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

Discrete Applied Mathematics 161 (2013) 1776–1783 Contents lists available at SciVerse ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam A linear time algorithm for computing a minimum paired-dominating set of a convex bipartite graph B.S Panda ∗ , D Pradhan Computer Science and Application Group, Department of Mathematics, Indian Institute of Technology Delhi, Hauz Khas, New Delhi 110 016, India article info Article history: Received 26 July 2010 Received in revised form December 2011 Accepted 17 April 2012 Available online 18 May 2012 Keywords: Paired-domination Perfect matching Convex bipartite graphs abstract 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 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 paireddomination 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 Section of 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 Section 4, we propose a linear time algorithm to compute a minimum paired-dominating set of a convex bipartite graph Preliminaries For a graph G = (V , E ), the sets N (v) = {u ∈ V (G)|uv ∈ E } and N [v] = N (v) ∪ {v} denote the neighborhood and the closed neighborhood of a vertex v , respectively The degree of a vertex v is |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 doi:10.1016/j.dam.2012.04.014 B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 1777 Fig 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 | = nx , |Y | = ny , and n = nx + ny 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] Let σ = (x1 , x2 , , xnx , y1 , y2 , , yny ) be a convex ordering of a convex bipartite graph G = (X , Y , E ) For each xi ∈ X , let minnbr (xi ) (resp maxnbr (xi )) be the minimum (maximum) indexed vertex with respect to σ adjacent to xi An ordering α = (x1 , x2 , , xnx , y1 , y2 , , yny ) of a convex bipartite graph G = (X , Y , E ) is called a lexicographic convex (lex-convex) ordering if α is a convex ordering and for xi and xj , i < j if either minnbr (xi ) < minnbr (xj ) or minnbr (xi ) = minnbr (xj ) and maxnbr (xi ) ≤ maxnbr (xj ) For an ordering σ = (v1 , v2 , , ), let σ −1 (x) = i if x = vi , ≤ i ≤ n The ordering σ = (x1 , x3 , x5 , x4 , x2 , x6 , x7 , x8 , x9 , y1 , , y11 ) is a lex-convex ordering of the convex bipartite graph G of Fig Let σ = (x1 , x2 , , xnx , y1 , y2 , , yny ) be an ordering of X ∪ Y of a bipartite graph G = (X , Y , E ) For xi ∈ X , let Nj (xi ) = N (xi ) ∩ {yj+1 , yj+2 , , yny } and dj (xi ) = |Nj (xi )| We denote M (yj ) to be the vertex x ∈ N (yj ) such that dj (x) = max{dj (x′ )|x′ ∈ N (yj )} For yi ∈ Y , let Nj (yi ) = N (yi ) ∩ {xj+1 , xj+2 , , xnx } and dj (yi ) = |Nj (yi )| We denote M (xj ) to be the vertex y ∈ N (xj ) such that dj (y) = max{dj (y′ )|y′ ∈ N (xj )} Let (x1 , x2 , , xnx , y1 , y2 , , yny ) be a lex-convex ordering of a convex bipartite graph G = (X , Y , E ) For xi ∈ X , ≤ i ≤ nx , we use the following notations to describe our algorithms • yi1 = minnbr (xi ) and NNi (yi1 ) = {xk |xk yi1 ̸∈ E , k > i} (the set of non-neighbors of yi1 which succeed xi in σ ) • M (xi ) = N (xi ) ∩ (∪x∈NNi (yi1 ) maxnbr (x)) and B(xi ) = {x ∈ NNi (yi1 )|N (xi ) ∩ {maxnbr (x)} ̸= ∅}  −1 (xi )), if M (xi ) = ∅; • m(i) = σmin{(kmaxnbr |y ∈ M (x )}, otherwise k i • D is an array such that D(v) = or D(v) = for each v ∈ V • N D (xi ) = {yk ∈ N (xi )|D(yk ) = 0}, BD (xi ) = {x ∈ B(xi )|D(x) = 0} and T (xi ) = ∪x∈BD (xi ) maxnbr (x) • ρ(N D (xi )) = min{k|yk ∈ N D (xi )} and ρ(T (xi )) = min{k|yk ∈ T (xi )} The following lemma follows from the definition of lex-convex ordering Lemma 