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Tic-tac-toe ($t^3$) can be played on a graph.
It is very similar to tic-tac-toe

  • 2 players; one X, one O
  • X starts, then O, repeat
  • the player, who can mark 3 adjacent vertices ('tics') in the shape of a 3-path ($P_3$), wins

The structure of the graphs allowing a winning strategy is known. Basically, all subgraphs allowing a winning strategy have been found - see On which simple graphs (no loops, no multi-edges) is tic-tac-toe winnable, where is it a draw?.

The $t^2$ game (2 adjacent vertices needed) is very easy; it is winnable, if the graph includes a $P_3$ subgraph (path with 3 vertices and 2 edges).
This is a bit boring as then only the singleton and $P_2$ graph lead to a draw...

Let's look here at the the game on a graph the player wins that places three Xs or Os in a triangle aka 3-cycle

On which graphs is the $t^3$-variation t-t-triangle aka cyc-cac-cycle winnable (i.e., has a winnable strategy)?

References

Michael T
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2 Answers2

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This question reminded me of a different subject of graph theory. It will not give a complete characterisation of winnable graphs, but might give more intuition about the problem.

Namely you can look at graphs where, even if both players play stupidly, there will never be a draw. If we make the claim even stronger, we could ignore that this is a turn-based game, and look for graphs where every 2-coloring of the vertices has a monochromatic triangle (= 3 equally-coloured vertices). This is called an 'arrowing' property and is denoted by $G \rightarrow (3,3)_v$. Now it is the study of Ramsey theory and Folkman numbers to find small graphs that arrow $(3,3)_v$.

For example $K_5 \rightarrow (3,3)_v$ is the smallest example, and so is obviously every graph that contains $K_5$ as a subgraph. But it is known that there are also $K_5$-free graphs that arrow $(3,3)_v$. This is the smallest example.
One can go further and also forbid $K_4$ as a subgraph. The smallest example is then this graph.

Now forbidding $K_3$ is obviously impossible since we want a monochromatic triangle, but forbidding diamonds appears to be still possible. The smallest known example is this graph. Since all your current families contain diamonds, this is certainly a graph that is missing.
Avoiding $C_4$ is also possible. I believe that avoiding arbitrary long cycles might be possible, but I don't know of examples yet.

Let me further note that all of this can be extended to more colours (so more players), but that the known graphs get very big very fast. It can also be extended to larger monochromatic subgraphs being desired. For example, there exist $K_5$-free graphs $G$ for which $G\rightarrow (4,4)_v$.
Also, checking if a graph arrows $(3,3)_v$ is co-NP complete, so there is a chance the simple collection of families you are looking for does not exist.

See for example: https://arxiv.org/abs/2110.03121

Steven
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  • hi Steven, would this be a simple example: a graph that is not bipartite, 2 linked vertices held by one player win? - This game can't result in a draw... – Michael T Mar 28 '25 at 10:13
  • I don't understand which graph you are trying to describe. You need at least 5 vertices for a draw to be impossible. – Steven Mar 28 '25 at 10:19
  • sorry, I did not make myself clear; I was changing the game to 'a player wins, if 2 linked vertices are held by one player'; this game can not have a draw unless the graph is bipartite. Actually, I would re-model to a collaborative game 'both players win, if no 2 vertices are held by the same player' - this game can not be won, unless the graph is bipartite. Would you agree? – Michael T Mar 28 '25 at 10:53
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    If it is a collaborative game without alternating turns, then yes, 'winnable' is the same is as 2-colourable, so bipartite. If turn order should be respected, then even some bipartite graphs are not winnable. – Steven Mar 28 '25 at 12:17
  • Hi Steven, agree; ad you second sentence, e.g., star graphs... isn't that true for any bipartite graph that is not balanced in the 2 parts? – Michael T Mar 28 '25 at 13:05
  • It is true for any bipartite graph where there is no subdivision into 2 (almost-)balanced parts, yes. But I don't see how this is contributing to the original question. – Steven Mar 30 '25 at 08:50
  • Hi Steven, sorry for digressing; just checked it: the (4,1,4) linked necklaces of triangles (https://houseofgraphs.org/graphs/53156) which is in the update of the other answer is isomorphic to a subgraph of https://houseofgraphs.org/graphs/51236. Many thanks!! Great additional check. – Michael T Mar 30 '25 at 12:07
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Below I have started collecting subgraphs that allow a winning strategy. Thanks for complementing the list to 'Especially Lime'!

Am I missing any other graphs that allow winning in cic-cac-cycle?

Subgraphs allowing a winning strategy

  • the wheel graph, $W_5$, with 5 vertices (and 4 spokes) and 4 triangles sharing a single 'hub vertex' aka $K_{1,2,2}$, a complete tripartite graph - the winning strategy starts in the hub
  • the 'double diamond' graph (D,D) with 2 diamond graphs linked at the axes; in Mathematica, it is called GraphData[{"Matchstick", {10,57}}] and it is in HoG at https://houseofgraphs.org/graphs/21218 - the winning strategy starts in the center
  • the family of '2 diamonds on a chain' graphs, which consist of 2 diamond graphs linked by a 'chain of triangles' (aka 'triangular snake graph'); the first one, (D,CT_k,D) with $k=1$ is shown below, has just one triangle between the 2 diamonds but any number $k\geq 0$ of triangles is possible ($k=0$ is the 'double diamond') - the winning strategy starts at one of the inner vertices of the 'diamond's axis'
  • the family of 'diamond necklace' graphs consists of a triangle necklace of 4 triangles linked to one diamond graph at a vertex on the diamond's axis (shown below as (TN_4, D)); more triangles can be linked in the chain (TN_k, $k\geq 4$); also chains of triangles can be added: (TN_k1, CT_k2, D) with $k1\geq 4$ and $k2\geq 0$ - the winning strategy starts at the 'diamond's axis' linked to the 'necklace' or 'chain of triangles'
  • finally, there is the family of linked 'necklaces of triangles' (TN_k1, CT_k2, TN_k3) with $k1, k3\geq 4$ and $k2\geq 0$; shown below is (TN_4, CT_0, TN_4)=(TN_4, TN_4)

graphs allowing winning strategy


References

Michael T
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    In the fourth picture, you could replace the right part by another copy of the left part, and/or insert a chain of triangles between the two parts. – Especially Lime Mar 27 '25 at 15:16
  • Hi Lime, great!! So, in total, if D='diamond graph' and $T_n$='closed cycle of triangles' ($n\geq 4$), we need DD, DT or TT and can link these directly or via a path of triangles; PLUS, we have the wheel... what else is missing? – Michael T Mar 27 '25 at 15:39
  • It's basically a degenerate case of something you already have, but for DD linked by a path of triangles, it can just be one D linked to the other side of itself. – Especially Lime Mar 28 '25 at 08:36
  • Hi Lime, I may misunderstand; I don't see 'one D with both 'sides' linked by a 'path of triangles'' work. – Michael T Mar 28 '25 at 08:49
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    Apologies, I think I made a mistake. – Especially Lime Mar 28 '25 at 09:24