A linear time algorithm for computing a minimum paired-dominating set of a convex bipartite graph

A linear time algorithm for computing a minimum paired-dominating set of a convex bipartite graph

journal homepage:www.elsevier.com/locate/dam

A linear time algorithm for computing a minimum paired-dominating set of a convex bipartite graph

Computer Science and Application Group, Department of Mathematics, Indian Institute of Technology Delhi, Hauz Khas, New Delhi 110 016, India

a r t i c l e i n f o

Article history:

Received 26 July 2010

Received in revised form 3 December 2011

Accepted 17 April 2012

Available online 18 May 2012

Keywords:

Paired-domination

Perfect matching

Convex bipartite graphs

a b s t r a c t

A set D of vertices of a graph G= (V,E)is a dominating set of G if every vertex in V\D has at least one neighbor in D A dominating set D of G is a paired-dominating set of G if the induced subgraph, G[D], has a perfect matching The paired-domination problem is for a given graph

G and a positive integer k to answer if G has a paired-dominating set of size at most k The

paired-domination problem is known to be NP-complete even for bipartite graphs In this paper, we propose a linear time algorithm to compute a minimum paired-dominating set

of a convex bipartite graph

©2012 Elsevier B.V All rights reserved

1 Introduction

A set D of vertices of a graph G= (V,E)is a dominating set of G if every vertex in V \D is adjacent to a vertex in D A matching M in G is a set of edges in G such that no two edges in M are adjacent A matching M of G is a perfect matching if

every vertex of G is incident to an edge of M A dominating set D of G is a paired-dominating set of G if the induced subgraph

G[D]has a perfect matching Graphs without isolated vertices have paired-dominating sets, since the vertices incident to edges of any maximal matching form such a set [7] The paired-domination problem is for a given graph G and a positive integer k to answer if G has a paired-dominating set of size at most k.

The paired-domination problem is known to be NP-complete for general graphs [7] and remains NP-complete even for various restricted families of graphs such as bipartite graphs, chordal graphs, and split graphs [3] However, the paired-domination problem can be solved in polynomial time for various special classes of graphs such as trees [9], interval graphs [3], block graphs [3], circular-arc graphs [5], permutation graphs [6], series–parallel graphs [1], and strongly chordal graphs [4

Hung et al [8] claimed to compute a minimum paired-dominating set of a convex bipartite graph in polynomial time However, we show in Section3of this paper that the set computed by their algorithm is not a dominating set and hence

is not a minimum paired-dominating set of the input graph In Section4, we propose a linear time algorithm to compute a minimum paired-dominating set of a convex bipartite graph

2 Preliminaries

For a graph G = (V,E), the sets N(v) = {u ∈V(G)|u ∈E}and N[v] =N(v) ∪ {v}denote the neighborhood and the

closed neighborhood of a vertexv, respectively The degree of a vertexvis|N(v)|and is denoted by d(v) For S⊆V , let G[S]

denote the subgraph induced by G on S A graph G= (V,E)is said to be bipartite if V(G)can be partitioned into two disjoint

∗Corresponding author Tel.: +91 11 26591448; fax: +91 11 26581005.

E-mail addresses:bspanda@maths.iitd.ac.in, bspanda1@gmail.com (B.S Panda), maz058251@maths.iitd.ac.in (D Pradhan).

0166-218X/$ – see front matter © 2012 Elsevier B.V All rights reserved.

Fig 1 A convex bipartite graph G.

sets X and Y such that every edge joins a vertex in X to another vertex in Y A partition(X,Y)of V is called a bipartition.

A bipartite graph with bipartition(X,Y)of V is denoted by G= (X,Y,E) Let n and m denote the number of vertices and number of edges of G, respectively For a bipartite graph G= (X,Y,E), let|X| =n x, |Y| =n y , and n=n x+n y

A bipartite graph G= (X,Y,E)is called convex bipartite if the vertices of Y can be ordered in such a way that for every

x∈X , the neighbors of x appear consecutively in the ordering of Y Y is said to exhibit the property of convexity Such an

ordering is called a convex ordering of G A convex ordering of a convex bipartite graph G can be computed in linear time [2 Letσ = (x1,x2, ,x n x,y1,y2, ,y n y)be a convex ordering of a convex bipartite graph G= (X,Y,E) For each x i ∈X ,

let minnbr(x i)(resp maxnbr(x i)) be the minimum (maximum) indexed vertex with respect toσadjacent to x i An ordering

α = (x1,x2, ,x n x,y1,y2, ,y n y)of a convex bipartite graph G= (X,Y,E)is called a lexicographic convex (lex-convex)

ordering ifαis a convex ordering and for x i and x j, i<j if either minnbr(x i) <minnbr(x j)or minnbr(x i) =minnbr(x j)and

maxnbr(x i) ≤maxnbr(x j) For an orderingσ = (v1, v2, , vn), letσ−1(x) =i if x= vi,1≤i≤n.

