9
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You can skip the back story and directly jump to the question.

Capitalism has arrived in suburban Chesster the community most famed for The Game, and with a vengeance. Rents have trebled in less than a year and your little community of scientists and artists already had to make hard choices and let all the pawns go from your flat share. Sharing a flat amongst free spirits is difficult at the best of times. Here is a warning example of a dysfunctional commune:

enter image description here

A list of things that contributed to the eventual dissolution of this share and a slew of most acrimonious law suits:

Wrong composition

  1. There is a pawn in it
  2. There is no king
  3. There are too many knights
  4. There are the right number of bishops but they are are both light-squared

Wrong arrangements (incomplete)

  1. Nb1 is isolating themselves and not interacting with others while
  2. Q and the other two N's are stalking two or even three of their flat mates

A healthy flat share looks more like this:

enter image description here

This model community consists of the ideal numbers of

  1. zero pawns
  2. one king
  3. one queen
  4. two rooks
  5. a pair of bishops
  6. two knights

Their arrangement is perfect in that every one shows an interest in exactly one of the others and is courted by exactly one other. Further, this chain of attraction runs through all in one single cycle. Otherwise there is a distinct possibility of it all breaking up into warring factions.

But, alas, a 64 square flat is out of reach in these trying times. To make ends meet you and your friends must move to a much smaller place while somehow meeting all the constraints laid out above.

Question

What is the smallest rectangular board such that one can place a half set (i.e. all white or all black) of proper pieces (no pawns) such that each piece attacks exactly one other, each is attacked by exactly one other and the ensuing daisy chain runs though every piece in one go? Please consider smallest both in terms of shorter edge and in terms of longer edge (two answers). EDIT: (This was in the backstory all the time but I omitted putting it in the summary here. Apologies.) The bishops should form a proper pair of light-squared and dark-squared specimen.

Ties are broken first by the length of the other edge and then by the number of distinct up to symmetry solutions.

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  • $\begingroup$ Nice idea for a puzzle, I wonder why we didn't have this type of puzzles before. $\endgroup$
    – Anonymous
    May 8 at 14:30
7
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3 x 7:

\begin{matrix}&K&.&B&.&.&Q&.\\&B&.&N&R&.&.&.\\&.&R&.&.&.&.&N\\\end{matrix}

4 x 6:

\begin{matrix}&.&Q&N&.&.&.\\&.&.&.&.&B&.\\&N&.&B&.&.&R\\&K&.&.&R&.&.\\\end{matrix}

5 x 5:

\begin{matrix}&.&R&.&.&.\\&B&B&.&.&.\\&.&.&R&.&K\\&.&N&.&N&.\\&.&Q&.&.&.\\\end{matrix}

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4
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What I believe to be two 4×7 solutions, built around the same core:

enter image description here

5×6 and 3×8, for good measure:

enter image description here
enter image description here

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  • 1
    $\begingroup$ A good start. One thing: I put it in the backstory but forgot to mention in the short version: The bishops should be a pair, i.e. one light squared one dark squared. Also, the 5x6 seems to be one rook short. $\endgroup$
    – loopy walt
    May 9 at 5:06

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