain solutions (generally the best ones), deposit pheromones on the edges of their solutions. Also, old pheromone trails have their potency decreased during this stage to prevent early convergence to a suboptimal solution. After a certain time period, ants will converge to a near-optimal path through the graph, just like real ants converge onto a food source. The following is pseudo-code give for the ACO algorithm: Function ACO_metaheuristic while(!termination_condition_met) { constructSolutions(); daemonActions(); pheromoneUpdate(); } I will refer to one pass through this loop as an iteration of the ACO algorithm.\n\n2.6.1\n\nTermination Condition\n\nThe termination condition signifies when the algorithm should finish. Since the ACO algorithm attempts to improve on previous iterations, there are many termination conditions that could be applicable. In fact, for all metaheuristics, there is no general termination condition . Therefore, there have been many different termination strategies developed. One possible termination condition is to bound the running time of the algorithm with a time limit, and finish after a certain amount of time has elapsed. Another possible condition is to terminate after a solution has been generated within a percentage deviation from a given optimum value. This approach requires knowledge of the problem before running the ACO algorithm, which may not always\n\n15\n\nbe available. Another possible termination condition involves setting a bound for a maximum number of iterations completed without an improvement in solution quality. If a significant number of iterations complete without improvement to the best solution found, it is likely the algorithm has converged to a solution. All of these are possible termination conditions for algorithms attempting to solve a TSP.\n\n2.6.2\n\nConstruction Stage\n\nIn this stage, each individual ant in the algorithm constructs a candidate solution to the problem. An ant is a problem specific data structure that can generate and retain the data of a completed candidate solution. The number of ants in an ACO algorithm run determines how many candidate solutions are constructed in every iteration. An important feature of the construction stage is that each individual solution can be constructed independantly, which means there is no communication between ants when they generate a solution. Ants are randomly placed at a starting node in the search graph, and then nodes are added to the candidate solution using a stochastic function until a complete tour has been generated. The stochastic function, is controlled by two variables, alpha and beta. Alpha is how heavily the pheromone trails are weighted in choosing the next node. Beta is how strong the orignal heuristic underlying the ACO algorithm is weighted in choosing the next node. For the function to work properly, it requires at least two matrices, one which has the heuristic values between nodes (the heuristic matrix) and one that stores the pheromone values\n\n16\n\nbetween nodes (the pheromone matrix). Now the stochastic function for applying ACO to a TSP can be generated. The heuristic matrix for a TSP stores the distances between cities. An ant k will rank the probability of choosing city y from current city x with the following function:\n\nIn the above function, denotes alpha, N denotes the set of unvisited cities for ant k and is the pheromone matrix. xy can be calculated a priori with the following equation: xy = 1/dxy where dxy signifies the distance between cities x and y. The summation at the bottom sums the remaining probabilites of the non-visited cities. This is done so that the probabilities correspond to a traditional sample space. In other words , where p(y) is the corresponding probability for visiting the\n\ncity. Each ant ranks the remaining cities using this stochastic function, chooses one city by generating a random number between (0,1), and then repeats until all the cities in the TSP have been visited.\n\n2.6.3\n\nPheromone Update\n\nThis stage updates the pheromone matrix using information generated from the construction stage. There are two stages to this stage, evaporation and intensification. In the evaporation stage, every pheromone edge is decremented by some value rho. Rho is always defined between (0,1). So every edge in the pheromone matrix, xy, is\n\n17\n\nmultiplied by rho to decrement its potency. Since early developed trails may be part of poor solutions, the process of evaporation allows these trails to be forgotten over time. The next stage, intensification, lays down new pheromones in a manner determined by the pheromone updating system. There has been much research into how exactly the pheromones should be updated, including differences in tours to use, relative strength of the tours, and general application of pheromone parameters. Some examples of ACO systems include classic AS , MAX-MIN , Elitist , and Rank-Based . We use the MAX-MIN system when developing the pheromone update stage on the GPU. The MAX-MIN system introduces two new parameters, max and min. The parameter max is an upper bound on the amount of pheromone an edge can have. When the pheromone matrix is first created, all the values in the solution space are intialized with value max. This allows for a more complete exploration of the search space, since initially all the edges are equally attractive with respect to their pheromone values. The value of min is a lower bound on the amount of pheromone deposited on an edge. By forcing all edges to keep at least a minimum pheromone value, it keeps all the edges in the search space to help prevent them from not being explored in later iterations. The MAX-MIN algorithm only deposits pheromone on the best tour from each iteration. The value of the pheromones it deposits are quantitively derived from the tours solution quality.\n\n18\n\n2.6.4\n\nDaemon Actions\n\nThe part of the algorithm named daemon actions refers to all functions that are not neccessarily part of the traditional ACO algorithm. This step can occur before or after the pheromone updating stage. Some packages may want to output statistical information on the algorithms performance during each iteration. Some ACO implementations may be used in conjunction with other optimization techniques, such as a local search . They may achieve better results using such techniques. All these types of activities fall under the category of daemon actions. In the implementations, we disable the daemon actions section of the ACOTSP package since it is not critical to the execution of the algorithm.\n\n2.7\n\nParallel ACO\n\nThe Max-Min ant system has complexity O(mn2) for TSP, where m is the number of ants and n is the number of cities in the problem . This is because the complexity of constructing each tour is approximately O(n2). This must be done m times since a tour must be constructed for every ant. This leads to a complexity of O(mn2) for the construction stage. All the other functions in the algorithm have smaller computational requirements than the construction stage. Evaluating the values of the tours is complexity O(mn), since each ant must sum every element in its constructed tour to get the tour length. The pheromone evaporation function is complexity O(n2), since every edge in the pheromone matrix must be updated, and there are n2 edges. The\n\n19\n\npheromone intensification function is a linear time function with complexity O(n), there will be one computation for each edge in the best tour, and the number of edges is equal to the number of cities in the tour. However, all of the other functions are much less than the O(mn2) complexity of the construction stage, and since Big O is an approximation, they are discarded in the overall complexity. Even though there seems to be a lot of work, all these functions put together are much less computation than a factorial O(n!) number of computations. However, the performance can be improved. Let p denote a number of processors available to run the algorithm in parallel. With perfect scaling among processors a theoretical runtime of O(m/pn2) can be achieved on each processor, if the work of constructing the tours is split evenly among the processors. Even taking into account the overhead from parallel computing, good results can still be achieved. There have been many different methods of parallelizing ACO algorithms . A set of ants sharing the same pheromone matrix and the same heuristic matrix is a colony. One way to parallelize the code is to split ants in the same colony up among several processors. Some ACO implementations, run multiple colonies with different parameters to diversify the search space even more. Another possibility involves decomposing the TSP domain. With this method, each processor p looks at generating a best path through a subset of cities, and then the segments are joined together into a tour. The theoretical complexity of the construction step on each processor would then be O(mq2), where q is the cardinality of the largest subset of cities. It is impossible to\n\n20\n\nguarantee that the optimal solution is found with this type of decomposition, since edges are removed, reducing the search space. The hope is that it is still possible to find good solutions much quicker with this type of decomposition. The city distribution and the size of the subsets is a big factor in how effective this method may be. Acan does an external memory approach which manages to get good results. The approach selects segments from already constructed good solutions, and then allows ants to construct the rest of the graph. However, constructing the rest of the graph after a segment is the same as searching a path in a subset of the cities for the TSP.\n\n21\n\nChapter 3 GPU COMPUTING Today, Central Processing Units, or CPUs, cannot handle all the computational requirements of modern day computer graphics. Almost all modern day computers have a Graphics Processing Unit (GPUs). GPUs are designed to have a large number of simple homogenous processors that can run many threads in parallel. This makes them a good choice for certain algorithms that can be phrased in a data-parallel fashion, where a GPU can yield a much higher price/performance than a comparably priced CPU. GPUs are also seen as useful for programs with high arithmetic intensity . That means programs that perform large amounts of arithmetic operations in contrast to memory operations will especially benefit from GPU implementation. Before the Compute Unified Device Architecture , (CUDA) several researchers attempted to do general purpose GPU (GPGPU) programming using programming languages and toolkits that were designed for graphics, such as OpenGL . However, with the advent of CUDA, programming GPUs to take advantage of their power for general-purpose applications has become much easier. This chapter explains development using CUDA and considerations for running on a modern day GPU architecture.\n\n22\n\n3.1\n\nSIMT Model\n\nCUDA programs have special functions, called kernels, that run on the GPU. Kernels are labeled with the __global__ keyword. They are called from the CPU side and are passed in two extra variables, the dimensions of the grid and the dimensions of the block. The grid variable is a one to three dimensional space that specifies the layout of the thread blocks, and the block variable is a one to three dimensional space that specifies the layout of threads within the blocks. For a more detailed account of this, it is recommended to read the NVIDIA CUDA Programming Guide . The variability in dimensions creates a large possibility of possible indexing options for the programmer. CUDA uses a Single Instruction Multiple Thread (SIMT) architecture, which is similar to the classic Single Instruction Multiple Data (SIMD) architecture . The SIMD architecture involves concurrently performing the same arthimetic instructions across an array of data. Typically, this makes it useful for applications that perform many of the same arithmetic calculations on different data, such as graphics processing or particle simulation. SIMT differentiates itself from SIMD for two reasons. One, the programming is done from a different perspective: Instead of programming the entire width of the data, SIMT can program individual threads as well as coordinated threads. Also, when there is a divergence between threads, which means that threads are\n\n23\n\n24\n\nunderstanding of thread execution, it is possible to achieve more on the GPU, such as task parallelism . Guevara et. al., manage to merge two different kernel tasks into one with proper indexing and memory strucuturing. This allowed for even increased performance by taking advantage of the massive amount of thread parallelism available. While it helps to think of CUDA from a SIMD perspective for performance reasons, CUDA maintains a high level of versatility in the types of applications that can be executed on a GPU.\n\n3.2\n\nGPU Memory\n\n25\n\n26\n\nthread. Using too much can limit multiprocessor occupancy, which may increase execution time. Grauer-Gray et. Al conducted a study observing the effects of multiprocessor occupancy and the tradeoffs in such situations. Threads should read memory in a coalesced manner. This means that all threads in a warp should access consective data elements in global memory. Shared memory is also more effective when accessed in a coalesced manner. Siegel et. al. suggest allocating memory in a structure of arrays (SoA) rather than an array of structures (AoS), so related data is contiguous in memory. They show a 30-50% improvement of performance between such memory schemes. The difference between SoA and AoS is illustrated in the figure below, where x, y, and z are data elements in a structure. In the first array, the data is structured in a SoA while in the second it is structured in a AoS.\n\n## Figure 3: An example of coalesced versus uncoalesced memory access.\n\n27\n\n3.2.1\n\nMemory Transfer\n\nAnother important aspect of memory coalescing is minimizing the amount of memory communication between the GPU and host. Since a memory transfer can take a significant amount of time, minimizing the amount of transfers is crucial to achieving good performance. For example, structures with pointers inside them cannot be transferred to and from the GPU as is, since pointers on the GPU cannot point into memory locations on the host device. It would be neccessary to individually copy every array from each struct to the GPU in such a case. This could lead to many memory copies per kernel run, which is highly inefficient. In my first CUDA implementation of ACO for the Travelling Salesman Problem, I solve this problem by consolidating the initial array allocation into one large array, and then having the structures point into the single array. Using this technique, the structures maintain the same functionality as the host side while minimizing memory transfers to one. The initial ACO implementation uses the technique of copying data between the host and the GPU, since it does not require an unmanageable amount of memory transfers. In my later implementation, I remove all memory transfers but one.\n\n3.3\n\nSynchronization Issues\n\nEven coarse-grained algorithms, which are especially suitable for parallization because they require little communication between threads, may need to synchronize at certain points. The easiest way to handle this problem is to transfer the memory back\n\n28\n\nto the host and handle the synchronization with CPU code. However, this may not always be ideal, especially if this causes a large amount of memory transfers between the host and the GPU. CUDA suports synchronization only within thread blocks. This is accomplised with the use of barriers, where all threads will stop computing and wait at a barrier until all the threads in a block have reached the same point. Synchronization in between thread blocks can be done with the use of multiple kernel calls. In subsequent kernels, a parallel reduction technique can be used to merge data from other thread blocks.\n\n3.4\n\n## CUDA for Evolutionary Computing\n\nMany researchers have begun to realize the benefits of using CUDA for paralleling their applications and reducing execution time. In the field of evolutionary computing, which the ACO algorithm falls under, several algorithms have shown notable gains using GPUs. For example, Tsutsui and Fujimoto showed a 3-12x speedup when using CUDA to parallelize a genetic algorithm. Franco, Krasnogor, and Bacardit achieved an improvement of over 50x when applying CUDA to the BioHEL evolutionary learning system. These types of performance gains are good and make using GPUs an interesting option for these types of algorithms.\n\n29\n\nChapter 4 INITIAL IMPLEMENTATION My initial implementation of the CUDA ACO algorithm for TSP consisted of porting only the construction stage of the algorithm, because it is where the majority of the computation occurs. Every ant in the ACO algorithm (which each creates one tour) is run as a thread within a CUDA kernel.\n\n4.1\n\n## Analyzing the Original Code\n\nMy initial parallel ACO implementation was done using CUDA and the ACOTSP package , a well-known ACO implementation. The ACOTSP package is a C implementation of the ACO algorithm used for solving Travelling Salesman Problems (TSP). The ACOTSP package was chosen because it is a highly-tuned sequential version of ACO and because it has many features, such as supporting multiple pheromone update functions and providing flexibility through command line options. It also provided a proven sequential version of code to compare my parallel implementation. Before porting the code, the relative computation time of the ACOTSPs functions was analyzed using the gprof profiler . It was found that the majority of\n\n30\n\ncomputation was spent in the construction phase of the algorithm, which took at least 80% of the execution time. After disabling functions to compute and output various non-essential statistics pertaining to the algorithms performance, analysis of the results found that the construction phase took over 90% of the execution time. Therefore the construction phase of the package was ported to CUDA. This allowed for improved performance while maintaining the ability to use different pheromone update functions as desired. Speedups of up to 10x were achieved with my initial parallel implementation over the sequential one, just from performing the construction stage in parallel.\n\n4.2\n\nTSPLIB\n\nThe experiments in this paper use problems exclusively from TSPLIB, a widely known library of TSP problems . The ACOTSP package supports parsing of the TSP files in this library. Another useful feature of TSPLIB is that the optimal solutions of the problems are documented within the library. Due to this being a popular source for TSPs, many other research papers explore the TSP instances defined within TSPLIB. Only strongly connected instances of TSP were chosen from the TSPLIB, though both symmetric and asymmetric TSPs were analyzed.\n\n31\n\n4.3\n\n## Issues in GPU Programming\n\nThis section describes some of the issues that arose when converting the sequential construction stage to run on the GPU.\n\n4.3.1\n\nMemory Coalescing\n\nA big issue with porting to CUDA is having memory coalesced for transfers between the host and the GPU. Since a memory transfer can take a significant amount of time, minimizing the amount of transfers is crucial to getting good performance. The ACOTSP package uses a C structure called ant_struct, which contains two pointers to arrays. One array holds the solution constructed for each ant, the other array is a flag array to determine what cities have been visited during the construction step. Transferring these structures as is would require an unreasonable amount of transfers, since each pointer needs to be transferred individually. Structures with pointers inside them cannot be transferred to and from the GPU as is, since pointers on the GPU cannot point into memory locations on the host device. Many memory transfers per iteration would lead to poor performance. To solve this problem, all fields in the structures were allocated as large consecutive arrays. Then, when the structures were created, a portion of this array was assigned to the pointer in the respective structure. Figure 4 below helps to illustrate this process. Then, the arrays can be copied in one memory transfer operation, and the proper sections can be referred to using offsets. This technique was also used for\n\n32\n\narrays of floats. This allowed for efficient memory transfers while maintaining compatibility with the original functions. Note that if improper indexing occurs, incorrect results can be obtained with buffer overflows. However, bugs such as these would generally lead to segmentation faults as well, so these bugs are easy to identify and correct.\n\nFigure 4: Shows how the memory from a structure may be mapped into a single array for use on a GPU.\n\n4.3.2\n\nAn important issue with the parallel ACO algorithm is that it is impossible to program the kernel to guarantee thread convergence. This is because of the stochastic nature of the algorithm. Since each ant/thread will have a different partial solution in memory, the future cities it will be checking are different for each thread. Then when accessing the probabilities to access each city from the memory, there is no guarantee\n\n33\n\nthose values are physically close in memory. One thread may be looking at one edge on the complete other side of the total matrix. Since the probability matrix is of size n2, where n is the number of cities in the TSP, accessing the edges in different areas of the matrix will lead to thread divergence. Therefore, thread divergence is likely unavoidable between the threads.\n\n4.3.3\n\n## Random Number Generation\n\nA key part of the ACO algorithm is the usage of a Random Number Generator (RNG) to select cities in the construction step. Random number generation in CUDA cannot have threads modifying the same seed to avoid race conditions. Also, every thread must have an original seed, otherwise each thread will calculate the same values. This is because if given the same starting seed a RNG will generate the same sequence of numbers. A good random number generator must also have a good distribution of values within the bounds given. The original ACOTSP package RNG used a common seed, which was not a problem with sequential code but did not work with a parallel implementation. For my initial implementation, a Linear Congruential Generator with a Combined Tausworthe Generator (as described in GPU Gems 3) was used . When the research began, there were not many RNG packages for GPU computing. In the second and multi-GPU implementations, I used NVIDIAs recently developed RNG package .