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– Re: how many different methods can we find for this question

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If you take the sum of all but the top two values (out of 26), that sum does not reach 100M. So it might be much faster, knowing that one or more of, say, the top four values are required in the final, to run the combinatorial on only the remaining lower (22!!!) values, and then run a short routine to check all those sums with the handful of larger ones. There are a lot of combinations; anything cleverer that can reduce that load is a real time saver.

Craig

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