• 沒有找到結果。

# On mixed domination problem in graphs

N/A
N/A
Protected

Share "On mixed domination problem in graphs"

Copied!
22
0
0

(1)

## On mixed domination problem in graphs

(2)

### On mixed domination problem in graphs

I

James K. Lana,∗, Gerard Jennhwa Changa,b,c

aDepartment of Mathematics, National Taiwan University, Taipei 10617, Taiwan

bTaida Institute for Mathematical Sciences, National Taiwan U., Taipei 10617, Taiwan

cNational Center for Theoretical Sciences, Taipei Oﬃce, Taiwan

Abstract

A mixed dominating set of a simple graph G = (V, E) is a subset D⊆ V ∪ E such that every vertex or edge not in D is adjacent or incident to at least one vertex or edge in D. The mixed domination problem is to determine a minimum mixed dominating set of G. This paper studies mixed domination in graphs from an algorithmic point of view. In particular, a linear-time labeling algorithm for the mixed domination problem in cacti is presented.

In addition, we ﬁx an incomplete proof of the NP-completeness of the mixed domination problem in split graphs [22]. Finally, we establish a primal-dual algorithm for the mixed domination problem in trees.

Keywords: Mixed domination, Cactus, Tree, Algorithm, NP-complete

1. Introduction

All graphs in this paper are simple, i.e., ﬁnite, undirected, loopless and without multiple edges. Domination is a core NP-complete problem in graph theory and combinatorial optimization. It has many applications in the real world such as location problems, sets of representatives, social network the- ory, etc; see [14] for more interesting applications. A dominating set of a graph G = (V, E) is a subset D of V such that every vertex not in D is adjacent to at least one vertex in D. The domination number γ(G) of G is

IThis research was partially supported by the National Science Council of the Republic of China under grants NSC100-2811-M-002-146 and NSC98-2115-M-002-013-MY3.

Corresponding author.

Email addresses: drjamesblue@gmail.com (James K. Lan), gjchang@math.ntu.edu.tw (Gerard Jennhwa Chang)

(3)

the minimum size of a dominating set of G. The domination problem is to ﬁnd a minimum dominating set of G.

The notion of dominating (or covering, interchangeably) vertices or edges by other vertices or edges has been widely studied in the literature. Tra- ditional (vertex) domination problem asks for dominating vertices by other vertices. Covering edges by vertices leads to the vertex cover problem. Cov- ering vertices by edges results in the edge cover problem. When edges are to be dominated by edges, we obtain the edge domination problem.

The extension of the above notion is naturally considered. Namely, dom- inating vertices and edges by other vertices and edges is studied [3, 4, 19, 20, 22]. Speciﬁcally, given a graph G = (V, E), a vertex is said to mixed dominate itself, its neighbors and all edges incident to it; an edge uv is said to mixed dominate u and v and all edges incident to u or v. A mixed domi- nating set of G is a subset D ⊆ V ∪ E such that every element in V ∪ E is mixed dominated by some element in D. In other words, every vertex and every edge not in D is adjacent or incident to at least one element in D.

The mixed domination number γm(G) of G is the minimum size of a mixed dominating set of G. The mixed domination problem is to ﬁnd a minimum mixed dominating set of G.

Mixed domination was introduced in early papers, and a mixed dominat- ing set is called a total cover in [3, 4, 19]. In this paper, we choose to follow the terminology of [14, 22]. Note that the meaning of mixed dominating set (used in this paper and [14, 22]) is distinct from that in papers or books such as [1, 8]. A practical application of mixed domination was introduced in [22]

as placing phase measurement units (PMUs) at selected vertices or edges to monitor theirs mixed neighbors’ state variables in an electric power system.

The domination problem is NP-complete for general graphs [13], and remains NP-complete for restricted classes of graphs such as bipartite graphs [6, 12], comparability graphs [12], chordal graphs [7, 9], planar graphs [13]

and split graphs [6]. Eﬃcient algorithms have been found in trees and interval graphs [8]. On the other hand, it has been proved that the mixed domination problem is NP-complete for general graphs [18], planar bipartite graphs [19]

and chordal graphs [16]. However, little work has been published on the existence of eﬃcient algorithms to this problem. Majumdar [18] gave the ﬁrst linear-time algorithm for the mixed domination problem in trees. Recently, Zhao et al. [22] gave a linear-time labeling algorithm for trees, and showed that the mixed domination problem remains NP-complete for split graphs.

Unfortunately, we found that there is a ﬂaw in the algorithm of Zhao et al.

(4)

and the NP-completeness for split graphs is incomplete. In addition, more eﬃcient algorithms for the mixed domination problem are still unknown and desired.

The total graph T (G) of a graph G has V (G)∪ E(G) as its vertex set and two vertices of T (G) are adjacent if and only if they are adjacent or incident in G. It is clear that the mixed domination number of a graph is equal to the domination number of its total graph. Thus, one way to solve the mixed domination problem in G is to solve the domination problem in T (G).

However, few graph classes are known to admit eﬃcient algorithms for the domination problem. Even if a new graph class F is found, it is still diﬃcult to ﬁnd graph class whose total graph is F. Moreover, the transformation between the graph and its total graph may increase the time complexity of the algorithm.

