Chapter 5Coloring of Graphs

5.1 Vertex Coloring and Upper Bound Definition:

A k-coloring of a graph G is a labeling f:V(G)S, where |S|=k (or S=[k]).

The labels are colors; the vertices of one color form a color class.

A k-coloring is proper if adjacent vertices have different labels.

A graph is k-colorable if it has a proper k-coloring.

The chromatic number (G) is the least k such that G is k-colorable.

Remark In a proper coloring, each color class is an

independent set. Graphs with loops are uncolorable; multiple

edges are irrelevant. So we consider simple graph only.

Definition A graph G is k-chromatic if (G)=k. A proper k-coloring of a k-chromatic graph is

an optimal coloring. If (H)< (G) = k for all proper subgraph H

of G is color-critical or k-critical. The clique number of a graph G, written

(G), is the max size of clique in G.

Proposition 5.1.7 For every graph G, (G)(G) and

(G)n(G)/(G). (G) is the max independent set of G.

(G) may exceed (G). r 2, C2r+1 join Ks

(G)=s+2 (G)=s+3

C5Ks

Cartesian Product The Cartesian product of G and H,

written GH, is the graph with vertex set V(G)V(H) specified by putting (u,v) adjacent to (u’,v’) iff (1) u=u’ and vv’E(H), (2) v=v’ and uu’E(G).b

x

c

y

z

a

(x,c)

(z,c)

(y,a)G

H

GH

Proposition 5.1.11 (GH)=max{(G), (H)}

(GH) max{(G), (H)} = k f(u,v) = {g(u)+h(v)} mod k

2

1

3

2

1

1

G

H

GH

312

31

2

312

Upper Bounds Greedy Coloring:

A vertex ordering v1, v2,…, vn of V(G), assign vi the smallest-indexed color not already used on its lower-indexed neighbors.

(G) (G) + 1 Using greedy coloring to prove.

If a graph G has degree sequence d1d2 …dn, then (G) 1+maxi min{di, i-1}

Apply greedy coloring to the vertices in the non-increasing order of degree. Vi has at most min{di, i-1} earlier neighbors. We assign the color to vi at most 1+min{di,i-1}. So we maximize over i to obtain upper bound.

Lemma 5.1.18 If H is a k-critical graph, then (H) k-1

Let x be a vertex of G. H-x is k-1-colorable. If d(x)<k-1, then N(x) cannot use all k-1 color and x can be assigned the rest color, contradiction.

Theorem: If G is a graph, then (G) 1+maxHG(H)

Let k= (G), and let H’ be a k-critical subgraph of G.

(G)-1 = (H’)-1 (H’) maxHG (H)

Gallai-Roy-Vitaver Theorem If D is an orientation of G with longest

path length l(D), then (G)1+l(D). Furthermore, equality holds for some orientation of G.

Proof: Let D be an orientation of G. Let D’ be a

maximal sub-digraph of D containing no cycle.

We assign color along the longest path by increase 1.

Every path can be assigned in increasing number.

So we use 1+ l(D) colors.1

6

53

1

42

Proof(cont): Let e is an edge in D not in D’. Since D’+e

forms a cycle and all path are in increasing order. So the two ends of e cannot be the same color.

1

6

53

1

42

Proof(cont): To prove the second statement, we

construct D* st. l(D*) (G)-1. Let f be optimal coloring of G. We set

uv a orient u to v in D* iff f(u)<f(v). So the max path length l(D*) (G)-1.

Brooks’ Theorem If G is a connected graph other than a

complete graph or an odd cycle, then (G) (G).

Proof: Let G be a connected graph. Let k= (G)

and k 3. Since k=1 is a K2, k=2 is a odd cycle or bipartite.

Our aim is to order the vertices st. each has at most k-1 lower neighbors.

Proof(cont): If G is not k-regular. We choose a vertex

with degree not k as vn. Choose one neighbor of as vi. So every vi has at least higher neighbors.

Thus vi has at most k-1 lower neighbors. So it can be colored by k colors.

nkinv

Proof(cont): G is k-regular and G has a cut-vertex x. Every component of G-x union x with edges

between them can be colored by k colors.

x

Proof(cont): G is k-regular. We assume that G is 2-

connected. Choose a vn has neighbors v1,v2 such that

v1v2 and G-{v1,v2} is connected. We can order G-{v1,v2} with 3,…,n. Every vi before vn has at most k-1 lower

neighbor and v1,v2 receive the same color. So greedy coloring also uses k colors.

