Author: Shep
Date: 00:43:50 01/27/00
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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. But the original question was "how many pieces of the same type can be on the board at the same time", and as I showed, you are not getting anything by promoting more than 8 pawns (W and B combined) because for each new piece you get by promotion you have to capture two pieces, reducing the overall number of pieces by 1. But let's just try how many knights we can get with this method (correction to my argumentation in above post): As I said, 8 pawns can promote without any trouble, so we have 8+4=12 knights. For each extra knight (= promoted pawn), a side has to capture 2 opponent pieces. Since we do not want to reduce the number of knights, they have to be non-knights. Each side has 2R/2B/1Q that can be legally captured, so each side can promote two additional pawns (using the 2-by-1 capture method I described) for a total of 4 new knights. Adding that to the 12 above, we get 16. The same argument holds for rooks and bishops. For queens, we can modify the argument (in short): 8+2=10 initially, then we have 2R+2N+2B to capture, so each side can promote 3 additional pawns, yielding 6 extra queens for a total of 16 as well. Summing up we have: max |Q| = max |N| = max |B| = max |R| = 16. q.e.d. --- Shep
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