Asteroidal

Triple-Free

Graphs

Derek

Corneil

Stephan

Olariu

Lorna

Stewart

Dept.

of

Computer

Science

Dept.

of

Computer

Science

Dept.

of

Computing

Science

University

of

Toronto

Old

Dominion

University

University

of

Alberta

Toronto

Ontario

Canada

Norfolk

Virginia

U.S.A.

Edmonton

Alberta

Canada

Abstract

An

independent

set

of

three

vertices

such

that

each

pair

is

joined

by

a

path

that

avoids

the

neighborhood

of

the

third

is

called

an

asteroidal

triple

A

graph

is

asteroidal

triple-free

-LRB-

AT-free

for

short

-RRB-

if

it

contains

no

asteroidal

triples

The

motivation

for

this

investigation

is

provided

in

part

by

the

fact

that

the

asteroidal

triple-free

graphs

provide

a

common

generalization

of

interval

permutation

trapezoid

and

cocomparability

graphs

The

main

contribution

of

this

work

is

to

investigate

and

reveal

fundamental

structural

properties

of

AT-free

graphs

Specifically

we

show

that

every

AT-free

graph

contains

a

dominating

pair

that

is

a

pair

of

vertices

such

that

every

path

joining

them

is

a

dominating

set

in

the

graph

We

then

provide

characterizations

of

AT-free

graphs

in

terms

of

dominating

pairs

and

minimal

triangulations

Subsequently

we

state

and

prove

a

decomposition

theorem

for

AT-free

graphs

An

assortment

of

other

properties

of

AT-free

graphs

are

also

provided

These

properties

generalize

known

structural

properties

of

interval

permutation

trapezoid

and

cocomparability

graphs

Introduction

The

original

motivation

for

this

work

was

provided

by

the

linear

structure

that

is

apparent

in

various

families

of

graphs

such

as

interval

graphs

permutation

graphs

trapezoid

graphs

and

cocomparability

graphs

Somewhat

surprisingly

the

linearity

of

the

interval

permutation

trapezoid

and

cocomparability

graphs

is

described

in

terms

of

different

and

seemingly

ad-hoc

properties

of

each

of

these

classes

of

graphs

For

example

in

the

case

of

interval

graphs

the

linearity

property

is

traditionally

expressed

in

terms

of

a

linear

order

on

the

set

of

maximal

cliques

-LSB-

-RSB-

For

permutation

graphs

the

linear

behavior

is

explained

in

terms

of

the

underlying

partial

order

of

dimension

two

-LSB-

-RSB-

For

cocomparability

graphs

the

linear

behavior

is

expressed

in

terms

of

the

well-known

linear

structure

of

comparability

graphs

-LSB-

-RSB-

and

so

on

Our

intention

is

to

provide

a

unifying

look

at

these

classes

in

the

hope

of

identifying

the

agent

responsible

for

their

linear

behavior

Before

proceeding

it

is

perhaps

appropriate

to

recall

a

few

definitions

A

graph

is

an

interval

graph

if

its

vertices

can

be

put

in

a

one-to-one

correspondence

with

a

set

of

intervals

on

the

real

line

in

such

a

way

that

two

vertices

are

adjacent

if

and

only

if

the

corresponding

intervals

overlap

A

graph

is

a

comparability

graph

if

the

edges

may

be

given

a

transitive

orientation

A

cocomparability

graph

is

the

complement

of

a

comparability

graph

A

graph

that

is

at

the

same

time

a

comparability

and

a

cocomparability

graph

is

said

to

be

a

permutation

graph

-LSB-

-RSB-

Figure

A

graph

G

A

trapezoid

representation

R

consists

of

two

parallel

lines

-LRB-

denoted

and

-RRB-

and

some

trapezoids

with

two

endpoints

lying

on

and

the

other

two

lying

on

A

graph

G

is

a

trapezoid

graph

if

it

is

the

intersection

graph

of

such

a

representation

Specifically

the

vertices

of

G

are

in

one-to-one

correspondence

with

the

trapezoids

in

R

and

two

vertices

in

G

are

adjacent

if

and

only

if

their

corresponding

trapezoids

intersect

If

the

trapezoids

degenerate

with

the

endpoints

on

-LRB-

resp

-RRB-

coinciding

-LRB-

i.e.

the

trapezoids

become

lines

-RRB-

then

the

intersection

graph

is

a

permutation

graph

Similarly

if

the

intervals

on

are

the

mirror

image

of

the

intervals

on

then

the

intersection

graph

is

an

interval

graph

The

reader

is

referred

to

Figure

for

an

illustration

of

these

notions

It

is

shown

in

-LSB-

-RSB-

that

permutation

graphs

and

interval

Figure

Trapezoid

interval

and

permutation

models

of

the

graph

in

Figure

graphs

are

strictly

contained

in

trapezoid

graphs

Furthermore

trapezoid

graphs

are

strictly

contained

in

cocomparability

graphs

-LSB-

-RSB-

Cocomparability

graphs

and

thus

trapezoid

permutation

and

interval

graphs

are

perfect

in

the

sense

of

Berge

-LSB-

-RSB-

i.e.

for

every

induced

subgraph

the

chromatic

number

equals

the

clique

number

The

trapezoid

representation

that

provides

the

common

thread

with

interval

and

permutation

graphs

also

indicates

that

in

some

sense

the

graphs

can

only

grow

linearly

For

example

referring

to

the

graph

in

Figure

which

is

at

the

same

time

an

interval

trapezoid

permutation

and

cocomparability

graph

we

can

add

a

new

vertex

adjacent

to

one

of

the

vertices

or

without

destroying

membership

in

any

of

these

families

however

when

looking

at

various

intersection

models

of

G

featured

in

Figure

it

seems

as

though

we

can

not

add

a

new

vertex

adjacent

to

without

destroying

membership

in

each

family

More

than

three

decades

ago

Lekkerkerker

and

Boland

-LSB-

-RSB-

set

out

to

identify

the

property

that

prevented

a

chordal

graph

namely

a

graph

in

which

every

cycle

of

length

at

least

four

has

a

chord

from

growing

in

three

directions

at

once

For

this

purpose

they

defined

an

asteroidal

triple

to

be

an

independent

set

of

three

vertices

such

that

each

pair

of

vertices

is

joined

by

a

path

that

avoids

the

neighborhood

of

the

third

For

an

illustration

the

reader

is

referred

to

Figure

featuring

various

instances

of

asteroidal

triples

Lekkerkerker

and

Boland

-LSB-

-RSB-

demonstrated

the

importance

of

asteroidal

triples

in

the

following

theorem

Theorem

A

graph

is

an

interval

graph

if

and

only

if

it

is

chordal

and

asteroidal

triple-free

Figure

Various

examples

of

asteroidal

triples

Thus

it

appears

that

the

condition

of

being

asteroidal

triple-free

-LRB-

AT-free

for

short

-RRB-

prohibits

a

chordal

graph

from

growing

in

three

directions

at

once

The

top

three

graphs

in

Figure

are

examples

of

chordal

graphs

that

are

not

interval

graphs

Later

Golumbic

et

al.

-LSB-

-RSB-

showed

that

cocomparability

graphs

-LRB-

and

thus

permutation

and

trapezoid

graphs

-RRB-

are

also

AT-free

Subsequently

it

was

shown

that

the

perfect

AT-free

graphs

strictly

contain

the

cocomparability

graphs

-LSB-

-RSB-

Since

the

complements

of

odd

chordless

cycles

of

size

greater

than

are

AT-free

the

AT-free

graphs

need

not

be

perfect

However

an

easy

argument

shows

that

the

celebrated

Strong

Perfect

Graph

Conjecture

is

true

for

asteroidal

triple-free

graphs

-LSB-

-RSB-

More

than

two

decades

ago

Gallai

-LSB-

-RSB-

in

his

monumental

work

on

comparability

graphs

obtained

the

first

characterization

of

AT-minimal

graphs

-LRB-

i.e.

graphs

that

contain

an

asteroidal

triple

and

are

minimal

with

this

property

-RRB-

in

terms

of

fifteen

families

of

subgraphs

Actually

Gallai

's

list

is

not

complete

Since

he

was

only

interested

in

graphs

with

no

induced

all

the

AT-minimal

graphs

containing

a

are

missing

from

-LSB-

-RSB-

For

a

full

list

of

AT-minimal

graphs

the

interested

reader

is

referred

to

-LSB-

-RSB-

After

Gallai

's

paper

little

work

has

been

done

on

AT-free

graphs

The

main

contribution

of

this

work

is

to

provide

a

number

of

structural

results

concerning

asteroidal

triple-free

graphs

To

anticipate

our

main

are

We

show

that

every

connected

AT-free

graph

has

a

dominating

pair

that

is

a

pair

of

vertices

such

that

every

path

joining

them

is

a

dominating

set

undefined

terms

the

reader

is

referred

to

Section

We

provide

properties

of

dominating

pairs

in

AT-free

graphs

related

to

the

concept

of

connected

domination

and

diameter

We

provide

a

characterization

of

AT-free

graphs

in

terms

of

dominating

pairs

We

provide

a

characterization

of

AT-free

graphs

in

terms

of

minimal

triangulations

We

provide

a

decomposition

theorem

for

AT-free

graphs

The

remainder

of

this

work

is

organized

as

follows

Section

provides

background

material

along

with

definitions

of

technical

terms

used

throughout

the

paper

In

Section

we

study

the

existence

of

dominating

pairs

in

connected

AT-free

graphs

In

Section

we

discuss

properties

of

dominating

pairs

in

the

context

of

connected

domination

and

show

that

some

dominating

pair

achieves

the

diameter

of

the

graph

In

Section

we

offer

two

characterizations

of

AT-free

graphs

Specifically

we

first

provide

a

characterization

of

AT-free

graphs

in

terms

of

dominating

pairs

next

we

provide

a

characterization

of

AT-free

graphs

in

terms

of

their

minimal

triangulations

In

Section

we

show

that

an

AT-free

graph

may

be

extended

to

another

AT-free

graph

by

adding

a

new

degree

vertex

adjacent

to

each

vertex

in

an

appropriate

dominating

pair

This

result

leads

to

a

decomposition

theorem

for

AT-free

graphs

whereby

an

AT-free

graph

is

reduced

to

a

single

vertex

by

a

sequence

of

contractions

In

Section

we

show

that

in

AT-free

graphs

of

diameter

greater

than

the

sets

of

vertices

that

can

be

in

dominating

pairs

are

restricted

to

two

disjoint

sets

thus

confirming

the

intuition

about

the

linear

structure

of

this

class

of

graphs

Finally

Section

offers

concluding

remarks

and

poses

some

open

problems

Preliminaries

All

graphs

in

this

paper

are

finite

with

no

loops

or

multiple

edges

We

use

standard

graph-theoretic

terminology

compatible

with

-LSB-

-RSB-

to

which

we

refer

the

reader

for

basic

definitions

As

usual

we

shall

write

G

-LRB-

V

E

-RRB-

to

denote

a

graph

G

with

vertex-set

V

and

edge-set

E

The

complement

of

a

graph

G

is

a

graph

G

having

the

same

vertex-set

as

G

vertices

u

and

v

are

adjacent

in

G

if

and

only

if

they

are

non-adjacent

in

G.

For

a

vertex

x

in

G

NG

-LRB-

x

-RRB-

denotes

the

set

of

all

the

vertices

adjacent

to

x

in

G

we

assume

that

adjacency

is

non-reflexive

and

so

x

NG

-LRB-

x

-RRB-

The

degree

of

vertex

x

in

the

graph

G

denoted

by

dG

-LRB-

x

-RRB-

is

the

cardinality

of

NG

-LRB-

x

-RRB-

A

vertex

x

will

be

said

to

be

pendant

if

its

degree

is

We

let

N

G

-LRB-

x

-RRB-

stand

for

the

set

of

all

the

vertices

adjacent

to

x

in

the

complement

G

of

G

The

notation

will

be

shortened

to

N

-LRB-

x

-RRB-

d

-LRB-

x

-RRB-

and

N

-LRB-

x

-RRB-

respectively

whenever

the

context

permits

If

H

is

a

subset

of

the

vertex-set

V

of

G

then

GH

will

denote

the

subgraph

of

G

induced

by

H.

Occasionally

if

no

confusion

is

possible

we

shall

use

H

as

a

shorthand

for

GH

A

path

is

a

sequence

vp

of

distinct

vertices

of

G

with

vi

E

for

all

i

-LRB-

i

p

-RRB-

A

chord

in

a

path

vp

is

an

edge

vivj

with

i

and

j

differing

by

more

than

A

cycle

of

length

p

is

a

sequence

vp

of

distinct

vertices

of

G

such

that

vi

E

for

all

i

-LRB-

i

p

-RRB-

and

E

We

let

Pn

and

Cn

denote

the

chordless

path

and

cycle

with

n

vertices

respectively

Unless

stated

otherwise

all

paths

in

this

work

will

be

assumed

to

be

chordless

A

path

joining

vertices

x

and

y

is

termed

an

x

y-path

A

vertex

u

misses

a

path

ss

if

u

is

adjacent

to

no

vertex

on

ss

otherwise

u

intercepts

ss

A

set

D

of

vertices

of

graph

G

is

said

to

be

dominating

if

every

vertex

outside

D

is

adjacent

to

some

vertex

in

D

A

pair

-LRB-

u

v

-RRB-

of

vertices

is

termed

a

dominating

pair

if

all

u

v-paths

are

dominating

Among

dominating

sets

D

that

induce

connected

subgraphs

of

G

one

is

often

interested

in

those

that

have

minimum

cardinality

In

the

remainder

of

this

paper

such

a

dominating

set

will

be

referred

to

as

a

mccds

Given

a

connected

graph

G

-LRB-

V

E

-RRB-

the

distance

dG

-LRB-

u

v

-RRB-

-LRB-

or

d

-LRB-

u

v

-RRB-

for

short

-RRB-

between

vertices

u

and

v

is

the

length

of

a

shortest

path

in

G

joining

u

and

v

The

diameter

of

G

is

defined

as

diam

-LRB-

G

-RRB-

maxu

dG

-LRB-

u

v

-RRB-

Two

vertices

u

and

v

such

that

d

-LRB-

u

v

-RRB-

diam

-LRB-

G

-RRB-

are

said

to

achieve

the

diameter

Dominating

Pairs

in

AT-free

Graphs

The

main

purpose

of

this

section

is

to

prove

a

fundamental

domination-related

property

of

AT-free

graphs

To

state

this

property

recall

that

a

pair

of

vertices

-LRB-

x

y

-RRB-

is

a

dominating

pair

in

a

graph

G

if

all

x

y-paths

in

G

are

dominating

sets

As

it

turns

out

connected

AT-free

graphs

always

contain

dominating

pairs

Although

it

is

straightforward

to

see

that

interval

permutation

trapezoid

and

cocomparability

graphs

all

contain

dominating

pairs

it

is

somewhat

surprising

that

up

to

now

this

property

had

not

been

noticed

for

these

classes

of

graphs

Throughout

this

section

we

assume

a

connected

AT-free

graph

G

-LRB-

V

E

-RRB-

along

with

an

arbitrary

vertex

x

of

G.

