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  • VFP Unit Matrix Multiply problem on the iPhone

    - by Ian Copland
    Hi. I'm trying to write a Matrix3x3 multiply using the Vector Floating Point on the iPhone, however i'm encountering some problems. This is my first attempt at writing any ARM assembly, so it could be a faily simple solution that i'm not seeing. I've currently got a small application running using a maths library that i've written. I'm investigating into the benifits using the Vector Floating Point Unit would provide so i've taken my matrix multiply and converted it to asm. Previously the application would run without a problem, however now my objects will all randomly disappear. This seems to be caused by the results from my matrix multiply becoming NAN at some point. Heres the code IMatrix3x3 operator*(IMatrix3x3 & _A, IMatrix3x3 & _B) { IMatrix3x3 C; //C++ code for the simulator #if TARGET_IPHONE_SIMULATOR == true C.A0 = _A.A0 * _B.A0 + _A.A1 * _B.B0 + _A.A2 * _B.C0; C.A1 = _A.A0 * _B.A1 + _A.A1 * _B.B1 + _A.A2 * _B.C1; C.A2 = _A.A0 * _B.A2 + _A.A1 * _B.B2 + _A.A2 * _B.C2; C.B0 = _A.B0 * _B.A0 + _A.B1 * _B.B0 + _A.B2 * _B.C0; C.B1 = _A.B0 * _B.A1 + _A.B1 * _B.B1 + _A.B2 * _B.C1; C.B2 = _A.B0 * _B.A2 + _A.B1 * _B.B2 + _A.B2 * _B.C2; C.C0 = _A.C0 * _B.A0 + _A.C1 * _B.B0 + _A.C2 * _B.C0; C.C1 = _A.C0 * _B.A1 + _A.C1 * _B.B1 + _A.C2 * _B.C1; C.C2 = _A.C0 * _B.A2 + _A.C1 * _B.B2 + _A.C2 * _B.C2; //VPU ARM asm for the device #else //create a pointer to the Matrices IMatrix3x3 * pA = &_A; IMatrix3x3 * pB = &_B; IMatrix3x3 * pC = &C; //asm code asm volatile( //turn on a vector depth of 3 "fmrx r0, fpscr \n\t" "bic r0, r0, #0x00370000 \n\t" "orr r0, r0, #0x00020000 \n\t" "fmxr fpscr, r0 \n\t" //load matrix B into the vector bank "fldmias %1, {s8-s16} \n\t" //load the first row of A into the scalar bank "fldmias %0!, {s0-s2} \n\t" //calulate C.A0, C.A1 and C.A2 "fmuls s17, s8, s0 \n\t" "fmacs s17, s11, s1 \n\t" "fmacs s17, s14, s2 \n\t" //save this into the output "fstmias %2!, {s17-s19} \n\t" //load the second row of A into the scalar bank "fldmias %0!, {s0-s2} \n\t" //calulate C.B0, C.B1 and C.B2 "fmuls s17, s8, s0 \n\t" "fmacs s17, s11, s1 \n\t" "fmacs s17, s14, s2 \n\t" //save this into the output "fstmias %2!, {s17-s19} \n\t" //load the third row of A into the scalar bank "fldmias %0!, {s0-s2} \n\t" //calulate C.C0, C.C1 and C.C2 "fmuls s17, s8, s0 \n\t" "fmacs s17, s11, s1 \n\t" "fmacs s17, s14, s2 \n\t" //save this into the output "fstmias %2!, {s17-s19} \n\t" //set the vector depth back to 1 "fmrx r0, fpscr \n\t" "bic r0, r0, #0x00370000 \n\t" "orr r0, r0, #0x00000000 \n\t" "fmxr fpscr, r0 \n\t" //pass the inputs and set the clobber list : "+r"(pA), "+r"(pB), "+r" (pC) : :"cc", "memory","s0", "s1", "s2", "s8", "s9", "s10", "s11", "s12", "s13", "s14", "s15", "s16", "s17", "s18", "s19" ); #endif return C; } As far as i can see that makes sence. While debugging i've managed to notice that if i were to say _A = C prior to the return and after the ASM, _A will not necessarily be equal to C which has only increased my confusion. I had thought it was possibly due to the pointers I'm giving to the VFPU being incrimented by lines such as "fldmias %0!, {s0-s2} \n\t" however my understanding of asm is not good enough to properly understand the problem, nor to see an alternative approach to that line of code. Anyway, I was hoping someone with a greater understanding than me would be able to see a solution, and any help would be greatly appreciated, thank you :-)

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  • How to round CGFloat

    - by Johannes Jensen
    I made this method + (CGFloat) round: (CGFloat)f { int a = f; CGFloat b = a; return b; } It works as expected but it only rounds down. And if it's a negative number it still rounds down. This was just a quick method I made, it isn't very important that it rounds correctly, I just made it to round the camera's x and y values for my game. Is this method okay? Is it fast? Or is there a better solution?

