Domination in distance-hereditary graphs
Maw-Shang Chang
a,Shaur-Ching Wu
a,Gerard J. Chang
b;∗,
Hong-Gwa Yeh
caDepartment of Computer Science and Information Engineering, National Chung Cheng University,
Min-Hsiun, Chiayi 621, Taiwan
bDepartment of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan c Department of Applied Mathematics, National University of Kaoshiung 811, Taiwan
Received 10 June 1997; revised 24 May 1999; accepted 2 October 2000
Abstract
The domination problem and its variants have been extensively studied in the literature. In this paper we investigate the domination problem in distance-hereditary graphs. In particular, we give a linear-time algorithm for the domination problem in distance-hereditary graphs by a labeling approach. We actually solve a more general problem,called the L-domination problem, which also includes the total domination problem as a special case. ? 2002 Elsevier Science B.V. All rights reserved.
Keywords: Domination; Distance-hereditary graph; Neighborhood; Twin; Leaf; Labeling
1. Introduction
All graphs in this paper are :nite,undirected,without loops and multiple edges. The distance dG(u; v) between two vertices u and v of a connected graph G is the
minimum length of a u–v path in G. A graph is distance-hereditary if each pair of vertices are equidistant in every connected induced subgraph containing them. Proper-ties of and optimization problems in distance-hereditary graphs have been extensively studied during the past two decades. In this paper,we study the domination problem in distance-hereditary graphs. We actually solve a more general problem,called the L-domination problem,which also includes the total domination as a special case.
The concept of domination can be used to model many location problems in operations research. In a graph G=(V; E); a dominating set is a subset D of V such that every vertex
Supported partly by the National Science Council under grants NSC85-2121-M194-020 and
NSC85-2121-M009-024.
∗Corresponding author. Tel.: +886-3-573-1945; fax: +886 3 572 4679.
E-mail addresses: [email protected] (M.-S. Chang),[email protected] (G.J. Chang), [email protected] (H.-G. Yeh).
0166-218X/02/$ - see front matter ? 2002 Elsevier Science B.V. All rights reserved. PII: S0166-218X(00)00330-9
in V − D is adjacent to at least one vertex in D. A dominating set D is independent, connected, total if the subgraph G[D] induced by D has no edges,is connected,and has no isolated vertices. The domination problem is one of :nding a minimum-sized dominating set of a given graph. The independent (connected, total) domination prob-lem is one of :nding a minimum-sized independent (connected,total) dominating set of a given graph. The domination problem and its variants have been extensively studied in the literature. For more information,readers are referred to [4,9–11].
It is well-known that the decision version of the domination problem and many of its variants are NP-complete for general graphs. Polynomial-time algorithms exist for the connected domination problem [6],the weighted connected domination prob-lem [19],the connected r-domination probprob-lem [2],the domination clique probprob-lem [7], and the weighted k-domination and the weighted k-dominating clique problems [20] in distance-hereditary graphs. The main purpose of this paper is to solve the dom-ination problem for distance-hereditary graphs in linear time. We employ a label-ing method which is powerful and widely used in many domination problems,see [3,5,12,14,15,17,18]. Our labeling method is mainly for the domination problem al-though the total domination is a by product. This coincides with the result by Kratsch and Stewart [13] that total domination can be reduced to domination for graph classes closed under adding false twins. We note that Nicolai and Szymczak [16] indepen-dently gave a linear-time algorithm for the domination problem in distance-hereditary graphs. Their method in fact solves the r-domination problem on homogeneous graphs, a supper class of distance-hereditary graphs,in O(|V ||E|) time.
In Section 2,we de:ne notation and introduce a characterization of distance-hereditary graphs that is useful in our algorithm. In Section 3,we introduce a more general concept, L-domination,which includes domination and total domination as special cases. Section 4 gives reduction lemmas from which a linear-time algorithm for the L-domination problem in distance-hereditary graphs is established. Concluding remarks are made in Section 5.