The orderingσ = (x1,x3,x5,x4,x2,x6,x7,x8,x9,y1, ,y11)is a lex-convex ordering of the convex bipartite graph G

ofFig 1

Letσ = (x1,x2, ,x n x,y1,y2, ,y n y)be an ordering of X ∪Y of a bipartite graph G = (X,Y,E) For x i ∈ X ,

let N j(x i) = N(x i) ∩ {y j+1,y j+2, ,y n y}and d j(x i) = |N j(x i)| We denote M(y j)to be the vertex x ∈ N(y j)such that

d j(x) =max{d j(x′)|x′∈N(y j)} For y i∈Y , let N j(y i) =N(y i) ∩ {x j+1,x j+2, ,x n x}and d j(y i) = |N j(y i)| We denote M(x j)

to be the vertex y∈N(x j)such that d j(y) =max{d j(y′)|y′∈N(x j)}

Let(x1,x2, ,x n x,y1,y2, ,y n y)be a lex-convex ordering of a convex bipartite graph G = (X,Y,E) For x i ∈ X ,

1≤i≤n x, we use the following notations to describe our algorithms

• y i1=minnbr(x i)and NN i(y i1) = {x k|x k y i1 ̸∈E,k>i}(the set of non-neighbors of y i1which succeed x iinσ)

• M(x i) =N(x i) ∩ (∪x∈NN i(yi1)maxnbr(x))and B(x i) = {x∈NN i(y i1)|N(x i) ∩ {maxnbr(x)} ̸= ∅}

• m(i) = σ− 1 (maxnbr(x i)), if M (x i) = ∅ ;

min {k|y k∈ M (x i)}, otherwise.

• D is an array such that D(v) =1 or D(v) =0 for eachv ∈V

• N D(x i) = {y k∈N(x i)|D(y k) =0},B D(x i) = {x∈B(x i)|D(x) =0}and T(x i) = ∪x∈B D(i)maxnbr(x)

• ρ(N D(x i)) =min{k|y k∈N D(x i)}andρ(T(x i)) =min{k|y k∈T(x i)}

The following lemma follows from the definition of lex-convex ordering

Lemma 2.1 Letσ = (x1,x2, ,x n x,y1,y2, ,y n y)be a lex-convex ordering of X ∪Y of a convex bipartite graph G = (X,Y,E) If x a y d,x b y c∈E with x a,x b∈X,y c,y d∈Y and a<b,c<d, then x a y c∈E.

3 Counterexample

In this section, we first introduce the algorithm proposed by Hung et al [8] and then give a counterexample to show the incorrectness of the algorithm To introduce the algorithm, we need to state some notations and terminologies which were introduced in [8

Let G= (X,Y,E)be a convex bipartite graph To be consistent with the notations used by Hung et el [8], we denote

x i by i and y j by j for 1 ≤ i ≤ n xand 1 ≤ j ≤ n y Denote[i,j]by the set of consecutive integers{i,i+1, ,j} Thus,

X= [1, |X|]and Y = [1, |Y|] Call[i,j]an integer interval starting from i and ending at j By definition of a convex bipartite graph, the neighbor of a vertex x in X can be represented as an interval I x = [x.begin,x.end] So the neighbors of vertices

of X can be represented by a set of intervals which is called the interval representation I(X)of X For an interval I x∈I(X),

the smallest integer and largest integer in I x are called the leftmost integer and rightmost integer of I x, respectively Further,

interval I x= [x.begin,x.end]is said to be dominated by integerℓif x.begin6ℓ6x.end.

Partition X into k disjoint clusters X1,X2, ,X k such that u.begin = v.begin if u andvare in the same cluster and

a.begin<b.begin if a ∈X i and b∈X j for i< j Sort the vertices of X i,1≤i≤k, such that a precedes b for a,b∈X iand

a.end6b.end.

Let X i be a cluster of X Define min(X i)and max(X i)to be two vertices in X isuch that min(X i).end6a.end6max(X i).end

for a∈X i Further, let Imin(X i) = [min(X i).begin,min(X i).end]and let Imax(X i) = [max(X i).begin,max(X i).end] The vertices

of Y in Imin(X i)are called common neighbors of vertices of X i Let X1,X2, ,X k be the disjoint sorted clusters of X Define

I (X) = {I (X)|16i6k}and I (X) = {I (X)|16i6k}

Fig 2 Clusters, interval representation I(X)of X, Imin(X), and Imax(X)of I(X)for the convex bipartite graph in Fig 1.

The following algorithm due to Hung et al [8] attempts to compute a minimum paired-dominating set of G.

Algorithm 1: MPD(G)

Partition X into k disjoint clusters X1,X2, ,X k;

Compute the interval representation I(X)of X , and construct Imin(X)and Imax(X)from I(X);

Call Procedure GD-Y on Imin(X)to findYD , and call Procedure GD-X on Imax(X)to findXD;

if (| Y D| =max{| X D|, | Y D|}) then

LetYD= {y1,y2, ,y|Y D|}, and let y i,16i6| Y D|, be the rightmost integer of the interval I x i in Imin(X),

where x i is a vertex of X ;

MPD= ∪16i6|Y D|{y i,x i};

else

LetXD= {x1,x2, ,x|X D|}, and let y i =x i.begin for 16i6| X D|

MPD= ∪16i6|X

D|{y i,x i};

end

The procedures GD-X and GD-Y due to Hung et al [8] computeXDandYD, respectively They claimed thatXDis a minimum

subset of X which dominates X , and YD is a minimum subset of Y which dominates Y

Procedure GD-Y: Given a set Imin(X), Procedure GD-Y computesYDusing a greedy technique as follows: Initially, letYD= ∅,

let Imin = Imin(X), and let s(Imin)be the interval in Iminwith the least rightmost integer Let z be the rightmost integer of

s Imin)and I z be the set of intervals in Imindominated by z Let YD= Y D∪ {z}and let Imin=Imin\I z Repeat the process until