In this paper, we explore eﬃcient algorithms for the mixed domination problem in graphs. In Section 3, a linear-time labeling algorithm for the mixed domination problem in cacti is presented. A cactus can be viewed as an extension of trees. In Section 4, we point out the proof of the NP- completeness for the mixed domination problem in split graphs in [22] is incomplete, and provide a new proof to this result. In Section 5, a linear- time labeling algorithm based on the primal-dual approach for the mixed domination problem in trees is introduced. The presented algorithm can serve as an alternative solution to the mixed domination problem in trees.

2. Definitions

Let G = (V, E) be a graph with vertex set V and edge set E. The degree deg(v) of a vertex v in G is the number of incident edges to v. An isolated vertex is a vertex v with deg(v) = 0. Deﬁne ε ∈ V ∪ E to be an element in G. For a vertex v in G, denote the open neighborhood by N (v) = { u ∈ V : uv ∈ E } and the closed neighborhood by N[v] = N(v) ∪ { v }. Let E(v) = { e ∈ E : v ∈ e } and E[v] = { v } ∪ E(v). The closed mixed neighborhood of an element ε in G is denoted by

Nm[ε] =

{ N (ε)∪ E[ε], if ε ∈ V ; E[u]∪ E[v], if ε = uv ∈ E.

Then for elements ε1 and ε2 in G, ε1 is mixed dominated by ε2 if and only if ε1 ∈ Nm2], and they are mixed neighbors of each other. A set D is a mixed dominating set of G if and only if D∩ Nm[ε]̸= ∅ for every element ε in G.

(5)

The subgraph of G induced by S ⊆ V is the graph G[S] with vertex set S and edge set {uv ∈ E : u, v ∈ S}. In a graph G = (V, E), the deletion of S ⊆ V from G, denoted by G − S, is the graph G[V \ S]; the deletion of F ⊆ E from G, denoted by G − F , is the graph (V, E \ F ). For an element ε in G, we write G− ε for G − { ε }. The union of two graphs G1 and G2 is the graph G1∪ G2 with vertex set V (G1)∪ V (G2) and edge set E(G1)∪ E(G2).

The length of a path is the number of edges in the path. The distance d(u, v) from vertex u to vertex v in G is the minimum length of a path from u to v;

d(u, v) = ∞ if there is no path from u to v.

A forest is a graph without cycles. A tree is a connected forest. A pendant vertex in a graph is a vertex with degree one (the pendant vertex is also called a leaf in a tree), and a pendant edge is an edge incident to a pendant vertex.

A penultimate vertex is a vertex all of whose neighbors except possibly one are pendant vertices. A path of order n is denoted by Pn; a cycle of order n is denoted by Cn.

In a graph G = (V, E), a stable set is a set of pairwise nonadjacent vertices and a clique is a set of pairwise adjacent vertices. A vertex v is a cut-vertex if the number of connected components is increased after removing v. A block of a graph is a maximal connected subgraph without any cut-vertex.

An end-block of a graph is a block containing at most one cut-vertex of the graph. A block graph is a graph whose blocks are cliques. A cactus is a connected graph whose blocks are either an edge or a cycle. Alternatively, a cactus is a connected graph in which two cycles have at most one vertex (cut-vertex) in common. A cactus is a tree if all the blocks are edges.

3. Linear-time labeling algorithm for mixed domination in graphs Labeling techniques are widely used in the literatures for solving the domination problem and its variants [5, 9, 10, 11, 15, 17, 21, 22]. Among them, the FBR labeling technique solves a slightly more general domination problem, which can be formulated as follows. For a graph G = (V, E), V is partitioned into three disjoint sets F, B and R, where F, B and R are called free set, bound set and required set, respectively. An optional domi- nating set of G (with respect to F, B and R) is a subset D ⊆ V such that R ⊆ D and D dominates B. Note that optional domination is the ordinary domination when specifying B = V and F = R =∅.

Based on this technique, Cockayne, Goodman and Hedetniemi [11] gave the ﬁrst linear-time algorithm for the domination problem in trees. Laskar et

(6)

al. [17] further partitioned V into F ·∪B ·∪R1·∪R2 to ﬁnd a subset D⊆ V such that R1 ∪ R2 ⊆ D and dominates B ∪ R1 for the total domination problem in trees. Zhao et al. [22] gave a linear-time labeling algorithm for the mixed domination problem in trees by partitioning V ∪ E = F ·∪B ·∪R, specifying B = V ∪ E, F = R = ∅ and replacing domination with mixed domination.

In this section, we employ the FBR labeling technique to ﬁnd a minimum mixed dominating set in a cactus. Unlike the labeling method used in [22], V ∪ E is partitioned into four disjoint sets F, B, B and R in our algorithm, where B is called strictly bound set. More speciﬁcally, given a graph G = (V, E), an FBR assignment L is a mapping that assigns each element ε in G a label1

L(ε)∈

{ { F, B, B, R} , if ε ∈ V ; { F, B, R } , if ε∈ E.

Namely, any element in G can be assigned a label in { F, B, B, R}, except that only vertices can be assigned a label B. Thus, V ∪ E = F ·∪B ·∪B·∪R.

An FBR mixed dominating set of G (with respect to L) is a subset D ⊆ V ∪E such that R⊆ D and

• for every element ε ∈ B, D ∩ Nm[ε]̸= ∅, and

• for every vertex v ∈ B, D∩ E[v] ̸= ∅.

