# Short proofs of coloring theorems on planar graphs

###### Abstract

A recent lower bound on the number of edges in a -critical -vertex graph by Kostochka and Yancey yields a half-page proof of the celebrated Grötzsch Theorem that every planar triangle-free graph is 3-colorable. In this paper we use the same bound to give short proofs of other known theorems on 3-coloring of planar graphs, among whose is the Grünbaum-Aksenov Theorem that every planar with at most three triangles is -colorable. We also prove the new result that every graph obtained from a triangle-free planar graph by adding a vertex of degree at most four is -colorable.

## 1 Introduction

Graphs considered in this paper are simple, i.e., without loops or parallel edges. For a graph , the set of its vertices is denoted by and the set of its edges by .

An *embedding* of a graph in a surface is an injective
mapping of
to a point set in and to non-self-intersecting curves in
such that
(a) for all and , is never an interior point of ,
and is an endpoint of if and only if is a vertex of , and
(b) for all , and can intersect only in vertices of .
A graph is *planar* if it has an embedding in the plane.
A graph with its embedding in the (projective) plane is a *(projective) plane* graph.
A cycle in a graph embedded in is *contractible* if it splits into
two surfaces where one of them is homeomorphic to a disk.

A *(proper) coloring* of a graph is a mapping from
to a set of colors such that whenever .
A graph is *-colorable* if there exists a coloring of using at most colors.
A graph is *-critical* if is not -colorable but every proper subgraph
of is -colorable.
By definition, if a graph is not -colorable then it contains
a -critical subgraph.

Dirac [12] asked to determine the minimum number of edges in a -critical graph. Ore conjectured [22] that an upper bound obtained from Hajós’ construction is tight. More details about Ore’s conjecture can be found in [18][Problem 5.3] and in [20]. Recently, Kostochka and Yancey [20] confirmed Ore’s conjecture for and showed that the conjecture is tight in infinitely many cases for every . In [19] they gave a 2.5-page proof of the case :

###### Theorem 1 ([19]).

If is a -critical -vertex graph then

Theorem 1 yields a half-page proof [19] of the celebrated Grötzsch Theorem [14] that every planar triangle-free graph is 3-colorable. This paper presents short proofs of some other theorems on -coloring of graphs close to planar. Most of these results are generalizations of Grötzsch Theorem.

###### Theorem 2 ([2, 17]).

Let be a triangle-free planar graph and be a graph such that for some edge of . Then is 3-colorable.

###### Theorem 3 ([17]).

Let be a triangle-free planar graph and be a graph such that for some vertex of degree 3. Then is 3-colorable.

###### Theorem 4.

Let be a triangle-free planar graph and be a graph such that for some vertex of degree 4. Then is 3-colorable.

Theorems 2 and 4 yield a short proof of the following extension theorem that was used by Grötzsch [14].

###### Theorem 5.

Let be a triangle-free planar graph and be a face of of length at most 5. Then each 3-coloring of can be extended to a 3-coloring of .

An alternative statement of Theorem 2 is that each coloring of two vertices of a triangle-free planar graph by two different colors can be extended to a 3-coloring of . Aksenov et al. [3] extended Theorem 2 by showing that each proper coloring of each induced subgraph on two vertices of extends to a 3-coloring of .

###### Theorem 6 ([3]).

Let be a triangle-free planar graph. Then each coloring of two non-adjacent vertices can be extended to a 3-coloring of .

We show a short proof of Theorem 6.

Another possibility to strengthen Grötzsch’s Theorem is to allow at most three triangles.

###### Theorem 7 ([1, 4, 15]).

Let be a planar graph containing at most three triangles. Then is 3-colorable.

The original proof by Grünbaum [15] was incorrect and a correct proof was provided by Aksenov [1]. A simpler proof was given by Borodin [4], but our proof is significantly simpler.

Youngs [30] constructed triangle-free graphs in the projective plane that are not 3-colorable. Thomassen [25] showed that if is embedded in the projective plane without contractible cycles of length at most 4 then is 3-colorable. We slightly strengthen the result by allowing two contractible 4-cycles or one contractible 3-cycle.

###### Theorem 8.

Let be a graph embedded in the projective plane such that the embedding has at most two contractible cycles of length 4 or one contractible cycle of length three such that all other cycles of length at most 4 are non-contractible. Then is 3-colorable.

It turned out that restricting the number of triangles is not necessary. Havel conjectured [16] that there exists a constant such that if every pair of triangles in a planar graph is at distance at least then is 3-colorable. The conjecture was proven true by Dvořák, Král’ and Thomas [13].

Without restriction on triangles, Steinberg conjectured [23] that every planar graph without 4- and 5-cycles is 3-colorable. Erdős suggested to relax the conjecture and asked for the smallest such that every planar graphs without cycles of length 4 to is 3-colorable. The best known bound for is 7 [9]. A cycle is triangular if it is adjacent to a triangle other than . In [6], it is proved that every planar graph without triangular cycles of length from to is -colorable, which implies all results in [7, 8, 9, 10, 11, 21, 26, 27, 28, 29].

