1 Introduction
Variants of domination in graphs form a rich area of graph theory, with many useful and interesting concepts, results, and challenging problems [33, 32, 35]. In this paper we consider a family of generalizations of classical domination and total domination known as domination and total domination. Given a graph and a positive integer , a dominating set in is a set such that every vertex has at least neighbors in , and a total dominating set in is a set such that every vertex has at least neighbors in . The domination and the total domination problems aim to find the minimum size of a dominating, resp. total dominating set, in a given graph. The notion of domination was introduced by Fink and Jacobson in 1985 [25] and studied in a series of papers (e.g., [29, 45, 14, 22, 24]) and in a survey Chellali et al. [13]. The notion of total domination was introduced by Kulli in 1991 [44] and studied under the name of tuple total domination by Henning and Kazemi in 2010 [34] and also in a series of recent papers [42, 56, 46, 1]. The terminology “tuple total domination” was introduced in analogy with the notion of “tuple domination”, introduced in 2000 by Harary and Haynes [31].^{1}^{1}1A set of vertices is said to be a tuple dominating set if every vertex of is adjacent or equal to at least vertices in . The redundancy involved in domination and total domination problems makes them useful in various applications, for example in forming sets of representatives or in resource allocation in distributed computing systems (see, e.g., [33]). However, these problems are known to be NPhard [40, 56] and also hard to approximate [18].
The domination and total domination problems are NPhard not only for general graphs but also in the class of chordal graphs. More specifically, the problems are NPhard in the class of split graphs [45, 56] and, in the case of total domination, also in the class of undirected path graphs [46]. We consider domination and total domination in another subclass of chordal graphs, the class of proper interval graphs. A graph is an interval graph if it has an intersection model consisting of closed intervals on a real line, that is, if there exist a family of intervals on the real line and a onetoone correspondence between the vertices of and the intervals of such that two vertices are joined by an edge in if and only if the corresponding intervals intersect. A proper interval graph is an interval graph that has a proper interval model, that is, an intersection model in which no interval contains another one. Proper interval graphs were introduced by Roberts [60], where it was shown that they coincide with the unit interval graphs, that is, interval graphs having an intersection model consisting of intervals of unit length. Various characterizations of proper interval graphs have been developed in the literature (see, e.g., [26, 39, 28, 52]) and several lineartime recognition algorithms are known, which in case of a yes instance also compute a proper interval model (see, e.g., [20] and references cited therein).
The usual domination and total domination problems (that is, when ) are known to be solvable in linear time in the class of interval graphs (see [36, 12, 6] and [43, 58, 59, 12, 10], respectively). Furthermore, for each fixed integer , the domination and total domination problems are solvable in time in the class of interval graphs where is the order of the input graph. This follows from recent results due to Kang et al. [41], building on previous works by BuiXuan et al. [8] and Belmonte and Vatshelle [3]. In fact, Kang et al. studied a more general class of problems, called domination problems, and showed that every such problem can be solved in time in the class of vertex interval graphs, where is a parameter associated to the problem (see Corollary 3.2 in [41] and the paragraph following it). The value of parameter for domination and total domination equals , yielding the claimed time complexity.
1.1 Our Results and Approach
To the best of our knowledge, the only known polynomialtime algorithms for domination and total domination for a general (fixed) in the class of interval graphs follow from the abovementioned work of Kang et al. [41] and run in time . We significantly improve the above result for the case of proper interval graphs. We show that for each positive integer , the domination and total domination problems are solvable in time in the class of proper interval graphs. Except for , this significantly improves on the best known running time.
Our approach is based on a reduction showing that for each positive integer , the total domination problem on a given proper interval graph can be reduced to a shortest path computation in a derived edgeweighted directed acyclic graph. A similar reduction works for domination. The reductions immediately result in algorithms with running time . We then show that with a suitable implementation the running time can be improved to . The algorithms can be easily adapted to the weighted case, at no expense in the running time.
An extended abstract of this work appeared in the Proceedings of ISCO 2018 [16].
1.2 Related Work
We now give an overview of related work and compare our results with the most related other results, in addition to those due to Kang et al. [41], which motivated this work.
