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[MATH] of [MATH] is an induced subgraph of [MATH] formed by a biconnected component [MATH] of [MATH] extended by paths attached to vertices of the biconnected component [MATH] . Each vertex of [MATH] has at most one path attached to it. Each vertex at which a path is attached is a separation vertex in [MATH] and not ad... |
Definitions and Basic Terminology We consider graphs [MATH] , where [MATH] is the set of vertices and [MATH] is the set of edges . We distinguish between undirected graphs with edge sets [MATH] and directed graphs with edge sets [MATH] Graph [MATH] is a subgraph of [MATH] if [MATH] and [MATH] . It is an induced subgrap... |
A sequence of [MATH] vertices [MATH] [MATH] [MATH] for [MATH] , is an undirected path of length [MATH] , if [MATH] for [MATH] . The vertices [MATH] and [MATH] are the end vertices of undirected path [MATH] . The sequence [MATH] is a directed path of length [MATH] , if [MATH] for [MATH] . Vertex [MATH] is the start vert... |
An undirected graph [MATH] is connected if there is a path between every pair of vertices. The distance [MATH] between two vertices [MATH] in a connected undirected graph [MATH] is the smallest integer [MATH] such that there is a path of length [MATH] between [MATH] and [MATH] . A connected component of an undirected g... |
[MATH] of [MATH] is an induced biconnected subgraph of [MATH] such that there is no biconnected induced subgraph [MATH] of [MATH] with [MATH] and [MATH] |
Definition 2.1 (Resolving set) Let [MATH] be a connected undirected graph. A vertex set [MATH] is a resolving set for [MATH] if for every vertex pair [MATH] |
[MATH] , there is a vertex [MATH] such that [MATH] The set [MATH] is a minimum resolving set for [MATH] , if there is no resolving set [MATH] for [MATH] with [MATH] . A connected undirected graph [MATH] has metric dimension |
[MATH] if [MATH] is the smallest positive integer such that there is a resolving set for [MATH] of size [MATH] Definition 2.2 Let [MATH] be a connected undirected graph. |
1. (leg, root, leaf, hooked leg, ordinary leg) A path [MATH] [MATH] , of [MATH] is a leg , if vertex [MATH] has degree one, the vertices [MATH] have degree [MATH] , and vertex [MATH] has degree [MATH] in [MATH] Vertex [MATH] is called the root of [MATH] . Vertex [MATH] is called the leaf of [MATH] A leg is called a hoo... |
2. (bridge) An edge [MATH] is called a bridge if [MATH] is not connected and if [MATH] is not an edge between two vertices of one and the same leg. |
3. (extended biconnected component (EBC)) biconnected component [MATH] of [MATH] extended by the subgraphs of [MATH] induced by the vertices of the hooked legs with roots in [MATH] is an extended biconnected component (EBC) of [MATH] |
4. (component) Every subgraph induced by the vertices of an ordinary leg, every subgraph induced by the two vertices of a bridge, and every EBC is called a component of [MATH] |
5. (amalgamation vertex) Separation vertices of [MATH] that belong to at least two components, i.e. separation vertices without the degree two vertices of the legs and roots of the hooked legs are called amalgamation vertices |
Every undirected graph [MATH] can be decomposed into legs, bridges and EBCs. This decomposition is a unique and edge-disjoint partition of [MATH] |
Definition 2.3 (EBC-tree) Let [MATH] be a connected undirected graph. 1. The EBC-tree [MATH] for [MATH] is a tree with two types of nodes called c-nodes (nodes for the components of [MATH] ) and a-nodes (nodes for the amalgamation vertices of [MATH] ). [MATH] has a c-node for every component of [MATH] . The vertex set ... |
