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Subject: Re: Is this where the 174 bit minimal figure comes from?

Author: José de Jesús García Ruvalcaba

Date: 08:59:48 05/19/99

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On May 19, 1999 at 11:55:50, José de Jesús García Ruvalcaba wrote:

>On May 19, 1999 at 11:32:51, Heiner Marxen wrote:
>
>>On May 18, 1999 at 08:02:09, Peter Klausler wrote:
>>
>>>On May 17, 1999 at 21:15:59, Dann Corbit wrote:
>>>
>>>>You would also need to store e.p. square, if any, I think. Ten bits more
>>>>[5ep+4castle+1stm]gives 174 bits.
>>>
>>>Two bits are all you need for the en passant file, actually.
>>>There can be at most four pawns of the opponent on his
>>>fourth rank next to a pawn of the player (on his fifth).
>>
>>Wrong.  There can be 5 such pawns.  E.g. 8/8/8/8/PpPPpPPp/8/8/8 b -
>>Even when we know already that there is an e.p. possible, we still
>>have to distinguish up to at least 5 cases.
>
>	Your position has five pawns on the fifth rank next to three of the opponent in
>the fourth rank. You and Peter are counting different things: he is counting the
>pawns in the fourth rank next to opponent pawns on their fifth rank (so for your
>position that count is three, less than four).
>	It is not neccesary to distinguish which pawns can make an en-passant capture,
>it is enough to know pawn can be taken en-passant. That is the reason Peter
>counts the pawns on the fourth rank (which can be captured en-passant) and not
>those which can make the capture.
>José.

	Sorry! I mis-read your position. That happens to me when I do not set up the
board and pieces (especially in the morning, here it is 11:00, too early for
me!).
José.



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