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What is the largest set of hexominoes that can be found in which no two of them are such that one can be converted into the other by cutting out one of its component squares (thus obtaining a pentomino) and glueing it elsewhere?

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  • 3
    $\begingroup$ It's definitely a question for you @hexomino $\endgroup$ – trolley813 Apr 5 at 22:07
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    $\begingroup$ Seriously, it's probably easily solvable via graph theory (vertices = hexominoes, edges connect the hexominoes which can be converted this way, and the problem is to find maximum subgraph in which no two vertices are connected with an edge). $\endgroup$ – trolley813 Apr 5 at 22:13
  • $\begingroup$ Please clarify whether reflections are considered the same hexomino. $\endgroup$ – msh210 Apr 5 at 23:19
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The largest set contains

Seven hexominoes

One such set is shown here in this ultra-fancy graphic:

 +---+---+---+---+---+---+   +---+
 |   |   |   |   |   |   |   |   |
 +---+---+---+---+---+---+   +---+
                             |   |
 +---+---+---+---+   +---+---+---+
 |   |   |   |   |   |   |   |   |
 +---+---+---+---+   +---+---+---+
 |   |       |   |   |   |
 +---+       +---+   +---+   +---+
                             |   |
 +---+---+---+           +---+---+
 |   |   |   |           |   |   |
 +---+---+---+---+---+   +---+---+
         |   |   |   |   |   |   |
 +---+   +---+---+---+   +---+---+
 |   |                       |   |
 +---+---+       +---+---+   +---+
 |   |   |       |   |   |
 +---+---+---+   +---+---+---+
     |   |   |       |   |   |
     +---+---+   +---+---+---+
         |   |   |   |   |
         +---+   +---+---+

Identified by the pentominoes they contain:

I, L, N, W, F/U, P/Y, V/Z

Proof that this is optimal:

There are five hexominoes that contain only one pentomino: the four shown above for the I, L, N and W pentominoes, and a fifth that also contains only the L pentomino. Thus, a set of 8 mutually independent hexominoes must include four that contain more than one pentomino. There are 8 remaining pentominoes, so these four hexominoes must contain exactly two pentominoes each, and one of them must include the X pentomino, which can be found in only two hexominoes. One of those also contains the F and P pentominoes, the other the T and Y pentominoes. We are therefore unable to complete a set of 8 mutually independent hexominoes.

| improve this answer | |
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    $\begingroup$ I don't disagree with the result, but apart from stating the answer, there seems to be precious little corroborating evidence in this answer. $\endgroup$ – Bass Apr 6 at 7:53
  • $\begingroup$ @Bass See edit. $\endgroup$ – Daniel Mathias Apr 6 at 11:02
  • $\begingroup$ I have no idea why, but the 'ultra-fancy graphic' has the illusion that the dashes are somehow misaligned with one another. $\endgroup$ – Cloudy7 Apr 7 at 23:34
  • $\begingroup$ @Cloudy7 The horizontal stroke of the plus sign is not aligned with the minus sign, disrupting our perception of what would otherwise be a straight line. $\endgroup$ – Daniel Mathias Apr 7 at 23:41
  • $\begingroup$ @DanielMathias That must be it. My brain wants to create this sort of bridge effect, where the center dash dips below the rest. $\endgroup$ – Cloudy7 Apr 7 at 23:45

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