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ELSEVIER

Discrete Mathematics 186 (1998) 15-24

DISCRETE

MATHEMATICS

Extremal graphs in some coloring problems

R. Balakrishnan*, R. Sampathkumar, V. Yegnanarayanan

Department of Mathematics, Annamalai University, Annamalainagar 608 002, India

Received 15 July 1994; revised 26 June 1996; accepted 24 February 1997

Abstract

For a simple graph G with chromatic number x(G), the Nordhaus-Gaddum inequalities give

upper and lower bounds for z(G)•(G ¢) and z(G)+ x(GC). Based on a characterization by Fink

of the extremal graphs G attaining the lower bounds for the product and sum, we characterize

the extremal graphs G for which A(G)B(G c) is minimum, where A and B are each of chromatic

number, achromatic number and pseudoachromatic number. Characterizations are also provided

for several cases in which A(G)+ B(G c) is minimum. (~) 1998 Elsevier Science B.V. All rights

reserved

1. Introduction

Throughout this paper, we consider only finite simple graphs. The notation and

terminology are as in [1]. We recall some of these below: Pv is the path of length

v - 1; Cv the cycle of length v; Kv the complete graph on v vertices; K(nl ..... n~) the

complete m-partite graph with ni vertices in the ith part, 1 <<. i <~m; K,n(n) the complete

m-partite graph with n vertices in each part; G v H the join of G and H; W~ the join

of Cv-l and K1; G c the complement of G; z(G) the chromatic number of G and co(G)

the number of components of G.

An achromatic k-coloring of a graph G is a proper vertex k-coloring of G in which

every pair of distinct colors is represented by some edge. The maximum k for which

G has an achromatic k-coloring is the achromatic number ~(G) of G. A pseudoachro-

matic k-coloring of G is a k-coloring of G in which every pair of distinct colors

is represented by some edge. The maximum k for which G has a pseudoachromatic

k-coloring is the pseudoachromatic number ~b(G) of G. It is clear that for any graph

G, z(G)<~7(G)<<.@(G). The celebrated Nordhaus-Gaddum [5] inequalities are:

v <<. z(G)x(G c) <~ L((v + 1)/2)2j

* Corresponding author.

0012-365X/98/$19.00 Copyright (~) 1998 Elsevier Science B.V. All rights reserved

Pll S0012-365X(97)002 1 6-1

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R. Balakrishnan et al./Discrete Mathematics 186 (1998) 15-24

and

I2x/~l <~ z(G) + z(G ~) <~ v + 1,

where v = IV(G)[.

Also, for the sum, the following upper bounds are known [2]:

x(G)+~(G~)<~v+ 1 and

¢(G)+~(G~)<~ V4v/3].

Fink [3] characterized graphs G for which z(G)z(G~)=v and those for which

z(G)+ z(G ¢) = [2vG]. His solutions of these graph equations are based on the con-

sideration of the set Tl(V;x, y) of graphs with v vertices, where x + y - 1 <<, v <~xy. The

set Tl(v;x,y) is defined as follows: Consider a rectangular array T with x rows and

y columns and place at most one dot in each of the xy entries of T according to the

following scheme: Place a dot in each entry of the first row and first column of T. This

accounts for x + y - 1 dots. Also place v - (x + y - 1) dots in any of the remaining

entries of T. Then a graph G in Tl(V;x,y) with v vertices is formed by taking the v

dots of T as the vertices of G and defining adjacency in G as follows: (i) Any two

dots in the same column are adjacent, (ii) no two dots in the same row are adjacent,

and (iii) any two dots which belong to distinct rows and columns of G may or may

not be adjacent. The two results of Fink are the following:

(i) z(G)z(G~)= v iff GETl(v;v/d,d), where d is a positive divisor of v.

(ii) z(G)+ z(GC)= [2vG] iff GETl(V;x,y) where x + y= FZx/-~ 1.

It is clear that for any graph GETl(V;x,y), z(G)=x,

Tl(v; y,x).

