Author: Sune Fischer
Date: 11:04:28 09/10/02
Go up one level in this thread
On September 10, 2002 at 13:22:59, Uri Blass wrote: >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. The examples are just meant to be hard to solve for the algorithm, to see if it works even in the extreme conditions. The position is not likely to occur of course, but if it can solve that, it can solve anything. >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. If you use bitboards there is a chance you have something that can produce these things quite fast. Imagine for instance a blocked pawn position, we want to know if the king can get through. The bitboard might look like this: 11111111 11111111 11111111 00001000 00010000 11111111 11111111 11111111 can he walk from e2 to g6? Turns out he can, but a rook or bishop of wrong color might not be able to help him, this is what the floodfill algorithm can tell you (with a few modifications). To me it looks like a tool we need to identify some of those fortress positions! -S. >Uri
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