proof of Vizing’s theorem (for graphs)

proof of Vizing’s theorem (for graphs)

Proof of Vizing’s theorem (for graphs)

We must prove, for any integer

ρ

, the following: if

G

is a graph all of
whose vertices have

valency

⩽

ρ

, then its edges can be colored (with

adjacent

edges receiving different colors) in no more than

ρ

+

1

colors.

This form of stating the

theorem

allows us to do

induction

on the
number of edges. A graph with zero edges, for

instance

, certainly doesn’t
need more than

ρ

+

1

colors. Now assume (for any given

ρ

) that
all graphs with

m

edges can be thus colored. For any graph

G

with

m

+

1

edges we choose an edge to remove, color the remaining

m

edges, and
try putting the edge back.

Another way of looking at this: suppose there was (for a certain

ρ

) a
graph

G

*

with

m

*

edges that could not be colored thus. There would then
be a largest number

m

,

bounded

by

0

⩽

m

<

m

*

, such that all graphs with

m

edges can still be colored thus. Then there would also be a

G

with

m

+

1

edges that cannot. Choose one of its edges and proceed as above. This time,
succeeding in

coloring

it after all would prove by

contradiction

that there
was no such

m

and hence no such

G

*

.

Either way we have our

G

where after picking an edge we can color everything
except that edge, and fitting it in too proves the theorem. Let the edge to
fit run between

v

and

w

1

. Some terms:

•

With a palette of

ρ

+

1

edge colors, and no more than

ρ

edges
at any vertex, each vertex has at least one

missing color

.

•

Let a

Kempe chain

H

⁢

(

a

,

b

)

be any

connected component

of the

subgraph

formed by all edges colored

a

or

b

. This is either a
cycle of even length, or an

open path

terminating at two vertices
where one of

a

or

b

is missing.

Step 1

. If there is a color missing both at

v

and at

w

1

, use
that color for the edge

v

w

1

and we’re finished.

Else, let color

a

be missing at

v

(but present at

w

1

), and color

b

1

be missing at

w

1

(but present at

v

). If the Kempe chain of
colors

a

and

b

1

that terminates at

v

is

disjoint

from the one
terminating at

w

1

, swap the colors in the latter chain, then use

a

for edge

v

w

1

and be done.

Else it’s the same

H

⁢

(

a

,

b

1

)

chain, in which case we proceed as follows.

Find the

b

1

-colored edge from

v

, call it

v

w

2

. Steal color

b

1

from it and give it to edge

v

w

1

, now

v

w

2

is the problem edge that needs a color.

Go back to Step 1 calling it

Step 2

, replacing

subscripts

1

by

2

in all text. This involves a color

b

2

that was missing at

w

2

(

a

remains the missing color at

v

). If

v

and

w

2

are awkward too,
being in the same

H

⁢

(

a

,

b

2

)

chain

find the

b

2

-colored edge

v

w

3

, give its color to

v

w

2

making

v

w

3

the problem:

Next time round (

Step 3

) increase the subscripts again.
And if still necessary, transfer the color again.

Sooner or later (if we don’t get to bail out at one of the Steps) we will run
out of colors, in the following sense: the color missing at some

w

j

+

1

is not some new color

b

j

+

1

but a color

b

i

we’ve seen before (it can’t
be

a

, that’s present at

w

j

+

1

). And not only is

i

<

j

+

1

but also

i

<

j

because

b

j

is the color we just stole from

w

j

+

1

.

A chain

H

⁢

(

a

,

b

i

)

runs from

v

to

w

i

+

1

and all its vertices
except

w

i

+

1

have color

b

i

. Now

w

j

+

1

isn’t

w

i

+

1

(because

i

<

j

) nor one of the other vertices of the chain (because
it doesn’t have color

b

i

). So

w

j

+

1

isn’t on this

H

⁢

(

a

,

b

i

)

.

w

k

+

1

does have color

a

so it’s on some other

H

′

⁢

(

a

,

b

i

)

disjoint
from

H

⁢

(

a

,

b

i

)

. This could be as short as a single

a

-colored edge or
longer, it doesn’t matter.

Swap the colors in that

H

′

⁢

(

a

,

b

i

)

. Now use

a

to finally color

v

w

j

+

1

(merging

H

⁢

(

a

,

b

i

)

and

H

′

⁢

(

a

,

b

i

)

in the process).

This is in essence the proof usually given, as taught to students and found for instance in

references

[FW77] and [Wil02] of the

http://planetmath.org/node/6930

parent entry.

Title

proof of Vizing’s theorem (for graphs)

Canonical name

ProofOfVizingsTheoremforGraphs

Date of creation

2013-03-22 15:10:42

Last modified on

2013-03-22 15:10:42

Owner

marijke (8873)

Last modified by

marijke (8873)

Numerical id

6

Author

marijke (8873)

Entry type

Proof

Classification

msc 05C15

Related topic

KempeChain

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