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Is the following conjecture true or false? (Hopefully it is true — I need it as a lemma.)

For every undirected graph $G=(V,E)$ there exist three pairwise disjoint sets of vertices $V_1,V_2,V_3$ (whose union is not necessarily $V$) such that for every $i\in\{1,2,3\}$ and for every cycle $C\in G$, $C\cap V_i\neq \emptyset$.

That is, each $V_i$ hits all cycles in $G$, and the $V_i$ are pairwise disjoint.

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  • $\begingroup$ The desired result in the second paragraph is untrue if $G$ is not connected (take $G$ to be the union of a cycle with some disconnected vertices). $\endgroup$ Jul 29, 2013 at 14:40
  • $\begingroup$ If $G$ is a clique on at least 7 vertices then one of the classes must have 3 vertices and so contain a triangle disjoint from the remaining two classes. Have I understood the question correctly? $\endgroup$
    – Ben Barber
    Jul 29, 2013 at 14:41
  • $\begingroup$ The vertex sets $V_1,V_2,V_3$ do not necessarily form a partition of $V$ (I clarified this in the question). So when $G$ is a cycle with some disconnected vertices, I just need each $V_i$ to be a vertex of the cycle. If $G$ is a clique, I would just take three different vertices. $\endgroup$
    – Ariel
    Jul 29, 2013 at 14:46
  • $\begingroup$ Thank you for the correction. I still think cliques are a problem: if any one $V_i$ misses any 3 vertices of a clique then won't it be missing the corresponding triangle? $\endgroup$
    – Ben Barber
    Jul 29, 2013 at 14:51

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This is false for large complete graphs: eventually one of the $V_i$ will miss three vertices, and will not meet the corresponding triangle.

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    $\begingroup$ Even any graph containing $K_4$ fails. $\endgroup$ Jul 29, 2013 at 15:20

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