Re^4: Algorithm to convert combinations to bitstring (size)

tye,
Ok, I think what you are saying is count in base-N even though you don't need all the bits? If that's the case then yes, it may make things easier and the extra space shouldn't be that much more.

Cheers - L~R

Comment on Re^4: Algorithm to convert combinations to bitstring (size)

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Re^5: Algorithm to convert combinations to bitstring (size) by tye (Sage) on Oct 18, 2006 at 18:50 UTC
That was my alternative (in an update) if K is small (or near N). That uses log2( N**K ) bits. But what I described above is using base 2. It uses only N bits no matter how large (or small) K is, and the translation doesn't depend on K either. It can describe any subset of your superset, not just subsets of size K. - tye	[reply]


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