PREPRINT

國立臺灣大學 數學系 預印本 Department of Mathematics, National Taiwan University

www.math.ntu.edu.tw/~mathlib/preprint/2013-10.pdf

On the algorithmic complexity of k-tupletotal domination

James K. Lan and Gerard Jennhwa Chang

May 29, 2013

On the algorithmic complexity of k-tuple total

dominationI

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

aDepartment of Mathematics, National Taiwan University, Taipei 10617, TaiwanbTaida Institute for Mathematical Sciences, National Taiwan University, Taipei 10617, Taiwan

cNational Center for Theoretical Sciences, Taipei Office, Taiwan

Abstract

For a fixed positive integer k, a k-tuple total dominating set of a graph Gis a subset D ⊆ V (G) such that every vertex of G is adjacent to at leastk vertices in D. The k-tuple total domination problem is to determine aminimum k-tuple total dominating set of G. This paper studies k-tuple totaldomination from an algorithmic point of view. In particular, we present alinear-time algorithm for the k-tuple total domination problem for graphs inwhich each block is a clique, a cycle or a complete bipartite graph, whichinclude trees, block graphs, cacti and block-cactus graphs. We also establishNP-completeness of the k-tuple total domination problem in undirected pathgraphs.

Keywords: k-Tuple total domination, Total domination, Block graph,Cactus, Algorithm, NP-complete, Undirected path graph

1. Introduction

All graphs in this paper are simple, i.e., finite, undirected, loopless andwithout multiple edges. Domination is a well studied subject in graph theoryand combinatorial optimization as it has many applications in the real worldsuch as location problems, sets of representatives, social network theory, etc

IThis research was partially supported by the National Science Council of the Republicof China under grants NSC100-2811-M-002-146 and NSC101-2115-M-002-005-MY3.

∗Corresponding author.Email addresses: drjamesblue@gmail.com (James K. Lan),

gjchang@math.ntu.edu.tw (Gerard Jennhwa Chang)

Preprint submitted to Elsevier May 29, 2013

and the literature on this topic has been surveyed and detailed in books[2, 5, 6]. A dominating set of a graph G is a subset D ⊆ V (G) such thatevery vertex not in D has at least one neighbor in D. The dominationnumber γ(G) of G is the minimum cardinality of a dominating set of G. Thedomination problem is to find a minimum dominating set of a graph.

The idea of dominating all vertices of the graph, rather than merelydominating vertices outside the set, is considered by Cockayne, Dawes andHedetniemi [3]. A total dominating set of a graph G is a subset D ⊆ V (G)such that every vertex of G has at least one neighbor in D. Every graphwithout isolated vertices has a total dominating set, since the vertex setV (G) is such a set. The total domination number γt(G) of G is the minimumcardinality of a total dominating set of G. Total domination is now a well-studied topic in graph theory; see the recent survey paper [7] for more details.

Another variation of domination, the k-tuple domination was introducedby Harary and Haynes [4]. For a fixed positive integer k, a k-tuple dominatingset of a graph G is a subset D ⊆ V (G) such that the closed neighborhoodof every vertex of G has at least k vertices in D. The k-tuple dominationnumber γ×k(G) of G is the minimum cardinality of a k-tuple dominatingset of G. The k-tuple domination number is only defined for graphs withminimum degree at least k − 1. The special case when k = 1 is the usualdomination.

Motivated by the concept of k-tuple domination, Henning and Kazemi [8]considered the following generalization of total domination. For a fixedpositive integer k, a k-tuple total dominating set of a graph G is a subsetD ⊆ V (G) such that every vertex of G has at least k neighbors in D. Everygraph with minimum degree at least k admits a k-tuple total dominating set.The k-tuple total domination number γ×k,t(G) of G is the minimum cardi-nality of a k-tuple total dominating set of G. The k-tuple total dominationproblem is to determine a minimum k-tuple total dominating set of a graph.Since a (k + 1)-tuple dominating set is also a k-tuple total dominating setand a k-tuple total dominating set is also a k-tuple dominating set, we haveγ×k(G) ≤ γ×k,t(G) ≤ γ×(k+1)(G) for graphs with minimum degree at leastk. The 1-tuple total domination is the well-studied total domination. The2-tuple total domination is called the double total domination in the litera-ture. Authors in [8, 9, 10] have established bounds on the number γ×k,t(G)in terms of different graph invariants.

On the complexity side of the k-tuple total domination problem, Prad-han [14] showed that the k-tuple total domination problem is NP-complete

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for bipartite graphs and for split graphs (and thus the chordal graphs). Thisproblem remains NP-complete for doubly chordal graphs, another subclass ofchordal graphs [14]. Apart from these, Pradhan also proposed some hardnessresults and approximation algorithms for this problem. On the other hand,Pradhan proved that the k-tuple total domination problem for chordal bipar-tite graphs is a subproblem of the k-tuple domination problem for stronglychordal graphs, which is solvable in polynomial time [13]. Therefore the k-tuple total domination problem for chordal bipartite graphs is polynomiallysolvable.

In this paper, we explore efficient algorithms for the k-tuple total domi-nation problem in graphs. In particular, we present a linear-time algorithmfor the k-tuple total domination problem in graphs in which each block is aclique, a cycle or a complete bipartite graph. This class of graphs includetrees, block graphs, cacti and block-cactus graphs. Since the k-tuple totaldomination is a generalization of the total domination, our algorithms coverthe partial results in [1, 7, 12]. Moreover, we also show that the k-tupletotal domination problem remains NP-complete for undirected path graphs,another subclass of chordal graphs.

2. Preliminaries

Let G = (V,E) be a graph with vertex set V and edge set E. For avertex v, the open neighborhood is the set NG(v) = {u ∈ V : uv ∈ E } andthe closed neighborhood is NG[v] = NG(v) ∪ { v }. The degree degG(v) of avertex v in G is the number of edges incident to v. When the graph G isclear from the context, it is dropped from notations. The minimum degreeamong the vertices of G is denoted by δ(G). An isolated vertex is a vertex vwith deg(v) = 0. A leaf of a graph is a vertex with degree one. The subgraphof 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]. For a vertex v in G, we write G− vfor G− { v }.

