Computer Chess Club Archives


Search

Terms

Messages

Subject: Re: pre-chess

Author: Marc Bourzutschky

Date: 17:51:49 05/16/04

Go up one level in this thread


On May 16, 2004 at 20:40:35, Norm Pollock wrote:

>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
>

There is only 1 way for the rook on h1, otherwise you will be double counting
positions where only the two rooks are swapped.  It looks like you have a
similar issue with your other calculation, which gives a factor of 2 overall.

>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



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.