Author: David Eppstein
Date: 11:03:01 08/19/99
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On August 18, 1999 at 20:25:06, KarinsDad wrote: >Doesn't 2 * 2^162 + 16 * 2^161 > 2^165? >>So even ignoring the cases you can't handle, you still have 166 bits required. >>Why do you say the maximum required is 162? > >The cases are not part of the same position, but rather templates of given >position types. For example, you could split positions into 17 cases: ones with >16 pawns, one with 15 pawns, ..., ones with zero pawns. Perhaps you misunderstood my question. Any count of the number of bits required to store a position has to include the number of bits used to tell which case you are in. Otherwise I could just divide into 2^180 cases and say that I store the position in 0 bits. It seems that when you are saying you require at most 162 bits (except for the cases you can't handle) that you are not counting this part of the representation.
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