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Subject: Re: Maximum # of N-B-R-Q for black and white (again (!) corrected proof)

Author: Shep

Date: 04:05:30 01/27/00

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On January 27, 2000 at 03:43:50, Shep wrote:

>On January 26, 2000 at 15:07:41, J. Wesley Cleveland wrote:
>
>
>>
>>You are doing an extra capture here. e.g. d4 c5 d5 nc6 dc6 and now both the
>>white pawn at c6 and the black d pawn can queen. I believe (but have not tried)
>>that with 8 captures you can have all 16 pawns promote.
>
>Of course you can, just as I outlined above.

My apologies, I didn't read your post with the required amount of attention. :)
You are right.

So once more (I hope it's finally right this time):

Let's first assume that no pawn captures another pawn (since that would not
change the number of additional BNRQ).
It suffices if 8 additional captures are made in the way you described it.
Without loss of generality, we can assume 4 white pawns and 4 black pawns
advance by capturing. Since each side has 2+2+1=5 pieces that can be captured
(i.e. if we're talking about promoting to Bishop, there are 2N+2R+1Q), this is
possible without reducing the number of pieces to find max() for.

Consequently, all 16 pawns can promote without a hitch, yielding

max |B| = max |N| = max |R| = 20
max |Q| = 18

q.e.d.

---
Shep



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