POWERFUL AND MAXIMAL RATIONAL METRIC DIMENSION OF A WHEEL

The rational distance from the vertex u to the vertex v in a graph G, denoted by d(v/u), is defined as the average distances from the vertex u to the closed neighbors of v if u 6= v, else it is 0. A subset S of vertices of G is called rational resolving set of G if for every pair u,v of distinct vertices in V − S, there is a w ∈ S such that d(u/w) 6= d(v/w) in G. In this paper powerful and maximal rational resolving sets are introduced and minimum cardinality of such sets are computed for the wheel graphs.


INTRODUCTION
Let G(V, E) be a connected simple finite graph. A path P a,b from the vertex a to b is an alternating sequence of distinct vertices and edges starting with a and ending with b in such a way each edge lies between its end vertices. The number of edges in a path P a,b is called the length of the path and is denoted by l(P a,b ). The distance between two vertices a and b in G, denoted by d G (a, b) (or simply d(a, b)), is the minimum length of a path between a and b. That is, d G (a, b) = min{l(P a,b )}. The number of edges incident with a vertex v of G is the degree of the vertex v in G and is denoted by deg G (v) or simply deg (v). Further, the closed neighborhood set of a vertex v ∈ V , denoted by N [v], is defined as The notion of rational distance is introduced in [12]. The rational distance from the vertex u to the vertex v ∈ V , denoted by d(v/u), is defined as The rational resolving sets are defined in [11] and studied by various authors in [6,7,8,9,10,12]. We recall that a subset S ⊆ V is a resolving set of G if for each pair u, v ∈ V there exists a vertex w ∈ S such that d(v, w) = d(u, w). The metric dimension of G, denoted by dim(G), is the minimum cardinality of a resolving set of G. A resolving set with minimum cardinality is called a metric basis. The concept of metric dimension was introduced by F. Harary and R. A.
The complement of a minimum dominating set is also a dominating set. But, the complement of a rational resolving set of minimum cardinality need not be a rational resolving set. For example every rational resolving set of a triangle should include at least 2 vertices of it and hence its complement is not. of B. Sooryanarayana and Suma A. S [20]. In this paper, we obtain similar results on rational resolving sets of a wheel.

RATIONAL DISTANCES IN A WHEEL
Since the diameter of the graph W 1,m is 2, it follows that ) whenever m ≥ 5. If m = 3, then the components of Γ(c 0 /S) and Γ(v i /S) are in 0, 3 4 for 0 ≤ i ≤ 2. If m = 4, then Γ(v i /S) ∈ 0, 1, 5 4 for 0 ≤ i ≤ 3. We recall the following results in [11] for immediate reference.
Theorem 2.1 ( [11]). For any integer n ≥ 3,  Throughout this paper, let R(G) be the collection of all rational resolving sets of the graph G.
Then R(G) is super hereditary, that is for every S ∈ R(G), the set T ∈ R(G) whenever S ⊆ T .

GAP IN A WHEEL
We first define the gap between two vertices in G with respect to a set S ∈ R(G).
Definition 3.1. Let G be a graph and S ∈ R(G). Let x, y ∈ V and S = V (G) − S. An S path between x and y in G is an xy-path of G containing all its internal vertices in S. The gap between  18 , v 23 } with Γ(c 0 /S) = 9 5 , 9 5 , 9 5 , 9 5 , 9 5 , 9 5 .
x and y with respect to S, denoted by g S (x, y), is defined as the minimum number of vertices of S in an S path (if it exists) between x and y, else it is 0.
Example: Consider the graph G of Figure 2.
We now begin with the following lemma which we often use in the proof of next theorems.
Let us suppose to the contrary that the condition (i) fails. Then either g T (a, b) ≥ 6 for some a, b ∈ S or there are vertices a, b, c, d ∈ S with |{a, b} ∪ {c, d}| ≥ 3 such that g T (a, b) = 5 and g T (c, d) = 5.
In this case, there are at least six consecutive rim vertices, contradiction to the fact that S ∈ R(W 1,m ).
In this case, there are five consecutive rim vertices, to a, and there are five consecutive rim vertices In case if the condition (ii) fails, then there are three vertices a, b, c in S such that g T (a, b) ∈ {3, 4, 5} and, g T (b, c) ≥ 3 or g T (a, c) ≥ 3. Without loss of generality, we take g T (b, c) ≥ 3.
Then there are three consecutive rim vertices v 1 , v 2 , v 3 in W 1,m with v 1 adjacent to b. Also there are three consecutive rim vertices u 1 , u 2 , u 3 with u 1 adjacent to b. But then, (which corresponds to the vertex b), a contradiction to the fact that S ∈ R(W 1,m ). Hence the conditions (i) and (ii) hold. Now to prove the converse part, let S be a k-element subset of the vertex set of W 1,m containing at least three rim vertices such that every pair of vertices in it satisfies the conditions (i) and (ii) of the theorem. We now show that S ∈ R(W 1,m ) by the method of contradiction. If S does In this case u and v are the rim vertices of W 1,m and w is adjacent to u and v (∵ m = 4).
In this case, w 1 is also adjacent to u and v, and w 1 is the rim vertex, and hence m = 4, a contradiction.
the cycle C m of G), a contradiction to the fact that m ≥ 9.
From the above two sub cases we see that only possibility is d(u/w i ) = d(v/w i ) = 7/4 for every w i ∈ S other than w whenever d(u/w) = d(v/w) = 1. But then, for the vertices s 1 , s 2 in S nearer to w in the cycle C m of W 1,m we see that g T (s 1 , w) ≥ 3 and g T (w, s 2 ) ≥ 3, a contradiction to the assumption of the condition (ii).
In this case neither u nor v is non-adjacent to w, for every w ∈ S. We first show that d(u/w) = d(v/w) = 7 4 for all w ∈ S. For this, let us assume to the contrary that d(u/w ) = d(v/w ) = 3/2 for some w ∈ S.
If possible, suppose to contrary that d(u/w 1 ) = d(v/w 1 ) = 3/2 for some w 1 ∈ S − {w }. Then w 1 can not be in a shortest uw -path or vw -path. Hence w 1 should be in a uv-path of C m not This is possible only if m = 8, a contradiction to the fact that m ≥ 9. Hence the claim.
which is a contradiction to the assumption of condition (ii) of the lemma.
Thus, we have arrived at the conclusion that d(u/w) = d(v/w) = 7/4 for all w ∈ S. That is This is possible only if one of the following hold.
(2) The vertex u is in the center of the S ∪ {c 0 } path between w 1 and w 2 for some w 1 , w 2 ∈ S, and, v is in the center of S ∪ {c 0 } path between w 3 and w 4 for some In either of the above possibilities we arrive at a contradiction to the assumption of condition (i). Hence the lemma.

POWERFUL RATIONAL METRIC DIMENSION
A rational resolving set S ∈ R(G) is called powerful if S ∈ R(G). The least cardinality of a powerful rational resolving set (if it exists) of G is called powerful rational metric dimension of G and is denoted by rmd p (G).
In this section we determine powerful rational metric dimension of a Wheel. We prove the claim by contradiction. Suppose that S / ∈ R(W 1,m ). Then there are two vertices and Let d(u/a i ) = l i , i = 1, 3. Then we have the following possibilities.
In this case, u, v ∈ {a 2 , v 0 } and hence n = 4, a contradiction to the fact that n ≥ 10. Since l 1 = 7/4, u = a i for some i ≥ 4 or i ≤ −2. Since l 3 = 7/4, u = a i for some i ≥ 6 or i ≤ 0. These two together imply u = a i for some i ≥ 6 or i ≤ −2. In either of the cases, d(u/a 2 ) = d(v/a 2 ) = 7/4, a contradiction to equation (2).
Therefore, in either of the cases, l 3 = 7/4, a contradiction.
Other cases follows by symmetry. Hence the Claim.
Therefore, by the above Claim, S ∈ R(W 1,m ), a contradiction to the fact that S is of minimum cardinality in R(W 1,m ). Hence the lemma. Proof. For 4 ≤ m ≤ 7, it is easy to see that a set S containing exactly two adjacent rim vertices of W 1,m is in R(W 1,m ) and is an rmd-set of W 1,m . Further, S also contains two adjacent rim vertices and hence by the super hereditary property of R(G) it follows immediately that S ∈ R(W 1,m ).
When m = 8, every rmd-set S of W 1,m is a 2-element set of its rim vertices v i , v j such that

MAXIMAL AND FOUL RATIONAL METRIC DIMENSION
A rational resolving set S ∈ R(G) is called maximal if S / ∈ R(G). The least cardinality of a maximal rational resolving set of G is called maximal rational metric dimension of G and is denoted by rmd m (G).
A subset S ⊆ V (G) is called foul rational resolving set if S / ∈ R(G) and S / ∈ R(G). The least cardinality of a foul rational resolving set of G is called foul rational metric dimension of G and is denoted by rmd f (G).
LetR(G) and ¬R(G) be the set of all maximal and foul rational resolving sets of G, respectively. In this section we determine maximal and foul rational metric dimension of a Wheel.   , then length of a longest path on C m between v i and v j to be at least 7 (so m ≥ 9 by condition (iii) as j = i + 1), or there are two longest paths of length 6 between v i and v j in C m (so m = 12). In either of the cases m ≥ 9, a contradiction to , v i+2 (mod m) }, j = i + k ≡ i − l (mod m) for some k, l ≥ 5, and hence m ≥ 10, again a contradiction. Proof. Let a = min{ j : j > i, v j ∈ S} and b = min{k : k > a, v k ∈ S}. If i + 3 / ∈ {a, b}, then the Hence, rmd m (W 1,m ) ≤ |S | ≤ (|S| − 1) + 4 = |S| + 3.
S satisfies all the conditions of Lemma 3.2 and hence S ∈ R(W 1,m ). Further,    Proof. Let S be an element of minimum cardinality inR(W 1,m ). Then S ∈ R(W 1,m ). For m = 3, ). Hence, S should contain at least 6 vertices. Therefore, |S| ≥ 6. On the other hand it is easy to see, by Lemma 3.2, that the and for all other pairs a, b ∈ V − S we get g S∪{c 0 } (a, b) = 0). Therefore, |S| = 6 for 9 ≤ m ≤ 12.
Let us now consider the cases m ≥ 13. In these cases, as S / ∈ R(W 1,m ), by Lemma 3.2 we see that S shall contain a 6-element proper subset T which is of the form , v i+k+4 } for some k ≥ 6. For the minimality of |S|, we consider the second option (which selects 6 vertices out of 7). Without loss of generality, we    In this case, a = 10 or 11 if l = 3 or 4 respectively, and g S∪{c 0  Hence the Claim.

ACKNOWLEDGMENT
Authors are very much thankful to the Management and the Principal of Dr. Ambedkar Institute of Technology, Bengaluru, for their constant support and encouragement during the preparation of this paper. Also special thanks to anonymous referees for their valuable suggestions for the improvement of the paper.

CONFLICT OF INTERESTS
The author(s) declare that there is no conflict of interests.