proof of chromatic number and girth

proof of chromatic number and girth

Let

α

⁢

(

G

)

denote the size of the largest

independent set

in

G

, and

χ

⁢

(

G

)

the

chromatic number

of

G

. We want to show
that there is a graph

G

with girth larger than

ℓ

and

χ

⁢

(

G

)

>

k

, for any

ℓ

,

k

>

0

.

We first prove the following claim.

Claim:

Given a

positive integer

ℓ

and a positive real
number

t

<

1

/

ℓ

, for all sufficiently large

n

, there is a
graph

G

on

n

vertices satisfying properties

1.

the number of cycles of length at most

ℓ

is less than

n

/

2

,

2.

α

⁢

(

G

)

<

3

⁢

n

1

-

t

⁢

log

⁡

n

.

Proof of claim:

Let

G

be a

random graph

on

n

vertices,
in which each pair of vertices joint by an edge independently with

probability

p

=

n

t

-

1

. Let

X

be the number of cycles of length
at most

ℓ

in

G

. The

expected value

of

X

is

E

⁢

[

X

]

=

∑

i

=

3

ℓ

n

⁢

(

n

-

1

)

⁢

⋯

⁢

(

n

-

i

+

1

)

2

⁢

i

⁢

p

i

<

∑

i

=

3

ℓ

(

n

⁢

p

)

i

2

⁢

i

<

∑

i

=

3

ℓ

n

t

⁢

i

<

ℓ

⁢

n

t

⁢

ℓ

By

Markov inequality

,

Pr

⁡

(

X

≥

n

/

2

)

<

2

⁢

ℓ

⁢

n

t

⁢

ℓ

-

1

,

we have

Pr

⁡

(

X

≥

n

/

2

)

→

0

as

n

→

∞

, since

t

⁢

ℓ

<

1

.

On the other hand, let

y

=

⌈

(

3

⁢

log

⁡

n

)

/

p

⌉

, and

Y

be
the number of independent sets of size

y

in

G

. By Markov
inequality again,

Pr

⁡

(

α

⁢

(

G

)

≥

y

)

=

Pr

⁡

(

Y

≥

1

)

≤

E

⁢

[

Y

]

.

However,

E

⁢

[

Y

]

=

(

n

y

)

⁢

(

1

-

p

)

y

⁢

(

y

-

1

)

/

2

Using the inequalities,

(

n

y

)

<

n

y

and

(

1

-

p

)

≤

e

-

p

, we get

Pr

⁡

(

α

⁢

(

G

)

≥

y

)

<

(

n

⁢

e

-

p

⁢

(

y

-

1

)

/

2

)

y

Our choice of

y

guarantees that

n

⁢

e

-

p

⁢

(

y

-

1

)

/

2

<

β

<

1

for some

β

, and

y

→

∞

as

n

approaches
infinity. Therefore,

Pr

⁡

(

α

⁢

(

G

)

≥

y

)

→

0

,

as

n

→

∞

.

We can thus find

n

0

such that for all

n

>

n

0

, both

Pr

⁡

(

X

≥

n

/

2

)

and

Pr

⁡

(

α

⁢

(

G

)

≥

y

)

are strictly less than

1

/

2

. For all

n

>

n

0

,

Pr

⁡

(

X

<

ℓ

⁢

and

⁢

α

⁢

(

G

)

<

y

)

>

1

-

Pr

⁡

(

X

≥

ℓ

)

-

Pr

⁡

(

α

⁢

(

G

)

≥

y

)

>

1

.

Therefore there exists a graph that satisfies the two properties
in the claim. This ends the proof of the claim.

Let

G

be a graph that satisfies the two properties in the
claim. Remove a vertex from each cycle of length at most

ℓ

in

G

. The resulting graph

G

′

has girth larger than

ℓ

, more
than

n

/

2

vertices, and

α

⁢

(

G

′

)

≤

α

⁢

(

G

)

. Since

http://planetmath.org/node/6037

χ

⁢

(

G

′

)

⁢

α

⁢

(

G

′

)

≥

|

G

′

|

, we have

χ

⁢

(

G

′

)

≥

n

/

2

3

⁢

n

1

-

t

⁢

log

⁡

n

=

n

t

6

⁢

log

⁡

n

We can pick sufficiently large

n

such that

χ

⁢

(

G

′

)

is larger
than

k

. Then the chromatic number of

G

′

is larger than

k

and
girth is larger than

ℓ

.

Reference

: N. Alon and J. Spencer,

The

probabilistic method

, 2nd, John Wiley.

Title

proof of chromatic number and girth

Canonical name

ProofOfChromaticNumberAndGirth

Date of creation

2013-03-22 14:31:02

Last modified on

2013-03-22 14:31:02

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

9

Author

kshum (5987)

Entry type

Proof

Classification

msc 05C38

Classification

msc 05C15

Classification

msc 05C80

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