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Crossposted from MO.

Wondering about the existence of graph gadget related to coloring (or 3-coloring) odd hole free graphs.

Let $G$ be simple $k$-chromatic connected graph with two vertices $u,v$.

Is it possible $G$ to satisfy:

  1. All induced $uv$ paths have odd order (even number of edges).
  2. In all proper $k$ colorings, $u$ and $v$ have distinct colors
  3. (optional) $G$ doesn't contain induced $C_{2n+1}$ for $n>1$

If this is possible, there is reduction $F$ to odd hole free $F'$.

Replace an edge $u'v'$ by the gadget $G$ where $u'=u,v'=v$ and the rest vertices of $G$ are new vertices.

According to graphclasses coloring odd hole free is NP hard and 3-coloring is unknown.

Computer search suggest small gadgets don't exist (modulo errors).

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One can extract an argument that this cannot work from the paper found by OP in the MO thread. Suppose $G=(V,E)$ is as required, and $c:V\to[k]$ is a $k$-coloring. By the assumption, $c(u)\neq c(v)$. Consider the (bipartite) subgraph $H$ induced by $\{x\in V\ |\ c(x)\in\{c(u),c(v)\}\}$.

If $u$ and $v$ are in the same connected component of $H$, pick any shortest path in $H$ between $u$ and $v$; it is an induced path in $G$, with colors alternating between $c(u),c(v)$, and must have an odd number of edges because the colors at its ends differ. This contradicts the assumption.

So $u,v$ are in different connected components; but then one can toggle the coloring of one of these components to obtain a coloring $c'$ with $c'(u)=c'(v)$, contradiction.

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  • $\begingroup$ Thank you. I allow induced triangles in all cases, does your answer still apply? $\endgroup$ – joro Jul 13 '15 at 11:58
  • $\begingroup$ Check the MO comments about "even pairs" and "merging even pairs preserves the chromatic number", this might disprove it. $\endgroup$ – joro Jul 13 '15 at 12:13
  • $\begingroup$ reverse image - I have updated to get rid of it (and deleted some obsolete comments) $\endgroup$ – Klaus Draeger Jul 13 '15 at 14:16
  • $\begingroup$ Are you saying that if $u,v$ is an even pair, in all colorings $c(u)=c(v)$? I believe this follows from preserving the chromatic number when merging $u,v$. $\endgroup$ – joro Jul 14 '15 at 6:06
  • $\begingroup$ Why non-existence of even path in the bipartite subgraph means no such in the WHOLE graph? A path may pass thru a vertex outside the bipartite graph or am I missing something? $\endgroup$ – joro Jul 14 '15 at 6:27

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