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Primitive root of a prime number n modulo n in C++
In this problem, we are given a prime number N. our task is to print the primitive root of prime number N modulo N.
Primitive root of prime number N is an integer x lying between [1, n-1] such that all values of xk (mod n) where k lies in [0, n-2] are unique.
Let’s take an example to understand the problem,
Input: 13 Output: 2
To solve this problem, we have to use mathematical function called Euler’s Totient Function.
Euler’s Totient Function is the count of numbers from 1 to n which are relatively prime to the number n.
A number i is relatively prime if GCD (i, n) = 1.
In solution, if the multiplicative order of x modulo n is equal to Euler’s Totient Function, then the number is primitive root otherwise not. We will check for all relative primes.
Note: Euler’s Totient Function of a prime number n=n-1
The below code will show the implementation of our solution,
Example
#include<bits/stdc++.h>
using namespace std;
bool isPrimeNumber(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 GeneratePrimes(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 findPrimitiveRoot(int n) {
unordered_set<int> s;
if (isPrimeNumber(n)==false)
return -1;
int ETF = n-1;
GeneratePrimes(s, ETF);
for (int r=2; r<=ETF; r++){
bool flag = false;
for (auto it = s.begin(); it != s.end(); it++){
if (power(r, ETF/(*it), n) == 1){
flag = true;
break;
}
}
if (flag == false)
return r;
}
return -1;
}
int main() {
int n= 13;
cout<<" Smallest primitive root of "<<n<<" is "<<findPrimitiveRoot(n);
return 0;
}Output
Smallest primitive root of 13 is 2
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