Note that in the EBC-tree all leaves are c-nodes and there is no edge between two a-nodes and no edge between two c-nodes. All ordinary legs are represented by leaves, all bridges are represented by inner c-nodes, and all EBCs are represented by leaves or inner c-nodes. |
Definition 2.4 (DEBC-tree) Let [MATH] be a connected undirected graph. 1. For the EBC-tree [MATH] for [MATH] and a node [MATH] let [MATH] be the directed EBC-tree (DEBC-tree) with root [MATH] that is defined as follows: [MATH] contains exactly one directed edge for every undirected edge of [MATH] such that for every no... |
2. For a node [MATH] , let [MATH] be the subtree of [MATH] induced by all nodes [MATH] for which there is a directed path from [MATH] to [MATH] in [MATH] . The root of [MATH] is [MATH] |
3. For a subtree [MATH] let [MATH] be the set of c-nodes of [MATH] and [MATH] . Then [MATH] is the subgraph of [MATH] induced by the vertices of [MATH] |
[MATH] is the subgraph of [MATH] represented by [MATH] . It is not necessary to refer to the a-nodes of [MATH] , because the vertices of [MATH] that are represented by the a-nodes are also represented by the c-nodes since for every a-node [MATH] there is a c-node [MATH] such that \textnu [MATH] |
Note that the EBC-tree and the DEBC-tree of [MATH] can be constructed in linear time with the help of any linear time algorithm for finding the biconnected components and bridges of [MATH] |
Computing the metric dimension based on a graph decomposition Without loss of generality we will use from now on the following assumptions: |
1. [MATH] is a connected undirected, but not biconnected graph. 2. [MATH] is the DEBC-tree for [MATH] with root [MATH] 3. [MATH] is the set of a-nodes of [MATH] and [MATH] is the set of c-nodes of [MATH] |
4. Root [MATH] is an a-node. 5. Root [MATH] has at least two children (because [MATH] is not biconnected). First we will describe the general idea of how to compute the metric dimension of [MATH] . The idea is based on dynamic programming. |
Property 3.1 For every subtree [MATH] [MATH] , of [MATH] we compute an information [MATH] satisfying the following properties: 1. |
For every a-node [MATH] with children [MATH] [MATH] , the information [MATH] can efficiently be computed from [MATH] 2. For every c-node [MATH] with children [MATH] [MATH] , the information [MATH] can efficiently be computed from [MATH] and [MATH] |
3. The metric dimension of [MATH] can efficiently be computed from [MATH] These properties allow an efficient bottom-up processing of [MATH] as follows: We start by computing [MATH] for every leaf [MATH] of [MATH] . Since the leafs are c-nodes without children we only need the subgraph [MATH] of [MATH] . For every inne... |
Before we define [MATH] we need a few more definitions. Definition 3.2 (Gate Vertex) Let [MATH] be a set of vertices. A vertex [MATH] is an [MATH] -gate of [MATH] , if there is a vertex [MATH] , such that for all [MATH] the equation [MATH] holds. Vertex [MATH] is called an out-vertex for [MATH] -gate [MATH] |
Intuitively this definition means that [MATH] is an [MATH] -gate if there is a vertex [MATH] that has a shortest path to any vertex [MATH] that passes [MATH] |
Observation 3.3 Let [MATH] and [MATH] be an [MATH] -gate of [MATH] . Then there is an out-vertex [MATH] adjacent to [MATH] Observation 3.4 |
Let [MATH] [MATH] and [MATH] be an [MATH] -gate of [MATH] . If [MATH] and [MATH] are two out-vertices for [MATH] -gate [MATH] with the same distance to [MATH] , i.e. [MATH] , then both vertices [MATH] and [MATH] have the same distance to all vertices of [MATH] . In this case [MATH] is not a resolving set for [MATH] . C... |
