Hardness and structural results for half-squares of restricted tree convex bipartite graphs

04/16/2018
by   Hoang-Oanh Le, et al.
0

Let B=(X,Y,E) be a bipartite graph. A half-square of B has one color class of B as vertex set, say X; two vertices are adjacent whenever they have a common neighbor in Y. Every planar graph is a half-square of a planar bipartite graph, namely of its subdivision. Until recently, only half-squares of planar bipartite graphs, also known as map graphs (Chen, Grigni and Papadimitriou [STOC 1998, J. ACM 2002]), have been investigated, and the most discussed problem is whether it is possible to recognize these graphs faster and simpler than Thorup's O(n^120)-time algorithm (Thorup [FOCS 1998]). In this paper, we identify the first hardness case, namely that deciding if a graph is a half-square of a balanced bisplit graph is NP-complete. (Balanced bisplit graphs form a proper subclass of star convex bipartite graphs.) For classical subclasses of tree convex bipartite graphs such as biconvex, convex, and chordal bipartite graphs, we give good structural characterizations of their half-squares that imply efficient recognition algorithms. As a by-product, we obtain new characterizations of unit interval graphs, interval graphs, and of strongly chordal graphs in terms of half-squares of biconvex bipartite, convex bipartite, and of chordal bipartite graphs, respectively. Good characterizations of half-squares of star convex and star biconvex bipartite graphs are also given, giving linear-time recognition algorithms for these half-squares.

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1 Introduction

The square of a graph , denoted , is obtained from by adding new edges between two distinct vertices whenever their distance is two. Then, is called a square root of . Given a graph , it is NP-complete to decide if is the square of some graph (MotwaniS ), even for a split graph (LauC ).

Given a bipartite graph , the subgraphs of the square induced by the color classes and , and , are called the two half-squares of (ChenGP98 ; ChenGP02 ).

While not every graph is the square of a graph and deciding if a graph is the square of a graph is hard, every graph is half-square of a bipartite graph: if is the bipartite graph with , then clearly . So one is interested in half-squares of special bipartite graphs. Note that is the subdivision of , hence every planar graph is half-square of a planar bipartite graph.

Let be a class of bipartite graphs. A graph is called half-square of if there exists a bipartite graph in such that . Then, is called a half-root of . With this notion, the following decision problem arises.

Half-Square Of Instance: A graph . Question: Is half-square of a bipartite graph in , i.e., does there exist a bipartite graph in s.t. ?

In this paper, we discuss Half-Square Of   for several restricted bipartite graph classes .

Previous results and related work.  Half-squares of bipartite graphs have been introduced in ChenGP98 ; ChenGP02 in order to give a graph-theoretical characterization of the so-called map graphs. A map graph is the (point-)intersection graph of simply-connected and interior-disjoint regions of the Euclidean plane. More precisely, a map of a graph is a function taking each vertex to a closed disc homeomorph (the regions) in the plane, such that all , , are interior-disjoint, and two distinct vertices and of are adjacent if and only if the boundaries of and intersect. A map graph is one having a map. It turns out that map graphs are exactly half-squares of planar bipartite graphs (ChenGP98 ; ChenGP02 ). As we have seen at the beginning, every planar graph is a map graph but not the converse; map graphs may have arbitrary large cliques. As such, map graphs form an interesting and important graph class. The main problem concerning map graphs is to recognize if a given graph is a map graph. In Thorup , Thorup shows that Half-Square Of Planar, that is, deciding if a graph is a half-square of a planar bipartite graph, can be solved in polynomial time 111

Thorup did not give the running time explicitly, but it is estimated to be roughly

with being the vertex number of the input graph; cf. ChenGP02 .. Very recently, in MnichRS , it is shown that Half-Squares Of Outerplanar and Half-Square Of Tree are solvable in linear time. Other papers deal with solving hard problems in map graphs include Chen ; DemaineFHT ; DemaineH ; EickmeyerK17 ; FominLS . Some applications of map graphs have been addressed in ChenHK99 . The paper Brandenburg discussed a relation between map graphs and -planar graphs, an interesting topic in graph drawing.