\n\n34\n\n4.4\n\nAbstraction of Implementation\n\nMy program uses parallelism at level of each individual ant. That is, each ant is given its own thread to run on a stream processor. Each stream processor is part of a multiprocessor that executes a thread block. For performance reasons, I ran thread blocks of 256 threads (and therefore 256 ants). The probability matrix is copied to the GPU before each iteration is run. Each ant constructs its own solution to the TSP problem on the GPU, and then the GPU returns all the tours to the CPU for completion of the algorithm. Figure 5 shows the high-level structure of the CUDA version of the algorithm, which is similar to the ACO pseudo-code shown earlier. Before the main loop, structures and arrays are allocated to hold the pheromones and tours constructed by each individual ant. The loop in Figure 5 repeats until the time limit set for the algorithm has expired. My implementation uses time as the termination condition. Initialize_GPU_memory_and_constants(); while( !termination_condition() ) { copy_data_to_GPU(); // executes construction stage for each ant in colony runTSP(); copy_data_from_GPU(); daemon_actions(); // any supported pheromone update function // supported by the ACOTSP package pheromone_update(); } Figure 5: Pseudo-code of the algorithm when running the construction stage on the GPU.\n\n35\n\n4.5\n\nPlatform\n\nThe experiments were performed using three different GPUs, the TeslaC870 GPU, TeslaC2050 GPU, and the Tesla C1060 GPU. The Tesla line of cards is NVIDIAs first offering of general-purpose computing GPUs. The C870 is the first offering, which only supports CUDA compute capability 1.0 and has 128 stream processors split across 16 multiprocessors with 1.5GB of memory. The C1060 has 240 stream processors split among 30 multiprocessors and 4GB of memory. The C2050 is based on NVIDIAs new Fermi architecture, and has 448 stream processors among 14 multiprocessors and 3GB of memory. It supports compute capability 2.0. I use Xeon processors to execute the original sequential version of the ACOTSP package. Results were normalized using the Intel Xeon E5335 2.0Ghz processor paired with 2GB memory for the C870, and a Xeon E5530 2.4Ghz processor with 24GB of memory for the C2050 and C1060.\n\n4.6\n\n## Experiments and Results\n\nI varied two variables in our experiments: problem size and the size of the ant colony (i.e., the number of ants used to find tours). For each of these combinations, ten trials was performed on each platform. While the original ACOTSP package has support for several different types of pheromone update functions, I used only the max-min variation for my experiments. Even though my version works with the other types of pheromone update functions, I found the difference in speed and solution\n\n36\n\nquality was consistent between the different pheromone functions. For the ACO parameters, Alpha was set to 1, Beta was set to 2, and Rho was set to 0.5. The program had a max memory usage of 1GB. For my graphs, I test each problem/ant combination for a timed interval, and then average the results. Speedups were calculated by enumerating the number of iterations per second and then comparing the execution times for the same number of iterations between the sequential and parallel versions. Figure 6 shows my results for the symmetric problems. The results show a definite improvement in execution time when using our CUDA implementation over the sequential CPU implementation on large sized graphs and a high number of ants. As I increase the problem size the performance gap between the CPU implementation and the CUDA implementation also increases. Since the construction step in the ACO algorithm is general and applicable to many pheromone update functions, these speedups are achievable for several variations of the ACO algorithm.\n\n37\n\nFigure 6: Speedup gained from running the initial implementation over the sequential algorithm on different problem sizes. Each graph represents a different number of ants.\n\n4.7\n\n## Asymmetric TSP Results\n\nThe original ACOTSP package supported only symmetric TSP graphs. However, there are many asymmetric graph problems in TSPLIB. The package was then modified to support running on asymmetric TSPs. First the parser was changed to\n\n38\n\nsupport TSPLIBs file format for asymmetric problems. Different functions that assumed symmetry in the programs matrices, such as pheromone evaporation, were modified to support asymmetric problems. The results achieved for asymmetric graphs are posted below in Figure 7. I used the same methodology for experiments as the symmetric graphs.\n\nFigure 7: Speedup gained from running the initial implementation over the sequential algorithm on asymmetric problems. Each graph represents a different number of ants.\n\n39\n\n4.8\n\nAnalysis of Results\n\nI highlight some of the results show in Figures 6 and 7. For example, the C2050 does not always outperform the older cards, especially at smaller problem sizes of under 1000 cities. A possibility for this is that the extra computational power of the newer card is not being fully utilized at these sizes. However, for large city sizes, e.g., over 1000, the C2050 gets significant performance improvements over the sequential code, and its speedups are much greater than the other two cards. For symmetric problems and the smaller ant colony size of 1024, our GPUs only achieve a performance improvement of up to 3x. For larger city sizes and older GPU models, I see a slowdown over the sequential version. I am currently investigating this anomaly. However, for symmetric problems and ant colony sizes above 1024, I consistently see good speedups especially for the newer GPUs based on the Fermi architecture. I can achieve speedups of almost 11x compared to a highly-tuned sequential version of ACO. For all the ATSP problem sizes, the speedups are relatively similar between the three cards. Overall, the newer card Fermi-based GPU performs better than the older GPUs. A possible reason the Fermi-based GPUs may achieve such better performance at larger city sizes is they may handle thread divergence better. The Fermi cards have a new memory architecture that works better with divergent memory accesses between threads. For larger problem sizes, it becomes much more likely that threads will become divergent between memory accesses because there is a much larger memory\n\n40\n\nspace, and each thread can be accessing information anywhere in the probability matrix.\n\n41\n\nChapter 5 SECOND IMPLEMENTATION By porting the construction step to a GPU, there was definitely an improvement in performance. However, it is possible that even better speedup can be achieved. If an estimate that only 10% of the execution time is done sequentially outside the construction stage, the execution time is bounded by a maximum 10x speedup if Amdahls Law is applied. While the sequential execution time is really between 510%, the rest of the algorithm can still be optimized for a significant overall speedup. By porting the rest of the algorithm to the GPU there is a two-fold benefit. First of all, the parallelism in the remaining functions in the algorithm can be further exploited. For example, each ant can calculate its own tour value in parallel. Secondly, all but one memory transfer between the GPU and CPU can be eliminated. Only the best tour would ever need to be transferred back and forth. Both these factors should increase the overall performance of the ACO algorithm.\n\n5.1\n\n## Calculating the Best Tour\n\nOne part of the ACO algorithm that could be ported is the ability to find the best tour. To find the overall best tour on the GPU, we can have each ant must evaluate its\n\n42\n\nown tour value in parallel. This involves running a kernel that simply sums the distances values in each tour. The next step is to somehow compare these tour values in parallel to calculate the best one. To do this a parallel reduction step is necessary. This is somewhat tricky to do in our case for two reasons. First, indexing for the tour values must be preserved to be able to link it to the corresponding tour it represents. Secondly, there is no thread synchronization between blocks, which means multiple kernels must be used. To solve the problem of index preservation, a second array that holds the original index values of the tour values is used. Then when two tour values in the parallel reduction step are swapped, the original index values are swapped as well. In this manner the index of the tour the value corresponds to can be tracked. To solve the problem of thread synchronization, the reduction step first runs only between threads in each block. After that kernel completes, another kernel is executed that performs a reduction step on the previous kernels results to find the best overall tour.\n\n5.2\n\nPheromone Updating\n\nThe MAX-MIN update system was used on the GPU for updating the pheromones in each iteration; this algorithm uses the best ant from each iteration to update the values. Stutlze achieves good performance using this system and shows it is a good update strategy. When evaporating pheromones and maintaining trail bounds on the GPU,\n\n43\n\nk blocks of size of 32 are run, where k is the number of cities to the next power of 2 divided by 32. A global id is then generated by multiplying the block id times 32 plus the thread id within the block. Threads with global ids greater than the number of cities are ignored. Each thread updates n edges, where n is equal to the number of cities. For the global update function we again run block sizes of 32, and k blocks. In this function, every edge updates one edge along the best path.\n\n5.3\n\n## Second Implementation Results\n\nFigure 8 shows the speedup over the CPU implementation when running the whole algorithm on the C2050 GPU for a number TSP problems and utilizing a variety of ants counts ranging from 256 to 8192. Using 256 ants is interesting because at this number only one thread block is running on the GPU for the construction stage. This means only one multiproccesor is executing on the GPU, yet the performance achieved is very similar to the performance of the CPU version. The results show a definite improvement in execution time when using a CUDA implementation over the sequential CPU implementation on large sized input sets and when utilizing a large number of ants; the trial with the largest speedup of almost 16X uses the fnl4461 TSP city set as input with 8192 ants. The results show that when increasing the problem size, the performance gap between the CPU implementation and the CUDA implementation also increases. Also, performance may be limited by divergence between the threads. Due to the stochastic nature of the algorithm, threads\n\n44\n\nrunning in the same warp may be reading from completely different parts of the probability matrix.\n\nFigure 8: Speedup gained from the second implementation over the sequential implementation. Each graph represents a different number of ants.\n\n5.4\n\nSolution Quality\n\nSolution quality was tested to make sure the implementations were providing good solutions. Solution quality between the first and second implementation was similar.\n\n45\n\nFor testing solution quality, ten trials were run and capped the execution time at 5 minutes per a trial. Values of alpha = 1, beta = 2, and rho = 0.5 were used. Each trial ran with 1024 ants in testing the quality of the solutions. To compute how accurate the experimental results were, RPD, or the relative percent deviation was used. RPD is calculated using the following formula:\n\nThe costactual is the optimal tour value for the problem and the costbest is the tour value found by the ACO algorithm. The ACO algorithm was run for four different problems and the following results were found:\n\nTable 1: A table showing the max and average RPD results for the second implementation on different problem sizes. Both the max RPD and the average RPD are considered. The max RPD signifies the RPD for the best solution found out of all the trials. The average RPD is done using the average of the solutions found from all our trials for that problem. The results are\n\n46\n\nconsistent with the RPD achieved in both the sequential and the initial implementations for the same amount of iterations.\n\n47\n\nChapter 6 MULTI-GPU IMPLEMENTATION A possible improvement to my original implementations was a multi-GPU implementation. The idea behind a multi-GPU approach is to split up the work among several GPUs and then combine the results together. The initial implementation I developed of a multi-GPU approach was to split the work up of the ants. This meant that each GPU would run its own collection of ants. The main idea here is to see if the program benefits from further parallelism among ants. This approach could also be easily extended to a multi-colony approach. This would require giving each GPU kernel its own pheromone matrix. Then problem parameters, such as alpha, beta, rho, the number of neighbors, etc. could be different for each of the GPUs. Another approach to a multi-GPU implementation would be to split up the domain space. This would mean each GPU would create segments of a candidate solution and work on a subset of the cities. This could improve execution performance, since each GPU would be splitting the work. This would also allow execution of bigger problem sizes. Note, after reaching a TSP size of about 8000 cities, a 1GB GPU would runs of memory. However, by splitting the cities up into subsets, and finding a path for each subset on the GPU, bigger problem sizes may be possible. This is a big factor, since\n\n48\n\nbigger problems are where performance is more of an issue. This type of decomposition may have a negative effect on the solution quality, since it reduces the search space by not searching edges that exist between the subsets of cities.\n\n6.1\n\nWhen running a multi-GPU program, each kernel call must be on a separate CPU thread. This is because every time a kernel function is called, the thread calling it will block until the kernel terminates. Therefore multiple CPU threads are required to run multiple kernel calls. The CPU threading is OS dependent, but all that is needed is the ability to run CPU threads and then wait for them to terminate collectively. To distinguish between multiple GPUs, every GPU installed on a machine is designated by an integer value. There exist CUDA functions that allow for managing which GPU to execute the kernels on. When creating a multi-GPU implementation, every GPU needs its own device pointers for its problem dependent memory allocations. We handle this by encapsulating all the memory requirements for a kernel into a structure. In an initialization stage, determine the problem information each GPU will need and create the appropriate structure. Then pass the appropriate structure to each thread, let each thread initialize the GPU it will execute the kernel on, and then allocate the memory with the structure. Then each GPU should be able to run the correct kernel call.\n\n49\n\n6.2\n\nMulti-GPU Results\n\nWe implemented a multi-GPU version that splits up both the ants and the cities. The algorithm evenly splits the ants evenly among the GPUs. It also naively splits the cities among the GPUs based on their sequential location in the input file. So if there are n cities and x amount of GPUs, the first n/x cities in the input file are put on the first GPU, the next n/x on the second GPU, and so on until all the cities have been assigned to a GPU. The best paths within each subset of cities are then connected to form the whole tour after the algorithm runs. The existence of an edge between the subsets can be guaranteed because the graph is strongly connected, allowing for the construction of a complete tour. Scott Grauer-Gray, a graduate student at University of Delaware, assisted in the coding and development of the multi-GPU implementation. Scott developed the majority of the code that splits the work along the cities. This implementation should have the benefit of increased execution time and allow for testing of larger problem sizes. In our experiments, we run each trial for two minutes. The same values of alpha, beta, and rho are used as in our previous experiments (1, 2, and 0.5 respectively). The GPUs used were all Tesla C2050s. The following execution time results were obtained:\n\n50\n\nFigure 9: A graph showing the number of iterations per a minute the multi-GPU version runs at for different number of GPUs.\n\nThe 1 GPU test speed is similar to the results in the previous single GPU implementation. The speedup results show an approximate speedup of 4x every time we doubled the number of GPUs. This is because there is a twofold benefit from the optimization one doubling for splitting the cities, another for splitting the ants. However, solution quality must be analyzed here, because we reduce the solution space by creating subsets of cities, possibly removing good solutions. RPD was computed as described before and these results were achieved:\n\n51\n\nTable 2: Shows the RPD for the multi-GPU implementation on a trial run of different problems. There results were achieved after only one trial and as such requires much more testing before being conclusive. These results show degradation in the solution quality with our multi-GPU ACO algorithm. The gaps in final solution quality between the GPUs may increase as the algorithm is run for longer periods of time.\n\n6.3\n\n## Improving the Multi-GPU Implementation\n\nThis implementation, while using a nave split and therefore suffering losses in solution quality, lays a framework for future work. More elaborate domain decomposition can be performed using a modified external memory approach, and splitting the cities up as segments of an already found good solution. This builds upon work done in previous external memory implementations as described in section 2.5 . This implementation also supports running different parameters and using a multi-colony approach such as done by Bai et al. . Also, Weiss , who implements an ACO algorithm for data mining on GPUs, mentions a multi-GPU approach as part of his future work section. This section is mainly to show that a\n\n52\n\nmulti-GPU approach can give even further performance improvement, especially in cases where a large number of ants are needed or the domain can be split in an intelligent way.\n\n53\n\nChapter 7 RELATED AND FUTURE WORK This chapter highlights other work published related to my research and plans for how to proceed with future research on the topic.\n\n7.1\n\nRelated Work\n\nThere have been promising results of parallel ACO implementations for specific problems. For TSP, Randall and Lewis developed a parallel version on a MIMD (Multiple Instruction Multiple Data) machine with MPI (Message Passing Interface). They proposed that the communication was slowing the implementation down and it would have benefited from a shared memory model over a distributed one. Shared memory model implementations have been developed using OpenMP for several different applications of ACO. For example, Delisle et al. implemented an OpenMP ACO algorithm to solve an industrial scheduling problem. Their implementation provided a 2x speedup with 2 processors and up to a 5x speedup with 16 processors. Bui et al. proposed a parallel ACO algorithm for the maximum clique problem. Their results ranged from a 40% performance increase on two\n\n54\n\nprocessors and up to 6x faster with an eight-core system. Both papers show the potential for a SIMD shared memory implementation of ACO algorithms. Recently, there have been a few papers on solving the TSP problem using CUDAimplementations of ACO. Bai et al. developed an implementation using MMAS (Max-Min Ant System). They achieve a 2.3x speedup, even though they performed a much larger workload on the GPU than the sequential version. Fu et al. published a recent paper using ACO to solve TSP problems. They achieve significant results by improving an ACO MATLAB implementation on TSP problems. They used the Jacket toolbox to improve over the sequential MATLAB implementation. Finally, the AntMiner system implements a parallel ACO algorithm for a data mining problem on GPUs using CUDA. Weiss achieves very good results in his implementation, gaining up to 100x performance in some cases. The fact that he achieves great results applying ACO to another problem on GPUs shows that many other problems may benefit from this type of platform/algorithm combination.\n\n7.2\n\nFuture Work\n\nMy analysis of the ACO algorithm on the TSPs looks at improving performance by exploiting parallelism inherent in the algorithm. While I achieve a significant increase, performance could still be improved. A more complex domain decomposition as described in Chapter 6 could be another research possibility. However, another possible avenue of research is applying parallel ACO algorithms to other NP-Hard\n\n55\n\nproblems. In the related work above, there were several papers that did just this and achieved good results. One possibility I looked at was applying a parallel ACO algorithm to the problem of instruction scheduling.\n\n7.2.1\n\n## ACO Instruction Scheduling\n\nIf a search algorithm can be used for solving the instruction scheduling problem, that means someone can probably apply ACO to solving instruction scheduling. The ACO algorithm could use the dependency time between instructions as the heuristic part of the probability equation. Then the instructions currently available for execution would be the local nodes used for picking the next element in the partial schedule. Pheromones would be laid down on the dependency edges between the instructions. There are several challenges to implementing parallel ACO for instruction scheduling. The algorithm would need to incorporate validating which edges are possible to select between iterations. The ACO algorithm would need to be given the amount of computing resources in the instruction pipeline. Also, evaluation of the instruction set could be a challenge. Different architectures may have different processing times for instructions, so the best way to evaluate the solutions would be an actual execution of the reordered instructions. If a large number of solutions was generated, this could be a very costly operation.