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  • How to embed AsciiMathML in Google Sites?

    - by Joannes Vermorel
    We would need to embed mathematical formulas through AsciiMathML into Google Sites pages (internal wiki for a research team). I am stuck with the limitation of Google Sites. Any idea how to do that? (ps: I have finally found a poorly practical work-around, but better ideas would still be appreciated)

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  • Draw fitted line (OpenCV)

    - by Sunny
    I'm using OpenCV to fit a line from a set of points using cvFitLine() cvFitLine() returns a normalized vector that is co-linear to the line and a point on the line. See details here Using this information how can I get the equation of a line so that I can draw the line?

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  • How to convert latitude or longitude to meters?

    - by Adam Taylor
    Hi, If I have a latitude or longitude reading in standard NMEA format is there an easy way / forumla to convert that reading to meters, which I can then implement in Java (J9)? Edit: Ok seems what I want to do is not possible /easily/, however what I really want to do is: Say I have a lat and long of a way point and a lat and long of a user is there an easy way to compare them to decide when to tell the user they are within a /reasonably/ close distance of the way point? I realise reasonable is subject but is this easily do-able or still overly maths-y? Thanks, Adam

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  • Calculating determinant by hand

    - by ldigas
    Okey, this is only half programming, but let's see how you are on terms with manual calculations. I believe many of you did this on your university's while giving "linear systems" ... the problem is it's been so long I can't remember how to do it any more. I know quite a few algorithms for calculating determinants, and they all work fine ... for large systems, where one would never try to do it manually. Unfortunatelly, I'm soon going on an exam, where I do have to calculate it manually, up to the system of 5. So, I have a K(omega) matrix that looks like this: [2-(omega^2)*c -4 2 0 0] [-2 5-(omega^2)*c -4 1 0] [1 -4 6-(omega^2)*c -4 1] [0 1 -4 5-(omega^2)*c -2] [0 0 2 -4 2-(omega^2)*c] and I need all the omegas which satisfy the det[K(omega)]=0 criteria. What would be a good way to calculate it so it can be repeated in a manual process ?

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  • What is O(n log n) or O(n log(log n))

    - by Mark Tomlin
    What does O, if indeed it is a Oh (As in the letter O) not the number Zero (0) mean? I think the n would be number, but I'm not sure as I'm not a 'real' computer programmer, just a hobbyist. And log would be logarithmic function, but I only know that because of smarter people then I have told me this, while never really explaining what a logarithm is. So please, in plain English, explain what this is, and the differences between the two (such as their applications.

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  • java cosine similarity problem

    - by agazerboy
    Hi again :) I developed some java program to calculate cosine similarity on the basis of TF*IDF. It worked very well. But there is one problem.... :( for example: If I have following two matrix and I want to calculate cosine similarity it does not work as rows are not same in length doc 1 1 2 3 4 5 6 doc 2 1 2 3 4 5 6 7 8 5 2 4 9 if rows and colums are same in length then my program works very well but it does not if rows and columns are not in same length. Any tips ???

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  • Calculating the square of BigInteger

    - by brickner
    Hi, I'm using .NET 4's System.Numerics.BigInteger structure. I need to calculate the square (x^2) of very large numbers. If x is a BigInteger, What is the time complexity of: x*x; or BigInteger.Pow(x,2); ? If it's worse than O(n^2), do you have a better implementation? Maybe something like Schönhage–Strassen algorithm?

