Size of edgecritical uniquely 3colorable planar graphs ^{1}^{1}1Supported by 973 Program of China 2013CB329601, 2013CB329603, National Natural Science Foundation of China Grant 61309015 and National Natural Science Foundation of China Special Equipment Grant 61127005.
Abstract
A graph is uniquely kcolorable if the chromatic number of is and has only one coloring up to permutation of the colors. A uniquely colorable graph is edgecritical if is not a uniquely colorable graph for any edge . Mel’nikov and Steinberg [L. S. Mel’nikov, R. Steinberg, One counterexample for two conjectures on three coloring, Discrete Math. 20 (1977) 203206] asked to find an exact upper bound for the number of edges in a edgecritical 3colorable planar graph with vertices. In this paper, we give some properties of edgecritical uniquely 3colorable planar graphs and prove that if is such a graph with vertices, then , which improves the upper bound given by Matsumoto [N. Matsumoto, The size of edgecritical uniquely 3colorable planar graphs, Electron. J. Combin. 20 (3) (2013) P49]. Furthermore, we find some edgecritical 3colorable planar graphs which have vertices and edges.
keywords:
planar graph; unique coloring; uniquely colorable planar graph; edgecriticalMsc:
05C151 Introduction
A graph is uniquely kcolorable if and has only one coloring up to permutation of the colors, where the coloring is called a unique coloring. In other words, all colorings of induce the same partition of into independent sets. In addition, uniquely colorable graphs may be defined in terms of their chromatic polynomials, which initiated by Birkhoff Birkhoff1912 () for planar graphs in 1912 and, for general graphs, by Whitney Whitney1932 () in 1932. Because a graph is uniquely colorable if and only if its chromatic polynomial is . For a discussion of chromatic polynomials, see Read Read1968 ().
Let be a uniquely colorable graph, is edgecritical if is not uniquely colorable for any edge . Uniquely colorable graphs were defined and studied firstly by Harary and Cartwright Harary1968 () in 1968. They proved the following theorem.
Theorem 1.1.
(Harary and Cartwright Harary1968 ()) Let be a uniquely colorable graph. Then for any unique coloring of , the subgraph induced by the union of any two color classes is connected.
As a corollary of Theorem 1.1, it can be seen that a uniquely colorable graph has at least edges. Furthermore, if a uniquely colorable graph has exactly edges, then is edgecritical. There are many references on uniquely colorable graphs. For example see Chartrand and Geller Chartrand1969 (), Harary, Hedetniemi and Robinson Harary1969 () and Bollobás Bollob¨¢s1978 ().
Chartrand and Geller Chartrand1969 () in 1969 started to study uniquely colorable planar graphs. They proved that uniquely 3colorable planar graphs with at least 4 vertices contain at least two triangles, uniquely 4colorable planar graphs are maximal planar graphs, and uniquely 5colorable planar graphs do not exist. Aksionov Aksionov1977 () in 1977 improved the low bound for the number of triangles in a uniquely 3colorable planar graph. He proved that a uniquely 3colorable planar graph with at least 5 vertices contains at least 3 triangles and gave a complete description of uniquely 3colorable planar graphs containing exactly 3 triangles.
For an edgecritical uniquely colorable planar graph , if , then it is easy to deduce that is tree and has exactly edges. If , then is a maximal planar graph and has exactly edges by Euler’s Formula. Therefore, it is sufficient to consider the size of uniquely colorable planar graphs. We denote by the set of all edgecritical uniquely colorable planar graphs and by the upper bound of the size of edgecritical uniquely colorable planar graphs with vertices.
In 1977 Aksionov Aksionov1977 () conjectured that . However, in the same year, Mel’nikov and Steinberg Mel'nikov1977 () disproved the conjecture by constructing a counterexample , which has 16 vertices and 30 edges. Moreover, they proposed the following problems:
Problem 1.2.
(Mel’nikov and Steinberg Mel'nikov1977 ()) Find an exact upper bound for the number of edges in a edgecritical 3colorable planar graph with vertices. Is it true that for any ?
Recently, Matsumoto Matsumoto2013 () constructed an infinite family of edgecritical uniquely 3colorable planar graphs with vertices and edges, where . He also gave a nontrivial upper bound for .
