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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 dimension of some convex polytopes q Jozef Kratica a,⇑, Vera Kovacˇevic´-Vujcˇic´ b, Mirjana Cˇangalovic´ b, Milica Stojanovic´ b a b
Mathematical Institute, Serbian Academy of Sciences and Arts, Kneza Mihaila 36/III, 11000 Belgrade, Serbia Faculty of Organizational Sciences, University of Belgrade, Jove Ilic´a 154, 11000 Belgrade, Serbia
a r t i c l e
i n f o
Keywords: Minimal doubly resolving set Strong metric dimension Convex polytopes
a b s t r a c t In this paper we consider two similar optimization problems on graphs: the strong metric dimension problem and the problem of determining minimal doubly resolving sets. We prove some properties of strong resolving sets and give an integer linear programming formulation of the strong metric dimension problem. These results are used to derive explicit expressions in terms of the dimension n, for the strong metric dimension of two classes of convex polytopes Dn and T n . On the other hand, we prove that minimal doubly resolving sets of Dn and T n have constant cardinality for n > 7. Ó 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 widely investigated [3–8]. It arises in many diverse areas including network discovery and verification [3], geographical routing protocols [9], the robot navigation, connected joints in graphs, chemistry, etc. Given a simple connected undirected graph G ¼ ðV; EÞ, where V ¼ f1; 2; . . . ; ng; jEj ¼ m; dðu; v Þ denotes the distance between vertices u and v, i.e. the length of a shortest u 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 set B ¼ fx1 ; x2 ; . . . ; xk g of G is a resolving set of G if every two distinct vertices of G are resolved by some vertex of B. Given a vertex t, the k-tuple rðt; BÞ ¼ ðdðt; x1 Þ; dðt; x2 Þ; . . . ; dðt; xk ÞÞ is called the vector of metric coordinates of t with respect to B. A metric basis of G is a resolving set of the minimum cardinality. The metric dimension of G, denoted by bðGÞ, is the cardinality of its metric basis. 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 doubly resolve vertices u, v of G if dðu; xÞ dðu; yÞ – dðv ; xÞ dðv ; yÞ. A vertex set D ¼ fx1 ; x2 ; . . . ; xl g of G is a doubly resolving set of G if every two distinct vertices of G are doubly resolved by some two vertices of D. The minimal doubly resolving set problem consists of finding a doubly resolving set of G with the minimum cardinality, denoted by wðGÞ. Note that if x, y doubly resolve u, v then dðu; 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 resolving set and bðGÞ 6 wðGÞ. The strong metric dimension problem (SMDP) was introduced by Sebö and Tannier [10] and further investigated by Oellermann and Peters-Fransen [11]. Recently, the strong metric dimension of distance hereditary graphs has been studied by May and Oellermann [12]. A vertex w strongly resolves two vertices u and v if u belongs to a shortest v w path or v belongs to 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 resolved by some vertex of S. A strong metric basis of G is a strong resolving set of the minimum cardinality. Now, the strong metric q
This research was partially supported by Serbian Ministry of Education and Science under the grant no. 174033.
⇑ Corresponding author.
E-mail addresses: [email protected] (J. Kratica), [email protected] (V. Kovacˇevic´-Vujcˇic´), [email protected] (M. Cˇangalovic´), [email protected] (M. Stojanovic´). 0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.03.047
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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 w strongly resolves vertices u and v then w also resolves these vertices. Hence every strong resolving set is a resolving set and bðGÞ 6 sdimðGÞ. All three previously defined problems are NP-hard in general case. The proofs of NP-hardness are given for the metric dimension problem in [13], for the minimal doubly resolving set problem in [14] and for the strong metric dimension problem 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 a graph with VðG0 Þ ¼ VðGÞ such that S strongly resolves G0 , and if for all vertices v we have rG ðv ; SÞ ¼ r G0 ðv ; SÞ, then G ¼ G0 . If S is a resolving set or a doubly resolving set, then set S need not uniquely determine G. The next example illustrates previously defined invariants bðGÞ; wðGÞ and sdimðGÞ. Example 1. Consider graph G1 of Fig. 1. It is easy to check that bðG1 Þ; wðG1 Þ and sdimðG1 Þ are all different. Namely, set fA; Bg is a minimal resolving set, set fA; B; E; Fg is a minimal doubly resolving set, while fA; B; Eg is a minimal strong resolving set of G1 . Therefore, bðG1 Þ ¼ 2; wðG1 Þ ¼ 4 and sdimðG1 Þ ¼ 3. For some simple classes of graphs it is possible to determine bðGÞ; wðGÞ and sdimðGÞ explicitly: the complete graph K n with n vertices has bðK n Þ ¼ wðK n Þ ¼ sdimðK n Þ ¼ n 1, the cycle C n with n vertices has bðC n Þ ¼ 2; wðC n Þ ¼ 3, while sdimðC n Þ ¼ dn=2e. 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 graphs Sn ; T n and U n , which are combinations of two graphs of convex polytopes. It has been proved that all these classes have the metric dimension equal to 3. In this paper we study the minimal cardinality of doubly resolving sets and the strong metric dimension of one representative of each group: Dn and T n . We prove that wðDn Þ ¼ 3 and
wðT n Þ ¼
3; n – 7; 4; n ¼ 7;
while strong metric dimension depends of the dimension n. Due to different structures of Dn and T n , the proofs are quite different. The paper is organized as follows. In Section 2 we prove some theoretical properties of strong resolving sets of an arbitrary graph and give an integer linear programming formulation of the strong metric dimension problem. In Sections 3 and 4 we get explicit expressions for wðDn Þ; sdimðDn Þ; wðT n Þ and sdimðT n Þ. 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; v Þ 6 dðu; v Þ for each w such that fw; ug 2 E and (ii) dðu; wÞ 6 dðu; v Þ 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 P 1 between a and v containing u, or (b) there exists a shortest path P 2 between a and u containing v.
Fig. 1. Graph G1 .
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If (a) is fulfilled, then denote by u0 a vertex of path P 1 which is adjacent to u and belongs to its part between a and u. As each part of P1 represents also a shortest path between the corresponding vertices of P1 , then, consequently, dðu0 ; v Þ ¼ 1 þ dðu; v Þ. But, since vertex u0 satisfies condition (i), it follows that dðu0 ; v Þ 6 dðu; v Þ, which is a contradiction. Starting from (b), in a similar way we can also reach a contradiction. Consequently, vertex a cannot strongly resolve vertices 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; v Þ dðw; v Þj 6 1 that dðw; v Þ ¼ 1 þ dðu; v Þ. It means that u belongs to a shortest path between w and v, and, therefore, w strongly resolves u and v. In the case when condition (ii) is not satisfied for a vertex w adjacent to v, using similar considerations, we also conclude 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) of Lemma 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 S which 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; v Þ ¼ DiamðGÞ; u 2 S or
v 2 S. Proof. If dðu; v Þ ¼ DiamðGÞ, then, obviously, u and h
v
satisfy conditions (i) and (ii) and, according to Property 1, u 2 S or
v 2 S.
2.2. Integer linear programming formulation In the literature there exist two integer linear programming (ILP) formulations of the metric dimension problem [5,6] and one ILP formulation of the minimal doubly resolving set problem [14]. To our knowledge, the ILP formulation (3)–(5) of the strong 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 length dðu; v Þ of a shortest u v path for all u; v 2 V, using any shortest path algorithm. The coefficient matrix A is defined as follows:
8 > < 1; dðu; iÞ ¼ dðu; v Þ þ dðv ; iÞ; Aðu;v Þ;i ¼ 1; dðv ; iÞ ¼ dðv ; uÞ þ dðu; iÞ; > : 0; otherwise;
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 ¼
1; i 2 S; 0;
ð2Þ
i R S:
The ILP model of the strong metric dimension problem can now be formulated as:
min
n P
yi
ð3Þ
i¼1
subject to: n P
Aðu;v Þ;i yi P 1;
1 6 u < v 6 n;
ð4Þ
i¼1
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 that dðu; iÞ ¼ dðu; v Þ þ dðv ; iÞ or dðv ; iÞ ¼ dðv ; uÞ þ dðu; iÞ. It follows that Aðu;v Þ;i ¼ 1, and consequently constraints (4) are satisfied. Constraints (5) are obviously satisfied by definition of variables yi .
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(() 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 exists i 2 f1; 2; . . . ; ng, such that Aðu;v Þ;i yi P 1, which implies yi ¼ 1 (i.e. i 2 S) and Aðu;v Þ;i ¼ 1 (i.e. dðu; iÞ ¼ dðu; v Þ þ dðv ; iÞ or dðv ; iÞ ¼ dðv ; uÞ þ dðu; iÞ). It follows that the set S is a strong resolving set of G. h n Note that ILP model (3)–(5) has only n variables and linear constraints. 2 Remark 2. According to Proposition 1, the objective function (3) represents the cardinality of a strong resolving set, and therefore 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
VðDn Þ ¼ fai ; bi ; ci ; di ji ¼ 0; 1; . . . ; n 1g; EðDn Þ ¼ fai aiþ1 ; di diþ1 ; ai bi ; bi ci ; ci di ; biþ1 ci ji 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, and similarly 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 the same holds for the cardinality of minimal doubly resolving sets of Dn . On the other hand, we prove that for n P 10 the strong metric 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 ; . . . ; xk g # V is a doubly resolving set of G if and only if for every p; q 2 V there exists i 2 f2; 3; . . . ; kg such that
dðp; xi Þ dðp; x1 Þ – dðq; xi Þ dðq; x1 Þ:
ð6Þ
For each v 2 V let r0 ðv ; DÞ ¼ ðdðv ; x2 Þ dðv ; x1 Þ; . . . ; dðv ; xk Þ dðv ; x1 ÞÞ. According to Proposition 2, D is a doubly resolving set if and only if vectors r0 ðv ; DÞ are different for all v 2 V. Theorem 1. For every convex polytope Dn it follows that wðDn Þ ¼ 3. Proof. Since wðGÞ P bðGÞ for every graph G and it has been proved in [19] that bðDn Þ ¼ 3, it follows that wðDn Þ P 3. We only need 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 total enumeration that the sets in Table 1 are doubly resolving sets of Dn . For n P 14 we will prove that
D¼
fa0 ; ak ; d0 g; n ¼ 2k; fa0 ; ak ; d3 g; n ¼ 2k þ 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 ; d0 g. 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 .
