Given a prime number n, the task is to find its primitive root under modulo n. 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 range[0, n-2] are different. Return -1 if n is a non-prime number.
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 range[0, n-2]. If all these values are different, then return r, else continue for next value of r. If all values of r are tried, return -1.
An efficient solution is based on below fact.
If the multiplicative order of a number r modulo n is equal to Euler Totient Function Φ(n) (Note that Euler Totient Function for a prime n is n-1), then it is a primitive root [Source : Wiki]
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 root for a prime number but we are only concerned for smallest one.If you want to find all roots then continue the process till p-1 instead of breaking up on finding first primitive root.
# Python3 program to find primitive root
# of a given number n
from math import sqrt
# Returns True if n is prime
def isPrime( n):
# Corner cases
if (n <= 1): return False if (n <= 3): return True # This is checked so that we can skip # middle five numbers in below loop 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 """ Iterative Function to calculate (x^n)%p in O(logy) */""" def power( x, y, p): res = 1 # Initialize result x = x % p # Update x if it is more # than or equal to p while (y > 0):
# If y is odd, multiply x with result
if (y & 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
# Utility function to store prime
# factors of a number
def findPrimefactors(s, n) :
# Pr the number of 2s that divide n
while (n % 2 == 0) :
n = n // 2
# n must be odd at this po. So we can
# skip one element (Note i = i +2)
for i in range(3, int(sqrt(n)), 2):
# While i divides n, pr i and divide n
while (n % i == 0) :
n = n // i
# This condition is to handle the case
# when n is a prime number greater than 2
if (n > 2) :
# Function to find smallest primitive
# root of n
def findPrimitive( n) :
s = set()
# Check if n is prime or not
if (isPrime(n) == False):
# Find value of Euler Totient function
# of n. Since n is a prime number, the
# value of Euler Totient function is n-1
# as there are n-1 relatively prime numbers.
phi = n – 1
# Find prime factors of phi and store in a set
# Check for every number from 2 to phi
for r in range(2, phi + 1):
# Iterate through all prime factors of phi.
# and check if we found a power with value 1
flag = False
for it in s:
# Check if r^((phi)/primefactors)
# mod n is 1 or not
if (power(r, phi // it, n) == 1):
flag = True
# If there was no power with value 1.
if (flag == False):
# If no primitive root found
# Driver Code
n = 761
print(“Smallest primitive root of”,
n, “is”, findPrimitive(n))
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
Smallest primitive root of 761 is 6
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Improved By : SHUBHAMSINGH10