Fit the most pieces on a chess board.
- You are only allowed to place Rooks, Knights and Bishops.
- The number of rooks, knights and bishops must be approximately equal (specifically, the number of rooks, knights and bishops must all be either $k$ or $(k+1)$ for some $k$).
- Every piece has to be attacked by exactly one other piece.
I don't know what the answer to this problem is. Highest legit answer wins, internet bonus points if you can prove that you found the optimum.
Answer
POST MORTEM EDIT (well, post-tick, anyway) There was way too much flexibility left in the earlier solution, I just had to keep wiggling everything until I got a bishop on the other long diagonal too. Here it is:
9 knights, 9 bishops and 8 rooks, for a total of 26 pieces.
Please do double check it; I checked it several times already, but wishful thinking often makes it impossible to see one's own mistakes.
POST POST MORTEM EDIT: Here’s a completely different approach with the same result (now with much more symmetry)
9 knights, 9 bishops, 8 rooks = 26 pieces.This time much easier to confirm: the rooks attack everything except the very corners, nothing else attacks anything, except the corner bishops that attack each other.
The original accepted answer is below the line.
Remorselessly stealing Daniel Duque's incredible solution, which had just enough wiggle room to add another knight:
25 pieces (9 knights, 8 bishops, 8 rooks)
Here's another with the same number of pieces, but with a lot more room for improvement, it seems (I hope there aren't any mistakes, I'm getting pretty cross-eyed here..)
9 knights, 8 bishops, 8 rooks = 25 pieces
(There are lots of promising lookin variations going on in here, please feel free to use this one in your own answer)
Here's my first (original) attempt with
20 pieces: 7 bishops, 7 knights and 6 rooks. (with some wiggle room left in the bottom middle)
Like so:
As you can probably tell, squeezed everything together by starting with a greedy approach with the difficult pieces on the top side.
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