2. Preliminaries
Suppose G = (V; E) is a graph. For any vertex v ∈ V; the neighborhood of v is NG(v) = {u ∈ V : uv ∈ E}
and the closed neighborhood of v is NG[v] = NG(v) ∪ {v}. For a subset X of V; the
neighborhood of X is NG(X ) =v∈XNG(v) and the closed neighborhood of X is
NG[X ] =v∈XNG[v]. We use N(v) for NG(v); N[v] for NG[v]; N(X ) for NG(X ); and
N[X ] for NG[X ] if there is no ambiguity.
Let G[X ] denote the subgraph of G induced by X ⊆ V and G−x denote the subgraph induced by V − {x} in G. The degree of a vertex v is deg(v) = |N(v)|. An isolated vertex is a vertex of degree zero and a leaf is a vertex of degree one. Two vertices u and v are called false twins if N(u) = N(v) and true twins if N[u] = N[v]. Note that, by the de:nition,true twins are adjacent and false twins are not.
An ordering v1; v2; : : : ; vn of V is called a one-vertex-extension ordering of G if vi
is a vertex of degree at most one or is a twin of some vertex vj in Gi= G[Vi] for
16i6n; where Vi= {v1; v2; : : : ; vi}. For convenience,we denote NGi(x) as Ni(x) and
NGi[x] as Ni[x] for all x ∈ Vi.
Theorem 1 (Bandelt and Mulder [1] and Hammer and MaLray [8]). A graph is
distance-hereditary if and only if it has a one-vertex-extension ordering.
A one-vertex-extension ordering of a distance-hereditary graph can be generated in O(n + m) time,where n is the number of vertices and m is the number of edges (see [8]). In the following section,we assume a one-vertex-extension ordering of a given distance-hereditary graph has been constructed.
For our convenience,if vi and vj are true twins in Gi with Ni[vi] = Ni[vj] = {vi; vj};
we shall view vi as a leaf adjacent to vj in Gi. Our algorithm needs the following
lemma which is easily seen.
Lemma 2. Suppose v1; v2; : : : ; vn is a one-vertex-extension ordering of G. If vi and vj
are twins in Gi; then v1; : : : ; vj−1; vi; vj+1; : : : ; vi−1; vj; vi+1; : : : ; vn is also a
one-vertex-extension of G.
3. L-domination and basic assumptions
In order to solve the domination problem for distance-hereditary graphs,we use a labeling method to design an algorithm in which a one-vertex-extension ordering is used. At each iteration,the algorithm decides if a vertex vi is to be included in a
minimum solution according to its label. It then deletes vi and updates the label of its
twin or its unique neighbor. During the iterations,some information of deleted vertices are still kept in the remaining graph. For this purpose,we introduce the following general concept called L-domination.
Suppose G = (V; E) is a graph whose vertex set V is a subset of a ground set V
which is a supper set of V . Furthermore,each vertex v in G is associated with a regular label L(v)=S(v) or a conditional label L(v)=S(v)∗p(v); where S(v) ⊆{{0}; {1}; {0; 1}} and p(v) ∈ V− V with the property that p(x) = p(y) for any two distinct vertices
x and y. An L-dominating set of G is a subset D ⊆ V such that (LD) holds when v
has a regular label L(v) = S(v); and (LD) holds or p(v) ∈ D when v has a conditional label L(v) = S(v) ∗ p(v).
(LD) For each A ∈ S(v); there exists some x ∈ D such that dG(v; x) ∈ A:
In other words,if L(v)={{0}}; then v ∈ D; if L(v)={{1}}; then v has a neighbor in D; if L(v) = {{0; 1}}; then either v ∈ D or v has a neighbor in D; and if L(v) = {{0; 1}} ∗ p(v); then p(v) ∈ D or v ∈ D or v has a neighbor in D; ; : : : ; etc. When V = V and
each vertex has a regular label {{0; 1}} (respectively, {{1}}),each vertex either is in D or has a neighbor in D (respectively,has a neighbor in D) and so L-domination is
just the usual domination (respectively,total domination). The L-domination number L(G) is the minimum size of an L-dominating set of G. The L-domination problem
is to determine L(G) of a graph G.
