Resolving Share and Topological Index
aa r X i v : . [ m a t h . C O ] A ug RESOLVING SHARE AND TOPOLOGICAL INDEX
MUHAMMAD SALMAN, IMRAN JAVAID ∗ , MUHAMMAD ANWAR CHAUDHRY Abstract.
An atom a of a molecular graph G uniquely determines (resolves) apair ( a , a ) of atoms of G if the distance between a and a is different from thedistance between a and a . In this paper, we quantify the involvement of eachatom a of G in uniquely determining (resolving) a pair ( a , a ) of atoms of G ,which is called the resolving share of a for the pair ( a , a ). Using this quantity,we define a distance-based topological index of a molecular graph, which reflectsthe topology of that molecular graph according to the resolvability behavior ofeach of its atom, and is called the resolving topological index. Then we computethe resolving topological index of several molecular graphs. Introduction
A major part of the current research in mathematical chemistry, chemical graphtheory and quantative structure-activity-property relationship QSAR/QSPR studiesinvolves topological indices. Topological indices are numerical identities derivedin an unambiguous manner from a molecular graph [22, 29]. These indices aregraph invariants which usually characterize the topology of that molecular graph.Some major classes of topological indices such as distance-based topological indices,connectivity topological indices and counting related polynomials and indices ofgraphs have found remarkable employment in several chemistry fields.The first non-trivial distance-based topological index was Wiener index, intro-duced by Wiener in 1947 [30]. To explain various chemical and physical propertiesof atoms, molecules, and to correlate the structure of molecules to their biologicalactivity, Wiener index plays a significant role [19]. Caused by this usefulness ofthe Wiener index, the research interest in Wiener index and related distance-basedindices is still considerable. In the last twenty years, surprisingly a large number ofmodifications and extensions of the Wiener index such as Schultz index
M T I ( G ),proposed by Schultz [27]; Szeged index S z ( G ), proposed by Gutman [7]; revisedWiener or revised Szeged index S z ∗ ( G ) proposed by Randi´c [21]; modified Wiener Key words and phrases. resolving set, resolving share, average resolving share, resolving topo-logical index2010
Mathematics Subject Classification. ∗ Corresponding author: [email protected]. ndex for trees m W ( T ), proposed by Nikoli´c et al. [18]; another class of modifiedWiener indices m W λ ( T ), proposed by Gutman et al. [8]; Harary index H ( G ), pro-posed by Plavsi´c et al. [20] and Baladan index J ( G ), proposed by Baladan [2] and byRandi´c [23], to name a few, was put forward and studied. An extensive bibliographyon this matter can be found in the reviews [6, 15].The problems on distance in graphs continues to seek the attention of scientistsboth as theory and applications. Among these problems, the most famous problemin graphs, which plays a vital role to uniquely distinguish all the vertices of a graph,is resolvability. Roughly speaking, by resolvability in a graph G , we mean that anytwo vertices of G in the pair ( u, v ) are said to be uniquely distinguished (representedor resolved) by a vertex w of G if the distance between u and w is distinct from thedistance between v and w . The vertices of G needed to distinguish all the pairs ofthe vertices of G form a remarkable set known as a resolving set for G , and it has asignificant application in pharmaceutical chemistry.A fundamental problem in pharmaceutical chemistry is to find out the uniquerepresentations of chemical compounds in several molecular structures (graphs).The intention behind uniquely representing the chemical compounds is to determinewhether any two compounds in the structure share the same functional group ata particular position. This comparative statement plays a fundamental part indrug discovery whenever it is to be determined whether the features of a chemicalcompound are responsible for its pharmacological activity [12, 13]. The solutionof this fundamental problem was addressed by the concept of resolving set. Aminimum resolving set is, in fact, the set of those few atoms in a molecular graphwhich determine the unique representations of the chemical compounds. Now, aquestion with remarkable interest arises that how much an atom of a moleculargraph partake in uniquely representing any pair of chemical compounds? . Precisely,in a graph G , how much a vertex of G involves itself to resolve any pair of verticesof G ? In this paper, we fix this problem by defining the amount of resolving doneby an atom (vertex) to represent (resolve) every pair of compounds of a moleculargraph, and is called the resolving share of that atom. With the help of resolvingshare of each atom, a numeric identity is associated with the molecular graph, calledthe resolving topological index, which reflects the total amount of resolving done bythe atoms in that molecular graph.Under a “molecular graph” we understand a simple graph, representing the atomskeleton of molecules (chemical compounds). Thus the vertices of a molecular graphrepresents the atoms and edges the atom-atom bonds. Let G be a non-trivial con-nected graph with vertex set V ( G ) and edge set E ( G ). We write u ∼ v if twovertices u and v are adjacent (form an edge) in G and write u v if they arenon-adjacent (do not form an edge). The join of two graphs G and G , de-noted by G + G , is a graph with vertex set V ( G ) ∪ V ( G ) and an edge set ( G ) ∪ E ( G ) ∪ { u ∼ v | u ∈ V ( G ) ∧ v ∈ V ( G ) } . The distance , d ( u, v ), betweentwo vertices u and v of G is defined as the length of a shortest u − v path in G ,where length is the number of edges in the path. The diameter of G , denoted by diam ( G ), is the maximum distance between any two vertices of G . We refer [4] forthe general graph theoretic notations and terminologies not described in this paper.A vertex u of G resolves two distinct vertices v and w of G if d ( v, u ) = d ( w, u ).A set R ⊆ V ( G ) is called a resolving set for G if every two distinct vertices of G areresolved by some elements of R . Such a set R with minimum cardinality is calleda metric basis , or simply a basis of G and that minimum cardinality is called the metric dimension of G , denoted by dim( G ) [3]. Obviously, the metric dimension ofa graph G is a topological index that suggests those minimum number of vertices of G which uniquely determine all the vertices of G by their shortest distances to thechosen vertices.The concept of resolving set was first introduced in the 1970s, by Slater [28] and,independently, by Harary and Melter [9]. Slater described the usefulness of this ideainto long range aids to navigation [28]. Moreover, this concept has some applicationsin chemistry for representing chemical compounds [12, 13] and in problems of patternrecognition and image processing, some of which involve the use of hierarchical datastructures [16]. Other applications of this concept to navigation of robots in networksand other areas appear in [3, 14]. In recent years, a considerable literature regardingthis notion has developed (see [1, 3, 5, 10, 11, 14, 16, 17, 24, 25, 26]).2. Resolving Share
In this section, we define the concept of resolving share and investigate somebasic results which later help in defining and computing a distance-based topologicalindex. We begin with the following useful preliminaries: V p denotes the collectionof all (cid:0) n (cid:1) pairs of the vertices of a graph G . For any vertex w of G , let V i ( w ) = { v ∈ V ( G ) − { w } | d ( v, w ) = i } be the distance neighborhood of w for 1 ≤ i ≤ diam ( G ),and the partition Π w = { V i ( w ) ; 1 ≤ i ≤ diam ( G ) } be the distance partition of the set V ( G ) − { w } with reference of w . By Π w − { x } ,we mean that the vertex x is not lying in any partite set of the distance partitionΠ w . By Π x = Π y , we mean that V i ( x ) = V i ( y ) for all 1 ≤ i ≤ diam ( G ). Definition 2.1.
Let G be a connected graph. For any pair ( u, v ) ∈ V p , let R ( u, v ) = { x ∈ V ( G ) | x resolves u and v } be the resolving neighborhood of the pair ( u, v ) .Then for any w ∈ V ( G ) , the quantity r w ( u, v ) = ( | R ( u,v ) | if u and v are resolved by w , otherwiseis called the resolving share of w for the pair ( u, v ) . v v v v v v v v G1: u u u v v v v G : Figure 1.
In the graph G , the resolving share of the vertex v iszero for the pair ( v , v ); is for the pair ( v , v ); is for the pair( v , v ) and is for the pair ( v , v ). Remarks 2.2. ( i ) The resolving neighborhood of a pair ( u, v ) ∈ V p is the class ofall those vertices whose resolving share for the pair ( u, v ) is same. ( ii ) For w ∈ V ( G ) and ( u, v ) ∈ V p , ≤ r w ( u, v ) ≤ . ( iii ) r u ( u, v ) = 0 = r v ( u, v ) . Lemma 2.3.
For a pair ( u, v ) ∈ V p and for a vertex w ∈ V ( G ) −{ u, v } , r w ( u, v ) = 0 if and only if both u and v belong to the same partite set of Π w .Proof. ( ⇒ ) r w ( u, v ) = 0 implies that d ( u, w ) = d ( v, w ) = i for some 1 ≤ i ≤ diam ( G ). It follows that both u and v belong to the same partite set V i ( w ) ∈ Π w .( ⇐ ) If u, v ∈ V i ( w ) ∈ Π w for some 1 ≤ i ≤ diam ( G ), then w does not resolve u and v , and hence r w ( u, v ) = 0. (cid:3) Lemma 2.4.