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J. Kratica et al. / Applied Mathematics and Computation 218 (2012) 9790–9801 Table 1 Minimal doubly resolving sets of Dn . n
D
7 9 11 13 n ¼ 2k n ¼ 2k þ 1, n R f7; 9; 11; 13g
{a0 ; a3 ; d1 } {a0 ; a4 ; d6 } {a0 ; a5 ; d2 } {a0 ; a6 ; d9 } fa0 ; ak ; d0 g fa0 ; ak ; d3 g
8 i ¼ 0; > < ð0; k; 3Þ; rðai ; DÞ ¼ ði; k i; i þ 2Þ; 1 6 i 6 k; > : ð2k i; i k; 2k i þ 3Þ; k þ 1 6 i 6 2k 1; 8 i ¼ 0; > < ð1; k þ 1; 2Þ; rðbi ; DÞ ¼ ði þ 1; k i þ 1; i þ 1Þ; 1 6 i 6 k; > : ð2k i þ 1; i k þ 1; 2k i þ 2Þ; k þ 1 6 i 6 2k 1; ði þ 2; k i þ 1; i þ 1Þ; 0 6 i 6 k 1; rðci ; DÞ ¼ ð2k i þ 1; i k þ 2; 2k i þ 1Þ; k 6 i 6 2k 1; ði þ 3; k i þ 2; iÞ; 0 6 i 6 k 1; rðdi ; DÞ ¼ ð2k i þ 2; i k þ 3; 2k iÞ; k 6 i 6 2k 1; Then the representations r0 are:
8 i ¼ 0; > < ðk; 3Þ; r 0 ðai ; DÞ ¼ ðk 2i; 2Þ; 1 6 i 6 k; > : ð2i 3k; 3Þ; k þ 1 6 i 6 2k 1; 8 i ¼ 0; > < ðk; 1Þ; r 0 ðbi ; DÞ ¼ ðk 2i; 0Þ; 1 6 i 6 k; > : ð2i 3k; 1Þ; k þ 1 6 i 6 2k 1; ðk 2i 1; 1Þ; 0 6 i 6 k 1; r 0 ðci ; DÞ ¼ ð2i 3k þ 1; 0Þ; k 6 i 6 2k 1; ðk 2i 1; 3Þ; 0 6 i 6 k 1; r 0 ðdi ; DÞ ¼ ð2i 3k þ 1; 2Þ; k 6 i 6 2k 1: Case 2: n is odd. In this case, we can write n ¼ 2k þ 1; k P 3. Let D ¼ fa0 ; ak ; d3 g. The representations of the vertices of Dn are:
8 ði; k i; 6 iÞ; 0 6 i 6 3; > > > > < ði; k i; i 1Þ; 4 6 i 6 k; rðai ; DÞ ¼ > k þ 1 6 i 6 k þ 4; > ð2k i þ 1; i k; i 1Þ; > > : ð2k i þ 1; i k; 2k i þ 7Þ; k þ 5 6 i 6 2k; 8 ði þ 1; k i þ 1; 5 iÞ; 0 6 i 6 3; > > > > < ði þ 1; k i þ 1; i 2Þ; 4 6 i 6 k; rðbi ; DÞ ¼ > ð2k i þ 2; i k þ 1; i 2Þ; k þ 1 6 i 6 k þ 4; > > > : ð2k i þ 2; i k þ 1; 2k i þ 6Þ; k þ 5 6 i 6 2k; 8 ði þ 2; k i þ 1; 4 iÞ; 0 6 i 6 3; > > > > < ði þ 2; k i þ 1; i 2Þ; 4 6 i 6 k 1; rðci ; DÞ ¼ > ð2k i þ 2; i k þ 2; i 2Þ; k 6 i 6 k þ 3; > > > : ð2k i þ 2; i k þ 2; 2k i þ 5Þ; k þ 4 6 i 6 2k; 8 ði þ 3; k i þ 2; 3 iÞ; 0 6 i 6 3; > > > > < ði þ 3; k i þ 2; i 3Þ; 4 6 i 6 k 1; rðdi ; DÞ ¼ > ð2k i þ 3; i k þ 3; i 3Þ; k 6 i 6 k þ 3; > > > : ð2k i þ 3; i k þ 3; 2k i þ 4Þ; k þ 4 6 i 6 2k:
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Then the representations r 0 are:
8 ðk 2i; 6 2iÞ; 0 6 i 6 3; > > > < ðk 2i; 1Þ; 4 6 i 6 k; r 0 ðai ; DÞ ¼ > ð2i 3k 1; 2i 2k 2Þ; k þ 1 6 i 6 k þ 4; > > : ð2i 3k 1; 6Þ; k þ 5 6 i 6 2k; 8 ðk 2i; 4 2iÞ; 0 6 i 6 3; > > > < ðk 2i; 3Þ; 4 6 i 6 k; r 0 ðbi ; DÞ ¼ > ð2i 3k 1; 2i 2k 4Þ; k þ 1 6 i 6 k þ 4; > > : ð2i 3k 1; 4Þ; k þ 5 6 i 6 2k; 8 ðk 2i 1; 2 2iÞ; 0 6 i 6 3; > > > < ðk 2i 1; 4Þ; 4 6 i 6 k 1; r 0 ðci ; DÞ ¼ > ð2i 3k; 2i 2k 4Þ; k 6 i 6 k þ 3; > > : ð2i 3k; 3Þ; k þ 4 6 i 6 2k; 8 ðk 2i 1; 2iÞ; 0 6 i 6 3; > > > < ðk 2i 1; 6Þ; 4 6 i 6 k 1; r 0 ðdi ; DÞ ¼ > ð2i 3k; 2i 2k 6Þ; k 6 i 6 k þ 3; > > : ð2i 3k; 1Þ; k þ 4 6 i 6 2k: 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 sdimðDn Þ ¼ 2n for n odd and n P 5, and sdimðDn Þ ¼ 5n=2 for n even and n P 10. Next we prove several lemmas which support the proof of Theorem 2. Lemma 2. For n ¼ 2k þ 1 and n P 5, if S is a strong resolving set of Dn , then jSj P 2n. Proof. Let us consider pairs of vertices ðai ; dkþi Þ; ðbi ; ckþi Þ for all i ¼ 0; 1; . . . ; n 1. It is easy to see that dðai ; dkþi Þ ¼ dðbi ; ckþi Þ ¼ k þ 3. As DiamðDn Þ ¼ k þ 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 jSj P 2n. h Lemma 3. For n ¼ 2k þ 1 and n P 5, set S ¼ fai ; ci ji 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 u and v. For each pair of vertices from V n S, the corresponding vertex of S which strongly resolves this pair is given in Table 3. Let us prove that, for each i; j 2 f0; 1; . . . ; n 1g such that i þ 1 6 j 6 k þ i, vertices bi and bj are strongly resolved by vertex cj . Really, it is easy to check that dðbi ; ciþ1 Þ ¼ 3 and dðbi ; cj Þ ¼ j i þ 3 for i þ 2 6 j 6 k þ i. Also, dðbi ; biþ1 Þ ¼ 2 and dðbi ; bj Þ ¼ j i þ 2 for i þ 2 6 j 6 k þ i, and, consequently, dðbi ; cj Þ ¼ dðbi ; bj Þ þ 1 for i þ 1 6 j 6 k þ i. As dðbj ; cj Þ ¼ 1, it means that vertex bj belongs to a shortest path between bi and cj . Therefore, cj strongly resolves bi and bj for i þ 1 6 j 6 k þ i. Similarly, for k þ i þ 1 6 j 6 n 1 it can be proved that bi and bj are strongly resolved by vertex cj1 . In this case, dðbi ; cnþi2 Þ ¼ 3; dðbi ; cj1 Þ ¼ n j þ i þ 3 for i þ k þ 1 6 j 6 n þ i 2; dðbi ; bnþi1 Þ ¼ 2 and dðbi ; bj Þ ¼ n j þ i þ 2 for i þ k þ 1 6 j 6 n þ i 2. Since dðbi ; cj1 Þ ¼ dðbi ; bj Þ þ 1 for i þ k þ 1 6 j 6 n 1, and cj1 is adjacent to v j , then bj belongs to a shortest path between bi and cj1 , i.e. cj1 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 jSj P 5n=2. Proof. Since dðai ; dkþi Þ ¼ k þ 2 ¼ DiamðDn Þ, according to Property 2, ai 2 S or dkþi 2 S for i ¼ 0; 1; . . . ; n 1, which means that S should contain at least n a-vertices or d-vertices. Also, dðbi ; bkþi Þ ¼ dðci ; ckþi Þ ¼ k þ 2 ¼ DiamðDn Þ 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 , where j R fi 2; i 1; i; i þ 1g satisfy conditions: (a) dðu; cj Þ 6 dðbi ; cj Þ for all vertices u adjacent to bi , and (b) dðv ; bi Þ 6 dðbi ; cj Þ for all vertices v adjacent to cj .