Overview. Several results on the complexity of domination and total domination problems were established in the literature. For every , the domination problem is NPhard in the classes of bipartite graphs [2] and split graphs [45]. The problem is solvable in linear time in the class of graphs every block of which is a clique, a cycle or a complete bipartite graph (including trees, block graphs, cacti, and blockcactus graphs) [45], and, more generally, in any class of graphs of bounded cliquewidth [53, 21] (see also [17]). For every positive integer , the total domination problem is NPhard in the classes of split graphs [56], doubly chordal graphs [56], bipartite graphs [56], undirected path graphs [46], and, for , also in the class of bipartite planar graphs [1]. The problem is solvable in linear time in the class of graphs every block of which is a clique, a cycle, or a complete bipartite graph [46], and, more generally, in any class of graphs of bounded cliquewidth [53, 21], and in polynomial time in the class of chordal bipartite graphs [56]. domination and total domination problems were also studied with respect to their (in)approximability properties, both in general [18] and in restricted graph classes [2], as well as from the parameterized complexity point of view [37, 9].
Besides domination and total domination, other variants of domination problems solvable in polynomial time in the class of proper interval graphs (or in some of its superclasses) include tuple domination for all [48] (see also [47] and, for , [57]), connected domination [59], independent domination [23], paired domination [15], efficient domination [11], liar’s domination [54], restrained domination [55], eternal domination [5], power domination [49], outerconnected domination [51], Roman domination [50], Grundy domination [7], etc.
Comparison. Bertossi [4] showed how to reduce the total domination problem in a given interval graph to a shortest path computation in a derived edgeweighted directed acyclic graph satisfying some additional constraints on pairs of consecutive arcs. A further transformation reduces the problem to a usual (unconstrained) shortest path computation. Compared to the approach of Bertossi, our approach exploits the additional structure of proper interval graphs in order to gain generality in the problem space. Our approach works for every and is also more direct, in the sense that the (usual or total, unweighted or weighted) domination problem in a given proper interval graph is reduced to a shortest path computation in a derived edgeweighted directed acyclic graph in a single, unified step.
The works of Liao and Chang [48] and of Lee and Chang [47] consider various domination problems in the class of strongly chordal graphs (and, in the case of [48], also dually chordal graphs). While the class of strongly chordal graphs generalizes the class of interval graphs, the domination problems studied in [48, 47] all deal with closed neighborhoods, and for those cases structural properties of strongly chordal and dually chordal graphs are helpful for the design of lineartime algorithms. In contrast, domination and total domination are defined via open neighborhoods and results of [48, 47] do not seem to be applicable or easily adaptable to our setting.
Structure of the paper. In Section 2, we describe the reduction for the total domination problem. The specifics of the implementation resulting in improved running time are given in Section 3. In Section 4, we discuss how the approach can be modified to solve the domination problem. Extensions to the weighted cases are presented in Section 5. We conclude the paper with some open problems in Section 6.
In the rest of the section, we fix some definitions and notation. Given a graph and a set , we denote by the subgraph of induced by and by the subgraph induced by . For a vertex in a graph , we denote by the set of neighbors of in . Note that for every graph , the set is a dominating set, while has a total dominating set if and only if every vertex of has at least neighbors. For notions not defined here, we refer the reader to [61, 19].
2 The Reduction for Total Domination
Let be a positive integer and a given proper interval graph. We may assume that is equipped with a proper interval model where for all . We may also assume that no two intervals coincide. Moreover, since in a proper interval model the order of the left endpoints equals the order of the right endpoints, we assume that the intervals are sorted increasingly according to their left endpoints, i.e., . We use notation if and say in this case that is to the left of and is to the right of . Also, we write if . Given three intervals , we say that interval is between intervals and if . We say that interval intersects interval if .
Our approach can be described as follows. Given , we compute an edgeweighted directed acyclic graph (where the superscript “” means “total” and is the constant specifying the problem) and show that the total domination problem on can be reduced to a shortest path computation in . In what follows, we first give the definition of digraph and illustrate the construction on an example (Example 2.1). Next we explain the intuition behind the reduction and, finally, prove the correctness of the reduction.
To distinguish the vertices of from those of , we refer to them as nodes. Vertices of are typically denoted by or , and nodes of by . Each node of is a sequence of intervals from the set , where , are two new, “dummy” intervals such that , , , and . We naturally extend the linear order on to the whole set . We say that an interval is associated with a node of if it appears in sequence . Given a node of , we denote the set of all intervals associated with by . The first and the last interval in with respect to ordering of are denoted by and , respectively. A sequence of intervals from is said to be increasing if .
The node set of is given by , where:

