Re^2: Algorithm to convert combinations to bitstring (size)

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tye,
N = 26 always and I will have 25 different bitstrings for K = 1 .. 25. I am not sure I understand what you mean by your 1 bit per superset and then set the bit if that member is in the subset.

ABCDE = 00000
AC    = 10100
DE    = 10100 + 00011 = 10111 ?
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Cheers - L~R

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Re^3: Algorithm to convert combinations to bitstring (size) by tye (Sage) on Oct 18, 2006 at 14:38 UTC
26 bits isn't very many so A=20=1, B=21=2, ..., Z=2**25=0x2000000. ABZ=0x2000003. - tye	[reply]
Re^4: Algorithm to convert combinations to bitstring (size) by Limbic~Region (Chancellor) on Oct 18, 2006 at 14:44 UTC
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	[reply]
Re^5: Algorithm to convert combinations to bitstring (size) by tye (Sage) on Oct 18, 2006 at 14:50 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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