Calculating nCr for large values of n...??

i was reading this tutorial to calculate nCr for large n values , but i couldn’t get that how to calculate

A-1 % M …?? the way it is told in this tutorial is not clear to me ,… plzz help me…!!


I encountered nCr for the first time on GCJ 08, Round 3 Problem D … link to analysis.

The first key idea is that of Lucas’ Theorem.

Lucas’s Theorem reduces nCr % M to

(n0Cr0 M) (n1Cr1 M) … (nkCrk % M)


(nknk-1…n0) is the base M representation of n
(rkrk-1…r0) is the base M representation of r

Note, if any of the above terms is zero because ri > ni or any other degeneracy, then the binomial coefficient nCr % M = 0
This means that any of the terms in the expansion of niCri is not divisible by M. But this is only half the job done.

Now you have to calculate nCr % M (ignoring subscripts for brevity) for some 0 ≤ r ≤ n < M

There are no ways around it, but to calculate

[ n! / ( r! (n-r)! ) ] % M

Without loss of generality, we can assume r ≤ n-r

Remember, you can always do the Obvious. Calculate the binomial and then take a modulo. This is mostly not possible because the binomial will be too large to fit into either int or long long int (and Big Int will be too slow)

This can then be simplified by using some clever ideas from Modular Arithmetic.

If A*B M = 1, A and B are modular multiplicative inverses of each other. For brevity, we say B M = A-1 % M

It is not always possible to calculate modular multiplicative inverses. If A and M are not co-prime, finding a B will not be possible.
For example, A = 2, M = 4. You can never find a number B such that 2*B % 4 = 1

Most problems give us a prime M. This means calculating B is always possible for any A < M.
For other problems, look at the decomposition of M. In the codesprint problem you mentioned

142857 = 33 * 11 * 13 * 37

You can find the result of nCr % m for each m = 27, 11, 13, 37. Once you have the answers, you can reconstruct the answer modulo 142857 using Chinese Remainder Theorem. These answers can be found by Naive Methods since, m is small.

I have also seen problems where M is a product of large primes, but square free. In these cases, you can calculate the answers modulo the primes that M is composed of using modular inverses (a little more about that below), and reconstruct the answer using CRT.

I am yet to see a problem where M is neither, but if it is. I do not know if there is a way to calculate binomial coefficients generally (since you cannot calculate modular inverses, and neither can you brote force). I can dream of a problem where there are powers of small primes, but square-free larger ones for a Number Theory extravaganza.

There is one other way to calculate nCr for any M which is small enough (say M ≤ 5000) or small n and r (say r ≤ n ≤ 5000) by using the following recursion with memoization

nCr = n-1Cr + n-1Cr-1

Since there are no divisions involved (no multiplications too) the answer is easy and precise to calculate even if the actual binomials would be very large.

So, back to calculating

[ n! / ( r! (n-r)! ) ] % M, you can convert it to

n * (n-1) … * (n-r+1) * r-1 * (r-1)-1 … * 1

Of course, each product is maintained modulo M.

This may be fast enough for problems where M is large and r is small.

But sometimes, n and r can be very large. Fortunately, such problems always have a small enough M :smiley:

The trick is, you pre-calculate factorials, modulo M and pre-calculate inverse factorials, modulo M.

fact[n] = n * fact[n-1] M ifact[n] = modular_inverse(n) * ifact[n-1] M

Modular Multiplicative Inverse for a prime M is in fact very simple. From Fermat’s Little Theorem

AM-1 % M = 1

Hence, A * AM-2 % M = 1
Or in other words,

A-1 M = AM-2 M, which is easy (and fast) to find using repeated squaring.

There is one last link I wish to paste to make this complete. Modular inverses can also be found using the Extended Euclid’s Algorithm. I have only had to use it once or twice among all the problems I ever solved.