Euler’s Totient Function
Euler’s Totient function Φ (n) for an input n is the count of numbers in {1, 2, 3, …, n-1} 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 nΦ
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 the 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 <iostream>using namespace std; // 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++) cout << "phi("<<n<<") = " << phi(n) << endl; return 0; } // This code is contributed by SHUBHAMSINGH10 |
C
// A simple C program to calculate Euler's Totient Function#include <stdio.h>// Function to return gcd of a and bint gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);}// A simple method to evaluate Euler Totient Functionint 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 functionint 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 Functionimport 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 |
Python3
# A simple Python3 program # to calculate Euler's # Totient Function# Function to return# gcd of a and bdef gcd(a, b): if (a == 0): return b return gcd(b % a, a)# A simple method to evaluate# Euler Totient Functiondef phi(n): result = 1 for i in range(2, n): if (gcd(i, n) == 1): result+=1 return result# Driver Codefor n in range(1, 11): print("phi(",n,") = ", phi(n), sep = "") # This code is contributed# by Smitha |
C#
// A simple C# program to calculate// Euler's Totient Functionusing System;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() { for (int n = 1; n <= 10; n++) Console.WriteLine("phi(" + n + ") = " + phi(n)); }}// This code is contributed by nitin mittal |
PHP
<Φphp// PHP program to calculate // Euler's Totient Function// Function to return // gcd of a and bfunction gcd($a, $b){ if ($a == 0) return $b; return gcd($b % $a, $a);}// A simple method to evaluate// Euler Totient Functionfunction phi($n){ $result = 1; for ($i = 2; $i < $n; $i++) if (gcd($i, $n) == 1) $result++; return $result;}// Driver Codefor ($n = 1; $n <= 10; $n++) echo "phi(" .$n. ") =" . phi($n)."\n";// This code is contributed by Sam007Φ> |
Javascript
<script>// Javascript program to calculate // Euler's Totient Function// Function to return // gcd of a and bfunction gcd(a, b){ if (a == 0) return b; return gcd(b % a, a);}// A simple method to evaluate// Euler Totient Functionfunction phi(n){ let result = 1; for (let i = 2; i < n; i++) if (gcd(i, n) == 1) result++; return result;}// Driver Codefor (let n = 1; n <= 10; n++) document.write(`phi(${n}) = ${phi(n)} <br>`);// This code is contributed by _saurabh_jaiswal</script> |
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. The time complexity of the gcd function is O(h) where “h” is the number of digits in a smaller number of given two numbers. Therefore, an upper bound on the time complexity of the above solution is O(N^2 log N) [HowΦ there can be at most Log10n digits in all numbers from 1 to n]
Auxiliary Space: O(log N)
Below is a Better Solution. The idea is based on Euler’s product formula which states that the value of totient functions is below the product overall 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 the implementation of Euler’s product formula.
C++
// C++ program to calculate Euler's // Totient Function using Euler's// product formula#include <bits/stdc++.h>using namespace std;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 -= result / n; //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n //if n is a prime number return (int)result;} // Driver codeint main(){ int n; for(n = 1; n <= 10; n++) { cout << "Phi" << "(" << n << ")" << " = " << phi(n) <<endl; } return 0;}// This code is contributed by koulick_sadhu |
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 -= result / n; //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n //if n is a prime number return (int)result;}// Driver program to test above functionint 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 formulaimport 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 -= result / n; //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n //if n is a prime number 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 formuladef 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 // n #Since in the set {1,2,....,n-1}, all numbers are relatively prime with n #if n is a prime number return int(result) # Driver program to test above functionfor n in range(1, 11) : print("phi(", n, ") = ", phi(n)) # This code is contributed# by Nikita Tiwari. |
C#
// C# program to calculate Euler's Totient// Function using Euler's product formulausing System;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 *= (float)(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 -= result / n; //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n //if n is a prime number return (int)result; } // Driver Code public static void Main() { int n; for (n = 1; n <= 10; n++) Console.WriteLine("phi(" + n + ") = " + phi(n)); }}// This code is contributed by nitin mittal. |
PHP
<Φphp// PHP program to calculate // Euler's Totient Function // using Euler's product formulafunction phi($n){ // Initialize result as n $result = $n; // Consider all prime factors // of n and for every prime // factor p, multiply result // with (1 - 1/p) for ($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 / $p)); } } // If n has a prime factor greater // than sqrt(n) (There can be at-most // one such prime factor) if ($n > 1) $result -= $result / $n; //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n //if n is a prime number return intval($result);}// Driver Codefor ($n = 1; $n <= 10; $n++)echo "phi(" .$n. ") =" . phi($n)."\n"; // This code is contributed by Sam007Φ> |
Javascript
// Javascript program to calculate // Euler's Totient Function // using Euler's product formulafunction phi(n){ // Initialize result as n let result = n; // Consider all prime factors // of n and for every prime // factor p, multiply result // with (1 - 1/p) for (let p = 2; p * p <= n; ++p) { // Check if p is // a prime factor. if (n % p == 0) { // If yes, then update // n and result while (n % p == 0) n /= p; result *= (1.0 - (1.0 / p)); } } // If n has a prime factor greater // than sqrt(n) (There can be at-most // one such prime factor) if (n > 1) result -= result / n; //Since in the set {1,2,....,n-1}, all numbers are relatively prime with n //if n is a prime number return parseInt(result);}// Driver Codefor (let n = 1; n <= 10; n++) document.write(`phi(${n}) = ${phi(n)} <br>`); // This code is contributed by _saurabh_jaiswal |
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
Time Complexity: O(√n log n)
Auxiliary Space: O(1)
We can avoid floating-point calculations in the 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 Φn).