Imin= ∅ OutputYD

Procedure GD-X: The strategy for finding XDuses the following greedy technique: Initially, letXD= ∅,Imax=Imax(X), and let

s Imax)be the interval in Imaxwith the least rightmost integer s Let I′= {I i∈Imax|s Imax) ⊂I i} If I′̸= ∅, then let z∈X such

that I z is the interval with the largest rightmost integer among I′; otherwise, let z =s Let XD= X D∪ {z} LetˆI= {I i ∈Imax|I i

is dominated by z and the rightmost integer of I i larger than the rightmost integer of I z} For I i∈ ˆI, let I˜i = [z+1,i end], i.e I˜i

is obtained from I iby removing[i begin,z] Let˜I= ∪I

i∈ˆI{I˜i}and Imax=Imax− {s Imax)} −I′− (ˆI∪ ˜I) Repeat until Imax = ∅ OutputXD

Consider the convex bipartite graph G= (X,Y,E)ofFig 1.Fig 2also contains the clusters of X , interval representation

I(X)of X,Imin(X), and Imax(X)of I(X) The procedure GD-Y outputsYD = {y3,y8,y11}, and the procedure GD-X outputs



X D= {x3,x6,x8} Now| Y D| =max{| Y D|, | X D|} So the if case needs to be considered.

For y3,x1and x5are the vertices such that y3is the rightmost integer of I x1and I x5in Imin(X) First assume that x1is chosen

by the algorithm For y8,x6is the vertex such that y8is the rightmost integer of I x6in Imin(X) Hence x6will be chosen by the algorithm MPD(G)to be paired with y8 Again for y11,x8and x9are the vertices such that y11is the rightmost integer of I x8 and I x9 in Imin(X) Hence either x8or x9will be chosen by the algorithm to pair with y11 Hence the algorithm MPD(G)can return the set{x1,y3,x6,y8,x8,y11}or the set{x1,y3,x6,y8,x9,y11}as the minimum paired-dominating set of G However, none of these sets are paired-dominating sets of G as y4is not dominated by any of these two sets

If x5 is chosen by the algorithm, then algorithm MPD(G) can return the set {x5,y3,x6,y8,x8,y11} or the set {x5,y3,x6,y8,x9,y11}as the minimum paired-dominating set of G However, none of these sets is a paired-dominating set of G as y4is not dominated by any of these sets

Hence Algorithm MPD(G) fails to produce a minimum paired-dominating set of G Therefore, Algorithm MPD(G)is incorrect

Fig 3 A convex bipartite graph with lex-convex ordering.

4 Our algorithm

Given a lex-convex orderingσ = (x1,x2, ,x n x,y1,y2, ,y n y)of a convex bipartite graph G, we present an algorithm

to compute a minimum paired-dominating set of G Our algorithm processes x1,x2, ,x n xin this order and takes some

decision while processing each vertex x i,1 ≤ i ≤ n x The algorithm maintains two arrays, namely, D and L Initially

L(v) =D(v) =0 for allv ∈X∪Y At any stage, D(v) =1 ifvis dominated by the so far constructed paired-dominating

set, and L(v) =2 if the vertexvhas been selected in the paired-dominating set The algorithm is presented below

Algorithm 2: MPDS(G)

Obtain a lex-convex orderingσ = (x1,x2, ,x n x,y1,y2, ,y n y)of G and assign D(v) =0 and L(v) =0 for all

v ∈X∪Y ;

for i=1 to n xdo

if (D(x i) ==0) then

Case 1: N D(x i) ̸= ∅andM(x i) = ∅

L(maxnbr(x i)) =2; L(M(yρ(ND(i)))) =2;

D(v) =1 for allv ∈ (N[maxnbr(x i)] ∪N[M(yρ(ND(i)))]);

Case 2: N D(x i) ̸= ∅andM(x i) ̸= ∅

L(y m(i)) =2;

x j′′ =M(y m(i))if m(i) < ρ(N D(x i)); else x j′′ =M(yρ(ND(i)));

L(x j′′) =2;

D(v) =1 for allv ∈ (N[y m(i)] ∪N[x j′′]);

Case 3: N D(x i) = ∅andM(x i) = ∅

L(maxnbr(x i)) =2;L(M(maxnbr(x i))) =2;

D(v) =1 for allv ∈ (N[maxnbr(x i)] ∪N[M(maxnbr(x i))]);

Case 4: N D(x i) = ∅andM(x i) ̸= ∅

L(y m(i)) =2;L(M(y m(i))) =2;

D(v) =1 for allv ∈ (N[y m(i)] ∪N[M(y m(i))]);

else

Case 5: N D(x i) ̸= ∅and B D(x i) = ∅

Let x p=M(yρ(ND(i)));

L(x p) =2;

y j′′=maxnbr(x p)if N(x p) ∩T(x p) = ∅; else y j′′=yρ(T(p));

L(y j′′) =2;

D(v) =1 for allv ∈ (N[y j′′] ∪N[x p]);

Case 6: N D(x i) ̸= ∅and B D(x i) ̸= ∅

L(yρ(T(i))) =2;

x j′′ =M(yρ(T(i)))if(ρ(T(x i))) < (ρ(N D(x i))); else x j′′ =M(yρ(ND(i)));

L(x j′′) =2;

D(v) =1 for allv ∈ (N[yρ(T(i))] ∪N[x j′′]);

end

end

Output (PDS= {v|L(v) =2});

Illustration of the algorithm MPDS(G):

We illustrate algorithm MPDS(G)for the graph G ofFig 3and describe the results of different iterations inTable 1 When

D(x i) =1 and N D(x i) = ∅, then the algorithm does not include anything in the minimum paired-dominating set of G So we have skipped such an iteration i.