We present the following result in the direction towards Steinberg’s conjecture with a Havel-type constraint on triangles. As a free bonus, the graph can be in the projective plane instead of the plane.

###### Theorem 9.

Let be a 4-chromatic projective planar graph where every vertex is in at most one triangle. Then contains a cycle of length 4,5 or 6.

There are numerous other results on the Three Color Problem in the plane. See a recent survey [5] or a webpage maintained by Montassier http://janela.lirmm.fr/~montassier/index.php?n=Site.ThreeColorProblem.

The next section contains proofs of the presented theorems and Section 3 contains constructions showing that some of the theorems are best possible.

## 2 Proofs

*Identification* of non-adjacent vertices and in a graph
results in a graph obtained from by adding a new vertex
adjacent to every vertex that is adjacent to at least one of and .

The following lemma is a well-known tool to reduce the number of 4-faces. We show its proof for the completeness.

###### Lemma 10.

Let be a plane graph and be a 4-face in such that . Let be obtained from by identifying and where . If the number of triangles increases in both and then there exists a triangle for some and . Moreover, contains vertices and not in such that and are paths in . Indices are modulo 4. See Figure 1.

###### Proof.

Let and be as in the statement of the lemma. Since the number of triangles increases in there must be a path in where . Similarly, a new triangle in implies a path in where . By the planarity of , and are not disjoint. Without loss of generality assume . This results in triangle and paths and . Note that and do not have to be distinct. See Figure 1(b). ∎

###### Proof of Theorem 2.

Let be a smallest counterexample and be a plane triangle-free graph such that for some edge . Let have vertices and edges and have faces. Note that has vertices and edges. By the minimality of , is 4-critical. So Theorem 1 implies .

CASE 1: has at most one 4-face. Then and hence . By this and Euler’s Formula applied on we have , i.e., . This contradicts Theorem 1.

CASE 2: Every 4-face of contains both and and there are at least two such 4-faces and . If there exists then has degree two in which contradicts the 4-criticality of .

Let be obtained from by identification of and into a new vertex . If is not triangle-free then there is a path in where . Since must cross and , we may assume that and . However, . This contradicts the existence of . Hence is triangle-free. Let . By the minimality of , there exists a 3-coloring of . This contradicts that is not -colorable since can be extended to by letting .

CASE 3: has a 4-face with vertices in the cyclic order where is neither nor . Since is triangle-free, neither nor are edges of . Lemma 10 implies that either and or and can be identified without creating a triangle. Without loss of generality assume that , obtained by from identification of and to a new vertex , is triangle-free. Let . By the minimality of , there is a 3-coloring of . The 3-coloring can be extended to by letting which contradicts the 4-criticality of . ∎

###### Proof of Theorem 4.

Let be a smallest counterexample and be a plane triangle-free graph such that for some vertex of degree 4. Let have vertices and edges and have faces. Then has vertices and edges. By minimality, is 4-critical. So Theorem 1 implies .

CASE 1: has no 4-faces. Then and hence . By this and Euler’s Formula applied to , we have , i.e., . This contradicts Theorem 1.

CASE 2: has a 4-face with vertices in the cyclic order. Since is triangle-free, neither nor are edges of and Lemma 10 applies. Without loss of generality assume that obtained from by identification of and is triangle-free.

By the minimality of , the graph obtained from by identification of and satisfies the assumptions of the theorem and hence has a 3-coloring. Then also has a 3-coloring, a contradiction. ∎

###### Proof of Theorem 5.

Let the 3-coloring of be .

CASE 1: is a 4-face where are its vertices in cyclic order.

CASE 1.1: and . Let be obtained from by adding a vertex adjacent to and . Since satisfies the assumptions of Theorem 4, there exists a 3-coloring of . In any such 3-coloring, and . Hence by renaming the colors in we obtain an extension of to a 3-coloring of .

By symmetry, the other subcase is the following.

CASE 1.2: and . Let be obtained from by adding the edge . Since satisfies the assumptions of Theorem 2, there exists a 3-coloring of . In any such 3-coloring, and hence . By renaming the colors in we obtain an extension of to a 3-coloring of .

CASE 2: is a 5-face where are its vertices in cyclic order. Observe that up to symmetry there is just one coloring of . So without loss of generality assume that and .

Let be obtained from by adding a vertex adjacent to and . Since satisfies the assumptions of Theorem 4, there exists a 3-coloring of . Note that in any such 3-coloring and . Hence by renaming the colors in we can extend to a 3-coloring of . ∎

###### Proof of Theorem 6.

Let be a smallest counterexample and let be the two non-adjacent vertices colored by . If then the result follows from Theorem 2 by considering graph obtained from by adding the edge . Hence assume that .

CASE 1: has at most two 4-faces. Let be a graph obtained from by identification of and . Any 3-coloring of yields a 3-coloring of where and are colored the same. By this and the minimality of we conclude that is 4-critical. Let have edges, vertices and faces.

