Claw-freeness, 3-homogeneous subsets of a graph and a reconstruction problem

# Claw-freeness, 3-homogeneous subsets of a graph and a reconstruction problem

## Abstract

We describe , the class of graphs such that and its complement are claw-free. With few exceptions, it is made of graphs whose connected components consist of cycles of length at least 4, paths, and of the complements of these graphs. Considering the hypergraph made of the -element subsets of the vertex set of a graph on which induces a clique or an independent subset, we deduce from above a description of the Boolean sum of two graphs and giving the same hypergraph. We indicate the role of this latter description in a reconstruction problem of graphs up to complementation.

pouzet@univ-lyon1.fr

sikaddour@univ-lyon1.fr

nicolas.trotignon@liafa.jussieu.fr

Done under the auspices of the French-Tunisian CMCU “Outils mathématiques pour l’Informatique” 05S1505

## 1Results and motivation

Our notations and terminology mostly follow . The graphs we consider in this paper are undirected, simple and have no loop. That is a graph is a pair , where is a subset of , the set of -element subsets of . Elements of are the vertices of and elements of its edges. We denote by the vertex set of and by its edge set. We look at members of as unordered pairs of distinct vertices. If is a subset of , the pair is the graph induced by on . The complement of is the simple graph whose vertex set is and whose edges are the unordered pairs of nonadjacent and distinct vertices of , that is , where . We denote by the complete graph on vertices and by the graph made of a vertex linked to a . The graph is called a claw, the graph

a co-claw.

In , Brandstädt and Mahfud give a structural characterization of graphs with no claw and no co-claw; they deduce several algorithmic consequences (relying on bounded clique width). We will give a more precise characterization of such graphs.
We denote by the graph on vertices made of a bounded by three (cf. Figure 1) and by the -element cycle, . We denote by the Paley graph on vertices (cf. Figure 1). Note that is isomorphic to its complement , to the line-graph of and also to , the cartesian product of by itself (see  page 30 if needed for a definition of the cartesian product of graphs, and see  page 176 and  page 28 for a definition and basic properties of Paley graphs). Figure 1

Given a set of graphs, we denote by the class of graphs such that no member of is isomorphic to an induced subgraph of . Members of , resp. are called triangle-free, resp. claw-free graphs.

The main result of this note asserts:

As an immediate consequence of Theorem , note that the graphs and are the only members of which contain a and a with no vertex in common. Note also that and are very important graphs for the study of how maximal cliques and stable sets overlap in general graphs. See the main theorem of , see also . Also, in , page 31, a list of all self-complementary line-graphs is given. Apart from , they are all induced subgraphs of .
From Theorem we obtain a characterization of the Boolean sum of two graphs having the same -homogeneous subsets. For that, we say that a subset of vertices of a graph is homogeneous if it is a clique or an independent set (note that the word homogeneous is used with this meaning in Ramsey theory; in other areas of graph theory it has other meanings, several in fact). Let be the hypergraph having the same vertices as and whose hyperedges are the -element homogeneous subsets of . Given two graphs and on the same vertex set , we recall that the Boolean sum of and is the graph on whose edges are unordered pairs of distinct vertices such that if and only if . Note that is the symmetric difference of and . The graph is also called the symmetric difference of and and denoted by in . Given a graph with vertex set , the edge-graph of is the graph whose vertices are the edges of and whose edges are unordered pairs such that , for three distinct elements such that is not an edge of . Note that the edge-graph is a spanning subgraph of , the line-graph of , not to be confused with it.

Claw-free graphs and triangle-free graphs are related by means of the edge-graph construction. Indeed, as it is immediate to see, for every graph , we have:

Our characterization is this:

As a consequence, if the graph satisfying Property (1) is disconnected, then contains no -element cycle, moreover, if contains no -element cycle then each connected component of is a cycle of even length, or a path, in particular is bipartite.

The implication in Theorem ? follows immediately from Theorem . Indeed, suppose that Property (2) holds, that is and are bipartite, then from Formula () and from the fact that and , , are respectively isomorphic to and to , we have:

From Theorem , Property (3) holds. The other implications, obtained by more straigthforward arguments, are given in subSection 2.3.

This leaves open the following:

A partial answer, motivated by the reconstruction problem discussed below, is given in . We mention that two graphs and as above are determined by the graphs induced on the connected components of and on a system of distinct representatives of these connected components (Proposition 10 ).

