Euler’s Totient function ? (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 :
?(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.
// 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 |
// 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;
} |
// 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 |
# 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 |
// 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 |
<Φphp // 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( $n ). "\n" ;
// This code is contributed by Sam007 Φ> |
<script> // Javascript program to calculate // Euler's Totient Function // Function to return // gcd of a and b function gcd(a, b)
{ if (a == 0)
return b;
return gcd(b % a, a);
} // A simple method to evaluate // Euler Totient Function function phi(n)
{ let result = 1;
for (let i = 2; i < n; i++)
if (gcd(i, n) == 1)
result++;
return result;
} // Driver Code for (let n = 1; n <= 10; n++)
document.write(`phi(${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) [How? 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 ?(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++ 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 code int 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 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 function int main()
{ int n;
for (n = 1; n <= 10; n++)
printf ( "phi(%d) = %d\n" , n, phi(n));
return 0;
} |
// 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 -= 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. |
# 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 / / 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 for n in range ( 1 , 11 ) :
print ( "phi(" , n, ") = " , phi(n))
# This code is contributed # by Nikita Tiwari. |
// 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 -= 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. |
<Φphp // 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 -= $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 Code for ( $n = 1; $n <= 10; $n ++)
echo "phi(" . $n . ") =" . phi( $n ). "\n" ;
// This code is contributed by Sam007 Φ> |
// Javascript program to calculate // Euler's Totient Function // using Euler's product formula function phi(n)
{ // Initialize result as n
let result = n;
// Consider all prime factors
// of n and for every prime
// factor p, multiply result
// with (1 - 1/p)
for (let p = 2; p * p <= n; ++p)
{
// Check if p is
// a prime factor.
if (n % p == 0)
{
// If yes, then update
// n and result
while (n % p == 0)
n /= p;
result *= (1.0 - (1.0 / p));
}
}
// If n has a prime factor greater
// than sqrt(n) (There can be at-most
// one such prime factor)
if (n > 1)
result -= result / n;
//Since in the set {1,2,....,n-1}, all numbers are relatively prime with n
//if n is a prime number
return parseInt(result);
} // Driver Code for (let n = 1; n <= 10; n++)
document.write(`phi(${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 ?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 the above algorithm.
// 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 code int 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 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;
} |
// 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 |
# 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 |
// 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 |
<Φphp // 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( $n ), "\n" ;
// This code is contributed // by ajit Φ> |
// Javascript program to calculate // Euler's Totient Function function phi(n)
{ // Initialize
// result as n
let result = n;
// Consider all prime
// factors of n and subtract
// their multiples from result
for (let p = 2;
p * p <= n; ++p)
{
// Check if p is
// a prime factor.
if (n % p == 0)
{
// If yes, then
// update n and result
while (n % p == 0)
n = parseInt(n / p);
result -= parseInt(result / p);
}
}
// If n has a prime factor
// greater than sqrt(n)
// (There can be at-most
// one such prime factor)
if (n > 1)
result -= parseInt(result / n);
return result;
} // Driver Code for (let n = 1; n <= 10; n++)
document.write(`phi(${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 is , i.e the number p itself, so subtracting 1 from p
Examples :
[Tex]\\[/Tex] [Tex]\\[/Tex]
2) For two prime numbers a and b
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) = ab - (a + b - 1)[/Tex] , removing all number with 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 numbers from 1 to Total multiples of Removing these multiples as with them [Tex]\\[/Tex]Example : p = 2, k = 5, = 32Multiples of 2 (as with them ) = 32 / 2 = 16 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32} [Tex]\phi(p ^ k) = p ^ k - p ^ {k - 1}[/Tex]
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]Normal Case : , [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 = 6 n = 8factors = {1, 2, 4, 8}n = = 1 + 1 + 2 + 4 = 8 n = 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 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
https://cp-algorithms.com/algebra/phi-function.html
http://mathcenter.oxford.memory.edu/site/math125/chineseRemainderTheorem/