# Tiling a Hexagon with Diamonds

Why are Round Diamonds priced higher than Fancy Shapes?
Why are Round Diamonds priced higher than Fancy Shapes?

Here’s my attempt to prove it.. It seemed impossible until I finally exploited a trick.

I start from a valid configuration where there’s only one change possible (rotating the 3 semilines in the middle: any other change would at same time change number of diamonds an create triangles.)

Once you do that change you are free to undo it (useless, I’ll mark it blue)
or to do other 3 changes (in red). You immediately note that you can do that “change” only on points that have lines placed like the middle of the first move, or the middle of the initial cube.

Once you do your 2nd move you’ll be unable to undo the first move (grey now) because doing so would create triangles and other shapes.

(Assuming my first move was a clockwise rotation of 1/6 round, my undo is a 1/6 anti-clockwise)

Diamond and Dot | Pinnacle Hexagon Patterns

Basically you can just check that the only possible moves are rotations of group of tiles made by 3 diamonds (1 for each orientation) (you can check all possible moves on a 2x2x2 “cube” and see that is true).

Hence you note also that rotation keeps number of diamonds for each orientation the same.

There’s a little piece missing of the proof:
I did not show that starting from my first cube I can do all possible tilings, that’s because rotations have “inter-dependencies” and I don’t know if at some point “I’ll get stuck” without more possible moves.

I’m too sleepy for that proof, but I developed another proof method I’ll let you the pleasure to use it:

Extruding columns starting from an “empty” cube:

You see you cannot extrude a column to a length greater than the preceding columns (there are 2 directions to check for preceding columns) because you’ll get triangles.

You have now a way to compute really all possible tilings. You start with the backmost column, and once decided a height you can extrude the 2 neighbour to any height lower or equal to the backmost column. After that you can do the same for the next 3 columns.

Hexagon Engagement Rings: The Complete Guide

There is no dependency on rotations here. You choose a number, and then you can choose again same number or a lower number. That’s much easier but have some help from imagination (3rd dimension in a problem that has 2 dimensions).

Well, probably that’s not a formal proof. But helps imagination you have 2 ways to attack the problem, and probably those can be worked around for a formal proof. But I think is more interesting the intuition than the proof. Without some intuition there will never be some proof.

The key seems always to be the same. Starting from a trivial configuration the only possible moves incidentally preserves the number of diamonds for each configuration.

P.S:

I have never seen that puzzle before. Hope you like my first answer in puzzling exchange.

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