Tiling hexagonal patterns

[ralphmerridew]

Imagine a finite number (divisible by 3) of hexagons arranged in a fixed pattern. That pattern is to be divided into a mixture of straight trihexes and triangular trihexes.

True or false:
If there are two tilings of a particular pattern, then the number of straight trihexes will be even for both or odd for both.

(Appears to be true for small cases, but I can't find a proof or counterexample.)

[L'lanmal]

Here's an approach. It isn't complete though.

Label each hex with an integer row and column number (as an ordered pair) from the underlying planar hex tessilation.

The sum of the coordinates of the three hexes making up a straight trihex, in any of the three orientations, is (0,0) (mod 3).

The sum of coordinates of any upward facing triangle is (1,1) (mod 3).
The sum of coordinates of any downward facing triangle is (2,2) (mod 3).

It is therefore impossible to lower the count of upward or downward facing triangles by 1 while raising the straight trihex count by 1. Or vice-versa.

It is possible for 2 straight trihexes to replace an upward and a downward triangle. This is ok, as it does not change the parity of the straight trihex count.

It remains to show that you cannot retile to change one of the triangle counts by a larger odd number at once, such as by three while adjusting the straight trihex count by three in the opposite direction.

I think if you can do this, you can reduce more complex changes to this case (such as replacing two upward facing triangles with one downward and one straight) by adding disjoint 2x3 regions and toggling between two straight and two triangler divisions in each.

[Zahariel]

Your claim is untrue. Consider the universe consisting of an upward pointing triangle of 10 hexes (like a bowling pin arrangement), with the center hex removed. This shape can be tiled by either 3 straight tri-hexes or 3 upward tri-hexes. I haven't been able to come up with a counterexample involving a solid universe; either the minimum size for a counterexample is pretty big, or it's actually true for solid universes but the proof relies on a hexagonal version of Pick's theorem.

[ralphmerridew]

Nicely done!