When L(v) is a regular (conditional) label, L is (is not) the same as S(v) and so is (is not) considered as a set. In this paper,we always use “A ∈ L(v)” as an abbreviation for “L(v) is a regular label and A ∈ L(v)” and “L(v) ⊆ S” for “L(v) is a regular label and L(v) ⊆ S”.
Lemma 3. Suppose v is a vertex such that {A; B} ⊆ S(v); where A is a proper subset
of B. If L is obtained from L by replacing S(v) with S(v) − {B}; then L-domination
is the same as L-domination.
Proof. The lemma follows from the fact that (LD) holds for A ∈ S(v) implies that
(LD) holds for B ∈ S(v).
By Lemma 3,we only need consider ∅; {{0}}; {{1}}; {{0; 1}}; {{0}; {1}} for S(v).
Lemma 4. Suppose vertex v has a conditional label S(v) ∗ p(v) but that neither
S(v) = {{1}} with N(v) = ∅ nor S(v) = {{0}; {1}}. If L is obtained from L by
relabeling v with the regular label S(v); then L(G) = L(G).
Proof. Any L-dominating set of G is clearly an L-dominating set of G. So,
L(G)6
L(G). Conversely,suppose D is an L-dominating set of G. If p(v) ∈ D; then D is
also an L-dominating set of G. If p(v) ∈ D; then D= (D − {p(v)}) ∪ {v} (when {1} ∈ S(v)) or D= (D − {p(v)}) ∪ {v} (when S(v) = {{1}} and v ∈ N(v) = ∅) is
an L-dominating set of G such that |D|6|D|. So, L(G)6L(G). The lemma then
follows.
According to Lemma 4,the only possible conditional labels are {{1}} ∗ p(v) with N(v) = ∅ and {{0}; {1}} ∗ p(v). We shall assume that originally the graph has no isolated vertex and so has no conditional label {{1}} ∗ p(v). Our lemmas in this paper will ensure that there is no {{1}} ∗ p(v) to be created. Hence we in fact assume that the only conditional label is {{0}; {1}}∗p(v). Thus,for each vertex v; we may assume that
L(v) is one of the following : ∅; {{0}}; {{1}}; {{0; 1}}; {{0}; {1}}; and {{0}; {1}} ∗ p(v):
4. The algorithm
Now,we are ready to develop reduction lemmas as the basis of the algorithm for the L-domination problem in distance-hereditary graphs. Lemmas 5–11 are also valid for a general graph G.
Lemma 5. Suppose L(x)={{0}; {1}}∗p(x) and L(y)={{0}; {1}}∗p(y) for an edge
xy. If L is obtained from L by relabeling x and y with {{0}}; then
L(G) = L(G).
Proof. Any L-dominating set of G is clearly an L-dominating of G. So,
L(G)6L(G).
Conversely,suppose D is an L-dominating set of G. By the de:nition of L-domination, either x ∈ D or p(x) ∈ D; also,either y ∈ D or p(y) ∈ D. In any case, D =
(D − {p(x); p(y)}) ∪ {x; y} is an L-dominating set of G such that |D|6|D|. So,
L(G)6L(G). The lemma then follows.
Throughout this paper,by deleting=adding sets A from=to a regular (conditional) label S(v) (S(v) ∗ p(v)) we mean relabeling v with the regular label S(v) − {A}/S(v) ∪ {A}. After adding A to S(v); by Lemma 3,we can then delete any proper superset B of A from S(v) ∪ {A}. Note that when A is added to a conditional label {{0}; {1}} ∗ p(x); the resulting label is always the regular label {{0}; {1}}.