CASE 2: has at least three 4-faces. Let be a 4-face with vertices in the cyclic order. Since is triangle-free, neither nor are edges of . Hence Lemma 10 applies.

Without loss of generality let from Lemma 10 be triangle-free. By the minimality of , has a 3-coloring where unless . Since , without loss of generality and . Moreover, the same cannot happen to from Lemma 10, hence contains a triangle. Thus contains a path where , and also contains a 5-cycle (see Figure 2). By Theorem 5, is a 5-face. Hence is a 2-vertex incident with only one 4-face.

By symmetric argument, is also a 2-vertex incident with one 4-face and 5-face. However, has at least one more 4-face where identification of vertices does not result in the edge , a contradiction to the minimality of . ∎

###### Proof of Theorem 8.

Let be a minimal counterexample with edges, vertices and faces. By minimality, is a 4-critical and has at most two 4-faces or one 3-face. From embedding, . By Euler’s formula, . Hence , a contradiction to Theorem 1. ∎

Borodin used in his proof of Theorem 7 a technique called *portionwise coloring*.
We avoid it and build the proof on the previous results arising
from Theorem 1.

###### Proof of Theorem 7.

Let be a smallest counterexample. By minimality, is 4-critical and every triangle is a face. By Theorem 5 for every separating 4-cycle and 5-cycle , both the interior and exterior of contain triangles.

CASE 1: has no 4-faces. Then and by Euler’s Formula , i.e., . This contradicts Theorem 1.

CASE 2: has a 4-face such that . By the minimality, and are both 3-faces and hence has 4 vertices, 5 edges and it is 3-colorable.

CASE 3: For every 4-face , neither nor are edges of . By Lemma 10, there exist paths and .

CASE 3.1: contains a 3-prism with one of its 4-cycles being a 4-face. We may assume that this face is our and , see Figure 1(b). Theorem 5 implies that one of , is a 4-face. Without loss of generality assume that is a 4-face. Let be obtained from by identification of and to a new vertex . Since is not 3-colorable, it contains a 4-critical subgraph . Note that contains triangle that is not in but is not in since in . By the minimality of , there exists another triangle that is in but not in . By planarity, . Hence there is a vertex such that and are neighbors of .

By considering identification of and and by symmetry, we may assume that there is a vertex such that and are neighbors of . By planarity we conclude that . This contradicts the fact that has at most three triangles. Therefore is 3-prism-free.

CASE 3.2: contains no 3-prism with one of its 4-cycles being a 4-face. Then , see Figure 1(a). If then is triangle-free and Theorem 3 gives a 3-coloring of , a contradiction. Similarly, .

Suppose that if a 4-face. Let be obtained from by adding edge . If the number of triangles in is at most three, then has a 3-coloring by the minimality of . Let be a 3-coloring of such that if and . Since the neighbors of in are neighbors of in , is a 3-coloring, a contradiction. Therefore has at least four triangles and hence contains a vertex adjacent to and . Since , the only possibility is . Having edge results in a 3-prism being a subgraph of which is already excluded. Hence is not a face and by symmetry is not a face either.

Since neither nor is a face, each of them contains a triangle in its interior. Since we know the location of all three triangles, Theorem 5 implies that is a 5-face. It also implies that the common neighbors of and are exactly and , and the common neighbors of and are exactly and . Without loss of generality, let be the outer face of .

Let be obtained from the 4-cycle and its interior by adding edge . The edge is in only two triangles, and there is only one triangle in the interior of the 4-cycle. Hence by the minimality of , there exists a 3-coloring of .

Let be obtained from the 4-cycle and its interior by adding edge . By the same argument as for , there is a 3-coloring of of .

Rename the colors in so that , and . Then is a 3-coloring of , a contradiction. ∎

###### Proof of Theorem 9.

Let be a 4-chromatic projective plane graph where every vertex is in at most one triangle and let be 4-,5- and 6-cycle free. Then contains a 4-critical subgraph . Let have edges, vertices and faces. Since is also 4-,5- and 6-cycle-free and every vertex is in at most one triangle, we get . By Euler’s formula, . Hence , a contradiction to Theorem 1. ∎

## 3 Tightness

Theorem 2 is best possible because there exists an infinite family [24] of -critical graphs that become triangle-free and planar after removal of just two edges. See Figure 3. Moreover, the same family shows also the tightness of Theorem 7, since the construction has exactly four triangles.

Aksenov [1] showed that every plane graph with one 6-face and all other faces being 4-faces has no 3-coloring in which the colors of vertices of form the sequence . This implies that Theorem 5 is best possible. It also implies that Theorems 4 and 6 are best possible. See Figure 4 for constructions where coloring of three vertices or an extra vertex of degree 5 force a coloring of a 6-cycle.

Theorem 8 is best possible because there exist embeddings of in the projective plane with three 4-faces or with two 3-faces and one 6-face.

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