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On the number of labeled graphs in hereditary classes The speed of hereditary properties Let K be the class of all complete graphs and O be the class of all empty (edgeless) graphs Theorem. A hereditary class X is finite if and only if neither K nor O is a subclasses of X. Definition. An infinite hereditary class X is constant if there is a constant N such that for every n1 the number of n-vertex graphs in X is bounded by N. Let S be the class of stars, i.e. graphs of the form K1,p, and all their induced subgraphs E1 be the class of graphs with at most one edge For any hereditary class X, denote by Xn the set of n-vertex graphs in X Xcomp the class of complements of graphs in X. For any hereditary class X, |Xn|=|Xncomp|. |Sn|=n+1 |E1n|=n(n-1)/2 + 1 On the number of labeled graphs in hereditary classes The speed of hereditary properties Theorem. An infinite hereditary class X is constant if and only if none of the following classes is a subclass of X : S, E1,Scomp,E1comp. Proof. None of the four classes is constant. Therefore if X contains at least one of them, then X is not constant. Conversely, assume X contains none of the four classes. In order to prove that X is constant, we will show that Y:=X-(K O) is a finite set. Suppose by contradiction that Y contains infinitely many graphs. Then by Ramsey’s Theorem in graphs in Y there are either arbitrarily large independent sets or arbitrarily large cliques. Suppose the first is true, i.e. for any k, there is a graph G in Y with an independent set of size k. Since G contains at least one edge, it must contain an induced subgraph of the form K1,s+Ok-s for some s1. Since G belongs to Y, it also belongs to X, and therefore, all induced subgraphs of G belong to X (because X is hereditary). In particular, all graphs of the form K1,s+Ok-s for some s1 belong to X, and since none of these graphs is empty or complete, all of them belong to Y. Among these graphs there must exist either graphs with arbitrarily large value of k-s (in which case E1 X) or with arbitrarily large value of s (in which case S X), since k can be arbitrarily large. A contradiction. Similarly, a contradiction arises if graphs in Y contain arbitrarily large cliques. Inclusion relationship between hereditary classes Jean, Michel An interval graph is a comparability graph. J. Combinatorial Theory 7 1969 189--190 Fishburn, Peter C. An interval graph is not a comparability graph. J. Combinatorial Theory 8 1970 442--443. Inclusion relationship between hereditary classes Comparability graphs Permutation graphs Triangle-free graphs Bipartite graphs Planar graphs Forests Inclusion relationship between hereditary classes Theorem. Free(M) Free(N) if and only if for every graph GN there is a graph HM such that H is an induced subgraph of G. Proof. Let Free(M) Free(N) and assume to the contrary that a graph G N contains no induced subgraphs from M. Then by definition G Free(M), implying that G Free(N), a contradiction. Conversely, suppose for every graph G N there is a graph H M such that H is an induced subgraph of G, but Free(M) is not a subclass of Free(N). Then there is a graph F Free(M)-Free(N). Since F Free(N) it contains a graph G N as an induced subgraph. Therefore, F contains H M as an induced subgraph, which contradicts the fact F Free(M). C4 2K 2 Hereditary classes of graphs C6 Chordal (triangulated) graphs = Free(C4,C5,C6,…) non-triangulated graphs triangulated graph Co-chordal graphs (complements to chordal graphs) = Free(C4 , C5 , C6 ,...) Free(C4 , C5 , C6 ,...) Free(C4 , C5 , C6 ,...) Free(C4 , C4 , C5 , C5 , C6 , C6 ,...) Free(C4 , C4 , C5 ) Hereditary classes of graphs Lemma. The vertex set of any graph GFree(C4,2K2,C5) can be partitioned into a clique and an independent set Proof. Let C be a maximal clique in G such that G-C has as few edges as possible. Our goal is to show that S=V(G)-C is an independent set. Assume by contradiction that S contains an edge ab. Then w.l.o.g. NC(a)NC(b), since otherwise a C4 arises. Observe that C must contain a vertex z non-adjacent to b, since otherwise C is not a maximal clique. On the other hand, if C contains two vertices y and z non-adjacent to b, then a,b,y,z induce a 2K2. Therefore, C=NC(b){z}. Suppose S contains a vertex x adjacent to z. Then x is non-adjacent to b (otherwise NC(b)NC(x)=C contradicting the maximality of C), which implies that x is adjacent to a (otherwise a,b,x,z induce a 2K2). Since C is a maximal clique, it must contain a vertex y non-adjacent to x. If y is adjacent to a, then y,a,x,z induce a C4, and if y is not adjacent to a, then y,b,a,x,z induce a C5. This contradiction shows that S contains no vertices adjacent to z. But then C’=(C-{z}){b} is a maximal clique such that G-C’ has fewer edges than G-C, contradicting the choice of C. Therefore, S is an independent set. Hereditary classes of graphs Definition. A graph G is a split graph of the vertex set of G can be partitioned into a clique and an independent set. We proved that Free(C4,2K2,C5) Split Graphs Split Graphs Free(C4,2K2,C5) because neither C4, nor 2K2, nor C5 are split Theorem. A graph G is a split graph if and only if G Free(C4,2K2,C5) 1977 Peter L. Hammer and Stefan Foldes Peter L. Hammer (December 23, 1936 - December 27, 2006) was an American mathematician native to Romania. He contributed to the fields of operations research and applied discrete mathematics through the study of pseudo-Boolean functions and their connections to graph theory and data mining. Threshold graphs