That is, D contains all required elements, and every bound element not in D is adjacent or incident to at least one element in D, and every strictly bound vertex not in D is incident to at least one edge in D. Free elements need not be mixed dominated, but can be included in D to mixed dominate elements in B∪B. The FBR mixed domination number γL(G) is the minimum size of an FBR mixed dominating set in G; such set is called a γL-set of G. Notice that if we specify B =∅, then FBR mixed domination is the optional mixed domination used in [22]; if we specify that B = V ∪ E and F = B = R =∅, then γL(G) = γm(G). Thus an algorithm for computing the value of γL(G) is suﬃcient to compute the value of γm(G).

Denote R = B ∪ B. Let G be a graph with an FBR assignment L.

Suppose ε is an element in G and ε ∈ R. It is clear that ε must be included in any FBR mixed dominating set of G. Hence, for every element ε ∈ Nm[ε]∩B we have D∩ Nm]̸= ∅, where D is a minimum FBR mixed dominating set

1For simplicity, the terms F, B, B, R represent sets and labels interchangeably.

(7)

of G. Moreover, if ε is an edge, then for every vertex v ∈ Nm[ε]∩ B we have D∩ E[v] ̸= ∅. Therefore these elements no longer need to retain its label.

Release is a procedure to relabel these elements with F .

Procedure Release(ε)

1 if ε is a vertex then

2 L(ε)← F for each ε ∈ Nm[ε]∩ B;

3 else // ε is an edge

4 L(ε)← F for each ε ∈ Nm[ε]∩ R;

3.1. Finding a γL-set in a tree and a cycle

In this subsection, we present algorithms to ﬁnd a minimum FBR mixed dominating set in trees and cycles. The presented algorithm will be used for ﬁnding a minimum FBR mixed dominating set in a cactus.

Theorem 3.1. Suppose G is a graph with an FBR assignment L. Suppose C is an end-block of G and |C| = 2. Let x be the unique cut-vertex of C and let y be the pendant vertex of C. Let G = G− y and let Gc denote the graph which results from G by relabeling x with c∈ { F, B, B, R}. Let L and L′′

be those restrictions of L on G and G′′, respectively.

(1) If y ∈ F and xy ∈ F , then γL(G) = γL(G).

(2) If y ∈ F and xy ∈ B, then γL(G) = γL(GB).

(3) If y ∈ B and xy ̸∈ R, then γL(G) = γL(GR).

(4) If y ∈ B and xy ̸∈ R, then γL(G) = γL′′(G′′), where G′′ results from G by relabeling y and xy with F and R, respectively.

(5) Suppose y ∈ R or xy ∈ R. Let S = { y, xy } ∩ R. Then γL(G) = γL′′(G′′) +|S|, where G′′ results from G by releasing ε for each ε∈ S and then deleting y.

Proof. We shall only provide the proofs of cases (2), (4) and (5); other cases are similar to obtain.

(2) Suppose D is a γL-set of GB. Since x∈ B, there exists an element ε∈ E[x] in GB such that ε∈ D. Then D is also an FBR mixed dominating set of G, since y is free and xy is mixed dominated by ε. Thus γL(GB) γL(G). Conversely, suppose D is a γL-set of G. By the minimality of D, x, y

(8)

and xy cannot concurrently be included in D. Let ε be y or xy. If ε ∈ D, then D = D− { ε } ∪ { x } is also an FBR mixed dominating set of GB and

|D| = |D|. Thus γL(G) ≥ γL(GB).

(4) Suppose D is a γL′′-set of G′′. Since xy ∈ R in G′′, xy∈ D. Then D is an FBR mixed dominating set of G, since D∩ E[y] ̸= ∅. Thus γL′′(G′′) γL(G). Conversely, suppose D is a γL-set of G. Since y ∈ B, D∩ E[y] = D∩ { y, xy } ̸= ∅. By the minimality of D, y and xy cannot be both included in D. If y ∈ D, then D = D−{ y }∪{ xy } is also an FBR mixed dominating set of G′′ and |D| = |D|. Thus, γL(G)≥ γL′′(G′′).

(5) Suppose D is a γL′′-set of G′′. Clearly D ∪ S is an FBR mixed dominating set of G. Thus γL′′(G′′) +|S| ≥ γL(G). Conversely, suppose D is a γL-set of G. For each ε∈ S, we have ε ∈ D. We only show the case y ∈ R and xy ̸∈ R; other cases are similar to obtain. Clearly, we have |S| = 1 and y ∈ D. Since xy ̸∈ R, by the minimality of D, x and xy cannot be both in D. If x ∈ D, then D − { y } is clearly an FBR mixed dominating set of G′′. If xy ∈ D, then D − { y, xy } ∪ { x } is an FBR mixed dominating set of G′′. If x ̸∈ D and xy ̸∈ D, then since all the elements in Nm[y]∩ B have been relabeled with F in G′′, D− { y } is an FBR mixed dominating set of G′′. In either case, γL(G)− |S| ≥ γL′′(G′′).

Based on the above theorem, we design a procedure, called MixDomLeaf, to handle the case of an end-block with cardinality two. The procedure will be used in algorithms for the FBR mixed domination problem in trees and cacti.

