1 Introduction
We consider only simple, undirected and finite graphs. A graph is said to be a threshold graph if it does not contain a pair of edges such that ; or equivalently, is free [1]. A graph is said to be covered by the graphs if . A graph is said to have a threshold cover of size if it can be covered by threshold graphs. The threshold dimension of a graph is defined to be the smallest integer such that has a threshold cover of size . Mahadev and Peled [12] give a comprehensive survey of threshold graphs and their applications.
Chvátal and Hammer [1] showed that the fact that a graph has a threshold cover of size is equivalent to the following: there exist linear inequalities on
variables such that the characteristic vector of a set
satisfies all the inequalities if and only if is an independent set of (see [13] for details). They further defined the auxiliary graph (defined in Section 2) corresponding to a graph and showed that any threshold cover of must have size at least . This gave rise to the question of whether there exist any graph that does not have a threshold cover of size . Cozzens and Leibowitz [4] showed the existence of such graphs. In particular, they showed that for every , there exists a graph such that but has no threshold cover of size . The question of whether such graphs exist for seems to have been intensely studied but remained open for a decade (see [11]). Ibaraki and Peled [7] showed that if is a split graph or if contains at most two nontrivial components, then if and only if has a threshold cover of size 2. They further conjectured that every graph satisfying has a threhold cover of size 2. If the conjecture were to be true, it would have the important consequence that graphs having a threshold cover of size 2 can be recognized in polynomial time. In contrast, Yannakakis [16] showed that it is NPcomplete to recognize graphs having a threshold cover of size 3. Cozzens and Halsey [3] studied some properties of graphs having a threshold cover of size 2 and showed that it can be decided in polynomial time whether the complement of a bipartite graph has a threshold cover of size 2. Finally, in 1995, Raschle and Simon [13] proved the conjecture of Ibaraki and Peled by extending the methods in [7]: they showed that every graph whose auxiliary graph is bipartite has a threshold cover of size 2. This proof is very technical and involves the use of a number of complicated reductions and previously known results. In particular, they construct a set of edges that have a “threshold completion” by finding a 2colouring of that is socalled “free”, where (a colouring of is free, if for each colour class , there is no cyclical sequence of vertices in such that if and only ifis odd). It is then shown that this reduces to finding a 2colouring of
which is free. This further reduces to finding a socalled “free” 2colouring of which further reduces to finding a socalled “double free 2colouring” of . The most intricate part is the proof of correctness of an algorithm that computes this particular kind of 2colouring of .The paper of Raschle and Simon also gives an algorithm that checks whether a graph has a threshold cover of size 2 and outputs two threshold graphs that cover in case it has. If the input graph does not have a threshold cover of size 2, the algorithm detects an odd cycle in the auxiliary graph . This odd cycle gives edges in , where is odd, such that the edges , for , and the edges , can never both belong to any threshold subgraph of (because their endpoints induce a , or in ). In this way, the algorithm provides an easily verifiable “certificate” for the fact that there does not exist two threshold graphs that cover . If does have a threshold cover of size 2, then the two threshold graphs returned by the algorithm that cover form an easily verifiable certificate for that fact. Such algorithms are called certifying algorithms [8].
Since as noted above, an odd cycle in the auxiliary graph corresponds to a structure present in that serves as an “obstruction” to it having a threshold cover of size 2, the result of Raschle and Simon can also be seen as a “forbidden structure characterization” of graphs having a threshold cover of size 2. That is, a graph has a threshold cover of size 2 if and only if the said obstruction is not present in . In fact, this is the only known forbidden structure characterization for this family of graphs. Such characterizations exist for many different classes of graphs — for example, interval graphs [9] and circulararc graphs [5].
In this paper, we propose a completely different and selfcontained proof for the theorem of Raschle and Simon that a graph can be covered by two threshold graphs if and only if is bipartite. Our proof is short and direct, and also gives rise to a simpler (although having the same asymptotic worst case running time of ) certifying recognition algorithm for graphs having a threshold cover of size 2.
Note that faster algorithms for determining if a graph has a threshold cover of size 2 are known. After the algorithm of Raschle and Simon [13], Sterbini and Raschle [15] used some observations of Ma [10] to construct an algorithm for the problem. But this algorithm is not a certifying algorithm in the sense that if the input graph does not have a threshold cover of size 2, it does not produce an obstruction in that prevents it from having a threshold cover of size 2. Note that there is an obvious way to make this algorithm a certifying algorithm: if the algorithm answers that the input graph does not have a threshold cover of size 2, run a secondary algorithm that constructs and finds an odd cycle in it (this odd cycle can serve as a certificate). But a naive implementation of the secondary algorithm will have worstcase running time , and it is not clear if it can be implemented to run in time .
