Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 28, Number 3, June 2023
Page(s) 221 - 222
DOI https://doi.org/10.1051/wujns/2023283221
Published online 13 July 2023

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

For graphs G and H, an embedding of G into H is an injection ϕ:V(G)V(H) such that ϕ(a)ϕ(b)E(H) whenever abE(G). A packing of p graphs G1,G2,,Gp into H is a p-tuple Φ=(ϕ1,ϕ2,,ϕp) such that, for i=1,2,,p, ϕi is an embedding of Gi into H and the p sets ϕi(E(Gi)) are mutually disjoint. When all Gi are isomorphic to G, we call it a k-parking of G. {L-End} A bipartite graph G with the vertex partition X=(X1,X2) is denoted as G(X1, X2) or G(X). For a packing Φ=(ϕ1,ϕ2,,ϕp) of G1(X), G2(X),,Gp(X) into a bipartite graph H(Y), we mean that Φ is a packing such that ϕi(Xj)Yj, i=1, 2,, p, j=1,2.

Packing problems are central to combinatorics. Many classical problems can be stated as packing problems, such as Mantel's Theorem which can be formulated by saying that if G is an n-vertex graph with less than (n2)-n24 edges, then the two graphs K3 and G can be packed into Kn. The packing problem has received a lot of attention. Many interesting results and elegant proofs of these results were obtained. For a survey, see Refs.[1,2]. Among the best known packing problems, the famous tree packing conjecture of Gyráfás and Lehel has driven a large amount of research in the area.

Conjecture 1 (Gyráfás and Lehel [3]) Given nN and trees T1,, Tn with Ti having order i, the graphs T1,,Tn can be packed into complete graph Kn.

A packing of many of the small trees from Conjecture 1 was obtained by Bolloba˙s [4], who showed that one can pack T1,,Tn/2 into Kn and that a better bound would follow from a famous conjecture of Erdo¨s. In a similar direction, Hobbs, Bourgeois and Kasiraj made the following conjecture.

Conjecture 2 (Hobbs, Bourgeois and Kasiraj[5]) Any sequence of trees T2,,Tn , with Ti having order i, can be packed into Kn-1,n/2.

The conjecture has been verified for several very special classes of trees. Hobbs, Bourgeois and Kasiraj[5] proved that any two trees of order m and n with m<n can be packed into a complete bipartite graph Kn-1,n/2. Yuster[6] proved that any sequence of trees T2,,Ts , s<5/8n can be packed into Kn-1,n/2. Motivated by these results, Wang proposed the following conjecture.

Conjecture 3 (Wang [7]) For each tree T of order n, there is a k-packing of T in some complete bipartite graph Bn+k-1(X,Y).

This conjecture is true for k=2 and k=3 (see Theorem 1 and Theorem 2).

Theorem 1[8] Let S(U0,U1) and T(V0,V1) be two trees of order n with |Ui|=|Vi| (i=0,1). Then there exists a complete bipartite graph Bn+1(X0, X1) such that there is a packing of S(U0,U1) and T(V0,V1) in Bn+1(X0,  X1).

Theorem 2[7] For each tree T of order n, there is a 3-packing of T in some complete bipartite graph Bn+2(X,Y).

In this paper we prove the following theorem.

Theorem 3   For each tree T of order n, whose bipartite vertex classes are of size k1 and k2, there is a k-packing of T in some complete bipartite graph Bn+m(X,Y), where m=(h1-k1)+(h2-k2), h1=|X| and h2=|Y|.

1 Proof of Theorem 3

We recall the following lemma due to Yuster[6].

Lemma 1[6] Let H be a bipartite graph with vertex classes H1 and H2 of sizes h1 and h2 , respectively, h1h2. Let T be a tree whose bipartite vertex classes are of size k1 and k2. If k1h1 and k2h2 and e(H)k2h1+k1h2+k1+k2-h1-h2-k1k2, then H contains a subgraph isomorphic to T.

Proof of Theorem 3   Let T be a tree of order n, whose bipartite vertex classes are of size k1 and k2, where k1+k2=n. Let Bn(X, Y) be a complete bipartite graph of order n with vertex partition sets X and Y of sizes k1 and k2, respectively. Now we add some vertices into X and Y such that |X|=h1, |Y|=h2, h1h2 and (h1-k1)(h2-k2)+(h1-k1)+(h2-k2)(k-1)(n-1). So we get a complete bipartite graph Bn+m(X,Y) of order n+m, where m=(h1-k1)+(h2-k2). Clearly, Bn+m(X,Y) contains a copy of T. Suppose that we have already packed k-1 copies of T in Bn+m(X,Y). Let H be the spanning subgraph of Bn+m(X,Y) which contains all the edges that do not appear in the packing. It is easy to see that e(H)=h1h2-(k1+k2-1)(k-1). Since (h1-k1)(h2-k2)+(h1-k1)+(h2-k2)(k-1)(n-1), we have e(H)k2h1+k1h2+k1+k2-h1-h2-k1k2. By Lemma 1, we find a copy of T in H, and add T to the packing. So there is a k-packing of T in the complete bipartite graph Bn+m(X,Y).

The proof is completed.

References

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