# Divisor graphs have arbitrary order and size

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1 Divisor graphs have arbitrary order and size arxiv:math/ v1 [math.co] 20 Jun 2006 Le Anh Vinh School of Mathematics University of New South Wales Sydney 2052 Australia Abstract A divisor graph G is an ordered pair (V,E) where V and for all u v V, uv E if and only if u v or v u. A graph which is isomorphic to a divisor graph is also called a divisor graph. In this note, we will prove that for any n 1 and 0 m ( n then there exists a divisor graph of order n and size m. We also present a simple proof of the characterization of divisor graphs which is due to Chartran, Muntean, Saenpholpant and Zhang. 1 Introduction The notion of divisor graph was first introduced by Singh and Santhosh [2]. A divisor graph G(V ) is an ordered pair (V, E) where V and for all u v V, uv E if and only if u v or v u. A graph which is isomorphic to a divisor graph is also called a divisor graph. The main result of this note is the following theorem. Theorem 1 For any n 1 and 0 m ( n then there exists a divisor graph of order n and size m. To prove Theorem 1, we need the following characterization of divisor graphs is due to Chartran, Muntean, Saenpholpant and Zhang [1]. Theorem 2 A graph G is divisor graph if and only if there is an orientation D of G such that if (x, y), (y, z) are edges of D then so is (x, z). The proof of Theorem 2 in [1] is by induction on the order of the graph. For the completeness of this note, we will present a simple (and direct) proof of this theorem in Section 3. From Theorem 2, we introduce the definition of a divisor digraph which will be usefull in the proof of Theorem 1. Definition 1 A digraph G is a divisor digraph if and only if (x, y), (y, z) are edges of G then so is (x, z). It is clear that if G is a divisor digraph then the graph obtained by ignoring the direction of edges of G is a divisor graph. 1

2 2 Proof of Theorem 1 Suppose that G = (V, E) is a graph with vertex set V = {v 1,..., v n } and size m. The degree of a vertex v i is the number of edges of G incident with v i. Let d 1 d 2... d n be the vertex degrees in non-increasing order and let f G = {i : d i i}. Let e i be the number of vertices with degree at least i, that is e i = {j : d j i}. Then we have e 1 e 2... e n. Moreover, we also have We have the following lemmas. d i = {j : e j i}. (1) Lemma 1 Let n 1 = d 1... d n 1 be a sequence of natural numbers and let e i = {j : d j i}. Suppose that d i = 2m, ( and d i = (e i 1) (3) for all 1 t f = {i : d i i}. Then there exists a divisor graph of order n and size m. Proof We construct a digraph G with vertex set v 1,...,v n as follows. For 1 i f, then (v i, v j ) is an edge of G for i + 1 j d i + 1. We first show that G is a graph of size m. It suffices to show that deg(v i ) = d i for all i (where deg(v i ) is the number of incident edges of vertex v i, regardless of their directions). We have three cases. 1. Suppose that 1 i f. Then it is clear from the construction that deg(v i ) = d i. 2. Suppose that i = f + 1. Then d i = d f+1 f (by the definition of f). From (3), we have d j = e j 1 for 1 j f. If d f+1 < f then e f f, or f d f = e f 1 < f, which is a contradiction. Hence d f+1 = f. For 1 j f then d j + 1 d f + 1 f + 1 = i. So (v j, v i ) is an edge of G and deg(v i ) = f. Thus, deg(v i ) = d i. 3. Suppose that i > f + 1. Then (v j, v i ) is an edge of G if and only if 1 j f and j + 1 i d j + 1. For j > f we have d j d f+1 < f + 1 i. This implies that deg(v i ) = {1 j d : d j i 1} = {j : d j i 1} = e i 1 for all i > f + 1. From (1), we have d i = e i 1 for 1 i f. For i > f + 1 then d j i 1 or e j i only if j f (since d f+1 = f < i 1). Hence e i 1 = {j : d j i 1} = {1 j f : d j i 1} = {1 j f : e j 1 i 1} = {j : e j i} = d i. Thus, deg(v i ) = e i 1 = d i for all i > f

3 Therefore, we have deg(v i ) = d i for 1 i n. This implies that G has order n and size m. Now, we will show that G is a divisor digraph. Suppose that (v i, v j ) and (v j, v k ) are two edges of G. Then from the above construction, 1 i, j d and k d j + 1 d i + 1. Thus (v i, v k ) is also an edge of G. This implies that G is a divisor digraph. Let H be the graph obtained from G by ignoring the direction of edges of G. Then H is a divisor graph of order n and size m. This concludes the proof of the lemma. Lemma 2 Let n 1 = d 1... d n 1 be a sequence of natural numbers and set e i = {j : d j i}. Suppose that and d i = 2m < n(n 1), (4) d i = (e i 1) (5) for all 1 t f = {i : d i i}. Then there exists a sequence n 1 = d 1... d n 1 of natural numbers such that d i = 2(m + 1), (6) and d i = (e i 1) (7) for all 1 t f = {i : d i i}, where e i = {j : d j i} for 1 i n. Proof If f n 1 then n 1 d 1... d n 1 d f f = n 1. Hence d 1 =.... = d n 1 = n 1, and n 1 e i = n(n 1). We have e i n for 1 i n 1, so e 1 =... = e n 1 = n. Hence d i n 1 for 1 i n or d i = n(n 1), which is a contradiction. Thus, f < n 1. Let g be the smallest index such that d i < n 1. Then we have 2 g f + 1 (since d f+1 f < n 1). We have two cases. 1. Suppose that 2 g f. Set h = d g + 2, d g = d g + 1, d h = d h + 1 and d i = d i for i g, h. We have (6) holds since d i = 2 + d i = 2(m + 1). 3

4 Recall that from the proof of Lemma 1, we have e i 1 if i f d i = f if i = f + 1 e i 1 if i > f + 1. (8) Since d g = h 2 g and g f, we have e g = h 1. This implies that d h 1 g and d h < g. Besides, d 1 =... = d g 1 = n 1 so d h d n = e n 1 g 1. Hence d h = g 1, d g = h 1 and d h = g. Therefore, we have e g = e g + 1, e h = e h + 1 and e i = e i for i g, h. We have d h = g 1 f 1 so h > f + 1. Hence d f+1 = d f+1 < f + 1 or f f. And we have (7) holds for 1 t f since (5) holds for 1 t f f. 2. Suppose that g = f + 1. Then from (8) we have d g = f = g 1. Set h = f + 2, d g = d g + 1, d h = d h + 1 and d i = d i for i g, h. Then it is clear that (6) holds. Besides, we have d 1 =... = d g 1 = n 1 so g 1 = d g... d n = e n 1 g 1. Hence d g =... = d n = g 1. We have d g = d h = g < h, so f = f+1, e g = e g+2 = g+1 and e i = e i for i g. From (5), we have d i = (e i 1) for 1 t f. We only need to check for t = f (= f + 1 = g). We have g f f g d i = g + d i = e g 1 + (e i 1) = (e i 1). Thus, (7) holds for 1 t f. This concludes the proof of the lemma. We are now ready to prove Theorem 1. From Lemma 1, we start with the sequence (n 1, 1,..., 1) to obtain a divisor graph of size n 1. Then apply Lemma 1 and Lemma 2 inductively to obtain divisor graphs of order n,..., ( n. To construct a divisor graph of order n and size m with m < n 1, we choose a vertex and join it with m other vertices. Thus, there exists a divisor graph of order n and size m for any n and 0 m ( n. This concludes the proof of the theorem. Remark 1 An interesting and open question is to find necessary and sufficient conditions for a non-increasing sequence n 1 d 1... d n 1 such that there exists a divisor graphs with degree sequence (d 1,...,d n ). 4

5 3 Proof of Theorem 2 Suppose that G is a divisor graph. Then there exists a set V of positive integer such that G G(V ). We give an orientation on each edge (i, j) of G as follows (i, j) E(G), i j if and only if i j. Suppose that (x, y), (y, z) are edges of D. Then x y and y z. Hence x z and (x, z) is an edge of G. Now suppose that there exists an orientation D of G such that if (x, y), (y, z) are edges of D then so is (x, z). We will show that G is a divisor graph. We will give an explicit labelling for G. We start with any vertex of G and label it by {a 1 } (a list of one symbol). Suppose that we have labelled k vertices of G and we have used a 1,..., a l symbols (each vertex is labelled by a list of symbols and we will update this list in each step). We choose any unlabelled vertex, says v. Consider two sets D I (v) = {u V (G) (u, v) E(D)}, D O (v) = {u V (G) (v, u) E(D)}. We label v by L v = {a l+1 }. For each u D I (v) and u was labelled by a list L u then we add this list into the list L v to have a new list L v for v. And for each u D O (v) which was labelled by a list L u then we add the new list L v into L u to have a new list for u. For each updated vertex u, we consider the set D O (u). For each w D O (u) which was labelled by a list L w, we add the new list L u into the list L w to have a new list for w. We keep doing until we have no vertex to update or we come back to some vertex which we met along the way. But in the latter case, we have a sequence of vertices, says w 1,...,w t such that w 1 w 2... w t w 1 in D. This implies that w 1 w 2, w 2 w 3,..., w t w 1. Hence w 1 =... = w t, which is a contradiction. Thus the process must be stopped. We repeat the process until all the vertices of G have been labelled by lists of symbols. Suppose that we have used r symbols a 1,...,a r. We choose r distinct primes p 1,..., p r and for each vertex v V (G) which is labelled by a list {a i1,..., a iv } {a 1,...,a r } then we label it by the number n(v) = p i1...p iv. From the construction above, if (u, v) is an edge of G then either L u L v or L v L u. This implies that either u v or v u. Hence G is a divisor graph. This concludes the proof of the theorem. 4 Acknowlegement I would like to thank Professor Ping Zhang for sending me reference papers [1, 3]. 5

6 References [1] G. Chartrand, R. Muntean, V. Saenpholphat, and P. Zhang, Which graphs are divisor graphs?, Congr. Numer. 151 (2001) [2] G. S. Singh and G. Santhosh, Divisor Graphs - I, Preprint. [3] R. Gera, V. Saenpholphat and P. Zhang, Divisor graphs with triangles, Congr. Numer. 165 (2003) [4] C. Pomerance, On the longest simple path in the divisor graph, Congr. Numer. 40 (1983)

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