SIGNED DOMINATION NUMBER AND MINUS ...shodhganga.inflibnet.ac.in/bitstream/10603/21209/9...domination number and signed domination number, in which we will use some of the results

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S I G N E D D O M I N AT I O N N U M B E R A N D M I N U S

D O M I N AT I O N N U M B E R O F T H E

G E N E R A L I Z E D P E T E R S E N G R A P H

32

2.1 introduction 33

2.1 introduction

The Petersen graph is one of the most useful graph in the theory

of graphs because it serves as a useful example and counterex-

ample for many problems in graph theory. The Petersen graph

is named for Julius Petersen, who in 1898 constructed it to be

the smallest bridgeless cubic graph with no three-edge-coloring.

In 1993, D. A. Holton from University of Otago, New Zealand

and J. Sheehan from University of Aberdeen published a book

[19] on Petersen graph. In 1950, H. S. M. Coxeter [5] introduced

the generalized Petersen graph family, and in 1969, M. Watkins

[26] studied the properties of generalized Petersen graphs.

For any two positive integers n and k, the generalized Pe-

tersen graph P(n, k) has its vertex set V(P(n, k)), which is the

union of the sets

U = {u1, u2, u3, . . . , un} and W = {w1, w2, w3, . . . , wn}

2.1 introduction 34

and its edge set is given by

E = {wiwi+1, wiui, uiui+k}, 1 ≤ i ≤ n,

where subscripts are reduced modulo n. The Petersen Graph

P(5, 2) is shown in the Figure 2.1.

Figure 2.1: The Petersen graph

Properties of generalized Petersen graphs:

1. P(n, k) and P(n, n− k) are isomorphic.

2. The generalized Petersen graph P(n, k), for n ≥ 3 and

1 ≤ k ≤ ⌊(n − 1)/2⌋ is a graph consisting of an inner

star polygon {n, k} (circulant graph Cn(k)) and an outer

regular polygon {n} (cycle graph Cn) with corresponding

vertices in the inner and outer polygons connected with

edges. P(n, k) has 2n nodes and 3n edges.

3. It is bipartite if and only if n is even and k is odd.

2.1 introduction 35

The Figure 2.2 shows a generalized Petersen graph P(8, 3).

Figure 2.2: P(8, 3)

Generalized Petersen graphs have a vast number of uses

and applications in telecommunication networks, VLSI design,

parallel and distributed computing and the main interest in gen-

eralized Petersen graphs lies in the role they play in the design

of networks. In this chapter, we fined signed domination num-

ber of the generalized Petersen graph P(n, k) for k ∈ {1, 2, 3},

sharp lower bounds for odd k ≥ 5, upper bound for even k and

some general results for cubic graphs. Also, a relation between

signed domination function and an efficient dominating set

have been established and minus domination number of P(n, 1)

have been calculated. Moreover we have shown that P(n, 2)

does not have any efficient minus dominating function.

2.1 introduction 36

Here we are listing some of the preliminary results of minus

domination number and signed domination number, in which

we will use some of the results in this chapter.

Theorem 2.1.1. [9] A minus (signed) dominating function f on a

graph G is minimal if and only if for every vertex v ∈ V(G) with

f (v) ≥ 0, there exist a vertex u ∈ N[v] with f (N[u]) = 1.

Theorem 2.1.2. [9] If T is a tree of order n ≥ 4, then γ(T) −

γ−(T) ≤ (n− 4)/5.

Theorem 2.1.3. [9] If G is a graph with maximum degree ∆(G) ≤ 5,

then γ−(G) ≥ 0.


then γ−(G) ≥ 1.

Theorem 2.1.5. [20] If G is a bipartite graph of order n, then

γ−(G) ≥ 4(√

n + 1− 1)− n.


then γs(G) ≥ n/3.

Theorem 2.1.7. [9] For every r-regular graph G of order n, γs(G) ≥

n/(r + 1) and this bound is sharp.

2.2 an embedding theorem 37

Theorem 2.1.8. [18] For every k-regular graph G of order n with k

odd, γs(G) ≥ 2n/(k + 1) and this bound is sharp.

Theorem 2.1.9. [6] Let G be a graph on n vertices. Then γs(G) = n

if and only if every vertex is either an end-vertex or adjacent to an

end vertex.

2.2 an embedding theorem

Before going to the main part, we will prove the result that a

graph G with SDF f can be embedded into a graph H having

MSDF.

Theorem 2.2.1. Let G be a graph with SDF f . Then, either f is a

MSDF or there exists a supergraph H of G with MSDF F such that

F(u) = f (u) for all u ∈ V(G).

Proof. If f is a MSDF then nothing to prove. So, let us assume f

is not a MSDF. If f (v) = +1; then by adding a pendent vertex

v′

corresponding to each vertex v in G we obtain a supergraph

H(say). Now, define F on H such that F(u) = f (u), for all

u ∈ V(G) and F(v′) = +1, for all v

′ ∈ (V(H) \ V(G)). Firstly,


we will show that F so defined is a SDF of H.

If v ∈ (V(H) \ V(G)) then v is a pendent vertex, in H. There-

fore,

F(N[v]) = F(v) + ∑u∈N(v)

F(u)

= F(v′) + f (v)

= 1 + 1

≥ 2.

If v ∈ V(G) then,

F(N[v]) = F(v) + ∑u∈N(v)

F(u)

≥ 1 i f f (v) = −1

≥ 2 i f f (v) = +1.

Thus, F(N[v]) ≥ 1, for all v ∈ V(H), whence F is a SDF of

H. Next, we will show that F is indeed a MSDF of H. On the

contrary, suppose F is not a MSDF on H. Then, by definition,

there exists a SDF F′on H such that wt(F

′) ≤ wt(F) and F

′(v) =


f (v), for all v ∈ V(G). This implies that there exists a vertex

v′ ∈ (V(H) \V(G)) such that F

′(v

′) = −1, whence,

F′(N[v

′]) = F

′(v

′) + F

′(v)

= F′(v

′) + f (v)

= −1 + 1

= 0.

This contradicts the fact that F′

is a SDF. Thus, F must be a

MSDF of H.