In this paper, we determine completely the extremal graphs G for which

A(G)B(GC)=v, where A,BE{x,a,~k} and the extremal graphs G for which z(G)+

~b(G ~) = f2vG1. As a consequence, we have improved the lower bound z(G) + ~(G ¢) >~

F2v~] by 1 for graphs G with at least 11 vertices. As v is a positive integer, there

exists a positive integer m such that m 2 <~v<<,m 2 +2m. Therefore we have:

g(GC)=y and that G~E

2m if v = m 2,

F2x/~]= 2m+l if m2+l<~v<~m2+m,

2m+2 if m2+m+l<~v<~m2+2m.

For v E {m 2, m e + m - 1, m 2 + m, m 2 + 2m}, we have completely determined the solution

set for the graph equation z(G) + ~(G c) = f2v~]. For the other values of v, the deter-

mination of the extremal graphs for the above graph equation appears to be difficult.

2. Some preliminary results

Let GETl(v;x,y) and let there be a dot in the entry (i,j) of Tl. Then we denote the

corresponding vertex by uij when regarded as a vertex of G and by vji when regarded

as a vertex of G c.

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R. Balakrishnan et al. / Discrete Mathematics 186 (1998) 15-24

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Lemma 1. Let x>~3 and y>~2 be any two positive integers and let GCTl(Xy;x,y).

Then ~(G c) = y iff G ~ yKx.

Proof. Clearly, G contains yK~. Suppose that ~(GC)=y and G~yKx. Then in G ~,

there exist two vertices, say, v(i and vkt, i¢k, j ~ I, such that (qj, vkt)~E(GC). Now

color the vertices of G c as follows: Color both the vertices ~j and vkl by cv+l and for

the remaining vertices color, t~n by cm. As x ~> 3, this coloring yields an achromatic

(y + 1)-coloring for G c, contradicting the fact that ~(G ~) = y.

Conversely, if G-~ yKx, then G c ~Ky(x) and therefore ct(G ~) --y.

Next we consider the case when x = 2.

Lemma 2. Let y>.2 be any positive integer and let GETI(2y;2,y). Then ct(GC)=y

iff G is a disjoint union of complete regular bipartite graphs.

Proof. Assume that a(GC)=y. We claim that for any four vertices ult, u2k, Ull and

u2t, the subgraph induced by them in G is isomorphic either to 2K2 or to C4. Otherwise,

the subgraph induced by them in both G as well as G c is P4. Suppose it is ullu2tulkUzk

in G. For i = 1,2 and j = 1,2 ..... y, color vji by c/ except the vertices vkl and vt2, to

which we assign the color Cy+l. This yields an achromatic (y + 1)-coloring for G c, a

contradiction•

We now obtain a new graph H from G as follows: Let V(H)= {ubu2,...,Uy} with

ui joined to uj in H iff the subgraph induced by {Uli, U2i, Ulj, U2j} in G is isomorphic

to C4. Now to show that each component of G is isomorphic to a complete regular

bipartite graph, it is enough to show that H is a disjoint union of complete graphs. If

this is not so, there must exist three vertices urn, Un and Us such that UmUs, UnusCE(H)

and u,~un fIE(H). Now color the vertices of G c as follows: Color vii by cj, where

(j,i)¢(m,2),(s,2),(n,2) and color Vm2, Vs2 and Dn2 by Cr+l, Cm and Cs respectively.

This yields an achromatic (y + 1)-coloring for G c, a contradiction.

I

Conversely, assume that G~-(.Ji_lKr,,r. Since c~(HIVH2)=e(HI)+e(H2), we

• =~i=l e(2Kr~)=~i=lri=Y • [5

have ~(GC)=~(2Kr, V " V2Kr,) - t l

Next, we consider graphs GETl(xy-1;x,y). As there exists exactly one gap in

Tl(xy- 1;x,y), without loss of generality, we can assume that the lone gap occurs at

the position (x, y) of Tl(Xy- 1,x, y).

