On some subclasses of interval catch digraphs

A digraph G = ( V, E ) is an interval catch digraph if for each vertex v ∈ V , one can associate an interval on real line and a point within it (say ( I v , p v ) ) in such a way that uv ∈ E if and only if p v ∈ I u . It was introduced by Maehara in 1984. It has many applications in real world situations like networking and telecommunication. In his introducing paper Maehara proposed a conjecture for the characterization of central interval catch digraph (where p v is the mid-point I v for each v ∈ V ) in terms of forbidden subdigraphs. In this paper, we disprove the conjecture by showing counter examples. Also we characterize this digraph by deﬁning a suitable mapping from the vertex set to the real line. We study oriented interval catch digraphs and characterize an interval catch digraph when it is a tournament. Finally, we characterize a proper interval catch digraph and establish relationships between these digraph classes.


Introduction
Intersection graphs have many important applications in problems related to real-world situations. Because of its diverse applications in network science, different kinds of intersection graphs were introduced in modeling various geometric objects. Among them, interval graph is the most important one. A simple graph G = (V, E) is an interval graph if one can map each vertex into an interval on the real line so that any two vertices are adjacent if and only if their corresponding intervals intersect. In 1984 Maehara [11] introduced an analogous concept for directed graphs (in short, digraphs). He defined a catch digraph of F as a digraph G = (V, E) in which uv ∈ E if and only if u = v and p v ∈ S u where F = {(S u , p u ) | u ∈ V } is a family of pointed sets (a set with a point within it) in a Euclidean space. The digraph G is said to be represented by F . Later on, interval catch digraphs (in brief, ICD) was also studied by Prisner [14] in 1989 where S u is represented by an interval I u in the real line. A digraph G = (V, E) is unilaterally connected if for each pair of distinct vertices u, v ∈ V , there is a directed path from u to v or from v to u (or both). The underlying graph of a digraph G = (V, E) is an undirected graph U (G) = (V , E ) where V = V and E = {uv | uv ∈ E or vu ∈ E}. An oriented graph G = (V, E) is a digraph with no (directed) cycle of length two, i.e., if uv ∈ E, u, v ∈ V , then vu / ∈ E. A tournament is an oriented graph G = (V, E) where every pair of distinct vertices are adjacent, i.e., for any distinct pair u, v ∈ V , either uv ∈ E or vu ∈ E (but not both).
In our paper, we study some natural subclasses of ICD, namely central interval catch digraph (in brief, central ICD), oriented interval catch digraph (in brief, oriented ICD) and proper interval catch digraph (in brief, proper ICD). A central ICD is an ICD where the points p u are the center points of the intervals I u . This digraph was introduced by Maehara [11] in name of "interval digraph" 1 and he proposed a conjecture for characterization of central ICD in terms of forbidden subdigraphs. We disprove the conjecture by showing counter examples. Also we characterize this digraph by defining a suitable mapping from the vertex set to the real line. We prove that an oriented ICD is acyclic and study various properties of it. Then we characterize adjacency matrix of an oriented ICD when it is a tournament. A proper ICD is an ICD where no interval contains other properly. We obtain characterization of the adjacency matrix of a proper ICD. In conclusion we discuss relationships between these classes of digraphs. Henceforth undirected graphs will be called simply graphs. For some recent applications of ICD and in general, catch digraphs one may consult [4,5,6,13].

Preliminaries
A proper interval graph G is an interval graph in which there is an interval representation of G such that no interval is a proper subinterval of other. A unit interval graph is an interval graph in which there is an interval representation of G such that all intervals have the same length. Let v ∈ V for any graph G = (V, E). Then the set The reduced graphG is obtained from G by merging vertices having same closed neighborhood. G(n, r) is a graph with n vertices x 1 , x 2 , . . . , x n such that x i is adjacent to x j if and only if 0 < |i − j| ≤ r, where r < n is a positive integer [12]. Among many characterizations [8,9] of proper interval graph we list the following which will serve our purpose. 1. G is a proper interval graph.

G is a unit interval graph.
3.G is an induced subgraph of G(n, r) for some positive integers n, r with n > r. 4. There is an ordering of V such that for all v ∈ V , elements of N [v] are consecutive (the closed neighborhood condition). 5. There exists an ordering '<' on V such that u < v < w and uw ∈ E imply uv and vw ∈ E.
(umbrella property) The following characterization is known for interval catch digraphs.
Theorem 2.2. [14] Let G = (V, E) be a simple digraph. Then G is an ICD if and only if there exists an ordering " < " of V such that for x < y < z ∈ V, xz ∈ E =⇒ xy ∈ E and zx ∈ E =⇒ zy ∈ E. (2.1) A (0, 1)-matrix is said to satisfy consecutive 1's property for rows if its columns can be permuted in such a way that the 1's in each row occur consecutively. The augmented adjacency matrix A * (G) of a digraph G is obtained from the adjacency matrix A(G) of G by replacing 0's by 1's along the principal diagonal. It follows from (2.1) that with respect to the ordering of vertices of G = (V, E) described in Theorem 2.2, A * (G) satisfy consecutive 1's property for rows. We call this ordering an ICD ordering of V . This ordering is not unique for an ICD. For a simple digraph is a pointed interval representation of an ICD G, then the points can be made distinct by slight adjustment and the increasing ordering of these points is an ICD ordering of the corresponding vertices. In the following we frequently use the above conditions equivalent to (2.1) without explicitly mentioning this every time.

