完全圖的極大路徑填充

國 立 交 通 大 學

應 用 數 學 系

碩士論文

完全圖的極大路徑填充

The maximal P

k+1

-packings of K

n

研 究 生：徐瑩晏

指導教授：傅恆霖 教授

完全圖的極大路徑填充

The maximal P

k+1

-packings of K

n

研 究 生：徐瑩晏 Student：Ying-Yen Hsu

指導教授：傅恆霖 Advisor：Hung-Lin Fu

國

立 交 通 大 學

應

用 數 學 系

碩

士 論 文

Submitted to Department of Applied Mathematics College of Science

National Chiao Tung University in Partial Fulfillment of the Requirements

for the Degree of Master

in

Applied Mathematics June 2010

Hsinchu, Taiwan, Republic of China

完全圖的極大路徑填充

研究生：徐瑩晏 指導教授：傅恆霖 教授

國 立 交 通 大 學

應 用 數 學 系

摘要

圖

G的H-填充是一個蒐集一些G中邊兩兩互相不同的

子圖的集合

℘ = {H

, H

, … , H

}，其中每一個子圖H

都和

H

同構。我們將那些沒有被

H

用到的邊集合所導出的子圖稱

為此填充的殘留。若殘留的部分找不到一個子圖和

H同構的

話，則稱此填充為極大填充。每一個極大填充不一定擁有

同樣多的基數。令

S(G; H)為一蒐集圖G所有極大H-填充的

基數的集合。如果圖

G是一個有 n 個點的完全圖，則將

S(K

; H)簡化為S(n; H)。

在此篇論文中，我們將探討

S(n; P

)，其中P

是一

條有

k + 1個點的路徑。值得注意的是，一個K

的極大

P

-填充若其基數為

S(n; P

)中的最小值，則此填充為K

的最

小

P

-填充且其殘留恰會是一個限制子圖為

P

的極圖。

反之，一個

K

的極大

P

-填充若其基數為

S(n; P

)中的最

大值，則此填充為

K

的最大

P

-填充。因此，我們有了以

下結果：當

k = 3,4,5,6時，我們確立了S(n; P

)。

中華民國一百年六月

The maximal P

_k+1

-packings of K

_n

Student: Ying-Yen Hsu

Advisor: Hung-Lin Fu

Department of Applied Mathematics National Chiao Tung University

June, 2010

Abstract

An H-packing P = {H1, H2, . . . , Hs} of a graph G is a set of edge-disjoint

subgraphs of G in which each subgraph Hi is isomorphic to H. The leave L

of P is the subgraph induced by the set of edges of G that does not occur in any Hi. P is a maximal H-packing if L contains no subgraph that is

isomorphic to H. Let S(G; H) denote the set of all possible cardinality of P such that P is a maximal H-packing of G. In case that G is the complete graph of order n, we use S(n; H) to denote S(Kn; H) for convenience.

In this thesis, we focus on the study of S(n; Pk+1) where Pk+1 is a path

with k + 1 vertices. Notice that the leave of the packing which attends min S(n; Pk+1) is the extremal graph which forbids Pk+1 and the packing

which attends max S(n; Pk+1) is a maximum packing of Kn with Pk+1’s.

