Given an integer n, find the value of Euler's Totient Function, denoted as Φ(n). The function Φ(n) represents the count of positive integers less than or equal to n that are relatively prime to n.
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.
If n is a positive integer and its prime factorization is; n = p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_k^{e_k}
Where p_1, p_2, \ldots, p_k are distinct prime factors of n, then:
\phi(n) = n \cdot \left(1 - \frac{1}{p_1}\right) \cdot \left(1 - \frac{1}{p_2}\right) \cdot \ldots \cdot \left(1 - \frac{1}{p_k}\right).
Examples:
Input: n = 11
Output: 10
Explanation: From 1 to 11, 1,2,3,4,5,6,7,8,9,10 are relatively prime to 11.
Input: n = 16
Output: 8
Explanation: From 1 to 16, 1,3,5,7,9,11,13,15 are relatively prime to 16.
[Naive Approach] Iterative GCD Method
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++ #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 etf(int n) { int result = 1; for (int i = 2; i < n; i++) if (gcd(i, n) == 1) result++; return result; } // Driver Code int main() { int n = 11; cout << etf(n); return 0; } 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); } // Function to compute Euler's Totient Function static int etf(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 = 11; System.out.println(etf(n)); } } # 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 etf(n): result = 1 for i in range(2, n): if gcd(i, n) == 1: result += 1 return result # Driver Code if __name__ == "__main__": n = 11 print(etf(n))
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 etf(int n) { int result = 1; for (int i = 2; i < n; i++){ if (gcd(i, n) == 1) result++; } return result; } // Driver Code static void Main(){ int n = 11; Console.WriteLine(etf(n)); } } // 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 etf(n) { let result = 1; for (let i = 2; i < n; i++) { if (gcd(i, n) === 1) result++; } return result; } // Driver Code let n = 11; console.log(etf(n)); Time Complexity: O(n log n)
Auxiliary Space: O(log min(a,b)) where a,b are the parameters of gcd function.
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.
1) Initialize result as n
2) Consider every number 'p' (where 'p' varies from 2 to Φ(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.
C++ #include <iostream> using namespace std; int etf(int 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 = 11; cout << etf(n); return 0; } class GfG { static int etf(int n) { int result = n; // Consider all prime factors of n // and subtract their multiples // from result for (int p = 2; p * p <= n; ++p) { if (n % p == 0) { 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 = 11; System.out.println(etf(n)); } } def etf(n): result = n # Consider all prime factors of n # and subtract their multiples # from result p = 2 while p * p <= n: if n % p == 0: while n % p == 0: n //= p result -= 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 -= result // n return result # Driver Code if __name__ == "__main__": n = 11 print(etf(n))
using System; class GfG { static int etf(int 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 void Main() { int n = 11; Console.WriteLine(etf(n)); } } function etf(n) { let result = n; // Consider all prime factors of n // and subtract their multiples // from result for (let p = 2; p * p <= n; ++p) { if (n % p === 0) { while (n % p === 0) n = Math.floor(n / p); result -= Math.floor(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 -= Math.floor(result / n); return result; } // Driver Code const n = 11; console.log(etf(n)); Time Complexity: O(√n)
Auxiliary Space: O(1)
Some Interesting Properties of Euler's Totient Function
1) For a prime number p, \phi(p) = p - 1
Proof :
\phi(p) = p - 1 , where p is any prime number
We know that gcd(p, k) = 1 where k is any random number and k \neq p
\\Total number from 1 to p = p
Number for which gcd(p, k) = 1 is 1, i.e the number p itself, so subtracting 1 from p \phi(p) = p - 1
Examples :
\phi(5) = 5 - 1 = 4\\\phi(13) = 13 - 1 = 12\\\phi(29) = 29 - 1 = 28.
2) For two prime numbers a and b \phi(a \cdot b) = \phi(a) \cdot \phi(b) = (a - 1) \cdot (b - 1) , used in RSA Algorithm
Proof :
Let a and b be distinct primes.
Then:
Total numbers from 1 to ab: ab
Multiples of a: b numbers
Multiples of b: a numbers
Common multiple (i.e., double-counted): only ab
So, numbers not coprime to ab:
a+b−1
Then,
ϕ(ab) = ab − (a + b − 1) = ab − a − b + 1 = (a−1)(b−1)
Hence,
ϕ(ab) = ϕ(a)* ϕ(b)
Examples :
- ϕ(5⋅7) = ϕ(5)⋅ϕ(7) = (5−1) (7−1) = 4⋅6 = 24
- ϕ(3⋅5) = ϕ(3)⋅ϕ(5) = (3−1) (5−1) = 2⋅4 = 8
- ϕ(3⋅7) = ϕ(3)⋅ϕ(7) = (3−1) (7−1) = 2⋅6 = 12
3) For a prime number p and integer k ≥ 1:
ϕ(pk) = pk−pk−1
Proof :
\phi(p^k) = p ^ k - p ^{k - 1} , where p is a prime number\\Total numbers from 1 to p ^ k = p ^ k Total multiples of p = \frac {p ^ k} {p} = p ^ {k - 1}
Removing these multiples as with them gcd \neq 1\\
Examples :
p = 2, k = 5, p ^ k = 32
Multiples of 2 (as with them gcd \neq 1) = 32 / 2 = 16\\\phi(p ^ k) = p ^ k - p ^ {k - 1}.
4) Special Case : gcd(a, b) = 1
\phi(a \cdot b) = \phi(a) \cdot \phi(b) \cdot \frac {1} {\phi(1)} = \phi(a) \cdot \phi(b).
Examples :
Special Case:
gcd(a, b) = 1, \phi(a \cdot b) = \phi(a) \cdot \phi(b)\phi(2 \cdot 9) = \phi(2) \cdot \phi(9) = 1 \cdot 6 = 6\\\phi(8 \cdot 9) = \phi(8) \cdot \phi(9) = 4 \cdot 6 = 24\\\phi(5 \cdot 6) = \phi(5) \cdot \phi(6) = 4 \cdot 2 = 8
Normal Case:
gcd(a, b) \neq 1, \phi(a \cdot b) = \phi(a) \cdot \phi(b) \cdot \frac {gcd(a, b)} {\phi(gcd(a, b))}\\\phi(4 \cdot 6) = \phi(4) \cdot \phi(6) \cdot \frac {gcd(4, 6)} {\phi(gcd(4, 6))}= 2 \cdot 2 \cdot \frac{2}{1}= 2 \cdot 2 \cdot 2 = 8\\\phi(4 \cdot 8) = \phi(4) \cdot \phi(8) \cdot \frac {gcd(4, 8)} {\phi(gcd(4, 8))} = 2 \cdot 4 \cdot \frac{4}{2} = 2 \cdot 4 \cdot 2 = 16\\\phi(6 \cdot 8) = \phi(6) \cdot \phi(8) \cdot \frac {gcd(6, 8)} {\phi(gcd(6, 8))} = 2 \cdot 4 \cdot \frac{2}{1} = 2 \cdot 4 \cdot 2 = 16.
5) Sum of values of totient functions of all divisors of n is equal to n.
Example :
n = 6 , factors = {1, 2, 3, 6}
n = \phi(1) + \phi(2) + \phi(3) + \phi(6) = 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
aΦ(n) Φ 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 Φ 1 (mod p)
Related Article:
Euler’s Totient function for all numbers smaller than or equal to n
Optimized Euler Totient Function for Multiple Evaluations
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