Chapter V DIVISOR GRAPHS

i J ' Chapter V DIVISOR GRAPHS

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..

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 5.2.2 [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

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 5.2.3 [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 5.2.4 [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 5.2.1 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. 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 5.2.6 [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 5.2.7 [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). 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 5.2.8 [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 5.2.9 [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

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 5.2.10 [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 5.2.11 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

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 5.2.12 [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 5.2.11, 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 5.2.13 [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 5.2.3 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 5.2.4 If G is a divisor graph so is mg for m ~ 2. 0 Theorem 5.2.14 [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. 0 Corollary 5.2.5 If G is a divisor graph so is G + K n for any n. 0 Corollary 5.2.6 The complete bipartite graphs K m n are divisor graphs. o Corollary 5.2.7 Wheels W n are divisor graphs for all even n (n > 2). Theorem 5.2.15 [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. 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 5.2.16 [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 5.2.1. 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. 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.

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 5.2.1 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. 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 5.2.2.! / 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. 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 5.2.2 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 2.------_a Cax P2 Figure 5.2.3

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 [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 5.2.4. 0