In a graph, an independent set is a set of pairwise nonadjacent vertices; aclique is a set of pairwise adjacent vertices. A tree is a connected graph with-out cycles. A vertex v is a cut-vertex if the number of connected componentsis increased after removing v. A block of a graph is a maximal connectedsubgraph without any cut-vertex. An end-block of a graph is a block con-taining at most one cut-vertex of the graph. A block graph is a graph whose

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blocks are cliques. A cactus is a connected graph whose blocks are either anedge or a cycle. A cactus is a tree if all the blocks are edges. A block-cactusgraph is a graph whose blocks are cliques or cycles.

A graph is an intersection graph if there is a correspondence between itsvertices and a family of sets (the intersection model) such that two distinctvertices are adjacent in the graph if and only if their two corresponding setshave a nonempty intersection. A graph G is chordal if G has no inducedcycle with length greater than 3. It is well-known that a graph is chordalif and only if it is the intersection graph of some subtrees of a certain tree.If these subtrees are paths, this chordal graph is called an undirected pathgraph.

3. Block-wise approach for k-tuple total domination

The main result of this section is an algorithm for the k-tuple total dom-ination problem in graphs. Actually our algorithm solves a slightly moregeneral problem, which will be formulated as “L-domination”.

3.1. Labeling method for k-tuple total domination

Labeling techniques are widely used in the literatures for solving the domi-nation problem and its variants [2, 11, 12, 13]. For k-tuple total domination,we employ the following labeling method which is similar to that in [11].Given a graph G, labeling L is a mapping that assigns each vertex v in G atwo-tuple label L(v) = (L1(v), L2(v)), where L1(v) ∈ {B,R }, and L2(v) isa nonnegative integer. Here a vertex v with L1(v) = R is called a requiredvertex ; a vertex v with L1(v) = B is called a bound vertex. An L-dominatingset of G is a subset D ⊆ V (G) such that

• if L1(v) = R, then v ∈ D, and

• for each v ∈ V (G), |NG(v) ∩D| ≥ L2(v).

That is, D contains all required vertices, and for each vertex v of G, v isadjacent to at least L2(v) vertices in D. The L-domination number γL(G)is the minimum cardinality of an L-dominating set in G, such set is called aγL-set of G. Clearly G has an L-dominating set if and only if L2(v) ≤ degG(v)for all v ∈ V (G), and we call such L a proper labeling. Notice that if L(v) =(B, k) for all v ∈ V (G), then γL(G) = γ×k,t(G). Thus an algorithm for γL(G)

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gives γ×k,t(G). In the following, we give a general approach (block-wise) tofind a minimum L-dominating set in a graph.

Suppose G is a graph with a proper labeling L = (L1, L2). Let C bean end-block of G and x be its unique cut-vertex. Let G′ denote the graphwhich results from G by deleting all vertices only in C. Suppose ℓ is anonnegative integer such that ℓ ≤ degC(x). The following notations will beused throughout the rest of this section.

• Lℓ : the restriction of L on C with the modification on Lℓ2(x) = ℓ.

• LℓR : the same as Lℓ except for the modification on LℓR1 (x) = R.

Since the value L2(x) may be greater than the number of neighbors of x in C,we need to set the L2 value of x as the minimum between L2(x) and degC(x)before evaluating the cardinality of a γL-set of C. Thus, for convenience, set

α = min{L2(x), degC(x)}.

Note that one always have γL0(C) ≤ γLℓ(C) ≤ γLℓR(C). Especially, substi-tuting α for ℓ gives

γL0(C) ≤ γLα(C) ≤ γLαR(C). (1)

For a vertex v of G with a proper labeling L, denote by D∗v(G,L) a min-

imum L-dominating set of G such that v has the most neighbors in thisset among all minimum L-dominating sets of G, i.e., |NG(v) ∩D∗

v(G,L)| ≥|NG(v) ∩D| for any minimum L-dominating set D of G. Fig. 1 (b) illustratesan example of D∗

v(G,L), while the minimum L-dominating set formed by theshaded vertices in Fig. 1 (a) cannot be selected as a D∗

v(G,L). The construc-tion and correctness of the algorithm is based on the following theorem.

(B,0)

(B,1)(B,1)

v

(a) (b)

(B,2) (B,0)

(B,0)

(B,1)(B,1)

v

(B,2) (B,0)

Figure 1: Minimum L-dominating sets of a graph.

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Theorem 1. Suppose G is a graph with a proper labeling L = (L1, L2).Suppose C is an end-block of G and x is its unique cut-vertex. Let G′ =(G − C) ∪ { x } and L′ be the restriction of L on G′ with modifications asdescribed below. Let α = min{L2(x), degC(x)}, s = |NC(x) ∩D∗

x(C,L0)| and

t = max{L2(x)− s− degG′(x), 0}.

(1) If γL0(C) = γLαR(C), then γL(G) = γL′(G′)+γLαR(C)−1, where L′1(x) =

R, L′2(x) = L2(x)− α.

(2) If γL0(C) = γLα(C) < γLαR(C), then γL(G) = γL′(G′) + γLα(C), whereL′2(x) = L2(x)− α.

(3) Suppose γL0(C) < γLα(C).

(3.1) If L1(x) = R or there exists y ∈ NG(x) such that L2(y) = degG(y),then γL(G) = γL′(G′) + γLℓR(C) − 1, where L′

1(x) = R, L′2(x) =

L2(x)− ℓ and ℓ = s+ t.(3.2) if L2(y) < degG(y) for all y ∈ NG(x), then γL(G) = γL′(G′) +

γLℓ(C), where L′2(x) = L2(x)− ℓ and ℓ = s+ t.

