3

I saw this puzzle and a solution sometime in the previous century, I don't remember where, and I don't remember the solution.

Place the 16 white men on an otherwise empty chessboard so that the following conditions are satisfied.

  1. Each man attacks one and only one other man.
  2. Each man is attacked by one and only one other man.
  3. The bishops are on opposite colors.
  4. One pawn on each file, no pawns on the first or last rank.
  5. There is one continuous cycle of attack, meaning that the men can be numbered from 1 to 16 so that #1 attacks #2, which attacks #3, which attacks #4, . . ., which attacks #15, which attacks #16, which attacks #1.

What's the solution?

Partial solutions (satisfying the first three conditions, but with doubled pawns or with two or more cycles of attack), will be appreciated and upvoted, but I'm hoping to accept an answer satisfying all five conditions.

1
  • Interesting puzzle. If I had to guess I'd think it's not even possible... But if it is a puzzle, presumably there must be some solution. – koedem Jan 3 at 0:53
5

Best solution

If there is nothing wrong with my rather spaghettoid script (admittedly, a big if) then there is no complete solution. The best my computer finds is two cycles. It finds essentially three such solutions plus a few minor variations. Interestingly, the split is always 3-13.

[FEN "1Q6/3R4/NB2P3/1P3P2/2P3P1/1N1P3P/P4KB1/R7 w - - 0 1"]

.

[FEN "R7/BP6/2PK4/6R1/2N3BP/1N1P2P1/P3PP2/Q7 w - - 0 1"]

.

[FEN "6R1/6BP/1N4P1/P4P2/BP2P1NK/2PP4/7Q/3R4 w - - 0 1"]

Strategy / computer implementation

From the assumptions follow a handful of constraints that enable an exhaustive search.

  1. There are no more than 4 chains of Ps. That is because Ps can only attack other Ps or Rs or Ns and each separate chain has a P at the base that is not available to be attacked by another P. Each chain is capped by a R or N.
  2. The major pieces each have a rank to themselves. That is because there is at least a P on their file, so they must spend their attack on that file. The Q, in particular, must also have her diagonals to herself.
  3. The Q must be attacked by a N.
  4. At least 1 chain of Ps must be capped off by a N. There are at least two Ps on the same rank and their respective chains cannot be both capped by Rs because then 1 chain would have to cross the private rank of the other chains R.

Using these constraints makes the code rather verbose and ugly but keeps the number of candidates below a million or so.

Relaxing the rules

If we lift the one pawn per file rule it becomes untractable for my computer program, but we can solve manually

[FEN "6R1/1R6/1BP3BN/3P2P1/2P2P2/1P4P1/P1N2K/Q7 w - - 0 1"]

Last word: The i files

We can actually achieve a full solution with no doubled pawns by ever so slightly bending the rules. All we need to do is rip out the f file but relax we are putting it right back: as the new i file. Job done:

enter image description here

3
  • Thank you! Can you get a solution with one cycle if you allow doubled pawns? – bof Jan 6 at 22:25
  • 3
    @bof my computer can't but I can ;-) – Paul Panzer Jan 7 at 5:11
  • Well done. On the assumption that your computer program is correct and I've been suffering from a false memory of having seen a solution with one cycle and no doubled pawns, I'm awarding the check mark. Of course, I will have to move it if a solution to the puzzle as I stated it ever turns up. – bof Jan 7 at 6:24
2

This has several issues. It's missing a pawn on c and a bishop on the black diagonal, and the pawn on d doesn't attack anything, as well as nothing attacks the rook on a3, but this is as close as I can apparently get. Maybe someone can work off this and do better.

However, there's a single cycle from Ra3 to d6. Anything else I try ends up with at least one piece attacking multiple other pieces, and I still can't get all the pieces on the board without several of these multiple attacks.

I put baby the queen in the corner to attempt to reduce the directions of her influence. Then I tried to put things in around her. I've been doing this manually, so I obviously haven't managed to try all combinations, but this methodology seems to have some merit, until it hits a wall.

[FEN "5R2/N5P1/1P1P3P/PB2P3/5P2/R7/4KN2/7Q w - - 0 1"]

There are so very many issues with this puzzle:

  1. The queen can only be attacked by a knight.
  2. The queen and both rooks require a whole rank to themselves, leaving only 5 ranks for the other 13 pieces.
  3. The queen needs the equivalent of at least one full diagonal.
  4. Pawns can't attack bishops or the king.
  5. Knights, bishops, and rooks can't attack like pieces.
  6. Because there's so many pawns, most major pieces are likely going to have to attack one.
1
  • @RewanDemontay, no. Unfortunately, I couldn't find places to put the one bishop or the last pawn that kept the cycle going without violating a bunch of puzzle rules. – computercarguy Jan 7 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.