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Rules:

  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 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?)

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    puzzlesland.com/blokus Might be a useful tool for answering this question. Not sure if you've already looked there or not.
    – Jacob Shetler
    Commented Mar 10, 2020 at 13:17
  • @Jacob No - I’d not seen that website before. Thanks! I’ll take a look later.
    – DonielF
    Commented Mar 10, 2020 at 13:45
  • @Jacob So that is helpful, but it A) only deals with a four-player game, and B) doesn’t demonstrate whether the standard 20x20 grid is the absolute minimum even for four players, where the theoretical minimum is 19x19. It at least sets an upper limit on the minimum for four players, but is that consistent that the theoretical minimum isn’t big enough?
    – DonielF
    Commented Mar 10, 2020 at 15:56
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    Since this has less to do with rules/mechanics and moreso with solving a mathematical puzzle, I'd try posting this over at puzzling.stackexchange.com. They love problems like this.
    – JRodge01
    Commented Mar 11, 2020 at 12:16
  • Good call, @JRodge01! I managed to squeeze the two-player set onto the 14x14 board, and it was definitely a blast.
    – Bass
    Commented Mar 11, 2020 at 20:55

1 Answer 1

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The 21 pieces together cover 89 squares (12x5 + 5x4 + 2x3 + 2 + 1).

Two sets will cover 178 and four sets 356. That leaves plenty of room on a 14x14 (196 squares) or 20x20 (400 squares) board.

Since 89 is prime, 178 and 356 don't make 'nice' rectangles. However, you can fit 2 sets on a 12x15 or 10x18 board (180 squares), and 4 sets on a 21x17 board (357 squares).

Examples of all the above can be seen at the site Puzzlesland's Blokus Discoveries page

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