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  • Is there an easily available implementation of erf() for Python?

    - by rog
    I can implement the error function, erf, myself, but I'd prefer not to. Is there a python package with no external dependencies that contains an implementation of this function? I have found http://pylab.sourceforge.net/packages/included_functions.htmlthis but this seems to be part of some much larger package (and it's not even clear which one!). I'm sorry if this is a naive question - I'm totally new to Python.

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  • java random percentages

    - by erw
    I need to generate n percentages (integers between 0 and 100) such that the sum of all n numbers adds up to 100. If I just do nextInt() n times, each time ensuring that the parameter is 100 minus the previously accumulated sum, then my percentages are biased (i.e. the first generated number will usually be largest etc.). How do I do this in an unbiased way?

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  • Fastest method to define whether a number is a triangular number

    - by psihodelia
    A triangular number is the sum of the n natural numbers from 1 to n. What is the fastest method to find whether a given positive integer number is a triangular one? I suppose, there must be a hidden pattern in a binary representation of such numbers (like if you need to find whether a number is even/odd you check its least significant bit). Here is a cut of the first 1200th up to 1300th triangular numbers, you can easily see a bit-pattern here (if not, try to zoom out): (720600, '10101111111011011000') (721801, '10110000001110001001') (723003, '10110000100000111011') (724206, '10110000110011101110') (725410, '10110001000110100010') (726615, '10110001011001010111') (727821, '10110001101100001101') (729028, '10110001111111000100') (730236, '10110010010001111100') (731445, '10110010100100110101') (732655, '10110010110111101111') (733866, '10110011001010101010') (735078, '10110011011101100110') (736291, '10110011110000100011') (737505, '10110100000011100001') (738720, '10110100010110100000') (739936, '10110100101001100000') (741153, '10110100111100100001') (742371, '10110101001111100011') (743590, '10110101100010100110') (744810, '10110101110101101010') (746031, '10110110001000101111') (747253, '10110110011011110101') (748476, '10110110101110111100') (749700, '10110111000010000100') (750925, '10110111010101001101') (752151, '10110111101000010111') (753378, '10110111111011100010') (754606, '10111000001110101110') (755835, '10111000100001111011') (757065, '10111000110101001001') (758296, '10111001001000011000') (759528, '10111001011011101000') (760761, '10111001101110111001') (761995, '10111010000010001011') (763230, '10111010010101011110') (764466, '10111010101000110010') (765703, '10111010111100000111') (766941, '10111011001111011101') (768180, '10111011100010110100') (769420, '10111011110110001100') (770661, '10111100001001100101') (771903, '10111100011100111111') (773146, '10111100110000011010') (774390, '10111101000011110110') (775635, '10111101010111010011') (776881, '10111101101010110001') (778128, '10111101111110010000') (779376, '10111110010001110000') (780625, '10111110100101010001') (781875, '10111110111000110011') (783126, '10111111001100010110') (784378, '10111111011111111010') (785631, '10111111110011011111') (786885, '11000000000111000101') (788140, '11000000011010101100') (789396, '11000000101110010100') (790653, '11000001000001111101') (791911, '11000001010101100111') (793170, '11000001101001010010') (794430, '11000001111100111110') (795691, '11000010010000101011') (796953, '11000010100100011001') (798216, '11000010111000001000') (799480, '11000011001011111000') (800745, '11000011011111101001') (802011, '11000011110011011011') (803278, '11000100000111001110') (804546, '11000100011011000010') (805815, '11000100101110110111') (807085, '11000101000010101101') (808356, '11000101010110100100') (809628, '11000101101010011100') (810901, '11000101111110010101') (812175, '11000110010010001111') (813450, '11000110100110001010') (814726, '11000110111010000110') (816003, '11000111001110000011') (817281, '11000111100010000001') (818560, '11000111110110000000') (819840, '11001000001010000000') (821121, '11001000011110000001') (822403, '11001000110010000011') (823686, '11001001000110000110') (824970, '11001001011010001010') (826255, '11001001101110001111') (827541, '11001010000010010101') (828828, '11001010010110011100') (830116, '11001010101010100100') (831405, '11001010111110101101') (832695, '11001011010010110111') (833986, '11001011100111000010') (835278, '11001011111011001110') (836571, '11001100001111011011') (837865, '11001100100011101001') (839160, '11001100110111111000') (840456, '11001101001100001000') (841753, '11001101100000011001') (843051, '11001101110100101011') (844350, '11001110001000111110') For example, can you also see a rotated normal distribution curve, represented by zeros between 807085 and 831405?

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  • Function to convert a z-score into a percentage

    - by Daniel
    Google doesn't want to help! I'm able to calculate z-scores, and we are trying to produce a function that given a z-score gives us a percent of the population in a normal distribution that would be under that z-score. All I can find are references to z-score to percentage tables. Any pointers?

