Download Hereditary classes of graphs

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 n1 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 s1. 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 s1 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 GN there is a graph HM
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 GFree(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 UV,
the characteristic vector of U is a binary vector (1,…,n) such
that i=1 if viU 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 aixib
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+a4x4b such that 0-1 solutions of
this inequality are precisely independent sets of one of these graphs. Then
a1+a3b, a2+a4b, 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 aixib
are precisely the characteristic vectors of independent sets in G. Let H be a
subgraph of G induced by a set SV(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 vV(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 GN there
is a graph HM 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 k3, co-C2k+1 contains C4 as an induced subgraph