On bipartite powers of bigraphs

The notion of graph powers is a well-studied topic in graph theory and its applications. In this paper, we investigate a bipartite analogue of graph powers, which we call bipartite powers of bigraphs . We show that the classes of bipartite permutation graphs and interval bigraphs are closed under taking bipartite power. We also show that the problem of recognizing bipartite powers is NP-complete in general.


Introduction
For a positive integer k, the kth power G k of a graph G has the same vertex set as G, and two vertices are adjacent in G k if and only if their distance in G is at most k.If H k = G, then H is called a kth root of G. (In general, kth roots of a graph are not unique and do not necessarily exist.)The concept of graph powers has been extensively studied in graph theory and its algorithmic applications (see Prisner (1995) and the references therein).It is known that several important graph classes are closed under the power operation.That is, for some graph class C, G ∈ C implies G k ∈ C as well.The recognition problem has also been studied for graph powers.It was shown that recognizing graph powers is NP-complete in general while it is solvable in polynomial time for some special graph classes (Chang et al. (2006); Farzad et al. (2009); Kearney and Corneil (1998); Lau (2006); Lau and Corneil (2004); Le and Nguyen (2010); Lin and Skiena (1995); Motwani and Sudan (1994)).
A graph is bipartite if its vertices can be partitioned into two parts in such a way that the endpoints of every edge belong to different parts.Bipartite graphs are also known as bigraphs.It is known that the class of bigraphs coincides with the class of graphs with no cycle of odd length.If a bigraph is connected and has at least three vertices, then clearly its kth power (for k ≥ 2) has a cycle of length three.Hence, the class of bigraphs is not closed under the power operation.Furthermore, any non-trivial subclass of the class of bigraphs is not closed under the power operation.Chandran et al. (2011) introduced a bipartite analogue of graph powers.For odd k ∈ Z + , the kth bipartite power G [k] of a bigraph G has the same vertex set as G, and two vertices are adjacent in G [k]  if and only if their distance in G is at most k and odd.It is easy to see that G [k] is also bipartite.They showed that for any tree T and for any odd k, T [k] is a chordal bipartite graph.They used this fact to prove that chordal bipartite graphs can have arbitrarily large boxicity.Recently, Chandran and Mathew (2012) have strengthened the closure-property result by showing that the class of chordal bipartite graphs is closed under the bipartite power operation.
In this paper, we complement the known closure-property results by proving that some subclasses of the class of chordal bipartite graphs are closed under the bipartite power operation.Note that the closure property of a graph class C under the bipartite power operation is not implied by the closure property of a superclass of C, and vice versa.We also focus on the computational complexity of the problem for recognizing the kth bipartite power graphs.We prove that given a bipartite graph G, the problem of determining whether there exists a bipartite graph H such that G = H [k] is NP-complete for any fixed odd k ≥ 3.

Preliminaries
In this paper, graphs are finite, simple, and undirected.We denote by N G (v) the neighborhood of a vertex v in a graph G.We denote by dist G (u, v) the distance between u and v in G.
A graph A bigraph is an interval bigraph if it has a bi-interval representation.An interval bigraph is a unit interval bigraph if it has a unit bi-interval representation.A bigraph is a chordal bipartite graph if every induced cycle is of length four.We denote the classes of bipartite permutation graphs, unit interval bigraphs, interval bigraphs, and chordal bipartite graphs by BP, UIB, IB, and CB, respectively.The following relations among these classes are known.

Closure properties of bipartite powers of bigraphs
It is known that several important (non-bipartite) graph classes are closed under the power operation.For example, the classes of unit interval graphs, interval graphs, strongly chordal graphs are closed under the kth power for any k (Chen and Chang (2001); Dahlhaus and Duchet (1987); Lubiw (1987); Raychaudhuri (1987Raychaudhuri ( , 1992))).On the other hand, few things are known on bipartite powers of bigraphs.To the best of our knowledge, the following is the only known closure property.
Theorem 3.1 (Chandran and Mathew (2012)) If G is a chordal bipartite graph, then so is G [k] for any odd k ∈ Z + .
In this section, we show that the closure property holds also for smaller graph classes.
We note that the closure property is not hereditary for the graph classes.That is, Theorem 3.1 does not imply Theorem 3.2, and Theorem 3.2 does not imply Theorem 3.3.

