domination parameters ofhypercubes

Journal of the Indian Math. Soc. Vol. 65, Nos. 1-4 (1998), 31-38

DOMINATION PARAMETERS OF HYPERCUBES S.

ARUMUGAM (Received:

and R.

August

KALA

10, 1995)

Let G be a connected graph. Let Y, Yi, Yr, Ye and Yp denote respectively the domination number, the independent domination number, the total domination

number,

domination

number of G. The /I-cube Qn is the graph whose vertex set is

the connected

the set of all /I-dimensional and only n = 2k

if they

differ

domination

boolean in exactly

vectors, one

number

two vertices

coordinate.

and the perfect being joined if

We prove

that if

I, then y(Qn) = y,{Qn) = Yp(Qn) = 2n - k We also prove that y(Qs) = 7, y(Q6) = 12, y(Q7) = 16 and y(Qn) S 2" - 3 for all 11 ? 7. We de-

termine the values ofYi' Y" Yeoand for these parameters for 11 7.

?

I

I.

Yp

for Qn when

11

S 6 and obtain bounds

INTRODUcrlON

By a graph we mean a finite, undirected, connected graph without loops or multiple edges. Terms not defined here are used in the sense of Harary [4]. Let G = (V, E) be a graph. Let S \: V· S is called a dominating set if every vertex in V - S is adjacent to at least one vertex in S. A dominating set S is called a perfect dominating set if every vertex in V - S is adjacent to exactly one vertex in S [2]. A dominating set S is called an independent dominating set if no two vertices of S are adjacent. A dominating set S is called a total dominating set if the subgraph induced by S has no isolated vertices. A dominating set S is called a connected dominating set if the subgraph induced by S is connected. The domination number Yof G is defined to be the minimum cardinality of a dominating set in G. In a similar way, we define the perfect domination number the independellt domination number Yi, the total domination number YI and the connected domination number Ye. A domatic partition of G is a partition of V(G), all of whose classes are dominating sets in G. The maximum number of classes of a domatic partition of G is called the domatic number of G and is denoted by d(G) [3].

Yp'

The n-cube Qn is the graph whose vertex set is the set of all n-dimensional boolean vectors, two vertices being joined if and only if they differ in exactly one coordinate. We observe that QI = K2 and Qn = Qn _ 1 X K2 if n ~ 2. The n-cube has applications in coding theory. It admits of a decomposition into hamiltonian © Indian Mathematical

Society, 1998

32

S. ARUMUGAM

AND R. KALA

cycles if n is even and into a perfect matching and hamiltonian cycles if n is odd. It has been successfully employed in the architecture of massively parallel computers. A dominating set in Qn can be interpreted as a set of processors from which information can be passed on to all the other processors. Hence the determination of the domination parameters of QIl is a significant unsolved problem [6]. Walikar et al. ([ 1], page 51) have conjectured that y(QIl) = 2n - 2 for II ~ 3. PK. Jha [5] has proved the following theorem thus disproving the above conjecture. THEOREM 2n

-II

+

1.1 [5]. Let y(Qn) denote the domination number of Qn' Then

2" :::;y(Q,,) :::; 1 2L logz ("

. If n = 2k

-

1,

then the two bounds coincide and hence

+ I)J

y(Qn) = 2" - k. These bounds are correct also for y;(Qn)' His proof is algorithmic in nature. In this paper we independently prove that if n = 2k

-

1, then y(Qn) = Yi(Qn) = yp(Q,,) = 2"-

y(Q6) = 12, y(Q7) = 16 and y(Q,,):::; 2"-

3

k.

We also prove that y(Qs) = 7,

for n ~ 7. We determine the values of

YI and Yc for n :::;6 and obtain bounds for the same when n ~ 7. We need the following theorems.

yp' y;,

THEOREM 1.2 ([ 1], p. 56). 1'1.,we havey~p/(I'1.+ 1). THEOREM

1.3 ([3]).

