Total proper connection of graphs1footnote 11footnote 1Supported by NSFC No.11371205 and 11531011, and PCSIRT.

# Total proper connection of graphs111Supported by NSFC No.11371205 and 11531011, and PCSIRT.

Hui Jiang, Xueliang Li, Yingying Zhang
Center for Combinatorics and LPMC-TJKLC
Nankai University, Tianjin 300071, China
E-mail: jhuink@163.com; lxl@nankai.edu.cn;
zyydlwyx@163.com
###### Abstract

A graph is said to be total-colored if all the edges and the vertices of the graph is colored. A path in a total-colored graph is a total proper path if any two adjacent edges on the path differ in color, any two internal adjacent vertices on the path differ in color, and any internal vertex of the path differs in color from its incident edges on the path. A total-colored graph is called total-proper connected if any two vertices of the graph are connected by a total proper path of the graph. For a connected graph , the total proper connection number of , denoted by , is defined as the smallest number of colors required to make total-proper connected. These concepts are inspired by the concepts of proper connection number , proper vertex connection number and total rainbow connection number of a connected graph . In this paper, we first determine the value of the total proper connection number for some special graphs . Secondly, we obtain that for any -connected graph and give examples to show that the upper bound is sharp. For general graphs, we also obtain an upper bound for . Furthermore, we prove that for a connected graph with order and minimum degree . Finally, we compare with and , respectively, and obtain that for any nontrivial connected graph , and that and can differ by for .

Keywords

: total-colored graph, total proper connection, dominating set

AMS subject classification 2010

: 05C15, 05C40, 05C69, 05C75.

## 1 Introduction

In this paper, all graphs considered are simple, finite and undirected. We refer to the book [2] for undefined notation and terminology in graph theory. A path in an edge-colored graph is a proper path if any two adjacent edges differ in color. An edge-colored graph is proper connected if any two vertices of the graph are connected by a proper path of the graph. For a connected graph , the proper connection number of , denoted by , is defined as the smallest number of colors required to make proper connected. Note that if and only if is a complete graph. The concept of was first introduced by Borozan et al. [3] and has been well-studied recently. We refer the reader to [1, 5, 8, 11] for more details.

As a natural counterpart of the concept of proper connection, the concept of proper vertex connection was introduced by the authors [7]. A path in a vertex-colored graph is a vertex-proper path if any two internal adjacent vertices on the path differ in color. A vertex-colored graph is proper vertex connected if any two vertices of the graph are connected by a vertex-proper path of the graph. For a connected graph , the proper vertex connection number of , denoted by , is defined as the smallest number of colors required to make proper vertex connected. Especially, set for a complete graph . Moreover, we have if is a noncomplete graph.

Actually, the concepts of the proper connection and proper vertex connection were motivated from the concepts of the rainbow connection and rainbow vertex connection. For details about them we refer to a book [10] and a survey paper [9]. Here we only state the concept of the total rainbow connection of graphs, which was introduced by Liu et al. [12] and also studied in [6, 13]. A graph is total-colored if all the edges and vertices of the graph are colored. A path in a total-colored graph is a total rainbow path if all the edges and internal vertices on the path differ in color. A total-colored graph is total rainbow connected if any two vertices of the graph are connected by a total rainbow path of the graph. For a connected graph , the total rainbow connection number of , denoted by , is defined as the smallest number of colors required to make total rainbow connected. Motivated by the concept of the total rainbow connection, now for the proper connection and proper vertex connection we introduce the concept of the total proper connection. A path in a total-colored graph is a total proper path if any two adjacent edges on the path differ in color, any two internal adjacent vertices on the path differ in color, and any internal vertex of the path differs in color from its incident edges on the path. A total-colored graph is total proper connected if any two vertices of the graph are connected by a total proper path of the graph. For a connected graph , the total proper connection number of , denoted by , is defined as the smallest number of colors required to make total proper connected. It is easy to obtain that if and only if is a complete graph, and if is not complete. Moreover,

 tpc(G)≥max{pc(G),pvc(G)}.                      (∗)

We can also extend the definition of the total proper connection to that of the total proper -connection in a similar way as the definitions of the proper -connection , proper vertex -connection and total rainbow -connection , which were introduced by Borozan et al. in [3], the present authors in [7] and Liu et al. in [12], respectively. However, one can see that when is larger very little have been known. Almost all known results are on the case for . So, in this paper we only focus our attention on the total proper connection of graphs, i.e., for the case .