and are sequences of intervals of length one.^{2}^{2}2This assures that the intervals and are well defined also for , in which case both are equal to .

is the set of socalled small nodes. Set consists exactly of those increasing sequences of (not necessarily consecutive) intervals from such that:

,

for all , and

every interval such that intersects at least intervals from the set .


is the set of socalled big nodes. Set consists exactly of those increasing sequences of (not necessarily consecutive) intervals from of length such that:

for all and

every interval such that intersects at least intervals from the set .

The arc set of is given by , where:

Set
consists exactly of those ordered pairs
such that:
and ,

every interval such that intersects at least intervals from ,

if is a big node, then the rightmost intervals associated with pairwise intersect, and

if is a big node, then the leftmost intervals associated with pairwise intersect.


Set consists exactly of those ordered pairs such that and there exist intervals in such that and .
To every arc of we associate a nonnegative length , defined as follows:
() 
The length of a directed path in is defined, as usual, as the sum of the lengths of its arcs.
Example 2.1.
Consider the problem of finding a minimum total dominating set in the graph given by the proper interval model depicted in Figure 1. Using the reduction described above, we obtain the digraph depicted in Figure 1 along with the length function on arcs, where, for clarity, nodes of are identified with the corresponding strings of indices . We also omit in the figure the (irrelevant) nodes that do not belong to any directed path from to . There is a unique shortest path in , namely . The path corresponds to , the only minimum total dominating set in .
Lemma 2.2.
Given a graph and a positive integer , let be a total dominating set in , and let be a component of . Then .
Proof.
Let . Since is a total dominating set, has at least neighbors in . However, since is a component of , all the neighbors of in are in fact in . Thus, . ∎
The following proposition establishes the correctness of the reduction.
Proposition 2.3.
Given a proper interval graph and a positive integer , let be the directed graph constructed as above. Then has a total dominating set of size if and only if has a directed path from to of length .
Before giving a detailed proof of Proposition 2.3, we explain the intuition behind the reduction. The subgraph of induced by a minimum total dominating set may contain several connected components. These components as well as vertices within them are naturally ordered from left to right. Moreover, since each connected subgraph of a proper interval graph has a Hamiltonian path, the nodes of correspond to paths in , see condition (2) for small nodes or condition (1) for big nodes. Since each vertex of has at least neighbors in the total dominating set, each component has at least vertices. Components with at least vertices give rise to directed paths in consisting of big nodes and arcs in . Each component with less than vertices corresponds to a unique small node in , which can be seen as a trivial directed path in . The resulting paths inherit the lefttoright ordering from the components. Any two consecutive paths (with respect to this ordering) are joined in by an arc in . Moreover, is joined to the leftmost node of the leftmost path by an arc in and, symmetrically, the rightmost node of the rightmost path is joined to by an arc in . Adding such arcs yields a directed path from to of the desired length.
The above process can be reversed. Given a directed path in from to , a total dominating set in of the desired size can be obtained as the set of all vertices corresponding to intervals in associated with the internal nodes of . Note that our construction of the graph implies that such a set is indeed a total dominating set in . For example, condition (3) from the definition of arcs in guarantees that the vertex corresponding to the rightmost interval associated with (where ) is dominated. The condition is related to the fact that in proper interval graphs the neighborhood of a vertex represented by an interval splits into two cliques: one for all intervals containing and another one for all intervals containing .
The digraph has nodes and arcs and can be, together with the length function on its arcs, computed from directly from the definition in time . A shortest directed path (with respect to ) from to all nodes reachable from in can be computed in polynomial time using any of the standard approaches, for example using Dijkstra’s algorithm. Actually, since is acyclic, a dynamic programming approach along a topological ordering of can be used to compute shortest paths from in linear time (in the size of ). Proposition 2.3 therefore implies that the total domination problem is solvable in time in the class of vertex proper interval graphs.
