Chapter 5 Coloring 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 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, C 2r+1 join K s  (G)=s+2 (G)=s+3 C5C5 KsKs

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 GHGH

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

Upper Bounds Greedy Coloring: A vertex ordering v 1, v 2, …, v n of V(G), assign v i 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 d 1 d 2  … d n, then (G)  1+max i min{d i, i-1} Apply greedy coloring to the vertices in the non-increasing order of degree. V i has at most min{d i, i-1} earlier neighbors. We assign the color to v i at most 1+min{d i,i-1}. So we maximize over i to obtain upper bound.

Lemma 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+max HG (H) Let k= (G), and let H ’ be a k-critical subgraph of G. (G)-1 = (H ’ )-1  (H ’ )  max HG (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

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

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 K 2, 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 v n. Choose one neighbor of as v i. So every v i has at least higher neighbors. Thus v i has at most k-1 lower neighbors. So it can be colored by k colors.

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 v n has neighbors v 1,v 2 such that v 1  v 2 and G-{v 1,v 2 } is connected. We can order G-{v 1,v 2 } with 3, …,n. Every v i before v n has at most k-1 lower neighbor and v 1,v 2 receive the same color. So greedy coloring also uses k colors.

Proof (cont) : We now proof that every 2-connected k-regular graph with k  3 has such triple v 1, v 2, v n. Choose a vertex x. If  (G-x)=2, let v 1 =x, v 2 with distance 2 from x (G is not complete), and v n 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 v 1,v 2 of x in two such blocks are nonadjacent. k  3 so G-{v 1,v 2 } is connected. v n =x v2v2 v1v1

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 {v 1,v 2, …,v n }, add U={u 1,u 2, …,u n } and w. add edges to make u i to adjacent to N G (v i ), and finally let N(w)=U. v1v1 v1v1 v2v2 u2u2 u1u1 w v2v2

Theorem: from a k-chromatic triangle-free graph G, Mycielski ’ s construction produces a k+1-chromatic triangle free graph G ’. V(G)={v 1,v 2, …,v n } is triangle free, U ={u 1,u 2, …,u n } is an independent set, and w cannot be contained in any triangle. So the triangles only can occur on some u i with two neighbors of v i, but it contains a triangle in G. Thus G ’ is triangle-free.

let  (G) = k. G ’ can be assigned in k+1 colors by set u i the same color with v i, 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.

We want to change all color of g(v i ) into g(u i ) st. g is still proper. If v i, v j adjacent, since v i also adjacent to u j, so v i has different color from v j. 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. 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.

Definition: A complete multipartite graph is a simple graph G whose vertices can be partition into K n1, K n2, … K nk. Where u  v iff u, v belong to different K ni. We written it as K n1,n2, …,nk. Every component in is a complete graph. Turán graph T n,r : A complete r-partite graph and the vertices number m of every part is  n/r   m   n/r .

Lemma: Among simple r-partite graphs with n vertices, the Turán graph is the unique graph with the most edges. 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, T n,r has the maximum of edges. 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 x  V(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) T n,r has the most edges within all r-colorable graphs.

Color-Critical Graphs Proposition: (a) for v  V(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 e  E(G), every proper k-1-coloring of G-e gives the same color to the two ends of e. (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. (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. let X 1, X 2, …, X k and Y 1, Y 2, …, Y k be the partition of X and Y by color classes. If there are no edges between X i and Y j, then X i  Y j is an independent set in G.

Recall that p : for every subgraph of K n,n with more than (k-1)n edges has a matching at least k. Construct bipartite H with vertices X 1, …,X k and Y 1, …,Y k, and edges when no edge on G between X i and Y j. 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 X i  Y j 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. 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 x  y. And there is a S-lobe H st.  (H+xy)=k. Let S={x,y} is a cutset of G with x  y, H 1, H 2, …, H t are S-lobes of G. Each H i is k-1-colorable. If x  y, then we must assign distinct color to x, y in each H i.

thus we can find a coloring st. x, y in assigned in the same color in every H i. Then G is k-1-colorable, contradiction. Now we prove second statement. If for all H i,  (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 K 4 -subdivision. induction on n(G) When n(G)=4, the graph is K 4 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 x  y, 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 K 4 -subdivision in H ’ +xy.

we replace xy into xy-path in other S- lobe other than H ’, it is a K 4 -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 ) shows that x has 3 vertex-disjoint paths connecting to cycle C, it forms a K 4 -subdivision.

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

Counting Proper Colorings Definition: Given k  N and a graph G, the value  (G;k) is the number of proper coloring using at most k colors. Examples:  (K n ;k) = k(k-1)(k-2) … (k-n+1),  ( ;k) = k n.

Proposition 5.3.3: If T is a tree with n vertices, then  (T;k) =k(k-1) n-1. choosing a vertex as a root and considering the coloring from it, then …. k-1 k

Proposition 5.3.4: let x (r) =x(x-1) … (x-r+1). If p r (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). when using r colors in a proper coloring, it will partition V(G) into r independent sets, which can happen in p r (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 k n.

Example: C 4, p 1 =0, p 2 =1, p 3 =2, p 4 =1  (C 4 ;k)=1  k(k-1)+2  k(k-1)(k-2)+1  k(k-1)(k-2)(k-3) =k(k-1)(k 2 -3k+3)

Theorem: if G is a simple graph and e  E(G), then  (G;k)=  (G-e;k)-  (G  e;k). 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  (G  e;k). Example:  (C 4 ;k)=  (P 3 ;k)-  (K 3 ;k)=k(k-1)(k 2 -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), …. we use induction on e(G). e(G)=0 holds.  (G-e;k):k n - [e(G)-1]k n-1 + a 2 k n-2 - … +(-1) i a i k n-i … -  (G  e;k): -( k n-1 - b 1 k n-2 + … +(-1) i-1 b i-1 k n-i … ) =  (G;k): k n - e(G)k n-1 +(a 2 +b 1 )k n-2 + … +(-1) i (a i +b i-1 )k n-i …

Theorem: let c(G) denote the number of components of a graph G. Given a set S  E(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:

Example: Like the theorem of “ exclusion and inclusion ”.  (G;k)=k 4 - 5k k 2 - (2k 2 +8k 1 ) + 5k - k =- = - … = = - ()

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

Chordal Graphs Definition: a vertex of G is simplicial if its neighborhood in G is a clique. A simplicial elimination ordering is a order v n, …,v 1 for deleting such that when deleting v i, v i is a simplicial vertex of the remaining graph induced by {v 1, …,v i }.

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 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 H S Q G’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. H S Q G’G’ T x u v HQ

Let G ’ = S  H, we can use induction hypothesis that (whether G ’ is a clique or not) there is a u  S has a simplicial vertex z  V(H) farthest from it. Since N G (z)  V(G ’ ), z is also simplicial in G. z is what we want. H S Q G’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 H  G. 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.

Theorem: chordal graphs are perfect. 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={v 1,v 2, …,v n }. For v i, the neighbors of v i among {v 1, …,v i-1 } forms a clique. We apply greedy coloring here. If v i uses color k, then 1, …,k-1 appear on earlier neighbors of v i, and we have a clique with size k. The obtain a clique whose size equals the number of color used.