Bipartite powers of interval bigraphs
Now we prove Theorem 3.2.Let G = (U, V ; E) be an interval bigraph, and This extension of intervals is inspired by the proof of Agnarsson et al. (2000), who showed that the kth power of an interval graph is also an interval graph.Note that Thus we have the theorem by the two lemmas.
In what follows, we assume u ∈ U and v ∈ V .We also assume that G is connected.It is easy to see that if Theorem 3.2 is valid for connected graphs, then it is also valid for disconnected graphs.
Proof: Without loss of generality, we assume u ≤ v .By this assumption and Hence, we assume there is no such index i.Then, the following claim holds.Claim 3.6 There is some index j such that I v ⊆ I vj .
Proof (Claim 3.6): Since I v does not intersect with any I ui , we can partition U ∩ V (P ) into two parts , it holds that r v < u 2d .Hence, both L(P ) and R(P ) are non-empty since u ∈ L(P ) and u 2d ∈ R(P ).Since P is a connected subgraph of G, there is a vertex v j that has neighbors both in L(P ) and in R(P ).Then, the corresponding interval I vj contains I v (see Fig. 1). 2 Since G is connected, v has a neighbor u in G. Thus, we have Therefore, the lemma holds. 2 Fig. 1: The bridge interval Iv j between L(P ) and R(P ).

Bipartite powers of bipartite permutation graphs
Next we prove Theorem 3.3.An asteroidal triple is a set of three pairwise non-adjacent vertices such that there is a path between any pair of them avoiding the neighbors of the third.If a graph has no asteroidal triple, then it is AT-free.We use the following characterization of bipartite permutation graphs.
Theorem 3.7 (Hell and Huang ( 2004)) A bipartite graph is a permutation graph if and only if it is ATfree.
By the theorem above, it is sufficient to show the following lemma.Our proof is inspired by the proof of Raychaudhuri (1987), who proved that the kth power of an AT-free graph is also AT-free.
Lemma 3.8 If G is bipartite and AT-free, then so is G [k] for any odd k ∈ Z + .
Proof: Let v 1 , v 2 , v 3 be any pairwise non-adjacent triple of vertices in G. Without loss of generality, assume that any v 1 -v 2 path in G passes through a neighbor of v 3 .We shall prove that any v 1 -v 2 path in G [k] passes through a neighbor of v 3 .
Let P = (u 1 , u 2 , . . ., u p ) be a v 1 -v 2 path in G [k] such that u 1 = v 1 and u p = v 2 .For 1 ≤ i ≤ p − 1, let P (u i , u i+1 ) denote a shortest u i -u i+1 path in G. Replacing every edge {u i , u i+1 } in P by the corresponding path P (u i , u i+1 ), we obtain a walk W P in G. Clearly, W P contains a v 1 -v 2 path in G, and hence, Without loss of generality, we assume dist G (u i , w) is odd, and dist G (w, u i+1 ) is even.Then, G has a v 3 -u i+1 path with length dist G (w, u i+1 ) + 1, which is odd and at most k.Since G is bipartite, G cannot have any v 3 -u i+1 path of even length.Hence dist G (v 3 , u i+1 ) is odd and at most k.This implies that 4 Recognizing bipartite powers of bigraphs is hard Powers of graphs are studied also in computational complexity aspects.Motwani and Sudan (1994) first showed that given a graph G, it is NP-complete to decide whether there exists a graph H such that H 2 = G.Following their result, several related results have been obtained (Chang et al. (2006); Farzad et al. (2009); Kearney and Corneil (1998); Lau (2006); Lau and Corneil (2004); Le and Nguyen (2010); Lin and Skiena (1995)).Recently, Le and Nguyen (2010) have shown the following hardness result.
Theorem 4.1 (Le and Nguyen (2010)) For any fixed k ≥ 2 and given a graph G, it is NP-complete to determine whether there exists a graph H such that H k = G.
In this section, we prove that the problem of recognizing bipartite powers is also hard.
Theorem 4.2 For any fixed odd k ≥ 3 and given a bipartite graph G, it is NP-complete to determine whether there exists a bipartite graph H such that The problem is clearly in NP.Thus, it is sufficient to show that the problem is NP-hard.We reduce SET SPLITTING, a well-known NP-complete problem, to ours.
Problem: SET SPLITTING (Garey and Johnson, 1979, SP4) Instance: Collection C of non-empty subsets of a finite set S.
Question: Is there a partition of S into two subsets S 1 and S 2 such that no subset in C is entirely contained in either S 1 or S 2 ?
Without loss of generality, we assume that every element of S belongs to at least one set in C.