The following observation is easily obtained.

Proposition 2. Suppose G is a graph with a proper labeling L. Let D be aγL-set of G. Then |D ∩ C| ≥ γL0(C).

Lemma 3. Suppose G is a graph with a proper labeling L. Let C be anend-block of G, x be its unique cut-vertex and G′ = (G − C) ∪ {x }. Let Dbe a γL-set of G. Let α = min{L2(x), degC(x)}. Let DC be a γLα-set or aγLαR-set of C and D = (D − C) ∪DC. Then

∣∣NG′(x) ∩D∣∣ ≥ L2(x)− α and∣∣NC(x) ∩D

∣∣ ≥ α. In other words,∣∣NG(x) ∩D

∣∣ ≥ L2(x).

Proof. By the definition of L-dominating set, |NG(x) ∩D| ≥ L2(x) and∣∣NC(x) ∩D∣∣ ≥ α. If L2(x) ≤ degC(x), i.e., α = L2(x), then

∣∣NG′(x) ∩D∣∣ ≥

0 = L2(x) − α. If L2(x) > degC(x), i.e., α = degC(x), then α = degC(x) ≥|NC(x) ∩D|, as x can have at most degC(x) neighbors in D. Hence |NG′(x)∩D| = |NG′(x)∩D| = |NG(x)∩D| − |NC(x)∩D| ≥ L2(x)− α. The assertionholds clearly.

Proof of Theorem 1 (1). Let D′ be a γL′-set of G′ and DC be a γLαR-set ofC. Since L′

1(x) = LαR1 (x) = R, x ∈ D′∪DC . Also that |NG(x) ∩ (D′ ∪DC)| =

|NG(x) ∩D′| + |NG(x) ∩DC | ≥ L′2(x) + α = L2(x), by the assumption. We

have D′ ∪DC is an L-dominating set of G and hence γL(G) ≤ |D′ ∪DC | =|D′|+ |DC | − 1 = γL′(G′) + γLαR(C)− 1.

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Conversely, suppose D is a γL-set of G. Let DC be a γLαR-set of C and letD = (D − C) ∪DC . By the assumption that γL0(C) = γLαR(C) and Eq. (1)and Proposition 2, |D| ≥

∣∣D∣∣. Note that no matter what L1(x) might be, x is

always be included in D. By Lemma 3,∣∣NG(x) ∩D

∣∣ ≥ L2(x) and thus D is

also an L-dominating set of G. Moreover,∣∣NG′(x) ∩D

∣∣ ≥ L2(x)−α = L′2(x)

and∣∣NC(x) ∩D

∣∣ ≥ α. Thus D∩G′ is an L′-dominating set of G′ and D∩Cis an LαR-dominating set of C. Hence γL(G) = |D| ≥

∣∣D∣∣ =∣∣D ∩G′

∣∣ +∣∣D ∩ C∣∣− 1 ≥ γL′(G′) + γLαR(C)− 1.

Note that under the assumptions of Theorem 1 (2), x must be a boundvertex, i.e., L1(x) = B. The argument is similar to that of Theorem 1 (1)and therefore is omitted.

Proposition 4. Suppose G is a graph that is also a block. Suppose L is aproper labeling on G. Let x be a vertex of G. Let s = |NG(x) ∩ D∗

x(G,L0)|and j be a nonnegative integer such that s+ j ≤ degG(x). Then γLs+j(G) =γLs(G) + j.

Proof. LetDs be a γLs-set of G andD = Ds∪J , where J consists of j verticesof NG(x) \ Ds. Clearly D is an Ls+j-dominating set of G and γLs+j(G) ≤|D| = |Ds|+ j = γLs(G) + j holds.

Conversely, suppose D is a γLs+j -set of G and we have |NG(x) ∩D| ≥s + j. Since D must contain an L0-dominating set of G, there exists asubset J ⊆ V (G) such that D \ J is an L0-dominating set of G. Clearly,|J | ≤ γLs+j(G) − γL0(G). Also, by the definition of s, γL0(G) = γLs(G) and|NG(x) ∩ (D \ J)| ≤ s. This indicates that |J | ≥ j. Therefore, γLs+j(G) ≥γL0(G) + j = γLs(G) + j.

Proof of Theorem 1 (3). We first consider (3.1) and assume that L1(x) =R. It is easy to check that LℓR is a proper labeling on V (C). Let D′ be aγL′-set of G′ and DC be a γLℓR-set of C. Since L′

1(x) = LℓR1 (x) = R, x ∈ D′∪

DC . Also by assumptions, we have |NG(x) ∩ (D′ ∪DC)| = |NG(x) ∩D′| +|NG(x) ∩DC | ≥ L2(x). Thus D

′∪DC is an L-dominating set of G and henceγL(G) ≤ |D′ ∪DC | = |D′|+ |DC | − 1 = γL′(G′) + γLℓR(C)− 1.

Conversely, suppose D is a γL-set of G. Let DC be a γLℓR-set of C.Case 1: |NC(x) ∩D| ≤ ℓ. Let D = (D − C) ∪ DC . We now claim that

|D ∩ C| ≥ γLℓR(C). If this claim holds, then we have |D| ≥∣∣D∣∣. First

consider the case of t = 0. Since ℓ = s + t, we have γLℓR(C) = γLsR(C).Since x can have at most s neighbors in a γL0R-set of C, γLsR(C) = γL0R(C).

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By Proposition 2 and the assumption that L1(x) = R, |D ∩ C| ≥ γL0(C) =γL0R(C) = γLℓR(C) holds.

Now suppose t > 0. From the definition of t, we have L2(x) = ℓ+degG′(x).Since D contains at least L2(x) vertices of NG(x), D must contain at least ℓvertices ofNC(x), i.e., |NC(x) ∩D| ≥ ℓ, for otherwise we have |NG′(x) ∩D| >L2(x) − ℓ = degG′(x), a contradiction. By the assumption that L1(x) = R,D ∩ C is also an LℓR-dominating set of C. Hence |D ∩ C| ≥ γLℓR(C). Wetherefore have this claim.