Lemma 3. Let x>,4, y>~2 be any two positive integers and let GETI(Xy- 1;x,y).

Then e(GC)=y tff G~-(y-1)KxUKx_j or rKxUH(x,y,r), with y>~r + 2, where

H(x,y,r) is the graph of Fig. 1. (Kx-i V {wi} is displayed by Kx-i = {wi}.)

Proof. Suppose that ~(GC)=y. By definition of T1,G contains (y-1)Kx UKx-1 as a

subgraph. If G ~ (y - 1 )Kx UKx-1, then there exist two vertices, say, q~/ and vkl, i ~ k,

j ¢ l, such that (v~j, vkt)~E(GC).

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R. Balakrishnan et al./Discrete Mathematics 186 (1998) 15-24

Kx- 1

I

Kx- 1

• ... •

• . . . •

Kx- 1

Kx- 1

• ... •

n

Wy-1- r

Fig. I. H(x,y,r)

Claim. Both the pairs (i,j) and (k,l) of vertices t~.j and vkt are in the set

S-- {(1,x), (2,x) ..... (y - 1,x), (y, 1),(y,2) ..... (y,x- 1)}.

Suppose not. Then at least one of (i,j) and (k, l) is not in S, say, (i,j). Now color the

vertices of G ¢ as follows: Color both the vertices t~ 7 and vkt by Cy+l and the remaining

vertices Vmn by Cm. As x~>4, this coloring yields an achromatic (y÷ 1)-coloring for

G c, a contradiction.

Hence, by the above claim (i,j)= (i,x) and (k, l)= (y, l) or (i,j)= (y,j) and (k, l)=

(k,x). Without loss of generality, assume the first possibility. We claim that for

any s, l<<.s<~y-1, the vertex V~x is either adjacent or nonadjacent to all of

Vyl, Vy2 ..... Vy(x_l) in G c. Otherwise, there exist two vertices Vyp and Vyq, 1 <~ p, q ~x- 1

and p ¢ q, such that V~x is joined to Vyp and is not joined to Vyq in G c. We color the ver-

tices of G c as follows: Color both Vsx and Vyq by Cy+l and color the remaining vertices

Vmn by Cm. This coloring is an achromatic (y + 1)-coloring for G c, a contradiction.

Permute the rows of Tl(Xy-1;y,x), if necessary, so that every vertex in

{Vlx, Vex ..... Vrx} is adjacent in G ° to all the vertices in Y={Vyl,Vy2 ..... Vy(x_l)} and

every vertex in {V(r+ 1)x,t~r+2)x ..... V(y--1)x} is nonadjacent in G c to all the vertices of

Y. Clearly, such an r exists. Hence, G ~ rKx U H(x, y, r). The existence of nonadjacent

vertices qx and Vyt in G c implies that y- 1 -r~>l i.e., y>Jr+2.

We now prove the converse. If G ~ (y - 1 )Kx UKx-a, then G ° ~-Ky_l(X) V Kx¢_l, and

therefore ~(G °) = y. If G ~- rKx U H(x, y, r) with y >~ r + 2, consider G o in T1 (xy - 1;

y,x). Clearly z(GO)=y and hence ~(G°)>~y. Suppose that ~(G¢)>y and cg is any

optimal achromatic coloring of G °. As the subgraph induced by C1 ={~1:1 <<.i<~y}

in G c is Ky, ~(v/1)¢c-g(Vjl) for i¢j. Assume that ~(t~-l)=Ci. Consider the vertex

vii with 1 ~<i ~<y- 1 and 2 <<.j~x- 1. Since v, 7 is adjacent to all the vertices in C1

except t~l, cg(qj) is either ci or a new color, say, Cy+l. Suppose it is Cy+l, then as

both the vertices v~l and vq are adjacent to all the vertices of G O except the vertices

in Ri = (ql,Vi2 ..... g;x}, the colors ci and Cy+l occur only in c'g'(Ri). But then there is

no edge with color ends ci and Cy+l in cg, a contradiction. Hence, ~(t~).)=ci. Now

consider Vy2, Vy3 ..... Vy(x-1). These vertices receive either the color Cy or a new one.