THEOREM

1.4 ([7]).

of the dimensioll 2k the dOlllatic number

-

For any graph G of order p alld maximum degree

For any graph G with p vertices d(G) :::;p/y(G). Let k be a positive integer. Theil the graph of the cube

1 and the graph of the cube of the dimellsioll 2k have both 2k.

We use the following notation. By (0) we mean the boolean vector I :::;il < i2< ... < ik:::; II, we denote by (it> i2, ... coordinates

it> i2, ...

ik

with all coordinates O. If the II-tuple having 1 in the

, ik)

and 0 elsewhere. We denote by S" the set of all vertices

i

in Q" having I in exactly coordinates. We observe that IS,I = (~IJ and each vertex of Sj dominates vertices in Si _ 1 and II - vertics in Sj + I.

i

i

2. THEOREM

2.1.

DOMINATION

IN QIl

y(Qs) = 7

Proof: S = {(O), (I, 2), (I, 3), (I, 4, 5), (2, 3, 4), (2, 3, 5), (2, 3, 4, 5)} is a dominating set of Qs and hence y(Qs):::;7. Now let D be any dominating set of Qs. We claim that IDI ~ 7. We consider the following cases.

33

DOMINATION PARAMETERS OF HYPERCUBES CASE 1. If

D

(0) E

S,nD=S2nD=0,

I(S4nD)I~2

and

I(S, u S3) 1'0 DI ~ 3. If I(S, u S3) 1'0 DI = 3. Now, IS3 1'0 DI = 1 or IS, Now, suppose

1'0

S,

If

and

if

SlnD1=0

I(S3nD)I=4

then

S2nD1=0

then

and

I(S, u S3) 1'0 DI > 3, then IDI ~ 7. Suppose 1'0 DI = 1 and IS3 1'0 DI = 2, then at least one

if IS,

of S2 is not dominated

element

I(S3nD)I~4

then

ID)I~7.

hence

by any element

of D. Hence

IS,

1'0

DI = 2 and

DI = 3 and IS3 1'0 DI = 0 and in both cases IDI ~ 7. 1'0

D = 0 and S2 1'0 D

1=

0. If IS2 1'0 DI ~ 3, then obviously

IDI ~ 7. If IS2 1'0 DI = 2, then IS3 1'0 DI ~ 3 and if IS3 1'0 DI = 3, then IS4 1'0 DI ~ 1.

1, then IS3nDI~4

If IS2nDI=

and if IS3nDI=4,

then IS4nDI~

1. Thus

IDI~7. Now,

S,

suppose

1'0

D

0

1=

S3 1'0 D

and

1=

If I(S, u S2) 1'0 D I = 2, then

0.

IS3 1'0 DI ~ 2 and if IS3 1'0 DI = 2, then IS4 1'0 DI ~ 2. If

I(SI u S2) 1'0 D I = 3,

IS3nDI~2

and IS4nDI~

then

either

CASE 2. Proof

is similar

Then IS,

to case 1, by symmetry.

1'0

1'0

If

D

I(S, u S4) 1'0 DI =:3

1

~

then

then

IS2n DI ~ 2

I(S2 u S3) 1'0 DI ~ 3.

If

IS3n DI ~ 2.

and

I(S, u S4) 1'0 DI = 5,

2. Hence in all cases IDI ~ 7 so that y(Q5) = 7.

THEOREM 2.2.

Proof:

e D.

DI ~ 1 and IS41'0 DI ~ 1. If I(SI u S4) 1'0 DI = 2 then IS2n DI ~ 2 and

I(S, u S4) 1'0 DI = 4 I(S2u S3)

I(S3 u S4) 1'0 DI ~ 2 or I(S3 u S4) 1'0 DI = 1

IDI~ 7.

(0) e D and (1, 2, 3,4,5)

IS3 1'0 DI ~ 2.

or

2, 3, 4, 5) E D.

(I,

CASE 3.

IS4n DI ~ 2

1.

