1. Each player has the twelve pentominoes, five tetrominoes, two trominoes, one domino, and lone square. These may be flipped and rotated in any manner.
  2. The board on which these are to be placed is square.
  3. A player’s first piece starts in any one corner of the board.
  4. All subsequent pieces must be placed such that all of one player’s pieces are touching only at the corners.
  5. There are no restrictions on how one player’s pieces may touch another player’s.

In the base game, the goal is to place as many of your pieces as possible while minimizing the amount of pieces your opponents can play. But here I am curious about a different question: what is the minimum-sized board needed to contain all pieces from every player?

Each player’s pieces add up to 89 squares, so two players, for instance, jointly have 178 squares to place. In theory, then, the minimum sized board must be the smallest square at least as large as this number, 14x14=196. However, because the pieces are rigid and because there are restrictions on how pieces may be placed, it’s surprisingly difficult to try to make this work.

In practice, is a 14x14 board sufficient to place all tiles of two players, given these rules? If not, what is the minimum?

Does this generalize to n players, that the smallest square larger than 89n, or the second-smallest or whatever, is sufficient to contain all of their pieces? (For example, this page provides a 20x20 solution for four players, but is 19x19 sufficient? Is it consistent that the theoretical minimum isn’t sufficient, or is sufficient?)


2 Answers 2


Wow, that sure is tight.

enter image description here

This pattern is symmetrical through the centre (or a 180 degree rotation), so I've only marked half of the pieces. The marked green pieces are the pentominoes (and the two 3s), which should help with any double checking.

And yes, even though it doesn't look like it at the first glance, the I and T pentominoes are legally connected to each other :-)

This took surprisingly long to find and (especially) double check, so I have nothing for the general case with n players.

  • 2
    $\begingroup$ Impressive! You made it look really easy, and quick. I had not yet begun to code a computer solution, on the grounds that it would take several light years to run unless I could come up with some clever optimisation. $\endgroup$ Commented Mar 11, 2020 at 20:59
  • 1
    $\begingroup$ @WeatherVane After posting, I followed the link in the comments of the cross-post, and looks like the computer solution is already done. Yeah, those guys even chopped a line off the edge of the board for good measure. Bloody show-offs :-) $\endgroup$
    – Bass
    Commented Mar 11, 2020 at 21:03
  • $\begingroup$ As 13x14 instead. $\endgroup$ Commented Mar 11, 2020 at 21:05

Bass, in a comment on their answer, provides a link to a site where someone has posted a solution for 4 players on a 19x19 board. But the solution at that site does not have rotational symmetry, which seems a shame. Here is a solution with rotational symmetry.

  • 1
    $\begingroup$ This is wonderful! $\endgroup$
    – athin
    Commented Sep 3, 2020 at 0:53

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