minimal doubly resolving sets and the strong metric dimension of some convex polytopes

Applied Mathematics and Computation 218 (2012) 9790–9801

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

Minimal doubly resolving sets and the strong metric dimensionof some convex polytopes q

Jozef Kratica a,⇑, Vera Kovacevic-Vujcic b, Mirjana Cangalovic b, Milica Stojanovic b

a Mathematical Institute, Serbian Academy of Sciences and Arts, Kneza Mihaila 36/III, 11000 Belgrade, Serbiab Faculty of Organizational Sciences, University of Belgrade, Jove Ilica 154, 11000 Belgrade, Serbia

a r t i c l e i n f o a b s t r a c t

Keywords:Minimal doubly resolving setStrong metric dimensionConvex polytopes

0096-3003/$ - see front matter � 2012 Elsevier Inchttp://dx.doi.org/10.1016/j.amc.2012.03.047

q This research was partially supported by Serbia⇑ Corresponding author.

E-mail addresses: [email protected] (J. Kratic

In this paper we consider two similar optimization problems on graphs: the strong metricdimension problem and the problem of determining minimal doubly resolving sets. Weprove some properties of strong resolving sets and give an integer linear programming for-mulation of the strong metric dimension problem. These results are used to derive explicitexpressions in terms of the dimension n, for the strong metric dimension of two classes ofconvex polytopes Dn and Tn. On the other hand, we prove that minimal doubly resolvingsets of Dn and Tn have constant cardinality for n > 7.

� 2012 Elsevier Inc. All rights reserved.

1. Introduction

The metric dimension problem (MDP), introduced independently by Slater [1] and Harary and Melter [2], has been widelyinvestigated [3–8]. It arises in many diverse areas including network discovery and verification [3], geographical routing pro-tocols [9], the robot navigation, connected joints in graphs, chemistry, etc. Given a simple connected undirected graphG ¼ ðV ; EÞ, where V ¼ f1;2; . . . ;ng; jEj ¼ m; dðu;vÞ denotes the distance between vertices u and v, i.e. the length of a shortestu � v path. A vertex x of the graph G is said to resolve two vertices u and v of G if dðu; xÞ– dðv; xÞ. A vertex setB ¼ fx1; x2; . . . ; xkg of G is a resolving set of G if every two distinct vertices of G are resolved by some vertex of B. Given a vertext, the k-tuple rðt;BÞ ¼ ðdðt; x1Þ; dðt; x2Þ; . . . ; dðt; xkÞÞ is called the vector of metric coordinates of t with respect to B. A metric basisof G is a resolving set of the minimum cardinality. The metric dimension of G, denoted by bðGÞ, is the cardinality of its metricbasis.

Caceres et al. [4] define the notion of a doubly resolving set as follows. Vertices x, y of the graph G (n P 2) are said to dou-bly resolve vertices u, v of G if dðu; xÞ � dðu; yÞ– dðv ; xÞ � dðv; yÞ. A vertex set D ¼ fx1; x2; . . . ; xlg of G is a doubly resolving set of Gif every two distinct vertices of G are doubly resolved by some two vertices of D. The minimal doubly resolving set problem consistsof finding a doubly resolving set of G with the minimum cardinality, denoted by wðGÞ. Note that if x, y doubly resolve u, v thendðu; xÞ � dðv ; xÞ – 0 or dðu; yÞ � dðv ; yÞ – 0, and hence x or y resolves u, v. Therefore, a doubly resolving set is also a resolvingset and bðGÞ 6 wðGÞ.

The strong metric dimension problem (SMDP) was introduced by Sebö and Tannier [10] and further investigated by Oel-lermann and Peters-Fransen [11]. Recently, the strong metric dimension of distance hereditary graphs has been studied byMay and Oellermann [12]. A vertex w strongly resolves two vertices u and v if u belongs to a shortest v �w path or v belongsto a shortest u�w path. A vertex set S of G is a strong resolving set of G if every two distinct vertices of G are strongly resolvedby some vertex of S. A strong metric basis of G is a strong resolving set of the minimum cardinality. Now, the strong metric

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n Ministry of Education and Science under the grant no. 174033.

a), [email protected] (V. Kovacevic-Vujcic), [email protected] (M. Cangalovic), [email protected] (M. Stojanovic).

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dimension of G, denoted by sdimðGÞ is defined as the cardinality of its strong metric basis. It is easy to see that if a vertex wstrongly resolves vertices u and v then w also resolves these vertices. Hence every strong resolving set is a resolving set andbðGÞ 6 sdimðGÞ.

All three previously defined problems are NP-hard in general case. The proofs of NP-hardness are given for the metricdimension problem in [13], for the minimal doubly resolving set problem in [14] and for the strong metric dimension prob-lem in [11]. The first metaheuristic approaches to all three problems have been proposed in [15,16,14].

If S is a strong resolving set of G, then set frðv; SÞjv 2 Vg uniquely determines graph G in the following sense. If G0 is agraph with VðG0Þ ¼ VðGÞ such that S strongly resolves G0, and if for all vertices v we have rGðv ; SÞ ¼ rG0 ðv ; SÞ, then G ¼ G0. IfS is a resolving set or a doubly resolving set, then set S need not uniquely determine G. The next example illustrates previ-ously defined invariants bðGÞ;wðGÞ and sdimðGÞ.

Example 1. Consider graph G1 of Fig. 1. It is easy to check that bðG1Þ;wðG1Þ and sdimðG1Þ are all different. Namely, set fA;Bg isa minimal resolving set, set fA;B; E; Fg is a minimal doubly resolving set, while fA;B; Eg is a minimal strong resolving set ofG1. Therefore, bðG1Þ ¼ 2; wðG1Þ ¼ 4 and sdimðG1Þ ¼ 3.