For a pair ( u, v ) ∈ V p and for all w ∈ V ( G ) − { u, v } , r w ( u, v ) = 0 ifand only if Π u − { v } = Π v − { u } .Proof. ( ⇒ ) Suppose that r w ( u, v ) = 0 for all w ∈ V ( G ) − { u, v } . This implies that d ( u, w ) = d ( v, w ) for all w ∈ V ( G ) − { u, v } . Contrarily assume that Π u − { v } 6 =Π v − { u } . It follows that there exists an element x in V ( G ) − { u, v } such that x liesin a partite set, say V i ( u ), of Π u − { v } and x lies in a partite set, say V j ( v ) ( j = i ),of Π v − { u } , and vice-versa. Thus d ( x, u ) = i = d ( x, v ) or d ( x, v ) = j = d ( x, u ), acontradiction. Hence Π u − { v } = Π v − { u } .( ⇐ ) Suppose that Π u − { v } = Π v − { u } . Assume contrarily that r w ( u, v ) = 0. Itfollows that w ∈ R ( u, v ) and hence d ( u, w ) = d ( v, w ). Thus, there exists a partiteset in Π u − { v } which is not equal to any members of Π v − { u } , a contradiction.Hence r w ( u, v ) = 0. (cid:3) emma 2.5. For a pair ( u, v ) ∈ V p and for a vertex w ∈ V ( G ) , r w ( u, v ) = if andonly if w ∈ { u, v } and Π u − { v } = Π v − { u } .Proof. ( ⇒ ) Suppose that r w ( u, v ) = . Then, clearly, w ∈ { u, v } because | R ( u, v ) | =2 and w resolves u and v . In fact, R ( u, v ) = { u, v } in this case. It follows that d ( u, x ) = d ( v, x ) for all x ∈ V ( G ) −{ u, v } , which concludes that Π u −{ v } = Π v −{ u } .( ⇐ ) Suppose that Π u − { v } = Π v − { u } . It follows that d ( u, z ) = d ( v, z ) forall z ∈ V ( G ) − { u, v } , which implies that the only vertices that resolves the pair( u, v ) are the vertices in the pair. So R ( u, v ) = { u, v } , and hence for w ∈ { u, v } , r w ( u, v ) = . (cid:3) Example 2.6.
Consider the graph G of Figure 1 with vertex set V ( G ) = U = { u , u , u } ∪ V = { v , v , v , v } . Let ( u, v ) ∈ V p be any pair of vertices of G . Thennote that, ( i ) Π u − { v } = Π v − { u } for either u, v ∈ U or u, v ∈ V ; ( ii ) if u ∈ U and v ∈ V , then R ( u, v ) = { u } ∪ V and both u, v belong to the same partite set of Π x for all x ∈ U − { u } . Hence, by previous three lemmas, we have r w ( u, v ) = if w ∈ { u, v } , for either u, v ∈ U or u, v ∈ V, if w ∈ { u } ∪ V, for u ∈ U and v ∈ V, otherwise . Remark 2.7.
For each pair ( u, v ) ∈ V p , R ( u, v ) ∩ R = ∅ for any resolving set R fora graph G . The following useful result for finding a resolving set for G was proposed byChartrand et al. in 2000. Lemma 2.8. [3]
Let R be a resolving set for a graph G and ( u, v ) ∈ V p . If d ( u, w ) = d ( v, w ) for all w ∈ V ( G ) − { u, v } , then u or v is in R . Lemma 2.9.
Let a pair ( u, v ) ∈ V p and R be any resolving set for G . (1) If r w ( u, v ) = 0 for all w ∈ V ( G ) − { u, v } , then u or v is in R . (2) If r w ( u, v ) = , then u or v is in R .Proof. (1) By Lemma 2.4, Π u −{ v } = Π v −{ u } , which implies that d ( u, w ) = d ( v, w )for all w ∈ V ( G ) − { u, v } , and hence Lemma 2.8 yields the result.(2) By Lemma 2.5, Π u − { v } = Π v − { u } and w ∈ { u, v } . In fact R ( u, v ) = { u, v } and d ( u, w ) = d ( v, w ) for all w ∈ V ( G ) − { u, v } . Hence, the result follows by Lemma2.8. (cid:3) Let | G | denotes the order of a graph G . The following assertion is directly followsfrom the definition of the resolving share. Proposition 2.10.
For a pair ( u, v ) ∈ V p and for each vertex w ∈ V ( G ) , r w ( u, v ) = | G | if and only if R ( u, v ) = V ( G ) . emma 2.11. For a pair ( u, v ) ∈ V p , if r w ( u, v ) = | G | for each w ∈ V ( G ) , then thedistance between u and v is odd.Proof. If the distance between u and v is even, i.e. , d ( u, v ) = 2 k for k ≥
1, then thereexists a vertex x in G such that d ( u, x ) = k = d ( x, v ), and hence x R ( u, v ). (cid:3) Theorem 2.12.
Let G be a graph with diam ( G ) = 2 . Then there are at most (cid:22)(cid:16) | G | (cid:17) (cid:23) pairs ( u, v ) in V p for which r w ( u, v ) = | G | for each w ∈ V ( G ) .Proof. Let ( u, v ) ∈ V p for which r w ( u, v ) = | G | for each w ∈ V ( G ). Then, R ( u, v ) = V ( G ), by Proposition 2.10, and since diam ( G ) = 2 so d ( u, v ) = 1, by Lemma 2.11.It follows that u ∈ V ( v ) ∈ Π v and v ∈ V ( u ) ∈ Π u . Moreover, V ( u ) and V ( v ) aredisjoint subsets of V ( G ), because if there is an element x in V ( u ) ∩ V ( v ), then x does not resolve u and v and hence R ( u, v ) = V ( G ). Also, V ( u ) ∪ V ( v ) = V ( G ).Otherwise, there exists an element y in V ( G ) − { u, v } such that u, v ∈ V ( y ), whichimplies that u and v are not resolved by y yielding R ( u, v ) = V ( G ). Hence V ( u )and V ( v ) form a partition of V ( G ). Further, for any a, b ∈ V ( u ) (or a, b ∈ V ( v )), a and b have the same distance from u (or v ) and hence R ( a, b ) = V ( G ). It followsthat the number of pairs ( u, v ) for which R ( u, v ) = V ( G ) is bounded above by | V ( u ) || V ( v ) | ≤ $(cid:18) | G | (cid:19) % . (cid:3) Lemma 2.13.