One may read Theorem 2.2.1 to have a measure that gives

an idea about how close a given SDF of G to being a MSDF by

minimizing the order and size of the host graph that contains

the labelled graph (G, f ) as an induced subgraph and extension

of f as a MSDF F of H. For specific graphs G, it may lead

further to develop an algorithm to possibly yield a MSDF of G

itself.

2.3 efficient dominating set and sign dominating function 40

2.3 efficient dominating set and sign dominating

function

For any MSDF f of a graph G, let Pf = {v ∈ V : f (v) = 1} and

M f = {v ∈ V : f (v) = −1}. In view of definition of SDF, one

can observe that if G is a cubic graph and f is a MSDF on G

then M f is a 2-packing set. Rubalcaba et al. [23] proved that, for

a regular graph G, ρ(G) = γ(G) if and only if G has an efficient

dominating set, where ρ(G) is an 2-packing number of G. In

this section, we give a relation between efficient dominating set

and MSDF, by first proving the following lemma.

Lemma 2.3.1. Let f be a SDF of a graph G having the property

f (N[u]) = ∑∀v∈N[u]

f (v) =

1 i f deg(u) is even;

2 i f deg(u) is odd.

Then f is a MSDF of G.

Proof. We will prove the result by contradiction, so let us sup-

pose if possible f is not a MSDF. Then there exists a SDF


g : V → {−1, 1} s.t. wt(g) < wt( f ), with at least one vertex v in

G satisfying −1 = g(v) < f (v) = 1. Without loss of generality,

assume that g(v) = −1 and g(u) ≤ f (u) for all v 6= u ∈ V.

By hypothesis, f (N[v]) is either 1 or 2 for all v ∈ V. Hence, if

N[v] = {v, v1, v2, . . . , vk} then we get

g(N[v]) = g(v) + g(v1) + g(v2) + · · ·+ g(vk)

≤ g(v) + f (v1) + f (v2) + · · ·+ f (vk)

= −1 + f (v) + f (v1) + f (v2) + · · ·+ f (vk)− f (v)

= −1 + f (N[v])− f (v)

= −1 + f (N[v])− 1

≤ 0,

which is a contradiction to the assumption that g is a SDF.

Therefore, f must be a MSDF.

Theorem 2.3.2. Let G be a graph with minimum degree 2 and maxi-

mum degree 3. If there exists an efficient dominating set D then, there

exists a MSDF f such that M f = D.


Proof. Let D be an efficient dominating set of G. Then, every

vertex u ∈ V(G) is adjacent to exactly one vertex of D, i.e.,

|N[u] ∩ D| = 1 and D is an independent set in G. Now, define

a function f : V → {−1, 1} such that,

f (u) =

−1 i f u ∈ D;

1 otherwise.

If u ∈ D, then u must be adjacent to at least two vertices of

V − D, as G is a connected graph with minimum degree 2 and

maximum degree 3 together with an independent set D. Also

f (vi) = +1, ∀ vi ∈ V − D, thus we have

f (N[u]) = ∑v∈N[u]

f (v) = f (u) + ∑v∈N(u)

f (v)

= −1 + 1 + 1

=

1 i f deg(u) is 2;

2 i f deg(u) is 3.


Now if u ∈ V − D then u is adjacent to exactly one vertex v1

of D and at least one vertex v2 of V − D. Also f (v1) = −1 and

f (v2) = +1, thus we have,

f (N[u]) = ∑vi∈N[u]

f (vi) = f (u) + ∑vi∈N(u)

f (vi)

= 1− 1 + 1

=

1 i f deg(u) is 2;

2 i f deg(u) is 3.

Therefore, f is a SDF. By making use of Lemma 2.3.1 it

follows that f is a MSDF.

The Figure 2.3 is an example of a graph with M f = D

Figure 2.3: The vertices of M f are marked in dark.

2.4 sign domination number of p(n, k) 44

2.4 sign domination number of p(n, k)

In [10], Ebrahimi et al. proved the necessary and sufficient

condition for the generalized Petersen graph to have an efficient

dominating set.

Lemma 2.4.1. [10] If P(n, k) has an efficient dominating set, then

γ(P(n, k)) = n2 and n ≡ 0(mod 4).

Theorem 2.4.2. [10] The generalized Petersen graph P(n, k) has an

efficient dominating set if and only if n ≡ 0(mod 4) and k is odd.

In view of Theorem 2.3.2 one can conclude,

Lemma 2.4.3. If P(n, k) has an efficient dominating set, then |M f | =

n2 and n ≡ 0(mod 4).

Theorem 2.4.4. In generalized Petersen graph P(n, k), |M f | = n2 , if

and only if n ≡ 0(mod 4) and k is odd.

In the following section we will calculate sign domination

number for three classes of generalized Petersen graphs P(n, k),

k = 1, 2, 3.


2.4.1 Let k=1

The following theorem determines the value of γs(P(n, 1)).

Theorem 2.4.5. If n ≥ 3, then

γs(P(n, 1)) =

2n− 2(n2 − 1) i f n ≡ 2(mod 4);

2n− 2⌊n2⌋ otherwise,

where ⌊x⌋ denotes the largest integer not larger than the real number

x.

Proof. If f is a MSDF of any graph G then

γs(G) = |Pf | − |M f | = n− 2|M f |.

Firstly, we will show that, for P(n, 1),

|M f | =

n2 − 1 i f n ≡ 2(mod 4);

⌊n2⌋ otherwise.

In order to show the lower bound, it is sufficient to find a SDF

g with |Mg| having the value mentioned in the above Lemma


2.4.3. Let l = ⌊n4⌋ and n ≡ r(mod 4), that is, n = 4l + r. Define g

such that

Mg =

{w4i+1, u4i+3 : 0 ≤ i ≤ l − 1}, i f r = 0, 1, 2;

{w4i+1, u4i+3 : 0 ≤ i ≤ l − 1} ∪ {wn−2}, i f r = 3.

And let Pg = V(P(n, 1)) \Mg.

It is easy to see that g is a SDF with

|Mg| =

n2 − 1 i f n ≡ 2(mod 4);

⌊n2⌋ otherwise.

For the upper bound, it is obvious that |M f | ≤ n2 . For n ≡

2(mod 4) there does not exist any efficient dominating set. There-

fore, |M f | ≤ n2 − 1. From the above two inequalities we get

|M f | =

n2 − 1 i f n ≡ 2(mod 4);

⌊n2⌋ otherwise.

