Thinking about graphs with 4 neighbours to every vertex, how about playing Go on a Sierpińsky triangle:

Sure, want to play a game?

Somehow, I think it would be a bit more aesthetically pleasing if the three outer edges formed a circle.

Sure, but not really if I have to manually draw the stones each time.

Perhaps some kind programmer could add it to the abstract boards available at our local variant server?

While it is a planar graph, I feel like the connectivity is bit better intuitively understood in 3 dimensions.

The outer three edges are depicted as much longer than the rest. If we folded each outer large triangle up, leaving the center large triangle flat as the base of the pyramid, we could then shrink the outer edges to be the same length as the rest.

However, leaving it as a basic polyhedron (with all straight edges) may give the false impression that disconnected points that are connected due to their physical closeness in 3D space. So, instead I’m envisioning the upright large triangles as bowing outward, so that only their uppermost tips get that close to each other.

Great idea! I’ve made a pull request: add sierpinsky triangle board by merowin · Pull Request #213 · govariantsteam/govariants · GitHub

However currently it doesn’t include the three “long” outer edges. Our program currently draws all edges in a straight line between the coordinates of the intersections, so I’ve left the curved connections out so far. How important is it to include those?

Without those edges, the adjacent points have the wrong number of liberties. EDIT: I mean, I guess that’s obvious, but since the goal seems to be “every point has 4 liberties”, it might be important to include them.

Essential

Wonderful. Distance between points seems more complex as with a grid.

9x13+5=122 intersections, between a 9x9 and a 13x13

I now think that not having them connected is fine, if they are connected, it should be done with three more fractal triangles, as in the 3D one above (which is still a planar graph, but it doesn’t look as nice on a plane)

I’m not a magician though, so don’t expect too much now In particular, I have no idea how the four fractal triangles should be embedded into the plane …

^ this

Think of it as a kind of stereographic projection: put a lamp just above the top (empty) triangle, then the graph would be the shadow that is projected on the ground plane.

alternative way of thinking of it:

Cut through the top 3 corners, separating the top 3 triangles

from each other, but not from the bottom 3 triangles.

Pull those triangles away from the center, so they end falling flat away from the center.

```
BBBBBB
bBBBBb
BBBB
bBBb
BB
AA
aAAa
AAAA
aAAAAa
AAAAAA
CCCCC DDDDD
CCCC DDDD
CCC DDD
CC DD
C D
```

(I don’t know what’s causing the DD to be red: That’s not intentional by me.)

Move the 6 outer corners, so the 3 outer triangles are back to touching each other.

On a general note, I think that the number of liberties of nodes is not a great indicator for how easy it is to surround territory (say e.g. for a local part of the board). I believe it is more complex than that, and I’m thinking that the (local) vertex-connectivity may be a better indicator, i.e. telling us how many stones are needed to cut off a part of the board (= “surround territory”, kind-of). In other words, identifying the local bottleneck intersections wrt. connectivity.

In the Sierpińsky triangle graph, even though all vertices (except the three corners) have four liberties, local parts can be separated with as little as 3 stones.

Also the liberties of chains behave much differently, adding a stone may not even increase your liberties. E.g. this 2-stone chain has only 4 liberties:

So in terms of strategy, I’m thinking that the bottleneck vertices may be more valuable outposts than others.

I dunno. Intermediate control 2 bottlenecks (and 2 Intermediates)… (but bottleneck 4 Intermediates). At this point i really can’t say.

The difference of amount of liberties by chains of same size does inspire me.