Definition. Given a graph G with n vertices V={v1,…,vn} and a subset UV, the characteristic vector of U is a binary vector (1,…,n) such that i=1 if viU and i=0 otherwise. Definition. A graph G with n vertices V={v1,…,vn} is a threshold graph if there exist real numbers a1,…,an and b such that 0-1 solutions of i aixib are precisely the characteristic vectors of independent sets in G. Observation 1. The following four graphs are not threshold graphs: v1 v4 v2 v3 v1 v4 v2 v3 v1 v4 v2 v3 Assume there is an equality a1x1+a2x2+a3x3+a4x4b such that 0-1 solutions of this inequality are precisely independent sets of one of these graphs. Then a1+a3b, a2+a4b, a1+a4>b, a2+a3>b, i.e. i ai 2b and i ai > 2b. A contradiction. Threshold graphs Observation 2. If G is a threshold graph, then every induced subgraph of G is a threshold graph. Proof. Let a1,…,an and b be real numbers such that the 0-1 solutions of i aixib are precisely the characteristic vectors of independent sets in G. Let H be a subgraph of G induced by a set SV(G), then the independent sets of H are precisely the 0-1 solutions of a x i:vi S i i b Corollary. The class of threshold graphs is hereditary and it is a subclass of Free(2K2,C4,P4). Theorem (Hammer, Chvatal,1977). A graph is threshold if and only if it is 2K2,C4,P4-free. Corollary. A graph is threshold if and only if its complement is threshold. G’ P4-free graphs x y u Complement reducible graphs (cographs) v Definition. A graph G is a cograph if and only if every induced subgraph of G with at least 2 vertices is either disconnected or the complement to a disconnected graph. Theorem. A graph G is P4-free if and only if G is a cograph. Proof. Every cograph is P4-free since neither P4 nor its complement is disconnected. Conversely, let G be a P4-free graph. We will show by induction on n=|V(G)| that either G or its complement is disconnected. Let vV(G) and G’=G-v. By induction we can assume w.l.o.g. that G’ is disconnected (otherwise we can consider the complement of G). We also may assume that • v has a non-neighbor u in G’, since otherwise the complement of G is disconnected. • v has a neighbor in each connected component of G’, since otherwise G is disconnected. Since u and x belong to the same connected component, there must be a path connecting them. But then G contains a P4, which contradicts our assumption. Threshold graphs Corollary. Every threshold graph has either an isolated vertex (i.e. a vertex of degree 0) or a dominating vertex (i.e. a vertex adjacent to all other vertices of the graph). Corollary. There are 2n-1 pairwise non-isomorphic threshold graphs with n vertices. Problem. Find all connected (P4,K3)-free planar graphs Solution. Let G be a connected (P4,K3)-free planar graph. Since G is connected and P4-free, its complement is disconnected. In other words, the vertices of G can be partitioned into at least 2 parts A and B so that every vertex of A is adjacent to every vertex of B. Then both A and B are independent sets, since otherwise a triangle arises. Therefore, G is a complete bipartite graph. Assuming w.l.o.g. that |A||B| we conclude that |A|<3, since otherwise G is not planar (contains K3,3). Therefore, G is either A 1 1 B 2 K1,n n 2 or K2,n n Perfect Graphs A vertex coloring is an assignment of colors to the vertices such that no edge connects vertices of the same color. The minimum number of colors needed to color a graph G is the chromatic number of G. The size of a maximum clique in G is the clique number of G. Observation. The chromatic number of a graph is greater than or equal to its clique number. There are graphs for which the chromatic number equals the clique number Examples: Bipartite graphs, split graphs. Moreover, the chromatic number equals the clique number not only for any bipartite or split graph G, but also for any induced subgraph of G, since the classes of bipartite and split graphs are hereditary. Definition. A graph G is perfect if the chromatic number equals the clique number for any induced subgraph of G. Perfect graphs Chordal graphs Comparability graphs Triangle-free graphs Split graphs Permutation graphs Bipartite graphs Planar graphs Cographs Forests Claude Berge (June 5, 1926 – June 30, 2002) was a French mathematician, recognized as one of the modern founders of combinatorics and graph theory. He is particularly remembered for his famous conjectures on perfect graphs Perfect Graphs The complement of C5 is not a perfect graph C5 is not a perfect graph The complement of C7 is not a perfect graph C7 is not a perfect graph C2k+1 is not a perfect graph The complement of C2k+1 is not a perfect graph Weak Berge Conjecture. A graph G is perfect if and only if its complement is perfect. This conjecture was solved by László Lovász (born March 9, 1948 in Budapest, Hungary) is a mathematician, best known for his work in combinatorics) in 1972 Strong Berge Conjecture. A graph G is perfect if and only if it contains neither odd cycles of length at least 5 nor their complements as induced subgraphs. This conjecture was solved by Chudnovsky, Robertson, Seymour, Thomas in 2002 Chudnovsky, Maria; Robertson, Neil; Seymour, Paul; Thomas, Robin (2006). "The strong perfect graph theorem". Annals of Mathematics 164 (1): 51–229. Theorem. Chordal graphs are perfect Theorem. Free(M) Free(N) if and only if for every graph GN there is a graph HM such that H is an induced subgraph of G. Chordal graphs = Free(C4,C5,C6,…) Perfect graphs = Free(C5, co-C5, C7, co-C7, C9, co-C9,…) For each k3, co-C2k+1 contains C4 as an induced subgraph