Therefore,

γs(P(n, 1)) =

2n− 2(n2 − 1) i f n ≡ 2(mod 4);

2n− 2⌊n2⌋ otherwise,


which completes the proof.

The Figure 2.4 shows the assignment of the vertices of M f

and Pf of P(8, 1) and P(10, 1).

Figure 2.4: P(8, 1) and P(10, 1)

2.4.2 Let k=2

Before going to the main theorem of this section we will prove

some lemmas for P(n, 2).

Let l = ⌊n7⌋ and n ≡ r(mod 7). i.e., n = 7l + r.

Lemma 2.4.6. If f is a MSDF of P(n, 2) where n ≥ 7 and n 6= 9, 12

then M f * W and M f * U.

Proof. We will prove the result by contradiction. Suppose f is

a MSDF of P(n, 2) and M f ⊆ W(or M f ⊆ U). Since P(n, 2) is a


cubic graph, W (or U) induces a cycle of length n (if n is even

then U induces two cycles of length n2 ). Also M f is a 2-packing

therefore, d(vi, vj) ≥ 3, ∀ vi, vj ∈ M f , whence |M f | = ⌊n3⌋. Now,

define a function g on P(n, 2), in such a manner that

Mg =

S i f r = 0, 1, 2;

S ∪ {wn−2} i f r = 3;

S ∪ {wn−3} i f r = 4;

S ∪ {wn−4} i f r = 5;

S ∪ {wn−5} ∪ {un−2} i f r = 6,

where S = {w7i+1, u7i+4, u7i+5 : 0 ≤ i ≤ l − 1}. And let Pg =

V(P(n, 2)) \Mg.

It is easy to verify that g is a SDF with

|Mg| =

3l = ⌊3n7 ⌋ i f r = 0, 1, 2;

3l + 1 = ⌊3n7 ⌋ i f r = 3, 4;

3l + 1 = ⌊3n7 ⌋ − 1 i f r = 5;

3l + 2 = ⌊3n7 ⌋ i f r = 6.


|Mg| =

⌊3n7 ⌋ − 1 i f n ≡ 5(mod 7);

⌊3n7 ⌋ otherwise.

This gives |M f | < |Mg|, for all n ≥ 7 and n 6= 9, 12. However,

this contradicts our assumption that f is a MSDF. Hence the

result.

If f is a MSDF of P(9, 2) and P(12, 2), then clearly |M f |

corresponding to P(9, 2) and P(12, 2) is 3 and 4 respectively.

Hence there exists a MSDF f such that M f ⊂ W or M f ⊂ U.

The Figure 2.5 shows the assignment of +1 and −1 to vertices

of P(n, 2).

Lemma 2.4.7. If f is a MSDF of P(n, 2) then V(P(n, 2)) \N[M f ] 6=

φ.

Proof. We will prove this lemma by contradiction. Let us assume

that v is either in M f or in N(M f ), for all v ∈ V. Consider a

vertex wi /∈ M f then by assumption, wi ∈ N(M f ). This means

one of ui, wi−1, wi+1 is in M f . Now we have three cases:

Case 1: Suppose wi+1 ∈ M f . Since M f is a 2-packing then

all the vertices at distance 2 from wi+1 belongs to Pf , whence


n = 7r

(a)

w0

wn−

1

n = 7r + 1

(b)

w0

wn−

1

n = 7r + 4

(e)

w0

wn−

3

n = 7r + 5

(f)

w0

wn−

4

n = 7r + 6

(g)

w0

wn−

5

un−

2

Figure 2.5


f (wi−1) = f (wi) = f (wi+2) = f (wi+3) = f (ui−1) = f (ui) =

f (ui+1) = f (ui+2) = f (ui+3) = +1.

Since wi+3 ∈ N[M f ] then, f (wi+4) = −1 and we get f (wi+5) =

f (wi+6) = f (ui+4) = f (ui+5) = f (ui+6) = +1. But then ui+2, ui+3

/∈ N[M f ] (see Figure 2.6(a)), a contradiction to the assumption

that v ∈ N[M f ] for all v ∈ V.

Case 2: Next suppose wi−1 ∈ M f . Proof of this case is given by

similar arguments those used in case 1 as wi−1 and wi+1 are

symmetric to wi (see Figure 2.6(b)).

Case 3: Suppose ui ∈ M f . Then f (wi) = f (wi−1) = f (wi−2) =

f (wi+1) = f (wi+2) = f (ui−2) = f (ui−4) = f (ui+2) = f (ui+4) =

+1. Since wi+1 ∈ N[M f ] then, f (ui+1) = −1, f (ui+3) = f (ui+5) =

f (ui−1) = f (ui−3) = f (wi+3) = +1, this gives wi−1, wi+2 /∈

N[M f ] (see Figure 2.6(c)), a contradiction. Hence in above

three cases, we arrive at a contradiction to the assumption

that v ∈ N[M f ], for all v ∈ V, thus our assumption is wrong.

Hence V(P(n, 2)) \ N[M f ] 6= φ.

In Figure 2.6 the vertices of M f are in dark, grey vertices in

Pf and the vertices in doted box are in V(P(n, 2)) \ N[M f ].


Figure 2.6

Lemma 2.4.8. If f is a MSDF of P(n, 2) then

|M f | ≥

⌊3n7 ⌋ − 1, i f n ≡ 5(mod 7);

⌊3n7 ⌋, otherwise.

Proof. In order to prove this lemma it is sufficient to give a SDF

with cardinality M f equal to the expression given in the lemma

. Let g be a function from V(P(n, 2)) to the set {−1,+1} as


defined in the Lemma 2.4.6 then one can easily verify that g is

a SDF of P(n, 2) with

|Mg| =

⌊3n7 ⌋ − 1 i f n ≡ 5(mod 7);


Let V′(i, t) = {wj, uj : i ≤ j ≤ i + t − 1} ⊆ V. With this

notation, we prove the next lemma, as follows.

Lemma 2.4.9. If f is a MSDF of P(n, 2) then |M f ∩V′(i, 7)| ≤ 3.

Proof. Since all the vertices at distance 2 from wi+3 lie in V′(i, 7).

