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;

Here is the link: http://e-maxx.ru/algo/reverse_element

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

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)

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

ithat is not greater thanm. In other words, .Thus , and so . That is,

r[i] is the modular inverse ofimodulom.There are some massive holes (mathematically) there, but should be easily patched up; but you want a simple explanation, not a completely correct one.

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!`

,Find the modular inverse of

N!. This is a result of Fermat Little Theorem and it can be computed inO(logP)Now we can calculate the inverse of each factorial backwards in

O(N).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:

Here is my code using this method: 35525761

Can you explain how the ifak[k] = ifak[k+1]*(k+1)%P works? More specifically, how modpow(fak[N],P-2) and ifak[k+1]*(k+1)%P produce the same result?

Notice that the inverse of factorials is computed in reverse order. You start by correctly computing the inverse factorial of $$$n$$$ which is $$$n!^{p - 2} mod(p)$$$:

`ifak[n] = modpow(fak[n], p - 2)`

Now that we now the inverse of $$$n!$$$ we can deduce smaller inverse in a similar fashion as we compute the factorials themselves

So we deduce $$$k!^{-1}$$$ from $$$(k+1)!^{-1}$$$ easily using:

`ifak[k] = ifak[k + 1] * (k + 1) % p`

.Not sure what do you mean by this.

`modpow(fak[N],P-2)`

isonlyused to compute the inverse of $$$N!$$$.`ifak[k+1]*(k+1)%P`

is used to compute the remaining factorial inverses. $$$0 \le k \lt N$$$What I meant earlier is that we could calculate inverse of each

x!for x from 0 to Nseparately using`modpow(fak[x],P-2)`

or we could do it faster with the iterative version, and I wanted to know how using both of them would give the same result.With the proof for the iterative step, it is very clear now,

Thank you for this wonderful explanation!