If I(S, u S2) 1'0 DI = 4, then either and IS5 1'0 DI = 1. Hence

IS3 1'0 DI = 1 and

If

then

I

y(Q6) = 12.

S = {(O), (1, 3), (2, 3), (l, 2, 4), (l, 5, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6). set of Q6 and

(l, 2, 4, 5), (l, 2, 4, 6), (3, 4, 5, 6), (1, 2, 3, 5, 6)} is a dominating hence

Y(Q6)

$

12. Proceeding

as

in Theorem

y(Q6) = 12 and we omit the details. THEOREM 2.3.

2.1,

it can

be

proved

that

I

y(Q7) = 16.

Proof S = {(O), (l, 2, 7), (l, 3, 4 ), (l, 5, 6), (2, 3, 5), (2, 4, 6), (3, 6, 7), (4,5.7). (I. 2,3,6), (l, 2, 4, 5), (1, 3,5,7), (1, 4, 6,7), (2, 3,4,7), (2, 5, 6. 7). (3. 4, 5, 6). (I. 2. 3, 4, 5, 6, 7)} is a dominating set of Q7 so that y(Q7) $ 16. Alsu by Theorm

1.2 y(Q7) ~ 16 and hence y(Q7) = 16.

COROLLARY 2.4.

y(Q7) = Yp(Q7) = 16.

I

34

ARUMUGAMANDR. KALA

S.

COROLLARY2.5. Proof:

Y(Q,,)

Since y(G x

THEOREM2.6. Proof'

K2)

If n =

$ 2" - 3for all n ~ 7. $ 2y(G), the result follows.

2k -

1, then y(Q,,) = 2" -

k.

It follows from Theorem 1.2 that y(Q,,) ~ 2" - k. Also it follows from

Theorem 1.3 and Theorem 1.4 that y(Q,,) COROLLARY2.7. Proof'

If n =

i-I,

$ 2" - k.

I

then Yi(Q,,)= yp(Q,,) = 2" -

k.

Since Q" is n-regular, any y-set is independent and perfect. 3.

We observe that (1,2,3,

I

PERFECTDOMINATIONIN Q" = 1,

yp(QI)

= Yp(Q3) = 2. Also {(O), (4), (1, 2, 3),

yp(Q2)

4)} is a perfect dominating set of Q4 so that

THEOREM3.1.

Yp(Qs)

I

YP(Q4)

=y(Q4) = 4.

= 8.

Proof: S = {(O), (1), (5), (1,5), (2, 3,4), (1,2,3,4), is a perfect dominating set of

Qs

and hence

Yp(Qs)

(2, 3, 4, 5), (I, 2, 3, 4, 5)}

$ 8.

Now, let D be any perfect dominating set of Qs. We claim that (0) and (1, 2, 3, 4, 5)

E

D. Suppose (0)

(2:

D. Then IS) n DI = 1, IS2n DI = 2 and IS3 n DI = 2.

The remaining elements of S3 cannot be dominated by elements of S4 retaining the property of perfectness. Hence (0) E D. Similarly (1, 2,3,4,5) If ISI

n DI = 0 then

n DI = 0 and the

IS2

E D.

elements of S3 cannot dominate the

elements of S2 retaining perfectness. Hence ISl

n DI ~ 1. Similarly

n DI ~ 1.

IS4

If IS! n DI = 1 = IS4n DI, then IS2n DI = IS3n DI = 0, and hence D is not a dominating set. A similar contradiction arises if IS!

n DI = 1 and IS4n DI = 2 or

if IS! n DI = 2 and IS4n DI = 1. Also if ISl n DI = 1 and IS4 n DI ~ 3 or if

n DI = 1 and IS) n DI ~ 3 then IDI ~ 8. Hence we may assume that IS! n DI ~ 2 and IS4n DI ~ 2 and in this case also IDI~ 8. Hence Yp(Qs) = 8 I

IS4

THEOREM3.2.

Yp(Q6)

= 16.

