More on total monochromatic connection of graphs1footnote 11footnote 1Supported by NSFC No.11371205 and 11531011, and PCSIRT.

# More on total monochromatic connection of graphs1

## Abstract

A graph is said to be total-colored if all the edges and the vertices of the graph are colored. A total-coloring of a graph is a total monochromatically-connecting coloring (TMC-coloring, for short) if any two vertices of the graph are connected by a path whose edges and internal vertices on the path have the same color. For a connected graph , the total monochromatic connection number, denoted by , is defined as the maximum number of colors used in a TMC-coloring of . Note that a TMC-coloring does not exist if is not connected, in which case we simply let . In this paper, we first characterize all graphs of order and size with and , respectively. Then we determine the threshold function for a random graph to have , where is a function satisfying . Finally, we show that for a given connected graph , and a positive integer with , it is NP-complete to decide whether .

Keywords

: total-colored graph, total monochromatic connection, random graphs, NP-complete

AMS subject classification 2010

: 05C15, 05C40, 05C75, 05C80, 68Q17.

## 1 Introduction

In this paper, all graphs are simple, finite and undirected. We refer to the book [2] for undefined notation and terminology in graph theory. Throughout this paper, let and denote the order (number of vertices) and size (number of edges) of a graph, respectively. Moreover, a vertex of a connected graph is called a leaf if its degree is one; otherwise, it is an internal vertex. Let and denote the number of leaves and the number of internal vertices of a tree , respectively, and let is a spanning tree of and is a spanning tree of for a connected graph . Note that the sum of and is for any connected graph of order . A path in an edge-colored graph is a monochromatic path if all the edges on the path have the same color. An edge-coloring of a connected graph is a monochromatically-connecting coloring (MC-coloring, for short) if any two vertices of the graph are connected by a monochromatic path of the graph. For a connected graph , the monochromatic connection number of , denoted by , is defined as the maximum number of colors used in an MC-coloring of . An extremal MC-coloring is an MC-coloring that uses colors. Note that if and only if is a complete graph. The concept of was first introduced by Caro and Yuster [6] and has been well-studied recently. We refer the reader to [4, 10] for more details.

In [11], the authors introduced the concept of total monochromatic connection of graphs. A graph is said to be total-colored if all the edges and the vertices of the graph are colored. A path in a total-colored graph is a total monochromatic path if all the edges and internal vertices on the path have the same color. A total-coloring of a graph is a total monochromatically-connecting coloring (TMC-coloring, for short) if any two vertices of the graph are connected by a total monochromatic path of the graph. For a connected graph , the total monochromatic connection number, denoted by , is defined as the maximum number of colors used in a TMC-coloring of . Note that a TMC-coloring does not exist if is not connected, in which case we simply let . An extremal TMC-coloring is a TMC-coloring that uses colors. It is easy to check that if and only if is a complete graph. Actually, these concepts are not only inspired by the concept of monochromatic connection number but also by the concepts of monochromatic vertex connection number and total rainbow connection number of a connected graph. For details about them we refer to [5, 12, 13, 14]. From the definition of the total monochromatic connection number, the following results follow immediately.

###### Proposition 1.

[11] If is a connected graph and is a connected spanning subgraph of , then .

###### Theorem 1.

[11] For a connected graph , .

In particular, if is a tree. The authors [11] also showed that there are dense graphs that still meet this lower bound.

###### Theorem 2.

[11] Let be a connected graph of order . If satisfies any of the following properties, then .

The complement of is -connected.

is -free.

.

.

has a cut vertex.

Moreover, the authors [11] gave an example to show that the lower bound is not always attained.

###### Example 1.

[11] Let be a complete multipartite graph with and . Then .

Let be a connected graph and be an extremal TMC-coloring of that uses a given color . Note that the subgraph formed by the edges and vertices with color is a tree where the color of each internal vertex is [11]. Now we define the color tree as the tree formed by the edges and vertices with color , denoted by . If has at least two edges, the color is called nontrivial; otherwise, is trivial. We call an extremal TMC-coloring simple if for any two nontrivial colors and , the corresponding trees and intersect in at most one vertex. If is simple, then the leaves of must have distinct colors different from color . Moreover, a nontrivial color tree of with edges and internal vertices is said to waste colors. For the rest of this paper we will use these facts without further mentioning them. In addition, we list a helpful lemma below.

###### Lemma 1.

[11] Every connected graph has a simple extremal TMC-coloring.

This paper is organized as follows. In Section , we characterize all graphs with , respectively. In Section , we show that for any function satisfying , if , where , then is a sharp threshold function for the property ; if , then is a sharp threshold function for the property . In Section , we prove that for a given connected graph , and a positive integer with , it is NP-complete to decide whether .

## 2 Characterization of graphs with small or large tmc

In this section, we characterize all graphs with , respectively. We call a connected graph unicyclic, bicyclic, or tricyclic if or , respectively. Let denote the set of the trees with , where . Note that if is a connected graph with , then is either a path or a cycle.

###### Theorem 3.

Let be a connected graph. Then if and only if is a path.