The main result obtained in this thesis is that we determine S(n; Pk+1) for

誌謝

首先，我想感謝我的指導教授-傅恆霖老師，這兩年來，

不論在課業或者研究上都給我很多指導和關心，使我能順利解

決難題。當我家裡發生變故時，也時常給予關心和問候，讓我

感覺很溫暖。許多時候，老師也會和我們分享做人處事的道理，

使我受益良多。

另外，也非常感謝兩位口委-黃國卿老師和史青林老師，

大老遠的來到交大替我口試，也提供不少建議，我的論文才能

更加完善。同時也感謝系上的老師們，這兩年來不僅在課業上

給予我指導，也教會我許多課外知識，謝謝可愛的老師們。

再來我想感謝各位學長姐們，尤其是黃明輝學長，在我的

研究題目上，提供我不少想法，也時常與我討論我的研究內容，

因為有他，我的論文才能突破瓶頸。也謝謝惠蘭學姐、敏筠學

姐、軒軒學長、Robin 學長、智懷學長、惠娟學姐總是不厭其

煩的教我許多東西，在論文寫作、口試簡報上也給我很多經驗

指導。

謝謝育慈、瑜堯、侖欣、小吵和志嘉學長，以及我的最佳

戰友-周小泡，每當遇到困難時，你們總是給予許多幫助：和

我討論 latex 的使用、指導我的英文文法、照顧生病的我、和

我一起走過瓶頸…等。因為有你們的陪伴和鼓勵，我的研究生

活才能如此多采多姿，謝謝你們給我很多快樂有趣的回憶。

最後謝謝我的家人，總是在背後默默支持我、鼓勵我，讓

我能夠無憂無慮的完成我的學業。謝謝你們。

Contents

Abstract (in Chinese) i

Abstract (in English) ii

Acknowledgement iii

Contents iv

List of Figures v

1 Introduction and Preliminary 1

1.1 Motivation . . . 1 1.2 Preliminaries . . . 1

1.3 Known results . . . 3

2 Maximum and minimum Pk+1-packing of Kn 4

3 The Spectrum of the maximal Pk+1-packing of Kn 9

List of Figures

1 Pk+1-decomposition of Kk,k. . . 5 2 P5-decomposition of K4,4,1 and P6-decomposition of K5,5,2. . . 6

1

Introduction and Preliminary

1.1

Motivation

The notation of ”graph” was first mentioned by Euler at around 1736 in which

he solved the well-known K¨onigsberg seven bridges. But, it was until 50 years

ago, ”graph theory” found its importance in computer sciences. Since then, graph

models were utilized in solving many discrete type problems, networking, schedul-ing, designs, . . . , etc. Especially, in recent decode, it was used in dealing several

problems in computational molecular biology including DNA sequencing. Without

a doubt, it is one of the most important branch of mathematics in 20th _{and 21}st

centuries.

Graph decomposition is one of the most popular topics studied in Graph Theory.

Among many reasons in its applications ”decomposing graphs into cliques” is most remarkable one since it is equivalent to obtaining combinatorial designs. Therefore,

this topic attracts many researchers from many aspects of combinatorial theory, graph theorists, combinatorialists and also coding theorists.

1.2

Preliminaries

A graph G = (V, E) consists of a vertex set V (G), an edge set E(G) and a

relation that associates with each edge two vertices called its endpoints. If uv is a edge, we say u and v adjacent, and u (or v) incident to the edge. If there are more

than one edge in the same pair of endpoints, these edges are called multiple edges. A loop is an edge which has the same endpoints. If a graph contains no multiple

edges and loop, we call the graph a simple graph. All of the graphs considered in this thesis are simple graphs. For graph terminologies, we refer to [4].

The order of G, denoted by |G|, is the number of vertices of G. The size of G, denoted by kGk, is the number of edges of G. Consider v ∈ V (G). The degree of

v means the number of vertices adjacent to v. The complement of G is denoted

by G where V (G) = V (G) and E(G) = {uv | u, v ∈ V (G) and uv /∈ E(G)}. The

union of the graphs G1 and G2 is denoted by G1∪ G2 where

if E(G1) ∩ E(G2) = ∅. The union of t copies of the same graph G is denoted by

Gt.

A path Pn is a simple graph where V (Pn) = {v0, v1, . . . , vn−1} and E(Pn) = {vivi+1 | i = 0, 1, . . . , n − 2}. If there exists a path from u to v for all u, v ∈ V (G), then G is connected. Let C be a connected subgraph of G. For all u ∈ V (G)\V (C),

if we can’t find v ∈ V (C) such that there exist a path from u to v, then C is a component of G. If for all u, v ∈ V (G), uv ∈ E(G), then we say G is a complete

graph. Let Kn denoted the complete graph of order n. A complete multigraph

λKn, is a complete graph Kn, in which every edge is taken λ times. The complete

m-partite graph Kn1,n2,...,nm is a simple graph has m partite sets Vi with order

ni, 1 ≤ i ≤ m, respectively and two vertices are adjacent if and only if they are

belonged to distinct partite sets. If n1 = n2 = . . . = nm = n in Kn1,n2,...,nm,

than the graph is denoted simply by Km(n). Let X = {x0, x1, . . . , xn1−1} and

Y = {y0, y1, . . . , yn2−1} be the partite sets of Kn1,n2, and X = {x0, x1, . . . , xn1−1},

Y = {y0, y1, . . . , yn2−1} and Z = {z0, z1, . . . , zn3−1} be the partite sets of Kn1,n2,n3

in later sections unless we give the other definition. The index of x, y or z will always be taken mod n1, n2 or n3, respectively. The bipartite difference of an edge

xiyj in Kn1,n2 as the value j − i ( mod n2)

If a graph H satisfies V (H) ⊆ V (G) and E(H) ⊆ E(G), then we say H is a

subgraph of G, denoted by H ⊆ G. Consider E0 ⊆ E, an edge-induced subgraph H

of G is defined by H = (V0, E0) where V0 = {v ∈ V | v is a endpoint of some e ∈

E0}.