Lemma 6. Suppose xy is an edge and x has a regular label L(x) containing {0}. If
L is obtained from L by deleting {1} and {0; 1} from L(y); then L(G) = L(G).
Proof. Since {0} ∈ L(x) = L(x); x ∈ D for any L-dominating or L-dominating set D
of G. The condition (LD) for vertex y and set A={1} or {0; 1} is then redundant. So, any L-dominating set of G is also an L-dominating set of G. Thus, L(G)6L(G).
Conversely,suppose D is an L-dominating set of G. If p(y) ∈ D; then D is also an L-dominating set of G. If p(y) ∈ D; then D=(D−{p(y)})∪{y} is an L-dominating
set of G such that |D|6|D|. So,
L(G)6L(G). The lemma then follows.
Lemma 7. If L(x) = {{0}} and L(z) is a regular label containing neither {1} nor
{0; 1} for all z ∈ N(x); then L(G) = L(G − x) + 1.
Proof. If D is an L-dominating set of G − x; then D ∪ {x} is clearly an L-dominating
set of G. So, L(G)6L(G − x) + 1. Conversely,suppose D is an L-dominating set of
G. Since {0} ∈ L(x); we have x ∈ D. Also,since L(z) is a regular label containing neither {1} nor {0; 1} for all z ∈ N(x); D= D − {x} is an L-dominating set of G − x
such that |D|6|D| − 1. So,
L(G − x)6L(G) − 1. The lemma then follows.
Lemma 8. If x and y are two distinct vertices such that N(x) ⊆ N(y) and L(x) = ∅;
then L(G) = L(G − x).
Proof. Since L(x) = ∅; any L-dominating set of G − x is an L-dominating set of G.
So, L(G)6L(G − x). Conversely,suppose D is an L-dominating set of G. If x ∈
D; then D is also an L-dominating set of G − x. If x ∈ D; since N(x) ⊆ N(y) and L(x) = ∅; D= (D − {x}) ∪ {y} is an L-dominating set of G − x such that |D|6|D|.
Lemma 9. Suppose x is a leaf adjacent to y.
1. Suppose L(x) = {{0; 1}} and L(y) = {{1}}. If L is obtained from L by relabeling
y with {{0}; {1}} ∗ x; then L(G) = L(G − x).
2. Suppose {1} ∈ L(x) or L(x) = {{0; 1}} with L(y) = {{1}}. If L is obtained from
L by adding {0} to L(y); then L(G) = L(G).
3. If L(x) = {{0}; {1}} ∗ p(x) and L(y) ⊆{{0; 1}}; then L(G) = L(G − x) + 1.
4. Suppose L(x) = {{0}; {1}} ∗ p(x) and L(y) = {{1}}. If L is obtained from L by
relabeling y with {{0; 1}}; then L(G) = L(G − x) + 1.
Proof. (1) For any L-dominating set D of G − x; either x = p(y) ∈ D or {y; z} ⊆ D
for some z ∈ N(y) − {x}. In either case, D is an L-dominating set of G. So,
L(G)6L(G − x). Conversely,suppose D is an L-dominating set of G. Also,
either x ∈ D or {y; z} ⊆ D for some z ∈ N(y) − {x}. Hence D is an L-dominating set
of G − x. So, L(G − x)6L(G). (1) of the lemma then follows.
(2) Any L-dominating set D of G is also an L-dominating set of G; since we only
add {0} to L(y). So, L(G)6L(G). Conversely,suppose D is an L-dominating set of
G. Consider the following two cases.
Case 1. y ∈ D. It is easy to see that D is also an L-dominating set of D if
L(y) is a regular label or L(y) is a conditional label with D containing a neighbor of y. Suppose L(y) is a conditional label but D does not contain a neighbor of y. By de:nition, p(y) ∈ D. Hence, D= (D − {p(y)}) ∪ {x} is an L-dominating set of G
such that |D|6|D|.
