Vertex colouring and chromatic
polynomials
for Master Course MSM380
Dong Fengming
National Institute of Education
Nanyang Technological University
Contents
1 Vertex Colourings
1
1.1
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Graphs G with χ(G) = v(G) − 1 . . . . . . . . . . . . . . . . . . .
4
1.3
Graphs G with χ(G) = v(G) − 2 (Optional) . . . . . . . . . . . .
5
1.4
Brooks Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.5
Extension of Brooks Theorem . . . . . . . . . . . . . . . . . . . .
13
1.6
Critical graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
1.7
Cliques and chromatic numbers . . . . . . . . . . . . . . . . . . .
17
1.8
Perfect graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
Exercises of Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
2 Planar Graphs
25
2.1
Four-Colour Problem . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.2
Plane Graphs and Euler’s Theorem . . . . . . . . . . . . . . . . .
28
2.2.1
Euler’s formula . . . . . . . . . . . . . . . . . . . . . . . .
29
2.2.2
Maximal plane graphs . . . . . . . . . . . . . . . . . . . .
33
Planar Graphs and Kuratowski’s Theorem . . . . . . . . . . . . .
34
2.3.1
Subdivisions of K5 and K(3, 3) . . . . . . . . . . . . . . .
35
2.3.2
Kuratowski’s Theorem . . . . . . . . . . . . . . . . . . . .
37
Four-colour problem in terms of graphs . . . . . . . . . . . . . . .
46
2.3
2.4
II
CONTENTS
III
2.4.1
Dual of a plane multigraph . . . . . . . . . . . . . . . . . .
46
2.4.2
From face colouring to vertex colouring . . . . . . . . . . .
47
2.4.3
Proof of Five-colour Theorem . . . . . . . . . . . . . . . .
48
Exercises of Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
3 Chromatic polynomials of Graphs
54
3.1
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
3.2
The Deletion-Contraction Formula . . . . . . . . . . . . . . . . .
58
3.3
Basic properties of chromatic polynomials . . . . . . . . . . . . .
63
3.4
The coefficients of Chromatic polynomials . . . . . . . . . . . . .
68
3.5
Connectivity and chromatic polynomials . . . . . . . . . . . . . .
73
Exercises of Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
Chapter 1
Vertex Colourings
1.1
Definition
Example 1.1.1
Given three colours (e.g. red, yellow and green), assign these
colours to the vertices of the graph in Figure 1.1, one colour per vertex, so that
adjacent vertices are assigned different colours.
v
.....
...............
...... ... .......
...... .... ............
.
.
.
.
.
.
.......
.....
.......
.......
.......
.......
...
......
.......
...
.......
.
.
.
.
.
.
..........................................................................................................................
.........
...
.
.
.
.
.
.
.........
.
.
...
.
.
.
.........
.
.
.
.
.
.........
.....
......... .................
....
v
v
v
v
Figure 1.1
Solution. One way of assigning three colours red (R), yellow (Y) and green (G)
to the vertices of the graph is shown in Figure 1.2, one colour per vertex, so that
adjacent vertices are assigned different colours:
Rv
................
...... .. .......
....... ..... .............
.
.
.
.
.
.
.......
.......
.....
.......
.......
.......
...
......
.......
...
......
.
.
.
.
.
.
.
............................................................................................................................
.........
....
.
.
.
.
.
.........
.
...
.
.
.
.
.
.........
.
.
.
.........
.......
......... ................
.....
Y v
vG
v
R
Figure 1.2
1
vY
2
k-vertex-colouring
Definition 1.1.1
Let G be a graph and k be any positive integer. A k-vertex-
colouring of G is a way of assigning k colours c1 , c2 , · · · , ck to the vertices of G, one
colour per vertex, so that adjacent vertices are assigned different colours.
If G has a k-vertex-colouring, then G is said to be k-vertex-colourable.
Note. If there is no danger of confusion, ‘k-vertex-colouring’ is simply called
‘k-colouring’ and ‘k-vertex-colourable’ is simply called ‘k-colourable’.
Question 1.1.1
(i) What graphs are 1-colourable?
(ii) Find some graphs which are 2-colourable?
(iii) Show that any path Pn , where n ≥ 2, is 2-colourable but not 1-colourable.
Question 1.1.2
Show that any even cycle Cn , where n is even, is 2-colourable.
Question 1.1.3
(i) Show that C5 is not 2-colourable but 3-colourable.
(ii) Show that odd cycle Ck , where k ≥ 3 is odd, is not 2-colourable but 3colourable.
On 2-colourable graphs, we have the following important result:
For any graph G, the following statements are equivalent:
(i) G is a bipartite graph;
(ii) G has no odd cycles; and
(iii) G is 2-colourable.
Question 1.1.4
Is the Petersen graph 3-colourable?
Question 1.1.5
For each of the following statements, judge whether it is true
or false:
(i) If G is k-colourable, then G is also (k − 1)-colourable.
(ii) If G is k-colourable, then Kk+1 is not a subgraph of G.
(iii) If Kk+1 is not a subgraph of G, then G is k-colourable.
Chromatic number
Definition 1.1.2
The minimum number k for which G is k-colourable is called the chromatic number of G and is denoted by χ(G).
