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I have ten copies of each of the twelve pentominoes. Can I use all of them to completely tile twelve 5 x 10 rectangles?

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  • $\begingroup$ Are the rectangles adjacent, so one of the pentominoes can overlap two, or are they separate? $\endgroup$
    – Stevo
    May 22 at 2:58
  • $\begingroup$ @Stevo They are separte. No tile can bec part of two rectangles. $\endgroup$ May 22 at 9:36

1 Answer 1

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It is easier to halve the size of the puzzle - using 5 sets of pentominoes to cover 6 of those rectangles. If that can be solved, you can simply use two copies of that as a solution.

I'll restrict things even further by assuming that each rectangle has no duplicate pieces. So each rectangle is covered using a single pentomino set with two pieces omitted. If the six omitted pairs of pieces form a single pentomino set, then the six rectangles are covered by exactly five sets.

So now I just have to look at ways of covering a single 5x10 rectangle with one set of pentominoes, and see which pentomino pairs can be omitted.

As there is no tag, I used a computer for this. It turns out that

you can omit almost any pair of pentominoes and still be able to cover the rectangle. The only pair for which this is not possible is the P and F pentomino pair.
So I split the pentominoes into six pairs but avoiding the (P,F) pair, and then tile rectangles while omitting each pair in turn.

Here is one such solution:

enter image description here

P.S.: If you omit I&L, I&P, Y&P, or N&P from a pentomino set then the 5x10 rectangle has a unique solution up to rotation/reflection.

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  • $\begingroup$ How about constructing a solution with the opposite of 'no repeated pieces in each rectangle' ie one with the maximum repeated pieces? Score the solution by "the number of distinct pieces in each rectangle". Your tiling scores the maximum, 120, what is the lowest score? $\endgroup$ May 23 at 5:52
  • $\begingroup$ @theonetruepath An interesting question. I don't see any useful approach to solving that problem. $\endgroup$ May 23 at 7:52
  • $\begingroup$ I don't either, at least not one involving large amounts of brute force. I got it down to a score of 46 with a few minutes of existing programs that aren't particularly suited to the task... $\endgroup$ May 23 at 8:03
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    $\begingroup$ Edo Timmermans got it to 39, Col George Sicherman to 30 and I found a 28, current best score. $\endgroup$ May 26 at 15:51

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