In the current work, we show that a graph has a threshold cover of size 2 if and only if its auxiliary graph is bipartite using a technique called the lexicographic method which was introduced by Hell and Huang [6]. Hell and Huang demonstrated how this method can lead to shorter proofs and simpler recognition algorithms for certain problems that can be viewed as orienting the edges of a graph satisfying certain conditions — for example, they showed how this method can lead to simpler characterization proofs and recognition algorithms for comparability graphs, proper interval graphs and proper circulararc graphs. The method starts by taking an arbitrary ordering of the vertices of the graph. It then prescribes choosing the lexicographically smallest (with respect to the given vertex ordering) edge to orient and then orienting it in one way or the other, along with all the edges whose orientations are forced by it. Hell and Huang showed that the lexicographic approach makes it easy to ensure that the orientation so produced satisfies the necessary conditions, if such an orientation exists. We adapt this technique to the problem of generating two threshold graphs that cover a given graph, if two such graphs exist. This shows that the applicability of the lexicographic method may not be limited to only problems involving orientation of edges. However, it should be noted that in our proof, we start with a LexBFS ordering of the vertices of the graph instead of an arbitrary ordering. It is an ordering of the vertices that gives the order in which a LexBFS, or Lexicographic Breadth First Search, a graph searching algorithm that was introduced by Rose, Tarjan and Lueker [14], may visit the vertices of the graph. A LexBFS ordering always gives an order in which a breadthfirst search can visit the vertices of the graph, but has some additional properties. LexBFS can be implemented to run in time linear in the size of the input graph and Rose, Tarjan and Lueker originally used this algorithm to construct a lineartime algorithm for recognizing chordal graphs. Later, LexBFS based algorithms were discovered for the recognition of many different graph classes (see [2] for a survey).
2 Preliminaries
Let be any graph. Two edges are said to form a pair of cross edges in if . If form a pair of cross edges in , we say that the set is a crossing set in (such a set is called an in [13]). It is easy to see that threshold graphs are exactly the graphs that contain no pairs of cross edges, or equivalently no crossing set.
For a graph , the auxiliary graph is defined to be the graph with and form a pair of cross edges in . We shall refer to the vertex of corresponding to an edge alternatively as or , depending upon the context. The following lemma is just a special case of the observation of Chvátal and Hammer [1] that a graph cannot have a threshold cover of size less than .
Lemma 1
If a graph has a threshold cover of size two then is bipartite.
Proof
Let be covered by two threshold graphs and . By the definition of , if then . The fact that and are threshold subgraphs of then implies that neither nor can contain both the edges and . We therefore conclude that the sets and are both independent sets in . Since is covered by and , we have that . Thus, forms a bipartition of into two independent sets. This completes the proof.∎
Our goal is to provide a new proof for the following theorem of Raschle and Simon [13].
Theorem 2.1
A graph can be covered by two threshold graphs if and only if is bipartite.
By Lemma 1, it is enough to prove that if is bipartite, then can be covered by two threshold graphs. In order to prove this, we find a specific 2coloring of the nontrivial components of using the lexicographic method of Hell and Huang [6].
Let be an ordering of the vertices of . Given two element subsets and of , where and , is said to be lexicographically smaller than , denoted by , if for some , and for all . In the usual way, we let denote the fact that either or . For a set , we abbreviate to just . Note that the relation (“is lexicographically smaller than”) that we have defined on element subsets of is a total order. Therefore, given a collection of element subsets of , the lexicographically smallest one among them is welldefined.
3 Proof of Theorem 2.1
Assume that is bipartite. Let denote a LexBFS ordering of the vertices of . The following observation states a wellknown property of LexBFS orderings [2].
Observation 1
For , if , and , then there exists such that , and .
We shall now construct a partial 2coloring of the vertices of using the colors . Notice that choosing a color for any vertex in a component of fixes the colors of all the other vertices in that component. Recall that every vertex of is a twoelement subset of . For every nontrivial component of , perform the following operation: Choose the lexicographically smallest vertex in (with respect to the ordering ) and assign the color 1 to it. This fixes the colors of all the other vertices in . Note that after this procedure, every vertex of that is in a nontrivial component has been colored either 1 or 2. For , let is colored . Further, let denote the set of all isolated vertices in . Clearly, is exactly the set of uncolored vertices of and we have . Consider the subgraphs and of . We claim that and are two threshold graphs that cover . Clearly ; so it only remains to be proven that both and are threshold graphs. Note that for any edge , and .
Observation 2
If form a pair of cross edges in , then exactly one of the following is true:

and , or

and .
Therefore, and cannot be present together in either or .
Proof
As form a pair of cross edges in , the vertices and are adjacent in . Therefore one of them will be colored 1 and the other 2 in the partial 2coloring of . This implies that one of belongs to and the other to . Since and , and cannot be both present in either or .∎
For , let , and and , . By Observation 2, it can be seen that the crossing sets in are exactly the elements of . Define and . Notice that in order to show that both and are threshold graphs, we only need to prove that . We shall first show that . Suppose not. Let be the lexicographically smallest element in .
Lemma 2
.
Proof
Suppose for the sake of contradiction that . By definition of , we can assume without loss of generality that , and . Since , we have that , which implies that . By the definition of , we have that belongs to a nontrivial component of and has been colored 2. Therefore is not the lexicographically smallest vertex in . Let be the lexicographically smallest vertex in ( is defined below). Then we have and by our construction the vertex must have received color 1. Let be a path in between and , where for , . Note that is odd, for each even and for each odd , where .
We claim that for each even and for each odd , where . We prove this by induction on . The case where is trivial as and . So let us assume that . Consider the case where is odd. As is even, by the induction hypothesis we have, . As and , by Observation 2, we have that . Now as form a pair of cross edges in and the same observation then implies that . Similarly, as and , we have . Again, as form a pair of cross edges in and we have . The case where is even is also similar and hence the claim.
By the above claim, . Since , . Recalling that , we now have that . Since , we have that , which is a contradiction.∎
Lemma 3
.
Proof
Suppose for the sake of contradiction that . By definition of , we can assume without loss of generality that , and . Recall that . As , the vertex belongs to a nontrivial component of . Then there exists a neighbor of in such that . By Observation 2, implies . Further, and implies that . Now form a pair of cross edges in . Since , we now have by Observation 2 that and . This implies that is in a nontrivial component of . Similarly, as and we have that . Now form a pair of cross edges in . Since , we have by Observation 2 that and . This implies that is in a nontrivial component of .
We now prove two claims using the fact that is a LexBFS ordering of .
Claim 1. is not possible.
Suppose not. Note that and . Then by Observation 1, there exists such that , and . Now form a pair of cross edges in . By Observation 2, implies that . As , and (recall that ) we then have that . Further, implies that which is a contradiction.
The next claim is symmetric to the claim above, so the proof is omitted.
Claim 2. is not possible.
As , must have received color 2 in the partial 2coloring of . This means that is not the lexicographically smallest vertex in the component . Let be the lexicographically smallest vertex in . Then we have and by our construction, the vertex must have received color 1. Let be a path in between and , where for , . Note that is odd, for each even and for each odd , where .
Claim 3. for each even and for each odd , where .
We prove this by induction on . The case is trivial as and . So let us assume that . Consider the case where is odd. As is even, we have by the induction hypothesis that . As and , by Observation 2, we have . Now since form a pair of cross edges in (recall that ) and , the same observation then implies that . Similarly, as and , we have that . Now since form a pair of cross edges in and , we can deduce as before that . The case where is even can be proved in the same way. Hence the claim.
Recall that , is in a nontrivial component of , and it has color 2 in the partial 2coloring of . Therefore, there exists a lexicographically smallest vertex in which has been colored 1. Clearly, . Let , be a path in between and , where for , . Note that is odd, for each even and for each odd , where .
The following claim is symmetric to Claim 3, and hence we omit the proof.
Claim 4. for each even and for each odd , where .
Recall that . By Claim 3, and , implying that . As we then have . Similarly, as , , and by Claim 3, we have and (as ), we have . We get the final contradiction from the following claim.
Claim 5. Either or .
Suppose . By Claim 3, we then have . Now, since , we have , and we are done. So we shall assume that . By Claim 3, we now have that , implying that . If , then we have , which implies that , proving the claim. So we shall assume that . Therefore, since , we have and . Thus we have , implying that .∎
Lemma 4
.
Proof
Suppose for the sake of contradiction that . Then there exists such that . Consider an element . We can assume without loss of generality that , . As , it belongs to a nontrivial component of . Therefore there exists a neighbor of in such that . Therefore by Observation 2, we have that . As , where , by the same observation we then have . Now if , then the fact that , and implies that which is a contradiction to our earlier observation that . Therefore . As and , it then follows that which again contradicts the fact that . This completes the proof.∎
4 A Certifying Algorithm
Our proof of Theorem 2.1 gives an algorithm which when given a graph as input, either constructs two threshold graphs that cover , or produces an odd cycle in as a certificate that cannot be covered by two threshold graphs.
Algorithm 2ThresholdCover
Input: A graph .
Output:
If has a threshold cover of size 2, two threshold graphs such that they cover , otherwise the auxiliary graph and an odd cycle in it.