Case 2. y ∈ D. In this case,we have that {1} ∈ L(x) and then L(x)={{0; 1}}. By the assumption, L(y) = {{1}}. Thus, L(y) = ∅ or {{0; 1}} or {{0}; {1}} or {{0}; {1}} ∗ p(y). Since x is only adjacent to y; x ∈ D and so D = (D − {x}) ∪ {y} is an
L-dominating set of G such that |D|6|D|.
In any case, L(G)6L(G). (2) of the lemma then follows.
(3) For any L-dominating set D of G − x; D∪ {p(x)} is clearly an L-dominating
set of G. So, L(G)6L(G − x) + 1. Conversely,suppose D is an L-dominating set of
G. There are three possible cases: p(x) ∈ D but x ∈ D; p(x) ∈ D and x ∈ D; p(x) ∈ D but {x; y} ⊆ D. For these cases, D= D − {p(x)}; D= (D − {p(x); x}) ∪ {y}; D=
D − {x}; respectively,are L-dominating sets of G − x such that |D|6|D| − 1. So,
L(G − x)6L(G) − 1. (3) of the lemma then follows.
(4) Suppose D is an L-dominating set of G − x. If y ∈ D; then z ∈ D for
some z ∈ N(y) − {x} and so D ∪ {p(x)} is an L-dominating set of G. If y ∈
D; then D∪ {x} is clearly an L-dominating set of G. So,
L(G)6L(G − x) + 1.
Conversely,suppose D is an L-dominating set of G. There are three possible cases: p(x) ∈ D but x ∈ D; p(x) ∈ D and x ∈ D; p(x) ∈ D but {x; y} ⊆ D. For these cases, D= D − {p(x)}; D= (D − {p(x); x}) ∪ {y}; D= D − {x}; respectively,are
L-dominating sets of G − x such that |D|6|D| − 1. So,
L(G − x)6L(G) − 1. (3)
of the lemma then follows.
(1) If L(x)={{0}; {1}}∗p(x) and L(y)={{0}; {1}}∗p(y); then L(G)=L(G−x)+1.
(2) Suppose L(x) is a regular label containing {1} or {0; 1} and L(y) = ∅. If L is
obtained from L by deleting {1} and {0; 1} from L(x) and adding {1} to L(y); then L(G) = L(G).
Proof. (1) For any L-dominating set D of G−x, D∪{p(x)} is clearly an L-dominating
set of G. So, L(G)6L(G − x) + 1. Conversely,suppose D is an L-dominating set
of G. If x ∈ D,then p(x) ∈ D and so D= D − {p(x)} is an L-dominating set of
G − x such that |D| = |D| − 1. If y ∈ D,then p(y) ∈ D and so D= D − {p(y)} is
an L-dominating set of G − y such that |D| = |D| − 1. Since x and y are false twins,
if either x ∈ D or y ∈ D,then we can interchange the roles of x and y to say that G − x has an L-dominating set D of size |D| = |D| − 1. So,we may assume x ∈ D
and y ∈ D. In this case, D= D − {x; p(x)} is an L-dominating set of G − x such that |D|6|D| − 1. So,
L(G − x)6L(G) − 1. (1) of the lemma then follows.
(2) For any L-dominating set D of G,since L(y) is a regular label containing {1}, z ∈ D for some z ∈ N(y) = N(x) and so D is an L-dominating set of G.
So, L(G)6L(G). Conversely,suppose D is an L-dominating set of G. Consider the
following two cases.
Case 1. D∩N(x)=D∩N(y) = ∅. It is easy to verify that D is also an L-dominating
set of G if L(y) is a regular label or L(y) is a conditional label with y ∈ D. Suppose L(y) is a conditional label and y ∈ D. Then p(y) ∈ D and hence D= (D − {p(y)}) ∪ {y} is an L-dominating set of G such that |D| = |D|.