Suppose some vertex Vyj, 2<<.j<~x- I receives Cy+l, then as cg is achromatic, there

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R. Balakrishnan et al./ Discrete Mathematics 186 (1998) 15-24

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must be an edge in G c with color ends Cy and Cy+l. This implies that either Cy or

Cy+l belongs to (g({Vlx,~x ..... V~x}). But then there will be an edge with color ends

either Cy and Cy or cy+a and Cy+~, a contradiction. Hence, cg({vv2 ..... Vy(x-~)})= {cy}.

Moreover, as the subgraph induced by C~ = {vi~: 1 <<.i<~y- 1} in G ~ is Ky-1, any two

distinct vertices of Cx cannot receive the color Cy+~ and hence there exists exactly one

vertex, say, q:~ in Cx with color Cv+l. But then, in G c there is no edge with color ends

ci and Cy+l, a contradiction. Thus, ~(GC)= y. []

Finally, we consider the case when x = 3.

Lemma 4. Let y~2 be any positive integer and let GcTl(3y-1;3,y).

~(G~)=y iffG~(y-1)K3 UK2, (y-2)K3U(2K2VK~) or (y-2)K3 U Ws.

Then

Proof. Assume that ~(G c) = y.

Claim 1. For icy, j C y and i C j, GC[ { Vil, Vi2, vi3; Vjl, Vj2, vj3 } ] ~ g3,3 .

If this claim were not true, then, without loss of generality assume that (vil,vj2)~

E(GC). Color t~l and vj2 by Cy+l and for the remaining vertices, color ~, by Cm. This

yields an achromatic (y + 1 )-coloring for G c, a contradiction.

Claim 2. For icy, (Vyl,Vi3)¢E(G c) iff (Vy2,~3)EE(GC).

First let (Vyl, Oi3)EE(GC). If (Vy2, vi3)~E(GC), then color both ~v2 and v/3 by Cy+l and

for the remaining vertices, color v~n by Cm. This yields an achromatic (y + 1)-coloring

for G c, a contradiction. Hence (vy2, t~3)EE(GC). The converse part follows by a similar

argument.

Claim 3. For icy, Ht = GC[ { ~il, ~i2, ~i3 ; Vyl,Vy2 } ] '~ K2,3, C4 U KI or 2K2 U K1.

Suppose not, then, by Claim 2, H r is one of the following graphs of Fig. 2, in which

the coloring is marked only for the vertices ~, q2, vi3, Vy~ and Vy2. For the rest of the

vertices Vmn, color them by Cm. This results in an achromatic (y + 1)-coloring in each

of the respective cases, a contradiction.

Claim 4. For i 7 ~ y, j ¢ y and i 7~j, at least one of Gi = GC[{Vil, vi2, vi3; Vyl, Vy2}] and

Gj = GC[ { Vjl, vj2, vj3; Vyl, Vy2 } ] is isomorphic to/(2,3.

Otherwise, by Claim 3, the following three cases arise. In each of these cases first

color the vertex Vrnn by Cm.

Case 4a: Both Gi and Gj are isomorphic to C4UK1, Recolor vi3,vj3 and Vy2,

respectively, by cy, Cy+x and Cy+l.

Case 4b: Both Gi and Gj are isomorphic to 2/£2 U KI. Recolor vii,vii and Vyl,

respectively, by Cy, cy+l and cj.

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R. Balakrishnan et al. / Discrete Mathematics 186 (1998) 15-24

c i

X<

Cy+l Cy

Cy+l c i Cy+l c i c i Cy+l c i

Y

Cy Cy+l

c.

1

Cy Cy+l

Cy+ 1 c i c i c i Cy+l

Cy Cy+l Cy+l Cy

C°

1

Fig. 2.

Case 4c: Gi is isomorphic to Ca UKI and Gj is isomorphic to 2K2 UKI. Recolor as

in case 4a.

