Author: Uri Blass
Date: 10:22:59 09/10/02
Go up one level in this thread
On September 09, 2002 at 19:01:17, Gerd Isenberg wrote: >On September 09, 2002 at 18:39:51, Vincent Diepeveen wrote: > >>On September 09, 2002 at 16:13:00, Gerd Isenberg wrote: >> >>Hello, >> >>Just to be clear is the next path a legal one: >> isconnected(d4,d8) ==> >> d4 e3 d1 c1 b2 a3 a4 a5 b6 c6 d6 e6 f7 e8 d8 > >Yes, Vincent, exactly. > >With the following bitboard > >0x18201e010911221c > >0 0 0 1 1 0 0 0 18 >0 0 0 0 0 1 0 0 20 >0 1 1 1 1 0 0 0 1e >1 0 0 0 0 0 0 0 01 >1 0 0 1 0 0 0 0 09 >1 0 0 0 1 0 0 0 11 >0 1 0 0 0 1 0 0 22 >0 0 1 1 1 0 0 0 1c How do you get that bitboard? I do not understand why do you choose 1 for d4 and e3. d4 and e3 are controlled by black pawns so they cannot be squares that white piece is going to go. I can think of a similiar path as a path for the white king. Another question is how do you plan to know the squares that are safe for the white king in the diagram. Evaluating this seems to me the main expensive problem and not calculating if 2 squares are connected. Uri
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