An H-packing of a graph G is a set P = {H1, H2, . . . , Hs} such that Hi is isomorphic to H of G for i = 1, 2, . . . , s where Hi and Hj are edge-disjoint for all i 6= j. The leave L of a packing P is the subgraph induced by the set of edges

of G that does not occur in any Hi. If L contains no edges, then G is said to be H-decomposable, denoted by H | G. A packing P is said to be maximal if the

leave of P contains no subgraph that is isomorphic to H. The size of a packing P, denoted by |P|, is the cardinality of P. P is a maximum maximal packing (or simply maximum packing) if |P| ≥ |P0| for all other maximal packing P0_{. On the} other hand, P is a minimum maximal packing (or simply minimum packing) if

set of all sizes such that there exists a maximal packing with this size. Clearly,

max S(G; H) is the size of the maximum packing, and min S(G; H) is the size of the minimum packing. In case that G is the complete graph of order n, we use

S(n; H) to denote S(Kn; H) for convenience.

1.3

Known results

The problem of path decompositions of complete graphs was first mentioned in

[7]. Earlier results on this topic are on the case when the paths have same size, and such that each vertex belongs to exactly l of these paths [9, 10, 11]. Tarsi proves

that if n be odd or λ even, and M = m1, m2, . . . , msa sequence of natural numbers with mi ≤ n − 3 and P mi = λn(n−1)₂ , then there exists a PM-decomposition [15]. He also proves the necessary and sufficient condition for the existence of a Pm

-decomposition of a λKn is λn(n − 1) ≡ 0 (mod 2m) and n ≥ m + 1 [15]. Recently,

Bryant [3] proves that Tarsi’s result is also true for any positive integers n, λ and sequence m1, m2, . . . , ms.

There are numerous papers written on packing problem. The maximum number of Kk-packing of Kn had solved only in these cases k = 3 [17] and k = 4 [1]. The

maximal Ck-packing of Kn had solved only when k = 3, k = 4 [18] and k = 5 [14]. Roditty proved the conjecture saying that max S(n; T ) = b n₂
/hc (h is the number of edges of T ) for all trees on at most 7 vertices [12, 13]. And then Caro and Yuster proved that conjecture for any trees if n ≥ n0(T ) [5]. In 1990, Fu, Huang and Shiue

[8] find the spectrum S(n; Sq) where Sq means the star with q edges. Chen, Fu and Huang studied the (P3 ∪ P2)-packing of G different from K1,1,3c+1 with |G| ≥ 5, kGk ≥ 6 and δ(G) ≥ 2 [6].

In this thesis, we study the problem of packing Pk+1 into Kn, n ≥ k + 1. In

Section 2, we present the maximum and minimum size of Pk+1-packing and then

2

Maximum and minimum P

-packing of K

Review that if P is a Pk+1-packing of Kn, P is said to be maximal if there is no path Pk+1 ∈ P such that {P/ k+1} ∪ P is also a packing. In this section, we find

the maximum and minimum number of the element in S(n; Pk+1).

Bryant [3] showed the following theorem.

Theorem 2.1. Let n, λ and s be positive integers and let m1, m2, . . . , ms be a

sequence of positive integers. there exist s pairwise edge-disjoint paths of lengths m1, m2, . . . , ms in λKn if and only if mi ≤ n − 1 for i = 1, 2, . . . , s and m1+ m2+ . . . + ms ≤ λn(n−1)₂ .

From theorem 1, we have the corollary.

Corollary 2.2. There exists a maximum (maximal) Pk+1-packing of Kn.

More-over, the size of the packing is b n₂
/kc, i.e., max S(n; Pk+1) = b n₂
/kc.

Now, we consider the minimum packing. Obviously, a minimum packing has a maximum number of edges of leave. Note that, P is maximal if and only if L

contains no Pk+1. Therefore, we consider the maximum number of edges of leave

which contains no Pk+1 for the minimum Pk+1-packing problem. For a given graph

F , ext(n; F ) denote the maximum number of edges of a graph of order n not containing F as a subgraph.

