www.elsevier.com/locate/disc
-labeling number of trees
夡
Chin-Lin Shiue
a, Hung-Lin Fu
baDepartment of Applied Mathematics, Chung Yuan Christian University, Chung Li 32023, Taiwan bDepartment of Applied Mathematics, National Chiao Tung University, Hsin Chu 30050, Taiwan
Received 7 January 2005; received in revised form 5 May 2006; accepted 25 June 2006 Available online 29 September 2006
Abstract
In this paper, we prove that the-labeling number of trees T, Tr/2n where n = |E(T )| and r is the radius of T. This improves the known result TeO(√nlog n)tremendously and this upper bound is very close to the upper bound Tn conjectured by Snevily. Moreover, we prove that a tree with n edges and radius r decomposes Ktfor some t(r + 1)n2+ 1.
© 2006 Elsevier B.V. All rights reserved.
Keywords:-labeling number; Tree decomposition
1. Introduction and preliminaries
Throughout this paper, all graphs we consider are finite and simple, i.e., multiple edges and loops are not allowed. For terms and notations used in this paper we refer to the textbook by West[10]. Let G be a graph. For any two vertices u and v, the distance from u to v, denoted by d(u, v), is the least length of a u, v-path. If G has no such path, then d(u, v)= ∞. The eccentricity of a vertex u, written e(u), is maxv∈V (G)d(u, v). The radius of a graph G is
minu∈V (G)e(u). Clearly, a tree of n edges has radius at mostn/2.
Let G be a graph with q edges. An injective function f : V (G) → S, S is a set of nonnegative integers, has been called a vertex labeling, a valuation, or a vertex numbering of G. For convenience, we denote the set{0, 1, 2, . . . , q} by[0, q]. A vertex labeling f of G is called a -labeling if f is an injection from V (G) into [0, q] such that the values |f (u) − f (v)| for the q pairs of adjacent vertices u and v are distinct. A -labeling is also known as a graceful labeling. If f is a-labeling of G such that there exists an integer so that each edge uv ∈ E(G) either f (u) < f (v) or
f (v) < f (u), then f is called an -labeling of G. Clearly, if G has an -labeling, then G must be a bipartite graph.
The following theorem is folklore now.
Theorem 1.1 (Rosa[7]). Let G be a graph with q edges, and let G have an-labeling. Then, the complete graph
K2pq+1can be decomposed into isomorphic copies of G, where p is an arbitrary positive integer.
夡Research supported in part by NSC 91-2115-M-033-001. E-mail address:[email protected](H.-L. Fu).
0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.06.016
To obtain an-labeling of a graph is not easy at all. Due to the close relation with graph decomposition, Snevily[9]
introduced the following notion.
H is called a “host” graph of G if H has an-labeling and H can be decomposed into copies of G. The -labeling
number of G is defined by G= min{t: ∃ a “host” graph H of G with |E(H )| = t|E(G)|}.
Then Snively conjectured that G<+ ∞ in[9]. Later, the conjecture was proved by El-Zanati et al.[1]. For trees, we can do even better. By the fact that an n-cube has an-labeling[9]for each n1 and an n-cube can be decomposed into copies of an arbitrary tree with n edges[2], we conclude that T2n−1. Furthermore, Shiue [8] proved that
TeO(√nlog n). Note that Snevily[9]conjectured that Tn. Clearly, this can be proved by showing that Kn,ntree
decomposition conjecture holds and Kn,nhas an-labeling.
Kn,n tree decomposition conjecture (Ringel[6]). For each tree T with n edges, Kn,ncan be decomposed into
iso-morphic copies of T.
In this paper, we manage to prove that a special regular bipartite graph H withr/2n2edges can be decomposed into isomorphic copies of arbitrary tree T with n edges, where r is the radius of T. Moreover, by showing this special bipartite graph has an-labeling, we conclude that the -labeling number of T, Tr/2n. Since r n, this improves the exponential upper bound for Tto a quadratic upper bound, and it is very close to the upper bound Tn especially when r is a constant.
2. The main result
We start with introducing the special bipartite graphs mentioned in Section 1. A bipartite graph defined on A∪ B where A∩ B = ∅, A = {a0, a1, . . . , al−1}, and B = {b0, b1, . . . , bl−1} is called an (n, l)-crown if for each i ∈ [0, l − 1],
ai is adjacent to bj, j ∈ {i, i + 1, . . . , i + n − 1} (mod l). Clearly, an (n, l)-crown is an n-regular bipartite graph with
2l vertices and nl edges.