For

a

vertex

u

of

G

we

let

D

-LRB-

u

x

-RRB-

denote

the

set

of

vertices

that

intercept

all

u

x-paths

In

this

terminology

a

pair

of

vertices

-LRB-

u

v

-RRB-

is

a

dominating

pair

whenever

D

-LRB-

u

v

-RRB-

V

We

are

now

in

a

position

to

state

the

main

result

of

this

section

Theorem

Every

connected

asteroidal

triple-free

graph

contains

a

dominating

pair

The

conclusion

of

Theorem

is

implied

by

the

following

stronger

result

Theorem

Let

x

be

an

arbitrary

vertex

in

a

connected

asteroidal

triple-free

graph

G.

Either

-LRB-

x

x

-RRB-

is

a

dominating

pair

or

else

for

a

suitable

choice

of

vertices

y

and

z

in

N

-LRB-

x

-RRB-

-LRB-

y

x

-RRB-

or

-LRB-

y

z

-RRB-

is

a

dominating

pair

Our

proof

of

Theorem

relies

on

a

number

of

intermediate

results

about

connected

AT-free

graphs

that

we

present

next

Claim

Let

u

v

and

w

be

arbitrary

vertices

of

G

If

u

D

-LRB-

v

x

-RRB-

w

D

-LRB-

u

x

-RRB-

and

u

and

w

are

not

adjacent

then

w

D

-LRB-

v

x

-RRB-

Proof

Suppose

that

w

misses

some

v

x-path

ss

v

vk

x.

Let

j

be

the

largest

subscript

for

which

u

is

adjacent

to

some

vertex

vj

of

ss

since

u

D

-LRB-

v

x

-RRB-

such

a

subscript

must

exist

But

now

w

misses

the

u

x-path

u

vj

vj

vk

x

contradicting

that

w

D

-LRB-

u

x

-RRB-

Figure

Illustrating

unrelated

vertices

We

say

that

vertices

u

and

v

are

unrelated

with

respect

to

x

if

u

D

-LRB-

v

x

-RRB-

and

v

D

-LRB-

u

x

-RRB-

The

reader

is

referred

to

Figure

for

an

illustration

The

paths

confirming

that

vertices

u

and

v

are

unrelated

with

respect

to

x

are

drawn

in

heavy

lines

In

the

remainder

of

this

section

we

shall

use

unrelated

as

a

shorthand

for

unrelated

with

respect

to

x

We

further

assume

that

F

is

an

arbitrary

connected

component

of

N

-LRB-

x

-RRB-

Claim

F

contains

no

unrelated

vertices

Proof

If

u

and

v

are

unrelated

vertices

in

F

then

the

connectedness

of

F

implies

that

fu

v

xg

is

an

asteroidal

triple

Claim

If

u

and

v

are

vertices

in

F

and

if

v

D

-LRB-

u

x

-RRB-

then

D

-LRB-

u

x

-RRB-

ae

D

-LRB-

v

x

-RRB-

Proof

From

Claim

it

follows

that

u

D

-LRB-

v

x

-RRB-

Let

w

be

an

arbitrary

vertex

in

D

-LRB-

u

x

-RRB-

If

w

and

u

are

not

adjacent

then

Claim

guarantees

that

w

D

-LRB-

v

x

-RRB-

if

w

and

u

are

adjacent

then

clearly

w

F

If

w

misses

some

v

x-path

then

in

particular

v

and

w

are

not

adjacent

Thus

with

ss

standing

for

some

u

x-path

missed

by

v

ss

-LSB-

fwg

contains

a

w

x-path

missed

by

v

But

now

v

and

w

are

unrelated

contradicting

Claim

Consequently

w

D

-LRB-

v

x

-RRB-

and

D

-LRB-

u

x

-RRB-

D

-LRB-

v

x

-RRB-

the

inclusion

is

strict

since

v

D

-LRB-

u

x

-RRB-

A

vertex

y

in

F

is

called

special

if

D

-LRB-

u

x

-RRB-

D

-LRB-

y

x

-RRB-

for

all

vertices

u

in

F

The

following

statement

provides

a

characterization

of

special

vertices

Claim

A

vertex

y

in

F

is

special

if

and

only

if

F

D

-LRB-

y

x

-RRB-

Proof

First

if

the

vertex

y

is

special

then

for

every

vertex

v

in

F

D

-LRB-

v

x

-RRB-

D

-LRB-

y

x

-RRB-

In

particular

v

D

-LRB-

v

x

-RRB-

implying

that

F

D

-LRB-

y

x

-RRB-

Conversely

suppose

that

F

D

-LRB-

y

x

-RRB-

Let

u

be

an

arbitrary

vertex

in

F

and

let

w

be

an

arbitrary

vertex

in

D

-LRB-

u

x

-RRB-

If

w

belongs

to

F

then

since

F

D

-LRB-

y

x

-RRB-

w

D

-LRB-

y

x

-RRB-

if

w

does

not

belong

to

F

then

u

and

w

are

not

adjacent

and

Claim

guarantees

that

w

D

-LRB-

y

x

-RRB-

confirming

that

D

-LRB-

u

x

-RRB-

D

-LRB-

y

x

-RRB-

Since

u

is

arbitrary

the

claim

follows

Claim

Every

component

of

N

-LRB-

x

-RRB-

contains

a

special

vertex

Proof

As

before

let

F

be

an

arbitrary

component

of

N

-LRB-

x

-RRB-

Choose

a

vertex

y

in

F

with

D

-LRB-

y

x

-RRB-

ae

D

-LRB-

t

x

-RRB-

for

no

vertex

t

in

F

If

y

is

not

special

then

by

Claim

we

find

a

vertex

v

in

F

with

v

D

-LRB-

y

x

-RRB-

By

Claim

D

-LRB-

y

x

-RRB-

ae

D

-LRB-

v

x

-RRB-

contradicting

our

choice

of

y.

Claim

Let

v

be

an

arbitrary

vertex

in

N

-LRB-

x

-RRB-

n

F

Either

v

D

-LRB-

w

x

-RRB-

for

all

vertices

w

in

F

or

else

v

D

-LRB-

w

x

-RRB-

for

all

vertices

w

in

F

Proof

Suppose

not

for

a

suitable

choice

of

vertices

w

and

in

F

we

have

v

D

-LRB-

w

x

-RRB-

and

v

D

-LRB-

x

-RRB-

Let

ss

stand

for

a

x-path

missed

by

v

and

let

stand

for

a

w

entirely

within

F

But

now

ss

-LSB-

contains

a

w

x-path

missed

by

v

contrary

to

our

assumption

Claim

Let

v

be

a

vertex

in

N

-LRB-

x

-RRB-

n

F

If

F

D

-LRB-

v

x

-RRB-

then

for

a

special

vertex

u

in

F

u

D

-LRB-

v

x

-RRB-

Proof

Write

U

fu

F

j

u

D

-LRB-

v

x

-RRB-

g

Since

F

D

-LRB-

v

x

-RRB-

U

is

nonempty

Choose

a

vertex

u

in

U

such

that

D

-LRB-

u

x

-RRB-

ae

D

-LRB-

u

x

-RRB-

for

no

vertex

u

in

U

If

u

is

not

special

then

by

Claim

there

exists

some

vertex

w

in

FnD

-LRB-

u

x

-RRB-

By

Claim

D

-LRB-

u

x

-RRB-

ae

D

-LRB-

w

x

-RRB-

by

our

choice

of

u

w

must

belong

to

F

n

U

This

however

implies

that

w

D

-LRB-

v

x

-RRB-

Since

w

D

-LRB-

u

x

-RRB-

Claim

implies

that

u

D

-LRB-

w

x

-RRB-

Since

u

and

w

are

not

adjacent

Claim

guarantees

that

u

D

-LRB-

v

x

-RRB-

which

is

the

desired

contradiction

Call

a

vertex

u

of

N

-LRB-

x

-RRB-

strong

if

N

-LRB-

x

-RRB-

ae

D

-LRB-

u

x

-RRB-

It

is

easy

to

verify

that

if

u

is

a

strong

vertex

then

-LRB-

u

x

-RRB-

is

a

dominating

pair

in

G.

From

now

on

we

shall

tacitly

assume

that

N

-LRB-

x

-RRB-

contains

no

strong

vertices

A

pair

-LRB-

y

z

-RRB-

of

vertices

in

distinct

components

of

N

-LRB-

x

-RRB-

is

an

admissible

pair

if

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

ae

D

-LRB-

t

x

-RRB-

-LSB-

D

-LRB-

x

-RRB-

for

no

vertices

t

in

distinct

components

of

N

-LRB-

x

-RRB-

Notice

that

if

N

-LRB-

x

-RRB-

is

connected

Claim

implies

that

N

-LRB-

x

-RRB-

contains

a

special

vertex

which

by

virtue

of

Claim

is

strong

We

shall

therefore

assume

that

N

-LRB-

x

-RRB-

is

disconnected

Now

the

absence

of

strong

vertices

in

N

-LRB-

x

-RRB-

guarantees

the

existence

of

admissible

pairs

As

it

turns

out

admissible

pairs

play

a

crucial

role

in

our

arguments

We

now

study

some

of

their

properties

Claim

Let

Y

and

Z

be

two

distinct

components

of

N

-LRB-

x

-RRB-

and

let

vertices

y

in

Y

and

z

in

Z

be

an

admissible

pair

Then

Y

D

-LRB-

z

x

-RRB-

and

Z

D

-LRB-

y

x

-RRB-

Proof

Assume

Z

ae

D

-LRB-

y

x

-RRB-

then

in

particular

z

D

-LRB-

y

x

-RRB-

To

see

that

D

-LRB-

z

x

-RRB-

D

-LRB-

y

x

-RRB-

note

that

for

an

arbitrary

vertex

w

in

D

-LRB-

z

x

-RRB-

w

D

-LRB-

y

x

-RRB-

whenever

w

Z

and

that

by

virtue

of

Claim

w

D

-LRB-

y

x

-RRB-

whenever

w

Z

Since

y

is

not

strong

we

find

a

vertex

in

N

-LRB-

x

-RRB-

nD

-LRB-

y

x

-RRB-

But

now

either

-LRB-

z

-RRB-

or

-LRB-

y

-RRB-

contradicts

our

choice

of

-LRB-

y

z

-RRB-

To

see

this

note

that

if

belongs

to

Y

then

by

Claim

D

-LRB-

y

x

-RRB-

ae

D

-LRB-

x

-RRB-

and

so

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

D

-LRB-

y

x

-RRB-

ae

D

-LRB-

x

-RRB-

D

-LRB-

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

If

does

not

belong

to

Y

then

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

D

-LRB-

y

x

-RRB-

D

-LRB-

x

-RRB-

-LSB-

D

-LRB-

y

x

-RRB-

Since

does

not

belong

to

D

-LRB-

y

x

-RRB-

the

inclusion

is

strict

The

fact

that

Y

D

-LRB-

z

x

-RRB-

follows

by

a

similar

argument

Claim

If

-LRB-

y

z

-RRB-

is

an

admissible

pair

then

N

-LRB-

x

-RRB-

ae

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

Proof

We

assume

without

loss

of

generality

that

vertices

y

and

z

belong

to

distinct

connected

components

Y

and

Z

of

N

-LRB-

x

-RRB-

respectively

If

the

claim

is

false

we

find

a

vertex

w

in

N

-LRB-

x

-RRB-

n

-LRB-

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

-RRB-

Clearly

w

D

-LRB-

y

x

-RRB-

and

w

D

-LRB-

z

x

-RRB-

We

claim

that

w

does

not

belong

to

Y

-LSB-

Z

-LRB-

-RRB-

If

the

vertex

w

belongs

to

Y

then

by

Claim

D

-LRB-

y

x

-RRB-

ae

D

-LRB-

w

x

-RRB-

and

since

w

D

-LRB-

z

x

-RRB-

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

ae

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

contradicting

that

-LRB-

y

z

-RRB-

is

an

admissible

pair

The

proof

of

the

fact

that

w

Z

is

similar

and

thus

omitted

Since

G

is

AT-free

it

is

easy

to

verify

that

no

distinct

vertices

t

in

N

-LRB-

x

-RRB-

are

pairwise

unrelated

with

respect

to

x

-LRB-

-RRB-

Further

we

claim

that

for

a

suitable

choice

of

vertices

u

and

v

in

N

-LRB-

x

-RRB-

u

D

-LRB-

y

x

-RRB-

n

-LRB-

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

-RRB-

and

v

D

-LRB-

z

x

-RRB-

n

-LRB-

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

-RRB-

-LRB-

-RRB-

To

justify

-LRB-

-RRB-

observe

that

by

-LRB-

-RRB-

y

z

and

w

belong

to

distinct

components

of

N

-LRB-

x

-RRB-

Since

-LRB-

y

z

-RRB-

is

an

admissible

pair

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

and

therefore

the

required

vertex

u

exists

A

similar

argument

asserts

the

existence

of

vertex

v.