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  • Polynomial division overloading operator (solved)

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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  • Modeling distribution of performance measurements

    - by peterchen
    How would you mathematically model the distribution of repeated real life performance measurements - "Real life" meaning you are not just looping over the code in question, but it is just a short snippet within a large application running in a typical user scenario? My experience shows that you usually have a peak around the average execution time that can be modeled adequately with a Gaussian distribution. In addition, there's a "long tail" containing outliers - often with a multiple of the average time. (The behavior is understandable considering the factors contributing to first execution penalty). My goal is to model aggregate values that reasonably reflect this, and can be calculated from aggregate values (like for the Gaussian, calculate mu and sigma from N, sum of values and sum of squares). In other terms, number of repetitions is unlimited, but memory and calculation requirements should be minimized. A normal Gaussian distribution can't model the long tail appropriately and will have the average biased strongly even by a very small percentage of outliers. I am looking for ideas, especially if this has been attempted/analysed before. I've checked various distributions models, and I think I could work out something, but my statistics is rusty and I might end up with an overblown solution. Oh, a complete shrink-wrapped solution would be fine, too ;) Other aspects / ideas: Sometimes you get "two humps" distributions, which would be acceptable in my scenario with a single mu/sigma covering both, but ideally would be identified separately. Extrapolating this, another approach would be a "floating probability density calculation" that uses only a limited buffer and adjusts automatically to the range (due to the long tail, bins may not be spaced evenly) - haven't found anything, but with some assumptions about the distribution it should be possible in principle. Why (since it was asked) - For a complex process we need to make guarantees such as "only 0.1% of runs exceed a limit of 3 seconds, and the average processing time is 2.8 seconds". The performance of an isolated piece of code can be very different from a normal run-time environment involving varying levels of disk and network access, background services, scheduled events that occur within a day, etc. This can be solved trivially by accumulating all data. However, to accumulate this data in production, the data produced needs to be limited. For analysis of isolated pieces of code, a gaussian deviation plus first run penalty is ok. That doesn't work anymore for the distributions found above. [edit] I've already got very good answers (and finally - maybe - some time to work on this). I'm starting a bounty to look for more input / ideas.

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  • Maths Question: number of different permutations

    - by KingCong
    This is more of a maths question than programming but I figure a lot of people here are pretty good at maths! :) My question is: Given a 9 x 9 grid (81 cells) that must contain the numbers 1 to 9 each exactly 9 times, how many different grids can be produced. The order of the numbers doesn't matter, for example the first row could contain nine 1's etc. This is related to Sudoku and we know the number of valid Sudoku grids is 6.67×10^21, so since my problem isn't constrained like Sudoku by having to have each of the 9 numbers in each row, column and box then the answer should be greater than 6.67×10^21. My first thought was that the answer is 81! however on further reflection this assume that the 81 number possible for each cell are different, distinct number. They are not, there are 81 possible numbers for each cell but only 9 possible different numbers. My next thought was then that each of the cells in the first row can be any number between 1 and 9. If by chance the first row happened to be all the same number, say all 1s, then each cell in the second row could only have 8 possibilites, 2-9. If this continued down until the last row then number of different permutations could be calculated by 9^2 * 8^2 * 7^2 ..... * 1^2. However this doesn't work if each row doesn't contain 9 of the same number. It's been quite a while since I studied this stuff and I can't think of a way to work it out, I'd appreciate any help anyone can offer.

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  • Game enemy move towards player

    - by Chris
    I'm creating a game in c++ and OpenGL and want an enemy to move towards the player. I tried finding the length of the hypotenuse between the player and the enemy. how could i get the enemy to move down that line?

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  • correct fisheye distortion

    - by Will
    I have some points that describe positions in a picture taken with a fisheye lens. I've found this description of how to generate a fisheye effect, but not how to reverse it. How do you calculate the radial distance from the centre to go from fisheye to rectilinear? My function stub looks like this: Point correct_fisheye(const Point& p,const Size& img) { // to polar const Point centre = {img.width/2,img.height/2}; const Point rel = {p.x-centre.x,p.y-centre.y}; const double theta = atan2(rel.y,rel.x); double R = sqrt((rel.x*rel.x)+(rel.y*rel.y)); // fisheye undistortion in here please //... change R ... // back to rectangular const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta)); fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y); return ret; }

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  • Polynomial division overloading operator

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) { polinom Result, temp; Iter i = poly.begin(); constIter j = P.poly.begin(); if (poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); *this = *this - Result; } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); temp = Result * P; *this = *this - temp; } else break; } } return Result; } The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Update Because no one seems to understand how i thought the algorithm, i'll explain: If the dividend contains only one term, we simply insert the quotient in Result, then we multiply it with the divisor ans subtract it from the first polynomial which stores the remainder. If the polynomial we do this until the second polynomial( P in this case) becomes bigger. I think this algorithm is called long division, isn't it? So based on these, can anyone help me with overloading the / operator correctly for my class? Thanks!