Proof. If is a path, then . Hence it remains to verify the converse. Let be a connected graph with . By Theorem 1, we get that and then as . Since is a connected graph, it follows that and . Thus is a path. ∎

###### Theorem 4.

Let be a connected graph. Then if and only if or is a cycle except for .

Proof. If or is a cycle except for , then by Theorem 2. Conversely, let be a connected graph with . First, we have by Theorem 1. Since and , it follows that or . If , then and so . Otherwise, from Theorem 3 we have that is a cycle and since . ∎

###### Theorem 5.

Let be a connected graph. Then if and only if or , where ; see Figure 1.

Proof. If or , where , then has a cut vertex and so by Theorem 2. Hence it remains to verify the converse. Let be a connected graph with . First, we have by Theorem 1. Since and , it follows that or . If , then and so . If , then we have from Theorem 4 and so is a unicyclic graph with ; see Figure 1. If , then we have from Theorems 3 and 4. However, , a contradiction.∎

###### Theorem 6.

Let be a connected graph. Then if and only if , or , where ; see Figure 2.

Proof. It is easy to verify the sufficiency by Theorem 2. Next we just need to prove the necessity. Let be a connected graph with . First, we have by Theorem 1. Since and , it follows that or . If , then and so . If , we have that from Theorem 5 and so is a unicyclic graph with ; see in Figure 2. Similarly, from Theorems 4 and 5, we have that if and then is a bicyclic graph with except for since ; see in Figure 2. If , we have that from Theorems 3 and 4. ∎

Recall that if and only if . In fact, there does not exist a graph such that . We are given a connected graph with and a simple extremal TMC-coloring of . Then there must be two nonadjacent vertices in a nontrivial color tree. Since every nontrivial color tree wastes at least 2 colors, we get that . Hence in the following, we characterize all graphs having . Let be nonadjacent edges of , where . Given a graph , let denote the graph obtained from by deleting the edges of .

###### Theorem 7.

Let be a connected graph. Then if and only if .

Proof. Clearly, . Then . Conversely, let be a connected graph with . We are given a simple extremal TMC-coloring of . Suppose that consists of nontrivial color trees, denoted by . Since each nontrivial color tree wastes at least two colors, it follows that and . Thus, .∎

###### Theorem 8.

Let be a connected graph. Then if and only if is either or .

Proof. Note that and then . Note that is a spanning subgraph of . Then by Proposition 1. Now we just need to prove that . Let be a simple extremal TMC-coloring of . Suppose that consists of nontrivial color trees. Since there are two pairs of nonadjacent vertices in two nontrivial color trees or in a common nontrivial color tree, it wastes at least three colors and then . Hence .

Now it remains to verify the converse. Let be a connected graph with . We are given a simple extremal TMC-coloring of . Suppose that consists of nontrivial color trees, denoted by . Since each nontrivial color tree wastes at least two colors, we get that and . Thus, is a spanning subgraph of . From Theorem 7, it can be checked that is either or . ∎

###### Theorem 9.

Let be a connected graph. Then if and only if .

Proof. Clearly, and . Thus we have that . If , there are three pairs of nonadjacent vertices and let be a simple extremal TMC-coloring of . Suppose that consists of nontrivial color trees. Then it wastes at least 4 colors and so . Since is a spanning subgraph of , by Proposition 1. Thus we get that . Similarly, it can be verified that .

Conversely, let be a connected graph with . We are given a simple extremal TMC-coloring of . Suppose that consists of nontrivial color trees, denoted by . Since each nontrivial color tree wastes at least two colors, we get the following two cases.

Case 1. .

Then or . If , then is a spanning subgraph of . From Theorems 7 and 8, we obtain that is either or . If , then is a spanning subgraph of . From Theorems 7 and 8, we get that .

Case 2. .

Then . From Theorem 8, and have not a common leaf. Thus is a spanning subgraph of . Since and , we have that . ∎

## 3 Random graphs

Let denote the random graph with vertices and edge probability [1]. For a graph property and for a function , we say that satisfies almost surely if the probability that satisfies tends to as tends to infinity. We say that a function is a sharp threshold function for the property if there are two positive constants and so that satisfies almost surely for all and almost surely does not satisfy .

Let and be two graphs on vertices. A property is said to be monotone if whenever and satisfies , then also satisfies . It is well-known that all monotone graph properties have sharp threshold functions; see [3] and [8]. For any graph with vertices and any function , having is a monotone graph property (adding edges does not destroy this property), so it has a sharp threshold function. In the following, we establish a sharp threshold function for the graph property .

###### Theorem 10.

Let be a function satisfying . Then

 p=⎧⎨⎩f(n)+nloglognn2if lnlogn≤f(n)<12n(n−1)+n, where l∈R+,lognnif f(n)=o(nlogn). (1)

is a sharp threshold function for the property .

###### Remark 1.

Note that if , then is a complete graph and . Hence we only concentrate on the case .

Before proving Theorem 10, we need some lemmas.

