Given any simple graph G = (V, E), not necessarily finite, and a ground set X, a set-indexer

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1 Chapter 2 Topogenic Graphs Given any simple graph G = (V, E), not necessarily finite, and a ground set X, a set-indexer of G is an injective set-valued function f : V (G) 2 X such that the induced edge function f L : E(G) 2 X, defined by f L (uv) = f(u) f(v) = (f(u) f(v)) (f(u) f(v)) for each uv E(G), is also injective. A set-indexer f of G is topogenic, if the family f(v (G)) f L (E(G)) is a topology on X, where f(v (G)) = {f(u) : u V (G)} and f L (E(G)) = {f L (e) : e E(G)}. In particular, if f(v (G)) f L (E(G)) = 2 X, then f is called a graceful topogenic set-indexer. In this chapter, we investigate some foundational results on topogenic and graceful topogenic set-indexers of graphs and also establish the necessary conditions for topogenic graphs and graceful topogenic graphs and analyze the relationship between topogenic graphs and existing various categories of set-valued graphs. We also establish the existence of non-topogenic graphs and identify certain classes of graphs that admit topogenic set-indexers and graceful topogenic set-indexers. 18

2 Topogenic Graphs Introduction By assigning subsets of a set to the vertices and the symmetric difference of sets associated with end vertices of an edge to the corresponding edge, Acharya [1] gave a new direction to the very topic of graph labeling. For a (p, q)-graph G = (V, E) and a non-empty set X of cardinality n, he defined a set-indexer of G as an injective setvalued function (or, equivalently, set-labeling ) f : V (G) 2 X such that the induced edge-function f : E(G) 2 X defined by f (uv) = f(u) f(v) for all uv E(G) is also injective, where 2 X is the set of all subsets of X and denotes the binary operation of taking the symmetric difference of pairs of subsets of X. He [1] called a graph G, (not necessarily finite), set-graceful if it admits a set-graceful labeling (or, graceful set-indexer) f, which is defined as a set-indexer such that f (E(G)) := {f (uv) : uv E(G)} = 2 X. Also, he [1, 4] defined G to be set-sequential if G admits a set-sequential labeling, which is a bijection f : V (G) E(G) 2 X such that f(uv) = f(u) f(v) for all uv E(G). Acharya [1] showed that every graph possesses a set-indexer f : V (G) 2 X such that f(v ) := {f(v) : v V (G)} is a

3 Topogenic Graphs 20 topology on the ground set X; he called such a set-indexer, a topological set-indexer (or, a T -set-indexer) of G. In particular, a set-graceful graph G is topologically set-graceful (or, T-set-graceful ) if G admits a graceful set-indexer f such that f(v ) is a topology on X; such setindexers are called graceful topological set-indexers of G. Hence, he [1] defined the topological number (or, simply, the T -number) t(g) of G as the smallest cardinality of a ground set X with respect to which G has a T -set-indexer. As we have seen, there are a number of ways to associate a topology with a given graph (see [1], [2], [5], [11]) or a graph with a given topology (see [5], [14], [19]). The depth and extent of intimacy of the properties of topologies on a given ground set X with that of the structure of a graph associated with a topology on X, or vice versa, could be a matter of investigation for a particular concept and may be driven by the interplay between topologies and what are known as knowledge structures [see [6], [8] and [13]]. A knowledge structure is defined as an ordered pair (Q, K) in which Q is a finite set, whose elements are called questions or items, and K is a family of subsets of Q, called knowledge states, such that Q, K. One of the fundamental question is : What is the special nature of

4 Topogenic Graphs 21 a knowledge structure when K is a topology on Q? Every knowledge structure (Q, K) can be associated with a directed graph D K (Q) by suitably orienting the 2-section hypergraph K (2) of K : q 1 q 2 if and only if the question q 2 has arisen while attempting to answer the question q 1. This motivates us to investigate the following new type of set-valuation of a graph. 2.2 Topogenic graphs Definition A graph G, not necessarily finite, is called topogenic if there exists a non-empty ground set X and an injective setassignment f : V (G) 2 X, such that (i) the induced edge function f : E(G) 2 X defined by f (uv) = f(u) f(v), uv E(G), where denotes the binary operation of taking symmetric difference of the subsets of X, is also injective, and (ii) f(v (G)) f (E(G)) = τ f is a topology on X; such a set-assignment f, if it exists, is called a topogenic set-indexer of G. Figure 2.1 is an example of a topogenic set-indexer of C 4, the quadrilateral. It is important to notice first that not every set-indexer of a graph need

5 Topogenic Graphs 22 Figure 2.1: A topogenic set-indexer of C 4 be topogenic, a fact illustrated in Figure 2.2 for K 4 {e}, the graph obtained by deleting an arbitrary edge from the complete graph K 4 of order 4. However, the set-valuation in which the vertices of degree three in this graph are assigned the sets and {2} respectively and the vertices of degree two are assigned the sets {1} and X = {1, 2, 3} respectively we get a topogenic set-indexer of K 4 {e}. Thus, topogenic graphs may have non-topogenic set-indexers. Also, we established that there are graphs which are not topogenic and hence the following observation is important to be noted. Observation For a topogenic set-indexer f of a graph G, if it exists, it is possible that f(u) = f (xy) for a vertex u and an edge xy of G. The very first basic question is whether every graph is topogenic. That

6 Topogenic Graphs 23 Figure 2.2: A set-indexer of K 4 e, which is not topogenic. is, given any graph G = (V, E), can we invariably find a ground set X and a set-indexer f of G such that f(v (G)) f (E(G)) is a topology on X? Towards gaining some insight in answering this question, we first give some examples of topogenic graphs and discuss their existence. Example The only graph of order 1 is the trivial graph K 1 ; it has a unique topogenic set-indexer that assigns the empty set, to its unique vertex. Example There are two graphs of order 2, viz., the complete graph K 2 of order two and the disconnected graph K 1 K 1 = K2, the complement of K 2 ; each of these graphs has the unique topogenic setindexer obtained by assigning the empty set to one vertex and the entire ground set X to the other; the optimal choice of the ground set to accomplish such a labeling is X = {x}.

7 Topogenic Graphs 24 Example There are four graphs of order 3, viz., G 1 = K3, G 2 = K1 K 2, G 3 = K1,2 = P3 and G 4 = K3. Consider the totally disconnected graph G 1 = K3 with its vertices labeled as u 1, u 2, u 3, the ground set X = {1, 2} and the set-labeling f 1 defined by letting f 1 (u 1 ) =, f 1 (u 2 ) = {1} and f 1 (u 3 ) = {1, 2} = X. Also, f 1 (V (K 3 )) f1 (E(K 3)) = f 1 (V (K 3 )) = {, {1}, {1, 2} = X}, which is a topology on X. Thus, by definition, G 1 = K3 is topogenic. In G 2, let the vertex of K 1 be assigned the empty set, and the two vertices of the component K 2 be assigned the sets {1} and {1, 2}, respectively. For the resulting set-indexer, say f 2, of G 2 we have f 2 (V (G 2 )) f2 (E(G 2)) = {, {1}, {2}, {1, 2} = X} to be a topology on X and hence f 2 is a topogenic set-indexer of the graph. Next, in the 2-star G 3, assign to its central vertex and let the two pendant vertices be assigned the sets {1} and {1, 2}, respectively. The resulting set-indexer of G 3, say f 3, may similarly be verified to be a topogenic set-indexer of the graph. Lastly, consider the set-valuation f 4. Let, f 4 (V (G 4 )) = {, {1}, {2}} then, f4 (E(G 4)) = {, {1}, {2}, X = {1, 2}}. Now, f 4 (V (G 4 )) f4 (E(G 4)) = {, {1}, {2}, {1, 2}}, a topology on X. Hence, f 4 is a topogenic set-indexer of G 4.

8 Topogenic Graphs 25 Figure 2.3: Topogenic labeling of K 1,5. Theorem For every positive integer n, there exists a connected topogenic graph of order n. Proof. Let G n be the (n 1)-star whose vertices are labeled u 1, u 2,..., u n so that u 1 is the central vertex of the star. Let X = {1, 2,..., n 1} and define f : V (G n ) 2 X such that f(u 1 ) =, f(u i ) = {1, 2,..., i 1}, i {2, 3..., n}. Then, f (E(G)) = {{1}, {1, 2},..., {1, 2,..., n 1}}, hence, f is a topogenic set-indexer of G n. Figure 2.3 illustrates the theorem for n = 5. Example A graph G, not necessarily finite, is set-graceful if it admits a set-graceful labeling (or, graceful set-indexer) f, which is defined as a set-indexer such that f (E(G)) = 2 X. As shown in

9 Topogenic Graphs 26 Figure 2.4: A graceful set-indexer of K 2 K 4, which is not topogenic. [1], any set-graceful labeling f of G with the property that f(v (G)) it is easy to see that f(v (G)) f (E(G)) is a topology on X, that is, G is topogenic. Note that the condition f(v (G)) is not necessary for f to be a set-graceful labeling of a graph G, as the instance displayed in Figure 2.4 illustrates. Theorem Every set-graceful graph G, having a graceful setindexer f with f(v (G)) is topogenic. However, a topogenic graph need not be set-graceful, as indicated by the topogenic set-indexed cycle C 4 of length 4 displayed in Figure 2.1, which is not a graceful set-indexer.

10 Topogenic Graphs 27 Example As in [1], a set-graceful graph G = (V, E) is topologically set-graceful (or, T-set-graceful ) if G admits a graceful setindexer f such that f(v (G)) is a topology on X; such set-indexers are called graceful T -set-indexers of G. Next, G is set-sequential if G admits a set-sequential labeling, which is a bijection f : V (G) E(G) 2 X such that f(uv) = f(u) f(v) for all uv E(G). Hence, let (G, f) be any graph G together with any one of its set-sequential labelings f. Consider the new graph H = G + K 1, V (K 1 ) = {w}, in which w is labeled with the empty set,. Let f denote the set-labeling of H such that f restricted to V (G) in H is precisely f and f (w) =. One can then easily see that H is a topogenic graph in the sense that (H, f ) is a topogenic set-labeled graph. Existence of set-sequential graphs is the only necessary condition to make this example non-vacuous, and this has been well established already in [1], [2], [4] and [12]. We have observed that not every set-graceful labeling of a graph need be topogenic. Also, a graph that is not even set-graceful may be topogenic. Figure 2.5 gives an example of a topogenic graph that is not set-graceful. We will now set out to find non-topogenic graphs. Towards this end, we need to prepare an adequately sound conceptual foundation.

11 Topogenic Graphs 28 Figure 2.5: A graph that is not set-graceful but having a topogenic set-indexer Topogenic strength of a graph Consider a topogenic set-indexer f : V (G) 2 X of a (p, q)-graph G = (V, E) and let τ f = f(v (G)) f (E(G)). The number of distinct f-open sets, viz., τ f, is called the topogenic strength of f over G and if G is finite, the minimum (respectively maximum) of τ f taken over all possible topogenic set-indexers f of G is denoted ϱ 0 (G) (respectively ϱ 1 (G)). Because of the injectivity of f and f, we must have ϱ 0 (G) f(v (G)) f (E(G)) ϱ 1 (G) p + q k, (1) where k is the number of vertices of G that are adjacent to the vertex w for which f(w) = (such a vertex w exists since τ f is a topology on X). Moreover, p ϱ 0 (G). Further, since / f (E(G)), q ϱ 0 (G) 1 or, equivalently, q + 1 ϱ 0 (G). Thus, for a topogenic (p, q)-graph G, p ϱ 0 (G) and q + 1 ϱ 0 (G). These observations are summarized in

12 Topogenic Graphs 29 the following lemma. Lemma For any topogenic (p, q)-graph G, max{p, q + 1} ϱ 0 (G) ϱ 1 (G) p + q δ, (2) where δ = δ(g) is the minimum vertex degree in G. Theorem For a topogenic path P n, n ϱ 0 2n 2. Proof. By Lemma 2.2.9, we have max{p, q + 1} ϱ 0 (G) p + q δ. For the path P n, we have p = n, q = n 1 and δ = 1. Thus, we have n ϱ 0 2n 2. Theorem For a topogenic cycle C n, n + 1 ϱ 0 2n 2. Proof. By Lemma 2.2.9, we have max{p, q + 1} ϱ 0 (G) p + q δ. For the cycle C n, we have p = n, q = n and δ = 2. Thus, we have n + 1 ϱ 0 2n 2. Theorem For a topogenic complete bipartite graph K m,n, n m, mn + 1 ϱ 0 m(n + 1). Proof. This follows from Lemma and the fact that for K m,n, we have, p = m + n, q = mn and δ = n. Theorem For a topogenic complete graph K p, ϱ 0 = ϱ 1 = p 2 p+2 2.

13 Topogenic Graphs 30 Proof. Suppose that, K p is topogenic and f is a topogenic set-indexer of K p. Since δ = p 1, we have p(p 1) ϱ 0 and ϱ 1 p + p(p 1) 2 (p 1). Hence, p(p 1) ϱ 0 and ϱ 1 p(p 1) ϱ 0 = ϱ 1 = p2 p+2 2, as claimed. Hence, the question arises to find values of p for which K p is indeed topogenic. We have already seen that K p is topogenic for values of p {1, 2, 3}. What about K p for values of p 4? An answer to this question is attempted in the following subsection Topogenic complete graphs K 1, K 2 and K 3 are topogenic [see Examples 2.2.2, and 2.2.4]. In this section, we aim to establish some more interesting results on topogenic complete graphs. First, we prove that K 4 is not topogenic. Theorem K 4 is not topogenic. Proof. If possible, let K 4 be topogenic, which implies there exists a topogenic set-indexer f of K 4 with respect to some non-empty ground set, say X, so that τ f = f(v (K 4 )) f (E(K 4 )) is a topology on

14 Topogenic Graphs 31 X. Then, by Theorem , ϱ 0 (K 4 ) = 7. Since f is injective, the empty set, cannot be obtained as a symmetric difference of two nonempty sets. Hence, empty set, should necessarily be assigned to a vertex. That is, f(v (K 4 )). Hence, let f(v (K 4 )) = {, V 1, V 2, V 3 }, where V 1, V 2, V 3 are non-empty subsets of X. Then, f (E(K 4 )) = {V 1, V2, V3, V1 V2, V 1 V3, V 2 V3 }. Since τ f is a topology on X, the entire set X must be an element of τ f. There arise two cases namely, X = V i for some i {1, 2, 3}, or X = V i Vj for some distinct i, j {1, 2, 3}. We prove the theorem in two steps. In step 1, we prove that X / f(v (K 4 )) and in step 2, we show that X / f (E(K 4 )); contradicting our assumption that τ f a is topology on X. Step 1 Without loss of generality, let V 3 = X. Then V 1, V 2 can be such that V 1 V 2 = V 3 or V 1 V 2 V 3. Case 1: V 1 V 2 = V 3 Then, V 1, V 2 might be such that V 1 V 2 = or V 1 V 2. Now, let V 1 V 2 =. Since V 1 V 2 = V 3, V 1 V2 = V 1 V 2 = V 3. Since f(v (K 4 )), there exists an edge with the label V 3 = V3, which contradicts the injectivity of f.

15 Topogenic Graphs 32 Therefore, V 1 V 2. Let V 1 V 2 = A X so that A. Since V 1, V 2 τ f, and τ f being a topology on X, V 1 V 2 = A, must be an element in τ f. We claim that, A V 1 and A V 2. If A = V 1 then, V 1 V 2 which gives V 1 V 2 = V 2. But V 1 V 2 = V 3 by assumption; which leads to a contradiction to the injectivity of f. A similar contradiction arises when A = V 2. Now, V 1 V2 = V 1 V 2 A = V 3 A; V 1 V3 = V 3 V 1 ; V 2 V3 = V 3 V 2. Hence, by the choice of V 1, V 2, V 3 and from the expressions for V i Vj for distinct i, j {1, 2, 3} we conclude that A V i for all i {1, 2, 3} and A V i Vj for all distinct i, j {1, 2, 3}. Thus, A / τ f, contradicting the hypothesis that τ f is a topology on X. Hence, condition in Case 1 is absurd. Case 2: V 1 V 2 V 3 Here, arises two possibilities namely, either V 1 V 2 = or V 1 V 2. Let V 1 V 2 =. Then, V 3 (V 1 V 2 ) = B, a non-empty subset of X and V 1 V 2 B = V 3. Now, V 1 V2 = V 1 V 2 τ f ; V 1 V3 = V 3 V 1 = V 2 B; V 2 V3 = V 3 V 2 = V 1 B. Again, (V 1 V3 ) (V 2 V3 ) = B τ f, since (V 1 V3 ), (V 2 V3 ) f (E(K 4 )) τ f. But by the choice of V 1, V 2, V 3 and from the expressions for V i Vj for distinct i, j {1, 2, 3}, it is clear that B V i for all i {1, 2, 3} and

16 Topogenic Graphs 33 B V i Vj for all distinct i, j {1, 2, 3}. This leads to a contradiction to our derivation that B τ f. Therefore, V 1 V 2. Now, let V 1 V 2 = C, a non-empty subset of X. Then, since τ f is a topology on X, C must be in τ f. We claim that C V 1 and C V 2. If C = V 1 then V 1 V 2 which gives V 1 V 2 = V 2. But V 1 V 2 V 3 by assumption. Then, V 2 V 3 and hence V 3 V 2 = K, a non-empty subset of X. Now, V 3 V2 = V 3 V 2 = K. Since K, V 1 τ f, K V 1 must be in τ f. Since K is neither contained in V 1 nor in V 2 ; and V 3 K we get K V 1 V i for all i {1, 2, 3}. Now, V 1 V2 = V 2 V 1 K V 1 ; V 1 V3 = V 3 V 1 K V 1 and V 2 V3 = V 3 V 2 K V 1. Hence, K V 1 V i Vj for all i, j {1, 2, 3}. That is, K V 1 / τ f, a contradiction to the fact that τ f is a topology. Hence, C V 1. A similar contradiction arises when C = V 2. Further, V 1 V2 = V 1 V 2 C; and since V 1 V 2 V 3, V 1 V3 = V 3 V 1, and V 2 V3 = V 3 V 2. We observe that C V i for any i {1, 2, 3} and C V i Vj for all distinct i, j {1, 2, 3}. This is again a contradiction to the fact that C τ f. Thus, X V i for all i {1, 2, 3}. From Case 1 and Case 2 we have shown that X / f(v (K 4 )).

17 Topogenic Graphs 34 Step 2 Therefore, X = V i Vj for some distinct i, j {1, 2, 3}. Without loss of generality, assume X = V 1 V2. Then, V 1 V 2 = X and V 1 V 2 =. Since for every i = 1, 2, 3, V i X, V 3 must have a non-empty intersection with at least one of the sets V 1, V 2 for, if not V 1 V 2 cannot be X. Without loss of generality, assume that V 1 V 3. Then, V 1 V 3 must be in τ f. Clearly V 1 V 3 ; V 1 V 3 V 2 ; and hence V 1 V 3 V i Vj for distinct i, j {1, 2, 3}. Hence, V 1 V 3 should necessarily be either V 1 or V 3. We claim that V 1 V 3 V 3 and V 1 V 3 V 1. Case 1 If possible, let V 1 V 3 = V 3. Then, V 3 V 1 and hence V 1 V 3 = H( ). Now, V 1 V3 = V 1 V 3 = H τ f. Since H, V 2 τ f, H V 2 must be in τ f. Assume, H V 2 f(v (K 4 )) which implies, H V 2 = V i for some i {1, 2, 3}. Now, by the choice of H we have, H V 2 V 1 since V 1 V 2 = ; H V 2 V 2 since H V 2 = ; H V 2 V 3 since V 2 V 3 = and H V 3. Hence, H V 2 / f(v (K 4 )). Then, H V 2 must be in f (E(K 4 )). But V 1 V3 = V 1 V 3 = H; V 2 V3 = V 2 V 3 ; V 1 V2 = X show that H V 2 V i Vj where i, j {1, 2, 3}. That is, H V 2 / f (E(K 4 )), a contradiction to the fact that τ f is a topology on X. Hence, V 1 V 3 V 3.

18 Topogenic Graphs 35 Case 2 If possible, let V 1 V 3 = V 1. Then, V 1 V 3, and we have, V 3 V 1 = S, a non-empty subset of X. Since S is not a subset of V 1 and V 1 V 2 = X; S V 2. Now, S V 2 implies V 2 S = T, a non-empty subset of X which gives V 2 = T S. Moreover, V 3 V 2 = S and T V 3 =. Consider V 2 V3 = (V 2 V 3 ) (V 2 V 3 ) = X S = (V 1 V 2 ) S = (V 1 T S) S = V 1 T τ f. Now, τ f being a topology, (V 1 T ), V 2 τ f implies that (V 1 T ) V 2 = T V 2 = T must be in τ f. Then, T = V i for some i = 1, 2, 3 or T = V i Vj for some distinct i, j {1, 2, 3}. But T V 1 since V 1 V 2 = ; T V 2 since V 2 = T S; T V 3 since V 3 T =. Hence, T V i for all i = 1, 2, 3. Moreover, V 1 V2 = X; V 1 V3 = V 3 V 1 = S; and V 2 V3 = V 1 T. That is, T V i Vj for all distinct i, j {1, 2, 3}. Thus, T / τ f, a contradiction to the fact that τ f is a topology. Hence, V 1 V 3 V 1. Thus, we have V 1 V 3 / τ f, a contradiction to the fact that τ f is a topology on X. Therefore, X V i Vj for distinct i, j {1, 2, 3}. Hence, the result follows by the principle of contraposition. Theorem K 5 is not topogenic.

19 Topogenic Graphs 36 Proof. If possible, let K 5 be topogenic. That is, there exists a topogenic set-indexer f with respect to some non-empty ground set say, X, so that τ f = f(v (K 5 )) f (E(K 5 )) is a topology on X. Then, by Theorem , ϱ 0 (K 5 ) = 11. Since the empty set, cannot be obtained as a symmetric difference of any two non-empty sets, it is necessary that f(v (K 5 )). Now, let f(v (K 5 )) = {, V 1, V 2, V 3, V 4 } where V 1, V 2, V 3, V 4 are non-empty subsets of X. Then, f (E(K 5 )) = {V 1, V2, V3, V4, V1 V2, V 1 V3, V 1 V4, V 2 V3, V 2 V4, V 3 V4 }. Since τ f is a topology on X, the entire set X must be an element of τ f. There arise two cases for X, namely X = V i for some i {1, 2, 3, 4}, or X = V i Vj for some distinct i, j {1, 2, 3, 4}. We establish the theorem in the following two steps. Step 1 Suppose X = V i for some i {1, 2, 3, 4}. Without loss of generality, assume that X = V 4. Then, V 1, V 2, V 3 can be such that V 1 V 2 V 3 = V 4 or V 1 V 2 V 3 V 4. Let V 1 V 2 V 3 = V 4. If the sets V 1, V 2, V 3 are pairwise disjoint then, V 1 V2 = V 1 V 2 and V 4 V3 = V 4 V 3 = X V 3 = V 1 V 2, a contradiction to the injectivity of f. Therefore, at least two of the sets V 1, V 2, V 3 must have a common element. Without loss of generality, suppose V 1 V 2 = A. Since τ f is

20 Topogenic Graphs 37 a topology on X, we have A must be in τ f. Therefore, A = V i for some i {1, 2, 3} or A = V i Vj for some distinct i, j {1, 2, 3}, for neither A = V 4 = X nor A = V 4 Vi = X V i = V c i for all i {1, 2, 3}. So let A = V 3. Then, V 1 V 2 V 3 = V 1 V 2 A = V 1 V 2 = V 4. Then, V 1 V4 = (V 1 V 4 ) (V 1 V 4 ) = V 1 V 2 V 1 = V 2 (V 2 V 1 ) = V 2 A = V 2 V 3. But V 2 V3 = V 2 V 3, a contradiction to the injectivity of f. Therefore, A = V 1 or V 2. Let A = V 1. Then, V 1 V 2 = V 2. Now, V 2 V 3 since otherwise, if V 2 V 3 = then, V 1 V 2 V 3 = V 2 V 3 and by assumption V 2 V 3 = V 4. Then, V 2 V3 = V 2 V 3 = V 4. But V 4 = V4, a contradiction to the injectivity of f. Also V 3 V4 = V c 3 and V 2 V4 = V c 2. Since V c 3, V c 2 τ f, V c 3 V c 2 must be in τ f. But by the choices of V 1, V 2, V 3, V 4 we have V c 3 V c 2 V i for all i {1, 2, 3, 4} and V c 3 V c 2 V i Vj for all distinct i, j {1, 2, 3, 4}. Hence, V c 3 V c 2 / τ f, a contradiction to the fact that τ f is a topology on X. Hence, A V 1. Analogously we can show that A V 2. That is, A V i for all i {1, 2, 3, 4}. Hence, A being an element of τ f, A = V i Vj for some i, j {1, 2, 3, 4}. But V 1 V2 = V 1 V 2 A; V 1 V3 = (V 1 V 3 ) (V 1 V 3 ); V 3 V2 = (V 3 V 2 ) (V 3 V 2 ) and

21 Topogenic Graphs 38 hence none of the sets V i Vj for distinct i, j {1, 2, 3, 4} equals A. Hence, A = V 1 V 2 / τ f, again a contradiction to the fact that τ f is a topology on X. The above analysis implies that V 1 V 2 V 3 X. Then, V 1 V 2 V 3 X. Let V 1 V 2 V 3 = B. Since τ f being a topology, B must be in τ f. Clearly, B V i and B V i for any i {1, 2, 3} and not a subset of the union of any pair of them. Hence, B = V i Vj for some distinct i, j {1, 2, 3}; without loss of generality, let B = V 1 V2. Then, V 1 V 2 V 3 = B = V 1 V2 = (V 1 V 2 ) (V 1 V 2 ) B, which is preposterous. Thus, it follows that X V i for all i {1, 2, 3, 4}. Step 2 Let X = V i Vj for some distinct i, j {1, 2, 3, 4}. Without loss of generality, assume that X = V 1 V2. Then, V 1 V 2 = X and V 1 V 2 =, for if V 1 V 2, then V 1 V2 = V 1 V 2 V 1 V 2 X. Then, since for every i = 1, 2, 3, 4; V i X, V 3 and V 4 have non-empty intersection with at least one of the sets V 1 and V 2. Without loss of generality, assume that C = V 1 V 3. Then, C must be in τ f and C, V 2. But none of the sets V i Vj for distinct i, j {1, 2, 3, 4} can be the set C. Therefore, C should necessarily be V 1, V 3 or V 4. Now, let C = V 4. It can be shown that V 1 V 3 cannot be equal to

22 Topogenic Graphs 39 V i for all i {1, 2, 3, 4} and also V 1 V 3 V i Vj for all distinct i, j {1, 2, 3, 4}. Hence C V 4. Therefore C = V 1 or V 3. We claim that C V 1 and C V 3. Now, the claim can be proved by invoking the proof given in Case 1 and Case 2 in Step 2 of Theorem Therefore, C / τ f, a contradiction to the fact that τ f is a topology on X. Hence, X V i Vj for distinct i, j {1, 2, 3, 4}. Theorem K 6 is topogenic. Proof. Let V (K 6 ) = {v 1, v 2, v 3, v 4, v 5, v 6 }. Let X = {1, 2, 3, 4}. Define f : V (K 6 ) 2 X such that f(v 1 ) =, f(v 2 ) = {1, 2, 3, 4}, f(v 3 ) = {1, 2, 3}, f(v 4 ) = {1, 2, 4}, f(v 5 ) = {1, 3, 4}, f(v 6 ) = {2, 3, 4}. Then, f (E) = {{1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}. Then, f(v (K 6 )) f (E(K 6 )) = 2 X. The topogenic labeling of K 6 is given in Figure 2.6. Finding the total number of labeled topologies T (n), one can define on a set X of cardinality n is still an open question. Also, there is no known simple formula giving T (n) for at least some specific values of n. For small values of n, may be found; for example, T (1) = 1, T (2) = 4,

23 Topogenic Graphs 40 Figure 2.6: A graceful topogenic set-indexer of K 6. and T (3) = 29. For n 4, the calculations are complicated. The online encyclopedia of Sloane [16] gives the value of T (n) for 1 n 14. An approach towards the determination of T (n) is explained in [15] and [18]. If t(n, k) denote the set of all labeled topologies on X having k open sets, 2 k 2 n, and T (n, k) = t(n, k), Sharp [15], Stanley [17] and Stephen [18] proved the following Theorems. Theorem [15, 18] For n 3, T (n, k) = 0 for 3.2 n 2 < k < 2 n. The following Theorem gives the number of topologies for large k. We denote (n) k, the number n(n 1)... (n k + 1). Theorem [17] For n 5, (i) T (n, 3.2 n 2 ) = (n) 2 ;

24 Topogenic Graphs 41 (ii) T (n, 5.2 n 3 ) = (n) 3 ; (iii) T (n, 7.2 n 4 ) = 9(n) (n) 5 ; (iv) T (n, 9.2 n 4 ) = 5(n) 5 6 ; (v) T (n, 15.2 n 5 ) = (n) 5 ; (vi) T (n, 17.2 n 5 ) = (n) 5 12 ; and (vii) T (n, 2 n 1 ) = (n) 4 + (n) 3 + (n) 2 2. Since the existence of topologies of order 3.2 n 2, 5.2 n 3, 7.2 n 4, 9.2 n 4, 17.2 n 5, 15.2 n 5, 2 n 1 on a set X of cardinality n is proved, to characterize topogenic complete graphs, it is worth examining whether there exists a topogenic set-indexer for complete graph with t(n, k) where k = 3.2 n 2, 5.2 n 3, 7.2 n 4, 9.2 n 4, 17.2 n 5, 15.2 n 5, 2 n 1. Since we have already proved the results for K p ; p 6, we can assume that n 5. Lemma K p, p 7 does not admit a topology of cardinality 3.2 n 2 for any n, where n 5. Proof. If possible, suppose K p admits such a topology. Then ϱ 0 (K p ) = 3.2 n 2

25 Topogenic Graphs 42 p2 p = 3.2 n 2 p 2 p + 2 = 3.2 n 1 The solution to this equation satisfies the polynomial congruence p 2 p+2 0 (mod 48), and this polynomial congruence has no solution. Lemma K p, p 7 does not admit a topology of cardinality 5.2 n 3 for any n, where n 5. Proof. If possible, suppose K p admits such a topology. Then ϱ 0 (K p ) = 5.2 n 3 p2 p = 5.2 n 3 p 2 p + 2 = 5.2 n 2 The solution to this equation satisfies the polynomial congruence p 2 p+2 0 (mod 40), and this polynomial congruence has no solution. Lemma K p, p 7 does not admit a topology of cardinality 9.2 n 4 for any n, where n 5. Proof. If possible, suppose K p admits such a topology. Then ϱ 0 (K p ) = 9.2 n 4

26 Topogenic Graphs 43 p2 p = 9.2 n 4 p 2 p + 2 = 9.2 n 3 The solution to this equation satisfies the polynomial congruence p 2 p+2 0(mod 36), and this polynomial congruence has no solution. Lemma K p, p 7 does not admit a topology of cardinality 17.2 n 5 for any n, where n 5. Proof. If possible, suppose K p admits such a topology. Then ϱ 0 (K p ) = 17.2 n 5 p2 p = 17.2 n 5 p 2 p + 2 = 17.2 n 4 The solution to this equation satisfies the polynomial congruence p 2 p+2 0(mod 34), and this polynomial congruence has no solution. Lemma K p, p 7 does not admit a topology of cardinality 15.2 n 5 for any n, where n 5. Proof. If possible, suppose K p admits such a topology. Then ϱ 0 (K p ) = 15.2 n 5

27 Topogenic Graphs 44 p2 p = 15.2 n 5 p 2 p + 2 = 15.2 n 4 The solution to this equation satisfies the polynomial congruence p 2 p+2 0(mod 30), and this polynomial congruence has no solution. Invoking Lemma , Lemma , Lemma , Lemma and Lemma , we have Theorem K p, p 7 is not topogenic with respect to the topologies of any of the cardinality 3.2 n 2, 5.2 n 3, 9.2 n 4, 17.2 n 5, 15.2 n 5 on a set X of cardinality n, where n 5. Thus, we strongly believe that K p is topogenic if and only if p {1, 2, 3, 6}. Hence, Conjecture 1. The complete graph K p is topogenic if and only if p {1, 2, 3, 6}. The following is an interesting problem to be investigated. Problem 1: Given a complete graph K p determination of the minimum number of edges {e i } to be deleted so that K p {e i } is topogenic is an open problem.

28 Topogenic Graphs Topogenic index of a graph The topogenic index of a graph G is defined as the least cardinality of a ground set X such that there is a topology τ on X which acts as a topogenic set-indexer of a graph H with least order that contains G as an induced subgraph; this number is denoted as Υ(G). If G is a topogenic graph, then Υ(G) is just the least cardinality of a ground set X such that there is a topology τ on X which acts as a topogenic setindexer of G. Finding good bounds for Υ(G) seems to be a challenging problem for topogenic graphs. Lemma If a (p, q)-graph is topogenic with respect to a set X of cardinality n, then 1 p + q 2 n+1 1. Proof. For a topogenic (p, q)-graph, we have 1 p 2 n and 0 q 2 n 1. Therefore, we have 1 p + q 2 n + 2 n 1 which implies 1 p + q 2 n+1 1. Remark Both the bounds for the inequality 1 p + q 2 n+1 1 are attainable, as the lower bound is attained by K 1 and the upper bound is attained by K 1,2 n 1. Theorem For a topogenic (p, q)-graph G, the topogenic index Υ(G) log( p+q+1 2 ) log 2.

29 Topogenic Graphs 46 Proof. By Lemma , for a (p, q)-graph G we have, p + q 2 X +1 1 p + q X +1 log(p + q + 1) ( X + 1)(log 2) log(p + q + 1) log 2 log 2 log( p+q+1 2 ) log 2 Thus, Υ(G) log( p+q+1 2 ) log 2. X Υ(G) Remark As the topogenic strength ϱ 0 (K 2 ) = 2, ϱ 0 (K 3 ) = 4 and ϱ 0 (K 6 ) = 16 we have Υ(K 2 ) = 1, Υ(K 3 ) = 2 and Υ(K 6 ) = More classes of topogenic graphs Theorem The star K 1,n is topogenic for any positive integer n. Proof. Let V (K 1,n ) = {u, v 1, v 2,..., v n } with u as the central vertex. Let X = {1, 2, 3,..., n}. Define f : V (K 1,n ) 2 X such that f(u) =, f(v 1 ) = X, f(v i ) = {1, 2,..., i 1}, 2 i n. Then, f (uv i ) = f(u) f(v i ) = f(v i ) for 1 i n. Therefore, f (E(K 1,n )) = f(v (K 1,n )). Then, f is injective and f(v (K 1,n ))

30 Topogenic Graphs 47 f (E(K 1,n )) = f(v (K 1,n )) is a topology on X. Figure 2.3 illustrates Theorem for K 1,5. Theorem The complete bipartite graph K m,n is topogenic for all positive integers m and n. Proof. Let {A, B} be the bipartition of K m,n, where A = m and B = n. Let A = {u 0, u 1, u 2,..., u m 1 } and B = {v 1, v 2, v 3,..., v n }. Then, V (K m,n ) = A B. Choose X = {1, 2, 3,..., m + n 1} as the ground set. Let f : V (K m,n ) 2 X be defined by f(u 0 ) =, f(u i ) = {1, 2,..., i}, 1 i m 1, f(v j ) = {m, m + 1,..., m + j 1}, 1 j n. Now, f(u 1 ) f(u 2 ) f(u 3 ) f(u m 1 ), f(v 1 ) f(v 2 ) f(v 3 ) f(v n ), f(u i ) f(v j ) = f(v (K m,n )), 1 i m 1, 1 j n. Now, f(u i ) f(v j ) = f (u i v j ) f (E(K m,n )), 1 i m 1 and 1 j n and f (u 0 v j ) = f(v j ) f(v (K m,n )), 1 j n. Moreover, f (u i v j ) f (u i+1 v j ) and f (u i v j ) f (u i v j+1 ). Hence, both f and f are injective and f(v (K m,n )) f (E(K m,n )) forms a topology on X. Hence K m,n is topogenic for all positive integers m and n.

31 Topogenic Graphs 48 Figure 2.7: A topogenic set-indexer of K 2,3 Figure 2.7 illustrates Theorem , for K 2,3. Theorem The tripartite graph K 1,m,n is topogenic for all positive integers m and n. Proof. Let {A, B, C} be a tripartition of K 1,m,n, where A = 1, B = m and C = n. Let A = {u}, B = {v 1, v 2, v 3,..., v m } and C = {w 1, w 2, w 3,..., w n }. Then, V (K 1,m,n ) = A B C. Choose X = {1, 2, 3,..., m + n} as the ground set and let f : V (K 1,m,n ) 2 X be defined by f(u) =, f(v i ) = {1, 2,..., i}, 1 i m, f(w j ) = {m + 1, m + 2,..., m + j}, 1 j n. Let f be the induced edge function. Now, f(v 1 ) f(v 2 ) f(v 3 ) f(v m ), f(w 1 ) f(w 2 ) f(w 3 ) f(w n ) and f(v i ) f(w j ) = f(v (K 1,m,n )), 1 i m, 1 j n.

32 Topogenic Graphs 49 Figure 2.8: A topogenic set-indexer of K 1,3,3 Further, f(v i ) f(w j ) = f (v i w j ) f (E(K 1,m,n )), 1 i m, 1 j n, f (v i w j ) f (v i+1 w j ), f (v i w j ) f (v i w j+1 ). f (uv i ) = f(v i ) f(v (K 1,m,n )), 1 i m and f (uw j ) = f(w j ) f(v (K 1,m,n )), 1 j n. Hence, both f and f are injective and f(v (K 1,m,n )) f (E(K 1,m,n )) forms a topology on X, whence K 1,m,n is topogenic for all positive integers m and n. Figure 2.8 illustrates Theorem , for K 1,3,3. Theorem The path P n for 2 n 14 admits topogenic set-indexer.

33 Topogenic Graphs 50 Proof. The topogenic labeling of P n for 2 n 14 are given in Table 2.1, Table 2.2, Table 2.3 and Table 2.4. f 2 : V (P 2 ) 2 X ; X = {1} V (P 2 ) f 2 (V (P 2 )) v 1, v 2 f 2 (v 1 ) =, f 2 (v 2 ) = {1} f 3 : V (P 3 ) 2 X ; X = {1, 2} V (P 3 ) f 3 (V (P 3 )) v 1, v 2, v 3 f 3 (v 1 ) = {1} f 3 (v 2 ) = f 3 (v 3 ) = {1, 2} f 4 : V (P 4 ) 2 X ; X = {1, 2, 3} V (P 4 ) f 4 (V (P 4 )) v 1, v 2, v 3, v 4 f 4 (v 1 ) = {1, 2, 3}, f 4 (v 2 ) =, f 4 (v 3 ) = {1}, f 4 (v 4 ) = {1, 2} f 5 : V (P 5 ) 2 X ; X = {1, 2, 3} V (P 5 ) f 5 (V (P 5 )) v 1, v 2, v 3, v 4, v 5 f 5 (v 1 ) = {2}, f 5 (v 2 ) = {1, 2, 3}, f 5 (v 3 ) =, f 5 (v 4 ) = {1}, f 5 (v 5 ) = {1, 2} f 6 : V (P 6 ) 2 X ; X={1,2,3} V (P 6 ) f 6 (V (P 6 )) v 1, v 2, v 3, v 4, v 5, v 6 f 6 (v 1 ) = {2, 3}, f 6 (v 2 ) = {3}, f 6 (v 3 ) =, f 6 (v 4 ) = {1}, f 6 (v 5 ) = {2}, f 6 (v 6 ) = {1, 3} Table 2.1: Topogenic set-indexers of P n ; 2 n 6 Given any graph G, its shadow graph Sh(G) is obtained by the adjunction of a new vertex v for each vertex v in G and then joining v to every neighbour of v in G (see [7]). Theorem The shadow graph Sh(K 1,n ) is topogenic for every positive integer n.

34 Topogenic Graphs 51 f 7 : V (P 7 ) 2 X ; X={1,2,3,4} V (P 7 ) f 7 (V (P 7 )) v 1, v 2, v 3, v 4, v 5, v 6, v 7 f 7 (v 1 ) = {2, 3}, f 7 (v 2 ) = {3}, f 7 (v 3 ) = {1, 2, 3, 4} f 7 (v 4 ) =, f 7 (v 5 ) = {1}, f 7 (v 6 ) = {2}, f 7 (v 7 ) = {1, 3} f 8 : V (P 8 ) 2 X ; X={1,2,3,4} V (P 8 ) f 8 (V (P 8 )) v 1, v 2, v 3, v 4, v 5, v 6, v 7, v 8 f 8 (v 1 ) = {2, 3}, f 8 (v 2 ) = {3}, f 8 (v 3 ) = {1, 2, 3, 4} f 8 (v 4 ) =, f 8 (v 5 ) = {1}, f 8 (v 6 ) = {2}, f 8 (v 7 ) = {1, 3}, f 8 (v 8 ) = {1, 2} f 9 : V (P 9 ) 2 X ; X={1,2,3,4} V (P 9 ) f 9 (V (P 9 )) v 1, v 2, v 3, v 4, v 5, v 6, v 7, v 8, v 9 f 9 (v 1 ) = {1}, f 9 (v 2 ) = {2}, f 9 (v 3 ) = {1, 2, 3} f 9 (v 4 ) = {1, 3}, f 9 (v 5 ) = {3}, f 9 (v 6 ) = {1, 3, 4}, f 9 (v 7 ) = {2, 3}, f 9 (v 8 ) =, f 9 (v 9 ) = {1, 4} f 10 : V (P 10 ) 2 X ; X = {1, 2, 3, 4} V (P 10 ) f 10 (V (P 10 )) v 1, v 2, v 3, f 10 (v 1 ) = {1}, f 10 (v 2 ) = {2}, f 10 (v 3 ) = {3} v 4, v 5, v 6, f 10 (v 4 ) = {4}, f 10 (v 5 ) = {1, 3}, f 10 (v 6 ) = {2, 4}, v 7, v 8, f 10 (v 7 ) = {2, 3, 4}, f 10 (v 8 ) = {1, 4}, v 9, v 10 f 10 (v 9 ) = {1, 2, 4}, f 10 (v 10 ) = Table 2.2: Topogenic set-indexers of P n ; 7 n 10 Proof. Let V (K 1,n ) = {u, v 1, v 2, v 3,..., v n } where u is the central vertex of K 1,n. Let u be the shadow vertex of u and v i be the shadow vertex of v i, 1 i n. Then, V (Sh(K 1,n )) = {u, v 1, v 2, v 3,..., v n, u, v 1, v 2,..., v n}. Choose X = {1, 2, 3,..., 2n + 1} as the ground set and define f : V (Sh(K m,n )) 2 X such that f(u) =, f(u ) = {1}, f(v i ) = {1, 2,..., i + 1}, 1 i n, f(v i ) = {1, 2,..., n + i + 1}, 1 i n. Now, f(u ) f(v 1 ) f(v 2 ) f(v n ) f(v 1) f(v 2) f(v n),

35 Topogenic Graphs 52 f 11 : V (P 11 ) 2 X ; X={1,2,3,4} V (P 11 ) f 11 (V (P 11 )) v 1, v 2, v 3, f 11 (v 1 ) =, f 11 (v 2 ) = {1}, f 11 (v 3 ) = {2} v 4, v 5, v 6, v 7 f 11 (v 4 ) = {3}, f 11 (v 5 ) = {4}, f 11 (v 6 ) = {1, 3}, f 11 (v 7 ) = {2, 4}, v 8, v 9, v 10, v 11 f 11 (v 8 ) = {2, 3, 4}, f 11 (v 9 ) = {1, 4}, f 11 (v 10 ) = {1, 2, 4}, f 11 (v 11 ) = {1, 2} f 12 : V (P 12 ) 2 X ; X={1,2,3,4} V (P 12 ) f 12 (V (P 12 )) v 1, v 2, v 3, f 12 (v 1 ) =, f 12 (v 2 ) = {1}, f 12 (v 3 ) = {2} v 4, v 5, v 6, v 7 f 12 (v 4 ) = {3}, f 12 (v 5 ) = {4}, f 12 (v 6 ) = {1, 3}, f 12 (v 7 ) = {2, 4}, v 8, v 9, v 10, f 12 (v 8 ) = {2, 3, 4}, f 12 (v 9 ) = {1, 4}, f 12 (v 10 ) = {1, 2, 4}, v 11, v 12 f 12 (v 11 ) = {1, 2}, f 12 (v 12 ) = {1, 3, 4} Table 2.3: Topogenic set-indexers of P n ; n = 11, 12 f (uv i ) = f(v i ) f(v ), 1 i n, f (uv i ) = f(v i ) f(v ), 1 i n, and f (u v i ) = f(v i ) f(u ) f(v i ). Hence, both f and f are injective and f(v ) f (E) forms a topology on X. Hence, the shadow graph of K 1,n is topogenic. Figure 2.9 illustrates Theorem , for Sh(K 1,4 ). For arbitrary positive integers r and s, the bistar T r,s is the tree of diameter 3 with the central edge uv such that there are r pendant vertices adjacent to u and s pendant vertices adjacent to v. Theorem The bistar T r,s is topogenic for all positive integers r, s.

36 Topogenic Graphs 53 f 13 : V (P 13 ) 2 X ; X={1,2,3,4} V (P 13 ) f 13 (V (P 13 )) v 1, v 2, v 3 f 13 (v 1 ) =, f 13 (v 2 ) = {1}, f 13 (v 3 ) = {2} v 4, v 5, v 6 f 13 (v 4 ) = {3}, f 13 (v 5 ) = {4}, f 13 (v 6 ) = {1, 3} v 7, v 8, v 9 f 13 (v 7 ) = {2, 4}, f 13 (v 8 ) = {2, 3, 4}, f 13 (v 9 ) = {1, 4} v 10, v 11, f 13 (v 10 ) = {1, 2, 4}, f 13 (v 11 ) = {1, 2} v 12, v 13 f 13 (v 12 ) = {1, 3, 4}, f 13 (v 13 ) = {1, 2, 3} f 14 : V (P 14 ) 2 X ; X={1,2,3,4} V (P 14 ) f 14 (V (P 14 )) v 1, v 2, v 3 f 14 (v 1 ) =, f 14 (v 2 ) = {1}, f 14 (v 3 ) = {2} v 4, v 5, v 6 f 14 (v 4 ) = {3}, f 14 (v 5 ) = {4}, f 14 (v 6 ) = {1, 3} v 7, v 8, v 9 f 14 (v 7 ) = {2, 4}, f 14 (v 8 ) = {2, 3, 4}, f 14 (v 9 ) = {1, 4} v 10, v 11, f 14 (v 10 ) = {1, 2, 4}, f 14 (v 11 ) = {1, 2} v 12, v 13, v 14 f 14 (v 12 ) = {1, 3, 4}, f 14 (v 13 ) = {1, 2, 3}, f 14 (v 14 ) = {3, 4} Table 2.4: Topogenic set-indexers of P n ; n = 13, 14 Proof. Let V (T r,s ) = {u, u 1, u 2,..., u r, v, v 1, v 2,..., v s } where u and v are the central vertices and u i is adjacent to u, 1 i r and v j is adjacent to v, 1 j s. Let m and n be the consecutive integers such that 2 m 1 r 2 n 2. Choose a set X of cardinality n + s and B a subset of X with cardinality n so that X = {1, 2, 3,..., n + s} and B = {1, 2, 3,..., n} and let f : V (T r,s ) 2 X be defined by f(u) = B, f(u i ) = B i, 1 i r, B i 2 B {B, } and for i j, B i B j, f(v) =, f(v j ) = B {n + 1, n + 2,..., n + j}, 1 j s. Hence, both f and f are injective. Also, f(v ) f (E) = {2 B, B {n + 1}, B {n + 1, n + 2},..., B {n + 1, n + 2, n + 3,..., n + s}},

37 Topogenic Graphs 54 Figure 2.9: A topogenic set-indexer of Sh(K 1,4 ) which is a topology on X. Hence, T r,s is topogenic for every integer r, s 1. Figure 2.10 illustrates Theorem , for T 5,4. Theorem P 2 +K m is topogenic for every positive integer m. Proof. Let V (K m ) = {v 1, v 2,..., v m } and V (P 2 ) = {u 1, u 2 }. Then, V (P 2 + K m ) = V (K m ) V (P 2 ). Let X = {1, 2, 3,..., m + 1} be the ground set and define f : V (P 2 + K m ) 2 X by letting f(v i ) = {1, 2,..., i}, 1 i m, f(u 1 ) =, f(u 2 ) = {m + 1}. By the definition of f, both f and f are injective. Now,

38 Topogenic Graphs 55 Figure 2.10: A topogenic set-indexer of T 5,4 f(v 1 ) f(v 2 ) f(v 3 ) f(v m ); f(v i ) f(u 2 ) = f(v ), 1 i m; f(v i ) f(u 2 ) = f (v i u 2 ) f (E), 1 i m and f(u 1 v i ) = f(v i ) f(v ), 1 i m. Also, f(v ) f (E) forms a topology on X and P 2 + K m is topogenic. Figure 2.11 illustrates Theorem , for P 2 + K 4. Theorem P 3 +K m is topogenic for every positive integer m. Proof. Let V (K m ) = {v 1, v 2,..., v m } and V (P 3 ) = {u 1, u 2, u 3 }. Then V (P 3 + K m ) = V (K m ) V (P 3 ). Without loss of generality, choose X = {1, 2, 3,..., m + 2}. Let f : V (P 3 + K m ) 2 X be defined by f(v i ) = {1, 2,..., i}, 1 i m; f(u 2 ) =

39 Topogenic Graphs 56 Figure 2.11: A topogenic set-indexer of P 2 + K 4 f(u 1 ) = {m + 1}; f(u 3 ) = {m + 1, m + 2}. Then, both f and f are injective. Moreover, f(v 1 ) f(v 2 ) f(v 3 ) f(v m ) and f(u 1 ) f(u 3 ), f(v i ) f(u j ) = f(v ), 1 i m, j = 1, 2, 3, f(v i ) f(u j ) = f (v i u j ) f (E), 1 i m, j = 1, 3 and f(u 2 v i ) = f(v i ) f(v ), 1 i m. Therefore, f(v ) f (E) forms a topology on X. Hence, P 3 + K m is topogenic. Figure 2.12 illustrates Theorem , for P 3 + K 3. In [3], Acharya discussed about the graphs whose subdivision graph is bipartite self-complementary and proved that there exists exactly seven such graphs, which is depicted in Figure 2.13.

40 Topogenic Graphs 57 Figure 2.12: A topogenic set-indexer of P 3 + K 3 Figure 2.13: Graphs whose subdivision graphs are bipartite self-complementary

41 Topogenic Graphs 58 Theorem The graph G whose subdivision graph is bipartite self-complementary is topogenic if and only if it is not isomorphic to K 4. Proof. The topogenic labeling of graphs, whose subdivision graphs are bipartite self-complementary and is not isomorphic to K 4 is shown in Figure Converse follows from Theorem Figure 2.14: Topogenic bipartite self-complementary graphs The following problem is worth of further investigation. Problem 2: Characterize topogenic graphs, in particular topogenic

42 Topogenic Graphs 59 trees. Establish bounds for Υ(G). 2.3 Graceful topogenic graphs We have already mentioned that the condition f(v (G)) is not necessary for f to be a set-graceful labeling of a graph G. It is an important open problem to determine set-graceful graphs G which admit graceful set-indexers f for which / f(v (G)), since such set-graceful labelings f do not render f(v (G)) f (E(G)) to be a topology on X. This observation makes the following new definition relevant to the study of topogenic graphs. Definition A graph G = (V, E) is a graceful topogenic graph if it admits a graceful topogenic set-indexer, in the sense that it is a set-indexer f : V (G) 2 X of G such that f(v (G)) f (E(G)) = 2 X. Observation Every topologically set-graceful graph has a graceful topogenic set-indexer. Proof. Suppose G is topologically set-graceful. Then, there exists a graceful set-indexer f of G with respect to a non-empty ground set X, giving f(v (G)) as a topology on X. Therefore, f(v (G)) and

43 Topogenic Graphs 60 f (E(G)) = 2 X. Then, f(v (G)) f (E(G)) = 2 X, the discrete topology on X. Remark A set-sequential labeling f cannot be topogenic due to the absence of empty set, in the entire set-labeling. However, given a set-sequential graph (connected or disconnected) by augmenting a new isolated vertex, say v and by assigning the empty set to v, we get a disconnected graph with a graceful topogenic set-indexer. That is, if G is a set-sequential graph then G {v} is a graceful topogenic graph. A graceful topogenic set-indexer need not be a graceful set-indexer as illustrated in the case of path P 3, displayed in Figure 2.14; it is well known that P 3 is not set-graceful. Hence, we have Theorem Every set-graceful graph G having a graceful setindexer f with f(v ) is graceful topogenic. Figure 2.15: A graceful topogenic set-indexer of P 3. Proposition For every positive integer n, there exists a connected graceful topogenic graph of order n.