For some simple classes of graphs it is possible to determine bðGÞ;wðGÞ and sdimðGÞ explicitly: the complete graph Kn withn vertices has bðKnÞ ¼ wðKnÞ ¼ sdimðKnÞ ¼ n� 1, the cycle Cn with n vertices has bðCnÞ ¼ 2;wðCnÞ ¼ 3, whilesdimðCnÞ ¼ n=2d e.

The metric dimension of some nontrivial classes of plane graphs has been determined in [4,13,17–23]. Specially, in [19]the group of convex polytopes Dn;Rn and Q n, introduced in [24] has been considered, while [20] studies the group of graphsSn; Tn and Un, which are combinations of two graphs of convex polytopes. It has been proved that all these classes have themetric dimension equal to 3.

In this paper we study the minimal cardinality of doubly resolving sets and the strong metric dimension of one represen-tative of each group: Dn and Tn. We prove that wðDnÞ ¼ 3 and

wðTnÞ ¼3; n – 7;4; n ¼ 7;

�

while strong metric dimension depends of the dimension n. Due to different structures of Dn and Tn, the proofs are quitedifferent.

The paper is organized as follows. In Section 2 we prove some theoretical properties of strong resolving sets of an arbi-trary graph and give an integer linear programming formulation of the strong metric dimension problem. In Sections 3 and 4we get explicit expressions for wðDnÞ; sdimðDnÞ;wðTnÞ and sdimðTnÞ. Finally, Section 5 summarizes conclusions of the paper.

2. Some new theoretical considerations of strong resolving sets in an arbitrary graph

2.1. Theoretical properties

Lemma 1. Let u;v 2 V ; u – v , and

(i) dðw;vÞ 6 dðu;vÞ for each w such that fw;ug 2 E and(ii) dðu;wÞ 6 dðu;vÞ for each w such that fv ;wg 2 E.

Then, there does not exist vertex a 2 V ; a – u; v , that strongly resolves vertices u and v.

Proof. Let us suppose that there exists vertex a 2 V ; a – u;v , that strongly resolves vertices u and v. It means that

(a) there exists a shortest path P1 between a and v containing u, or(b) there exists a shortest path P2 between a and u containing v.

Fig. 1. Graph G1.

9792 J. Kratica et al. / Applied Mathematics and Computation 218 (2012) 9790–9801

If (a) is fulfilled, then denote by u0 a vertex of path P1 which is adjacent to u and belongs to its part between a and u. As eachpart of P1 represents also a shortest path between the corresponding vertices of P1, then, consequently, dðu0;vÞ ¼ 1þ dðu;vÞ.But, since vertex u0 satisfies condition (i), it follows that dðu0;vÞ 6 dðu;vÞ, which is a contradiction.

Starting from (b), in a similar way we can also reach a contradiction. Consequently, vertex a cannot strongly resolvevertices u and v. h

Remark 1. In the case when condition (i) is not satisfied for a vertex w adjacent to u, it follows from jdðu;vÞ � dðw;vÞj 6 1that dðw;vÞ ¼ 1þ dðu;vÞ. It means that u belongs to a shortest path between w and v, and, therefore, w strongly resolves uand v. In the case when condition (ii) is not satisfied for a vertex w adjacent to v, using similar considerations, we also con-clude that w strongly resolves u and v.

Property 1. If S � V is a strong resolving set of graph G, then, for every two vertices u;v 2 V which satisfy conditions (i) and (ii) ofLemma 1, u 2 S or v 2 S.

Proof. Let us suppose that u R S and v R S. Then, it follows, directly from Lemma 1, that there does not exist a vertex from Swhich strongly resolves vertices u and v, which is in contradiction with the assumption that S is a strong resolving set.

Let DiamðGÞ denote the diameter of graph G, i.e. the maximal distance between two vertices in G.

Property 2. If S � V is a strong resolving set of graph G, then, for every two vertices u;v 2 V such that dðu;vÞ ¼ DiamðGÞ; u 2 S orv 2 S.

Proof. If dðu;vÞ ¼ DiamðGÞ, then, obviously, u and v satisfy conditions (i) and (ii) and, according to Property 1, u 2 S orv 2 S. h

2.2. Integer linear programming formulation

In the literature there exist two integer linear programming (ILP) formulations of the metric dimension problem [5,6] andone ILP formulation of the minimal doubly resolving set problem [14]. To our knowledge, the ILP formulation (3)–(5) of thestrong metric dimension problem is new.

Given a simple connected undirected graph G ¼ ðV ; EÞ, where V ¼ f1;2; . . . ;ng; jEj ¼ m, it is easy to determine the lengthdðu; vÞ of a shortest u � v path for all u;v 2 V , using any shortest path algorithm. The coefficient matrix A is defined asfollows:

Aðu;vÞ;i ¼1; dðu; iÞ ¼ dðu;vÞ þ dðv; iÞ;1; dðv ; iÞ ¼ dðv ;uÞ þ dðu; iÞ;0; otherwise;

8><>: ð1Þ

where 1 6 u < v 6 n; 1 6 i 6 n. Variable yi described by (2) determines whether vertex i belongs to a strong resolving set S.

yi ¼1; i 2 S;

0; i R S:

�ð2Þ

The ILP model of the strong metric dimension problem can now be formulated as:

minPni¼1

yi ð3Þ

subject to:

Pni¼1

Aðu;vÞ;i � yi P 1; 1 6 u < v 6 n; ð4Þ

yi 2 f0;1g; 1 6 i 6 n: ð5Þ

The following proposition shows that each solution of (4) and (5) defines a strong resolving set S of G, and vice versa.

Proposition 1. S is a strong resolving set of G if and only if constraints (4) and (5) are satisfied.