\n\n56\n\nThere have already been several ACO implementations of instruction scheduling done using CUDA. One implementation done by Wang et al. manages to achieve better performance using a hybrid MAX-MIN ACO algorithm rather than just list scheduling. ACO has also been done for other types of scheduling problems. Socha et al. manages to get good performance for scheduling university courses using an MAX-MIN ACO algorithm as well.\n\n57\n\nChapter 8 CONCLUSION The research shows promising results. I achieve up to 10x speedup when porting the construction stage of the algorithm, and up to 16x speedup when porting the whole ACO algorithm while maintaining comparable solution quality per an iteration. The multi-GPU version achieves an additional speedup of 4x every time the number of GPUs is doubled, though solution quality suffers. These types of speedups are especially relevant to larger problem sizes, where each iteration of the ACO algorithm takes a large amount of time. I also provide a strong framework for future work in the topic. I proposed several ways for improving the multi-GPU implementation, and laid some groundwork for applying the ACO algorithm to the Instruction Scheduling problem. Further optimization may also be possible in the CUDA kernel. The results also suggest that other metaheuristic algorithms may also benefit from GPU implementations as well.\n\n58\n\nBIBLIOGRAPHY Thomas Stuetzle. ACOTSP, Version 1.0. 2004. Susan L. Graham, Peter B. Kessler, and Marshall K. McKusick. gprof: a call graph execution profiler. SIGPLAN Not. 39, 4 (April 2004), 49-57. M. Dorigo and T. Sttzle, The Ant Colony Optimization Metaheuristic: Algorithms, Applications, and Advances. Handbook of Metaheuristics, 2002. NVIDIA CUDA Programming Guide 3.1. 2010. Stuart J. Russell and Peter Norvig. 2003. Artificial Intelligence: A Modern Approach (2 ed.). Pearson Education. Marco Dorigo and Thomas Sttzle. 2004. Ant Colony Optimization. Bradford Co., Scituate, MA, USA. M. Dorigo and L.M. Gambardella. Ant Colony System: A Cooperative Learning Approach to the Traveling Salesman Problem. IEEE Transactions on Evolutionary Computation, 1, 1, (1997), 53-66. M.Dorigo, V. Maniezzo and A. Colorni. Ant System: Optimization by a colony of cooperating agents, IEEE Transactions on Systems, Man, and Cybernetics-Part B, 26(1), 29-41, 1996. T. Sttzle and H. H. Hoos, MAX-MIN Ant System. Future Generation Computer Systems. 16(8):889--914,2000.\n\n59\n\n B. Bullnheimer, R. F. Hartl and C. Strauss, A New Rank Based Version of the Ant System: A Computational Study. Central European Journal for Operations Research and Economics, 7(1):25-38, 1999. A. Acan, An external memory implementation in ant colony optimization, in Fourth International Workshop on Ant Colony Optimization and Swarm Intelligence ANTS 2004, Lecture Notes in Computer Science, 2004, pp. 73-82. G. Reinelt. TSPLIBA traveling salesman problem library. ORSA J. Comput. 3, (1991), 376-384. S. Tsutsui and N. Fujimoto. 2009. Solving quadratic assignment problems by genetic algorithms with GPU computation: a case study. In Proceedings of the 11th Annual Conference Companion on Genetic and Evolutionary Computation Conference: Late Breaking Papers (GECCO '09). ACM, New York, NY, USA, 25232530. Mara A. Franco, Natalio Krasnogor, and Jaume Bacardit. 2010. Speeding up the evaluation of evolutionary learning systems using GPGPUs. In Proceedings of the 12th annual conference on Genetic and evolutionary computation (GECCO '10). CUDA Toolkit 3.2 Math Library Performance. 2011. L. Howes and D. Thomas. Efficient Random Number Generation and Application Using CUDA. In GPU Gems 3, Hubert Nguyen, chapter 37, Addison Wesley. S. Janson, D. Merkle, and M. Middendorf. In Parallel Metaheuristics, chapter Parallel Ant Colony Algorithms, John Wiley & Sons, 2005, 171-201.\n\n60\n\n Marcus Randall and Andrew Lewis. 2002. A parallel implementation of ant colony optimization. J. Parallel Distrib. Comput. 62, 9 (September 2002), 1421-1432. P. Delisle, M. Krajecki, M. Gravel, and C. Gagn. Parallel Implementation of an Ant Colony Optimization Metaheuristic with OpenMP. In International conference of parallel architectures and complication techniques (PACT), Proceedings of the third European workshop on OpenMP (EWOMP 2001), pages 8-12, Barcelona, Spain, September 8-9 2001. Thang N. Bui, ThanhVu Nguyen, and Joseph R. Rizzo, Jr.. 2009. Parallel shared memory strategies for ant-based optimization algorithms. In Proceedings of the 11th Annual conference on Genetic and evolutionary computation (GECCO '09). ACM, New York, NY, USA, 1-8. Hongtao Bai, Dantong OuYang, Ximing Li, Lili He, and Haihong Yu. 2009. MAX-MIN Ant System on GPU with CUDA. In Proceedings of the 2009 Fourth International Conference on Innovative Computing, Information and Control (ICICIC '09). IEEE Computer Society, Washington, DC, USA, 801-804. Jie Fu, Lin Lei, and Guohua Zhou. A parallel Ant Colony Optimization algorithm with GPU-acceleration based on All-In-Roulette selection. In Advanced Computational Intelligence (IWACI), 2010 Third International Workshop on, 260-264. Gang Wang, Wenrui Gong, and Ryan Kastner. 2005. Instruction scheduling using MAX-MIN ant system optimization. In Proceedings of the 15th ACM Great Lakes symposium on VLSI (GLSVLSI '05). ACM, New York, NY, USA, 44-49.\n\n61\n\n Marisabel Guevara, Chris Gregg, Kim Hazelwood, Kevin Skadron. \"Enabling Task Parallelism in the CUDA Scheduler,\" in Proceedings of the Workshop on Programming Models for Emerging Architectures (PMEA). Raleigh, NC. September 2009, pages 69-76. S. Grauer-Gray, J. Cavazos. Optimizing and Auto-tuning Belief Propagation on the GPU. In The 23rd International Workshop on Languages and Compilers for Parallel Computing (LCPC 2010) . John Nickolls, Ian Buck, Michael Garland, and Kevin Skadron. 2008. Scalable Parallel Programming with CUDA. Queue 6, 2 (March 2008), 40-53. Abid M. Malik, Tyrel Russell, Michael Chase, and Peter Beek. 2008. Learning heuristics for basic block instruction scheduling. Journal of Heuristics 14, 6 (December 2008), 549-569. Jakob Siegel, Juergen Ributzka, and Xiaoming Li. 2009. CUDA Memory Optimizations for Large Data-Structures in the Gravit Simulator. In Proceedings of the 2009 International Conference on Parallel Processing Workshops (ICPPW '09). IEEE Computer Society, Washington, DC, USA, 174-181. R.M.Weiss, Gpu-accelerated ant colony optimization, in: W.M.W. Hwu (Ed.), GPU Computing Gems: Emerald Edition, Applications of GPU Computing, Morgan Kaufmann, 2011, pp. 325340. Eliot Moss, Paul Utgoff, John Cavazos, Carla Brodley, David Scheeff, Doina Precup, and Darko Stefanovic;. 1998. Learning to schedule straight-line code. In\n\n62\n\nProceedings of the 1997 conference on Advances in neural information processing systems 10 (NIPS '97), Michael I. Jordan, Michael J. Kearns, and Sara A. Solla (Eds.). MIT Press, Cambridge, MA, USA, 929-935. The Traveling Salesman Problem: A Case Study in Local Optimization, D. S. Johnson and L. A. McGeoch, Local Search in Combinatorial Optimization, E. H. L. Aarts and J. K. Lenstra (editors), John Wiley and Sons, Ltd., 1997, pp. 215-310. Russell, Tyrell. Learning Instruction Scheduling Heuristics from Optimal Data. Masters thesis, University of Waterloo. 2006. Michael Sipser (1997). Introduction to the Theory of Computation. PWS Publishing. Sean Luke, 2009, Essentials of Metaheuristics, Lulu, available for free at http://cs.gmu.edu/~sean/book/metaheuristics/ 2001. Introduction to Algorithms (2nd ed.). MIT Press, Cambridge, MA, USA. Mark Harris. Mapping Computational Concepts to GPUs. In GPU Gems 2, Hubert Nguyen, chapter 31, Addison Wesley. Mason Woo, Jackie Neider, Tom Davis, Dave Shreiner, OpenGL Programming Guide: The Official Guide to Learning OpenGL, Version 1.2, Addison-Wesley Longman Publishing Co., Inc., Boston, MA, 1999 Krzysztof Socha, Joshua Knowles, and Michael Sampels. 2002. A MAX-MIN Ant System for the University Course Timetabling Problem. In Proceedings of the\n\n63\n\nThird International Workshop on Ant Algorithms (ANTS '02), Marco Dorigo, Gianni Di Caro, and Michael Sampels (Eds.). Springer-Verlag, London, UK, 1-13.\n\n64" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90403813,"math_prob":0.95879716,"size":84626,"snap":"2019-26-2019-30","text_gpt3_token_len":17269,"char_repetition_ratio":0.19799815,"word_repetition_ratio":0.04467666,"special_character_ratio":0.24199419,"punctuation_ratio":0.2835145,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.95299995,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-19T21:16:30Z\",\"WARC-Record-ID\":\"<urn:uuid:b315a25d-fab6-4821-9620-8c36eafeebba>\",\"Content-Length\":\"367022\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50f13322-302e-4c98-9412-b37c705b46a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:0283703b-d622-40ac-be55-fd7ca71b1175>\",\"WARC-IP-Address\":\"151.101.250.152\",\"WARC-Target-URI\":\"https://id.scribd.com/document/82803617/A-Parallel-Ant-Colony-Optimization-Algorithm-for-the-Salesman-Problem\",\"WARC-Payload-Digest\":\"sha1:ZST55NDQO4EXMU3U3UAINNIWV63NOLCH\",\"WARC-Block-Digest\":\"sha1:TILHCSLBY2DQMSAJEIBFFHPX7BNRTGZT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526359.16_warc_CC-MAIN-20190719202605-20190719224605-00203.warc.gz\"}"}
https://vrcacademy.com/formulas/z-test-two-proportions/	[ "## Two sample proportion test\n\nSuppose we want to compare two distinct populations $A$ and $B$ with respect to possessions of certain attribute among their members. Suppose take samples of sizes $n_1$ and $n_2$ from the population A and B respectively.\n\nLet $X_1$ and $X_2$ be the observed number of successes i.e., number of units possessing the attributes, from the two samples respectively.\n\nThe hypothesis testing problem can be setup as:\n\nSituation Hypothesis Testing Problem\nSituation A : $H_0: p_1=p_2$ against $H_a : p_1 < p_2$ (Left-tailed)\nSituation B : $H_0: p_1=p_2$ against $H_a : p_1 > p_2$ (Right-tailed)\nSituation C : $H_0: p_1=p_2$ against $H_a : p_1 \\neq p_2$ (Two-tailed)\n\n## Formula\n\nThe test statistic for testing $H_0$ is\n\n### $Z = \\frac{(\\hat{p}_1-\\hat{p}_2)-(p_1-p_2)}{SE(\\hat{p}_1-\\hat{p}_2)}$\n\nwhere\n\n• $\\hat{p}_1=\\frac{X_1}{n_1}$ be the observed proportion of successes in the sample from population $A$,\n\n• $\\hat{p}_2=\\frac{X_2}{n_2}$ be the observed proportion of successes in the sample from population $B$.\n\n• $SE(\\hat{p}_1-\\hat{p}_2) = \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n_1}+\\frac{\\hat{p}(1-\\hat{p})}{n_2}}$ is the standard error of the difference between two proportions,\n\n• $\\hat{p} =\\dfrac{X_1 +X_2}{n_1 + n_2}$ is the pooled estimate of sample proportion.\n\nThe test statistic $Z$ follows standard normal distribution $N(0,1)$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77963597,"math_prob":0.99988854,"size":1333,"snap":"2020-24-2020-29","text_gpt3_token_len":459,"char_repetition_ratio":0.13844997,"word_repetition_ratio":0.13218391,"special_character_ratio":0.34133533,"punctuation_ratio":0.09163347,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000002,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-05T19:00:46Z\",\"WARC-Record-ID\":\"<urn:uuid:9e891c66-4cad-4f45-a0c0-5a6518d82d8b>\",\"Content-Length\":\"46376\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8591afba-08eb-41ef-aee5-336d7f4bee4b>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad642457-551d-4c0d-81ed-538e9a369df7>\",\"WARC-IP-Address\":\"104.237.2.19\",\"WARC-Target-URI\":\"https://vrcacademy.com/formulas/z-test-two-proportions/\",\"WARC-Payload-Digest\":\"sha1:CNXUR7TPCT4MC5W7OB6AAB2E243HYNGA\",\"WARC-Block-Digest\":\"sha1:S4ZKVU7NDAMTMHCDCHXTE6GWAEZDUQFO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655888561.21_warc_CC-MAIN-20200705184325-20200705214325-00028.warc.gz\"}"}
https://www.physicsforums.com/threads/having-trouble.361718/	[ "# Having trouble\n\nthe problem says:\n\nfind and state the component form of the following vectors.\nc= 3i + 4j P(0,0),Q(5,-2)\n\nfor some reason, i think i ended up doing unneccessary work. I found the Work of the problem instead, which is 7. Is work the same thing as component form? If not, what am I supposed to be looking for?\n\nRelated Precalculus Mathematics Homework Help News on Phys.org\nMark44\nMentor\nthe problem says:\n\nfind and state the component form of the following vectors.\nc= 3i + 4j P(0,0),Q(5,-2)\n\nfor some reason, i think i ended up doing unneccessary work. I found the Work of the problem instead, which is 7. Is work the same thing as component form? If not, what am I supposed to be looking for?\nWork? How does work enter into a problem that asks for the compenents?\n\nIf v = ai + bj, the horizontal component is a and the vertical component is b.\n\nIf v is the vector from P(a, b) to Q(c, d), v = (c - a)i + (d - b)j and its components are c -a and d - b." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94718456,"math_prob":0.8844116,"size":608,"snap":"2020-45-2020-50","text_gpt3_token_len":181,"char_repetition_ratio":0.12251656,"word_repetition_ratio":0.87931037,"special_character_ratio":0.29276314,"punctuation_ratio":0.15646258,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99097466,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T21:33:34Z\",\"WARC-Record-ID\":\"<urn:uuid:3fd2e918-788f-4acb-84df-9b2777ab136a>\",\"Content-Length\":\"64795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ab44d97-108f-4b50-aa79-1bae52584de9>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c53d968-5b76-48e4-886f-54d71f719837>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/having-trouble.361718/\",\"WARC-Payload-Digest\":\"sha1:5STXLGZRLZ5XHE4CPOEUY27DZUYS6QW7\",\"WARC-Block-Digest\":\"sha1:RHV4LNGRTZTAJARIJ57ZX6P6EIYURF5O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141716970.77_warc_CC-MAIN-20201202205758-20201202235758-00368.warc.gz\"}"}
https://dukarahisi.com/topic-7-application-of-simple-statistics-geography-form-3/	[ "Home GEOGRAPHY TOPIC 7: APPLICATION OF SIMPLE STATISTICS \| GEOGRAPHY FORM 3\n\n# TOPIC 7: APPLICATION OF SIMPLE STATISTICS \| GEOGRAPHY FORM 3\n\n2708\n4", null, "#### b. Inferential statistics it draws conclusions from data that are subject to random variation (e.g., observational errors, sampling variation). Descriptive statistics are most often concerned with two sets of properties of a distribution (sample or population).\n\nSTATISTICAL DATA\nData refers to the actual pieces of information collected through your study. For example, if you ask five of your friends how many pets they own, they might give you the following data: 0, 2, 1, 4, 18. Also data defined as facts or figures from which conclusions may be drawn. Datum is the singular form of the noun data.\n\n#### Occupational Prestige (12-96) Ratio Indicates a difference, the direction of the difference, the amount of the difference in equal intervals, an absolute zero Temperature in Kelvin Income Years of schooling\n\nVARIABLE\nA variable is anything or characteristic that data may have, or an attribute which changes in value under given conditions. Variables include population size, age, sex, altitude, temperature and time.\n\n#### For example the higher the attitude the lower the temperature and vise versa, for that reason increase or decrease of temperature depends on attitude.\n\nDATA PRESENTATION\nData presentation refers to the process of organizing data and presenting them into different forms such as line graphs, pie chart, bar proportional diagrams, polygons and others.\n\nGRAPHICAL DATA\nAfter data have been collected, the next step is to present the data in different ways and forms. Some of the forms in which the data may be presented include charts, graphs, lists, diagrams, tables, essays, graphs, histograms, and even sketches.\n\nLINE (LINEAR) GRAPHS\nLine graphs have unique properties that distinguish them from other graphs. The properties of line graphs are as follows: General procedure to present data using line graphs\na. Get the data needed for plotting the graph.\nb. Identify the independent and dependent variable. Statistically, the independent variables are placed on the x-axis while the dependent variables are placed on the y-axis.\nc. Decide on the vertical scale depending on the graph space and values of the independent variable available.\nd. Decide on the horizontal spacing of the graph according to graph space available.\ne. Draw and divide the vertical and horizontal axes depending on the respective scales.\nf. Plot and join the points to get the graph.\ng. Write the title of the graph you have drawn.\nh. Indicate the scale of the graph.\ni. Show the key for the graph where necessary\n\nLine graphs can be sub-divided into the following categories.\na. Simple line graphs\nb. Group (comparatives) line graphs\nc. Compound line graphs\nd. Divergent line graphs\n\nSimple line graph Construction procedure:\nUse the following table which shows the average monthly temperature recorded in a certain weather station: Average monthly temperature for station X Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec Temp (°C) 23 24 26 28 29 28 26 26 26 27 26 25\n\n#### The following procedures may be used: 1. Identify the variables. The dependent variable is temperature and the independent variable is months. 2. Determine a vertical scale. 3. Determine the horizontal scale (y-axis) depending on the available space 4. Draw both axes and label them: y-axis for temperature and x-axis for months. 5. Plot the points and join them by a smooth line to make a curve. 6. Insert the title and scale. The following is a simple line graph showing monthly temperature for station X. Average monthly temperature for Station X Source: Hypothetical data Scale • Vertical – 1cm:3°C • Horizontal – 1cm: 1month\n\n1. They are easy to draw, read and interpret.\n2. They show specific values of data\n3. They show patterns in data clearly\n4. They enable the viewer to make predictions about the results of data.\n5. It is easy to read the exact values against plotted points on straight line graphs.\n\n1. They limit presentation of only one data or item over time.\n2. One can change the data of a line graph by not using consistent scales on the axis.\n3. They can give a wrong impression on the continuity of data even when there are periods when data is not available.\n4. They do not give a clear visual impression of the actual quantities.\n\nGroup (comparative) line graph\nA group line graph is the graph that involves drawing more than one line on the same graph. It shows the relationship between sets of similar statistics for two or more items. A group line graph is also known as Comparative line graph, Composite line graph, multiple line graph, Polygraph. Usefulness of a group line graph\n• Comparing different values or trends in two or more data variables.\n• Examining the possibility of a relationship existing between the distributions of a number of variables over time.\n• Comparing the distribution of the same variable at different places.\n\nConstruction:\nThe method of drawing a group line graph is the same as for a simple line graph. Therefore, to draw each single line in a group line graph, follow similar steps used for construction of the simple line graph.\n\n#### The following things should be considered before drawing the graph: 1. The lines drawn should not be uniform in colour, thickness, general appearance, etc 2. The number of lines that a graph can accommodate should not exceed five (5). The following table shows banana production (in tonnes) by three villages in Ingwe Division, Tarime district. These data have been used to plot the group (comparative) line graph as shown below: Village/Year Geisangora Itiryo Bungurere Nyansincha 2000 10 15 25 25 2001 20 10 15 20 Source: Hypothetical data Maize production by three villages between 2000 and 2002 Scale: Vertical scale: 1cm to 5 tones Horizontal scale: 2 cm to 1 year\n\n1. The quantity of each component is shown clearly by different line shadings.\n2. Gives comparative analysis of data\n3. It saves time and space since all the line graphs are drawn as a group.\n\n1. The lines can be overcrowded and hence become difficult to read and interpret if many data are involved. 2. It does not give a clear visual impression of actual quantities. Compound line graph A compound line graph is used to analyse the total and the individual inputs of the specific commodities or economic sectors. The graph involves drawing two or more lines, each line corresponding to one item in a different year or region. The items are differentiated from each other or one another by shading differently. Construction: The table below is used for construction of the graph. The table contains hypothetical figures for mineral exports between 2010 and 2012. Year/Mineral Diamond Gold Tanzanite 2010 10,000 16,000 20,000 2011 20,000 25,000 32,000 2012 25,000 35,000 40,000\n\nProcedure:\n• Simplify the data to make the presentation work easy by dividing each value by 1000. Year /Mineral Diamond Gold Tanzanite 2010 10 16 20 2011 20 25 32 2012 25 35 40\n• Add the values for each year to get the cumulative export\n• Plot the values for mineral exports against years on a graph. Usually the line graph for data with the highest values is drawn first.\n• Draw the second line graph above the first one to show the next component. To get the values for plotting the second line graph, add the values of the first\n• Draw the line graph for the last item (diamond) above that of the second item.\n• Shade the component parts between the line graphs using different shadings.\n• Label the axes, show the key and indicate the scale used to construct the graph.\n\n1. Total values are shown clearly shown\n2. It gives good visual impression which encourage understanding and interpreter\n3. Combining all graphs in one saves space.\n\n1. Graph construction is difficult and time-consuming.\n2. It involves a lot of calculations which are difficult and time-consuming.\n3. Reading and interpretation the value is difficult\n\nDivergent line graph\nAre graphs which represents negative (minus value) and positive (plus value) around a mean. They are loss and gain graphs which show divergence or variation between export and import or profit and loss etc. The mean is represented by zero axis drawn horizontally across the graph paper. Year Yield (tonnes) 2012 1000 2013 1500 2014 500 2015 3000\n\nConstruction\n• Sum up the values of all items or commodities. 