Hence, either wi+3 ∈ N[M f ] or wi+3 ∈ V \N[M f ]. Now we have

the following cases:

Case 1: Suppose wi+3 ∈ N[M f ], and for this we have the follow-

ing four subcases:

subcase 1.1: If wi+3 ∈ M f then

f (wi+1) = f (wi+2) = f (wi+4) = f (wi+5) = f (ui+1)

= f (ui+2) = f (ui+3) = f (ui+4) = f (ui+5) = +1.


Clearly, remaining vertices of V′(i, 7) are {wi, ui, wi+6, ui+6}.

Also,

{wiui, wi+6ui+6} ∈ E(P(n, 2)).

Therefore, at most two negative vertices are in M f \ {wi+3},

whence |M f ∩V′(i, 7)| ≤ 3 (see Figure 2.7(a)).

subcase 1.2: If wi+2 ∈ M f then

f (wi) = f (wi+1) = f (wi+3) = f (wi+4) = f (ui)

= f (ui+1) = f (ui+2) = f (ui+3) = f (ui+4) = +1,

so remaining vertices of V′(i, 7) are {wi+5, ui+5, wi+6, ui+6}.

Also,

{wi+5ui+5, wi+6ui+6, wi+5wi+6} ∈ E(P(n, 2)).

Therefore, at most two negative vertices are in M f \ {wi+2} (see

Figure 2.7(b)), whence

|M f ∩V′(i, 7)| ≤ 3.

subcase 1.3: If wi+4 ∈ M f then similar proof can be given as in

subcase 1.2 as wi+2 and wi+4 are symmetric to the vertex wi+3


(see Figure 2.7(c)).

subcase 1.4: If ui+3 ∈ M f then

f (wi+1) = f (wi+2) = f (wi+3) = f (wi+4)

= f (wi+5) = f (ui+1) = f (ui−1) = f (ui+5) = f (ui+7) = +1.

Clearly remaining vertices of V′(i, t) are {wi, ui, ui+2, ui+4, wi+6,

ui+6}.

Since

{wiui, uiui+2, ui+2ui+4, ui+4ui+6, ui+6wi+6} ∈ E(P(n, 2)),

it follows that at most two negative vertices are in M f \ {ui+3}

(see Figure 2.7(d)). Consequently

|M f ∩V′(i, 7)| ≤ 3.


From all the subcases of case 1 we obtained that, if wi+3 ∈

N[M f ], then |M f ∩V′(i, 7)| ≤ 3.

Case 2: If wi+3 ∈ V \ N[M f ] then

f (wi+2) = f (wi+3) = f (ui+3) = f (wi+4) = +1,

whence the remaining vertices of V′(i, t) are

{wi, ui, wi+1, ui+1, ui+2, ui+4, ui+5, wi+5, wi+6, ui+6}.

Since {wiui, uiui+2, ui+2ui+4, ui+4ui+6, ui+6wi+6, wi+1ui+1, wiwi+1,

wi+6wi+5, ui+5wi+5} ∈ E(P(n, 2)) whence V′(i, 7) \N[wi+3] have

at most three vertices in M f (see Figure 2.7(e)). i.e., |M f ∩

V′(i, 7)| ≤ 3.

From the above analysis of both the cases, we conclude that

|M f ∩V′(i, 7)| ≤ 3. Hence the result.

In the Figure 2.7 the vertices of M f are in dark, grey vertices

in Pf and the vertices in doted box are the possible vertices in

M f .


Figure 2.7



|M f | ≤

⌊3n7 ⌋ − 1, i f n ≡ 5(mod 7);

⌊3n7 ⌋, otherwise.

Proof. Let f be a MSDF of P(n, 2) then we have following cases

to show the desired result:

Case 1: If n ≡ 0(mod 7) then by Lemma 2.4.9 we have

|M f | =n7−1

∑i=0

|M f ∩V′(7i, 7)| ≤ n

7× 3 =

3n

7= ⌊3n

7⌋.

Case 2: If n ≡ 1, 2, 3, 4, 6(mod 7) then by Lemma 2.4.9 we have

7|M f | =n−1

∑i=0

|M f ∩V′(7i, 7)| ≤ n× 3 = 3n,

implies |M f | ≤ ⌊3n7 ⌋.

Case 3: If n ≡ 5(mod 7) then there exists a positive integer l

such that n = 7l + 5, which gives

|M f | ≤ ⌊3n

7⌋ = ⌊3(7l + 5)

7⌋ = 3l + 2.


Now suppose |M f | = 3l + 2, then |V ′(7i, 7) ∩ M f | = 3, for all

0 ≤ i ≤ l − 1, and |V ′(7l, 5) ∩ M f | = 2. Following subcases

arise:

Subcase 3.1: If {wn−1, wn−4} ⊆ M f then w1, w0, wn−2, wn−3,

wn−5, wn−6, u1, u0, un−1un−2, un−3, un−4, un−5 and un−6 are all in

Pf .

Again since

|V ′(7i, 7) ∩M f | = 3, f or every 0 ≤ i ≤ l − 1,

then w7i, w7i+1, u7i, u7i+1 will be in Pf . Hence

w7l−7 = wn−12, w7l−6 = wn−11, u7l−7 = un−12, u7l−6 = un−11

are in Pf . Further, wn−6, un−6, wn−12, wn−11, un−12, un−11 are in Pf ,

the remaining vertices of V′(7(l − 1), 7) \ Pf = V

′(7l − 5, 4) in

which maximum two vertices are in M f , therefore, |V ′(7(l −

1), 7) ∩M f | ≤ 2 (see Figure 2.8), which is a contradiction to the

hypothesis that |V ′(7i, 7)∩M f | = 3, for all 0 ≤ i ≤ l− 1. Hence

{wn−1, wn−4} * M f .