Proof(cont): We now proof that every 2-connected k-

regular graph with k3 has such triple v1, v2, vn.

Choose a vertex x. If (G-x)=2, let v1=x, v2 with distance 2 from

x (G is not complete), and vn is their common neighbor.

Proof(cont): If (G-x)=1, since G has no cut-vertex, x has

a neighbor in every leaf block. Neighbors v1,v2 of x in two such blocks are

nonadjacent. k 3 so G-{v1,v2} is connected.

vn=x

v2v1

5.2 Structure of k-chromatic Graphs

Bound (G) (G) is bad. (G), (G), (G) over all graphs is

approximating to 2ln n, 2ln n, n/(2ln n) (in 8.5), so n/(G) is a good bound.

Graphs with Large Chromatic Number

Definition: A simple graph G, Mycielski’s

construction produces a simple graph G’containing G. Beginning with G having vertex {v1,v2,…,vn}, add U={u1,u2,…,un} and w. add edges to make ui to adjacent to NG(vi), and finally let N(w)=U.

v1 v1

v2 u2

u1

w

v2

Theorem: from a k-chromatic triangle-free graph G, Mycielski’s construction produces a k+1-chromatic triangle free graph G’.

<pf> V(G)={v1,v2,…,vn} is triangle free, U ={u1,u2,…,un} is an independent set, and w cannot be contained in any triangle.

So the triangles only can occur on some ui with two neighbors of vi, but it contains a triangle in G.

Thus G’ is triangle-free.

<pf(cont)> let (G) = k. G’ can be assigned in k+1 colors by set ui

the same color with vi, and w the color k+1. Hence (G’) (G) +1.

We now want to prove that (G) < (G’) . If a proper coloring g on G’ using k colors,

let g(w)=k -> U using colors {1,2,…,k-1}, V(G) may use k colors.

<pf(cont)> We want to change all color of g(vi) into

g(ui) st. g is still proper. If vi, vj adjacent, since vi also adjacent to uj,

so vi has different color from vj. So G is k-1-colorable.

Extremal Problems and Turán’s Theorem

Proposition 5.2.5: every k-chromatic graph with n vertices at least edges.

<pf> there are pairs of colors meaning the two ends of a edge. If (i,j) does not exist, color i and j can be merge into one color.

Maximization is more interesting.

2

k

2

k

Definition: A complete multipartite graph is a simple

graph G whose vertices can be partition into Kn1, Kn2,… Knk. Where uv iff u, v belong to different Kni.

We written it as Kn1,n2,…,nk. Every component in is a complete graph.

Turán graph Tn,r: A complete r-partite graph and the vertices

number m of every part is n/r m n/r.

G

Lemma: Among simple r-partite graphs with n vertices, the Turán graph is the unique graph with the most edges.

<pf> We consider complete r-partite only. let G be a r-partite graph other than Turán

graph with most edges. We chose v from largest class (size i) and

move it to the smallest class (size j), i-1>j. We loss j edges and gain i-1 edges, so we

have more edges than G, contradiction.

Theorem: Among the n-vertex simple graphs with no r+1-clique, Tn,r has the maximum of edges.

<pf> if we can prove that the maximum is achieved by an r-partite graph. Then we can use earlier lemma to prove Turán graph is the maximum.

We want to construct a r-partite graph H from graph G with at least as many edges.

Prove by induction on r: When r=1 ,then no edges. Consider r>1. Let G as an n-vertex graph with no r+1-

clique, and xV(G) be a vertex of degree k=(G).

G’ is the induced subgraph of G, and V(G’)=N(x). Since G has no r+1 clique, G’ has no r-clique.

Applying induction hypothesis, there exists a r-1-partite H’ st. e(H’)e(G’).

Let H be the graph formed by joining N(x) and S = V(G) - N(x), H is r-partite.

e(G) e(G’) + (n-k)k e(H’) + (n-k)k = e(H) Tn,r has the most edges within all r-

colorable graphs.

Color-Critical Graphs Proposition:

(a) for vV(G), there is a proper k-coloring of G in which the color on v appears on v only, and other k-1 appear on N(v).

(b) for eE(G), every proper k-1-coloring of G-e gives the same color to the two ends of e.

<pf>(a) giving proper k-1-coloring on G-v and color k to v forms a proper k-coloring on G. N(v) must use k-1 colors, otherwise G is k-1-colorable.

