On the bcontinuity of the lexicographic product of graphs
^{†}^{†}thanks: All authors are members of the ParGO Research Group  Parallelism, Graphs and Optimization.
This work was partially supported by CNPq and CAPES, Brazil.
Abstract
A bcoloring of the vertices of a graph is a proper coloring where each color class contains a vertex which is adjacent to each other color class. The bchromatic number of is the maximum integer for which has a bcoloring with colors. A graph is bcontinuous if has a bcoloring with colors, for every integer in the interval . It is known that not all graphs are bcontinuous. Here, we investigate whether the lexicographic product of bcontinuous graphs and is also bcontinuous. Using homomorphisms, we provide a new lower bound for , namely , where , and prove that if is bcontinuous for every positive integer , then admits a bcoloring with colors, for every in the interval . We also prove that is bcontinuous, for every positive integer , whenever a sparse graph, and we give further results on the bspectrum of , when is chordal.
Keywords:
bchromatic numberbcontinuitybhomomorphism chordal graphssparse graphs∎
1 Introduction
Given a simple graph ^{1}^{1}1The graph terminology used in this paper follows BM08 ()., and a function , we say that is a proper coloring of with colors if for every . A value is called color , while the subset is called color class . Graph colorings are a very useful model for situations in which a set of objects is to be partitioned according to some prescribed rules. For example, problems of scheduling Werra.85 (), frequency assignment Gamst.86 (), register allocation Chow.Hennessy.84 (); Chow.Hennessy.90 (), and the finite element method Saad.96 (), are naturally modelled by colourings. In these applications, one is interested in finding a proper coloring with the smallest number of colors. This motivates the definition of the chromatic number of , denoted , the smallest integer for which has a proper coloring with colors. Deciding if a graph admits a proper colouring with colours is an NPcomplete problem, even if is not part of the input Hol81 (). The chromatic number is also hard to approximate: for all , there is no algorithm that approximates the chromatic number within a factor of unless P = NP Has96 (); Zuc07 ().
One approach to obtain proper colorings of a graph is to use coloring heuristics. Consider a proper coloring of graph , for which we want to reduce the number of colors. A vertex in color class is called a bvertex of color if has at least one neighbor in color class , for every . If color has no vertices, we may recolor each in color class with some color that does not appear in the neighborhood of . In this way, we eliminate color , and obtain a new coloring for that uses colors. The procedure may be repeated until we reach a coloring in which every color contains a vertex. Such a coloring is called a coloring. Clearly, the described procedure cannot decrease the number of colors used in a proper coloring of with colors. Therefore, we are actually interested in investigating the worstcase scenario for the described procedure. This motivates the definition of the chromatic number of a graph , denoted , being the largest such that has a coloring with colors.
This concept was introduced by Irving and Manlove in IM99 (), where they prove that determining the chromatic number of a graph is an NPcomplete problem. In fact, it remains so even when restricted to bipartite graphs KTV02 (), connected chordal graphs HLS11 (), and line graphs CLMSSS15 ().
In KTV02 () it is proved that , the graph obtained from by removing a perfect matching, admits colorings only with or colors. And in BCF07 (), the authors prove that, for every finite , there exists a graph that admits a bcoloring with colors if and only if . These facts motivate the definition of bspectrum and of bcontinuous graphs, introduced in BCF03 (). The bspectrum of a graph , denoted by , is the set containing every positive value for which admits a bcoloring with colors; and is said to be bcontinuous if contains every integer in the closed interval . In the same article, they prove that interval graphs are continuous. This result was generalized for chordal graphs independently in F04 () and KKV.04 (). Other examples of bcontinuous graphs are cographs and sparse graphs BDM+09 (), as well as the more general class of tidy graphs VBK10 (); Kneser graphs for JaOm09 (); regular graphs with girth at least 6 and not containing cycles of length 7 BK.12 (); and, more recently, graphs with girth at least 10 SLS.16 ().
Given graphs and , the lexicographic product of by is the graph , where and if and only if either and or . Intuitively, is the graph obtained from and by replacing each vertex of with a copy of and adding every possible edge between two copies of if and only if the corresponding vertices of are adjacent. For every , we denote the copy of related to in by . The chromatic number of the lexicographic product of graphs was studied in JaPe12 (). In LVS15 (), which was coauthored by the authors, they considered the following question, which we continue to investigate in this paper.