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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 ; cj Þ ¼ j þ 3 for 2 6 j 6 k 1 and dðb0 ; cj Þ ¼ n j þ 2 for k 6 j 6 n 3. Let us prove the condition (b). Vertices adjacent to vertex cj are bj ; bjþ1 and dj . For 2 6 j 6 k 1 the corresponding distances are dðbj ; b0 Þ ¼ dðdj ; b0 Þ ¼ j þ 2 ¼ dðb0 ; cj Þ 1; dðbjþ1 ; b0 Þ ¼ j þ 3 ¼ dðb0 ; cj Þ and, consequently, the condition (b) is satisfied. If k 6 j 6 n 3 then dðbjþ1 ; b0 Þ ¼ dðdj ; b0 Þ ¼ n j þ 1 ¼ dðb0 ; cj Þ 1; dðbj ; b0 Þ ¼ n j þ 2 ¼ dðb0 ; cj Þ, and (b) is satisfied. Due to the symmetry of b-vertices and cvertices, 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 ; dkþi Þ and at least k c-vertices, and, therefore, jSj P 2n þ 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 jSj P 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; i þ 1g, and all bt ; t R fj 1; j; j þ 1; j þ 2g. As S should also contain n vertices from pairs fai ; dkþi g, then jSj P 3n 8. Since n P 16; 3n 8 P 5n=2, and jSj P 5n=2. h Lemma 5. For n ¼ 2k and n P 16, set S ¼ fbi ; cj ; dj ji 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 each i 2 f0; 1; . . . ; k 1g and each j 2 fi þ 1; i þ 2; . . . ; n 1g vertex bi strongly resolves vertices ai and aj . Really, it is easy to check that dðbi ; aj Þ ¼ j i þ 1 and dðai ; aj Þ ¼ j i for i þ 1 6 j 6 i þ k, and dðbi ; aj Þ ¼ n ðj iÞ þ 1 and dðai ; aj Þ ¼ n ðj iÞ for i þ k þ 1 6 j 6 n 1. Since dðbi ; ai Þ ¼ 1, then dðbi ; aj Þ ¼ dðbi ; ai Þ þ dðai ; aj Þ for i þ 1 6 j 6 n 1, i.e. vertex ai belongs to a shortest path between bi and aj and, therefore, bi strongly resolves ai and aj . When i 2 fk; k þ 1; . . . ; n 2g and j 2 fi þ 1; i þ 2; . . . ; n 1g, then dðai ; bkþi Þ ¼ k þ 1; dðai ; aj Þ ¼ j i and dðaj ; bkþi Þ ¼ k þ i j þ 1. As dðai ; bkþi Þ ¼ dðai ; aj Þ þ dðaj ; bkþi Þ, then aj belongs to a shortest path between ai and bkþi , i.e. bkþi strongly resolves ai and aj . In a similar way all other vertices w from Table 4 can be verified. h
Table 2 The strong metric dimension of Dn . n
sdim Dn
Strong metric basis
6 8 n ¼ 2k; n R f6; 8g n ¼ 2k þ 1
13 19 5n=2 2n
{b3 ; b4 ; b5 ; c2 ; c3 ; c4 ; c5 ; d0 ; d1 ; d2 ; d3 ; d4 ; d5 } {b4 ; b5 ; b6 ; b7 ; c0 ; c2 ; c3 ; c4 ; c5 ; c6 ; c7 ; di ; ði ¼ 0; 1; . . . ; 7Þ} fbi ; cj ; dj ji ¼ 0; 1; . . . ; k 1; j ¼ 0; 1; . . . ; n 1g {ai ; ci ji ¼ 0; 1; . . . ; 2kg
Table 3 Cases of Lemma 3. u,
v
bi ; bj 06i
w cj , j 6 k þ i cj1 , j P k þ i þ 1 ai ci
Table 4 Cases of Lemma 5. u; v
w
ai ; aj 06i
bi ,i 6 k 1 bkþi , i P k bjk cj
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Now we can give a proof of Theorem 2: Proof. It follows directly from Lemmas 2 and 3 that for n ¼ 2k þ 1; sdimðDn Þ ¼ 2n. Also, according to Lemmas 4 and 5, for n ¼ 2k and n P 16; sdimðDn Þ ¼ 5n=2. In order to determine sdimðDn Þ for n ¼ 10; 12; 14, we used the integer programming model (3)–(5), defined in Section 2. In all cases, CPLEX has obtained the strong metric basis with the cardinality 5n=2 and of the same type as set S in Lemma 5. Therefore, sdimðDn Þ 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. sdimðDn Þ < 5n=2 (see Table 2). 4. Convex polytopes T n The graph of convex polytope T n ; n P 5 (Fig. 3), introduced in [21], consists of 4n 3-sided faces, n 4-sided faces and a pair of n sided faces, and it is obtained by the combination of the graph of convex polytope Rn [25] and the graph of an antiprism An [24]. We have
VðT n Þ ¼ fai ; bi ; ci ; di ji ¼ 0; 1; . . . ; n 1g;
EðT n Þ ¼ fai aiþ1 ; bi biþ1 ; ci ciþ1 ; di diþ1 ; ai bi ; bi ci ; ci di ; aiþ1 bi ; ciþ1 di j i ¼ 0; 1; . . . ; n 1g;
where indices are taken modulo n. It has been proved in [21] that the metric dimension of T n 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 T n ; n > 7. On the other hand, we prove that for n P 5 the strong metric dimension of T n is equal to 2n for n odd, and 5n=2 for n even. 4.1. Minimal doubly resolving sets of T n Theorem 3. For every convex polytope T n it follows
wðT n Þ ¼
3; n – 7; 4; n ¼ 7:
Proof. Since wðGÞ P bðGÞ for every graph G and in [21] is proved that bðT n Þ ¼ 3 it follows that wðT n Þ P 3. In order to prove the assertion of the theorem, we only need to prove that for n P 9 there exists a doubly resolving set of T n 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 T n .
Table 5 Minimal doubly resolving sets of T n . n
D
5 7 8 n ¼ 2k, n – 8 n ¼ 2k þ 1, n R f5; 7g
{a0 ; a2 ; c3 } {a0 ; a1 ; a3 ; c4 } {a0 ; a4 ; d1 } fa0 ; bk1 ; d2k2 g fa0 ; ak ; ckþ2 g
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D¼
fa0 ; bk1 ; d2k2 g; n ¼ 2k; n ¼ 2k þ 1 fa0 ; ak ; ckþ2 g;
is a doubly resolving set of T n ; 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 ; bk1 ; d2k2 g. We will show that D is a doubly resolving set of T n . To this end, we give representations of all vertices of T n :
8 0 6 i 6 k 1; > < ði; k i; i þ 3Þ; rðai ; DÞ ¼ ð2k i; i k þ 1; 2k i þ 1Þ; k 6 i 6 2k 2; > : ð1; k; 3Þ; i ¼ 2k 1; 8 ði þ 1; k i 1; i þ 3Þ; 0 6 i 6 k 2; > > > > < ðk; 0; k þ 1Þ; i ¼ k 1; rðbi ; DÞ ¼ > ð2k i; i k þ 1; 2k iÞ; k 6 i 6 2k 2; > > > : ð1; k; 2Þ; i ¼ 2k 1; 8 ði þ 2; k i; i þ 2Þ; 0 6 i 6 k 2; > > > > < ðk þ 1; 1; kÞ; i ¼ k 1; rðci ; DÞ ¼ > ð2k i þ 1; i k þ 2; 2k i 1Þ; k 6 i 6 2k 2; > > > : ð2; k þ 1; 1Þ; i ¼ 2k 1; 8 ði þ 3; k i; i þ 2Þ; 0 6 i 6 k 2; > > > > < ðk þ 2; 2; k 1Þ; i ¼ k 1; rðdi ; DÞ ¼ > ð2k i þ 1; i k þ 3; 2k i 2Þ; k 6 i 6 2k 2; > > > : ð3; k þ 1; 1Þ; i ¼ 2k 1: Then the representations r0 are:
8 0 6 i 6 k 1; > < ðk 2i; 3Þ; 0 r ðai ; DÞ ¼ ð2i 3k þ 1; 1Þ; k 6 i 6 2k 2; > : ðk 1; 2Þ; i ¼ 2k 1; 8 ðk 2i 2; 2Þ; 0 6 i 6 k 2; > > > > < ðk; 1Þ; i ¼ k 1; r 0 ðbi ; DÞ ¼ > ð2i 3k þ 1; 0Þ; k 6 i 6 2k 2; > > > : ðk 1; 1Þ; i ¼ 2k 1; 8 ðk 2i 2; 0Þ; 0 6 i 6 k 2; > > > > < ðk; 1Þ; i ¼ k 1; r 0 ðci ; DÞ ¼ > ð2i 3k þ 1; 2Þ; k 6 i 6 2k 2; > > > : ðk 1; 1Þ; i ¼ 2k 1; 8 ðk 2i 3; 1Þ; 0 6 i 6 k 2; > > > > < ðk; 3Þ; i ¼ k 1; r 0 ðdi ; DÞ ¼ > ð2i 3k þ 2; 3Þ; k 6 i 6 2k 2; > > > : ðk 2; 2Þ; i ¼ 2k 1: Case 2: n is odd. In this case, we can write n ¼ 2k þ 1; k P 3. Let D ¼ fa0 ; ak ; ckþ2 g. The representations of vertices of T n are:
8 ði; k i; k þ iÞ; > > > < ði; k i; k i þ 4Þ;
0 6 i 6 2;
3 6 i 6 k; > ð2k i þ 1; i k; k i þ 4Þ; k þ 1 6 i 6 k þ 2; > > : ð2k i þ 1; i k; i k 1Þ; k þ 3 6 i 6 2k; 8 ði þ 1; k i; k þ iÞ; 0 6 i 6 1; > > > < ði þ 1; k i; k i þ 3Þ; 2 6 i 6 k 1; rðbi ; DÞ ¼ > ð2k i þ 1; i k þ 1; k i þ 3Þ; k 6 i 6 k þ 2; > > : ð2k i þ 1; i k þ 1; i k 1Þ; k þ 3 6 i 6 2k; rðai ; DÞ ¼
J. Kratica et al. / Applied Mathematics and Computation 218 (2012) 9790–9801
8 ði þ 2; k i þ 1; k þ i 1Þ; > > > < ði þ 2; k i þ 1; k i þ 2Þ; rðci ; DÞ ¼ > ð2k i þ 2; i k þ 2; k i þ 2Þ; > > : ð2k i þ 2; i k þ 2; i k 2Þ; 8 ði þ 3; k i þ 1; k þ iÞ; > > > > > ði þ 3; k i þ 1; k i þ 2Þ; > > > < ðk þ 2; 3; 3Þ; rðdi ; DÞ ¼ > ð2k i þ 2; i k þ 3; k i þ 2Þ; > > > > > ð2k i þ 2; i k þ 3; i k 1Þ; > > : ð3; k þ 2; k 1Þ;
9799
0 6 i 6 1; 2 6 i 6 k 1; k 6 i 6 k þ 2; k þ 3 6 i 6 2k; 0 6 i 6 1; 2 6 i 6 k 2; i¼k1 k 6 i 6 k þ 1; k þ 2 6 i 6 2k 1; i ¼ 2k:
0
Then the representations r are:
8 ðk 2i; kÞ; 0 6 i 6 2; > > > < ðk 2i; k 2i þ 4Þ; 3 6 i 6 k; r 0 ðai ; DÞ ¼ > ð2i 3k 1; k þ 3Þ; k þ 1 6 i 6 k þ 2; > > : ð2i 3k 1; 2i 3k 2Þ; k þ 3 6 i 6 2k; 8 ðk 2i 1; k 1Þ; 0 6 i 6 1; > > > < ðk 2i 1; k 2i þ 2Þ; 2 6 i 6 k 1; r 0 ðbi ; DÞ ¼ > ð2i 3k; k þ 2Þ; k 6 i 6 k þ 2; > > : ð2i 3k; 2i 3k 2Þ; k þ 3 6 i 6 2k; 8 ðk 2i 1; k 3Þ; 0 6 i 6 1; > > > < ðk 2i 1; k 2iÞ; 2 6 i 6 k 1; r 0 ðci ; DÞ ¼ > ð2i 3k; kÞ; k 6 i 6 k þ 2; > > : ð2i 3k; 2i 3k 4Þ; k þ 3 6 i 6 2k; 8 ðk 2i 2; k 3Þ; 0 6 i 6 1; > > > > > ðk 2i 2; k 2i 1Þ; 2 6 i 6 k 2; > > > < ðk þ 1; k þ 1Þ; i ¼ k 1; r 0 ðdi ; DÞ ¼ > ð2i 3k þ 1; kÞ; k 6 i 6 k þ 1; > > > > > ð2i 3k þ 1; 2i 3k 3Þ; k þ 2 6 i 6 2k 1; > > : ðk 1; k 4Þ; i ¼ 2k: 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 T n