Case 2. D ∩ N(x) = D ∩ N(y) = ∅. In this case, {1} ∈ L(x) and so L(x) = {{0; 1}} and x ∈ D. Similarly, L(y)={{0; 1}} or {{0}} or {{0}; {1}} or {{0}; {1}}∗p(y),and so y ∈ D or p(y) ∈ D. Choose a vertex z ∈ N(x) = N(y). Then D= (D − {x}) ∪ {z}
or D= (D − {x; p(y)}) ∪ {z; y} is an L-dominating set of G such that |D|6|D|.
In any case, L(G)6L(G) (2) of the lemma then follows.
Lemma 11. Suppose x and y are true twins such that N[x] = N[y] = {x; y}.
(1) If L(x) = ∅ and L(y) = {{1}}; then L(G) = L(G − x).
(2) Suppose L(x)={{1}} and L(y)=∅. If L is obtained from L by relabeling y with {{0; 1}}; then L(G) = L(G − x).
(3) Suppose L(x) = {{0; 1}} or {{1}} and L(y) = {{0}; {1}} ∗ p(y). If L is obtained
from L by relabeling y with {{0}; {1}}; then L(G) = L(G − x).
(4) If L(x) and L(y) are both {{0; 1}} or {{1}} except that L(x) = {{0; 1}} and L(y) = {{1}}; then L(G) = L(G − x).
Proof. (1) Since L(x) = ∅,any L-dominating set of G − x is an L-dominating set
of G. So, L(G)6L(G − x). Conversely,suppose D is an L-dominating set of G. If
x ∈ D,then D is also an L-dominating set of G − x. Now suppose x ∈ D. If y ∈ D, then D= (D − {x}) ∪ {z} is an L-dominating set of G − x such that |D|6|D|,where
z ∈ N(y)−{x}. Suppose y ∈ D. Then,either L(y) is a conditional label with p(y) ∈ D or L(y) is a regular label with {0} ∈ L(y). By assumption, {1} ∈ L(y) if L(y) is a
regular label. Thus D= (D − {x}) ∪ {y} is an L-dominating set of G − x such that |D|6|D|. So,
L(G − x)6L(G). (1) of the lemma then follows.
(2) For any L-dominating set Dof G−x,since L(y)={{0; 1}},we have D∩N(x)=
D∩(N[y]−{x}) = ∅ and so Dis also an L-dominating set of G. So,
L(G)6L(G−x).
Conversely,suppose D is an L-dominating set of G. Since L(x) = {{1}},we have D ∩ N(x) = ∅. Then D= D (when x ∈ D) or D= (D − {x}) ∪ {y} (when x ∈ D)
is an L-dominating set of G − x with |D|6|D|. So,
L(G − x)6L(G). (2) of the
lemma then follows.
(3) For any L-dominating set D of G − x,since {0} ∈ L(y), y ∈ D and so
D is an L-dominating set of G. So, L(G)6L(G − x). Conversely,suppose D is
an L-dominating set of G. Then either z1= p(y) ∈ D with some z2 ∈ D ∩ N[x],or
p(y) ∈ D with some z1= y, z2∈ D ∩ N(y). In any case, D= (D − {z1, z2}) ∪ {y; z}
is an L-dominating set of G − x such that |D|6|D| for some z ∈ N(y) − {x}. So,
L(G − x)6L(G). (3) of the lemma then follows.
(4) For any L-dominating set D of G − x,since L(y) = {{0; 1}} or {{1}}, D
contains some vertex in N[y] − {x} and so D is also an L-dominating set of G. So,
L(G)6L(G − x). Conversely,suppose D is an L-dominating set of G. If x ∈ D,then
D is also an L-dominating set of G − x. Suppose x ∈ D. Consider the following three cases.
Case 1. y ∈ D. In this case D= (D − {x}) ∪ {z} is an L-dominating set of G − x
with |D|6|D| for some z ∈ N(y) − {x}.
