1 Introduction
All graphs considered in this paper are undirected, finite and simple. We follow [2] for notation and terminology not described here. Let be a graph. We use to denote the vertexset, edgeset, the number of vertices, number of edges, and the diameter of , respectively.
A path of an edgecolored graph is said to be a rainbow path if no two edges on the path have the same color. The graph is called rainbow connected if every pair of distinct vertices of is connected by a rainbow path in . An edgecoloring of a connected graph is called a rainbow connection coloring if it makes the graph rainbow connected. For a connected graph , the rainbow connection number of is defined as the smallest number of colors that are needed in order to make rainbow connected. These concepts on the rainbow connection of graphs were introduced by Chartrand et al. [10] in 2008. The reader is referred to [22, 23, 24] for details.
Inspired by the rainbow connection colorings and proper colorings in graphs, Borozan et al. [3] in 2012, and Andrews et al. [1] independently, introduced the concept of properpath coloring. Let be a nontrivial connected graph with an edgecoloring. A path in is called a proper path if no two adjacent edges of the path receive the same color. An edgecoloring of a connected graph is called a properpath coloring if every pair of distinct vertices of are connected by a proper path in . If colors are used, then is called a properpath coloring. An edgecolored graph is proper connected if any two vertices of are connected by a proper path. For a connected graph , the minimum number of colors that are needed in order to make proper connected is called the proper connection number of , denoted by . For more details, we refer to [14, 16, 19, 12] and a dynamic survey [20].
Czap et al. [11] in 2018 introduced the concept of conflictfree connection of graphs. An edgecolored graph is called conflictfree connected if each pair of distinct vertices is connected by a path which contains at least one color used on exactly one of its edges. This path is called a conflictfree path, and this coloring is called a conflictfree connection coloring of . The conflictfree connection number of a connected graph , denoted by , is defined as the smallest number of colors needed to color the edges of such that is conflictfree connected. More results can be found in [6, 7, 8].
Note that a rainbow path has the maximum number of colors allowed on the path. Inspired by the concept of rainbow connection, we would like to ask what about a path uses the minimum number of colors ? This will be the minimum cost when we try to establish a point to point communication. As mentioned by Caro and Yuster [5], it is a natural opposite counterpart of the rainbow connection colorings. At first, we introduce the notion colordistance in the following.
Let be a nontrivial connected graph associated with an edgecoloring , . Let and be two vertices of the edgecolored graph . The colordistance between and , denoted by , is the minimum number of distinct colors that a path can have. It is easy to see that ; if and only if ; ; . Then is a metric space. We can also define colordistance between two sets of vertices: . Naturally, the colordiameter of , denoted by , is the maximum colordistance among all pairs of vertices. We have . The question is, when we fix the colordiameter, how colorful can a coloring be ? That is to say, we already have , and then we tend to maximize the number of distinct colors . We call a color connection coloring if . The color connection number of a connected graph , denoted by , is the maximum number of colors that are needed to make a color connection coloring. We call an extremal color connection coloring if uses colors.
If , then is exactly the monochromatic connection number introduced by Caro and Yuster [5] in 2011. More results on the monochromatic connection of graphs can be found in [25, 4, 13, 15, 26].
Now, for a given a graph we have got four parameters , , and . The question is that for a given graph, weather there exist efficient ways to compute these parameters ? Chakraborty et al. [9] first solved this problem for the rainbow connection number . They proved that computing is NPhard. However, for a long time it has been tried to fix the computational complexity for , and of a graph. Till now it has not been solved, yet. Only the complexity results for the strong version, i.e., the strong proper connection number [17] and the strong conflictfree connection number [18], were determined to be NPhard. In this paper, we prove that to compute each of , and of a graph is NPhard. This solves a long standing problem in this field, asked in many talks of workshops and papers; see [20, 21, 25, 7, 8, 18].
The paper is organized as follows. In Sections 2,3 and 4, we prove the NPhardness for the proper connection number, conflictfree connection number and monochromatic connection number, respectively. In the last section, Section 5, we work on the color connection number and give a lineartime algorithm to compute the color connection number of trees.
2 NPhardnes for the proper connection number
The classical 3Satisfiability problem (3SAT for short) is wellknown to be NPcomplete. Its variant NotAllEqual 3 Satisfiability problem (NAE3SAT for short) was also proved to be NPcomplete by Schaefer’s dichotomy theorem [27]. For ease of reading, we state the variant problem as follows.
The NAE3SAT problem:
INSTANCE: A boolean formula in 3 conjunctive normal form (3CNF for short) such that each clause is made up of three distinct literals.
QUESTION: Is there a satisfiable assignment for such that in each clause, at least one literal is equal to true and at least one literal is equal to false ?