Procedure MixDomLeaf(G, C)

1 let x be the unique cut-vertex of C and y be its neighbor;

2 if L(y) = R or L(xy) = R then

3 S ← { ε: ε ∈ { y, xy } , ε ∈ R };

4 D← D ∪ S;

5 foreach ε∈ S do Release(ε);

6 else

7 if L(y) = B then L(x)← R;

8 if L(y) = B then D ← D ∪ { xy } and Release(xy);

9 if L(y) = F and L(xy) = B then L(x)← B;

10 G← G − y;

Next, we present an algorithm, called MixDomTree to ﬁnd a minimum FBR mixed dominating set in a tree.

(9)

Algorithm 1. MixDomTree (Finding a γL-set of a tree) Input: A tree T = (V, E) with an FBR assignment L.

Output: A minimum FBR mixed dominating set D of T .

1 T ← T ;

2 D← ∅;

3 while |T| > 1 do

4 let C be an end-block of T;

5 MixDomLeaf(T, C);

6 end

7 let the only left vertex of T be x;

8 if x̸∈ F then D ← D ∪ { x };

Theorem 3.2. Algorithm MixDomTree finds a minimum FBR mixed domi- nating set in a tree in linear-time.

Proof. The correctness comes from Theorem 3.1 and the fact that every induced subgraph of a tree of size greater than one has an end-block with cardinality two. Besides, MixDomTree runs in linear-time, since it visits each vertex once, and all of the statements within which can be executed in at most O(deg(x)) time, where x is the neighbor of a pendant vertex y.

Remark. Algorithm MixDomTree is a one stage algorithm and it can also ﬁnd a minimum FBR mixed dominating set of a forest without modifying the algorithm. We notice that the labeling algorithm for the mixed domination problem in trees in [22] is a two-stage algorithm and it has a ﬂaw shown as follows. Let T be a rooted tree with V (T ) = { x, y, z }, E(T ) = { xy, xz } and root=x. Let all the elements of T be labeled F except that y is labeled R and xz is labeled B. The algorithm will execute Stage B, return D = { y } and then stop, where D is the output of the algorithm. However, D is not a minimum optional mixed dominating set because xz is not mixed dominated by any element in D. To ﬁx this ﬂaw, modify Stage B as follows:

D← D ∪ R. If there exists an element ε ∈ B such that D ∩ Nm[ε] =∅, then L(root) = R, D ← D ∪ {root} and stop; Otherwise, stop. See [22] for further details.

Now we consider ﬁnding a minimum FBR mixed dominating set of a cycle. The idea is to cut the cycle into a path and then use MixDomTree.

(10)

Let ε be an element in C. Denote the cycle C with element ε relabeled with c∈ { F, B, B, R} by Cεc.

Algorithm 2. MixDomCYC (Finding a γL-set of a cycle) Input: A cycle C with an FBR assignment L.

Output: A minimum FBR mixed dominating set D of C.

1 D← ∅;

2 if ∃ an element ε ∈ R then

3 Release(ε) and D ← MixDomTree(C − ε) ∪ { ε };

4 else

5 if ∃ an element ε ∈ B or B then

6 if ε∈ B then S ← Nm[ε] else S ← E[ε];

7 Umin ← ∞;

8 foreach εi in S do

9 Ui ← MixDomCYC(CεRi);

10 if |Ui| < |Umin| then min ← i;

11 D← Umin;

Based on Theorem 3.2, the construction and correctness of MixDomCYC is straightforward, and the proof is omitted. For the time complexity, since Mix- DomTree is linear and MixDomCYC makes at most ﬁve calls to MixDomTree (the number of mixed neighbors of an element in a cycle), it is clear that MixDomCYC is linear.

3.2. Finding a γL-set in cacti

In this subsection, we present an algorithm to ﬁnd a minimum FBR mixed dominating set in a cactus. The construction and correctness of the algorithm is based on the following theorem. Note that for a cycle C with an FBR assignment L and a vertex v in C, γL(CvF)≤ γL(CvB)≤ γL(CvB) γL(CvR) < γL(CvF) + 1 holds, where L is the same as L with the above modiﬁcations on v.

Theorem 3.3. Suppose K is a cactus with an FBR assignment L. Suppose C is an end-block of K and |C| ≥ 3. Let x be the unique cut-vertex of C and y, z be the two neighbors of x on C. Let K denote the cactus which results from K by deleting all vertices only in C. Let Kc denote the cactus which results from K by relabeling x with c ∈ { F, B, B, R}. Let L and L′′ be those restrictions of L on K and C, respectively.

(1) If x∈ R, then γL(K) = γL(K) + γL′′(C)− 1.

(11)

(2) Suppose x̸∈ R and γL′′(C) < γL′′(CxR).

(2.1) If γL′′(C) < γL′′(CxyR) and γL′′(C) < γL′′(CxzR), then γL(K) = γL(KF ) + γL′′(C).

(2.2) If γL′′(C) = γL′′(CxyR) or γL′′(C) = γL′′(CxzR), then γL(K) = γL(K′′)+

γL′′(CeR), where K′′ results from K by releasing e, where e = xy or xz in which γL′′(C) = γL′′(CeR), and then deleting all vertices only in C.

(3) Suppose x∈ B or B and γL′′(CxF) < γL′′(C) = γL′′(CxR). Then γL(K) = γL(K) + γL′′(CxF).

(4) Suppose x̸∈ R and γL′′(CxF) = γL′′(C) = γL′′(CxR).

(4.1) If γL′′(C) < γL′′(CxyR) or γL′′(C) < γL′′(CxzR) or γL′′(C) = γL′′(CxyR) = γL′′(CxzR) = γL′′(C − x) then γL(K) = γL(KR ) + γL′′(CxR)− 1.