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  • division problems

    - by David
    This line of code: System.out.println ("aray[j], "+aray[j]+", divided by sum, "+sum+", equals: aray[j]/sum: "+ aray[j]/sum) ; is yeilding this line of text: aray[j], 21, divided by sum, 100, equals: aray[j]/sum: 0 why is it doing this? (everything is right eccept that the answer should be .21)

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  • License key pattern detection?

    - by Ricket
    This is not a real situation; please ignore legal issues that you might think apply, because they don't. Let's say I have a set of 200 known valid license keys for a hypothetical piece of software's licensing algorithm, and a license key consists of 5 sets of 5 alphanumeric case-insensitive (all uppercase) characters. Example: HXDY6-R3DD7-Y8FRT-UNPVT-JSKON Is it possible (or likely) to extrapolate other possible keys for the system? What if the set was known to be consecutive; how do the methods change for this situation, and what kind of advantage does this give? I have heard of "keygens" before, but I believe they are probably made by decompiling the licensing software rather than examining known valid keys. In this case, I am only given the set of keys and I must determine the algorithm. I'm also told it is an industry standard algorithm, so it's probably not something basic, though the chance is always there I suppose. If you think this doesn't belong in Stack Overflow, please at least suggest an alternate place for me to look or ask the question. I honestly don't know where to begin with a problem like this. I don't even know the terminology for this kind of problem.

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  • log (1-var) operation in C

    - by heike
    I am trying to code this algorithm. I am stuck in the part of log((1.0-u)/u))/beta; As I understand, I can not get the result of this in C, as it will always return me with negative value log (returning imaginary value). Tried to print the result of log(1-5) for instance, it gives me with Nan. How can I get the result of double x = (alpha - log((1.0-u)/u))/beta then? Would appreciate for any pointers to solve this problem. Thank you

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  • PHP: Trying to come up with a "prev" and "next" link

    - by fwaokda
    I'm displaying 10 records per page. The variables I have currently that I'm working with are.. $total = total number of records $page = whats the current page I'm displaying I placed this at the top of my page... if ( $_GET['page'] == '' ) { $page = 1; } //if no page is specified set it to `1` else { $page = ($_GET['page']); } // if page is specified set it Here are my two links... if ( $page != 1 ) { echo '<div style="float:left" ><a href="index.php?page='. ( $page - 1 ) .'" rev="prev" >Prev</a></div>'; } if ( !( ( $total / ( 10 * $page ) ) < $page ) ) { echo '<div style="float:right" ><a href="index.php?page='. ( $page + 1 ) .'" rev="next" >Next</a></div>'; } Now I guess (unless I'm not thinking of something) that I can display the "Prev" link every time except when the page is '1'. How can make it where the "Next" link doesn't show on the last page though?

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  • Populate array from vector

    - by Zag zag..
    Hi, I would like to populate an 2 dimensional array, from a vector. I think the best way to explain myself is to put some examples (with a array of [3,5] length). When vector is: [1, 0] [ [4, 3, 2, 1, 0], [4, 3, 2, 1, 0], [4, 3, 2, 1, 0] ] When vector is: [-1, 0] [ [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4] ] When vector is: [-2, 0] [ [0, 0, 1, 1, 2], [0, 0, 1, 1, 2], [0, 0, 1, 1, 2] ] When vector is: [1, 1] [ [2, 2, 2, 1, 0], [1, 1, 1, 1, 0], [0, 0, 0, 0, 0] ] When vector is: [0, 1] [ [2, 2, 2, 2, 2], [1, 1, 1, 1, 1], [0, 0, 0, 0, 0] ] Have you got any ideas, a good library or a plan? Any comments are welcome. Thanks. Note: I consulted Ruby "Matrix" and "Vector" classes, but I don't see any way to use it in my way... Edit: In fact, each value is the number of cells (from the current cell to the last cell) according to the given vector. If we take the example where the vector is [-2, 0], with the value *1* (at array[2, 3]): array = [ [<0>, <0>, <1>, <1>, <2>], [<0>, <0>, <1>, <1>, <2>], [<0>, <0>, <1>, *1*, <2>] ] ... we could think such as: The vector [-2, 0] means that -2 is for cols and 0 is for rows. So if we are in array[2, 3], we can move 1 time on the left (left because 2 is negative) with 2 length (because -2.abs == 2). And we don't move on the top or bottom, because of 0 for rows.

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  • Mysterious combination

    - by pstone
    I decided to learn concurrency and wanted to find out in how many ways instructions from two different processes could overlap. The code for both processes is just a 10 iteration loop with 3 instructions performed in each iteration. I figured out the problem consisted of leaving X instructions fixed at a point and then fit the other X instructions from the other process between the spaces taking into account that they must be ordered (instruction 4 of process B must always come before instruction 20). I wrote a program to count this number, looking at the results I found out that the solution is n Combination k, where k is the number of instructions executed throughout the whole loop of one process, so for 10 iterations it would be 30, and n is k*2 (2 processes). In other words, n number of objects with n/2 fixed and having to fit n/2 among the spaces without the latter n/2 losing their order. Ok problem solved. No, not really. I have no idea why this is, I understand that the definition of a combination is, in how many ways can you take k elements from a group of n such that all the groups are different but the order in which you take the elements doesn't matter. In this case we have n elements and we are actually taking them all, because all the instructions are executed ( n C n). If one explains it by saying that there are 2k blue (A) and red (B) objects in a bag and you take k objects from the bag, you are still only taking k instructions when 2k instructions are actually executed. Can you please shed some light into this? Thanks in advance.

