BOUNDS ON THE CHROMA.TIC AND ACHROMATIC NUMBERS
OF COMPLIMENTARY GRAPHS
by
Ram Prakash Gupta
University of North Carolina
Institute of Statistics Mimeo Series No. 577
April 1968
This research was supported by the National
Science Foundation Grant No. GP-5790.
DEPARTMENT OF STATISTICS
University of North Carolina
Chapel Hill, N. C.
ABSTRACT
In the present note, exact upper bounds on the sums of chromatic
and achromatic numbers of complimentary graphs are determined Which prove
in particular a conjecture by Hedetniemi (1966) and imply a bound due to
Nordhaus and Gaddum (1956).
1.
INTRODUCTION
The graphs considered below are assumed to be non-null, finite, undirected
which are to have no loops and no multiple edges.
A graph G consists of a set V(G) of vertices together with a set E(G)
of unordered pairs [u,v] of distinct vertices u, v
E(G) are called edges of the graph G.
u and v are said to be adjacent.
vertices.
E
V(G).
The elements of
If [u,v] is an edge of G, the vertices
The order of a graph G is the number of its
For any S S V(G), the s~bgraph G' of G, induced. by S, is defined
as follows:
[u,v]
€
E(G).
V(G')
=S
and for any u, v
I
S, [u,v] • E(G') if and only if
Two graphs G and G are called complimentary if they have the
same set of vertices and any two vertices are adjacent in one of G or G but
not in both, i.e., V(G)
= V(G)
Consider a graph G and let
and [u,v] E E(G) if and only if [u,v] ~ E(G).
a , ••• , ~ represent k distinct colors.
2
Any function f which assigns to each vertex v of G a unique color
~,
f(v) « (~, ~, ... , ~) is called colorins or more specifically a k-coloring
of G.
If f(v) =
a,
we say that v is colored
a
or that v is an a-vectex.
any S S V(G), f(S) denotes the set of colors (f(v)/v
I
S).
For
Any k-coloring
f of G induces a decomposition of V(G),
where Vi is precisely the set of all ai-vectices.
Conversely, any decom-
position (1.1) of V(G) induces a k-coloring f of G such that f(v)
ever v e Vi' l.s i .s k.
= 0i
when-
Thus, there is a natural 1-1 correspondence between
k-colorings of G and decompositions of V(G) into k mutually disjoint sets.
In the present note, we shall consider k-colorings f of a graph G with
(1.1) as the induced decomposition of V(G) which satisfy one or both of the
following two conditions:
2
(R)
For any two vertices u and v of G, [u, v] e E(G) implies
feu) 1= fey) or, equivalently, for each i, 1 ~ i ~ k, u,
V
€
Vi
implies u and v are not adjacent.
(c)
For any two colors a i and a j , 1
[u, v]
E
E(G) with feu) = a
each i and j, 1
~
i < j
i < j < k, there is an edge
~
i
and fey) = a
S k,
there exist
j
or, equivalently, for
vertices u E Vi' v E Vj
such that u and v are adjacent.
A k-coloring f of G is called
re~J
pseu.io-coDJIllete or complete
according as it satisfies the condition (R), (C) or both (R) and (C),
respectively.
The
chromatic number of G, denoted by X(G), is the minimum
number k for which a regular k-coloring of G exists.
The pseudo-achromatic
number of G, denoted byts{G), is defined to be the maximum number k for
which a pSE!l'do-complete
k~coloring
of G exists •
Finally, the achromatic
number of G, denoted by t(G), is the maximum number k for which a complete
k-coloring of G exists.
From the definitions, it is obvious that for any
graph G, t(G) < t (G). Also, if X(G) = k and f is any regular k-coloring
- s
of G, then it is easily seen that f must also be pseudo-complete so that
clearly XiG) S "(G).
(1.2)
Hence, for any graph G, we always have
X,.{G) S "(G) ~
"8 (G).
Further, there exist graphs G for which the strict inequalities X{G) < "(G)
and/or "(G) <
"8 (G) may hold.
For instance, if G is the graph of Figure 1,
then, it is easily verified that X(G)
Figure 1.
= 2,
"(G)
= 3 and
"s(G)
= 4.
