1 Introduction
Let be a -matrix. We denote and the th row and the th column of matrix . Let and for each . Two rows and are disjoint if there is no such that . We say that is contained in if for each such that also . We say that and are nested if is contained in or is contained in . Finally, we say that and start (resp. end) in the same column if (resp. ), and we say and start (end) in different columns otherwise. We say a -matrix has the consecutive ones property for the rows (for short, C1P) if there is permutation of the columns of such that the 1’s in each row appear consecutively. Tucker characterized all the minimal forbidden submatrices for the C1P, later known as Tucker matrices. For the complete list of Tucker matrices, see [5], where a graphic representation of them can be found in Figure 3.
We say a -matrix is nested if it has the consecutive ones property for the rows (C1P) and every two rows are either disjoint or nested. We say a -matrix is 2-nested if it has the C1P for the rows and there is a partition of the rows such that each submatrix obtained is nested.
All graphs in this work are simple, undirected, with no loops and no multiple edges. The pair is a split partition of a graph if is a partition of the vertex set of and the vertices of (resp. ) are pairwise adjacent (resp. nonadjacent), and we denote it . A graph is a split graph if it admits some split partition. Let be a split graph with split partition , , and . Let and be linear orderings of and , respectively. Let be the matrix defined by if is adjacent to and , otherwise.
A split graph is nested (resp. 2-nested) if there is a linear ordering of , such that the associated matrix is nested (resp. 2-nested) and if its columns are ordered as in then the ones in each row occur in consecutive columns.
Circle graphs [2] are intersection graphs of chords in a circle. These graphs were characterized by Bouchet [1] in 1994 by forbidden induced subgraphs under local complementation. However, no complete characterizations of circle graphs by forbidden induced subgraphs of the graph itself are known. It follows from the definition that nested and 2-nested graphs are common subclasses of circle graphs. Furthermore, nested and 2-nested graphs are also a superclass of threshold graphs (see Golumbic [4] for more details on these definitions).
The problem of characterizing 2-nested graphs by minimal forbidden induced subgraphs arises as a natural subproblem in our ongoing efforts to obtain the same kind of characterization of those split graphs that are circle graphs. We started by considering a split graph such that is minimally non-circle. Since comparability graphs are a subclass of circle graphs, in particular is not a comparability graph. Notice that permutation graphs are those comparability graphs for which their complement is also a comparability graph. It is easy to prove that permutation graphs are precisely those circle graphs having a circle model with an equator. See Gallai [3] for the complete list of minimal forbidden subgraphs of comparability graphs. Using the list of minimal forbidden subgraphs of comparability graphs and the fact that is also a split graph, we conclude that contains either a tent, a 4-tent, a co-4-tent or a net as a subgraph. We first considered the case in which contains an induced tent as a subgraph, thus reaching a problem when trying to give a circle model for . Once analyzed the compatibilities between the vertices in the complete and independent partitions of such a graph, it arises that there is exactly one subclass –which we denoted – of independent vertices for which both endpoints of each vertex could be entirely drawn in two distinct areas of the circle model, when for every other vertex there is a unique possible placement. Hence, for the subgraph induced by taking the tent graph union the subclass to admit a circle model, the subclass must be partitioned into two disjoint subsets such that, for each subset, every pair of vertices are either disjoint or nested, thus leading to the definition of 2-nested graphs.
2 Nested matrices
We begin by giving the following characterization of nested matrices.
Theorem 1.
A -matrix is nested if and only if it contains no as a submatrix (see Figure 1).
Proof.
Since no Tucker matrix has the C1P and the rows of are neither disjoint nor nested, no nested matrix contains a Tucker matrix or as submatrices. Conversely, as each Tucker matrix contains as a submatrix, every matrix containing no as a submatrix is a nested matrix. ∎
Corollary 2.
A split graph is nested if and only if it contains no induced gem.
3 2-nested matrices
We define the following matrices, since they play an important role in the sequel.
, for any odd
.Theorem 3.
A -matrix is 2-nested if and only if there is a linear ordering of the columns such that the matrix with its columns ordered according to does not contain any Tucker matrix, or , , for every odd as a configuration.
We define the auxiliary graph where the vertex set has one vertex for each row in , and two vertices and in are adjacent if and only if the rows and are neither disjoint nor nested. By abuse of language, will refer to both the vertex in and the row of . In particular, the definitions given in the introduction apply to the vertices in ; i.e., we say two vertices and in are nested (resp. disjoint) if the corresponding rows and are nested (resp. disjoint). And two vertices and in start (resp. end) in the same column if the corresponding rows and start (resp. end) in the same column. It follows from the definition of 2-nested matrices that is a 2-nested matrix if and only if there is a bicoloring of the auxiliary graph or, equivalently, if is bipartite (i.e., does not contain cycles of odd length).
Proof.
Since admits a C1P, then contains no Tucker matrices. Moreover, if contains , or for some odd , since the corresponding subgraphs in of every such matrix induces an odd cycle, then it follows that does not admit a proper 2-coloring and this results in a contradiction. Therefore, does not contain any , or for any odd as a configuration.