Next

we

claim

that

y

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

and

z

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

-LRB-

-RRB-

To

see

this

note

that

if

y

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

then

our

choice

of

w

guarantees

that

y

and

w

are

unrelated

Therefore

it

must

be

that

z

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

for

otherwise

y

z

and

w

would

be

pairwise

unrelated

contradicting

-LRB-

-RRB-

Consider

the

vertex

v

specified

in

-LRB-

-RRB-

since

z

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

and

v

D

-LRB-

z

x

-RRB-

n

-LRB-

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

-RRB-

Claim

implies

that

z

and

v

are

adjacent

But

now

fy

v

wg

is

an

asteroidal

triple

this

follows

since

y

and

w

are

unrelated

and

both

v

w

and

v

y

are

unrelated

by

-LRB-

-RRB-

and

Claim

Along

similar

lines

one

can

prove

that

z

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

Thus

must

hold

Further

we

claim

that

u

Y

and

v

Z

-LRB-

-RRB-

By

-LRB-

-RRB-

y

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

by

-LRB-

-RRB-

u

D

-LRB-

y

x

-RRB-

n

-LRB-

D

-LRB-

z

x

-RRB-

-LSB-

D

-LRB-

w

x

-RRB-

-RRB-

It

follows

that

u

and

y

are

adjacent

for

otherwise

we

contradict

Claim

The

fact

that

v

Z

is

proved

similarly

To

complete

the

proof

of

Claim

we

first

observe

that

-LRB-

-RRB-

-LRB-

-RRB-

and

Claim

combined

guarantee

that

u

D

-LRB-

v

x

-RRB-

and

v

D

-LRB-

u

x

-RRB-

and

so

u

and

v

are

unrelated

Similarly

by

-LRB-

-RRB-

-LRB-

-RRB-

and

Claim

the

vertices

u

and

w

are

unrelated

as

are

v

and

w

But

now

the

vertices

u

v

and

w

are

pairwise

unrelated

contradicting

-LRB-

-RRB-

With

this

the

proof

of

Claim

is

complete

Proof

of

Theorem

If

N

-LRB-

x

-RRB-

is

empty

then

-LRB-

x

x

-RRB-

is

a

dominating

pair

If

N

-LRB-

x

-RRB-

is

nonempty

but

contains

a

strong

vertex

y

then

clearly

-LRB-

x

y

-RRB-

is

a

dominating

pair

Otherwise

let

-LRB-

y

z

-RRB-

be

an

admissible

pair

we

assume

without

loss

of

generality

that

y

and

z

belong

to

distinct

connected

components

Y

and

Z

of

N

-LRB-

x

-RRB-

respectively

By

Claim

Claim

and

Claim

we

find

special

vertices

y

in

Y

and

z

in

Z

such

that

y

D

-LRB-

z

x

-RRB-

and

z

D

-LRB-

y

x

-RRB-

Put

differently

y

and

z

are

unrelated

Furthermore

since

y

and

z

are

special

we

have

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

D

-LRB-

y

x

-RRB-

-LSB-

D

-LRB-

z

x

-RRB-

implying

that

-LRB-

y

z

-RRB-

is

also

an

admissible

pair

Finally

we

claim

that

-LRB-

y

z

-RRB-

is

a

dominating

pair

in

G.

By

Claim

any

vertex

v

that

misses

some

y

z

path

must

be

in

N

-LRB-

x

-RRB-

-LRB-

Observe

that

v

and

x

are

distinct

since

every

y

z

path

contains

at

least

one

vertex

in

N

-LRB-

x

-RRB-

-RRB-

Since

y

and

z

are

unrelated

y

misses

some

z

x-path

ss

and

z

misses

some

y

x-path

But

now

we

have

reached

a

contradiction

fy

z

vg

is

an

asteroidal

triple

To

see

this

note

that

by

assumption

v

misses

some

y

z

path

in

addition

y

misses

the

z

v-path

ss

-LSB-

fvg

and

z

misses

the

y

v-path

-LSB-

fvg

It

is

perhaps

interesting

to

note

that

Claim

suggests

the

following

characterization

of

AT-free

graphs

The

proof

is

immediate

and

left

to

the

reader

Theorem

A

graph

G

is

AT-free

if

and

only

if

for

every

vertex

x

of

G

no

component

F

of

N

-LRB-

x

-RRB-

contains

unrelated

vertices

Distance

Properties

of

Dominating

Pairs

The

purpose

of

this

section

is

to

examine

various

distance

properties

featured

by

dominating

pairs

in

connected

AT-free

graphs

Specifically

we

study

the

maximum

distance

between

vertices

of

a

dominating

pair

as

well

as

the

relationship

between

dominating

pairs

and

minimum

cardinality

connected

dominating

sets

To

begin

we

state

a

property

of

connected

AT-free

graphs

that

will

be

used

throughout

this

section

Claim

A

connected

asteroidal

triple-free

graph

G

is

a

clique

if

and

only

if

it

contains

no

non-adjacent

dominating

pair

Proof

The

only

if

part

is

trivial

To

prove

the

if

part

note

that

if

G

is

not

a

clique

then

for

some

vertex

x

of

G

N

-LRB-

x

-RRB-

is

nonempty

By

Theorem

there

exist

vertices

y

z

N

-LRB-

x

-RRB-

such

that

either

-LRB-

x

y

-RRB-

is

a

dominating

pair

or

failing

this

-LRB-

y

z

-RRB-

is

a

dominating

pair

In

the

latter

case

the

vertices

y

and

z

belong

to

distinct

connected

components

of

N

-LRB-

x

-RRB-

and

consequently

must

be

non-adjacent

In

the

remainder

of

this

section

we

assume

a

connected

AT-free

graph

G

which

is

not

a

clique

Claim

guarantees

that

we

can

find

a

non-adjacent

dominating

pair

-LRB-

x

-RRB-

in

G.

Let

F

be

the

connected

component

of

N

-LRB-

x

-RRB-

containing

and

let

Y

be

the

set

of

vertices

y

in

F

for

which

-LRB-

x

y

-RRB-

is

a

dominating

pair

in

G

A

vertex

a

in

F

n

Y

is

called

an

attractor

if

Y

ae

D

-LRB-

a

x

-RRB-

Claim

F

contains

no

attractors

Proof

If

the

statement

is

false

then

the

set

A

of

attractors

in

F

n

Y

is

nonempty

Let

a

be

a

vertex

in

A

for

which

D

-LRB-

a

x

-RRB-

ae

D

-LRB-

a

x

-RRB-

for

no

vertex

a

in

A

We

claim

that

-LRB-

a

x

-RRB-

is

a

dominating

pair

in

G

If

the

statement

is

false

we

find

a

vertex

t

that

misses

some

a

x-path

ss

However

-LRB-

i

-RRB-

t

A

by

our

choice

of

a

and

Claim

combined

-LRB-

ii

-RRB-

t

Y

because

Y

ae

D

-LRB-

a

x

-RRB-

-LRB-

iii

-RRB-

t

N

-LRB-

x

-RRB-

n

F

for

otherwise

t

would

miss

a

x-path

Such

a

path

is

contained

in

the

concatenation

of

ss

with

a

a

path

in

F

-LRB-

iv

-RRB-

t

F

n

-LRB-

A

-LSB-

Y

-RRB-

Since

Y

ae

D

-LRB-

a

x

-RRB-

t

must

be

adjacent

to

every

vertex

in

Y

implying

that

t

belongs

to

A

contrary

to

-LRB-

i

-RRB-

The

next

result

concerns

the

maximum

distance

between

vertices

in

a

dominating

pair

Theorem

In

every

connected

asteroidal

triple-free

graph

some

dominating

pair

achieves

the

diameter

Our

proof

of

Theorem

relies

on

the

following

intermediate

results

Lemma

Let

G

be

a

connected

AT-free

graph

and

let

vertices

x

and

a

of

G

be

such

that

d

-LRB-

x

a

-RRB-

diam

-LRB-

G

-RRB-

If

-LRB-

x

y

-RRB-

is

a

dominating

pair

with

vertex

y

in

N

-LRB-

x

-RRB-

then

there

exists

a

vertex

z

such

that

-LRB-

x

z

-RRB-

is

a

dominating

pair

and

d

-LRB-

x

z

-RRB-

diam

-LRB-

G

-RRB-

Proof

Clearly

we

may

assume

that

d

-LRB-

x

a

-RRB-

Let

Y

be

the

set

of

vertices

y

in

N

-LRB-

x

-RRB-

such

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

We

assume

that

a

does

not

belong

to

Y

for

otherwise

there

is

nothing

to

prove

Observe

that

Y

is

contained

in

the

component

of

N

-LRB-

x

-RRB-

containing

a

otherwise

d

-LRB-

x

y

-RRB-

and

d

-LRB-

x

a

-RRB-

since

every

path

joining

x

and

y

must

dominate

a.

By

virtue

of

Claim

a

can

not

be

an

attractor

we

find

a

vertex

y

in

Y

such

that

y

D

-LRB-

a

x

-RRB-

In

particular

a

and

y

are

non-adjacent

Consider

an

arbitrary

shortest

x

y-path

ss

-LRB-

x

y

-RRB-

x

uk

y

Since

-LRB-

x

y

-RRB-

is

a

dominating

pair

a

must

be

adjacent

to

some

vertex

uj

Since

a

and

y

are

non-adjacent

j

k

But

now

diam

-LRB-

G

-RRB-

d

-LRB-

x

a

-RRB-

d

-LRB-

x

uj

-RRB-

d

-LRB-

x

y

-RRB-

diam

-LRB-

G

-RRB-

implying

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

with

d

-LRB-

x

y

-RRB-

diam

-LRB-

G

-RRB-

-RRB-

This

completes

the

proof

of

Lemma

Proof

of

Theorem

Let

vertices

x

and

a

be

such

that

d

-LRB-

x

a

-RRB-

diam

-LRB-

G

-RRB-

Let

C

be

the

connected

component

of

N

-LRB-

x

-RRB-

containing

a

We

may

assume

that

x

is

in

no

dominating

pair

involving

a

vertex

in

N

-LRB-

x

-RRB-

otherwise

we

are

done

by

Lemma

By

the

proof

of

Theorem

there

exists

a

dominating

pair

-LRB-

y

z

-RRB-

with

vertices

y

and

z

belonging

to

distinct

components

of

N

-LRB-

x

-RRB-

We

observe

that

precisely

one

of

y

and

z

belongs

to

C

otherwise

d

-LRB-

y

z

-RRB-

and

we

are

done

-LRB-

To

see

this

note

that

if

neither

of

y

and

z

is

in

C

then

a

must

be

adjacent

to

a

neighbor

of

x

therefore

diam

-LRB-

G

-RRB-

d

-LRB-

a

x

-RRB-

and

d

-LRB-

y

z

-RRB-

diam

-LRB-

G

-RRB-

implying

that

-LRB-

y

z

-RRB-

is

a

dominating

pair

of

distance

diam

-LRB-

G

-RRB-

-RRB-

Assume

without

loss

of

generality

that

y

belongs

to

C

and

that

z

belongs

to

some

component

C

-LRB-

C

-RRB-

of

N

-LRB-

x

-RRB-

If

there

exists

a

shortest

z

y-path

ss

-LRB-

z

y

-RRB-

z

uk

y

such

that

a

is

adjacent

to

uj

for

some

j

k

then

diam

-LRB-

G

-RRB-

d

-LRB-

x

a

-RRB-

d

-LRB-

x

uj

-RRB-

d

-LRB-

z

uj

-RRB-

d

-LRB-

z

y

-RRB-

diam

-LRB-

G

-RRB-

and

-LRB-

y

z

-RRB-

is

the

required

dominating

pair

Otherwise

y

is

the

only

vertex

on

ss

adjacent

to

a

and

diam

-LRB-

G

-RRB-

d

-LRB-

x

a

-RRB-

d

-LRB-

a

z

-RRB-

d

-LRB-

y

z

-RRB-

diam

-LRB-

G

-RRB-

Therefore

d

-LRB-

a

z

-RRB-

diam

-LRB-

G

-RRB-

and

the

conclusion

follows

by

Lemma

Thus

in

a

connected

AT-free

graph

some

dominating

pair

achieves

the

diameter

We

now

consider

shortest

dominating

paths

and

their

relation

to

connected

dominating

sets

A

minimum

cardinality

connected

dominating

set

-LRB-

mccds

for

short

-RRB-

in

a

graph

G

-LRB-

V

E

-RRB-

is

a

minimum

cardinality

set

S

V

which

is

a

dominating

set

and

which

induces

a

connected

subgraph

of

G

In

the

remainder

of

this

section

we

shall

find

it

convenient

to

make

use

of

a

special

notation

that

we

now

introduce

When

referring

to

a

path

ss

we

shall

denote

by

ss

y

the

path

obtained

from

ss

by

removing

y

one

of

its

endpoints

Similarly

we

let

ss

x

denote

the

path

obtained

from

ss

by

the

addition

of

x

as

a

new

endpoint

Theorem

In

every

connected

asteroidal

triple-free

graph

some

mccds

induces

a

path

Proof

Let

G

be

a

connected

AT-free

graph

and

let

D

be

an

arbitrary

mccds

and

let

-LRB-

x

y

-RRB-

be

an

arbitrary

dominating

pair

in

G

We

note

that

if

fx

yg

ae

D

then

D

induces

a

path

-LRB-

-RRB-

This

follows

from

the

fact

that

every

x

y-path

ss

in

D

is

a

connected

dominating

set

implying

that

D

ss

Next

we

claim

that

if

x

D

-LRB-

or

y

D

-RRB-

then

some

mccds

induces

a

path

-LRB-

-RRB-

To

justify

-LRB-

-RRB-

assume

without

loss

of

generality

that

x

D.

By

-LRB-

-RRB-

we

may

assume

that

y

D.

Let

Y

consist

of

all

the

vertices

in

D

adjacent

to

y

Since

D

is

connected

we

find

a

path

ss

joining

x

and

a

vertex

in

Y

such

that

all

vertices

in

ss

are

in

DnY

Clearly

D

ss

else

ss

y

is

a

dominating

path

of

cardinality

at

most

j

D

j.

Thus

-LRB-

-RRB-

must

hold

By

-LRB-

-RRB-

and

-LRB-

-RRB-

combined

we

may

assume

that

neither

x

nor

y

belongs

to

D

We

may

further

assume

that

a

shortest

x

y-path

contains

at

least

two

edges

else

j

D

j

and

the

claim

is

trivial

Thus

j

D

j

Let

X

and

Y

be

the

sets

of

vertices

in

D

adjacent

to

x

and

y

respectively

Observe

that

X

and

Y

must

be

disjoint

for

otherwise

with

w

standing

for

an

arbitrary

vertex

in

X

Y

fx

w

yg

induces

a

dominating

path

and

there

is

nothing

to

prove

Connectedness

of

D

guarantees

the

existence

of

vertices

in

X

in

Y

and

of

an

ss

in

D

all

of

whose

internal

vertices

are

in

Dn

-LRB-

X

-LSB-

Y

-RRB-

We

claim

that

j

Dnss

j

-LRB-

-RRB-

To

see

that

this

is

the

case

observe

that

if

D

ss

then

we

are

done

if

j

Dnss

j

then

ss

x

y

is

a

dominating

path

of

cardinality

at

most

j

D

j.