2.1 Let σ = (x1 , x2 , , xnx , y1 , y2 , , yny ) be a lex-convex ordering of X ∪ Y of a convex bipartite graph G = (X , Y , E ) If xa yd , xb yc ∈ E with xa , xb ∈ X , yc , yd ∈ Y and a < b, c < d, then xa yc ∈ E 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 xi by i and yj by j for ≤ i ≤ nx and ≤ j ≤ ny 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 Ix = [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 Ix ∈ I (X ), the smallest integer and largest integer in Ix are called the leftmost integer and rightmost integer of Ix , respectively Further, interval Ix = [x.begin, x.end] is said to be dominated by integer ℓ if x.begin ℓ x.end Partition X into k disjoint clusters X1 , X2 , , Xk such that u.begin = v.begin if u and v are in the same cluster and a.begin < b.begin if a ∈ Xi and b ∈ Xj for i < j Sort the vertices of Xi , ≤ i ≤ k, such that a precedes b for a, b ∈ Xi and a.end b.end Let Xi be a cluster of X Define min(Xi ) and max(Xi ) to be two vertices in Xi such that min(Xi ).end a.end max(Xi ).end for a ∈ Xi Further, let Imin (Xi ) = [min(Xi ).begin, min(Xi ).end] and let Imax (Xi ) = [max(Xi ).begin, max(Xi ).end] The vertices of Y in Imin (Xi ) are called common neighbors of vertices of Xi Let X1 , X2 , , Xk be the disjoint sorted clusters of X Define Imin (X ) = {Imin (Xi )|1 i k} and Imax (X ) = {Imax (Xi )|1 i k} 1778 B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 Fig Clusters, interval representation I (X ) of X , Imin (X ), and Imax (X ) of I (X ) for the convex bipartite graph in Fig 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 , , Xk ; 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 find YD , and call Procedure GD-X on Imax (X ) to find XD ; if (|YD | = max{|XD |, |YD |}) then Let YD = {y1 , y2 , , y|YD | }, and let yi , i |YD |, be the rightmost integer of the interval Ixi in Imin (X ), where xi is a vertex of X ; MPD = ∪16i6|YD | {yi , xi }; else Let XD = {x1 , x2 , , x|XD | }, and let yi = xi begin for i |XD |; MPD = ∪16i6|XD | {yi , xi }; end The procedures GD-X and GD-Y due to Hung et al [8] compute XD and YD , respectively They claimed that XD is 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 computes YD using a greedy technique as follows: Initially, let YD = ∅, let Imin = Imin (X ), and let s(Imin ) be the interval in Imin with the least rightmost integer Let z be the rightmost integer of s(Imin ) and I z be the set of intervals in Imin dominated by z Let YD = YD ∪ {z } and let Imin = Imin \ I z Repeat the process until Imin = ∅ Output YD Procedure GD-X: The strategy for finding XD uses the following greedy technique: Initially, let XD = ∅, Imax = Imax (X ), and let s(Imax ) be the interval in Imax with the least rightmost integer s Let I ′ = {Ii ∈ Imax |s(Imax ) ⊂ Ii } If I ′ ̸= ∅, then let z ∈ X such that Iz is the interval with the largest rightmost integer among I ′ ; otherwise, let z = s Let XD = XD ∪ {z } Let ˆI = {Ii ∈ Imax |Ii is dominated by z and the rightmost integer of Ii larger than the rightmost integer of Iz } For Ii ∈ ˆI, let I˜i = [z + 1, i.end], i.e I˜i is obtained from Ii by removing [i.begin, z ] Let ˜I = ∪I ∈ˆI {I˜i } and Imax = Imax − {s(Imax )} − I ′ − (ˆI ∪ ˜I ) Repeat until Imax = ∅ i Output XD Consider the convex bipartite graph G = (X , Y , E ) of Fig Fig also contains the clusters of X , interval representation I (X ) of X , Imin (X ), and Imax (X ) of I (X ) The procedure GD-Y outputs Y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 , x1 and x5 are the vertices such that y3 is the rightmost integer of Ix1 and Ix5 in Imin (X ) First