By the choice of D, x is included in D. Also that∣∣NC(x) ∩D

∣∣ ≥ ℓ

and thus D ∩ C is an LℓR-dominating set of C. By the assumption that|NC(x) ∩D| ≤ ℓ, we have

∣∣NG′(x) ∩D∣∣ = |NG(x) ∩D| − |NC(x) ∩D| ≥

L2(x)− ℓ = L′2(x), and thus D∩G′ is also an L′-dominating set of G′. Hence

γL(G) = |D| ≥∣∣D∣∣ = ∣∣D ∩G′

∣∣+ ∣∣D ∩ C∣∣− 1 ≥ γL′(G′) + γLℓR(C)− 1.

Case 2: |NC(x) ∩D| > ℓ. For easy writing, set r = |NC(x) ∩D|. ThenD ∩ C is an LrR-dominating set of C. Hence |D ∩ C| ≥ γLrR(C). LetD = (D − C) ∪DC ∪ Y , where Y is a subset of NG′(x) of cardinality r − ℓ.Note that Y may contain vertices that are already in D. Clearly |DC ∪ Y | =γLℓR(C) + r − ℓ. We now show that |D ∩ C| ≥ |DC ∪ Y |. Since r > ℓ ≥ s,by Proposition 4, γLrR(C) = γLℓR(C) + r − ℓ and thus |D ∩ C| ≥ γLrR(C) =γLℓR(C) + r − ℓ = |DC ∪ Y | holds. Consequently, we have |D| ≥ |DC |.

By the choice of D, x is included in D. Also that∣∣NC(x) ∩D

∣∣ ≥ ℓ and

thus D ∩ C is an LℓR-dominating set of C. Now we shall show that D ∩ G′

is an L′-dominating set of G′. From the definition of ℓ and t, if t = 0 thenL2 − ℓ = L2 − s ≤ degG′(x); if t > 0 then L2 − ℓ = degG′(x). In bothcases, we have L2− ℓ ≤ degG′(x). Since D contains at least L2(x) vertices ofNG(x), |NG′(x) ∩D| ≥ L2(x) − r. Let A be a vertex subset of NG′(x) ∩ Dof cardinality L2(x)− r. Now we pick arbitrarily exactly r − ℓ vertices fromNG′(x) \ A to form the vertex set Y . Note that Y is well-defined as wehave shown that L2 − ℓ ≤ degG′(x). Since NG′(x) ∩D = (NG′(x) ∩D) ∪ Y ,we have

∣∣NG′(x) ∩D∣∣ ≥ |A| + |Y | = L2(x) − ℓ = L′

2(x). Consequently,

γL(G) = |D| ≥∣∣D∣∣ = ∣∣D ∩G′

∣∣+ ∣∣D ∩ C∣∣− 1 ≥ γL′(G′) + γLℓR(C)− 1.

Now we assume that L1(x) = B. If x has a neighbor y such that L2(y) =deg(y), then x must be included in any L-dominating set of G. Let L be thesame as L except for the modification on L1(x) = R. In this case, we haveγL(G) = γL(G). Thus the assertion holds clearly.

The argument of (3.2) is similar to (3.1) and therefore is omitted.

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3.2. Algorithm

We are now in the position to present our algorithm, called kTTDom,to determine a minimum L-dominating set of a graph. In our algorithm, weassume kTTDomB is a subroutine that can find a minimum L-dominatingset of each end-block C of a graph.

Algorithm: kTTDom (A block-wise approach for finding a γL-set ingraphs. kTTDomB is a subroutine we assume it can find aγL-set of each end-block of the graph)

Input: A graph G with a proper labeling L = (L1, L2).Output: A minimum L-dominating set D of G.Method:G′ ← G;D ← ∅;while G′ ̸= ∅ do

if G′ is a block thenD ← D ∪ kTTDomB(G′, L);G′ ← ∅;

elselet C be an end-block of G′ and x be its unique cut-vertex;α← min {L2(x), degC(x) };U0 ← kTTDomB(C,L0);UR ← kTTDomB(C,LαR);if |U0| = |UR| then

D ← D ∪ UR;L1(x)← R;L2(x)← L2(x)− α;

elseif L1(x) = B and ∃ y ∈ NG(x) s.t. L2(y) = degG(y) then

L1(x)← R;s← |NC(x) ∩D∗

x(C,L0)|;

ℓ← s+max{L2(x)− s− degG′(x), 0};D ← D ∪ kTTDomB(C,Lℓ);L2(x)← L2(x)− ℓ;

G′ ← (G′ − C) ∪ { x };end

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Theorem 5. Algorithm kTTDom finds a minimum L-dominating set ofa graph G in linear-time if kTTDomB and D∗

x(C,L0) take linear time to

compute for each end-block C of G with cut-vertex x.

Proof. The correctness comes from Theorem 1. For the time complexity,since kTTDom calls at most three times of kTTDomB and computesD∗

x(C,L0) at most once for each end-block C of the graph, it is clear that

kTTDom is linear by the assumptions.

4. L-domination for some classes of graphs

In this section we present linear-time algorithms for finding a minimumL-dominating set for complete graphs, cycles and complete bipartite graphs.In addition, the computations ofD∗

v(G,L) for the mentioned classes of graphsare also discussed. Combing with the algorithm presented in the previoussection, we obtain a linear-time algorithm for the k-tuple total dominationproblem in graphs whose blocks are cliques, cycles or complete bipartitegraphs. These include block graphs, cacti and block-cactus graphs.