(4.2) If γL′′(C) = γL′′(CxyR) = γL′′(CxzR) = γL′′(C − x) + 1, then γL(K) = γL(K′′) + γL′′(C − x), where K′′ results from K by deleting all elements in C\ Nm[x], and then relabels y and z with F .

Proof. We shall only provide the proofs of cases (2) and (4); other cases are similar to obtain.

(2) In this case, we give a proof for the case x∈ B; the argument can be applied to cases x ∈ F and x ∈ B. Note that γL′′(CxF) = γL′′(C) < γL′′(CxR).

(2.1) Let D1 and D2 be γL-set of KF and γL′′-set of C, respectively.

Clearly, D1∪ D2 is an FBR mixed dominating set of K and thus γL(KF ) + γL′′(C)≥ γL(K). Conversely, suppose D is a γL-set of K. We have two cases.

Case 1 : x ∈ D. Since γL′′(C) < γL′′(CxR), |D ∩ C| > γL′′(C). Let C be a γL′′-set of C and let D = (D − C) ∪ { x } ∪ C. Then |D| = |D| and D is a FBR mixed dominating set of K. Clearly, D− C and C are FBR mixed dominating sets of KF and C, respectively. We have γL(K) γL(KF ) + γL′′(C).

Case 2 : x̸∈ D. Since x ∈ B, there exists an element ε ∈ D ∩ Nm[x].

Subcase 2-1 : ε ∈ C. Suppose ε = xy or xz. Since γL′′(C) < γL′′(CεR),

|D ∩ C| > γL′′(C). Let C be a γL′′-set of C. Then D = (D− C) ∪ { x } ∪ C is also a γL-set of K in which x ∈ D. Thus, we go back to Case 1 and γL(K) ≥ γL(KF ) + γL′′(C) holds. Now suppose ε is a vertex in C. Then clearly D − C and D ∩ C are FBR mixed dominating sets of KF and C, respectively. We have γL(K)≥ γL(KF ) + γL′′(C).

Subcase 2-2 : ε̸∈ C. In this case, we have |D ∩ C| ≥ γL′′(C). Let C be a γL′′-set of C and let D = (D− C) ∪ C. Now we have a γL-set of K in which

(12)

x is mixed dominated by some element in C. As in the previous subcase, γL(K)≥ γL(KF ) + γL′′(C).

(2.2) W.l.o.g. assume that e = xy, i.e., γL′′(C) = γL′′(CxyR). Let D1 and D2 be γL-set of K′′and γL′′-set of CxyR, respectively. Then D1∪D2is an FBR mixed dominating set of K, since the elements that have been relabeled with F in K′′ are mixed dominated by e. Hence γL(K′′) + γL′′(CxyR) ≥ γL(K).

Conversely, suppose D is a γL-set of K. Since γL′′(C) < γL′′(CxR), we have

|D ∩ C| ≥ γL′′(C). Let C be a γL′′-set of CxyR, which is also a γL′′-set of C.

Let D = (D− C) ∪ { x } ∪ C. Then D is also a γL-set of K. D− C and C are FBR mixed dominating sets of K′′ and CxyR, respectively. We have γL(K)≥ γL(K′′) + γL′′(CxyR).

(4.1) Let D1 and D2 be γL-set of KR and γL′′-set of CxR, respectively.

Then x ∈ D1∩ D2. Clearly D1 ∪ D2 is an FBR mixed dominating set of K and we have γL(KR ) + γL′′(CxR)− 1 ≥ γL(K). Conversely, suppose D is a γL-set of K. Two cases.

Case 1 : x∈ D. Then (D−C)∪{x} and D∩C are FBR mixed dominating sets of KR and CxR, respectively. We have γL(K)≥ γL(KR ) + γL′′(CxR)− 1.

Case 2 : x ̸∈ D. We claim that |D ∩ C| ≥ γL′′(C). Suppose on the contrary that |D ∩ C| = γL′′(C)− 1. Suppose γL′′(C) < γL′′(CeR) for some e ∈ { xy, xz }. Then (D ∩ C) ∪ { e } results in a γL′′-set of C, contradicts to the assumption of γL′′(C) < γL′′(CeR). Now suppose γL′′(C) = γL′′(CxyR) = γL′′(CxzR) = γL′′(C−x). Note that |D ∩ C| = γL′′(C)−1 implies γL′′(C−x) ≤ γL′′(C)− 1, contradicts to the assumption of γL′′(C) = γL′′(C − x). Hence we have the claim. Now let C be a γL′′-set of CxR, which is also a γL′′-set of C. Let D = (D− C) ∪ C. We have |D| = |D| and D is also a γL-set of K in which x∈ D. By Case 1, we have the desired result.

(4.2) Let D1 and D2 be γL-set of K′′ and γL′′-set of C − x, respectively.

Clearly D1∪D2 is an FBR mixed dominating set of K and we have γL(K′′)+

γL′′(C − x) ≥ γL(K). Conversely, suppose D is a γL-set of K. Two cases:

Case 1 : |D ∩ C| ≥ γL′′(C). Let D′′ be a γL′′-set of C − x, and let D = (D − C) ∪ { x } ∪ D′′. Since γL′′(C) = γL′′(C − x) + 1, |D| = |D| and thus D is also a γL-set of K. Clearly, D ∩ K′′ and D ∩ (C − x) are FBR mixed dominating sets of K′′ and C − x, respectively. Thus γL(K) γL(K′′) + γL′′(C − x).