Thus

-LRB-

-RRB-

must

hold

By

-LRB-

-RRB-

we

write

fzg

Dnss

Since

the

path

ss

x

is

of

cardinality

j

D

j

we

find

a

vertex

u

that

misses

ss

x.

Similarly

since

the

path

ss

y

is

of

cardinality

j

D

j

we

find

a

vertex

v

that

misses

ss

y

The

following

are

easily

seen

ffl

u

v

and

uy

vx

are

edges

-LRB-

otherwise

we

contradict

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

-RRB-

ffl

u

and

v

are

not

adjacent

-LRB-

else

fu

x

is

an

AT

in

G

-RRB-

ffl

u

z

v

z

and

both

uz

vz

are

edges

-LRB-

otherwise

we

contradict

that

D

is

a

connected

dominating

set

-RRB-

ffl

are

both

edges

-LRB-

if

is

not

an

edge

then

fu

vg

is

an

AT

-RRB-

We

claim

that

fu

z

vg

is

a

mccds

To

see

this

let

w

be

a

vertex

that

misses

the

path

induced

by

fu

z

vg

Since

D

is

dominating

w

must

be

adjacent

to

some

vertex

on

ss

But

now

it

is

easy

to

confirm

that

fu

v

wg

is

an

AT

Next

we

show

that

Theorem

can

be

strengthened

To

make

this

statement

precise

we

shall

refer

to

a

mccds

that

induces

a

path

as

a

path-mccds

Theorem

In

every

connected

asteroidal

triple-free

graph

the

endpoints

of

some

path-mccds

are

a

dominating

pair

Our

proof

of

Theorem

relies

on

the

following

technical

result

Lemma

Let

G

be

a

connected

asteroidal

triple-free

graph

and

let

ss

-LRB-

x

a

-RRB-

be

a

path-mccds

in

G

with

endpoints

a

and

x

If

x

belongs

to

a

dominating

pair

involving

a

vertex

in

N

-LRB-

x

-RRB-

then

there

exists

a

vertex

y

in

N

-LRB-

x

-RRB-

such

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

and

some

x

y-path

is

a

mccds

Proof

Write

ss

-LRB-

x

a

-RRB-

x

uk

a

We

may

assume

that

k

Let

C

be

the

component

of

N

-LRB-

x

-RRB-

containing

a.

Observe

that

every

vertex

that

forms

a

dominating

pair

with

x

must

belong

to

C.

To

clarify

this

suppose

such

a

vertex

t

belongs

to

a

component

C

distinct

from

C.

Then

since

the

path

ss

-LRB-

x

a

-RRB-

is

dominating

t

is

adjacent

to

implying

that

d

-LRB-

x

t

-RRB-

k

and

there

is

nothing

to

prove

Let

Y

be

the

set

of

all

special

vertices

in

C

It

is

easy

to

see

that

x

forms

a

dominating

pair

with

every

vertex

in

Y

Thus

we

may

assume

that

a

Y

Note

that

if

some

vertex

in

Y

is

adjacent

to

uj

with

j

k

then

we

are

done

otherwise

a

is

an

attractor

contradicting

Claim

This

completes

the

proof

of

Lemma

Proof

of

Theorem

For

convenience

we

inherit

the

notation

of

Lemma

We

may

assume

that

x

is

in

no

dominating

pair

involving

a

vertex

in

N

-LRB-

x

-RRB-

otherwise

we

are

done

by

Lemma

By

the

proof

of

Theorem

there

exists

a

dominating

pair

-LRB-

y

z

-RRB-

with

y

and

z

in

distinct

components

of

N

-LRB-

x

-RRB-

We

observe

that

precisely

one

of

the

vertices

y

and

z

belongs

to

C

otherwise

d

-LRB-

y

z

-RRB-

k

and

we

are

done

Assume

without

loss

of

generality

that

y

C

and

that

z

belongs

to

a

component

C

distinct

from

C.

Note

that

since

ss

-LRB-

x

a

-RRB-

is

dominating

z

is

adjacent

to

Thus

y

is

adjacent

to

a

and

to

no

other

vertex

on

ss

-LRB-

x

a

-RRB-

for

otherwise

d

-LRB-

y

z

-RRB-

k

We

claim

that

at

least

one

of

the

paths

ss

-LRB-

x

a

-RRB-

x

y

or

ss

-LRB-

x

a

-RRB-

x

z

is

dominating

Observe

that

both

these

paths

are

of

length

k

and

each

of

them

is

anchored

at

a

vertex

belonging

to

a

dominating

pair

Therefore

once

we

establish

this

claim

the

conclusion

of

Theorem

follows

from

Lemma

If

neither

of

these

paths

is

dominating

then

ffl

there

exists

a

vertex

v

missing

ss

-LRB-

x

a

-RRB-

x

y

trivially

both

vx

and

vz

are

edges

ffl

there

exists

a

vertex

w

missing

ss

-LRB-

x

a

-RRB-

x

z

trivially

both

wx

and

wy

are

edges

But

now

we

have

reached

a

contradiction

fa

w

zg

is

an

AT

and

the

proof

of

the

theorem

is

complete

Two

Characterizations

of

AT-free

Graphs

The

goal

of

this

section

is

to

offer

two

characterizations

of

AT-free

graphs

To

motivate

our

first

characterization

notice

that

Theorems

and

do

not

lead

to

a

characterization

of

AT-free

graphs

For

example

vertices

achieving

the

diameter

in

the

constitute

a

dominating

pair

Furthermore

if

we

add

a

universal

vertex

to

an

arbitrary

graph

we

obtain

a

graph

that

has

a

dominating

pair

consisting

of

the

universal

vertex

and

any

other

vertex

Clearly

any

attempt

to

provide

a

characterization

of

AT-free

graphs

involving

dominating

pairs

must

not

only

be

based

on

induced

subgraphs

it

must

also

restrict

the

types

of

dominating

pairs

For

example

the

graph

contains

an

AT

yet

every

induced

subgraph

has

a

dominating

pair

The

first

goal

is

to

provide

a

characterization

of

AT-free

graphs

based

on

dominating

pairs

As

indicated

previously

such

a

result

must

restrict

the

types

of

dominating

pairs

In

particular

we

impose

an

adjacency

condition

on

G

with

dominating

pair

-LRB-

x

y

-RRB-

whereby

the

connected

component

of

G

n

fxg

containing

y

has

a

dominating

pair

-LRB-

y

-RRB-

with

adjacent

to

x

As

illustrated

in

Figure

the

graph

fails

this

criterion

Here

-LRB-

x

y

-RRB-

is

a

dominating

pair

in

the

graph

yet

neither

-LRB-

y

-RRB-

nor

-LRB-

y

-RRB-

is

a

dominating

pair

in

the

graph

obtained

by

removing

vertex

x

We

begin

by

stating

a

simple

property

of

vertices

in

AT-free

graphs

which

is

of

independent

interest

Claim

Let

u

v

and

y

be

vertices

in

a

connected

asteroidal

triple-free

graph

such

that

v

D

-LRB-

u

y

-RRB-

If

D

-LRB-

u

y

-RRB-

D

-LRB-

v

y

-RRB-

then

for

some

vertex

w

in

D

-LRB-

u

y

-RRB-

v

and

w

are

unrelated

with

respect

to

y.

Proof

Let

ss

be

a

u

y-path

missed

by

v.

Let

w

be

an

arbitrary

vertex

in

the

set

D

-LRB-

u

y

-RRB-

n

D

-LRB-

v

y

-RRB-

Since

w

does

not

belong

to

D

-LRB-

v

y

-RRB-

w

misses

some

v

y-path

Since

w

belongs

to

D

-LRB-

u

y

-RRB-

w

intercepts

ss

and

moreover

ss

-LSB-

fwg

contains

a

chordless

w

y-path

missed

by

v

confirming

that

v

and

w

are

unrelated

with

respect

to

y.

Let

ss

uk

and

vl

be

two

paths

We

shall

refer

to

the

path

ui

with

i

k

as

a

prefix

of

ss

A

vertex

w

is

a

cross

point

of

ss

and

Figure

does

not

satisfy

the

spine

property

if

w

ui

vj

and

the

four

vertices

ui

vj

ui

and

vj

are

all

defined

and

distinct

For

later

reference

we

now

investigate

properties

of

asteroidal

triples

Let

G

be

a

graph

containing

an

AT

Choose

an

induced

subgraph

H

of

G

with

the

least

number

of

vertices

such

that

some

triple

fx

y

zg

is

an

AT

in

H.

Let

ss

-LRB-

x

y

-RRB-

ss

-LRB-

x

z

-RRB-

and

ss

-LRB-

y

z

-RRB-

be

paths

in

H

demonstrating

that

fx

y

zg

is

an

AT

In

the

following

we

write

ss

-LRB-

x

y

-RRB-

x

uk

y

ss

-LRB-

x

z

-RRB-

x

vl

z

and

ss

-LRB-

z

y

-RRB-

z

wt

y.

Clearly

the

choice

of

H

guarantees

that

x

y

and

z

have

degree

at

most

Claim

No

pair

of

paths

among

ss

-LRB-

x

y

-RRB-

ss

-LRB-

x

z

-RRB-

and

ss

-LRB-

y

z

-RRB-

has

a

cross

point

Proof

Suppose

that

the

paths

ss

-LRB-

x

y

-RRB-

and

ss

-LRB-

x

z

-RRB-

have

a

cross

point

w

such

that

w

ui

vj

Observe

that

the

definition

of

a

cross

point

guarantees

that

i

and

j

Our

choice

of

the

paths

ss

-LRB-

x

y

-RRB-

and

ss

-LRB-

x

z

-RRB-

guarantees

that

i

k

and

that

j

l

But

now

in

H

H

n

fvj

y

misses

the

x

z-path

ui

vj

vj

z

and

x

misses

the

y

z-path

y

uk

ui

vj

vj

z.

Thus

fx

y

zg

is

an

AT

in

H

contradicting

the

minimality

of

H.

Claim

Let

i

be

the

largest

subscript

for

which

there

exists

a

subscript

j

such

that

ui

vj

and

ui

vj

Then

i

j

and

ut

vt

for

all

t

i.

Proof

Since

y

are

z

are

distinct

and

the

subscript

i

in

the

statement

of

the

claim

always

exists

Since

by

Claim

ui

can

not

be

a

cross

point

we

must

have

ui

vj

Let

t

be

the

least

value

for

which

ui?t

vj?t

We

may

assume

that

such

a

t

exists

for

otherwise

there

is

nothing

to

prove

Clearly

implies

that

t

minfi

j

Consequently

we

can

remove

vertex

vj?t

from

H

while

still

ensuring

that

fx

y

zg

is

an

AT

in

the

remaining

graph

This

contradiction

completes

the

proof

of

the

claim

Theorem

There

exist

unique

vertices

in

H

such

that

-LRB-

i

-RRB-

The

unique

path

between

x

and

is

a

prefix

of

both

ss

-LRB-

x

y

-RRB-

and

ss

-LRB-

x

z

-RRB-

-LRB-

ii

-RRB-

The

unique

path

between

y

and

is

a

prefix

of

both

ss

-LRB-

y

x

-RRB-

and

ss

-LRB-

y

z

-RRB-

-LRB-

iii

-RRB-

The

unique

path

between

z

and

is

a

prefix

of

both

ss

-LRB-

z

x

-RRB-

and

ss

-LRB-

z

y

-RRB-

Proof

Claim

guarantees

that

one

can

associate

with

x

a

unique

vertex

corresponding

to

the

largest

subscript

for

which

ui

vi

Put

differently

the

path

x

ui

in

H

is

the

common

prefix

of

both

ss

-LRB-

x

y

-RRB-

and

ss

-LRB-

x

z

-RRB-

In

a

perfectly

similar

way

one

can

define

vertices

and

As

it

turns

out

vertices

have

a

number

of

interesting

properties

We

present

some

of

them

next

Claim

The

vertices

and

are

either

all

distinct

or

else

they

coincide

Proof

Suppose

that

exactly

two

of

the

vertices

coincide

Symmetry

allows

us

to

assume

that

Write

ui

and

wt?k

i

Since

-LRB-

-RRB-

can

not

be

a

cross

point

of

ss

-LRB-

x

z

-RRB-

and

ss

-LRB-

z

y

-RRB-

we

must

have

vi

wt?k

i

Now

an

argument

similar

to

that

of

the

proof

of

Claim

guarantees

that

the

subpaths

of

ss

-LRB-

x

z

-RRB-

and

ss

-LRB-

z

y

-RRB-

between

z

and

coincide

a

contradiction

Claim

and

the

minimality

of

H

combined

imply

the

following

result

Corollary

Vertices

and

coincide

if

and

only

if

H

is

isomorphic

to

the

graph

in

Figure

Figure

Illustrating

Claim

Claim

Vertex

is

distinct

from

x

if

and

only

if

dH

-LRB-

x

-RRB-

Furthermore

if

and

are

distinct

and

x

then

is

an

edge

Proof

First

observe

that

if

x

then

by

Claim

dH

-LRB-

x

-RRB-

Conversely

if

vertices

x

and

are

distinct

then

ss

-LRB-

x

y

-RRB-

and

ss

-LRB-

x

z

-RRB-

have

at

least

one

edge

in

common

confirming

that

dH

-LRB-

x

-RRB-

To

settle

the

second

part

of

the

claim

assume

that

ui

with

i

Since

are

distinct

ui

misses

the

path

ss

-LRB-

y

z

-RRB-

and

thus

fui

y

zg

is

an

AT

in

H

n

fxg

The

conclusion

follows

For

reasons

that

will

become

clear

later

we

shall

say

that

a

connected

graph

H

with

a

dominating

pair

satisfies

the

spine

property

if

for

every

non-adjacent

dominating

pair

-LRB-

ff

-RRB-

in

H

there

exists

a

neighbor

of

such

that

-LRB-

-RRB-

is

a

dominating

pair

of

the

connected

component

of

H

n

f

g

containing

We

are

now

in

a

position

to

state

the

first

main

result

of

this

section

Theorem

-LRB-

The

Spine

Theorem

-RRB-

A

graph

G

is

asteroidal

triple-free

if

and

only

if

every

connected

induced

subgraph

H

of

G

satisfies

the

spine

property

Proof

of

Theorem

To

settle

the

only

if

part

we

observe

that

if

G

is

a

clique

then

there

is

nothing

to

prove

We

may

therefore

assume

that

G

is

not

a

clique

Claim

guarantees

that

some

induced

subgraph

H

of

G

contains

a

nonadjacent

dominating

pair

-LRB-

ff

-RRB-

Let

C

denote

the

connected

component

of

H

n

f

g

that

contains

Let

A

denote

N

-LRB-

-RRB-

C

We

choose

a

vertex

in

A

such

that

D

-LRB-

-RRB-

ae

D

-LRB-

t

-RRB-

for

no

vertex

t

in

A

We

claim

that

-LRB-

ff

-RRB-

is

a

dominating

pair

in

C

-LRB-

-RRB-

To

see

that

-LRB-

-RRB-

holds

suppose

that

a

vertex

t

in

C

misses

some

ff

path

Observe

that

t

must

belong

to

A

for

otherwise

this

path

extends

to

an

ff

path

in

H

missed

by

t

contradicting

that

-LRB-

ff

-RRB-

is

a

dominating

pair

Our

choice

of

guarantees

that

D

-LRB-

-RRB-

D

-LRB-

t

-RRB-

By

Claim

we

find

a

vertex

w

in

D

-LRB-

-RRB-

such

that

t

and

w

are

unrelated

with

respect

to

Note

that

w

belongs

to

A

otherwise

the

t

path

missed

by

w

would

extend

to

an

ff

path

missed

by

w

But

now

w

and

t

are

in

the

same

component

of

N

-LRB-

-RRB-

and

are

unrelated

with

respect

to

contradicting

Claim

This

completes

the

proof

of

the

only

if

part

To

prove

the

if

part

let

H

be

an

induced

subgraph

of

G

with

the

least

number

of

vertices

in

which

some

set

fx

y

zg

is

an

AT

Since

H

is

connected

but

not

a

clique

Claim

guarantees

the

existence

of

a

non-adjacent

dominating

pair

-LRB-

a

b

-RRB-

in

H.

