Cmjv04i03p0143

J. Comp. & Math. Sci. Vol.4 (3), 143-151 (2013)

Independence Related Parameters and Edge Removal D. K. THAKKAR1 and D. D. PANDYA2 1

Department of Mathematics, Saurashtra University, Rajkot, Gujarat, INDIA. 2 Mathematics Department, L. E. College, Morbi, Gujarat, INDIA. (Received on: May 9, 2013) ABSTRACT Independence, Independent Domination and Vertex Covering are Parameters related to Domination. In this paper we establish necessary and sufficient conditions under which these parameters change when an edge is removed from the graph. In particular We prove that the vertex covering number of the graph does not increase when an edge is removed from the graph. we give some related examples.

Keywords: Independent Dominating Set, Independent Domination Number, Minimal Independent Dominating Set, Independence Set, Maximum Independence Set, Vertex Covering Set, Minimum Vertex Covering set.

AMS subject classification (2000): 05c69.

PRELIMINARIES AND NOTATIONS If G is a graph then V(G) or V will denote the set of vertices of the graph G And E(G) or E will denote the set of edges of the graph G. If u and v are the end vertices of an edge e then we will write e=uv or e=vu. The sub graph obtained by removing an edge e=uv will be denoted as G - {e} or G {uv}.The sub graph obtained by removing a vertex v from the graph G will be denoted as G - {v}.we will consider only simple and undirected graphs in this paper.

Vertex covering Definition 1.1 Vertex Covering Set[6] Let G be a graph. A set S ⊂V (G) is said to be Vertex Covering Set of the graph G if every edge has at least one end point in S. Definition 1.2 Minimum Vertex Covering Set[6] A Vertex Covering Set with minimum cardinality is called minimum vertex covering set. It is also called γ cr Set. Definition 1.3 Vertex Covering

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Number[6] The Vertex Covering Number of the graph G is the Cardinality of any minimum Vertex Covering Set of the graph G. It is denoted by α 0 (G) or simply α0 . >>It is useful to partition the Edge set of G into two sets according to how their Removal affects γ cr (G) .

Ecr − = {e ∈ E (G ) / γ cr (G − e) < γ cr (G )}

Theorem 1.5: An Edge − e = uv ∈ E cr ⇔ There is a minimum vertex covering Set S of G such that All neighbours of u except possibly v are in S and All neighbours of v except possibly u are in S. Proof: ⇒ Suppose e = uv ∈ E cr −

Ecr 0 = {e ∈ E (G ) / γ cr (G − e) = γ cr (G )}

Let S1 be a minimum vertex covering set of

It is Obvious that E = Ecr − ∪ Ecr 0

Covering Set of G. Therefore u∉S1 & v∉S1 .

>>It is useful to partition the Vertex set of G into two sets according to how their Removal affects γ cr (G) .

Let S = S1 ∪{u} . Then because of condition given , S is a minimum vertex Covering Set of G. Now if we consider S then u ∈ S . Let w be

G − {uv} .Now S1 can not be a Vertex

Vcr − = {v ∈ V (G ) / γ cr (G − v ) < γ cr (G )}

any neighbour of u in G − {uv} . Since S1 is

Vcr = {v ∈ V (G ) / γ cr (G − v ) = γ cr (G )}

a Vertex Covering Set of G − {uv} , w∈S1

0

It is Obvious that V = Vcr − ∪ Vcr 0 The following theorem says that the vertex covering number of the graph does not increase when an edge is removed from graph.

and hence w∈ S . So, All neighbours of u except v are in S & All neighbours of v are in S.

⇐ Conversely

Theorem 1.4 Let G be a graph and e be an

Let S be a minimum Vertex Covering Set of G such that u ∈ S . Let

Proof: Let S be a minimum vertex covering set of G. It is obvious that S is also a vertex

except possible v are in S. It follows that every neighbour of u in G − {uv} is in S. If v ∉ S then every neighbour of v in

edge of G. Then α 0 (G − e) < α 0 (G )

covering

set

of

G − {e} Therefore

α 0 (G − e) < S = α 0 (G ) .

This proves the theorem.

█

S1 = S − {u} . Since every neighbour of u

G − {uv} is in S1 .Thus S1 is Vertex Covering Set in G − {uv} . █

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Theorem 1.6: An Edge e = uv ∈ E cr 0 ⇔ For every minimum Vertex Covering Set S of G There is a neighbour of u different from v which lies outside of S OR There is a neighbour of v different from u which lies outside of S. Proof: ⇐ Suppose Condition holds.

Let T be a minimum Vertex Covering Set of G − {uv} .

Vertex Covering Set of G − {uv} which implies that γ cr (G − uv) < γ cr (G ) Which is a Contradiction.

