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

4.1

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.

For example, Φ(4) = 2, Φ(3) = 2 and Φ(5) = 4. There are 2 numbers smaller or equal to 4 that are relatively prime to 4, 2 numbers smaller or equal to 3 that are relatively prime to 3. And 4 numbers smaller than or equal to 5 that are relatively prime to 5.

We have discussed different methods for computation of Φ(n) in previous post.

How to compute Φ for all numbers smaller than or equal to n?
Example:

Input: n = 5
Output: Totient of 1 is 1
        Totient of 2 is 1
        Totient of 3 is 2
        Totient of 4 is 2
        Totient of 5 is 4 

We strongly recommend you to minimize your browser and try this yourself first.

A Simple Solution is to call Φ(i) for i = 1 to n.

An Efficient Solution is to use idea similar to Sieve of Eratosthenes to precompute all values/. The method is based on below product formula.

eulersproduct

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.

Below is complete algorithm:

1) Create an array phi[1..n] to store Φ values of all numbers 
   from 1 to n.  

2) Initialize all values such that phi[i] stores i.  This
   initialization serves two purposes.
   a) To check if phi[i] is already evaluated or not. Note that
      the maximum possible phi value of a number i is i-1.
   b) To initialize phi[i] as i is a multiple in above product
      formula. 

3) Run a loop for p = 2 to n
    a) If phi[p] is p, means p is not evaluated yet and p is a 
       prime number (slimier to Sieve), otherwise phi[p] must 
       have been updated in step 3.b
    b) Traverse through all multiples of p and update all 
       multiples of p by multiplying with (1-1/p).

4) Run a loop from i = 1 to n and print all Ph[i] values. 

Below is C++ implementation of above algorithm.

// C++ program to compute Totient function for
// all numbers smaller than or equal to n.
#include<iostream>
using namespace std;

// Computes and prints totien of all numbers
// smaller than or equal to n.
void computeTotient(int n)
{
    // Create and initialize an array to store
    // phi or totient values
    long long phi[n+1];
    for (int i=1; i<=n; i++)
        phi[i] = i; // indicates not evaluated yet
                    // and initializes for product
                    // formula.

    // Compute other Phi values
    for (int p=2; p<=n; p++)
    {
        // If phi[p] is not computed already,
        // then number p is prime
        if (phi[p] == p)
        {
            // Phi of a prime number p is
            // always equal to p-1.
            phi[p] = p-1;

            // Update phi values of all
            // multiples of p
            for (int i = 2*p; i<=n; i += p)
            {
               // Add contribution of p to its
               // multiple i by multiplying with
               // (1 - 1/p)
               phi[i] = (phi[i]/p) * (p-1);
            }
        }
    }

    // Print precomputed phi values
    for (int i=1; i<=n; i++)
      cout << "Totient of " << i << " is "
           << phi[i] << endl;
}

// Driver program to test above function
int main()
{
    int n = 12;
    computeTotient(n);
    return 0;
}

Output:

Totient of 1 is 1
Totient of 2 is 1
Totient of 3 is 2
Totient of 4 is 2
Totient of 5 is 4
Totient of 6 is 2
Totient of 7 is 6
Totient of 8 is 4
Totient of 9 is 6
Totient of 10 is 4
Totient of 11 is 10
Totient of 12 is 4

The same solution can be used when we have large number queries for computing totient function.

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

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