Definition 3.5 [MATH] -resolving set, non-gate- [MATH] -resolving set) Let [MATH] 1. [MATH] -resolving set for [MATH] is a resolving set [MATH] for [MATH] with [MATH] |
2. minimum [MATH] -resolving set for [MATH] is a resolving set [MATH] for [MATH] with [MATH] such that there is no [MATH] -resolving set [MATH] for [MATH] with [MATH] |
3. non-gate- [MATH] -resolving set for [MATH] is a [MATH] -resolving set [MATH] for [MATH] with [MATH] and [MATH] is not an [MATH] -gate in [MATH] |
4. minimum non-gate- [MATH] -resolving set for [MATH] is a [MATH] -resolving set [MATH] for [MATH] with [MATH] and [MATH] is not an [MATH] -gate in [MATH] , such that there is no non-gate- [MATH] -resolving set [MATH] for [MATH] with [MATH] |
Note that a minimum [MATH] -resolving set is not necessarily a minimum resolving set. It is possible that no minimum resolving set contains [MATH] . Also a minimum non-gate- [MATH] -resolving set is not necessarily a minimum [MATH] -resolving set. It is possible that in every minimum [MATH] -resolving set [MATH] vertex... |
Lemma 3.6 Let [MATH] . Let [MATH] a minimum resolving set for [MATH] [MATH] a minimum [MATH] -resolving set for [MATH] , and [MATH] a minimum non-gate- [MATH] -resolving set for [MATH] , then [MATH] and [MATH] |
Proof [MATH] : If [MATH] contains vertex [MATH] [MATH] is already a minimum [MATH] -resolving set. If [MATH] the set [MATH] is a [MATH] -resolving set. |
[MATH] : If [MATH] is not an [MATH] -gate, [MATH] is already a minimum non-gate- [MATH] -resolving set. If [MATH] is an [MATH] -gate there are out-vertices [MATH] for [MATH] . From Observation 3.4 we know that these out-vertices are on a shortest path between [MATH] and the out-vertex with longest distance to [MATH] , ... |
[MATH] Figure shows that these bounds are tight. Lemma 3.7 Let [MATH] be a separation vertex and [MATH] [MATH] , be the vertex sets of the connected components of [MATH] . Let [MATH] be a resolving set for [MATH] . Then there is at most one [MATH] such that [MATH] |
Proof Assume there are two sets [MATH] such that [MATH] and [MATH] . Consider a vertex [MATH] adjacent to [MATH] and a vertex [MATH] adjacent to [MATH] . For any [MATH] we have [MATH] . This contradicts the assumption that [MATH] is a resolving set. |
[MATH] Lemma 3.8 Let [MATH] be a separation vertex and [MATH] [MATH] , be the vertex sets of the connected components of [MATH] such that if [MATH] in every resolving set [MATH] for [MATH] there is a vertex [MATH] and a vertex [MATH] . Let [MATH] with [MATH] . If [MATH] is a resolving set for [MATH] then [MATH] is reso... |
Proof Consider two vertices [MATH] that are separated by [MATH] . We will show, that there is a vertex [MATH] [MATH] , that separates [MATH] and [MATH] |
Assume [MATH] and [MATH] are both in the same set [MATH] [MATH] . Then, by assumption and Lemma 3.7 , there is a vertex [MATH] [MATH] . Since every path from [MATH] to [MATH] or [MATH] to [MATH] contains vertex [MATH] and [MATH] separates [MATH] and [MATH] , vertex [MATH] does the same. |
Assume [MATH] and [MATH] are in different sets. Without loss of generality let [MATH] [MATH] [MATH] , and [MATH] Case 1: [MATH] If [MATH] then by assumption [MATH] has at least three components. By Lemma 3.7 there is a component [MATH] [MATH] [MATH] , such that [MATH] . Let [MATH] , then we have |