The rest of this paper is organized as follows: In Section , we mainly determine the value of for some special graphs, and moreover, we present some preliminary results. In Section , we obtain that for any -connected graph and give examples to show that the upper bound is sharp. For general graphs, we also obtain an upper bound for . In Section , we prove that for a connected graph with order and minimum degree . In Section , we compare with and , respectively, and obtain that for any nontrivial connected graph , and that and can differ by for .

## 2 Preliminary results

In this section, we present some preliminary results on the total proper connection number and determine the value of when is a nontrivial tree, a complete bipartite graph and a complete multipartite graph.

###### Proposition 1.

If is a nontrivial connected graph and is a connected spanning subgraph of , then . In particular, for every spanning tree of .

###### Proposition 2.

Let be a connected graph of order that contains a bridge. If is the maximum number of bridges incident with a single vertex in , then .

Let denote the maximum degree of a connected graph . We have the following.

###### Theorem 1.

If is a tree of order , then .

Proof. Since each edge in is a bridge, we have by Proposition 2. Now we just need to show that . Let be the vertex with maximum degree and denote its neighborhood. Take the vertex as the root of . Define a total-coloring of with colors in the following way: Let be a vertex in . If , color and its incident edges with distinct colors from , and with the color from for . If , there exists a father of , say . Let denote the neighborhood of . Color the edges with distinct colors from , and the vertex with the color from for .

For any two vertices and in , let be a path from to , where . Next we shall show that there is a total proper path between and . If and are edge-disjoint, then ; otherwise, we walk from along to the earliest common vertex, say , and then switch to and walk to , i.e., . Thus, , and therefore, . ∎

The consequence below is immediate from Proposition 1 and Theorem 1.

###### Corollary 1.

For a nontrivial connected graph ,

 tpc(G)≤min{Δ(T)+1: T is a spanning tree of G}.

A Hamiltonian path in a graph is a path containing every vertex of and a graph having a Hamiltonian path is a traceable graph. We get the following result.

###### Corollary 2.

If is a traceable graph that is not complete, then .

Let denote a complete bipartite graph, where . Clearly, and if . For , we have the result below.

###### Theorem 2.

For , we have .

Proof. Let the bipartition of be and , where and . Since is not complete, it suffices to show that . Now we divide our discussion into two cases.

Case 1. .

We first give a total-coloring of with colors. Color the vertex and the edge with color , the vertex and the edge with color , and all the other edges and vertices with color . Then we show that there is a total proper path between any two vertices of . It is clear that and are total proper connected by an edge if they belong to different parts of the bipartition. Next we consider that and are in the same part of the bipartition. For , we have . For , if one of them is , then ; otherwise, .

Case 2. .

Similarly, we first give a total-coloring of with colors. Color the vertices and edges of the cycle starting from in turn with the colors . For and , color with color , with color , and all the other edges and vertices with color . Now we show that there is a total proper path between any two vertices of . It is clear that and are total proper connected by an edge if they belong to different parts of the bipartition. For , we have . For , we have . It can be checked that and are total proper connected in all other cases.

Therefore, the proof is complete.∎

Since any complete multipartite graph has a spanning complete bipartite subgraph, we obtain the following corollary.

###### Corollary 3.

If is a complete multipartite graph that is neither a complete graph nor a tree, then .

## 3 Connectivity

In this section, we first prove that for any -connected graph . Also we show that this upper bound is sharp by presenting a family of a -connected graphs. Finally, we state an upper bound of for general graphs.

Given a colored path between any two vertices and , we denote by the color of the first edge in the path, i.e., , and by the last color, i.e., . Moreover, let be the color of the first internal vertex in the path, i.e., , and be the last color, i.e., . If is just the edge , then , and .

###### Definition 1.

Let be a total-coloring of that makes total proper connected. We say that has the strong property if for any pair of vertices , there exist two total proper paths between them (not necessarily disjoint) such that and for , and both and are -sets.

Let be a connected graph and be a spanning subgraph of . We say that is a spanning minimally -connected subgraph of if the removal of any edge from would leave -connected.

###### Theorem 3.

Let be a -connected graph. Then and there exists a total-coloring of with colors such that has the strong property.

Proof. Let be a spanning minimally -connected subgraph of . We apply induction on the number of ears in an ear-decomposition of . The base case is that is simply a cycle . Obviously, and for . Next define a total-coloring of with colors by

 c(vivi+1)=⎧⎨⎩1,if i is odd,1≤i≤2k−1 for n=2k or n=2k+12,if i is even,2≤i≤n for n=2k or n=2k+14,if i=2k+1 for n=2k+1 (1)

and

 c(vi)=⎧⎨⎩3,if i is odd,1≤i≤2k−1 for n=2k or n=2k+14,if i is even,2≤i≤2k for n=2k or n=2k+11,if i=2k+1 for n=2k+1. (2)

Clearly, the total-coloring makes have the strong property.