Example 1.1.2
(i) χ(Nn ) = 1, where Nn is the empty graph on n vertices.
(ii) χ(G) ≥ 2 for every graph G which is not empty.
(iii) χ(Pn ) = 2 for every path Pn on n vertices, where n ≥ 2.
(vi) χ(T ) = 2 for every tree T with at least two vertices.
Theorem 1.1.1
Let G be a nonempty graph. Then χ(G) = 2 if and only if G is bipartite.
i.e., χ(G) ≥ 3 if and only if G contains an odd cycle.
Theorem 1.1.2
(i) If H is a subgraph of G, then χ(H) ≤ χ(G).
(ii) χ(Kn ) = n for all n ≥ 1.
(iii) If G contains Kn as a subgraph, then χ(G) ≥ n.
(vi) If G has G1 , G2 , · · · , Gk as its components, then
χ(G) = max{χ(Gi ) : i = 1, 2, · · · , k}.
Note that the complement of a graph G, denoted by G, is defined to be the
graph with vertex set V (G) and edge set
{xy : xy ∈
/ E(G), x, y ∈ V (G)} .
Question 1.1.6
Determine χ(Cn ) for n = 3, 4, 5.
Question 1.1.7
Show that χ(Cn ) = dn/2e for all n ≥ 4.
Question 1.1.8
Determine χ(Pn ) for n ≥ 3.
1.2
Graphs G with χ(G) = v(G) − 1
In this section, we shall study the structures of graphs G with χ(G) = n − 1,
where n is the order of G.
Theorem 1.2.1
Let G be a graph of order n. Then
(i) χ(G) ≤ n;
(ii) χ(G) = n if and only if G ∼
= Kn ;
(iii) χ(G) = n − 1 if and only if G ∼
6= Kn and Kn−1 is a subgraph of G.
Proof. We just prove (iii).
(⇐) Suppose G ∼
6= Kn and G contains a Kn−1 . By (ii), we have χ(G) < n.
Also, G contains a Kn−1 , implying that χ(G) ≥ n − 1. Thus χ(G) = n − 1.
(⇒) Suppose χ(G) = n − 1. By (ii), we have that G ∼
6= Kn . It remains to show
that Kn−1 is a subgraph of G.
Suppose on the contrary that Kn−1 is not a subgraph of G. Then G−x ∼
6= Kn−1
for every x ∈ V (G).
Since G ∼
6= Kn , there exist u, v ∈ V (G) such that uv ∈
/ E(G). Then xy ∈ E(G)
for any x, y ∈ V (G) \ {u, v}; otherwise, [{u, v, x, y}] is 2-colourable, implying that
χ(G) ≤ n − 2, a contradiction. Hence G − {u, v} ∼
= Kn−2 .
Note that G − u ∼
6= Kn−1 and G − v ∼
6= Kn−1 . Since G − {u, v} ∼
= Kn−2 , there
exist u0 , v 0 ∈ V (G) \ {u, v} such that uu0 , vv 0 ∈
/ E(G).
If u0 = v 0 , then [{u, v, u0 }] is 1-colourable, implying that χ(G) ≤ n − 2, a
contradiction.
If u0 6= v 0 , then [{u, v, u0 , v 0 }] is 2-colourable, implying that χ(G) ≤ n − 2, a
contradiction too.
Therefore Kn−1 is a subgraph of G.
1.3
2
Graphs G with χ(G) = v(G) − 2 (Optional)
For any graph G, a set U ⊆ V (G) is called a cover of E(G) (or a vertex cover of
G) if every edge of G is incident with a vertex in U . The covering number of
G is defined to be the minimum |U | over all covers U of G.
Question 1.3.1
(i) Determine the covering number of Kn .
(ii) Determine the covering number of K(p, q), where p ≥ q ≥ 1.
(iii) Determine the covering number of Pn , where n ≥ 3.
(iv) Determine the covering number of Cn , where n ≥ 3.
Recall that a matching of G is a subset M of E(G) such that no two edges
of M are incident with a common vertex.
Question 1.3.2
Is it true that for any matching M of G, the covering number
of G is at least |M |?
The matching number of G is defined to be the maximum |M | over all
matchings M of G.
Question 1.3.3
(i) Determine the matching numbers of Kn , where n ≥ 3.
(ii) Determine the matching number of K(p, q), where p ≥ q ≥ 1.
(iii) Determine the matching numbers of Pn , where n ≥ 3.
(iv) Determine the matching numbers of Cn , where n ≥ 4.
It is not difficult to show that
For any graph G,
covering number of G ≥ matching number of G.
Question 1.3.4
Find a graph such that it covering number is equal to its
matching number.
Question 1.3.5
Find a graph such that it covering number is larger than its
matching number.
König’s Theorem, namely Theorem ??, shows that a bipartite graph has the
same matching number and covering number.
For any two disjoint graphs G and H, define the graph G + H as follows:
V (G + H) = V (G) ∪ V (H)
and
E(G + H) = E(G) ∪ E(H) ∪ {xy : x ∈ V (G), y ∈ V (H)}.
Question 1.3.6
Is it true that Kn+1 = K1 + Kn−1 ?
