Find the number of primitive roots modulo prime in C++.

In this problem, we are given a prime number N. Our task is to find the number of primitive roots modulo prime.

Primitive Root of a number − It is a number (r) smaller than N which has all values of r^x(mod N) different for all X in range [0, n-2].

Let’s take an example to understand the problem,

Input : N = 5
Output : 2

Solution Approach

A simple solution to the problem is based on a trial method. We will check for all numbers from 2 to (N-1) for the conditions with x ranging from [0, n-2] and break if a value is found that satisfies the condition.

This solution is naive and is easy to implement but the time complexity of the solution is of the order of N². This might lead to long runs in case of a large value of N.

So, a more efficient solution to the problem is by using the Euler Totient function φ(N)

So, for a number r to be the primitive root of N. Its multiplicative order with modulo N is equal to φ(N). Here are steps to follow −

We need to find all prime factors of (N-1) for the prime number N. Then calculate all powers using (N-1) / prime factors. Then check the value of prime numbers power modulo n. If its is never 1 then i is the primitive root. The first value which we see will be the returned value as there can be more than one primitive root for a number but we need only the smallest one.

Example

Let’s take an example to understand the problem

#include<bits/stdc++.h>
using namespace std;
int calcPowerMod(int x, unsigned int y, int p){
   int modVal = 1;
   x = x % p;
   while (y > 0){
      if (y & 1)
         modVal = (modVal*x) % p;
      y = y >> 1;
      x = (x*x) % p;
   }
   return modVal;
}
void findAllPrimeFactors(unordered_set<int> &s, int n){
   while (n%2 == 0){
      s.insert(2);
      n = n/2;
   }
   for (int i = 3; i*i <= n; i = i+2){
      while (n%i == 0){
         s.insert(i);
         n = n/i;
      }
   }
   if (n > 2)
      s.insert(n);
}
int findSmallestPrimitiveRoot(int n){
   unordered_set<int> primes;
   int phi = n-1;
   findAllPrimeFactors(primes, phi);
   for (int r=2; r<=phi; r++){
      bool flag = false;
      for (auto it = primes.begin(); it != primes.end(); it++){
         if (calcPowerMod(r, phi/(*it), n) == 1){
            flag = true;
            break;
         }
      }
      if (flag == false)
         return r;
   }
   return -1;
}
int main(){
   int n = 809;
   cout<<"The smallest primitive root is "<<findSmallestPrimitiveRoot(n);
   return 0;
}

Output

The smallest primitive root is 3

sudhir sharma

Updated on 24-Jan-2022 13:34:13

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