In an ear-decomposition of , let be the last ear with at least one internal vertex since is assumed to be minimally -connected. And denote by the graph after removal of the internal vertices of . Let and be the vertices of and then . By induction hypothesis, there exists a total-coloring of with colors such that is total proper connected with the strong property. We give such a total-coloring to . Then there exist two total proper paths and from to such that and for , and both and are -sets. Let . Color the edge with the color from , and then total-properly color from to so that and . If , it will become clear that this is the easier case, and so we consider the case that in the following.

Without loss of generality, suppose that . We will show that is total proper connected with the strong property under this coloring. For any two vertices of , there exist two total proper paths connecting them with the strong property by induction hypothesis. Since forms a total proper connected cycle, any two vertices in this cycle also have the desired paths. Assume that and . Next we will show that there are two total proper paths from to with the strong property.

Since , there exist two total proper paths and starting at and ending at with the strong property. Analogously, there exist two total proper paths and starting at and ending at with the strong property. Since these paths have the strong property, suppose that and are total proper paths. If , then and are the desired pair of paths. Thus, assume that .

Then there exists a total proper walk for some (suppose ). If is a path, then and are the desired two paths. Otherwise, let denote the vertex closest to on which is in . Now consider the path . If is a total proper path, then and are the desired two paths, and so we suppose that . Since and are total proper paths, , and . Then . Let and . Obviously, and are two total proper paths. Note that . Thus, and have the strong property. Since by Proposition 1, we have and there exists a total-coloring of with 4 colors such that has the strong property. This completes the proof of Theorem 3. ∎

In order to show that the bound obtained in Theorem 3 is sharp, we give a family of -connected graphs with (see Figure 1).

###### Proposition 3.

Let be the graph obtained from an even cycle by adding two ears which are as long as their interrupting segments respectively, such that each segment has () edges. Then .

Before proving Proposition 3, we give the following fact.

###### Fact 1.

Let . If there exists a total-coloring of with three colors such that there are two total proper paths and where and , then .

Proof of Proposition 3: Since by Theorem 3, we just need to prove that . Assume that there is a total-coloring of with colors such that is total proper connected. Label the segments and some vertices of as in Figure 1, where and are the neighbours of the vertex for .

Firstly, we shall show that the segments and are two total proper paths. If one of them is not, say , then there is no total proper path in from to or from to (say from to ). Hence there exists a total proper path through connecting and , suppose (this assumption, as opposed to using any of or , does not lose any generality). Next we consider the total proper path between and . Then there must exist a total proper path using the segments or . If there is a total proper path , then . Thus the total proper path between and is unique, i.e., , and then . However, we can not find a total proper path from to , a contradiction. If there is a total proper path , then . Thus the total proper path connecting and is unique, i.e., . Then and are two total proper paths in which is an even cycle of length , which contradicts Fact 1. Hence there is no total proper path from to , a contradiction. Therefore, the segments and are two total proper paths.

Secondly, we will show that at least one of or must be total proper (and similarly, at least one of or ). Suppose both and are not total proper. Then and are total proper connected by a path through or , say . However, we can not find a total proper path connecting and in a similar discussion above, which is impossible. Thus, suppose and are total proper without loss of generality.

Finally, we know that at least one of the paths and must be not total proper by Fact 1. As we have shown, the only place which we can not go through is at the intersections, and so assume that the path is not total proper, where and . In the following, we consider the total proper path from to and divide our discussion into two cases:

Case 1. is or .

Between and , there must exist a total proper path . If is , then . Hence, there is only one total proper path from to ; otherwise and are two total proper paths in the cycle for a contradiction. Then it follows that . Similarly, we can deduce that there is no total proper path connecting and , which is impossible. If is , then . In a similar discussion, we obtain that the total proper path from to is unique, i.e., . Then and are two total proper paths in , a contradiction. If is or , then and are two total proper paths in , which again contradicts Fact 1.

Case 2. is .

Consider the total proper path from to , where . If is , then we can prove that this subcase could not happen in a similar way as Case 1. If is , then . From to , there is only one total proper path since we can not go through . However, and are two total proper paths in for a contradiction.

The proof is thus complete. ∎

###### Remark 1.

Remember that for a -connected graph , we have that the proper connection number ; see [3]. But, if we consider a -connected bipartite graph , then we have that . That means that the bipartite property can lower down the number of color by 1. However, from Proposition 3 we see that the bipartite property cannot play a role in general to lower down the number of colors for the total proper connection number, since the graphs in Proposition 3 are bipartite but their total proper connection numbers reach the upper bound .

