Unless there’s some clever structure to the optimal position, this is a problem best solved by computer search. Potentially it could make a good Project Euler problem, potentially not (depending on how much computation the best human-writable program requires to do it).
Also, this post gave the false impression that there was a known answer. That irritated me when I read the comments.
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.
Unless there’s some clever structure to the optimal position, this is a problem best solved by computer search. Potentially it could make a good Project Euler problem, potentially not (depending on how much computation the best human-writable program requires to do it).
Also, this post gave the false impression that there was a known answer. That irritated me when I read the comments.
I think I can prove 53 is the maximum. And the solution does have a nice structure.
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.
What about the three sets? And four?
Cool! Using the bishops trick, any N-sets for N>2 reduce to N*64 − 8
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?
I mean that the following position:
achieves 55 connections, by using mixed colors. Unless my arithmetic is failing me again.
Edit to add: And unless arithmetic is failing me again, the following has 56 connections:
Yes, I know now what you mean. It is a bit different puzzle now. The color of a piece maters this way
Do you think your solution is optimal?
I meant that with mixed colors, we can have the following:
which unless my arithmetic fails me again, is 55 connections.