Our aim is to reduce the NAE3SAT problem to the problem of deciding if for a graph . We then obtain our main result of this section.
Theorem 2.1
For a graph , deciding if is NPcomplete. In particular, computing is NPhard.
Proof. At first, we show that deciding if is in NP. Actually, Edmonds and Manoussakis showed that to check if a 2edgecoloring of a graph makes the graph proper connected is polynomialtime solvable; see [21]. From their result, one can see that given a yesinstance of a 2edgecolored connected graph, it is polynomialtime checkable if the graph is proper connected. Then we prove that the problem is NPcomplete by reducing the NAE3SAT problem to it.
Given a 3CNF on variables , the clause is made up of three distinct literals for each . We construct a graph as follows:
First, we construct the variable garget for a variable for each (see Figure 1). Assume that the variable appears in clauses of . For each , we start from a path . Next we add pendent vertices as follows: each of the pendent vertices and is adjacent to , the pendent vertex is adjacent to , the pendent vertex is adjacent to , and each of the pendent vertices and is adjacent to . Then we add a new vertex , and a set of edges . Now we get the graph . The variable garget is obtained by adding a set of edges to the union of the graphs .
Second, we construct the graph represent for the clause for each , see Figure 2. We start from the vertices and such that there are three edgedisjoint paths , and between them. Next we add a new vertex such that is adjacent to . Then we add two pendent vertices and such that each of and is adjacent to . Then we add three vertices such that is adjacent to , is adjacent to , and is adjacent to . If a variable is positive such that in the clause for some , then we add an edge for some . If a variable is negative such that in the clause for some , then we identity the vertices and for some . We do this similarly for the literals and .
Third, we add some extra vertices and edges to complete the construction of the graph . We construct a complete graph on the set of vertices . For each , we add a new vertex such that is adjacent to , and add two pendent vertices and such that each of and is adjacent to . Next we add a set of edges . Now we get the graph .
In the following, we only need to show that if and only if is satisfiable.
Suppose . Let be a properpath coloring of using 2 colors.
Claim. The three entries , and of share a same color for each .
Proof of the Claim: Assume, to the contrary, that the three entries , and of do not share a same color. Notice that is the only path between the vertices and (see Figure 1). Then we can assume that the edge is colored with 1 and the edge is colored with 2. Similarly, we can assume that the edges , are colored with 1 and the edges , are colored with 2. We distinguish the following two cases to complete the proof.
Case 1. The two entries and share a same color.
Without loss of generality, we can assume that the entries , are colored with 1, and the third entry is colored with 2. Since there exists a proper path between vertices and , we can assume that the proper path is . Since there is a proper path between vertices and , it follows that the edge is colored with 1. Notice that there exists a proper path between vertices and , then the edge is colored with 2. There is a proper path between vertices and , then the edge is colored with 1. Also there is a proper path between vertices and , then the edge is colored with 1. However, there is no proper path between vertices and . Thus, is not a properpath coloring, a contradiction.
Case 2. The two entries and do not share a same color.
Without loss of generality, we assume that the entries is colored with 1, and the other two entries , are colored with 2. Then the proper path between vertices and is . Since there is a proper path between vertices and , it follows that the edge is colored with 1. Notice that there exists a proper path between vertices and , then the edge is colored with 2. Suppose that the edge is colored with 2. Then the edge is colored with 1. There is no proper path between vertices and , then is not a properpath coloring, a contradiction. We have that the edge is colored with 1. Since there is a proper path between vertices and , the edge is colored with 1. Also, there is a proper path between vertices and , then the edge is colored with 1. However, there is no proper path between vertices and . Thus, is not a properpath coloring, a contradiction.
As a result, the three entries , and of share a same color for each .
Now we have that the edges in the edge set share a same color for each . If the edge is colored with 1, then we set for each . If the edge is colored with 2, then we set for each . Let be two vertices, be a path from to such that is in a graph for some . Then the length of is even. Thus, two of the four edges should be colored with 1, and the other two edges should be colored with 2 for each . Then admits a NotAllEqual satisfiable assignment for the clause . As a result, we get a satisfiable assignment for .