Question 1.3.7
Is it true that K(p, q) = Np + Nq , where Np is the null graph
of order p?
Let χ(G) = n − 2, where n = v(G). It is clear that Kn−1 is not a subgraph of
G. If Kn−2 is a subgraph of G, then G − {u, v} ∼
= Kn−2 for some u, v ∈ V (G) and
G contains either a subgraph K3 or a matching with two edges.
If Kn−2 is not a subgraph of G, what is the structure of G? The following
result answers this question.
Let G be a graph order n, where n ≥ 3. If χ(G) = n − 2, then
is a subgraph of G or G ∼
= Kn−5 + C5 .
Theorem 1.3.1
either Kn−2
Proof. Assume that χ(G) = n − 2 and Kn−2 is not a subgraph of G. We just
need to show that G ∼
= Kn−5 + C5 under this condition.
Since Kn−2 is not a subgraph of G, the covering number of G is at least 3.
(Why?)
Since χ(G) = n − 2, the matching number of G is less than 3. (Why?)
Then, by Theorem ??, G is not bipartite. Thus, G contains odd cycles.
Claim 1: G contains no k-cycles with k ≥ 6.
Otherwise, the matching number of G is at least 3, a contradiction.
Claim 2: G contains no 3-cycles.
Suppose on the contrary that G contains a 3-cycle xyzx. Then [{x, y, z}] is
1-colourable, implying that G − {x, y, z} ∼
= Kn−3 . (Why?)
Since Kn−2 is not a subgraph of G and G − {x, y, z} ∼
= Kn−3 , there exists
x0 , y 0 , z 0 in G − {x, y, z} such that x and x0 are not adjacent, y and y 0 are not
adjacent and z and z 0 are not adjacent.
It is impossible that x0 = y 0 = z 0 ; otherwise, [{x, y, z, x0 }] is 1-colourable,
implying that χ(G) ≤ n − 3 (Why?), a contradiction.
It is impossible that |{x0 , y 0 , z 0 }| = 3; otherwise, the matching number of G is
at least 3, a contradiction.
Thus, |{x0 , y 0 , z 0 }| = 2. Assume that x0 = y 0 and z 0 6= x0 . But, then both
[{x, y, x0 }] and [{z, z 0 }] are 1-colourable, implying that χ(G) ≤ n − 3, a contradiction. Hence Claim 2 holds.
Claim 3: G ∼
= C5 + Kn−5 .
Since G contains odd cycles, by Claims 1 and 2, G contains a 5-cycle.
We must have e(G) = 5; otherwise, either G contains a 3-cycle or the matching
number of G is at least 3, a contradiction.
Since G contains a 5-cycle and e(G) = 5, Claim 3 holds.
This completes the proof.
2
To end this section, we propose the following problem.
Problem 1.1
Let G be a graph of order n with χ(G) = n − 3, where n ≥ 4.
Determine the structure of G if G contains no subgraph Kn−3 .
The following problem is more general and very difficult. It is suitable for
graduate students to try some special cases, e.g., r = 3, 4.
Problem 1.2
Let G be a graph of order n with χ(G) = n − r, where 3 ≤ r ≤
n − 1. Determine the structure of G if G contains no subgraph Kn−r .
1.4
Brooks Theorem
For graph G, let
∆(G) = max{d(v) : v ∈ V (G)}.
∆(G) is called the maximum vertex degree of G.
Theorem 1.4.1
For any graph G, χ(G) ≤ ∆(G) + 1.
Question 1.4.1
Prove Theorem 1.4.1.
Observe that the following two types of connected graphs have the property
that χ(G) = ∆(G) + 1:
• All complete graphs;
• All odd cycles.
Brooks in 1941 showed that these are the only connected graphs having the
property that χ(G) = ∆(G) + 1. In other words, if a connected graph G is not
complete nor an odd cycle, then χ(G) ≤ ∆(G).
In the following, we shall apply the idea in a proof given by Lovász in 1975 to
find a very result, which is very useful in the proof of Brooks’ Theorem.
Theorem 1.4.2
For any graph G, χ(G) ≤ k if there exists U ⊆ V (G) such
that
(i) [U ] is k-colourable and G − U is connected, and
(ii) d(x) ≤ k for every x ∈ V (G) \ U and d(w) ≤ k − 1 for some w ∈ V (G) \ U .
Proof. Assume that m = |V (G)| − |U |. Since G − U is connected, we are able to
label all vertices in G − U by x1 , x2 , · · · , xm such that
(i) w is labeled by xm ;
(ii) every xi , i < m, is adjacent to some xj with j > i.
Let c1 , c2 , · · · , ck be k different colours. We now assign these k colours to
vertices in G such that any two adjacent vertices are assigned different colours.
Since [U ] is k-colourable, we can find a proper k-colouring in [U ] by these k
colours.
Then colour vertices x1 , x2 , · · · , xm .
For any i with 1 ≤ i < m, assume that only xi , xi+1 , · · · , xm have not been
assigned colours yet. Since d(xi ) ≤ k and xi xj ∈ E(G) for some j > i, at most
k − 1 neighbours of xi have been assigned colours, and so some colour, say c1 , is
not assigned to any neighbour of xi . Then we can assign c1 to xi .
