On (k, d)-multiplicatively indexable graphs

Transcription

1 Chapter 3 On (k, d)-multiplicatively indexable graphs A (p, q)-graph G is said to be a (k,d)-multiplicatively indexable graph if there exist an injection f : V (G) N such that the induced function f : (E(G)) = {k, k + d,..., k + (q 1)d} defined by f (uv) = f(u)f(v) for every uv E(G). If f(g) = {1,,..., p} then G is said to be a (k, d)-strongly-multiplicatively indexable graph. The corresponding labelings are said to be a (k, d)-multiplicative indexer and a (k, d)-strongly multiplicative indexer respectively. The first section deals with some properties of (k, d)-multiplicatively indexable graphs. Then we find a necessary and sufficient condition for the complete graph C 3 to be (k, d) multiplicatively indexable. Moreover we can embed every graph into a (k, 1)-multiplicatively indexable graph. In the second section we give in some cases for which strongly (k, d)-multiplicatively indexable graphs does not exists. Also we show that there exists strongly (k, 1)-multiplicatively indexable graphs for all k > 1 and strongly (k, )-multiplicatively indexable graphs for all k even. Finally we characterize some particular strongly (k, d)-multiplicatively indexable graphs. 57

2 Chapter Introduction B.D. Acharya and S.M. Hegde,[AH91a] defined (p, q)-graph G = (V, E) multiplicative if there exist an injection f : V (G) N such that f (uv) = f(u)f(v) for every uv E(G) are all distinct. L. W. Beineke and S.M. Hegde,[BH01] defined (p, q)-graph G = (V, E) strongly multiplicative if there exist a bijection f : V (G) {1,,..., p} such that f (uv) = f(u)f(v) for every uv E(G) are all distinct. In this chapter we generalize the multiplicative graphs into (k, d)-multiplicatively indexable graphs (k,d)-multiplicatively indexable graphs and strongly multiplicative graphs into (k, d)-strongly multiplicatively indexable graphs. The notion of (k, d)-multiplicative indexers is closely related to the other known numberings such as strongly multiplicative labelings and strong indexers, [AH91b] [BH01]. Here we carried out a separate study of (k, d)-multiplicatively indexable graphs. Most of the theorems and properties given by B.D. Acharya and S.M. Hegde,[AH90] will be of immense use in our study here. Also we try to reformulate some results into our terminology. Such an attempt is given in the following few theorems. Definition A (p, q)-graph G is said to be a (k, d) - multiplicatively indexable graph if there exist an injection f : V (G) N such that the induced function f (E(G)) = {k, k + d,..., k + (q 1)d} defined by f (uv) = f(u)f(v) for every uv E(G). If f(g) = {1,,..., p} then G is said to be a strongly (k, d)- strongly multiplicatively indexable graph. The corresponding labelings are called

3 Chapter 3 59 a (k, d)-multiplicative indexer and a (k, d)-strongly multiplicative indexer respectively. Figure 3.1 gives two (k, d)-multiplicatively indexable graphs and Figure 3. gives two (k, d)-strongly multiplicatively indexable graphs. Figure 3.1: Figure 3.:

4 Chapter (k, d)-multiplicatively indexable graphs. Theorem Let G be a (p, q)-graph. Then G is (k, d) - multiplicatively indexable if and only if it is (m k, m d)-multiplicatively indexable for all non - negative integer m. Proof. Let G be (k, d) - multiplicatively indexable graph. Hence there exist a multiplicative indexer f : V (G) N such that f is injective and f (E(G)) = {k, k + d,..., k + (q 1)d}. Now define a multiplicative labelling f 1 : V (G) N as, f 1 (u i ) = mf(u i ) for every u i V (G), m > 0. Then f 1 (u iu j ) = f 1 (u i )f 1 (u j ) = mf(u i )mf(u j ) = m f (u i u j ) Hence f (E(G)) = {m k, m (k + d),..., m (k + (q 1)d)}. Therefore f 1 is (m k, m d)-multiplicative indexer for m > 0. Conversely let G is (m k, m d)-multiplicatively indexable. m = 1 we get G is (k, d)-multiplicatively indexable. Corollary 3... Taking If f is a (k, d)-multiplicative indexer and f 1 is a (m k, m d)-multiplicative indexer on G, then f 1 (e) = m q f (e) f 1 (e) = m f (e) for every e E(G). Theorem Let G = (V, E) be a (p, q)-graph which admits a (k, d)-multiplicative indexer, f where k and d are not simultaneously even. Let X = {u V (G) : f(u) is odd}. Then the number of edges with both ends in X is (i) [ ] q+1 if k and d are both odd.

