Lets say you have two players, Red and Blue, that alternate filling an arbitrarily large hexagonal grid of tessellated hexagons with pieces of their color. Hexagons can either be filled or empty. A piece maybe be placed on an empty hexagon (filling it) as long as that empty hexagon meets AT LEAST ONE of these conditions:

  1. No more than two filled adjacencies.
  2. The filled adjacencies contain an equal number of Red and Blue pieces.

Is there a way to fill up the grid completely?

  • 1
    $\begingroup$ Ah, the AT LEAST ONE part is tripping people up, haha $\endgroup$
    – justhalf
    Apr 1 at 23:50

1 Answer 1


I claim that the answer is


and here's why:

You can start by constructing a 'honeycomb' pattern like this. This is easy to do while never filling a cell with more than 2 adjacent cells: just go row-by-row, for instance.
honeycomb pattern - a 3-colored hex grid in red, blue, and white, with no two adjacent cells sharing a color
Once you've done this, we can just fill all the white cells however we want! All of them are next to 3 red cells and 3 blue cells.

The only potential problem with this strategy is the border. An empty cell on the border of the grid but not in the corner is still okay - that cell will just have two red and two blue cells next to it. This takes care of everything but the corners.

If we have an empty cell in the corner, this strategy breaks: it'll have 3 adjacent filled cells, so it will fail both conditions. However, we can get around this problem: no matter the grid size, we can align our honeycomb pattern so it covers all of the corners. (The way this is done is based on the grid size modulo 3, so I have provided 3 examples to demonstrate.) three honeycombs, tri-coloring hexagons with a repeating red/blue/white pattern

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    $\begingroup$ You are so smart! Wow, trying this out for myself now $\endgroup$ Apr 1 at 15:29

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