Question 1
Is it true that is bcontinuous whenever and are?
We mention that a similar question is answered in the negative for the cartesian product and strong product of graphs. For the cartesian product, it suffices to observe that the cube , which is known to be nonbcontinuous, can also be viewed as the cartesian product of by . As for the strong product, consider and , . Observe that is isomorphic to , and as mentioned before, the spectrum of is .
In JaPe12 (), the authors show that , while in GS75 () it is shown that . Therefore, if is bcontinuous, then the closed interval must be contained in the bspectrum of . In LVS15 (), the authors show that this is the case when .
Theorem 1.1 (Lvs15 ())
If and are bcontinuous and , then .
In Section 2, we apply the concept of bhomomorphism to show that is in fact at least , where . This improves the lower bound given in JaPe12 (). We also show that if is bcontinuous for every integer , then contains every integer in the closed interval , which by what is said before, contains the interval . This shows that an important step to answer Question 1 is to first answer the particular case below. Observe that this is complementary to Theorem 1.1.
Question 2
Let be a bcontinuous graph and be any positive integer. Is bcontinuous?
In LVS15 (), the authors have answered this question positively for chordal graphs, and in Section 3, we show that this is also the case for sparse graphs. Below, we show that if the roles of and are reversed, then the answer is yes.
Theorem 1.2
If is a continuous graph, then is continuous, for every positive integer .
Proof
Denote by . First observe that . Now, take any coloring of with colors, . Then, for some , is colored with colors. Observe as well that the colors of only occur in . Therefore, restricted to is a coloring with colors, and since is continuous, it can be turned into a coloring with colors, to produce a coloring of with colors.
In Section 4, we further investigate the bspectrum of , when is chordal. We prove that if is chordal, is bcontinuous and is an integer in the interval such that does not have a bcoloring with colors, then is in the open interval , where and . If this interval is empty, it means that is bcontinuous. Therefore, a good question is the following:
Question 3
What are the graphs and such that is chordal, is bcontinuous and ?
2 bHomomorphism
In this section, we investigate the concept of bhomomorphism. This will help us obtaining a new lower bound for , and showing that contains the integers of the interval , whenever is bcontinuous and is bcontinuous, for every . We refer the reader to HN.book () for an overview on graph homomorphisms.
Given any function and a subset , we denote by the set . Given graphs and , and a function , we say that is a homomorphism from to if for every ; and that is a bhomomorphism from to if it is a homomorphism from to and, for every , there exists such that . If such a function exists, we write . Note that is surjective by definition.
Proposition 1
If and , then .
Proof
Let be homomorphisms from to and to , respectively, and let (composition function of and ). It is known that is a homomorphism from to , therefore we just need to prove that it is also a bhomomorphism. So, let be any vertex. Then, there exists such that ; in turn, there exists such that . Therefore, is such that , as desired.
Lemma 1
If , then and , .
Proof
Let be a bhomomorphism from to and define as . We prove that is a bhomomorphism. First, note that if is an edge in , then either and , in which case is an edge in , or and , in which case and is an edge in . Therefore, is a homomorphism.
Now, denote by and by . Consider , and let be such that . We want to prove that . Note that since is a homomorphism. Therefore, it remains to prove that , i.e., we need to prove that for each , there exists such that . So, consider any . Then, either and , or and . If the latter occurs, by the choice of , there must exist such that ; hence, is the desired vertex. If the former occurs, because is surjective, there must exist such that . Then, and , as desired.
Now we want to prove the second part of the lemma. So now let denote , respectively, and consider defined as . If , then either and , or and . If the former occurs, then is also an edge in ; and if the latter occurs, then and again is also an edge in . Therefore, is an homomorphism and it remains to show that it is a bhomomorphism.
So, consider any , and let be such that . We prove that . For this, consider any . Then one of two cases occurs. If and , then we know that and clearly . So consider and . Then, by the choice of , there must exist such that . Therefore, is such that .
It is easy to verify that the following holds.
Proposition 2
if and only if has a bcoloring with colors.
The next lemma and corollary show the importance of Question 2.
Lemma 2
Consider any , and let , , and . Then, the following hold:

;

; and

.
Proof
Recall that JaPe12 (); this explains the second inequality in (1). Also, recall that GS75 (), which explains the first equality and first inequality in (2). Now, consider any positive integer , and let be a proper coloring of . Note that the colors used in any pair of distinct copies of in are disjoint. Therefore we get and . This explains the equality in (1) and the second equality in (2). Also, because for every graph , we get: . This explains the last inequality in (2).
The next corollary easily follows from Lemma 2.
Corollary 1
If is bcontinuous, for every positive integer , and is bcontinuous, then (below, denotes )
Given a graph , let denote the value . In JaPe12 (), the authors prove that if , then . Therefore, for any and , we have that , while . This tells us that can be arbitrarily larger than . Nevertheless, we give an example where is strictly larger than , showing that our lower bound improves the best previously known lower bound for , namely JaPe12 (). For this, consider the tree obtained from the , , by adding one pendant leaf at , one at , two at and two at . It is not hard to verify that . On the other hand, one can verify that the precoloring of can be completed into a bcoloring of with 7 colors, thus showing that .
Corollary 1 shows the importance of knowing the value . In IM99 (), the authors introduce an upper bound for . Observe that if has a bcoloring with colors, then has at least vertices with degree at least , namely the bvertices. Therefore, if is the largest for which has at least vertices with degree at least , then . Observe that can be easily computed by ordering the vertices of according to their degrees in a nonincreasing way. As a consequence of the following proposition, we get that the distance is at most .
Proposition 3
Let be any graph and let be any positive integer. Then,
Proof
Denote by and by . First, we prove that there are at least vertices of degree at least in . For this, let be the subset of vertices of with degree at least . By the definition of , there are at least such vertices. Also, for each and each , we have . Therefore, the set contains the desired vertices.
Now, we prove that there are at most vertices of degree at least , which implies that cannot exceed . For this, just consider any such that . Since , we get that . Therefore, each such vertex in defines a vertex of with degree at least . By the definition of , one can see that there are at most such vertices in , which implies that there are at most vertices of degree at least in .
In the following sections, we investigate the answer to Question 2 restricted to some known bcontinuous classes.
3 sparse graphs
In this section, we prove the following theorem.
Theorem 3.1
Let be a sparse graph and be any positive integer. Then, is bcontinuous.
We mention that in BDM+09 (), the authors prove that sparse are bcontinuous. Our proof generalizes theirs, since it also holds when . It is also worth mentioning that tidy graphs, a superclass of sparse graphs, are also bcontinuous VBK10 (). A good question is whether our result can be generalized to tidy graphs.
Before we proceed, we need some definitions. Consider a graph ; we say that is complete if contains every possible edge, and that is empty if is empty. Let ; the subset is called a clique if is complete, and it is called a stable set if is empty. A matching in is a collection of pairwise nonadjacent edges, while an antimatching in is a matching in (complement graph of ). Given disjoint subsets of vertices , we say that is complete to if every possible edge between and exists in , and that is anticomplete to if is complete to in .
Given disjoint graphs and , the union of and is the graph , while the join of and is obtained from their union by adding every possible edge between and . Finally, let and be the complete and empty graphs on vertices, respectively, , and add either a matching or an antimatching between and . Also, let be any subset disjoint from both and . The spider operation applied to and is obtained by adding every possible edge between the sets and . The obtained graph is called a spider and we say that is the head of the spider. If , we say that is a spider with empty head. The following decomposition theorem is an important tool in our proof.
Theorem 3.2 (Ho95 (); JaOl95 ())
If is a nontrivial sparse graph, then exactly one of the following holds:

is the union of two sparse graphs; or

is the join of two sparse graphs; or

is a spider whose head is either empty or a sparse graph.
Given a sparse graph , let be a tuple, where is a rooted tree having as internal vertices, is a function , and associates to each leaf of a subset , disjoint from for every , and such that is either a clique, or a stable set, or induces a spider with empty head. We say that is a primeval decomposition of if can be constructed from by searching the tree in an upward way and applying the operation defined by the label on the respective internal node of . We suppose that has a minimal number of internal vertices, i.e., if is such that , and is adjacent to leaves , then at least one between and is not empty as otherwise the decomposition obtained from by removing and relating to is also a primeval decomposition of . A similar argument can be done when and are complete. For each node , we denote by the tuple restricted to the subtree of rooted at . Observe that itself is a primeval decomposition of , the subgraph formed by . We mention that if is an internal node or is not a spider, then is a module in , i.e., for every , we get that is either complete or anticomplete to .
In our proof, we start with a bcoloring of and iteratively try to remove a fixed color, namely color 1, from some , where is a leaf in the primeval decomposition of . While doing this, we allow for the bvertices to lose color 1 in their neighborhoods. Therefore, if at the end no more vertex is colored with 1, then the obtained coloring is a bcoloring of with colors. However, removing color 1 from a leaf is not always possible. When this happens, we restrict our attention to a subgraph of such that restricted to is also a bcoloring with colors. If eventually we arrive at a clique and color 1 cannot be removed, we get that ; hence no bcoloring with colors can exist and we are done. Before we proceed to our proof, we need some further definitions that tell us what is an acceptable coloring and how we can reduce the subgraph being investigated.
Consider any graph and let be a proper coloring of . We say that is a vertex in if and is adjacent to a vertex colored with , for every color distinct from and ; and we say that is a miss1bcoloring of if there exists a vertex colored with , for every color distinct from 1. Note that vertices are not necessarily nonadjacent to ; therefore any bvertex of satisfies the condition, i.e., a bcoloring with colors is also a miss1bcoloring. In addition, note that a miss1bcoloring where no vertex is colored with color 1 is a bcoloring with colors. In our proof, starting with a bcoloring with colors, we manipulate miss1bcolorings until color 1 disappears, or until we can prove that .
Now, consider a minimal primeval decomposition of . Given a leaf of , let be the parent node of in . We know that is an internal node labeled with one of the operations in , which are binary operations. This means that has exactly one other child different from , say , and that, when constructing , the subgraph is operated with the subgraph by applying the operation . We introduce some reduction operations on that allow us to restricted our attention to a subgraph of . Below, we show that can be easily adapted to a minimal primeval decomposition of . We implicitly use these decompositions in the proof.