Proof. ()) Suppose that S is a strong resolving set. Then for each u;v 2 V ; u – v there exists i 2 S (i.e. yi ¼ 1), such thatdðu; iÞ ¼ dðu;vÞ þ dðv ; iÞ or dðv ; iÞ ¼ dðv ;uÞ þ dðu; iÞ. It follows that Aðu;vÞ;i ¼ 1, and consequently constraints (4) are satisfied.Constraints (5) are obviously satisfied by definition of variables yi.


(() According to (2), S ¼ fi 2 f1;2; . . . ;ngjyi ¼ 1g. If constraints (4) are satisfied then for each 1 6 u < v 6 n there existsi 2 f1;2; . . . ;ng, such that Aðu;vÞ;i � yi P 1, which implies yi ¼ 1 (i.e. i 2 S) and Aðu;vÞ;i ¼ 1 (i.e. dðu; iÞ ¼ dðu;vÞ þ dðv ; iÞ ordðv ; iÞ ¼ dðv ;uÞ þ dðu; iÞ). It follows that the set S is a strong resolving set of G. h

Note that ILP model (3)–(5) has only n variables and n2

� �linear constraints.

Remark 2. According to Proposition 1, the objective function (3) represents the cardinality of a strong resolving set, andtherefore each optimal solution of model (3)–(5) defines a strong metric basis of graph G.

3. Convex polytopes Dn

The graph of convex polytope Dn;n P 5 (Fig. 2), introduced in [24], consists of 2n 5-sided faces and a pair of n sided faces.We have

VðDnÞ ¼ fai; bi; ci;diji ¼ 0;1; . . . ;n� 1g;EðDnÞ ¼ faiaiþ1;didiþ1; aibi; bici; cidi; biþ1ciji ¼ 0;1; . . . ;n� 1g;

where indices are taken modulo n. In further considerations vertices ai; i ¼ 0;1; . . . ;n� 1 will be refered as a-vertices, andsimilarly other vertices will be called b-vertices, c-vertices and d-vertices.

It has been proved in [19] that the metric dimension of Dn is constant and equal to 3. We will prove in this section that thesame holds for the cardinality of minimal doubly resolving sets of Dn. On the other hand, we prove that for n P 10 the strongmetric dimension of Dn is equal to 2n for n odd, and 5n=2 for n even.

3.1. Minimal doubly resolving sets of Dn

In the proof of Theorem 1 we will use the following:

Proposition 2 [14]. A subset D ¼ fx1; x2; . . . ; xkg# V is a doubly resolving set of G if and only if for every p; q 2 V there existsi 2 f2;3; . . . ; kg such that

dðp; xiÞ � dðp; x1Þ– dðq; xiÞ � dðq; x1Þ: ð6Þ
For each v 2 V let r0ðv ;DÞ ¼ ðdðv; x2Þ � dðv ; x1Þ; . . . ; dðv ; xkÞ � dðv; x1ÞÞ. According to Proposition 2, D is a doubly resolving
set if and only if vectors r0ðv ;DÞ are different for all v 2 V .

Theorem 1. For every convex polytope Dn it follows that wðDnÞ ¼ 3.

Proof. Since wðGÞP bðGÞ for every graph G and it has been proved in [19] that bðDnÞ ¼ 3, it follows that wðDnÞP 3. We onlyneed to prove that for each n there exists a doubly resolving set of Dn with cardinality 3. For n 6 13 we have proved by totalenumeration that the sets in Table 1 are doubly resolving sets of Dn. For n P 14 we will prove that

D ¼fa0; ak;d0g; n ¼ 2k;

fa0; ak;d3g; n ¼ 2kþ 1

�

is a doubly resolving set of Dn.We have two cases:Case 1: n is even. In this case, we can write n ¼ 2k; k P 3. Let D ¼ fa0; ak; d0g. We will show that D is a doubly resolving

set of Dn. To this end we give the representations of all vertices of Dn:

Fig. 2. The graph of convex polytope Dn.

Table 1Minimal doubly resolving sets of Dn .

n D

7 {a0 ; a3; d1}9 {a0 ; a4; d6}11 {a0 ; a5; d2}13 {a0 ; a6; d9}n ¼ 2k fa0; ak; d0gn ¼ 2kþ 1, n R f7;9;11;13g fa0; ak; d3g


rðai;DÞ ¼ð0; k;3Þ; i ¼ 0;ði; k� i; iþ 2Þ; 1 6 i 6 k;

ð2k� i; i� k;2k� iþ 3Þ; kþ 1 6 i 6 2k� 1;

8><>:

rðbi;DÞ ¼ð1; kþ 1;2Þ; i ¼ 0;ðiþ 1; k� iþ 1; iþ 1Þ; 1 6 i 6 k;

ð2k� iþ 1; i� kþ 1;2k� iþ 2Þ; kþ 1 6 i 6 2k� 1;

8><>:

rðci;DÞ ¼ðiþ 2; k� iþ 1; iþ 1Þ; 0 6 i 6 k� 1;ð2k� iþ 1; i� kþ 2;2k� iþ 1Þ; k 6 i 6 2k� 1;

�

rðdi;DÞ ¼ðiþ 3; k� iþ 2; iÞ; 0 6 i 6 k� 1;ð2k� iþ 2; i� kþ 3;2k� iÞ; k 6 i 6 2k� 1;

�

Then the representations r0 are:

r0ðai;DÞ ¼ðk;3Þ; i ¼ 0;ðk� 2i;2Þ; 1 6 i 6 k;

ð2i� 3k;3Þ; kþ 1 6 i 6 2k� 1;

8><>:

r0ðbi;DÞ ¼ðk;1Þ; i ¼ 0;ðk� 2i;0Þ; 1 6 i 6 k;

ð2i� 3k;1Þ; kþ 1 6 i 6 2k� 1;

8><>:

r0ðci;DÞ ¼ðk� 2i� 1;�1Þ; 0 6 i 6 k� 1;ð2i� 3kþ 1;0Þ; k 6 i 6 2k� 1;