In this paper, we give some properties of edgecritical uniquely 3colorable planar graphs with vertices and improve the upper bound of given by Matsumoto Matsumoto2013 () to , where . Moreover, we give some edgecritical 3colorable planar graphs which have vertices and edges. It follows that the conjecture of Mel’nikov and Steinberg Mel'nikov1977 () is false because if .
2 Notation
Only finite, undirected and simple graphs are considered in this paper. For a planar graph , , and are the sets of vertices, edges and faces of , respectively. We denote by and the minimum degree and maximum degree of graph . The degree of a vertex , denoted by , is the number of neighbors of in . The degree of a face , denoted by , is the number of edges in its boundary, cut edges being counted twice. When no confusion can arise, and are simplified by and , respectively. A face is a face if and a kface if . The similar notation is used for cycles. We denote by the set of vertices of with degree and by the set of vertices of with degree at least , where . The similar notation is used for the set of faces of .
A wheel is the graph consists of a single vertex and a cycle with vertices together with edges from to each vertex of . A planar (resp. outerplanar) graph is maximal if is not planar (resp. outerplanar) for any two nonadjacent vertices and of . Let and be two disjoint subset of , we use to denote the number of edges of with one end in and the other in . In particular, if or , we simply write or for , respectively. To contract an edge of a graph is to delete the edge and then identify its ends. The resulting graph is denoted by . Two faces and of are adjacent if they have at least one common edge. A cycle is said to be a separating cycle in if the removal of disconnects the graph .
A kcoloring of is an assignment of colors to such that no two adjacent vertices are assigned the same color. Naturally, a coloring can be viewed as a partition of , where denotes the set of vertices assigned color , and is called a color class of the coloring for any . Two colorings and of are said to be distinct if they produce two distinct partitions of into color classes. A graph is kcolorable if there exists a coloring of , and the chromatic number of , denoted by , is the minimum number such that is colorable.
The notations and terminologies not mentioned here can be found in Bondy2008 ().
3 Properties of edgecritical uniquely colorable planar graphs
Let be a 3colorable planar graph and be a 3coloring of . It is easy to see that the restriction of to is a 3coloring of , where . For convenience, we also say is a 3coloring of . If there exists a 3coloring of such that , then we say that can be extended to a 3coloring of .
Theorem 3.1.
Let be a uniquely 3colorable planar graph. Then if and only if is 3colorable for any edge .
Proof Suppose that , then, by definition, has at least two distinct 3colorings for each . Since is uniquely 3colorable, we conclude that there exists a 3coloring of such that . Hence is 3colorable.
Conversely, suppose that . Then there exists an edge such that is also a uniquely 3colorable planar graph. Obviously, for any unique 3coloring of , we have . So is not 3colorable. This establishes Theorem 3.1. ∎
The following result is obtained by Theorem 3.1.
Corollary 3.2.
Let and . If is incident with exactly one 4face and all other faces incident with are triangular, then is even.
Proof Suppose that the result is not true. Let be the neighbors of and and be the vertices of the 4face. Then the graph contains a wheel. Hence is not 3colorable, a contradiction with Theorem 3.1. ∎
Theorem 3.3.
Suppose that and is a subgraph of . If is uniquely 3colorable, then we have
 (i)

;
 (ii)

For any vertex , .
Proof (i) Suppose that , then there exists an edge such that is also uniquely 3colorable. Let be a unique 3coloring of . Since , then has a 3coloring which is distinct from . Note that , we have . Thus, can be extended to a 3coloring of . So has two distinct 3colorings and , which contradicts .
(ii) Suppose that there exists a vertex such that . Let be a unique 3coloring of and be the three neighbors of in . Then their exist at least two vertices among and receive the same color. We assume w.l.o.g. that . Since , then has a 3coloring which is distinct from . Note that is uniquely 3colorable and , we have . Thus, can be extended to a 3coloring of . This is a contradiction. ∎
Corollary 3.4.
Suppose that contains a sequence of triangles satisfying and have a common edge, where and . Let and be the vertices in and , respectively, then and .
Proof Let be the neighbors of in . Since the subgraph of consists of triangles is uniquely 3colorable, by Theorem 3.3, we know that is not adjacent to or in . Thus, . Similarly, since the subgraph of consisting of triangles is uniquely 3colorable, we have . ∎
By Corollary 3.4, we have the following result.