Figure 2.8

Subcase 3.2: If {wn−1, wn−5} ⊆ M f then w1, w0, wn−2, wn−3,

wn−4, wn−6, wn−7, u1, u0, un−1, un−2, un−3, un−4, un−5, un−6 and un−7

are all in Pf . As discussed in the subcase 3.1, each |V ′(7i, 7) ∩

M f | = 3, for every 0 ≤ i ≤ l − 1, then

w7l−7 = wn−12, w7l−6 = wn−11, u7l−7 = un−12, u7l−6 = un−11

are in Pf . Since wn−6, un−6, wn−7, un−7, wn−12, wn−11, un−12 and

un−11 are in Pf , the remaining vertices of V′(7(l − 1), 7)) \ Pf =

V′(7l − 5, 3) in which maximum two vertices are in M f (see

Figure 2.9), i.e.,

|V ′(7(l − 1), 7) ∩M f | ≤ 2,


which is a contradiction to the hypothesis that |V ′(7i, 7) ∩

M f | = 3, for all 0 ≤ i ≤ l − 1. Hence

{wn−1, wn−5} * M f .

Figure 2.9

Subcase 3.3: If {wn−1, un−4} ⊆ M f , then w1, w0, wn−2, wn−3,

wn−4, wn−5, wn−6, u1, u0, un−1, un−2, un−3, un−6, and un−8 are in Pf .

By subcase 3.1 and 3.2 if wn−1 ∈ M f then

{wn−12, wn−11, un−12, un−11} ⊂ Pf .

Clearly, remaining vertices of V′(7(l − 1), 7)) \ Pf = V

′(7l −

5, 4) \ {un−8} in which maximum two vertices are in M f (see

Figure 2.10), therefore,

|V ′(7(l − 1), 7) ∩M f | ≤ 2.


If un−5 ∈ M f then

|V ′(7(l − 1), 7) ∩M f | ≤ 1.

By which we arrive at a contradiction to the hypothesis that

|V ′(7i, 7) ∩M f | = 3, for all 0 ≤ i ≤ l − 1. Hence

{wn−1, un−4} * M f .

Figure 2.10

Subcase 3.4: If {wn−1, un−5} ⊆ M f , then proof of this subcase

is similar to that as given in subcase 3.3 (see Figure 2.11). Hence

{wn−1, un−5} * M f .


Figure 2.11

Subcase 3.5: If {wn−2, wn−5} ⊆ M f , then proof of this sub-

case follows from the subcase 3.1 as the vertices wn−1 and wn−5

are symmetric to wn−3 (see Figure 2.12). Hence

{wn−2, wn−5} * M f .

Figure 2.12

Subcase 3.6: If {wn−2, un−5} ⊆ M f then w0, wn−1, wn−3, wn−4,

wn−5, wn−6, wn−7, u0, un−1, un−2, un−3, un−4, un−7 and un−9 are


all in Pf . Since |V ′(7i, 7)∩M f | = 3, for every 0 ≤ i ≤ l− 1, then

w7i, u7i will be in Pf . Hence

w7l−7 = wn−12, u7l−7 = un−12 are in Pf .

Again, since wn−6, wn−7, un−7, un−9, wn−12, and un−12 are in Pf ,

the remaining vertices of V′(7(l − 1), 7) \ Pf = {un−6, un−8,

un−10, un−11, wn−8, wn−9, wn−10, wn−11} in which maximum two

vertices are in M f , i.e., |V ′(7(l − 1), 7) ∩ M f | ≤ 2 (see Figure

2.13) which is a contradiction to the hypothesis that |V ′(7i, 7) ∩

M f | = 3, for all 0 ≤ i ≤ l − 1. It follows that

{wn−2, un−5} * M f .

Figure 2.13

Subcase 3.7: If {un−1, un−2} ⊆ M f then w1, w0, wn−1, wn−2,

wn−3, wn−4, u3, u2, u1, u0, un−3, un−4, un−5 and un−6 are in Pf .


Since w0, w1, u0, u1 are in Pf , as discussed in subcase 3.1, {w7i,

w7i+1, u7i, u7i+1} are in Pf and therefore, |V ′(7(l− 1), 7)∩M f | ≤

2 (see Figure 2.14) which is a contradiction to the hypothesis

that |V ′(7i, 7) ∩ M f | = 3, for all 0 ≤ i ≤ l − 1. If wn−5 ∈ M f

then |V ′(7(l − 1), 7) ∩M f | ≤ 1. Hence {un−1, un−2} * M f .

Figure 2.14

Subcase 3.8: If {un−1, un−4} ⊆ M f then w1, w0, wn−1, wn−2,

wn−3, wn−4, wn−5, wn−6, u3, u1, un−3, un−5, u0, un−2, un−6 and

un−8 has to be in Pf . Since w0, w1, u0, u1 are in Pf . Then, by giv-

ing same argument as given in subcase 3.1, w7i, w7i+1, u7i, u7i+1,

u7i+2 and u7i+3 are in Pf , whence we get |V ′(7(l− 1), 7)∩M f | ≤

2 (see Figure 2.15) which is a contradiction to the hypoth-

esis that |V ′(7i, 7) ∩ M f | = 3, for all 0 ≤ i ≤ l − 1. Hence

{un−1, un−4} * M f .


Figure 2.15

Subcase 3.9: If {un−2, un−3} ⊆ M f then w0, wn−1, wn−2, wn−3,

wn−4, wn−5, u2, u1, u0, un−1, un−4, un−5, un−6 and un−7 has to be

in Pf . Since w0, u0, u1, u2, are in Pf . Then, following the similar

argument as given in subcase 3.1, we get |V ′(7i, 7) ∩ M f | = 3,

for every 0 ≤ i ≤ l − 1. Than

w7l−7 = wn−12, u7l−7 = un−12, u7l−6 = un−11, u7l−5 = un−10

are in Pf . Since un−6, un−7, wn−12, un−12, un−11 and un−10 are in

Pf , the remaining vertices of V′(7(l− 1), 7)) \ Pf = {wn−6, wn−7,

wn−8, wn−9, wn−10, wn−11, un−8, un−9} in which maximum two

vertices are in M f , therefore, |V ′(7(l − 1), 7) ∩ M f | ≤ 2 (see

Figure 2.16), which is a contradiction as |V ′(7i, 7) ∩ M f | = 3,

for all 0 ≤ i ≤ l − 1. Hence

{un−2, un−3} * M f .


Figure 2.16

Subcase 3.10: If {un−3, un−4} ⊆ M f , then similar proof can

be given as given in the subcase 3.9, which gives

{un−3, un−4} * M f .


follows from the arguments analogous to those used in subcase

3.3. Hence

{wn−5, un−2} * M f .