<pf>(b) if some k-1-coloring of G-e gave distinct colors to the two ends of e, then G is k-1-colorable.

The join of two color critical graphs is still color-critical.

Lemma: let G be a graph with (G)>k, and let X,Y be a partition of V(G). If G[X] and G[Y] are k-colorable, then the edge cut [X,Y] has at least k edges.

<pf> let X1, X2,…, Xk and Y1, Y2,…, Yk be the partition of X and Y by color classes. If there are no edges between Xi and Yj, then XiYj is an independent set in G.

Recall that p.121 3.1.29: for every subgraph of Kn,n with more than (k-1)n edges has a matching at least k.

Construct bipartite H with vertices X1,…,Xk and Y1,…,Yk, and edges when no edge on G between Xi and Yj.

Let |[X,Y]|<k, then H has more than (k-1)k edges, so it has a matching at least k (perfect matching).

We assign the same color to each matching pair. Since XiYj matching in H means that they are independent set in G, the coloring is proper.

Theorem: Every k-critical graph is k-1-edge-connected.

<pf>using earlier lemma, proved.

Definition: let S be a set of vertices of G. an S-lobe of

G is a induced subgraph whose vertex set is S union one of the component of G-S.

Proposition: if G is k-critical, then G has no cutset {x,y} with xy. And there is a S-lobe H st. (H+xy)=k.

Let S={x,y} is a cutset of G with xy, H1, H2,…, Ht are S-lobes of G. Each Hi is k-1-colorable. If xy, then we must assign distinct color to x, y in each Hi.

<pf(cont)> thus we can find a coloring st. x, y in assigned in the same color in every Hi.

Then G is k-1-colorable, contradiction. Now we prove second statement. If for all Hi, (H+xy) < k, then G is k-1-

colorable.

Forced Subdivision Definition: An H-subdivision is a graph

obtained from a graph H by successive edge subdivisions.

Theorem: every graph with chromatic number at least 4 contains a K4-subdivision.

<pf> induction on n(G) When n(G)=4, the graph is K4 itself. Consider n(G)>4. G has a 4-critical subgraph H.

H has no cut-vertex. If (H)=2, let cutset S={x,y} and not xy, then

there is a S-lobe H’ st. (H’+xy)4. Since n(H’+xy)<n(G), we can apply the

induction hypothesis to obtain K4-subdivision in H’+xy.

<pf(cont)> we replace xy into xy-path in other S-lobe other than H’, it is a K4-subdivision in G. Such path exists since x, y connect to every S-lobe.

If (H)=3, choose a vertex x. Since H-x is 2-connected, H-x contains a cycle C. And Since H is 3-connected, the Fan lemma (theorem 4.2.23) shows that x has 3 vertex-disjoint paths connecting to cycle C, it forms a K4-subdivision.

Enumerative Aspects Counting Proper Colorings Chordal Graph A Hint of Perfect Graphs

Counting Proper Colorings Definition:

Given kN and a graph G, the value (G;k) is the number of proper coloring using at most k colors.

Examples: (Kn;k) = k(k-1)(k-2)…(k-n+1), ( ;k) = kn.

nK

Proposition 5.3.3: If T is a tree with n vertices, then (T;k) =k(k-1)n-1.

<pf> choosing a vertex as a root and considering the coloring from it, then ….

k-1

k

k-1k-1

k-1 k-1 k-1

Proposition 5.3.4: let x(r)=x(x-1)…(x-r+1). If pr(G) denotes the number of partitions of V(G) into r nonempty independent sets, then (G;k) = , which is the polynomial in k of degree n(G).

<pf>when using r colors in a proper coloring, it will partition V(G) into r independent sets, which can happen in pr(G) ways.

When k colors available, k(r) ways to choose colors.

The way to partition V(G) into n(G) independent sets is only 1, it leads to the leading term kn.

)(

1 )()(Gn

r rr kGp

Example: C4, p1=0, p2=1, p3=2, p4=1 (C4;k)=1k(k-1)+2k(k-1)(k-2)+1k(k-1)(k-

2)(k-3) =k(k-1)(k2-3k+3)

)(

1 )()(Gn

r rr kGp

Theorem: if G is a simple graph and eE(G), then (G;k)= (G-e;k)- (Ge;k).

<pf> if the proper coloring (G-e;k) assigns the two ends of e distinct color, then the coloring is also proper in (G;k).

If the two ends of e are assigned in the same color in (G-e;k), the number is the same with (Ge;k).