A quite natural problem, related to the study of Ramsey numbers for triples, is this:

An asymptotic lower bound of the size of in terms of was established by A.W. Goodman .

The motivation for Theorem ? (and thus Theorem ) originates in a reconstruction problem on graphs that we present now. Considering two graphs and on the same set of vertices, we say that and are isomorphic up to complementation if is isomorphic to or to the complement of . Let be a non-negative integer, we say that and are -hypomorphic up to complementation if for every -element subset of , the graphs and induced by and on are isomorphic up to complementation. Finally, we say that is -reconstructible up to complementation if every graph which is -hypomorphic to up to complementation is in fact isomorphic to up to complementation. The following problem emerged from a question of P.Ille :

It is immediate to see that if the conclusion of the problem above is positive for some , then is distinct from and and, with a little bit of thought, that if then (see Proposition 4.1 of ). With J. Dammak, G. Lopez  and  we proved that the conclusion is positive if:

1. or

2. and .

We do not know if in (ii) the condition can be dropped. For , we checked that the conclusion holds if and noticed that for larger values of it could be negative or extremely hard to obtain, indeed, a positive conclusion would imply that Ulam’s reconstruction conjecture holds (see Proposition 19 of ).

The reason for which Theorem ? plays a role in that matter relies on properties of incidence matrices.

Given non-negative integers , , let be the by incidence matrix of ’s and ’, the rows of which are indexed by -element subsets of , the colums are indexed by the -element subsets of , and where the entry is if and is otherwise.

Let and be the column vector associated to the graph . The matrix product where the computation is made in the two elements field is if and only if the number of edges of and have the same parity for all ’s, a condition satisfied if and are -hypomorphic up to complementation and or . According to R.M. Wilson , the dimension (over ) of the kernel of is if and that is is the constant matrix or , and thus is equal to or to . If , the dimension is and the kernel consists of (the colum matrices of) complete bipartite graphs and their complement . If we add the fact that and have the same -homogeneous subsets then, according to Theorem ?, is claw and co-claw free. If , it follows readily that is either the empty graph or the complete graph. Hence is equal to or to . If , it turns out that two graphs and which are -hypomorphic up to complementation are -hypomorphic up to complementation, which amounts to the fact that and have the same -homogeneous subsets, thus in the case , and are equal up to complementation. Indeed, a famous Gottlieb-Kantor theorem on incidence matrices () asserts that the matrix has full row rank over the field of rational numbers provided that . From which follows that:

Up to now, Wilson theorem has not been applied successfully to the cases and . Instead, efforts concentrated on the structure of pairs of -hypomorphic graphs and with the same -homogeneous subsets. The form of their Boolean sum as given in (3) of Theorem ? was the first step of a description. With that in hands, it was shown in  that the additional hypothesis that and are -hypomorphic to complementation for some , , was enough to ensure that and are isomorphic up to complementation.

## 2Proofs

Let be a graph. For an unordered pair of distinct vertices, we set if and otherwise. Let ; we denote by and the neighborhood and the degree of (that is and ). For , we set .

### 2.1Proof of Theorem .

Trivially, the graphs described in Theorem belong to . We prove the converse.

The diamond is the graph on four vertices with five edges. We say that a graph contains a graph when has an induced subgraph isomorphic to .

Since  is very difficult to find, we include a short proof. Checking that a line-graph of a triangle-free graph contains no claw and no diamond is a routine matter. Conversely, let be graph with no claw and no diamond. A theorem of Beineke  states that there exists a list of nine graphs such any graph that does not contain a graph from is a line-graph. One of the nine graphs is the claw and the eight remaining ones all contain a diamond. So, for some graph . Let be the graph obtained from by replacing each connected component of isomorphic to a triangle by a claw. So, . We claim that is triangle-free. Else let be a triangle of . From the construction of , there is a vertex in the connected component of that contains . So we may choose with a neighbor in . Now the edges of and one edge from to induce a diamond of , a contradiction.

Let be in the class .

(1) We may assume that and are connected.