Suppose that is satisfiable. Given a NotAllEqual satisfiable assignment on the variables , we color the graph as follows:
If , then is colored with 1. For each , the edges , , , , , are colored with 1, and the remaining edges in the graph is colored with 2. If , then is colored with 2. For each , the edges , , , , , are colored with 2, and the remaining edges in the graph is colored with 1. For each , the edge shares a same color with the edge , and the edges , receive the distinct color. If a variable is positive in the clause for some , then the edge shares a same color with the edge and the edge receives a distinct color. If a variable is negative such that in the clause for some , then the edge and the edge receive distinct colors. The coloring is similar for the literals and . Without loss of generality, we can assume that the edges and are colored with 1, the edge is colored with 2, since is satisfiable. Then the edges , , , , , , are colored with 1, the edges , , , , , are colored with 2. Let be a vertex in . Then there exists a unique vertex such that is adjacent to . If the edge is colored with 1, then we color the edge with 2. If the edge is colored with 2, then we color the edge with 1. Let be two vertices. If the edges and share a same color 1, then we color the edge with 1. If the edges and share a same color 2, then we color the edge with 2. The remaining edges can be arbitrarily colored.
One can verify that there exists a proper path between every pair of vertices in the graph . Thus, . The proof is now complete.
3 NPhardness for the conflictfree connection number
Let be the subgraph of induced by the set of cutedges of , and let . The following result was obtained in [11].
Lemma 3.1
In the following, our aim is to reduce the NAE3SAT problem to the problem of deciding whether or for the case when . We then know that deciding whether or is NPcomplete by Theorem 3.4. At first, we need the following lemma.
Lemma 3.2
Let be a graph such that contains two edgedisjoint Hamilton paths and . The edges of is colored with 1 and the edges of is colored with 2. Then between any pair of vertices of , there exists a monochromatic color 1 path, a monochromatic color 2 path, a conflictfree path with color 1 used only once and a conflictfree path with color 2 used only once.
Proof. Let and be two vertices of . Clearly, and lie on , then the path is a monochromatic color 1 path. Similarly, the path is a monochromatic color 2 path. Now consider the path . Let be the edge incident with on the path . Then contains two component and ( lies on ). Since is a Hamilton path, there exists an edge on such that . Now we have that the path is a conflictfree path with color 1 used only once. Similarly, there exists a conflictfree path with color 2 used only once.
Next we need the lemma given by M Ji et al. [18] as following:
Lemma 3.3
[18] Given a connected graph and a coloring : of , determining whether or not is, respectively, conflictfree connected, conflictfree vertexconnected, strongly conflictfree connected under the coloring can be done in polynomialtime.
Our main result of this section is as follows.
Theorem 3.4
Let be a graph such that . Then deciding whether or 3 is NPcomplete. In particular, computing is NPhard.
Proof. As a result of Lemma 3.3, the problem is in NP. Then we prove that the problem is NPcomplete by reducing the NAE3SAT problem to it. Given a 3CNF on variables , the clause is made up of three distinct literals for each . We construct a graph as follows.
First, we construct the variable garget for a variable for each . We start from a complete graph on four vertices . Next we add two pendent vertices such that each of the vertices is adjacent to . Then we get by adding a cycle of length 5.
Second, we construct the graph representing for the clause for each . We start from a complete graph on four vertices . Next we add two pendent vertices and such that each of and is adjacent to . Let the resulting graph be . Then we add two copies and of . Finally, we get by adding the set of edges to the union of , and . If a literal is positive in the clause for some , then is adjacent to . If a literal is negative in the clause for some , then is adjacent to . We do this similarly for the literals and .
Third, we add some extra vertices and edges to complete the construction of the graph . We construct a complete graph on vertices . Next we add two vertices and such that is a path of length two. Then the resulting graph is .
In the following, we only need to show that if and only if is satisfiable.
Suppose . Notice that is the only path between the vertices and (see Figure 4). Then we can assume that the edge is colored with 1 and the edge is colored with 2. Similarly, we can assume that the edges , are colored with 1 and the edges , are colored with 2. We can assume that is colored with 1 and is colored with 2, since is the only path between vertices and (see Figure 3). There exists a monochromatic color 1 path between vertices and , since color 1 has been used at least twice on the conflictfree path between vertices and . Similarly, there exists a monochromatic color 2 path between and . The graph has exact three entries, then these three edges can not have a same color; otherwise, we can not get both a monochromatic color 1 path and a monochromatic color 2 path between and .
Claim. There exists no monochromatic path going through some .
Proof of the Claim: Suppose that there exists a monochromatic color 1 path going through some two entries of . Then the third entry should be colored with 2. There exists a monochromatic color 1 path and a monochromatic color 2 path between vertices and , since any path between and goes through the vertex . Similarly, there exists a monochromatic color 1 path and a monochromatic color 2 path between vertices and ; there exists a monochromatic color 1 path and a monochromatic color 2 path between vertices and . Then we have that the edge is colored with 2. Next, we color the edge with 1 and the edge with 2. Hence, there exists no monochromatic color 1 path going through for any . Similarly, there exists no monochromatic color 2 path going through some .