[EQUATION] Case 2: [MATH] Then we have for any [MATH] [EQUATION] [MATH] Lemma 3.9 Let [MATH] be a separation vertex and [MATH] [MATH] , be the vertex sets of the connected components of [MATH] such that if [MATH] in every resolving set [MATH] for [MATH] there is a vertex [MATH] and a vertex [MATH] . Let [MATH] [MATH] ,... |
[MATH] is a minimum [MATH] -resolving set for [MATH] and there is at most one [MATH] such that [MATH] is an [MATH] -gate in [MATH] |
Proof [MATH] : Let [MATH] be a resolving set for [MATH] . We will show that (1) for every [MATH] the set [MATH] is an [MATH] -resolving set for [MATH] and (2) there is at most one index [MATH] such that [MATH] is an [MATH] -gate in [MATH] |
(1): Consider two vertices [MATH] . Obviously [MATH] is a resolving set for those pairs of vertices that are separated by a vertex [MATH] Assume that for a pair of vertices [MATH] there is no vertex in [MATH] that separates them, i.e. [MATH] are separated by a vertex [MATH] [MATH] . Since all paths from [MATH] to [MATH... |
[EQUATION] Therefore [MATH] implies [MATH] , i.e. [MATH] separates [MATH] and [MATH] (2): Assume there are two indices [MATH] [MATH] , for which [MATH] is an [MATH] -gate in [MATH] and an [MATH] -gate in [MATH] . Then there are out-vertices [MATH] for [MATH] -gate [MATH] in [MATH] and out-vertices [MATH] for [MATH] -ga... |
[EQUATION] This contradicts the assumption that [MATH] is a resolving set for [MATH] , see also Figure [MATH] : Let [MATH] be an [MATH] -resolving set for [MATH] and let there be at most one [MATH] such that [MATH] is an [MATH] -gate in [MATH] . We will proof that [MATH] is a resolving set for [MATH] |
Let [MATH] . Obviously [MATH] separates all pairs of vertices that are in the same set [MATH] . Consider two vertices [MATH] that are in different set. Without loss of generality let [MATH] and [MATH] [MATH] [MATH] and [MATH] the component in which [MATH] is not an [MATH] -gate for [MATH] . This implies that there is a... |
Now we will show the minimality. Assume that [MATH] is a minimal resolving set for [MATH] and [MATH] is not a minimal [MATH] -resolving set, i.e. there is set [MATH] with [MATH] that contains [MATH] and resolves all vertices from [MATH] . Consider the set [MATH] . As shown above [MATH] is a resolving set for [MATH] and... |
Assume that [MATH] is a minimal [MATH] -resolving set and there is at most one [MATH] such that [MATH] is an [MATH] -gate in [MATH] , but [MATH] is not minimal, i.e. there is a resolving set [MATH] for [MATH] with [MATH] . This implies that there is at least one [MATH] such that [MATH] is a resolving set for [MATH] and... |
[MATH] Lemma 3.10 Let [MATH] be a separation vertex and [MATH] [MATH] , the vertex sets of the connected components of [MATH] such that if [MATH] in every resolving set [MATH] for [MATH] there is a vertex [MATH] and a vertex [MATH] . Let [MATH] . If [MATH] is a minimum resolving set for [MATH] , then [MATH] is a minimu... |
Proof Let [MATH] be a minimum resolving set and [MATH] an [MATH] -resolving set. From Lemma 3.8 we know that [MATH] , therefore [MATH] . Assume there is a miniumum [MATH] -resolving set [MATH] with [MATH] . Then, by Lemma 3.8 [MATH] is a resolving set for [MATH] and we have [MATH] . This contradicts the assumption that... |
[MATH] Now we define [MATH] [MATH] , as introduced at the beginning of the section. Definition 3.11 1. Let [MATH] be an a-node. We define [MATH] , where [MATH] is the size of a minimum non-gate- \textnu [MATH] -resolving set for [MATH] and [MATH] is the size of a minimum \textnu [MATH] -resolving set for [MATH] |
