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 is the implementation of above idea.
C++
#include<bits/stdc++.h>
using namespace std;
int GCD ( int a , int b )
{
if (b == 0 )
return a;
return GCD( b , a%b ) ;
}
int multiplicativeOrder( int A, int N)
{
if (GCD(A, N ) != 1)
return -1;
unsigned int result = 1;
int K = 1 ;
while (K < N)
{
result = (result * A) % N ;
if (result == 1)
return K;
K++;
}
return -1 ;
}
int main()
{
int A = 4 , N = 7;
cout << multiplicativeOrder(A, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int GCD( int a, int b) {
if (b == 0 )
return a;
return GCD(b, a % b);
}
static int multiplicativeOrder( int A, int N) {
if (GCD(A, N) != 1 )
return - 1 ;
int result = 1 ;
int K = 1 ;
while (K < N) {
result = (result * A) % N;
if (result == 1 )
return K;
K++;
}
return - 1 ;
}
public static void main(String args[]) {
int A = 4 , N = 7 ;
System.out.println(multiplicativeOrder(A, N));
}
}
|
Python3
def GCD (a, b ) :
if (b = = 0 ) :
return a
return GCD( b, a % b )
def multiplicativeOrder(A, N) :
if (GCD(A, N ) ! = 1 ) :
return - 1
result = 1
K = 1
while (K < N) :
result = (result * A) % N
if (result = = 1 ) :
return K
K = K + 1
return - 1
A = 4
N = 7
print (multiplicativeOrder(A, N))
|
C#
using System;
class GFG {
static int GCD( int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
static int multiplicativeOrder( int A, int N)
{
if (GCD(A, N) != 1)
return -1;
int result = 1;
int K = 1;
while (K < N)
{
result = (result * A) % N;
if (result == 1)
return K;
K++;
}
return -1;
}
public static void Main()
{
int A = 4, N = 7;
Console.Write(multiplicativeOrder(A, N));
}
}
|
PHP
<?php
function GCD ( $a , $b )
{
if ( $b == 0 )
return $a ;
return GCD( $b , $a % $b ) ;
}
function multiplicativeOrder( $A , $N )
{
if (GCD( $A , $N ) != 1)
return -1;
$result = 1;
$K = 1 ;
while ( $K < $N )
{
$result = ( $result * $A ) % $N ;
if ( $result == 1)
return $K ;
$K ++;
}
return -1 ;
}
$A = 4; $N = 7;
echo (multiplicativeOrder( $A , $N ));
?>
|
Javascript
<script>
function GCD(a, b) {
if (b == 0)
return a;
return GCD(b, a % b);
}
function multiplicativeOrder(A, N) {
if (GCD(A, N) != 1)
return -1;
let result = 1;
let K = 1;
while (K < N) {
result = (result * A) % N;
if (result == 1)
return K;
K++;
}
return -1;
}
let A = 4, N = 7;
document.write(multiplicativeOrder(A, N));
</script>
|
Output :
3
Time Complexity: O(N)
Space Complexity: O(1)
Reference: https://en.wikipedia.org/wiki/Multiplicative_order
Last Updated :
14 Mar, 2023
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