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Let's have an 17x17 grid. We can fill this grid with 96 trominoes of three different colors, 32 trominoes of each color. On this particular grid the empty single square is the position A1. By visual inspection we see only 6 trominoes that do not form a pair (a pair is a 2x3 or3x2 rectangle); I marked these with red color. The rest of the trominoes form 45 pairs. Two trominoes of the same color are not allowed to touch anywhere side to side. Can you put the trominoes on this grid with only 10 pairs? You can put the empty single square anywhere on the grid.

dec19

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    $\begingroup$ Downvoting this because the OP has been posting essentially the same puzzle three times with minor parameter differences, and they didn't even check if their puzzle has a solution or not so I believe the puzzles are produced with very low effort. $\endgroup$
    – Bubbler
    Dec 21, 2020 at 1:50
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    $\begingroup$ I have something to say about the images too. I guess it took a lot of time to draw the tilings with pencil and paper, but it's not the right kind of effort, since using any kind of image or spreadsheet software would give much clearer images. $\endgroup$
    – Bubbler
    Dec 21, 2020 at 2:00

1 Answer 1

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I left the empty square at A1. I found 25000 tilings that work for A1, out of an estimated 500 000 give or take a few hundred thousand.

The rectangles are marked with lighter colours.

enter image description here

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  • $\begingroup$ I assume you found this with a computer program - could you share the code in your answer? Surround it with three backticks (`) for code formatting. $\endgroup$
    – bobble
    Dec 21, 2020 at 1:33
  • $\begingroup$ Not sure it's a good idea to drop 32k lines of pretty much unmaintainable code in here, most of it is not applicable and even if I did the huge job of separating out the applicable stuff, it would be pretty much unreadable. I've been developing the code since it started life as Pascal in 1981 or so, now it's C and not the good, modular sort. $\endgroup$
    – theonetruepath
    Dec 21, 2020 at 5:35
  • $\begingroup$ Ah, just leave as-is then $\endgroup$
    – bobble
    Dec 21, 2020 at 5:41

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