Central interval catch digraphs
In 1984, Maehara posed the following conjecture: Maehara's conjecture [11]: If a digraph G has no induced subdigraph isomorphic to one of the digraphs in Figure 1 and A * (G) satisfy the consecutive 1's property for rows, then G is a central ICD.  where i 1 and i 2 be the least and the highest numbers such that is an interval and c i be its center point. We arrange vertices of G according to the increasing order of their center points. It is easy to check G satisfy (2.1) with respect to this ordering. On contrary, let us assume for some i < j, i 1 > j 1 and i 2 > j 2 . Then Example 3.2. Consider the digraphs G k = (V k , E k ) with vertex set V k and edge set E k for k = 1, 2, 3, 4 in Figure 2. Note that G 4 is the digraph (a) in Figure 1. From Theorem 2.1 one can easily verify that v 1 < v 2 < v 3 < v 4 and its reverse ordering are the only possible ICD ordering of V k for each graph G k (i.e., for which A * (G k ) satisfies the consecutive 1's property for rows). Now this augmented adjacency matrix (in Figure 2) shows a contradiction to (3.1) for i = 2, j = 3 for each k. Thus these graphs are not central ICD.
In the following we characterize central ICD. Let R + be the set of all positive real numbers.
(i.e., every out-neighbor distance is less than every non-out-neighbor distance from a vertex) and only if c j ∈ I i and assume that v i 's are ordered according to increasing sequence of c i 's (without loss of generality, we also assume that c i 's are distinct). Define a labeling f : This vertices are ordered according to increasing order of their labels.
We consider following cases: Case 2. i < k < j or, j < k < i. These cases are not possible by (2.1) as G is an ICD.
Conversely, suppose G satisfies (3.2) with a labeling f . Let us arrange vertices according to increasing order of their labels. For each i = 1, 2, . . . , n, define c i = f (v i ). Let i 1 and i 2 be the least and the highest numbers such that As c i is the center point of I i for each i ∈ {1, . . . , n}, it is sufficient to prove that G is an ICD.
We verify (2.1) to show that G is an ICD. Let i < j < k and Example 3.4. Consider the digraph D 1 in Figure 5. It can be easily checked that Interestingly a family of (undirected) graphs satisfying the condition analogous to (3.2) becomes a well known class of graphs, namely, proper interval graphs.
Proof. Let G = (V, E) be a proper interval graph. By Theorem 2.1, the reduced graphG = (Ṽ ,Ẽ) of G is an induced subgraph of G(n, r) = (V n , E ) for some n, r ∈ N with n > r (see [12]).
if u is a k th copy among z copies of y j (for any but a fixed permutation of them). We arrange the vertices of V according to the increasing order of vertices inṼ keeping copies of same vertices together.
with a labeling f . We arrange the vertices of G according to the increasing order of their labels.
Thus G satisfies umbrella property and hence by Theorem 2.1, G becomes a proper interval graph.