In each of the above cases, these recolorings together with Claim 1 result in an

achromatic (y + 1)-coloring for G c, a contradiction.

The above claims ensure that G~(y-1)K3[3K2,

(y -- 2)K3 L3 W5.

We leave the converse as an exercise to the reader.

(y-2)K3[3(2KzVKI)

or

[]

3. Extremal graphs for the products

Theorem 5. For a graph G on v vertices, z(G)~(G °) = v iff G is either isomorphic

to dKv/d, where d is a positive divisor of v, or each component of G is isomorphic to

a complete regular bipartite graph.

Proof. If z(G)ct(GC)=v, then x(G)z(GC)=v and therefore z(Ge)=cz(GC). Fink's

result quoted in Section 1 implies that G is a graph in Tl(v; rid, d), where d is a

positive divisor of v. As G C 2"1 (v; v/d, d), G c C T1 (v; d, v/d). If d = 1, then G is Kv and

if d = v then G is vK1. So assume that 2 <~d <-N v/2. Suppose that 2 <~d <~v/3, then by

Lemma 1, G-~ dKv/d. Suppose that d = v/2, then by Lemma 2, each component of

G is isomorphic to a complete regular bipartite graph. The converse follows from

Lemmas 1 and 2. []

Corollary 5.1. For a graph G on v vertices, ~(G)ct(G c) = v iff G is &omorphic to

either Kv, K~, 2Kv/2, or Kv/2,v/2.

Proof. ct(G)ct(G c) = v implies that z(G)~(G c) = v and ~(G)x(G¢)= v. By Theorem 5,

both G and G c are in {H: H is either isomorphic to dK~/d, where d is a positive

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R. Balakrishnan et al./Discrete Mathematics 186 (1998) 15-24

21

H 1 H 2 H 3 H 4

Fig. 3.

divisor of v, or each component of H is isomorphic to a complete regular bipartite

graph}. Hence G is isomorphic to either K,.,KC,2Kv/2, or Kv/2,~/2. The converse holds

trivially. []

Since ~(K,,/2,~,/2)= 2 and ~(K~,/2,~,/2)= (v/2)+ 1, we have

Corollary 5.2. For a graph G on v vertices, qJ(G)~(G~)=v iff G is isomorphic to

either K~., K c or 2Kv/2.

Corollary 5.3. For a graph G on v vertices, ~(G)~p(GC)=v iff G is isomorphic to

either Kv or K c.

Corollary 5.4. For a graph G on v vertices, z(G)O(G ~) = v iff G is isomorphic to

either Kv, K c or Kv/2,v/2.

Proof. By Theorem 5, G is either isomorphic to dKv/d, where d is a positive divisor

of v, or each component of G is isomorphic to a complete regular bipartite graph.

Since for 2<~d<~(v/2), tP(Kd(v/d))>d=a(Kd(v/d)), we have d= 1 or v. If d= 1 or

v then G is either Kv or K~. If every component of G is isomorphic to a complete

regular bipartite graph, then by hypothesis ~9(G c) = ct(G c) = v/z(G)= v/2. If re(G)~>2,

then G c contains two vertex disjoint copies of Kv/2. Now color the vertices of one

copy Kv/2 of G c by cl, c2 ..... cv/2 and color all the vertices of the other copy Kv/2 of

G ~ by c(r/2)+l. This shows that ~b(GC)>~(v/2)+ 1. Hence ~o(G)= l, i.e., G~Kv/2,v/2.

The converse holds trivially. []

4. Extremal graphs for the sums

We begin with the case when B=~k in A(G)+B(GC).

Theorem 6. For a graph G on v vertices, z(G)+~O(GC) = I2x/~] iff G is one of the

following graphs: K. with n= 1,2,3; Kn,. with n=2,3,4,5; K.,.+I with n=2,3; P.