Finally, for xm , since d(xm ) < k, some colour is not assigned to any neighbour
of xm . Then we can assign this colour to xm .
Hence G has a proper k-colouring.
For any connected graph G, if ∆(G) ≤ k and d(w) ≤ k − 1
Corollary 1.4.1
for some w ∈ V (G), then χ(G) ≤ k.
Lemma 1.4.1
2
2
Let G be a 2-connected graph of order n with 3 ≤ d(x) ≤ n − 2
for some x ∈ V (G). Then there exist u, v, w ∈ V (G) such that
(i) uv ∈
/ E(G);
(ii) uwv is a path in G;
(iii) G − {u, v} is connected.
Proof. Choose a vertex x in G with 3 ≤ d(x) ≤ n − 2.
Case 1: G − x is 2-connnected.
Since d(x) ≤ n − 2, there exists a path xyz with xz ∈
/ E(G). Since G − x is
2-connnected, G − {x, z} is connected. Let u be x, w be y and v be z. Note that
u, v, w satisfy conditions (i), (ii) and (iii).
Case 2: G − x is not 2-connnected.
Note that G − x has at least two end-blocks G1 and G2 , where a block of a
connected graph H is called an end-block of H if this block contains only one
cut-point of H, and a vertex z of H is called a cut-point if H − z is disconnected.
Since G is 2-connected, x has a neighbor yi in every end-block Gi of G − x
such that yi is not a cut-point of G − x.
As Gi is a block, it is 2-connected and so Gi − yi is connected. Since yi is
not a cut-point of G − x, i = 1, 2, G − {x, y1 , y2 } is connected. Since d(x) ≥ 3,
G − {y1 , y2 } is also connected.
Now let w be x, u be y1 and v be y2 . Then u, v, w satisfy conditions (i), (ii)
2
and (iii).
Now we introduce Brooks’ Theorem.
Theorem 1.4.3 (Brooks, 1941)
Let G be a connected graph. If G is not a
complete graph nor an odd cycle, then χ(G) ≤ ∆(G).
Proof. Let ∆(G) = k. If G is not k-regular, then, by Corollary 1.4.1, χ(G) ≤ k.
Assume that G be k-regular. If k = 2, then G is an even cycle and so χ(G) =
2 = k. Now assume that k ≥ 3. Since G is not complete, k ≤ v(G) − 2.
If G is not 2-connected, then, by Corollary 1.4.1, χ(Gi ) ≤ k for each block Gi
of G, since ∆(Gi ) ≤ k and Gi is not k-regular. Then χ(G) ≤ k, by the result that
a graph is k-colourable if every block of this graph is k-colourable. (Why?)
Thus it remains the case that G is 2-connected. By Lemma 1.4.1, there exists
u, v, w ∈ V (G) which satisfy conditions (i), (ii) and (iii) in Lemma 1.4.1.
The remaining part of the proof is similar to the proof of Theorem 1.4.2.
Label all vertices in the subgraph G − {u, v} by x1 , x2 , · · · , xm , where m =
v(G) − 2, such that
(i) w is labeled by xm ;
(ii) every xi , i < m, is adjacent to some xj with j > i.
Let c1 , c2 , · · · , ck be k different colours. We now assign these k colours to
vertices in G such that any two adjacent vertices are assigned different colours.
Assign u and v by the same colour, say c1 . Then colour x1 , x2 , · · · , xm by this
order.
Let 1 ≤ i ≤ m − 1 and assume that only vertices xi , xi+1 , · · · , xm have not
been assigned colours yet. Since d(xi ) ≤ k and xi xj ∈ E(G) for some j > i, at
most k − 1 neighbours of xi have been assigned colours, and so some colour, say
c1 , is not assigned to any neighbour of xi . Then we can assign c1 to xi .
Finally, for xm , since d(xm ) = k and u, v are two neighbours of xm which are
assigned the same colour, namely c1 , some colour is not assigned to any neighbour
of xm . Then we can assign this colour to xm .
2
Hence G has a proper k-colouring.
Question 1.4.2
Use Brooks’ Theorem to prove that the chromatic number of
the Petersen graph is 3.
Question 1.4.3
Determine χ(G1 ) by Brooks’ Theorem, where G1 is the graph
in Figure 1.3.
v
v
v
.......................................................................................................................................................................
....
.
...
....
...
...
....
...
...
....
...
...
....
.
.
....
...
.
....
...
...
....
...
...
....
...
....
...
.
.
....
...
..
....
...
.... .....
...
.... ..
...
...............................................................................................................
v
v
Figure 1.3
Question 1.4.4
Determine χ(G2 ) by applying Brooks’ Theorem, or otherwise,
where G2 is the graph in Figure 1.4.
v
.
.............
.... .. ........
.... ... .. ....
.... .... ..... ........
.
.
.
.... ... .... ........
.
....
.
.................................................................................................................................................
........ .. .
...
... ......................
..
...................................... ....
... ......... ..................
............ ...........
..
....... ......
................. ...
.... ............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.... .......... ............. ....
.
... ..
.. .......
.......................
.... .... ...........................