Throughout the rest of this section, suppose G is a graph with a properlabeling L = (L1, L2). The aim is to find a minimum L-dominating set D ofG. Define

R̃ = { v ∈ V : L1(v) = R or ∃u ∈ NG(v) s.t. L2(u) = degG(u) } . (2)

And let |R̃| = r. By the definition of L-dominating set, all vertices of R̃ mustbe included in D.

4.1. Complete graphs

Suppose G = (V,E) is a complete graph with n vertices. Let V \ R̃ ={ v1, v2, . . . , vn−r }. If vi ̸∈ D and vj ∈ D with L2(vi) < L2(vj) for some

vi, vj ∈ V \ R̃, then (D − vj) ∪ { vi } is also a minimum L-dominating

set of G. This indicates that we shall choose vertices in V \ R̃ with L2

value as small as possible. Suppose vRmax (resp. vmax) has the maximum

L2 value among all vertices in R̃ (resp. V \ R̃). It is the case that D =

R̃ ∪ S, where S consists of the smallest s vertices of V \ R̃, where s =max {L2(v

Rmax)− r + 1, L2(vmax)− r, 0 }. However, some vertex in S may

have L2 value no less than r + s. In that case, one can simply add anothervertex of V \ R̃ to S. In other words, choose the smallest s + 1 vertices of

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V \ R̃ and R̃ to form a minimum L-dominating set of G. The process of thealgorithm is described as follows.

Algorithm: kTTDomKn (Finding a γL-set of a complete graph)Input: A complete graph G = (V,E) with a proper labeling L = (L1, L2).Output: A minimum L-dominating set D of G.Method:D ← ∅;R̃← { v ∈ V : L1(v) = R or ∃u ∈ NG(v) s.t. L2(u) = degG(u) };r ← |R̃|;v0 ← ∅ ; // pseudo vertex

L2(v0)← 0;

let v1, v2, . . . , vn be a vertex ordering of V such that vi ∈ V \ R̃ for1 ≤ i ≤ n− r, L2(v1) ≤ · · · ≤ L2(vn−r) and L2(vn−r+1) ≤ · · · ≤ L2(vn);s← max {L2(vn)− r + 1, L2(vn−r)− r, 0 };if L2(vs) ≥ r + s then

s← s+ 1;

D ← R̃ ∪ { vi ∈ V \ R̃ : 0 ≤ i ≤ s };

Theorem 6. Algorithm kTTDomKn finds a minimum L-dominating setfor a complete graph in linear time.

Proof. The correctness is clear and is omitted. The time complexity is boundby the computation of the vertex ordering of V \ R̃. Note that L is a proper

labeling, L2(v) ≤ degG(v) < n for all v ∈ V \ R̃. Since each L2(v) is aninteger in the range 0 to n and there are at most n integers need to sort,one can use linear-time sorting algorithms, for examples, Counting sort, toobtain the vertex ordering.

Consider the computation of D∗x(G,L) of a complete graph G for some

fixed vertex x. Since each pair of vertices of G is adjacent, any minimumL-dominating set D of G has the property that |NG(x) ∩D| is maximum,and can be selected as D∗

x(G,L). Thus D∗x(G,L) can be found in linear time.

4.2. Trees

Since each end-block of a tree T of order at least 2 is a clique of cardinality2, a solution to find a minimum L-dominating set of a tree T is to usekTTDom and kTTDomKn. However, since tree has a simple structure,

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we provide an alternative algorithm to find a minimum L-dominating setin trees. The presented algorithm is actually the simplified version of thecombination of kTTDom and kTTDomKn for the input graph to be trees,and will also be used later as a subroutine to find a minimum L-dominatingset for cycles, cacti and block-cactus graphs.

Given a tree T of n vertices, it is well-known that T has a vertex orderingv1, v2, . . . , vn such that vi is a leaf of Gi = G[vi, vi+1, . . . , vn] for 1 ≤ i ≤ n−1.This ordering can be found in linear-time by using, for example, the breadth-first-search (BFS) algorithm.

The algorithm visits vertices of T along the tree ordering v1, v2, . . . , vn. Ineach iteration, processing a leaf vi of a tree T , which is adjacent to a uniquevertex u. The label of vi is used to possibly relabel u. After v is visited,vi is ignored from T and a new tree T ′ is obtained. A linear-time labelingalgorithm for finding a γL-set in trees is shown as follows.

Algorithm: kTTDomT (Finding a γL-set of a tree)Input: A tree T of n vertices with a tree ordering v1, v2, . . . , vn and a

proper labeling L = (L1, L2).Output: A minimum L-dominating set D of T .Method:D ← ∅;for i← 1 to n do

T ′ ← T [vi, . . . , vn];let u be the parent of vi in T ′ (regard u as vi if i = n);if L2(vi) = 1 then

L1(u)← R;

if L1(vi) = R or L2(u) = degT ′(u) then // vi ∈ R̃L2(u)← max {L2(u)− 1, 0 };D ← D ∪ { vi };

end

The construction and correctness of the algorithm is based on the follow-ing lemma.

Lemma 7. Suppose T is a tree of order at least 2 with a proper labelingL = (L1, L2). Let v be a leaf adjacent to u in T

(1) If L2(v) = 1, then γL(T ) = γL′(T ), where L′ is the same as L except forthe modification on L′

1(u) = R.

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(2) If v ∈ R̃, then γL(T ) = γL′(T − v) + 1, where L′ is the restriction of Lon T − v with the modification on L′

2(u) = max {L2(u)− 1, 0 }(3) If L(v) = (B, 0) and v ̸∈ R̃, then γL(T ) = γL′(T − v), where L′ is the

restriction of L on T − v.

Proof. (1) By the definition of L-domination, γL(T ) ≤ γL′(T ) holds clearly.Now suppose D is a γL-set of T . Since L2(v) = 1, u must be included in D.Thus D is also an L′-dominating set of T . Hence we have γL′(T ) ≤ γL(T ).

