Question

We discuss the Euclidean algorithm that finds the greatest common divisor of 2 numbers u and v. We want to extend and compute the gcd of n integers gcd⁡(u1,u2,….un). One way to do it is to assume all numbers are non-negative, so if only one of if uj≠0 it is the gcd. Otherwise replace uk by uk mod uj  for all k≠j where uj is the minimum of the non-zero elements (u’s). The algorithm can be made significantly faster if one consider the identity:

gcdu1,u2….un=gcd⁡(u1,gcdu2,u3,..un).

Then we may calculate gcdu1,u2….un with following algorithm:

            Step 1. set d←un, j←n-1

            Step 2. if d≠1 and j>0 setgcduj,dand j=j-1 and repeat Step 2.

Else d=gcd(u1,u2,…un).

What is the smallest value of un such that the calculation of: gcd(u1,u2,..., un)
by steps Step 1 and Step 2 (given above) requires N divisions, if Euclid’s algorithm is used throughout? Assume that N ≥ n.

Answer 1

This java code can be used to find the gcd of array of elements:

CODE:

public class GCD{
static int gcd(int x,int y) //GCD of two numbers
{
if (x==0)
return y;
return gcd(y%x,x);
}
static int findgcd(int arr[], int n)//Function to find gcd of array of numbers. Ex:gcd⁡(u_1,u_2,…,u_n).
{
int result = arr[0];
for (int i=1;i<n;i++){
result=gcd(arr[i],result);

if(result==1)
{
return 1;
}
}
return result;
}
public static void main(String args[])
{
int arr[] = {2,4,6,26,45}; //Input elemets
//you can write a loop to get the inputs as per user requirement
int n=arr.length;
System.out.println(findgcd(arr,n));//Array elements and array length are passed as arguments
}
}