Finally, we prove an upper bound of for general graphs.

###### Theorem 4.

Let be a connected graph and denote the maximum degree of a vertex which is an endpoint of a bridge in . Then if and otherwise.

In order to prove Theorem 4, we need a lemma below. Let denote the set of colors presented on the vertex and edges incident to .

###### Lemma 1.

Let be a graph obtained from a block with by adding nontrivial blocks and pendant edges at for . Consider as the maximum degree of a vertex which is an endpoint of a bridge in . Then .

Proof. Let and . We give a total-coloring of using as follows.

Step 1. If is a trivial block, then we give a total-coloring with 3 colors to such that are different from each other; otherwise, we give a total-coloring with 4 colors to that makes it have the strong property by Theorem 3. Let modulo for .

Step 2. For , if , then we give a total-coloring with 4 colors from to each uncolored nontrivial block at , denoted by , that makes each of them have the strong property by Theorem 3; afterwards if , then color uncolored pendant edges at , denoted by , with distinct colors from and then color each pendant vertex using for .

Next we show that is total proper connected under the coloring . If is a nontrivial block, then each pair of the vertices in has two total proper paths between them with the strong property. It will become clear that this is the easier case so we consider the case that is a trivial block. Let and be two vertices of . It is obvious that there exists a total proper path connecting them if both belong to the same block. Suppose that and for . If and , then there exist two paths and from to with the strong property. We know that is a total proper path for some (suppose ). Similarly, there exists a total proper path from to where is a total proper path connecting and . Thus, we can find a total proper path between and . If and , then is a total proper path under the coloring . For the other cases, it can be checked that there exists a total proper path connecting and in a similar way. Therefore, . ∎

Now we are ready to prove Theorem 4.

Proof of Theorem 4: Let be the blocks of and denote the block graph of with vertex set . Now, we consider a breadth-first search tree (BFS-tree) of with root and suppose that the blocks have an order . Let and . We will give a total-coloring using in the following.

We give a total-coloring to and its neighbor blocks of in a similar way as in Lemma 1. Then we can get that is total proper connected if there are no more blocks in . Hence, suppose that there are uncolored blocks in . We extend our coloring from in a Breadth First Search way until there is no more blocks in , i.e., if has uncolored neighbor blocks, we give a total-coloring to its uncolored neighbor blocks of in a similar way as Step 2; otherwise, consider .

Now we prove that is total proper connected. Let and be two vertices in . It is obvious that there exists a total proper path between them if both belong to the same block. Suppose that and . Let denote the path from to in the BFS-tree . Then we can find a total proper path from to traversing the blocks on under the coloring . Therefore, if and otherwise. ∎

## 4 Minimum degree

In this section, we prove the following result concerning the minimum degree.

###### Theorem 5.

Let be a connected graph of order with minimum degree , then .

Given a graph , a set is called a two-step dominating set of if every vertex in which is not dominated by has a neighbor that is dominated by . Moreover, a two-step dominating set is called a two-way two-step dominating set if every pendant vertex of is included in , and every vertex in has at least two neighbors in , where denotes the set of all vertices at distance exactly from . Further, if is connected, is called a connected two-way two-step dominating set of .

###### Lemma 2.

[4] Every connected graph of order and minimum degree has a connected two-way two-step dominating set of size at most .

Proof of Theorem 5: The proof goes similarly as that of the main result in [11] by Li et al.

We are given a connected graph of order with minimum degree . The assertion can be easily verified for and so suppose . Let denote a connected two-way two-step dominating set of and . Then we have by Lemma 2. Let is a neighbor of in for and is a neighbor of in for .

Case 1. For each vertex , its neighbors in has at least one common neighbor in , i.e., .

We consider a spanning subgraph of (see Figure 2, where denotes the union of graphs each of which is isomorphic to the graph and similarly for and ). Next, we give a total-coloring to using . For the edges and vertices of , let be a spanning tree of . Then by Theorem 1, can be total-colored using such that for each edge , the colors of and are different from each other. We color in such a way and the edges of with any used colors (denote this coloring of by ). For the other edges and vertices in , color them as depicted in Figure 2.

Since each pair of vertices has a total proper path connecting them such that and , it suffices to show that is total proper connected in the assumption that . Take any two vertices and in . If , then has a neighbor in and similarly has a neighbor in . Hence, if , is a total proper path; otherwise, is a total proper path where is another neighbor of in . It is easy to check that and are total proper connected in all other cas