(2) By the Case (1), we can assume L1(u) = R if L2(v) = 1 withoutloss of generality. Suppose D′ is a γL′-set of T − v. Set D = D′ ∪ { v }.Since L′ is the restriction of L on T − v with the modifications on L′

2(u)and L2(u) ≤ L′

2(u) + 1, we have |NT (u) ∩D| ≥ L2(u). If L2(v) = 1, thenby assumption that L1(u) = R and thus u ∈ D′. Therefore D is also anL-dominating set of T . Thus γL(T ) ≤ |D| = |D′|+ 1 = γL′(T − v) + 1.

Conversely, suppose D is a γL-set of T . By the assumption that v ∈ R̃, vmust be included in D. Set D′ = D \ { v }. As L′ is the restriction of L onT − v with the modification on L′

2(u), |NT−v(x) ∩D′| ≥ L2(u)− 1 = L′2(u).

Thus D′ is an L′-dominating set of V (T ′). Hence γL′(T − v)+1 ≤ |D′|+1 =|D| = γL(T ).

(3) Suppose D′ is a γL′-set of T − v. Since L′ is the restriction of L onT − v and |N(v) ∩ D′| ≥ 0 = L2(v), it is clear that D′ is an L-dominatingset of T . Thus γL(T ) ≤ |D′| = γL′(T − v).

Conversely, suppose D is a γL-set of T . If v ̸∈ D, then D is also anL′-dominating set of T − v. Now suppose v ∈ D. If |NT−v(u) ∩ (D − v)| ≥L′

2(u), then we are done. For otherwise, pick a vertex w ∈ NT−v(u) \D and

set D′ = (D − v) ∪ {w }. Since v ̸∈ R̃, we have L2(u) < degT (u) and thus|NT−v(u) ∩D′| ≥ L′

2(u). Therefore D′ is also an L′-dominating set of T − v.Hence γL′(T − v) ≤ |D′| = |D| = γL(T ).

Theorem 8. Algorithm kTTDomT finds a minimum L-dominating set fora tree in linear-time.

4.3. Cycles

Suppose G = (V,E) is a cycle. We will use kTTDomT as a subroutineto find a minimum L-dominating set of G. Basically, the idea is to pick aparticular vertex v∗ of G, cut the cycle at v∗ to get a path Pv∗ , and then applykTTDomT to obtain a minimum L′-dominating set of Pv∗ . The principleof picking is to choose a vertex that must be included in D and is shown as

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follows. If R̃ ̸= ∅, then we know that all vertices of R̃ must be included inD. Thus pick arbitrarily a vertex from R̃ as v∗. Now assume R̃ = ∅, i.e.,L1(v) = B and L2(v) < 2 for all v ∈ V . If L2(v) = 0 for all v ∈ V , thenclearly D = ∅. Otherwise there must exist some vertex v that has L2(v) = 1.This indicates that at least one of the two neighbors of v, say u and w, mustbe included in D. In this case, we can pick v∗ from u or w by testing thecardinality of the minimum L′-dominating set of C, where L′ is the same asL except for the modification on L1(u) = R or L1(w) = R.

After picking v∗, we need to modify the labels of its neighbors accordingly.Let u∗ and w∗ be the two neighbors of v∗ on C. Since v∗ must be includedin D, the L2 values of u∗ and w∗ should be decreased by 1. Let Pv∗ be thepath of C − v∗ and LP be the restriction of L on Pv∗ with the modificationsas described below.

Case 1: L2(v∗) = 2. Clearly u∗ and w∗ must be included in D. Thus just

set the LP1 values of u∗ and w∗ as R.

Case 2: L2(v∗) = 1. So at least one of u∗ and w∗ must be included in

D. In this case, we test the cardinality of the minimum LQ-dominating setof Pv∗ to decide which one of u∗ and w∗ should be chosen, where LQ is thesame as LP except for the modifications on LQ

1 (u∗) = R or LQ

1 (w∗) = R.

Case 3: L2(v∗) = 0. Then none has to be changed in this case. The

detailed algorithm is shown as in Algorithm kTTDomCYC.The correctness is clear and therefore is omitted. The time complexity is

clearly linear, since kTTDomT is linear and kTTDomCYC calls at mosttwo times of kTTDomT.

Theorem 9. Algorithm kTTDomCYC finds a minimum L-dominating setin a cycle in linear time.

Now consider the computation of D∗x(G,L) of a cycle G for some fixed

vertex x. Let y and z be the neighbors of x in G. By the definition ofD∗

x(G,L), |NG(x) ∩D∗x(G,L)| ≤ 2. If |NG(x) ∩D∗

x(G,L)| = 2, then y, z ∈D∗

x(G,L) and |D∗x(G,L)| = γL′(G), where L′ is the same as L with the

modifications on L1(y) = R and L1(z) = R. If |NG(x) ∩D∗x(G,L)| = 1 and

suppose y ∈ D∗x(G,L), then |D∗

x(G,L)| = γL′(G), where L′ is the same asL with the modifications on L1(y) = R. Thus one can find D∗

x(G,L) byexamining among all possible combinations of modifications on L′

1(v) = Rfor all neighbors v of x with the condition that γL′(G) = γL(G). Since findinga γL-set of a cycle can be done in linear-time, the computation of D∗

x(G,L)clearly can be done in linear time, too.