An auxiliary lemma
Before presenting the reduction, we show a useful lemma.In their proof of NP-hardness, Le and Nguyen (2010) used the following tail structure presented first by Motwani and Sudan (1994) and generalized later by Lau (2006).Note that for two sets A and B, we mean by A ⊂ B that A ⊆ B and A = B; that is, A is properly included in B.
Lemma 4.3 (Lau (2006)) Let G be a connected graph with {v 1 , . . ., v k+1 } ⊂ V (G) where N G (v 1 ) = {v 2 , . . ., v k+1 }, and The tail structure played an important role in the proofs of NP-hardness of recognizing powers of graphs (Lau (2006); Le and Nguyen (2010); Motwani and Sudan (1994)).Although we cannot use the structure directly, we show the following similar structure for bipartite powers.
Lemma 4.4 Let k ≥ 3 be odd, and G be a connected bipartite graph with {v 0 , . . ., To prove Lemma 4.4, we need some definitions.
Proof (Lemma 4.4): First, observe that the vertices v 0 and v 1 are not bipartite universal in G.
Since v 0 is not bipartite universal in G = H [k] and G is connected, there is a vertex u ∈ V (H) such that dist H (v 0 , u) is finite and greater than k.This implies that N 2i+1 Next we apply a similar argument to v 3).Thus the first property holds.The second property immediately follows from the first one.2 ...

Reduction
Let (S, C) be an instance of SET SPLITTING, where S = {u 1 , . . ., u n } is the ground set, and C = {c 1 , . . ., c m } is a set of non-empty subsets of S. For any fixed odd k ≥ 3, we construct a graph G from (S, C).The vertex set V (G) of G consists of the following vertices: • Element vertices u i for all element u i ∈ S; • Subset vertices c 0 j , . . ., c k+1 j for all c j ∈ C; and • Partition vertices S 0 1 , . . ., S k−2 1 and S 0 2 , . . ., S k−2
The edge set E(G) of G consists of the following edges: For each subset c j ∈ C, • c p j is adjacent to c q j if p ≡ q (mod 2), • c p j is adjacent to c q j if p + q ≤ k − 4 and p ≡ q (mod 2); • c p j is adjacent to c q j if c j ∩ c j = ∅ and p + q = k − 2, • c p j is adjacent to all u i ∈ S if p is even and 0 ≤ p ≤ k − 3, • c p j is adjacent to S q 1 and S q 2 if p + q ≤ k − 2 and p ≡ q (mod 2), For each element u i ∈ S, • u i is adjacent to S 0 1 , S 2 1 , . . ., S k−3 1 and S 0 2 , S 2 2 , . . ., S k−3 2 ; For each i ∈ {1, 2}, • S p i is adjacent to S q i if p ≡ q (mod 2), • S p 1 and S q 2 are adjacent if p + q ≤ k − 4 and p ≡ q (mod 2).

Equivalence
Now we prove that (S, C) has a desired partition if and only if there exists a bipartite graph H such that H [k] = G.
Lemma 4.5 If there exists a partition of S into two disjoint subsets S 1 and S 2 such that each subset in C intersects both S 1 and S 2 , then there exists a bipartite graph H such that G = H [k] .
• For each c j ∈ C, {c i j , c i+1 j } ∈ E(H) for 0 ≤ i ≤ k, and {c 0 j , u i } ∈ E(H) for each u i ∈ c j .
• For each u i ∈ S, {u i , S 0 j } ∈ E(H) if u i ∈ S j .
It is a routine exercise to verify that G = H [k] . 2 Lemma 4.6 If there is a bigraph H such that H [k] = G, then there exists a partition of S into two disjoint subsets S 1 and S 2 such that each subset in C intersects both S 1 and S 2 .
Proof: We construct a partition (S 1 , S 2 ) of S such that S 1 = {u i | {u i , S 0 1 } ∈ E(H)} and S 2 = S \ S 1 .We show that (S 1 , S 2 ) is a good partition; that is, c j ∩ S i = ∅ for each c j ∈ C and i ∈ {1, 2}.To this end, we show that 1. for any c j ∈ C and ∈ {1, 2}, some u i ∈ c j has S 0 as a neighbor in H, and