2. Let [MATH] be a c-node with father [MATH] in [MATH] . We define [MATH] , where [MATH] is the size of a minimum non-gate- \textnu [MATH] -resolving set for [MATH] and [MATH] is the size of a minimum \textnu [MATH] -resolving set for [MATH] |
To get familiar with this definition we will investigate the smallest possible values for [MATH] and [MATH] . For an arbitrary node [MATH] with father [MATH] we have [MATH] , i.e. the [MATH] -th component of [MATH] is less or equal than the [MATH] -th component of [MATH] , since [MATH] is a subgraph of [MATH] . Therefo... |
For an a-node [MATH] that has only leafs as children the graph [MATH] consists of EBCs and paths, that are connected by the separation vertex \textnu [MATH] . Note that if [MATH] has exactly one child [MATH] the graph [MATH] is not an ordinary leg, since this contradicts the decomposition of [MATH] into EBCs, ordinary ... |
Observation 3.12 Let [MATH] be a minimum resolving set for [MATH] . For any a-node [MATH] the subgraph [MATH] contains at least one resolving node, i.e. [MATH] |
We will now show that this definition satisfys the properties in 3.1 Theorem 3.13 For every a-node [MATH] with children [MATH] [MATH] [MATH] can be computed from [MATH] |
Proof [MATH] : If [MATH] has exactly one child [MATH] then [MATH] . Since \textnu [MATH] and [MATH] is the only child of [MATH] , we can follow that [MATH] . Therefore a minimum non-gate- \textnu [MATH] -resolving set for [MATH] is also a minimum non-gate- \textnu [MATH] -resolving set for [MATH] . The same holds for a... |
[MATH] : Let [MATH] [MATH] . Then [MATH] with 1. [MATH] 2. [MATH] Let [MATH] be a minimum non-gate- \textnu [MATH] -resolving set for [MATH] and thus [MATH] . Then every [MATH] is also a minimum \textnu [MATH] -resolving set and there is no [MATH] [MATH] , such that [MATH] is a \textnu [MATH] -gate in [MATH] . With the... |
In a minimum \textnu [MATH] -resolving set [MATH] for [MATH] there is at most one index [MATH] such that \textnu [MATH] is a [MATH] -gate in [MATH] , otherwise there would be two vertices [MATH] such that for every [MATH] there is a shortest path to [MATH] via \textnu [MATH] , i.e. [MATH] is not a resolving set (see Le... |
[MATH] Theorem 3.14 For every c-node [MATH] with father [MATH] and children [MATH] [MATH] [MATH] can be computed from [MATH] and [MATH] |
To proof this theorem, we need the following lemma: Lemma 3.15 Let [MATH] be a c-node with father [MATH] and children [MATH] [MATH] . Let [MATH] with \textnu [MATH] . Let [MATH] [MATH] , and [MATH] [MATH] is a \textnu [MATH] -resolving set for [MATH] if and only if |
1. [MATH] is a resolving set for [MATH] and 2. [MATH] is a \textnu [MATH] -resolving set for [MATH] and 3. For every [MATH] vertex \textnu [MATH] is not an [MATH] -gate in [MATH] or not an [MATH] -gate in [MATH] |
Proof [MATH] ”: Let [MATH] be a \textnu [MATH] -resolving set for [MATH] 1. We show that [MATH] is a resolving set for [MATH] . Let [MATH] . Pair [MATH] is either resolved by a vertex [MATH] , or by a vertex [MATH] , or by a vertex [MATH] [MATH] [MATH] . Since every path from a vertex of [MATH] to [MATH] or to [MATH] c... |
2. We show that [MATH] is a \textnu [MATH] -resolving set for [MATH] . Since \textnu [MATH] by definition, we just have to show that [MATH] is a resolving set. Let [MATH] . Pair [MATH] is either resolved by a vertex [MATH] or by a vertex [MATH] [MATH] . Since every path from a vertex in [MATH] to [MATH] contains vertex... |
3. We show that for every [MATH] vertex \textnu [MATH] is not an [MATH] -gate in [MATH] or not an [MATH] -gate in [MATH] . Assume there is an index [MATH] such that vertex \textnu [MATH] is an [MATH] -gate in [MATH] and an [MATH] -gate in [MATH] . From Observation 3.3 we know that there is an out-vertex [MATH] with res... |