Our results.  We identify the first class of bipartite graphs for which Half-Square Of   is NP-hard. Our class is a subclass of the class of the bisplit bipartite graphs and of star convex bipartite graphs (all terms are given later). For some other subclasses of tree convex bipartite graphs, such as star convex and star biconvex, convex and biconvex, and chordal bipartite graphs, we give structural descriptions for their half-squares, that imply polynomial-time recognition algorithms:

  • Recognizing half-squares of balanced bisplit graphs (a proper subclass of star convex bipartite graphs) is hard, even when restricted to co-bipartite graphs;

  • Half-squares of star convex and star biconvex can be recognized in linear time;

  • Half-squares of biconvex bipartite graphs are precisely the unit interval graphs;

  • Half-squares of convex bipartite graphs are precisely the interval graphs;

  • Half-squares of chordal bipartite graphs are precisely the strongly chordal graphs.

2 Preliminaries

Let be a graph with vertex set and edge set . A stable set (a clique) in is a set of pairwise non-adjacent (adjacent) vertices. The complete graph on vertices, the complete bipartite graph with vertices in one color class and vertices in the other color class, the cycle with vertices are denoted , and , respectively. A is also called a triangle, a complete bipartite graph is also called a biclique, a complete bipartite graph is also called a star.

The neighborhood of a vertex in , denoted by , is the set of all vertices in adjacent to ; if the context is clear, we simply write . A universal vertex in is one with , i.e., is adjacent to all other vertices in .

For a subset , is the subgraph of induced by , and stands for . We write for bipartite graphs with a bipartition into stable sets and . For subsets , we denote for the bipartite subgraph of induced by .

We will consider half-squares of the following well-known subclasses of bipartite graphs: Let be a bipartite graph.

  • is -convex if there is a linear ordering on such that, for each , is an interval in . Being -convex is defined similarly. is convex if it is -convex or -convex. is biconvex if it is both -convex and -convex. We write Convex and Biconvex to denote the class of convex bipartite graphs, respectively, the class of biconvex bipartite graphs.

  • is chordal bipartite if has no induced cycle of length at least six. Chordal Bipartite stands for the class of chordal bipartite graphs.

  • is tree -convex if there exists a tree such that, for each , induces a subtree in . Being tree -convex is defined similarly. is tree convex if it is tree -convex or tree -convex. is tree biconvex if it is both tree -convex and tree -convex. When is a star, we also speak of star convex and star biconvex bipartite graphs.

    Tree Convex and Tree Biconvex are the class of all tree convex bipartite graphs and all tree biconvex bipartite graphs, respectively, and Star Convex and Star Biconvex are the class of all star convex bipartite graphs and all star biconvex bipartite graphs, respectively.

It is known that Biconvex Convex Chordal Bipartite Tree Biconvex Tree Convex. All inclusions are proper; see Spinrad ; Liu for more information on these graph classes.

Given a graph , we often use the following two kinds of bipartite graphs associated to :

Definition 1

Let be an arbitrary graph.

  • The bipartite graph with is the subdivision of .

  • Let denote the set of all maximal cliques of . The bipartite graph with is the vertex-clique incidence bipartite graph of .

Note that the subdivision of a planar graph is planar, and subdivisions and vertex-clique incidence graphs of triangle-free graphs coincide.

Proposition 1

Every graph is half-square of its vertex-clique incidence bipartite graph. More precisely, if is the vertex-clique incidence bipartite graph of , then . Similar statement holds for subdivisions.

Proof

For distinct vertices , if and only if for some , if and only if and are adjacent in . That is, . ∎

3 Recognizing half-squares of balanced bisplit graphs is hard

A graph is a split graph if there is a partition of its vertex set into a clique and stable set . Recall that a biclique is a complete bipartite graph. Following the concept of split graphs, we call a bipartite graph bisplit if it can be partitioned into a biclique and a stable set. In this section, we show that Half-Square Of Balanced Bisplit is NP-hard. Balanced bisplit graphs form a proper subclass of bisplit graphs, and are defined as follows.

