New puzzle! Place a number of knights and bishop on a chessboard so that none of them threaten each other (simple enough). The challenge is to maximize the product (number of knights) x (number of bishops). Don't post your solution yet, just post the product you got. I'll post the product I got later (I don't want to give any hints).
Originally posted by Jirakon1024.
New puzzle! Place a number of knights and bishop on a chessboard so that none of them threaten each other (simple enough). The challenge is to maximize the product (number of knights) x (number of bishops). Don't post your solution yet, just post the product you got. I'll post the product I got later (I don't want to give any hints).
For some reason, I'm not sure whether you're actually serious or just being smart. So let me reword it: Place a number of bishops and knights on a chessboard so that none of them stand on a square that another piece is threatening, and so that the product (number of knights on the board)x(number of bishops on the board) is maximized.
Originally posted by JirakonWith the constraint that no piece can observe another, I get a maximum product of 112.
For some reason, I'm not sure whether you're actually serious or just being smart. So let me reword it: Place a number of bishops and knights on a chessboard so that none of them stand on a square that another piece is threatening, and so that the product (number of knights on the board)x(number of bishops on the board) is maximized.
Originally posted by CZekeI'd bet $$ that 112 is the maximum.
[Never mind, I thought this was an improvement, but it wasn't. I still think 112 can be improved on, though.
Side note: You can't put two FEN boards in one post. That seems dumb somehow.]
Putting another B on the board requires 3 Ns to leave (at minimum), which results in a lower product. Removing a B from the board only allows 2 more Ns to be placed (at maximum), which also results in a lower product.
5 Bishops require 23 (!) N's for their product to exceed 112. The closest I could get was 21 N's.
Originally posted by SwissGambitwhy didn't you fill the board with knights on all 32 white squares?
I'd bet $$ that 112 is the maximum.
Putting another B on the board requires 3 Ns to leave (at minimum), which results in a lower product. Removing a B from the board only allows 2 more Ns to be placed (at maximum), which also results in a lower product.
5 Bishops require 23 (!) N's for their product to exceed 112. The closest I could get was 21 N's.
32 x 7 = 224