Φ(5) = 4 gcd(1, 5) is 1, gcd(2, 5) is 1, gcd(3, 5) is 1 and gcd(4, 5) is 1 Φ(6) = 2 gcd(1, 6) is 1 and gcd(5, 6) is 1,
We have discussed different methods to compute Euler Totient function that work well for single input. In problems where we have to call Euler’s Totient Function many times like 10^5 times, simple solution will result in TLE(Time limit Exceeded). The idea is to use Sieve of Eratosthenes.
Find all prime numbers upto maximum limit say 10^5 using Sieve of Eratosthenes.
To compute Φ(n), we do following.
- Initialize result as n.
- Iterate through all primes smaller than or equal to square root of n (This is where it is different from simple methods. Instead of iterating through all numbers less than or equal to square root, we iterate through only primes). Let the current prime number be p. We check if p divides n, if yes, we remove all occurrences of p from n by repeatedly dividing it with n. We also reduce our result by n/p (these many numbers will not have GCD as 1 with n).
- Finally we return result.
10 12 30 40 32 60 72 100
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