Proof' S = ((O), (1), (2), (6), (1, 2), (1, 6), (2, 6), (1, 2, 6), (3, 4, 5), (1, 3, 4, 5), (2, 3, 4, 5), (3, 4, 5, 6), (1,2, 3,4,5), (1, 3, 4,5,6), (2, 3,4,5,6), (1, 2, 3, 4, 5, 6)} is a perfect dominating set of

Q6'

Hence

Now, let D be any perfect dominating set of proved that (0) E D, (1,2,3,4,5,6) THEOREM3.3. Proof:

Yp(Qll)

Q6'

YP(Q6)

$ 16.

As in Theorem 3.1, it can be

E D and IDI~ 16.

I

$ 2,,-3 for all n ~ 7.

Follows from Corollary 2.4.

I

DOMINATION

4.

PARAMETERS

INDEPENDENT

35

OF HYPERCUBES

DOMINATION

IN Q"

We observe that Yi(Qd= 1, Yi(Q2)=Yi(Q)) =2. Also

(CO),

(1, 2), (1, 3,4),

(2, 3, 4)} is an independent dominating set of Q4 and hence y(Q4) = Yi(Q4)= 4. 4.1.

THEOREM

Yi(QS)

= 8.

Proof: S = ((O), (1,5), (2, 3), (1,2,4), (1, 3, 4), (2, 4,5), (3, 4, 5), (1,2,3,5)} is an independent dominating set of Qs. Hence Yi(QS)~ 8. Let

Now, suppose there exists an independent dominating set D such that IDI = 7. Obviously, DnX"I'0 and X=SOUS2US4 and Y=SI uS)uSs.

n Y"I' '0. Since IDI = 7, we n Xl = 4 and ID n y] = 3.

D ID

CASE

may assume

without loss of generality

that

1. (0) i: D.

If IS] n DI = IS) n DI =

ISs

n DI = I then IS4n DI = 0 so that IS2n DI = 4 and

in this case at least one element of S) is not dominated by D. If IS] n DI = 1 and IS) n DI = 2, then at least one element of S2 is not dominated. If IS] n DI = 2 and IS) n DI = 1, then at least one element of S4 is not dominated. If IS] n DI = 3 then IS) n DI = be dominated.

ISs

n DI = 0 and hence all elements of S4 cannot

2. (0) E D.

CASE

Hence IS] n DI = O. If IS) n DI = 2 and

ISs

n DI = 1 then IS4n DI = 0 and

IS2n DI = 3 and at least one element of S2 is not dominated. If IS) n DI = 3 and ISs

n DI = 0, at least one element of S2 is not dominated. Hence Yi(QS)= 8. 4.2.

THEOREM

•

Yi(Q6)= 12.

Proof: S = {(1), (2, 3), (4, 5), (5, 6), (1, 2, 5), (1, 3, 5), (2,4, 6), (3,4, 6), (1,2,3,4), (1, 2,3,6), (1, 4, 5, 6), (2, 3, 4, 5, 6)} is an independent dominating set of Q6· Hence Yi(Q6)=y(Q6) = 12. I 4.3.

THEOREM

Proof: x = (XI>

X2,'

Let

x

.•

,X,,-l)

Yi(Q,,)~ 2" - 2for all n ~ 3. be

a maximum

E X, let x' =

can be easily verified that Yi(Q,,)~ 2,,-2 . 5.

Y

independent

(XI> XI>X2,'

.• ,

set of Q" _ I' For X,,_I)'

each

Let Y= {x'ix E X}. It

is an independent dominating set of Q" so that

I

TOTAL DOMINATION

IN Q"

36

S. ARUMUGAM AND R. KALA y/(Q5) = 8.

THEOREM 5.1.

Proof

5 = {(O), (1), (5), (3, 4), (2, 3, 4), (1,2, 3, 5), (1,2,4,5),

is a total

dominating

If

If

151

151 155

n DI ~

1. If

= 2 then

n DI

= 3 or 4 then

Thus in all cases CASE 2. Proof

(l, 2, 3, 4, 5)

let D be any tota

n DI = 1 then n DI ~ 3 and 153 n DI ~ 3. n DI ~ 3 and if n DI = 3, then 1(53 u 54) n DI ~:

151

n DI

152

n DI = 1.