1000 + 1500 + 500 + 3000 = 6000 • Calculate the arithmetic mean (average) of the values.\n• Calculate the deviation from the mean of each value as shown in the table below. Deviation from the mean value Year X X – 2012 1000 -500 2013 1500 0 2014 500 -1000 2015 3000 +1500\n• Plot the graph using the values of deviation from the mean; and remember to include the title and scale of the graph.\n\n1. It clearly shows how items fluctuate from the mean.\n2. It compares the values of the items and hence facilitates a sound conclusion.\n3. It shows both the positive (profit) and negative (loss) phenomena. 4. It is easy to construct, read and interpret.\n\n1. It involves many calculations and hence time-consuming.\n2. It might be difficult to interpret if one lacks statistical skills.\n3. It is applicable for only one item per graph.\n\nBAR GRAPHS\nAre graphs drawn to show variation of distribution of items by means of bars. The bars should be separated from one another by a space. A bar graph is also called bar chart or columnar graph. Types of bar graphs:\na. Simple bar graphs\nb. Group or comparative bar graphs\nc. Compound bar graphs\nd. Divergent bar graphs\n\nSimple bar graph\nA simple bar graph is drawn to show a single item per bar and represents simple data. Consider the data in the table below which shows the value of sisal exported by Tanzania between 1900 and 1993: Year Sisal export (Tsh ‘000) 1990 106126 1991 107430 1992 142601 1993 161180 1994 202425\n\nConstruction:-\n1. Choose the appropriate scale.\n2. Draw the axes and insert the bars. All the bars must have the same width and spacing.\n4. Insert vertical and horizontal scales and the title. Tanzania sisal export Scale: 1 cm to 50,000 tonnes\n\nAdvantages of a simple bar graph\n1. It is simple to construct, read and interpret.\n2. It has a good visual impression.\n3. It can be used to compare how the amount of an item varies from time to time.\n\nDisadvantages of a simple bar graph\n1. It is limited to only one item or commodity and hence not suitable for massive data.\n2. Not suitable for continuous data such as temperature.\n\nGroup (comparative) bar graph\nA comparative bar graph consists of several bars drawn side by side on the same chart for the purpose of comparison. The technique involves grouping of bars in a chart. The graph can be used to show how production of certain commodities varies each year.\n\nConstruction:\nThe procedure for construction of the comparative bar graph is similar to that of drawing the simple bar graph except that the simple bar graph contains a single bar while the comparative bar graph comprises of multiple bars. Consider the data in the table below, showing agricultural production in metric tonnes.\n\n#### Year/Commodity 1986 1987 1988 Sorghum 1200 5000 8000 Tea 9000 7000 6000 Tobacco 3000 5000 4000 The graph for the data is as shown below. Group (comparative) bar graph showing crop yields in ‘000 kg (1986-1988)\n\nAdvantages of a group bar graph\n1. The total values are expressed well for illustration of points.\n2. It is easy to construct, read and interpret.\n3. The importance of each component is shown clearly.\n\nDisadvantages of a group bar graph\n1. It is difficult to compare the totals of each item/component.\n2. Trends such as fall and rise cannot be shown easily.\n\nCompound (divided) bar graph\nThis is a method of data presentation that involves construction of bars which are divided into segments to show both the individual and cumulative values of items. The length of each segment represents the contribution of an individual item in the total length while that of the whole bar represents the total (cumulative) value of the different items in each group.\n\nConstruction\n• Get the data needed for presentation. For example, consider the table below, which shows the number of tourists who visited the named Tanzania National Parks from 1998 to 2002. Year /park 1998 1999 2000 2002 2003 Manyara 120,000 160,000 172,000 170,000 203,000 Serengeti 175,000 160,000 148,000 185,010 201,000 Tarangire 29,000 30,000 54,100 79,000 102,000 Mikumi 100,000 110,000 111,000 150,000 183,400\n• Simplify the data (to make the presentation work easy) by dividing each value by 10,000. Then add the values to get the total for each year. The simplified data are as shown in the table below.\n• Determine the scale of the bar length based on the highest total value. In this case, the highest total value is 68 (20 + 20 + 10 + 18).\n• Decide on the bar spacing, for example, 1 cm apart.\n• Draw the axes and label them.\n• Start by drawing bars that represent the highest values.\n• The first sets of bars to be drawn are those that represent the highest values. On top of these, the second highest segments are drawn. The last segments to be drawn are those with the lowest values in general.\n• To make it easy to follow the rise and fall of individual values, a soft line could be drawn across bars to separate individual segments.\n• Colour or shade the segments to improve the appearance and simplify interpretation.\n• Inset the scales, key and title. Compound (divided) bar graphs showing tourist visits in 0’000 (1998-2002)\n\nAdvantages of compound (divided) bar graph\n1. It is easy to read and interpret as the totals are clearly shown.\n2. It gives a clear visual impression of the total values.\n3. It clearly shows the rise and fall in the grand total values.\n\nDisadvantages of compound (divided) bar graph\n1. The values of individual segments above the first set are difficult to establish because they don’t start at zero. To get the correct values of the top segments, you have to add the figures, which is difficult for someone not well equipped with statistical skills.\n2. The graph is very difficult to construct and interpret.\n3. It is not easy to represent a large number of components as this would involve very long bars with many segments.\n\nDivergent bar graph\nA divergent bar graph is a graph which shows the fluctuation of individual items from the mean.\n\nConstruction:\n1. Calculate the arithmetic mean (average) of the items.\n2. Subtract the mean from each item.\n3. Draw the graph using the resulting values.\n4. Insert the scale and title of the graph. The data below show the enrolment of Form One students at Mara Secondary School from 1980–1985. Study the table and present the data by a divergent bar graph. Year Number of students 1980 100 1981 150 1982 175 1983 200 1984 225 1985 300 Procedure: • Find the arithmetic mean:\n• Subtract the mean from each item: Year Number of students X – 1980 100 -92 1981 150 -42 1982 175 -17 1983 200 8 1984 225 33 1985 300 108\n• Choose a suitable scale and construct the graph using the obtained values (X – ). A divergent bar graph showing student enrolment (1980-1985)\n\n1. Fluctuation in values, which helps to detect the problem in general terms, is shown.\n2. It is important for comparison of positives and negatives.\n3. Profit (success) or loss (failure) can easily be deduced.\n4. They are simple to construct, read and interpret.\n\n1. Graph construction is time-consuming since it involves many steps.\n2. The calculations involved may be difficult to someone who is poor at mathematics.\n3. It is limited to analysis of only one variable.\n\nDivided circles (pie charts)\nA divided circle is also known as pie chart, circle chart or pie graph. The chart involves dividing the circle into “pie slices” to represent and show relative sizes of data. The size of each slice or segment is always proportional to the value it represents.\n\nDivided circles can appear in two forms:\na. Simple divided circles.\nb. Proportional divided circles. A simple divided circle involves a single set of data whereas the proportional divided circle involves more than one set of data such that the circles will be proportional to the total quantity that each circle represents.\n\nSimple divided circle Construction:\n• Obtain the data to work on. Study this hypothetical record showing enrolment of Form One students in selected Secondary Schools in Tarime District: A table showing student enrolment in selected schools in Tarime District Name of school Number of students Nyansincha 85 Bungurere 80 Nyanungu 78 Magoto 78 Tarime 65 Nyamongo 70 Total 456\n• Calculate the total number of students as shown in the table.\n• Calculate the angle in a circle that would represent the number of students enrolled in each school. For example, 85 out of 456 students enrolled in Nyansincha Secondary School will be represented in the circle by a segment with an angle of 85/456 ×630 = 67 degrees. This will give the following results. Name of school Number of students Degrees Nyansincha 85 67° Bungurere 80 63° Nyanungu 78 62° Magoto 78 62° Tarime 65 51° Nyamongo 70 55° Total 456 360°\n• Draw a circle of a reasonable size.\n• Using a protractor, draw a radius from the 6 o’clock mark to the centre of the circle.\n• Starting with the largest segment representing a specific component, measure and draw its angle from the centre of the circle.\n• Do the same for other components in ascending order.\n• Divide a circle into segments according to the sizes of the angles.\n• Shade the segments and write the title and key of the drawn graph. Student enrolment in selected Secondary Schools in Tarime District\n\n1. It is easy to compare components as they are represented by angles.\n2. Analysis and interpretation of data is easy.\n3. It is easy to assess the proportion of individual components against the total.\n4. Construction of this graphical representation is relatively simple.\n5. It is easy to determine the value of each component since it is indicated on each segment.\n6. Visual impression of the individual components is clear and facilitates the understanding of the information in the data.\n\n1. It is time-consuming because it involves a lot of calculations.\n2. The represented actual values remain hidden as the values shown on the faces of the segments may be in percentages.\n3. Where the range of data is large and involves small and big values, accurate construction of the chart is difficult.\n4. When the values of data set vary slightly, it is difficult to visualize the proportional differences between values (as it is the case in the pie chart above).\n\nThe Importance of Statistics to the User Statistics is important in geography because of the following reasons:\n1. It enables the geographers to handle large sets of data and summarize them in a way that can be easily understood.\n2. Statistics is very useful for planning at local and national levels. For example, statistics on census can be used to plan for social services.\n3. It can also enable the geographers to make comparisons between geographical phenomena, e.g. to compare the amount of rainfall and agriculture production or population distribution in different regions, etc. 4. Statistics translates data into mathematical ways which make the application of quantitative techniques possible.\n5. It enables the geographers to store the information in forms of numbers, graphs, tables, charts, etc.\n6. Statistics give precise rather than generalized information. This offers a lot of satisfaction to the user.\n\nSUMMARIZATION OF MASSIVE DATA\nThe massive data collected from the field have to be summarized so as to make it easy to read, interpret and apply. The massive data can be summarized by the following ways:\n1. Frequency distribution A frequency distribution shows a summarized grouping of data divided into mutually exclusive classes and the number of occurrences in a class. It is a way of showing unorganized data e.g. to show results of an election, income of people for a certain region, sales of a product within a certain period, student loan amounts, etc.\n\n#### These are 0.5 below or above the apparent limits. From the above summarized data, other measures of statistics can be deduced. Such measures include the measures of central tendency, measures of dispersion (variability), measures of relationship (correlation) and measures of relative position.\n\nMETHODS OF PRESENTING SIMPLE AND MIXED DATA\nMeasures of central tendency (averages) A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. As such, measures of central tendency are sometimes called measures of central location.\n\n#### Disadvantages 1. It is greatly affected by extreme values of the data 2. It cannot be obtained if a single observation (item) is missing 3. It is not appropriate in some distributions\n\nMedian\nThe median is the middle score for a set of data that has been arranged in order of magnitude.\n\n#### So, if we look at the example below: 65, 55, 89, 56, 35, 14, 56, 55, 87, 45 We again rearrange the data in order of magnitude (smallest first): 14, 35, 45, 55, 55, 56, 56, 65, 87, 89 Only now we have to take the 5th and 6th score in our data set and average them to get a median of 55.5.\n\n1. It is easy to calculate and understand\n2. It can also be calculated in qualitative data\n3. It is appropriate for skewed distribution\n4. It is not affected by all extreme observations. Hence, it is a better average than the arithmetic mean when extreme observations are present.\n5. The values of a median can be obtained graphically.\n\n1. It is not suitable for further mathematical treatment.\n2. It is not rigidly defined.\n3. It is based on all values or observations.\n4. Compared to mean, median is more affected by fluctuation of sampling.\n5. In case of ungrouped data, rearrangement of values in order of magnitude becomes necessary.\n\nMode\nThe mode is the most frequent score in a data set. It represents the highest bar in a bar chart or histogram. You can, therefore, sometimes consider the mode as being the most popular option. An example of a mode is presented below:\n\n1. It is simple to compute.\n2. It is easy to understand and calculate. In some cases it can be located merely by inspection. The value of the mode can be obtained graphically from the histogram.\n3. It gives a rough idea of the differences of the data set.\n4. It is the only average that can be used when the data is not numerical.\n\n1. It is not rigidly defined; hence it is unstable for large samples.\n2. It is independent of sample size except under special circumstances.\n3. It is not based on all the values of the data.\n4. Mode is not suitable for further mathematical treatment.\n5. As compared to mean, mode is affected to a great extent by the fluctuation of sampling.\n6. There may be more than one mode (as is the case in the previous graph).\n7. There may be no mode at all if none of the data are the same.\n8. It may not accurately represent the data.\n\nThe Significance of Mean, Mode and Median\nMeasures of central tendency are very useful in statistics. Their importance is because of the following reasons:\n\n#### Measures of central tendency can be calculated from grouped data, for example: Calculation of measures of central tendency for grouped data Study the frequency distribution table below: Calculation from the table: Assumed mean (A) = 12 Note: we find the class interval by using the class limits as follows: i = upper class limit – lower class limit + 1 Importance of statistics Helps in the comparison of different geographical phenomena for example climate, population, commodity and production Used to summarize raw and bulk data for easy interpretative and visual explanation It facilitates land use planning Helps resources allocation and provision of social services for example food, health, water, education. Makes it easy to compare data Its knowledge simplifies research activities\n\n1.", null, "Cassandra\n\nI do not know if it’s just me or if everybody else encountering issues with your site.\nIt seems like some of the text in your content are running off the screen. Can someone else please provide feedback and let me\nknow if this is happening to them as well? This might be a issue with my browser because\nI’ve had this happen before. Thanks\n\n2. excellent points altogether, you just won a new reader.\nWhat might you recommend in regards to your\npost that you made some days ago? Any certain?\n\n3. Hey just wanted to give you a brief heads up and let you know a" ]	[ null, "https://dukarahisi.com/wp-content/uploads/2020/10/STATISTICS-640x321.jpg", null, "https://secure.gravatar.com/avatar/2f9a2a3f5db82d09fc45a27ccabfe831", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89418346,"math_prob":0.9539852,"size":34624,"snap":"2021-43-2021-49","text_gpt3_token_len":7930,"char_repetition_ratio":0.14006355,"word_repetition_ratio":0.032360833,"special_character_ratio":0.2387939,"punctuation_ratio":0.12315343,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9814268,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T09:18:47Z\",\"WARC-Record-ID\":\"<urn:uuid:428ad30e-f1f5-45d2-9353-46011d670676>\",\"Content-Length\":\"207133\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8162c85-ef84-454a-915b-e8e609f54e06>\",\"WARC-Concurrent-To\":\"<urn:uuid:4317bad1-3871-490f-aad5-ae38b1270922>\",\"WARC-IP-Address\":\"104.21.77.232\",\"WARC-Target-URI\":\"https://dukarahisi.com/topic-7-application-of-simple-statistics-geography-form-3/\",\"WARC-Payload-Digest\":\"sha1:BM76EVE373GDJOUJOX6S2TWQOFNQL72C\",\"WARC-Block-Digest\":\"sha1:47HBZD4USOTSBAY7SXJL3AKXPP6TZOS7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585916.29_warc_CC-MAIN-20211024081003-20211024111003-00063.warc.gz\"}"}
http://mfat.imath.kiev.ua/article/?id=686	[ "Open Access\n\n# On exit space extensions of symmetric operators with applications to first order symmetric systems\n\n### Abstract\n\nLet $A$ be a symmetric linear relation with arbitrary deficiency indices. By using the conceptof the boundary triplet we describe exit space self-adjointextensions $\\widetilde A^\\tau$ of $A$ in terms of a boundary parameter $\\tau$. We characterize certain geometrical properties of $\\widetilde A^\\tau$ and describe all $\\widetilde A^\\tau$ with ${\\rm mul}\\, \\widetilde A^\\tau=\\{0\\}$. Applying these results to general (possibly non-Hamiltonian) symmetric systems $Jy'- B(t)y=\\Delta(t)y, \\; t \\in [a,b\\rangle,$ we describe all matrix spectral functions of theminimally possible dimension such that the Parseval equality holdsfor any function $f\\in L_\\Delta^2([a,b \\rangle)$.\n\nKey words: Symmetric relation, exit space extension, boundary triplet, first order symmetric system, spectral function.\n\n### Article Information\n\n Title On exit space extensions of symmetric operators with applications to first order symmetric systems Source Methods Funct. Anal. Topology, Vol. 19 (2013), no. 3, 268-292 MathSciNet MR3136731 zbMATH 1289.47046 Milestones Received 21/03/2013; Revised 02/04/2013 Copyright The Author(s) 2013 (CC BY-SA)\n\n### Authors Information\n\nV. I. Mogilevskii\nDepartment of Mathematical Analysis, Lugans'k Taras Shevchenko National University, 2 Oboronna, Lugans'k, 91011, Ukraine\n\n### Citation Example\n\nV. I. Mogilevskii, On exit space extensions of symmetric operators with applications to first order symmetric systems, Methods Funct. Anal. Topology 19 (2013), no. 3, 268-292.\n\n### BibTex\n\n@article {MFAT686,\nAUTHOR = {Mogilevskii, V. I.},\nTITLE = {On exit space extensions of symmetric operators with applications to first order symmetric systems},\nJOURNAL = {Methods Funct. Anal. Topology},\nFJOURNAL = {Methods of Functional Analysis and Topology},\nVOLUME = {19},\nYEAR = {2013},\nNUMBER = {3},\nPAGES = {268-292},\nISSN = {1029-3531},\nMRNUMBER = {MR3136731},\nZBLNUMBER = {1289.47046},\nURL = {http://mfat.imath.kiev.ua/article/?id=686},\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67754304,"math_prob":0.96625215,"size":1953,"snap":"2023-40-2023-50","text_gpt3_token_len":564,"char_repetition_ratio":0.11441765,"word_repetition_ratio":0.09411765,"special_character_ratio":0.29493088,"punctuation_ratio":0.18911175,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9828336,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T14:32:49Z\",\"WARC-Record-ID\":\"<urn:uuid:84f7eb15-0000-4ddc-bb3f-877a04c80ffe>\",\"Content-Length\":\"34068\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b19d279b-68fc-4001-8a67-0b365588649d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a5aa9eaa-fe9d-445b-a690-c1a0e236b423>\",\"WARC-IP-Address\":\"194.44.31.54\",\"WARC-Target-URI\":\"http://mfat.imath.kiev.ua/article/?id=686\",\"WARC-Payload-Digest\":\"sha1:3XYRSCY435KOWWUKCC7DVRT56SMU6GRQ\",\"WARC-Block-Digest\":\"sha1:ASG4EDUKELMHHKMLVKT27OV6CPCRSB7Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510516.56_warc_CC-MAIN-20230929122500-20230929152500-00502.warc.gz\"}"}
