Author: Dann Corbit
Date: 11:07:55 11/03/99
Go up one level in this thread
On November 03, 1999 at 08:44:12, Guido wrote: >In the previous message I calculated the number of different positions with 32 >regular pieces (without captures, promotions, castling, ep) obtaining the value: > >Total dispositions: (48!/(32!*8!))^2 / 64 = 2.14 10^40 ===> 134 bits > >I think that the number is wrong and that the correct answer is 1/4 of this >value, so 132 bits are sufficient to represent all the positions. A new help for >the goal of a maximum of 160 bits to represent every chess legal position! > >The reason is that each bishop can occupy only 32 squares, and therefore we have >to exclude as illegal all the positions where both white bishops or both black >bishops occupy squares of the same colour. The complete symmetry of the problem >should guarantee the correctness of my statement. You can have many bishops on the board. I believe that the maximum would be 10 bishops, with up to 9 of a single color. >The problem become very complicated with promotions and I don't try here to >solve! Perhaps it could be convenient in the calculations to consider bishops on >white squares and bishops on black squares as two different pieces, with the >possibilities for each bishop of occupying only 32 squares. But a pawn promoted >can always choose only between 4 pieces depending on the colour of the square on >which it is promoted. Any scheme to number all positions must take into account any legal board position. Otherwise it cannot be used and its value is limited >In another successive message KarinsDad said: >" >The only illegal positions that can occur in the variable length code method >(that I perceive off the top of my head and assuming the decoder can check for >illegality of too many pieces and too many pieces of a given type) are: side not >to move king in check; side to move king in check by more than 2 pieces or in >check by more than 1 knight; pawn misplacement based on number of promotions >(i.e. with no captures and no promotions, all pawns must be on their starting >column); bishop misplacement based on number of promotions; piece location where >it could not be (say for example the king bishop realised when none of the king >pawns have moved). >" > >I would specify that IMO are also illegal all the positions where 2 identical >pieces give check, not necessarily 2 knights, specifying that in this case: > >- Queen and bishop are considered identical if queen acts as a bishop (on >diagonals) >- Queen and rook are considered identical if queen acts as a rook (on rows or >columns) > >There is one only exception to the second statement: it is legal a double check >of queens, rooks or queen-rook in the only situation in which the capture move >of a pawn determines its promotion (queen or rook). The pawn, doing so, could >discover the enemy king to an attack along the column where the pawn was before >the move. > >It was interesting to calculate how many positions for a double check and other >reasons are illegal in the very famous KDDKDD ending. > >Regards >Guido
This page took 0.01 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.