�

r0ðdi;DÞ ¼ðk� 2i� 1;�3Þ; 0 6 i 6 k� 1;ð2i� 3kþ 1;�2Þ; k 6 i 6 2k� 1:

�

Case 2: n is odd. In this case, we can write n ¼ 2kþ 1; k P 3. Let D ¼ fa0; ak; d3g. The representations of the vertices of Dn

are:

rðai;DÞ ¼

ði; k� i;6� iÞ; 0 6 i 6 3;

ði; k� i; i� 1Þ; 4 6 i 6 k;

ð2k� iþ 1; i� k; i� 1Þ; kþ 1 6 i 6 kþ 4;

ð2k� iþ 1; i� k;2k� iþ 7Þ; kþ 5 6 i 6 2k;

8>>>><>>>>:

rðbi;DÞ ¼

ðiþ 1; k� iþ 1;5� iÞ; 0 6 i 6 3;

ðiþ 1; k� iþ 1; i� 2Þ; 4 6 i 6 k;

ð2k� iþ 2; i� kþ 1; i� 2Þ; kþ 1 6 i 6 kþ 4;

ð2k� iþ 2; i� kþ 1;2k� iþ 6Þ; kþ 5 6 i 6 2k;

8>>>><>>>>:

rðci;DÞ ¼

ðiþ 2; k� iþ 1;4� iÞ; 0 6 i 6 3;

ðiþ 2; k� iþ 1; i� 2Þ; 4 6 i 6 k� 1;

ð2k� iþ 2; i� kþ 2; i� 2Þ; k 6 i 6 kþ 3;

ð2k� iþ 2; i� kþ 2;2k� iþ 5Þ; kþ 4 6 i 6 2k;

8>>>><>>>>:

rðdi;DÞ ¼

ðiþ 3; k� iþ 2;3� iÞ; 0 6 i 6 3;

ðiþ 3; k� iþ 2; i� 3Þ; 4 6 i 6 k� 1;

ð2k� iþ 3; i� kþ 3; i� 3Þ; k 6 i 6 kþ 3;

ð2k� iþ 3; i� kþ 3;2k� iþ 4Þ; kþ 4 6 i 6 2k:

8>>>><>>>>:



r0ðai;DÞ ¼

ðk� 2i;6� 2iÞ; 0 6 i 6 3;ðk� 2i;�1Þ; 4 6 i 6 k;

ð2i� 3k� 1;2i� 2k� 2Þ; kþ 1 6 i 6 kþ 4;ð2i� 3k� 1;6Þ; kþ 5 6 i 6 2k;

8>>><>>>:

r0ðbi;DÞ ¼

ðk� 2i;4� 2iÞ; 0 6 i 6 3;ðk� 2i;�3Þ; 4 6 i 6 k;

ð2i� 3k� 1;2i� 2k� 4Þ; kþ 1 6 i 6 kþ 4;ð2i� 3k� 1;4Þ; kþ 5 6 i 6 2k;

8>>><>>>:

r0ðci;DÞ ¼

ðk� 2i� 1;2� 2iÞ; 0 6 i 6 3;ðk� 2i� 1;�4Þ; 4 6 i 6 k� 1;ð2i� 3k;2i� 2k� 4Þ; k 6 i 6 kþ 3;ð2i� 3k;3Þ; kþ 4 6 i 6 2k;

8>>><>>>:

r0ðdi;DÞ ¼

ðk� 2i� 1;�2iÞ; 0 6 i 6 3;ðk� 2i� 1;�6Þ; 4 6 i 6 k� 1;ð2i� 3k;2i� 2k� 6Þ; k 6 i 6 kþ 3;ð2i� 3k;1Þ; kþ 4 6 i 6 2k:

8>>><>>>:

Since in both cases vectors r0 are mutually different we conclude that the given sets are doubly resolving. h

3.2. The strong metric dimension of Dn

Theorem 2. For any convex polytope Dn it follows that sdimðDnÞ ¼ 2n for n odd and n P 5, and sdimðDnÞ ¼ 5n=2 for n even andn P 10.

Next we prove several lemmas which support the proof of Theorem 2.

Lemma 2. For n ¼ 2kþ 1 and n P 5, if S is a strong resolving set of Dn, then jSjP 2n.

Proof. Let us consider pairs of vertices ðai; dkþiÞ; ðbi; ckþiÞ for all i ¼ 0;1; . . . ;n� 1. It is easy to see thatdðai; dkþiÞ ¼ dðbi; ckþiÞ ¼ kþ 3. As DiamðDnÞ ¼ kþ 3, then, according to Property 2, ai 2 S or dkþi 2 S for i ¼ 0;1; . . . ;n� 1,and bi 2 S or ckþi 2 S for i ¼ 0;1; . . . ;n� 1. It means that jSjP 2n. h

Lemma 3. For n ¼ 2kþ 1 and n P 5, set S ¼ fai; ciji ¼ 0;1; . . . ;n� 1g is a strong resolving set of Dn.

Proof. We need to prove that for every two vertices u;v R S; u – v , there exists a vertex w 2 S which strongly resolves uand v. For each pair of vertices from V n S, the corresponding vertex of S which strongly resolves this pair is given in Table3. Let us prove that, for each i; j 2 f0;1; . . . ; n� 1g such that iþ 1 6 j 6 kþ i, vertices bi and bj are strongly resolved by vertexcj. Really, it is easy to check that dðbi; ciþ1Þ ¼ 3 and dðbi; cjÞ ¼ j� iþ 3 for iþ 2 6 j 6 kþ i. Also, dðbi; biþ1Þ ¼ 2 anddðbi; bjÞ ¼ j� iþ 2 for iþ 2 6 j 6 kþ i, and, consequently, dðbi; cjÞ ¼ dðbi; bjÞ þ 1 for iþ 1 6 j 6 kþ i. As dðbj; cjÞ ¼ 1, it meansthat vertex bj belongs to a shortest path between bi and cj. Therefore, cj strongly resolves bi and bj for iþ 1 6 j 6 kþ i.