https://blog.mathnasium.com/ask-education-what-is-infinity-part-2/	[ "# Ask Education: What is Infinity? (Part 2)\n\nBy Mathnasium \| March 30, 2022\n\nOur expert team of math educators and enthusiasts has spent over 40 years developing and refining the most powerful teaching methods and materials into the comprehensive, industry-leading Mathnasium Method™. These “Ask Education” features are a way for the Education team members to share their knowledge and love of math with our curious readers and fans.\n\nHi, Mathnasium! What is infinity?\n\n~ Saakshi K.\n\nIn our previous “What is infinity?” blog post (part 1), we looked at the concept of infinity. We established that infinity is not a number and does not behave the same way numbers do; it is greater than any real number and is the idea of real numbers being endless.\n\n# Infinity Applied to the Real World\n\nAs we navigate the real world, opportunities arise to observe the properties of infinity. The following are some scenarios of infinity in action.\n\nQuestion: Infinitely many students wait in line outside a classroom. Another student joins the line. How many students are now waiting in line?\n\nAnswer: One plus an infinite amount of students is still an infinite amount of students. Any number added to infinity still leaves us with infinity!\n\nQuestion: Infinitely many students join the line of already infinitely many students waiting outside a classroom. How many students are now waiting in line?\n\nAnswer: We still have infinitely many students in line!\n\nQuestion: Three siblings decide to leave the line of infinitely many students. How many students are left in line?\n\nAnswer: Taking a finite amount from an infinite amount still leaves us with infinite students.\n\nOne popular demonstration of infinity is the Hilbert’s Hotel Paradox, which illustrates that a hotel with infinite rooms, even if they’re all occupied, can still accommodate an endless number of guests.\n\nMany misconceptions arise about infinity because it does not behave as real numbers do. Remember that:\n\nInfinity is endless. We cannot travel far enough to get there since there is no end. We do not need to define any end to infinity, which makes it simpler to work with than things with an end.\n\nInfinity just IS — it is not growing or getting larger. It already is.\n\n# Infinite Sets\n\nOnce understanding of this concept grows, infinity can become captivating. For example, did you know there are types of infinity?\n\nNot all infinities are the same. The two main types of infinity are countable and uncountable. Countable means we could count all the numbers in the set in an infinite amount of time, whereas uncountable means we could not.\n\nAn infinite set is any set that has an endless number of elements, for example, all integers {…, -2, -1, 0, 1, 2, …}, all even numbers {…, -4, -2, 0, 2, 4, …}, all odd numbers {…, -5, -3 -1, 1, 3, 5…}, and all square numbers {0, 1, 4, 9, 16, …}. These are all countable infinite sets because we can pair each number in the set with a natural number (known as a bijection).\n\nInteresting fact: There are exactly the same amount of numbers between 0 and 1 as there are between 0 and 2. Isn’t that incredible?\n\nDespite containing infinite numbers, the sizes of infinite sets can differ. Uncountable infinite sets, however, contain so many elements that we cannot make a pairing (bijection) with the set of natural numbers.\n\nInteresting fact: Cantor’s Diagonal demonstrates the idea of uncountable infinity, where there are infinitely many real numbers and infinitely many positive integers, but the number of items in the set of real numbers is bigger than the number of items in the set of positive integers.\n\nWe know this idea of something having no end can be challenging to process. It remains unproven whether the universe is infinite or finite in size. Our human intuition about magnitude makes it hard to fathom the enormity of infinity, let alone its properties. But the beauty of mathematics and our universe is that they open us to think beyond the limits of our minds.\n\nIf you would like to learn more about infinity or discuss any other mathematical concepts, reach out to your nearest Mathnasium Learning Center. They would be happy to talk to you!\n\nReaders: Do YOU have a math-related question you’d like our education team to answer? Submit it at: http://bit.ly/AskMathnasiumEducation." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9416777,"math_prob":0.7528769,"size":4202,"snap":"2022-05-2022-21","text_gpt3_token_len":912,"char_repetition_ratio":0.15721773,"word_repetition_ratio":0.025423728,"special_character_ratio":0.21251784,"punctuation_ratio":0.13032886,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98269856,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T23:11:10Z\",\"WARC-Record-ID\":\"<urn:uuid:68e086cf-a9c6-4258-9c6c-1b5589e3c865>\",\"Content-Length\":\"41201\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7f0ca6b-4276-4047-99d6-4477b847685c>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3f14e52-be16-46ee-9c30-9c2a8dfce858>\",\"WARC-IP-Address\":\"52.23.39.62\",\"WARC-Target-URI\":\"https://blog.mathnasium.com/ask-education-what-is-infinity-part-2/\",\"WARC-Payload-Digest\":\"sha1:U2RZ767ZIZTQSNXQAQF2XKVQXMKEH2FB\",\"WARC-Block-Digest\":\"sha1:MWS3S3XGSVPTXYJ33RVZRCIQUIRW5XRM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662541747.38_warc_CC-MAIN-20220521205757-20220521235757-00476.warc.gz\"}"}
https://syfexirakenoge.abcdfestivalgoa.com/how-to-work-with-data-probability-grade-4-book-24771wj.php	[ "Last edited by Akirg\nWednesday, May 6, 2020 \| History\n\n5 edition of How To Work with Data & Probability, Grade 4 found in the catalog.", null, "# How To Work with Data & Probability, Grade 4\n\n## by TEACHER CREATED RESOURCES\n\nWritten in English\n\nSubjects:\n• Education / Teaching Methods & Materials / Mathematics,\n• Teaching Methods & Materials - Mathematics,\n• Education,\n• Education / Teaching\n\n• The Physical Object\nFormatPaperback\nNumber of Pages48\nID Numbers\nOpen LibraryOL11817670M\nISBN 101420637401\nISBN 109781420637403\nOCLC/WorldCa59282260\n\nThe probability is the same because there are 3 odd numbers and 3 even numbers. 5. What is the probability of the spinner landing on red? 2 out of 4 6. What is the probability of the spinner landing on orange? 1 out of 4 7. What is the probability of the spinner landing on a primary color? 3 out of 4 8. Free Printable Math Worksheets for Grade 4. This is a comprehensive collection of free printable math worksheets for grade 4, organized by topics such as addition, subtraction, mental math, place value, multiplication, division, long division, factors, measurement, fractions, and decimals. They are randomly generated, printable from your.\n\n5th grade probability Skill: Introduction to probability. How likely is it that snow will fall in August? Remember there are possibilities between definite and impossible. This math worksheet introduces your child to probability with common sense statements to plot on a line. Probability lines help visualize answers. Probability scale 0 to 1. Grade Probability Game: Investigating 8-Sided Dice Roll and Illustration writing math and science differently; this activity will open your students' minds and encourages them to think critically at math and strengthen his or her math knowledge of probability.\n\nYou might also like\nday with Josephine and her friends\n\nday with Josephine and her friends\n\nLand use planning and pollution control.\n\nLand use planning and pollution control.\n\nWest Point\n\nWest Point\n\nA looking-glass for Elder Clarke and Elder Wightman, and the church under their care.\n\nA looking-glass for Elder Clarke and Elder Wightman, and the church under their care.\n\nPricing Practices of Miss Mary Maxim Ltd.\n\nPricing Practices of Miss Mary Maxim Ltd.\n\nHalf broke horses\n\nHalf broke horses\n\ngrey widow-maker\n\ngrey widow-maker\n\nThe impact of the credit crunch on small business\n\nThe impact of the credit crunch on small business\n\nIn her chosen field\n\nIn her chosen field\n\nA beginning teachers guide to special educational needs\n\nA beginning teachers guide to special educational needs\n\nNew Solar Homes Partnership new construction home buyers market research report\n\nNew Solar Homes Partnership new construction home buyers market research report\n\nLondon street finder\n\nLondon street finder\n\n### How To Work with Data & Probability, Grade 4 by TEACHER CREATED RESOURCES Download PDF EPUB FB2\n\nBelow, you will find a wide range of our printable worksheets in chapter Probability of section Data, Graphs, and worksheets are appropriate for Fourth Grade have Grade 4 book many worksheets covering various aspects of this topic, possible outcomes, organized lists, making predictions, fractions and probability, combinations, and many more.\n\nChances Are - predict the likelihood of events using a circle graph with percentages as a model ; Coin Toss - toss enough coins to make a prediction about probability (maximum number of tossesbut you can keep tossing to get a larger data set) ; Hand Squeeze - (a data collection and analysis class experiment) - Pass a \"hand squeeze\" around a circle and measure the amount of time that it.\n\nIt’s a safe bet that our probability worksheets will help fourth grade students make heads and tails of probability. These fourth grade worksheets use visual aides, word problems, and kid-centric themes to keep kids engaged.\n\nKids will also gain extra practice with fractions as they solve problems and summarize information. Tapping and Hugs for Managing Strong Emotions. Physical touch through tapping and hugging can be a wonderful way to regulate emotions.\n\nIn this social emotional learning activity, your child will explore the power of tapping or hugging to help them calm down and relieve feelings of stress or anxiety. 4th grade. Matching and Displaying Data. Emily Thulier from Bruce-monroe Elementary School. Location: 4th Grade Math Description: This unit teaches the basic concepts involved in creating bar graphs, line graphs, line plots, circle Grade 4 book, and other common graph forms.\n\nThe lessons tea. 57 rows Do You Wanna Bet. Your Chance to Find Out About Probability: Cuschman, Jean: 7–. Probability (Grade 4) Give students practice with probability, graphs, and tables.\n\nIn this math worksheet, students read tally tables and a bar graph to compare the number of items and determine which scenario is most or least likely. This page contains all our printable worksheets in section Data, Graphs, and Probability of Fourth Grade you scroll down, you will see many worksheets for organizing data, data: interpret and graph, probability, and more.\n\nA brief description of the worksheets is on each of the worksheet widgets. Improve your math knowledge with free questions in \"Understanding probability\" and thousands of other math skills. Probability worksheets for kids from grade 4 and up include probability on single coin, two coins, days in a week, months in a year, fair die, pair of dice, deck of cards, numbers and more.\n\nMutually exclusive and inclusive events, probability on odds and other challenging probability worksheets are useful for grade 6 and up students. Test (KTCA) grade 4/5: Probability Test grade 5 Probability Test grade 4 Note – I spelled “sundae” wrong in the above tests – hope you caught that one 🙂 Slightly updated tests and an example of a teachers copy (with answers): Probability Test grade 4 Probability Test grade 4 teacher copy.\n\n/ 4 4 2 3 IntroDuCtIon Data Management and Probability, Grades 4 to 6 is a practical guide that teachers will find useful in helping students to achieve the curriculum expectations outlined for Grades 4 to 6 in the Data Management and Probability strand of The Ontario Curriculum, Grades 1–8: Mathematics, This guide provides teachers.\n\nCard Spinner Experiment (Basic) Try this spinner experiment to test the mathematical and experimental probability of spinning diamonds, spades, clubs, or hearts on a spinner.\n\n4th through 7th Grades. View PDF. Coin Flip Experiment (Basic) Try this coin-flipping experiment to test your hypothesis on probability.\n\nExample: the chances of rolling a \"4\" with a die. Number of ways it can happen: 1 (there is only 1 face with a \"4\" on it). Total number of outcomes: 6 (there are 6 faces altogether). So the probability = 1 6. Everyday Mathematics is divided into Units, which are divided into Lessons.\n\nIn the upper-left corner of the Study Link, you should see an icon like this: The Unit number is the first number you see in the icon, and the Lesson number is the second number. In this case, the student is working in Unit 5, Lesson 4.\n\nContinuous Probability Distributions. When you work with continuous probability distributions, the functions can take many forms. These include continuous uniform, exponential, normal, standard normal (Z), binomial approximation, Poisson approximation, and.\n\nTIPS4Math Grade 4 Probability Overall Expectations Students will: • Collect and organize discrete primary data and display the data using charts and graphs, including stem File Size: KB. Probability (5th grade) What's the chance of getting heads in a coin toss.\n\nThis math worksheet introduces your child to probability with common sense questions and probability. Explore what probability means and why it's useful. Explore what probability means and why it's useful. If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains. Probability, Fourth 4th Grade Math Standards, 4th Grade Data Analysis, Grade Level Help, Internet 4 Classrooms Internet resources: teachers, students, children, parents Questions to test your ability to work out the probability of picking specific cards from a pack of playing cards.\n\nIf you are using this Practice Book with another curriculum, use the tables of pages grouped by skill (iii–x) to assign pages based on the skills they address, rather than in order by page number. Bridges in Mathematics Grade 4 Practice Book Blacklines The Math Learning Center, PO BoxSalem, Oregon Tel.\n\n1 –Improve your math knowledge with free questions in \"Find the probability\" and thousands of other math skills.Learn fourth grade math—arithmetic, measurement, geometry, fractions, and more.\n\nThis course is aligned with Common Core standards. If you're seeing this message, it means we're having trouble loading external resources on our website." ]	[ null, "https://covers.openlibrary.org/b/id/2802021-M.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8994285,"math_prob":0.6914912,"size":8541,"snap":"2020-45-2020-50","text_gpt3_token_len":1806,"char_repetition_ratio":0.14923275,"word_repetition_ratio":0.051862672,"special_character_ratio":0.20875776,"punctuation_ratio":0.12059012,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97506833,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T08:14:21Z\",\"WARC-Record-ID\":\"<urn:uuid:f0fe6702-d832-40c0-ae5f-c1d672d2006f>\",\"Content-Length\":\"22867\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2058841a-06f3-49ce-ba14-780dbd027dfd>\",\"WARC-Concurrent-To\":\"<urn:uuid:62c483f0-fa8d-4e98-a04f-795651975d00>\",\"WARC-IP-Address\":\"172.67.200.4\",\"WARC-Target-URI\":\"https://syfexirakenoge.abcdfestivalgoa.com/how-to-work-with-data-probability-grade-4-book-24771wj.php\",\"WARC-Payload-Digest\":\"sha1:DADW7X42227UC5H33PWPBFM444MOJAGB\",\"WARC-Block-Digest\":\"sha1:PE4PVPARNY4AHLFORRTHLKMAXKPOIO7A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107888402.81_warc_CC-MAIN-20201025070924-20201025100924-00271.warc.gz\"}"}
https://sanj.ink/posts/2016-06-28-scalaz-try-operations.html	[ "If you are looking to use scalaz to get some additional functionality for your vanilla `scala.util.Try` class, then you’ve got a couple of options. This can be confusing at first because you might not know which import to use.\n\n### 1. Functions that accept a Try instance\n\nTo import only functions that must be supplied a Try instance use:\n\n``import scalaz.std.`try`._``\n\nThis will give you functions of the form:\n\n``````def cata[A, B](t: Try[A])(success: A => B, failure: Throwable => B): B\n\ndef toDisjunction[A](t: Try[A]): Throwable \\/ A\n\ndef fromDisjunction[T <: Throwable, A](d: T \\/ A): Try[A]\n\ndef toValidation[A](t: Try[A]): Validation[Throwable, A]\n\ndef toValidationNel[A](t: Try[A]) : ValidationNel[Throwable, A]\n\ndef fromValidation[T <: Throwable, A](v: Validation[T, A]) : Try[A]``````\n\nExample:\n\n``cata(Try(..))(..)``\n\nTo get a pimped up version of Try use:\n\n``import scalaz.syntax.std.`try`._``\n\nThis will give you functions directly on your Try instance:\n\n``````final def cata[B](success: A => B, failure: Throwable => B): B\n\nfinal def toDisjunction: Throwable \\/ A\n\nfinal def toValidation: Validation[Throwable, A]\n\nfinal def toValidationNel: ValidationNel[Throwable, A]``````\n\nExample:\n\n``Try(..).cata(..)``" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6205255,"math_prob":0.81190175,"size":1228,"snap":"2022-27-2022-33","text_gpt3_token_len":324,"char_repetition_ratio":0.17320262,"word_repetition_ratio":0.057142857,"special_character_ratio":0.26547232,"punctuation_ratio":0.22846442,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98330873,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T15:15:26Z\",\"WARC-Record-ID\":\"<urn:uuid:058d3d1e-604f-4841-85a0-c88065f619a5>\",\"Content-Length\":\"7382\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aaa92c72-c78d-46de-8e47-e094c2ea1e66>\",\"WARC-Concurrent-To\":\"<urn:uuid:29c8efd5-b30b-452d-b529-474597edd751>\",\"WARC-IP-Address\":\"128.199.88.145\",\"WARC-Target-URI\":\"https://sanj.ink/posts/2016-06-28-scalaz-try-operations.html\",\"WARC-Payload-Digest\":\"sha1:TLNRWR5R7OTHS6SLXZ22UFXFXEKJ542G\",\"WARC-Block-Digest\":\"sha1:3G4GYWQCOQRPGQCKWXHSJ5CMNICIWLUE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103556871.29_warc_CC-MAIN-20220628142305-20220628172305-00723.warc.gz\"}"}
https://libguides.polk.edu/c.php?g=519822&p=3555355	[ "Skip to main content\nIt looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.\n\n# MAT0018 Additional resources - Oliver: Solving Equations\n\nAdditional resources that may prove useful to MAT0018 Developmental I math students\n\n## Solving Equations - Primer\n\n• Definitions:\nq A Term is a constant or the product of a constant and one or more variables.\n\nq If a term contains a variable it is called a variable term.\n\nq A number multiplied by the variable is called a coefficient.\n\nq A term with no variable is called a constant term.\n\n• Equations can take 4 forms in this course.\n\n1. x + a = b\n\n2. ax = b\n\n3. ax + b = c\n\n4. ax + b = cx + d\n\n## Survey\n\n• Were there any of these videos that need improvement?\nSurvey\nEquation type 1: x + 2 = 5: 1 votes (33.33%)\nEquation type 2: 3x = 12: 1 votes (33.33%)\nEquation type 3: 2x + 2 = 12: 1 votes (33.33%)\nEquation type 4: 3x + 2 = 2x - 1: 0 votes (0%)\nTotal Votes: 3\n\nPolk State College is committed to equal access/equal opportunity in its programs, activities, and employment. For additional information, visit polk.edu/equity." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9182947,"math_prob":0.9904421,"size":973,"snap":"2021-04-2021-17","text_gpt3_token_len":266,"char_repetition_ratio":0.118679054,"word_repetition_ratio":0.039106146,"special_character_ratio":0.28057554,"punctuation_ratio":0.14563107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97028464,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T19:31:52Z\",\"WARC-Record-ID\":\"<urn:uuid:09faf308-78e6-4e0f-82dd-c8d9606e31d7>\",\"Content-Length\":\"53474\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1d8c6a62-5541-48ec-b58e-fff5f66a09f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:9412b069-ddde-4fd3-8d2b-1146e805f9ba>\",\"WARC-IP-Address\":\"54.88.35.99\",\"WARC-Target-URI\":\"https://libguides.polk.edu/c.php?g=519822&p=3555355\",\"WARC-Payload-Digest\":\"sha1:UPSGW6W5KH3FJHASXLQZG55YYMUPSD4D\",\"WARC-Block-Digest\":\"sha1:QWU5UVQ5B3XTK7HCRHBJDIDI4ODNYIFC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038064898.14_warc_CC-MAIN-20210411174053-20210411204053-00136.warc.gz\"}"}
https://experts.mcmaster.ca/display/publication220960	[ "Interlayer coupling andc-axis quasiparticle transport in high-Tccuprates Academic Article", null, "•\n• Overview\n•\n• Research\n•\n• Identity\n•\n• Additional Document Info\n•\n• View All\n•\n\nabstract\n\n• The c-axis quasiparticle conductivity shows different behavior depending on the nature of the interlayer coupling. For coherent coupling with a constant hopping amplitude $t_{\\perp}$, the conductivity at zero frequency and zero temperature $\\sigma(0,0)$ depends on the direction of the magnetic field, but it does not for angle-dependent hopping $t(\\phi)$ which removes the contribution of the nodal quasiparticles. For incoherent coupling, the conductivity is also independent of field direction and changes only when paramagnetic effects are included. The conductivity sum rule can be used to determine the admixture of coherent to incoherent coupling. The value of $\\sigma(0,0)$ can be dominated by $t_{\\perp}$ while at the same time $t(\\phi)$ dominates the temperature dependence of the superfluid density.\n\npublication date\n\n• February 1, 2001" ]	[ null, "https://experts.mcmaster.ca/images/individual/uriIcon.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82864517,"math_prob":0.95128214,"size":821,"snap":"2019-43-2019-47","text_gpt3_token_len":175,"char_repetition_ratio":0.13341494,"word_repetition_ratio":0.0,"special_character_ratio":0.20097442,"punctuation_ratio":0.073529415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9710097,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T23:05:14Z\",\"WARC-Record-ID\":\"<urn:uuid:335fd408-f835-4557-8a10-a6ee616771c2>\",\"Content-Length\":\"21430\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97d08048-8008-48e7-8f31-e0e32e8dc984>\",\"WARC-Concurrent-To\":\"<urn:uuid:52fa2bea-cc73-40aa-9967-39fa857d57eb>\",\"WARC-IP-Address\":\"130.113.213.148\",\"WARC-Target-URI\":\"https://experts.mcmaster.ca/display/publication220960\",\"WARC-Payload-Digest\":\"sha1:SE6NNPHDAP6HMT6ALSRCDGDECGHMO35G\",\"WARC-Block-Digest\":\"sha1:PHPA5KPJCSD4SSGPTZSQV36WZGF5XZ5G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986700435.69_warc_CC-MAIN-20191019214624-20191020002124-00514.warc.gz\"}"}
https://statstuneup.com.au/correlation/	[ "## Before You Watch\n\nThis video will describe the concept of comparing two continuous numerical variables and formally testing whether responses to one variable are associate with responses from the other variable. This builds on the Random Variables video where variables type is described, as well as the Turning Data Into Information- Two Variables video where exploratory visual comparisons are described.\n\nReminder: a numerical variable is a variable where the possible responses are numbers which reflect the quantity, or amount, of something. Take for example height, in centimeters, or weight, in kilograms.\n\n...\n\n## Now What?\n\nThis video introduced how to determine if two numerical variables are related to each other, and whether that relationship is statistically significant or not. You can look to further develop your understanding of this concept by looking to the Other Links listed below. Alternatively, how to determine if two categorical variables are related is covered in Association Between Two Categorical Variables: Chi-Squared Test.\n\n## But When Am I Going to Use This?\n\nStatistics is essential in the study of systems of situations where there is an unpredictable random element. This includes a huge number of situations, such as any system that involves living things, which always have a degree of unpredictability. In fact, any studies that involve people involve statistics: for example, medical processes, education and economics. Other areas statistics can be applied to include quality control, stars, nature, or how wear and tear affects machinery." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87008256,"math_prob":0.89775777,"size":593,"snap":"2023-40-2023-50","text_gpt3_token_len":102,"char_repetition_ratio":0.15110357,"word_repetition_ratio":0.0,"special_character_ratio":0.16694772,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95995986,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T09:21:56Z\",\"WARC-Record-ID\":\"<urn:uuid:fd66bd17-1cfb-4a3e-9201-c0e0b931efb8>\",\"Content-Length\":\"273050\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:207e0754-d593-4b27-b6db-6b0bfe32e657>\",\"WARC-Concurrent-To\":\"<urn:uuid:ffcbca9a-c0fa-45ba-b795-daca89983481>\",\"WARC-IP-Address\":\"104.18.153.16\",\"WARC-Target-URI\":\"https://statstuneup.com.au/correlation/\",\"WARC-Payload-Digest\":\"sha1:NX2EPG22QF57MYB6VIM3P5D66XG3ECTS\",\"WARC-Block-Digest\":\"sha1:I7YTF24IBSJ3PCSTRAKY2O24E2MOBKMW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100381.14_warc_CC-MAIN-20231202073445-20231202103445-00263.warc.gz\"}"}
