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I was trying to solve the following problem using Burnside's lemma, but when testing the solution for a $12\times 12$ matrix involving $20$ possibilities for each matrix element, it had an awful run time.

Does this problem have a better solution than the Burnside's lemma, so it can give the result with a better time complexity?

The problem:

Suppose that a matrix is equivalent to another one if it can be obtained by swapping the rows and/or columns of the other one. The goal is to find the number of inequivalent $h\times w$ matrices with each of its elements having $s$ possible choices.

For example, 

1 5    
0 0

would be equivalent to itself and to:

0 0
1 5

and would also be equivalent to:

0 0 
5 1

and to:

5 1
0 0
  • 0
    what does swapping rows and columns mean? – 2017-01-07
  • 0
    It might be useful to consider that this swapping preserves the row sums and column sums. – 2017-01-07
  • 0
    This question recently appeared at this [MSE link](http://math.stackexchange.com/questions/2056708/). – 2017-01-07

0 Answers 0