We, next, prove that algorithm MPDS(G)finds a minimum paired-dominating set of G.

3 x6;D(x6)=1 Case 5 x7y10

The following lemma easily follows by observation of algorithm MPDS(G)

Lemma 4.1 Let x i be the considered vertex for 1≤i≤n x at some point of the algorithm MPDS(G) Then

(i) L(x j) =0 or 2 for all 1≤j≤i−1;

(ii) D(v) =1 for allv ∈ ∪i−1

j=1N[x j].

Let S⊆V and M be a perfect matching of G[S] For U⊆S, we say U is perfect with respect to M if M∩E(G[U])is a perfect

matching of G[U] For each 1≤i≤n x , let D i = {v|L(v) =2}when x i has just been considered In particular, D0 = ∅ At the termination of the algorithm MPDS(G), we have D n x =PDS.Lemma 4.1(ii) implies that PDS is a paired-dominating set Therefore, to prove that PDS is a minimum dominating set, it is sufficient to prove that there is a minimum paired-dominating set D∗of G such that PDS⊆D∗and PDS is perfect with respect to a perfect matching of G[D∗] We prove this by

induction on i in the following lemma.

Lemma 4.2 For each 0≤i≤n x , there is a minimum paired-dominating set D∗of G such that D i ⊆D∗and D i is perfect with respect to a perfect matching of G[D∗].

Proof We prove this by induction i, that is, for every 0≤i≤n x , there is a minimum paired-dominating set D∗of G such that D i ⊆D∗and D i is perfect with respect to a perfect matching of G[D∗] The basis step is trivial Let i=0 Then D0 = ∅

and D0is contained in any minimum paired-dominating set of G Assume that there is a minimum paired-dominating set

D′of G such that D i−1⊆D′and D i−1is perfect with respect to a perfect matching of G[D′] Next we show the existence of a

minimum paired-dominating set D∗∗of G containing D i and D i is perfect with respect to a perfect matching of G[D∗∗] Note

that D is an array used in algorithm MPDS(G)such that D(v) =1 or D(v) =0 for eachv ∈V According to the algorithm

MPDS(G), we discuss the following cases

Case 1: D(x i) =0,N D(x i) ̸= ∅andM(x i) = ∅

Clearly D i = D i−1∪ {maxnbr(x i),M(yρ(ND(i)))} Let y j = maxnbr(x i),x j′′ = M(yρ(ND(i)))and j′ = ρ(N D(x i)) Suppose

y a∈D′is the vertex which dominates x i and let y a be paired with x b in D′ ByLemma 4.1(i), it is clear that y a̸∈D i−1 Since

D i−1is perfect with respect to a perfect matching of G[D′],x b̸∈D i−1 As y a x i ∈E,a≤j SinceM(x i) = ∅,N i(y a) ⊆N i(y j)

First assume that b<i Then x b y i1 ∈E byLemma 2.1 If N j′(x b) = ∅, then(D′\ {y a,x b}) ∪ {y j,x j′′}is a required minimum

paired-dominating set of G If N j′(x b) ̸= ∅, then byLemma 2.1, x b y j′ ∈E and hence N j′(x b) ⊆N j′(x j′′)by choice of x j′′ Since

j′<j and x j′′ y j′ ∈E, we have x j′′ y j∈E Let D′′= (D′\ {y a,x b}) ∪ {y j,x j′′} Then D′′is a required minimum paired-dominating

set of G.

Next assume that b ≥ i If x b y j′ ∈ E, then N j′(x b) ⊆ N j′(x j′′)by the choice of x j′′ and hence(D′\ {x b,y a}) ∪ {x j′′,y j}is

the required minimum paired-dominating set of G So assume that x b y j′ ̸∈ E Since D(y j′) = 0, let x c ∈ D′be the vertex

that dominates y j′ and be paired with y d Clearly byLemma 4.1, x c ̸∈D i−1and y d̸∈D i−1as D i−1is perfect with respect to a

perfect matching of G[D′] Let i1 = σ− 1(minnbr(x i)) If d< i1, then N i(y d) = ∅; otherwise byLemma 2.1, x i y d ∈E which

contradicts the choice of a So d≥i1 By the choice of x j′′ , we have N j′(x c) ⊆ N j′(x j′′) Let x r ∈ N i(y d) Then it is clear that

σ−1(minnbr(x r)) ≥i1 AsM(x i) = ∅,x r y j ∈E So N i(y d) ⊆ N i(y j) Now(D′\ {x c,y d}) ∪ {x j′′,y j}is a required minimum

paired-dominating set of G.