3
From the above observations, it is evident that the concept of pseudoachromatic number of a graph as defined here is a proper generalization of
the concept of achromatic number of a
gr~ph,
introduced by Hedetniemi [1].
Let G and G be complimentary graphs defined on a set of p vertices.
Then it is known [2] that
~
(1.3)
S X(G)
+ x'(G}
sp
+ 1.
In the present note, we determine the following upper bounds:
leG) + ts(G) S p + 1,
ts(G) + ts(G) S
(~p).
The bounds (1.4) and (1.5) are shown to be exact.
As a corollary to (1.4),
we obtain the bound
(1.6)
x( G) + t(G}
Sp
+ 1,
conjectured by Hedetriemi [1] and also the upper bound in (1.3) due to
Nordhaus and Gaddum [2].
From (1.5), we obtain the (exact) upper bound
t(G) + t(G) S
(1.7)
(~p)
which answers a qUestion by Hedetniemi [1] in the negative.
2.
In the following,
B<X1.NDS
IA I denotes, as usua1, the number of elements in the
set A; [x] denotes the integral part of the number x and (x) is the smallest
integer greater than or equal to x.
we first prove the following
Theorem 2.1.
If G and G are complimentary graphs of order p, then
X,(G) + t s (G) S p + 1,
x·(G) + t(G) S p + 1,
X(G) + ,X(G) S p + 1.
4
For any p ~ 1, the bounds (2. 1 ), (2.2) and (2.3) are attainable.
Proof:
We shall first prove (2.1).
The bounds (2.2) and (2.3) then
follow inmediately from (2.1) and (1.2).
Let G and G' be complimentary graphs of order p, and let t (G) = k.
s
Obviously, 1 =:: k =:: p. If k = 1, then since clearly X(G) =:: p, we have the
inequality (2.1).
We may therefore assume that k
> 1. Now, consider
any
pseudo-complete k-coloring of G' and let
F
= V1UV2U •••
be the induced decomposition of V(G).
UVk , VrnVj =., i
j,
Let G denote the sUbgraph of G
induced by the set of vertices V(Gr ) -.
vl Uv2U
(2.4)
V(G) = V(G')
r = 1, 2, ••• , k.
r
••• UV
r and let IV(Gr )
1= p r ,
We assert that x(Gr ) =:: Pr - r + 1.
To prove the assertion, it is evidently sufficient to show that G
r
possesses a (Pr - r + l)-coloring f r which is regular.
we may define f
Now, for r
= 1,
by assining Pl distinct colors to the vertices of G so
l
that any two vertices are colored differently. Let us assume as induction
l
hypothesis that we have already defined a (fixed) regular (Pr - r + l}coloring f r of G for some r, 1 =:: r < k. Now consider the graph G + •
r
r l
Since the decomposition (2. 4 ) is induced by a pseudo-complete coloring of
G,
by definition it is clear that there exist vertices v
l e Vl ' v2
E
V2 '
••• , V E V such that for some vertices u i • V +1 (1 < i < r) which need
r
r
r
-not all be distinct, [vi,ui ] € E(G) so that clearly [vi'u i ] ;. E(Gr + ).
l
Now, we define a coloring f r +l of Gr +l by using the coloring f r of Gr.
If
each of the colors fr(V ), f (v ), ••• , fr(V ) is also assigned by f to
l
r
r 2
r
some vertex in V(Gr ) - (Vl'v , ... ,vr ), then evidently the number of colors
2
actually used by f r does not exceed IV(Gr ) - (Vl ' V2 ' ... , vr ) I
In this case, we define f +
= pr
- r.
r l as follows: assign Ivr+ll =Pr+l-Pr new colors
5
to the vertices in Vr +l and let fr+l(V) = frey) for the rest of the vertices
v eV(Gr +1).
And" if there is a color ex among f r (vl )" f r (v2 )" ... " f r (vr )
which is not assigned by f r to any vertex in V(Gr ) - (Vl "v2" ••• "Vr ) so that
all the Cl-vertices Vi "vi" ... " V. (1 =:s t =:s r)" say" are among
1 2
J. t
v l " v2" ... " vr " then" we define f r +l as follows: assign Pr+l - Pr new
colors to the vertices in Vr+1; delete the color
f r +l (vi )
j
= f r +l (uij )"
1 =:s j =:s t" where
U
ij
E
a by putting
Vr +l is such that
] ~ E(Gr +l ); for all other vertices veV('\.+l)" let fr+l(V) = freY).
ij
In either case" clearly" f r +l is a (Pr+l - r+r + l)-coloring of Gr +l " and
[vi/ u
it is easily verified that f + is regular.
r l
by finite induction.