Definition 2

A bipartite graph is called balanced bisplit if it satisfies the following properties:

  • ;

  • there is partition such that is a biclique;

  • there is partition such that the edge set of is a perfect matching.

Note that by (i) and (iii), , and by (ii) and (iii), every vertex in is universal in .

In order to prove the NP-hardness of Half-Square Of Balanced Bisplit, we will reduce the following well-known NP-complete problem Edge Clique Cover to it.

Edge Clique Cover Instance: A graph and a positive integer . Question: Do there exist cliques in such that each edge of is contained in some of these cliques?

Edge Clique Cover is NP-complete Holyer ; KouSW ; Orlin , even when restricted to co-bipartite graphs LeP . (A co-bipartite graph is the complement of a bipartite graph.)

Theorem 3.1

Half-Square Of Balanced Bisplit is NP-complete, even when restricted to co-bipartite graphs.

Proof

It is clear that Half-Square Of Balanced Bisplit is in NP, since guessing a bipartite-half root with , verifying that is balanced bisplit, and that can obviously be done in polynomial time. Thus, by reducing Edge Clique Cover to Half-Square Of Balanced Bisplit, we will conclude that Half-Square Of Balanced Bisplit is NP-complete.

Let be an instance of Edge Clique Cover. Note that we may assume that , and that is connected and has no universal vertices. We construct an instance of Half-Square Of Balanced Bisplit as follows: is obtained from by adding a set of new vertices, , and all edges between vertices in and all edges with , . Thus, , and the new vertices in are exactly the universal vertices of . Clearly, can be constructed in polynomial time , and in addition, if is co-bipartite, then is co-bipartite, too. We now show that Edge Clique Cover if and only if Half-Square Of Balanced Bisplit.

First, suppose that the edges of can be covered by cliques in . We are going to show that is half-square of some balanced bisplit bipartite graph. Consider the bipartite graph (see also Figure 1) with

In particular, . The edge set is as follows:

  • has all edges between and , i.e., is a biclique,

  • has edges , . Thus, the edge set of forms a perfect matching, and

  • has edges , , , whenever is contained in clique , .

Figure 1: The balanced bisplit graph proving ; is adjacent to if and only if .

Thus, is a balanced bisplit graph. Moreover, by the construction of , we have in :

  • is a clique (as is a biclique),

  • every vertex is adjacent to all vertices (recall that is connected, so every is in some , and is a common neighbor of and ), and

  • no two distinct vertices have, in , a common neighbor in . So and are adjacent in if and only if and have a common neighbor in , if and only if and belong to a clique in , if and only if and are adjacent in .

That is, .

Conversely, suppose for some balanced bisplit graph with and partitions and as in Definition 2. We are going to show that the edges of can be covered by cliques. As is a biclique, all vertices in are universal in . Hence

because no vertex in is universal in (recall that has no universal vertex). Therefore (recall that )

Note that, as is a balanced bisplit graph, . Write and observe that no two vertices in have a common neighbor in . Thus, for each edge in , and have a common neighbor in . Therefore, the cliques in , , induced by the neighbors of in , cover the edges of . ∎

Theorem 3.1 indicates that recognizing half-squares of restricted bipartite graphs is algorithmically much more complex than recognizing squares of bipartite graphs; the latter can be done in polynomial time (Lau ).

Observe that balanced bisplit graphs are star convex: Let be a bipartite graph with the properties in Definition 2. Fix a vertex and consider the star . Since every vertex is adjacent to , induces a substar in . Note, however, that the hardness of Half-Square Of Balanced Bisplit does not imply the hardness of Half-Square Of Star Convex. This is because the proof of Theorem 3.1 strongly relies on the properties of balanced bisplit graphs.