151

Now,

D.

(0) ~

CASE 1.

and

ytCQ5) ~ 8.

set of Q5' We claim that IDI ~ 8.

dominating

Then

set of Q5' Hence

CASE 3.

152

152

n DI ~ 3. If 15, n DI = 5 then

152

n DI ~ 3.

IDI ~ 8.

5) ~ D.

(1,2,3,4,

is similar

152

to Case

1.

(0) E D and (1, 2, 3, 4, 5) E D.

n DI ~

1 and 154

n DI ~ 1.

Then

15,

If 1(5,

u 54) n DI

= 2 then

1(52

u 53)

If 1(5,

u 54) n DI = 3 then

1(52

u 53) nDI

~ 3.

If 1(5,

u 54) n DI

1(52

u 53) n DI

~ 2.

= 4 then

Thus IDI ~ 8 . Hence y,(Q5)

nDI ~ 4.

= 8.

y/(Q6) = 14.

THEORM 5.2.

Proof (2,4,5,6),

5 = {(O), (1), (2), (6), (1, 2), (1, 6), (3, 4, 5), (2, 3, 5, 6), (2, 3, 4, 6) (1, 3,4,5), (1, 2, 3, 4,6), (1, 2, 3, 5, 6), (1, 2, 4, 5, 6») is a Iota

dominating

set of Q6' Hence YtCQ6) ~ 14. Proceeding

verified

that IDI ~ 14 for any total dominating

COROLLARY 5.3.

Proof

5.1, it can hi

set D of Q6'

Y,(Q,,) ~ 2" - 2 - 2/1 - 4 for all /I ~ 7.

Since y/(Q,,) ~ 2y(Q/I_ 1) the result follows.

6. We observe

Proof

5=

dominating

CONNECTED DOMINATION IN Qr

that Yc(Q:) = 1, Yc(Q2) = 2 and Y,(Q}) = 4.

THEOREM 6.1.

Yc(Q4)

= 6.

{CO), (1), (4), (2, 4), (2,3,4),

set of Q4 and

hence

CASE 1. 151

e

D.

n DI ~

1.

(0)

(1, 2, 3, 4») is a connecte( let D be any connecte(

Y,·(Q4) ~ 6. Now,

set of Q4' We claim that IDI ~ 6.

dominating

Then

as in Theorm

I

DOMINATION PARAMETERS OF HYPERCUBES

3,

If ISI n DI ~ IS4

n DI

=

If ISI

n DI = 4,

then

IS2

n

DI ~

2

if IS2 n DI =

and

2

then

37 IS3 n DI ~

2

and

1. then IS2

n DI ~ 2.

Thus IDI ~ 6. CASE

2. (1,2,3,4)

Proof is similar CASE

3.

i: D.

to case 1.

(1, 2,3,4)

(0) E D and

E D.

Then ISI n DI ~ 1 and IS3 n DI?- 1. If I(SI I(SI

U S3)

U S3) n

n DI

DI = 2, then the induced

subgraph

(D) is not connected

and if

~ 3, then IDI ~ 6.

~u~~=6.

•

THEOREM 6.2.

= 10.

Ye(QS)

S = {(O), (1), (2), (1, 2), (1, 5), (1, 3, 5), (3, 4, 5). (l, 3, 4, 5),

Proof:

(2, 3, 4, 5), (1, 2, 3, 4, 5)} Ye(QS)

~ 10. Now,

is a connected

let D be a con;,ected

dominating

dominating

set of Qs. Hence

set of Qs. We claim

that

IDI ~ 10.