assume that x1 is chosen by the algorithm For y8 , x6 is the vertex such that y8 is the rightmost integer of Ix6 in Imin (X ) Hence x6 will be chosen by the algorithm MPD(G) to be paired with y8 Again for y11 , x8 and x9 are the vertices such that y11 is the rightmost integer of Ix8 and Ix9 in Imin (X ) Hence either x8 or x9 will 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 y4 is 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 y4 is 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 B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 1779 Fig A convex bipartite graph with lex-convex ordering Our algorithm Given a lex-convex ordering σ = (x1 , x2 , , xnx , y1 , y2 , , yny ) of a convex bipartite graph G, we present an algorithm to compute a minimum paired-dominating set of G Our algorithm processes x1 , x2 , , xnx in this order and takes some decision while processing each vertex xi , ≤ i ≤ nx The algorithm maintains two arrays, namely, D and L Initially L(v) = D(v) = for all v ∈ X ∪ Y At any stage, D(v) = if v is dominated by the so far constructed paired-dominating set, and L(v) = if the vertex v has been selected in the paired-dominating set The algorithm is presented below Algorithm 2: MPDS(G) Obtain a lex-convex ordering σ = (x1 , x2 , , xnx , y1 , y2 , , yny ) of G and assign D(v) = and L(v) = for all v ∈ X ∪ Y; for i = to nx if (D(xi ) == 0) then Case 1: N D (xi ) ̸= ∅ and M (xi ) = ∅ L(maxnbr (xi )) = 2; L(M (yρ(N D (xi )) )) = 2; D(v) = for all v ∈ (N [maxnbr (xi )] ∪ N [M (yρ(N D (xi )) )]); Case 2: N D (xi ) ̸= ∅ and M (xi ) ̸= ∅ L(ym(i) ) = 2; xj′′ = M (ym(i) ) if m(i) < ρ(N D (xi )); else xj′′ = M (yρ(N D (xi )) ); L(xj′′ ) = 2; D(v) = for all v ∈ (N [ym(i) ] ∪ N [xj′′ ]); Case 3: N D (xi ) = ∅ and M (xi ) = ∅ L(maxnbr (xi )) = 2; L(M (maxnbr (xi ))) = 2; D(v) = for all v ∈ (N [maxnbr (xi )] ∪ N [M (maxnbr (xi ))]); Case 4: N D (xi ) = ∅ and M (xi ) ̸= ∅ L(ym(i) ) = 2; L(M (ym(i) )) = 2; D(v) = for all v ∈ (N [ym(i) ] ∪ N [M (ym(i) )]); else Case 5: N D (xi ) ̸= ∅ and BD (xi ) = ∅ Let xp = M (yρ(N D (xi )) ); L(xp ) = 2; yj′′ = maxnbr (xp ) if N (xp ) ∩ T (xp ) = ∅; else yj′′ = yρ(T (xp )) ; L(yj′′ ) = 2; D(v) = for all v ∈ (N [yj′′ ] ∪ N [xp ]); Case 6: N D (xi ) ̸= ∅ and BD (xi ) ̸= ∅ L(yρ(T (xi )) ) = 2; xj′′ = M (yρ(T (xi )) ) if (ρ(T (xi ))) < (ρ(N D (xi ))); else xj′′ = M (yρ(N D (xi )) ); L(xj′′ ) = 2; D(v) = for all v ∈ (N [yρ(T (xi )) ] ∪ N [xj′′ ]); end end Output (PDS = {v|L(v) = 2}); Illustration of the algorithm MPDS(G): We illustrate algorithm MPDS(G) for the graph G of Fig and describe the results of different iterations in Table When D(xi ) = and N D (xi ) = ∅, 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 1780 B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 Table Minimum paired-dominating set of the Fig Iteration i Considering vertex v with D(v) Case of algorithm applied Edge selected x1 ; D(x1 ) = x3 ; D(x3 ) = x6 ; D(x6 ) = x9 ; D(x9 ) = x12 ; D(x12 ) = x14 ; D(x14 ) = x16 ; D(x16 ) = x19 ; D(x19 ) = x22 ; D(x22 ) = Case Case Case Case Case Case Case Case Case x1 y3 x3 y5 x7 y10 x10 y14 x13 y17 x14 y20 x17 y23 x21 y28 x23 y31 The following lemma easily follows by observation of algorithm MPDS(G) Lemma 4.1 Let xi be the considered vertex for ≤ i ≤ nx at some point of the algorithm MPDS(G) Then (i) L(xj ) = or for all ≤ j ≤ i − 1; (ii) D(v) = for all v ∈ ∪ij− =1 N [xj ] 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 ≤ i ≤ nx , let Di = {v|L(v) = 2} when xi has just been considered In particular, D0 = ∅ At the termination of the algorithm MPDS(G), we have Dnx = PDS Lemma 4.1(ii) implies that PDS is a paired-dominating set Therefore, to prove that PDS is a minimum paired-dominating set, it is sufficient to prove that there is a minimum paireddominating 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 ≤ i ≤ nx , there is a minimum paired-dominating set D∗ of G such that Di ⊆ D∗ and Di is perfect with respect to a perfect matching of G[D∗ ] Proof We prove this by induction i, that is, for every ≤ i ≤ nx , there is a minimum paired-dominating set D∗ of G such that Di ⊆ D∗ and Di is perfect with respect to a perfect matching of G[D∗ ] The basis step is trivial Let i = Then D0 = ∅ and D0 is contained in any minimum paired-dominating set of G Assume that there is a minimum paired-dominating set D′ of G such that Di−1 ⊆ D′ and Di−1 is 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 Di and Di 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) = or D(v) = for each v ∈ V According to the algorithm MPDS(G), we discuss the following cases Case 1: D(xi ) = 0, N D (xi ) ̸= ∅ and M (xi ) = ∅ Clearly Di = Di−1 ∪ {maxnbr (xi ), M (yρ(N D (xi )) )} Let yj = maxnbr (xi ), xj′′ = M (yρ(N D (xi )) ) and j′ = ρ(N D (xi )) Suppose ya ∈ D′ is the vertex which dominates xi and let ya be paired with xb in D′ By Lemma 4.1(i), it is clear that ya ̸∈ Di−1 Since Di−1 is perfect with respect to a perfect matching of G[D′ ], xb ̸∈ Di−1 As ya xi ∈ E , a ≤ j Since M (xi ) = ∅, Ni (ya ) ⊆ Ni (yj ) First assume that b < i Then xb yi1 ∈ E by Lemma 2.1 If Nj′ (xb ) = ∅, then (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a required minimum paired-dominating set of G If Nj′ (xb ) ̸= ∅, then by Lemma 2.1, xb yj′ ∈ E and hence Nj′ (xb ) ⊆ Nj′ (xj′′ ) by choice of xj′′ Since j′ < j and xj′′ yj′ ∈ E, we have xj′′ yj ∈ E Let D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } Then D′′ is a required minimum paired-dominating set of G Next assume that b ≥ i If xb yj′ ∈ E, then Nj′ (xb ) ⊆ Nj′ (xj′′ ) by the choice of xj′′ and hence (D′ \ {xb , ya }) ∪ {xj′′ , yj } is the required minimum paired-dominating set of G So assume that xb yj′ ̸∈ E Since D(yj′ ) = 0, let xc ∈ D′ be the vertex that dominates yj′ and be paired with yd Clearly by Lemma 4.1, xc ̸∈ Di−1 and yd ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Let i1 = σ −1 (minnbr (xi )) If d < i1 , then Ni (yd ) = ∅; otherwise by Lemma 2.1, xi yd ∈ E which contradicts the choice of a So d ≥ i1 By the choice of xj′′ , we have Nj′ (xc ) ⊆ Nj′ (xj′′ ) Let xr ∈ Ni (yd ) Then it is clear that σ −1 (minnbr (xr )) ≥ i1 As M (xi ) = ∅, xr yj ∈ E So Ni (yd ) ⊆ Ni (yj ) Now (D′ \ {xc , yd }) ∪ {xj′′ , yj } is a required minimum paired-dominating set of G Case 2: D(xi ) = 0, N D (xi ) ̸= ∅, and M (xi ) ̸= ∅ Let a be the minimum index such that ya ∈ D′ and ya dominates xi and ya be paired with xb in D′ By Lemma 4.1, ya ̸∈ Di−1 and xb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Let j = m(i) and j′ = ρ(N D (xi )) Then we have a < j; otherwise let xp be the vertex such that j = σ −1 (maxnbr (xp )) and let ya′ ∈ D′ dominate xp in D′ So, a′ < j As yi1 = minnbr (xi ), i1 < a′ Hence i1 < a′ < j This implies xi ya′ ∈ E and contradicts the fact that a is the minimum index such that ya ∈ D′ and ya dominates xi Claim 1: Ni (ya ) ⊆ Ni (yj ) Let xl ∈ Ni (ya ) First assume that xl yi1 ∈ E, where yi1 = minnbr (xi ) If xl yj ̸∈ E, then xl will precede xi So xl yj ∈ E Next assume that xl yi1 ̸∈ E If xl yj ̸∈ E, then m(i) ≤ a < j, which is a contradiction to the fact that j = m(i) So xl yj ∈ E Hence Ni (ya ) ⊆ Ni (yj ) and the claim is proved