Indeed, we will show in the next section that half-squares of star-convex bipartite graphs can be recognized in polynomial time.

4 Half-squares of star convex and star biconvex bipartite graphs

We need more notations for stating our results. Let and be two (vertex-disjoint) graphs. For a vertex , we say that the graph with

  • vertex set and

  • edge set

is obtained from by substituting the vertex of by the graph . Thus, substituting a vertex of by the graph results in the graph obtained from and by adding all edges between and .

Recall that a split graph is one, whose vertex set can be partitioned into a clique and a stable set. In a graph, a connected component is big if it has at least two vertices.

Lemma 1

Let be a star convex bipartite graph with an associated star . Then

  • has at most one big connected component and the big connected component has a universal vertex.

  • is obtained from a split graph by substituting vertices by cliques.

Proof

Let be the center vertex of the star . Note that, if has at least two vertices, then must contain .

(i): If for all , then is clearly edgeless, and (i) trivially holds. So, assume that

is not empty. Write

Then and . Moreover, for every and , and have no common neighbor in . Thus, consists of the big component induced by in which is a universal vertex and many one-vertex components.

(ii): Partition into and with

Then, clearly,

Let (otherwise, (ii) obviously holds), and write for some . Note that every vertex has degree one since . Thus, consists of vertex-disjoint stars at center vertices , . For each , fix a vertex . (See Figure 2.)

Figure 2: Proof of Lemma 1 (ii) illustrated.

Then, as no two of have a common neighbor in ,

Thus, is obtained from the split graph by substituting by clique with vertex set , . ∎

The following facts show that the reverse statements in Lemma 1 hold true, too.

Fact 1

Let have at most one big connected component, and let the big connected component have a universal vertex. Then is half-square of a star convex bipartite graph.

Proof

It is obvious that graphs having no edges are half-squares of star convex bipartite graphs. So, let be the set of universal vertices of the big component of , let be the set of isolated vertices of , and let be set of the remaining vertices of , . Thus, is the big component of .

If , let with and . If , construct a bipartite graph as follows.

  • , where and is the edge set (possibly empty) of ,

  • is a biclique, the edge set of is the perfect matching ,

  • the remaining edges of are between and . Two vertices and are adjacent in if and only if is an endvertex of the edge in . That is, is the subdivision of .

It is not difficult to verify, by construction, that . Moreover, is star convex: Fix a vertex and let be the star with edge set . Clearly, for every , forms a substar in . ∎

Fact 2

Graphs obtained from split graphs by substituting vertices by cliques are half-squares of star convex bipartite graphs.

Proof

We first show that split graphs are half-squares of star convex bipartite graphs. Let be a split graph with a partition of its vertex set into clique and stable set . Construct a bipartite graph as follows.

  • ,

  • .

By construction, . Moreover, is star convex with the associated star .

Now, if is obtained from by substituting by clique , then clearly is the half-square of the bipartite graph obtained from by substituting vertices by stable sets . Obviously, is star convex with the same star associated to . ∎

By Lemma 1 and Facts 1 and 2, we obtain:

Theorem 4.1

A graph is a half-square of a star convex bipartite graph if and only if

  • it has at most one big connected component and the big connected component has a universal vertex, or

  • it is obtained from a split graph by substituting vertices by cliques.

In case of star biconvex bipartite graphs, we obtain:

Theorem 4.2

A graph is a half-square of a star biconvex bipartite graph if and only if it has at most one big connected component and the big connected component is obtained from a split graph having a universal vertex by substituting vertices by cliques.

Proof

The necessity part follows directly from Theorem 4.1.

For the sufficiency part, we first show that split graphs in which the big connected component has a universal vertex are half-squares of star biconvex bipartite graphs. Let be a split graph with a partition of its vertex set into clique and stable set . Let be the set of all isolated vertices of (thus, is the big component of ). Construct a bipartite graph as follows.

  • ,

  • .