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R. Balakrishnan et aL /Discrete Mathematics 186 (1998) 15 24

with n=3,5; nK~ with n=2,3; K(2,2,3); K(1,3,3); K2UK1; C4UK~; K3,3UK~;

K2,3-e; K3,4-e; Ws; K~VP4; K~V2K2; Hi with i= 1,2,3,4 as in Fig. 3.

Proof. Assume that z(G)+~b(GC)- - r2v~l. By Nordhaus-Gaddum inequality z(G)+

z(G~)>~ r2x/~]. As z(G~)<<,~b(G¢), our hypothesis implies that z(G)+z(G ~) = [2v/~]

and z(G ~) = O(G~). Now by Fink's result (ii) quoted in the Introduction, G is a graph

in Tl(V; x,y) with x+y= r2x/~]. Further, GE Tl(V; x,y) implies that ;t(G)=x and

z(G¢)=y. We claim that v+ 1 <x+2y. If not, v÷ 1 >>,x+2y. In this case, color the

vertices of the first column of G c in Tl(v; y,x) with Cl,C2 ..... Cy in order, then color

the rest of the vertices of the first row of Tl(v; y,x) with Cy+l and the remaining

vertices of G c with colors cl,c2 ..... Cy such that, for every color ci, 1 <~i<~y, there

exist at least two vertices in G c with color ci. (As v- (x + y- 1)~> y, such a color-

ing is possible.) This yields a pseudoachromatic (y + 1)-coloring for G ~, a contradic-

tion. Now v+ 1 <x+2y implies that v+ 1 ~<4xF and hence v~<13. A search for all

possible pairs (x,y) with x+y= [2x/~], v+ 1 <x+2y, v~<13 and xy~v results in

the collection of graphs that are listed in the statement. The converse can be easily

verified. []

As a consequence of the above theorem, for any graph G with at least 11 vertices,

0(G) + ~J(G c)/> c~(G) + 0(G c)/> z(G) + 0(G c)/> F2 7 ÷ 1.

We now consider the case when B(G c) = ~(G~).

Theorem 7. For a graph G on m 2 vertices, z(G)+~(GC)=2m (= [2x/~]) iff G is

isomorphic to mKm or C4.

Proof. If z(G)+~(G¢)=2m, then by Fink's result (ii), GETl(m2;x,y) with

x+ y=2m. As xy>>,m 2, (x,y)=(m,m). Also ct(GC)=x(GC)=y=m. Ifm~>3, then

by Lemma 1, G TM mKm. If m = 2, then by Lemma 2, G is either 2K2 or C4. If m = 1,

then G is Ks. The converse holds trivially. []

Theorem 8. For a graph G on m2+m-1

is one of the following graphs: mKmUKm-1, m~>4; (m-1)Km+lUKm, m~>3;

rKmUH(m,m+l,r), m>~r+l and m~>4; rKm+lUH(m+l,m,r), m>>.r+2 and

m>>.3;3K3UK2;2K3U(2K2VK1); 2K3UWs; 2K2UK1; P3UK2; C4UK1; Ps; K2.3-e;

K2,3;K3 U K2; (2K2 V K1) and Ws.

vertices, z(G)+c~(Ge)=2m+ I iff G

Proof. If z(G) + c~(G c) = 2m + 1, then G E 7'1 (m 2 + m - 1; x, y) with x + y = 2m + 1.

Since xy>>.m 2 +m - 1, solving for x and y results in (x,y) to be either (m,m+ 1) or

(m+ 1,m).

Case 1: (x,y)=(m,m+ l)

We have c~(GC)=x(GC)=y=m+ 1. If m~>4, then by Lemma 3, G~-mKmUKm-~

or rKmUH(m,m+l,r), with m>~r+l. If m=3, then by Lemma 4, G~-3K3UK2,

2K3U(2K2VK1) or 2K3UWs. If m=2, then G~-2K2UK1; P3UK2; C4UKI; Ps;

K2.3 - e or K2,3. Note that m ~ 1.

