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Subject: Re: pre-chess

Author: Norm Pollock

Date: 17:40:35 05/16/04

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On May 16, 2004 at 20:21:38, Marc Bourzutschky wrote:

>On May 16, 2004 at 19:56:08, Norm Pollock wrote:
>
>>On May 16, 2004 at 17:27:10, Marc Bourzutschky wrote:
>>
>>>On May 16, 2004 at 17:12:12, Dieter Buerssner wrote:
>>>
>>>>On May 16, 2004 at 14:28:48, Marc Bourzutschky wrote:
>>>>
>>>>>Max Euwe: 4,147,200
>>>>>Noam Elkies: 8,294,400
>>>>>Paul Epstein: 5,317,600
>>>>>Marc Bourzutschky: 5,149,368
>>>>
>>>>Dieter Bürßner: 4,665,582
>>>>
>>>>Idea: 2880 positions per side, of which 2694 have no castling possibilities.
>>>>
>>>>x = 2880^2-2694^2/2
>>>>
>>>>I fear, I thought too simple,
>>>>Dieter
>>>
>>>If instead of 2694 in your formula you use 2508 you get the Bourzutschky result.
>>> The difference is that 2508 is the number of positions where neither the
>>>position itself, nor the mirrored position, has castling rights...
>>
>>Fwiw, I disagree with the explanation of 2508.
>>
>>I think the 2508 is just the number of positions that do NOT have castling
>>positions. I calculate 372 castling positions of the 2880 possible positions,
>>therefore 2508 in my calculations is the number of positions without castling
>>rights.
>>
>>x = 2880^2 - (2508^2/2) = 5,149,368 is still correct.
>>
>>It says take all the positions of both sides then remove the duplicate of those
>>positions where neither side had castling rights.
>>
>>-- Norm
>
>I only calculate 186 castling positions for white, assuming the white king has
>to be on e1.  In addition, your argument does not seem quite right, because even
>if one side has castling rights the symmetry is broken, even if the other side
>does not have castling rights...

Would you show me your analysis for the 186 castling positions you calculated
for white, assuming the white king to be on e1? I want to see where I differ
from you. It's a factor of 2 so it should be easy to spot. I'm putting my
analysis for 372 castling positions below.

For ----K--R, there are 3*3 ways for 2 bishops of opp color, 1 way for the king,
2 ways for the castle at h1, 4*3/2 ways for 2 knights, 2 ways for the queen and
1 way for the 2nd castle.
Sub-result= 3*3*1*2*(4*3/2)*2*1 = 216

For R---K---, there are 4*2 ways for 2 bishops of opp color, 1 way for the king,
2 ways for castle at a1, 4*3/2 ways for 2 knights, 2 ways for the queen and 1
way for the 2nd castle.
Sub-result= 4*2*1*2*(4*3/2)*2*1 = 192

For R---K--R (to remove duplicates from 2 sub-results), 2 ways for castle at a1,
1 way for castle at h1, 1 way for king at e1, 3*2 ways for 2 bishops of opp
color, 3*2/2 ways for 2 knights, and 1 way for the queen.
Sub result= 2*1*1*3*2*(3*2/2)*1 = 36

Result = 216 + 192 - 36 = 372

-Norm



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