Case 2 : |D ∩ C| = γL′′(C)− 1. In this case, we have x ̸∈ D, xy ̸∈ D, xz ̸∈

D, and at least one of x, xy and xz must be mixed dominated by an element in D− C. We claim that there must exist an edge e such that e ∈ D − C and x ∈ e. Suppose the claim is not true. Then D ∩ C induces an FBR mixed

(13)

dominating set of CxF, contradicts to the assumption of γL′′(CxF) = γL′′(C).

As a result, D∩ K′′ and D∩ (C − x) are clearly FBR mixed dominating sets of K′′ and C− x, respectively. Thus γL(K)≥ γL(K′′) + γL′′(C− x).

Based on the Theorem 3.3, we design a procedure, called MixDomC3, to handle the case of an end-block with cardinality larger than two.

Procedure MixDomC3(G, C)

1 let x be the unique cut-vertex of C and y, z be its neighbors;

/* Compute the following sets only when they are needed. */

U ← MixDomCYC(C), UR← MixDomCYC(CxR), UF ← MixDomCYC(CxF), Uxy← MixDomCYC(CxyR), Uxz ← MixDomCYC(CxzR), UP ← MixDomTree(C − x);

2 if L(x) = R then D← D ∪ U; /* Case 1 */

3 else if |U| < |UR| then

4 if |U| < |Uxy| and |U| < |Uxz| then /* Case 2.1 */

5 L(x)← F and D ← D ∪ U;

6 else /* Case 2.2 */

7 choose e∈ { xy, xz } such that |U| = |Ue|;

8 Release(e) and D ← D ∪ Ue;

9 else

10 if L(x)̸= F and |UF| < |U| then /* Case 3 */

11 D← D ∪ UF;

12 else if |U| < |Uxy| or |U| < |Uxz| or |U| = |UP| then /* Case 4.1 */

13 L(x)← R and D ← D ∪ UR;

14 else /* Case 4.2 */

15 L(y)← F , L(z) ← F , D ← D ∪ UP and Case← 4.2;

16 if Case = 4.2 then

17 G← G − { ε ∈ C : ε ̸∈ Nm[x]};

18 else

19 G← (G − C) ∪ { x };

We are now ready to present our algorithm, called MixDomCAC, to de- termine a minimum FBR mixed dominating set in a cactus. Our algorithm takes MixDomTree and MixDomCYC as subroutines, which ﬁnds a minimum FBR mixed dominating sets of a tree and a cycle, respectively.

(14)

Figure 1: Mixed domination in a cactus.

Algorithm 3. MixDomCAC (Finding a γL-set of a cactus) Input: A cactus K = (V, E) with an FBR assignment L.

Output: A minimum FBR mixed dominating set D of K.

1 K ← K;

2 D← ∅;

3 while |K| ̸= ∅ do

4 if K is a block then

5 if |K| < 3 then

6 D← D ∪ MixDomTree(K) and stop;

7 else

8 D← D ∪ MixDomCYC(K) and stop;

9 let C be an end-block of K;

10 if |C| = 2 then MixDomLeaf(K,C) else MixDomC3(K,C);

11 end

It is well-known that cacti can be recognized in linear-time [2, 15]. For each end-block of a cactus, Algorithm MixDomCAC calls MixDomTree at most one time and calls MixDomCYC at most ﬁve times. Since MixDomTree and MixDomCYC are linear, algorithm MixDomCAC is clearly linear. Fig. 1 shows an example of a minimum mixed dominating set in a cactus.

Theorem 3.4. Algorithm MixDomCAC finds a minimum FBR mixed domi- nating set in a cactus in linear-time.

(15)

4. NP-completeness of mixed domination in split graphs

In this section, we study the complexity of the mixed domination problem:

MIXED DOMINATION

INSTANCE: A graph G = (V, E) and positive integers k.

QUESTION: Does G have a mixed dominating set of size ≤ k?

In [22], Zhao et al. showed that the mixed domination problem remains NP-complete for split graphs. A split graph is a graph whose vertex set is the disjoint union of a clique and a stable set. Unfortunately, the proof in [22] is incomplete. In the following, we indicate the ﬂaw and present a counterexample to their proof.

To show the NP-completeness of MIXED DOMINATION, Zhao et al. [22]

adopted the reduction from a well-known NP-complete problem, namely the vertex cover problem for general graphs. A vertex cover of a graph G = (V, E) is a subset C ⊆ V such that for every edge uv ∈ E we have u ∈ C or v ∈ C.

The vertex cover problem is to ﬁnd a minimum vertex cover of G.

Let the decision version of the vertex cover problem be denoted by VER- TEX COVER. The transformation is constructed as follows. Given an in- stance of VERTEX COVER, construct the graph G = (V, E) with V = V ∪ E and E = { uv : u ̸= v, u, v ∈ V } ∪ { ve: v ∈ V, e ∈ E, v ∈ e }. Then showed that:

G has a vertex cover with cardinality k if and only if the split graph G has a mixed dominating set of size k +n−k

2

⌉. (∗)

For the suﬃciency part of (∗), suppose the split graph G has a mixed dom- inating set D of size k +n−k

2

. Zhao et al. [22] assumed that D contains no edge between V and E by claiming that if there exists some edge ve∈ D such that v ∈ V and e ∈ E in G, then edge ve can be replaced by v in D without changing the size of D. However, the claim is not always true. Take Fig. 2 for example. Suppose G has a mixed dominating set D = { ae, cf, d } of size three. It is clear that D ={ a, c, d } is not a mixed dominating set of G, since the edge bf is not mixed dominated by any element in D. Since the claim is not always true, the proof is incomplete.