3.4. Hence

{wn−5, un−1} * M f .




3.6. Hence

{wn−4, un−1} * M f .

Subcase 3.14: If {un−5, un−4} ⊆ M f , then proof of this subcase


3.7. Hence

{un−5, un−4} * M f .

Subcase 3.15: If {un−5, un−2} ⊆ M f , then proof of this subcase


3.8. Hence

{un−5, un−2} * M f .

From the foregoing analysis of subcases 3.1 to 3.15 we have

seen that if |V ′(7l, 5) ∩M f | = 2, then

|V ′(7(l − 1), 7) ∩M f | ≤ 2,


which is a contradiction as |V ′(7i, 7) ∩M f | = 3, for all 0 ≤ i ≤

l − 1. It follows that

|V ′(7l, 5) ∩M f | = 1.

Thus, if n ≡ 5(mod 7) then, |M f | = 3l + 1, i.e., |M f | = ⌊3n7 ⌋ − 1,

which is the desired result.

As an immediate consequence of Lemma 2.4.8 and 2.4.10, we

have the following:


|M f | =

⌊3n7 ⌋ − 1 i f n ≡ 5(mod 7);


Theorem 2.4.12. If f is a MSDF of P(n, 2) then

γs(P(n, 2)) =

2n− 2(⌊3n7 ⌋ − 1) i f n ≡ 5(mod 7);

2n− 2⌊3n7 ⌋ otherwise.

Proof. The result follows immediately from the Lemma 2.4.11

and the fact that if f is a MSDF of a graph G then γs(G) =

|V(G)| − 2|M f |.


2.4.3 Let k=3

In this section we give results for γs(P(n, 3)).

Theorem 2.4.13. If f is a MSDF of P(n, 3), for n ≥ 7, then

γs(P(n, 3)) =

2n− 2⌊n2⌋ i f n ≡ 0, 3(mod 4) or n = 9;

2n− 2(⌊n2⌋ − 1) otherwise.

Proof. In order to prove the theorem it is enough to show that

|M f | =

⌊n2⌋ i f n ≡ 0, 3(mod 4) or n = 9;

⌊n2⌋ − 1 otherwise.

Let l = ⌊n4⌋ and n ≡ r(mod 4). Then, n = 4l + r. Firstly, for the

lower bound, we define a SDF g for each case.

Mg =

S i f r = 0;

S \ {un−2} i f r = 1, and n 6= 9;

S \ {un−3} ∪ {wn−2} i f r = 2;

S ∪ {wn−2} i f r = 3;

{w1, u3, u5, w7} i f n = 9,


where S = {w4i+1, u4i+3 : 0 ≤ i ≤ l − 1}. Consider Pg =

V(P(n, 3)) \Mg. One can easily verify that g is an SDF with

|Mg| =

2l = n2 i f r = 0 and n = 9;

2l = ⌊n2⌋ − 1 i f r = 1 and n 6= 9;

2l − 1 = (n2 )− 1 i f r = 2;

2l + 1 = ⌊n2⌋ i f r = 3.

Therefore, |Mg| ≤ |M f |.

Now, we shall show that in each case g is a MSDF. Using Lemma

2.4.3, we have

|Mg| ≤ ⌊n

2⌋.

So it will cover the cases when n ≡ 0, 3 (mod 4) and n = 9. Next,

when n ≡ 1, 2 (mod 4), by Theorem 2.4.2, P(n, k) has an efficient

dominating set if and only if n ≡ 0 (mod 4) and k is odd. This

means that for n ≡ 1, 2 (mod4), there does not exist a MSDF f

such that all vertices belong to N[M f ]. This means that there

exists a subset S ⊂ V(P(n, 3)) such that v /∈ N[M f ] for all v ∈ S.

Now we have two cases.

Case 1: If w0 ∈ S. Then w1, wn−1 and u0 are in Pf . Further we


have following five subcases and we will study each subcase

separately.

Subcase 1.1: If N(w0) ∈ S. Then |S| ≥ 4.

Subcase 1.2: If {wn−1, w0} ⊂ S, then either u3 or un−3 has to be

in M f since u0 ∈ N(M f ). Suppose u3 ∈ M f then u9, u6, un−3,

w6, w4, w3 and w2 are in Pf . Since w1 and w2 belong to N(M f ),

u2 and u1 has to be in M f , also u8, u7, u5, u4, un−1, un−4, un−5,

w5, wn−2 and wn−1 are in Pf . Thus one can observe that w5, w4

are in S (see Figure 2.17). Similar arguments may be used to

show that wn−5 and wn−6 are in S when un−3 ∈ M f . Therefore,

|S| ≥ 4.

Figure 2.17

In Figure 2.17 the vertices of M f are in dark, grey vertices in

Pf and the vertices in doted box are in S.


Subcase 1.3: If {w1, w0} ⊂ S then proof of this subcase is similar

to that of above subcase 1.2 (see Figure 2.18).

Figure 2.18

Subcase 1.4: If {w0, u0} ⊂ S. Then u3, un−3, w1 and wn−1

are in Pf . Since w1 ∈ N(M f ), then either w2 or u1 has to be in

M f , clearly, for this subcase we need to consider following two

subcases:

Subcase 1.4.1: Suppose w2 ∈ M f . Then u5, u2, u1, w4, un−1 and

w3 are in Pf . By the similar arguments as given in subcase 1.1

wn−2, u4, w6, un−4 and wn−6 has to be in M f . This means u10, u7,

un−2, w7, w5, w8, u9, u6, wn−3, wn−4, un−5, wn−5, wn−7, un−7, un−6,

wn−8 and un−9 are in Pf (see Figure 2.19), whence the vertices

u3 and un−3 are in S. Therefore, |S| ≥ 4.

Subcase 1.4.2: If u1 ∈ M f . Then u7, w4, u4, w2, un−2, wn−2 and

un−5 are in Pf . By the similar arguments as given in subcases


Figure 2.19

1.1 un−1, w3, w6, u8, w10, wn−3, wn−6, un−8 and wn−10 are in M f .