44 Topogenic Graphs 61 Figure 2.16: Illustration of Proposition Proof. Let G n be the (n 1)-star whose vertices are labeled u 1, u 2,..., u n, where u 1 being the central vertex. Let k be the smallest positive integer such that n 2 k. Choose a set X of cardinality k, say X = {1, 2,..., k}. Define f : V 2 X such that f(u 1 ) = X, f(u i ) = X i ; X i 2 X X for 2 i n. The resulting graph is a graceful topogenic graph. Figure 2.17 illustrates Proposition In this figure p = 5 and so k = 3. Invoking Lemma 2.2.9, for a topogenic (p, q)-graph G, max{p, q + 1} ϱ 0 (G) ϱ 1 (G) p + q δ, (1) where δ = δ(g) is the minimum vertex degree in G.

45 Topogenic Graphs 62 The topogenic strength of a graceful topogenic graph is 2 n, where n is the cardinality of the ground set X. Thus from (1) we have, 2 n p+q δ or 2 n + δ p + q (2) For a non-trivial connected graph G, δ(g) 1. Hence we have 2 n +1 p + q. This shows that, for a graceful topogenic connected (p, q)-graph, p + q 2 n for any positive integer n. Hence, it is worthwhile to analyze the bounds for order and size of a graceful topogenic graph with respect to a set X of cardinality n. Clearly p + q 2 n+1. Now, since / f (E(G)) and all the nonempty elements of f(v (G)) may be repeated in f (E(G)) (for example, complete graphs, star) we get that p + q 2 n + 2 n 1 = 2 n+1 1 (3) From (2) and (3) we get 2 n + δ p + q 2 n+1 1. Thus, we have Theorem A necessary condition for a (p, q)-graph G to admit a graceful topogenic set-indexer with respect to a set X of cardinality n

46 Topogenic Graphs 63 is 2 n + δ p + q 2 n+1 1 where δ = δ(g) is the minimum degree of G. Remark Both the bounds of the inequality in Theorem are sharp as shown by the following examples. Example Consider the labeling of K 3 given in Figure 2.17 (a). Here X = 2, 2 n = 4, δ = 2. Therefore, 2 n + δ = 6 = p + q. 2. Consider the labeling of K 1,7 given in Figure 2.17 (b). Here X = 3, 2 n+1 1 = 15 and p + q = 15. Theorem A 2-regular graph with 2 k vertices is not graceful topogenic. Proof. Assume, to the contrary, that G be a 2-regular graph with 2 k vertices which is graceful topogenic with respect to a set X of cardinality n. Then by Theorem we have 2 n k + 2 k 2 n+1 1 = 2 n k+1 2 n+1 1. But there exists no n N satisfying this inequality.

47 Topogenic Graphs 64 Figure 2.17: Graphs attaining bounds of the inequality in Theorem Theorem K 4, the complete graph on four vertices is not graceful topogenic. Proof. K 4 is a 3-regular graph on 4 vertices. Therefore, by Theorem 2.3.4, we have 2 n n+1 1 Now, 2 n = n 2 and 10 2 n+1 1 = n 3. There exists no n satisfying this inequality. Theorem K 5, the complete graph on five vertices is not graceful topogenic. Proof. K 5 is a 4-regular graph on 5 vertices. Therefore, by Theorem 2.3.4, we have 2 n n+1 1. Now, 2 n = n 3 and 15 2 n+1 1 = n 3. Therefore, X = 3. Now, 2 3 = 8, but

48 Topogenic Graphs 65 q = 10. Hence, to maintain the injectivity of f we must have n 4, a contradiction. Hence, K 5 is not graceful topogenic Graceful topogenic complete graphs We have seen already, the complete graphs K 1 and K 2 are topogenic. Moreover, K 1 and K 2 are graceful topogenic. Now, we shall show that the complete graph K 3 is also graceful topogenic with the following vertex assignments. Let V (K 3 ) = {u 1, u 2, u 3 } and X = {1, 2}. Define f : V (K 3 ) 2 X defined by f(u 1 ) =, f(u 2 ) = {1} and f(u 3 ) = {2}. Then, f (E(K 3 )) = {{1}, {2}, {1, 2} = X}. Thus, f(v (K 3 )) f (E(K 3 )) = {, {1, 2}, {1}, {2}} = 2 X. By Theorem and Theorem 2.3.8, K 4 and K 5 are not graceful topogenic. The topogenic set-indexer of K 6 defined in Theorem is a graceful topogenic labeling. Figure 2.6 illustrates a graceful topogenic labeling of K 6. Remark From Theorem , if K p is graceful topogenic with respect to a ground set X of cardinality m, then p 2 p + 2 = 2 m+1. This implies that p = 1, 2, 3, 6. Hence, we characterize gracefully topogenic complete graphs.

49 Topogenic Graphs 66 Theorem The complete graph K p is graceful topogenic if and only if p {1, 2, 3, 6}. Conjecture 2. The complete graphs K p is topogenic if and only if K p is graceful topogenic More classes of graceful topogenic graphs Theorem The star K 1,n is graceful topogenic for every positive integer n. Proof. Let V (K 1,n ) = {u, v 1, v 2,..., v n } where u is the central vertex. Choose a set X = {1, 2,..., k} where k is the smallest positive integer such that n 2 k. Define f : V (K 1,n ) 2 X so that f(u) = X; f(v 1 ) = ; f(v i ) = X i, (i = 2, 3,..., n), where X i X and X i X j if i j. Then, f(v (K 1,n )) f (E(K 1,n )) = 2 X, the discrete topology on X. Theorem The complete bipartite graph K 2,m, where m = t 2 Cj t 1, for any t N, t > 2 is graceful topogenic. j=0 Proof. Let K 2,m = (S, T ) where S = {u 1, u 2 } and T = {v 1, v 2,..., v m }. Choose a set X so that X = t where t > 2. Fix an element x X. Define f : V (K 2,m ) 2 X such that

50 Topogenic Graphs 67 f(u 1 ) = ; f(u 2 ) = X, f(v i ) = X i, (i = 1, 2,..., m) where X i X with x X i and X i X j for i j. Then, f(v (K 2,m )) f (E(K 2,m )) = 2 X, the discrete topology on X. Figure 2.18 illustrate Theorem when t = 4. Figure 2.18: Graceful topogenic labeling of K 2,7 Theorem The totally disconnected graph K m is graceful topogenic if and only if m = 2 k for some k. Proof. Let K m be graceful topogenic with respect to the set X. Then there exists f : V 2 X such that f(v (K m )) f (E(K m )) = 2 X. But E(K m ) =. Hence, we must have f(v (K m )) = 2 X. Therefore, m = f(v (K m )) = 2 k for some k. To prove the converse assume that m = 2 k for some k. Choose X =

51 Topogenic Graphs 68 {1, 2,..., k}. Now, assign the elements of 2 X to the vertices in an injective manner. Then, f(v (K m )) = 2 X and hence K m is graceful topogenic. 2.4 References 1. Acharya, B. D., Set-valuations of graphs and their applications, MRI Lecture Notes in Applied Mathematics, No.2, Mehta Research Institute of Mathematics and Mathematical Physics, Allahabad, 1983, Acharya, B. D., Set-indexers of a graph and set-graceful graphs, Bull. Allahabad Math. Soc., 16(2001), Acharya, B. D., There are exactly seven graphs whose subdivision graphs are bipartite self-complementary, Nat. Acad. Sci. Letters, Vol.11, No.5, 1988, Acharya, B. D. and Hegde, S. M., Set-sequential graphs, Nat. Acad. Sci. Letters, 8(12), 1985, Acharya, B. D., Germina, K. A., Princy, K. L. and Rao, S. B., Topologically set-graceful graphs, Research Report

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