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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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  • Required Working Precision for the BBP Algorithm?

    - by brainfsck
    Hello, I'm looking to compute the nth digit of Pi in a low-memory environment. As I don't have decimals available to me, this integer-only BBP algorithm in Python has been a great starting point. I only need to calculate one digit of Pi at a time. How can I determine the lowest I can set D, the "number of digits of working precision"? D=4 gives me many correct digits, but a few digits will be off by one. For example, computing digit 393 with precision of 4 gives me 0xafda, from which I extract the digit 0xa. However, the correct digit is 0xb. No matter how high I set D, it seems that testing a sufficient number of digits finds an one where the formula returns an incorrect value. I've tried upping the precision when the digit is "close" to another, e.g. 0x3fff or 0x1000, but cannot find any good definition of "close"; for instance, calculating at digit 9798 gives me 0xcde6 , which is not very close to 0xd000, but the correct digit is 0xd. Can anyone help me figure out how much working precision is needed to calculate a given digit using this algorithm? Thank you,

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  • Encoding / Error Correction Challenge

    - by emi1faber
    Is it mathematically feasible to encode and initial 4 byte message into 8 bytes and if one of the 8 bytes is completely dropped and another is wrong to reconstruct the initial 4 byte message? There would be no way to retransmit nor would the location of the dropped byte be known. If one uses Reed Solomon error correction with 4 "parity" bytes tacked on to the end of the 4 "data" bytes, such as DDDDPPPP, and you end up with DDDEPPP (where E is an error) and a parity byte has been dropped, I don't believe there's a way to reconstruct the initial message (although correct me if I am wrong)... What about multiplying (or performing another mathematical operation) the initial 4 byte message by a constant, then utilizing properties of an inverse mathematical operation to determine what byte was dropped. Or, impose some constraints on the structure of the message so every other byte needs to be odd and the others need to be even. Alternatively, instead of bytes, it could also be 4 decimal digits encoded in some fashion into 8 decimal digits where errors could be detected & corrected under the same circumstances mentioned above - no retransmission and the location of the dropped byte is not known. I'm looking for any crazy ideas anyone might have... Any ideas out there?

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  • Percentage calculation around 0.5 (0.4 = -20% and 0.6 = +20%)

    - by Micheal
    I'm in a strange situation where I have a value of 0.5 and I want to convert the values from 0.5 to 1 to be a percentage and from 0.5 to 0 to be a negative percentage. As it says in the title 0.4 should be -20%, 0.3 should be -40% and 0.1 should be -80%. I'm sure this is a simple problem, but my mind is just refusing to figure it out :) Can anyone help? :)

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  • Code Golf: Evaluating Mathematical Expressions

    - by Noldorin
    Challenge Here is the challenge (of my own invention, though I wouldn't be surprised if it has previously appeared elsewhere on the web). Write a function that takes a single argument that is a string representation of a simple mathematical expression and evaluates it as a floating point value. A "simple expression" may include any of the following: positive or negative decimal numbers, +, -, *, /, (, ). Expressions use (normal) infix notation. Operators should be evaluated in the order they appear, i.e. not as in BODMAS, though brackets should be correctly observed, of course. The function should return the correct result for any possible expression of this form. However, the function does not have to handle malformed expressions (i.e. ones with bad syntax). Examples of expressions: 1 + 3 / -8 = -0.5 (No BODMAS) 2*3*4*5+99 = 219 4 * (9 - 4) / (2 * 6 - 2) + 8 = 10 1 + ((123 * 3 - 69) / 100) = 4 2.45/8.5*9.27+(5*0.0023) = 2.68... Rules I anticipate some form of "cheating"/craftiness here, so please let me forewarn against it! By cheating, I refer to the use of the eval or equivalent function in dynamic languages such as JavaScript or PHP, or equally compiling and executing code on the fly. (I think my specification of "no BODMAS" has pretty much guaranteed this however.) Apart from that, there are no restrictions. I anticipate a few Regex solutions here, but it would be nice to see more than just that. Now, I'm mainly interested in a C#/.NET solution here, but any other language would be perfectly acceptable too (in particular, F# and Python for the functional/mixed approaches). I haven't yet decided whether I'm going to accept the shortest or most ingenious solution (at least for the language) as the answer, but I would welcome any form of solution in any language, except what I've just prohibited above! My Solution I've now posted my C# solution here (403 chars). Update: My new solution has beaten the old one significantly at 294 chars, with the help of a bit of lovely regex! I suspected that this will get easily beaten by some of the languages out there with lighter syntax (particularly the funcional/dynamic ones), and have been proved right, but I'd be curious if someone could beat this in C# still. Update I've seen some very crafty solutions already. Thanks to everyone who has posted one. Although I haven't tested any of them yet, I'm going to trust people and assume they at least work with all of the given examples. Just for the note, re-entrancy (i.e. thread-safety) is not a requirement for the function, though it is a bonus. Format Please post all answers in the following format for the purpose of easy comparison: Language Number of characters: ??? Fully obfuscated function: (code here) Clear/semi-obfuscated function: (code here) Any notes on the algorithm/clever shortcuts it takes.