Case 2: D(x i) =0,N D(x i) ̸= ∅, andM(x i) ̸= ∅

Let a be the minimum index such that y a∈D′and y a dominates x i and y a be paired with x b in D′ ByLemma 4.1, y a̸∈D i−1

and x b̸∈D i−1as D i−1is perfect with respect to a perfect matching of G[D′] Let j=m(i)and j′= ρ(N D(x i)) Then we have

a < j; otherwise let x p be the vertex such that j = σ− 1(maxnbr(x p))and let y a′ ∈ D′dominate x p in D′ So, a′ < j As

y i1 = minnbr(x i),i1 < a′ Hence i1 <a′ <j This implies x i y a′ ∈ E and contradicts the fact that a is the minimum index

such that y a∈D′and y a dominates x i

Claim 1: N i(y a) ⊆N i(y j)

Let x l ∈N i(y a) First assume that x l y i1 ∈E, where y i1 =minnbr(x i) If x l y j̸∈E, then x l will precede x i So x l y j∈E Next

assume that x l y i1 ̸∈E If x l y j̸∈E, then m(i) ≤a< j, which is a contradiction to the fact that j=m(i) So x l y j∈E Hence

N(y) ⊆N(y)and the claim is proved

Depending on j<j′or j′≤j, we need to discuss the following two cases.

Subcase 2.1: j<j′

Clearly D i=D i−1∪ {y j,x j′′}, where x j′′ =M(y j) Let y l∈N j(x b) Since a<j<l and y a x b∈E, we have x b y j∈E and hence

by the choice of x j′′ , we get N j(x b) ⊆N j(x j′′)

Let D′′= (D′\ {y a,x b}) ∪ {y j,x j′′} Then D′′is a required minimum paired-dominating set of G.

Subcase 2.2: j′≤j.

In this subcase, D i=D i−1∪ {y j,x j′′}, where x j′′ =M(y j′) Depending on a≤j′or a>j′, we have two subcases

Subcase 2.2(a): a≤j′

Let y l∈N j′(x b) Since a≤j′<l and x b y a∈E, by property of convexity, we have x b y j′ and hence by choice of x j′′, we get

N j′(x b) ⊆N j′(x j′′) Then D′′= (D′\ {y a,x b}) ∪ {y j,x j′′}is a required minimum paired-dominating set of G.

Subcase 2.2(b): j′<a.

If b<i, then x b y j′∈E byLemma 2.1and hence by the choice of x j′′,N j′(x b) ⊆N j′(x j′′) Then D′′= (D′\ {y a,x b}) ∪ {y j,x j′′}

is a required minimum paired-dominating set of G If b≥i and x b y j′ ∈E, then D′′= (D′\ {y a,x b}) ∪ {y j,x j′′}is a required

minimum paired-dominating set of G So assume that x b y j′ ̸∈E Since D(y j′) =0, let x c∈D′dominate y j′and be paired with

y d in D′ Clearly byLemma 4.1, x c̸∈D i−1and y d̸∈D i−1as D i−1is perfect with respect to a perfect matching of G[D′]

Depending on c < i or c ≥ i, we need to consider two cases First assume that c < i Again we have either d≤ j′or

j′<d≤j or j<d.

So, first assume that d≤j′ Then N i(y d) = ∅; otherwise, byLemma 2.1, x i y d∈E since d≤j′<a This is a contradiction

to the fact that a is the minimum index such that y a ∈D′and y a dominates x i Hence we have N i(y d) = ∅ ⊂ N i(y j) Since

x c∈N(y j′), we have N j′(x c) ⊆N j′(x j′′)by choice of x j′′

Next assume that j′<d≤j Then x i y d∈E Clearly d>a by choice of a Now let x l∈N i(y d) Let y i1 =minnbr(x i) If x l y i1 ∈

E and x l y j̸∈E, then x l will precede x i So, x l y j∈E If x l y i1 ̸∈E and x l y j̸∈E, then m(i)(=min{k|y k∈M(x i)}) <j, which is a

contradiction to the fact that j=m(i) This implies x l y j∈E and hence N i(y d) ⊆N i(y j) Therefore D′′= (D′\{y d,x c})∪{y j,x j′′}

is a required minimum paired-dominating set of G.

Next assume that j < d If N d(x b) ̸= ∅, then x b y d ∈ E Then D′′ = (D′\ {x c,y a}) ∪ {x j′′,y j}is a required minimum

paired-dominating set of G If N d(x b) = ∅, then N j′(x b) ⊆ N j′(x c) ⊆N j′(x j′′) So D′′ = (D′\ {y a,x b}) ∪ {y j,x j′′}is a desired

minimum paired-dominating set of G.

Now suppose that c≥i If d≤j′, then x i y d∈E which contradicts the choice of a So d>j′ Therefore, either j′<d≤j

or j<d.

First assume that j′ < d≤j Then x i y d ∈E Clearly d> a by choice of a Now let x l ∈ N i(y d) Let y i1 = minnbr(x i) If

x l y i1 ∈E and x l y j ̸∈E, then x l will precede x i So, x l y j∈ E If x l y i1 ̸∈E and x l y j̸∈ E, then m(i)(=min{k|y k∈M(x i)}) < j,

which is a contradiction to the fact that j = m(i) This implies that x l y j ∈ E and hence N i(y d) ⊆ N i(y j) Therefore

D′′= (D′\ {y d,x c}) ∪ {y j,x j′′}is a required minimum paired-dominating set of G.