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  • How to know if two stocks move togheter?

    - by Damiano
    Hello, I have two stocks with their prices, example: STOCK1: 10.56 11.23 12.32 8.90 STOCK2: 1.26 5.80 3.26 10.3 I only found Pearson correlation, but, is there another method to know if two stocks move togheter? (esample: co-integration??) Thank you so much!

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  • Fixing VBSCRIPT inaccurate mathematical results due to rounding

    - by jay
    Try running this in a .VBS file MsgBox(545.14-544.94) You get a neat little answer of 0.199999999999932! This rounding issue also occurs unfortunately in Sin(2 * pi) since VB can only ever see the (user defined) variable pi as accurate as 3.14159265358979. Is rounding it manually (and loosing accuracy) the only way to improve the result? What is the most effective way of dealing with this kind of problem?

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  • Test if a number is fibonacci

    - by VaioIsBorn
    I know how to make the list of the Fibonacci numbers, but i don't know how can i test if a given number belongs to the fibonacci list - one way that comes in mind is generate the list of fib. numbers up to that number and see if it belongs to the array, but there's got to be another, simpler and faster method. Any ideas ?

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  • Orcad / Matlab: How to plot the roots of a polynomial

    - by Tom
    Hi everyone, Im trying to plot the roots of a polynomial, and i just cant get it. First i create my polynomial p5 = [1 0 0 0 0 -1] %x^5 - 1 r5 = roots(p5) stem (p5) Im using the stem function, but I would like to remove the stems, and just get the circle around the roots. Is this possible, is stem the right command? Thanks in advance, PS: This is not homework, but very close, will tag it if requested.

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  • How to determine the modulus of a Float in Ada 95

    - by mat_geek
    I need to determine the amount left of a time cycle. To do that in C I would use fmod. But in ada I can find no reference to a similar function. It needs to be accurate and it needs to return a float for precision. So how do I determine the modulus of a Float in Ada 95? elapsed := time_taken mod 10.348; left := 10.348 - elapsed; delay Duration(left);

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  • List all possible combinations of k integers between 1...n (n choose k)

    - by Asaf R
    Hi, Out of no particular reason I decided to look for an algorithm that produces all possible choices of k integers between 1...n, where the order amongst the k integer doesn't matter (the n choose k thingy). From the exact same reason, which is no reason at all, I also implemented it in C#. My question is: Do you see any mistake in my algorithm or code? And, more importantly, can you suggest a better algorithm? Please pay more attention to the algorithm than the code itself. It's not the prettiest code I've ever written, although do tell if you see an error. EDIT: Alogirthm explained - We hold k indices. This creates k nested for loops, where loop i's index is indices[i]. It simulates k for loops where indices[i+1] belongs to a loop nested within the loop of indices[i]. indices[i] runs from indices[i - 1] + 1 to n - k + i + 1. CODE: public class AllPossibleCombination { int n, k; int[] indices; List<int[]> combinations = null; public AllPossibleCombination(int n_, int k_) { if (n_ <= 0) { throw new ArgumentException("n_ must be in N+"); } if (k_ <= 0) { throw new ArgumentException("k_ must be in N+"); } if (k_ > n_) { throw new ArgumentException("k_ can be at most n_"); } n = n_; k = k_; indices = new int[k]; indices[0] = 1; } /// <summary> /// Returns all possible k combination of 0..n-1 /// </summary> /// <returns></returns> public List<int[]> GetCombinations() { if (combinations == null) { combinations = new List<int[]>(); Iterate(0); } return combinations; } private void Iterate(int ii) { // // Initialize // if (ii > 0) { indices[ii] = indices[ii - 1] + 1; } for (; indices[ii] <= (n - k + ii + 1); indices[ii]++) { if (ii < k - 1) { Iterate(ii + 1); } else { int[] combination = new int[k]; indices.CopyTo(combination, 0); combinations.Add(combination); } } } } I apologize for the long question, it might be fit for a blog post, but I do want the community's opinion here. Thanks, Asaf

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