Conversely, let be a linear ordering of the columns such that the matrix does not contain any for any odd or Tucker matrices as configurations. Due to Tucker’s Theorem, since there are no Tucker submatrices in , the matrix has the C1P.
Towards a contradiction, suppose that the auxiliary graph is not bipartite. Hence there is an induced odd cycle in .
Suppose first that has an induced odd cycle of length 3, and suppose without loss of generality that the first rows of are those corresponding to the cycle . Since and are adjacent, both begin and end in different columns. The same holds for and , and and . We assume without loss of generality that the vertices start in the order of the cycle, in other words, that .
Since starts first, it is clear that , thus the column of is the same as the first column of the matrix .
Since has the C1P and and are adjacent, then . As stated before, starts before and thus . Hence, column is equal to the second column of .
The third column of is , for is adjacent to and , hence it is straightforward that .
To find the next column of , let us look at column . Notice that . Since is adjacent to and , and and both start after , then necessarily , and thus is equal to the fourth column of .
Finally, we look at the column . Notice that . Since has the C1P, and , then and , which is equal to last column of . Therefore we reached a contradiction that came from assuming that there is a cycle of length 3 in .
Suppose now that has an induced odd cycle of length . We assume without loss of generality that the first rows of are those in and that is ordered according to the C1P.
Remark 1.
Let be vertices in . If and are adjacent and starts before , then and , .
Remark 2.
If and for some , then for all , is nested in . The same holds if and . Since and , then and are not disjoint, thus necessarily is nested in . It follows from this argument that this holds for .
Notice that and are nonadjacent, hence they are either disjoint or nested. Using this fact and Remark 1, we split the proof into two cases.
Case 1.
and are nested
We may assume without loss of generality that is nested in , for if not, we can rearrange the cycle backwards as , , . Moreover, we will assume without loss of generality that both and start before . First, we need the following Claim.
Claim 1.
If and are nested, then is nested in , for .
Suppose first that and are disjoint, and towards a contradiction suppose that and are disjoint. In this case, . The contradiction is clear if . If instead and starts before , then for all , which contradicts the assumption that is nested in . Hence, necessarily is nested in and and are disjoint. This implies that and once more, for all , which contradicts the fact that is nested in .
Suppose now that is nested in . Towards a contradiction, suppose that is not nested in . Thus, and are disjoint since they are nonadjacent vertices in . Notice that, if is nested in , then and . Furthermore, since is adjacent to and nonadjacent to , then . This holds for every odd .
If , since is nested in , then , which results in a contradiction for and are adjacent.
Suppose that . If and are disjoint for all , then and are nonadjacent for is nested in , which results in a contradiction. Conversely, if and are not disjoint for some , then they are adjacent, which also results in a contradiction that came from assuming that and are disjoint. Therefore, since is nested in , and are nonadjacent and is adjacent to for all , then necessarily is nested in , which finishes the proof of the Claim.
Claim 2.
Suppose that and are nested. Then, if is nested in , then for all . If instead and are disjoint, then for all .
Recall that, by the previous Claim, since is nested in for all , in particular is nested in . Moreover, since and are adjacent, notice that, if is nested in , then , and if and are disjoint, then .
It follows from Remark 2 that, if , then is nested in for all , which contradicts the fact that and are adjacent. The proof of the first statement follows from applying this argument successively.
The second statement is proven analogously by applying Remark 2 if , and afterwards successively for all .
If and are disjoint, then we obtain first, by putting the first row as the last row, and considering the submatrix given by columns , , , , , (using the new ordering of the rows). If instead is nested in , then we obtain by taking the submatrix given by the columns , , , , , .
Case 2.
and are disjoint
We assume without loss of generality that and .
Claim 3.
If and are disjoint, then for all .
Notice first that, in this case, is nested in , for all . If not, then using Remark 2, we notice that it is not possible for the vertices to induce a cycle. This implies, in particular, that is nested in and thus . Furthermore, using this and the same remark, we conclude that for all , therefore proving Claim 3.
In this case, we obtain by considering the submatrix given by the columns , , , , , .∎
References
- [1] A. Bouchet, Circle graph obstructions, J. Combin. Theory B 60 (1994), 107–144.
- [2] S. Even, A. Itai, Queues, stacks and graphs, in: A. Kohavi, A. Paz (Eds.), Theory of Machines and Computations, Academic Press, New York (1971), 71–86.
- [3] T. Gallai, Transitiv orientierbare Graphen, Acta Mathematica Academiae Scientiarum Hungaricae 18 (1-2) (1967), 25–66.
- [4] M. C. Golumbic, Algorithmic graph theory and perfect graphs, Elsevier Science B.V., Amsterdam (2004).
- [5] A.C. Tucker, A structure theorem for the consecutive 1’s property, J. Combin. Theory 12 (B) (1972), 153–162.
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