So condition follows.

█

Lemma 1.7: Let G be a Graph and e=uv be an edge of Graph G. Then α 0 (G − v) <α 0 (G − uv) <α 0 (G ) .

Proof: We need to prove only that

α 0 (G − v) < α 0 (G − uv)

Claim: | T |= γ cr (G )

Let S be a minimum vertex covering set of

Suppose | T |< γ cr (G ) . Then T can not be Vertex Covering Set of G. Therefore u ∉T & v∉ T .

Case 1: u ∉S and v∉ S Obviously S is a vertex covering set of

Now Consider S = T ∪{u} .Obviously S is a minimum Vertex Covering Set of G. Since T is Vertex Covering Set of G − {uv} .

Every neighbour of u in G − {uv} is in T. Therefore every neighbour of u different from v in G lies in S. This Contradicts the hypothesis.

Therefore | T |= γ cr (G ) and e = uv ∈ E cr 0 .

⇒ Suppose e = uv ∈ Ecr 0 Let S be a minimum Vertex Covering Set of G. Then u ∈S or v ∈ S Suppose Every neighbour of u different from v is in S and Every neighbour of v different from u is in S. Then S − {u} is a

G − {uv}

G − {v}

Case 2: u ∈S and v∉ S Here also S is a vertex covering set of G − {v} .

Case 3: u ∉S and v∈ S If xy is an edge of G − {v} then x ≠ v and y ≠ v .

If x=u then uy is an edge of G − {uv} and therefore it has an end vertex in S. Similarly y=u then ux is an edge of G − {uv} and therefore it has an end vertex in S. Case 4: u ∈S and v∈ S Consider the set S1 = S − {v} . If xy is an edge of G − {v} then x ≠ v and y ≠ v . Then xy is an edge of G − {uv} not containing the

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vertex v. Since S is a vertex covering set of G − {uv} , x∈ S or y ∈ S Equivalently

x∈S1 or y∈S1 . Thus S1 is a vertex covering set of G − {v} .

Hence α 0 (G − v) < α 0 (G − uv) . █ Theorem 1.6 : Let G be a Graph and e=uv be an edge of Graph G. If uv∈ Ecr − then

u ∈Vcr − and v∈Vcr − . uv∈ Ecr − Then α 0 (G − uv) < α 0 (G ) . And we know that α 0 (G − v) < α 0 (G − uv) .

Therefore α 0 (G − v) < α 0 (G ) and Similarly α 0 (G − u ) < α 0 (G ) . Hence Proof: Suppose

u ∈Vcr − and v∈Vcr − .

Example 1.8: Consider the Graph P5 with vertex set S={a,b,u,v,c}.It can be verified that the edge uv ∈ Ecr 0 .

a b u vc

a bu v c

Example 1.9: Consider the Peterson graph. In Peterson graph all edges are in Ecr 0 . Here |S|=6. Consider the edge e=uv.

We can see that all neighbours of u are in S.

u

█

*Following Example shows that converse of above theorem is not always true.

v

Example 1.7: Consider the graph P6 (path with 6 vertices).Here u ∈Vcr − and v∈Vcr − but e = uv∉ Ecr −

u

Infact e = uv∈ Ecr 0 .

v u

v

Note: It may be noted that every vertex in Peterson graph is in Vcr − .

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D. K. Thakkar, et al., J. Comp. & Math. Sci. Vol.4 (3), 143-151 (2013)

Example 1.10: Consider the Graph C3 with vertex set S={a,u,v}. It can be verified that the edge uv ∈ Ecr − .

a

u

a

v

u

covering set S such that v ∈ S . (2) A vertex

v ∉ S for every minimum

vertex covering set S of G. It has been further proved that if 0 v∈Vcr then for every vertex u such that u is adjacent to v, u ∈Vcr − Infact every neighbour of v belongs to every minimum vertex covering set of G. That is N (v) is subset of intersection of all minimum vertex covering sets of G. Corollary 1.11: If v∈Vcr 0 then for every edge e = uv , e = uv∈ Ecr 0 0 Proof : Since v∈Vcr α 0 (G − v) = α 0 (G) . .

Now from Lemma 1.7

α 0 (G − v) = α 0 (G − uv) = α 0 (G)

Hence e = uv∈ Ecr 0 MAXIMUM INDEPENDENT SET

Definition 2.1 Independent Set[6] Let G be a graph. A set S ⊂V (G) is said to be Independent Set if any two distinct vertices of S are non-adjacent.