[MATH] ”: Let 1. [MATH] be a resolving set for [MATH] 2. [MATH] be a \textnu [MATH] -resolving set for [MATH] 3. for every [MATH] vertex \textnu [MATH] be not an [MATH] -gate in [MATH] or not an [MATH] -gate in [MATH] |
We show that [MATH] is a \textnu [MATH] -resolving set. [MATH] is a \textnu [MATH] -resolving set, if \textnu [MATH] and if for every pair [MATH] there is a vertex in [MATH] , that resolves [MATH] and [MATH] . Obviously \textnu [MATH] , because \textnu [MATH] , so we just have to show that [MATH] is a resolving set. We... |
(a) [MATH] [MATH] (b) [MATH] [MATH] [MATH] [MATH] (c) [MATH] (d) [MATH] and [MATH] [MATH] We will have a closer look at all these cases. |
(a) Let [MATH] . Pair [MATH] is either resolved by a vertex [MATH] , or by vertex \textnu [MATH] . Since [MATH] and every pair that is resolved by \textnu [MATH] is also resolved by a vertex in [MATH] \textnu [MATH] ), [MATH] resolves all pairs [MATH] , see Figure 7(a) |
(b) Let [MATH] and [MATH] . Pair [MATH] is either resolved by [MATH] or by [MATH] , because it is not possible that [MATH] and [MATH] have the same distance to both, [MATH] and [MATH] , what can be seen as follows: |
Let [MATH] [MATH] [MATH] [MATH] [MATH] [MATH] and [MATH] , (see Figure 7(b) ). Since [MATH] we can follow [MATH] . Assume that neither [MATH] nor [MATH] separate [MATH] , i.e. [MATH] and [MATH] . From [MATH] and [MATH] we get [MATH] and [MATH] . Finally we have [MATH] what implies [MATH] and thus [MATH] . This contradi... |
(c) Let [MATH] . Pair [MATH] is either resolved by a vertex [MATH] or by a vertex in [MATH] . Since [MATH] for every [MATH] (see Observation 3.12 ) and every pair that is resolved by \textnu [MATH] is also resolved by a vertex in [MATH] [MATH] resolves all pairs [MATH] , see Figure 7(c) |
(d) Let [MATH] and [MATH] . Since for every [MATH] vertex \textnu [MATH] is not an [MATH] -gate in [MATH] or not an [MATH] -gate in [MATH] , pair [MATH] is either resolved by a vertex [MATH] (see Observation 3.12 ) or by a vertex [MATH] \textnu [MATH] ), what can be seen as follows: |
Assume pair [MATH] cannot be resolves by a vertex [MATH] and cannot be resolved by a vertex [MATH] . Without loss of generality let \textnu [MATH] be no [MATH] -gate in [MATH] . Then there is a vertex [MATH] such that there is no shortest path from [MATH] to [MATH] via vertex \textnu [MATH] Therefore [MATH] . Since [MA... |
Vertex [MATH] does not resolve pair [MATH] either, that means [MATH] . It follows [MATH] , what contradicts the asumption. [MATH] |
Proof of Theorem 3.14 Graph [MATH] is composed by the graph [MATH] and the graphs [MATH] [MATH] . We compute [MATH] by computing a minimum-non-gate- \textnu [MATH] -resolving set [MATH] for [MATH] with [MATH] and a minimum \textnu [MATH] -resolving set [MATH] for [MATH] with [MATH] with the help of Lemma 3.15 |
Let [MATH] be a minimum-non-gate- \textnu [MATH] -resolving set for [MATH] and [MATH] be a minimum- \textnu [MATH] -resolving set for [MATH] [MATH] . To compute sets [MATH] and [MATH] and thus [MATH] and [MATH] we can do the following: |
For every subset [MATH] that contains vertices \textnu [MATH] [MATH] and resolves all pairs [MATH] we determine a resolving set [MATH] for [MATH] |
[MATH] contains the vertices in [MATH] and for every [MATH] either the vertices in [MATH] or in [MATH] . If vertex \textnu [MATH] is a [MATH] -gate in [MATH] then [MATH] contains the vertices in [MATH] else the vertices in [MATH] [MATH] is a \textnu [MATH] -resolving set for [MATH] (Lemma 3.15 ) and by Lemma 3.8 we get... |
