Cyclic attack puzzle 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.

Each man attacks one and only one other man.
Each man is attacked by one and only one other man.
The bishops are on opposite colors.
One pawn on each file, no pawns on the first or last rank.
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.
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 &quot;1Q6/3R4/NB2P3/1P3P2/2P3P1/1N1P3P/P4KB1/R7 w - - 0 1&quot;]

.
[FEN &quot;R7/BP6/2PK4/6R1/2N3BP/1N1P2P1/P3PP2/Q7 w - - 0 1&quot;]

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

Strategy / computer implementation
From the assumptions follow a handful of constraints that enable an exhaustive search.

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.
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.
The Q must be attacked by a N.
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 &quot;6R1/1R6/1BP3BN/3P2P1/2P2P2/1P4P1/P1N2K/Q7 w - - 0 1&quot;]

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: