greatest-common-divisor

Using Euclid Algorithm to find GCF(GCD)

由 淺唱寂寞╮ 提交于 2019-12-08 08:19:26
I am trying to write a function to find the gcd of 2 numbers, using Euclid's Algorithm which I found here . From the larger number, subtract the smaller number as many times as you can until you have a number that is smaller than the small number. (or without getting a negative answer) Now, using the original small number and the result, a smaller number, repeat the process. Repeat this until the last result is zero, and the GCF is the next-to-last small number result. Also see our Euclid's Algorithm Calculator. Example: Find the GCF (18, 27) 27 - 18 = 9 18 - 9 = 9 9 - 9 = 0 So, the greatest

time complexity of below gcd algorithm

由 人盡茶涼 提交于 2019-12-08 03:47:35
问题 I have been finding it difficult to calculate the time complexity of The binary GCD algorithm, also known as Stein's algorithm which is given to be O(n^2) where n is the number of bits in the larger of the two numbers. Shouldn't it be O(n) ? Algorithm is as below : 1.gcd(0, v) = v, because everything divides zero, and v is the largest number that divides v. Similarly, gcd(u, 0) = u. gcd(0, 0) is not typically defined, but it is convenient to set gcd(0, 0) = 0. 2.If u and v are both even, then

Using Euclid Algorithm to find GCF(GCD)

由 生来就可爱ヽ(ⅴ＜●) 提交于 2019-12-08 03:41:57
问题 I am trying to write a function to find the gcd of 2 numbers, using Euclid's Algorithm which I found here. From the larger number, subtract the smaller number as many times as you can until you have a number that is smaller than the small number. (or without getting a negative answer) Now, using the original small number and the result, a smaller number, repeat the process. Repeat this until the last result is zero, and the GCF is the next-to-last small number result. Also see our Euclid's

GCD and LCM relation

由 岁酱吖の 提交于 2019-12-07 05:49:45
问题 The following relation works only for two (3, 12) numbers, it fails to produce the right answer when used for three numbers (3,12,10) . Just wondering if is it my understanding or it is just for two numbers and for me same is true for Euclid algorithm as well. LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) 回答1: The analogous formulas to LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) with three variables are simply not valid, as your example with (3, 12, 10)

BigIntegers, gcd , modulus inverse to find public key

由 ≡放荡痞女 提交于 2019-12-06 13:42:46
So, Im using java to find the public key of a RSA password. Right now Im unsure what I'm doing and also if it's correct. I have this information for the public key. C = 5449089907 n = p*q = 8271344041 q = 181123 p = n/q = 45667 d = 53 phi(n) = (p-1)(q-1) = 8271117252 What complicates things are the BigIntegers, the numbers are way to huge for int and longs, so I have to use the clumsy BigIntegers. As far as I understand I have the following equation to solve. e*5198987987 - x*8271117252 = 1 I'm trying to use euklidske algorithm to solve it. In Java i think i can use the following method : I

Least Common Multiple of an array values using Euclidean Algorithm

由 五迷三道 提交于 2019-12-06 10:41:52
问题 I want to calculate the least common multiple of an array of values, using Euclideans algorithm I am using this pseudocode implementation: found on wikipedia function gcd(a, b) while b ≠ 0 t := b; b := a mod b; a := t; return a; My javascript implementation is such function smallestCommons(arr) { var gcm = arr.reduce(function(a,b){ let minNum = Math.min(a,b); let maxNum = Math.max(a,b); var placeHolder = 0; while(minNum!==0){ placeHolder = maxNum; maxNum = minNum; minNum = placeHolder%minNum;

Knuth the art of computer programming ex 1.1.8

由 自闭症网瘾萝莉.ら 提交于 2019-12-06 03:45:34
问题 I can't figure out what Knuth meant in his instructions for an exercise 8 from Chapter 1.1. The task is to make an efficient gcd algorithm of two positive integers m and n using his notation theta[j] , phi[j] , b[j] and a[j] where theta and phi are strings and a and b - positive integers which represent computational steps in this case. Let an input be the string of the form a^mb^n . An excellent explanation of Knuth's algorithm is given by schnaader here. My question is how this may be

Java: Get Greatest Common Divisor, which method is better?

由 こ雲淡風輕ζ 提交于 2019-12-05 18:48:13
From this question Java: get greatest common divisor In getting the gcd of any data type whether int , long , Integer , Long , which answer is better in terms of precision, speed, cpu usage, etc.? A: private static int gcdThing(int a, int b) { return BigInteger.valueOf(a).gcd(BigInteger.valueOf((b))).intValue(); } B: public int GCD(int a, int b) { return b==0 ? a : GCD(b, a%b); } Random r = new Random(); int[] ints = new int[500000]; for (int i = 0 ; i < ints.length ; i++) ints[i] = r.nextInt(); for (int i = 0 ; i < ints.length-1; i++) GCD(i,i+1); for (int i = 0 ; i < ints.length-1; i++)

GCD and LCM relation

由 爱⌒轻易说出口 提交于 2019-12-05 11:19:49
The following relation works only for two (3, 12) numbers, it fails to produce the right answer when used for three numbers (3,12,10) . Just wondering if is it my understanding or it is just for two numbers and for me same is true for Euclid algorithm as well. LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) The analogous formulas to LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) with three variables are simply not valid, as your example with (3, 12, 10) shows readily. The product of these three numbers is 360. The GCD is 1. The LCM is 60. It is our common

What algorithm does Python employ in fractions.gcd()?

由 吃可爱长大的小学妹 提交于 2019-12-05 01:56:04
I'm using the fractions module in Python v3.1 to compute the greatest common divisor. I would like to know what algorithm is used. I'm guessing the Euclidean method, but would like to be sure. The docs ( http://docs.python.org/py3k/library/fractions.html?highlight=fractions.gcd#fractions.gcd ) don't help. Can anybody clue me in? According to the 3.1.2 source code online , here's gcd as defined in Python-3.1.2/Lib/fractions.py : def gcd(a, b): """Calculate the Greatest Common Divisor of a and b. Unless b==0, the result will have the same sign as b (so that when b is divided by it, the result
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