Oriented interval catch digraph
We first note that the characterization of an oriented ICD is obvious. A simple digraph G = (V, E) is an oriented ICD if and only if G is oriented and there exists an ordering of vertices in V satisfying (2.1) or, equivalently, there exists an ordering of vertices in V such that A * (G) = (a i,j ) satisfies consecutive 1's property for rows and for each pair i = j, a i,j = 1 implies a j,i = 0. Now we study some important properties of oriented ICD. Proof. Let G be an oriented ICD. We first show that G has no induced directed cycles. Note that any induced subdigraph of an ICD is also an ICD by definition. Now it is easy to check that no induced directed cycles of length ≥ 3 satisfies (2.1) and so they are not ICD. Finally, since the digraph is oriented, there are no 2-cycles. Now we use induction on the length of the cycle. Since G has no induced directed cycles, in particular, G has no directed cycles of length 3. Suppose there are no directed cycles of length less than k and G has a directed cycle C of length k > 3. Since C is not induced, there must be a chord. Interestingly, every such arc (chord) forms a smaller directed cycle on one side of it along with some arcs of C. This contradicts the induction hypothesis.
The following corollary is immediate from the above theorem. In [11] Maehara proved that an acyclic ICD is a central ICD. Thus we have the following:    Now being an acyclic oriented graph, an oriented ICD has some rich properties. For example, a unilaterally connected oriented ICD has unique source (vertex of indegree zero) and unique sink (vertex of outdegree zero). Hence it possesses unique hamiltonian path (see [1]). In the following we characterize oriented ICDs which are tournaments.
A Ferrers digraph is a directed graph G = (V, E) whose successor sets are linearly ordered by inclusion, where the successor set of u ∈ V is its set of out-neighbors {v ∈ V | uv ∈ E}. A (0, 1)matrix M is a Ferrers matrix if 1's are clustered in a corner of M . A digraph G is Ferrers digraph if and only if there exists a permutation of vertices of G such that its adjacency matrix is a Ferrers matrix [2,7]. For a (0, 1)-matrix M , M denotes the matrix obtained from M by interchanging 0's and 1's. Theorem 4.6. Let G = (V, E) be an oriented ICD. Then G is a tournament if and only if there is an ordering of vertices of G with respect to which the augmented adjacency matrix A * (G) takes one of the following forms: where M is an upper triangular matrix with all entries on and above the principal diagonal are 1 and other enties are 0; N and P are lower triangular matrices with all entries on and below the principal diagonal are 1 and other entries are 0 and F is a Ferrers matrix.
Proof. Let G = (V, E) be an oriented ICD which is a tournament. We order the vertices of V according to the increasing value of the associated points in their corresponding intervals. Let {v 1 , . . . , v n } be the required ordering which is in fact, an ICD ordering, i.e., {v 1 , . . . , v n } satisfies (2.1). Hence A = A * (G) satisfies consecutive 1's property for rows with respect to this ordering. For each 1 ≤ i ≤ n, let (v i ) 1 and (v i ) 2 be the column numbers where first and last 1 occur in the i-th row. We denote the (i, j)-th entry of the matrix A by a i,j .
Subcase 1(a). If (v 2 ) 2 = 2 then 2 = (v 2 ) 2 = (v 2 ) 1 which implies the vertex v 2 has out degree 0. In this case we define l = 2 (note that the matrix M in the statement is of the order l × l).
Already we have shown that 2 ∈ S. This implies S = ∅. As |V | is finite S must have a maximum, say, l. First we will show that {i | 2 ≤ i ≤ l} ⊆ S. As 2 ∈ S, applying induction we assume {2, 3, . . . , j} ⊆ S. If j + 1 ≤ l, then we show that j + 1 ∈ S.
In this case a 1,j = 0 for all j such that 1 < j ≤ n. Since G is a tournament this implies a j,1 = 1 for all such j. Then all the entries in the first column of A are 1. Also since A is the augmented adjacency matrix of G, a j,j = 1 for all j = 1, 2, . . . , n. Moreover, A satisfies consecutive 1's property for rows. Thus we have a j,i = 1 for all i such that 1 ≤ i ≤ j, for all j = 1, 2, . . . , n. Finally since G is a tournament a i,j = 0 for all i < j, for all j = 1, 2, . . . , n. Therefore A is the lower triangular matrix P with all entries 1 on and below the principal diagonal.
The converse part is obvious from the structure of A * (G) as both the matrices described in the statement have consecutive 1's property for rows and for each pair i = j, a i,j = 0 if and only if a j,i = 1. Thus G is an ICD which is a tournament. [4,9]