Lemma 2.3. [2] If n = tk + r, 0 ≤ r < k, then ext(n; Pk+1) =

tk(k−1)

2 +

r(r−1) 2 .

Moreover, a graph G of order n has the edge number ext(n; Pk+1) if and only if

G has t + 1 connected components where one is Kr and the others are Kk, i.e.

G = Kt

k∪ Kr.

We have the corollary.

Corollary 2.4. Consider n = tk + r, 0 ≤ r < k. If P is a minimum (maximal)

Pk+1-packing of Kn, then |P| ≥ tk(t−1)

2 + tr.

Proof. If L is the leave of P, then kLk ≤ tk(k−1)₂ + r(r−1)₂ since P is maximal.

The number of edges of all Pk+1 of P is at least t(t−1)k

2

2 + tkr, and then |P| ≥

tk(t−1)

x₀ x₁ x₂ x_k-1 x_k-1

y₀ y₁ y₂ y_k-1 y_k-1 …

…

Figure 1: Pk+1-decomposition of Kk,k.

The following lemmas and Open problems are essential for finding the upper

bound of the size of minimum packing. Following the Lemma 2.3, we consider whether exists a Pk+1-packing of Kn such that L = Kkt ∪ Kr. Hence we need to know whether Kk,r, r ≤ k has Pk+1-decomposition.

Lemma 2.5. There exists a Pk+1-decomposition of Kk,k.

Proof. For 0 ≤ i ≤ k − 1, let

pi =  



yixiyi+1xi−1. . . yi+(k+1₂ −1)xi−(k+1₂ −1), if k is odd; and yixiyi+1xi−1. . . y_i+(k+2

2 −2)xi−( k+2

2 −2)yi+( k+2

2 −1), if k is even.

(see Fig. 1). By the fact that all edges of pi receive different bipartite labeling,

p1, p2, . . . pk−1 are edge-disjoint paths of length k. Let P = {pi | 0 ≤ i ≤ k − 1}, P is a Pk+1-decomposition of Kk,k.

Lemma 2.6. There exists no Pk+1-decomposition of Kk,r if 1 ≤ r < dk₂e or both k and r are odd.

Proof. It is clear for r < dk₂e. Each Pk+1 has one of its end vertices in X and the other one in Y when k is odd. Since all k vertices of X have odd degree r and

there are only r paths in decomposing Kk,r into Pk+1’s, we are done.

Lemma 2.7. There exists a Pk+1-decomposition of Kk,k,r if 1 ≤ r < dk₂e.

Proof. The Pk+1-decomposition of Kk,k,r is obtained from the decomposition of

Kk,k. Take a subgraph Kk,k of Kk,k,r, P = {pi | 0 ≤ i ≤ k − 1} is a decomposition of Kk,k where pi is defined in Lemma 2.5. For each pi, 0 ≤ i ≤ k − 1, delete the

x₀ x₁ x₂ x₃ x₄ y₀ y₁ y₂ y₃ y₄ x₀ x₁ x₂ x₃ x₄ y₀ y₁ y₂ y₃ y₄ z₀ z₁ x₀ x₁ x₂ x₃ y₀ y₁ y₂ y₃ x₀ x₁ x₂ x₃ y₀ y₁ y₂ y₃ z₀

Figure 2: P5-decomposition of K4,4,1 and P6-decomposition of K5,5,2.

last 2r edges and combine with the path having length 2r in Kk,k,r as following if

k is odd :    y_i+(k+1 2 −r)zr−1xi−( k+1 2 −(r−1))zr−2. . . yi+( k+1 2 −2)z1xi−( k+1 2 −1)z0yi, if r is even; x_i−(k+1 2 −r)zr−1yi+( k+1 2 −(r−1))zr−2. . . yi+( k+1 2 −2)z1xi−( k+1 2 −1)z0yi, if r is odd, or as following if k is even :          y_i+(k 2−r+1)zr−1xi−( k 2−r+1)zr−2. . . yi+( k 2−1)z1xi−( k 2−1)z0yi, if r is even; x_i−(k 2−r)zr−1yi+( k 2−(r−2))∪ y_i+(k 2−(r−2))zr−2xi−( k 2−(r−2))zr−3. . . yi+( k 2−1)z1xi−( k 2−1)z0yi, if r is odd.