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  • Best fit curve for trend line

    - by Dave Jarvis
    Problem Constraints Size of the data set, but not the data itself, is known. Data set grows by one data point at a time. Trend line is graphed one data point at a time (using a spline/Bezier curve). Graphs The collage below shows data sets with reasonably accurate trend lines: The graphs are: Upper-left. By hour, with ~24 data points. Upper-right. By day for one year, with ~365 data points. Lower-left. By week for one year, with ~52 data points. Lower-right. By month for one year, with ~12 data points. User Inputs The user can select: the type of time series (hourly, daily, monthly, quarterly, annual); and the start and end dates for the time series. For example, the user could select a daily report for 30 days in June. Trend Weight To calculate the window size (i.e., the number of data points to average when calculating the trend line), the following expression is used: data points / trend weight Where data points is derived from user inputs and trend weight is 6.4. Even though a trend weight of 6.4 produces good fits, it is rather arbitrary, and might not be appropriate for different user inputs. Question How should trend weight be calculated given the constraints of this problem?

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  • java jama array problem

    - by agazerboy
    Hi All, I asked a question before but duffymo said it is not clear so i am going to post it again here. I am using Jama api for SVD calculation. I know very well about jama and SVD. Jama does not work if your column are more than rows. I have this situation. What should I do?? any help? I can't transpose the matrix too as it can produce wrong results. Thanks. P.S: I am calculating LSI with the help of jama. I am going like column(docs) and rows ( terms )

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  • polynomial multiplication using fastfourier transform

    - by mawia
    i am going through the above topic from CLRS(CORMEN) (page 834) and I got stuck at this point. Can anybody please explain how the following expression, A(x)=A^{[0]}(x^2) +xA^{[1]}(x^2) follows from, n-1 ` S a_j x^j j=0 Where, A^{[0]} = a_0 + a_2x + a_4a^x ... a_{n-2}x^{\frac{n}{2-1}} A^{[1]} = a_1 + a_3x + a_5a^x ... a_{n-1}x^{\frac{n}{2-1}} WITH REGARDS THANKS

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  • Fast permutation -> number -> permutation mapping algorithms

    - by ijw
    I have n elements. For the sake of an example, let's say, 7 elements, 1234567. I know there are 7! = 5040 permutations possible of these 7 elements. I want a fast algorithm comprising two functions: f(number) maps a number between 0 and 5039 to a unique permutation, and f'(permutation) maps the permutation back to the number that it was generated from. I don't care about the correspondence between number and permutation, providing each permutation has its own unique number. So, for instance, I might have functions where f(0) = '1234567' f'('1234567') = 0 The fastest algorithm that comes to mind is to enumerate all permutations and create a lookup table in both directions, so that, once the tables are created, f(0) would be O(1) and f('1234567') would be a lookup on a string. However, this is memory hungry, particularly when n becomes large. Can anyone propose another algorithm that would work quickly and without the memory disadvantage?

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