Definition 2.1 Maximum Independent Set[6] An Independent Set with maximum cardinality is called maximum Independent Set.

v

*In [2] it has been proved that (1) A vertex v∈Vcr − ⇔ there is a minimum vertex

v∈Vcr 0 ⇔

147

█

Definition 2.3 Independence Number[6] The cardinality of a maximum Independent set is called Independence Number of the graph G and it is denoted by β0 (G) or simply β0 . >>It is useful to partition the Edge set of G into two sets according to how their Removal affects β0 (G) .

EI + = {e ∈ E (G ) / β 0 (G − e) > β 0 (G )} EI 0 = {e ∈ E (G ) / β 0 (G − e) = β 0 (G )} It is Obvious that E = E I + ∪ E I 0 >>It is useful to partition the Vertex set of G into two sets according to how their Removal affects β0 (G) .

VI − = {v ∈ V (G ) / β 0 (G − v ) < β 0 (G )} VI 0 = {v ∈ V (G ) / β 0 (G − v) = β 0 (G )} It is Obvious that V = VI − ∪ VI 0

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It may be noted that the independence number of the graph does not decrease when an edge is removed from the graph.

In the following theorem we prove the necessary and sufficient condition under which this number increases when an edge is removed.

Theorem 2.4: Let G be a graph and e=uv be an edge of graph G. Then

β0 (G − e) > β0 (G) ⇔

There is a maximum independence set S of G such that

u ∈ S , v ∉ S & N (v) ∩ S = {u}

Proof: ⇐ (sufficiency) Let S be a maximum independence set of G such that u ∈ S , v ∉ S & N (v) ∩ S = {u} .Now v is not adjacent to any vertex of S in graph G-{e} and therefore S ∪ {v} is an independence set in G-{e}.

Thus β0 (G − e)> | S ∪ {v}|> β 0 (G ) .

⇒ (Necessity) Suppose β 0 (G − e) > β 0 (G) Let S be any maximum independence set in G-{e}.Then S can not be independence set in G. This implies that u, v ∈ S .

Now Consider the set S-{v} in graph G. Since S is independence in G-{e}, v is adjacent to only one vertex u of S{v}.Therefore N (v) ∩ S − {v} = {u} .Also S − {v} is independence set in G-{e} because v ∉ S

If T is an independence set in G such that | T | > | S − {v}| then | T | > | S | .This implies that β0 (G)> | T |> β 0 (G − e) Which is a contradiction.

So, S-{v} is a maximum independence set in G. Thus condition is satisfied by S-{v}. This completes the proof. █ Theorem 2.5: Let G be a graph and e=uv be an edge of graph G then β0 (G − e) = β0 (G) ⇔ (for every maximum independence set S of G) one of following holds (1) u ∈ S , v ∉ S &| N (v) ∩ S | > 2 (2) u, v ∉ S Proof:

⇐ CASE

A: independence

Let set

S

be of

maximum G and

u ∈ S , v ∉ S &| N (v) ∩ S | > 2 .

Consider G-{e}.Now S is still independence set of G-{e}. Let T be maximum independence set of G-{e} with |T|>|S| Case (1) u ∈ T , v ∉ T T is maximum independence set of G with |T|>|S|. β0 (G )> | T |>| S |= β0 (G ) .

So, Which is a contradiction.

Case (2) u , v ∉ T .T is maximum independence set of G with |T|>|S|.

So, β 0 (G )> | T |>| S |= β 0 (G ) .Which is a contradiction Case (3) u , v ∈ T Consider T-{v}. So T{v} is maximum independence set of G with

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|T-{v}|(maximal)=|S|(maximum) Thus T-{v} is maximum independence set in G such that v ∉T − {v} and N (v) ∩ T − {v} contains exactly one vertex u , which is a contradiction. Therefore such a set T does not exist in G-{e}.

Therefore S is a maximum independence set n G-{e}.

CASE B: Suppose u, v ∉ S If then according to previous theorem, there must be a maximum S1 such that independence set

Corollary 2.6: Suppose G is a graph and e = uv be an edge of graph G. e = uv∈ EI + then u ∈VI 0 or v ∈VI 0

u ∈ S1 , v ∉ S1 & N (v) ∩ S1 = {u} .Which is

In [3] it has been proved that Suppose v ∈VI 0 then

not true in this Case.

Therefore β 0 (G − e) = β 0 (G) ⇒ (Necessity) Let S be a maximum independence Set in G such that β 0 (G − e) = β 0 (G) Suppose (1) u, v ∉ S then condition (2) is satisfied.

Suppose(2) u , v ∈ S . which is not possible as S is maximum independence set.

Suppose(3) u ∈ S , v ∉ S & N (v) ∩ S = {u} Here v is not adjacent to any vertex of S in G-{e}.Therefore S ∪ {v} is an independence set in G-{e}.

Thus | S ∪ {v}| > | S |

β0 (G − e) >| S ∪{v}| >| S |= β0 (G ) .