(There are examples for k = 4, r = 1 and k = 5, r = 2 in Fig. 2.) Let p0_i denote the new paths obtain from pi for 0 ≤ i ≤ k − 1. Clearly, the edges both in p0i and pi be

used only once since pi is a decomposition. The edges adjacent to zj, 0 ≤ j ≤ r − 1, are all used and only once. Next we consider the edges deleted above the edge set

of those edges is {x_i−(k+1 2 −(j+1))yi+( k+1 2 −j), yi+( k+1 2 −j)xi−( k+1 2 −j)| 0 ≤ i ≤ k − 1, 1 ≤ j ≤ r} if k is even or is {y_i+(k 2−j)xi−( k 2−j), xi−( k 2−j)yi+( k 2−(j−1)) | 0 ≤ i ≤ k − 1, 1 ≤ j ≤ r}

if k is odd. Since x_i−(k+1

2 −(j+1)) = xi+1−( k+1 2 −j) and yi+( k 2−(j−1)) = yi+1+( k 2−j) for

y₀ y₁ y₂ y₃ y₀ y₁ y₂ y₃

x₀ x₁ x₂ x₃ x₀ x₁ x₂ x₃ o

y0 y1 y2 y3

x₀ x₁ x₂ x₃ o₁ o₂

Figure 3: P5-decomposition of K5,4 and K6,4.

each Hamilton cycle, we can decomposition it into two path of length k, and so we

have 2r’s Pk+1. Let P0 be the set of all these paths and p0i for 0 ≤ i ≤ k − 1, then P0 _{is a P}

k+1-decomposition of Kk,k,r.

Open problem 1. Is Kk,k+r Pk+1-decomposable if 1 ≤ r < dk₂e?

Open problem 2. Is Kk,k,r Pk+1-decomposable if dk₂e ≤ r < k and both k and r are odd?

Open problem 3. Is Kk,r Pk+1-decomposable if dk₂e ≤ r < k and at least one of k, r is even?

Remark. In Truszczynski’s paper [16], he verified that Kk,r can decomposed into

Pk+1’s if r ≥ dk₂e and r is even. Hence the only unknown case in Open problem 3 is that k is even and r is odd. Note that the case k, r odd is not possible (Lemma 2.6).

When r = k − 1 or r = k − 2, the Open problem 3 can construct direct from the Pk-decomposition of Kk−1,k−1. (The figure 3 give the example for the construction

of K5,4 and K6,4.) If r = k − 1, let the vertex classes of Kk,r (i.e. Kk,k−1) as follow:

X = {x0, x1, . . . , xk−2, o}, Y = {y0, y1, . . . , yk−2} if k is odd. X = {x0, x1, . . . , xk−2}, Y = {y0, y1, . . . , yk−2, o} if k is even.

From Lemma 2.5, the Pk-decomposition of Kk−1,k−1 is {pi | 0 ≤ i ≤ k − 2}

where pi defined as in Lemma 2.5. Let p0i = pi ∪ {y_i+(k+1

2 −1)o} when k is odd

and p0_i = pi ∪ {xi−(k₂−1)o} when k is even for 0 ≤ i ≤ k − 2. Then p0i is a Pk+1 -decomposition of Kk,k−1.

If r = k −2, both k and r are even and let the vertex classes of Kk,r (i.e. Kk,k−2) be

The Pk−1-decomposition of Kk−2,k−2 is {pi | 0 ≤ i ≤ k − 3} where pi defined as in Lemma 2.5, and then {{o1yi} ∪ pi ∪ {yi+(k₂−1)o2} | 0 ≤ i ≤ k − 3} is a Pk+1-decomposition of Kk,k−2.

Now, we have a lemma about the size of minimum packing for some small n.

Corollary 2.8. If n = k + r, r < k, and either 1 ≤ r < dk₂e or both k and r are

odd, there is no minimum (maximal) packing of Kn which has size r.

Proof. From Lemma 2.6, we know that there is no Pk+1-decomposition of Kk,r.

Therefore, we can’t have a packing with the leave Kk ∪ Kr and then there is no

3

The Spectrum of the maximal P

-packing of

K

In this section, we consider all the size of maximal Pk+1-packings of Kn where k ≤ 6. From Corollary 2.2, we have max S(n; Pk+1) = b n₂
/kc. Now, we consider the min S(n; Pk+1).