B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 1781 Depending on j < j′ or j′ ≤ j, we need to discuss the following two cases Subcase 2.1: j < j′ Clearly Di = Di−1 ∪ {yj , xj′′ }, where xj′′ = M (yj ) Let yl ∈ Nj (xb ) Since a < j < l and ya xb ∈ E, we have xb yj ∈ E and hence by the choice of xj′′ , we get Nj (xb ) ⊆ Nj (xj′′ ) Let D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } Then D′′ is a required minimum paired-dominating set of G Subcase 2.2: j′ ≤ j In this subcase, Di = Di−1 ∪ {yj , xj′′ }, where xj′′ = M (yj′ ) Depending on a ≤ j′ or a > j′ , we have two subcases Subcase 2.2(a): a ≤ j′ Let yl ∈ Nj′ (xb ) Since a ≤ j′ < l and xb ya ∈ E, by property of convexity, we have xb yj′ and hence by choice of xj′′ , we get Nj′ (xb ) ⊆ Nj′ (xj′′ ) Then D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a required minimum paired-dominating set of G Subcase 2.2(b): j′ < a If b < i, then xb yj′ ∈ E by Lemma 2.1 and hence by the choice of xj′′ , Nj′ (xb ) ⊆ Nj′ (xj′′ ) Then D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a required minimum paired-dominating set of G If b ≥ i and xb yj′ ∈ E, then D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a required minimum paired-dominating set of G So assume that xb yj′ ̸∈ E Since D(yj′ ) = 0, let xc ∈ D′ dominate yj′ and be paired with yd in D′ Clearly by Lemma 4.1, xc ̸∈ Di−1 and yd ̸∈ Di−1 as Di−1 is 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 Ni (yd ) = ∅; otherwise, by Lemma 2.1, xi yd ∈ E since d ≤ j′ < a This is a contradiction to the fact that a is the minimum index such that ya ∈ D′ and ya dominates xi Hence we have Ni (yd ) = ∅ ⊂ Ni (yj ) Since xc ∈ N (yj′ ), we have Nj′ (xc ) ⊆ Nj′ (xj′′ ) by choice of xj′′ Next assume that j′ < d ≤ j Then xi yd ∈ E Clearly d > a by choice of a Now let xl ∈ Ni (yd ) Let yi1 = minnbr (xi ) If xl yi1 ∈ E and xl yj ̸∈ E, then xl will precede xi So, xl yj ∈ E If xl yi1 ̸∈ E and xl yj ̸∈ E, then m(i)(= min{k|yk ∈ M (xi )}) < j, which is a contradiction to the fact that j = m(i) This implies xl yj ∈ E and hence Ni (yd ) ⊆ Ni (yj ) Therefore D′′ = (D′ \{yd , xc })∪{yj , xj′′ } is a required minimum paired-dominating set of G Next assume that j < d If Nd (xb ) ̸= ∅, then xb yd ∈ E Then D′′ = (D′ \ {xc , ya }) ∪ {xj′′ , yj } is a required minimum paired-dominating set of G If Nd (xb ) = ∅, then Nj′ (xb ) ⊆ Nj′ (xc ) ⊆ Nj′ (xj′′ ) So D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a desired minimum paired-dominating set of G Now suppose that c ≥ i If d ≤ j′ , then xi yd ∈ 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 xi yd ∈ E Clearly d > a by choice of a Now let xl ∈ Ni (yd ) Let yi1 = minnbr (xi ) If xl yi1 ∈ E and xl yj ̸∈ E, then xl will precede xi So, xl yj ∈ E If xl yi1 ̸∈ E and xl yj ̸∈ E, then m(i)(= min{k|yk ∈ M (xi )}) < j, which is a contradiction to the fact that j = m(i) This implies that xl yj ∈ E and hence Ni (yd ) ⊆ Ni (yj ) Therefore D′′ = (D′ \ {yd , xc }) ∪ {yj , xj′′ } is a required minimum paired-dominating set of G Next assume that j < d If Nd (xb ) ̸= ∅, then xb yd ∈ E Then D′′ = (D′ \ {xc , ya }) ∪ {xj′′ , yj } is a required minimum paired-dominating set of G If Nd (xb ) = ∅, then Nj′ (xb ) ⊆ Nj′ (xc ) ⊆ Nj′ (xj′′ ) So D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a desired minimum paired-dominating set of G Case 3: D(xi ) = 0, N D (xi ) = ∅, and M (xi ) = ∅ It is obvious that Di = Di−1 ∪ {yj , xj′ }, where j = σ −1 (maxnbr (xi )) and xj′ = M (maxnbr (xi )) Suppose ya ∈ D′ is the vertex which dominates xi and let ya be paired with xb in D′ By Lemma 4.1, ya ̸∈ Di−1 and xb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Clearly