Find the number of primitive roots modulo prime
Given a prime . The task is to count all the primitive roots of .
A primitive root is an integer x (1 <= x < p) such that none of the integers x – 1, x2 – 1, …., xp – 2 – 1 are divisible by but xp – 1 – 1 is divisible by .
Examples:
Input: P = 3
Output: 1
The only primitive root modulo 3 is 2.
Input: P = 5
Output: 2
Primitive roots modulo 5 are 2 and 3.
Approach: There is always at least one primitive root for all primes. So, using Eulers totient function we can say that f(p-1) is the required answer where f(n) is euler totient function.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countPrimitiveRoots( int p)
{
int result = 1;
for ( int i = 2; i < p; i++)
if (__gcd(i, p) == 1)
result++;
return result;
}
int main()
{
int p = 5;
cout << countPrimitiveRoots(p - 1);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int __gcd( int a, int b)
{
if (a == 0 )
return b;
if (b == 0 )
return a;
if (a == b)
return a;
if (a > b)
return __gcd(a-b, b);
return __gcd(a, b-a);
}
static int countPrimitiveRoots( int p)
{
int result = 1 ;
for ( int i = 2 ; i < p; i++)
if (__gcd(i, p) == 1 )
result++;
return result;
}
public static void main (String[] args) {
int p = 5 ;
System.out.println( countPrimitiveRoots(p - 1 ));
}
}
|
Python3
from math import gcd
def countPrimitiveRoots(p):
result = 1
for i in range ( 2 , p, 1 ):
if (gcd(i, p) = = 1 ):
result + = 1
return result
if __name__ = = '__main__' :
p = 5
print (countPrimitiveRoots(p - 1 ))
|
C#
using System;
class GFG {
static int __gcd( int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return __gcd(a-b, b);
return __gcd(a, b-a);
}
static int countPrimitiveRoots( int p)
{
int result = 1;
for ( int i = 2; i < p; i++)
if (__gcd(i, p) == 1)
result++;
return result;
}
static public void Main (String []args) {
int p = 5;
Console.WriteLine( countPrimitiveRoots(p - 1));
}
}
|
PHP
<?php
function __gcd( $a , $b )
{
if ( $a == 0)
return b;
if ( $b == 0)
return $a ;
if ( $a == $b )
return $a ;
if ( $a > $b )
return __gcd( $a - $b , $b );
return __gcd( $a , $b - $a );
}
function countPrimitiveRoots( $p )
{
$result = 1;
for ( $i = 2; $i < $p ; $i ++)
if (__gcd( $i , $p ) == 1)
$result ++;
return $result ;
}
$p = 5;
echo countPrimitiveRoots( $p - 1);
?>
|
Javascript
<script>
function __gcd( a, b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return __gcd(a-b, b);
return __gcd(a, b-a);
}
function countPrimitiveRoots(p)
{
var result = 1;
for ( var i = 2; i < p; i++)
if (__gcd(i, p) == 1)
result++;
return result;
}
var p = 5;
document.write( countPrimitiveRoots(p - 1));
</script>
|
Time Complexity: O(p * log(min(a, b))), where a and b are two parameters of gcd.
Auxiliary Space: O(log(min(a, b)))
Last Updated :
29 Jun, 2022
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