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 NPcomplete 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 halfsquares 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 halfsquare of a bipartite graph: if is the bipartite graph with , then clearly . So one is interested in halfsquares of special bipartite graphs. Note that is the subdivision of , hence every planar graph is halfsquare of a planar bipartite graph.
Let be a class of bipartite graphs. A graph is called halfsquare of if there exists a bipartite graph in such that . Then, is called a halfroot of . With this notion, the following decision problem arises.
HalfSquare Of Instance: A graph . Question: Is halfsquare of a bipartite graph in , i.e., does there exist a bipartite graph in s.t. ?
In this paper, we discuss HalfSquare Of for several restricted bipartite graph classes .
Previous results and related work. Halfsquares of bipartite graphs have been introduced in ChenGP98 ; ChenGP02 in order to give a graphtheoretical characterization of the socalled map graphs. A map graph is the (point)intersection graph of simplyconnected and interiordisjoint 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 interiordisjoint, 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 halfsquares 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 HalfSquare Of Planar, that is, deciding if a graph is a halfsquare of a planar bipartite graph, can be solved in polynomial time ^{1}^{1}1
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 HalfSquares Of Outerplanar and HalfSquare 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 HalfSquare Of is NPhard. 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 halfsquares, that imply polynomialtime recognition algorithms:

Recognizing halfsquares of balanced bisplit graphs (a proper subclass of star convex bipartite graphs) is hard, even when restricted to cobipartite graphs;

Halfsquares of star convex and star biconvex can be recognized in linear time;

Halfsquares of biconvex bipartite graphs are precisely the unit interval graphs;

Halfsquares of convex bipartite graphs are precisely the interval graphs;

Halfsquares 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 nonadjacent (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 halfsquares of the following wellknown 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 vertexclique incidence bipartite graph of .
Note that the subdivision of a planar graph is planar, and subdivisions and vertexclique incidence graphs of trianglefree graphs coincide.
Proposition 1
Every graph is halfsquare of its vertexclique incidence bipartite graph. More precisely, if is the vertexclique 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 halfsquares 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 HalfSquare Of Balanced Bisplit is NPhard. 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 NPhardness of HalfSquare Of Balanced Bisplit, we will reduce the following wellknown NPcomplete 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 NPcomplete Holyer ; KouSW ; Orlin , even when restricted to cobipartite graphs LeP . (A cobipartite graph is the complement of a bipartite graph.)
Theorem 3.1
HalfSquare Of Balanced Bisplit is NPcomplete, even when restricted to cobipartite graphs.
Proof
It is clear that HalfSquare Of Balanced Bisplit is in NP, since guessing a bipartitehalf root with , verifying that is balanced bisplit, and that can obviously be done in polynomial time. Thus, by reducing Edge Clique Cover to HalfSquare Of Balanced Bisplit, we will conclude that HalfSquare Of Balanced Bisplit is NPcomplete.
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 HalfSquare 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 cobipartite, then is cobipartite, too. We now show that Edge Clique Cover if and only if HalfSquare Of Balanced Bisplit.
First, suppose that the edges of can be covered by cliques in . We are going to show that is halfsquare 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 , .
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 halfsquares 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 HalfSquare Of Balanced Bisplit does not imply the hardness of HalfSquare 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 halfsquares of starconvex bipartite graphs can be recognized in polynomial time.
4 Halfsquares of star convex and star biconvex bipartite graphs
We need more notations for stating our results. Let and be two (vertexdisjoint) 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 onevertex 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 vertexdisjoint stars at center vertices , . For each , fix a vertex . (See Figure 2.)
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 halfsquare of a star convex bipartite graph.
Proof
It is obvious that graphs having no edges are halfsquares 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 halfsquares of star convex bipartite graphs.
Proof
We first show that split graphs are halfsquares 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 halfsquare of the bipartite graph obtained from by substituting vertices by stable sets . Obviously, is star convex with the same star associated to . ∎
Theorem 4.1
A graph is a halfsquare 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 halfsquare 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 halfsquares 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 halfsquare 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
HalfSquare Of Star Convex and HalfSquare Of Star Biconvex can be solved in linear time. A star (bi)convex bipartite halfroot, if any, can be constructed in linear time.
5 Halfsquares of biconvex and convex bipartite graphs
In this section, we show that halfsquares of convex bipartite graphs are precisely the interval graphs and halfsquares 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 wellknown (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 halfsquares 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 halfsquares of a biconvex bipartite graph are free. ∎
Lemma 3

Every interval graph is halfsquare of a convex bipartite graph. More precisely, if is an interval graph and is the vertexclique 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 vertexclique 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 halfsquare of a biconvex bipartite graph if and only if it is a unit interval graph.
Proof
First, by Lemma 3 (ii), halfsquares of biconvex bipartite graphs are interval graphs, and then by Lemma 2, halfsquares of biconvex bipartite graphs are unit interval graphs.
Next we show that every unit interval graph is halfsquare of some biconvex bipartite graph. Let be a unit interval graph. Let be the vertexclique 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 nonneighbor with in the sorted list, say. In particular, belong to , but not; see also Figure 4.
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 nonadjacent to ), and there exists with (hence is nonadjacent 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 halfsquare of a biconvex bipartite graph. ∎
We next characterize halfsquares 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 halfsquares.
Theorem 5.2
A graph is a halfsquare of a convex bipartite graph if and only if it is an interval graph.
Proof
By Lemma 3 (i), interval graphs are halfsquares of convex bipartite graphs. It remains to show that halfsquares 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.
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
HalfSquare Of Convex and HalfSquare Of Biconvex can be solved in linear time. A (bi)convex bipartite halfroot, if any, can be constructed in linear time.
6 Halfsquares of chordal bipartite graphs
In this section, we show that halfsquares 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 wellknown (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 nonadjacent 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 nonadjacent to . That is, induce a in . ∎
Theorem 6.1
A graph is halfsquare of a chordal bipartite graph if and only if it is a strongly chordal graph.
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