a < j Since M (xi ) = ∅, Ni (ya ) ⊆ Ni (yj ) Since N D (xi ) = ∅, Nj (xb ) ⊆ Nj (xj′ ) Let D′′ = (D′ \ {xb , ya }) ∪ {yj , xj′ } Then D′′ is a required minimum paired-dominating set of G Case 4: D(xi ) = 0, N D (xi ) = ∅, and M (xi ) ̸= ∅ In this case, Di = Di−1 ∪ {yj , xj′ }, where j = m(i) and xj′ = M (yj ) Suppose that a is the minimum index such that ya ∈ D′ and ya dominates xi and let ya be paired with xb in D′ As D(xi ) = 0, by Lemma 4.1, ya ̸∈ Di−1 So xb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Clearly a < j; otherwise let xp be the vertex such that yj = maxnbr (xp ) and let ya′ ∈ D′ dominate xp in D′ So, a′ < j Again i1 < a′ Hence i1 < a′ < j This implies xi ya′ ∈ E and contradicts the choice of a Again let xk ya ∈ E where k > i Since j = m(i), it is clear that yj xk ∈ E and hence Ni (ya ) ⊆ Ni (yj ) Since N D (xi ) = ∅, Nj (xb ) ⊆ Nj (xj′ ) Let D′′ = (D′ \ {xb , ya }) ∪ {yj , xj′ } Then D′′ is a desired minimum paired-dominating set of G Case 5: D(xi ) = 1, N D (xi ) ̸= ∅, and BD (xi ) = ∅ Let j′ = ρ(N D (xi )) and xp = M (yj′ ) Let T (xp ) = {maxnbr (x)|x ∈ BD (xp )} We need to consider the following two subcases Subcase 5.1: N (xp ) ∩ T (xp ) = ∅ In this subcase, Di = Di−1 ∪ {xp , yj′′ }, where j′′ = σ −1 (maxnbr (xp )) Let xa ∈ D′ be the vertex which dominates yj′ and be paired with yb By Lemma 4.1, xa ̸∈ Di−1 and yb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Let xl ∈ Ni (yb ) be such that D(xl ) = Since BD (xi ) = ∅ and N (xp )∩ T (xp ) = ∅, xl ∈ Ni (yj′′ ) By the choice of xp , Nj′ (xa ) ⊆ Nj′ (xp ) So D′′ = (D′ \ {xa , yb }) ∪ {xp , yj′′ } is a desired minimum paired-dominating set of G Subcase 5.2: N (xp ) ∩ T (xp ) ̸= ∅ In this subcase Di = Di−1 ∪ {M (yj′ ), yj′′′ }, where j′′′ = ρ(T (xp )) Let xa ∈ D′ be the vertex which dominates yj′ and be paired with yb By Lemma 4.1, xa ̸∈ Di−1 and yb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Clearly j′ < j′′′ ≤ σ −1 (maxnbr (xp )) First assume that b < i1 Then Ni (yb ) = ∅; otherwise yi1 ̸= minnbr (xi ) Hence Ni (yb ) ⊆ Ni (yj′′′ ) 1782 B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 Next assume that b ≥ i1 Let xl ∈ Ni (yb ) Then either xl ∈ B(xi ), x ∈ B(xp ) or xl ∈ B(xi ) ∩ B(xp ) If xl ∈ B(xi ), then D(xl ) = since BD (xi ) = ∅ If x ∈ B(xp ), then D(x) = or xl yj′′′ ∈ E; otherwise we get a contradiction to the choice of j′′′ Hence Ni (yb ) ⊆ Ni (yj′′′ ) By the choice of xp , Nj′ (xa ) ⊆ Nj′ (xp ) Let D′′ = (D′ \ {xa , yb }) ∪ {M (yj ), yj′′′ } Then D′′ is a desired minimum paired-dominating set of G Case 6: D(xi ) = 1, N D (xi ) ̸= ∅, and BD (xi ) = ̸ ∅ Let T (xi ) = ∪x∈BD (xi ) maxnbr (x), j = ρ(T (xi )) and j′ = ρ(N D (xi )) Depending upon j < j′ or j′ ≤ j, we discuss the following two subcases Subcase 6.1: j < j′ In this case, Di = Di−1 ∪ {yj , xj′′ }, where xj′′ = M (yj ) Let xp ∈ BD (xi ) be the vertex such that j = σ −1 (maxnbr (xp )) and let ya ∈ D′ dominate xp and be paired with xb in D′ Clearly by Lemma 4.1, ya ̸∈ Di−1 and xb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Again a < j Let xl ∈ Ni (ya ) be such that xl is not dominated by Di−1 Then xl ∈ BD (xi ) If xl yj ̸∈ E, then ρ(T (xi )) = (min{k|yk ∈ T (xi )}) < j, which is a contradiction So xl yj ∈ E and Ni (ya ) ⊆ Ni (yj ) ∪ (∪v∈Di−1 Ni (v)) Let yl ∈ Nj (xb ) Then xb yj ∈ E since xb ya ∈ E with a < j < l Hence by the choice of xj′′ , we have Nj (xb ) ⊆ Nj (xj′′ ) Then D′′ = (D′ \ {xb , ya }) ∪ {xj′′ , yj } is a required minimum paired-dominating set of G Subcase 6.2: j′ ≤ j In this case, Di = Di−1 ∪{yj , xj′′ }, where xj′′ = M (yj′ ) Let xp ∈ BD (xi ) be the vertex such that j = σ −1 (maxnbr (xp )) and let a be the minimum index such that ya ∈ D′ and ya dominates xp and is paired with xb in D′ Clearly by Lemma 4.1, ya ̸∈ Di−1 and xb ̸∈ Di−1 as Di−1 is perfect with respect to a perfect matching of G[D′ ] Again a < j Then Ni (ya ) ⊆ Ni (yj ) ∪ (∪v∈Di−1 Ni (v)) Depending on a ≤ j′ or j′ < a, we have the following subcases Subcase 6.2(a): a ≤ j′ Let yl ∈ Nj′ (xb ) Then by the convexity property, xb yj′ ∈ E Hence by the choice of xj′′ , we have Nj′ (xb ) ⊆ Nj′ (xj′′ ) Then D′′ = (D′ \ {xb , ya }) ∪ {xj′′ , yj } is a required minimum paired-dominating set of G Subcase 6.2(b): j′ < a If b < i, then xb yj′ ∈ E by Lemma 2.1 and hence by the choice of xj′′ , Nj′ (xb ) ⊆ Nj′ (xj′′ ) Then D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } If b ≥ i and xb yj′ ∈ E, then D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } is a minimum paired-dominating set of G such that Di ⊆ D′′ So assume that xb yj′ ̸∈ E Since D(yj′ ) = 0, let xc ∈ D′ dominate yj′ and be paired with yd in D′ By Lemma 4.1, xc ̸∈ Di−1 and yd ̸∈ Di−1 as Di−1 is 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 Np (yd ) = ∅; otherwise by Lemma 2.1, xp yd ∈ E with d ≤ a This is a contradiction to the choice of a Hence we have Np (yd ) = ∅ ⊂ Np (yj ) If c ≥ p, then xp yd ∈ E This contradicts the choice of a Again if there is a vertex xr ∈ N (yd ) with i < r < p, then xr yj ∈ E; otherwise we get a contradiction to the choice of j So we have Ni (yd ) ⊆ Ni yj Again since xc ∈ N (yj′ ), we have Nj′ (xc ) ⊆ Nj′ (xj′′ ) by the choice of xj′′ Let D′′ = (D′ \ {xc , yd }) ∪ {xj′′ , yj } Clearly D′′ is a required minimum paired-dominating set of G Next assume that a < d ≤ j, then xi yd ∈ E Now let xl ∈ Ni (yd ) such that D(xl ) = Then xl yj ∈ E; otherwise we get a contradiction to the choice of j Hence Ni (yd ) ⊆ Ni (yj ) ∪ (∪v∈Di−1 Ni (v)) Then D′′ = (D′ \ {yd , xc }) ∪ {yj , xj′′ } is a required minimum paired-dominating set of G Next assume that j < d If Nd (xb ) ̸= ∅, then xb yd ∈ E Let D′′ = (D′ \ {xc , ya }) ∪ {xj′′ , yj } Clearly D′′ is a minimum paireddominating set of G such that Di ⊆ D′′ If Nd (xb ) = ∅, then Nj′ (xb ) ⊆ Nj′ (xc ) ⊆ Nj′ (xj′′ ) Let D′′ = (D′ \ {ya , xb }) ∪ {yj , xj′′ } 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 ordering σ = (x1 , x2 , , xnx , y1 , y2 , , yny ), it takes O(d(xi )) time to find the minnbr (xi ) and maxnbr (xi ) for each i, ≤ i ≤ nx So minnbr (x) and maxnbr (x) can be computed for all vertices in X in O(n + m) time Given a convex ordering σ = (x1 , x2 , , xnx , y1 , y2 , , yny ) of G, and (minnbr (xi ), maxnbr (xi )) for ≤ i ≤ nx , a lex-convex ordering α of G can be computed in O(n + m) time using a bucket sort We maintain an array A[1 ny ] of pointers such that A[i] points to the set Si = {x ∈ X |maxnbr (x) = yi } which is represented as a doubly linked list As each xi has a unique maximum neighbor, Si ∩ Sj = ∅ for i ̸= j We also maintain an array AD [1 ny ] of pointers such that AD [i] points to the set SiD = {x ∈ X |maxnbr (x) = yi , D(x) = 0} of X , which is represented as a doubly linked list We maintain an array AX [1 nx ] of pointers such that AX [i] points the cell containing xi which is present in the doubly linked list pointed to by AD [j] where yj = maxnbr (xi ) This is needed so that any xi can be deleted from the set Sj , where j = maxnbr (xi ) in O(1) time Initially, A = AD Note that the A, AD , and AX can be computed in O(n + m) time We also maintain an array C [1 nx ] such that C [i] = The algorithm needs to find m(i), M (yi ), and j = ρ(T (xi )), where T (xi ) = ∪x∈BD (xi ) maxnbr (x), B(xi ) = {x ∈ NNi (yi1 )| N (xi ) ∩ {maxnbr (x)} ̸= ∅}, and BD (xi ) = {x ∈ B(xi )|D(x) = 0} To compute m(i), we need to check first whether M (xi ) = ∅ This is done as follows: Let N (xi ) = {yi1 , yi2 , , yik } Scan the sets starting from Si2 to Sik till a vertex, say x, which is not a neighbor of yi1 is found If j is the minimum index such that Sij contains a vertex that is not a neighbor of yi1 , then m(i) = j If no such j is found, then M (xi ) = ∅ and we return m(i) = maxnbr (xi ) To check whether a vertex xj is a neighbor of yi1 in O(1) time, we use the array C Let N (yi1 ) = {xi1 , xi2 , , xir } Set C [ij ] = for j = 1, 2, , r This preprocessing takes O(d(yi1 )) time Now C [j] = if and only if xj is a neighbor of yi1 After the ith iteration, we compute N (xi+1 ) = {y(i+1)1 , y(i+1)2 , , y(i+1)k } B.S Panda, D Pradhan / Discrete Applied Mathematics 161 (2013) 1776–1783 1783 If yi1 ̸= y(i+1)1 , then we make C [ij ] = for j = 1, 2, , r so that at the start of the (i + 1)th iteration C [k] = for k = 1, 2, , nx If yi1 = y(i+1)1 , we leave C unchanged after the ith iteration and use it in the (i + 1)th iteration So checking whether M (xi ) = ∅ and computing m(i) takes O(d(yi1 )) time at the ith iteration To compute ρ(T (xi )), we use the array AD Note that S D [i] = {xj |maxnbr (xj ) = yi and D(xj ) = 0} Let N (xi ) = {yi1 , yi2 , , yik } Scan the sets starting from SiD2 to SiDk till a vertex, say x, which is not a neighbor of yi1 is found If j is the minimum index such that SiDj contains a vertex that is not a neighbor of yi1 , then return j If no such j is found, then BD (xi ) = ∅ So checking whether M (xi ) = ∅ and computing ρ(T (xi )) takes O(d(yi1 )) time at the ith iteration Whenever a vertex yi is selected in the paired-dominating set, D(xj ) becomes for all xj ∈ N (yi ) We reflect this in AD Using the pointer AX [j], we delete the vertex xj from the doubly linked list AD [k], where yk = maxnbr (xj ) So AD can be updated in O(d(yi )) time whenever yi is selected in the paired-dominating set To compute M (yj ), we maintain an array R[1 ny ] of pointers such that R[j] points to the set {(x, dj (x))|x ∈ N (yj )}, where dj (x) = |N (x) ∩ {yj+1 , yj+2 , , yny }| This set is implemented as a linked list such that each cell of the linked list contains two fields: one containing a neighbor of yj , say x and the other contains dj (x) To construct the array R, we first use an array DG[1 nx ] such that DG[i] = d(xi ) For all xi ∈ N (y1 ), decrement DG[i] by one and then form the linked list R[1] by inserting all the neighbors of yi and their current DG values Once the linked list R[i] is formed, decrement one from DG[k] for all xk ∈ N (yi+1 ) Now form the linked list R[i + 1] by inserting each neighbor of yi+1 along 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 yi and di (x), ≤ i ≤ ny As constructing R[i] takes O(d(yi )) time, the construction of R takes O(n + m) time Now M (yj ) is the vertex, say x, in R[j] having maximum dj (x) So M (yj ) can be computed in O(n + m) time for all j = 1, 2, , ny 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 of Lemma 4.2 and 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 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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,... represented as a doubly linked list As each xi has a unique maximum neighbor, Si ∩ Sj = ∅ for i ̸= j We also maintain an array AD [1 ny ] of pointers such that AD [i] points to the set SiD = {x ∈ X |maxnbr... 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