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The cographs are exactly the P_4-free graphs

Statement

A finite graph is a cograph if and only if it is $P_{4}$ -free.

Facts & Assumptions

Given: A finite graph $G$ .

[L1]

Every induced subgraph of a cograph is a cograph (Every induced subgraph of a cograph is a cograph).

[L2]

Every nontrivial cograph is disconnected or has disconnected complement (Every nontrivial cograph is disconnected or has disconnected complement).

[L3]

The four-vertex path $P_{4}$ has vertices $0, 1, 2, 3$ and edges $01, 12, 23$ , and its complement has edges $02, 03, 13$ . Hence both $P_{4}$ and $\overline{P_{4}}$ are connected (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices, Connected graphs and connected components defined by the existence of vertex paths, Graph isomorphisms, automorphisms and graph complements).

[L4]

Every nontrivial $P_{4}$ -free graph is disconnected or its complement is disconnected (Every nontrivial $P_{4}$ -free graph is disconnected or has disconnected complement).

[L5]

Connected components partition the vertex set, anticomponents do too, distinct components are anticomplete, and distinct anticomponents are complete (The connected components of a graph partition its vertex set and are its maximal connected subgraphs, The anticonnected components of $G$ are exactly the connected components of $\overline{G}$ , Distinct connected components are anticomplete, and distinct anticonnected components are complete).

[F1]

A graph is $P_{4}$ -free when it contains no induced copy of the path $P_{4}$ ( $H$ -free and $F$ -free graphs under the induced-subgraph convention, Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[F2]

Every induced subgraph of a $P_{4}$ -free graph is again $P_{4}$ -free, because an induced copy inside an induced subgraph is also an induced copy in the whole graph ( $H$ -free and $F$ -free graphs under the induced-subgraph convention, Subgraphs, induced subgraphs and spanning subgraphs).

Proof

technique · induction on the order of a $P_4$-free graph

1.1

Suppose first that $G$ is a cograph. If $G$ had an induced copy of $P_{4}$ on some vertex set $X$ , then [L1] would make $G [X]$ a cograph. But $G [X]$ is isomorphic to $P_{4}$ , and [L3] shows that both $P_{4}$ and its complement are connected, contradicting [L2]. Therefore every cograph is $P_{4}$ -free.

L1L2L3F1

1.2

For the converse, we prove by induction on $n : = ∣ V (G) ∣$ that every $P_{4}$ -free graph on $n$ vertices is a cograph. If $n \leq 1$ , then $G$ is the null graph or a one-vertex graph, and the recursive definition makes it a cograph.

F1base

2.1

Assume now that $n \geq 2$ and that every smaller $P_{4}$ -free graph is a cograph. Because $G$ is $P_{4}$ -free, [L4] gives that $G$ is disconnected or $\overline{G}$ is disconnected.

step 1.2L4ih

3.1

If $G$ is disconnected, choose a connected component $C$ of $G$ and let $D : = V (G) ∖ C$ . Then $C$ and $D$ are nonempty, [L5] makes them anticomplete, and [F2] makes both $G [C]$ and $G [D]$ smaller $P_{4}$ -free graphs. By the induction hypothesis they are cographs, so $G$ is their disjoint union and hence a cograph.

step 2.1L5F2ihchoose

3.2

If $\overline{G}$ is disconnected, choose an anticomponent $C$ of $G$ and let $D : = V (G) ∖ C$ . Again $C$ and $D$ are nonempty, [L5] makes them complete to one another, and [F2] makes $G [C]$ and $G [D]$ smaller $P_{4}$ -free graphs. By the induction hypothesis they are cographs, so $G$ is their complete connection and hence a cograph.

step 2.1L5F2ihchoose

4.1

Steps 3.1 and 3.2 close the induction, proving that every $P_{4}$ -free graph is a cograph. Together with step 1.1, this proves the equivalence.

step 1.1step 3.1step 3.2discharge-induction∎

Depends on

Used by

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Tero Harju, Lecture Notes on Combinatorial Structures in Graph Theory, Exercise 5.3 (standard reference, not scraped)
Maria Chudnovsky, The Erdos-Hajnal Conjecture - A Survey, Theorem 2.1 (standard reference, not scraped)