Primitive root of a prime number n modulo n
Given a prime number n, the task is to find its primitive root under modulo n. The primitive root of a prime number n is an integer r between[1, n-1] such that the values of r^x(mod n) where x is in the range[0, n-2] are different. Return -1 if n is a non-prime number.
Examples:
Input : 7
Output : Smallest primitive root = 3
Explanation: n = 7
3^0(mod 7) = 1
3^1(mod 7) = 3
3^2(mod 7) = 2
3^3(mod 7) = 6
3^4(mod 7) = 4
3^5(mod 7) = 5
Input : 761
Output : Smallest primitive root = 6
A simple solution is to try all numbers from 2 to n-1. For every number r, compute values of r^x(mod n) where x is in the range[0, n-2]. If all these values are different, then return r, else continue for the next value of r. If all values of r are tried, return -1.
An efficient solution is based on the below facts.
If the multiplicative order of a number r modulo n is equal to Euler Totient Function ?(n) ( note that the Euler Totient Function for a prime n is n-1), then it is a primitive root.
1- Euler Totient Function phi = n-1 [Assuming n is prime]
1- Find all prime factors of phi.
2- Calculate all powers to be calculated further
using (phi/prime-factors) one by one.
3- Check for all numbered for all powers from i=2
to n-1 i.e. (i^ powers) modulo n.
4- If it is 1 then 'i' is not a primitive root of n.
5- If it is never 1 then return i;.
Although there can be multiple primitive roots for a prime number, we are only concerned with the smallest one. If you want to find all the roots, then continue the process till p-1 instead of breaking up by finding the first primitive root.
C++
#include<bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1) return false ;
if (n <= 3) return true ;
if (n%2 == 0 || n%3 == 0) return false ;
for ( int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false ;
return true ;
}
int power( int x, unsigned int y, int p)
{
int res = 1;
x = x % p;
while (y > 0)
{
if (y & 1)
res = (res*x) % p;
y = y >> 1;
x = (x*x) % p;
}
return res;
}
void findPrimefactors(unordered_set< int > &s, int n)
{
while (n%2 == 0)
{
s.insert(2);
n = n/2;
}
for ( int i = 3; i <= sqrt (n); i = i+2)
{
while (n%i == 0)
{
s.insert(i);
n = n/i;
}
}
if (n > 2)
s.insert(n);
}
int findPrimitive( int n)
{
unordered_set< int > s;
if (isPrime(n)== false )
return -1;
int phi = n-1;
findPrimefactors(s, phi);
for ( int r=2; r<=phi; r++)
{
bool flag = false ;
for ( auto it = s.begin(); it != s.end(); it++)
{
if (power(r, phi/(*it), n) == 1)
{
flag = true ;
break ;
}
}
if (flag == false )
return r;
}
return -1;
}
int main()
{
int n = 761;
cout << " Smallest primitive root of " << n
<< " is " << findPrimitive(n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static boolean isPrime( int n)
{
if (n <= 1 )
{
return false ;
}
if (n <= 3 )
{
return true ;
}
if (n % 2 == 0 || n % 3 == 0 )
{
return false ;
}
for ( int i = 5 ; i * i <= n; i = i + 6 )
{
if (n % i == 0 || n % (i + 2 ) == 0 )
{
return false ;
}
}
return true ;
}
static int power( int x, int y, int p)
{
int res = 1 ;
x = x % p;
while (y > 0 )
{
if (y % 2 == 1 )
{
res = (res * x) % p;
}
y = y >> 1 ;
x = (x * x) % p;
}
return res;
}
static void findPrimefactors(HashSet<Integer> s, int n)
{
while (n % 2 == 0 )
{
s.add( 2 );
n = n / 2 ;
}
for ( int i = 3 ; i <= Math.sqrt(n); i = i + 2 )
{
while (n % i == 0 )
{
s.add(i);
n = n / i;
}
}
if (n > 2 )
{
s.add(n);
}
}
static int findPrimitive( int n)
{
HashSet<Integer> s = new HashSet<Integer>();
if (isPrime(n) == false )
{
return - 1 ;
}
int phi = n - 1 ;
