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Thomas comments on A puzzle - Less Wrong
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Comments (44)
Prove it, since your answer is correct, as far as I know.
Can you answer for the case of the two sets of pieces? Black and white, 32 of them. Again the color does not matter.
No, I'll leave it to you for now.
For two sets, the answer is 118.
119.
According to my proof, more than 128-10 is impossible. Are you sure?
Thanks! Clever trick with a rook and two bishops to reduce the length of the top boundary, I missed it.
Cebbs fxrgpu sbe n fvatyr frg: svefg, pbafvqre gjb xavtugf naq n xvat. Rnfl gb frr gurl unir ng yrnfg 4 serr pbaarpgvbaf, ab znggre ubj bgure cvrprf ner cynprq. Gura pbafvqre gur gbc obhaqnel (=cvrprf jvgu ng yrnfg bar serr hcjneq ovfubc-zbir naq ng yrnfg bar serr ebbx-zbir). Gur gbc obhaqnel pna'g or yrff guna 6 cvrprf, rnpu jvgu ng yrnfg bar serr pbaarpgvba, naq ng yrnfg bar bs gur obhaqnel cvrprf unf abguvat va gur ebjf nobir vg, fb vg unf ng yrnfg gjb serr pbaarpgvbaf. Nygbtrgure 64-(4+6+1)=53.
Can you give a position for 53 'bounds'?
Here
Then the clarification mentioned in a comment:
is false, because by turning the pawn of the second row, and the first pawn of the third row into black pieces, we go up to 55 connections from 53.
Makes no sense to me what you said. Can you please clarify?
What about the three sets? And four?
Cool! Using the bishops trick, any N-sets for N>2 reduce to N*64 - 8