If I had a grid of squares, they can be labeled with Cartesian coordinates such that given square $(x,y)$, you know it shares a boundary with squares $(x+1,y),(x-1,y),(x,y+1),(x,y-1)$.
Is there a way of labeling a tessellated hexagon grid, so that given any hexagon label you can work out it's neighbors?
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While I use redblobgames' hexagons page for all sorts of things hexagonal, there is a simple 'other way', very similar to the cartesian neighbour calculation.
The normal way we imagine bricks (stretcher bond) has the same relationship as a hex-grid.
Recognising that bricks on every other row have a horizontal offset, we count (along the rows) using +2 increments instead of +1.
This gives us our x values as following
3:0 2 4 6 8
2: 1 3 5 7
1:0 2 4 6 8
0: 1 3 5 7
012345678
Whereas y values are the same as cartesian. Now we can work out our neighbours:
For any brick/hex (x, y) where (x+y)%2==1 the neighbours are (x-2,y),(x+2,y),(x-1,y+1),(x+1,y+1),(x-1,y-1),(x+1,y-1)
Placing a hex (x,y) into an approximate cartesian position can be done by (x+y%2,y) (where % is the modulo operation)
If we want to label without using +2 increments, we can still do that, but we need to use the modulo operation for calculating neighbours.
3:0 1 2 3 4
2: 1 2 3 4
1:0 1 2 3 4
0: 1 2 3 4
0-1-2-3-4 y%2=1
-1-2-3-4- y%2=0
Now we can work out our neighbours: For any brick/hex (x, y) the neighbours are (x-1,y),(x+1,y),(x,y+1),(x+2(y%2)-1,y+1),(x,y-1),(x+2(y%2)-1,y-1).
Create the grid by drawing horizontal lines thru ALL hexagon centers and drawing vertical lines thru ALL hexagon centers. The neighbors are:
(+1,-1), (+2,0), (+1,+1), (-1,+1), (-2,0), (-1,-1)
Note: the x-axis is scaled by a factor of sqrt(3). This is ok. Cartesian coordinates can have an axis that is scaled.
Note: the x,y does not match the hexagon label. This is ok.
If we set the upper left hexagon with a coordinate of (0,0), the hexagon label is: Math.floor(x/2)+1 plus y+1
(0,0) => 0101 (2,0) => 0201 (4,0) => 0301 (1,1) => 0102 (3,1) => 0202 (5,1) => 0302