findPrimefactors(s, phi);
for ( int r = 2 ; r <= phi; r++)
{
boolean flag = false ;
for (Integer a : s)
{
if (power(r, phi / (a), n) == 1 )
{
flag = true ;
break ;
}
}
if (flag == false )
{
return r;
}
}
return - 1 ;
}
public static void main(String[] args)
{
int n = 761 ;
System.out.println( " Smallest primitive root of " + n
+ " is " + findPrimitive(n));
}
}
|
Python3
from math import sqrt
def isPrime( n):
if (n < = 1 ):
return False
if (n < = 3 ):
return True
if (n % 2 = = 0 or n % 3 = = 0 ):
return False
i = 5
while (i * i < = n):
if (n % i = = 0 or n % (i + 2 ) = = 0 ) :
return False
i = i + 6
return True
def power( x, y, p):
res = 1
x = x % p
while (y > 0 ):
if (y & 1 ):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
def findPrimefactors(s, n) :
while (n % 2 = = 0 ) :
s.add( 2 )
n = n / / 2
for i in range ( 3 , int (sqrt(n)), 2 ):
while (n % i = = 0 ) :
s.add(i)
n = n / / i
if (n > 2 ) :
s.add(n)
def findPrimitive( n) :
s = set ()
if (isPrime(n) = = False ):
return - 1
phi = n - 1
findPrimefactors(s, phi)
for r in range ( 2 , phi + 1 ):
flag = False
for it in s:
if (power(r, phi / / it, n) = = 1 ):
flag = True
break
if (flag = = False ):
return r
return - 1
n = 761
print ( "Smallest primitive root of" ,
n, "is" , findPrimitive(n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static bool isPrime( int n)
{
if (n <= 1)
{
return false ;
}
if (n <= 3)
{
return true ;
}
if (n % 2 == 0 || n % 3 == 0)
{
return false ;
}
for ( int i = 5; i * i <= n; i = i + 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false ;
}
}
return true ;
}
static int power( int x, int y, int p)
{
int res = 1;
x = x % p;
while (y > 0)
{
if (y % 2 == 1)
{
res = (res * x) % p;
}
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static void findPrimefactors(HashSet< int > s, int n)
{
while (n % 2 == 0)
{
s.Add(2);
n = n / 2;
}
for ( int i = 3; i <= Math.Sqrt(n); i = i + 2)
{
while (n % i == 0)
{
s.Add(i);
n = n / i;
}
}
if (n > 2)
{
s.Add(n);
}
}
static int findPrimitive( int n)
{
HashSet< int > s = new HashSet< int >();
if (isPrime(n) == false )
{
return -1;
}
int phi = n - 1;
findPrimefactors(s, phi);
for ( int r = 2; r <= phi; r++)
{
bool flag = false ;
foreach ( int a in s)
{
if (power(r, phi / (a), n) == 1)
{
flag = true ;
break ;
}
}
if (flag == false )
{
return r;
}
}
return -1;
}
public static void Main(String[] args)
{
int n = 761;
Console.WriteLine( " Smallest primitive root of " + n
+ " is " + findPrimitive(n));
}
}
|
Javascript
<script>
function isPrime(n) {
if (n <= 1)
return false ;
if (n <= 3)
return true ;
if (n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
}
function power(x, y, p) {
let res = 1;
x = x % p;
while (y > 0) {
if (y & 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
function findPrimefactors(s, n) {
while (n % 2 == 0) {
s.add(2);
n = n / 2;
}
for (let i = 3; i <= Math.sqrt(n); i = i + 2) {
while (n % i == 0) {
s.add(i);
n = n / i;
}
}
if (n > 2)
s.add(n);
}
function findPrimitive(n) {
let s = new Set();
if (isPrime(n) == false )
return -1;
let phi = n - 1;
findPrimefactors(s, phi);
for (let r = 2; r <= phi; r++) {
let flag = false ;
for (let it of s) {
if (power(r, phi / it, n) == 1) {
flag = true ;
break ;
}
}
if (flag == false )
return r;
}
return -1;
}
let n = 761;
document.write( " Smallest primitive root of " + n + " is " + findPrimitive(n));
</script>
|
Output:
Smallest primitive root of 761 is 6
Time Complexity : O(n^2 * logn)
Space Complexity : O(sqrt(n))
Last Updated :
08 Mar, 2023
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