We consider next the case that every 2*K*_{2} has 3 colors. Like the above result, the characterization is that the graph *G* must be “nearly bipartite”.

Define the following graphs and graph classes. Let 𝓗_{1} be the set of graphs that contain two adjacent vertices *x* and *y* such that every other edge is incident to *x* or *y*. Let 𝓗_{2} be the graphs that contain a triangle *x*, *y*, *z* such that every other vertex is a leaf with a neighbor in {*x*, *y*, *z*}.

Let *F*_{1} be the graph that is obtained from the disjoint union of *K*_{4} and *K*_{2} by identifying a vertex of each. Let *F*_{2} be the graph that is obtained from *K*_{4} − *e* by picking a vertex *x* of degree 3 and a vertex *y* of degree 2 and adding a leaf adjacent only to *x* and one only to *y*. Let *F*_{3} be the graph that is obtained from the disjoint union of *K*_{4} − *e* and *P*_{3} by identifying a minimum-degree vertex of each. Let *F*_{4} be the graph that is obtained from *P*_{6} : *v*_{1}, *v*_{2}, …, *v*_{6} by adding edge *v*_{3}*v*_{5}. We draw these in Figure 1.

Fig. 1 Coloring graphs so that every 2*K*_{2} receives 3 colors

For a graph *G*, we define the *reduced version* of *G* by considering every vertex in turn and, if it has more than one leaf neighbor, discarding all but one of these neighbors. It is immediate that a graph has a valid coloring if and only if its reduced version has, because we may assume all leaves at a vertex are the same color.

#### Theorem 4.2

*A graph G has a coloring where every* 2*K*_{2} *receives* 3 *colors if and only if*

*G is bipartite*;

*G is formed from the disjoint union of a bipartite graph H and K*_{3} *by identifying one vertex of each*;

*G is the disjoint union of graphs H and M where H* ∈ 𝓗_{1} ∪ 𝓗_{2} *and M* ∈ 𝓜 *(the family of graphs with matching number* 1*)*; *or*

*the reduced version of G is F*_{1}, *F*_{2}, *F*_{3}, *or F*_{4}, *or a subgraph thereof*.

#### Proof

These graphs each have a valid coloring. In case (i), color all vertices in one partite set red and give all remaining vertices unique colors. In case (ii), let *x* be the vertex that was identified. Color it and all other vertices in its partite set of *H* red, color its two neighbors in the *K*_{3} sapphire, and then give all remaining vertices unique colors. In case (iii), color the subgraph *H* as shown in the above picture and color the subgraph *M* monochromatically with color 4. In case (iv), color the graphs with three colors as shown in the above picture.

We turn to the proof that these are all the graphs. So assume *G* has a valid coloring.

#### Claim 4.3

*Graph G cannot have an odd cycle of length* 5 *or more*.

#### Proof

We claim first that there cannot be two consecutive vertices of the same color. Consider a portion of the cycle *abcdef* (where possibly *a* = *f*) where *c* and *d* have the same color, say 1. Then, by considering the pair *bc*, *de*, without loss of generality *b* has color 2 and *e* has color 3. By considering the pair *cd*, *ef*, the vertex *f* must have a color other than 1 and 3. By considering pair *bc*, *ef*, it follows that vertex *f* has color 2. By similar argument, *a* has color 3 (and in particular *a* ≠ *f*). But then pair *ab*, *ef* is a contradiction.

Now suppose there are vertices at distance three of the same color. Consider a portion of the cycle *abcdef* (where possibly *a* = *f*) where *b* and *e* have same color, say 1. Then by considering *bc*, *de*, without loss of generality *c* has color 2 and *d* has color 3. By considering pair *cd*, *ef*, vertex *f* must have a color from {1, 2, 3}. By the lack of consecutives of the same color, *f* cannot be color 1; by considering pair *bc*, *ef*, vertex *f* cannot be color 2. Therefore *f* has color 3. Similarly, vertex *a* has color 2. In particular *a* ≠ *f*, and so there is another vertex next to *f*, say *g*. Then by considering pair *bc*, *fg*, vertex *g* must have a color from {1, 2, 3}. But it can easily be checked that each choice leads to a contradiction.

So we have shown that there cannot be two consecutive vertices of the same color nor two vertices at distance three. Now consider again a portion of the cycle *abcdef* (where possibly *a* = *f*). There must be a duplicate color at distance two within *abcd*. Say *a* and *c* have color 1, with *b* of color 2 and *d* of color 3. Then consider the pair *bc*, *de*. Since vertices *b* and *d* have different colors, it must be that vertices *c* and *e* have the same color. Then *f* must have a color different from *c*, *d* and *e*. Further, it cannot have the same color as *b*, by the pair *ab*, *ef*. By repeated application of this, it follows that every alternate vertex on the cycle has color 1. In particular the cycle has even length. □

If the graph *G* is bipartite, we are done. So assume there is a triangle *T*.

#### Claim 4.4

*If T is properly colored*, *then we are in case (iii) or have reduced graph F*_{1} *or F*_{2}.