Corollary 3.5.
Suppose that has no separating 3cycles. Let be a subgraph of that consists of a sequence of triangles such that each has a common edge with for some , where . Then is a maximal outerplanar graph.
For a planar graph , if has no separating 3cycles, we call the subgraph in Corollary 3.5 a trianglesubgraph of . Note that a triangle is a trianglesubgraph of . Therefore, any has at least one trianglesubgraph. A trianglesubgraph of is maximal if there is no maximal outerplanar subgraph of such that . In other words, the graph consists of the longest sequence of triangles such that each has a common edge with for some .
Theorem 3.6.
Suppose that has no separating 3cycles. Let be a uniquely 3colorable subgraph and be any two maximal trianglesubgraphs of . If , , then we have
 (i)

and have at most one common vertex;
 (ii)

If and have a common vertex , then ; otherwise, ;
 (iii)

If and have a common vertex , and have a common vertex and , , then the union of and is uniquely 3colorable.
Proof Let be a unique 3coloring of .
(i) Suppose, to the contrary, that and have two common vertices and . Since , then and are not adjacent in both and . Otherwise, if , this contradicts Corollary 3.4; if , then is uniquely 3colorable but not edgecritical, a contradiction with Theorem 3.3. By the definition of a trianglesubgraph, we know that there exists a sequence of triangles in such that and have a common edge and , , where .
If . Let be a neighbor of in , then . Since , then has a 3coloring which is distinct from . Note that both and the subgraph of consists of triangles are uniquely 3colorable and . So , , namely . Therefore, can be extended to a 3coloring of which is distinct from . This contradicts .
If . Let be a neighbor of in satisfying . Since , then has a 3coloring which is distinct from . Since both and the subgraph of consists of triangles are uniquely 3colorable, we have . Therefore, can be extended to a 3coloring of which is distinct from . It is a contradiction.
(ii)Case 1. and have a common vertex .
Suppose that and are two edges with and . If there exists a vertex such that , we assume w.l.o.g. that , then . Since , has a 3coloring which is distinct from . Note that both and are uniquely 3colorable, we have and . Thus and then can be extended to a 3coloring of which is distinct from . If for any , then . Thus, we have either and , or and . Since has a 3coloring which is distinct from , and and are uniquely 3colorable, we have . Therefore, can be extended to a 3coloring of which is distinct from .
Case 2. and have no common vertex.
Suppose that and are 4 edges with and , . Then there exist two edges, say and , such that . By using a similar argument to Case 1, we can obtain a 3coloring of , which is distinct from and can be extended to a 3coloring of . It is a contradiction.
(iii) By definition of , there exists a sequence of triangles in such that and have a common edge and , , where .
Suppose that , . Let be an arbitrary neighbor of in . Then . Since , has a 3coloring which is distinct from . Note that and the subgraph of consists of triangles are uniquely 3colorable, we have and . Therefore, can be extended to a 3coloring of which is distinct from . This contradicts .
Suppose that , then their exists such that . We assume w.l.o.g. that . Let be a neighbor of in satisfying . If , then . Since , has a 3coloring which is distinct from . Note that and the subgraph of consists of triangles are uniquely 3colorable, we have , and . Thus, . Therefore, can be extended to a 3coloring of which is distinct from . This contradicts . If, then . Since , has a 3coloring which is distinct from . Since and the subgraph of consists of triangles are uniquely 3colorable, we have and . Thus, . This contradicts .
Suppose that . Using the fact that any coloring of two vertices with can be extended uniquely to a 3coloring of , we can obtain that the union of and is uniquely 3colorable, where . ∎
4 Size of edgecritical uniquely colorable planar graphs
In this section, we consider the upper bound of for edgecritical uniquely 3colorable planar graphs with vertices.
Suppose that and has no separating 3cycles. Let be all of the maximal trianglesubgraphs of . For two maximal trianglesubgraphs and having a common vertex , if there exists such that and satisfy the condition of Case (iii) in Theorem 3.6, namely and have a common vertex (say ), and have a common vertex (say ) and , then we say that and satisfy Property P. Let . (We will use such notation without mention in what follows.) Now we analyse the relationship between , the number of 4faces of , and . For a vertex , we use to denote the number of maximal trianglesubgraphs of that contain .