Further

let

ss

-LRB-

x

y

-RRB-

ss

-LRB-

x

z

-RRB-

and

ss

-LRB-

y

z

-RRB-

be

-LRB-

chordless

-RRB-

paths

in

H

demonstrating

that

fx

y

zg

is

an

AT

Claim

Vertices

a

and

b

are

distinct

from

x

y

z

and

Proof

To

begin

we

show

that

a

and

b

are

distinct

from

x

y

and

z.

Suppose

not

we

may

assume

without

loss

of

generality

that

a

x

Since

-LRB-

a

b

-RRB-

is

a

dominating

pair

b

must

belong

to

ss

-LRB-

y

z

-RRB-

Consider

the

x

b-path

contained

in

the

concatenation

of

ss

-LRB-

x

y

-RRB-

with

the

y

b

portion

of

ss

-LRB-

y

z

-RRB-

This

path

is

missed

by

z

unless

vertices

b

and

z

are

adjacent

A

mirror

argument

shows

that

b

and

y

are

also

adjacent

Since

by

assumption

H

satisfies

the

spine

property

and

vertices

a

and

b

are

nonadjacent

we

should

be

able

to

find

a

neighbor

of

b

such

that

-LRB-

a

-RRB-

is

a

dominating

pair

in

H

n

fbg

However

if

belongs

to

ss

-LRB-

x

y

-RRB-

then

z

misses

the

corresponding

a-path

if

belongs

to

ss

-LRB-

x

z

-RRB-

then

y

misses

a

a-path

The

fact

that

a

is

distinct

from

follows

by

an

identical

argument

whose

details

are

omitted

Claim

has

the

following

interesting

corollary

Claim

Each

pair

of

vertices

x

and

y

and

and

z

and

must

coincide

Proof

First

observe

that

the

vertices

are

distinct

for

otherwise

by

Corollary

H

is

isomorphic

to

the

graph

in

Figure

which

does

not

satisfy

the

spine

property

If

the

statement

is

false

then

we

may

assume

without

loss

of

generality

that

x

and

are

distinct

By

Claim

x

has

degree

in

H.

By

Claim

a

-LRB-

resp

b

-RRB-

is

distinct

from

both

x

and

implying

that

x

misses

some

a

b-path

a

contradiction

By

virtue

of

Claim

and

Claim

combined

x

y

and

z

have

degree

exactly

in

H

and

moreover

H

is

biconnected

Without

loss

of

generality

let

vertices

a

and

b

belong

to

ss

-LRB-

x

y

-RRB-

and

to

ss

-LRB-

x

z

-RRB-

respectively

Observe

that

vertices

a

and

y

must

be

adjacent

for

otherwise

the

a

b-path

through

x

is

missed

by

y.

Similarly

vertices

b

and

z

are

also

adjacent

else

the

a

b-path

through

x

is

missed

by

z.

Further

either

a

or

b

is

adjacent

to

x

for

if

not

then

the

a

b-path

through

y

and

z

is

missed

by

x.

Symmetry

allows

us

to

assume

without

loss

of

generality

that

a

and

x

are

adjacent

We

claim

that

vertices

b

and

x

are

adjacent

-LRB-

-RRB-

Since

vertices

a

and

b

are

not

adjacent

the

spine

property

guarantees

that

we

can

find

a

neighbor

of

a

such

that

-LRB-

b

-RRB-

is

a

dominating

pair

of

H

nfag

Clearly

can

not

be

x

if

b

and

x

are

not

adjacent

then

can

not

be

y.

Therefore

must

belong

to

ss

-LRB-

y

z

-RRB-

But

now

x

misses

the

b-path

containing

z

a

contradiction

Thus

-LRB-

-RRB-

must

hold

To

complete

the

proof

of

the

if

part

we

claim

that

-LRB-

b

y

-RRB-

is

a

dominating

pair

-LRB-

-RRB-

It

is

clear

that

once

-LRB-

-RRB-

is

proved

we

have

reached

a

contradiction

by

Claim

y

can

not

be

in

a

dominating

pair

To

prove

-LRB-

-RRB-

consider

a

vertex

c

that

misses

a

path

ss

joining

b

and

y

that

does

not

involve

a.

Trivially

c

must

belong

to

ss

-LRB-

y

z

-RRB-

But

now

fc

x

yg

is

an

AT

in

H

n

fag

To

see

this

note

that

ss

x

is

an

x

y-path

missed

by

c

the

y

c-path

consisting

of

the

portion

of

ss

-LRB-

y

z

-RRB-

from

y

to

c

is

missed

by

x

finally

ss

-LRB-

x

z

-RRB-

concatenated

with

the

c

z

portion

of

ss

-LRB-

y

z

-RRB-

contains

a

c

x-path

missed

by

y

This

completes

the

proof

of

Theorem

ff?fi

ff

fi

ff

fi

ff

fi

Figure

Illustrating

the

Spine

Theorem

Let

G

-LRB-

V

E

-RRB-

be

an

AT-free

graph

and

let

-LRB-

x

y

-RRB-

be

an

arbitrary

non-adjacent

dominating

pair

in

G.

Construct

a

sequence

xk

of

vertices

of

G

and

a

sequence

Gk

of

subgraphs

of

G

defined

as

follows

i

-RRB-

G

and

x

ii

-RRB-

for

all

i

-LRB-

i

k

-RRB-

xiy

E

and

xky

E

iii

-RRB-

for

all

i

-LRB-

i

k

-RRB-

let

Gi

stand

for

the

subgraph

of

Gi

induced

by

the

component

of

Gi

n

fxi

containing

y

iv

-RRB-

for

all

i

-LRB-

i

k

-RRB-

let

xi

be

a

vertex

in

Gi

adjacent

to

xi

and

such

that

-LRB-

xi

y

-RRB-

is

a

dominating

pair

in

Gi

The

existence

of

the

sequence

xk

is

guaranteed

by

the

Spine

Theorem

The

sequence

xk

y

will

be

referred

to

as

a

spine

of

G.

For

an

illustration

of

the

Spine

Theorem

the

reader

is

referred

to

Figure

The

sequence

of

graphs

featured

in

Figure

begins

with

a

graph

G

with

vertex-set

fa

b

c

d

e

x

yg

and

with

dominating

pair

-LRB-

x

y

-RRB-

The

sequence

continues

with

the

graph

G

n

fxg

with

dominating

pair

-LRB-

a

y

-RRB-

and

so

on

The

spine

of

the

graph

G

is

featured

in

heavy

lines

The

Spine

Theorem

could

lead

one

to

the

false

conclusion

that

a

graph

is

AT-free

if

and

only

if

there

is

a

path

xj

such

that

-LRB-

xi

xj

-RRB-

is

a

dominating

pair

in

G

n

xi

As

a

counterexample

consider

the

graph

in

Figure

In

this

graph

g

is

an

AT

Furthermore

-LRB-

-RRB-

is

a

dominating

pair

in

G

and

-LRB-

-RRB-

is

a

dominating

pair

in

G

n

-LRB-

Furthermore

-LRB-

-RRB-

is

a

dominating

pair

of

G

n

-RRB-

Note

however

that

the

induced

subgraph

G

n

has

a

dominating

pair

-LRB-

-RRB-

yet

G

n

has

no

dominating

pair

consisting

of

and

a

neighbor

of

Let

G

-LRB-

V

E

-RRB-

be

an

arbitrary

graph

A

triangulation

T

-LRB-

G

-RRB-

of

G

is

a

set

of

edges

such

that

the

graph

-LRB-

V

E

-LSB-

T

-LRB-

G

-RRB-

-RRB-

is

chordal

A

triangulation

T

-LRB-

G

-RRB-

is

minimal

when

no

proper

subset

of

T

-LRB-

G

-RRB-

is

a

triangulation

of

G.

Recently

M?ohring

-LSB-

-RSB-

proved

the

following

result

Proposition

If

G

is

an

asteroidal

triple-free

graph

then

for

every

minimal

triangulation

T

-LRB-

G

-RRB-

of

G

the

graph

-LRB-

V

E

-LSB-

T

-LRB-

G

-RRB-

-RRB-

is

an

interval

graph

The

second

main

goal

of

this

section

is

to

provide

a

characterization

of

AT-free

graphs

by

proving

the

converse

of

Proposition

Theorem

A

graph

G

is

asteroidal

triple-free

if

and

only

if

for

every

minimal

triangulation

T

-LRB-

G

-RRB-

of

G

the

graph

-LRB-

V

E

-LSB-

T

-LRB-

G

-RRB-

-RRB-

is

an

interval

graph

Our

arguments

rely

in

part

on

the

following

result

which

is

of

independent

interest

Lemma

Let

G

be

an

arbitrary

graph

and

let

H

-LRB-

V

-LRB-

H

-RRB-

E

-LRB-

H

-RRB-

-RRB-

be

an

induced

subgraph

of

G.

Let

T

-LRB-

H

-RRB-

be

an

arbitrary

minimal

triangulation

of

H

There

exists

Figure

a

triangulation

T

-LRB-

G

-RRB-

of

G

such

that

the

only

edges

in

T

-LRB-

G

-RRB-

joining

vertices

in

H

are

those

in

T

-LRB-

H

-RRB-

Proof

If

the

statement

is

false

then

we

select

a

triangulation

T

-LRB-

G

-RRB-

of

G

that

adds

as

few

new

edges

to

E

-LRB-

H

-RRB-

-LSB-

T

-LRB-

H

-RRB-

as

possible

Since

T

-LRB-

H

-RRB-

is

a

triangulation

of

H

some

edge

uv

with

both

u

and

v

in

H

present

in

T

-LRB-

G

-RRB-

but

not

in

T

-LRB-

H

-RRB-

must

be

the

unique

chord

of

a

having

-LRB-

at

least

-RRB-

one

vertex

outside

H.

Let

w

and

be

the

remaining

vertices

of

this

with

w

outside

of

H

The

triangulation

of

G

obtained

by

replacing

the

edge

uv

by

is

minimal

since

is

the

unique

chord

of

the

and

has

one

fewer

edges

added

to

E

-LRB-

H

-RRB-

-LSB-

T

-LRB-

H

-RRB-

contradicting

our

choice

of

T

-LRB-

G

-RRB-

Proof

of

Theorem

The

only

if

part

follows

trivially

from

Proposition

To

prove

the

if

part

let

G

be

a

graph

containing

an

AT

Choose

an

induced

subgraph

H

-LRB-

V

-LRB-

H

-RRB-

E

-LRB-

H

-RRB-

-RRB-

of

G

with

the

least

number

of

vertices

such

that

some

triple

fx

y

zg

is

an

AT

in

H.

Let

ss

-LRB-

x

y

-RRB-

ss

-LRB-

x

z

-RRB-

and

ss

-LRB-

y

z

-RRB-

be

paths

in

H

demonstrating

that

fx

y

zg

is

an

AT

and

write

ss

-LRB-

x

y

-RRB-

x

uk

y

ss

-LRB-

x

z

-RRB-

x

vl

z

and

ss

-LRB-

z

y

-RRB-

z

wt

y.

