The 1990 Game Boy game Daedalian Opus is essentially a series of 36 pentomino puzzles. In the first level, however, you only have three pentominos; the rest are introduced in the levels after that, one at a time. A good proportion of the shapes you have to fit the pentominos into are also plain rectangles. (See a tool-assisted speedrun of the game here).

That led me to the following problem. What is the smallest $k$ such that there is an ordering of the 12 distinct free pentominos where

  • the first $k$ can tile a $5×k$ rectangle exactly
  • the first $k+1$ can tile a $5×(k+1)$ rectangle exactly
  • and so on, to the first $12$ tiling a $5×12$ rectangle exactly (which can always be done)?

For example, if $k=9$ there would have to exist an ordering – say PINWVZXFTUYL – where

  • PINWVZXFTUY tiles a $5×11$ rectangle
  • PINWVZXFTU tiles a $5×10$ rectangle
  • PINWVZXFT tiles a $5×9$ rectangle

1 Answer 1


There is no way to tile a 2x5 rectangle with two distinct pentominos. But there are many ways to make such a sequence starting with three pentominos.

Here's an example, chosen for the unique tilings up to level 7:

FPU-LXTVINWYZ enter image description here

Here's a better example, with pic. I automated the bit I was doing manually (finding sets with one extra piece from the list of all possible pieces). The maximum is 8 pieces with unique tilings. 9 through 12 I just printed the first tiling.

LTY-PWZVXFINU enter image description here

While we're at it, the one sided pentomino case works too. There are six extra asymmetric pentominos (call them F', L', N', P', Y' and Z') for a total of 18. This time you can get all the way to a 5x12 with unique tilings.

PP'V-YNFZN'F'Z'XLIUTWL'Y' enter image description here


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