First we construct a new graph from with
, , , where in corresponds to in for any . The edges in are constructed by the following two steps.
Step 1: For every with , add the edge to if both and contain . (see e.g. Fig. 1)
Step 2: For every with , let contain and they appear in clockwise order around .
For any with and satisfying Property P, then add the edge to .
Let be the subgraph of with vertex set and edge set .
Then we add some edges in to such that the resulting graph, denoted by , is connected and has the minimum number of edges. Now the construction of the edges of the graph is completed. (see e.g. in Fig. 1, we first join the edges , and , then join the edges , , , and .)
Remark. By the definition of , if , then and must have a common vertex. For a edgecritical uniquely 3colorable graph , if for any , then the graph obtained by above construction is unique; otherwise, is not unique. Furthermore, we have Theorem 4.1.
Theorem 4.1.
Suppose that has no separating 3cycles. Let , , be all of the maximal trianglesubgraphs of , then is a simple planar graph and .
Proof By Corollary 3.5 and Theorem 3.6(i), we know that has no loops or parallel edges. So is a simple graph. Note that is a planar graph. For any with and any with and satisfying Property P, then is a planar graph and there exist no edges such that and , where and the subscripts are taken modulo . Now we prove that is a forest. If , then, by the definition of and Theorem 3.6(iii), we know that there exist and such that the graph is uniquely 3colorable. Thus, and does not satisfy Property P, namely . Therefore, is a forest. By the definition of , it is easy to see that is a tree. By the definition of , we can conclude that is a planar graph.
For any distinct faces and of , by the definition of , it can be seen that there exist two distinct 4faces of corresponding to and , respectively. Conversely, for any 4face of , let be all of the maximal trianglesubgraphs satisfying and have common edges, . Let be the common vertex of , because is tree, there exists a unique face of incident with . Thus, . ∎
Theorem 4.2.
Suppose that has no separating 3cycles. Let , , be a sequence of faces in such that and are adjacent, , . If and , , let be all of the vertices incident with the faces . Then and is uniquely 3colorable, where denotes the set of the vertices incident with .
Proof The proof is by induction on . Let and be the vertices incident with , but not incident with , , . If , since , by Theorem 3.6 (iii), we know that is uniquely 3colorable. Suppose that . By hypothesis, is uniquely 3colorable. Since , by Theorem 3.6 (i), we have , namely . Therefore, by Theorem 3.6 (iii), we obtain that is uniquely 3colorable. ∎
For a planar graph , let and be two cycles of . and are dependent if there exists a sequence of cycles of such that and have common edges and , where , . Obviously, if and have a common edge, then they are dependent.
Lemma 4.3.
Let be a planar graph, . If any cycle of is dependent with at most 3cycles for and with at most 3cycles for , then .
Proof The proof is by contradiction. Let be a smallest counterexample to the lemma, then satisfies the conditions of the lemma and . Suppose that is not connected, let be a connected component of . If and , it is easy to see that . This is a contradiction. Otherwise, we assume w.l.o.g. that . Since any cycle of is dependent with at most 3cycles for and with at most 3cycles for , the same is true of and . By the minimality of , we have . Furthermore, if , then ; otherwise, . Therefore, , a contradiction.
Suppose that is connected. If contains a cut vertex , let be the vertex sets of the connected components of , respectively, and , . Obviously, satisfies the conditions of the lemma. If , then, by the minimality of , ; otherwise, , . Therefore, , a contradiction.
Now we assume that is 2connected. If contains no 3faces, then . Thus . By Euler’s Formula, we have . This contradicts the choice of . If contains exactly one 3face, then contains at least one 5face. Thus, and then . If contains at least two 3faces, then each 3face is dependent with at least two 5faces for is 2connected. We claim that , namely .
For any face , we set the initial charge of to be . We now use the discharging procedure, leading to the final charge , defined by applying the following rule:
RULE. Each 3face receives from each dependent 5face.
For any face , if , since is dependent with at least two 5faces, then . If , then . If , then . If , then, by hypothesis, . Therefore,