what is the fastest way to find the gcd of n numbers?

Question

what is the fastest way to compute the greatest common divisor of n numbers?

Answer 1

Without recursion:

int result = numbers[0];
for(int i = 1; i < numbers.length; i++){
    result = gcd(result, numbers[i]);
}
return result;

For very large arrays, it might be faster to use the fork-join pattern, where you split your array and calculate gcds in parallel. Here is some pseudocode:

int calculateGCD(int[] numbers){
    if(numbers.length <= 2){
        return gcd(numbers);    
    }
    else {
        INVOKE-IN-PARALLEL {
            left = calculateGCD(extractLeftHalf(numbers));
            right = calculateGCD(extractRightHalf(numbers));
        }
        return gcd(left,right);
    }
}

Answer 2

You may want to sort the numbers first and compute the gcd recursively starting from the smallest two numbers.

Answer 3

C++17

I have written this function for calculating gcd of n numbers by using C++'s inbuilt __gcd(int a, int b) function.

int gcd(vector<int> vec, int vsize)
{
    int gcd = vec[0];
    for (int i = 1; i < vsize; i++)
    {
        gcd = __gcd(gcd, vec[i]);
    }
    return gcd;
}

To know more about this function visit this link .

Also refer to Dijkstra's GCD algorithm from the following link. It works without division. So it could be slightly faster (Please correct me if I am wrong.)

Answer 4

You should use Lehmer's GCD algorithm.

Answer 5

How about the following using Euclidean algorithm by subtraction:

function getGCD(arr){
    let min = Math.min(...arr); 
    let max= Math.max(...arr);
    if(min==max){
        return min;
    }else{
         for(let i in arr){
            if(arr[i]>min){
                arr[i]=arr[i]-min;
            }
        }
        return getGCD(arr);
    }
   
}

console.log(getGCD([2,3,4,5,6]))

The above implementation takes O(n^2) time. There are improvements that can be implemented but I didn't get around trying these out for n numbers.

Answer 6

Use the Euclidean algorithm :

function gcd(a, b)
while b ≠ 0
   t := b; 
   b := a mod b; 
   a := t; 
return a;

You apply it for the first two numbers, then the result with the third number, etc... :

read(a);
read(b);

result := gcd(a, b);
i := 3;
while(i <= n){
    read(a)
    result := gcd(result, a);
}
print(result);

Answer 7

Here below is the source code of the C program to find HCF of N numbers using Arrays.

#include<stdio.h>

int main()
{
    int n,i,gcd;
    printf("Enter how many no.s u want to find gcd : ");
    scanf("%d",&n);
    int arr[n];
    printf("\nEnter your numbers below :- \n ");
    for(i=0;i<n;i++)
    {
        printf("\nEnter your %d number = ",i+1);
        scanf("%d",&arr[i]);
    }
    gcd=arr[0];
    int j=1;
    while(j<n)
    {
       if(arr[j]%gcd==0)
       {
           j++;
       }
       else
       {
           gcd=arr[j]%gcd;
           i++;
       }
    }
    printf("\nGCD of k no.s = %d ",gcd);
    return 0;
}

For more refer to this website for further clarification.......

Answer 8

You can use divide and conquer. To calculate gcdN([]), you divide the list into first half and second half. if it only has one num for each list. you calculate using gcd2(n1, n2).

I just wrote a quick sample code. (assuming all num in the list are positive Ints)

def gcdN(nums):
    n = len(nums)
    if n == 0: return "ERROR"
    if n == 1: return nums[0]
    if n >= 2: return gcd2(gcdN(nums[:n//2]), gcdN(nums[n//2:]))

def gcd2(n1, n2):
    for num in xrange(min(n1, n2), 0, -1):
        if n1 % num == 0 and n2 % num == 0:
            return num

Answer 9

If you have a lot of small numbers, factorization may be actually faster.

//Java
int[] array = {60, 90, 45};
int gcd = 1;
outer: for (int d = 2; true; d += 1 + (d % 2)) {
    boolean any = false;
    do {
        boolean all = true;
        any = false;
        boolean ready = true;
        for (int i = 0; i < array.length; i++) {
            ready &= (array[i] == 1);
            if (array[i] % d == 0) {
                any = true;
                array[i] /= d;
            } else all = false;
        }
        if (all) gcd *= d;
        if (ready) break outer;
    } while (any);
}
System.out.println(gcd);

(works for some examples, but not really tested)

Answer 10

Here's a gcd method that uses the property that gcd(a, b, c) = gcd(a, gcd(b, c)).
It uses BigInteger's gcd method since it is already optimized.

public static BigInteger gcd(BigInteger[] parts){
    BigInteger gcd = parts[0];
    for(int i = 1; i < parts.length; i++)
        gcd = parts[i].gcd(gcd);
    return gcd;
}

Answer 11

//Recursive solution to get the GCD of Two Numbers

long long int gcd(long long int a,long long int b)<br>
{
   return b==0 ? a : gcd(b,a%b);
}
int main(){
  long long int a,b;
  cin>>a>>b;
  if(a>b) cout<<gcd(a,b);
  else cout<<gcd(b,a);
return 0;
}

Answer 12

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

class GCDArray{
    public static int [] extractLeftHalf(int [] numbers)
    {
        int l =numbers.length/2;
        int arr[] = Arrays.copyOf(numbers, l+1);
        return arr;
    }

    public static int [] extractRightHalf(int [] numbers)
    {
        int l =numbers.length/2;
        int arr[] = Arrays.copyOfRange(numbers,l+1, numbers.length);
        return arr;
    }

    public static int gcd(int[] numbers)
    {
        if(numbers.length==1)
            return numbers[0];
        else {
            int x = numbers[0];
            int y = numbers[1];
            while(y%x!=0)
            {
                int rem = y%x;
                y = x;
                x = rem;
            }
            return x;
        }
    }
    public static int gcd(int x,int y)
    {
            while(y%x!=0)
            {
                int rem = y%x;
                y = x;
                x = rem;
            }
            return x;

    }
    public static int calculateGCD(int[] numbers){
        if(numbers.length <= 2){
            return gcd(numbers);    
        }
        else {

                    int left = calculateGCD(extractLeftHalf(numbers));
                    int right = calculateGCD(extractRightHalf(numbers));

            return gcd(left,right);
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int arr[] = new int[n];
        for(int i=0;i<n;i++){
            arr[i]=sc.nextInt();
        }
        System.out.println(calculateGCD(arr));
    }
}

**

Above is the java working code ..... the pseudo code of which is already mention by https://stackoverflow.com/users/7412/dogbane

**

Answer 13

A recursive JavaScript (ES6) one-liner for any number of digits.

const gcd = (a, b, ...c) => b ? gcd(b, a % b, ...c) : c.length ? gcd(a, ...c) : Math.abs(a);

Answer 14

This is what comes off the top of my head in Javascript.

function calculateGCD(arrSize, arr) {
    if(!arrSize)
        return 0;
    var n = Math.min(...arr);
    for (let i = n; i > 0; i--) {
        let j = 0;
        while(j < arrSize) {
            if(arr[j] % i === 0) {
                j++;
            }else {
                break;
            }
            if(j === arrSize) {
                return i;
            }
        }
    }
}

console.log(generalizedGCD(4, [2, 6, 4, 8]));
// Output => 2

Answer 15

Here was the answer I was looking for. The best way to find the gcd of n numbers is indeed using recursion.ie gcd(a,b,c)=gcd(gcd(a,b),c). But I was getting timeouts in certain programs when I did this.

The optimization that was needed here was that the recursion should be solved using fast matrix multiplication algorithm.