Chapter V DIVISOR GRAPHS

Size: px

Start display at page:

Download "Chapter V DIVISOR GRAPHS"

Leon Armstrong
5 years ago
Views:

1 i J ' Chapter V DIVISOR GRAPHS

2 Divisor graphs 173 CHAPTER V DIVISOR GRAPHS 5.1 Introduction In this chapter we introduce the concept of a divisor graph. Adivisor graph 0(8) of a finite subset S of Zis the graph (V; E) where V= Sand uv E E if and only if either u divides v or v divides u. A graph Gis a divisor graph if it is isomorphic to the divisor graph 0(8) of some subset S of Z. We also give an equivalent definition of a divisor graph as follows. Agraph G is called a divisor graph if there exists 'a labeling / of its vertices with distinct integers such that for any two distinct vertices u and v, uv is an edge of G if and only if either / (u) divides / (v) or / (v) divides/(u). In this case we say that/is a divisor labeling of G. First, we give an example of a divisor graph. Consider the star KI,n' Take the set S = { 1, PI' P2'...,Pn} where Pi stands for the lu1 prime. Clearly KI,n == O(S). 5.2 On Divisor Graphs In this section we give some results on cycle related graphs, complete graphs and trees and also study some general results on divisor graphs..

3 Divisor graphs 174 Theorem 5.2.1[154] The cycles C 2n are divisor graphs for all n ~ 2. Proof. Let { v l1 v 2 ' v 3 '...,v 2n }, n ~ 2 be the vertex set of C 2n. When n= 2, the resulting graph is C 4 Let SI = {V l,v 2,V 3,V 4 } where VI = 2, v 2 = 12, v 3 = 3, v 4 = 18. It is easy to see that C 4 == O(SJ. Next, consider all even cycles of length greater than 4. Set V 2i _ l = Pi' i = 1, 2,..., n V 2i = PiPi+l' i = 1, 2,..., (n - 1) where Pi is the Z'th prime. Let S = { VI' V 2, V 3,..., v 2n } where the Vi'S are as defined above. Then, obviously C 2n == O(S) when n> 2. o Theorem [154] Odd cycles C 2n +Jor all n > '1 are not divisor graphs. Proof. On the contrary, let us assume that C 2n + l (n> 1) is a divisor graph. Let S = V (C 2n + l ) = { a p a 2, a 3,..., a 2n + l } where a i ':f. 0, aie Z, 1 ~ i ~ 2n + 1. Since C 2n + l is a divisor graph, note that a vertex a i in S can divide only the vertices immediately preceding it and succeeding it along the rim of the cycle. Without loss of generality, let us assume ~ I a 2 (a l divides a 2 ). The cycle being given by a l a 2 a 3 a 2n + 1 ~, by definition, a 2 1 a 3 or a 3 1 a 2 If a 2 1 a 3

4 Divisor graphs 175 then all a 3 implying the given cycle is a triangle. Hence a 3 1 a 2. Thus all a 2 and a 3 1 a 2 Similarly a 3 and as must divide a 4 Proceeding like this we get, a 2n - l and a 2n + l must divide a 2n. But then, either a 1 must divide a 2n + l in which case a l I a 2n (or) a 2n + l must divide a l in which case a 2n + l l a 2, both of which are contradictions. Thus C 2n + l for n> 1 is not a divisor graph. o Theorem [154] An induced subgraph of a divisor graph is a divisor graph. Proof. Let G be the divisor graph of the set 5 c Z. Let 51 ~ 5. Then the induced subgraph (8 1 ) is clearly the divisor graph of 51' D Theorem [154] Any graph with an induced subgraph which is an odd cycle of length greater than or equ8.l to 5 is not a divisor graph. Proof. By Theorem 5.2.2, it is clear that an odd cycle of length ~ 5 is not a divisor graph. Then, if we use Theorem 5.2.3, the result follows. D Corollary The Petersen graph is not a divisor graph. Proof. Since the Petersen graph has an induced subgraph which is a cycle of length 5, by Theorem 5.2.4, Petersen graph is not a divisor graph. D

Divisor graphs 176 Theorem 5.2.5 [154] Let G be a divisor graph containing both v and -v such that v E Z - { 0 }. Then G contains at least one triangle if <5 (G ) ~ 2. Proof.