.
.
.
.... .................. ........
. ....... ......
...
.......... ....
. .....
........
.... ......................
.
.
.
.
.. .
.
.
.
.
.
..
... ........
... ..
. ........... ....
.. ...
......................................................................................................................................................................................................................
..........
.
.
.
.
.
.
.
.....
.
.
.
.
.
.
.
.
.
.......... ..
...
.........
.....
.......... .....
... ....................
.... ....
..............
.. .
.... .......
.... .................... ..
... ................ .......
... ... ................ ................ ... ...
... .. ..
............
... .. .....
... .... ...... ................................... ........ ..... ...
.......... ........
........................
................
.............
...
v
v
v
v
v
v
Figure 1.4
1.5
Extension of Brooks Theorem
In this section, we shall apply the idea in Lovász’s proof of Brooks’ Theorem to
find some results more general than Brooks’ Theorem.
For any two vertices u, v in G, let G/uv denote the graph obtained from G
by contracting u and v, i.e., G/uv is the graph with
V (G/uv) = (V (G) \ {u, v}) ∪ {w}
and
E(G) = E[G − {u, v}] ∪ {wx : x ∈ V (G) \ {u, v}, xu ∈ E(G) or xv ∈ E(G)}.
Lemma 1.5.1
For any two vertices u, v in G, if uv ∈
/ E(G) and G/uv is k-
colourable, then G admits a proper k-colouring ψ with ψ(u) = ψ(v).
Proof. Let c1 , c2 , · · · , ck be different colours. Since G/uv is k-colourable, G/uv
admits a proper k-colouring φ : V (G/uv) → {c1 , c2 , · · · , ck }.
Let w be the new vertex in G/uv after contracting u and v. Now define a
k-colouring ψ of G:
ψ(x) =
φ(w),
φ(x),
x ∈ {u, v},
x ∈ V (G) \ {u, v}.
Since uv ∈
/ E(G), ψ is a proper k-colouring of G.
Theorem 1.5.1
2
For any graph G, χ(G) ≤ k if there exist V 0 ⊆ V (G) such
that
(i) d(x) ≤ k for every x ∈ V (G) \ V 0 ;
(ii) G − V 0 is connected;
(iii) there exists w ∈ V (G) \ V 0 such that [V 0 ]/uv is k-colourable for some
distinct u, v ∈ V 0 ∩ N (w).
Proof. By Lemma 1.5.1, it suffices to show that G/uv is k-colourable. We shall
apply Theorem 1.4.2 to show that G/uv is k-colourable. Let H denote G/uv.
Let w be the new vertex in G/uv after contracting u and v. So w is also
considered the new vertex in [V 0 ]/uv after contracting u and v. Since [V 0 ] is
k-colourable, [V 0 ]/uv is also k-colourable, as uv ∈
/ E(G).
0
Let U = V ([V ]/uv). It is clear that H − U ∼
= G − V 0 is connected. For
every x ∈ V (H) \ U , we have dH (x) ≤ dG (x) ≤ k. Since u, v ∈ NG (w), we have
dH (w) = dG (w) − 1 ≤ k − 1.
Hence H (i.e., G/uv) is k-colourable by Theorem 1.4.2.
Question 1.5.1
2
Let G be the graph in Figure 1.5.
(i) Can we conclude that χ(G) ≤ 3 by Brooks’ Theorem?
(ii) Prove that χ(G) = 3 by Theorem 1.5.1.
v
v
......................................................................................................................................................................................
... ..
...... ................
............
...
............
......
............
............
...
............
.
... ...
.
............
.
.
.
.
..
.
.
.
.
.
............ ..............
..
... ....
.
.
................
...
.
.
.
.
... ...
.
.
.
.
.
.
.
............
....
.
.
.
.
.
.
.
... ...
.
.
....
.
.
.
.
.
............
.
............
... ... .......................
............ .....
.. ..................
.................................................................................................................................................................................
....
... ...
..
....
... ...
...
....
... ...
....
...
.
.
... ..
.
.
.
....
.
... ...
....
...
... ...
....
...
....
... ...
....
...
... ...
.
.
.
....
......
..
....
......
.... .....
.....
.... ..
.... ...
.....
.............................................................................................................
v
v
v
v
v
Figure 1.5
Theorem 1.5.1 can be even extended to the following result by a similar proof.
Theorem 1.5.2
For any graph G and positive integer k, we have χ(G) ≤ k if
there exist U ⊆ V (G) such that
(i) d(x) ≤ k for every x ∈ V (G) \ U ;
(ii) [U ] has a k-colouring ψ;
(iii) each component of G − U contains a vertex w with either d(w) ≤ k − 1
or ψ(u) = ψ(v) for some distinct vertices u, v ∈ N (wi ) ∩ U .
1.6
2
Critical graphs
Example 1.6.1
Observe that χ(C5 ) = 3, but χ(C5 − v) = 2 for each vertex v
in C5 .
We say that C5 is 3-critical, because of the property in Example 1.6.1. In
general, a k-critical graph is defined below.
A graph G is called k-critical if χ(G) = k and χ(G − v) < k for each
vertex v of G.