Similarly, for kþ iþ 1 6 j 6 n� 1 it can be proved that bi and bj are strongly resolved by vertex cj�1. In this case,dðbi; cnþi�2Þ ¼ 3; dðbi; cj�1Þ ¼ n� jþ iþ 3 for iþ kþ 1 6 j 6 nþ i� 2; dðbi; bnþi�1Þ ¼ 2 and dðbi; bjÞ ¼ n� jþ iþ 2 foriþ kþ 1 6 j 6 nþ i� 2. Since dðbi; cj�1Þ ¼ dðbi; bjÞ þ 1 for iþ kþ 1 6 j 6 n� 1, and cj�1 is adjacent to v j, then bj belongs toa shortest path between bi and cj�1, i.e. cj�1 strongly resolves bi and bj.

In a similar way we can verify existance of vertex w for all other possible pairs of vertices from set V n S given in Table 3. h

Lemma 4. For n ¼ 2k and n P 16, if S is a strong resolving set of Dn, then jSjP 5n=2.

Proof. Since dðai; dkþiÞ ¼ kþ 2 ¼ DiamðDnÞ, according to Property 2, ai 2 S or dkþi 2 S for i ¼ 0;1; . . . ;n� 1, which means thatS should contain at least n a-vertices or d-vertices. Also, dðbi; bkþiÞ ¼ dðci; ckþiÞ ¼ kþ 2 ¼ DiamðDnÞ for i ¼ 0;1; . . . ; k� 1, and,therefore, S should contain at least k b-vertices and at least k c-vertices. We will prove that two vertices bi and cj, wherej R fi� 2; i� 1; i; iþ 1g satisfy conditions:

(a) dðu; cjÞ 6 dðbi; cjÞ for all vertices u adjacent to bi, and(b) dðv ; biÞ 6 dðbi; cjÞ for all vertices v adjacent to cj.


Without loss of generality we can assume that i ¼ 0 and j R f0;1;n� 2;n� 1g. It is easy to check that dðb0; cjÞ ¼ jþ 3 for2 6 j 6 k� 1 and dðb0; cjÞ ¼ n� jþ 2 for k 6 j 6 n� 3. Let us prove the condition (b). Vertices adjacent to vertex cj arebj; bjþ1 and dj. For 2 6 j 6 k� 1 the corresponding distances are dðbj; b0Þ ¼ dðdj; b0Þ ¼ jþ 2 ¼ dðb0; cjÞ � 1; dðbjþ1; b0Þ ¼jþ 3 ¼ dðb0; cjÞ and, consequently, the condition (b) is satisfied. If k 6 j 6 n� 3 then dðbjþ1; b0Þ ¼ dðdj;

b0Þ ¼ n� jþ 1 ¼ dðb0; cjÞ � 1; dðbj; b0Þ ¼ n� jþ 2 ¼ dðb0; cjÞ, and (b) is satisfied. Due to the symmetry of b-vertices and c-vertices, condition (a) also holds. Therefore, according to Property 1, bi 2 S or cj 2 S.

Now, let us consider three possible cases:

Case 1: Let bi 2 S for all i ¼ 0;1; . . . ;n� 1. Then S contains n b-vertices, n vertices from pairs ðai; dkþiÞ and at least k c-ver-tices, and, therefore, jSjP 2nþ k ¼ 5n=2.

Case 2: Let ci 2 S for all i ¼ 0;1; . . . ;n� 1. Using similar considerations as in Case 1, it can be also proved that jSjP 5n=2.Case 3: Let bi R S and cj R S for some i; j ¼ 0;1; . . . ;n� 1. Then, according to previous considerations, and the symmetry

between b-vertices and c-vertices, S should contain all cl; l R fi� 2; i� 1; i; iþ 1g, and allbt ; t R fj� 1; j; jþ 1; jþ 2g. As S should also contain n vertices from pairs fai; dkþig, then jSjP 3n� 8. Sincen P 16; 3n� 8 P 5n=2, and jSjP 5n=2. h

Lemma 5. For n ¼ 2k and n P 16, set S ¼ fbi; cj; djji ¼ 0;1; . . . ; k� 1; j ¼ 0;1; . . . ;n� 1g is a strong resolving set of Dn.

Proof. For each pair u;v 2 V n S a vertex w 2 S which strongly resolves this pair is given in Table 4. Let us prove that for eachi 2 f0;1; . . . ; k� 1g and each j 2 fiþ 1; iþ 2; . . . ;n� 1g vertex bi strongly resolves vertices ai and aj. Really, it is easy to checkthat dðbi; ajÞ ¼ j� iþ 1 and dðai; ajÞ ¼ j� i for iþ 1 6 j 6 iþ k, and dðbi; ajÞ ¼ n� ðj� iÞ þ 1 and dðai; ajÞ ¼ n� ðj� iÞ foriþ kþ 1 6 j 6 n� 1. Since dðbi; aiÞ ¼ 1, then dðbi; ajÞ ¼ dðbi; aiÞ þ dðai; ajÞ for iþ 1 6 j 6 n� 1, i.e. vertex ai belongs to a short-est path between bi and aj and, therefore, bi strongly resolves ai and aj. When i 2 fk; kþ 1; . . . ;n� 2g andj 2 fiþ 1; iþ 2; . . . ;n� 1g, then dðai; bkþiÞ ¼ kþ 1; dðai; ajÞ ¼ j� i and dðaj; bkþiÞ ¼ kþ i� jþ 1. As dðai; bkþiÞ ¼dðai; ajÞ þ dðaj; bkþiÞ, then aj belongs to a shortest path between ai and bkþi, i.e. bkþi strongly resolves ai and aj. In a similarway all other vertices w from Table 4 can be verified. h

