Euler’s Totient function Φ(n) for an input n is count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.

Examples:

Φ(1) = 1 gcd(1, 1) is 1 Φ(2) = 1 gcd(1, 2) is 1, but gcd(2, 2) is 2. Φ(3) = 2 gcd(1, 3) is 1 and gcd(2, 3) is 1 Φ(4) = 2 gcd(1, 4) is 1 and gcd(3, 4) is 1 Φ(5) = 4 gcd(1, 5) is 1, gcd(2, 5) is 1, gcd(3, 5) is 1 and gcd(4, 5) is 1 Φ(6) = 2 gcd(1, 6) is 1 and gcd(5, 6) is 1,

**How to compute Φ(n) for an input n?**

A **simple solution** is to iterate through all numbers from 1 to n-1 and count numbers with gcd with n as 1. Below is C implementation of the simple method to compute Euler’s Totient function for an input integer n.

## C++

// A simple C program to calculate Euler's Totient Function #include <stdio.h> // Function to return gcd of a and b int gcd(int a, int b) { if (a == 0) return b; return gcd(b%a, a); } // A simple method to evaluate Euler Totient Function int phi(unsigned int n) { unsigned int result = 1; for (int i=2; i < n; i++) if (gcd(i, n) == 1) result++; return result; } // Driver program to test above function int main() { int n; for (n=1; n<=10; n++) printf("phi(%d) = %d\n", n, phi(n)); return 0; }

## Java

// A simple java program to calculate // Euler's Totient Function import java.io.*; class GFG { // Function to return GCD of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b%a, a); } // A simple method to evaluate // Euler Totient Function static int phi( int n) { int result = 1; for (int i=2; i < n; i++) if (gcd(i, n) == 1) result++; return result; } // Driver code public static void main (String[] args) { int n; for (n=1; n<=10; n++) System.out.println("phi("+n+") = "+ phi(n)); } } // This code is contributed by sunnusingh

Output:

phi(1) = 1 phi(2) = 1 phi(3) = 2 phi(4) = 2 phi(5) = 4 phi(6) = 2 phi(7) = 6 phi(8) = 4 phi(9) = 6 phi(10) = 4

The above code calls gcd function O(n) times. Time complexity of the gcd function is O(h) where h is number of digits in smaller number of given two numbers. Therefore, an upper bound on time complexity of above solution is O(nLogn) [How? there can be at most Log_{10}n digits in all numbers from 1 to n]

Below is a **Better Solution**. The idea is based on Euler’s product formula which states that value of totient functions is below product over all prime factors p of n.

The formula basically says that the value of Φ(n) is equal to n multiplied by product of (1 – 1/p) for all prime factors p of n. For example value of Φ(6) = 6 * (1-1/2) * (1 – 1/3) = 2.

We can find all prime factors using the idea used in this post.

1) Initialize : result = n 2) Run a loop from 'p' = 2 to sqrt(n), do following for every 'p'. a) If p divides n, then Set: result = result * (1.0 - (1.0 / (float) p)); Divide all occurrences of p in n. 3) Return result

Below is C implementation of Euler’s product formula.

## C++

// C program to calculate Euler's Totient Function // using Euler's product formula #include <stdio.h> int phi(int n) { float result = n; // Initialize result as n // Consider all prime factors of n and for every prime // factor p, multiply result with (1 - 1/p) for (int p=2; p*p<=n; ++p) { // Check if p is a prime factor. if (n % p == 0) { // If yes, then update n and result while (n % p == 0) n /= p; result *= (1.0 - (1.0 / (float) p)); } } // If n has a prime factor greater than sqrt(n) // (There can be at-most one such prime factor) if (n > 1) result *= (1.0 - (1.0 / (float) n)); return (int)result; } // Driver program to test above function int main() { int n; for (n=1; n<=10; n++) printf("phi(%d) = %d\n", n, phi(n)); return 0; }

## Java

// Java program to calculate Euler's Totient // Function using Euler's product formula import java.io.*; class GFG { static int phi(int n) { // Initialize result as n float result = n; // Consider all prime factors of n and for // every prime factor p, multiply result // with (1 - 1/p) for (int p = 2; p * p <= n; ++p) { // Check if p is a prime factor. if (n % p == 0) { // If yes, then update n and result while (n % p == 0) n /= p; result *= (1.0 - (1.0 / (float) p)); } } // If n has a prime factor greater than sqrt(n) // (There can be at-most one such prime factor) if (n > 1) result *= (1.0 - (1.0 / (float) n)); return (int)result; } // Driver program to test above function public static void main(String args[]) { int n; for (n=1; n<=10; n++) System.out.println("phi("+n+") = " + phi(n)); } } // This code is contributed by Nikita Tiwari.

