Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 3, June 2024
Page(s) 239 - 241
DOI https://doi.org/10.1051/wujns/2024293239
Published online 03 July 2024

© Wuhan University 2024

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

We shall use standard graph theory notation. For any graph G, a neighbor of a vertex v is a vertex adjacent to v in G. NG(v) denotes the set of neighbors of a vertex v in G. The degree of v, denoted by degG(v), is |NG(v)|. Given a subset A of V(G), NG(v,A) is NG(v)A for the vertex vV(G). When the context is clear, the subscript G is omitted.

An endvertex is a vertex of degree 1 and non-endvertex is a vertex of degree >1. A node is a vertex adjacent to an endvertex. A supernode is a node x of G such that, with one exception, every neighbor of x is an endvertex. A star is a treewith only one non-endvertex.

An embedding of a graph 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 p-parking of G.

A k-partite graph G with the partition X=(X1,X2,,Xk) is denoted as G(X1,X2,,Xk) or G(X). In this case, it is said that G admits the partition X and |X|=k. If G admits two distinct partitions X and Y, then the notion that G(X)G(Y) is adopted here. If G and H admit the k-partitions X and Y, respectively, and ϕ is an embedding of G into H such that ϕ(Xi)Yi, then ϕ is restrained and this is denoted as ϕ:G(X)H(Y). A packing Φ=(ϕ1,ϕ2,,ϕp) of G1(X1),G2(X2),,Gp(Xp) into H(Y) is restrained if each embedding of Φ is restrained. A k-partite tree is a k-partite graph without cycles. Let T(X) denote the k-partite tree with X=(X1,X2,,Xk) as its k-partition and Bn(Y) denote the complete k-partite graph of order n with Y=(Y1,Y2,,Yk) as its k-partition.

Packing problems are central to combinatorics. Many exciting results and elegant proofs of these results were obtained [1-4]. For a survey, see Refs. [5,6]. Among the best-known packing problems, the famous Tree Packing Conjecture of Gyárfás and Lehel [7] has driven a large amount of research in the area. Bollobás [8] confirmed the "Tree Packing Conjecture" for many small trees. Motivated by the "Tree Packing Conjecture", Wang Hong made the following conjecture:

Conjecture[9] For each k-partite tree T(X) of order n, there is a restrained packing of two copies of T(X) into a complete k-partite graph Bn+m(Y), where m=k/2.

This conjecture has been verified in Ref. [10] for k=2. Recently, Sapozhnikov[11] proved this conjecture with k=3, which is stated as the following proposition:

Proposition 1[11] Let T(X) be a 3-partite tree of order n with partites X1,X2,X3. Then there exists a restrained 2-packing of T(X) into a complete3-partite graph Bn+1(Y1,Y2,Y3).

In this paper, we prove that the conjecture is true for k=4.

1 Main Results

In the following lemmas, we assume that T(X) is a counter-example of Theorem 1 of minimum order n.

Lemma 1[9] The endvertices of T(X) are adjacent to the same node if they are in the same partite.

Lemma 2[9] If v is a node of T(X) adjacent to endvertices in a partite X, then |X| is odd and deg (v,X)=(|X|+1)/2.

Theorem 1   For each 4-partite tree T(X) of order n with partition X=(X1,X2,X3,X4), there is a restrained 2-packing of T(X) into some complete 4-partite graph Bn+2(Y).

Proof   We assume that T(X) is a counter-example of Theorem 1 of minimum order n with partition X=(X1,X2,X3,X4). Then |Xi|>1,i=1,2,3,4, since Theorem 1 holds if |Xi|=1 for some i clearly.

If T(X) has exactly one node v, then T(X) is a star. So, for some i, deg (v,Xi)=|Xi|>1. By Lemma 2, we have deg (v,Xi)=(|Xi|+1)/2=|Xi|, a contradiction. Therefore, there are at least two nodes in T(X). By observing a longest path of T(X), there exist at least two supernodes in T(X).

Let v be a supernode of T(X) and u be the only one non-endvertex adjacent to v. Without loss of generality, we may assume that vX1. Let Vi=N(v,Xi), i=2,3,4. Then there is at least one of Vi, i=2,3,4, which is non-empty. Without loss of generality, we may assume that V2. By Lemma 1, we can see that all the endvertices of X2 are in the set V2. Let V=X2\V2. Then, X2=V2V and |V2|=|V|+1 by Lemma 2. So we may assume that V2={v1,v2,,vm} and V={vm+1,vm+2,,v2m-1}.

Consider the graph T(X')=T(X)\(X2{v}), where X'=(X1',X3',X4') with X1'=X1\{v},X3'=X3,X4'=X4. So T(X') is a 3-partite tree with order n', where n'=n-1-|X2|. By Proposition 1, there exists a restrained 2-packing Φ'=(ϕ1',ϕ2') of T(X') into a complete3-partite graph Bn'+1(Y1,Y3,Y4) with |Xi'||Yi|,i=1,3,4.

Suppose that there is ij such that Vi,Vj. Since u be the only one non-endvertex adjacent to v, without loss of generality, we may assume uXj and j=2. Now add two vertices x1, x2 to Y1 and a partite set Y2 to Bn'+1(Y1,Y3,Y4) such that |Y2|=|X2|. LetY2={y1,y2,,y2m-1}. Then the packing Φ' can be extended to T(X) as follows: define Φ(x)=Φ'(x) for xT(X'),ϕ1(v)=x1,ϕ2(v)=x2. Define Φ(X2) as follows:

ϕ 1 ( v i ) = y i for i=1,2,,2m-1, ϕ2(v1)=y1,ϕ2(vi)=y2m-i+1 for i=2,3,,2m-1.

Thus, we extend the Φ' to T(X) so that a restrained 2-packing Φ of T(X) into the Bn+2(Y) is obtained, where Y=(Y1{x1,x2},Y2,Y3,Y4), a contradiction.

Therefore, there exists only one nonempty Vi, which implies that all the endvertices adjacent to v are in Vi and uVi. Without loss of generality, we may assume that i=2 and u=v1. Now add a vertex x to Y1 and add a partite set Y2 to Bn'+1(Y1,Y3,Y4) such that |Y2|=|X2|+1. Let Y2={y1,y2,,y2m}. Then we extend the Φ' to T(X) as follows:

Define Φ(x)=Φ'(x) for xT(X') and ϕ1(v)=ϕ2(v)=x. Define Φ(X2) as follows: ϕ1(vi)=yi for i=1,2,,2m-1,ϕ2(vi)=y2m-i+1 for i=1,2,,2m-1. Thus, we extend Φ' to T(X) so that a restrained 2-packing Φ of T(X) into the Bn+2(Y) is obtained, where Y=(Y1{x},Y2,Y3,Y4), a contradiction.

The proof is completed.

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