We give a direct construction to prove the Open problem 1, 2 and 3 are right

for k ≤ 6, and then have a minimum Pk+1-packing of Kn.

Lemma 3.1. The Open problem 1, 2 and 3 are true for k ≤ 6. That is,

1. There is a Pk+1-decomposition of Kk,k+r if 1 ≤ r < dk₂e.

2. There is a Pk+1-decomposition of Kk,k,r if dk₂e ≤ r < k and both k and r are odd.

3. There is a Pk+1-decomposition of Kk,r if dk₂e ≤ r < k and at least one of k, r is even.

Proof. Consider the cases k = 3, 4, 5 and 6, respectively.

Case 1. k = 3

The Open problem 2 can’t occur in these case. When k = 3, there are only

K3,2 and K3,4 be considered in Open problem 3 and Open problem 1, respectively.

From the Remark of Open problem 3, we can construct a P4-decomposition of K3,2

direct. Decompose K3,4 by two K3,2’s, then we are done by the P4-decomposition of K3,2.

Case 2. k = 4

The Open problem 2 can’t occur in the case. In Open problem 3, we only

con-sider the graphs K4,2 and K4,3. We have a direct construction of P5-decomposition

of K4,2 and K4,3 from the Remark of Open problem 3. K4,5 is the only case in

Open problem 1. Since K4,5 can decomposed to two graphs K4,3 and K4,2, we are

done.

When k = 5, the graph in Open problem 2 is only K5,5,3. Let

P = {j0z0j1z2j2z1, j2z0j3z1j4z2 | j = x, y} ∪ {y1x1z1x0z2x3, x1y2x0y3x4z0}

∪ {yixiyi+1xi−1yi+2xi−2| i = 0, 2, 3} ∪ {x2y1z1y0z2y3, z0y4x4y0x3y1},

then P is a P6-decomposition of K5,5,3. Consider K5,4, the only graph in Open

problem 3, has the P6-decomposition by the Remark of the Open problem 3. There

are two kind of graphs, K5,6 and K5,7, in Open problem 1 when k = 5. Let

p1 = x1y0x3y1x0y2, p2 = x1y3x3y4x2y5, p3 = x3y2x1y1x4y0, p4 = x3y5x1y4x4y3, p5 =

x4y2x2y0x0y4 and p6 = x4y5x0y3x2y1, then {pi | 1 ≤ i ≤ 6} is a P6-decomposition of K5,6. Let p0i = yixiyi+1xi−1yi+2x4, 1 ≤ i ≤ 3, p04 = x3y4x1y5x2y6, p05 = x1y6x3y5x0y4,

p0₆ = x0y1x3y2x4y5 and p07 = x2y4x4y6x0y0. Then {p0i | 1 ≤ i ≤ 7} is a P6 -decomposition of K5,7.

Case 4. k = 6

There is no graph in Open problem 2 and, we have three graphs K6,3, K6,4 and

K6,5 in Open problem 3 when k = 6. Clearly,

{x0y0x1y2x5y1x3, x1y1x2y0x3y2x4, x2y2x0y1x4y0x5}

is a P7-decomposition of K6,3. K6,4 and K6,5 have the P7-decomposition from the

Remark of Open problem 3. Final, the graphs in Open problem 1 are K6,7 and K6,8. Similar to case 2, K6,7 and K6,8have the P7-decomposition by use P7-decomposition

of K6,3 and K6,4.

Accordingly we have the theorem.

Theorem 3.2. Consider k ≤ 6, n = tk + r ≥ k + 1, 0 ≤ r < k. There exist a

minimum (maximal) Pk+1-packing P of Kn which with size

|P| =   

r + 1, if t = 1 and either 1 ≤ r < dk

2e or k and r are odd; tk(t−1)

2 + tr, otherwise.

Proof. Let L denote the leave of P. Note that, if we can find a minimum packing such that the leave with size tk(k−1)₂ +r(r−1)₂ , then |P| = ((tk+r)(tk+r−1)₂ − tk(k−1)₂ −

r(r−1)

2 )/k =

tk(t−1)

First, consider t = 1 (i.e. n = k + r) and either 1 ≤ r < dk₂e or both k and r are odd. tk(t−1)₂ + tr = r when t = 1. We have |P| 6= r from Corollary 2.8, and then |P| > r by Corollary 2.4. Consider n = 4 when k = 3, n = 5 when k = 4,

n = 6, 7, 8 when k = 5 and n = 7, 8 when k = 6.