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R. Balakrishnan et al. / Discrete Mathematics 186 (1998) 15-24

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Case 2: (x,y)=(m+ 1,m)

If m>~3, then by Lemma 3, G~=(m-1)Km+lUKm or rKm+lUH(m+l,m,r),

m>~r+2. If m=2 then by Lemma 4, G~K3UK2; (2K2 VK1) or Ws.

It is easy to verify the converse. [~

Theorem 9. For a 9raph G on m 2 +m vertices, z(G)+et(GC)=2m+ 1 iff G & one of

the followin9 9raphs: mKm+l with m~>2; (m+ 1)Kin with m~>3; K3,3; C4UK2; 3K2;

K2 and K~.

ProoL If )~(G) + ~(G c) = 2m + 1, then G E/'1 (m 2 + m;x, y) with x + y = 2m + 1. Since

xy>~m 2 +m, solving for x and y we get (x,y) to be either (m+ 1,m) or (m,m+ 1).

Case 1: (x,y)=(m+ 1,m)

If rn>~2, then by Lemma 1, G~-mKm+l. If m= 1, then G is K2.

Case 2: (x,y)=(m,m+ 1)

If m~>3, then by Lemma 1, G~-(m+ 1)Kin. If m=2, then by Lemma 2, G-~K3,3,

C4 LIK2 or 3K2. If m = 1, then G is K~.

Once again, it is easy to check the converse. []

Theorem 10. For a 9raph G on m 2 +2m vertices, z(G)+~(GC)=2m+ 2 (fiG is one

of the followin9 9raphs: mKm+2; mKm+l UKm with m~>3; rKm+l UH(m+ 1,m+ 1,r)

with m>~3 andm~r + l; (m+2)Km with m~>3; 2K3 UK2; K3U(2K2 VK1); K3UWs;

K2UKI; P3; 4K2; 2K2UC4; K2UK3,3; 2C4; K4.4 andK~.

ProoL If z(G) + ~(G c) = 2m + 2, then G E T1 (m 2 4- 2m; x, y) with x + y = 2m + 2. Since

xy>~m2+2m, we see that (x,y) to be either (m+2,m), (m+ 1,m+ 1) or (re, m+2).

Case 1: (x,y)=(m+2,m)

If m ~> 2, then by Lemma 1, G ~- mKm+2. If m = 1, then G ~ K3.

Case 2: (x,y)=(m+ 1,m+ 1)

If m~>3, then by Lemma 3, G~-mKm+l UKm or rKm+l UH(m+ 1,m+ 1,r) with

m >~ r + 1. If m = 2, then by Lemma 4, G ~ 2K3 U 1£2, 1(3 U (21£2 V K1 ) or /£3 U Ws.

If m = 1, then G ~ K2 U KI or P3.

Case 3: (x,y)=(m,m+ 2)

If m~>3, then by Lemma 1, G~(m+2)Km. If m=2, then by Lemma 2, G-~4K2;

2K2UC4; K2UK3,3; 2C4; K4,4. Ifm=l, then G is K~. []

From Theorems 7-10, one can deduce that for any graph G on v vertices,

vE{mZ,mZ+m- 1,m 2 +m,m 2 +2m}, m~>3, ~(G)+ c~(GC)>~ I2v/Vl + 1.

Acknowledgements

This research was supported by Project 401-1 of the Indo-French Centre for the

Promotion of Advanced Research (Centre Franco-Indien Pour la Promotion de la

Recherche Advance).

Page 10

24 R. Balakrishnan et al./Discrete Mathematics 186 (1998) 15-24

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Academic Publishers, London, 1990.

[3] H.J. Fink, On the chromatic numbers of a graph and its complement, in: P. Erdos, G. Katona (Eds.),

Theory of Graphs (Proc. Colloquium Tihany, 1966) Akademiai, Kiodo, Budapest, 1968, pp. 99-113.

[4] F. Harary, S. Hedetniemi, The achromatic number of a graph, J. Combin. Theory 8 (1970) 154-161.

[5] E.A. Nordhaus, J.W. Gaddum, On complementary graphs, Am. Math. Monthly 63 (1956) 175-177.

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