Table 3Cases of Lemma 3.

u, v w

bi; bj cj , j 6 kþ i0 6 i < j 6 n� 1 cj�1, j P kþ iþ 1bi;dj ai

0 6 i; j 6 n� 1di; dj ci

0 6 i < j 6 n� 1

Table 2The strong metric dimension of Dn .

n sdim Dn Strong metric basis

6 13 {b3; b4; b5; c2; c3; c4; c5;d0; d1; d2;d3;d4;d5}8 19 {b4; b5; b6; b7; c0; c2; c3; c4; c5; c6; c7; di; ði ¼ 0;1; . . . ;7Þ}n ¼ 2k; n R f6;8g 5n=2 fbi; cj; djji ¼ 0;1; . . . ; k� 1; j ¼ 0;1; . . . ;n� 1gn ¼ 2kþ 1 2n {ai; ciji ¼ 0;1; . . . ;2kg


u;v w

ai; aj bi ,i 6 k� 10 6 i < j 6 n� 1 bkþi , i P kai; bj

0 6 i 6 n� 1 bj�k

k 6 j 6 n� 1bi; bj cj

k 6 i < j 6 n� 1


Now we can give a proof of Theorem 2:

Proof. It follows directly from Lemmas 2 and 3 that for n ¼ 2kþ 1; sdimðDnÞ ¼ 2n. Also, according to Lemmas 4 and 5, forn ¼ 2k and n P 16; sdimðDnÞ ¼ 5n=2. In order to determine sdimðDnÞ for n ¼ 10;12;14, we used the integer programmingmodel (3)–(5), defined in Section 2. In all cases, CPLEX has obtained the strong metric basis with the cardinality 5n=2 and ofthe same type as set S in Lemma 5. Therefore, sdimðDnÞ ¼ 5n=2 also holds for n ¼ 10;12;14. h

Let us mention that for n ¼ 6 and n ¼ 8, CPLEX has obtained values 13 and 19, respectively, i.e. sdimðDnÞ < 5n=2 (see Table 2).

4. Convex polytopes Tn

The graph of convex polytope Tn; n P 5 (Fig. 3), introduced in [21], consists of 4n 3-sided faces, n 4-sided faces and a pairof n sided faces, and it is obtained by the combination of the graph of convex polytope Rn [25] and the graph of an antiprismAn [24].

We have

VðTnÞ ¼ fai; bi; ci; diji ¼ 0;1; . . . ;n� 1g; EðTnÞ ¼ faiaiþ1; bibiþ1; ciciþ1; didiþ1; aibi; bici; cidi; aiþ1bi; ciþ1di j i ¼ 0;1; . . . ;n� 1g;

where indices are taken modulo n.It has been proved in [21] that the metric dimension of Tn is constant and equal to 3. In this section we will prove that the

same holds for the cardinality of minimal doubly resolving sets of Tn;n > 7. On the other hand, we prove that for n P 5 thestrong metric dimension of Tn is equal to 2n for n odd, and 5n=2 for n even.

4.1. Minimal doubly resolving sets of Tn

Theorem 3. For every convex polytope Tn it follows

wðTnÞ ¼3; n – 7;4; n ¼ 7:

�

Proof. Since wðGÞP bðGÞ for every graph G and in [21] is proved that bðTnÞ ¼ 3 it follows that wðTnÞP 3. In order to provethe assertion of the theorem, we only need to prove that for n P 9 there exists a doubly resolving set of Tn of cardinality 3.Other cases have been resolved by total enumeration, and results are displayed in Table 5.

In the sequel we shall prove that

Fig. 3. The graph of convex polytope Tn.

Table 5Minimal doubly resolving sets of Tn .

n D

5 {a0; a2; c3}7 {a0; a1; a3; c4}8 {a0; a4; d1}n ¼ 2k, n – 8 fa0; bk�1;d2k�2gn ¼ 2kþ 1, n R f5;7g fa0; ak; ckþ2g


D ¼fa0; bk�1; d2k�2g; n ¼ 2k;fa0; ak; ckþ2g; n ¼ 2kþ 1

�

is a doubly resolving set of Tn; n P 9.We have two cases:Case 1: n is even. In this case, we can write n ¼ 2k; k P 3. Let D ¼ fa0; bk�1; d2k�2g. We will show that D is a doubly

resolving set of Tn. To this end, we give representations of all vertices of Tn:

rðai;DÞ ¼ði; k� i; iþ 3Þ; 0 6 i 6 k� 1;

ð2k� i; i� kþ 1;2k� iþ 1Þ; k 6 i 6 2k� 2;

ð1; k;3Þ; i ¼ 2k� 1;

8><>:

rðbi;DÞ ¼

ðiþ 1; k� i� 1; iþ 3Þ; 0 6 i 6 k� 2;

ðk;0; kþ 1Þ; i ¼ k� 1;

ð2k� i; i� kþ 1;2k� iÞ; k 6 i 6 2k� 2;

ð1; k;2Þ; i ¼ 2k� 1;

8>>>><>>>>:

rðci;DÞ ¼

ðiþ 2; k� i; iþ 2Þ; 0 6 i 6 k� 2;

ðkþ 1;1; kÞ; i ¼ k� 1;

ð2k� iþ 1; i� kþ 2;2k� i� 1Þ; k 6 i 6 2k� 2;

ð2; kþ 1;1Þ; i ¼ 2k� 1;

8>>>><>>>>:

rðdi;DÞ ¼

ðiþ 3; k� i; iþ 2Þ; 0 6 i 6 k� 2;

ðkþ 2;2; k� 1Þ; i ¼ k� 1;