#### Proof

Say triangle *T* has vertex *a* of color 1, vertex *b* of color 2, and vertex *c* of color 3. Suppose there is another color present, say vertex *d* has 4. Then consider an edge incident with *d*. If it goes to triangle *T*, we have a contradiction. So say it is *de*. By considering the pairs formed by *de* and each edge of *T*, it follows that *e* cannot have a new color, nor can it have color from {1, 2, 3}. Thus *e* has color 4. That is, the edges incident with the vertices not colored from {1, 2, 3} are monochromatically colored; and indeed must induce a component from 𝓜 colored 4.

So consider the vertices with color from {1, 2, 3}. We cannot have an edge disjoint from the triangle, since if it is monochromatic one can pair it with an edge of *T* that includes that color, and if proper one can pair it with the edge of *T* that is identically colored. So all such edges have (at least) one end in the triangle. If the component containing *T* is in 𝓗_{1}, then we are done. So assume the component containing *T* is not in 𝓗_{1}. That is, every vertex of *T* has a neighbor outside the triangle. Consider the possibilities.

One possibility is that the three vertices of *T* have a common neighbor *v*. By the above, *v* must have one of their colors, say 1. If not just *K*_{4}, there is another vertex, say *w*. This vertex cannot be incident with either *a* or *v*, so say incident with vertex *c*. Then *w* must be color 2 and must be a leaf. One can check that all remaining edges must be similarly incident with *c*. Further, the edge *va* being monochromatic implies that there is no monochromatic edge of color 4. That is, the reduced version of *G* is *F*_{1} or *K*_{4}.

A second possibility is that *a*, *b* have a common neighbor *v* while *c* has neighbor *w*. Suppose that *w* has color 3. Then considering the pair *va*, *cw* it follows that *v* must be color 2, but then the pair *vb*, *cw* is not validly colored. Thus vertex *w* has color 1, say. By considering the pair *va*, *cw*, it follows that the vertex *v* has color 2. Any other neighbor *x* of *a* must have color 2 because of *ax*, *cw* but not color 2 because of *vb*, *ax*, so *a* has no other neighbor. Any other neighbor *y* of *b* must have color 3 because of the pair *av*, *by*. And, any other neighbor *z* of *c* must have color 1 because of the pair *bv*, *cz*. Further, the edge *vb* being monochromatic means that there is no monochromatic edge of color 4. That is, the reduced version of *G* is *F*_{2} or *F*_{2} with one/both of the leaves deleted.

A third possibility is that *a*, *b*, and *c* have only distinct neighbors outside the triangle. That is, the component containing *T* is in 𝓗_{2}. □

So we can assume that there is no properly colored triangle *T*.

#### Claim 4.5

*If triangle T contains two colors*, *then we are in case (ii) or have reduced graph F*_{3} *or F*_{4}.

#### Proof

Say *T* has two vertices of color 1, say *a* and *b*, and the remaining vertex *c* has color 2. By the conditions and the lack of properly colored triangles, there is no other common neighbor of *a* and *c*, nor of *b* and *c*.

Assume first that there is no edge disjoint from *T*. Then the component containing *T* is in 𝓗_{2} and we are in case (iii) (with *M* being null) unless *a* and *b* have another common neighbor, say *d*. The vertex *d* must have a different color, say 3. Then the component containing *T* is in 𝓗_{1} (albeit with an alternative coloring) unless *c* has another neighbor, say *e*. The vertex *e* must have color 3. Then neither *a* nor *b* can have any other neighbors, and we are in case *F*_{2} minus the leaf incident with the vertex of degree 4.

So assume second there is an edge *ds* disjoint from *T*. Since one can pair it with *ab*, edge *ds* must be properly colored and neither end is color 1. Since one can pair it with *ac*, one end must have color 2. Further, no isolated vertex *w* of *G* – *T* can have color 1 or 2, since *V*(*T*) ∪ {*w*} contains a 2*K*_{2}. In particular, *G* – *T* is bipartite with bipartition (*D*,*S*), where *d* and every other vertex in *D* has color 2, while *s* and every other vertex in *S* has color distinct from {1, 2}.

Note that none of {*a*, *b*, *c*} can have a neighbor in *D*. Therefore, if neither *a* nor *b* has a neighbor in *S*, then we are in case (ii). So assume one of them, say *a*, has a neighbor *e* in *S*, say of color 3. Note that possibly *e* = *s*.

Then consider any vertex *v* other than {*a*, *b*, *c*, *d*, *e*}. Suppose it has a color not in {1, 2, 3}. By potential pairing with *ae* or *ac*, it follows that all neighbors of *v* have color 1 (so in particular *v* ≠ *s*) and then there is a contradiction from pairing with *ds*. That is, all vertices other than *a*, *b* have colors from {2, 3}.

It follows that the nontrivial component in *G* − {*a*, *b*} is a star. Now, note that if both *b* and *c* have degree 2, we are in case (ii) of the theorem. So either *b* is adjacent to *d*, or *c* is adjacent to *s*. In the first case we get the reduced graph *F*_{3} or possibly with edge removed so that it is just *K*_{4} − *e*∪ *K*_{2} (in which case *G* is covered by case (iii)). In the second case we get *F*_{4}. And one can check that all remaining vertices are clones of existing leaves. □

Finally suppose that all triangles are monochromatic. Then there can be only one. Indeed it must be an isolated component, while the remainder of the graph is bipartite, so that we are in case (iii).

This completes the proof of Theorem. □

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