Euler's Totient for First n Natural Numbers

Euler's Totient function φ(n) for an input n is the 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.

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
Explanation:
1 has 1 coprime number → φ(1) = 1
2 has 1 coprime number (1) → φ(2) = 1
3 has 2 coprime numbers (1,2) → φ(3) = 2
4 has 2 coprime numbers (1,3) → φ(4) = 2
5 has 4 coprime numbers (1,2,3,4) → φ(5) = 4

Input: n = 12
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
Explanation:
1 has 1 coprime number → φ(1) = 1 → {1}
2 has 1 coprime number → φ(2) = 1 → {1}
3 has 2 coprime numbers → φ(3) = 2 → {1,2}
4 has 2 coprime numbers → φ(4) = 2 → {1,3}
5 has 4 coprime numbers → φ(5) = 4 → {1,2,3,4}
6 has 2 coprime numbers → φ(6) = 2 → {1,5}
7 has 6 coprime numbers → φ(7) = 6 → {1,2,3,4,5,6}
8 has 4 coprime numbers → φ(8) = 4 → {1,3,5,7}
9 has 6 coprime numbers → φ(9) = 6 → {1,2,4,5,7,8}
10 has 4 coprime numbers → φ(10) = 4 → {1,3,7,9}
11 has 10 coprime numbers → φ(11) = 10 → {1,2,3,4,5,6,7,8,9,10}
12 has 4 coprime numbers → φ(12) = 4 → {1,5,7,11}

[Naive Approach] Iterative GCD Method - O(√n) time and O(1) space

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.

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[Efficient Approach] Sieve of Eratosthenes - O(n log(log n)) time and O(n) space

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

Below is the complete algorithm: 

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

Step 2. Initialize all values such that phi[i] stores i. This initialization serves two purposes.

• To check if phi[i] is already evaluated or not. Note that the maximum possible phi value of a number i is i-1.

• To initialize phi[i] as i is multiple in the above product formula.

Step 3. Run a loop for p = 2 to n

• If phi[p] is p, means p is not evaluated yet and p is a prime number (similar to Sieve), otherwise phi[p] must have been updated in step 3.b

• Traverse through all multiples of p and update all multiples of p by multiplying with (1-1/p).

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

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...

Time Complexity: O(n log(log n))
Auxiliary Space: O(n)

Prime Factorization - O(sqrt(n)*log(n)) time and O(1) space

The sieve-based solution is useful when we have a large number of queries for computing the totient function.

For numbers upto n, we can use Prime Factorization.

This approach computes φ(n) using prime factorization instead of iterating over all numbers.

For example, φ(12) = { (2^(2-1)) x (2-1) } x { (3^(1-1)) x (3-1) } =4

Note that φ(n) = n - 1 if n is prime.

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 o...