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Algorithm: kTTDomCYC (Finding a γL-set of a cycle)Input: A cycle G = (V,E) with a proper labeling L = (L1, L2).Output: A minimum L-dominating set D of G.Method:D ← ∅;R̃← { v ∈ V : L1(v) = R or ∃u ∈ NG(v) s.t. L2(u) = 2 };if R̃ ̸= ∅ then

pick an arbitrarily vertex in R̃ as v∗;else if ∃ v s.t. L2(v) = 1 then

let u,w be the two neighbors of v;Uu ← kTTDomCYC(G,L′), where L′ ← L with L′

1(u)← R;Uw ← kTTDomCYC(G,L′), where L′ ← L with L′

1(w)← R;if |Uu| ≤ |Uw| then D ← Uu else D ← Uw;

else stop ;let the two neighbors of v∗ be u∗ and w∗;let Pv∗ be the path G− v∗ and LP be the restriction of L on Pv∗ ;LP

2 (u∗)← max{L2(u

∗)− 1, 0};LP

2 (w∗)← max{L2(w

∗)− 1, 0};if L2(v

∗) = 2 thenLP

1 (u∗)← LP

1 (w∗)← R;

D ← kTTDomT(Pv∗ , LP ) ∪ {v∗};

else if L2(v∗) = 1 then

Uu∗ ← kTTDomT(Pv∗ , LQ), where LQ ← LP with LQ

1 (u∗)← R;

Uw∗ ← kTTDomT(Pv∗ , LQ), where LQ ← LP with LQ

1 (w∗)← R;

if |Uu∗| ≤ |Uw∗| then D ← Uu∗ ∪ {v∗} else D ← Uw∗ ∪ {v∗};else

D ← kTTDomT(Pv∗ , LP ) ∪ {v∗};

4.4. Complete bipartite graphs

Suppose G = (A ∪ B,E) is a complete bipartite graph whose vertex set

is a disjoint union of two independent sets A and B. Let r1 = |A ∩ R̃| andr2 = |B ∩ R̃|. The argument is similar to that of complete graphs. Let amax

(resp. bmax) be a vertex in A that has the maximum L2 value among allvertices of A (resp. B). Since D must contain at least L2(amax) verticesof B, it is the case that we shall choose the smallest max{L2(amax) − r2, 0}(resp. max{L2(bmax) − r1, 0}) vertices of B \ R̃ (resp. A \ R̃). The processof the algorithm is described as follows.

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Algorithm kTTDomKmn (Finding a γL-set of a complete bipartitegraph)Input: A complete bipartite graph G whose vertex set is a disjoint union of

two independent sets A and B, and a proper labeling L = (L1, L2).Output: A minimum L-dominating set D of G.Method:D ← ∅;R̃← { v ∈ V : L1(v) = R or ∃u ∈ NG(v) s.t. L2(u) = degG(u) };r1 ← |A ∩ R̃|;r2 ← |B ∩ R̃|;let a1, a2, . . . , a|A| and b1, b2, . . . , b|B| be vertex orderings of A and B,

respectively, such that ai ∈ A \ R̃ for 1 ≤ i ≤ |A| − r1, bi ∈ B \ R̃ for1 ≤ i ≤ |B| − r2, L2(a1) ≤ · · · ≤ L2(a|A|−r1), L2(a|A|−r1+1) ≤ · · · ≤ L2(a|A|),L2(b1) ≤ · · · ≤ L2(b|B|−r2), and L2(b|B|−r2+1) ≤ · · · ≤ L2(b|B|);a0 ← ∅; b0 ← ∅ ; // pseudo vertices

sa ← max{L2(a|A|)− r2, L2(a|A|−r1)− r2, 0};sb ← max{L2(a|B|)− r1, L2(a|B|−r1)− r1, 0};D ← R̃ ∪ { ai ∈ A \ R̃ : 0 ≤ i ≤ sa } ∪ { bi ∈ B \ R̃ : 0 ≤ i ≤ sb };

Theorem 10. Algorithm kTTDomKmn finds a minimum L-dominatingset in a complete bipartite graph in linear time.

Proof. The correctness is clear and is omitted. The time complexity is linear,since the vertex orderings of A\R̃ and B\R̃ can be found by using linear-timesorting algorithms.

The computation of D∗x(G,L) of a complete bipartite graph G for some

fixed vertex x can be found in linear time as any minimum L-dominating setD of G has the property that |NG(x) ∩D| is maximum, and can be selectedas D∗

x(G,L).It is well-known that block graphs, cacti, block-cactus graphs can be

recognized in linear-time. By Theorems 5, 6, 9, and 10, one can immediatelyhave the following result.

Theorem 11. Algorithm kTTDom finds a minimum L-dominating set inlinear time for graphs in which each block is a clique, a cycle or a completebipartite graph, including block graphs, cacti and block-cactus graphs.

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5. NP-completeness result

In this section, we study the complexity of the k-tuple total dominationproblem:

k-TUPLE TOTAL DOMINATION (kTTD)INSTANCE: A graph G = (V,E) and positive integers k and s.QUESTION: Does G have a k-tuple total dominating set of size ≤ s?

It has been proved that kTTD is NP-complete for bipartite graphs andsplit graphs [14], in which the reductions are mainly from the well-knownvertex cover problem. Pradhan also showed that kTTD is NP-complete fordoubly chordal graphs, a subclass of chordal graphs. In this section, we showthat kTTD remains NP-complete for undirected path graphs, another sub-class of chordal graphs. The argument is similar to that in [12], where the re-duction is from another well-known NP-complete problem, the 3-dimensionalmatching problem. For the sake of completeness, we will describe the reduc-tion completely.

3-DIMENSIONAL MATCHING (3DM)INSTANCE: Disjoint sets X, Y and Z, each of cardinality q, and a set M ⊆X × Y × Z of triples having cardinality p.QUESTION: Is there a set of q triples in M such that each element of X ∪Y ∪ Z is contained in exactly one of these triples?

Theorem 12. For any fixed positive integer k, kTTD is NP-complete forundirected path graphs.

Proof. Obviously kTTD belongs to NP, since it is easy to verify a “yes”instance of kTTD in polynomial time. Consider an instance of 3DM. Let

X = {xr : 1 ≤ r ≤ q } , Y = { ys : 1 ≤ s ≤ q } , Z = { zt : 1 ≤ t ≤ q } ,

and a subset

M = {mi = (xr, ys, zt) : xr ∈ X, ys ∈ Y and zt ∈ Z for 1 ≤ i ≤ p }

of triples X × Y × Z. Now we construct a clique tree T having 6p+ 3q + 1cliques from which we will obtain an undirected path graph. The vertices ofthe tree T , which are represented by sets, are explained below.

