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Re^3: Speeding permutation countingby Limbic~Region (Chancellor) |
on Jul 19, 2007 at 20:46 UTC ( [id://627617]=note: print w/replies, xml ) | Need Help?? |
BrowserUk,
What I am suggesting is that if you know the number of 1s and 0s in all strings in position 1, you should be able to calculate the number of '11', '00', '10', and '01' at position 1 without comparing any of the strings. This would still have to be done for all strings at all positions - I am just trying to eliminate the comparisons. Cheers - L~R
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