Next assume that j < d If N d(x b) ̸= ∅, then x b y d ∈ E Then D′′ = (D′\ {x c,y a}) ∪ {x j′′,y j}is a required minimum

paired-dominating set of G If N d(x b) = ∅, then N j′(x b) ⊆ N j′(x c) ⊆N j′(x j′′) So D′′ = (D′\ {y a,x b}) ∪ {y j,x j′′}is a desired

minimum paired-dominating set of G.

Case 3: D(x i) =0,N D(x i) = ∅, andM(x i) = ∅

It is obvious that D i = D i−1∪ {y j,x j′}, where j= σ− 1(maxnbr(x i))and x j′ = M(maxnbr(x i)) Suppose y a ∈ D′is the

vertex which dominates x i and let y a be paired with x b in D′ ByLemma 4.1, y a̸∈D i−1and x b̸∈D i−1as D i−1is perfect with

respect to a perfect matching of G[D′] Clearly a<j SinceM(x i) = ∅,N i(y a) ⊆N i(y j) Since N D(x i) = ∅,N j(x b) ⊆N j(x j′)

Let D′′= (D′\ {x b,y a}) ∪ {y j,x j′} Then D′′is a required minimum paired-dominating set of G.

Case 4: D(x i) =0,N D(x i) = ∅, andM(x i) ̸= ∅

In this case, D i=D i−1∪ {y j,x j′}, where j=m(i)and x j′ =M(y j) Suppose that a is the minimum index such that y a∈D′

and y a dominates x i and let y a be paired with x b in D′ As D(x i) = 0, byLemma 4.1, y a ̸∈ D i−1 So x b ̸∈ D i−1as D i−1is

perfect with respect to a perfect matching of G[D′] Clearly a<j; otherwise let x p be the vertex such that y j=maxnbr(x p)

and let y a′ ∈ D′dominate x p in D′ So, a′ < j Again i1 < a′ Hence i1 < a′ < j This implies x i y a′ ∈ E and contradicts

the choice of a Again let x k y a ∈ E where k >i Since j = m(i), it is clear that y j x k ∈ E and hence N i(y a) ⊆ N i(y j) Since

N D(x i) = ∅,N j(x b) ⊆N j(x j′) Let D′′= (D′\ {x b,y a}) ∪ {y j,x j′} Then D′′is a desired minimum paired-dominating set of G.

Case 5: D(x i) =1,N D(x i) ̸= ∅, and B D(x i) = ∅

Let j′= ρ(N D(x i))and x p=M(y j′) Let T(x p) = {maxnbr(x)|x∈B D(x p)} We need to consider the following two subcases

Subcase 5.1: N(x p) ∩T(x p) = ∅

In this subcase, D i =D i−1∪ {x p,y j′′}, where j′′ = σ− 1(maxnbr(x p)) Let x a∈D′be the vertex which dominates y j′and

be paired with y b ByLemma 4.1, x a̸∈D i−1and y b̸∈D i−1as D i−1is perfect with respect to a perfect matching of G[D′] Let

x l∈N i(y b)be such that D(x l) =0 Since B D(x i) = ∅and N(x p) ∩T(x p) = ∅,x l∈N i(y j′′) By the choice of x p,N j′(x a) ⊆N j′(x p)

So D′′= (D′\ {x a,y b}) ∪ {x p,y j′′}is a desired minimum paired-dominating set of G.

Subcase 5.2: N(x p) ∩T(x p) ̸= ∅

In this subcase D i = D i−1∪ {M(y j′),y j′′′}, where j′′′ = ρ(T(x p)) Let x a ∈D′be the vertex which dominates y j′and be

paired with y b ByLemma 4.1, x a̸∈D i−1and y b̸∈D i−1as D i−1is perfect with respect to a perfect matching of G[D′] Clearly

j′<j′′′≤ σ− 1(maxnbr(x p)) First assume that b<i1 Then N i(y b) = ∅; otherwise y i ̸=minnbr(x i) Hence N i(y b) ⊆N i(y j′′′)

Case 6: D(x i) =1,N (x i) ̸= ∅, and B (x i) ̸= ∅.

Let T(x i) = ∪x∈B D(i)maxnbr(x),j = ρ(T(x i))and j′ = ρ(N D(x i)) Depending upon j < j′or j′ ≤ j, we discuss the

following two subcases

Subcase 6.1: j<j′

In this case, D i =D i−1∪ {y j,x j′′}, where x j′′ =M(y j) Let x p∈B D(x i)be the vertex such that j= σ− 1(maxnbr(x p))and

let y a ∈D′dominate x p and be paired with x b in D′ Clearly byLemma 4.1, y a ̸∈D i−1and x b̸∈D i−1as D i−1is perfect with

respect to a perfect matching of G[D′] Again a<j Let x l∈N i(y a)be such that x l is not dominated by D i−1 Then x l∈B D(x i) If

x l y j̸∈E, thenρ(T(x i)) = (min{k|y k∈T(x i)}) <j, which is a contradiction So x l y j∈E and N i(y a) ⊆N i(y j) ∪ (∪v∈ i−1N i(v))

Let y l ∈N j(x b) Then x b y j ∈E since x b y a ∈ E with a< j <l Hence by the choice of x j′′ , we have N j(x b) ⊆ N j(x j′′) Then

D′′= (D′\ {x b,y a}) ∪ {x j′′,y j}is a required minimum paired-dominating set of G.

Subcase 6.2: j′≤j.

