Euler?s Totient Function
Euler?s Totient function ?(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 :
?(1) = 1
gcd(1, 1) is 1
?(2) = 1
gcd(1, 2) is 1, but gcd(2, 2) is 2.
?(3) = 2
gcd(1, 3) is 1 and gcd(2, 3) is 1
?(4) = 2
gcd(1, 4) is 1 and gcd(3, 4) is 1
?(5) = 4
gcd(1, 5) is 1, gcd(2, 5) is 1,
gcd(3, 5) is 1 and gcd(4, 5) is 1
?(6) = 2
gcd(1, 6) is 1 and gcd(5, 6) is 1,
How to compute ?(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 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 <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; } |
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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 |
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Python3
# A simple Python3 program� # to calculate Euler's� # Totient Function ��# Function to return # gcd of a and b def gcd(a, b): ������if (a == 0): ��������return b ����return gcd(b % a, a) ��# A simple method to evaluate # Euler Totient Function def phi(n): ������result = 1����for i in range(2, n): ��������if (gcd(i, n) == 1): ������������result+=1����return result ��# Driver Code for n in range(1, 11): ����print("phi(",n,") = ",� �����������phi(n), sep = "") �������������# This code is contributed # by Smitha |
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C#
// A simple C# program to calculate // Euler's Totient Function using 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 |
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PHP
<?php // PHP program to calculate� // Euler's Totient Function ��// Function to return� // gcd of a and b function gcd($a, $b) { ����if ($a == 0) ��������return $b; ����return gcd($b % $a, $a); } ��// A simple method to evaluate // Euler Totient Function function phi($n) { ����$result = 1; ����for ($i = 2; $i < $n; $i++) ��������if (gcd($i, $n) == 1) ������������$result++; ����return $result; } ��// Driver Code for ($n = 1; $n <= 10; $n++) ����echo "phi(" .$n. ") =" . phi($n)."\n"; ��// This code is contributed by Sam007 ?> |
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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 Log10n 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 ?(n) is equal to n multiplied by product of (1 ? 1/p) for all prime factors p of n. For example value of ?(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 <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; } |
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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. |
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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. |
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C#
// C# program to calculate Euler's Totient // Function using Euler's product formula using 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 *= (float)(1.0 - (1.0 / (float)n)); ����������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. |
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PHP
<?php // PHP program to calculate� // Euler's Totient Function� // using Euler's product formula function 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 *= (1.0 - (1.0 / $n)); ������return intval($result); } ��// Driver Code for ($n = 1; $n <= 10; $n++) echo "phi(" .$n. ") =" . phi($n)."\n"; ������// This code is contributed by Sam007 ?> |
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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 the implementation of above algorithm.
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 function int main() { ����int n; ����for (n = 1; n <= 10; n++) ��������printf("phi(%d) = %d\n", n, phi(n)); ����return 0; } |
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Java
// Java program to calculate� // Euler's Totient Function import 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 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 ajit |
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Python3
# Python3 program to calculate� # Euler's Totient Function def 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 Code for n in range(1, 11): ����print("phi(",n,") =", phi(n)); ������# This code is contributed� # by mits |
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C#
// C# program to calculate� // Euler's Totient Function using System; ��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 Code static public void Main () { ����int n; ����for (n = 1; n <= 10; n++) ��������Console.WriteLine("phi(" + n +� ������������������������������") = " + ������������������������������phi(n)); } } ��// This code is contributed� // by akt_mit |
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PHP
<?php // PHP program to calculate� // Euler's Totient Function ��function 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 $result; } ��// Driver Code for ($n = 1; $n <= 10; $n++) ����echo "phi(", $n,") =",� ����������phi($n), "\n"; ������// This code is contributed� // by ajit ?> |
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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, ?(p) is p-1. For example ?(5) is 4, ?(7) is 6 and ?(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 ?(ab) = ?(a) * ?(b). For example ?(5) is 4 and ?(6) is 2, so ?(30) must be 8 as 5 and 6 are relatively prime.
3) For any two prime numbers p and q, ?(pq) = (p-1)*(q-1). This property is used in RSA algorithm.
4) If p is a prime number, then ?(pk) = pk ? pk-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 ?(1) + ?(2) + ?(3) + ?(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?(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 ?(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
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