Question 1.6.1
Is C4 2-critical? Why? In general, is an even cycle graph a
2-critical graph?
Question 1.6.2
Show that K1 and K2 are the only 1-critical and 2-critical
graphs.
Question 1.6.3
Show that Cn is a 3-critical graph for every odd integer n with
n ≥ 3.
Question 1.6.4
Is Kn is n-critical? Why?
Question 1.6.5
Determine χ(G − v) for v ∈ V (G) if G is k-critical.
Question 1.6.6
Show that if G is 3-critical, then G is an odd cycle.
Question 1.6.7
Show that the graph in Figure 1.6 is 4-critical.
v
.....
...... ......
....... .......
........ ............
.
.
.
... ....
.
.... ...
... .....
.... ..
... .....
.... ..
... .....
.... .....
.
.
.
... .....
.. ...
.
.
.
....
...
.
..
.
.
....
.
.
.
...
.
..
....
.
.
.
.
.
...
....
.
..
.
.
.
.
.
.
....
.
...
..
.
.
.
....
.
.
.........
.............
.
..
.
.
.
.
.
.
....
.
.
.
.
.
.
.
.
.
......
......... ..........
...
..
....
.
.
.
.
.
.
.
.
.
.
.
.
...........
......
.
..
..
.
.
.
.
.
.
.
.
.
.
.
...... ........
...
.. ........
.
.
.
.
.
...... .....
..
.. .......
.
.
.
.
.
.
...... .....
...
.. .......
.
.
...... ....
.
..
...... ....
.... ......
...... ....
...
.... .....
...... ....
............... ...............................
.
...............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
...... ....
.
.
................
......
.
.........
.
.
.
.
.
.
.
.
........
.
.
.
.
.
.
.
.
.
.
.
................
.......... ......................
.
.
.
.
.
.
.
.
.................................
...........................
.
.
.
.......................................................................................................................................................................................................................................
v
v
v
v
v
v
Figure 1.6
For a connected graph G and S ⊂ V (G), S is a cut-set of G if G − S is
disconnected. A cut-set S of G is called a complete cut-set of G if [S] is a
complete graph.
It is clear that every cut-point x of G forms a cut-set {x}. Thus, if a connected
graph contains no cut-set, then it is 2-connected.
Theorem 1.6.1 (Dirac, 1952)
Let G be a k-critical graph. Then
(i) d(x) ≥ k − 1 for every x ∈ V (G);
(ii) G is 2-connected;
(iii) G has no complete cut-set.
Question 1.6.8
Prove Theorem 1.6.1.
Question 1.6.9
Let G be the Petersen graph. Note that χ(G) = 3. Show that
(i) G is not 3-critical; and
(ii) G has the three properties in Theorem 1.6.1 for k = 3.
Question 1.6.10
Show that if G is k-critical, then G + K1 is (k + 1)-critical.
Actually, we have the following more general result.
Theorem 1.6.2
If G and H are s-critical and t-critical respectively, then G+H
is (s + t)-critical.
Question 1.6.11
Prove Theorem 1.6.2.
Question 1.6.12
Apply Theorem 1.6.2, or otherwise, to construct 4-critical
graphs rather than K4 .
1.7
Cliques and chromatic numbers
For any graph G, a complete subgraph of G is called a clique of G. The clique
number of G, denoted by ω(G), is defined to be the largest integer k such that
G has a clique isomorphic to Kk .
Lemma 1.7.1
Let G be a non-empty graph of order n.
(i) If G is disconnected, then ω(G) = max{ω(Gi ) : Gi is a component of G}.
(ii) 2 ≤ ω(G) ≤ χ(G) ≤ n;
(iii) ω(G) = n if and only if G ∼
= Kn ;
(vi) ω(G) = n − 1 if and only if χ(G) = n − 1.
Question 1.7.1
Prove Lemma 1.7.1 (vi).
Question 1.7.2
Determine ω(Cn ), where n ≥ 3.
Question 1.7.3
If ω(G) < χ(G) = v(G) − 2, then G ∼
= C5 + Kn−5 , where
n = v(G).
Note that χ(G) ≥ ω(G) for all graphs G. But Exercise 1.7.2 shows that
χ(G) > ω(G) for some graphs G. In this section, we shall show that for any
integer k ≥ 2, there exist graphs Gk with ω(G) = 2 such that χ(G) = k.
For any graph G, we construct the graph γ(G) as follows:
Step 1 : Assume that V (G) = {v1 , v2 , · · · , vn }.
Step 2 : Construct γ(G) with vertex set V (G) ∪ {v, u1 , u2 , · · · , un } and edge
set
[
E(G) ∪ {vui : 1 ≤ i ≤ n} ∪
{ui vj : vi vj ∈ E(G), 1 ≤ j ≤ n} .
1≤i≤n
For any i : 1 ≤ i, j ≤ n, where i 6= j, the graph [v, ui , vi , vj ] is shown in Figure
1.7:
vi
v
vj
vi
v
v
...........................................................
.
.........
.........
.........
.........
.
.
.
.
.
.
.
....
...
...
..
...
..
...
..
...
..
ui v
vj
v
ui ......v
.....
...
..
..
....
.....