By construction, . Moreover, is star biconvex with the associated stars and , where is a universal vertex of (hence for all ).

Now, if is obtained from by substituting vertices by cliques , then clearly is the half-square of the bipartite graph obtained from by substituting vertices by stable sets . Obviously, is star biconvex with the same star associated to and the star , where is a vertex in ( is a universal vertex of ). ∎

By definition, if is obtained from a graph by substituting vertex by clique with , then is a module in , that is, every vertex in outside is adjacent to all or to none vertices in . Now, note that all maximal clique modules of a given graph can be computed in linear time (see, e.g, (McConnell03, , Corollary 7.4)). Note also that split graphs can be recognized in linear time (cf. Golumbic ), and a partition into a clique and a stable set of a given split graph can be computed in linear time (HeggernesK07 ). Thus, Theorems 4.1 and 4.2 and their proofs imply:

Corollary 1

Half-Square Of Star Convex and Half-Square Of Star Biconvex can be solved in linear time. A star (bi)convex bipartite half-root, if any, can be constructed in linear time.

5 Half-squares of biconvex and convex bipartite graphs

In this section, we show that half-squares of convex bipartite graphs are precisely the interval graphs and half-squares of biconvex bipartite graphs are precisely the unit interval graphs.

Recall that is an interval graph if it admits an interval representation , such that two vertices in are adjacent if and only if the corresponding intervals intersect. Let be an interval graph. It is well-known (FulkersonG ; GilmoreH ) that there is a linear ordering of the maximal cliques of , say , such that every vertex of belongs to maximal cliques that are consecutive in that ordering, that is, for every vertex of , there are indices and with

If and are distinct maximal cliques of , then and are both not empty, that is, for every , there are vertices and such that .

Recall that unit interval graphs are those interval graphs admitting an interval representation in which all intervals have the same length. It is well known (Roberts ) that a graph is a unit interval graphs if and only if it has an interval representation in which no interval is properly contained in another interval (a proper interval graph), if and only if it is a -free interval graph.

Lemma 2

The half-squares of a biconvex bipartite graph are -free.

Proof

Let be a biconvex bipartite graph. By symmetry we need only to show that is -free. Suppose, by contradiction, that induce a in with edges and . Let be a common neighbor of and , be a common neighbor of and , and be a common neighbor of and . Then, are pairwise distinct and induce, in , a subdivision of ; see also Figure 3. This is a contradiction because biconvex bipartite graphs do not contain an induced subdivision of the . Thus, the half-squares of a biconvex bipartite graph are -free. ∎

Figure 3: The subdivision of is convex, but not biconvex.
Lemma 3

  • Every interval graph is half-square of a convex bipartite graph. More precisely, if is an interval graph and is the vertex-clique incidence bipartite graph of , then and is -convex.

  • If is -convex, then is an interval graph.

Proof

(i): Let be an interval graph, and let be the vertex-clique incidence bipartite graph of . Since each appears in the interval in , is -convex. Moreover, by Proposition 1, .

(ii): This is because admits a linear ordering such that, for each , is an interval in . This collection is an interval representation of because and are adjacent in if and only if . ∎

Theorem 5.1

A graph is half-square of a biconvex bipartite graph if and only if it is a unit interval graph.

Proof

First, by Lemma 3 (ii), half-squares of biconvex bipartite graphs are interval graphs, and then by Lemma 2, half-squares of biconvex bipartite graphs are unit interval graphs.

Next we show that every unit interval graph is half-square of some biconvex bipartite graph. Let be a unit interval graph. Let be the vertex-clique incidence bipartite graph of . By Lemma 3 (i), and is -convex. We now are going to show that is -convex, too.

Consider a linear order in , , such that each is contained in exactly the cliques , . Let be lexicographically sorted according . We claim that is -convex with respect to this ordering. Assume, by a contradiction, that some has neighbors and non-neighbor with in the sorted list, say. In particular, belong to , but not; see also Figure 4.

Figure 4: Assuming , and , but .