ð2k� iþ 1; i� kþ 3;2k� i� 2Þ; k 6 i 6 2k� 2;

ð3; kþ 1;1Þ; i ¼ 2k� 1:

8>>>><>>>>:


r0ðai;DÞ ¼ðk� 2i;3Þ; 0 6 i 6 k� 1;

ð2i� 3kþ 1;1Þ; k 6 i 6 2k� 2;

ðk� 1;2Þ; i ¼ 2k� 1;

8><>:

r0ðbi;DÞ ¼

ðk� 2i� 2;2Þ; 0 6 i 6 k� 2;

ð�k;1Þ; i ¼ k� 1;

ð2i� 3kþ 1;0Þ; k 6 i 6 2k� 2;

ðk� 1;1Þ; i ¼ 2k� 1;

8>>>><>>>>:

r0ðci;DÞ ¼

ðk� 2i� 2; 0Þ; 0 6 i 6 k� 2;

ð�k;�1Þ; i ¼ k� 1;

ð2i� 3kþ 1;�2Þ; k 6 i 6 2k� 2;

ðk� 1;�1Þ; i ¼ 2k� 1;

8>>>><>>>>:

r0ðdi;DÞ ¼

ðk� 2i� 3;�1Þ; 0 6 i 6 k� 2;

ð�k;�3Þ; i ¼ k� 1;

ð2i� 3kþ 2;�3Þ; k 6 i 6 2k� 2;

ðk� 2;�2Þ; i ¼ 2k� 1:

8>>>><>>>>:

Case 2: n is odd. In this case, we can write n ¼ 2kþ 1; k P 3. Let D ¼ fa0; ak; ckþ2g. The representations of vertices of Tn

are:

rðai;DÞ ¼

ði; k� i; kþ iÞ; 0 6 i 6 2;ði; k� i; k� iþ 4Þ; 3 6 i 6 k;

ð2k� iþ 1; i� k; k� iþ 4Þ; kþ 1 6 i 6 kþ 2;ð2k� iþ 1; i� k; i� k� 1Þ; kþ 3 6 i 6 2k;

8>>><>>>:

rðbi;DÞ ¼

ðiþ 1; k� i; kþ iÞ; 0 6 i 6 1;ðiþ 1; k� i; k� iþ 3Þ; 2 6 i 6 k� 1;ð2k� iþ 1; i� kþ 1; k� iþ 3Þ; k 6 i 6 kþ 2;ð2k� iþ 1; i� kþ 1; i� k� 1Þ; kþ 3 6 i 6 2k;

8>>><>>>:


rðci;DÞ ¼

ðiþ 2; k� iþ 1; kþ i� 1Þ; 0 6 i 6 1;ðiþ 2; k� iþ 1; k� iþ 2Þ; 2 6 i 6 k� 1;ð2k� iþ 2; i� kþ 2; k� iþ 2Þ; k 6 i 6 kþ 2;ð2k� iþ 2; i� kþ 2; i� k� 2Þ; kþ 3 6 i 6 2k;

8>>><>>>:

rðdi;DÞ ¼

ðiþ 3; k� iþ 1; kþ iÞ; 0 6 i 6 1;ðiþ 3; k� iþ 1; k� iþ 2Þ; 2 6 i 6 k� 2;ðkþ 2;3;3Þ; i ¼ k� 1ð2k� iþ 2; i� kþ 3; k� iþ 2Þ; k 6 i 6 kþ 1;ð2k� iþ 2; i� kþ 3; i� k� 1Þ; kþ 2 6 i 6 2k� 1;ð3; kþ 2; k� 1Þ; i ¼ 2k:

8>>>>>>>><>>>>>>>>:


r0ðai;DÞ ¼

ðk� 2i; kÞ; 0 6 i 6 2;ðk� 2i; k� 2iþ 4Þ; 3 6 i 6 k;

ð2i� 3k� 1;�kþ 3Þ; kþ 1 6 i 6 kþ 2;ð2i� 3k� 1;2i� 3k� 2Þ; kþ 3 6 i 6 2k;

8>>><>>>:

r0ðbi;DÞ ¼

ðk� 2i� 1; k� 1Þ; 0 6 i 6 1;ðk� 2i� 1; k� 2iþ 2Þ; 2 6 i 6 k� 1;ð2i� 3k;�kþ 2Þ; k 6 i 6 kþ 2;ð2i� 3k;2i� 3k� 2Þ; kþ 3 6 i 6 2k;

8>>><>>>:

r0ðci;DÞ ¼

ðk� 2i� 1; k� 3Þ; 0 6 i 6 1;ðk� 2i� 1; k� 2iÞ; 2 6 i 6 k� 1;ð2i� 3k;�kÞ; k 6 i 6 kþ 2;ð2i� 3k;2i� 3k� 4Þ; kþ 3 6 i 6 2k;

8>>><>>>:

r0ðdi;DÞ ¼

ðk� 2i� 2; k� 3Þ; 0 6 i 6 1;ðk� 2i� 2; k� 2i� 1Þ; 2 6 i 6 k� 2;ð�kþ 1;�kþ 1Þ; i ¼ k� 1;ð2i� 3kþ 1;�kÞ; k 6 i 6 kþ 1;ð2i� 3kþ 1;2i� 3k� 3Þ; kþ 2 6 i 6 2k� 1;ðk� 1; k� 4Þ; i ¼ 2k:

8>>>>>>>><>>>>>>>>:

Since in both cases vectors r0 are mutually different we conclude that the given sets are doubly resolving. h

4.2. The strong metric dimension of Tn

Theorem 4. For any convex polytope Tn and n P 5 it follows that sdimðTnÞ ¼ 2n for n odd and sdimðTnÞ ¼ 5n=2 for n even.Next we prove several lemmas which support the proof of Theorem 4.

Lemma 6. For n ¼ 2kþ 1 and n P 5, if S is a strong resolving set of Tn, then jSjP 2n.