Example: (C4 ;k)= (P3;k)- (K3;k)=k(k-1)(k2-3k+3)

Theorem: the chromatic polynomial (G;k) of a simple graph G has degree n(G), with integer coefficients alternating sign and beginning 1, -e(G),….

<pf>we use induction on e(G). e(G)=0 holds. (G-e;k):kn- [e(G)-1]kn-1 + a2kn-2-…+(-1)iaikn-i… -(Ge;k): -( kn-1 - b1kn-2+…+(-1)i-1bi-1kn-i…) =(G;k): kn - e(G)kn-1 +(a2+b1)kn-2+…+(-1)i

(ai+bi-1)kn-i…

Theorem: let c(G) denote the number of components of a graph G. Given a set SE(G) of edges in G, let G(S) denote the spanning subgraph of G with edge set S. Then the number (G;k) of proper k-chromatic of G is given by:

)(

))((||)1();(GES

SGcS kkG

Example: Like the theorem of “exclusion and

inclusion”. (G;k)=k4 - 5k3 + 10k2 - (2k2+8k1) + 5k - k

= -

=

-

…==

-

( )

)(

))((||)1();(GES

SGcS kkG

<pf> Multiple edges do not effect the theorem. When all edges have been deleted or

contracted, the graph remains isolated vertices. The remaining vertices corresponding to components of G(S);

So the term is kc(G(S)), and the sign is changed by contracting edge, so the contribution is positive iff |S| is even.

)(

))((||)1();(GES

SGcS kkG

Chordal Graphs Definition: a vertex of G is simplicial if

its neighborhood in G is a clique. A simplicial elimination ordering is a

order vn,…,v1 for deleting such that when deleting vi, vi is a simplicial vertex of the remaining graph induced by {v1,…,vi}.

Definition: A chord of a cycle C is an edge not in C

whose end points lie in C. A chordless cycle in G is a cycle of length

at least 4 that has no chord. A graph G is chordal if it is simple and has

no chordless cycle.

Lemma: for every vertex x in a chordal graph G, there is a simplicial vertex of G among the vertices farthest from x in G

<pf> induction on n(G), when n=1 trivial. If x is adjacent to all other vertices, then G-x

has a simplicial vertex y. And y in G is also simplicial since x adjacent to every vertex.

We consider the rest case.

Let T be the set of farthest points from x, H is a component of G[T].

Let S be the set of vertices in G-T having neighbors in V(H).

And let Q be the component of G-S contains x.

We claim that S is a clique

HSQ

G’

T

x

If not, there exist u,v S such that u,v being not adjacent.

u,v have neighbors in H, and u,v have neighbors in Q. So there is a uv-path through H, and a uv-path through Q.

If u and v non-adjacent, then there is a chordless cycle, contradiction.

So S is a clique.HS

Q

G’

T

x

u

v

HQ

Let G’= SH, we can use induction hypothesis that (whether G’ is a clique or not) there is a uS has a simplicial vertex z V(H) farthest from it. Since NG(z)V(G’), z is also simplicial in G.

z is what we want.

HSQ

G’

T

x

Theorem: a simple graph has a simplicial elimination ordering iff it is a chordal graph.

->: let G be a graph with simplicial elimination ordering. Let C be a cycle in G of length at least 4.

When we first deleted a vertex v from C. Since the neighborhood of v in rest graph is

a clique, the edge join the two neighbors of v in C is a chord of C, so no chordless cycle.

v

<-: by earlier lemma, every chordal graph has a simplicial vertex. Since every induced subgraph of a chordal graph is a chordal graph, proved.

A Hint of Perfect graphs Definition:

A graph G is perfect if (H)=(H) for every induced subgraph HG.

The clique cover number (G) of a graph G is the minimum number of cliques in G needed to cover V(G); note that (G) =( ).

A family of graphs G is heredictary if every induced subgraph of a graph in G is also a graph in G.

More detail in section 8.1.

G

Theorem: chordal graphs are perfect. <pf>every induced subgraph of chordal graph

is chordal. We only need to prove that (G)=(G) when G is chordal.

We have known that G has a simplicial elimination ordering, the reverse of the ordering={v1,v2,…,vn}.

For vi, the neighbors of vi among {v1,…,vi-1} forms a clique. We apply greedy coloring here.

If vi uses color k, then 1,…,k-1 appear on earlier neighbors of vi, and we have a clique with size k.

The obtain a clique whose size equals the number of color used.