Run the LexBFS algorithm on (starting from an arbitrarily chosen vertex) to produce a LexBFS ordering of .

Construct the auxiliary graph .

Initialize and belongs to a trivial component of .

While there exist uncolored vertices in a nontrivial component of , do

Choose the lexicographically smallest vertex in and assign the color 1 to it.

Complete the 2coloring of by doing a BFS starting from the vertex . If an odd cycle is detected, return along with the cycle and exit. Otherwise update is colored 1 in , is colored 2 in .


Output and .
Correctness of the algorithm follows from the proof of Theorem 2.1. The LexBFS on can be done in time and the remaining steps in time. As contains at most vertices and at most edges, the running time of this algorithm is .
5 The Chain Subgraph Cover Problem
A bipartite graph is called a chain graph if it does not contain a pair of edges whose endpoints induce a in . A collection of chain graphs is said to be a chain subgraph cover of a bipartite graph if it is covered by . The problem of deciding whether a bipartite graph can be covered by chain graphs, i.e. whether has a chain subgraph cover, is known as the chain subgraph cover (CSC) problem. He showed that 3CSC is NPcomplete and pointed out that using the results of Ibaraki and Peled [7], the CSC problem can be solved in polynomial time as it can be reduced to the problem of determining whether a split graph can be covered by two threshold graphs. Ma and Spinrad [11] note that a direct implementation of this approach to the 2CSC problem only gives an algorithm and instead propose an algorithm for the problem. This algorithm works by reducing the CSC problem to the problem of deciding whether a partial order has DushnikMiller dimension at most 2. Note that this algorithm does not produce a directly verifiable certificate, such as a forbidden structure in the graph, in case the input graph does not have a 2chain subgraph cover. Our algorithm can be easily modified to make it an certifying algorithm for deciding if an input bipartite graph has a 2chain subgraph cover as explained below. In fact, the only modification that is needed is to change the definition of so that two edges of are adjacent in if and only if they induce in . As shown below, we can start with an arbitrary ordering of vertices in this case, i.e. we do not need to run the LexBFS algorithm to produce a LexBFS ordering of the input graph as the first step.
Let be a bipartite graph. We now redefine the meaning of the term “cross edges”. Two edges are now said to be cross edges if and only if , and . Note that the meaning of the auxiliary graph now changes, but our proof that if and only if there exists two graphs , each containing no cross edges, such that still works almost verbatim—the only change to be made is that all occurrences of expressions of the form “” (where and are some two vertices in ) in the proof of Theorem 2.1 have to be now read as “ and belong to different parts of the bipartition and ”. As noted before, while adapting the proof, we could let the ordering on be any arbitrary ordering. In that case, we cannot use Observation 1 and any argument that uses it. Note that Observation 1 is used only in the proof of Lemma 3; in particular, only in Claims 1 and 2, which in turn are used only in Claim 5. Remove Claims 1 and 2 and replace the proof of Claim 5 with the following proof.
Claim 5. Either or .
As and , we have . From Claim 3, we have . Then, where implies that . As , we can conclude that and . Since and , we further have that and . Therefore we get,
(1) 
From Claim 3, we have . Now where implies that . Since and , we have . As we can conclude that and . Since and , we further have that and . Therefore we get,
(2) 
If and , we get by (1) and (2) that , which is a contradiction. Therefore, either or . If , then by (1), we get , and we are done. Similarly, if , then by (2), we have , again we are done. This proves the claim.
Thus Algorithm 2ThresholdCover can be modified into a certifying recognition algorithm for deciding if a bipartite graph has a 2chain subgraph cover by just changing the definition of . Moreover, this algorithm can choose any arbitrary ordering of the vertices of the input graph to start with and hence does not require the implementation of the LexBFS algorithm. Note that we do not know the answer to the following question: Would Algorithm 2ThresholdCover correctly decide whether the input graph has a threshold cover of size 2 even if it lets be an arbitrary ordering of ?
6 Conclusion
Chvátal and Hammer [1] showed that the problem of deciding whether an input graph has a threshold cover of size at most is NPcomplete, when is part of the input. Yannakakis [16] observes that a bipartite graph has a chain subgraph cover if and only if the split graph obtained from by making every pair of vertices in adjacent to each other has a threshold cover of size . He notes that therefore, his proof of the NPcompleteness of the 3CSC problem implies that the problem of deciding if an input graph has a threshold cover of size at most 3 is also NPcomplete.
We believe that our result demonstrates once again the power of the lexicographic method in yielding short and elegant proofs for certain kinds of problems that otherwise seem to need more complicated proofs. Further research could establish the applicability of the method to a wider range of problems.
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