Clearly

the

choice

of

H

guarantees

that

x

y

and

z

have

degree

at

most

Our

plan

is

to

exhibit

a

minimal

triangulation

T

-LRB-

H

-RRB-

of

H

that

results

in

a

noninterval

graph

H

-LRB-

V

-LRB-

H

-RRB-

E

-LRB-

H

-RRB-

-LSB-

T

-LRB-

H

-RRB-

-RRB-

For

this

purpose

let

and

be

the

vertices

specified

in

Theorem

and

consider

the

triangulation

T

-LRB-

H

-RRB-

of

H

returned

by

the

following

procedure

Step

If

then

set

T

-LRB-

H

-RRB-

and

return

Step

Let

F

be

the

graph

obtained

from

H

by

removing

vertices

x

y

z

and

by

adding

the

edges

-LRB-

in

case

x

-RRB-

uk

-LRB-

in

case

y

-RRB-

and

vl

-LRB-

in

case

z

-RRB-

Let

T

-LRB-

F

-RRB-

be

an

arbitrary

triangulation

of

F

Return

T

-LRB-

H

-RRB-

T

-LRB-

F

-RRB-

-LSB-

yuk

ywt

zvl

-LRB-

in

case

x

one

adds

the

edge

instead

of

the

edges

and

etc

-RRB-

Now

Claim

along

with

an

easy

ad-hoc

argument

shows

that

T

-LRB-

H

-RRB-

is

a

minimal

triangulation

of

H

and

that

fx

y

zg

is

still

an

AT

in

the

graph

H

-LRB-

V

-LRB-

H

-RRB-

E

-LRB-

H

-RRB-

-LSB-

T

-LRB-

H

-RRB-

-RRB-

By

Lemma

there

must

exists

some

triangulation

T

-LRB-

G

-RRB-

of

G

such

that

H

is

an

induced

subgraph

of

G

-LRB-

V

E

-LSB-

T

-LRB-

G

-RRB-

-RRB-

The

conclusion

follows

Augmenting

AT-free

Graphs

The

purpose

of

this

section

is

twofold

First

we

exhibit

a

structural

property

of

AT

free

graph

that

naturally

allows

one

to

stretch

an

AT-free

graph

to

a

new

AT-free

graph

Next

we

provide

a

decomposition

theorem

for

AT-free

graphs

To

begin

we

address

the

issue

of

creating

new

AT-free

graphs

out

of

old

ones

Specifically

we

show

how

to

augment

an

arbitrary

AT-free

graph

G

to

obtain

a

new

AT-free

graph

This

augmentation

will

be

accomplished

by

finding

a

particular

dominating

pair

-LRB-

x

y

-RRB-

and

by

adding

new

vertices

and

adjacent

to

x

and

y

respectively

This

augmentation

of

G

again

confirms

our

intuition

about

the

linear

structure

of

AT-free

graphs

since

the

dominating

pair

-LRB-

x

y

-RRB-

has

been

stretched

to

a

new

dominating

pair

-LRB-

-RRB-

Figure

Illustrating

pokable

and

unpokable

vertices

In

preparation

for

stating

the

first

main

result

of

this

section

we

need

to

define

a

few

terms

A

vertex

v

of

an

AT-free

graph

G

is

called

pokable

if

the

graph

obtained

from

G

by

adding

a

pendant

vertex

adjacent

to

v

is

AT-free

For

example

referring

to

Figure

vertex

u

is

pokable

since

the

addition

of

a

pendant

vertex

does

not

create

an

AT

in

the

graph

At

the

same

time

vertex

v

is

unpokable

for

the

addition

of

the

vertex

creates

the

AT

fa

b

A

dominating

pair

-LRB-

x

y

-RRB-

is

referred

to

as

pokable

if

both

x

and

y

are

pokable

For

further

reference

we

take

note

of

the

following

simple

observation

whose

proof

is

routine

Observation

A

vertex

v

of

an

asteroidal

triple-free

graph

G

is

unpokable

if

and

only

if

there

exist

non-adjacent

vertices

u

and

w

in

G

and

induced

paths

v

up

u

missed

by

w

v

wq

w

missed

by

u

as

well

as

a

u

w-path

that

does

not

contain

v

The

notation

of

Observation

will

be

used

whenever

we

are

in

the

presence

of

an

unpokable

vertex

v

We

are

now

in

a

position

to

make

the

previous

discussion

precise

Theorem

Every

connected

asteroidal

triple-free

graph

contains

a

pokable

dominating

pair

Proof

The

theorem

is

trivial

for

cliques

We

shall

therefore

assume

that

G

is

not

a

clique

Now

Claim

guarantees

the

existence

of

a

non-adjacent

dominating

pair

-LRB-

x

-RRB-

in

G.

Let

F

be

the

connected

component

of

N

-LRB-

x

-RRB-

containing

and

let

Y

stand

for

the

set

of

vertices

y

in

F

for

which

-LRB-

x

y

-RRB-

is

a

dominating

pair

in

G

The

conclusion

of

Theorem

is

implied

by

the

following

technical

result

that

will

be

proved

later

Lemma

Y

contains

a

pokable

vertex

To

see

how

Theorem

follows

from

Lemma

let

be

a

pokable

vertex

in

Y

and

let

X

denote

the

set

of

vertices

in

the

same

component

of

N

-LRB-

-RRB-

as

x

for

which

-LRB-

-RRB-

is

a

dominating

pair

Clearly

x

belongs

to

X

and

so

X

is

not

empty

By

applying

Lemma

again

with

as

the

anchor

we

find

a

pokable

vertex

in

X

The

proof

of

Theorem

is

established

by

noting

that

-LRB-

ff

-RRB-

is

the

desired

pokable

dominating

pair

Proof

of

Lemma

The

proof

is

by

induction

on

the

number

of

vertices

in

G.

Assume

the

lemma

true

for

all

connected

AT-free

graphs

with

fewer

vertices

than

G

We

now

present

various

facts

that

are

used

in

the

proof

Claim

Let

v

be

an

unpokable

vertex

in

Y

Then

all

vertices

ui

and

wj

-LRB-

i

p

j

q

-RRB-

belong

to

F

Proof

Without

loss

of

generality

let

i

be

the

smallest

subscript

for

which

ui

lies

outside

F

Trivially

ui

must

belong

to

N

-LRB-

x

-RRB-

Since

w

can

not

miss

the

v

x-path

v

ui

x

and

since

w

is

adjacent

to

no

vertex

on

the

path

v

ui

it

follows

that

w

belongs

to

N

-LRB-

x

-RRB-

Similarly

since

u

can

not

miss

the

v

x-path

v

wq

w

x

and

since

u

is

adjacent

to

no

vertex

on

the

path

v

wq

it

follows

that

u

belongs

to

N

-LRB-

x

-RRB-

But

now

fu

v

wg

is

an

AT

contradicting

G

being

AT-free

It

is

important

to

note

that

by

virtue

of

Claim

Lemma

is

established

as

soon

as

we

exhibit

a

vertex

in

Y

that

is

pokable

in

the

subgraph

of

G

induced

by

F

If

F

and

Y

coincide

then

by

the

induction

hypothesis

such

a

vertex

must

exist

Therefore

from

now

on

we

shall

assume

that

F

n

Y

-LRB-

-RRB-

Let

Yk

-LRB-

k

-RRB-

be

the

connected

components

of

the

subgraph

of

G

induced

by

Y

Claim

Let

t

be

a

vertex

in

F

n

Y

If

some

vertex

z

in

Yi

satisfies

z

D

-LRB-

t

x

-RRB-

then

Yi

ae

D

-LRB-

t

x

-RRB-

Proof

If

the

claim

is

false

then

we

find

vertices

z

in

Yi

such

that

z

D

-LRB-

t

x

-RRB-

and

D

-LRB-

t

x

-RRB-

Since

Yi

is

a

connected

subgraph

of

G

there

exists

a

chordless

path

z

sr

joining

z

and

in

G

with

all

internal

vertices

in

Yi

Let

j

be

the

smallest

subscript

for

which

sj

D

-LRB-

t

x

-RRB-

Since

D

-LRB-

t

x

-RRB-

such

a

subscript

must

exist

But

now

in

G

sj

and

sj

are

non-adjacent

and

sj

misses

some

t

x-path

while

sj

intercepts

all

such

paths

It

follows

that

sj

misses

a

sj

x-path

a

contradiction

since

sj

belongs

to

Y

Claim

Y

induces

a

disconnected

subgraph

of

G.

Proof

First

we

claim

that

j

Y

j

-LRB-

-RRB-

If

-LRB-

-RRB-

is

false

then

Y

Let

U

stand

for

the

set

of

all

vertices

in

F

adjacent

to

Note

that

-LRB-

-RRB-

along

with

the

connectedness

of

F

guarantee

that

U

is

nonempty

But

now

for

every

u

in

U

Y

ae

D

-LRB-

u

x

-RRB-

Thus

u

is

an

attractor

contradicting

Claim

Therefore

-LRB-

-RRB-

holds

Note

that

by

virtue

of

-LRB-

-RRB-

it

makes

sense

to

talk

about

Y

being

disconnected

in

the

complement

We

now

continue

the

proof

of

Claim

If

Y

-LRB-

-RRB-

and

the

connectedness

of

F

imply

the

existence

of

a

vertex

z

in

Y

adjacent

to

some

vertex

t

in

F

n

Y

Note

in

particular

that

z

belongs

to

D

-LRB-

t

x

-RRB-

and

so

by

Claim

Y

ae

D

-LRB-

t

x

-RRB-

However

now

t

is

an

attractor

a

contradiction

With

this

the

proof

of

Claim

is

complete

Claim

Let

v

be

an

unpokable

vertex

in

Y

Then

ffl

for

all

i

-LRB-

i

p

-RRB-

v

belongs

to

D

-LRB-

ui

x

-RRB-

and

ffl

for

all

j

-LRB-

j

q

-RRB-

v

belongs

to

D

-LRB-

wj

x

-RRB-

Proof

Since

v

is

adjacent

to

it

follows

that

v

D

-LRB-

x

-RRB-

Let

i

be

the

smallest

subscript

for

which

v

does

not

belong

to

D

-LRB-

ui

x

-RRB-

Let

ss

be

a

ui

x-path

missed

by

v.

Note

that

w

must

intercept

ss

for

otherwise

w

would

miss

a

v

x-path

contained

in

fv

uig

-LSB-

ss

However

now

fu

v

wg

is

an

AT

The

proof

that

v

belongs

to

D

-LRB-

wj

x

-RRB-

follows

by

a

mirror

argument

For

every

i

-LRB-

i

k

-RRB-

let

Ti

stand

for

the

set

of

vertices

t

in

F

n

Y

with

the

property

that

Yi

ae

D

-LRB-

t

x

-RRB-

By

renaming

the

Yi

's

if

necessary

we

ensure

that

j

j

j

j

j

Tk

j

Claim

Every

vertex

in

is

adjacent

to

all

vertices

in

Proof

The

statement

is

vacuously

true

if

is

empty

Now

assume

that

is

nonempty

and

let

t

be

a

vertex

in

non-adjacent

to

some

z

in

Since

by

Claim

t

can

not

be

an

attractor

we

find

a

subscript

j

-LRB-

j

-RRB-

such

that

for

some

in

Yj

does

not

belong

to

D

-LRB-

t

x

-RRB-

jTjj

implies

that

there

must

exist

a

vertex

in

Tj

n

such

that

z

does

not

belong

to

D

-LRB-

x

-RRB-

Note

that

t

does

not

belong

to

D

-LRB-

x

-RRB-

otherwise

by

Claim

z

would

belong

to

D

-LRB-

x

-RRB-

a

contradiction

Since

does

not

belong

to

D

-LRB-

t

x

-RRB-

in

particular

is

not

adjacent

to

t

Since

t

does

not

belong

to

D

-LRB-

x

-RRB-

there

exists

a

x-path

missed

by

t

But

now

-LSB-

contains

a

x-path

missed

by

t

contradicting

that

is

in

Y

We

now

continue

the

proof

of

Lemma

Let

Z

be

a

connected

component

of

the

subgraph

of

G

induced

by

By

the

induction

hypothesis

Z

contains

a

vertex

v

pokable

in

Z.

To

complete

the

proof

of

Lemma

we

only

need

show

that

v

is

also

pokable

in

F

Suppose

not

By

Claims

and

combined

all

the

vertices

ui

and

wj

-LRB-

i

p

j

q

-RRB-

belong

to

Y

To

clarify

this

last

point

note

that

by

Claim

and

Claim

at

most

and

belong

to

FnY

But

now

either

and

w

or

and

u

contradict

Claim

Further

since

Y

is

disconnected

in

the

complement

all

the

ui

's

and

wj

's

must

belong

to

In

fact

it

is

easy

to

confirm

that

all

the

ui

's

and

wj

's

must

belong

to

Z

a

contradiction

This

completes

the

proof

of

Lemma

Theorem

implies

the

following

results

that

are

interesting

in

their

own

right

Corollary

Every

asteroidal

triple-free

graph

is

either

a

clique

or

else

contains

two

non-adjacent

pokable

vertices

Corollary

-LRB-

The

Composition

Theorem

-RRB-

Given

two

asteroidal

triple-free

graphs

and

and

pokable

dominating

pairs

-LRB-

-RRB-

and

-LRB-

-RRB-

in

and

respectively

let

G

be

the

graph

constructed

from

and

by

identifying

vertices

and

Then

G

is

an

asteroidal

triple-free

graph

The

reader

is

referred

to

Figure

for

an

illustration

of

the

Composition

Theorem

We

now

show

that

the

existence

of

a

pokable

dominating

pair

in

a

connected

AT

free

graph

leads

to

a

natural

decomposition

scheme

In

preparation

for

stating

the

second

main

result

of

this

section

we

first

give

a

necessary

and

sufficient

condition

for

a

vertex

in

a

dominating

pair

to

be

pokable

Specifically

we

have

the

following

result

Claim

Let

G

be

a

connected

asteroidal

triple-free

graph

with

a

dominating

pair

-LRB-

x

y

-RRB-

Then

x

is

pokable

if

and

only

if

there

are

no

unrelated

vertices

with

respect

to

x.

Figure

Illustrating

the

Composition

Theorem

Proof

The

if

part

is

easily

seen

To

prove

the

only

if

part

consider

unrelated

vertices

u

and

v

with

respect

to

x

In

particular

we

find

a

v

x-path

missed

by

u

and

a

u

x-path

missed

by

v

Since

-LRB-

x

y

-RRB-

is

a

dominating

pair

u

and

v

intercept

every

path

joining

x

and

y.

Let

ss

be

such

a

path

and

let

and

be

vertices

on

ss

adjacent

to

u

and

v

respectively

Trivially

both

and

are

distinct

from

x

But

now

there

exists

a

u

v-path

in

G

that

does

not

contain

x

-LRB-

this

path

contains

vertices

and

a

subpath

of

ss

-RRB-

implying

that

x

is

not

pokable

Let

G

-LRB-

V

E

-RRB-

be

a

connected

AT-free

graph

with

at

least

two

vertices

and

let

-LRB-

x

y

-RRB-

be

a

pokable

dominating

pair

in

G.

Define

a

binary

relation

R

on

G

by

writing

for

every

pair

u

v

of

vertices

u

R

v

-LRB-

-RRB-

D

-LRB-

u

x

-RRB-

D

-LRB-

v

x

-RRB-

-LRB-

-RRB-

Clearly

R

is

an

equivalence

relation

let

Ck

-LRB-

k

-RRB-

be

the

equivalence

classes

of

G

R

A

class

Ci

is

termed

non-trivial

if

j

Ci

j

The

existence

of

nontrivial

equivalence

classes

with

respect

to

R

is

not

immediately

obvious

The

following

result

guarantees

that

non-trivial

equivalence

classes

always

exist

Claim

G

R

contains

at

least

one

non-trivial

equivalence

class

Proof

If

N

-LRB-

x

-RRB-

is

empty

then

C

-LRB-

y

-RRB-

V

and

there

is

nothing

to

prove

Otherwise

Claim

and

the

proof

of

Lemma

combined

guarantee

that

x

and

y

are

nonadjacent

Let

F

be

the

connected

component

of

N

-LRB-

x

-RRB-

containing

y

and

let

Y

stand

for

the

subset

of

F

consisting

of

all

the

vertices

that

are

in

a

dominating

pair

with

x.