5 Divisor graphs 176 Theorem [154] Let G be a divisor graph containing both v and -v such that v E Z - { 0 }. Then G contains at least one triangle if <5 (G ) ~ 2. Proof. Since <5 (G ) ~ 2, we have d (v ) ~ 2. Thus there exists at least one vertex u -:;e -v which is adjacent to v. Then u must be adjacent to -v also. i.e., v, -v, u form a triangle in G. o Theorem [154] Every graph is a subgraph of a divisor graph. Proof. Let a be a nonzero integer. Consider the set s={a, a 2, a 3,., an} where n E N. It follows that K n == 0(8). Since any graph on n vertices is a subgraph of K n, the proof is complete. 0 Let Hm,n denote the complete n-partite graph on n ~ 2 sets of m ~ 2 independent vertices. Also let [a 1, a 2, a 3,,an1denote the least common multiple of n positive integers a 1, a 2, a 3,,an' Theorem [154] H mn is a divisor graph., Proof. Let Pi denote the i th prime. Let 1 = [ PI' P2,P3'..., Pm 1and 2 s; k ~ n. Consider the sets

$Dilfisor graphs 177... V n = {PI L 1 L2... L n - 1, P2 L 1 L2 '" L n _ 1,..., Pm L 1 L... 2 L _ } n 1 Let S = V; U V 2 U... U Vn. We claim that Hm,n == O(S).$

6 Dilfisor graphs V n = {PI L 1 L2... L n - 1, P2 L 1 L2 '" L n _ 1,..., Pm L 1 L... 2 L _ } n 1 Let S = V; U V 2 U... U Vn. We claim that Hm,n == O(S). To prove this claim we observe that Pi does not divide Pj for I :;; j, 1 :s; I, j :s; m. Consider u = P r L 1 L 2... L j _ 1 E V j and v = Ps L 1 L 2 L j -1 E V j for 1 :s; r, s :s; m, r:;; 5, 2 ::;; j :s; n. If u I v then we get Prl Ps' a contradiction. If v Iu then also we get a similar contradiction. Hence u does not divide v and v does not divide u. Also it is easy to find the vertices X J yes such that either x I y or y I x. This completes the proof of the theorem. 0 Corollary 5.2;2 The cocktail party graph H 2,n is a divisor graph. 0 Theorem [154] mk n, the union of m copies of K n is a divisor graph for all m > 1. Proof. Recall that the complete graphs K n are divisor graphs. Let V j denote the vertex set of the j th copy of K n, 1 ::;; j::;; m. Let V j = { P j' P~, P~,..., pi }for 1 ::;; j::;; m where Pi denotes the i th prime.. Consider the set S = V; U V 2 U... U V m We show that mk n == 0(8). Let u E V; and v E V j, i :j; j, 1 ::;; i, j ::;; m. Then u = p7, 1 :s; s:s; n and v = P;, 1 :s; t:s; n.. If u Iv then we get Pi IPj' a contradiction. Similarly if vlu then pjl Pi' a contradiction.

DilJisor graphs 178 Hence u does not divide v and v does not divide u. It is easy to find the vertices x) yes such that either x I y or Y I x. This completes the proof of the theorem. CJ Theorem 5.2.

7 DilJisor graphs 178 Hence u does not divide v and v does not divide u. It is easy to find the vertices x) yes such that either x I y or Y I x. This completes the proof of the theorem. CJ Theorem [154] If F be the union of m complete graphs of different orders with a vertex in common then F is a divisor graph. Proof. Let K n. denote the complete graph of order n i, 1 ~ i ~ m, I having vertex set Vi' Let V i = { 1, Pi' P~,..., p'(-l } for 1 ~ i.~ m where Pi is the i th prime. Consider the set S = ~ U V 2 U... U V m We show that F == O(S). For i :;t.j, 1 ~ i,) ~ m, let U E v: and v E Vj' Suppose that u:;t. 1 and v:;t. 1. Then U = P:, for 1 ~ s ~ n i - 1 and v =p}, for 1 ~ t ~ n j - 1, i :;t. j, 1 ~ i,) ~ m. If u I v then Pi I Pj' and if v Iu then Pjl Pi' both are contradictions. Hence u does not divide v and v does not divide u. It is easy to find the vertices x) yes such that either x I y or y I x. This completes the proof of the theorem. CJ Notations Let G = (V, E ) be a divisor graph on n vertices. Let d j denote the number of divisors of i in V and m i, the number of multiples of i in V. Also let d (v ) denotes the degree of a vertex v in V. Let ~ = { U E V I u:;t. 0, - u ~ V }, V 2 = { U E V I u:;t. 0, - u E V }, Va == { }. Clearly ~, V 2, V 3 are mutually disjoint subsets of V and