This means u2, u5, wn−4, un−4, un−7, w12, w11, w9, w8, w7, w5,

wn−5, wn−7, wn−8, wn−9, wn−11, wn−12, u14, u13, u11, u10, u9, u6, u3,

un−3, un−5, un−6, un−9, un−10, un−11, un−13 and un−14 are in Pf

(see Figure 2.20). This implies that u7 and un−7 are in S. Hence

|S| ≥ 4.

Notice that if n = 9 then u7 = un−2 and un−7 = u2, whence

both are adjacent to u1 and u8 that is they are in M f . Therefore,

|S| = 2.

Figure 2.20


Subcase 1.5: If N(w0) * S. Then w1, wn−1 and u0 are in

Pf . Since u0 is in N(M f ), either u3 or un−3 has to be in M f . If

u3 ∈ M f , then u9, u6, un−3, w6, w4, w3 and w2 are in Pf . Then

following the same arguments as mentioned in the subcase

1.4.2, u1, un−1, wn−3, wn−6, un−8, wn−10, w5, w8, u10, and w12, are

in M f . This means that the vertices u15, u14, u13, u12, u11, u8, u7,

u5, u4, u2, un−2, un−4, un−5, un−6, un−7, un−9, un−10, un−11, un−13,

un−14, w13, w11, w10, w9, w7, wn−2, wn−4, wn−5, wn−7, wn−8, wn−9,

wn−11 and wn−12 are in Pf (see Figure 2.21). One can observe

that the vertices u9, w2 and wn−7 are in S. Thus |S| ≥ 4.

Figure 2.21

Case 2: If u0 ∈ S and N(u0) /∈ S. Then w0, u3, and un−3

are in Pf . Since w0 ∈ N[M f ] then either w1 or wn−1 has to

be in M f . Suppose w1 ∈ M f , then u4, u2, un−2, w3, w2 and

wn−1 are in Pf , again by the similar arguments w3 ∈ N[M f ].

Therefore, w4 ∈ M f . Continuing this process we obtain a set


of vertices {w4i, u4i+2 : ∀ 1 ≤ i ≤ l − 2} which has to be in

M f , and remaining other vertices are in Pf . Now if i = l − 2

then w4(l−2) = wn−9 and u4(n−2)+2 = un−7 but if i = l − 3

then u4(l−3)+2 = un−11. This implies that wn−6, wn−8, wn−7, un−6,

wn−4, un−9, un−8, un−5, un−4 and un−1 are in Pf (see Figure 2.22).

Therefore, u5, u2 and un−1 are in S. Hence |S| ≥ 4.

Figure 2.22

Since wn−6 ∈ N[M f ], wn−5 ∈ M f and wn−3 ∈ Pf , which

implies that un−3 ∈ S. Hence if u0 and un−3 are in S, |S| ≥ 4.

From the above analysis for each case (including subcases)

we obtain |S| ≥ 4. Now,

3|M f |+ |M f |+ |S| = 2n

|M f | =2n− |S|

4

≤ 2n− 4

4

≤⌊n

2

⌋

− 1

2.5 bounds for signed domination number of p(n, k) 77

Therefore, in both the cases, when n ≡ 1(mod 4) or n ≡ 2(mod 4)

|M f | ≤⌊

n2

⌋

− 1. Hence

|M f | =

⌊

n2

⌋

i f n ≡ 0, 3(mod 4) or n = 9;

⌊

n2

⌋

− 1 otherwise.

2.5 bounds for signed domination number of p(n, k)

In this section we will give the lower bound for γs(P(n, k)), for

odd k and upper bound for even k.

Lemma 2.5.1. For odd k γs(P(n, k)) ≥ 2n− 2⌊

n2

⌋

.

Proof. Proof is straight forward by using the Theorem 4.4, i.e.,

|M f | ≤ n2 , for odd k together with n ≡ 0 (mod 4).

Lemma 2.5.2. For even k γs(P(n, k)) ≤ 2n− 2⌊

n3

⌋

.

2.6 minus domination number of p(n, k) 78

Proof. In order to prove the Lemma it is sufficient to find an

SDF with |M f | = ⌊n3⌋. Let f : V(P(n, k))→ {1,−1} be an SDF

such that,

f (wi) =

−1 i f i ≡ 0(mod 3);

1 otherwise.

And

f (ui) = 1, ∀ 0 ≤ i ≤ n− 1.

Since 〈W〉 is a cycle of length n, it is easy to verify that f is a

SDF with |M f | = ⌊n3⌋. Hence, for even k,

γs(P(n, k)) ≤ 2n− 2⌊n

3

⌋

.

2.6 minus domination number of p(n, k)

This section is devoted to the exact minus domination number

of generalized Petersen graph P(n, 1). Further it is shown that


there does not exist any efficient minus dominating function

for P(n, 2).

For any MMDF f of a graph G, let Pf = {v ∈ V : f (v) = 1},

Z f = {v ∈ V : f (v) = 0} and M f = {v ∈ V : f (v) = −1}.

By Theorem 2.4.2 we know that P(n, k) has an efficient dom-

inating set if n ≡ 0(mod 4) and k is odd.

Lemma 2.6.1. If G has an efficient dominating set then there exists

an efficient minus dominating function (EMDF).

Proof. Let D be the EDS of G and f : V(G) → {−1, 0, 1} be

defined in such a manner that

Pf = D, Z f = V \ D and M f = ∅, then one can easily verify

that f is EMDF.

2.6.1 Let k=1

Theorem 2.6.2.

γ−(P(n, 1)) =

⌈

n2

⌉

+ 1 i f n ≡ 2(mod 4);

⌈

n2

⌉

otherwise.


Proof. For n ≡ 0, 1, 3(mod 4) the result holds due to Lemma

2.3.1 and Theorem 2.4.2. Now, if n ≡ 2(mod 4), then there are

two cases.

Case 1: If |M f | = ∅ then f will be dominating function as

proved by Ebrahim et al.[10], i.e.,

γ(P(n, 1)) =

⌈

n2

⌉

+ 1 i f n ≡ 2(mod 4);

⌈

n2

⌉

otherwise.

Hence the result.

