Search Results

Search found 9047 results on 362 pages for 'double math'.

Page 73/362 | < Previous Page | 69 70 71 72 73 74 75 76 77 78 79 80  | Next Page >

  • Histrogram matching - image processing - c/c++

    - by Raj
    Hello I have two histograms. int Hist1[10] = {1,4,3,5,2,5,4,6,3,2}; int Hist1[10] = {1,4,3,15,12,15,4,6,3,2}; Hist1's distribution is of type multi-modal; Hist2's distribution is of type uni-modal with single prominent peak. My questions are Is there any way that i could determine the type of distribution programmatically? How to quantify whether these two histograms are similar/dissimilar? Thanks

    Read the article

  • A Plot Graph .NET WindowsForm Component Free

    - by user255946
    Hello All, I'm searching for a plot .NET component to plot a 2D line chart, given an array of data. It will be used with WindowsForm (C#) and It will be very helpful if it could be freeware. It is for a scientific application. This is my first asked question in stackoverflow, and excuse me for my terrible English written.

    Read the article

  • Intersection of line and rectangle with maximum segment length

    - by Aarkan
    I have a vector represented by the slope m. Then there is rectangle (assume axis aligned), which is represented by top-left and bottom-right corner. Of course, there may be many lines with slope m and intersecting the given rectangle. The problem is to find out the line whose length of line intercept inside the rectangle is maximum among all such lines. i.e., if the line intersects rectangle at P1 and P2, then the problem is to find the equation of line for which length of P1P2 is maximum. I proceeded like this. Let the line is: y = m*x + c. Then find out the intersection with each side of rectangle and finding out the maxima for distance function between each pair of points. But it will only give me the length of line segment and there seem to be many corner cases to handle. Could anyone please suggest a better way to do this. Thanks in advance.

    Read the article

  • Basic C++ code for multiplication of 2 matrix or vectors (C++ beginner)

    - by Ice
    I am a new C++ user and I am also doing a major in Maths so thought I would try implement a simple calculator. I got some code off the internet and now I just need help to multiply elements of 2 matrices or vectors. Matrixf multiply(Matrixf const& left, Matrixf const& right) { // error check if (left.ncols() != right.nrows()) { throw std::runtime_error("Unable to multiply: matrix dimensions not agree."); } /* I have all the other part of the code for matrix*/ /** Now I am not sure how to implement multiplication of vector or matrix.**/ Matrixf ret(1, 1); return ret; }

    Read the article

  • 3D points to quaternions

    - by Hubrus
    For the simplicity, we'll consider two 3D points, that moves one relatively to other, in time. Let's say: at moment t0, we have P1(0,0,0) and P2(0,2,0) at moment t1, P1 is still (0,0,0) but P2 changed to (0,2,2). From what I've understood reading about quaternions, is that, at moment t0, Q1 (representing P1) and Q2 (representing P2) will be both (0, 0, 0, 0). But at the moment t1, Q2 will become something else (w, x, y, z). How do I calculate the Q2 at t1 moment? I've googled a lot on this subject, but I was able to find only rotation between quaternions. I will appreciate any guidance. Thanks!

    Read the article

  • Shortest distance between points on a toroidally wrapped (x- and y- wrapping) map?

    - by mstksg
    I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean rectangle, but when a point moves to the right boundary, it will appear at the left boundary (at the same y value), given by x_new = x_old % width Basically, points are plotted based on: (x_new, y_new) = ( x_old % width, y_old % height) Think Pac Man -- walking off one edge of the screen will make you appear on the opposite edge. What's the best way to calculate the shortest distance between two points? The typical implementation suggests a large distance for points on opposite corners of the map, when in reality, the real wrapped distance is very close. The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula. But that would involve many checks, calculations, operations -- some that I feel might be unnecessary. Is there a better way?

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • Wpf. Chart optimization. More than million points

    - by Evgeny
    I have custom control - chart with size, for example, 300x300 pixels and more than one million points (maybe less) in it. And its clear that now he works very slowly. I am searching for algoritm which will show only few points with minimal visual difference. I have link to component which have functionallity exactly what i need (2 million points demo): http://www.mindscape.co.nz/demo/SilverlightElements/demopage.html#/ChartOverviewPage I will be grateful for any matherials, links or thoughts how to realize such functionallity.

    Read the article

  • How to calculate the state of a graph?

    - by zcb
    Given a graph G=(V,E), each node i is associated with 'Ci' number of objects. At each step, for every node i, the Ci objects will be taken away by the neighbors of i equally. After K steps, output the number of objects of the top five nodes which has the most objects. Some Constrains: |V|<10^5, |E|<2*10^5, K<10^7, Ci<1000 My current idea is: represent the transformation in each step with a matrix. This problem is converted to the calculation of the power of matrix. But this solution is much too slow considering |V| can be 10^5. Is there any faster way to do it?

    Read the article

  • Scaling range of values with negative numbers

    - by Pradeep Kumar
    How can I scale a set of values to fit a new range if they include negative numbers? For example, I have a set of numbers (-10, -9, 1, 4, 10) which have to scaled to a range [0 1], such that -10 maps to 0, and 10 maps to 1. The regular method for an arbitrary number 'x' would be: (x - from_min) * (to_max - to_min) / (from_max - from_min) + to_min but this does not work for negative numbers. Any help is appreciated. Thanks!!

    Read the article

  • Dilemma with two types and operator +

    - by user35443
    I have small problem with operators. I have this code: public class A { public string Name { get; set; } public A() { } public A(string Name) { this.Name = Name; } public static implicit operator B(A a) { return new B(a.Name); } public static A operator+(A a, A b) { return new A(a.Name + " " + b.Name); } } public class B { public string Name { get; set; } public B() { } public B(string Name) { this.Name = Name; } public static implicit operator A(B b) { return new A(b.Name); } public static B operator +(B b, B a) { return new B(b.Name + " " + a.Name); } } Now I want to know, which's conversion operator will be called and which's addition operator will be called in this operation: new A("a") + new B("b"); Will it be operator of A, or of B? (Or both?) Thanks....

    Read the article

  • 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

    Read the article

  • 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)

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

< Previous Page | 69 70 71 72 73 74 75 76 77 78 79 80  | Next Page >