Else, up to symmetry, is disconnected. If contains a vertex of degree at least , then contains an edge (for otherwise there is a claw), so contains a triangle. This is a contradiction since by picking a vertex in another component we obtain a co-claw. So all vertices of are of degree at most . It follows that the components of are cycles (of length at least , or there is a co-claw) or paths, an outcome of the theorem. This proves (1).
(2) We may assume that and contain no induced path on six vertices.
Else has an induced subgraph that is either a path on at least vertices or a cycle on at least vertices. Suppose maximal with respect to this property. If then we are done. Else, by (1), we pick a vertex in with at least one neighbor in . From the maximality of , has a neighbor in the interior of some of . Up to symmetry we assume that v has a neighbor where . So contains an edge for otherwise induces a claw. If then must be adjacent to for otherwise there is a co-claw; so induces a claw. If then must be adjacent to for otherwise there is a co-claw, so from the symmetry between and we may rely on the previous case. If then must be adjacent to for otherwise there is a co-claw; so induces a claw. In all cases there is a contradiction. This proves (2).
(3) We may assume that and contain no .
Suppose that contains . Then, let be three disjoint edges of such that the only edges between them are . If , an outcome of the theorem is satisfied, so let be a seventh vertex of . We may assume that (else there is a co-claw). If then (else there is a co-claw) so is a claw. Hence . We have (or is a claw) and similarly . So is a co-claw. This proves (3).
(4) We may assume that and contain no diamond.
Suppose for a contradiction that contains a diamond. Then, contains a co-diamond, that is four vertices that induce only one edge, say . By (1), there is a path from to some vertex that has a neighbor in . We choose such a path minimal and we assume up to symmetry that the path is from .

If is adjacent to both then induces a co-claw unless is adjacent to , similarly is adjacent to , so induces a claw. Hence is adjacent to exactly one of , say to a. So, is an induced path and for convenience we rename its vertices . If has a neighbor in then, from the minimality of , this neighbor must be . So, induces a claw. Hence, has no neighbor in .

By (1), there is a path from to some vertex that has a neighbor in . We choose minimal with respect to this property. From the paragraph above, . Let (resp. ) be the neighor of in with minimum (resp. maximum) index. If then is a path on at least vertices a contradiction to (2). So, if then and symmetrically, , so is a claw. Hence . If then , where is the neighbor of along , is a claw. So, . So is a triangle. Hence , and , for otherwise there is a co-claw. Hence, form an induced of , a contradiction to (3). This proves (4).

Now is connected and contains no claw and no diamond. So, by Theorem ?, is the line-graph of some connected triangle-free graph . Symmetrically, is also a line-graph. These graphs are studied in .

If contains a vertex of degree at least then all edges of must be incident with , for else an edge non-incident with together with three edges of incident with and non-adjacent to form a co-claw in . So all vertices of have degree at most since otherwise, is a clique, a contradiction to (1). We may assume that has a vertex of degree for otherwise is a path or a cycle. Let be the neighbors of . Since has degree , all edges of must be incident with or for otherwise contains a co-claw.

If one of , say , is of degree , then and all edges of are incident with one of (or there is a co-claw). So is a subgraph of . So, since , is an induced subgraph of , an outcome of the theorem. Hence we assume that are of degree at most . If , then contains the pairwise non-adjacent edges say, and the edges are vertices of that induce an , a contradiction to (3). So, which means again that is a subgraph of .

### 2.2Ingredients for the proof of Theorem .

The proof of the equivalence between Properties (1) and (2) of Theorem ? relies on the following lemma.

Observe first that Property (b) is equivalent to the conjunction of the following properties:
: If is an edge of then iff .
and
: If is an edge of then iff .

. Let us show .
Let , then . By contradiction, we may suppose that (the other case implies thus is similar). Since and are edges of then . Let such that , . Then and thus iff .
If , is a homogeneous subset of . Since and have the same -element homogeneous subsets, is an homogeneous subset of . Hence, since , , thus , a contradiction.
If , then ; since it follows that is a homogeneous subset of . Consequently is a homogeneous subset of . Since , then , a contradiction.
The implication is similar.
. Let be a of . Suppose that is not a homogeneous subset of then we may suppose with and or and . In the first case , which contradicts Property , in the second case , which contradicts Property .
. First and . Let be two distinct elements of (respectively ). Then (respectively ). From we have . Then and are independent sets of . The proof that and are independent sets of is similar.
. This implication is trivial.

### 2.3Proof of Theorem .

Implication follows directly from implication of Lemma ?. Indeed, Property (c) implies trivially that and are bipartite.
. Suppose that and are bipartite. Let and be respectively a partition of and into independent sets. Note that , for . Let be two graphs with the same vertex set as such that and . Clearly . Thus . To conclude that Property holds, it suffices to show that and have the same -element homogeneous subsets, that is Property (a) of Lemma ? holds. For that, note that , , and and thus Property (c) of Lemma ? holds. It follows that Property (a) of this lemma holds.

