Multiplicative order

3.5

In number theory, given an integer A and a positive integer N with gcd( A , N) = 1, the multiplicative order of a modulo N is the smallest positive integer k with A^k( mod N ) = 1. ( 0 < K < N )

Examples:

Input : A = 4 , N = 7 
Output : 3
explanation :  GCD(4, 7) = 1  
               A^k( mod N ) = 1 ( smallest positive integer K )
               4^1 = 4(mod 7)  = 4
               4^2 = 16(mod 7) = 2
               4^3 = 64(mod 7)  = 1
               4^4 = 256(mod 7) = 4
               4^5 = 1024(mod 7)  = 2
               4^6 = 4096(mod 7)  = 1

smallest positive integer K = 3  

Input :  A = 3 , N = 1000 
Output : 100  (3^100 (mod 1000) == 1) 

Input : A = 4 , N = 11 
Output : 5 

IF we take a close look then we observe that we do not need to calculate power every time. we can be obtaining next power by multiplying ‘A’ with the previous result of a module .

Explanation : 
A = 4 , N = 11  
initially result = 1 
with normal                with modular arithmetic (A * result)
4^1 = 4 (mod 11 ) = 4  ||  4 * 1 = 4 (mod 11 ) = 4 [ result = 4]
4^2 = 16(mod 11 ) = 5  ||  4 * 4 = 16(mod 11 ) = 5 [ result = 5]
4^3 = 64(mod 11 ) = 9  ||  4 * 5 = 20(mod 11 ) = 9 [ result = 9]
4^4 = 256(mod 11 )= 3  ||  4 * 9 = 36(mod 11 ) = 3 [ result = 3]
4^5 = 1024(mod 5 ) = 1 ||  4 * 3 = 12(mod 11 ) = 1 [ result = 1]

smallest positive integer  5 

Run a loop from 1 to N-1 and Return the smallest +ve power of A under modulo n which is equal to 1.

Below C++ implementation of above idea.

// C++ program to implement multiplicative order
#include<bits/stdc++.h>
using namespace std;

// fuction for GCD
int GCD ( int a , int b )
{
    if (b == 0 )
        return a;
    return GCD( b , a%b ) ;
}

// Fucnction return smallest +ve integer that
// holds condition A^k(mod N ) = 1
int multiplicativeOrder(int A, int  N)
{
    if (GCD(A, N ) != 1)
        return -1;

    // result store power of A that rised to
    // the power N-1
    unsigned int result = 1;

    int K = 1 ;
    while (K < N)
    {
        // modular arithmetic
        result = (result * A) % N ;

        // return samllest +ve integer
        if (result  == 1)
            return K;

        // increment power
        K++;
    }

    return -1 ;
}

//driver program to test above function
int main()
{
    int A = 4 , N = 7;
    cout << multiplicativeOrder(A, N);
    return 0;
}

Output:

3

Time Complexity: O(N)

Reference : https://en.wikipedia.org/wiki/Multiplicative_order
This article is contributed by Nishant Singh . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:



3.5 Average Difficulty : 3.5/5.0
Based on 4 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.