- Each player has the twelve pentominoes, five tetrominoes, two trominoes, one domino, and lone square. These may be flipped and rotated in any manner.
- The board on which these are to be placed is square.
- A player’s first piece starts in any one corner of the board.
- All subsequent pieces must be placed such that all of one player’s pieces are touching only at the corners.
- 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 exactly two players?
Each player’s pieces add up to 89 squares, so two players 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?)