Author: Ralf Elvsén
Date: 18:00:46 01/28/02
Go up one level in this thread
On January 28, 2002 at 20:46:04, Ralf Elvsén wrote: >On January 28, 2002 at 13:26:02, Sune Fischer wrote: > >>On January 28, 2002 at 13:04:05, Sune Fischer wrote: >> >>This is what I get for half the rook, similar stuff is done for the backwards >>attacks using the reversed occupied bitboard. >> >>BITBOARD RookAttacksForward(BOARD &occupied, char square) >>{ >> BITBOARD a,x,y; >> >> x=occupied>>square; // shift board so square ends in lower left corner >> y=x&RANK1; // mask out ewerything but the first rank >> a=y^(y-2); // get the attacked bits on this rank >> y=x&FILE1; // now mask so we get a clean first file >> y=y^(y-2); // get the attacked squares >> a=a|(y&FILE1); // add the attacked bits after masking with the file again >> a=a<<square); // shift the attacked bits back (for ease of readability) >> >> return a; >>} >> >> >>..untested, but something along those lines. >>Not entirely sure how to do it for the bishop, I guess we just shift to the >>lower right corner when attacking north-west and lower left when attacking >>north-east. >> >>But what happens if x=0? >>Will y then be 0 or something strange, the bitboard is unsigned so it should >>remain 0 right? >> >>-S. > >I think one should change > >y = x&RANK1 > >to > >y = x|0x80 The above is not correct. The most significant bit in the rank is not nr 7 any longer. I should know better then improvise bit manipulation 3 am, sorry :) Ralf > >That way the most significant bit in the rank will be set and when >you do the x^(x-2)-thing you will get a correct answer. > >I don't understand how you can use the same "occupied" to get >the file attacks but it's late here... :) > >x will never be zero since there is slider at "square". > >Ralf
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.