https://numbermatics.com/n/987854/	[ "# 987854\n\n## 987,854 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 987854 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.\n\n987854 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.\n\n## Prime factorization of 987854:\n\n### 2 × 7 × 41 × 1721\n\nSee below for interesting mathematical facts about the number 987854 from the Numbermatics database.\n\n### Names of 987854\n\n• Cardinal: 987854 can be written as Nine hundred eighty-seven thousand, eight hundred fifty-four.\n\n### Scientific notation\n\n• Scientific notation: 9.87854 × 105\n\n### Factors of 987854\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 1771\n\n### Divisors of 987854\n\n• Number of divisors d(n): 16\n• Complete list of divisors:\n• Sum of all divisors σ(n): 1735776\n• Sum of proper divisors (its aliquot sum) s(n): 747922\n• 987854 is a deficient number, because the sum of its proper divisors (747922) is less than itself. Its deficiency is 239932\n\n### Bases of 987854\n\n• Binary: 111100010010110011102\n• Base-36: L68E\n\n### Squares and roots of 987854\n\n• 987854 squared (9878542) is 975855525316\n• 987854 cubed (9878543) is 964002784105511864\n• The square root of 987854 is 993.9084464879\n• The cube root of 987854 is 99.5934830119\n\n### Scales and comparisons\n\nHow big is 987854?\n• 987,854 seconds is equal to 1 week, 4 days, 10 hours, 24 minutes, 14 seconds.\n• To count from 1 to 987,854 would take you about four days!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 987854 cubic inches would be around 8.3 feet tall.\n\n### Recreational maths with 987854\n\n• 987854 backwards is 458789\n• 987854 is a Harshad number.\n• The number of decimal digits it has is: 6\n• The sum of 987854's digits is 41\n• More coming soon!\n\nMLA style:\n\"Number 987854 - Facts about the integer\". Numbermatics.com. 2022. Web. 8 August 2022.\n\nAPA style:\nNumbermatics. (2022). Number 987854 - Facts about the integer. Retrieved 8 August 2022, from https://numbermatics.com/n/987854/\n\nChicago style:\nNumbermatics. 2022. \"Number 987854 - Facts about the integer\". https://numbermatics.com/n/987854/\n\nThe information we have on file for 987854 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 987854, math, Factors of 987854, curriculum, school, college, exams, university, Prime factorization of 987854, STEM, science, technology, engineering, physics, economics, calculator, nine hundred eighty-seven thousand, eight hundred fifty-four.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8674316,"math_prob":0.87052137,"size":2768,"snap":"2022-27-2022-33","text_gpt3_token_len":723,"char_repetition_ratio":0.12626629,"word_repetition_ratio":0.03456221,"special_character_ratio":0.3240607,"punctuation_ratio":0.1663516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98285854,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T22:56:50Z\",\"WARC-Record-ID\":\"<urn:uuid:bda78d03-ffb0-485e-b55c-490cf921bf03>\",\"Content-Length\":\"18276\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:65b8c934-3fd7-44fa-97d7-b9d7d827a29b>\",\"WARC-Concurrent-To\":\"<urn:uuid:b88d83fc-779d-45fe-89f0-866c369a3d9c>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/987854/\",\"WARC-Payload-Digest\":\"sha1:VGIMAPVCWDEF4FXAQO5EPSV2KSUCBJFC\",\"WARC-Block-Digest\":\"sha1:4US6WU5J67LJLPUKCLLFFFBF3TYXBFNR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570879.1_warc_CC-MAIN-20220808213349-20220809003349-00677.warc.gz\"}"}
https://www.analystforum.com/t/share-valuation/2837	[ "", null, "# Share valuation\n\nFrom schweser, Given, EPS=\\$6/share Current dividend rate=\\$3/share Estimated growth rate=6% Required rate of return=10% One-year expected sale price of stock=\\$35/share How much would you be willing to pay for the stock? A)31.82 B)34.54 C)34.71 D)37.60 Using DDM I get a valuation of 3(1.06)/(.1-.06)=79.50?\n\nI get C… FV = 35 Pmt = 3 * (1.06) = 3.18 I/y = 10 n = 1 pv = 34.71\n\nB. (3+35)/1.1\n\nIt is D/1+k + P/1+k - DDM for a holding period of one year => 3.18+35/1.1 = \\$34.709 - C\n\n(3*1.06+35)/1.1=34.71 c\n\nthe answer is C nann_82 Wrote: ------------------------------------------------------- > It is D/1+k + P/1+k - DDM for a holding period of > one year => 3.18+35/1.1 = \\$34.709 - C thanks must have missed that\n\ni got C as well" ]	[ null, "https://analystforum-uploads.s3.dualstack.us-east-1.amazonaws.com/original/2X/8/8e7be8e6512cde25d070f18d332292fb5a3804d9.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82625973,"math_prob":0.9844177,"size":722,"snap":"2021-04-2021-17","text_gpt3_token_len":271,"char_repetition_ratio":0.14763232,"word_repetition_ratio":0.19354838,"special_character_ratio":0.52770084,"punctuation_ratio":0.14210527,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981306,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-20T21:45:50Z\",\"WARC-Record-ID\":\"<urn:uuid:67a3e60a-3c21-4008-8ea6-16d1313a9f7c>\",\"Content-Length\":\"22838\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0ee3ad3e-317f-47fa-99ec-c5fe44b81f77>\",\"WARC-Concurrent-To\":\"<urn:uuid:cbd4c6a6-810f-4223-9e4f-51af8e4272ed>\",\"WARC-IP-Address\":\"45.79.51.137\",\"WARC-Target-URI\":\"https://www.analystforum.com/t/share-valuation/2837\",\"WARC-Payload-Digest\":\"sha1:77XOD6KEBI5PLGRDIVNOSZHRDF6VGPIV\",\"WARC-Block-Digest\":\"sha1:P3W7645FSCHZFAZNA5BHBJPSLUBTUVCM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703522133.33_warc_CC-MAIN-20210120213234-20210121003234-00311.warc.gz\"}"}
https://learnetutorials.com/cpp-programming/programs/quadratic-equation-roots	[ "", null, "# C++ Program to find roots of a quadratic equation\n\nNovember 4, 2021, Learn eTutorial\n917\n\nHere we are discussing a C++ program to find all roots of a quadratic equation.\n\n## What is a quadratic equation?\n\nIn math, we define a quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax2 + bx + c, where a, b, and c are numbering and a cannot be 0.\n\nExample : 6x2 + 11x – 35 = 0.", null, "## How to find the roots of a quadratic equation?\n\nFor a quadratic equation ax2 + bx + c = 0, The roots are calculated using the formula,\n\nx = (-b ± √ (b² - 4ac) ) / 2a\n\nWhere,\n\n• a, b, and c are coefficients.\n• b2 - 4ac is called the discriminant.", null, "With respect to the value of discriminant the roots of the equation will be\n\n• If the discriminant of the equation is greater than zero, then the roots will be real and different\n• If the discriminant of the quadratic is Zero, then the roots will be real and equal.\n• If the discriminant of the equation is less than zero, then the roots are complex and different.\n\n## How to calculate the roots of a quadratic equation using C++ program?\n\nAsk the user to enter the value of three variables a, b, and c. First, calculate the value of the discriminant by using the formula, Discriminant = b² - 4ac. Store the value to variable discrim, which is of float type.\n\nFor the discriminant value, we have three conditions. as we discussed above which will be less than or equal to, or greater than zero. This can be done by using the `if….else if….else `statement.\n\nIf the value of discrim is greater than zero ( discrim > 0 ) calculate the roots by the formula (-b ± √ (b² - 4ac) )/2a. In this equation, the square root will find out using the function sqrt function. The sqrt() function in C++ is used for calculating the square root. it is defined in the cmath header of C++.\n\n• R1= (-b + √ (b² - 4ac) )/2a and\n• R2 = (-b - √ (b² - 4ac) )/2a\n• Print R1 and R2.\n\nElse if the discriminant value is equal to zero ( discrim = = 0 )\n\n• Root R1 = root R 2\n• R1 = -b / 2a\n• Print R1\n\nElse there are two parts real part and the imaginary part.\n\n• The real part is calculated by using the formula rpart = -b / 2a;\n• Imaginary part is calculated by using the formula ipart = i√- (b² - 4ac) )/2a\n• Print the imaginary part and the real part.\n\n### Algorithm\n\nStep 1: Call the header file `iostream.`\n\nStep 2: Call the header file `cmath.`\n\nStep 3: Use the `namespace std`\n\nStep 4: Open the main function `int main().`\n\nStep 5: Declare float variables; a, b, c, R1, R2, discrim, ipart, rpart\n\nStep 6: Print a message to enter coefficients a, b, and c;\n\nStep 7: Read the numbers into the variables a, b, and c.\n\nStep 8: Calculate the discriminant discrim = b2 - 4ac.\n\nStep 9: if the discrim >0\n\n• Then R1= (-b + sqrt(discrim )/2a and\n• R2 = (-b - sqrt(discrim )/2a\n• Print R1 and R2\n\nElse if discrim = 0\n\n• Then R1 = R2\n• R1 = -b / 2a\n\nElse if discrim < 0\n\n• Real part rpart = -b / 2a;\n• Imaginary part ipart = i√- sqrt(discrim)/2a\n• Print rpart and ipart\n\nStep 10: Exit\n\n## C++ Source Code\n\n``` ```#include <iostream>\n#include <cmath>\nusing namespace std;\n\nint main() {\n\nfloat a, b, c, R1, R2, discrim, rPart, iPart;\ncout << \"Enter coefficients a, b and c: \";\ncin >> a >> b >> c;\ndiscrim= bb - 4ac;\n\nif (discrim> 0) {\nR1 = (-b + sqrt(discrim)) / (2a);\nR2 = (-b - sqrt(discrim)) / (2a);\ncout << \"Roots are real and different.\" << endl;\ncout << \"R1 = \" << R1 << endl;\ncout << \"R2 = \" << R2 << endl;\n}\n\nelse if (discrim == 0) {\ncout << \"Roots are real and same.\" << endl;\nR1 = -b/(2a);\ncout << \"R1 = R2 =\" << R1 << endl;\n}\n\nelse {\nrPart = -b/(2a);\niPart =sqrt(-discrim)/(2a);\ncout << \"Roots are complex and different.\" << endl;\ncout << \"x1 = \" << rPart << \"+\" << iPart << \"i\" << endl;\ncout << \"x2 = \" << rPart << \"-\" << iPart << \"i\" << endl;\n}\n\nreturn 0;\n}```\n```\n\n## OUTPUT\n\n```Enter coefficients a, b and c: 5\n3\n6\nRoots are complex and different.\nx1 = -0.3+1.05357i\nx2 = -0.3-1.05357i\n```" ]	[ null, "https://learnetutorials.com/assets/uploads/course/c++blue.webp", null, "https://learnetutorials.com/assets/images/programs/quadratic-equation-roots.png", null, "https://learnetutorials.com/assets/images/programs/quadratic-polynomial-roots.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75402325,"math_prob":0.9994229,"size":2727,"snap":"2023-14-2023-23","text_gpt3_token_len":844,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.04693141,"special_character_ratio":0.35496882,"punctuation_ratio":0.17123288,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999844,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T04:51:24Z\",\"WARC-Record-ID\":\"<urn:uuid:6518fb2a-c336-49bc-acc4-3429068116d7>\",\"Content-Length\":\"65969\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3378637c-b69b-4e39-8f0d-4fd8e0b7f960>\",\"WARC-Concurrent-To\":\"<urn:uuid:4dbaf9cc-8ecb-4c27-9316-869ff99628e3>\",\"WARC-IP-Address\":\"162.159.136.54\",\"WARC-Target-URI\":\"https://learnetutorials.com/cpp-programming/programs/quadratic-equation-roots\",\"WARC-Payload-Digest\":\"sha1:CKTOB7ZAS7TSYFYHZJPAJFCJKIPEE2SV\",\"WARC-Block-Digest\":\"sha1:LO3YJKU7W5VNMIMZPEKO6S66SZ2EFLBF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647614.56_warc_CC-MAIN-20230601042457-20230601072457-00749.warc.gz\"}"}
https://metanumbers.com/131938	[ "# 131938 (number)\n\n131,938 (one hundred thirty-one thousand nine hundred thirty-eight) is an even six-digits composite number following 131937 and preceding 131939. In scientific notation, it is written as 1.31938 × 105. The sum of its digits is 25. It has a total of 3 prime factors and 8 positive divisors. There are 64,320 positive integers (up to 131938) that are relatively prime to 131938.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 6\n• Sum of Digits 25\n• Digital Root 7\n\n## Name\n\nShort name 131 thousand 938 one hundred thirty-one thousand nine hundred thirty-eight\n\n## Notation\n\nScientific notation 1.31938 × 105 131.938 × 103\n\n## Prime Factorization of 131938\n\nPrime Factorization 2 × 41 × 1609\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 131938 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 131,938 is 2 × 41 × 1609. Since it has a total of 3 prime factors, 131,938 is a composite number.\n\n## Divisors of 131938\n\n8 divisors\n\n Even divisors 4 4 4 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 202860 Sum of all the positive divisors of n s(n) 70922 Sum of the proper positive divisors of n A(n) 25357.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 363.233 Returns the nth root of the product of n divisors H(n) 5.20312 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 131,938 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 131,938) is 202,860, the average is 2,535,7.5.\n\n## Other Arithmetic Functions (n = 131938)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 64320 Total number of positive integers not greater than n that are coprime to n λ(n) 8040 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12292 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares\n\nThere are 64,320 positive integers (less than 131,938) that are coprime with 131,938. And there are approximately 12,292 prime numbers less than or equal to 131,938.\n\n## Divisibility of 131938\n\n m n mod m 2 3 4 5 6 7 8 9 0 1 2 3 4 2 2 7\n\nThe number 131,938 is divisible by 2.\n\n• Deficient\n\n• Polite\n\n• Square Free\n\n• Sphenic\n\n## Base conversion (131938)\n\nBase System Value\n2 Binary 100000001101100010\n3 Ternary 20200222121\n4 Quaternary 200031202\n5 Quinary 13210223\n6 Senary 2454454\n8 Octal 401542\n10 Decimal 131938\n12 Duodecimal 6442a\n20 Vigesimal g9gi\n36 Base36 2tsy\n\n## Basic calculations (n = 131938)\n\n### Multiplication\n\nn×y\n n×2 263876 395814 527752 659690\n\n### Division\n\nn÷y\n n÷2 65969 43979.3 32984.5 26387.6\n\n### Exponentiation\n\nny\n n2 17407635844 2296728657985672 303025785677313592336 39980616110693400745627168\n\n### Nth Root\n\ny√n\n 2√n 363.233 50.9085 19.0587 10.57\n\n## 131938 as geometric shapes\n\n### Circle\n\n Diameter 263876 828991 5.46877e+10\n\n### Sphere\n\n Volume 9.62051e+15 2.18751e+11 828991\n\n### Square\n\nLength = n\n Perimeter 527752 1.74076e+10 186589\n\n### Cube\n\nLength = n\n Surface area 1.04446e+11 2.29673e+15 228523\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 395814 7.53773e+09 114262\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.01509e+10 2.70672e+14 107727\n\n## Cryptographic Hash Functions\n\nmd5 420f3c51bce93da9132e86041fe2fdc2 ea1fec27b1dfcd171ab3c8887d699112c65a2ec0 6d4afe4c71c4a9cedaf05762f5f38eeae105b7d0495212a0127754c9e0101910 9c1bd71d0630fd60c662573685d95abbe3222564c448c9af09a792a9bc1b148387158cb9aa385860a7e58a29015e0df7e76ee41730dfcf2af4908f32dd5a5350 5a8ee8d9e7c72ba948276421f712b72c58c0f8bb" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.61647344,"math_prob":0.98233473,"size":4627,"snap":"2021-31-2021-39","text_gpt3_token_len":1626,"char_repetition_ratio":0.12005191,"word_repetition_ratio":0.031019202,"special_character_ratio":0.4586125,"punctuation_ratio":0.078580484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954401,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T05:14:04Z\",\"WARC-Record-ID\":\"<urn:uuid:f4a28720-92e5-4b4f-8d10-244aad6c2ae1>\",\"Content-Length\":\"39941\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76706dc1-bb66-4141-8330-0af8a0324898>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab0d5b22-87f2-4305-bb92-ca44ecdc6f24>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/131938\",\"WARC-Payload-Digest\":\"sha1:IYMP3RUQATNXPRRL43DP366E6KHI722E\",\"WARC-Block-Digest\":\"sha1:62Y6FW77ZOTEDR6WKUWBOKJJCX4EHERT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057504.60_warc_CC-MAIN-20210924050055-20210924080055-00470.warc.gz\"}"}
https://journals.tubitak.gov.tr/math/vol40/iss2/10/	[ "•\n•\n\n## Turkish Journal of Mathematics\n\n#### Article Title\n\nWeakly $2$-absorbing submodules of modules\n\n#### DOI\n\n10.3906/mat-1501-55\n\n#### Abstract\n\nLet $M$ be a module over a commutative ring $R.$ A proper submodule $N$ of $M$ is called weakly $2$-absorbing, if for $r,s\\in R$ and $x\\in M$ with $0\\neq rsx\\in N,$ either $rs\\in (N:M)$ or $rx\\in N$ or $sx\\in N.$ We study the behavior of $(N:M)$ and $\\sqrt{(N:M)},$ when $N$ is weakly $2$-absorbing. The weakly $2$-absorbing submodules when $R=R_1\\oplus R_2$ are characterized. Moreover we characterize the faithful modules whose proper submodules are all weakly $2$-absorbing.\n\n#### Keywords\n\nPrime submodule, $2$-absorbing submodule, weakly $2$-absorbing submodule, weakly prime submodule, weak prime submodule\n\n350\n\n364\n\nCOinS" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5159961,"math_prob":0.9988555,"size":987,"snap":"2023-40-2023-50","text_gpt3_token_len":339,"char_repetition_ratio":0.19125128,"word_repetition_ratio":0.0,"special_character_ratio":0.3100304,"punctuation_ratio":0.16585366,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99946433,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T07:17:35Z\",\"WARC-Record-ID\":\"<urn:uuid:3ac60e7f-5c75-4452-bf17-756210ecc1e0>\",\"Content-Length\":\"50175\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1063afe-e2c3-4c47-a06d-fcb2c653bf61>\",\"WARC-Concurrent-To\":\"<urn:uuid:70013b28-ca72-4000-9b07-1f7297927d24>\",\"WARC-IP-Address\":\"50.18.241.247\",\"WARC-Target-URI\":\"https://journals.tubitak.gov.tr/math/vol40/iss2/10/\",\"WARC-Payload-Digest\":\"sha1:EGCAOKF5CXS2ST6VUP2TWJVDBANVF447\",\"WARC-Block-Digest\":\"sha1:5YFG37GVACSJR4KHLGWWH2W2IK3UMIHI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506339.10_warc_CC-MAIN-20230922070214-20230922100214-00398.warc.gz\"}"}
https://mathoverflow.net/questions/69824/ramification-in-p-division-fields-associated-to-elliptic-curves-with-good-ordina	[ "# Ramification in p-division fields associated to elliptic curves with good ordinary reduction\n\nDear MO,\n\nLet $p$ be a prime and let $E/\\mathbb{Q}_p$ be an elliptic curve. Suppose that $E/\\mathbb{Q}_p$ has good ordinary reduction at $p$. In his 1972 paper Propriétés galoisiennes des points d'ordre fini des courbes elliptiques'' (more specifically, see Corollaire, in p. 274), Serre shows along the way that the inertia subgroup of $\\operatorname{Gal}(\\mathbb{Q}_p(E[p])/\\mathbb{Q}_p)$ is, with respect to a suitable basis of $E[p]$, isomorphic to either a matrix group of the form {$[\\ast\\ 0; 0\\ 1]$} or {$[\\ast\\ \\ast; 0\\ 1]$} as a subgroup of $\\operatorname{GL}(2,\\mathbb{F}_p)$. After this result, Serre remarks that he doesn't know of any simple criterion that would determine whether one is in the first case or the second case.\n\nQuestion: Nowadays, do we know of a criterion to tell whether one is in the first case or the second case?\n\nA more concrete question: Here is the particular example that I am working with: Let $E/\\mathbb{Q}$ be 1225h1'' in Cremona's tables, given by $$E : y^2 + xy + y = x^3 + x^2 - 8x + 6.$$ This curve has a rational $37$-isogeny and therefore $\\operatorname{Gal}(\\mathbb{Q}(E)/\\mathbb{Q})$ is a Borel subgroup of $\\operatorname{GL}(2,\\mathbb{F}_{37})$. The curve $E$ has good ordinary reduction at $p=37$ and I am trying to find out whether the ramification index of $37$ in the extension $\\mathbb{Q}(E)/\\mathbb{Q}$ is just $\\varphi(37)$ or rather $\\varphi(37)\\cdot 37$, where $\\varphi$ is the Euler phi function.\n\nThe $37$th division polynomial of $E/\\mathbb{Q}$ has degree $684$ and it factors (over $\\mathbb{Q}[x]$) as a product of $4$ polynomials of degrees $6$, $6$, $6$ and $666$, respectively. The extension of degree $666$ is, well, diabolically large and I can't find the ramification at $37$ computationally... or at least I don't know how to!\n\n• You won't need to factor your diabolic polynomial; the slopes of the Newton polygon in $\\mathbb{Q}_p[x]$ are enough to determine the ramification. In your case there are 666 unit roots and 18 of valuation $−1/18$. Jul 9, 2011 at 8:05\n• Chris, if I am not mistaken, the slopes of the Newton polygon will determine the valuations of the x-coordinates of the 37-torsion points. But even if the valuations of the roots of the polynomial of degree $666$ are $0$, that doesn't mean that the extension generated by the roots is unramified, does it? For instance, the slopes of $1+x+x^2$ are $[0,0]$ for $p=3$ but, of course, the prime $3$ ramifies in $\\mathbb{Q}(\\zeta_3)/\\mathbb{Q}$. Jul 9, 2011 at 14:20\n• Absolutely correct. I am embarrassed about my mistake. The only thing it shows is that the reduction is ordinary. Jul 10, 2011 at 15:18\n\nAssume $p \\ne 2$. The condition for the representation to be tamely ramified (i.e $* = 0$ in the upper right entry of the matrix) is that $j(E) \\equiv j_0 \\mod p^2$ where $j(E)$ is the $j$-invariant of $E$ and $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$. This is proved in Gross \"A tameness criterion for galois representations...\" Duke J. 61 (1990) on page 514. For $p=2$ you need the congruence modulo $8$. Serre gives an algorithm for computing $j_0$ in Lubin-Serre-Tate.\n• Thank you, Felipe! I will try to calculate $j_0$ in this case and report back. Jul 9, 2011 at 14:22\n• Ok, if I am understanding everything correctly, there is a typo in the very last line of Lubin-Serre-Tate. When $p=2$ and $\\lambda=1$, one should have $j_0 = 3375/31 = 3^3\\cdot 5^3/31$. The polynomial $\\phi_2(x,x^2)$ factors as $2^4(31x-3375)(3x^4 + 3x^3 + 82531x^2 + 26622000x + 2916000000)$, where $\\phi_2(x,y)$ is the classical modular polynomial for $N=2$. The only root in $\\mathbb{Q}_2$ congruent to $1$ mod $2$ is $3375/31$. Jul 12, 2011 at 4:09\n• @Álvaro: Careful, $s(x)$ is not $x^2$, only congruent to it modulo $2$. It is the Frobenius on Witt vectors which squares the Witt coordinates. I doubt there is a typo, Serre would have picked it up. Most of the corrections I have listed on that page were his. Jul 12, 2011 at 11:22" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8305109,"math_prob":0.99954677,"size":1827,"snap":"2023-40-2023-50","text_gpt3_token_len":570,"char_repetition_ratio":0.11684037,"word_repetition_ratio":0.04964539,"special_character_ratio":0.3223864,"punctuation_ratio":0.1027027,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99993193,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T04:16:29Z\",\"WARC-Record-ID\":\"<urn:uuid:37055377-1013-4371-afd7-439002e65ff3>\",\"Content-Length\":\"117387\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5afaf58f-395c-4f03-bd62-5ee1e54ee86a>\",\"WARC-Concurrent-To\":\"<urn:uuid:e3b12ddd-b354-46b6-bd7e-48435bd01842>\",\"WARC-IP-Address\":\"104.18.37.74\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/69824/ramification-in-p-division-fields-associated-to-elliptic-curves-with-good-ordina\",\"WARC-Payload-Digest\":\"sha1:3VOOIA72QIRDI5RT4U2UYX5GZRJTARWJ\",\"WARC-Block-Digest\":\"sha1:QXPMBYTOVOMR7TQMB43XVYOKTF7R4XGZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100264.9_warc_CC-MAIN-20231201021234-20231201051234-00137.warc.gz\"}"}
