Post Your Own Puzzles

Black 3



White 4

Same question as before. White can mate with this move. Where's White's invisible king? Send me a message with the solutions, rather than posting them here.

I PM'ed solutions to the last 3 puzzles.

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 Jirakon
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).

1024.

I think you did something wrong. The only way to get 1024 is to fill up the chessboard. How can no piece be threatening another piece?

Originally posted by Jirakon
I think you did something wrong. The only way to get 1024 is to fill up the chessboard. How can no piece be threatening another piece?

Easy. I used all white pieces. Pieces from the same army don't threaten one another.

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 Jirakon
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.

With the constraint that no piece can observe another, I get a maximum product of 112.

I just had a very quick go at this and I also got a product of 112.

I can do better, but I need a question answered first. Jirakon, do knights block bishops' paths? [EDIT: Never mind, stupid question. Then the knights would be threatened.]

Here's how everyone is getting 112:

[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.]

Originally posted by CZeke
[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.]

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.

Originally posted by SwissGambit
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.

why didn't you fill the board with knights on all 32 white squares?
32 x 7 = 224

Originally posted by Mephisto2
why didn't you fill the board with knights on all 32 white squares?
32 x 7 = 224

Because no piece can observe another.