In particular" .since Gk
The assertion is now proved
= G"
we have X(G) =:s p-k+l
from which we obtain (2.1) iDmediately.
To see that the bounds (2.1)-(2.3) are attainable" it is sufficient
to let G be a conplete graph of order p" i.e." G has p vertices each pair
of which is adjacent in G so that no two vertices are adjacent inG.
clearly leG)
hold·
= p"
'X(G)
= .(G) = .s(G)
Then"
= 1 and the equalities in (2.1)-(2.3)
This completes the proof of the theorem.
We shall now prove the following
Theorem 2.2:
If G and G are comp]jmentary graphs of order p" then
(2.5)
.s (G) + .s (G) =:s
(2.6)
.(G) + .s (G) =:s
(2.7)
.(G) + .(G)=:s
~ p}"
~ p}"
(~ p).
For any P ~ 1" the bounds (2.5)" (2.6) and (2.7) are attainable.
Proof:
We shall first prove (2.5).
follow 1Jrmediately from (2.5) and (1.2).
The bounds (2.6) and (2.7) then
6
Let G and a be complimentary graphs of order p, and let "'s(G) = k.
Obviously, 1 =s k =s p.
inequality (2.5).
=s p, we have the
I f k = 1, then since clearly "'s(G)
We may therefore assume that k
> 1. Now, consider
any
pseudo-complete' k-coloring of G and let
(2.8)
V(G)
= v(a) = V1LN2U... UVk,
vinvj
be the induced decomposition of Vea,. C'learly, Vi
= .,
F•
i
F j,
for i = 1,
2, ..., k.
Let the number of sets among Vl' V2' ... , Vk which consist of exactly one
vertex each be r where r ~ O. To be definite, we may assume that Vi = (vi ),
i = 1, 2, ... , r (if r
> 0). Then, the remaining sets Vr+l' ... , Vk consist
of at least two vertices each.
Since the sets Vi are mutually disjoint,
we have evidently
Now, let "s(G) = k' and f be any pseudo-complete k'-coloring of
Since the decomposition (2.8) is induced by a
ps~do-complete
G'.
coloring of
G, it is observed that each pair of vertices in(vl , v ' ••• , vrl must be
2
adjacent in G so that no two vertices in (Vl ' v2' ••• , Vr ) are adjacent in
a.
Therefore, since f is a pseudo-complete coloring of
G', it is easily seen
that there can be at most one color in (f(Vl ), f(v2 ), ••• , f(v,.)} which is
not assigned by f to any vertex in V(G) - (Vl '
2' ••• , vr ).
V
Hence, we
have evidently, k' =s Iv(a) - (Vl ' v2' ... , vrJl + 1, or
(2.10)
k'
=s p
-
r + 1.
We shall next prove the following inequality.
(2.11)
k' =s p -
~].
To this end, consider the sets of colors f(Vl ), f(V2 ), ... , f(Vk ).
If for each index i, 1 =s i =s k, we have either If(Vi) .1< IVi I or f(vi)nf(Vj)F.
for some j, j F i, then it is easily seen (by induction on k) that
e,
7
k
k
k
k
k' =Ii~l f(Vi)I:sli~l vii - (2 )= p -( 2) and hence, a fortiori. we have the
inequality (2.11).
We may therefore assume that f(Vl ), say, is such that
If(Vl ) I = Ivll = Pl and f(Vl )
n f(Vj )
= • for 2 :s j :s k.
Let, if possible,
there be another set f(V2 ), say, such thatlf(V2) ,I, = IV21, and f(V2 ) n f(V j ) = •
for j ~ 2. Now, since there exist vertices vl • Vl ' v2 E V2 such that
[v ,v ] E E(G) or [v ,v2 ]J E(G), it is seen that there can be no edge
l
l 2
[v,u] E E(G) with fey) = f(vl ) and feu) • f(v2 ). This, however, contradicts
the fact that f is a pseudo-complete coloring of G.