.
v
v
v
v
if vi vj ∈ E(G)
if vi vj ∈
/ E(G)
Figure 1.7
Example 1.7.1
The graph γ(K2 ) is shown in Figure 1.8.
K2
γ(K2 )
.....................................................vv
v1 ..v
2
.................................................................v v2
v1 ..v
.......
.......
=⇒
.......
....... ............
.......
...... .............
.
.
.......
.....
......
.......
.
.
.
.
.
......
...
...
...
....
...
.
.
...
..
.
...
.
....
...
.... ......
... ....
.....
.
u1 v
v u2
v
Figure 1.8
Lemma 1.7.2
3e(G).
v
For any graph G, γ(G) is of order 2v(G) + 1 and size v(G) +
Question 1.7.4
Draw the graphs γ(P3 ) and γ(K3 ).
Question 1.7.5
Prove Lemma 1.7.2.
If G is the null graph Nn , then γ(G) is a graph consisting of the star K1,n and
n isolated vertices. The following result concerns the case that G is not null.
Theorem 1.7.1
For any non-null graph G, ω(γ(G)) = ω(G) and χ(γ(G)) =
χ(G) + 1.
Proof. Here we prove that χ(γ(G)) = χ(G) + 1 and it is left to the reader to
show that ω(γ(G)) = ω(G).
Let V (G) = {v1 , v2 , · · · , vn } and
V (γ(G)) = V (G) ∪ {v, u1 , u2 , · · · , un }.
By definition, the edge set of γ(G) can be
E(G) ∪ {vui : 1 ≤ i ≤ n} ∪
[
{ui vj : vi vj ∈ E(G), 1 ≤ j ≤ n} .
1≤i≤n
Assume that χ(G) = k. Let ψ : V (G) → {c1 , c2 , · · · , ck } be a k-colouring of
G. Then we get a (k + 1)-colouring φ : V (γ(G)) → {c1 , c2 , · · · , ck , ck+1 } of γ(G),
defined by
φ(x) =
ψ(vi ),
ck+1 ,
x ∈ {vi , ui }, 1 ≤ i ≤ n;
x = v.
Thus χ(γ(G)) ≤ k + 1.
It remains to show that γ(G) is not k-colourable. Suppose on the contrary
that γ(G) has a k-colouring φ : V (γ(G)) → {c1 , c2 , · · · , ck }. Let φ(v) = ck . Then
G has a (k − 1)-colouring ψ : {v1 , v2 , · · · , vn } → {c1 , c2 , · · · , ck−1 }:
φ(vi ),
if φ(vi ) 6= ck ;
ψ(vi ) =
φ(ui ),
otherwise.
We can show that ψ is really a (k − 1)-colouring of G.
First, since φ(v) = ck and vui ∈ E(γ(G)) for all i = 1, 2, · · · , n, we have
φ(ui ) ∈ {c1 , c2 , · · · , ck−1 } and so ψ(vi ) ∈ {c1 , c2 , · · · , ck−1 } for all i = 1, 2, · · · , n,
by the definition of ψ.
Second, we shall show that ψ(vi ) 6= ψ(vj ) for all vi , vj with vi vj ∈ E(G). If
both φ(vi ) 6= ck and φ(vj ) 6= ck , then
ψ(vi ) = φ(vi ) 6= φ(vj ) = ψ(vj ).
Now consider the case that φ(vi ) = ck . As φ is a colouring of γ(G), φ(vj ) 6= ck .
By the definition of ψ,
ψ(vi ) = φ(ui ) 6= φ(vj ) = ψ(vj ).
Thus, ψ is a (k − 1)-colourable, contradicting the assumption that χ(G) = k.
Hence χ(γ(G)) ≥ k + 1, and therefore χ(γ(G)) = k + 1.
2
Consider the following sequence of graphs:
K2 , γ(K2 ), γ (2) (K2 ), · · · , γ (m) (K2 ), · · · ,
where γ (1) (G) = γ(G) and γ (m+1) (G) = γ(γ (m) (G)). By Theorem 1.7.1, we have:
Corollary 1.7.1
For any m ≥ 1,
χ(γ (m) (K2 )) = m + 2,
ω(γ (m) (K2 )) = 2.
Question 1.7.6
Show that ω(γ(G)) = ω(G) for any non-empty graph G.
Question 1.7.7
Is it true that ω(γ(G)) = ω(G) if G is an empty graph? Why?
Question 1.7.8
Find explicit expressions, in terms of k, for v(γ (k) (K2 )) and
e(γ (k) (K2 )).
1.8
Perfect graphs
For a graph G and S ⊆ V (G), let G[S] (or simply [S]) denote the graph with
vertex set S and edge set
{xy : xy ∈ E(G), x, y, ∈ S}.
Observe that [S] can actually be obtained from G by removing all vertices in
V (G) \ S. It is clear that if S = V (G), we have [S] ∼
= G.
A subgraph H of G is called an induced subgraph of G if H ∼
= [V (H)]. In
other words, H is an induced subgraph of G if every edge xy of G belongs to H
whenever x, y ∈ V (H).
Question 1.8.1
that H ∼
= C5 ?
