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Here are the one sided hexominoes arranged into 12 congruent shapes. But there are one or two flaws: The dark blue hexominoes, which are the symmetric ones, may not occur more than once each in a shape. And the other colours, which are in chiral pairs (ie left- and right-handed versions), also may not occur in the same shape. The flaws: Shapes 6 and 8 have two symmetrical hexominoes, shape 1 has two staircases, 5 has a matched pair, shape 8 has a matched pair as well as the two symmetric pieces. And shape 12 has two 'long N' shapes. Your task is a simple one - rearrange these hexominoes to eliminate these flaws. To make your task easier, there's only one solution so you can't stumble on the wrong one by mistake. Also each shape will have only one tiling. That's not a requirement, but you can use that fact to make things easier. These hexominoes are one-sided so you can't flip them.

TLDR version: Print these pieces in colour, cut them out, and arrange them, coloured side up, in twelve copies of the given shape with no colour repeated in a shape.

enter image description here

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  • $\begingroup$ Just to clarify, the goal is to rearrange the pieces from all twelve solutions combined and produce a new set of 12 solutions with the same pieces except that no two tiles within a solution are congruent and no solution has more than one tile with a mirror symmetry? $\endgroup$ Dec 20, 2017 at 5:35
  • $\begingroup$ Almost... except that "except that no two tiles within a solution are congruent" should be "except that no two tiles within a solution may be a chiral pair". Technically the left-and right-handed versions of a hexomino aren't congruent. $\endgroup$ Dec 20, 2017 at 6:01
  • $\begingroup$ I re-read your question, new answer: Change it to this: "The goal is to arrange the pieces from all twelve solutions combined and produce a new set of 12 solutions (all in the same outside shape) with the same pieces except that no solution has more than one symmetric hexomino and no solution has more than one of a mirror-symmetric pair" $\endgroup$ Dec 20, 2017 at 6:37

1 Answer 1

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The single possible tiling is:

60 hexomino layout

The method of solving is the same as for my answer to this similar problem, except that when picking sets of five hexominoes, colours that are already included aren't considered.

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  • $\begingroup$ I can see I'm going to have to make a much more difficult puzzle just for you... $\endgroup$ Dec 21, 2017 at 23:17
  • $\begingroup$ :) This was a nice variation on the other question. I'm sorry for the terse answer. After BmyGuest's comments on the other question I tried to tidy up my code and post it, but then this question came along, so I started working on that. I might post my code, but this will have to wait until after Christmas. My code takes longer than a second: This one takes 12s to find the combos and about a second to find the possible distributions. $\endgroup$
    – M Oehm
    Dec 22, 2017 at 6:33

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