# Get ring of tiles in hexagon grid

Thanks to this post: Hexagonal tiles and finding their adjacent neighbours, I'm able to collect adjacent tiles to a given tile. But I'm pretty much stuck on an algorithm that gives me only a "ring" of tiles specified by an offset. The algorithm given in that Stack Overflow post doesn't exactly care about the order in which it collects the tiles.

I know that with every offset 6 tiles are added.

• Offset 1 gives you 6 tiles (the first adjacent tiles).
• Offset 2 gives you 12.
• Offset 3 gives you 18, etc.

There is a constant growth of 6 with each offset. So I assume there should be a rule which adapts to these offsets. I can't exactly figure this one out. Anyone?

A hexagonal ring with the radius of N consists of 6 straight lines, each with length N - see my extremely crude example below :) For N=2: The arrows cover 2 hexes each.

I assume you have some functions which give you the neighbouring tile in a specific direction, like north(), southeast() etc. So your algorithm, in pseudocode, should be something like this:

var point = startingPoint.north(N)
for i = 0..N-1:
point = point.southeast(1);
for i = 0..N-1:
point = point.south(1);
for i = 0..N-1:
point = point.southwest(1);
for i = 0..N-1:
point = point.northwest(1);
for i = 0..N-1:
point = point.north(1);
for i = 0..N-1:
point = point.northeast(1);


Note that this should work also for edge cases N=1, returning 6 tiles, and N=0 returning an empty set.

I know the code isn't perfect :) There is some redundancy here. In my projects using regularly tiled maps (hexagonal or otherwise) I usually have an enum "Direction", which allows me to do this more smoothly:

var point = startingPoint.inDir(N, Direction.North)
var dir = Direction.SouthEast.
for d = 0..Direction.count():
for i = 0..N-1:
point = point.inDir(1, dir);
dir = nextDirection(dir);

• This should push me in the right direction. Thanks! Mar 18, 2013 at 14:34
• Note that the code sample will add duplicate points for the first five segments. However, it's a nice answer. Mar 18, 2013 at 14:38
• @Byte56 Yeah I figured. But at least I see the connection between shifts in direction! Mar 18, 2013 at 14:39
• @Byte56 Really? Hm. I tried to avoid that one... 0..N-1 gives 0..1 for N=2, so that's i=0 and i=1, which is 2 values. 2 values from each times 6 directions is 12 tiles, as it should be ...? Mar 18, 2013 at 15:10
• Nope. You're right. Since each loop is adding a point from the last loop I was off by one for the loops, my mistake. It's a clever algorithm. Mar 18, 2013 at 15:57

I have found this article to be a very good reference for hexagonal grid algorithms, and its section on "Distances" provides a method for determining the number of steps between two tiles. If you convert your axial coordinates (x-y) into cube coordinates (x-y-z), the distance is always equal to the largest of the coordinate offsets between the two tiles, or max(|dx|, |dy|, |dz|).

An exhaustive search of the whole grid for tiles at the desired distance is $O(n^2)$ with the grid dimensions, but it is a simple implementation that works well for small grids. 