If we take one P4 in K4 (i.e. r = 1), the other edges is also form a P4 clearly.

So K4 has a minimum P4-packing of size 2 (i.e. r + 1). We know that the size of

the maximal P5-packing of K5 more than r = 1, and let {ai | 0 ≤ i ≤ 4} be the

vertex set of K5. Since {a0a1a2a3a4, a0a2a4a1a3} is a minimum P5-packing of K5 with leave P3, we have a packing with size r + 1 = 2.

When k = 5, denote the vertex sets of K6 (i.e. r = 1), K7 (i.e. r = 2)

and K8 (i.e. r = 3) by {ai | 0 ≤ i ≤ 5}, {bi | 0 ≤ i ≤ 6} and {ci | 0 ≤

i ≤ 7}, respectively. {a5a0a1a3a4a2, a2a1a5a4a0a3} is a minimum P6-packing of K6, {b1b5b6b4b0b3, b3b6b2b0b1b4, b6b1b2b3b4b5} is a minimum P6-packing of K7, and {c1c4c2c5c3c6, c2c1c3c7c4c0, c3c0c5c1c6c7, c4c6c0c2c7c5} is a minimum P6-packing of K8. The size of these packing is r + 1.

Consider K7 and K8 with the vertex sets {ai | 0 ≤ i ≤ 6} and {bi | 0 ≤ i ≤ 7}, respectively. Since {a2a1a3a6a4a0a5, a3a0a2a6a5a1a4} is a minimum P7-packing of K7 and {b1b0b2b7b3b4b5, b1b6b5b7b4b0b3, b3b2b1b7b6b0b5} are a minimum P7-packing of

K8, we have packing with size r + 1. From above, we have |P| = r + 1 when

n = k + r and either 1 ≤ r < dk₂e or both k and r are odd.

Final, consider the other n. We have the proof by induction on n. If n =

k + r where k > r ≥ dk₂e and at least one of k, r is even, then there is a Pk+1

-decomposition of Kk,r from Lemma 3.1. In other words, we have a packing P

with leave L = Kk ∪ Kk ∪ Kr and |P| =

2k(2−1)

2 + 2r. If n = 2k, we have a

packing P with leave L = Kk ∪ Kk from Lemma 2.5 and then |P| = 2k(2−1)₂ . If

n = 2k + r, 1 ≤ r < dk₂e, we have a Pk+1-decomposition of Kk,k,r from Lemma 2.7.

That is, there is a minimum packing P with leave L = Kk ∪ Kk∪ Kr and then

|P| = 2k(2−1)₂ + 2r. If n = 2k + r where k > r ≥ dk₂e and both k and r are odd,

we have a Pk+1-decomposition of Kk,k,r from Lemma 3.1. Accordingly, there is

minimum packing P with leave L = Kk∪ Kk∪ Kr. In this case, |P| =

2k(2−1)

2 + 2r.

Suppose there exist a minimum packing of the complete graph with order

the complete graph with order n, there are four cases to be considered.

Case 1. n = tk + r where t ≥ 3 and 1 ≤ r < dk₂e.

Let G0 = G\{v1, v2, . . . , vk}, then G0 is a complete graph with order (t−1)k +r. Since (t − 1)k + r < n, by induction hypothesis, there exists a packing P0 of G0 with leave L0 where L0 = K_kt−1∪ Kr and |P0| = (t−1)k(t−2)₂ + (t − 1)r. Obviously P0 is also a packing of G. The leave of P0 in G has edges E(L0) ∪ {uvi|u ∈ V (G0), i = 1, 2, . . . , k}∪{vivj|1 ≤ i < j ≤ k}. Consider K(t−1)k+r,k = Kk,kt−2∪Kk+r,k, K(t−1)k+r,k can decomposed to (t − 1)k + r’s edge-disjoint paths of length k by Lemma 2.5 and Lemma 3.1. Let P00 be the set of these paths and P = P0∪ P00_{, then P is a} Pk+1-packing of G with leave L = L0 ∪ Kk. Clearly, P is a minimal Pk+1-packing of Kn. Note that, |P| = (t−1)k(t−2)₂ + (t − 1)r + (t − 1)k + r = tk(t−1)₂ + tr.

Case 2. n = tk where t ≥ 3.

The proof of this case is similar to case 1 (let r = 0) by use Lemma 2.5.