CASE 1 (0) Then

IS1

D.

n DI ~

If ISI n DI IS3

e

= 1,

n DI = 3 then

1. then

I(S4 u

IS2

n DI = 2, then If ISI n DI =

IS3

IS4 u

3,

n DI = 2, then

IS2

n DI = 2, then

IS4

IS3 n

DI ~ 3, and if

n DI ~ 3; if

IS2

n DI = 3, then

IS3 n

DI ~ 2 and if

IS2 n

DI ~ 3; if

DI = 3, then

IS3 n

DI ~ 2 and if

IS2 n

n DI ~ 2 and ISsn DI = 1.

If ISI n DI = 4, then IS3

n DI = 3, then

DI ~ 3.

then

IS4

IS2

Ss)n DI ~ 3.

If IS1 n DI = 2, then IS3

n DI ~ 3; if

IS2

n

DI ~ 3; if IS2

n DI ~ 2 and

ISs

n

DI

n DI = 3, then

IS3 n DI ~ 2 and

= I.

If ISI n DI = 5, then IS2 n DI ~ 3, and if IS2 n DI = 3 then IS3 n DI ~ 2. Thus IDI ~ 10. CASE

2. (1,2,3,4,5)

Proof

is similar

CASE

3.

to Case

(0) E D and

e D. I.

(1, 2, 3,4,5)

Then ISI n DI ~ 1 and IS4n If I(SI If I(SI

U Sol) n DI = 2, then U S.~)nDI = 3, then

E D.

DI ~ I. I(S2 IS2 n

U S3)

n DI ~ 8.

DI ~ 2 and

IS3

n DI ~ 3.

if

38

S. ARUMUGAM AND R. KALA

U S4) n DI = 4 then IS2 n DI ~ 2 and IS3 n DI ~ 2. I(SI U S4) n DI = 5, then IS2 n DI ~ 1 and IS3 n DI ~ 2.

If I(SI If

Hence

Yc(Qs)

Thus IDI ~ 10.

I

= 10.

THEOREM 6.3.

Yc(Q6)

=

18.

Proof s= {(O), (2), (3), (1, 3), (2, 3), (2, 6), (1, 4, 5), (1, 5, 6), (2, 4, 6), (4,5,6), (1,2,4,5), (1, 4,5,6), (2, 4, 5, 6), (3,4,5,6), (1, 2, 3, 4, 6), (1,2,3,5,6), (2,3,4,5,6), (1,2, 3,4, 5, 6)} is a connected dominating set of Q6 and hence Yc(Q6) ~ 18. Proceeding as in Theorem 6.2, it can be proved that Yc(Q6) = 18 and we omit the details.

I

THEOREM 6.4.

Proof:

Yc(Qn) ~ 2n - 2 + 4 for all n ~ 7.

Since Yc(Qn) ~ 2Yc(Qn _ I) the result follows from Theorem 6.3

I

REFERENCES [1]

B.D. Acharya, H.B. Walikar and E. Sampathkumar, Recent developments in the theory of domination in graphs, Mehta R~search Institute, Allahabad, MRI Leetllre Notes in Math., I (1979).

[2]

EJ. Cockayne, BL Hartnell, S.T. Hedetniemi and R. Laskar, Perfect domination in graphs, JClSS (to appear).

[3] E.J. Cockayne and S.T. Hedetniemi, Towards a theory of domination in graphs, Networks, 7 (1977), 247-261. [4]

F. Harary, Graph Theory, Addison-Wesley, Reading, Mass, 1969.

[5]

P.K. Jha, Hypercubes, median graphs and products of graphs: Some algorithmic and combinatorial results, Ph.D. dissertation, Department of Computer Science, Iowa State University, 1990.

[6]

R. Laskar and H.B.Walikar. On domination related concepts in graph theory, Proceedings of the international symposium, Indian Statistical Institute, Calcutta, 1980, Lecture notes in Mathematics, No. 885, Springer-Verlag, Berlin, 1981,308--320.

[7]

B. Zelinka, Domination numbers of cube graphs, Math Slovaea, 32(2), (1982),117-119.

Department of Mathematics Manonmaniam Sundaranar University Tirunelveli-627 012, India·