In this case, D i=D i−1∪{y j,x j′′}, where x j′′ =M(y j′) Let x p∈B D(x i)be the vertex such that j= σ− 1(maxnbr(x p))and let a

be the minimum index such that y a∈D′and y a dominates x p and is paired with x b in D′ Clearly byLemma 4.1, y a̸∈D i−1and

x b̸∈D i−1as D i−1is perfect with respect to a perfect matching of G[D′] Again a<j Then N i(y a) ⊆N i(y j) ∪ (∪v∈ i−1N i(v))

Depending on a≤j′or j′<a, we have the following subcases.

Subcase 6.2(a): a≤j′

Let y l ∈N j′(x b) Then by the convexity property, x b y j′ ∈E Hence by the choice of x j′′ , we have N j′(x b) ⊆N j′(x j′′) Then

D′′= (D′\ {x b,y a}) ∪ {x j′′,y j}is a required minimum paired-dominating set of G.

Subcase 6.2(b): j′<a.

If b<i, then x b y j′∈E byLemma 2.1and hence by the choice of x j′′,N j′(x b) ⊆N j′(x j′′) Then D′′= (D′\ {y a,x b}) ∪ {y j,x j′′}

If b≥i and x b y j′ ∈E, then D′′ = (D′\ {y a,x b}) ∪ {y j,x j′′}is a minimum paired-dominating set of G such that D i ⊆D′′ So

assume that x b y j′ ̸∈E Since D(y j′) =0, let x c ∈D′dominate y j′ and be paired with y d in D′ ByLemma 4.1, x c ̸∈D i−1and

y d̸∈D i−1as D i−1is perfect with respect to a perfect matching of G[D′]

First assume that either d ≤j′ < a or j′ < d≤a If c < p, then N p(y d) = ∅; otherwise byLemma 2.1, x p y d ∈E with

d≤a This is a contradiction to the choice of a Hence we have N p(y d) = ∅ ⊂N p(y j) If c≥p, then x p y d∈E This contradicts

the choice of a Again if there is a vertex x r ∈ N(y d)with i < r < p, then x r y j ∈ E; otherwise we get a contradiction

to the choice of j So we have N i(y d) ⊆ N i y j Again since x c ∈ N(y j′), we have N j′(x c) ⊆ N j′(x j′′)by the choice of x j′′ Let

D′′= (D′\ {x c,y d}) ∪ {x j′′,y j} Clearly D′′is a required minimum paired-dominating set of G.

Next assume that a<d≤j, then x i y d∈E Now let x l ∈N i(y d)such that D(x l) =0 Then x l y j∈E; otherwise we get a

contradiction to the choice of j Hence N i(y d) ⊆N i(y j) ∪ (∪v∈ i−1N i(v)) Then D′′ = (D′\ {y d,x c}) ∪ {y j,x j′′}is a required

minimum paired-dominating set of G.

Next assume that j<d If N d(x b) ̸= ∅, then x b y d∈E Let D′′= (D′\ {x c,y a}) ∪ {x j′′,y j} Clearly D′′is a minimum

paired-dominating set of G such that D i ⊆D′′ If N d(x b) = ∅, then N j′(x b) ⊆N j′(x c) ⊆N j′(x j′′) Let D′′ = (D′\ {y a,x b}) ∪ {y j,x j′′}

Clearly D′′is a minimum paired-dominating set of G. 

Next we show how algorithm MPDS(G)can be implemented in linear time

Given a convex orderingσ = (x1,x2, ,x n x,y1,y2, ,y n y), it takes O(d(x i))time to find the minnbr(x i)and maxnbr(x i)

for each i,1≤i≤n x So minnbr(x)and maxnbr(x)can be computed for all vertices in X in O(n+m)time Given a convex orderingσ = (x1,x2, ,x n x,y1,y2, ,y n y)of G, and(minnbr(x i),maxnbr(x i))for 1≤i≤n x, a lex-convex orderingαof

G can be computed in O(n+m)time using a bucket sort

We maintain an array A[1 n y]of pointers such that A[i]points to the set S i = {x ∈ X|maxnbr(x) = y i}which is

represented as a doubly linked list As each x i has a unique maximum neighbor, S i ∩S j = ∅for i ̸= j We also maintain

an array A D[1 n y]of pointers such that A D[i]points to the set S D

i = {x ∈ X|maxnbr(x) = y i,D(x) = 0}of X , which is represented as a doubly linked list We maintain an array A X[1 n x]of pointers such that A X[i]points the cell containing x i which is present in the doubly linked list pointed to by A D[j]where y j =maxnbr(x i) This is needed so that any x ican be

deleted from the set S j , where j=maxnbr(x i)in O(1)time Initially, A=A D Note that the A,A D , and A Xcan be computed in

O(n+m)time We also maintain an array C[1 n x]such that C[i] =0

The algorithm needs to find m(i),M(y i), and j= ρ(T(x i)), where T(x i) = ∪x∈B D(i)maxnbr(x),B(x i) = {x ∈NN i(y i1)|

N(x i) ∩ {maxnbr(x)} ̸= ∅}, and B D(x i) = {x∈B(x i)|D(x) =0} To compute m(i), we need to check first whetherM(x i) = ∅ This is done as follows:

Let N(x i) = {y i1,y i2, ,y i k} Scan the sets starting from S i2to S i k till a vertex, say x, which is not a neighbor of y i1is

found If j is the minimum index such that S i j contains a vertex that is not a neighbor of y i1, then m(i) = j If no such j is

found, thenM(x i) = ∅and we return m(i) = maxnbr(x i) To check whether a vertex x j is a neighbor of y i1in O(1)time,

we use the array C Let N(y i1) = {x i1,x i2, ,x i r} Set C[i j] =1 for j=1,2, ,r This preprocessing takes O(d(y i1))time

Now C[j] =1 if and only if x j is a neighbor of y i After the ith iteration, we compute N(x i+1) = {y(i+ 1 ),y(i+ 1 ), ,y(i+ 1 )}

If y i1 ̸= y(i+1)1, then we make C[i j] = 0 for j = 1,2, ,r so that at the start of the(i+1)th iteration C[k] = 0 for

k=1,2, ,n x If y i1=y(i+1)1, we leave C unchanged after the ith iteration and use it in the(i+1)th iteration So checking whetherM(x i) = ∅and computing m(i)takes O(d(y i1))time at the ith iteration.

To computeρ(T(x i)), we use the array A D Note that S D[i] = {x j|maxnbr(x j) = y i and D(x j) = 0} Let N(x i) = {y i1,y i2, ,y i k} Scan the sets starting from S i D

2 to S i D k till a vertex, say x, which is not a neighbor of y i1 is found If j is the minimum index such that S D

i j contains a vertex that is not a neighbor of y i1, then return j If no such j is found, then

B D(x i) = ∅ So checking whetherM(x i) = ∅and computingρ(T(x i))takes O(d(y i1))time at the ith iteration Whenever a vertex y i is selected in the paired-dominating set, D(x j)becomes 1 for all x j∈N(y i) We reflect this in A D Using the pointer

A X[j], we delete the vertex x j from the doubly linked list A D[k], where y k = maxnbr(x j) So A D can be updated in O(d(y i))

time whenever y iis selected in the paired-dominating set

To compute M(y j), we maintain an array R[1 n y]of pointers such that R[j]points to the set{(x,d j(x))|x∈N(y j)}, where

d j(x) = |N(x) ∩ {y j+1,y j+2, ,y n y}| This set is implemented as a linked list such that each cell of the linked list contains

two fields: one containing a neighbor of y j , say x and the other contains d j(x) To construct the array R, we first use an array

DG[1 n x]such that DG[i] =d(x i) For all x i∈N(y1), decrement DG[i]by one and then form the linked list R[1]by inserting

all the neighbors of y i and their current DG values Once the linked list R[i]is formed, decrement one from DG[k]for all

x k∈ N(y i+1) Now form the linked list R[i+1]by inserting each neighbor of y i+1along with their corresponding current

DG values In this process the array R of linked lists is constructed Note that each cell of the linked list pointed to by R[i]

contains a neighbor, say x, of y i and d i(x), 1≤i≤n y As constructing R[i]takes O(d(y i))time, the construction of R takes

O(n+m)time Now M(y j)is the vertex, say x, in R[j]having maximum d j(x) So M(y j)can be computed in O(n+m)time

for all j=1,2, ,n y

It is easy to check that the other parts of the algorithm MPDS(G)can be implemented in O(n+m)time So algorithm MPDS(G)takes O(n+m)time

In view ofLemma 4.2and the above discussions, we have the following theorem

Theorem 4.3 A minimum paired-dominating set of a convex bipartite graph G= (X,Y,E)can be computed in O(n+m)time.

Acknowledgments

The authors would like to thank the anonymous referees for their detailed review and for their helpful and constructive comments leading to improvements in the presentation of the paper

The second author was supported by Council of Scientific & Industrial Research (CSIR), INDIA

References

[1] T Araki, A linear time algorithm for finding the minimum paired-dominating set in series–parallel graphs, Inf Process Soc Japan 04 (2007) 15–22.

[2] K.S Booth, G.S Lueker, Testing for the consecutive ones property, interval graphs, and graph planarity using PQ -tree algorithms, J Comput System Sci.

13 (1976) 335–379.

[3] L Chen, C Lu, Z Zeng, Labelling algorithms for paired-domination problems in block and interval graphs, J Comb Optim 19 (2008) 457–470 [4] L Chen, C Lu, Z Zeng, A linear-time algorithm for paired-domination problem in strongly chordal graphs, Inform Process Lett 110 (2009) 20–23 [5] T.C.E Cheng, L.Y Kang, C.T Ng, Paired domination on interval and circular-arc graphs, Discrete Appl Math 155 (2007) 2077–2086.

[6] T.C.E Cheng, L.Y Kang, E Shan, A polynomial-time algorithm for the paired-domination problem on permutation graphs, Discrete Appl Math 157 (2009) 262–271.

[7] T.W Haynes, P.J Slater, Paired-domination in graphs, Networks 32 (1998) 199–206.

[8] R.W Hung, C.H Laio, C.K Wang, Efficient algorithm for the paired domination problem in convex bipartite graphs, in: Proceedings of the International MultiConference of Engineers and Computer Scientists 2010, IMECS 2010, March 17–19, 2010, Hong Kong, China, volume I, 2010 See http://www.iaeng.org/publication/IMECS2010/IMECS2010_pp365-369.pdf.

[9] H Qiao, L Kang, M Cardei, D.Z Du, Paired-domination of trees, J Global Optim 25 (2003) 43–54.