###### Lemma 2.

[7] Let . Then

 Pr[G(n,p) is connected]→⎧⎨⎩ee−aif |a|=O(1),0a→−∞,1a→+∞. (2)
###### Lemma 3.

[1] (Chernoff Bound) If is binomial variable with expectation , and , then

 Pr[X<(1−δ)μ)]≤exp(−δ2μ2)

and

 Pr[X>(1+δ)μ)]≤exp(−δ2μ2+δ).
###### Lemma 4.

Let be a noncomplete connected graph of order with minimum degree . Then .

Proof. For a noncomplete graph , we have that whose proof is contained in the proof of Theorem 6 in [11]. Moreover, by Proposition 12 in [6]. Thus .∎

Proof of Theorem 10: We divide our proof into two cases according to the range of .

Case 1. , where .

We first prove that there exists a constant such that the random graph with almost surely has . Let

 C={5if nlogn≤f(n)<12n(n−1)+n,5lif f(n)=lnlogn, where 0

It is easy to check that is almost surely connected by Lemma 2. Let denote the expectation of the number of edges in . Then

 μ1=n(n−1)2⋅Cp=C2(n−1nf(n)+(n−1)loglogn).

Moreover from Lemma 3, it follows that . Suppose that . By Theorem 1 we have that for sufficiently large,

 tmc(G(n,Cp))≥|E(G(n,Cp))|−n+2+l(G)≥μ12−n+2+l(G)
 =C4(n−1nf(n)+(n−1)loglogn)−n+2+l(G)
 ≥54(n−1nf(n)+(n−1)loglogn)−n+2+2
 ≥f(n).

Thus, we conclude that holds with the probability at least .

Next we show that there exists a constant such that the random graph with almost surely has . Let and denote the expectation of the number of edges in . Then we have

 μ2=n(n−1)2⋅cp=12(n−1nf(n)+(n−1)loglogn).

Furthermore by Lemma 3, it follows that . If is not connected, then . Otherwise, let be the minimum degree of . Suppose that . From Lemma 4, we have that for sufficiently large,

 tmc(G(n,cp))≤|E(G(n,cp))|−n+δ+1+l(G)≤32μ2−n+δ+1+l(G)
 =34(n−1nf(n)+(n−1)loglogn)−n+δ+1+l(G)
 <34(n−1nf(n)+(n−1)loglogn)−n+n+1+n−1


Hence, we conclude that holds with the probability at least .

Case 2. .

Let and . By Lemma 2, we have that is almost surely connected and is almost surely not connected. It can be checked that almost surely holds in a similar way as Case 1. On the other hand, since is almost surely not connected, almost surely holds. ∎

## 4 Hardness result for computing tmc

Given a graph , a set is called a dominating set of if every vertex of not in has a neighbor in . If the subgraph induced by is connected, then is called a connected dominating set. The connected dominating number, denoted by , is the minimum cardinalities of the connected dominating sets of . Note that the sum of and is because a vertex subset is a connected dominating set if and only if its complement is contained in the set of leaves of a spanning tree. In this section, we mainly prove the following result.

###### Theorem 11.

The following problem is NP-complete: Given a connected graph and a positive integer , decide whether .

In order to prove Theorem 11, we need the lemma as follows.

###### Lemma 5.

[15] The first problem defined below is polynomially reducible to the second one:
Problem 1. Given a graph and a positive integer , decide whether there is a dominating set of size or less.
Problem 2. Given a connected graph with a cut vertex and a positive integer with , decide whether there is a connected dominating set of size or less.

Proof of Theorem 11: Given a connected graph with a cut vertex, and a positive integer . Note that if and only if by Theorem 2. Then Problem 2 can be polynomially reducible to Problem 3: given a connected graph with a cut vertex and a positive integer with , decide whether . Thus, Problem 1 can be reducible to Problem 3 by Lemma 5. Moreover, Problem 1 is known as a NP-complete problem in [9]. Hence the problem in Theorem 11 is NP-hard.

Next we prove that given a connected graph and a nonnegative integer , to decide whether is NP. Recall that a problem belongs to NP-class if given any instance of the problem whose answer is “yes”, there is a certificate validating this fact which can be checked in polynomial time. For any fixed integer , to prove the problem of deciding whether is NP, we choose a TMC-coloring of with colors as a certificate. For checking a TMC-coloring with colors, we only need to check that colors are used and for any two vertices and of , there exists a total monochromatic path between them. Notice that for any two vertices and of , there are at most paths of length , since if we let , then there are less than choices for each . Clearly, the path wastes at least colors. Then and so which implies that . Therefore, contains at most - paths of length at most . Then, check these paths in turn until one finds a path whose edges and internal vertices have the same color. It follows that the time used for checking is at most . Since is a fixed integer, we conclude that the certificate can be checked in polynomial time. Then the problem of deciding whether belongs to NP-class and so is the problem in Theorem 11.

Therefore, the proof is complete. ∎

###### Corollary 1.

Let be a connected graph. Then computing is NP-hard.

### Footnotes

1. Supported by NSFC No.11371205 and 11531011, and PCSIRT.

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