We will show in Section 3 that, with a careful implementation, a shortest path in can be computed without examining all the arcs of the digraph, leading to an improved running time of .
Proof of Proposition 2.3
We assume all notation up to Example 2.1 and, additionally, fix the following notation useful for both directions of the proof: for , we denote by the set of intervals in corresponding to vertices in .
First we establish the forward implication. Suppose that has a total dominating set of size . The components of can be naturally ordered from left to right, say as . To each component we will associate a path in defined as a sequence of nodes. The desired directed path from to in will be then obtained by combining the paths into a single sequence of nodes preceded by and followed by .
We say that a component is small if and big, otherwise. To every small component we associate a sequence , consisting of the intervals corresponding to the vertices of , ordered increasingly. We claim that is a small node of . By Lemma 2.2, has at least vertices, which, together with the fact that is small, implies property (1) of the definition of a small node. Property (2) follows from the fact that is a proper interval model of and is connected. To show that satisfies property (3) of the definition of a small node, consider an interval such that . Let be the vertex of corresponding to . Since is a total dominating set, vertex has at least neighbors in . Since , all the neighbors of in must belong to , more specifically, . It follows that intersects at least intervals from the set , establishing also property (3) and with it the claim that is a small node of . The path associated to component is the onenode path having as a node.
Let now be a big component and let . Then . Let be the intervals corresponding to the vertices of , ordered increasingly. For every , let denote the subsequence of these intervals of length starting at th interval, that is, . We claim that for each , sequence is a big node of . Property (1) follows from the fact that is a proper interval model of and is connected. To show that satisfies property (2) of the definition of a big node, consider an interval such that and let be the vertex of corresponding to . Since is a total dominating set, vertex has at least neighbors in . Since , all the neighbors of in must belong to . It follows that intersects at least intervals from the set . Suppose for a contradiction that intersects strictly less that intervals from the set . Since intersects at least intervals from the set , there is an interval, call it , in the set such that . Note that where and . Suppose first that , that is, for some . Since is a proper interval model of , conditions and imply that interval intersects each of the two intervals and (possibly ). Similarly, the fact that intersects both and implies that also intersects each of the intervals in the set ; in particular, intersects each of the intervals in the set , all of which are in . It follows that interval intersects at least intervals from the set , contradicting the assumption on . The case is symmetric to the case . This establishes property (2) and with it the claim that is a big node of . The path associated to component is defined as .
Let denote the sequence of nodes of obtained by combining the paths into a single sequence of nodes preceded by and followed by in the natural order . Since , the paths are pairwise nodedisjoint, and none of them contains or , path has at least nodes. Moreover, note that for each node of other than and , the vertices of corresponding to intervals associated with all belong to the same component of , call it .
We claim that is a path in , that is, that every two consecutive nodes of form an arc in . Consider a pair of consecutive nodes of . Clearly, . We consider three subcases depending on whether , , or .
Suppose first that . We claim that . Property (1) of the definition of the edges in clearly holds, as does (vacuously) property (3) (since ). To show property (2), consider an interval such that . Since is a total dominating set, interval intersects at least intervals from . Since and is the leftmost interval corresponding to a vertex of , it follows that intersects the leftmost intervals from . All these intervals belong to vertices from component , and therefore to . This establishes property (2). A similar argument shows that if , then in order to make sure that the vertex corresponding to has at least neighbors in , interval must intersect the intervals associated with immediately following in the sequence. This implies that the leftmost intervals associated with pairwise intersect, thus establishing property (4). It follows that , as claimed.
Suppose now that . We claim that . Property (1) of the definition of the edges in follows from the construction of and the fact that . Property (2) follows from the construction of together with the fact that is a total dominating set in . Property (3) is satisfied vacuously. A similar argument as in the case establishes property (4). It follows that , as claimed.