Case 2. y ∈ D and D ∩ (N(y) − {x}) = ∅. In this case, D= (D − {x}) ∪ {y} is an
L-dominating set of G − x with |D|6|D|.
Case 3. D ∩ (N[y] − {x}) = ∅. In this case, L(x) = {{0; 1}} and L(y) = {{0; 1}}. Thus, D= (D − {x}) ∪ {y} is an L-dominating set of G − x with |D|6|D|.
In any case, L(G − x)6L(G). (4) of the lemma then follows.
Based on these lemmas,we present the following algorithm for the L-domination problem in distance-hereditary graphs.
Algorithm LD-dh. Find the L-domination number of a distance-hereditary graph.
Input. A distance-hereditary graph G = (V; E) with label L(v) for all v ∈ V and a one-vertex-extension ordering v1; v2; : : : ; vn.
Output. L(G).
Method.
← 0; =∗ as a short notation for L(G) ∗=
update labels according to Lemmas 3 and 4;
for i = n to 1 step −1 do
{ x = vi;
Case 1. x is an isolated vertex in Gi,i.e.,Ni(x) = ∅.
if {1} ∈ L(x) then STOP since there is no feasible solution else ← + 1;
2.1 if N[x] = N[y] = {x; y} and L(x) = ∅ then interchange x and y by applying Lemma 2;
2.2 if L(x) and L(y) are conditional labels then L(x) ← L(y) ← {{0}}; 2.3 if {1} ∈ L(x) or L(x) = {{0; 1}} with L(y) = {{1}} then add {0} to L(y); 2.4 if L(y) contains {0} then delete {1} and {0; 1} from L(x);
2.5 if N[x] = N[y] = {x; y} and L(x) = ∅ then interchange x and y by applying Lemma 2;
2.6 if L(x) = {{0}} then delete {1} and {0; 1} from L(y) and ← + 1; 2.7 if L(x) = {{0; 1}} and L(y) = {{1}} then relabel y with {{0}; {1}} ∗ x; 2.8 if L(x) is a conditional label and L(y) ⊆{{0; 1}} then ← + 1;
2.9 if L(x) is a conditional label and L(y) = {{1}} then {relabel y with {{0; 1}}; ← + 1; }
Case 3. x and y are false twins in Gi with Ni(x) = Ni(y) = ∅.
3.1 if L(x) and L(y) are conditional labels then ← + 1 else
3.2 {assume L(x) = ∅ or L(x) is a regular label with L(y) = ∅ by interchanging x
and y if necessary;
3.3 if L(x) contains {1} or {0; 1} then delete {1} and {0; 1} from L(x) and add
{1} to L(y);
3.4 if L(x) = {{0}} then delete {1} and {0; 1} from L(z) for all z ∈ N(x) and
← + 1;}
Case 4. x and y are true twins in Gi with Ni[x] = Ni[y] = {x; y}.
4.1 if L(x) and L(y) are conditional labels then L(x) ← L(y) ← {{0}}; 4.2 if {0} ∈ L(x) then delete {1} and {0; 1} from L(y);
if {0} ∈ L(y) then delete {1} and {0; 1} from L(x); if {0} ∈ L(x) then delete {1} and {0; 1} from L(y);
4.3 assume one of the following cases holds by interchanging x and y if necessary; Case 4.1. L(x) = {{0}}: delete {1} and {0; 1} from L(z) for all z ∈ N(x) and
← + 1;
Case 4.2. L(x) = ∅ and L(y) = {{1}}: do nothing;
Case 4.3. L(x) = {{1}} and L(y) = ∅: relabel y with {{0; 1}};
Case 4.4. L(x) = {{0; 1}} or {{1}} and L(y) = {{0}; {1}} ∗ p(y): relabel y with
{{0}; {1}};
Case 4.5. L(x) and L(y) are {{0; 1}} or {{1}} except L(x) = {{0; 1}} and L(y) =
{{1}}: do nothing; }
Theorem 12. Algorithm LD-dh solves the L-domination problem for distance-hereditary
graphs in linear time.