Clearly

y

Y

and

so

Y

is

nonempty

If

F

contains

at

least

two

vertices

then

-LRB-

-RRB-

guarantees

that

Y

itself

contains

at

least

two

vertices

and

so

the

equivalence

class

containing

y

is

non-trivial

We

may

therefore

assume

that

F

fyg

Let

be

an

arbitrary

neighbor

of

y

in

N

-LRB-

x

-RRB-

Clearly

D

-LRB-

x

-RRB-

V

for

otherwise

if

some

vertex

z

does

not

belong

to

D

-LRB-

x

-RRB-

then

z

must

miss

the

y

x-path

consisting

of

y

and

x.

Consequently

the

equivalence

class

containing

y

is

non-trivial

and

the

proof

of

Claim

is

complete

Remark

In

fact

the

proof

of

Claim

also

tells

us

that

the

class

C

-LRB-

y

-RRB-

containing

y

is

always

non-trivial

as

long

as

the

original

graph

has

at

least

two

vertices

A

non-trivial

class

C

of

G

R

is

said

to

be

valid

if

C

induces

a

connected

subgraph

of

G

As

before

the

existence

of

valid

equivalence

classes

is

not

immediately

obvious

As

we

shall

prove

next

such

classes

always

exist

Specifically

we

propose

to

show

that

C

-LRB-

y

-RRB-

is

valid

As

it

will

turn

out

all

valid

classes

of

G

R

enjoy

very

interesting

properties

that

will

allow

us

to

select

an

arbitrary

one

for

the

purpose

of

decomposing

the

original

graph

This

freedom

of

choice

opens

the

door

to

parallel

decomposition

algorithms

for

AT-free

graphs

Claim

G

R

contains

at

least

one

valid

equivalence

class

Proof

Let

-LRB-

x

y

-RRB-

be

a

pokable

dominating

pair

in

G

If

N

-LRB-

x

-RRB-

is

empty

then

C

-LRB-

y

-RRB-

V

and

there

is

nothing

to

prove

We

may

therefore

assume

that

N

-LRB-

x

-RRB-

is

nonempty

As

before

we

may

also

assume

that

y

belongs

to

N

-LRB-

x

-RRB-

Let

F

be

the

connected

component

of

N

-LRB-

x

-RRB-

containing

y

let

Y

stand

for

the

subset

of

F

consisting

of

all

the

vertices

that

are

in

a

dominating

pair

with

x

and

let

C

-LRB-

y

-RRB-

be

the

equivalence

class

containing

y.

Notice

that

every

vertex

w

that

belongs

to

N

-LRB-

x

-RRB-

and

to

C

-LRB-

y

-RRB-

must

be

adjacent

to

all

the

vertices

in

F

In

particular

if

such

a

vertex

exists

then

C

-LRB-

y

-RRB-

which

by

Claim

is

non-trivial

must

be

connected

and

thus

valid

We

will

assume

therefore

that

N

-LRB-

x

-RRB-

and

C

-LRB-

y

-RRB-

are

disjoint

In

turn

this

implies

that

C

-LRB-

y

-RRB-

Y

Recall

that

by

Claim

Y

induces

a

disconnected

component

of

G

confirming

that

C

-LRB-

y

-RRB-

is

connected

as

a

subgraph

of

G

The

conclusion

follows

Let

S

be

a

set

of

vertices

of

G

The

graph

is

said

to

arise

from

G

by

an

S

contraction

if

contains

all

the

vertices

in

GnS

along

with

a

new

vertex

s

adjacent

in

to

all

the

vertices

in

G

n

S

that

were

adjacent

in

G

to

some

vertex

in

S

Our

next

result

states

a

fundamental

property

of

valid

equivalence

classes

namely

that

contracting

any

of

them

will

result

in

an

AT-free

graph

The

details

are

spelled

out

as

follows

Lemma

Let

G

be

an

asteroidal

triple-free

graph

and

let

C

be

an

arbitrary

valid

equivalence

class

of

G

R

The

graph

obtained

from

G

by

a

C-contraction

is

asteroidal

triple-free

Proof

Let

c

be

the

vertex

in

obtained

by

contracting

C.

To

begin

we

claim

that

there

are

no

vertices

u

v

in

such

that

fu

v

cg

is

an

AT

-LRB-

-RRB-

To

justify

-LRB-

-RRB-

note

that

if

ss

-LRB-

u

v

-RRB-

is

a

u

v-path

missed

by

c

then

the

same

path

is

missed

in

G

by

all

the

vertices

in

C.

Let

ss

-LRB-

u

c

-RRB-

be

a

u

c-path

in

missed

by

v.

Then

there

exists

a

vertex

in

C

such

that

v

misses

the

path

ss

-LRB-

u

c

-RRB-

c

Similarly

let

ss

-LRB-

v

c

-RRB-

be

a

u

c-path

in

missed

by

u

There

must

exist

a

vertex

in

C

such

that

v

misses

the

path

ss

-LRB-

u

c

-RRB-

c

Since

C

induces

a

connected

subgraph

of

G

there

exists

a

path

joining

and

all

of

whose

internal

vertices

are

in

C.

By

a

previous

observation

both

u

and

v

miss

this

path

Therefore

for

a

suitably

chosen

vertex

in

C

fu

v

is

an

AT

in

G

a

contradiction

Thus

-LRB-

-RRB-

must

hold

To

complete

the

proof

of

Theorem

let

fu

v

wg

be

an

arbitrary

AT

in

By

-LRB-

-RRB-

c

is

distinct

from

u

v

w.

Let

ss

-LRB-

u

v

-RRB-

ss

-LRB-

u

w

-RRB-

and

ss

-LRB-

v

w

-RRB-

be

paths

in

confirming

that

fu

v

wg

is

an

AT

If

c

belongs

to

of

these

paths

then

fu

v

wg

is

an

AT

in

G

We

may

therefore

assume

without

loss

of

generality

that

c

belongs

to

ss

-LRB-

u

v

-RRB-

Since

w

misses

ss

-LRB-

u

v

-RRB-

it

is

clear

that

w

is

adjacent

to

no

vertex

in

C

We

claim

that

there

exists

a

path

-LRB-

u

v

-RRB-

in

G

missed

by

w

This

path

contains

the

same

vertices

as

ss

-LRB-

u

v

-RRB-

outside

of

C.

Inside

C

it

contains

a

path

between

two

vertices

and

of

C

such

that

ffl

w

misses

a

u

consisting

of

a

subpath

of

ss

-LRB-

u

v

-RRB-

ffl

w

misses

a

v-path

consisting

of

the

remaining

vertices

in

ss

-LRB-

u

v

-RRB-

c

This

completes

the

proof

of

Lemma

The

example

in

Figures

and

shows

that

the

connectivity

of

the

equivalence

class

C

in

Lemma

is

required

if

we

are

to

guarantee

that

the

resulting

graph

is

AT-free

To

wit

the

graph

G

featured

in

Figure

is

AT-free

with

a

pokable

dominating

pair

-LRB-

x

e

-RRB-

The

contraction

of

the

equivalence

class

fa

bg

yields

the

graph

in

Figure

which

has

the

AT

For

the

reader

's

benefit

the

various

values

of

the

D

-LRB-

x

-RRB-

sets

along

with

the

equivalence

classes

corresponding

to

the

graph

in

Figure

are

summarized

in

Table

below

Let

C

-LRB-

y

-RRB-

be

the

equivalence

class

containing

y.

Let

be

the

graph

obtained

from

G

by

a

C

-LRB-

y

-RRB-

contraction

Recall

that

the

proof

of

Claim

guarantees

that

C

-LRB-

y

-RRB-

is

valid

and

so

Lemma

asserts

that

the

graph

is

also

AT-free

Let

be

the

vertex

of

obtained

by

contracting

C

-LRB-

y

-RRB-

We

now

show

that

in

fact

more

can

be

said

about

Specifically

we

have

the

following

result

Lemma

-LRB-

x

-RRB-

is

a

pokable

dominating

pair

in

Figure

A

graph

G

Equivalence

Class

D

-LRB-

x

-RRB-

x

fx

V

n

fc

d

eg

V

nfc

dg

a

b

V

nfdg

V

nfeg

w

c

d

e

V

Table

Illustrating

the

various

equivalence

classes

Proof

To

begin

we

establish

that

-LRB-

x

-RRB-

is

a

dominating

pair

in

For

this

purpose

suppose

that

there

exists

some

path

ss

-LRB-

x

-RRB-

joining

x

and

missed

by

a

vertex

w.

Clearly

w

is

adjacent

in

G

to

no

vertex

in

C

-LRB-

y

-RRB-

In

particular

w

is

not

adjacent

to

y

Since

C

-LRB-

y

-RRB-

is

valid

w

misses

in

G

a

y

x-path

consisting

of

all

the

vertices

in

ss

-LRB-

x

-RRB-

along

with

a

suitable

path

in

C

-LRB-

y

-RRB-

Therefore

-LRB-

x

-RRB-

must

be

a

dominating

pair

in

Next

we

show

what

both

x

and

are

pokable

vertices

of

Suppose

that

x

is

not

pokable

Now

Claim

guarantees

the

existence

of

unrelated

vertices

u

and

v

-LRB-

with

respect

to

x

-RRB-

This

in

turn

implies

the

existence

of

paths

ss

-LRB-

v

x

-RRB-

and

ss

-LRB-

u

x

-RRB-

in

missed

by

u

and

v

respectively

Since

-LRB-

x

-RRB-

is

a

dominating

pair

in

belongs

to

neither

of

these

paths

But

now

these

paths

must

have

been

paths

in

G

a

contradiction

Finally

suppose

that

is

not

pokable

By

virtue

of

Claim

this

implies

the

existence

of

vertices

u

and

v

and

of

paths

ss

-LRB-

v

-RRB-

and

ss

-LRB-

u

-RRB-

in

missed

by

u

and

fi

fl

Figure

The

graph

obtained

by

contracting

fa

bg

v

respectively

In

particular

neither

u

nor

v

is

adjacent

to

In

turn

this

implies

that

neither

u

nor

v

is

adjacent

to

a

vertex

in

C

-LRB-

y

-RRB-

But

now

in

G

there

exists

a

u

y-path

missed

by

v

and

a

v

y-path

missed

by

u

contradicting

that

y

is

pokable

This

completes

the

proof

of

Lemma

At

this

stage

the

reader

may

wonder

whether

the

class

C

-LRB-

y

-RRB-

is

the

only

one

whose

contraction

leaves

x

pokable

The

answer

is

provided

by

the

following

result

that

complements

Lemma

Lemma

Let

C

be

an

arbitrary

valid

equivalence

class

in

an

asteroidal

triplefree

graph

G

and

let

the

graph

obtained

from

G

by

a

C-contraction

If

C

is

distinct

from

C

-LRB-

x

-RRB-

and

C

-LRB-

y

-RRB-

then

-LRB-

x

y

-RRB-

is

a

pokable

dominating

pair

in

Proof

Let

c

be

the

vertex

of

obtained

by

the

C-contraction

By

assumption

c

is

distinct

from

y

and

from

x

We

begin

by

showing

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

in

Suppose

that

there

exists

some

path

ss

-LRB-

x

y

-RRB-

joining

x

and

y

in

missed

by

a

vertex

w.

Clearly

c

must

belong

to

ss

-LRB-

x

y

-RRB-

Notice

that

w

is

adjacent

in

G

to

no

vertex

in

C

Since

C

is

valid

w

misses

in

G

a

y

x-path

consisting

of

all

the

vertices

in

ss

-LRB-

x

y

-RRB-

c

along

with

a

suitable

path

in

C.

Thus

-LRB-

x

y

-RRB-

must

be

a

dominating

pair

in

Next

we

show

what

both

x

and

y

are

pokable

vertices

of

If

x

is

not

pokable

Claim

guarantees

the

existence

of

vertices

u

and

v

unrelated

with

respect

to

x

In

turn

this

implies

the

existence

of

paths

ss

-LRB-

v

x

-RRB-

and

ss

-LRB-

u

x

-RRB-

in

missed

by

u

and

v

respectively

Since

x

is

pokable

in

G

c

must

belong

to

-LRB-

at

least

-RRB-

one

of

these

paths

Symmetry

allows

us

to

assume

with

no

loss

of

generality

that

c

belongs

to

ss

-LRB-

u

x

-RRB-

The

fact

that

v

misses

ss

-LRB-

u

x

-RRB-

guarantees

that

v

is

adjacent

in

G

to

no

vertex

in

C

But

now

we

have

reached

a

contradiction

v

misses

a

u

x-path

in

G

consisting

of

all

the

vertices

of

ss

-LRB-

u

x

-RRB-

outside

C

along

with

a

suitably

chosen

path

in

C.

Thus

x

must

be

pokable

in

A

perfectly

similar

argument

whose

details

are

omitted

asserts

that

y

is

also

pokable

With

this

the

proof

of

Lemma

is

complete

Lemmas

and

combined

set

the

stage

for

a

decomposition

theorem

for

AT

free

graphs

Consider

an

AT-free

graph

G

-LRB-

V

E

-RRB-

and

let

-LRB-

x

-RRB-

be

a

pokable

dominating

pair

in

G.

Let

Gk

be

a

sequence

of

graphs

defined

as

follows

i

-RRB-

G

ii

-RRB-

For

all

i

-LRB-

i

k

-RRB-

let

Ri

be

the

equivalence

relation

defined

on

Gi

by

setting

uRiv

-LRB-

-RRB-

D

-LRB-

u

x

-RRB-

D

-LRB-

v

x

-RRB-

and

let

C

be

an

arbitrary

valid

equivalence

class

of

Gi

Ri

Let

Gi

be

the

graph

obtained

from

Gi

by

a

C-contraction

-LRB-

i.e.

Gi

contains

all

the

vertices

in

GinC

as

well

as

a

new

vertex

c

which

is

adjacent

to

all

vertices

in

GinC

that

were

adjacent

to

at

least

one

vertex

in

C

-RRB-

iii

-RRB-

Gk

consists

of

a

single

vertex

Such

a

sequence

Gk

is

called

involutive

The

reader

is

referred

to

Figure

featuring

the

first

five

graphs

in

an

involutive

sequence

of

the

given

graph

Note

that

in

the

transition

from

to

in

Figure

two

equivalence

classes

could

be

contracted

namely

fa

bg

and

fd

efghg

We

have

selected

to

contract

the

class

C

fa

bg

The

obvious

question

is

whether

every

connected

AT-free

graph

has

such

an

involutive

sequence

The

answer

to

this

fundamental

question

is

provided

by

the

following

result

Theorem

Every

connected

asteroidal

triple-free

graph

G

has

an

involutive

sequence

Proof

If

G

is

a

clique

then

there

is

nothing

to

prove

We

shall

therefore

assume

that

G

is

not

a

clique

By

Claim

and

the

proof

of

Lemma

we

find

a

nonadjacent

pokable

dominating

pair

-LRB-

x

-RRB-

in

G.