5 Chapter 3 61 (ii) [ q ] if k is even and d is odd, and (iii) q if k is odd and d is even. Proof. Since k and d are not simultaneously even, the number of odd numbers in f (E(G)) is precisely given by (i), (ii) and (iii) in the respective cases. Further an edge e of G has both ends in X if and only if f (e) is odd and hence the result follows. Corollary (mod) for every u V (G). Theorem Let G be a connected (k, k)-multiplicatively indexable graph. If with k is odd and d is even, Then f(u) Then for any (k, k)-multiplicative indexer f of G, if k divides f(u) for every u V (G) {v}, then f(v) = 1. Proof. Since f is a (k, k)-multiplicative indexer of G, then f (E(G)) = {jk : 1 j q}. Then there exists an edge xy E(G) with f (xy) = k. Assume k be a prime number. Now f (xy) = f(x)f(y), which implies either f(x) or f(y) equal to 1, since k divides f(u) for every u V (G) {v}. If k is not a prime say, k = k 1 k where k 1 < k and k < k, then there exists no u V (G) such that f(u) = k 1 or k. Hence either f(x) or f(y) is to be equal to 1. Since k divides f(u) for every u V (G) {v}, we have x = v so that f(v) = 1. Remark The converse of the above proposition is not always true. For example consider the figure 3.3 which is a (3, 3)- multiplicatively indexable graph. Here f(v (G)) = {1,, 3, 4, 5, 6, 9}

6 Chapter 3 6 and f (E(G)) = {3, 6, 9, 1, 15, 18}. Note that 1 f(v (G). However 3 does not divide f(u) for all u V (G) {v}. Figure 3.3: Remark However note that every (k, d)-multiplicatively indexable graph with k prime or square of a prime 1 should be assigned as a vertex labelling. Remark For every (k, d)-multiplicatively indexable graph with k and d relatively prime 1 f(v (G)) The following theorem gives a straight forward neessary and sufficient condition for the cycle to be (k, d)-multiplicatively indexable. Theorem The cycle C 3 is (k, d)-multiplicatively indexable if and only if there exists three positive integers a, b, c such that a < b < c and a is the harmonic mean of b and c where k = ab and d = a(b c). Proof. Suppose C 3 = (u v w u) is (k, d)-multiplicatively indexable. Let f : V (C 3 ) N defined by f(u) = a, f(v) = b and f(w) = c with a < b < c be a multiplicative indexer of G so that ab = k, bc = k + d

7 Chapter 3 63 and ca = k + d. Clearly a = bc b+c and d = a(b c). Conversely if there exist positive integers a, b, c such that a < b < c and a is the harmonic mean of b and c then f : V (C 3 ) N defined by f(u) = a, f(v) = b and f(w) = c is a (k, d)- multiplicative indexer of G where k = ab and d = a(b c). Theorem The cycle C 3 is (k, d)-multiplicatively indexable for some k, d > 0 then, d 0{modk}. Proof. Since C 3 is (k, d)- multiplicatively indexable then f (E(C 3 )) = {k, k + d, k + d} = {ab, bc, ca}. Without loss of generality let ab = k, bc = k + d and ca = k + d then abc = (k + d)(k + d) so that kc = k + 3kd + d hence d 0(modk). Theorem If C 3 is (k, d)-multiplicatively indexable for some k, d > 0 then 1 / f(v (C 3 )). Proof. Let V (C 3 ) = {u, v, w} Since C 3 is (k, d)-multiplicatively indexable, there exists f : V (C 3 ) N and f (E(C 3 )) = {k, k + d, k + d}. Let f(u) = 1 and f(v) = b and f(w) = c with b < c. Then f (E(C 3 )) = {b, c, bc}. Since b, c, bc are in arithmetic progression, we get b = c b 1+c.That is c = b Since b > 0 and b 1 we get c < 0 which is a contradiction. Hence 1 / f(v (C 3 )). Theorem The cycle C 4 is not (k, d)-multiplicatively indexable for any k, d > 0.