8 Divisor graphs 179 ~ U V 2 U V 3 = V. Let I ~ I = ~ and IV 2 I = n 2. Any vertex in G must be adjacent to all its divisors and multiples in V, except itself. Thus, d (u ) = d u + m u - 2 for all U E V 1, For any W E V 2, - w is a divisor of w and also a multiple of w. Hence d (w ) = d w + m w - 3 for all W E V 2. Also d ( )= n - 1. Now using these symbols we prove the following theorem. Theorem [154] Let G = (V, E) be a graph on n vertices and 8 edges. If G is a divisor graph, then 2& = L d u + L m u - 2n 1-3n 2 +(n -1) v E Vi U V~ V E Vi U V~ where ~ = I ~ I and n 2 = I V 2 I. Proof. Suppose G = (V, E ) is a divisor graph on n vertices and 8 edges. Let V; ={U E V IU"* 0, - U ~ V }, V 2 ={U E V IU:f. 0, - U E V }, V 3 = {O }. Then, 2& = I d(v) = I d(v) + Ld(v) + (n - 1) UEV UEV 1 UEV2 = L (d u + m u - 2) + I (d v + m v - 3) + (n - 1) UE Vi VE V 2 = L d v + I m v - 2n 1-3n 2 + (n - 1). UEViUV~ VEViUV~ o Theorem Let G be a divisor graph with no triangles. Then for every vertex v of G, the label I (v) is a common multiple or common

9 Divisor graphs 180 divisor of { 1(x) I x E No (v)} where No (v) denotes the neighbourhood of v and 1(v) denotes the number labeled at v. Proof. Let u be a vertex of G such that 1(u) is a multiple of 1(a) and a divisor of 1 (b) where a, b E No (u). Then 1(a) divides 1(b) and the three vertices u, a, b form a triangle in G. o Theorem [158] Every tree is a divisor graph. Proof. We proceed by induction on the order n. The result is true if n = 1. Assume that all trees of order k are divisor graphs. Let T be a tree of order k + 1. Let v be an end-vertex of T and let u be adjacent to v in T. Then T 1 = T - v is a tree of order k and is therefore a divisor graph. Let f be a divisor labeling of ~. Let No (v) denote the neighbourhood of v. By Theorem , we need only consider two cases. Case 1 f(u) is a common divisor of {f(x): x E No (u) } Define a labeling 9 of T as follows: 9 (x) =f(x) for all x E V(T ) 1 and 9 (v) = p f (u) where p is a prime not occurring as a factor of any labelf(x), x E V(T 1 ). Case 2 f(u) is a common multiple of {f(x): x E No (u) }. Define a labeling 9 of T as follows: 9 (x) = pf (x) for all x in T 1 where p is a prime not occurring as a factor of any labelf(x), x E V(T 1 ) and 9 (v) =f(u).

Divisor graphs 181 In both cases it is easy to see that 9 gives a divisor labeling of Tand this completes the proof. 0 Theorem 5.2.

10 Divisor graphs 181 In both cases it is easy to see that 9 gives a divisor labeling of Tand this completes the proof. 0 Theorem [158] If G 1 and G 2 are two divisor graphs then we can label the vertices of G 1 and G 2 so that for all x E V (G 1 ) and y E V (G 2 ), I (x) and I (y) are relatively prime where I (x) denotes the number labeled at x. Proof. Consider labelings of the divisor graphs G 1 and G 2 in which the label of every vertex is either a prime or a product of distinct primes. Also choose the labeling of G 2 such that, for all v E V (G 2 ), every prime factor of I (v) is greater than M where M=max { Pi I Pi is a prime factor of I (x), X E V (G 1 ) }. It follows that for all x E V (G 1 ) and y E V (G 2 ), I (x) and I (y) are relatively prime. 0 Corollary If G 1 and G 2 are two divisor graphs with disjoint vertex sets then G 1 u G 2 is also a divisor graph. 0 Corollary If G is a divisor graph so is mg for m ~ 2. 0 Theorem [158] If G and H are vertex disjoint divisor graphs then G + H is also a divisor graph. Proof. Let f and 9 be the divisor graph labelings of G and H respectively such that f (x) and 9 (y) are relatively prime for all x E V (G) and y E V (H ). We define a labeling function h of G+ H as follows.

Divisor graphs 182 h(v) = f(v) for all v E V(G) L1g(v) for all v E V(H) where L 1 is the least common multiple of {f(x): x E V(G)} It is easy to see that h is a divisor labeling of G + H.

