I am solving a programming problem which is stuck at calculating nCr
efficiently and at the same time avoiding overflows. I have made the following trivial simplifi
I had to solve this problem, too. What I did was use the fact that there are the same number of multiplications as divisions and bundled them together, taking one multiplication and one division at a time. It comes out as an integer at the end, but I use double for the intermediate terms and then round to the nearest integer at the end.
// Return the number of combinations of 'n choose k'
unsigned int binomial(unsigned int n, unsigned int k) {
unsigned int higher_idx;
unsigned int lower_idx;
if(k > n-k) {
higher_idx = k;
lower_idx = n - k;
} else {
higher_idx = n - k;
lower_idx = k;
}
double product = 1.0;
double factor;
unsigned int idx;
for(idx=n; idx>higher_idx; idx--) {
factor = (double)idx / double(lower_idx - (n - idx));
product *= factor;
}
return (unsigned int)(product + 0.5);
}