8 Chapter 3 64 Proof. If possible, assume C 4 is (k, d)-multiplicatively indexable for some k, d > 0 Let V (C 4 ) = {u 1, u, u 3, u 4 } and f : V (C 4 ) N by f(u 1 ) = a, f(u ) = x 1, f(u 3 ) = b, f(u 4 ) = x. Then f (E(C 4 )) = {ax 1, ax, bx 1, bx } Out of the 4 possible numberings of edge values, the only six nonisomorphic way of numbering of the edges are given below; ax 1, ax, bx 1, bx ax 1, ax, bx, bx 1 ax 1, bx 1, ax, bx ax 1, bx 1, bx, ax ax 1, bx, ax, bx 1 ax 1, bx, bx 1, ax. Now consider first way of numbering. Since they are in arithmetic progression we have a(x x 1 ) = b(x x 1 ) a = b, which is a contradiction to the fact that f is injective. A similar proof holds for the other way of numberings also. The above discussions lead to the following conjecture: Conjecture The cycle C n is (k, d)-multiplicatively indexable if and only if n 3. Theorem The star K 1,b is (k, d)-multiplicatively indexable. Proof. Let V (K 1,b )) = {u 0, u 1,..., u b }. Define f : V (K 1,b ) N by f(u 0 ) = 1 and f(u i ) = k + (i 1)d for 1 i b. It can be easily

9 Chapter 3 65 verified that f is a (k, d) multiplicative indexer of K 1,b. Corollary If k and d are relatively prime, the above (k, d)- multiplicative indexer is unique in the sense that 1 f(v (K 1,b )) and 1 should be assigned at the central vertex of the star. it is interesting to see that any (k, d)-multiplicatively indexable graphs together with a star could be used to generate an infinite ascending chain of (k, d)-multiplicatively indexable graphs each of which contains its predecessor as a subgraph, which leads to the following theorem. Theorem Let G = (V, E) be a (p, q)-graph which admits a (k, d)-multiplicative indexer f such that there exists exactly one pair of vertices u, v with f(v) = mf(u) for some non-zero positive integer m. Consider G 1, G,..., G n, n copies of G with (u i, v i ) belongs to V (G i ) such that f(v i ) = mf(u i ). Let H be the graph obtained by identifying u and v 1, u 3 and v,..., u n and v n 1. Then there exists a positive integer s such that H K 1,s is (k, d)-multiplicatively indexable. Proof. Let G admits a (k, d)-multiplicatively indexaer f. Let G 1, G,..., G n be n copies of G such that V (G 1 ) = {u 1 v 1, u 11, u 1,..., u 1(p ) } V (G ) = {u v, u 1, u,..., u (p ) } V (G n ) = {u n v n, u n1, u n,..., u n(p ) }.