11 Divisor graphs 182 h(v) = f(v) for all v E V(G) L1g(v) for all v E V(H) where L 1 is the least common multiple of {f(x): x E V(G)} It is easy to see that h is a divisor labeling of G + H. 0 Corollary If G is a divisor graph so is G + K n for any n. 0 Corollary The complete bipartite graphs K m n are divisor graphs. o Corollary Wheels W n are divisor graphs for all even n (n > 2). Theorem [158] The subdivision graph of a divisor graph is a divisor graph. Proof. Let G be a divisor graph and f denote its divisor labeling. Let w ll W 2,,w m denote the newly added vertices on the edges ell e 2,...,em of G. We define a labeling function 9 of the subdivision graph 5 (G ) as follows. 9 (Wi) = Pi for 1 ~ i ~ m where Pi is a prime such that Pi and f (vj) are relatively prime for all 1 ~ i ~ m, 1 ~ j ~ n. 9 (v j ) = k j f (v j ) for 1 ~ j ~ n where k j is the product of primes adjacent to v j in 5(G). It follows that 9 is a divisor labeling of 5 (G ). It is easy to prove the following lemma.

Divisor graphs 183 Lemma 5.2.1 Suppose u i is a product of midistinct primes for i = 1,2.

12 Divisor graphs 183 Lemma Suppose u i is a product of midistinct primes for i = 1,2. If u 1 and u 2 have k common prime factors where 0 ~ k < min { m 1, m 2 }, then u 1 does not divide u 2 and u 2 does not divide ~. o Theorem [158] The ladder graph L n is a divisor graph. ladder Ln' In -lj V 2i+ 1= P 2 i + 2 for 1 < i ~ -2-, U 2i = P2i + 1 for 1 < i ~l;j' where Pj denotes the j th prime. defined above. The labeling is illustrated in Figure We claim that L n == O(S). To prove this claim we observe that V2i + 1 'S and u 2j 's are primes. Hence we get the following results. ""'I does not divide V 2j,1 for j"ln; 1 "j, 0 " i, 1J. u2 ; does not

Divisor graphs 184 divide u 2j for i "j, 1 ~ i,j ~l;j. "2id does not divide V 2i + 1 for O~i5ln;1j, 15 j5l;j. does not divide ~j and u 2j Next we consider primes and product of primes.

13 Divisor graphs 184 divide u 2j for i "j, 1 ~ i,j ~l;j. "2id does not divide V 2i + 1 for O~i5ln;1j, 15 j5l;j. does not divide ~j and u 2j Next we consider primes and product of primes. U 2i + 1 'S and v 2j 's are products of distinct primes. By definition, the prime U 2i is not a factor of u 2j + 1 if j ~ i, (i - 1). Hence U 2i does not divide u 2j + 1 and.. (. 1) 1. lnj 0. In,-lj u 2j +1does not divide U 2i for J ~ I, I -, 5I 52' 5J ~ -2-. Similarly we get V 2i + 1 does not divide v 2j and v 2j does not divide V 2i + 1 for j " i, ( i + 1), 0 ~ i ~ ln; 1j, 1 ~ j ~l;j. For i "j, the prime u" is not a factor of v 2j. Hence U 2i does not divide v 2j and v 2j does not divide ~, for i "j, 1 ~ i, j ~l;j. A similar argument shows that "2,.1 does not divide u 2j + 1 and u 2j + 1 does not divide V 2i + 1 if i ~ j, O ~ I,}.. 5 In-2- -lj. Finally we consider products of primes. By definition, U 2i + 1 'S and v 2j 's are products of three distinct prime factors and both have at most two common factors. Hence by Lemma 5.2.1, U 2i + 1 does not divide "OJ and "2j does not divide u 2i, for 0 ~ i ~ln; 1j. 1 ~ j ~l;j Also if i ~ j, both U 2i + 1 and u 2j + 1 have at most one common factor.

14 Divisor graphs By Lemma 5.2.1, U 2i + 1 does not divide U 2j + 1 for i '" j, 185 O ~ 1,J~ " In-lj -2-. Bya similar argument, V 2i does not divide v 2j for i '" j, 1 ~ i, j ~ l;j..... ~ L s Figure Y I x.. S such that either x I y or Also it is easy to fmd the vertlces x, Y E This completes the proof of the theorem. o

Divisor graphs Theorem 5.2.17 [158] Prisms em x P n 186 are divisor graphs for all even positive integers m and all positive integers n. Proof.