Proposition 2.1. Let n and l be two positive integers such that n > 1 and n|l. Then an (n, l)-crown has an -labeling. Proof. Let G= (A, B) be an (n, l)-crown such that A = {a0, a1, . . . , al−1} and B = {b0, b1, . . . , bl−1} and ai is
adjacent to bj if and only if j ∈ {i, i + 1, . . . , i + n − 1}(mod l). First, we partition A into n(=l/k) sets such that
A0= {a0, a1, . . . , ak−1} and for each 1hn − 1, Ah= {at|t = l − h − j (n − 1), j = 0, 1, 2, . . . , k − 1}. Define a
vertex labeling f of G as follows:
1. f (x)= i if x = bi, i= 0, 1, 2, . . . , l − 1;
2. f (x)= nl − jn + j if x = aj, j= 0, 1, . . . , k − 1; and
3. f (x)= hl + j if x = atwhere t = l − h − jn + j, j = 0, 1, 2, . . . , k − 1.
Then, it is a routine matter to check that f is an injective function from V (G) into[0, nl] and can be chosen as l− 1. So, it is left to verify that {f (ax)− f (by)|ax ∈ A, by∈ B and axby∈ E(G)} = [1, |E(G)|] = [1, nl].
Let Eidenote the set of edges which are incident to the vertices in Ai, i=0, 1, 2, . . . , n−1, and let f (Ei)={f (ax)−
f (by)|ax∈ Ai, by ∈ B and axby ∈ Ei}. Now, by the definition of f, we have
f (E0)= k−1 j=0 {f (aj)− f (bi)|i = j, j + 1, j + 2, . . . , j + n − 1} = k−1 j=0 {nl − jn + j − i|i = j, j + 1, j + 2, . . . , j + n − 1} = k−1 j=0 [(l − 1)n − jn + 1, ln − jn] = [(n − 1)l + 1, nl]. (1)
Note that the vertex al−his adjacent to the n vertices bl−h, bl−h+1, . . . , bl−1, b0, b1. . . , bn−h−1for each 1hn − 1.
By the definition of f, we have f (Eh)= k−1 j=0 {f (at)− f (b)|b ∈ N(at), t= l − h − jn + j} = k−1 j=1 {hl + j − i|i = l − h − jn + j, l − h − jn + j + 1, . . . , l − h − jn + j + (n − 1)} ∪ {hl − i|i = l − h, l − h + 1, . . . , l − 1} ∪ {hl − i|i = 0, 1, . . . , n − h − 1} = k−1 j=1 [(h − 1)l + h + (j − 1)n + 1, (h − 1)l + h + jn] ∪ [(h − 1)l + 1, (h − 1)l + h] ∪ [lh + h − n + 1, lh] = [(h − 1)l + h + 1, (h − 1)l + h + kn − n] ∪ [(h − 1)l + 1, (h − 1)l + h] ∪ [lh + h − n + 1, lh] = [(h − 1)l + h + 1, lh + h − n] ∪ [(h − 1)l + 1, (h − 1)l + h] ∪ [lh + h − n + 1, lh] = [(h − 1)l + 1, hl], h = 1, 2, . . . , n − 1. (2)
By combining (1) and (2), we obtain f (E)= n−1 h=1 f (Eh) ∪ f (E0)= n−1 h=1 [(h − 1)l + 1, hl] ∪ [(n − 1)l + 1, nl] = [1, nl].
This concludes the proof.
For clearness, we give an example to show the idea of our-labeling.
Example. n= 5, l = 20.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 100 96 92 88 83 63 43 23 82 62 42 22 81 61 41 21 80 60 40 20
f (E0)= [81, 100], f (E1)= [1, 20], f (E2)= [21, 40], f (E3)= [41, 60],
f (E4)= [61, 80].
In order to find the-labeling number of a given tree T with n edges, it is left to prove that there exists an (n, l)-crown H such that n|l and H can be decomposed into isomorphic copies of T. First, we need a definition of bilabeling.
Let G= (A, B) be a bipartite graph with q edges. We call a function f : V (G) → [1, m] a bilabeling of G if f |A
and f|B are injective functions. We let the strength of f be max{f (x)|x ∈ A ∪ B}. A bilabeling of G = (A, B) is
a-bilabeling if {f (y) − f (x)| x ∈ A, y ∈ B and {x, y} ∈ E(G)} = [0, q − 1], where q = |E(G)|. Moreover, if a -bilabeling of G = (A, B) has strength |E(G)|, then the -bilabeling is in fact the -bilabeling of G or bigraceful labeling named by Ringel et al.[5]. Not surprisingly, they use this labeling to tackle the Kn,n tree decomposition
conjecture. In what follows, we shall use a-bilabeling to obtain our main result. First, we claim that a tree does have a-bilabeling with larger strength.