10 Chapter 3 66 Let u i v i V (G i ) such that f(v i ) = mf(u i ) and H be the graph obtained by by identifying u and v 1, u 3 and v,..., u n and v n 1. It is given that f is a (k, d)-multiplicative indexer on G = G 1. Let f(u 1 ) = a 1, f(v 1 ) = a p, f(u 11 ) = a, f(u 1 ) = a 3, f(u 1(p ) ) = a p 1 so that f(v 1 ) = mf(u 1 ). On G extend f as f(u ) = f(v 1 ) = mf(u 1 ), f(u 3 ) = mf(v ) = m f(u 1 ) and f(u i ) = mf(u 1i ), 1 i p. In general on G n, we have f(u n ) = f(v n 1 ) = m n 1 f(u 1 ), f(v n ) = mf(u n ) and f(u ni ) = mf(u n 1i ), 1 i p. Here all the edge labels of H are the terms of the arithmetic progression with first term k = minf (e), e E(G) and common difference d. Let the highest edge value of H be the r th term of the above arithmetic progression, then k + (r 1)d = m n 1 a i m n 1 a j = m n 1 a i a j where a i a j = maxf (e), e E(G). Thus we get k + (r 1)d = m n 1 [k + (q 1)d] so that r = m(n 1) [k+(q 1)d] k d + 1. In order to make the graph H to be (k, d)-multiplicatively indexable we add an isolated vertex w with f(w) = 1. There are qn edges of H. Hence for each number x with k < x < k + (r 1)d, x is an element of the arithmetic progression and x is not an edge label add a vertex w x such that f(w x ) = x.then the resulting graph obtained by the union of H and a star K 1,s where s = t qn is a (k, d)-multiplicatively indexable graph.

11 Chapter 3 67 The following figure 3.4 illustrates the theorem. Figure 3.4: We pose the following problem: Problem indexable graphs. Characterize the class of (k, d)-multiplicatively Theorem Every graph can be embedded into a (k, 1)- multiplicatively indexable graph. Proof. Let G be a graph with n vertices and let V (G) = {u 1, u,..., u n }. Let k < p <... < p n where each p i is a prime and let, f(u i ) = p i ; i n. Now add a vertex v, let f(v) = 1and join v to the vertices u 1 u,..., u n so that k, p,..., p n are the labels of the edges vu 1, vu,..., vu n. Now for each integer x with k < x < p n 1 p n and x is not an edge label, add a new vertex v x, let f(v x ) = x and join v x to v. The resulting graph G 1 is (k, 1)-multiplicatively indexable and G is an induced subgraph of G 1.

12 Chapter 3 68 Figure 3.5 gives the embedding of a graph with 5 vertices into a (6, 1)-multiplicatively indexable graph. Figure 3.5: 3.3 (k, d)-strongly multiplicatively indexable graphs In this section we establish some classes of graphs which do not admit a (k, d)-strongly multiplicative indexer for any particular values of k and d. Theorem The complete graph K n is not (k, d)-strongly multiplicatively indexable for n 3. Proof. Since K n for n > 5, f is not injective it is enough to check for K n for n =1,, 3, 4, 5. For K 1 and K the result is trivial. Consider K 3. In this case for any multiplicative labelling f : V (K 3 ) {1,, 3}

13 Chapter 3 69 the elements of f (E(G)) = {, 3, 6} and not in the form {k, k +d, k + d}. Hence K 3 is not (k, d)-strongly multiplicatively indexable. By a similar arguments one can show that K 4 and K 5 are not (k, d)-strongly multiplicatively indexable. Theorem There does not exists any (p, q)-graph which is (k, d)- strongly multiplicatively indexable graphs in the following cases: (1) For every k odd and d even () For every k even, d odd, and p d and d does not divide k (3) For every k odd and d odd and p d and d does not divide k. Proof. (1) Since k odd and d even f E(G)) consists only odd numbers. Hence there does not exists a vertex u of G such that f(u) is even so that f(v (G)) {1,,..., p} and f is not a (k, d)-strongly multiplicative indexer of G. () Let k be even, d odd, and p d and d does not divide k and f be a (k, d) strongly multiplicative indexer of G. Hence f (E(G)) = {k, k+d,..., k+(q 1)d}. Also f (e) k( mod d) for every e E(G). Since p d and d f(v (G)) then there exists an edge xy E(G) with either, f(x) = d or f(y) = d. In this case f (xy) 0(mod d) which is not possible since d does not divide k and f (xy) k(mod d) for every e E(G). Hence f is not a (k, d) strongly multiplicative indexer of G. (3) The proof of this case follows immediately from case (). Corollary For a star K 1,p 1 there exists a unique (, 1)- strongly multiplicative indexer which assigns 1 to the central vertex