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  • Calculating percent "x/y * 100" always results in 0?

    - by Patrick Beninga
    In my assignment i have to make a simple version of Craps, for some reason the percentage assignments always produce 0 even when both variables are non 0, here is the code. import java.util.Random; Header, note the variables public class Craps { private int die1, die2,myRoll ,myBet,point,myWins,myLosses; private double winPercent,lossPercent; private Random r = new Random(); Just rolls two dies and produces their some. public int roll(){ die1 = r.nextInt(6)+1; die2 = r.nextInt(6)+1; return(die1 + die2); } The Play method, this just loops through the game. public void play(){ myRoll = roll(); point = 0; if(myRoll == 2 ||myRoll == 3 || myRoll == 12){ System.out.println("You lose!"); myLosses++; }else if(myRoll == 7 || myRoll == 11){ System.out.println("You win!"); myWins++; }else{ point = myRoll; do { myRoll = roll(); }while(myRoll != 7 && myRoll != point); if(myRoll == point){ System.out.println("You win!"); myWins++; }else{ System.out.println("You lose!"); myLosses++; } } } This is where the bug is, this is the tester method. public void tester(int howMany){ int i = 0; while(i < howMany){ play(); i++; } bug is right here in these assignments statements winPercent = myWins/i * 100; lossPercent = myLosses/i* 100; System.out.println("program ran "+i+" times "+winPercent+"% wins "+ lossPercent+"% losses with "+myWins+" wins and "+myLosses+" losses"); } }

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  • Minimizing distance to a weighted grid

    - by Andrew Tomazos - Fathomling
    Lets suppose you have a 1000x1000 grid of positive integer weights W. We want to find the cell that minimizes the average weighted distance.to each cell. The brute force way to do this would be to loop over each candidate cell and calculate the distance: int best_x, best_y, best_dist; for x0 = 1:1000, for y0 = 1:1000, int total_dist = 0; for x1 = 1:1000, for y1 = 1:1000, total_dist += W[x1,y1] * sqrt((x0-x1)^2 + (y0-y1)^2); if (total_dist < best_dist) best_x = x0; best_y = y0; best_dist = total_dist; This takes ~10^12 operations, which is too long. Is there a way to do this in or near ~10^8 or so operations?

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  • Replace for loop with formula

    - by hamax
    I have this loop that runs in O(end - start) and I would like to replace it with something O(1). If "width" wouldn't be decreasing, it would be pretty simple. for (int i = start; i <= end; i++, width--) if (i % 3 > 0) // 1 or 2, but not 0 z += width; start, end and width have positive values

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  • Write number N in base M

    - by VaioIsBorn
    I know how to do it mathematically, but i want it now to do it in c++ using some easy algorithm. Is is possible? The question is that i need some methods/ideas for writing a number N in base M, for example 1410 in base 3: (14)10 = 2*(3^0) + 1*(3^1) + 1*(3^2) = (112)3 etc.

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  • Permutations with extra restrictions

    - by Full Decent
    I have a set of items, for example: {1,1,1,2,2,3,3,3}, and a restricting set of sets, for example {{3},{1,2},{1,2,3},{1,2,3},{1,2,3},{1,2,3},{2,3},{2,3}. I am looking for permutations of items, but the first element must be 3, and the second must be 1 or 2, etc. One such permutation that fits is: {3,1,1,1,2,2,3} Is there an algorithm to count all permutations for this problem in general? Is there a name for this type of problem? For illustration, I know how to solve this problem for certain types of "restricting sets". Set of items: {1,1,2,2,3}, Restrictions {{1,2},{1,2,3},{1,2,3},{1,2},{1,2}}. This is equal to 2!/(2-1)!/1! * 4!/2!/2!. Effectively permuting the 3 first, since it is the most restrictive and then permuting the remaining items where there is room.

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  • Solve Physics exercise by brute force approach..

    - by Nils
    Being unable to reproduce a given result. (either because it's wrong or because I was doing something wrong) I was asking myself if it would be easy to just write a small program which takes all the constants and given number and permutes it with a possible operators (* / - + exp(..)) etc) until the result is found. Permutations of n distinct objects with repetition allowed is n^r. At least as long as r is small I think you should be able to do this. I wonder if anybody did something similar here..

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  • Assigning XY positions to points based on a "weight" between them

    - by sanity
    I have a bunch of points in a graph, and for every pair of these points I have "weight" value indicating what their proximity should be, between -1 and 1. I want to choose XY coordinates for these points such that those that have a proximity of 1 are in the same position, and those with a proximity of -1 are distant from each-other. All points must reside within a bounded area. What algorithms should I investigate to achieve this?

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