## Python3

# Python 3 program to calculate # Euler's Totient Function # using Euler's product formula def phi(n) : result = n # Initialize result as n # Consider all prime factors # of n and for every prime # factor p, multiply result with (1 - 1/p) p=2 while(p*p<=n) : # Check if p is a prime factor. if (n % p == 0) : # If yes, then update n and result while (n % p == 0) : n = n // p result = result * (1.0 - (1.0 / (float) (p))) p = p + 1 # If n has a prime factor # greater than sqrt(n) # (There can be at-most one # such prime factor) if (n > 1) : result = result * (1.0 - (1.0 / (float)(n))) return (int)(result) # Driver program to test above function for n in range(1,11) : print("phi(",n,") = ",phi(n)) # This code is contributed # by Nikita Tiwari.

Output:

phi(1) = 1 phi(2) = 1 phi(3) = 2 phi(4) = 2 phi(5) = 4 phi(6) = 2 phi(7) = 6 phi(8) = 4 phi(9) = 6 phi(10) = 4

We can avoid floating point calculations in above method. The idea is to count all prime factors and their multiples and subtract this count from n to get the totient function value (Prime factors and multiples of prime factors won’t have gcd as 1)

1) Initialize result as n 2) Consider every number 'p' (where 'p' varies from 2 to √n). If p divides n, then do following a) Subtract all multiples of p from 1 to n [all multiples of p will have gcd more than 1 (at least p) with n] b) Update n by repeatedly dividing it by p. 3) If the reduced n is more than 1, then remove all multiples of n from result.

Below is C implementation of above algorithm.

// C program to calculate Euler's Totient Function #include <stdio.h> int phi(int n) { int result = n; // Initialize result as n // Consider all prime factors of n and subtract their // multiples from result for (int p=2; p*p<=n; ++p) { // Check if p is a prime factor. if (n % p == 0) { // If yes, then update n and result while (n % p == 0) n /= p; result -= result / p; } } // If n has a prime factor greater than sqrt(n) // (There can be at-most one such prime factor) if (n > 1) result -= result / n; return result; } // Driver program to test above function int main() { int n; for (n=1; n<=10; n++) printf("phi(%d) = %d\n", n, phi(n)); return 0; }

Output:

phi(1) = 1 phi(2) = 1 phi(3) = 2 phi(4) = 2 phi(5) = 4 phi(6) = 2 phi(7) = 6 phi(8) = 4 phi(9) = 6 phi(10) = 4

Let us take an example to understand the above algorithm.

n = 10. Initialize: result = 10 2 is a prime factor, so n = n/i = 5, result = 5 3 is not a prime factor. The for loop stops after 3 as 4*4 is not less than or equal to 10. After for loop, result = 5, n = 5 Since n > 1, result = result - result/n = 4

**Some Interesting Properties of Euler’s Totient Function**

**1)** For a prime number p, Φ(p) is p-1. For example Φ(5) is 4, Φ(7) is 6 and Φ(13) is 12. This is obvious, gcd of all numbers from 1 to p-1 will be 1 because p is a prime.

**2)** For two numbers a and b, if gcd(a, b) is 1, then Φ(ab) = Φ(a) * Φ(b). For example Φ(5) is 4 and Φ(6) is 2, so Φ(30) must be 8 as 5 and 6 are relatively prime.

**3)** For any two prime numbers p and q, Φ(pq) = (p-1)*(q-1). This property is used in RSA algorithm.

**4)** If p is a prime number, then Φ(p^{k}) = p^{k} – p^{k-1}. This can be proved using Euler’s product formula.

**5)** Sum of values of totient functions of all divisors of n is equal to n.

For example, n = 6, the divisors of n are 1, 2, 3 and 6. According to Gauss, sum of Φ(1) + Φ(2) + Φ(3) + Φ(6) should be 6. We can verify the same by putting values, we get (1 + 1 + 2 + 2) = 6.

**6)** The most famous and important feature is expressed in * Euler’s theorem* :

The theorem states that if n and a are coprime (or relatively prime) positive integers, then a^{Φ(n)}≡ 1 (mod n)

The RSA cryptosystem is based on this theorem:

In the particular case when m is prime say p, Euler’s theorem turns into the so-called * Fermat’s little theorem* :

a^{p-1}≡ 1 (mod p)

**7)** Number of generators of a finite cyclic group under modulo n addition is Φ(n).

**Related Article:**

Euler’s Totient function for all numbers smaller than or equal to n

Optimized Euler Totient Function for Multiple Evaluations

**References:**

http://e-maxx.ru/algo/euler_function

http://en.wikipedia.org/wiki/Euler%27s_totient_function

This article is contributed by **Ankur**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above