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 the implementation of the above algorithm.
C++
// C++ program to calculate Euler's// Totient Function#include <bits/stdc++.h>using namespace std;int phi(int n){ // Initialize result as n int result = 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 codeint main(){ int n; for(n = 1; n <= 10; n++) { cout << "Phi" << "(" << n << ")" << " = " << phi(n) << endl; } return 0;}// This code is contributed by koulick_sadhu |
C
// 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 functionint 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 Functionimport java.io.*;class GFG {static int phi(int n){ // Initialize result as n int result = 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 Codepublic 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 ajit |
Python3
# Python3 program to calculate # Euler's Totient Functiondef phi(n): # Initialize result as n result = n; # Consider all prime factors # of n and subtract their # multiples from result 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 = int(n / p); result -= int(result / 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 -= int(result / n); return result;# Driver Codefor n in range(1, 11): print("phi(",n,") =", phi(n)); # This code is contributed # by mits |
C#
// C# program to calculate // Euler's Totient Functionusing System;class GFG{ static int phi(int n){// Initialize result as nint result = n; // Consider all prime // factors of n and // subtract their // multiples from resultfor (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 Codestatic public void Main (){ int n; for (n = 1; n <= 10; n++) Console.WriteLine("phi(" + n + ") = " + phi(n));}}// This code is contributed // by akt_mit |
PHP
<Φphp// PHP program to calculate // Euler's Totient Functionfunction phi($n){ // Initialize // result as n $result = $n; // Consider all prime // factors of n and subtract // their multiples from result for ($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) $n = (int)$n / $p; $result -= (int)$result / $p; } } // If n has a prime factor // greater than sqrt(n) // (There can be at-most // one such prime factor) if ($n > 1) $result -= (int)$result / $n; return $result;}// Driver Codefor ($n = 1; $n <= 10; $n++) echo "phi(", $n,") =", phi($n), "\n"; // This code is contributed // by ajitΦ> |
Javascript
// Javascript program to calculate // Euler's Totient Functionfunction phi(n){ // Initialize // result as n let result = n; // Consider all prime // factors of n and subtract // their multiples from result for (let 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 = parseInt(n / p); result -= parseInt(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 -= parseInt(result / n); return result;}// Driver Codefor (let n = 1; n <= 10; n++) document.write(`phi(${n}) = ${phi(n)} <br>`); // This code is contributed // by _saurabh_jaiswal |
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
Time Complexity: O(√n log n)
Auxiliary Space: O(1)
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, 
Proof :
, where p is any prime numberWe know that
where k is any random number and
[Tex]\\[/Tex]Total number from 1 to p = p Number for which
, i.e the number p itself, so subtracting 1 from p
Examples :
[Tex]\\[/Tex]
[Tex]\\[/Tex]
2) For two prime numbers a and b
, used in RSA Algorithm
Proof :
, where a and b are prime numbers
,
[Tex]\\[/Tex]Total number from 1 to ab = ab Total multiples of a from 1 to ab =
=
Total multiples of b from 1 to ab =
=
Example:a = 5, b = 7, ab = 35Multiples of a =
= 7 {5, 10, 15, 20, 25, 30, 35}Multiples of b =
= 5 {7, 14, 21, 28, 35}
Can there be any double counting ?(watch above example carefully, try with other prime numbers also for more grasp)Ofcourse, we have counted
twice in multiples of a and multiples of b so, Total multiples = a + b - 1 (with which
with
[Tex]\phi(ab) = (a - 1) \cdot (b - 1)[/Tex]
Examples :
[Tex]\\[/Tex]
[Tex]\\[/Tex]
3) For a prime number p, 
Proof :
, where p is a prime number
Total multiples of
Removing these multiples as with them
= 32Multiples of 2 (as with them
Examples :
[Tex]\\[/Tex]
[Tex]\\[/Tex]
4) For two number a and b 
Special Case : gcd(a, b) = 1
Examples :
Special Case :,
[Tex]\\[/Tex]
[Tex]\\[/Tex]
,
[Tex]\\[/Tex]
[Tex]\\[/Tex]
[Tex]\\[/Tex]
5) Sum of values of totient functions of all divisors of n is equal to n.
Examples :
n = 6
factors = {1, 2, 3, 6}
n = 
= 1 + 1 + 2 + 2 = 6n = 8factors = {1, 2, 4, 8}n = 
= 1 + 1 + 2 + 4 = 8n = 10factors = {1, 2, 5, 10}n = 
= 1 + 1 + 4 + 4 = 10
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 aΦ(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 :
ap-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
https://cp-algorithms.com/algebra/phi-function.html
http://mathcenter.oxford.memory.edu/site/math125/chineseRemainderTheorem/
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
Please Login to comment...