Suppose now that . If , then we conclude that by symmetry with the case , . If , then we conclude that by symmetry with the case . Let now . If , then we can use similar arguments as in the case to show that . If , let be the index such that . The construction of implies that and are nodes of such that and for some where . The definitions of and now imply that , showing in particular that is an arc of .
This shows that is a directed path from to in , as claimed. Furthermore, the definition of the length function and the construction of imply that the length of equals the size of , which is .
Now we establish the backward implication. Suppose that has a directed path from to of length . Let be the set of all vertices such that the interval corresponding to is associated with some node of . We claim that is a total dominating set in of size .
Note that since neither of and is a big node, ; moreover, since condition in the definition of an arc in fails. Therefore, and has at least nodes. The set of nodes of can be uniquely partitioned into consecutive sets of nodes, say , such that each is the node set of a maximal subpath of such that . (Equivalently, the ’s are the vertex sets of the components of the undirected graph underlying the digraph .) Note that and . For all , let be the set of vertices of corresponding to intervals associated with nodes in .
We claim that for every and for every pair of intervals and we have and . This can be proved by induction on . If , then let be the unique arc of connecting a vertex in with a vertex in . By the definition of the ’s, we have and therefore . This implies that every interval associated with is smaller (with respect to ordering ) and disjoint from every interval associated with . Consequently, the definitions of , , , , and the properties of the arcs in imply that every interval in is smaller (with respect to ordering ) and disjoint from every interval in . The inductive step follows from the transitivity of the relation on in which interval is in relation with interval if and only if and .
The above claim implies that no edge of connects a vertex in with a vertex in whenever . More specifically, we claim that for every , the subgraph of induced by is a component of . This can be proved using the properties of the arcs in , as follows. Let . Suppose first that . Since every arc in connects a pair of big nodes, we infer that for some . Using property (2) in the definition of a small node, we infer that is connected. Therefore, since no edge of connects a vertex in with a vertex in for , we infer that is a component of , as claimed. Suppose now that , that is, . Let be the subpath of such that . Since consists only of big nodes and only of arcs in , we can use property (1) in the definition of a big node to infer that is connected. It follows that is a component of also in this case.
Let us now show that the size of equals the length of . This will imply that . For every , let be the subpath of of consisting of all the arcs of entering a node in . By construction, the paths are pairwise arcdisjoint and their union is . For every , let denote the the length of . The definitions of the ’s and of the length function imply that , where is the (unique) arc of such that and . Since , we have
It follows that
Furthermore, we have for all , which implies . Since the subgraph of induced by is a component of , we have , which is exactly the length of (this follows from the fact that the paths are pairwise arcdisjoint and their union is ). Therefore, .
It remains to show that is a total dominating set of , that is, that every vertex has at least neighbors in . Let , let be the interval corresponding to . We need to show that intersects at least intervals from the set . We consider two cases depending on whether or not.
Case 1. . In this case, and is associated with some node of . Let be such a node. Note that . By construction of , we have . Suppose first that is a small node. Then, condition in the definition of a small node implies that intersects at least intervals from the set , which is a subset of .
Suppose now that is a big node. There exists a unique such that . Note that is the node set of a subpath of , say , consisting only of big nodes and arcs in . Let be the intervals corresponding to the vertices in , ordered increasingly. The fact that all arcs of are in imply that where and . Moreover, there exists a unique index such that .
Suppose first that . Let be the (unique) arc of such that and . Since and , condition in the definition of an arc in guarantees that the intervals pairwise intersect. Clearly, this implies that interval intersects at least intervals from the set .
The case when is symmetric to the case and can be analyzed using condition in the definition of an arc in .
Suppose now that . Let and let . Then and is a big node of . Moreover,
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