Proof. The algorithm certainly runs in linear time. To see that it solves the L-domination
cases for various values of L(x) and L(y) at every iteration by indicating the corre-sponding lemma(s) each line of the algorithm uses.
Case 1. x is an isolated vertex in Gi. The correctness follows from de:nition.
Case 2. x is a leaf adjacent to y in Gi. If L(x) and L(y) are conditional labels,by
Lemma 5,we can change them to {{0}} as in line 2.2. Now,either L(x) or L(y) is regular. If {1} ∈ L(x) or L(x) = {{0; 1}} with L(y) = {{1}},then Lemma 9 (2) is applied to add {0} to L(y) as in line 2.3. Then Lemma 6 is applied to remove {1} and {0; 1} from L(x) when {0} ∈ L(y) as in line 2.4. Immediately after line 2.4 is executed,we have that L(x)=∅,or L(x)={{0}},or L(x)={{0; 1}} with L(y)={{1}}, or L(x) is a conditional label with {0} ∈ L(y). After the special treatments in lines 2.1 and 2.5,it is impossible that N[x]=N[y]={x; y} and L(x)=∅ and L(y) = ∅. Therefore, after line 2.5,there are six possibilities: (1) N[x] = N[y] = {x; y} and L(x) = N(y) = ∅, (2) N(y) contains some y = x and L(x) = ∅,(3) L(x) = {{0}},(4) L(x) = {{0; 1}}
and L(y) = {{1}},(5) L(x) is a conditional label and L(y) ⊆{{0; 1}},(6) L(x) is a conditional label and L(y) = {{1}}. For the :rst case,by de:nition,we need to only delete x from G. For the second case,since N(x) ⊆ N(y),we also delete x from G
by applying Lemma 8. The other four cases are handled in lines 2.6 to 2.9 by using Lemmas 6,7,9 (1),9 (3),9 (4),respectively.
Case 3. x and y are false twins in Gi with Ni(x) = Ni(y) = ∅. By Lemma 10 (1),
line 3.1 handles the case when L(x) and L(y) are conditional labels. By Lemma 2,we can interchange x and y,if necessary,to assume L(x) = ∅ or L(x) is a regular label with L(y) = ∅,as shown in line 3.2. By Lemma 10 (2),we can remove {1} and
{0; 1} from L(x),if necessary,as in line 3.3. Now,either L(x) = ∅ or L(x) = {{0}}.
If L(x) = {{0}},then Lemmas 6 and 7 are applied as in line 3.4. If L(x) = ∅,then Lemma 8 is applied and D remains unchanged.
Case 4. x and y are true twins in Gi with Ni[x] = Ni[y] = {x; y}. If L(x) and L(y)
are conditional labels,by Lemma 5,we can change them to {{0}} as in line 4.1. If
{0} ∈ L(x) or {0} ∈ L(y),then we can apply Lemma 6 to delete {1} and {0; 1} from
L(y) or L(x),respectively,as in line 4.2. Lemma 6 is applied three times to handle the case of L(x) = {{0}; {1}} ∗ p(x) and {0} ∈ L(y). Then,by applying Lemma 2 to interchange x and y if necessary,there are :ve possibilities,which are handled as in Cases 4.1–4.5 by using Lemmas 7 and 11.
5. Concluding remarks
The weighted connected domination problem is solvable in linear time for distance-hereditary graphs [19]. In this paper,we give a linear-time algorithm for the domi-nation and the total domidomi-nation problems in distance-hereditary graphs. On the other hand,the time complexity of the independent domination problem in distance-hereditary graphs is still unknown. It is desirable study the time complexity of the independent domination,the weighted domination,and the weighted total domination problems in distance-hereditary graphs.
Acknowledgements
The authors thank the referees for many useful suggestions.
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