Theorem 4. For any convex polytope T n and n P 5 it follows that sdimðT n Þ ¼ 2n for n odd and sdimðT n Þ ¼ 5n=2 for n even. Next we prove several lemmas which support the proof of Theorem 4. Lemma 6. For n ¼ 2k þ 1 and n P 5, if S is a strong resolving set of T n , then jSj P 2n. Proof. As dðai ; ckþi Þ ¼ dðbi ; dkþi Þ ¼ k þ 2 for i ¼ 0; 1; . . . ; n 1, and DiamðT n Þ ¼ k þ 2, then, according to Property 2, S should contain at least n a-vertices or c-vertices, and at least n b-vertices or d-vertices. Therefore, jSj P 2n. h Lemma 7. For n ¼ 2k þ 1 and n P 5, set S ¼ fai ; di ji ¼ 0; 1; . . . ; n 1g is a strong resolving set of T n . Proof. Let us prove that vertex dkþi strongly resolves pairs ðbi ; bj Þ; ðbi ; cj Þ and ðci ; cj Þ for each i; j; i – j. It is easy to see that, for each i 2 f0; 1; . . . ; n 1g; dðbi ; dkþi Þ ¼ k þ 2 ¼ DiamðT n Þ, 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 .
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v
bi ; bj 06i
w ckþi
ajkþ1 dik
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, belongs to at least one of previous paths and, consequently, dkþi strongly resolves bi and cj . Similarly, dðci ; dkþi Þ ¼ k þ 1 and the corresponding 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 T n , then jSj P 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 to Property 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, it can be easily proved that dðai ; dkþi1 Þ ¼ k þ 2 ¼ DiamðT n Þ and pairs ðai ; akþi Þ; ðai ; di1 Þ; ðdi ; dkþi Þ satisfy conditions (i) and (ii) of Lemma 1. Therefore, according to Properties 1 and 2, (a) (b) (c) (d)
ai ai ai di
2S 2S 2S 2S
or or or or
dkþi1 2 S; akþi 2 S; di1 2 S; dkþi 2 S.
As a consequence of (a)–(d) it follows that S should contain at least 3 vertices from set X i ¼ fai ; akþi ; di1 ; dkþi1 g. For example, if akþi R S, then, according to (a), dkþðkþiÞ1 ¼ di1 2 S, and according to (b), ai 2 S, and according to (c), dðkþiÞ1 2 S. Since for i ¼ 0; 1; . . . ; n 1, the number of all distinct sets X i is equal to n=2, then S should contain at least 3n=2 a-vertices or d-vertices. As S should also contain at least n b-vertices or c-vertices, it means that jSj P 5n=2. h Lemma 9. For n ¼ 2k and n P 5, set S ¼ fai ; ci ; dj ji ¼ 0; 1; . . . ; n 1; j ¼ 0; 1; . . . ; k 1g is a strong resolving set of T n . 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 each i 2 f0; 1; . . . ; n 1g and each j 2 fk; k þ 1; . . . ; n 1g vertex ajkþ1 strongly resolves bi and dj . It is easy to check that, for each j 2 fk; k þ 1; . . . ; n 1g; dðajkþ1 ; dj Þ ¼ k þ 2 ¼ DiamðT n Þ with the corresponding shortest paths between ajkþ1 and dj : ajkþ1 ; bjkþ1 ; bjkþ2 ; . . . ; bj ; cj ; dj ; dj ; cjþ1 ; bjþ1 ; bjþ2 ; . . . ; bjþk ; ajþkþ1 ¼ ajkþ1 . As each vertex bi ; i – j, belongs to one of these paths, ajkþ1 strongly resolves bi and dj . All other results from Table 6 can be 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 2k þ 1; sdimðT n Þ ¼ 2n. Also, according to Lemmas 8 and 9, for n ¼ 2k; sdimðT n Þ 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 and T n . We have proved that the cardinality of minimal doubly resolving sets is constant and equal to 3, except for T 7 . In order to
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study the strong metric dimension, we have established some general properties of strong resolving sets and introduced an integer linear programming formulation of the strong metric dimension problem. These results have been used to obtain expressions for the strong metric dimension of Dn and T n , which depends on the dimension n. Future work can be directed towards obtaining minimal doubly resolving sets and/or the strong metric dimension of some other challenging classes of graphs. References [1] P.J. Slater, Leaves of trees, Congr. Numerantium 14 (1975) 549–559. [2] F. Harary, R.A. Melter, On the metric dimension of a graph, Ars Combinatoria 2 (1976) 191–195. [3] Z. Beerliova, F. Eberhard, T. Erlebach, A. Hall, M. Hoffmann, M. Mihalák, L.S. Ram, Network discovery and verification, IEEE J. Selected Areas Commun. 24 (12) (2006) 2168–2181. [4] J. Caceres, C. Hernando, M. Mora, I.M. Pelayo, M.L. Puertas, C. Seara, D.R. Wood, On the metric dimension of Cartesian products of graphs, SIAM J. Disc. Math. 21 (2) (2007) 423–441. [5] G. Chartrand, L. Eroha, M.A. Johnson, O.R. Oellermann, Resolvability in graphs and the metric dimension of a graph, Discr. Appl. Math. 105 (2000) 99– 113. [6] J.D. Currie, O.R. Oellerman, The metric dimension and metric independence of a graph, J. Comb. Math. Comb. Comput. 39 (2001) 157–167. [7] M. Fehr, S. Gosselin, O.R. Oellermann, The metric dimension of Cayley digraphs, Discr. Math. 306 (1) (2006) 31–41. [8] C. Hernando, M. Mora, I.M. Pelayo, C. Seara, J. Caceres, M.L. Puertas, On the metric dimension of some families of graphs, Electr. Notes Discr. Math. 22 (2005) 129–133. [9] K. Liu, N. Abu-Ghazaleh, Virtual coordinate backtracking for void traversal in geographic routing, Lecture Notes Comput. Sci. 4104 (2006) 46–59. [10] A. Sebö, E. Tannier, On metric generators of graphs, Math. Oper. Res. 29 (2) (2004) 383–393. [11] O.R. Oellermann, J. Peters-Fransen, The strong metric dimension of graphs and digraphs, Discr. Appl. Math. 155 (2007) 356–364. [12] T. May, O.R. Oellermann, The strong dimension of distance hereditary graphs, J. Comb. Math. Comb. Comput. 76 (2011) 59–73. [13] S. Khuller, B. Raghavachari, A. Rosenfeld, Landmarks in graphs, Discr. Appl. Math. 70 (1996) 217–229. ˇ angalovic´, V. Kovacˇevic´-Vujcˇic´, Computing minimal doubly resolving sets of graphs, Comput. Oper. Res. 36 (7) (2009) 2149–2159. [14] J. Kratica, M. C [15] J. Kratica, V. Kovacˇevic´-Vujcˇic´, M. Cˇangalovic´, Computing strong metric dimension of some special classes of graphs by genetic algorithms, Yugoslav J. Oper. Res. 18 (2) (2008) 43–51. ˇ angalovic´, Computing the metric dimension of graphs by genetic algorithms, Comput. Optim. Appl. 44 (2009) 343– [16] J. Kratica, V. Kovacˇevic´-Vujcˇic´, M. C 361. [17] P.S. Buczkowski, G. Chartrand, C. Poisson, P. Zhang, On k-dimensional graphs and their bases, Periodica Math. Hung. 46 (1) (2003) 9–15. [18] J. Caceres, C. Hernando, M. Mora, I.M. Pelayo, M.L. Puertas, C. Seara, D.R. Wood, On the metric dimension of some families of graphs, Electr. Notes Disc. Math. 22 (2005) 129–133. [19] M. Imran, A.Q. Baig, A. Ahmad, Families of plane graphs with constant metric dimension, Utilitas Math., in press. [20] M. Imran, A.Q. Baig, M.K. Shafiq, A. Semanicˇová-Fenˇovcˇíková, Classes of convex polytopes with constant metric dimension, Utilitas Math., in press. [21] M. Imran, S. Ahtsham, U.H. Bokhary, A.Q. Baig, On families of convex polytopes with constant metric dimension, Comput. Math. Appl. 60 (9) (2010) 2629–2638. [22] I. Javaid, M.T. Rahim, K. Ali, Families of regular graphs with constant metric dimension, Utilitas Math. 75 (2008) 21–33. [23] I. Tomescu, I. Javaid, On the metric dimension of the Jahangir graph, Bull. Math. de la Soc. Sci. Math. de Roum. 50 (98) (2007) 371–376. [24] M. Bacˇa, Labellings of two classes of convex polytopes, Utilitas Math. 34 (1988) 24–31. [25] M. Bacˇa, On magic labellings of convex polytopes, Ann. Disc. Math. 51 (1992) 13–16.
