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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
19
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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