In the following, we give an alternative proof that MIXED DOMINA- TION is NP-complete for splits graphs. The proof involves a transformation from modified vertex cover, which may be deﬁned as follows:

MODIFIED VERTEX COVER

INSTANCE: A graph G = (V, E) of n vertices and a nonnegative integer k

(16)

G G a

b c d

a b c d

e=ab f=bc g=cd

Figure 2: A counterexample to the proof in [22]. The collection of shaded elements forms a mixed dominating set in the graph.

with n + k odd and at least one isolated vertices.

OUTPUT: “Yes” if G has a vertex cover of size k, “no” otherwise.

Theorem 4.1. MODIFIED VERTEX COVER is NP-complete.

Proof. Clearly MODIFIED VERTEX COVER belongs to NP. The correct- ness follows from the fact that C is a vertex cover of G if and only if it is a vertex cover of G∪ K1, and a trivial reduction from VERTEX COVER.

Theorem 4.2. ([22]) MIXED DOMINATION is NP-complete for split graphs.

Proof. Clearly MIXED DOMINATION belongs to NP. We construct a polynomial- time reduction from MODIFIED VERTEX COVER. Given an instance of MODIFIED VERTEX COVER, construct a graph G = (V, E) with V = V ∪ E and E ={ uv : u ̸= v, u, v ∈ V } ∪ { ve: v ∈ V, e ∈ E, v ∈ e }. Clearly the graph G can be constructed in linear time in|V | and |E|. We claim that G has a vertex cover of size k if and only if G has a mixed dominating set of size n+k2−1.

Assume that G has a vertex cover C of size k. Since n− k is odd, we can pair the vertices of V − C into n−k−12 pairs with one vertex left (choose to be an isolated vertex of G). These n−k−12 pairs form a set C of edges in G. It is easy to see that C∪ C form a mixed dominating set of G of size n+k2−1.

On the other hand, suppose G has a mixed dominating set D of size

n+k−1

2 . For any e = uv ∈ E ⊆ V with e ∈ D, either EG[u]∩ D ̸= ∅ or EG[v]∩ D ̸= ∅, for otherwise EG[u] ∩ D = EG[v]∩ D = ∅ imply that uv ∈ E is not mixed dominated by any element in D. In this case, we

(17)

may assume that EG[u]∩ D ̸= ∅ and so (D − { e }) ∪ { v } is also a mixed dominating set. Hence we may assume that D∩ E = ∅.

Now for any e ∈ E ⊆ V, either u ∈ D or ue ∈ D for some u ∈ e. We collect all of these u and ue to form a subset D ⊆ D and collect all of these u (includes the u with ue∈ D) to form a subset C ⊆ V . (If u ∈ D and ue ∈ D for some e∈ E and u ∈ e, then just replace u with v where e = uv.) Notice that C is a vertex cover of G, and |C| = |D|. In order to mixed dominate all edges whose end-vertices are in V − C, it is the case that |D − D| ≥ n−|C|−12 and so n+k2−1 =|D| = |D − D| + |D| ≥ n−|C|−12 +|C| which gives k ≥ |C|.

Adding enough vertices to C results a vertex cover of G of size k.

The theorem then follows from Theorem 4.1.

5. Primal-dual approach for the mixed domination problem in trees Although we have presented Algorithm 3 for ﬁnding a minimum mixed dominating set in a tree, it is still desire to design an algorithm without using the labeling method, as it needs extra space to store the labels for each element of a graph. In the following, we present a simple, nonlabeling algorithm that can compute a minimum mixed dominating set in a tree.

The most beautiful technique used in domination may be the primal-dual approach. In this technique, besides the original mixed domination problem, the following dual problem is also considered. In a graph G = (V, E), a 2- stable set is a subset S ⊆ V in which every two distinct vertices u and v have distance d(u, v) > 2. The 2-stability number s2(G) of G is the maximum size of a 2-stable set in G. Since we are handling both vertices and edges concurrently in mixed domination, a parameter which can be viewed as the

“mixed version” of 2-stable set is introduced. Let ε1 and ε2 be two elements in a graph G = (V, E). Deﬁne the distance between ε1 and ε2 by

dm1, ε2) =



d(ε1, ε2), if ε1, ε2 ∈ V ;

min{ d(x, y): x ∈ ε1, y ∈ ε2} , if ε1 ∈ E or ε2 ∈ E.

A 2-mix-stable set of a graph G = (V, E) is a subset S ⊆ V ∪ E in which every two distinct elements ε1, ε2 of S satisﬁes the 2-mix-stability:

(i) dm1, ε2) > 2, if ε1, ε2 ∈ V ;

(ii) dm1, ε2)≥ 2, if ε1 ∈ E or ε2 ∈ E. (2-mix-stability)

(18)

The 2-mix-stability number ms2(G) of G is the maximum size of a 2-mix- stable set in G. Clearly, a 2-stable set in a graph is also a 2-mix-stable set.