If the diameter of a graph G is one, then for every pair ( u, v ) ∈ V p , r w ( u, v ) ∈ { , } for each w ∈ V ( G ) .Proof. Since diam ( G ) = 1, so G is isomorphic to a complete graph, and for everytwo vertices u and v of a complete graph, Π u − { v } = Π v − { u } . Hence, the resultfollowed by Lemma 2.4 if w
6∈ { u, v } , or followed by Lemma 2.5 if w ∈ { u, v } . (cid:3) Theorem 2.14.
Let G be a graph. Then for every pair ( u, v ) ∈ V p and for each w ∈ V ( G ) , r w ( u, v ) = (cid:26) if w ∈ { u, v } , if w
6∈ { u, v } , if and only if diam ( G ) = 1 .Proof. Suppose that r w ( u, v ) = (cid:26) if w ∈ { u, v } , w
6∈ { u, v } , for every pair ( u, v ) ∈ V p and for each w ∈ V ( G ). It follows that R ( u, v ) = { u, v } and w ∈ R ( u, v ) or w R ( u, v ). We claim that diam ( G ) = 1. Suppose that ( u, v ) = k ≥ u, u , u , . . . , u k − , v be a u − v geodesic (shortest path) in G oflength k . This implies that d ( u , u ) = 1 and d ( u , v ) = k −
1, and hence u ∈ R ( u, v ).But u
6∈ { u, v } , a contradiction to the fact that R ( u, v ) = { u, v } . Consequently d ( u, v ) = 1 for every two distinct vertices u, v of G . So diam ( G ) = 1.The converse part of the theorem followed by Lemma 2.13. (cid:3) Resolving Topological Index
We define the average resolve share of each vertex of a graph and then by using itwe establish a distance based topological index of that graph. Further, we computethe resolving topological index of certain graphs.
Definition 3.1.
Let G be a connected graph. For any vertex w ∈ V ( G ) , let R ( w ) = { ( u, v ) ∈ V p | u and v are resolved by w } be the resolvent neighborhood of the vertex w . Then the quantity ar w ( G ) = 1 | R ( w ) | X ( u,v ) ∈ R ( w ) r w ( u, v ) is the average of the amount of resolving done by w in G , and is called the averageresolving share of w in G . Remark 3.2.
Since for each x ∈ V ( G ) −{ w } , ( w, x ) ∈ R ( w ) , so | R ( w ) | and ar w ( G ) will never zero for every w ∈ V ( G ) . Since P ( u,v ) ∈ R ( w ) r w ( u, v ) = | R ( w ) | a ⇔ r w ( u, v ) = a for all ( u, v ) ∈ R ( w ). So, wehave the following straightforward proposition: Proposition 3.3.
Let G be a graph. For any w ∈ V ( G ) , ar w ( G ) = a if and only if r w ( u, v ) = a for all ( u, v ) ∈ R ( w ) . Proposition 3.4.
Let G be a graph and w be any vertex of G . If each partite setof the distance partition Π w is a singleton set, then ar w ( G ) = 2 | G | ( | G | − X ( u,v ) ∈ V p r w ( u, v ) . Proof.
If each partite set of the distance partition Π w is a singleton set, then thevertices of each pair ( u, v ) ∈ V p are resolved by w . It follows that R ( w ) = V p andthe proof is complete. (cid:3) Definition 3.5.
Let G be a graph. Then the positive real number R ( G ) = X w ∈ V ( G ) ar w ( G ) is called the resolving topological index of G. Theorem 3.6.
The resolving topological index of the Petersen graph is . Figure 2.
The Petersen graph
Proof.
Let G be the Petersen graph. The vertices of G are the 2-element subsets ofthe set { , , , , } . Let the vertex set of G be { S ij = { i, j } ; 1 ≤ i < j ≤ } andtwo subsets will be connected by an edge if their intersection is the empty set (seeFigure 2). Let ( S ij , S kl ) be any pair of V p . Then either S ij ∩ S kl = ∅ , or S ij ∩ S kl isa singleton set. In the former case, i, j = k, l and d ( S ij , S kl ) = 1, and in the latercase, i = k , or i = l , or j = k , or j = l and d ( S ij , S kl ) = 2. Now, we discuss twocases. Case 1.
When S ij ∩ S kl = ∅ . Then for each element A ∈ V ( S ij ), A ∩ S kl = ∅ , and foreach element B ∈ V ( S kl ), B ∩ S ij = ∅ . It follows that each A and each B resolves S ij and S kl . Further, for each S ab ∈ V ( G ) − ( V ( S ij ) ∪ V ( S kl )), S ab ∩ S ij = ∅ as wellas S ab ∩ S kl = ∅ , and hence S ij , S kl ∈ V ( S ab ). Thus, the resolving neighborhood ofthe pair ( S ij , S kl ) is V ( S ij ) ∪ V ( S kl ) having six elements. Thus, for X ∈ V ( G ), byLemma 2.3, we have r X ( S ij , S kl ) = (cid:26) if X ∈ V ( S ij ) ∪ V ( S kl ) , . Case 2.
When S ij ∩ S kl = ∅ . Then ( i ) V ( S ij ) ∩ V ( S kl ) = { S ab } , where a, b = i, j, k, l ,and hence S ij , S kl belong to the same partite set V ( S ab ), ( ii ) | V ( S ij ) ∩ V ( S kl ) | = 3and for each A ∈ V ( S ij ) ∩ V ( S kl ), S ij , S kl belong to the same partite set V ( A ).Thus R ( S ij , S kl ) = V ( G ) − [( V ( S ij ) ∩ V ( S kl )) ∪ { S ab } ]. Hence, for X ∈ V ( G ), byLemma 2.3, we have r X ( S ij , S kl ) = (cid:26) if X V ( G ) − [( V ( S ij ) ∩ V ( S kl )) ∪ { S ab } ] , . Note that, for any 2-element subset X ∈ V ( G ), r X ( Y, Z ) = for all ( Y, Z ) ∈ R ( X ).Thus, by Proposition 3.3, ar X ( G ) = for all X ∈ V ( G ). It completes the proof. (cid:3) heorem 3.7. Let G be a path on n ≥ vertices. Then R ( G ) = n P i =1 2 n − n − i +4 n ( n − − n − i − if n is even , n − n − n ( n − + ⌊ n ⌋ P i =1 2 n − n − n ( i − n ( n − − n ( n − i − if n is odd . Proof.
Let G : v , v , . . . , v n be a path on n ≥ v i , v j ) ∈ V p . Then R ( v i , v j ) = ( V ( G ) if i + j is odd ,V ( G ) − { v i + j } if i + j is even . Hence for w ∈ V ( G ), r w ( v i , v j ) = i + j, and w = v i + j , n − for even i + j, and for all w ∈ V ( G ) − { v i + j } , n for odd i + j, and for all w ∈ V ( G ) . Now, sinceΠ v = (cid:26) { V j ( v ) ; 1 ≤ j ≤ n − i } for v ∈ { v i , v n − i +1 } , ≤ i ≤ ⌊ n ⌋ , { V j ( v ) ; 1 ≤ j ≤ i − } for v = v i and i = ⌈ n ⌉ for odd n. So Π v i and Π v n − i +1 contain ( i −
1) 2-element subsets of V ( G ) for each 1 ≤ i ≤ ⌊ n ⌋ and i = ⌈ n ⌉ . It follows that each v i and v n − i +1 do not resolve i − V p , andhence | R ( v i ) | = | R ( v n − i +1 ) | = (cid:18) n (cid:19) − ( i − . The resolvent neighborhood of each vertex of G consists of two types of pairs of V p :the pairs ( v a , v b ) for which a + b is even, we refer them the pairs of type-I; the pairs( v a , v b ) for which a + b is odd, we refer them the pairs of type-II. In above, we haveproved that the resolving share of the vertices of G for the pairs of type-I is n − andfor the pairs of type-II is n . We discuss the following two cases: Case 1. ( n is even) Since the sum a + b of two natural numbers a, b will be evenwhen both a and b are even, or both a and b are odd, and there are n even and n oddnatural numbers in the set { , , . . . , n } in this case. So V p contains 2 (cid:0) n (cid:1) = n ( n − (cid:0) n (cid:1) − n ( n −
2) = n pairs of type-II. Out of n ( n −
2) pairsof type-I, those n ( n − − ( i −
1) pairs belong to R ( v i ) and R ( v n − i +1 ) for whichthe sum of the indices of the vertices in the pairs is not equal to 2 i or 2( n − i + 1),respectively. Also, all the n pairs of type-II belong to R ( v i ) and R ( v n − i +1 ), where1 ≤ i ≤ n . It follows that for each 1 ≤ i ≤ n , ar v i ( G ) = 2 n ( n − − i − (cid:18) n ( n − n − − i − n − n (cid:19) = ar v n − i +1 ( G ) . ase 2. ( n is odd) Since there are n − even and n +12 odd natural numbers in theset { , , . . . , n } in this case. So V p contains (cid:0) n − (cid:1) + (cid:0) n +12 (cid:1) = ( n − pairs oftype-I and (cid:0) n (cid:1) − ( n − = ( n −
1) pairs of type-II. Out of ( n − pairs oftype-I, those ( n − − ( i −
1) pairs belong to R ( v i ) and R ( v n − i +1 ) for which thesum of the indices of the vertices in the pairs is not equal to 2 i or 2( n − i + 1),respectively. Also, all the ( n −
1) pairs of type-II belong to R ( v i ) and R ( v n − i +1 ),where 1 ≤ i ≤ ⌈ n ⌉ . It follows that for each 1 ≤ i ≤ ⌈ n ⌉ , ar v i ( G ) = 2 n ( n − − i − (cid:18) n − − i − n − n − n (cid:19) = ar v n − i +1 ( G ) . By adding the average resolving shares of all the vertices of G in the both abovecases, one can easily derive the stated resolving topological index. (cid:3) An even path in a graph G is such whose length is even, and a path is odd if itslength is odd. Theorem 3.8.
Let G be a cycle on n ≥ vertices. Then R ( G ) = ( n ( n − n − n +2 when n is even , nn − when n is odd . Proof.
We consider the following two cases:
Case 1. ( n in even) Let G : v , v , . . . , v n , v be a cycle on even n ≥ D be the diameter of G . Then for ( v i , v j ) ∈ V p ( i = j ), R ( v i , v j ) = ( V ( G ) when i + j is odd ,V ( G ) − { v i + j , v i + j + D } when i + j is even . Hence for w ∈ V ( G ), r w ( v i , v j ) = i + j, and w ∈ { v i + j , v i + j + D } , n − for even i + j, and for all w ∈ V ( G ) − { v i + j , v i + j + D } , n for odd i + j, and for all w ∈ V ( G ) . Since for each 1 ≤ i ≤ n , the distance partition Π v i = { V j ( v i ) ; 1 ≤ j ≤ D } contains ( D −
1) 2-element subsets of V ( G ), so each v i do not resolve D − V p , and hence | R ( v i ) | = (cid:18) n (cid:19) − ( D −
1) = n ( n − − D − . Note that, the resolvent neighborhood of each vertex of G consists of two typesof pairs of V p : the pairs ( v a , v b ) for which a + b is even, we refer them the pairs oftype-I; the pairs ( v a , v b ) for which a + b is odd, we refer them the pairs of type-II.In above, we have proved that the resolving share of the vertices of G for the pairsof type-I is n − and for the pairs of type-II is n . Since the sum a + b of two natural umbers a, b will be even when both a and b are even, or both a and b are odd, andthere are n even and n odd natural numbers in the set { , , . . . , n } in this case.So V p contains 2 (cid:0) n (cid:1) = n ( n −
2) pairs of type-I and (cid:0) n (cid:1) − n ( n −
2) = n pairsof type-II. Out of n ( n −
2) pairs ( v a , v b ) of type-I, those pairs belong to R ( v i ) forwhich i = a + b and i = a + b + D , and three are n ( n − − ( D −
1) such pairs, where1 ≤ i ≤ n . Also, all the n pairs of type-II belong to R ( v i ), where 1 ≤ i ≤ n . Itfollows that for each 1 ≤ i ≤ n , ar v i ( G ) = 2 n ( n − − D − (cid:18) n ( n − n − − D − n − n n (cid:19) = n − n − n + 2 , because D = n . It completes the proof. Case 2. ( n is odd) Let ( u, v ) be any pair of V p and let l be the length of the even u − v path in G . Then there exists a vertex x in G such that u, v ∈ V l ( x ) ∈ Π x . Itfollows, by Lemma 2.3, that r w ( u, v ) = (cid:26) w = x, n − for all w ∈ V ( G ) − { x } . Since the resolving share of each vertex w of G for every pair in V P (and hencefor every pair in the resolvent neighborhood of w ) is n − . So Proposition 3.3 yieldsthat ar w ( G ) = n − for all w ∈ V ( G ), and it concludes the proof. (cid:3) Theorem 3.9.
The resolving topological index of a complete graph on at least twovertices is | G | .Proof. Since diam ( G ) = 1, so Theorem 2.14 implies that for every pair ( u, v ) ∈ V p and for each w ∈ V ( G ), r w ( u, v ) = (cid:26) if w ∈ { u, v } , w
6∈ { u, v } . Hence R ( G ) = | G | because ar w ( G ) = for all w ∈ V ( G ). (cid:3) Theorem 3.10.
Let G be a complete k -partite graph K n ,n ,...,n k , where each n i ≥ , ≤ i ≤ k and k ≥ . Then R ( G ) = k X i =1 n i n i − n i + k X t =1 t = i n t − n i − n i + k X t =1 t = i n t n i + n t . .Proof. Let the partite sets of G are V i = { v i , v i , . . . , v in i } , where n i ≥ ≤ i ≤ k . Let ( x, y ) ∈ V p . If x, y ∈ V i for any 1 ≤ i ≤ k , then Π x − { y } = Π y − { x } . Itfollows, by Lemmas 2.4 and 2.5, that r w ( x, y ) = (cid:26) if w ∈ { x, y } , w
6∈ { x, y } . f x ∈ V i and y ∈ V j = i , where 1 ≤ i, j ≤ k . Then for each u ∈ V i − { x } , v ∈ V j − { y } and z ∈ V l (for all 1 ≤ l ≤ k and l = i, j ), x ∈ V ( v ) ∪ V ( u ) , y ∈ V ( u ) ∪ V ( v ) and x, y ∈ V ( z ). It follows that the resolving neighborhood of ( x, y ) is V i ∪ V j . Thus,by Lemma 2.3, r w ( x, y ) = ( n i + n j if w ∈ V i ∪ V j , . For each 1 ≤ i ≤ k and for any v ij ∈ V i (1 ≤ j ≤ n i ), R ( v ij ) = { ( v ij , v il ) , ( v ia , v tb ) ; 1 ≤ l = j ≤ n i , ≤ a ≤ n i , ≤ b ≤ n t , ≤ t = i ≤ k } . Since r v ij ( v ij , v il ) = 12 for all 1 ≤ l = j ≤ n i , and r v ij ( v ia , v tb ) = 1 n i + n t for all 1 ≤ a ≤ n i , ≤ b ≤ n t , ≤ t = i ≤ k. So for each 1 ≤ i ≤ k and for each 1 ≤ j ≤ n i , ar v ij ( G ) = 1 | R ( v ij ) | n i X l =1 l = j
12 + k X t =1 t = i n i X a =1 n t X b =1 n i + n t = ( n i −
1) + n i k X t =1 t = i n t − n i −
12 + n i k X t =1 t = i n t n i + n t . = n i k X t =1 t = i n t − − n i
12 + k X t =1 t = i n t n i + n t − . Now, by taking the summation of the average resolving shares of the vertices ofall the partite sets of V ( G ), we get the required result. (cid:3) Theorem 3.11.
For each wheel graph W n ( n ≥ , the resolving topological indexis ( n − n +8)2( n − n − .Proof. A wheel graph W n is the join of a cycle C n − and the vertex c (called thecentral vertex of the wheel). First note that, Π c = { V ( c ) = V ( C n − ) } and for each v ∈ V ( C n − ), Π v = { V ( v ) , V ( v ) } with | V ( v ) | = 3 and | V ( v ) | = n − x, y ) ∈ V p . Case 1.
When d ( x, y ) = 1. If y = c and x ∈ V ( C n − ), then each vertex u of C n − such that d ( u, x ) = 2 belongs to the resolving neighborhood of ( x, y ). Also, or each v ∈ V ( x ) − { c } , x and y belong to the same partite set of Π v . Thus R ( x, y ) = V ( x ) ∪ { x, y } , and hence Lemma 2.3 yields that r w ( x, y ) = (cid:26) w ∈ V ( x ) − { c } , n − otherwise . If both x and y belong to V ( C n − ), then x ∈ V ( y ) and y ∈ V ( x ). In this case, x, y ∈ V ( c ) and for each z ∈ V ( x ) ∩ V ( y ), x, y ∈ V ( z ). Also, for u ∈ V ( x ) − { y, c } and for v ∈ V ( y ) − { x, c } , x ∈ V ( v ) and y ∈ V ( u ). It follows that R ( x, y ) = { u, v, x, y } , and hence, by Lemma 2.3, we have r w ( x, y ) = (cid:26) w ∈ { c } ∪ ( V ( x ) ∩ V ( y )) , otherwise . Case 2.
When d ( x, y ) = 2, then x, y ∈ V ( C n − ) and x ∈ V ( y ) , y ∈ V ( x ). Thereare two subcases to discuss. Subcase 2.1.
When | V ( x ) ∩ V ( y ) | = 1. In this case, for each u ∈ V ( x ) − { c } andfor each v ∈ V ( y ) − { c } , x ∈ V ( v ) and y ∈ V ( u ). So R ( x, y ) = V ( x ) ▽ V ( y ) and | R ( x, y ) | = 6 (the symbol X ▽ Y denotes the symmetric difference of two sets X and Y ). Further, for each z V ( x ) ▽ V ( y ), x, y belong to the same partite set ofΠ z . It follows, by Lemma 2.3, that r w ( x, y ) = (cid:26) w V ( x ) ▽ V ( y ) , otherwise . Subcase 2.2.
When | V ( x ) ∩ V ( y ) | = 2. In this case, for u ∈ V ( x ) − ( V ( x ) ∩ V ( y ))and for v ∈ V ( y ) − ( V ( x ) ∩ V ( y )), x ∈ V ( v ) and y ∈ V ( u ). So R ( x, y ) = V ( x ) ▽ V ( y ) = { u, v, x, y } . Moreover, for each z
6∈ { u, v, x, y } , x, y belong to thesame partite set of Π z . It concludes, by Lemma 2.3, that r w ( x, y ) = (cid:26) w
6∈ { u, v, x, y } , otherwise . Since R ( c ) = { ( c, v ) ; v ∈ V ( C n − ) } and each v ∈ V ( C n − ) does not resolve thepairs ( a, b ) and ( x, y ) for all distinct a, b ∈ V ( v ) and for all distinct x, y ∈ V ( v ). Itfollows that | R ( c ) | = n − v ∈ V ( C n − ), | R ( v ) | = (cid:0) n (cid:1) − (cid:0) (cid:1) − (cid:0) n − (cid:1) =4 n −
13. For v ∈ V ( C n − ), let V ( v ) = { c, u, w } . Then out of 4 n −
13 pairs in R ( v ), ( i ) n − c, a ) , a ∈ V ( W n ) − V ( v ), and the resolving share of v forall these pairs is n − ; ( ii ) n − v, b ) , b ∈ V ( W n ) −{ c, v } , and theresolving share of v for 4 out of these n − and is for the remaining n − iii ) 2( n −
4) pairs are of the form ( u, d ) and ( w, d ) , d ∈ V ( W n ) − { c, u, v, w } ,and the resolving share of v for 4 out of these 2( n −
4) pairs is and is forthe remaining 2( n −
6) pairs. Also, the resolving share of the central vertex c foreach pair in R ( c ) is n − . Hence, by Proposition 3.3, ar c ( W n ) = n − and for each ∈ V ( C n − ), ar v ( W n ) = 14 n − (cid:18) ( n −
3) 1 n − n −
6) 16 + 4( 14 ) + 2( n −
6) 16 (cid:19) = n − n − n − n − . Hence R ( W n ) = 1 n − n − n − n − n − n −
13) = ( n − n + 8)6( n − n − . (cid:3) Theorem 3.12.
The resolving topological index of a friendship graph F n is n − n +4 n − n (3 n − ,where n ≥ .Proof. A friendship graph F n is the join K + G and having 2 n + 1 vertices, where K is a graph having only one vertex c (called the central vertex) and G is thegraph obtain by taking the union of n copies of the path P . For each u ∈ V ( G ),Π u = { V ( u ) , V ( u ) } with | V ( u ) | = 2 and | V ( u ) | = 2( n −
1) since the diameterof F n is two. Let v ∈ V ( F n ) − { u } . If v ∈ V ( u ) and v = c , then for the vertex x ∈ V ( u ) − { v } , u and v belong to the same partite set of Π x . Also, for each y ∈ V ( u ), v ∈ V ( y ) and u ∈ V ( y ). Hence, together with Lemma 2.3 and abovediscussion, we have r w ( u, v ) = (cid:26) w = x, n otherwise . If v ∈ V ( u ) and v = c , then u and v are the vertices of the same copy of P andΠ u − { v } = Π v − { u } . It follows, by Lemmas 2.4 and 2.5, that r w ( u, v ) = (cid:26) w
6∈ { u, v } , if w ∈ { u, v } . If v ∈ V ( u ), then for a ∈ V ( u ) − { c } and for b ∈ V ( v ) − { c } , u ∈ V ( b ) and v ∈ V ( a ). Also, for all x ( V ( u ) ∪ V ( v )) − { c } , u and v belong to the same partiteset of Π x . Thus R ( u, v ) = { a, b, u, v } , and hence r w ( u, v ) = (cid:26) w ∈ { c } ∪ ( V ( u ) ∪ V ( v )) , otherwise . For any u ∈ V ( G ), let V ( u ) = { a, c } . Then R ( u ) = V p − ( { ( a, c ) } ∪ { ( x, v ) ; x, v ∈ V ( u ) } )= { ( u, w ) , ( a, y ) , ( c, y ) ; w ∈ V ( F n ) − { u } , y ∈ V ( F n ) − { a, c, u }} . In R ( u ), ( i ) the number of pairs of the form ( u, w ) is 2 n , and the resolving share of u for the pair ( u, a ) is , for the pair ( u, c ) is n and for the remanding 2( n −
1) pairsis ; ( ii ) the number of pairs of the form ( a, y ) is 2( n − f u for all these pairs is ; ( iii ) the number of pairs of the form ( c, y ) is 2( n − u for all these pairs is n . Thus | R ( u ) | = 2(3 n − ar u ( F n ) = 12(3 n − (cid:18)
12 + 12 n + 2( n −
1) 14 + 2( n −
1) 14 + 2( n −
1) 12 n (cid:19) = ( n + 1)(2 n − n (3 n − . Since R ( c ) = { ( c, v ) ; v ∈ V ( G ) } and the resolving share of c for all these 2 n pairsis n . So, Proposition 3.3 concludes that ar c ( F n ) = n . It concludes that R ( F n ) = 12 n +( n −
1) ( n + 1)(2 n − n (3 n −
2) = 2 n − n + 4 n − n (3 n − . (cid:3) Concluding Remarks
We investigated the amount of the resolving done by a vertex v of a graph G forevery pair of vertices of G , called the resolving share of v , and then we establishedsome related results. We also quantified the average of the amount of resolving doneby v in G , and we called it the average resolving share of v . Using average resolvingshare of each vertex of G , we associated a distance-based topological index withthe graph G , which describes the topology of that graph with respect to the totalresolving done by each vertex of that graph, and we called this topological index, theresolving topological index. Then, by computing the resolving shares and averageresolving shares of all the vertices, we worked out the resolving topological indicesof certain well-known graphs such as the Petersen graph, paths, cycles, completegraphs, complete k -partite graphs, wheel graphs and friendship graph. The workdone in this paper is a revelation for the researchers working with resolvability todetermine, in different graphical structures, how they have the topology accordingto the resolving done by their vertices. References [1] S. Ahamd, M. A. Chaudhry, I. Javaid, M. Salman, On the metric dimension of the generalizedPetersen graphs,
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