Notice that is the only path between vertices and (see Figure 4). Then we can assume that the edge is colored with 1 and the edge is colored with 2. We have that there exists a monochromatic color 1 path and a monochromatic color 2 path between vertices and . Then the two edges and are colored with distinct colors by our Claim. If (, respectively) is incident with the entry (, respectively) of , then the edge (, respectively) and the edge (, respectively) share a same color to ensure the monochromatic paths between vertices and . Similar thing happens for the remaining two entries of . If is colored with 1, then we set . If is colored with 2, then we set . As a result, the variables admit a NotAllEqual satisfiable assignment of .
Suppose that is satisfiable. Given a NotAllEqual satisfiable assignment on variables , we color the graph as follows.
If , then the edges incident with the vertex are colored with 1 and the edges incident with the vertex is colored with 2 for each . If , then the edges incident with the vertex are colored with 2 and the edges incident with the vertex is colored with 1 for each . The three entries of do not share a same color, since is satisfiable for each . Without loss of generality, assume that the entry incident with and the entry incident with are colored with 1, and the entry incident with is colored with 2 for each . Then we color the two edges and with 2, and color the edge with 1 for each . Next the edges , , are colored with 1, and the edges , , are colored with 2 for each . The edge is colored with 1 and the edge is colored with 2 for each . The remaining parts consist of complete subgraphs of order at least 4. By induction, we know that a complete graph of order at least 4 contains two edgedisjoint Hamilton paths. For each complete subgraph, one Hamilton path is colored with 1, and the other is colored with 2.
Next we should verify that there exists a conflictfree path between every pair of vertices. Notice that there exist both a monochromatic color 1 path and a monochromatic color 2 path between and for each . Let be two vertices. Then there exists a conflictfree path between and by Lemma 3.2. Let be two vertices for some . Then there exists a conflictfree path between and by Lemma 3.2. Let be two vertices for some . Then there exists a conflictfree path between and by Lemma 3.2. Let , be two vertices for some . Then there exists a conflictfree path between and by Lemma 3.2. Let , be two vertices for some . Then there exists a conflictfree path between and by Lemma 3.2. Without loss of generality, for some , assume that the entry incident with and the entry incident with are colored with 1, and the entry incident with is colored with 2. Then we color the two edges and with 2, and color the edge with 1. Let , be two vertices. Then we can find a conflictfree path from to with color 2 which is the unique color such that edge is in . Let , be two vertices. Then we can find a conflictfree path from to with color 2 which is the unique color such that edge is in . Let , be two vertices. Then we can find a conflictfree path from to with color 1 which is the unique color such that edge is in . As a result, we have that . The proof is now complete.
4 NPhardness for the monochromatic connection number
In the last section we will deal with the complexity for the monochromatic connection number. The following result was obtained in [5].
Theorem 4.1
[5] Let be a connected graph with vertices and edges. If satisfies one of the following conditions, then .

is 4connected.

is trianglefree.

. In particular, this holds if or .

.

has a cut vertex.
Inspired by the concept of a monochromatic path which uses exact one color, we consider a path using at most distinct colors. Next, we will use the colordistance as a tool. Given a positive integer and a graph , if one asks for an edgecoloring of satisfying that for any pair of vertices of there exists a path in which uses at most distinct colors between the two vertices, then the maximum number of distinct colors for the edgecoloring is . We can consider in an opposite way which counts the total number of ”wasted” colors in the following lemma.
Lemma 4.2
Let be an extremal color connection coloring of . Then each color forms a forest. For a color , denote by the subgraph induced by the set of edges colored with . We call the color forest of the color . A color forest with edges is said to waste colors. Hence, is said to be nontrivial if and only if . Let be the total number of wasted colors. Then .
Proof. We claim that the subgraph induced by the set of edges colored with is a forest. Assume to the contrary, there is a cycle in . Then it is possible to choose any edge in this cycle and color it with a fresh color while still maintaining a color connecting coloring, a contradiction.
Next, the color connection number counts the number of distinct color forests. Hence, we have .
Lemma 4.3
For a connected graph , .
Proof. By definition, we know that if and only if is at most . Let be a connected induced subgraph of . We can construct a color connecting coloring of by adding new colors to a color connecting coloring of , and hence . Then we have by Lemma 4.2.
Next we consider the upper bound of . Let be a spanning tree of graph . Then can be calculated in Section 4. Hence, is an alternative upper bound. In general, we have by our Algorithms 1 and 2 in next section. Moreover, achieves its upper bound if and only if is a path.
Lemma 4.4
Let be a positive integer. Given a connected graph associated with an edgecoloring on vertices, one can decide if in polynomial time.
Proof. Let be the color set of . Clearly, . Next we choose a pair of vertices in arbitrarily. Let be a color set chosen from such that . In this step, we have exact
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