Case 2: If |M f | 6= ∅ then for any r- regular graph G

γ−(G) ≥⌈ |V(G)|

r + 1

⌉

. Now, if n = 4l + 2, l > 0, then γ−(P(n, 1)) ≥⌈

|V(P(4l+2,1))|r+1

⌉

=

⌈n2⌉ = 2l + 1, which implies that f (N[v]) = 1, ∀ v ∈ V(P(4l +

2, 1)).

If w1 ∈ M f then any one of w2, w0 or u1 must be in Z f as

f (N[w1]) = 1. Now we have following three subcases:

Subcase 2.1: Firstly, suppose u1 ∈ Z f . Then w0, w2, u2 and u0

have to be in Pf . Also f (N[w2]) = 1, then w3 has to be in Z f .


Now similar arguments may be used to show that {w4i+1, u4i+3}

all are in M f and {u4i+1, w4i+3} have to be in Z f , ∀ 0 ≤ i ≤ l− 1.

This implies that wn−5 and un−3 has to be M f and wn−3, un−5

are in Z f . Thus un−2 and wn−2 are in Pf and wn−1, un−1 are in

Z f (see Figure 2.23).This gives us f (N[un−1]) = 2 = f (N[wn−1])

a contradiction as f (N[v]) = 1, ∀ v ∈ V(P(4l + 2, 1)).

Figure 2.23

In Figure 2.23 the vertices of M f are in dark, Grey vertices

are in Z f and the vertices in white are in Pf .

Subcase 2.2: Next, if w2 ∈ Z f , then w3 u2 u1 and w0 are in Pf .

Since f (N[u2]) = 1, u3 and u4 has to be in M f and Z f respec-

tively and f (N[u3]) = 1. Now similar arguments may be used

to show that {w4i+1, u4i+3} are all in M f and {w4i+2, u4i} are in

Z f , ∀ 0 ≤ i ≤ l − 1. This implies that un−3 is in M f and un−2 is

in Z f , then wn−2 and un−1 are in Pf (see Figure 2.24). Thus, we


get f (N[wn−2]) = 2 = f (N[wn−1]) again a contradiction.

Figure 2.24

Subcase 2.3: If w0 ∈ Z f . As the vertices w0 and w2 are sym-

metric to w1, therefore, the proof of this subcase is similar to

the above subcase 2.

Hence in all the subcases we have seen that, if |M f rvert 6= ∅

then there exist at least two vertices with f (N[v]) = 2, therefore,

γ−(P(n, 1)) = 2l + 2 =⌈n

2

⌉

+ 1.

Hence the proof.

2.6.2 Let k=2

In this section we will show that there does not exist any effi-

cient minus dominating function for P(n, 2).


Theorem 2.6.3. For P(n, 2), there does not exist any efficient minus

dominating function.

Proof. By Theorem 4.2, we know that P(n, 2) does not have any

efficient dominating set, so if M f = ∅ then result holds trivially.

Let’s suppose M f 6= ∅. Now we have the following cases:

Case 1: Suppose w1 ∈ M f , then any one of w2 or u1 or w0 has

to be in Z f as f (N[w1]) = 1. Now the following three subcases

arise:

Subcase 1.1: Firstly, suppose u1 ∈ Z f . Then un−1, w2, w0 and u3

are in Pf . Also f (N[w2]) = 1, then either w3 or u2 has to be in

Pf , if w3 is in Pf then f (N[w3]) ≥ 2 (not possible). Therefore,

u2 is in Pf and w3 in Z f . Now, similar arguments may be used

to show that w4 is in M f and u4 is in Pf . Clearly, f (N[u2]) ≥ 2

which is a contradiction to the fact that f (N[v]) = 1 for all

v ∈ V(P(n, 2)).

Subcase 1.2: If w2 ∈ Z f , then w3, u2, u1 and w0 are in Pf . Since

f (N[u0]) = 1 then either u0 ∈ Z f together with un−2 ∈ M f or

u0 ∈ M f together with un−2 ∈ Z f . If u0 ∈ M f and un−2 ∈ Z f ,

then f (N[w0]) ≤ 0 (not possible). So u0 ∈ Z f and un−2 ∈


M f , then wn−1, wn−2, un−4 are in Pf . Now, it is easy to see

that f (N[wn−1]) ≥ 2, a contradiction to the assumption that

f (N[v]) = 1, for all v ∈ V(P(n, 2)).

Subcase 1.3: If w0 ∈ Z f , since the vertices w0 and w2 are sym-

metric to w1, therefore, the proof of this subcase is similar to

the above subcase 1.2.

Case 2: Next, suppose u1 ∈ M f then any one of u3 or un−1 or

w1 has to be in Z f as f (N[u1]) = 1. Now we have the following

three subcases:

Subcase 2.1: If w1 ∈ Z f , then u3, w2, w0 and un−1 are all in

Pf . Since f (N[w3]) = 1 then either w3 ∈ Z f or w3 ∈ M f . If

w3 ∈ M f , then f (N[u3]) ≤ 0 (not possible). Therefore, w3 ∈ Z f ,

w4 ∈ M f and w5, u4 and u5 ∈ Pf , this implies that f (N[u5]) ≥ 2,

which is a contradiction to the fact that f (N[v]) = 1, for all

v ∈ V(P(n, 2)).

Subcase 2.2: If u3 ∈ Z f , then w1, w3, u5, and un−1 are all in Pf .

Since f (N[w2]) = 1 then w2 ∈ Z f and u2 ∈ M f this implies that

w0 and u0 are in Pf . So f (N[w0]) ≥ 2, a contradiction to the fact

that f (N[v]) = 1 for all v ∈ V(P(n, 2)).

2.7 open problems 85

Subcase 2.3: If un−1 ∈ Z f , the proof is similar to the above

subcase 2.2.

In all the cases we have shown that if M f 6= ∅ then there

doesn’t exist any efficient minus dominating function for P(n, 2).

Hence the result.

2.7 open problems

We can see that the minus domination number and domination

number of P(n, 1) are equal. Hence, we propose the following:

1. Prove or disprove γ(P(n, k)) = γ−(P(n, k)) when k is odd.

2. Find the value for γ−(P(km, k)) when k ≥ 3, where m is

any positive integer.

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SIGNED DOMINATION NUMBER AND MINUS ...shodhganga.inflibnet.ac.in/bitstream/10603/21209/9...domination number and signed domination number, in which we will use some of the results