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Minimal doubly resolving sets and the strong metric dimension of some convex polytopes q Jozef Kratica a,⇑, Vera Kovacˇevic´-Vujcˇic´ b, Mirjana Cˇangalovic´ b, Milica Stojanovic´ b a b
Mathematical Institute, Serbian Academy of Sciences and Arts, Kneza Mihaila 36/III, 11000 Belgrade, Serbia Faculty of Organizational Sciences, University of Belgrade, Jove Ilic´a 154, 11000 Belgrade, Serbia
a r t i c l e
i n f o
Keywords: Minimal doubly resolving set Strong metric dimension Convex polytopes
a b s t r a c t In this paper we consider two similar optimization problems on graphs: the strong metric dimension problem and the problem of determining minimal doubly resolving sets. We prove some properties of strong resolving sets and give an integer linear programming formulation of the strong metric dimension problem. These results are used to derive explicit expressions in terms of the dimension n, for the strong metric dimension of two classes of convex polytopes Dn and T n . On the other hand, we prove that minimal doubly resolving sets of Dn and T n have constant cardinality for n > 7. Ó 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 widely investigated [3–8]. It arises in many diverse areas including network discovery and verification [3], geographical routing protocols [9], the robot navigation, connected joints in graphs, chemistry, etc. Given a simple connected undirected graph G ¼ ðV; EÞ, where V ¼ f1; 2; . . . ; ng; jEj ¼ m; dðu; v Þ denotes the distance between vertices u and v, i.e. the length of a shortest u 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 set B ¼ fx1 ; x2 ; . . . ; xk g of G is a resolving set of G if every two distinct vertices of G are resolved by some vertex of B. Given a vertex t, the k-tuple rðt; BÞ ¼ ðdðt; x1 Þ; dðt; x2 Þ; . . . ; dðt; xk ÞÞ is called the vector of metric coordinates of t with respect to B. A metric basis of G is a resolving set of the minimum cardinality. The metric dimension of G, denoted by bðGÞ, is the cardinality of its metric basis. 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 doubly resolve vertices u, v of G if dðu; xÞ dðu; yÞ – dðv ; xÞ dðv ; yÞ. A vertex set D ¼ fx1 ; x2 ; . . . ; xl g of G is a doubly resolving set of G if every two distinct vertices of G are doubly resolved by some two vertices of D. The minimal doubly resolving set problem consists of finding a doubly resolving set of G with the minimum cardinality, denoted by wðGÞ. Note that if x, y doubly resolve u, v then dðu; 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 resolving set and bðGÞ 6 wðGÞ. The strong metric dimension problem (SMDP) was introduced by Sebö and Tannier [10] and further investigated by Oellermann and Peters-Fransen [11]. Recently, the strong metric dimension of distance hereditary graphs has been studied by May and Oellermann [12]. A vertex w strongly resolves two vertices u and v if u belongs to a shortest v w path or v belongs to 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 resolved by some vertex of S. A strong metric basis of G is a strong resolving set of the minimum cardinality. Now, the strong metric q
This research was partially supported by Serbian Ministry of Education and Science under the grant no. 174033.
⇑ Corresponding author.
E-mail addresses: [email protected] (J. Kratica), [email protected] (V. Kovacˇevic´-Vujcˇic´), [email protected] (M. Cˇangalovic´), [email protected] (M. Stojanovic´). 0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.03.047
J. Kratica et al. / Applied Mathematics and Computation 218 (2012) 9790–9801
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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 w strongly resolves vertices u and v then w also resolves these vertices. Hence every strong resolving set is a resolving set and bðGÞ 6 sdimðGÞ. All three previously defined problems are NP-hard in general case. The proofs of NP-hardness are given for the metric dimension problem in [13], for the minimal doubly resolving set problem in [14] and for the strong metric dimension problem 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 a graph with VðG0 Þ ¼ VðGÞ such that S strongly resolves G0 , and if for all vertices v we have rG ðv ; SÞ ¼ r G0 ðv ; SÞ, then G ¼ G0 . If S is a resolving set or a doubly resolving set, then set S need not uniquely determine G. The next example illustrates previously defined invariants bðGÞ; wðGÞ and sdimðGÞ. Example 1. Consider graph G1 of Fig. 1. It is easy to check that bðG1 Þ; wðG1 Þ and sdimðG1 Þ are all different. Namely, set fA; Bg is a minimal resolving set, set fA; B; E; Fg is a minimal doubly resolving set, while fA; B; Eg is a minimal strong resolving set of G1 . Therefore, bðG1 Þ ¼ 2; wðG1 Þ ¼ 4 and sdimðG1 Þ ¼ 3. For some simple classes of graphs it is possible to determine bðGÞ; wðGÞ and sdimðGÞ explicitly: the complete graph K n with n vertices has bðK n Þ ¼ wðK n Þ ¼ sdimðK n Þ ¼ n 1, the cycle C n with n vertices has bðC n Þ ¼ 2; wðC n Þ ¼ 3, while sdimðC n Þ ¼ dn=2e. 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 graphs Sn ; T n and U n , which are combinations of two graphs of convex polytopes. It has been proved that all these classes have the metric dimension equal to 3. In this paper we study the minimal cardinality of doubly resolving sets and the strong metric dimension of one representative of each group: Dn and T n . We prove that wðDn Þ ¼ 3 and
wðT n Þ ¼
3; n – 7; 4; n ¼ 7;
while strong metric dimension depends of the dimension n. Due to different structures of Dn and T n , the proofs are quite different. The paper is organized as follows. In Section 2 we prove some theoretical properties of strong resolving sets of an arbitrary graph and give an integer linear programming formulation of the strong metric dimension problem. In Sections 3 and 4 we get explicit expressions for wðDn Þ; sdimðDn Þ; wðT n Þ and sdimðT n Þ. 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; v Þ 6 dðu; v Þ for each w such that fw; ug 2 E and (ii) dðu; wÞ 6 dðu; v Þ 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 P 1 between a and v containing u, or (b) there exists a shortest path P 2 between a and u containing v.
Fig. 1. Graph G1 .
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If (a) is fulfilled, then denote by u0 a vertex of path P 1 which is adjacent to u and belongs to its part between a and u. As each part of P1 represents also a shortest path between the corresponding vertices of P1 , then, consequently, dðu0 ; v Þ ¼ 1 þ dðu; v Þ. But, since vertex u0 satisfies condition (i), it follows that dðu0 ; v Þ 6 dðu; v Þ, which is a contradiction. Starting from (b), in a similar way we can also reach a contradiction. Consequently, vertex a cannot strongly resolve vertices 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; v Þ dðw; v Þj 6 1 that dðw; v Þ ¼ 1 þ dðu; v Þ. It means that u belongs to a shortest path between w and v, and, therefore, w strongly resolves u and v. In the case when condition (ii) is not satisfied for a vertex w adjacent to v, using similar considerations, we also conclude 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) of Lemma 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 S which 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; v Þ ¼ DiamðGÞ; u 2 S or
v 2 S. Proof. If dðu; v Þ ¼ DiamðGÞ, then, obviously, u and h
v
satisfy conditions (i) and (ii) and, according to Property 1, u 2 S or
v 2 S.
