Progressive Daedalian Opus

Question

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

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

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

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'