Then we have [MATH] with [MATH] and [MATH] with [MATH] [MATH] Theorem 3.16 The metric dimension of [MATH] can efficiently be computed from [MATH] |
Proof Let [MATH] be a minimum- \textnu [MATH] -resolving set for [MATH] . Since the conditions of Lemma 3.8 are given, we get that [MATH] is a resolving set for [MATH] and with Lemma 3.6 we get that [MATH] is a minimum resolving set for [MATH] . Thus, if [MATH] , then [MATH] is the metric dimension of [MATH] |
[MATH] Algorithm and Time Complexity Let [MATH] be a connected undirected graph with [MATH] and [MATH] . To compute a resolving set for [MATH] we first compute the DEBC-tree [MATH] for [MATH] . This can be done in [MATH] with the help of any linear-time-algorithm for finding the biconnected components and bridges of [M... |
Definition 4.1 An undirected graph [MATH] is (minimum) [MATH] -EBC-bounded for some positive integer [MATH] , if there is a (minimum) resolving set [MATH] for [MATH] such that every EBC of [MATH] contains at most [MATH] vertices of [MATH] [MATH] is called a (minimum) [MATH] -EBC-bounded-resolving set for [MATH] |
Let [MATH] and [MATH] be the class of graphs that are [MATH] -EBC-bounded and minimum- [MATH] -EBC-bounded, respectively. A set of graphs [MATH] is (minimum) EBC-bounded , if for every graph [MATH] there is a [MATH] such that [MATH] is (minimum) [MATH] -EBC-bounded. |
Corollary 4.2 The following problems can be solved in polynomial time for any fixed positive integer [MATH] 1. Given an undirected graph [MATH] . Is [MATH] |
2. Given a set [MATH] of EBC-bounded graphs. Find the smallest integer [MATH] such that [MATH] 3. Given an undirected graph [MATH] . Compute a minimum [MATH] -EBC-bounded-resolving set for [MATH] |
4. Given an undirected graph [MATH] . Compute a minimum resolving set for [MATH] and thus the metric dimension of [MATH] To solve these problems we use our algorithm with slight modifications. |
Instead of checking every subset [MATH] if it is resolving, we do the following: For the problems 1., 3. and 4. we only test those subsets with at most [MATH] vertices. For the problem 2. we run our algorithm for [MATH] and increase [MATH] successively by one until we get a resolving set. By doing these modifications t... |
Obviously it holds that [MATH] . Vice versa it holds that for all [MATH] there is a graph [MATH] such that [MATH] , see Figure . Moreover, the complexity of the following problems remain open: |
1. Given an undirected graph [MATH] a fixed positive integer [MATH] . Is [MATH] 2. Given an undirected graph [MATH] . Find the smallest integer [MATH] such that [MATH] |
The following problem, however, still remains NP-complete. The proof can be found in the full version of this paper. [MATH] -bounded BC Metric Dimension |
Given: An undirected graph [MATH] and a positive integer [MATH] such that there is a minimum resolving set [MATH] for [MATH] that contains at most [MATH] vertices from each biconnected component of [MATH] |
Question: Is the metric dimension of [MATH] at most [MATH] Theorem 4.3 [MATH] -bounded BC Metric Dimension is NP-complete for all positive integers [MATH] |
Proof We use a slight modification of the NP-completeness proof of Metric Dimension from Khuller et al. in , where they reduce from 3-SAT. Let [MATH] be a 3-SAT instance with [MATH] variables and [MATH] clauses. For each variable [MATH] they construct the gadget in Figure 9(a) and for each clause [MATH] the gadget in F... |
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