2.2. Integer linear programming formulation In the literature there exist two integer linear programming (ILP) formulations of the metric dimension problem [5,6] and one ILP formulation of the minimal doubly resolving set problem [14]. To our knowledge, the ILP formulation (3)–(5) of the strong 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 length dðu; v Þ of a shortest u v path for all u; v 2 V, using any shortest path algorithm. The coefficient matrix A is defined as follows:
8 > < 1; dðu; iÞ ¼ dðu; v Þ þ dðv ; iÞ; Aðu;v Þ;i ¼ 1; dðv ; iÞ ¼ dðv ; uÞ þ dðu; iÞ; > : 0; otherwise;
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 ¼
1; i 2 S; 0;
ð2Þ
i R S:
The ILP model of the strong metric dimension problem can now be formulated as:
min
n P
yi
ð3Þ
i¼1
subject to: n P
Aðu;v Þ;i yi P 1;
1 6 u < v 6 n;
ð4Þ
i¼1
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 that dðu; iÞ ¼ dðu; v Þ þ dðv ; iÞ or dðv ; iÞ ¼ dðv ; uÞ þ dðu; iÞ. It follows that Aðu;v Þ;i ¼ 1, and consequently constraints (4) are satisfied. Constraints (5) are obviously satisfied by definition of variables yi .
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(() 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 exists i 2 f1; 2; . . . ; ng, such that Aðu;v Þ;i yi P 1, which implies yi ¼ 1 (i.e. i 2 S) and Aðu;v Þ;i ¼ 1 (i.e. dðu; iÞ ¼ dðu; v Þ þ dðv ; iÞ or dðv ; iÞ ¼ dðv ; uÞ þ dðu; iÞ). It follows that the set S is a strong resolving set of G. h n Note that ILP model (3)–(5) has only n variables and linear constraints. 2 Remark 2. According to Proposition 1, the objective function (3) represents the cardinality of a strong resolving set, and therefore 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
VðDn Þ ¼ fai ; bi ; ci ; di ji ¼ 0; 1; . . . ; n 1g; EðDn Þ ¼ fai aiþ1 ; di diþ1 ; ai bi ; bi ci ; ci di ; biþ1 ci ji 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, and similarly 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 the same holds for the cardinality of minimal doubly resolving sets of Dn . On the other hand, we prove that for n P 10 the strong metric 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 ; . . . ; xk g # V is a doubly resolving set of G if and only if for every p; q 2 V there exists i 2 f2; 3; . . . ; kg such that
dðp; xi Þ dðp; x1 Þ – dðq; xi Þ dðq; x1 Þ:
ð6Þ
For each v 2 V let r0 ðv ; DÞ ¼ ðdðv ; x2 Þ dðv ; x1 Þ; . . . ; dðv ; xk Þ dðv ; x1 ÞÞ. According to Proposition 2, D is a doubly resolving set if and only if vectors r0 ðv ; DÞ are different for all v 2 V. Theorem 1. For every convex polytope Dn it follows that wðDn Þ ¼ 3. Proof. Since wðGÞ P bðGÞ for every graph G and it has been proved in [19] that bðDn Þ ¼ 3, it follows that wðDn Þ P 3. We only need 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 total enumeration that the sets in Table 1 are doubly resolving sets of Dn . For n P 14 we will prove that
D¼
fa0 ; ak ; d0 g; n ¼ 2k; fa0 ; ak ; d3 g; n ¼ 2k þ 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 ; d0 g. 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 .
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J. Kratica et al. / Applied Mathematics and Computation 218 (2012) 9790–9801 Table 1 Minimal doubly resolving sets of Dn . n
D
7 9 11 13 n ¼ 2k n ¼ 2k þ 1, n R f7; 9; 11; 13g
{a0 ; a3 ; d1 } {a0 ; a4 ; d6 } {a0 ; a5 ; d2 } {a0 ; a6 ; d9 } fa0 ; ak ; d0 g fa0 ; ak ; d3 g
8 i ¼ 0; > < ð0; k; 3Þ; rðai ; DÞ ¼ ði; k i; i þ 2Þ; 1 6 i 6 k; > : ð2k i; i k; 2k i þ 3Þ; k þ 1 6 i 6 2k 1; 8 i ¼ 0; > < ð1; k þ 1; 2Þ; rðbi ; DÞ ¼ ði þ 1; k i þ 1; i þ 1Þ; 1 6 i 6 k; > : ð2k i þ 1; i k þ 1; 2k i þ 2Þ; k þ 1 6 i 6 2k 1; ði þ 2; k i þ 1; i þ 1Þ; 0 6 i 6 k 1; rðci ; DÞ ¼ ð2k i þ 1; i k þ 2; 2k i þ 1Þ; k 6 i 6 2k 1; ði þ 3; k i þ 2; iÞ; 0 6 i 6 k 1; rðdi ; DÞ ¼ ð2k i þ 2; i k þ 3; 2k iÞ; k 6 i 6 2k 1; Then the representations r0 are:
8 i ¼ 0; > < ðk; 3Þ; r 0 ðai ; DÞ ¼ ðk 2i; 2Þ; 1 6 i 6 k; > : ð2i 3k; 3Þ; k þ 1 6 i 6 2k 1; 8 i ¼ 0; > < ðk; 1Þ; r 0 ðbi ; DÞ ¼ ðk 2i; 0Þ; 1 6 i 6 k; > : ð2i 3k; 1Þ; k þ 1 6 i 6 2k 1; ðk 2i 1; 1Þ; 0 6 i 6 k 1; r 0 ðci ; DÞ ¼ ð2i 3k þ 1; 0Þ; k 6 i 6 2k 1; ðk 2i 1; 3Þ; 0 6 i 6 k 1; r 0 ðdi ; DÞ ¼ ð2i 3k þ 1; 2Þ; k 6 i 6 2k 1: Case 2: n is odd. In this case, we can write n ¼ 2k þ 1; k P 3. Let D ¼ fa0 ; ak ; d3 g. The representations of the vertices of Dn are:
8 ði; k i; 6 iÞ; 0 6 i 6 3; > > > > < ði; k i; i 1Þ; 4 6 i 6 k; rðai ; DÞ ¼ > k þ 1 6 i 6 k þ 4; > ð2k i þ 1; i k; i 1Þ; > > : ð2k i þ 1; i k; 2k i þ 7Þ; k þ 5 6 i 6 2k; 8 ði þ 1; k i þ 1; 5 iÞ; 0 6 i 6 3; > > > > < ði þ 1; k i þ 1; i 2Þ; 4 6 i 6 k; rðbi ; DÞ ¼ > ð2k i þ 2; i k þ 1; i 2Þ; k þ 1 6 i 6 k þ 4; > > > : ð2k i þ 2; i k þ 1; 2k i þ 6Þ; k þ 5 6 i 6 2k; 8 ði þ 2; k i þ 1; 4 iÞ; 0 6 i 6 3; > > > > < ði þ 2; k i þ 1; i 2Þ; 4 6 i 6 k 1; rðci ; DÞ ¼ > ð2k i þ 2; i k þ 2; i 2Þ; k 6 i 6 k þ 3; > > > : ð2k i þ 2; i k þ 2; 2k i þ 5Þ; k þ 4 6 i 6 2k; 8 ði þ 3; k i þ 2; 3 iÞ; 0 6 i 6 3; > > > > < ði þ 3; k i þ 2; i 3Þ; 4 6 i 6 k 1; rðdi ; DÞ ¼ > ð2k i þ 3; i k þ 3; i 3Þ; k 6 i 6 k þ 3; > > > : ð2k i þ 3; i k þ 3; 2k i þ 4Þ; k þ 4 6 i 6 2k:
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Then the representations r 0 are:
8 ðk 2i; 6 2iÞ; 0 6 i 6 3; > > > < ðk 2i; 1Þ; 4 6 i 6 k; r 0 ðai ; DÞ ¼ > ð2i 3k 1; 2i 2k 2Þ; k þ 1 6 i 6 k þ 4; > > : ð2i 3k 1; 6Þ; k þ 5 6 i 6 2k; 8 ðk 2i; 4 2iÞ; 0 6 i 6 3; > > > < ðk 2i; 3Þ; 4 6 i 6 k; r 0 ðbi ; DÞ ¼ > ð2i 3k 1; 2i 2k 4Þ; k þ 1 6 i 6 k þ 4; > > : ð2i 3k 1; 4Þ; k þ 5 6 i 6 2k; 8 ðk 2i 1; 2 2iÞ; 0 6 i 6 3; > > > < ðk 2i 1; 4Þ; 4 6 i 6 k 1; r 0 ðci ; DÞ ¼ > ð2i 3k; 2i 2k 4Þ; k 6 i 6 k þ 3; > > : ð2i 3k; 3Þ; k þ 4 6 i 6 2k; 8 ðk 2i 1; 2iÞ; 0 6 i 6 3; > > > < ðk 2i 1; 6Þ; 4 6 i 6 k 1; r 0 ðdi ; DÞ ¼ > ð2i 3k; 2i 2k 6Þ; k 6 i 6 k þ 3; > > : ð2i 3k; 1Þ; k þ 4 6 i 6 2k: 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 sdimðDn Þ ¼ 2n for n odd and n P 5, and sdimðDn Þ ¼ 5n=2 for n even and n P 10. Next we prove several lemmas which support the proof of Theorem 2. Lemma 2. For n ¼ 2k þ 1 and n P 5, if S is a strong resolving set of Dn , then jSj P 2n. Proof. Let us consider pairs of vertices ðai ; dkþi Þ; ðbi ; ckþi Þ for all i ¼ 0; 1; . . . ; n 1. It is easy to see that dðai ; dkþi Þ ¼ dðbi ; ckþi Þ ¼ k þ 3. As DiamðDn Þ ¼ k þ 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 jSj P 2n. h Lemma 3. For n ¼ 2k þ 1 and n P 5, set S ¼ fai ; ci ji 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 u and v. For each pair of vertices from V n S, the corresponding vertex of S which strongly resolves this pair is given in Table 3. Let us prove that, for each i; j 2 f0; 1; . . . ; n 1g such that i þ 1 6 j 6 k þ i, vertices bi and bj are strongly resolved by vertex cj . Really, it is easy to check that dðbi ; ciþ1 Þ ¼ 3 and dðbi ; cj Þ ¼ j i þ 3 for i þ 2 6 j 6 k þ i. Also, dðbi ; biþ1 Þ ¼ 2 and dðbi ; bj Þ ¼ j i þ 2 for i þ 2 6 j 6 k þ i, and, consequently, dðbi ; cj Þ ¼ dðbi ; bj Þ þ 1 for i þ 1 6 j 6 k þ i. As dðbj ; cj Þ ¼ 1, it means that vertex bj belongs to a shortest path between bi and cj . Therefore, cj strongly resolves bi and bj for i þ 1 6 j 6 k þ i. Similarly, for k þ i þ 1 6 j 6 n 1 it can be proved that bi and bj are strongly resolved by vertex cj1 . In this case, dðbi ; cnþi2 Þ ¼ 3; dðbi ; cj1 Þ ¼ n j þ i þ 3 for i þ k þ 1 6 j 6 n þ i 2; dðbi ; bnþi1 Þ ¼ 2 and dðbi ; bj Þ ¼ n j þ i þ 2 for i þ k þ 1 6 j 6 n þ i 2. Since dðbi ; cj1 Þ ¼ dðbi ; bj Þ þ 1 for i þ k þ 1 6 j 6 n 1, and cj1 is adjacent to v j , then bj belongs to a shortest path between bi and cj1 , i.e. cj1 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 jSj P 5n=2. Proof. Since dðai ; dkþi Þ ¼ k þ 2 ¼ DiamðDn Þ, according to Property 2, ai 2 S or dkþi 2 S for i ¼ 0; 1; . . . ; n 1, which means that S should contain at least n a-vertices or d-vertices. Also, dðbi ; bkþi Þ ¼ dðci ; ckþi Þ ¼ k þ 2 ¼ DiamðDn Þ 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 , where j R fi 2; i 1; i; i þ 1g satisfy conditions: (a) dðu; cj Þ 6 dðbi ; cj Þ for all vertices u adjacent to bi , and (b) dðv ; bi Þ 6 dðbi ; cj Þ for all vertices v adjacent to cj .