Since and , we have . Since is not in , we therefore have

In particular, . Since and , we have

Hence . Now, by the maximality of the cliques, there exists with (hence is non-adjacent to ), and there exists with (hence is non-adjacent to and ; note that and are possible). But then , and induce a in , a contradiction.

Thus, we have seen that every unit interval graph is half-square of a biconvex bipartite graph. ∎

We next characterize half-squares of convex bipartite graphs as interval graphs. This is somehow surprising because the definition of being convex bipartite is asymmetric with respect to the two half-squares.

Theorem 5.2

A graph is a half-square of a convex bipartite graph if and only if it is an interval graph.

Proof

By Lemma 3 (i), interval graphs are half-squares of convex bipartite graphs. It remains to show that half-squares of convex bipartite graphs are interval graphs. Let be an -convex bipartite graph. By Lemma 3 (ii), is an interval graph. We now are going to show that is an interval graph, too.

Let be obtained from by removing all vertices with is properly contained in for some . Clearly, . We show that is -convex, hence, by Lemma 3 (ii), is an interval graph, as claimed. To this end, let such that, for every , is an interval in . (Recall that , hence is -convex.) For let , and sort increasing according . Then, for each , is an interval in : Assume, by contradiction, that there is some such that is not an interval in . Let be a leftmost and be a rightmost vertex in . By the assumption, there is some with . Then, as and are intervals in , must be a subset of ; see also Figure 5. Since but , must be a proper subset of , contradicting to the fact that in , no such pair of vertices exists in . Thus, is -convex.

Figure 5: Assuming , and is adjacent to and , but non-adjacent to .

Note that is indeed biconvex, hence, by Theorem 5.1, is even a unit interval graph. ∎

Since (unit) interval graph with vertices and edges can be recognized in linear time and all maximal cliques of an (unit) interval graph can be listed in the same time complexity (cf. Golumbic ), Theorems 5.2 and 5.1 imply:

Corollary 2

Half-Square Of Convex and Half-Square Of Biconvex can be solved in linear time. A (bi)convex bipartite half-root, if any, can be constructed in linear time.

6 Half-squares of chordal bipartite graphs

In this section, we show that half-squares of chordal bipartite graphs are precisely the strongly chordal graphs. Recall that a graph is chordal if it has no induced cycle of length at least four. It is well-known (see, e.g., Golumbic ; McKeyM ; Spinrad ) that a graph is chordal if and only if it admits a tree representation, that is, there exists a tree such that, for each vertex , is a subtree of and two vertices in are adjacent if and only if the corresponding subtrees in intersect. Moreover, the vertices of can be taken as the maximal cliques of the chordal graph (a clique tree). Recall also that a graph is strongly chordal if it is chordal and has no induced -sun, . Here, a -sun consists of a stable set and a clique and edges , , . (Indices are taken modulo .)

We first begin with the following fact.

Lemma 4

Let be a bipartite graph without induced and let . If contains an induced -sun, then contains an induced cycle of length .

Proof

We first show that

every clique in stems from a star in .

Suppose, by a contradiction, that there is some clique in such that, for any vertex , . Choose a vertex with is maximal. Let . Since is a clique in , there is a vertex adjacent to and some vertices in . Choose such a vertex with is maximal. By the choice of , there is a vertex . Again, since is a clique in , there is a vertex adjacent to both and . By the choice of , there is a vertex non-adjacent to . But then and induce a in , a contradiction. Thus, there must be a vertex such that .

Now, consider a -sun in with stable set , clique , and edges , , . Let such that , and let such that , . Since is adjacent, in the -sun, only to and , we have in , that is non-adjacent to . That is, induce a in . ∎

Theorem 6.1

A graph is half-square of a chordal bipartite graph if and only if it is a strongly chordal graph.

Proof

We first show that half-squares of chordal bipartite graphs are chordal. Let be a chordal bipartite graph. It is known that is tree convex (JiangLWX ; Lehel ). Thus, there is a tree