Proper interval catch digraph
Let G = (V, E) be a simple digraph. Then the augmented adjacency matrix A * (G) has a monotone consecutive arrangement (MCA) [2] if and only if it has independent row and column permutations such that 1's appear consecutively in each row and 1 1 ≤ 2 1 ≤ . . . ≤ n 1 and 1 2 ≤ 2 2 ≤ . . . ≤ n 2 where |V | = n and the values i 1 and i 2 denote the initial column and final column containing 1's in the i-th row. Now one can seperate the 1's and 0's by drawing upper and lower stairs (polygonal path from top left to bottom right) as in Figure 4 (right). In the following we give an adjacency matrix characterization of a proper ICD. Now for i = j suppose i 1 < j 1 and i 2 > j 2 hold simultaneously in A * (G). Then p i 1 / ∈ I j as i 1 < j 1 . Again i 1 < j 1 ≤ j implies p i 1 < p j 1 ≤ p j . Since p j ∈ I j , the interval I j lies entirely right to p i 1 . So we get a i < p i 1 < a j . As the intervals do not contain other properly we have b i < b j . Now as p i 2 ∈ I i and p i 2 > p j (for i 2 > j 2 ≥ j) we get a j ≤ p j < p i 2 < b i < b j . Then p i 2 ∈ I j which is a contraction as i 2 > j 2 . Hence i 2 ≤ j 2 . 3 Conversely, let G satisfy conditions (1) and (2). We permute the rows of A = A * (G) = (a i,j ) according to the non-decreasing order of i 1 's keeping the columns intact (see Figure 4). If there exist pair of rows i, j for which i 1 = j 1 , then place i-th row prior to j-th row when i 2 < j 2 . If i 1 = j 1 and i 2 = j 2 then keep i-th row prior to j-th row only when i < j in A. Let us call the modified matrix by B = (b i,j ). As A satisfies (2), after permuting only the rows of A it is easy to check that B satisfies MCA. Moreover permutation of rows of A does not affect the adjacency of the graph G, i.e., if i-th row of A is shifted to k-th row in B, then a i,j = 1 if and only if b k,j = 1. (5.1) Step 1. We show that the diagonal entries of the matrix B must be 1. Suppose the i-th row of B was the k-th row of A. If k = i then b i,i = 1 as a i,i = 1. Suppose k > i in A. For each j = 1, 2, . . . , i, j 1 ≤ j ≤ i. Now the k-th row of A moved upward to the i-th row position in B. This implies there exists at least one such j so that k 1 ≤ j 1 . Then k 1 ≤ i < k ≤ k 2 . Hence a k,i = 1 which implies b i,i = 1. Again if k < i in A then there exists at least one j ≥ i such that j 1 < k 1 or, j 1 = k 1 and j 2 < k 2 . Then by the condition (2), in either case, we have j 2 ≤ k 2 . Thus k 1 ≤ k < i ≤ j ≤ j 2 ≤ k 2 . This implies a k,i = 1 and hence b i,i = 1.
Step 2. Defining intervals I i = [l i , r i ] for the i-th row of B.
By (5.2), b i > a io and so b i ≥ a io + 1 > l i . Thus l i < r i for all i = 1, 2, . . . , n. Also since B satisfies MCA, one can check that l i < l j and r i < r j for all i < j. Therefore the set of intervals [l i , r i ] gives proper representation.
Step 3. Defining points p j for each column j of B (as well as of A). Let S j = {l i | i ≥ j, b i,j = 1}. Note that S j = ∅ as b j,j = 1. Assign p j = max {a j , max S j }. Now p j ≥ l j as l j ∈ S j . Suppose for some i ≥ j, b i,j = 1. Then i o ≤ j ≤ i e which implies a io ≤ a j . So l i < a io + 1 ≤ a j + 1 ≤ b j . Also since b j,j = 1 by Step 1, we have a j < b j by (5.2). Thus p j < b j . Hence p j ∈ [l j , r j ]. and uw ∈ E. Suppose u, v, w correspond to rows i, j, k respectively in A * (G) = (a i,j ), where i < j < k. Since the matrix is augmented, a i,i = 1. Also uw ∈ E implies a i,k = 1. Thus by consecutive 1's property for rows, we have a i,j = 1 which implies uv ∈ E. Now the digraph is oriented. Thus a j,i = a k,i = 0. Then j 1 > i ≥ i 1 . Thus by condition (2), j 2 ≥ i 2 . But a i,k = 1 and so i 2 ≥ k. Then j 2 ≥ k which implies a j,k = 1. Hence vw ∈ E. The other part can be proved similarly.
Conversely, let G = (V, E) be an oriented graph satisfying the given condition. By Theorem 2.2, it follows that G is an ICD and so it satisfies the condition (1) of Theorem 5.1. Suppose in A * (G) = (a i,j ), i 1 < j 1 for two distinct rows i, j. Now if we can show that a j,i 2 = 1, then it follows that i 2 ≤ j 2 as required. If i 2 ≤ j, then i 2 ≤ j 2 as j ≤ j 2 . Let i 2 > j. If i < j, then i < j < i 2 and a i,i 2 = 1. Then by the given condition (first part), we have a j,i 2 = 1. Now if i > j, then i 1 < j 1 ≤ j < i and a i,i 1 = 1. Thus by the given condition (second part), we have a j,i 1 = 1. But this implies j 1 ≤ i 1 which is a contradiction.
Remark 5.4. It follows from the proof of the converse part of the above theorem that in the case of a proper oriented ICD G, A * (G) itself satisfies MCA which may not be true for a proper ICD, in general.

Conclusion
In this paper we consider three subclasses of the class of ICD, namely central ICD, oriented ICD and proper ICD. We obtain characterizations for central ICD, oriented ICD when it is a tournament and proper ICD. In the following we show relationships between these digraphs in Figure  5 whose proofs follow from these characterization theorems. Several combinatorial optimization problems remain open for these digraphs, for example, to find the complete list of forbidden digraphs or to construct recognition algorithms for them.
Oriented ICD Proper oriented ICD Figure 5: Examples of relations between some subclasses of ICD