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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 ; cj Þ ¼ j þ 3 for 2 6 j 6 k 1 and dðb0 ; cj Þ ¼ n j þ 2 for k 6 j 6 n 3. Let us prove the condition (b). Vertices adjacent to vertex cj are bj ; bjþ1 and dj . For 2 6 j 6 k 1 the corresponding distances are dðbj ; b0 Þ ¼ dðdj ; b0 Þ ¼ j þ 2 ¼ dðb0 ; cj Þ 1; dðbjþ1 ; b0 Þ ¼ j þ 3 ¼ dðb0 ; cj Þ and, consequently, the condition (b) is satisfied. If k 6 j 6 n 3 then dðbjþ1 ; b0 Þ ¼ dðdj ; b0 Þ ¼ n j þ 1 ¼ dðb0 ; cj Þ 1; dðbj ; b0 Þ ¼ n j þ 2 ¼ dðb0 ; cj Þ, and (b) is satisfied. Due to the symmetry of b-vertices and cvertices, 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 ; dkþi Þ and at least k c-vertices, and, therefore, jSj P 2n þ 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 jSj P 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; i þ 1g, and all bt ; t R fj 1; j; j þ 1; j þ 2g. As S should also contain n vertices from pairs fai ; dkþi g, then jSj P 3n 8. Since n P 16; 3n 8 P 5n=2, and jSj P 5n=2. h Lemma 5. For n ¼ 2k and n P 16, set S ¼ fbi ; cj ; dj ji 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 each i 2 f0; 1; . . . ; k 1g and each j 2 fi þ 1; i þ 2; . . . ; n 1g vertex bi strongly resolves vertices ai and aj . Really, it is easy to check that dðbi ; aj Þ ¼ j i þ 1 and dðai ; aj Þ ¼ j i for i þ 1 6 j 6 i þ k, and dðbi ; aj Þ ¼ n ðj iÞ þ 1 and dðai ; aj Þ ¼ n ðj iÞ for i þ k þ 1 6 j 6 n 1. Since dðbi ; ai Þ ¼ 1, then dðbi ; aj Þ ¼ dðbi ; ai Þ þ dðai ; aj Þ for i þ 1 6 j 6 n 1, i.e. vertex ai belongs to a shortest path between bi and aj and, therefore, bi strongly resolves ai and aj . When i 2 fk; k þ 1; . . . ; n 2g and j 2 fi þ 1; i þ 2; . . . ; n 1g, then dðai ; bkþi Þ ¼ k þ 1; dðai ; aj Þ ¼ j i and dðaj ; bkþi Þ ¼ k þ i j þ 1. As dðai ; bkþi Þ ¼ dðai ; aj Þ þ dðaj ; bkþi Þ, then aj belongs to a shortest path between ai and bkþi , i.e. bkþi strongly resolves ai and aj . In a similar way all other vertices w from Table 4 can be verified. h
Table 2 The strong metric dimension of Dn . n
sdim Dn
Strong metric basis
6 8 n ¼ 2k; n R f6; 8g n ¼ 2k þ 1
13 19 5n=2 2n
{b3 ; b4 ; b5 ; c2 ; c3 ; c4 ; c5 ; d0 ; d1 ; d2 ; d3 ; d4 ; d5 } {b4 ; b5 ; b6 ; b7 ; c0 ; c2 ; c3 ; c4 ; c5 ; c6 ; c7 ; di ; ði ¼ 0; 1; . . . ; 7Þ} fbi ; cj ; dj ji ¼ 0; 1; . . . ; k 1; j ¼ 0; 1; . . . ; n 1g {ai ; ci ji ¼ 0; 1; . . . ; 2kg
Table 3 Cases of Lemma 3. u,
v
bi ; bj 06i
w cj , j 6 k þ i cj1 , j P k þ i þ 1 ai ci
Table 4 Cases of Lemma 5. u; v
w
ai ; aj 06i
bi ,i 6 k 1 bkþi , i P k bjk cj
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Now we can give a proof of Theorem 2: Proof. It follows directly from Lemmas 2 and 3 that for n ¼ 2k þ 1; sdimðDn Þ ¼ 2n. Also, according to Lemmas 4 and 5, for n ¼ 2k and n P 16; sdimðDn Þ ¼ 5n=2. In order to determine sdimðDn Þ for n ¼ 10; 12; 14, we used the integer programming model (3)–(5), defined in Section 2. In all cases, CPLEX has obtained the strong metric basis with the cardinality 5n=2 and of the same type as set S in Lemma 5. Therefore, sdimðDn Þ 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. sdimðDn Þ < 5n=2 (see Table 2). 4. Convex polytopes T n The graph of convex polytope T n ; n P 5 (Fig. 3), introduced in [21], consists of 4n 3-sided faces, n 4-sided faces and a pair of n sided faces, and it is obtained by the combination of the graph of convex polytope Rn [25] and the graph of an antiprism An [24]. We have
VðT n Þ ¼ fai ; bi ; ci ; di ji ¼ 0; 1; . . . ; n 1g;
EðT n Þ ¼ fai aiþ1 ; bi biþ1 ; ci ciþ1 ; di diþ1 ; ai bi ; bi ci ; ci di ; aiþ1 bi ; ciþ1 di j i ¼ 0; 1; . . . ; n 1g;
where indices are taken modulo n. It has been proved in [21] that the metric dimension of T n 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 T n ; n > 7. On the other hand, we prove that for n P 5 the strong metric dimension of T n is equal to 2n for n odd, and 5n=2 for n even. 4.1. Minimal doubly resolving sets of T n Theorem 3. For every convex polytope T n it follows
wðT n Þ ¼
3; n – 7; 4; n ¼ 7:
Proof. Since wðGÞ P bðGÞ for every graph G and in [21] is proved that bðT n Þ ¼ 3 it follows that wðT n Þ P 3. In order to prove the assertion of the theorem, we only need to prove that for n P 9 there exists a doubly resolving set of T n 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 T n .