So, Which is a contradiction.

So, u ∈ S , v ∉ S &| N (v) ∩ S | > 2 █

Proof: By above theorem ,There is a maximum independent set S such that

u ∈ S , v∉ S and N (v) ∩ S = {u}

By above remark, it is clear that v ∈VI .

Corollary 2.7: Let G be a Graph and e = uv be an edge. If e = uv∈ EI 0 then 0

u ∈VI 0 or v∈VI 0

(1)

If | N (v) ∩ S | > 2 then all edge

belongs to EI 0 (2)

If | N (v) ∩ S | = {u} then

e = uv∈ EI + .

Corollary 2.8: Suppose v ∈VI 0 then the edge e = uv∈ EI + ⇔ there is a maximum independence set S of G such that u ∈ S and N (v) ∩ S = {u} Note: If v ∈VI 0 then all edges incident to v are in EI 0 Corollary: Let G be a Graph and e=uv .

If β 0 (G − v) = β 0 (G ) then for every edge

uv, β 0 (G − uv) = β 0 (G ) INDEPENDENT DOMINATION

*In [3] it has been Proved that a vertex v ∈VI 0 ⇔ There is a maximum independent set S of G such that v ∉ S

Definition 3.1 Independent Dominating Set

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Let G be a graph. A set S ⊂V (G) is said to be Independent Dominating Set of the graph G if (1) S is Dominating Set and (2) S is an Independent Set. Definition 3.2 Minimum Independent Dominating Set An Independent Dominating Set with minimum cardinality is called minimum Independent Dominating set.

Definition 3.3 Independent Domination Number The Independent Domination Number of the graph G is the Cardinality of any minimum Independent Dominating Set of the graph G denoted by i (G) >>It is useful to partition the Edge set of G into two sets according to how their Removal affects i (G) .

Ei − = {e ∈ E (G ) / i(G − e) < i(G)} Ei 0 = {e ∈ E (G) / i(G − e) = i(G)} Ei + = {e ∈ E (G ) / i (G − e) > i(G )} It is Obvious that E = Ei − ∪ Ei 0 ∪ Ei + Theorem 3.4: An edge e = P1 P2 ∈ Ei

+

⇔

(for every γ i set S of G) P1 ∈ S , P2 ∉S and for every w ≠ P1 in N ( P2 ) , N ( w) ∩ S ≠ φ Proof:

⇐ Let S be a γ i set in G and condition satisfies.

Consider G – {e}. Now S is not independent dominating set in G – {e}.

Let T be γ i set in G – {e} with | T | < | S |

P1 ∈ T , P2 ∈ T . Now in G, T − {P2 } will be γ i set with | T − {P2 }| < | S | . (a)

which is a contradiction.

P1 ∉ T , P2 ∉ T .Now in G, T is γ i (b) set with | T | < | S | and P1 or P2 ∉T which contradicts our assumption.

P1 ∈ T , P2 ∉T .Now P2 is adjacent (c) to x ∈T .Consider G,Now T is independent dominating set in G also. But N ( x) ∩ S = φ because x ∈T . Which contradicts our assumption.

So, e = P1 P2 ∈ Ei +

⇒

An edge e = P1 P2 ∈ Ei

+

Let S be any γ i set of G. Then S can not be a dominating set in G – {e}.There are two possibilities.

P1 & P2 ∉ S .Then S is a dominating (1) set in G – {e}. Which is a contradiction.

P1 ∈ S , P2 ∉ S .Now S is not a (2) dominating set in G – {e}.Therefore P2 is not adjacent to any vertex of S. If w∈ N ( P2 ) & w ≠ P1 then w is adjacent to some vertex of S. Therefore N ( w) ∩ S ≠ φ █ Theorem 3.5: An Edge e = P1 P2 ∈ Ei − ⇔ (For any dominating set T of G with | T | < γ i (G ) ) P1 & P2 ∈T and they are the only non-isolated vertices in T. Proof:

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D. K. Thakkar, et al., J. Comp. & Math. Sci. Vol.4 (3), 143-151 (2013)

⇐ Let T be a set which satisfies given conditions in Theorem. Then T is a independent dominating set in G − {PP 1 2} Therefore i (G − PP 1 2 ) < | T |< i (G ) .

Therefore e = P1 P2 ∈ Ei

⇒

−

.

Suppose e = P1 P2 ∈ Ei

−

Let T be a minimum independent dominating set in G − {PP 1 2 } .Then T can not be independent dominating set in G. Since T must be a dominating set in G, it can not be independent set in G.

P1 & P2 ∈T and Therefore Obviously they are the only non-isolated vertices in T. █ Note: In Peterson graph, wheel graph all E ∈ Ei 0 .

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