greatest-common-divisor

I have an array of size n. I need to find the GCD of each element with a given number and if it's greater than 1, add it to another array. What's the fastest way to do this? int gcd(int a, int b) { if(b == 0) { return a; } else { return gcd(b, a % b); } } The following code uses the normal method that we humans use to calculate the GCD and is by far, according to me the fastest way to find GCD(HCF) of 2 numbers: #include <iostream> using namespace std; int main() { int N1; int N2; cin<<N1; cin<<N2; int small=N1; int big=N2; if (N1>N2){ N1=small; N2=big; } int P=1; int i=1; while (i<=N1){ if

I want to simplify a fraction in my application. The fraction is like, x/y where x and y are integers. I want to simplify the fraction to its simplest form. Can anyone please give me hints how to do it. Thanks in advance. Compute the greatest common divisor for x and y Divide both of them by the GCD Euclid's algorithm is an easy way to compute the GCD. Divide both by gcd(x,y) The Binary GCD algorithm is a fast way to compute the GCD on a computer. #include<iostream> using namespace std; struct fraction { int n1, d1, n2, d2, s1, s2; }; void simplification(int a,int b) { bool e = true; int t;

Does numpy have a gcd function somewhere in its structure of modules? I'm aware of fractions.gcd but thought a numpy equivalent maybe potentially quicker and work better with numpy datatypes. I have been unable to uncover anything on google other than this link which seems out of date and I don't know how I would access the _gcd function it suggests exists. Naively trying: np.gcd np.euclid hasn't worked for me... You can write it yourself: def numpy_gcd(a, b): a, b = np.broadcast_arrays(a, b) a = a.copy() b = b.copy() pos = np.nonzero(b)[0] while len(pos) > 0: b2 = b[pos] a[pos], b[pos] = b2,

Can someone give an example for finding greatest common divisor algorithm for more than two numbers? I believe programming language doesn't matter. Start with the first pair and get their GCD, then take the GCD of that result and the next number. The obvious optimization is you can stop if the running GCD ever reaches 1. I'm watching this one to see if there are any other optimizations. :) Oh, and this can be easily parallelized since the operations are commutative/associative. Michael Foukarakis The GCD of 3 numbers can be computed as gcd(a, b, c) = gcd(gcd(a, b), c) . You can apply the

I'm writing a mixed numeral class and need a quick and easy 'greatest common divisor' function. Can anyone give me the code or a link to the code? I'm tempted to vote to close -- it seems difficult to believe that an implementation would be hard to find, but who knows for sure. template <typename Number> Number GCD(Number u, Number v) { while (v != 0) { Number r = u % v; u = v; v = r; } return u; } In C++ 17 or newer, you can just #include <numeric> , and use std::gcd (and if you care, about the gcd, chances are pretty fair that you'll be interested in the std::lcm that was added as well). The

Suppose you have a list of floating point numbers that are approximately multiples of a common quantity, for example 2.468, 3.700, 6.1699 which are approximately all multiples of 1.234. How would you characterize this "approximate gcd", and how would you proceed to compute or estimate it? Strictly related to my answer to this question . You can run Euclid's gcd algorithm with anything smaller then 0.01 (or a small number of your choice) being a pseudo 0. With your numbers: 3.700 = 1 * 2.468 + 1.232, 2.468 = 2 * 1.232 + 0.004. So the pseudo gcd of the first two numbers is 1.232. Now you take

I would like to find the greatest common divisor using JavaScript. Anyone done that before and willing to share? Here is a recursive solution. var gcd = function(a, b) { if ( ! b) { return a; } return gcd(b, a % b); }; Our base case is when b is equal to 0 . In this case, we return a . When we're recursing, we swap the input arguments but we pass the remainder of a / b as the second argument. Taken from Wikipedia. Recursive: function gcd_rec(a, b) { if (b) { return gcd_rec(b, a % b); } else { return Math.abs(a); } } Iterative: function gcd(a,b) { a = Math.abs(a); b = Math.abs(b); if (b > a)