Question

I would like to be able to create multiple combinations that sum to 100%, given a defined number of “buckets” with a defined ‘difference factor’. In the below example, the difference is a factor of 20 to make it simple, but I will probably reduce it to 1 in the final solution.

For example, with 3 “buckets” A, B, C you could have:

A     100     80      80      60      60  ...   0
B     0       20      0       20      40  ...   0
C     0       0       20      20      0   ...   100

Each column is one combination (summing to 100) that I would like to store and do further calculations on.

This is a business problem and not homework.

Please help me come up with a solution. A brute force way would be to create a multi-dimension array for every possible combination, e.g. 100x100x100 and then go through each 1 million combination to see which ones sum to 100. However this looks like it will be way too inefficient.

Much appreciated. I hope I have explained clearly enough.

Answer 1

This problem is known as partitions rather than combinations, which is something different.

First off: the ‘difference factor’ just turns the problem from finding partitions of 100 to (in your example) finding partitions of 5 (then multiplying by 20).

Next up: If the number of buckets is constant, you can just do (pseudo code):

for i = 0 to n
  for j = 0 to n-i
    output (i, j, n-(i+j))

If the number of buckets is going to be dynamic, you’d have to be a bit cleverer, but this approach will basically work.