Table 5 Minimal doubly resolving sets of T n . n
D
5 7 8 n ¼ 2k, n – 8 n ¼ 2k þ 1, n R f5; 7g
{a0 ; a2 ; c3 } {a0 ; a1 ; a3 ; c4 } {a0 ; a4 ; d1 } fa0 ; bk1 ; d2k2 g fa0 ; ak ; ckþ2 g
9798
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D¼
fa0 ; bk1 ; d2k2 g; n ¼ 2k; n ¼ 2k þ 1 fa0 ; ak ; ckþ2 g;
is a doubly resolving set of T n ; 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 ; bk1 ; d2k2 g. We will show that D is a doubly resolving set of T n . To this end, we give representations of all vertices of T n :
8 0 6 i 6 k 1; > < ði; k i; i þ 3Þ; rðai ; DÞ ¼ ð2k i; i k þ 1; 2k i þ 1Þ; k 6 i 6 2k 2; > : ð1; k; 3Þ; i ¼ 2k 1; 8 ði þ 1; k i 1; i þ 3Þ; 0 6 i 6 k 2; > > > > < ðk; 0; k þ 1Þ; i ¼ k 1; rðbi ; DÞ ¼ > ð2k i; i k þ 1; 2k iÞ; k 6 i 6 2k 2; > > > : ð1; k; 2Þ; i ¼ 2k 1; 8 ði þ 2; k i; i þ 2Þ; 0 6 i 6 k 2; > > > > < ðk þ 1; 1; kÞ; i ¼ k 1; rðci ; DÞ ¼ > ð2k i þ 1; i k þ 2; 2k i 1Þ; k 6 i 6 2k 2; > > > : ð2; k þ 1; 1Þ; i ¼ 2k 1; 8 ði þ 3; k i; i þ 2Þ; 0 6 i 6 k 2; > > > > < ðk þ 2; 2; k 1Þ; i ¼ k 1; rðdi ; DÞ ¼ > ð2k i þ 1; i k þ 3; 2k i 2Þ; k 6 i 6 2k 2; > > > : ð3; k þ 1; 1Þ; i ¼ 2k 1: Then the representations r0 are:
8 0 6 i 6 k 1; > < ðk 2i; 3Þ; 0 r ðai ; DÞ ¼ ð2i 3k þ 1; 1Þ; k 6 i 6 2k 2; > : ðk 1; 2Þ; i ¼ 2k 1; 8 ðk 2i 2; 2Þ; 0 6 i 6 k 2; > > > > < ðk; 1Þ; i ¼ k 1; r 0 ðbi ; DÞ ¼ > ð2i 3k þ 1; 0Þ; k 6 i 6 2k 2; > > > : ðk 1; 1Þ; i ¼ 2k 1; 8 ðk 2i 2; 0Þ; 0 6 i 6 k 2; > > > > < ðk; 1Þ; i ¼ k 1; r 0 ðci ; DÞ ¼ > ð2i 3k þ 1; 2Þ; k 6 i 6 2k 2; > > > : ðk 1; 1Þ; i ¼ 2k 1; 8 ðk 2i 3; 1Þ; 0 6 i 6 k 2; > > > > < ðk; 3Þ; i ¼ k 1; r 0 ðdi ; DÞ ¼ > ð2i 3k þ 2; 3Þ; k 6 i 6 2k 2; > > > : ðk 2; 2Þ; i ¼ 2k 1: Case 2: n is odd. In this case, we can write n ¼ 2k þ 1; k P 3. Let D ¼ fa0 ; ak ; ckþ2 g. The representations of vertices of T n are:
8 ði; k i; k þ iÞ; > > > < ði; k i; k i þ 4Þ;
0 6 i 6 2;
3 6 i 6 k; > ð2k i þ 1; i k; k i þ 4Þ; k þ 1 6 i 6 k þ 2; > > : ð2k i þ 1; i k; i k 1Þ; k þ 3 6 i 6 2k; 8 ði þ 1; k i; k þ iÞ; 0 6 i 6 1; > > > < ði þ 1; k i; k i þ 3Þ; 2 6 i 6 k 1; rðbi ; DÞ ¼ > ð2k i þ 1; i k þ 1; k i þ 3Þ; k 6 i 6 k þ 2; > > : ð2k i þ 1; i k þ 1; i k 1Þ; k þ 3 6 i 6 2k; rðai ; DÞ ¼
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8 ði þ 2; k i þ 1; k þ i 1Þ; > > > < ði þ 2; k i þ 1; k i þ 2Þ; rðci ; DÞ ¼ > ð2k i þ 2; i k þ 2; k i þ 2Þ; > > : ð2k i þ 2; i k þ 2; i k 2Þ; 8 ði þ 3; k i þ 1; k þ iÞ; > > > > > ði þ 3; k i þ 1; k i þ 2Þ; > > > < ðk þ 2; 3; 3Þ; rðdi ; DÞ ¼ > ð2k i þ 2; i k þ 3; k i þ 2Þ; > > > > > ð2k i þ 2; i k þ 3; i k 1Þ; > > : ð3; k þ 2; k 1Þ;
9799
0 6 i 6 1; 2 6 i 6 k 1; k 6 i 6 k þ 2; k þ 3 6 i 6 2k; 0 6 i 6 1; 2 6 i 6 k 2; i¼k1 k 6 i 6 k þ 1; k þ 2 6 i 6 2k 1; i ¼ 2k:
0
Then the representations r are:
8 ðk 2i; kÞ; 0 6 i 6 2; > > > < ðk 2i; k 2i þ 4Þ; 3 6 i 6 k; r 0 ðai ; DÞ ¼ > ð2i 3k 1; k þ 3Þ; k þ 1 6 i 6 k þ 2; > > : ð2i 3k 1; 2i 3k 2Þ; k þ 3 6 i 6 2k; 8 ðk 2i 1; k 1Þ; 0 6 i 6 1; > > > < ðk 2i 1; k 2i þ 2Þ; 2 6 i 6 k 1; r 0 ðbi ; DÞ ¼ > ð2i 3k; k þ 2Þ; k 6 i 6 k þ 2; > > : ð2i 3k; 2i 3k 2Þ; k þ 3 6 i 6 2k; 8 ðk 2i 1; k 3Þ; 0 6 i 6 1; > > > < ðk 2i 1; k 2iÞ; 2 6 i 6 k 1; r 0 ðci ; DÞ ¼ > ð2i 3k; kÞ; k 6 i 6 k þ 2; > > : ð2i 3k; 2i 3k 4Þ; k þ 3 6 i 6 2k; 8 ðk 2i 2; k 3Þ; 0 6 i 6 1; > > > > > ðk 2i 2; k 2i 1Þ; 2 6 i 6 k 2; > > > < ðk þ 1; k þ 1Þ; i ¼ k 1; r 0 ðdi ; DÞ ¼ > ð2i 3k þ 1; kÞ; k 6 i 6 k þ 1; > > > > > ð2i 3k þ 1; 2i 3k 3Þ; k þ 2 6 i 6 2k 1; > > : ðk 1; k 4Þ; i ¼ 2k: 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 T n
Theorem 4. For any convex polytope T n and n P 5 it follows that sdimðT n Þ ¼ 2n for n odd and sdimðT n Þ ¼ 5n=2 for n even. Next we prove several lemmas which support the proof of Theorem 4. Lemma 6. For n ¼ 2k þ 1 and n P 5, if S is a strong resolving set of T n , then jSj P 2n. Proof. As dðai ; ckþi Þ ¼ dðbi ; dkþi Þ ¼ k þ 2 for i ¼ 0; 1; . . . ; n 1, and DiamðT n Þ ¼ k þ 2, then, according to Property 2, S should contain at least n a-vertices or c-vertices, and at least n b-vertices or d-vertices. Therefore, jSj P 2n. h Lemma 7. For n ¼ 2k þ 1 and n P 5, set S ¼ fai ; di ji ¼ 0; 1; . . . ; n 1g is a strong resolving set of T n . Proof. Let us prove that vertex dkþi strongly resolves pairs ðbi ; bj Þ; ðbi ; cj Þ and ðci ; cj Þ for each i; j; i – j. It is easy to see that, for each i 2 f0; 1; . . . ; n 1g; dðbi ; dkþi Þ ¼ k þ 2 ¼ DiamðT n Þ, 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 .
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v
bi ; bj 06i
w ckþi
ajkþ1 dik
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, belongs to at least one of previous paths and, consequently, dkþi strongly resolves bi and cj . Similarly, dðci ; dkþi Þ ¼ k þ 1 and the corresponding 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 T n , then jSj P 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 to Property 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, it can be easily proved that dðai ; dkþi1 Þ ¼ k þ 2 ¼ DiamðT n Þ and pairs ðai ; akþi Þ; ðai ; di1 Þ; ðdi ; dkþi Þ satisfy conditions (i) and (ii) of Lemma 1. Therefore, according to Properties 1 and 2, (a) (b) (c) (d)
ai ai ai di
2S 2S 2S 2S
or or or or
dkþi1 2 S; akþi 2 S; di1 2 S; dkþi 2 S.
As a consequence of (a)–(d) it follows that S should contain at least 3 vertices from set X i ¼ fai ; akþi ; di1 ; dkþi1 g. For example, if akþi R S, then, according to (a), dkþðkþiÞ1 ¼ di1 2 S, and according to (b), ai 2 S, and according to (c), dðkþiÞ1 2 S. Since for i ¼ 0; 1; . . . ; n 1, the number of all distinct sets X i is equal to n=2, then S should contain at least 3n=2 a-vertices or d-vertices. As S should also contain at least n b-vertices or c-vertices, it means that jSj P 5n=2. h Lemma 9. For n ¼ 2k and n P 5, set S ¼ fai ; ci ; dj ji ¼ 0; 1; . . . ; n 1; j ¼ 0; 1; . . . ; k 1g is a strong resolving set of T n . 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 each i 2 f0; 1; . . . ; n 1g and each j 2 fk; k þ 1; . . . ; n 1g vertex ajkþ1 strongly resolves bi and dj . It is easy to check that, for each j 2 fk; k þ 1; . . . ; n 1g; dðajkþ1 ; dj Þ ¼ k þ 2 ¼ DiamðT n Þ with the corresponding shortest paths between ajkþ1 and dj : ajkþ1 ; bjkþ1 ; bjkþ2 ; . . . ; bj ; cj ; dj ; dj ; cjþ1 ; bjþ1 ; bjþ2 ; . . . ; bjþk ; ajþkþ1 ¼ ajkþ1 . As each vertex bi ; i – j, belongs to one of these paths, ajkþ1 strongly resolves bi and dj . All other results from Table 6 can be 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 2k þ 1; sdimðT n Þ ¼ 2n. Also, according to Lemmas 8 and 9, for n ¼ 2k; sdimðT n Þ 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 and T n . We have proved that the cardinality of minimal doubly resolving sets is constant and equal to 3, except for T 7 . In order to
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