https://socketloop.com/tutorials/golang-shuffle-array-of-list/?utm_source=socketloop&utm_medium=tutesidebar	[ "# Golang : Shuffle array of list\n\nContinue from previous tutorial on how to shuffle an array of strings, to shuffle array of list is not really straight forward, but it can be done easily with `rand.Perm()` function. Use this code example to shuffle an array of list.\n\n`````` package main\n\nimport (\n\"fmt\"\n\"math/rand\"\n\"time\"\n)\n\nfunc shuffleList(start, end int) []int {\nif end < start {\nstart, end = end, start\n}\n\nlength := end - start\nrand.Seed(time.Now().UTC().UnixNano())\n\nlist := rand.Perm(length)\nfor index, _ := range list {\nlist[index] += start\n}\nreturn list\n}\n\nfunc main() {\n\nmapList := make([]map[int]int, 3)\n\nmapList = map[int]int{1: 1}\n\nmapList = map[int]int{2: 2}\n\nmapList = map[int]int{3: 3}\n\nshuffled := shuffleList(0, len(mapList))\n\nfmt.Printf(\"Original order : %v\\n\", mapList)\n\nfmt.Printf(\"Shuffled order : %v\\n\", shuffled)\n\n}\n``````\n\nSample output :\n\nOriginal order : [map[1:1] map[2:2] map[3:3]]\n\nShuffled order : [1 0 2]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66344714,"math_prob":0.99291635,"size":1964,"snap":"2023-40-2023-50","text_gpt3_token_len":541,"char_repetition_ratio":0.12346939,"word_repetition_ratio":0.006472492,"special_character_ratio":0.2790224,"punctuation_ratio":0.17128463,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97859,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T16:03:11Z\",\"WARC-Record-ID\":\"<urn:uuid:daaf7857-9a59-4f27-b426-50401f0bb4ac>\",\"Content-Length\":\"452989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37b4d86c-b5bc-47c0-bbc3-20fe74a30fd8>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c9387e8-c634-469c-a19e-8a68d3a3e4d5>\",\"WARC-IP-Address\":\"104.21.16.89\",\"WARC-Target-URI\":\"https://socketloop.com/tutorials/golang-shuffle-array-of-list/?utm_source=socketloop&utm_medium=tutesidebar\",\"WARC-Payload-Digest\":\"sha1:OT2QCWNLDBUAORQ57APVWUA75D2MRILY\",\"WARC-Block-Digest\":\"sha1:L7VV453THB6QSCL7MJPYODGEMMVVA6NW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100912.91_warc_CC-MAIN-20231209134916-20231209164916-00313.warc.gz\"}"}
https://www.onemathematicalcat.org/Math/Precalculus_obj/squareRootNeg.htm	[ " The Square Root of a Negative Number\n\n# THE SQUARE ROOT OF A NEGATIVE NUMBER\n\n• PRACTICE (online exercises and printable worksheets)\n\nSquare roots of negative numbers were introduced in a prior lesson:\nArithmetic with Complex Numbers in the Algebra II materials.\nFor your convenience, that material is repeated (and extended) here.\n\nFirstly, recall some information from beginning algebra:\n\n## The Square Root of a Nonnegative Real Number\n\nFor a nonnegative real number $\\,k\\,$, $$\\overbrace{\\ \\ \\ \\sqrt{k}\\ \\ \\ }^{\\text{the square root of \\,k\\,}} := \\ \\ \\text{the unique nonnegative real number which, when squared, equals \\,k\\,}$$\n\n(Note: when the distinction becomes important in higher-level mathematics, then $\\,\\sqrt{k}\\,$ is called the principal square root of $\\,k\\,$.)\n\nThus, the number that gets to be called the square root of $\\,k\\,$ satisfies two properties:\n\n• it is nonnegative (not negative; i.e., greater than or equal to zero)\n• when squared, it gives $\\,k\\,$\nFor example, $\\,\\sqrt{9} = 3\\,$ since the number $\\,3\\,$ satisfies these two properties:\n• $\\,3\\,$ is nonnegative\n• $\\,3\\,$, when squared, gives $\\,9\\,$: $\\,3^2 = 9\\,$\n\nSo, what is (say) $\\,\\sqrt{-4}\\$?\nThere does not exist a nonnegative real number which, when squared, equals $\\,-4\\$.\nWhy not?\nEvery real number, when squared, is nonnegative: for all real numbers $\\,x\\,$, $\\,x^2 \\ge 0\\,$.\nComplex numbers to the rescue!\n\n## The Square Root of a Negative Number\n\nComplex numbers allow us to compute the square root of negative numbers, like $\\,\\sqrt{-4}\\$.\nRemember the key fact: $\\,i:=\\sqrt{-1}\\,$, so that $\\,i^2=-1\\,$\n\nTHE SQUARE ROOT OF A NEGATIVE NUMBER\nLet $\\,p\\,$ be a positive real number, so that $\\,-p\\,$ is a negative real number. Then: $$\\overbrace{\\ \\ \\ \\sqrt{-p}\\ \\ \\ }^{\\text{the square root of negative \\,p\\,}} \\ :=\\ i\\,\\sqrt{\\vphantom{h}p}$$\n(Note: when the distinction becomes important in higher-level mathematics, then $\\,\\sqrt{-p}\\,$ is called the principal square root of $\\,-p\\,$.)\n\nObserve that $\\,i\\sqrt{\\vphantom{h}p}\\,$, when squared, does indeed give $\\,-p\\$: $$(i\\sqrt{\\vphantom{h}p})^2 \\ =\\ i^2(\\sqrt{\\vphantom{h}p})^2 \\ =\\ (-1)(p) \\ =\\ -p$$\n\nSome of my students like to think of it this way:\nYou can slide a minus sign out of a square root, and in the process, it turns into the imaginary number $\\,i\\,$!", null, "Here are some examples:\n\n• $\\sqrt{-4} = i\\sqrt{4} = i2 = 2i\\,$\n\nIt is conventional to write $\\,2i\\,$, not $\\,i2\\,$.\nWhen there's no possible misinterpretation (see below), write a real number multiplier before the $\\,i\\,$.\n\n• $\\sqrt{-5} = i\\sqrt{5}$\n\nIn this situation, it's actually better to leave the $\\,\\sqrt{5}\\,$ after the $\\,i\\,$.\nWhy? If you pull the $\\,\\sqrt{5}\\,$ to the front, it can lead to a misinterpretation, as follows:\nThe numbers $\\,\\sqrt{5}i\\,$ and $\\,\\sqrt{5i}\\,$ are different numbers—look carefully!\nIn $\\ \\sqrt{5}i\\$, the $\\,i\\,$ is outside the square root.\nIn $\\ \\sqrt{5i}\\$, the $\\,i\\,$ is inside the square root.\nUnless you look carefully, however, you might mistake one of these for the other.\nBy writing $\\,i\\sqrt{5}\\,$, you eliminate any possible confusion.\n\n• TWO DIFFERENT QUESTIONS; TWO DIFFERENT ANSWERS\n\nRecall these two different questions, with two different answers:\n\n• What is $\\,\\sqrt{4}\\,$?\nAnswer: By definition, $\\,\\sqrt{4} = 2\\,$.\nThe number $\\ \\sqrt{4}\\$ is the nonnegative number which, when squared, gives $\\,4\\,$.\n• What are the solutions to the equation $\\,x^2 = 4\\,$?\nAnswer: $\\,x= \\pm\\ 2\\,$\nThere are two different real numbers which, when squared, give $\\,4\\,$.\n\nSimilarly, there are two different questions involving complex numbers, with two different answers:\n\n• What is $\\,\\sqrt{-4}\\,$?\nAnswer: $\\sqrt{-4} = 2i\\,$\n• What are the (complex) solutions to the equation $\\,x^2 = -4\\,$?\nAnswer: $\\,x=\\pm \\sqrt{-4} = \\pm\\ 2i\\,$\nThere are two different complex numbers which, when squared, give $\\,-4\\,$.\nVerifying the second solution: $\\,(-2i)(-2i) = 4i^2 = 4(-1) = -4$\n\n## Square Roots of Products\n\nThe following statement is true: for all nonnegative real numbers $\\,a\\,$ and $\\,b\\,$, $\\,\\sqrt{ab} = \\sqrt{a}\\sqrt{b}\\$.\nFor nonnegative numbers, the square root of a product is the product of the square roots.\n\nDoes this property work for negative numbers, too?\nThe answer is NO, as shown next.\n\nCertainly, anything called ‘the square root of $\\,-4\\,$’ must have the property that, when squared, it equals $\\,-4\\$.\nUnfortunately, the following incorrect reasoning gives the square as $\\,4\\,$, not $\\,-4\\,$: $$\\text{? ? ? ? }\\ \\ \\ \\ (\\sqrt{-4})^2 \\ \\ =\\ \\ \\sqrt{-4}\\sqrt{-4} \\overbrace{\\ \\ =\\ \\ }^{\\text{this is the mistake}}\\ \\ \\sqrt{(-4)(-4)}\\ \\ =\\ \\ \\sqrt{16}\\ \\ =\\ \\ 4\\ \\ \\ \\ \\text{? ? ? ? }$$ Here is the correct approach: $$(\\sqrt{-4})^2 \\ \\ =\\ \\ \\sqrt{-4}\\sqrt{-4}\\ \\ =\\ \\ i\\sqrt{4}\\ i\\sqrt{4} \\ \\ =\\ \\ i^2(\\sqrt{4})^2 \\ \\ =\\ \\ (-1)(4) \\ \\ =\\ \\ -4$$\n\nSimilarly, $\\,\\sqrt{-5}\\sqrt{-3}\\,$ is NOT equal to $\\,\\sqrt{(-5)(-3)}\\,$.\nInstead, here is the correct simplication: $$\\sqrt{-5}\\sqrt{-3} \\ \\ =\\ \\ i\\sqrt{5}\\ i\\sqrt{3} \\ \\ =\\ \\ i^2\\sqrt{5}\\sqrt{3} \\ \\ =\\ \\ (-1)\\sqrt{(5)(3)} \\ \\ =\\ \\ -\\sqrt{15}$$ Be careful about this!\n\nMaster the ideas from this section" ]	[ null, "https://www.onemathematicalcat.org/Math/Precalculus_obj/graphics/slide_i_out.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8087557,"math_prob":1.0000036,"size":3519,"snap":"2020-34-2020-40","text_gpt3_token_len":1105,"char_repetition_ratio":0.15789473,"word_repetition_ratio":0.025089607,"special_character_ratio":0.37254903,"punctuation_ratio":0.23684211,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999654,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-20T13:52:11Z\",\"WARC-Record-ID\":\"<urn:uuid:79b44a04-37ee-4901-9a7d-655531297679>\",\"Content-Length\":\"57500\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e623ab74-7539-4551-9bda-4afdd83ebf74>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e0b134b-d290-457e-82d1-b49e18bd6f3f>\",\"WARC-IP-Address\":\"8.21.33.44\",\"WARC-Target-URI\":\"https://www.onemathematicalcat.org/Math/Precalculus_obj/squareRootNeg.htm\",\"WARC-Payload-Digest\":\"sha1:GGYY7CBXH6PA3PEEEDY3AR645A3LPVOY\",\"WARC-Block-Digest\":\"sha1:3GVIQ3VQO3E3ICMCUPC6K25TZYVEBC3P\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198213.25_warc_CC-MAIN-20200920125718-20200920155718-00189.warc.gz\"}"}
http://top.zhan.com/cihui/ielts-integrity.html	[ "# integrity\n\n• 单词助记\n• 考点解读\n• 单词详解\n\n## 扫码背更多单词", null, "4周搞定2000+托福核心词\n\nintegr ity\n\nintegral = integr(完整) + al(...的) = 构成整体所必需的;完整的\n\nintegration = integr(完整) + ation(表结果) = 集合，综合\n\n+ 表性质\n\naccreditation = ac(加强) + cred(相信) + ity(表性质) + ate(表动词) = 加强信用→委派\n\nacidity = acid(酸) + ity(表性质) = 酸的性质→酸性\n\n[ɪnˈtɛɡrɪtɪ]\n\n• 托福\n• 雅思\n• 剑雅词频:1 核心词\n\nintegrity\n\n常考释义\n\n• n. 诚实\n变形词\n• 复数 integrities\n\n• 英汉双解\n• 同反义词\n• n. 完好；完善；健全<正式>\n\n• n. 完整；完全；统一<正式>\n\n• n. 正直；诚实\n\n• 同义词\n\nn.诚实；正直; 统一；一致；联合;\n\n反义词\n\nn. 伪善；虚伪；矫饰;\n\n• 每日搜索热词排行榜", null, "" ]	[ null, "https://top-static.zhan.com/beikao/vocabulary/img/weixinCode2x.png", null, "https://top-static.zhan.com/beikao/images/seo-region/gov.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.67301184,"math_prob":0.999251,"size":323,"snap":"2020-10-2020-16","text_gpt3_token_len":249,"char_repetition_ratio":0.12852664,"word_repetition_ratio":0.0,"special_character_ratio":0.34365326,"punctuation_ratio":0.14492753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801868,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-02T21:08:10Z\",\"WARC-Record-ID\":\"<urn:uuid:f9934b60-6b0c-4713-b423-899995e0f015>\",\"Content-Length\":\"56962\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64bfa6c4-5d12-49cb-b6da-719735d64ff6>\",\"WARC-Concurrent-To\":\"<urn:uuid:6fafb6c8-776c-4994-b2b4-4f7df036091f>\",\"WARC-IP-Address\":\"47.103.175.190\",\"WARC-Target-URI\":\"http://top.zhan.com/cihui/ielts-integrity.html\",\"WARC-Payload-Digest\":\"sha1:R6BXTAMZ5S2KX4G33GMK6SLSK3SDKNSV\",\"WARC-Block-Digest\":\"sha1:FSQQ5YKNKW4KHJKFRFMWODQNMB6CTKYE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370508367.57_warc_CC-MAIN-20200402204908-20200402234908-00122.warc.gz\"}"}
https://www.enotes.com/homework-help/describe-how-find-inverse-function-y-3x-12-262861	[ "# Describe how to find the inverse of function y=3x+12?", null, "We need to find the inverse of y = 3x + 12. For this, express x in terms of y.\n\ny = 3x + 12\n\n=> 3x = y - 12\n\n=> x = y/3 - 4\n\nsubstitute x and y.\n\n=> y = x/3 - 4\n\nThe inverse of y = 3x + 12 is y = x/3 - 4\n\nApproved by eNotes Editorial Team" ]	[ null, "https://static.enotescdn.net/images/main/illustrations/illo-answer.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.722947,"math_prob":0.9999026,"size":395,"snap":"2021-43-2021-49","text_gpt3_token_len":183,"char_repetition_ratio":0.18670076,"word_repetition_ratio":0.9661017,"special_character_ratio":0.53417724,"punctuation_ratio":0.078431375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99430436,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T08:01:24Z\",\"WARC-Record-ID\":\"<urn:uuid:1ae45c10-523c-41b6-8713-dcf65f8ddf1e>\",\"Content-Length\":\"67685\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9396ab7-7335-481b-92be-a8b95aec983c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e328085-d084-4c36-9526-8540247f524f>\",\"WARC-IP-Address\":\"104.26.4.75\",\"WARC-Target-URI\":\"https://www.enotes.com/homework-help/describe-how-find-inverse-function-y-3x-12-262861\",\"WARC-Payload-Digest\":\"sha1:7WLM7XLABOEJSVSESYUMMWJL7WNAC62M\",\"WARC-Block-Digest\":\"sha1:WTN2U7BSIWAD4NDDC2TTVFX7A42II53M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585460.87_warc_CC-MAIN-20211022052742-20211022082742-00246.warc.gz\"}"}
https://www.clutchprep.com/chemistry/practice-problems/116522/an-aqueous-nacl-solution-is-made-using-126-g-of-nacl-diluted-to-a-total-solution	[ "# Problem: An aqueous NaCl solution is made using 126 g of NaCl diluted to a total solution volume of 1.15 L .Calculate the molarity of the solution.\n\n87% (80 ratings)\n###### Problem Details\n\nAn aqueous NaCl solution is made using 126 g of NaCl diluted to a total solution volume of 1.15 L .\n\nCalculate the molarity of the solution.\n\nFrequently Asked Questions\n\nWhat scientific concept do you need to know in order to solve this problem?\n\nOur tutors have indicated that to solve this problem you will need to apply the Molarity concept. You can view video lessons to learn Molarity. Or if you need more Molarity practice, you can also practice Molarity practice problems." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93222535,"math_prob":0.81803465,"size":475,"snap":"2021-04-2021-17","text_gpt3_token_len":103,"char_repetition_ratio":0.14225054,"word_repetition_ratio":0.0,"special_character_ratio":0.20210527,"punctuation_ratio":0.08791209,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98370147,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T01:55:58Z\",\"WARC-Record-ID\":\"<urn:uuid:b1f02a21-7f5e-484e-bbfc-219bb67ca2a8>\",\"Content-Length\":\"154230\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eae6b3e6-3ac2-436c-b56b-007b039956e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:b21382b1-7a8d-4c89-a48a-784f37f6435b>\",\"WARC-IP-Address\":\"54.159.163.191\",\"WARC-Target-URI\":\"https://www.clutchprep.com/chemistry/practice-problems/116522/an-aqueous-nacl-solution-is-made-using-126-g-of-nacl-diluted-to-a-total-solution\",\"WARC-Payload-Digest\":\"sha1:EKCJJA7FL3VTF7USYUTRR5R2KV4RVVTZ\",\"WARC-Block-Digest\":\"sha1:QHP7H2QTTW2DRCMU354SHMKIG6M7WWEQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038464146.56_warc_CC-MAIN-20210418013444-20210418043444-00479.warc.gz\"}"}
https://itprospt.com/num/17728130/2-52-35-what-is-the-range-of-the-function-g-x	[ "5\n\n# 2 +52 35_ What is the range of the function g(x)...\n\n## Question\n\n###### 2 +52 35_ What is the range of the function g(x)\n\n2 +52 3 5_ What is the range of the function g(x)", null, "", null, "#### Similar Solved Questions\n\n##### Find th cu-cm: 30,000 volume 'pasn H top unoue and closed base that minimizes the square 1 38 (8 pts) dimensionsvalue pressure the blood verify resulting sure Be the occur? dru ] (cc) ofa certain 0.16 blood 3 maximum centimeters the cubic Genti gptsyill (r)a Eeyofx what dosage aBesop - maximum: approximate At 2drug? (2 pts for pressure blood maximum the What\nFind th cu-cm: 30,000 volume 'pasn H top unoue and closed base that minimizes the square 1 38 (8 pts) dimensions value pressure the blood verify resulting sure Be the occur? dru ] (cc) ofa certain 0.16 blood 3 maximum centimeters the cubic Genti gptsyill (r)a Eeyofx what dosage aBesop - maximum...\n##### Suppose that Y, Yz; Yn denote random sample of size from population with an expurettial distribulion whose detsity is given by(1/0)e->fly)elsewhere.If Ya) min(Y , Yz, denotes the smallest-order statistic show that 6 = nY) is an unbiased estimator for 0 and find MSE(O).\nSuppose that Y, Yz; Yn denote random sample of size from population with an expurettial distribulion whose detsity is given by (1/0)e-> fly) elsewhere. If Ya) min(Y , Yz, denotes the smallest-order statistic show that 6 = nY) is an unbiased estimator for 0 and find MSE(O)....\n##### 2) Recall that for any quadratic equation of the form ax2 br + c = 0 with a,b,0 > 0, the quadratic formula gives us that the smaller root is: 6_ b2_c 11123Consider the quadratic equation 12 =0. Compute T1, using the formula above; in four-digit arithmetic with chopping:Suppose that the true four digit solution is â‚¬1 0.0054. Compute the relative and absolute error of your above approximationNow; rationalize the numerator of the expression for 11. Use this new expression to compute T1 for th\n2) Recall that for any quadratic equation of the form ax2 br + c = 0 with a,b,0 > 0, the quadratic formula gives us that the smaller root is: 6_ b2_c 11 123 Consider the quadratic equation 12 =0. Compute T1, using the formula above; in four-digit arithmetic with chopping: Suppose that the true fo...\n##### 5) (9 pts) Sketch the Lewis Dot Struclure for the sulfite ion; SO; ! Sketch all of the resonance fors and label the formal charges, if applicable\n5) (9 pts) Sketch the Lewis Dot Struclure for the sulfite ion; SO; ! Sketch all of the resonance fors and label the formal charges, if applicable...\n##### Consider the exponential function Q =ra ~. Letting 4 = In Q show that linear function of bv writing it in the formq b + mt. State the values of mandb:b =\nConsider the exponential function Q =ra ~. Letting 4 = In Q show that linear function of bv writing it in the formq b + mt. State the values of mandb: b =...\n##### Circular loop of wire, whose area is 0.20 m\" , lies so that an external, uniform magnetic field of magnitude 0.20 T is perpendicular to the face of the loop The magnetic field reverses its direction, and its magnitude changes to 0.10 Tin 1.0 \\$. Find the magnitude of the average induced emf during this period of time\ncircular loop of wire, whose area is 0.20 m\" , lies so that an external, uniform magnetic field of magnitude 0.20 T is perpendicular to the face of the loop The magnetic field reverses its direction, and its magnitude changes to 0.10 Tin 1.0 \\$. Find the magnitude of the average induced emf duri...\n##### 2. (10 points) Suppose thatf(,v 2) =r2 ytan z, 9(t) : (sin? t, cos 4,44) .points) Is f 0 @ well-defined? What about @0 f? Briefly justify your answers. points) Demonstrate your knowledge of the chain rule to find &n in terms of t. Lf you do not use the chain rule you will not receive credit for your work\n2. (10 points) Suppose that f(,v 2) =r2 ytan z, 9(t) : (sin? t, cos 4,44) . points) Is f 0 @ well-defined? What about @0 f? Briefly justify your answers. points) Demonstrate your knowledge of the chain rule to find &n in terms of t. Lf you do not use the chain rule you will not receive credit fo...\n##### Fhee transfarmtim matci ees Atenhrxs \" are one O onfo Write O \"no (n each cel( 'Qs' A = c[:]l:6Oxe-+o-Onlonte\nfhee transfarmtim matci ees Atenhrxs \" are one O onfo Write O \"no (n each cel( 'Qs' A = c[:]l: 6 Oxe-+o-Onl onte...\n##### Physicists describe certain properties, such as angular momentum and energy, as being conserved. What does this mean? Do these conservation laws imply that an individual object can never lose or gain angular momentum or energy? Explain your reasoning.\nPhysicists describe certain properties, such as angular momentum and energy, as being conserved. What does this mean? Do these conservation laws imply that an individual object can never lose or gain angular momentum or energy? Explain your reasoning....\n##### Question 10Wnat Is the molecular geometry araund the centra atom in PH3C)inearbentErIgonal pyramidaltetranedra\nQuestion 10 Wnat Is the molecular geometry araund the centra atom in PH3 C)inear bent ErIgonal pyramidal tetranedra...\n##### A pole-vaulter runs toward the launch point with horizontal momentum. Where does the vertical momentum come from as the athlete vaults over the crossbar?\nA pole-vaulter runs toward the launch point with horizontal momentum. Where does the vertical momentum come from as the athlete vaults over the crossbar?...\n##### QUESTIONPlease refer - the bear phylogeny shown beloyPolar Dear (RefhDear 0zPolar bear 01An erican black Dea\"Sloth bearSun bearAsiatic black bearSpectacied Dear 01Spectaclcd bear 02Bears highlighted in yellow hibernate Bears that are not highlighted do not hibernate_ Given this arrangement of descendants; which of the following would be the most parsimonious trait for ancestor exhibit?Tnis nor accurate it is just for the sake of this question:hibernatesdoes not hibernateBoth of the above ar\nQUESTION Please refer - the bear phylogeny shown beloy Polar Dear (Refh Dear 0z Polar bear 01 An erican black Dea\" Sloth bear Sun bear Asiatic black bear Spectacied Dear 01 Spectaclcd bear 02 Bears highlighted in yellow hibernate Bears that are not highlighted do not hibernate_ Given this arran...\nEntered Answer Preview 4/9 The answer above is NOT correct point) Determine whether the series bnverges,and if so find its sum 4*+3 gkH1 4/9 (Enter DNE if the sum does not exist ) Preview My Answers Submit Answers score Was recorded...\n##### Find the values of P, q and r Cari nilai-nilai p, 4 dan r. [4 markslmarkah] Find the value of m, if the composite index for the year 2015 based on the year 2013 is 12353 Cari nilai m, jika indeks gubahan 'pada tahun 2015 berasaskan tahun 2013 ialah 123.53. [3 marks/markah] Find the unit price of the electrical item for the vear 2015 if the unit price of the item for the year 2013 is RM425. Cari harga seunit barangan elektrik itu pada t telutz 2015 jika harga seunit barangan tersebut pada\nFind the values of P, q and r Cari nilai-nilai p, 4 dan r. [4 markslmarkah] Find the value of m, if the composite index for the year 2015 based on the year 2013 is 12353 Cari nilai m, jika indeks gubahan 'pada tahun 2015 berasaskan tahun 2013 ialah 123.53. [3 marks/markah] Find the unit price...\n##### After reading this section, write out the answers to these questions. Use complete sentences.When do you use the quadratic formula to solve a quadratic equation?\nAfter reading this section, write out the answers to these questions. Use complete sentences. When do you use the quadratic formula to solve a quadratic equation?...\n##### [3 Marks] Q 2) Evaluate:1) lim I,42) lim JIIS I+43) In the belowgraph (figure 1.1.2) _figure 1.1.2T(then find the value of the limit: lim f(z) I .\n[3 Marks] Q 2) Evaluate: 1) lim I,4 2) lim JIIS I+4 3) In the below graph (figure 1.1.2) _ figure 1.1.2 T( then find the value of the limit: lim f(z) I ...." ]	[ null, "https://cdn.numerade.com/ask_images/0bfa6e6c51da4d35a05574c3584b110c.jpg ", null, "https://cdn.numerade.com/previews/64e0386a-ad1e-4d27-8c8c-c19c7e2fda93_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78047913,"math_prob":0.9477327,"size":10011,"snap":"2022-27-2022-33","text_gpt3_token_len":2753,"char_repetition_ratio":0.10952333,"word_repetition_ratio":0.63564134,"special_character_ratio":0.26281092,"punctuation_ratio":0.12150434,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98971504,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T13:02:15Z\",\"WARC-Record-ID\":\"<urn:uuid:20b31540-91cd-4540-8434-3b064aabc0aa>\",\"Content-Length\":\"68137\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8264893f-a82e-45b7-831e-c7515e2a605c>\",\"WARC-Concurrent-To\":\"<urn:uuid:28ef1696-fc52-436f-b685-dc2d4ce77d6a>\",\"WARC-IP-Address\":\"104.26.7.163\",\"WARC-Target-URI\":\"https://itprospt.com/num/17728130/2-52-35-what-is-the-range-of-the-function-g-x\",\"WARC-Payload-Digest\":\"sha1:AUPJFRHZBO7VK3TX76GR2VJF6UELOAGW\",\"WARC-Block-Digest\":\"sha1:MEAKXVFDZOSNV64BVMCJZXQRVTLE2DIT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572908.71_warc_CC-MAIN-20220817122626-20220817152626-00091.warc.gz\"}"}