Consider

the

transition

from

Gi

to

Gi

for

some

i

-LRB-

i

k

-RRB-

Let

C

be

an

arbitrary

valid

equivalence

class

in

Gi

Ri

and

let

-LRB-

x

yi

-RRB-

be

a

pokable

dominating

pair

in

Gi

Define

yi

to

be

yi

in

case

C

is

distinct

from

C

-LRB-

yi

-RRB-

and

to

be

the

vertex

obtained

by

contracting

C

-LRB-

yi

-RRB-

otherwise

Clearly

Gi

is

connected

whenever

Gi

is

By

Lemmas

and

combined

Gi

is

AT-free

and

-LRB-

x

yi

-RRB-

is

a

pokable

dominating

pair

in

Gi

This

completes

the

proof

of

Theorem

We

close

with

the

obvious

question

can

such

an

involutive

sequence

be

constructed

efficiently

The

answer

to

this

question

seems

to

rely

on

an

efficient

algorithm

for

computing

the

D

sets

ffi

ffl

ffl

ffi

flffiffl

flffiffl

fiflffiffl

fffiflffiffl

Figure

Illustration

of

an

involutive

sequence

On

the

Linear

Structure

of

AT-free

Graphs

The

reader

may

recall

that

the

original

motivation

for

this

work

was

the

intuition

about

a

concept

that

we

vaguely

described

as

the

linear

structure

of

AT-free

graphs

The

purpose

of

this

section

is

to

make

this

a

little

more

precise

by

showing

that

in

the

case

of

AT-free

graph

of

a

diameter

larger

than

the

set

of

vertices

that

can

be

in

dominating

pairs

are

restricted

to

two

disjoint

sets

Specifically

we

have

the

following

result

Theorem

Let

G

be

a

connected

asteroidal

triple-free

graph

with

diameter

at

least

There

exist

nonempty

disjoint

sets

X

and

Y

of

vertices

of

G

such

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

if

and

only

if

x

X

and

y

Y

We

note

that

Theorem

is

best

possible

in

the

sense

that

for

AT-free

graphs

of

diameter

less

than

the

sets

X

and

Y

are

not

guaranteed

to

exist

To

wit

and

the

graph

below

provide

counterexamples

of

diameter

and

respectively

s

s

s

sss

Proof

of

Theorem

Let

-LRB-

-RRB-

be

a

dominating

pair

in

G

achieving

the

diameter

Let

Y

stand

for

the

set

of

all

the

vertices

y

in

G

such

that

-LRB-

y

-RRB-

is

a

dominating

pair

and

let

X

be

the

set

of

all

the

vertices

x

in

G

for

which

-LRB-

x

-RRB-

is

a

dominating

pair

We

propose

to

show

that

X

and

Y

are

the

sets

with

the

property

specified

in

Theorem

Our

proof

relies

on

a

number

of

intermediate

results

that

we

present

next

To

begin

we

note

that

X

and

Y

-LRB-

-RRB-

In

addition

by

Claim

both

X

and

Y

are

disconnected

in

G

-LRB-

-RRB-

Our

choice

of

and

guarantees

that

-LRB-

resp

-RRB-

are

adjacent

to

no

vertices

in

Y

-LRB-

resp

X

-RRB-

-LRB-

-RRB-

-LRB-

Otherwise

-LRB-

-RRB-

and

-LRB-

-RRB-

would

imply

that

d

-LRB-

-RRB-

-RRB-

Note

that

-LRB-

-RRB-

and

-LRB-

-RRB-

combined

guarantee

that

X

and

Y

are

disjoint

-LRB-

-RRB-

-LRB-

For

if

z

X

Y

then

in

particular

z

X

and

so

-LRB-

z

-RRB-

is

a

dominating

pair

By

-LRB-

-RRB-

there

exists

a

z

contained

in

Y

By

-LRB-

-RRB-

misses

this

path

-RRB-

Let

x

and

y

be

arbitrary

vertices

in

X

and

Y

respectively

We

claim

that

-LRB-

x

y

-RRB-

is

a

dominating

pair

-LRB-

-RRB-

-LRB-

To

justify

-LRB-

-RRB-

suppose

that

some

vertex

u

misses

an

x

y-path

ss

Observe

that

-LRB-

-RRB-

guarantees

the

existence

of

an

y-path

contained

in

ss

-LSB-

X

Since

-LRB-

y

-RRB-

is

a

dominating

pair

this

path

is

dominating

By

-LRB-

-RRB-

must

be

adjacent

to

a

vertex

of

ss

n

fxg

Thus

ss

-LSB-

contains

an

x

This

path

must

be

dominating

and

so

u

must

be

adjacent

to

A

perfectly

similar

argument

shows

that

u

is

adjacent

to

contradicting

that

and

achieve

the

diameter

-RRB-

Next

let

x

be

an

arbitrary

vertex

in

X

We

claim

that

if

-LRB-

x

z

-RRB-

is

a

dominating

pair

then

z

Y

-LRB-

-RRB-

-LRB-

Trivially

z

X

since

diam

-LRB-

G

-RRB-

x

and

z

are

not

adjacent

If

z

Y

there

exists

an

z-path

ss

missed

by

some

vertex

u.

Note

that

ss

-LSB-

X

contains

an

x

z-path

Since

by

assumption

-LRB-

x

z

-RRB-

is

a

dominating

pair

this

path

is

dominating

and

so

must

intercept

it

By

-LRB-

-RRB-

intercepts

ss

Since

-LRB-

-RRB-

is

a

dominating

pair

it

follows

that

u

is

adjacent

to

Trivially

u

is

not

adjacent

to

x

otherwise

the

path

u

x

which

is

dominating

implies

that

x

and

are

adjacent

and

so

d

-LRB-

-RRB-

Further

u

and

x

being

non-adjacent

guarantees

that

x

and

are

also

non-adjacent

else

u

misses

the

x

z-path

contained

in

ss

-LSB-

fxg

Now

-LRB-

-RRB-

guarantees

that

some

in

X

is

adjacent

to

both

and

x

Since

-LRB-

x

z

-RRB-

is

a

dominating

pair

u

must

be

adjacent

to

However

this

implies

that

d

-LRB-

-RRB-

a

contradiction

-RRB-

Let

y

be

an

arbitrary

vertex

in

Y

As

above

we

can

prove

that

if

-LRB-

y

z

-RRB-

is

a

dominating

pair

then

z

X

-LRB-

-RRB-

Finally

we

claim

that

if

-LRB-

v

w

-RRB-

is

a

dominating

pair

then

v

X

and

w

Y

-LRB-

-RRB-

Note

that

by

virtue

of

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

and

-LRB-

-RRB-

to

complete

the

proof

of

Theorem

we

only

need

to

establish

-LRB-

-RRB-

The

remainder

of

this

proof

is

devoted

to

establishing

-LRB-

-RRB-

By

-LRB-

-RRB-

-LRB-

-RRB-

and

-LRB-

-RRB-

the

only

way

-LRB-

-RRB-

can

fail

is

to

have

v

X

-LSB-

Y

and

w

X

-LSB-

Y

Let

F

be

the

component

of

N

-LRB-

-RRB-

that

contains

Y

-LRB-

Observe

that

diam

-LRB-

G

-RRB-

guarantees

that

Y

is

restricted

to

a

unique

component

of

N

-LRB-

-RRB-

-RRB-

We

claim

that

v

or

w

belongs

to

F

-LRB-

-RRB-

-LRB-

Consider

a

shortest

v

w-path

in

G.

By

assumption

this

path

is

dominating

and

so

both

and

must

intercept

it

Assume

without

loss

of

generality

that

intercepts

the

path

closer

to

w

than

at

a

vertex

t.

Trivially

is

adjacent

to

no

vertex

on

this

path

from

t

to

w

and

the

conclusion

follows

-RRB-

Let

H

be

the

component

of

N

-LRB-

-RRB-

that

contains

X.

By

virtue

of

-LRB-

-RRB-

we

may

assume

without

loss

of

generality

that

w

F

and

that

v

H

We

claim

that

w

is

an

attractor

in

F

-LRB-

-RRB-

-LRB-

To

begin

observe

that

can

miss

no

w

as

by

-LRB-

-RRB-

such

a

path

extends

to

a

w

v-path

missed

by

Similarly

no

vertex

y

Y

non-adjacent

to

can

miss

an

w

otherwise

y

would

miss

an

a

contradiction

Let

y

Y

be

a

vertex

that

misses

some

w

ss

By

the

previous

argument

y

and

are

adjacent

However

since

-LRB-

w

v

-RRB-

is

a

dominating

pair

y

must

intercept

any

w

v-path

contained

in

ss

-LSB-

H

implying

that

y

is

adjacent

to

some

neighbor

of

x

But

now

we

have

reached

a

contradiction

and

are

joined

by

a

path

of

length

-RRB-

With

this

the

proof

of

Theorem

is

complete

Concluding

Remarks

and

Open

Problems

Many

families

of

graphs

including

interval

graphs

permutation

graphs

trapezoid

graphs

and

cocomparability

graphs

demonstrate

a

type

of

linear

ordering

on

their

vertex

sets

It

is

precisely

this

linear

order

that

is

exploited

in

one

form

or

another

in

a

search

for

efficient

algorithms

for

these

classes

of

graphs

The

classes

mentioned

are

known

to

have

wide-ranging

practical

applications

In

addition

they

are

all

subfamilies

of

the

class

of

graphs

called

asteroidal

triple-free

graphs

-LRB-

AT-free

for

short

-RRB-

This

work

is

the

first

known

to

us

at

investigating

structural

properties

of

the

AT-free

graphs

In

this

direction

our

contributions

are

as

follows

We

showed

that

every

connected

AT-free

graph

has

a

dominating

pair

that

is

a

pair

of

vertices

such

that

every

path

joining

them

is

a

dominating

set

We

provided

properties

of

dominating

pairs

in

AT-free

graphs

related

to

the

concept

of

connected

domination

and

diameter

We

provided

a

characterization

of

AT-free

graphs

in

terms

of

dominating

pairs

We

provided

a

characterization

of

AT-free

graphs

in

terms

of

minimal

triangulations

preliminary

version

of

this

work

has

appeared

in

-LSB-

-RSB-

We

provided

a

decomposition

theorem

for

AT-free

graphs

Many

other

questions

are

still

open

For

example

it

is

well

known

-LSB-

-RSB-

that

cocomparability

graphs

have

a

linear

ordering

this

ordering

exemplifies

the

linear

structure

we

observe

in

interval

graphs

permutation

graphs

and

trapezoid

graphs

It

would

be

interesting

to

see

whether

the

AT-free

graphs

also

possess

some

linear

ordering

Such

an

ordering

could

conceivably

be

exploited

for

algorithmic

purposes

A

further

natural

question

to

ask

is

What

exactly

are

the

roles

of

dominating

pairs

and

pokable

vertices

in

the

subfamilies

of

AT-free

graphs

It

is

clear

that

the

extreme

vertices

of

any

intersection

representation

for

a

connected

graph

in

any

of

the

subfamilies

form

a

dominating

pair

Some

additional

partial

answers

to

this

question

have

been

given

in

a

slightly

different

setting

in

-LSB-

-RSB-

and

-LSB-

-RSB-

Investigating

further

properties

of

dominating

pairs

and

pokability

in

each

of

these

particular

families

promises

to

be

a

fruitful

area

for

further

research

Recently

M?ohring

-LSB-

-RSB-

has

added

to

the

understanding

of

the

linear

structure

of

AT-free

graphs

by

showing

that

the

pathwidth

of

an

AT-free

graph

equals

its

treewidth

Just

as

there

are

many

families

of

perfect

AT-free

graphs

one

would

expect

to

see

a

rich

hierarchy

of

families

of

non-perfect

AT-free

graphs

So

far

nothing

is

known

here

Since

perfect

AT-free

graphs

strictly

contain

cocomparability

graphs

it

would

be

interesting

to

study

the

perfect

AT-free

graphs

The

fastest

recognition

algorithm

known

to

us

runs

in

O

-LRB-

-RRB-

with

an

n-vertex

graph

as

input

Is

is

a

tantalizing

open

problem

to

produce

a

recognition

algorithm

that

is

more

efficient

perhaps

even

optimal

Acknowledgement

The

authors

are

indebted

to

two

anonymous

referees

for

an

exceptionally

thorough

reviewing

job

Their

constructive

comments

have

greatly

contributed

to

improving

the

presentation

D.G.

Corneil

and

L.

Stewart

wish

to

thank

the

Natural

Sciences

and

Engineering

Research

Council

of

Canada

for

financial

assistance

S.

Olariu

was

supported

in

part

by

the

National

Science

Foundation

under

grant

Last

but

not

least

the

authors

wish

to

thank

F.

Maffray

and

M.

Preissmann

of

Universit?e

de

Grenoble

for

providing

us

with

an

English

translation

of

Gallai

's

paper

Note

Very

recently

we

have

learned

that

Andreas

Parra

-LSB-

-RSB-

has

independently

obtained

a

result

similar

to

our

Theorem

However

we

have

not

seen

Parra

's

proof

References

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-RSB-

K.

A.

Baker

P.

C.

Fishburn

and

F.

S.

Roberts

Partial

orders

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dimension

two

Networks

-LRB-

-RRB-

-LCB-

-LSB-

-RSB-

J.

A.

Bondy

U.

S.

R.

Murty

Graph

Theory

with

Applications

North-Holland

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K.

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and

G.

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Lueker

Testing

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ones

property

interval

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-RRB-

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-LSB-

-RSB-

K.

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A

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Cheah

A

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University

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available

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A.

Kamula

Extensions

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interval

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Proc

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Graph

Theory

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Computing

Boca

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February

-LCB-

-LSB-

-RSB-

D.

G.

Corneil

S.

Olariu

L.

Stewart

Asteroidal

triple-free

graphs

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A.

Parra

private

communication

asteroidal triples
cocomparability graphs
dominating pairs
interval graphs
permutation graphs
trapezoid graphs

asteroidal triple-free graphs
graph decompositions
structural graph theory