If is a clique and , we say that is a creduction of if it is obtained from by removing every vertex of nonadjacent to . A primeval decomposition of can be obtained from by removing and the subtree rooted in from , and relating either with when , or with when , where is a spider with empty head and clique set ;

If is a stable set, we say that is an sreduction of if it is obtained from by removing every vertex of except one, say . A primeval decomposition of can be obtained from by relating to . Denote node by and note that it is a clique leaf node in the new decomposition;

If is a spider with partition , we say that is a preduction of . A primeval decomposition of can be obtained from by changing the label of to .
Note that the decomposition may not be a minimal decomposition anymore, but that it is not hard to obtain a minimal decomposition from simply by removing that have a common parent and relating to , whenever either are stable sets and , or are cliques and . Therefore, at all times we consider the decomposition being used to be a minimal decomposition. We also want the reader to remark that the creduction and preduction are not the same. The main difference is that, in the creduction, vertices from are removed, while in the preduction, it is that loses vertices. If is a creduction, or an sreduction, or a preduction of , we say that it is a reduction of . Observe that if is a reduction of , then is an induced subgraph of . In the proof, we iteratively reduce the graph. For this, if is the current subgraph being considered, we implicitly deal with a minimal primeval decomposition of and with a bcoloring of , and we say that a leaf node contains color 1 if .
Proof (of Theorem 3.1)
We construct a sequence of pairs such that:

is a miss1bcoloring of with colors, for every ;

For each , either is a reduction of , or and in there is either one less leaf or one less clique leaf containing color 1; and

Either color class 1 is empty in , or .
For the reader to better understand Condition (ii), we mention that sometimes we are able to decrease the number of clique leaves containing color 1, but without decreasing the number of leaves containing color 1. This is because a recoloring is made in a way that color 1 appears in a nonclique leaf that did not contain color 1 before.
Naturally, we start by setting to . We show how to construct the sequence and prove that if is a bcoloring of with colors (which is the case if ), then a bcoloring of with colors can be obtained. In fact, throughout the construction, whenever is a reduction of , we explain how to obtain a bcoloring of with colors, given a bcoloring of with colors. This gives us what we need. Because , we get that if , then and we are done (no bcolorings of with colors can exist).
Consider we are at step of the construction and let be any leaf of containing color (if no such leaf exists, we are done since there are no more vertices colored with ). If , by the definition of miss1bcoloring we get that each color class distinct from is also nonempty; hence either is the complete graph with vertices, in which case we are done, or is a spider with empty head. So suppose the latter occurs and let be the partition of such that is a clique and is a stable set of . Note that , and that , for every . Therefore, if , then no vertex of can be a vertex; but then we have at most vertices, namely the vertices in , contradicting the fact that is a miss1bcoloring of . Hence, either and we are done, or and a bcoloring with colors of can be easily obtained.
Therefore, suppose that , and let be the label of the parent node of in . Also, let be the subgraph operated with in the construction of , and denote by the graph .
First, suppose that is a stable set, and let be such that . We suppose that as otherwise we can treat as a clique. If there exists , we switch colors and in . We can repeat this argument for other vertices of containing color 1. Therefore, we can suppose that . In this case, we obtain an sreduction of by removing every vertex of , and we let be restricted to . Clearly (i) and (ii) hold. Also, if is a bcoloring of with colors, then by coloring with , for each , we obtain a bcoloring of with colors.
Now, suppose that is a spider with empty head. Let be the partition of where is a clique and is a stable set; denote by the subsets , respectively. Observe that . If , change the color of every colored with to ; such a color exists since is not complete to . If and there exists , then switch colors and in and proceed as before. Finally, if , then let be a preduction of obtained by removing , and let be equal to restricted to . One can verify that (i) and (ii) hold. Now, consider to be any vertex of (recall that since ). If is a bcoloring of with colors, then a bcoloring of with colors can be obtained by giving colors to for every .
Now, suppose that is a clique. If , let and let be obtained by switching colors 1 and in . Since ) is a module and is complete to for every , we know that color cannot lose all of its vertices, and that any is adjacent to the same set of colors (i.e., (i) holds). Also, because is minimal, we know that is not a clique; hence has one less clique leaf containing color 1, i.e., (ii) holds for . We can therefore suppose that . Let denote , and first suppose that there exists a color that appears in but does not appear in . Note that color cannot appear in because is a module in . If is obtained by switching colors and in , then (i) and (ii) hold.
Finally, suppose that is a clique, , and that