15 Divisor graphs Theorem [158] Prisms em x P n 186 are divisor graphs for all even positive integers m and all positive integers n. Proof. Let Vi)' 1 :::; i :::; m, 1 :::; j :::; n denote the vertices of the Pi j if i and j are both odd or both even Set vi) = product of the labels of the neighboursof ViP if i and j are of different parity, where Pij are distinct primes. Let S = { viji 1:::; i :::; m, 1 :::; j:::; n} where Vi) are as defined above. The labeling is illustrated in Figure ! / I J I PUP""P31 IL P'J'J Pl~'J'JP"J'33 / P"" I I I I P 31 P31 P'J" P33 P4'J P33 P~33P3~44 II'.. P~ P31P4"P: 51 P4'J P3~4'JPJs3 P 44 PSI PSIP S3 P4"p6'J \ P S3 P5~55PJ64 \ \ \ \ \ Ill. P Figure 5.2.2

Divisor graphs 187 We claim that C m x P n == O(S). To prove this claim we observe that if vpq and v rs are distinct primes then v pq does not divide v rs and v rs does not divide v pq.

16 Divisor graphs 187 We claim that C m x P n == O(S). To prove this claim we observe that if vpq and v rs are distinct primes then v pq does not divide v rs and v rs does not divide v pq. If v ij and v tw are products of primes, by Lemma 5.2.1, v ij does not divide v 1w and v t w does not divide vi}' From the definition it follows that if v ij is a prime, it is not a factor of v rs if v rs is not a neighbour of vi}' It is easy to find the vertices x, yes such that either x I y or y I x. This completes the proof of the theorem. 0 Lemma C 3 X P n is not a divisor graph for all integers n> 1. Proof. We first show that C 3 x P 2 is not a divisor graph. On the contrary, suppose that C 3 x P 2 is a divisor graph. Let V (C 3 X P 2 ) = { al' a 2, a 3, a 4, as, a 6 } where a j 7; 0, a j E Z. Then either a l divides a 2 or a 2 divides a l. (See Figure 5.2.3). a _a Cax P2 Figure 5.2.3

17 Divisor graphs 188 We consider these two cases. Case 1. a 1 divides a 2 In this case if a 6 divides a 1 then a 6 divides a 2, a contradiction. Hence a 1 divides a 6 Similarly we can show that ~ divides a 3, a 4 divides a 3, a 2 divides a 3, a 4 divides a 6, as divides a 6, and a 4 divides as' Finally, either a 2 divides as or as divides a 2 If a 2 divides as then since a 1 divides a 2 we get a 1 divides as, a contradiction. If as divides a 2 then since a 4 divides as we get a 4 divides a 2, which is impossible. Case 2. a 2 divides a 1 In this case if a 1 divides a 6 then we get a 2 divides a 6, a contradiction. Hence a 6 divides a 1 Similarly we can show that a 3 divides ai' a 3 divides a 4, a 3 divides a 2, a 6 divides a 4, a 6 divides as, and as divides a 4 Finally, either a 2 divides as or as divides a 2 If a 2 divides as then, since as divides a 4 we get a 2 divides a 4, a contradiction. If as divides a 2 then, since a 2 divides ~ we get as divides a 1, a contradiction. Since both these cases lead to contradictions C 3 x P 2 is not a divisor graph. But C 3 x P 2 is an induced subgraph of C 3 X P n for all integers n ~ 3. By Theorem 5.2.3, any induced subgraph of a divisor

Dillisor graphs graph is a divisor graph. Hence C 3 X P n 189 is not a divisor graph for all integers n ~ 3. This completes the proof. o Theorem 5.2.

18 Dillisor graphs graph is a divisor graph. Hence C 3 X P n 189 is not a divisor graph for all integers n ~ 3. This completes the proof. o Theorem [158] Prisms C m x P n are not divisor graphs for all odd integers m ~ 3 and all integers n> 1. Proof. In view of Lemma 5.2.2, we need only consider prisms C m x P n with odd integers m ~ 5 and all integers n > 1. Such a prism is a graph with an induced subgraph which is a cycle of length m. Then the theorem follows from Theorem

R(p, k) = the near regular complete k-partite graph of order p. Let p = sk+r, where s and r are positive integers such that 0 r < k.

R(p, k) = the near regular complete k-partite graph of order p. Let p = sk+r, where s and r are positive integers such that 0 r < k. MATH3301 EXTREMAL GRAPH THEORY Definition: A near regular complete multipartite graph is a complete multipartite graph with orders of its partite sets differing by at most 1. R(p, k) = the near regular