The proof of implication was given in Section 1. For the converse implication, let be a graph satisfying Property (3). It is clear from Figure 1 that is bipartite (vertical edges and horizontal edges form a partition). Since is isomorphic to , is bipartite too. Thus, if is isomorphic to an induced subgraph of , Property (2) holds. If not, we may suppose that the connected components of are cycles of even length or paths (otherwise, replace by ). In this case, is trivially bipartite. In order to prove that Property 2 holds, it suffices to prove that is bipartite too. This is a direct consequence of the following claim:

If is a colouring of , set defined by .

With this, the proof of Theorem ? is complete.

### 2.4A direct proof for (3) ⟹ (1) of Theorem .

In  we gave all possible decompositions of a graph satisfying (1) into a Boolean sum where and have the same -element homogeneous sets.

When , a decomposition can be given by a picture (see Figure 2). Figure 2

For the other cases, we introduce the following notation.

Let . Let be an -element set, be an enumeration of , and . Set , , . Let and be the graphs with vertex set and edge sets and respectively. Let for even, . For we give a picture (see Figure 3). For convenience, we set the graph with one vertex and we put . When is a graph of the form , , or , with , we put and . Figure 3

When is a cycle of even size , a decomposition can be given by and . When is a path of size , a decomposition can be given by and .

When the connected components of are cycles of even length or paths, we define and satisfying as follows: For each connected component of , is given by the previous step. For distinct connected components and of , , , (and ) if and only if and , or and .

When the connected components of are cycles of even length or paths, from , the previous step gives a pair , then a pair .

We thank S. Thomassé for his helpful comments. We thank the anonymous referee for his careful examination of the paper and his suggestions.

### References

1. L.W. Beineke, Characterizations of derived graphs, J. Combinatorial Theory 9 (1970) 129-135.
2. L.W. Beineke, Derived graphs with derived complements, In Lecture Notes in Math. (Proc. Conf., New York, 1970) pages 15-24. Springer (1971).
3. J.A. Bondy, U.S.R. Murty, Graph Theory, Basic Graph Theory, Graduate Texts in Mathematics, vol 244, Springer, 2008, 651 pp.
4. A. Brandstädt and S. Mahfud, Maximum weight stable set on graphs without claw and co-claw (and similar graph classes) can be solved in linear time, Information Processing Letters 84 (2002) 251-259.
5. J. Dammak, G. Lopez, M. Pouzet, H. Si Kaddour, Hypomorphy up to complementation, JCTB, Series B 99 (2009) 84-96.
6. J. Dammak, G. Lopez, M. Pouzet, H. Si Kaddour, Reconstruction of graphs up to complementation, in Proceedings of the First International Conference on Relations, Orders and Graphs: Interaction with Computer Science, ROGICS08, May 12-15 (2008), Mahdia, Tunisia, pp. 195-203.
7. X. Deng, G. Li, W. Zang, Proof of Chvátal’s conjecture on maximal stable sets and maximal cliques in graphs, JCTB, Series B 91 (2004) 301-325.
8. X. Deng, G. Li, W. Zang, Corrigendum to proof of Chvátal’s conjecture on maximal stable sets and maximal cliques in graphs, JCTB, Series B 94 (2005) 352-353.
9. A.W. Goodman, On sets of acquaintances and strangers at any party, Amer. Math. Monthly 66 (1959) 778-783.
10. A. Farrugia. Self-complementary graphs and generalisations: a comprehensive reference manual, Master’s thesis, University of Malta (1999).
11. A class of incidence matrices, Proc. Amer. Math. Soc. 17 (1966) 1233-1237.
D.H. Gottlieb,
12. P. Ille, personnal communication, September 2000.
13. F. Harary and C. Holzmann, Line graphs of bipartite graphs, Rev. Soc. Mat. Chile 1 (1974) 19-22.
14. W. Kantor, On incidence matrices of finite projective and affine spaces, Math.Zeitschrift 124 (1972) 315-318.
15. J.H. Van Lint, R.M. Wilson, A course in Combinatorics, Cambridge University Press (1992).
16. R.M. Wilson, A Diagonal Form for the Incidence Matrices of -Subsets -Subsets, Europ J. Combinatorics 11 (1990) 609-615.
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