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A subset of the nodes of a graph is a dominating set if every other node of is adjacent to some node in the subset. Let

\(\)

localid="1663236390542" \text{DOMINATING-SET}=\{langleG,k\rangle\midG\text{hasadominatingsetwith}k\text{nodes}\}

\(\)

Show that it is NP-complete by giving a reduction from VERTEX-COVER.

Short Answer

Expert verified


DOMINATING-SETisNP-complete.

Step by step solution

01

To Check node and Polynomial time

DOMINATING-SETis in :

Consider an instance$\langleG,k\rangle$ of theDOMINATING-SET and a covering D.

Check that each node ofG is adjacent to some node in D.

This can be done in polynomial time.

Therefore, DOMINATING-SETis in NP.

VERTEX-COVER$\leq_{mathrm{p}}$DOMINATING-SET

Now show that VERTEX-COVER reduces to DOMINATING-SET.

02

To Consider an instance angle

Consider an instance $langle(V,E),k\rangle$ofVERTEX-COVER

Construct an instance$\left\langle\left((V-S)\cupV^{\prime},E\cupE^{\prime}\right,k\right\rangle$

of DOMINATING-SETwhere$S\subseteqV$ are nodes of degree 0 .

For each edge $(u,v)\in , there are edges$(u,w)$ and$(w,v){\text{in}}E^{\prime}$ where $w\inv^{prime}$in new vertex corresponding torole="math" localid="1663233233403" (u,v)

Let$G=\langleV,E\rangle$and$G^{\prime}=\left\langle(V-S)\cupV^{\prime},E\cupE^{\prime}\right\rangle$

Suppose$(\langleV,E\rangle,k)$is inVERTEX-COVER .

There exits $C\subseteqV$of sizek where each edge $(u,v)\inE{\text{haseither}}u\inE{\text{haseither}}u\inC$or$V\inC$.

If $v\i(V-S)$then the degree ofv is one or more then there exists a node usuch that$(u,v)\inE$ which implies that at least one of uorv is in C. Thus, vis covered.

03

To Implies at least one

If$w\inV^{\prime}$then$\mathrm{w}$is adjacent to bothuand vwhere$(u,v)\inE$which implies that at least one ofuorv is inC . Thus, is covered.

In other direction, suppose that$\left\langle\left((V-S)\cupV^{\prime},E\cupE^{\prime}\right),k\right\rangle$is in DOMINATING-SET.

Then there exists $C\subseteq\left((V-S)\cupV^{\prime}\right)$of size k.

In such cases in which multiple such $C$exist, it can be said that at least one includes no vertices in $V^{prime}$.

This is always exists since $w\in\left(C\capV^{\prime}\right)$that corresponds to edge $(u,v)$covers only nodes u,v,wbut using instead of w Covers and possibly more.

Therefore, $C\subseteq(V-S)$and $C$is a vertex cover for G.

This is because Cis aDOMINATING-SET for implying that all nodes ofare $V^{\prime}$covered. Thus, every edge$(u,v)\inE{\text {has at least one of }u,v}$ in$c$.

DOMINATING-SETis NP-complete.

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