C 9 -3 14 13 -4 8 -5 -6 -7 12 11 10 0 -1 -2 Fig. 1.
Proposition 2.2. Let T be a tree with n > 0 edges. Then T has a-bilabeling with strength not greater than r/2n
where r is the radius of T.
Proof. Let c be a vertex in the center of T. Then we construct a rooted tree Tc with c as the root by assigning direction to the edges of T. Since T is a bipartite graph let T = (A, B) where c ∈ A. For convenience, we also let V = V (T )\{c}, |A| = s, and |B| = t. Then we have n = s + t − 1. If s = 1, then T is star, and it is easy to see that T has a bigraceful labeling. This completes the proof. Otherwise, we assume s > 1. In Tc, each vertex v in V has in-degree
one, we can denote the unique arc adjacent to v by ev. Hence, the arc set of Tc, E(Tc)= {ev| v ∈ V } and we partition
E(Tc)into two sets EA= {ev∈ E(Tc)| v ∈ A} and EB= {ev ∈ E(Tc)| v ∈ B}. Let Pvbe the unique path joint from
the root c to v for each vertex v∈ V . Then on the path Pv, the edges occur in EBand EAalternately and this implies
the following fact.
Fact 1. |E(Pv)∩ EA| = |E(Pv)|/2 and |E(Pv)∩ EB| = |E(Pv)|/2 for each v ∈ V .
In order to construct a-bilabeling, we first label the arcs of Tc. Since|E(Tc)| = n, we can define a bijective function
g mapping E(Tc)to[−(s − 2), 0] ∪ [s − 1, n − 1] by the following rules.
Rule 1: g(EA)= [−(s − 2), 0] and g(EB)= [s − 1, n − 1].
Rule 2: For each pair evand ev in the same arc set, if|E(Pv)| < |E(Pv )|, then g(ev) < g(ev ).
Rule 3: For each pair evand ev in the same arc set, let eu and eu be the previous arc of evand ev in Pvand Pv ,
respectively. Then g(ev) < g(ev )provided that g(eu) < g(eu ); otherwise, g(ev) > g(ev ).
SeeFig. 1for an example.
By Rule 1 of the definition of g and Fact 1, we have the following: Fact 2. e
u∈E(Pv)g(eu) >0 for each v∈ V .
Now, we are ready to define the desired vertex labeling f. Let f be a function mapping V (T ) to a set of nonnegative integers which is defined as follows:
1. f (c)= 1 and
2. f (v)=eu∈E(Pv)g(eu)+ 1 where v ∈ V .
SeeFig. 2for example.
Since the length of each path starting from the center is no more than the radius in a tree, we have
f (v)= eu∈E(pv) g(eu)+ 1 eu∈E(pv)∩EB g(eu)+ 1 |E(Pv)∩ EB|(n − 1) + 1 = |E(Pv)|/2(n − 1) + 1r/2(n − 1) + 1r/2n
21 1 10 7 20 6 9 4 3 2 16 13 12 14 15 21 Fig. 2. Claim 1. f is a bilabeling.
By Fact 2 and f (c)= 1, we have f (v)1 for each v ∈ V (T ). This implies that f maps V (T ) into a set of positive integers. It is left to claim that both f|Aand f|Bare injective. Clearly, by the definition of f and Fact 2, f (c)= f (v)
for each v∈ A\{c}. For any pair of vertices v and v both in A\{v} (or both in B), let Pv= c − v1− v2− · · · − vl(=v) and
Pv = c − v1 − v2 − · · · − vm (=v ).
For convenience, we also let evi= ei and evj = e jfor i= 1, 2, . . . l, j = 1, 2, . . . , m.
Case 1: l= m.
Suppose that g(ev) < g(ev ). Then g(ei) < g(e i)for i= l − 1, l − 2, . . . , 1 by Rule 3. Hence
f (v)= l i=1 g(ei)+ 1 < m j=1 g(e j)+ 1 = f (v ). Case 2: l= m.
Suppose that l < m. Then |E(Pv)| < |E(Pv )|, and we have g(ev) < g(ev )by Rule 2. Also, by Rule 3, we have
g(el−i) < g(e m−i)for i= 1, 2, . . . , l − 1. Hence, by Fact 2, we have
f (v)= l i=1 g(ei)+ 1 < m j=m−l+1 g(e j)+ 1 < ew∈E(Pu) g(ew)+ m j=m−l+1 g(e j)+ 1 = m j=1 g(e j)+ 1 = f (v ), where u= vm −l.
In any case, we have that both f|Aand f|Bare injective (SeeFig. 3for example). Therefore, we have the Claim. Claim 2. {f (y) − f (x)|x ∈ A, y ∈ B and {x, y} ∈ E(T )} = [0, n − 1].
For each arc ev= (u, v) ∈ E(Tc), that is, for each edge{u, v} ∈ E(T ), if ev∈ EA, then v∈ A and u ∈ B. Hence,
we have f (u)− f (v) = ⎡ ⎣ w∈Pu g(ew)+ 1 ⎤ ⎦ − ⎡ ⎣ w∈Pv g(ew)+ 1 ⎤ ⎦ = −g(ev).
Fig. 3.
Otherwise, u∈ A and v ∈ B, and we have f (v)− f (u) = ⎡ ⎣ w∈Pv g(ew)+ 1 ⎤ ⎦ − ⎡ ⎣ w∈Pu g(ew)+ 1 ⎤ ⎦ = g(ev).
This implies that
{f (y) − f (x)|x ∈ A, y ∈ B and {x, y} ∈ E(T )} = {−g(ev)|ev∈ EA} ∪ {g(ev)|ev∈ EB}
= {0, 1, 2, . . . , s − 2} ∪ {s − 1, s, s + 1, . . . , n − 1} (by Rule 1) = [0, n − 1].
Therefore, we have the proof.
Proposition 2.3. Let G be a bipartite graph with n edges. If G has a-bilabeling of strength m, then an (n, l)-crown
H can be decomposed into isomorphic copies of G for each lm.
Proof. Let H = (U, V ) be an (n, l)-crown, l m. Therefore, U and V are l-set, let them be {u0, u1, . . . , ul−1} and
{v0, v1, . . . , vl−1}, respectively. By the definition of an (n, l)-crown uivj ∈ E(H ) if and only if j = i, i + 1, . . . , i +
n− 1(mod l).
Now, let G= (A, B) be a bipartite graph with n edges where A = {a0, a1, . . . , as−1} and B = {b0, b1, . . . , bt−1}.
By the hypothesis, G has a-bilabeling f. For convenience, let f (ai)= siand f (bj)= tj, respectively. It is not difficult
to see the following bipartite graphs G1, G2, . . . ,and Gl are isomorphic to G where Gj = (Aj, Bj)is defined as
follows:
(i) Aj= {usi+j (mod l)|i ∈ Zs} and Bj= {vti +j (mod l)|i
∈ Z t}, and
(ii) usi+j (mod l)is adjacent to vti +j (mod l)if and only if aiand bi are adjacent in G.
It is left to show that G1, G2, . . . ,and Gl form a decomposition of H. Suppose not. Then, let uv ∈ E(Gj1)∩
E(Gj2), j1 = j2. This implies that we have two distinct edges x1y1 and x2y2 in E(G) such that f (x1)+ j1 ≡
f (x2)+ j2≡ (mod l) and f (y1)+ j1≡ f (y2)+ j2≡ (mod l).
Hence, f (y1)− f (x1)= f (y2)− f (x2). By the fact that f is a-bilabeling, f (y1)− f (x1)= f (y2)− f (x2)if and
only if x1y1= x2y2. This leads to a contradiction. Thus, G1, G2, . . . ,and Gldecompose H.
With the above propositions, we are able to obtain the following result.
Theorem 2.4. Let r be the radius of a tree T with n edges. Then T2rn.
Proof. By Proposition 2.2, T has a-bilabeling with strength r/2n. Since an (n, r/2n)-crown has an -labeling
(by Proposition 2.1) and an (n,r/2n)-crown can be decomposed into r/2n isomorphic copies of T (by Proposition 2.3), we conclude that Tr/2n.
Corollary 2.6. TO(n2).
Proof. This is a direct consequence of the fact that the radius of a tree with n edges is at mostn/2.
By combining Theorems 1.1 and 2.4, we also have a result which is a slight improvement of the research work obtained by Kézdy and Snevily[4].
Corollary 2.7. Let T be a tree with n edges and radius r, then T decomposes Kt for some t(r + 1)n2+ 1.
3. Concluding remark
As mentioned in Section 1, Snevily conjectured that the-labeling number of a tree with n edges is at most n. Note that there are trees which do not have-labelings. For these trees T, T2. We believe that Tn is quite possible. Hopefully, by finding a better-bilabeling with smaller strength, we can prove the conjecture in the near future. Acknowledgment
The authors with to extend their gratitude to the referees for their helpful comments in revision. References
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