### pranay2063's blog

By pranay2063, 4 years ago, ,

Hi all,

While searching about inverse modulo, i got to know about a concise algorithm to find inverse modulo of numbers in range[1...n) under modulo m.

Time complexity of this approach is O(n).

Implementation of the algorithm:

r[1] = 1; for (int i=2; i<n; ++i) r[i] = (m — (m/i) * r[m%i] % m) % m;

I am unable to understand the proof of the algorithm. It would be very helpful if anyone explains the same in a simple way.

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 » 4 years ago, # | ← Rev. 3 →   +1 m mod i = m — m/i*i m mod i = — m/i*i (mod m) // multiply by r[i] r[i] * (m mod i) = — m/i*i *r[i] (mod m) r[i] * (m mod i) = — m/i (mod m) // multiply by r[m % i] — reverse to (m mod i) (reverse modulo m) r[i] = — m/i * r[m % i] (mod m)
 » 4 years ago, # |   +1 This is the simplest I can think of:We want to prove that .Now multiply both sides by . We assume that already.And note that is the largest multiple of i that is not greater than m. In other words, .Thus , and so . That is, r[i] is the modular inverse of i modulo m.There are some massive holes (mathematically) there, but should be easily patched up; but you want a simple explanation, not a completely correct one.
 » 7 months ago, # |   +11 I was discussing with mgold earlier about another way to find the modular inverse of the first N numbers mod a prime number P.Find the factorial of the first N numbers, let's say fak[n] = n!, fak[0] = 1 for k=1...N: fak[k] = fak[k-1] * k % P Find the modular inverse of N!. This is a result of Fermat Little Theorem and it can be computed in O(logP) ifak[N] = modpow(fak[N], P - 2) Now we can calculate the inverse of each factorial backwards in O(N). for k=N-1...0: ifak[k] = ifak[k+1] * (k+1) % P Most of the times I need to pre calculate the inverse of each of the first N numbers just to calculate the inverse of the factorial. So I can get C(n,k) in O(1). In that case we are done.But If you actually need the inverse of the first N numbers you can compute them: for k=1...N: inv[k] = ifak[k] * fak[k-1] % P Here is my code using this method: 35525761