Proof. As dðai; ckþiÞ ¼ dðbi; dkþiÞ ¼ kþ 2 for i ¼ 0;1; . . . ;n� 1, and DiamðTnÞ ¼ kþ 2, then, according to Property 2, S shouldcontain at least n a-vertices or c-vertices, and at least n b-vertices or d-vertices. Therefore, jSjP 2n. h

Lemma 7. For n ¼ 2kþ 1 and n P 5, set S ¼ fai; diji ¼ 0;1; . . . ;n� 1g is a strong resolving set of Tn.

Proof. Let us prove that vertex dkþi strongly resolves pairs ðbi; bjÞ; ðbi; cjÞ and ðci; cjÞ for each i; j; i – j. It is easy to see that, foreach i 2 f0;1; . . . ;n� 1g; dðbi; dkþiÞ ¼ kþ 2 ¼ DiamðTnÞ, where the following paths represent the shortest paths between bi

and dkþi:

� bi; b1þi; . . . ; bkþi; ckþi; dkþi;� dkþi; ckþiþ1; bkþiþ1; bkþiþ2; . . . ; b2kþi ¼ bi;� bi; ci; c1þi; . . . ; ckþi; dkþi;� dkþi; ckþiþ1; ckþiþ2; . . . ; c2kþi; b2kþi ¼ bi.


u, v w

bi; bj ckþi

0 6 i < j 6 n� 1bi;dj

0 6 i 6 n� 1 aj�kþ1

k 6 j 6 n� 1di; dj di�k

k 6 i < j 6 n� 1


As each vertex bj; j – i, belongs to one of these paths, then dkþi strongly resolves bi and bj. Also, each vertex cj; j – i, be-longs to at least one of previous paths and, consequently, dkþi strongly resolves bi and cj. Similarly, dðci; dkþiÞ ¼ kþ 1 and thecorresponding shortest paths are:

� ci; c1þi; . . . ; ckþi; dkþi;� dkþi; ckþiþ1; ckþiþ2; . . . ; c2kþi ¼ ci.

Since each vertex cj; j – i, belongs to one of these paths, dkþi strongly resolves ci and cj. h

Lemma 8. For n ¼ 2k and n P 5, if S is a strong resolving set of Tn, then jSjP 5n=2.

Proof. It is easy to see that vertices bi and ckþi satisfy conditions (i) and (ii) of Lemma 1 in Section 2. Therefore, according toProperty 2, bi 2 S _ ckþi 2 S for i 2 f0;1; . . . ;n� 1g. It means that S should contain at least n b-vertices or c-vertices. Also, itcan be easily proved that dðai; dkþi�1Þ ¼ kþ 2 ¼ DiamðTnÞ and pairs ðai; akþiÞ; ðai; di�1Þ; ðdi; dkþiÞ satisfy conditions (i) and (ii) ofLemma 1. Therefore, according to Properties 1 and 2,

(a) ai 2 S or dkþi�1 2 S;(b) ai 2 S or akþi 2 S;(c) ai 2 S or di�1 2 S;(d) di 2 S or dkþi 2 S.

As a consequence of (a)–(d) it follows that S should contain at least 3 vertices from set Xi ¼ fai; akþi; di�1; dkþi�1g. Forexample, if akþi R S, then, according to (a), dkþðkþiÞ�1 ¼ di�1 2 S, and according to (b), ai 2 S, and according to (c), dðkþiÞ�1 2 S.Since for i ¼ 0;1; . . . ;n� 1, the number of all distinct sets Xi is equal to n=2, then S should contain at least 3n=2 a-vertices ord-vertices. As S should also contain at least n b-vertices or c-vertices, it means that jSjP 5n=2. h

Lemma 9. For n ¼ 2k and n P 5, set S ¼ fai; ci; djji ¼ 0;1; . . . ;n� 1; j ¼ 0;1; . . . ; k� 1g is a strong resolving set of Tn.

Proof. Vertices w 2 S which strongly resolve pairs u;v 2 V n S are given in Table 6. Let us show that, for eachi 2 f0;1; . . . ;n� 1g and each j 2 fk; kþ 1; . . . ;n� 1g vertex aj�kþ1 strongly resolves bi and dj. It is easy to check that, for eachj 2 fk; kþ 1; . . . ;n� 1g; dðaj�kþ1; djÞ ¼ kþ 2 ¼ DiamðTnÞ with the corresponding shortest paths between aj�kþ1 and dj:

� aj�kþ1; bj�kþ1; bj�kþ2; . . . ; bj; cj; dj;� dj; cjþ1; bjþ1; bjþ2; . . . ; bjþk; ajþkþ1 ¼ aj�kþ1.

As each vertex bi; i – j, belongs to one of these paths, aj�kþ1 strongly resolves bi and dj. All other results from Table 6 canbe proved in a similar way. h

Now we can give a proof of Theorem 4:

Proof. It follows directly from Lemmas 6 and 7that for n ¼ 2kþ 1; sdimðTnÞ ¼ 2n. Also, according to Lemmas 8 and 9, forn ¼ 2k; sdimðTnÞ ¼ 5n=2. h

5. Conclusions

In this paper we have studied minimal doubly resolving sets and the strong metric dimension of convex polytopes Dn andTn. We have proved that the cardinality of minimal doubly resolving sets is constant and equal to 3, except for T7. In order to


study the strong metric dimension, we have established some general properties of strong resolving sets and introduced aninteger linear programming formulation of the strong metric dimension problem. These results have been used to obtainexpressions for the strong metric dimension of Dn and Tn, which depends on the dimension n.

Future work can be directed towards obtaining minimal doubly resolving sets and/or the strong metric dimension ofsome other challenging classes of graphs.

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minimal doubly resolving sets and the strong metric dimension of some convex polytopes

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