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For each triple mi ∈M , there are six vertices depend only upon the tripleitself and not upon the elements within the triple:

{Aij , Bij , Cij , Dij }{Aij , Bij , Dij , Fij }{Cij , Dij , Gij }{Aij , Bij , Dij , Eij }{Aij , Eij , Hij }{Bij , Eij , Iij } for 1 ≤ i ≤ p, 1 ≤ j ≤ k.

These six vertices form the subtree corresponding to mi, which is illustratedin Fig. 2. Next there is a vertex for each element of X, Y and Z that dependsupon the triples of M to which each respective element belongs:

{Rr } ∪ {Ai1 . . . , Aik : xr ∈ mi } for all xr ∈ X,

{Ss } ∪ {Bi1 . . . , Bik : ys ∈ mi } for all ys ∈ Y,

{Tt } ∪ {Ci1 . . . , Cik : zt ∈ mi } for all zt ∈ Z.

Finally, the last vertex, the root of the tree T , which contains

{Aij , Bij , Cij : 1 ≤ i ≤ p, 1 ≤ j ≤ k } .

The arrangement of these vertices in the tree T is shown in Fig. 2. This thenresult in an undirected path graph G with vertex set

{Aij , Bij , Cij , Dij , Eij , Fij , Gij , Hij , Iij : 1 ≤ i ≤ p, 1 ≤ j ≤ k }∪{Rj, Sj, Tj : 1 ≤ j ≤ q }

of cardinality 9pk + 3q, where the undirected path in T corresponding to avertex v of G consists of those vertices (sets) containing v in the tree T .

Now we shall show that 3DM has a solution if and only if G has a k-tupletotal dominating set of cardinality 2kp + kq. Suppose 3DM has a solutionM ′ of cardinality q. Let

D = {Aij , Bij , Cij : mi ∈M ′, 1 ≤ j ≤ k }∪{Dij , Eij : mi ∈M \M ′, 1 ≤ j ≤ k } .

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It is straightforward to check that D is a k-tuple total dominating set of Gof cardinality 3kq + 2k(p− q) = 2kp+ kq.

Conversely, suppose D is a k-tuple total dominating set of G of cardi-nality 2kp + kq. Observe that for any i, the cardinality of a k-tuple totaldominating set of the subgraph induced by the vertex set {Aij ,Bij ,Cij ,Dij ,Eij ,Fij ,Gij ,Hij ,Iij : 1 ≤ i ≤ p, 1 ≤ j ≤ k} corresponding to mi is at least2k, and the only way is to choose Dij and Eij for all j, 1 ≤ j ≤ k. Anylarger k-tuple total dominating set might just as well consist of Aij , Bij andCij for all j, 1 ≤ j ≤ k, since none of the other possible vertices domi-nates any vertex outside of the subtree. Consequently, suppose D containsAij , Bij , Cij , 1 ≤ j ≤ k for ℓ mi’s, and Dij , Eij , 1 ≤ j ≤ k for p− ℓ other mi’s,and at least max{3k(q − ℓ), 0}Rr, Ss, Tt. Then we have

2kp+ kq = |D| ≥ 3kℓ+ 2k(p− ℓ) + 3k(q − ℓ) = 2kp+ 3kq − 2kℓ

and so ℓ ≥ q. Picking q triples mi for which Aij , Bij , Cij , 1 ≤ j ≤ k are in Dform a matching M ′ of cardinality q.

for for for

for

Figure 2: A transformation to an undirected path graph.

[1] G.J. Chang, Total domination in block graphs, Operations ResearchLetters 8 (1989) 53–57.

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[2] G.J. Chang, Algorithmic aspects of domination in graphs (1998) 339–405. In: D. Du and P.M. Pardalos (Eds), Handbook of CombinatorialOptimization.

[3] E. Cockayne, R. Dawes, S. Hedetniemi, Total Domination in Graphs,Networks 10 (1980) 211–219.

[4] F. Harary, T. Haynes, Double domination in graphs, Ars Combinatoria55 (2000) 201–213.

[5] T. Haynes, S. Hedetniemi, P. Slater, Domination in Graphs: AdvancedTopics, Monographs and Textbooks in Pure and Applied Mathematics,Marcel Dekker, 1998.

[6] T. Haynes, S. Hedetniemi, P. Slater, Fundamentals of domination inGraphs, Monographs and Textbooks in Pure and Applied Mathematics,Marcel Dekker, 1998.

[7] M.A. Henning, A survey of selected recent results on total dominationin graphs, Discrete Mathematics 309 (2009) 32–63.

[8] M.A. Henning, A.P. Kazemi, k-tuple total domination in graphs, Dis-crete Applied Mathematics 158 (2010) 1006–1011.

[9] M.A. Henning, A.P. Kazemi, k-tuple total domination in cross productsof graphs, Journal of Combinatorial Optimization 24 (2012) 339–346.

[10] M.A. Henning, A. Yeo, Strong transversals in hypergraphs and doubletotal domination in graphs, SIAM Journal on Discrete Mathematics 24(2010) 1336–1355.

[11] J.K. Lan, G.J. Chang, Algorithmic aspects of the k-domination problemin graphs, Discrete Applied Mathematics 161 (2013) 1513 – 1520.

[12] R. Laskar, J. Pfaff, S. Hedetniemi, S. Hedetniemi, On the algorithmiccomplexity of total domination, SIAM Journal on Algebraic and DiscreteMethods 5 (1984) 420–425.

[13] C.S. Liao, G.J. Chang, Algorithmic aspect of k-tuple domination ingraphs, Taiwanese Journal of Mathematics 6 (2003) 415–420.

[14] D. Pradhan, Algorithmic aspects of k-tuple total domination in graphs,Information Processing Letters 112 (2012) 816–822.

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