Hence, we must have
for every index i, 2:S i :s k, either If(Vi ) I < IVi I or f(Vi )
some j, j
k
~ i. Hence, as above,
"
"
we have k' - Pl
"k
n f(V j )
~ • for
k
=Ii~ f(Vi)I:sli~Vi,-{k~l) =
"
.. .
P-Pl-C[] whence we obtain (2.11) inmediately.
Now, we shall derive (2.5) from (2.9), (2.10)
and (2.11).
If
r ~ [~] + 1, then from (2.9) and (2.10), we obtain .s(G) + ts(G) =
k + k' :s P +
[p~r]
+ 1 :s
~ pl.
(The last inequality is obtained by substituting
r = [~] + 1 and some elementary simplification.)
(2.9)
If r :s [J] + 1, then from
and (2.11) we obtain similary .s(G) + .s(G) :s
~ pl.
This completes
the proof of (2.5).
It now remains to show that. for any p ~ 1, the bounds (2.5)-(2.7) are
attainable,. For p :s 3, this is obvious.
In general, it is evidently
sufficient if we show that the bound (2.7) is attainable for all p of the
form 3r+l, r=l, 2, ....
To this end, we construct below examples of graIbs
G of order 3r+l successively for r = 1, 2, ••• and define (2r+l)-colorings
r
f r and £r of Gr and Gr , respectively.
follows:
V(Gl ) = {vO,vl ,v2 'v },
3
For r = 1, we define Gl' fl' 1'1 as
E(Gl ) = ([vO'Y2]' [vl ,v ], [v2 ,v3 ]);
3
f l (vo) = f l (vl ) = ell' f l (v2 ) =~, f l (v3 ) =~; 11 (vO) = 0:1.' 1'1(vI) = ~,
£1(v ) = £1(v )
SUppose that we have already defined Gr , f r , 1'r for
2
3
=.,.
8
same r ~ 1.
Now, we define Gr +l , f r +l , f r +l as follows:
V(Gr ) U (V3r+l ' v3r+2' v3r+3 ),
E(Gr +l )
U ([v3r+2' v3i ] ,I i = 0, 1,' ... ,
r+1~
i
= 1,
= E(Gr )
= ~r+3'
f r +l (vi)
= fr(v i )
U ([v3r+l ' v3i - 1 1i
= 1,2, ••• ,r)
U ([v3r+3 ' v3i _l ], [v3r+3 ' v3i ] 1
2, ••• , r) U ([v3r+3 , v1 ]); f r +1 (v3r+1 )
f r +1 (v3r+3 )
V(Gr +l ) =
= f r +l (v3r+2) = CX2r+2'
for i ::: 0, 1, ••• , 3r;
f r +1 (v3r+l ) ::: ~+2' f r +1 (V3r+2) = f r +l (v3r+3 ) = Cl2r+3, f r +1 (Vi) = fr(Vi )
= 0,
It is easily checked that f r and f r are complete
(2r+l)-colorings of Gr and Gr , respectively. Hence, by definition, we
for i
1, ••• , 3r.
have t(Gr ) + t(Or)
~ 2(2r+1)
=
~3r+l)}.
But, tr'om (2.7), the equality
t(Gr ) + t (Or) =( ~3r+l)} must hold.
This completes the proof of the theorem.
Remark:
The research work presented in this note was motivated by
the suggestions made by Hedetniemi [1].
complimentary graphs G and
a of order p,
He conjectured that for any
we have: (1) t(G) +
and further, asked the question: (2) t(G) + t(G)
,(G) S p+l,
S p+2? Clearly, Theorem
2.1 yields a stronger result than (1) and Theorem 2.2 answers the question
(2) in the negative.
9
REFERENCES
[1]
Hedetniemi., S. T.,
Homomorphism of Graphs and Automata, Tech. Report,
The University of Michigan, Ann Arbor (1966), pp. 24-25.
[2]
Nordhaus, E. A. and Gaddum, J. W.,
On Complimentary Graphs, Amer.
M:Lth. Monthly, vol. 63 (1956), pp. 175-177.