Is there an induced subgraph H of the Petersen graph such
Question 1.8.2
that H ∼
= N5 ?
Is there an induced subgraph H of the Petersen graph such
A graph G is said perfect if χ(H) = ω(H) for every induced subgraph H in
G.
Example 1.8.1
Every complete graph is perfect.
Question 1.8.3 Show that
(i) every odd cycle Cr is not perfect, where r ≥ 5 is odd;
(ii) a graph is not perfect if it contains an odd cycle Cr as an induced
subgraph, where r ≥ 5 is odd;
(iii) every bipartite graph is perfect.
Lemma 1.8.1
The complement of a bipartite graph is perfect.
The proof of Lemma 1.8.1 is left to the reader (see Exercise 19 of Chapter 1).
Lemma 1.8.1 was generalized to any perfect graph by Lovász and Fulkerson.
Theorem 1.8.1 (Lovász and Fulkerson)
The complement of a perfect graph
2
is also perfect.
Strong Perfect Graph Conjecture (Berge, 1960):
Conjecture 1.1
A graph G is perfect if and only if neither G nor its comple-
ment G contains an induced odd cycle of length at least 5.
After 40 years, this conjecture was finally proved by M. Chudnovsky, N.
Robertson, P. Seymour and R. Thomas in 2002, and so now it can be called
the Strong Perfect Graph Theorem.
Theorem 1.8.2 (Strong Perfect Graph Theorem)
A graph G is perfect if
and only if neither G nor its complement G contains an induced odd cycle of length
at least 5.
Exercises of Chapter 1
1.1 Let G be a graph of order n. Show that n ≤ χ(G) × α(G), where α(G)
is the independence number of G.
1.2 Determine χ(Pn ) for n ≥ 3.
1.3 Show that χ(Cn ) = dn/2e for all n ≥ 4.
1.4 Determine n ≥ 3 such that Cn is critical.
1.5 Show that χ(G + H) = χ(G) + χ(H) for any graphs G and H.
1.6 Show that if G contains exactly one odd cycle, then χ(G) = 3.
1.7 Show that if all odd cycles of G have a vertex w in common, then
χ(G) = 3.
1.8 Let G be a graph which contains some odd cycles and S an independent
set of G. Show that if V (C) ∩ S 6= ∅ for every odd cycle C in G, then
χ(G) = 3.
1.9 Let G be a graph with χ(G) = 3. Show that G contains an independent
set S such that V (C) ∩ S 6= ∅ for every odd cycle C in G.
1.10 Prove that the 3-critical graphs are just the odd cycles.
1.11 Is the Petersen graph critical? Why?
1.12 Determine the chromatic number of the graph in Figure 1.9. Is it
critical?
v
v
.....................................................................................................................................................................................
.... ..
...... ................
............ .........
............
... ....
.
............
............
............
............
... ....
... ...
............ .......................
.. ...
... .....
.
.
.
.
.
.
.
.
.
.
... ...
.
.
.
.
...
.
.
....... ................
..
...
...
...
.........
.........
...
...
...
...
.........
.........
.
.
.
.
.
.
...
.
.
..
.
...
.
.
......... ..
.................
...
.
.
...
.
.........
.........
.
..
.
.
..
.
.
.
... ...
. .........
.
.
.
.
.
.
...
.
.
.
....
.
.........
...
...
.
.
.
.
.
.
.
..
.
.
.
.
.
...
.........
...
...
.
.
.
.
.
.
.
.
...
.
.
.
.
.
.
.........
...
........................
.. ....
... .....
.
.
.
.
.
.
.
... ...
.
.
.
.
.
..
.
.
............
.....
.
.
.
.
.
.
.
.
.
.
.
.. ...
.
.
.
.
.
.
.
............
.. ....
....
............
... ...
............
............ ..... ....
..... .......................
...................................................................................................................................................................................................
v
v
v
v
v
v
Figure 1.9
1.13 Determine the chromatic number of the graph in Figure 1.10. Is it
critical?
v
...
..................
....... .. ... .......
....... .. .. .......
....... ..... ..... ..............
.
.
.
.
.
.
.......
..
...
.......
...
.......
...
.......
.......
...
...
.......
.......
...
...
.......
.......
.
.
.
.
.
.
.
.
.
.
.......
...
...
.
.
.
.
.
.
.....
.
.
.
.
..
.....................................................................................................................................................................................
...
.
.
.
...
.
.
...
...
.
...
.
...
...
...
.
...
.
.
...
.
.
...
..
...
...
...
...
...
...
...
...
...
... ....
... ....
.
.
.
.. .
... ...
.......................................................................................................................
v
v
v
v
v
v
Figure 1.10
1.14 Let G be a graph obtained from K100 by deleting any 50 edges. Determine the range for ω(G).
1.15 Let G be a graph obtained from Kn by deleting any m edges. Show
that ω(G) ≥ n − m.
1.16 (∗) Let G be a graph obtained from K100 by deleting 99 edges. Determine the minimum value for ω(G).
1.17 Let G be a graph obtained from K(100, 100) by adding 100 edges.
Determine the range for ω(G).
1.18 Characterize G such that γ(G) is connected.
1.19 Show that the complement of a bipartite graph is perfect.