https://webpages.uncc.edu/brvainbe/mathematical-physics-seminar.html	[ "# Mathematical Physics/PDE Seminar\n\nIn Spring 2021, the Math.Physics/PDE seminar meets Tuesdays 5:00 -6:15 pm online on Zoom. For more information please contact Boris Vainberg\n\n Back to the main page of the Department of Mathematics and Statistics\n\nApril 20, 2021\n\nS. Molchanov will finish his talk on the sparse potentials\n\nApril 20, 2021\n\nS. Molchanov will continue his talk on the sparse potentials.\n\nApril 13, 2021\n\nS. Molchanov will give a talk “Introduction to the spectral theory of lattice Schrodinger operators with sparse potentials\n\nAbstarct. The talk will present several results. In particular, we will describe the essential spectrum of lattice Schrodinger operators with sparse potentials and the point component of the spectral measure of the Schrodinger operator outside of the spectrum of the Laplacian. We will also discuss the open problems and possible approaches to their solutions.\n\nApril 6, 2021\n\nS. Molchanov will continue to talk on “Introduction to the spectral theory of Schrodinger operators with sparse potentials”\n\nAbstract: Kapitsa pendulum. Example of a band-gap spectrum. Review of results on the 1-D theory (Pearson, Kiselev-Last Simon, Molchanov, Cook-Holt-Molchanov).\n\nS. Molchanov will give a talk “Introduction to the spectral theory of Schrodinger operators with sparse potentials”\n\nAbstract: In the classical spectral theory of the Schrodinger operators H=-Δ+V(x) (i.e. in the quantum mechanics) there are two fundamental models:\n\na) V(x) vanishes at infinity (may be, with additional restriction on the rate of the decay of the potential). Examples. Hydrogen atom H, here V(x)=-c/\|x\|, general atoms and molecules, scattering theory.\n\nb) Periodic potentials. Examples. Theory of the ideal crystals, their heat and electric conductivities, theory of metals and semiconductors.\n\nPeriodic potentials are also related to the problem of stability of the mechanical system (Kapitsa pendulum). What happens between these two classical cases? The simplest intermediate models are the operators with the sparse potentials where there are infinitely many elementary scatterers (bumps) such that the distances between the bumps are growing to infinity. These models demonstrate many new effects.\n\nMarch 23, 2021\n\nO. Safronov will continue to talk on “Discrete spectrum of a periodic Schrodinger operator perturbed by a decaying impurity potential”\n\nAbstract: First, we explain how compact operators naturally appear in problems where one studies the flow of eigenvalues through a fixed point λ. In particular, if H(t)=H-t V where H is a differential operator and V is a positive decaying function, then one can reduce the study of eigenvalues of H(t) to the study of the operator W(H- λ )^{-1}W where W^2=V. The latter operator turns out to be compact. After that we will discuss applications of this reduction principle to the case of a periodic operator H.\n\nMarch 16, 2021\n\nO. Safronov will give a talk “Discrete spectrum of a periodic Schrodinger operator perturbed by a decaying impurity potential”\n\nAbstract. Let H be a periodic Schrodinger operator and let V be a positive fast decaying function on R^d. We consider the family of operators H(t)= H-tV. We study the number N(t) of eigenvalues of H(t) in a fixed interval [a,b] consisting of regular points of H. We obtain an asymptotic formula for N(t) as t goes to infinity. However, the limit in this asymptotic formula is understood in some integral Cesaro-like sense." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8576553,"math_prob":0.91655606,"size":3372,"snap":"2021-21-2021-25","text_gpt3_token_len":804,"char_repetition_ratio":0.14133017,"word_repetition_ratio":0.11753372,"special_character_ratio":0.20670225,"punctuation_ratio":0.10483871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98424417,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-23T08:17:11Z\",\"WARC-Record-ID\":\"<urn:uuid:d5f26e8f-c7b0-48ee-87ad-384ee843f216>\",\"Content-Length\":\"55814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63a004fd-6023-45dc-a1d7-7f0dfafbd08a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c997f8df-4370-443b-b37e-282061657b97>\",\"WARC-IP-Address\":\"152.15.36.56\",\"WARC-Target-URI\":\"https://webpages.uncc.edu/brvainbe/mathematical-physics-seminar.html\",\"WARC-Payload-Digest\":\"sha1:N2KKJE6Q45C4DSZ4PRW6YAN3LHY7TULW\",\"WARC-Block-Digest\":\"sha1:2MWQQ7F7FO5AVKG6PDUQYX7B67BZSPEQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488536512.90_warc_CC-MAIN-20210623073050-20210623103050-00263.warc.gz\"}"}
https://unix.stackexchange.com/questions/386598/grep-patterns-from-file1-into-column-of-file2	[ "# grep patterns from file1 into column of file2\n\nI have two files:\n\n``````\\$ cat File1\nA\nB\nC\n``````\n``````\\$ cat File2\nA aaa B\nD bbb A\nB aaa h\n``````\n\nI would like to search patterns from `File1` into `File2`, a sort of what `grep -f File1 File2` would do, but searching the patterns reported in `File1` only in `\\$1` of `File2`\n\nSample output:\n\n``````\\$cat File3\nA aaa B\nB aaa h\n``````\n• Would it be acceptable to have a modified version of File1: `sed 's/^/^/' File1 > File1-anchored`? Then use that modified file as pattern file for `grep`. Aug 17, 2017 at 5:40\n\nWith `awk`:\n\n``````awk 'NR==FNR{a[\\$0]=NR; next} a[\\$1]' f1.txt f2.txt\n``````\n• `NR==FNR{a[\\$0]=NR; next}`: for first file (`f1.txt`) we are putting the record as key to an assiciative array with the corresponding record number as the value\n\n• `a[\\$1]`: for second file (`f2.txt`), the record is only printed if the first field is a key of array `a`\n\nExample:\n\n``````% cat f1.txt\nA\nB\nC\n\n% cat f2.txt\nA aaa B\nD bbb A\nB aaa h\n\n% awk 'NR==FNR{a[\\$0]=NR; next} a[\\$1]' f1.txt f2.txt\nA aaa B\nB aaa h\n``````\n\nUsing `join` command:\n\n``````join <(sort file1) <(sort file2)\n``````\n\nIf the files are sorted.\n\n``````join file1 file2\n``````\n\nWith `bash` or any shell that understands process substitution:\n\n``````\\$ grep -f <( awk '{ printf(\"^%s[[:blank:]]\\n\", \\$0) }' File1 ) File2\nA aaa B\nB aaa h\n``````\n\nThe idea here is to create the correct patterns for `grep -f File1` to work directly on `File2` by transforming each line in `File` from `something` to the regular expression `^something[[:blank:]]` (prefix it with a circumflex and suffix it with `[[:blank:]]`).\n\nThe circumflex anchors the pattern to the start of the line and `[[:blank:]]` forces a match against a space or a tab character.\n\nGNU `grep` may also read patterns from standard input:\n\n``````\\$ awk '{ printf(\"^%s[[:blank:]]\\n\", \\$0) }' File1 \| grep -f - File2\nA aaa B\nB aaa h\n``````\n\nThe `awk` command may be replaced by an equivalent `sed` command (if you prefer `sed` over `awk`):\n\n``````\\$ sed -e 's/^/^/' -e 's/\\$/[[:blank:]]/' File1 \| grep -f - File2\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71306956,"math_prob":0.54894733,"size":905,"snap":"2022-27-2022-33","text_gpt3_token_len":272,"char_repetition_ratio":0.13762486,"word_repetition_ratio":0.101265825,"special_character_ratio":0.31491712,"punctuation_ratio":0.113513514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9542288,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T16:31:54Z\",\"WARC-Record-ID\":\"<urn:uuid:ea7e392f-8ffb-4fbb-a042-df5eec3c399f>\",\"Content-Length\":\"244569\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7d161da-4649-4808-868d-4d6c707a0ee2>\",\"WARC-Concurrent-To\":\"<urn:uuid:070e6000-4477-4c01-bd7f-63310fadb882>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://unix.stackexchange.com/questions/386598/grep-patterns-from-file1-into-column-of-file2\",\"WARC-Payload-Digest\":\"sha1:BTNGGY6C5WADJCJKC7FEGTYT2TUQ365U\",\"WARC-Block-Digest\":\"sha1:OXWUYH7XF5GZX63HHP53527EO3CROA6F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103556871.29_warc_CC-MAIN-20220628142305-20220628172305-00527.warc.gz\"}"}
https://www.physicsforums.com/threads/math-joke-thread-lol.158839/page-2	[ "I once came across a number of spoof proofs the only one of which I remember is the:-\n\nProof by Deferrred Responsibility\n\nAt school: The proof of this is too complicated at your current stage. When you get to University you will be given a proof.\n\nAt University: You will know from the maths you did at school that it is true that..........\n\nhere is another limerick:\n$$\\frac{12 + 144 + 20 + 3\\sqrt{4}}{7} + 5\\times11 = 9^2 + 0$$\n\nit says:\nA dozen, a gross, and a score,\nPlues three times the square root of four,\nDivided by seven,\nPlues five times eleven,\nIs nine squared, and not a bit more.\n\nGib Z\nHomework Helper\nO adding to the spoof proofs\n\nProof By Imtimidation:\n\"Trivial\"\n\nI particularly like that one :D\n\nProof by Confusing Notation:\nGood when you have access to at least 4 languages and special symbols.\n\nProof by Example:\n\"Lets try when n=2, the other 253 cases are analogous..\"\n\nmathwonk\nHomework Helper\nthe one about the wife and mistress is funny, because true. that is really what mathematicians are like. but i suspect only a mathematician (or a wife of one) would know this.\n\nSGT\nAt the time of the Iron Curtain, two mathematicians try to leave Polland by stealing an airplane.\nThey enter the aircraft and one of them tries to undestand the functions of the instruments. The other worries:\n\"Hurry up, we will be caught!\"\n\"Calm down, I´m a simple Pole in a complex plane!\"\n\n$$\\frac{\\sinx}{n} = 6$$\n\n$$\\frac{\\sinx}{n} = 6$$\ni didnt get this one.\n\nedit: oh, i see. you wanted to write $$\\frac{\\sin x}{n} = 6$$\n\nThere is three erors in this post.\n\nI will finish this post before you can say Jack Rob" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9709237,"math_prob":0.9773867,"size":611,"snap":"2020-10-2020-16","text_gpt3_token_len":141,"char_repetition_ratio":0.14003295,"word_repetition_ratio":0.7256637,"special_character_ratio":0.26022914,"punctuation_ratio":0.20714286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9968211,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T17:00:57Z\",\"WARC-Record-ID\":\"<urn:uuid:09bb1fae-cce3-4538-ad7b-62e9f2ef9075>\",\"Content-Length\":\"91389\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:791aa284-c06e-45b6-b7c8-7e26fb7e340a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d43b834-d74d-45cb-9b24-6ee589aa405f>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/math-joke-thread-lol.158839/page-2\",\"WARC-Payload-Digest\":\"sha1:DP4IRRM6D6DLHYUU7JEZ6AHFGLRH5HCO\",\"WARC-Block-Digest\":\"sha1:Q45JUKOTYCAUXYYH6FSAD4NQJTESZ2SO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146414.42_warc_CC-MAIN-20200226150200-20200226180200-00090.warc.gz\"}"}
https://math.stackexchange.com/questions/2967855/a-non-orientable-surface-s-such-that-t-ps-is-time-type	[ "# A non-orientable surface $S$ such that $T_pS$ is time-type.\n\nI'm currently taking a course in Semi-Euclidean (Semi-Riemannian) Geometry and am given the task to give an example of a non-orientable surface $$S$$ in $$\\mathbb{R}^3_1$$ which is time-type.\n\nWith the Lorentzian dot product: $$\\langle u,v\\rangle = u_1v_1 + u_2v_2 - u_3v_3$$ We say that a vector is time-type whenever $$\\langle v,v\\rangle < 0$$. And we say that a surface is time-type iff the tangent plane to at any point in $$S$$ contains a time-type vector.\n\nI cannot think of an example that is non-orientable and time-type. We know the classic example of the mobius band which is non-orientable but there is at least one point where the tangent plane does not contain one such vector." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8908994,"math_prob":0.9984348,"size":682,"snap":"2019-13-2019-22","text_gpt3_token_len":190,"char_repetition_ratio":0.12241888,"word_repetition_ratio":0.0,"special_character_ratio":0.25953078,"punctuation_ratio":0.05925926,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99756056,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T11:49:49Z\",\"WARC-Record-ID\":\"<urn:uuid:0534953c-71e4-49fa-a231-bfe36997f020>\",\"Content-Length\":\"121923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2bd5e183-1fde-4b06-96b7-d014b23304cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9467a0e-5730-424e-a4ca-cab766231f98>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2967855/a-non-orientable-surface-s-such-that-t-ps-is-time-type\",\"WARC-Payload-Digest\":\"sha1:HSCIIODMRIGVBE35YMNW3SZ5IT2URXER\",\"WARC-Block-Digest\":\"sha1:SG2C36OLOVWT2SKWDEIAMDEK354LWH7Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259126.83_warc_CC-MAIN-20190526105248-20190526131248-00308.warc.gz\"}"}
http://www.webgenii.ca/generating-random-numbers-and-letters/	[ "", null, "# Generating Random Numbers and Letters\n\nNeed to generate a random number? Excel has two formula variations that can do that for you.\n\nThe first is RAND, it generates a random value between 0 and 1. But if you need larger values, try RANDBETWEEN.\n\nRANDBETWEEN generates random numbers between a bottom and top value that you specify.\n\nLastly, if you need to generate a random letter value – try this formula: =CHAR(RANDBETWEEN(65,90))\n\nIn this formula the CHAR function returns the character symbol specified by a number. Letters A-Z are between 65 and 90.", null, "" ]	[ null, "http://www.webgenii.ca/wp-content/uploads/2018/03/krissia-cruz-362015-unsplash-672x372.jpg", null, "http://assets.pinterest.com/images/pidgets/pinit_fg_en_rect_red_28.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8136835,"math_prob":0.9954001,"size":511,"snap":"2020-10-2020-16","text_gpt3_token_len":122,"char_repetition_ratio":0.13412228,"word_repetition_ratio":0.0,"special_character_ratio":0.22504893,"punctuation_ratio":0.11764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9613518,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-10T05:17:24Z\",\"WARC-Record-ID\":\"<urn:uuid:d266ae32-b32e-430c-a2d8-d16ae73af049>\",\"Content-Length\":\"85030\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aae419d7-f48c-4754-8f7d-e8a235ede6ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:cec61f6a-e076-4dbf-b521-01faa8ca268e>\",\"WARC-IP-Address\":\"69.49.101.51\",\"WARC-Target-URI\":\"http://www.webgenii.ca/generating-random-numbers-and-letters/\",\"WARC-Payload-Digest\":\"sha1:A5OKIGQG75LULDL6PKYCFCV37UK27Y6M\",\"WARC-Block-Digest\":\"sha1:7CUNCBYV2C74QOFZ2NAGZOII7VX6T65R\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371886991.92_warc_CC-MAIN-20200410043735-20200410074235-00545.warc.gz\"}"}
https://thekooshy.com/wap.html	[ "# Math help algebra 2\n\nThere is Math help algebra 2 that can make the technique much easier. Our website can solving math problem.\n\n## The Best Math help algebra 2", null, "Here, we will show you how to work with Math help algebra 2. In mathematics, the domain of a function is the set of all input values for which the function produces a result. For example, the domain of the function f(x) = x2 is all real numbers except for negative numbers, because the square of a negative number is undefined. To find the domain of a function, one must first identify all of the possible input values. Then, one must determine which input values will produce an undefined result. The set of all input values that produce a defined result is the domain of the function. In some cases, it may be possible to solve for the domain algebraically. For example, if f(x) = 1/x, then the domain is all real numbers except for 0, because division by 0 is undefined. However, in other cases it may not be possible to solve for the domain algebraically. In such cases, one can use graphing to approximate thedomain.\n\nFirst, when you multiply or divide both sides of an inequality by a negative number, you need to reverse the inequality sign. For example, if you have the inequality 4x < 12 and you divide both sides by -2, you would get -2x > -6. Notice that the inequality sign has been reversed. This is because we are multiplying by a negative number, so we need to \"flip\" the inequality around. Second, when solving an inequality, you always want to keep the variable on one side and the constants on the other side. This will make it easier to see what values of the variable will make the inequality true. Finally, remember that when solving inequalities, you are looking for all of the values that make the inequality true. This means that your answer will often be a range of numbers. For example, if you have the inequality 2x + 5 < 15, you would solve it like this: 2x + 5 < 15 2x < 10 x < 5 So in this case, x can be any number less than 5 and the inequality will still be true.\n\nIn mathematics, \"solving for x\" refers to the process of finding the value of an unknown variable in an equation. In most equations, the variable is represented by the letter \"x.\" Fractions can be used to solve for x in a number of ways. For example, if the equation is 2x + 1 = 7, one can isolated the x term by subtracting 1 from each side and then dividing each side by 2. This would leave x with a value of 3. In some cases, more than one step may be necessary to solve for x. For example, if the equation is 4x/3 + 5 = 11, one would first need to multiply both sides of the equation by 3 in order to cancel out the 4x/3 term. This would give 12x + 15 = 33. From there, one could subtract 15 from each side to find that x = 18/12, or 1.5. As these examples demonstrate, solving for x with fractions is a matter of careful algebraic manipulation. With a little practice, anyone can master this essential math skill.\n\nIn some cases, you may need to do a bit of research to find the answer. However, if you take your time and carefully read the question, you should be able to find the correct answer. With a little practice, you will be able to confidently answer math questions and improve your understanding of the subject.\n\nTrying to solve a binomial equation can be frustrating, especially when you don't have the right tools. A binomial solver can be a helpful tool for anyone struggling with this type of equation. Binomial equations are quadratic equations with twoterms, and they can be difficult to solve without the proper tools. The Binomial theorem is a formula that allows you to expand these equations, and a Binomial solver can help you to use this theorem to solve your equation. With a Binomial solver, you can input the coefficients of your equation and see the expanded form of the equation. This can be helpful in finding the roots of your equation. Binomial solvers are easy to use and can be a valuable tool for anyone struggling with binomial equations.\n\n## We solve all types of math troubles", null, "This is the best math-app I have ever used in my life! It processes your photo of the math problem quick and hassle-free. It also features a well-equipped calculator that works flawlessly. Would really recommend!\n\nOra Roberts", null, "This is an amazing app that has helped me through many of my math problems. This has been especially helpful whenever I would study for a math exam since it shows all the solving steps and helps me understand the problems properly.\n\nDior Stewart\n\nHow to solve an expression Online differential equation solver Cpm mathematics homework help Quadratic function to standard form solver Algebra 1 problem solving" ]	[ null, "https://thekooshy.com/PSY6320a40ff2880/author-profile.jpg", null, "https://thekooshy.com/PSY6320a40ff2880/testimonial-one.jpg", null, "https://thekooshy.com/PSY6320a40ff2880/testimonial-two.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9367688,"math_prob":0.99364686,"size":4535,"snap":"2022-40-2023-06","text_gpt3_token_len":1027,"char_repetition_ratio":0.12690355,"word_repetition_ratio":0.03117506,"special_character_ratio":0.22557883,"punctuation_ratio":0.10064935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99928945,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T00:22:02Z\",\"WARC-Record-ID\":\"<urn:uuid:69f5a0b2-ea2a-40ba-a967-6c0d978c9cf0>\",\"Content-Length\":\"21222\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3fd5f0c7-cc31-4f7a-981c-f827fc53ef64>\",\"WARC-Concurrent-To\":\"<urn:uuid:58d01a3d-911f-41bb-80fb-d96221658e29>\",\"WARC-IP-Address\":\"107.167.10.253\",\"WARC-Target-URI\":\"https://thekooshy.com/wap.html\",\"WARC-Payload-Digest\":\"sha1:5DKS4UV4LK53DM3YYNQ73X3FZPKOM3KA\",\"WARC-Block-Digest\":\"sha1:M6SBXDWX6RUYURE4KB74X4G3WQUEWATE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499831.97_warc_CC-MAIN-20230130232547-20230131022547-00405.warc.gz\"}"}