14 Chapter 3 70 and the numbers, 3,..., p to the vertices of unit degree. Theorem A star K 1,n is (k, k)-strongly multiplicatively indexable if n = k 1. In otherwords the maximum possible edges of a (k, k)-strongly multiplicatively indexable star is k 1. Proof. Since f(k 1,n )) = {1,,..., (n + 1)} and f (E(K 1,n )) = {k, k,..., nk}, we should assign the value k to the central vertex and the numbers 1,,..., n should assign to the pendant vertices. Hence n = k 1 and so the maximum possible edges of a (k, k)-strongly multiplicatively indexable star is k 1. Theorem Let k be a positive integer and let n = k. Then the graph G obtained from the star K 1,n by subdividing exactly one edge is (k, k)-strongly multiplicatively indexable. Proof. Let V (K 1,n ) = {u, u 1,..., u n where n = k, degu = n and let v be the vertex subdividing the edge uu 1. Then f : V (G) {1,,..., n} defined by f(u) = k, f(v) = 1 and f(u j ) = k if j = i j if 1 j k 1 j + 1 if k j k is a (k, k)-strongly multiplicative indexer G.

15 Chapter 3 71 The following figure 3.6 illustrates the theorem for k = 3. Figure 3.6: Remark In general the converse of the theorem is not true as the figure 3.7 illustrates. Let G be a (4, 4)-strongly multiplicatively indexable graph with f(v (G)) = {1,,..., 8} and f (E(G)) = {4, 8, 1, 16, 0, 4, 8}. Here n is not equal to k. However, if k is a prime number, the converse of the above theorem is true which is our next theorem. Figure 3.7:

16 Chapter 3 7 Theorem Let G be the graph obtained from the star K 1,n by subdividing exactly one edge. If k is a prime number then G is (k, k)-strongly multiplicatively indexable if and only if n = k. Proof. Let the graph obtained by joining one end vertex of K 1,n to an isolated vertex and assume G is (k, k)-strongly multiplicatively indexable. Hence f(v (G)) = {1,,..., (n + 1)} and f (E(G)) = {k, k, 3k,..., nk}. Assume k is a prime number, there exists an edge e = (uu 1 ) E(G). Since k is prime, one of f(u) = 1 or f(u 1 ) = 1 and the other is of labelling k. If n = k there exists an edge which is labelled as k. By theorem the possible way of labelling the consecutive numbers, 3,..., k 1 only to the vertices that are adjacent to u 1 and the value k can be assigned to the vertex which is adjacent to u. Let it be v so that f(v) = k. Since k is in f (E(G)), we should assign all the consecutive numbers from k + 1 to k 1 to the vertices that are adjacent to u 1. Then we have f(v (G)) = {1,,..., k } and f (E(G)) = {k, k,..., (k 1)k}. Hence the number of edges of G is n = k. Converse part follows from the theorem Theorem For all positive integer k > 1 there exists a connected (k, 1)-strongly multiplicatively indexable graph. Proof. We know that for m k, k! + m is divisible by m. Consider a graph G with k! + 1 vertices and k! edges. Let V (G) = {u 1, u,..., u k!+1 } and E(G) = {u 1 u i : k i k! + 1} {u j u k!+j : j j k 1}. Define f : V (G) {1,,..., k! + 1} as f(u i ) =

17 Chapter 3 73 i, 1 i k! + 1 and f induces f on E(G) as follows: f (u 1 u i ) = i, i k! + 1 and f (u j u k!+j ) = k! + j, k j k 1. Clearly j f (E(G)) = {k, k + 1,..., k! + 1, k! +,..., k! + k 1}. That is G is a connected (k, 1)-strongly multiplicatively indexable graph. Figure 3.8 illustrates the theorem for k = 7. Figure 3.8:

18 Chapter 3 74 Theorem For all positive even integer k there exists a connected (k, )-strongly multiplicatively indexable graph. Proof. Let k be an even number and G be a graph with k!+4 vertices. Let V (G) = {u, v, u 1, u,..., u m, v 1, v,..., v n } where m = k! k+4 and n = k 4 so that m + n + = k!+4. Let E(G) = E 1(G) E (G) E 3 (G) where E 1 (G) = {uu i, 1 i m}, E (G) = {vu k } and E 3 (G) = {u j1 v 1, u j v, u j3 v 3,..., u jn v n, 1 ji m}. Define f : V (G) {1,,..., k!+4 } by f(u) =, f(v) = 1, f(u i) = i+ k, 1 i m and f(v i ) = i +, 1 i n where n = k 4. Here we note that f(u 1 ) = 1 + k, f(u ) = + k, in particular f(u k ) = k + k = k. Similarly f(v 1 ) = 3, f(v ) = 4,..., f(v n ) = k. Since m + n + = k!+4 k!+6, we have the following relation given by 1 < 3 < k!+8 4 < k!+10 5 <... < k!+k k < k!+4. That is we are selecting certain values from the set {1,,..., k!+4 }. Let u j1, u j,..., u jn be the vertices such that f(u j1 ) = k!+6 3, f(u j) = k!+8 4,..., f(u jn) = k!+k k. Clearly the induced edge function f is injective, for if f (uu i ) = f(u)f(u i ) = i + k, hence f (E 1 (G)) = {k +, k + 4,..., k + m} where k + m = k! + 4. Also f (vu k ) = k so that f (E (G)) = {k}. Similarly f (u j1 v 1 ) = k! = k! + 6, f (u j v ) = k! = k! + 8,..., f (u jn v n ) = k!+k k k = k! + k so that f (E 3 (G)) = k! + 6, k! + 8,..., k! + k. Hence f (E ( G)) = {k, k +, k + 4,..., k! + 4, k! + 6, k! + 8,..., k! + k} so that G is a connected (k, )-strongly multiplicatively indexable graph.

19 Chapter 3 75 The following figure 3.9 illustrates the theorem for k = 10. Figure 3.9: Theorem A graph G is (, 4)-strongly multiplicatively indexable if and only if G = P 3 Proof. Let G is (, 4)-strongly multiplicative. Then f (e) (mod 4) for every e E(G). Hence 4 / f (E(G)) since G is connected and any edge with one end vertex labelled as 4 is 0(mod4). So G = P 3. Theorem (i) The graph G is (, 4)-strongly multiplicatively indexable if and only if G = P 3. (ii) The graph G is (, 3)-strongly multiplicatively indexable if and only if G = K.

20 Chapter 3 76 (iii) The graph G is (, 10)-strongly multiplicatively indexable if and only if G = K. (iv) The graph G is (3, 5)-strongly multiplicatively indexable if and only if G = K. Proof. Let G be (, 4)-strongly multiplicative indexer. Then f (e) (mod4) for every e E(G) so that 4 / f(v (G)). Thus V (G) = 3 and hence G = P 3.Conversely if G = P 3 = (v 1, v, v 3 )then f defined by f(v 1 ) = 1, f(v ) =, f(v 3 ) = 3 is a (, 4)-strongly multiplicative indexer of G. The proof of (ii), (iii) and (iv) are similar.

21 Chapter 3 77 References [AG96] [AH90] S. Arumugam and K.A. Germina. On indexable graphs. Discrete Mathematics, (161):85 89, B.D. Acharya and S. M. Hegde. Arithmetic graphs. Journal of Graph Theory, 14:75 99, [AH91a] B.D. Acharya and S. M. Hegde. On certain vertex valuations of graphs. Indian J. Pure Appl Math., : , [AH91b] B.D. Acharya and S.M. Hegde. Strongly indexable graphs. Discrete Mathematics, (93):13 19, [BH01] L.W. Beineke and S. M. Hegde. Strongly multiplicative graphs. Discussions Mathematicac, Graph Theory., 1:63 75, 001. [Bur06] [Gal05] [Gol80] David M. Burton, editor. Elementary Number Theory. TATA McGRAW-HILL,, 006. J.A. Gallian. A dynamic survey of graph labelling. The Electronic Journal of Combinatorics(DS6), pages 1 148, 005. Martin Charles Goloumbic. Algorithmic Graph Theory and Perfect Graphs. Academic Press Inc., New York, [Har7] Frank Harary. Graph Theory. Addision Wesley, Massachusetts, 197.