lately, I have encountered a problem that I struggle to find a satisfactory solution for. I need to prove that triangle-free 3-colorability is NP-complete. Therefore I assume the right way is to find an NP-complete problem and reduce it to my problem.

I tried it with general 3-colorability, but I was unable to create function F() that would reduce a general graph to a triangle-free graph without changing its colorability.

Is it a good direction? Does it make sense to reduce the 3-colorability problem?

If yes, I would appreciate any advice on how to finish the proof. If I am doing it wrong, I would appreciate any advice on the reduction.

  • 2
    $\begingroup$ Where have you encountered this problem? I am familiar with a paper that proves hardness of this problem by a reduction from 3-color. It is a bit challenging, perhaps there are easier reductions, I'm not sure though. $\endgroup$ Jan 6, 2021 at 17:22
  • $\begingroup$ Can you give a self-contained definition of the "triangle-free 3-colorability" problem? $\endgroup$
    – D.W.
    Jan 6, 2021 at 19:31
  • $\begingroup$ Presumably, 3-colorability of triangle-free graphs. $\endgroup$ Jan 6, 2021 at 20:29
  • $\begingroup$ Yes, as Yuval stated, triangle-free 3-colorability is simply the problem of 3-colorability on graphs that contain no triangles. $\endgroup$
    – kolomann
    Jan 6, 2021 at 21:00
  • $\begingroup$ The proof here is a one paragraph gadget reduction. $\endgroup$ Jan 7, 2021 at 10:23

1 Answer 1


We reduce from 3-coloring, following Kráľ, Kratochvíl, Tuza and Woeginger, Complexity of Coloring Graphs without Forbidden Induced Subgraphs. We will need to use a gadget $H$, with the following properties:

  • $H$ is a 3-colorable triangle-free graph.
  • There are two vertices $a,b \notin H$, not connected by an edge, such that any 3-coloring of $H$ gives $a,b$ different colors.

Given such a gadget $H$, here is how to reduce 3-coloring to triangle-free 3-coloring. We start with a graph $G$. We construct a new graph $G'$ as follows. The graph $G'$ will have all vertices of $G$, together with some new vertices. For each edge $e = (x,y) \in G$, there is a copy $H_e$ of $H$. We identify $a_e$ (the copy of $a$ in $H_e$) with $x$, and $b_e$ with $y$. This completes the construction.

The new graph is triangle-free. Indeed, suppose that it contained some triangle $u,v,w$. Since the vertices $u,v$ are connected, they must reside in some copy $H_e$ of $H$. Since $(a,b) \notin H$, at least one of these vertices doesn't belong to $V$, say $u \notin V$. Since all neighbors of $u$ are in $H_e$, it follows that $u,v,w$ all reside in $H_e$, contradicting the triangle-freeness of $H$.

If $G$ is 3-colorable, then we can easily complete a 3-coloring of $G$ to a 3-coloring of $G'$ by using a 3-coloring of $H$ (by symmetry, it doesn't matter what colors $a,b$ get, as long as the colors are different). Conversely, the restriction of any 3-coloring of $G'$ to $V$ is a 3-coloring of $G$.

To complete the proof, we need to describe the gadget $H$. Our starting point is any critically 4-colorable triangle-free graph $H'$, such as the Grötzsch graph; here critically 4-colorable means that the graph is not 3-colorable, but becomes 3-colorable if we remove any edge.

We take any edge $(a,c) \in H'$ and replace it by an edge $(b,c)$, where $b$ is a new vertex (in particular, $(a,b)$ is not an edge). The resulting graph $H$ is still triangle-free. Since $H'$ was critically 4-colorable, the graph $H$ is 3-colorable: take a 3-coloring of $H' \setminus \{(a,c)\}$, and color $b$ appropriately. Any 3-coloring of $H$ must give $a,b$ different colors, since otherwise its restriction to all vertices but $b$ would be a 3-coloring of $H'$.


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