The following lemma shows that in a graph, the cardinality of any 2-mix- stable set is no larger than the cardinality of any mixed dominating set.

Lemma 5.1. In a graph G = (V, E), if S is a 2-mix-stable set and D is a mixed dominating set, then |S| ≤ |D|.

Proof. Deﬁne a function f from S to D by mapping each element ε in S into some mixed neighbor of ε in D. We now claim that f is one-to-one. Suppose two distinct elements ε1 and ε2 of S are mapped into the same element bε. Case 1 : ε1, ε2 ∈ V . If bε∈ V , then clearly dm1, ε2)≤ d(ε1, bε) + d(ε2, bε) 2; if bε∈ E, then dm1, ε2)≤ 1. Case 2 : ε1 ∈ V, ε2 ∈ E. Then dm1, ε2) dm1, bε) ≤ 1. Case 3 : ε1, ε2 ∈ E. If bε ∈ V , then dm1, ε2) = 0; if bε ∈ E, then dm1, ε2) = 1. Since ε1 ̸= ε2 and S is a 2-mix-stable set, we have the claim and therefore |S| ≤ |D| holds.

Corollary 5.2. ms2(G)≤ γm(G) for any graph G.

It should be noticed that the inequality of Corollary 5.2 can be strict, as shown by the n-cycle Cn that

ms2(Cn) =



2n

5

, if n̸≡ 0 (mod 5);

2n

5

, if n≡ 0 (mod 5),

while γm(Cn) = ⌈2n

5

for n≥ 3.

For a tree T , an algorithm which outputs a mixed dominating set D and a 2-mix-stable set S with|D| ≤ |S| is designed. From Corollary 5.2, we have

|S| ≤ ms2(T )≤ γm(G) ≤ |D| ,

which implies that all inequalities are equalities. As a result, D is a minimum mixed dominating set and S is a maximum 2-mix-stable set.

The algorithm starts from any penultimate vertex v. At each iteration of the algorithm, penultimate v (resp. vu) is added to D if there exists a pendant vertex (resp. pendant edge) adjacent (resp. incident) to v that is not mixed dominated by any element in D. Besides, the pendant vertex (resp. pendant edge) is also put into S.

(19)

Algorithm 4. MixDomTreePD Input: A tree T .

Output: A minimum mixed dominating set D and a maximum 2-mix-stable set S of T .

1 T ← T ;

2 D← ∅;

3 S← ∅;

4 while |T| > 2 do

5 let v be a vertex in T with neighborhoods u, u1, . . . , ur such that degT(ui) = 1 for all 1≤ i ≤ r;

6 if ∃ ui such that Nm[ui]∩ D = ∅ then

7 D← D ∪ { v } and S ← S ∪ { ui};

8 else

9 if ∃ uiv such that Nm[uiv]∩ D = ∅ then

10 D← D ∪ { vu } and S ← S ∪ { uiv};

11 T ← T\ { u1, u2, . . . , ur};

12 end

13 if ∃ an element ε in T such that Nm[ε]∩ D = ∅ then

14 D← D ∪ { ε } and S ← S ∪ { ε };

Theorem 5.3. Algorithm MixDomTreePD finds a minimum mixed dominat- ing set D and a maximum 2-mix-stable set S of a tree T with |D| = |S|.

Proof. To verify the algorithm, it is suﬃcient to show that D is a mixed dominating set, S is a 2-mix-stable set and |D| ≤ |S|. By the construction of the algorithm, D is obviously a mixed dominating set.

We now show that |D| = |S| at the end of the algorithm. Suppose elements εD and εS are added to D and S after the i-th iteration of the algorithm, respectively. Then at the beginning of the i-th iteration of the algorithm, εS is clearly a new element to S. In addition, εD must not be in D, for otherwise, εS should not be added to S. Hence, after each iteration of the algorithm, new elements are added to D and S concurrently, and thus

|D| = |S| at the end of the algorithm.

Suppose at the end of the algorithm|D| = |S| = k. Let ε1, ε2, . . . , εk and s1, s2, . . . , sk be the ordered list of elements added to D and S, respectively.

Suppose on the contrary that S contains two elements si and sj with i < j such that si and sj violate the 2-mix-stability. Let T be the tree at the beginning of the iteration of adding εi and si to D and S, respectively. Then

In this talk, we introduce a general iterative scheme for ﬁnding a common element of the set of solutions of variational inequality problem for an inverse-strongly monotone mapping

In this paper, we have studied a neural network approach for solving general nonlinear convex programs with second-order cone constraints.. The proposed neural network is based on

For the proposed algorithm, we establish a global convergence estimate in terms of the objective value, and moreover present a dual application to the standard SCLP, which leads to

We propose a primal-dual continuation approach for the capacitated multi- facility Weber problem (CMFWP) based on its nonlinear second-order cone program (SOCP) reformulation.. The

* All rights reserved, Tei-Wei Kuo, National Taiwan University, 2005..

Given a graph and a set of p sources, the problem of finding the minimum routing cost spanning tree (MRCT) is NP-hard for any constant p &gt; 1 [9].. When p = 1, i.e., there is only

The MTMH problem is divided into three subproblems which are separately solved in the following three stages: (1) find a minimum set of tag SNPs based on pairwise perfect LD

▪ Can we decide whether a problem is “too hard to solve” before investing our time in solving it.. ▪ Idea: decide which complexity classes the problem belongs to