# crowded 5x5 chessboard

While working on an earlier chess piece placement problem, I happened to come across a silly (but somewhat interesting) configuration:

1. The board size is 5x5 squares
2. There are 16 pieces of the same colour:
• 4 rooks
• 4 bishops
• 8 knights
3. Each piece is attacked by exactly 1 other piece. (Maybe "protected" would be a better word, since the pieces are of the same colour. But you get the point.)

Puzzle 1: Can you fit all those pieces on such a tiny board while following the rules?

Puzzle 2: Which notable chess board property is also demonstrated by the answer?

(Disclaimer: I'm fairly confident that the solution to puzzle 1 is unique, but haven't brute-forced it. If I'm wrong, puzzle 2 may be unsolvable.)

• "Each piece is attacked by exactly 1 other piece" or "Each piece is attacked by at most 1 other piece"? – ibrahim mahrir Dec 6 '17 at 1:29
• @ibrahimmahrir, exactly 1 piece. Looks like you figured that out yourself with RaT’s help in the other comment chain :-) – Bass Dec 6 '17 at 6:24

## Solution

B N . N B
N . R . N
. R . R .
N . R . N
B N . N B

Image representation (thanks to @ibrahimmahrir):

This works because

each knight is defended by exactly one rook while not attacking anything, and the rooks and bishops cover each other in a perfect one-to-one correspondence.

## Methodology / ramblings

I swiftly realised that one of the keys to this puzzle is to use

symmetry. We can split the squares of the $5\times5$ chessboard into five symmetrical sets, labelling these sets A, B, C, D, E as follows (and ignoring the central square since it's not perfectly symmetric with anything):

A B C B A
B D E D B
C E . E C
B D E D B
A B C B A

It makes sense that a neat solution will involve all squares of the same letter being occupied by pieces of the same type (or unoccupied). Maybe not all solutions are symmetric (if there are more than one), but surely the one you have in mind is, since it should be elegant enough to find by hand.

After fiddling around a bit with trial and error, I started to consider it logically. The squares labelled E

can't be occupied by knights, since then the squares labelled A and C and D would have to be empty (each of these is covered by TWO knights), contradiction. They can't be occupied by bishops either, since then each bishop would be covered by TWO bishops. So let's try putting rooks on those squares (the other possibility is that they're empty).

Now the squares labelled D must be empty, since each one is covered by TWO rooks. So we have four A squares, eight B squares, and four C squares on which to fit four bishops and eight knights.

Putting the knights on the B squares works beautifully, and then the bishops can easily be slotted in to the A squares, giving the solution found above.

• I have the same configuration but hesitated because of this. – ibrahim mahrir Dec 6 '17 at 1:39
• @ibrahim Why should that make you hesitate? In this configuration, each piece is attacked by exactly one other piece, so we've solved the problem as stated. – Rand al'Thor Dec 6 '17 at 1:42
• I guess my understanding is that each piece should be attacked. The rooks are not attacked so I thought the rules aren't fullfilled. – ibrahim mahrir Dec 6 '17 at 1:44
• @ibrahim The rooks are attacked ... by each other. And thanks for the image! I was lazy and just did this in Notepad :-) – Rand al'Thor Dec 6 '17 at 1:45
• I'm guessing the puzzle 2 may have something to do with the centre square, which has the maximum of a certain property. – Apep Dec 6 '17 at 1:46

Given Rand al'Thor's solution to puzzle 1:

B  N  .  N  B
N  .  R  .  N
.  R  .  R  .
N  .  R  .  N
B  N  .  N  B

I believe that the answer to puzzle 2 is:

The centre square of the solution demonstrates the maximum number of pieces that can threaten/protect a single square on a chess board. All 16 pieces in the solution are threatening the centre square. It is impossible to add a 17th piece to threaten it without blocking another piece, even if the board is extended to the standard 8x8.

• Oh, nice! I hadn't even noticed they're all attacking the same square. – Rand al'Thor Dec 6 '17 at 13:25
• Yes, that's pretty much exactly how I would have put it. Interestingly, if you replace a couple of the rooks with a king and a queen, it becomes apparent that there are just enough pieces in the regular starting position to reach this maximum via a legal game. (I hope you agree that the tick needs to go to RaT's answer; upvoting this one too, of course.) – Bass Dec 7 '17 at 5:12