Case 3. n = tk + r where t ≥ 3, k > r ≥ dk₂e and both k and r are odd.

The case can be show as case 1. Note that, we consider K(t−1)k+r,k = Kk,kt−2∪ Kk−1,k ∪ Kr+1,k in this case. Since k and r are odd, r ≤ k − 2, there exist Pk+1

-decompositions of Kk−1,k and Kr+1,k by Lemma 3.1. The graph Kk,k has Pk+1

-decomposition from Lemma 2.5.

Case 4. n = tk + r where t ≥ 2, k > r ≥ dk₂e and at least one of k, r is even. The idea of case 4 is also like case 1. But in this case, we replaced K(t−1)k+r,k by K_k,kt−1∪ Kr,k and to complete it by Lemma 2.5 and Lemma 3.1.

Therefore, the proof concludes by mathematical induction.

In our study of the spectrum S(n; Pk), there are some special technique. The

main technique used in the section need switching some edges of the paths in a

given maximal Pk+1-packing of Kn with size s and the edges in the leave of the

packing to produce a new Pk+1-packing. The goal is causing the new leave contains

one only path of length k and add the path to the new Pk-packing, then we have

a new maximal Pk+1-packing of Kn with size s + 1. Let A = {n = tk + r | either r ≥ dk

2eand one of k, r is even or t > 1}. If n ∈ A,

then there exist a minimum Pk+1-packing P of Kn with the construction as in

Theorem 3.2. Note that the leave is L = Kt

k∪ Kr, that is, the graph induce by the packing from Kn is Kt(k),r. The way of the edge switching as the following step:

Step 1. Consider a subgraph H as Kk,r or Kk,k,r, 0 ≤ r < k, of Kk,...,k,r at a time.

Step 2. Take one or two paths from P which is also contain in H. Let P1 denote

the set of these paths.

Step 3. Choose k edges from L, and rearranging the k edges and the paths took in step 2 to produce a new Pk+1-packing P2.

Step 4. Let P0 = P \ P1 ∪ P2. Since L contains no Pk+1, the leave of P0 is also contains no Pk+1. Hence P0 is a maximal Pk+1-packing of Knwith size |P|+1.

Step 5. Repeat step 1 to 4, then we can have a maximal Pk+1-packing of Kn with

desired size s until s = b n₂
/kc.

Afterward we study the spectrum S(n; Pk), n = tk + r ≥ k + 1, 0 ≤ r < k for

k ≤ 6. In our way, the subgraph Kk,r or Kk,k,r be considered only in first took

and only took Kk,k later in step 1. Note that, all subgraphs considered are edge

disjoint. We only need verify the step 3 can always operate, then the following

Lemmas are done.

Lemma 3.3. Consider n = 3t + r ≥ 4, 0 ≤ r < 3. 1. S(n; P4) = {s | r + 1 ≤ s ≤ b n₂
/3c} = {2} if n = 4. 2. S(n; P4) = {s | 3t(t−1) 2 + tr ≤ s ≤ b n 2
/3c} if n ≥ 5.

Proof. First, consider n = 4. Since r = 1 and b n₂
/3c = 2 = r + 1 is the size of minimum P4-packing of K4, we are done.

Second, consider n ≥ 5. We have a minimum P4-packing P of Kn with leave L

as defined in the Theorem 3.2. Study the three cases. Case 1. n ≡ 0 ( mod 3).

Clearly, K3,3 is a subgraph of Kn. Recall that, y0x0y1x2 and y2x2y0x1 are paths of P. Let P0 = P \ {y0x0y1x2} ∪ {x1x0y1x2, y0x0x2x1}, then P0 is a maximal P4 -packing of Kn with leave L0. Note that y2x2y0x1 is still in P0 and there are three edges y0y1, y1y2 and y2y0 which in E(K3,3) not be used. Let P00 = P0\ {y2x2y0x1} ∪ {x2y2y0x1, x2y0y1y2}, then P00 is a maximal P4-packing of Kn with leave L00. If n = 6, there is no edge in L00 and then |P00| = |P0_{| + 1 = |P| + 2 = 5 = b} n

Reuse the subgraph K3,3 as well as the way of the edge switching when n ≥ 9, then

we have the maximal P4-packing with desired size until the size of leave less than k.

Case 2. n ≡ 2 ( mod 3).

Consider K3,2 ⊆ Kn with vertex classes X = {x0, x1, o} and Y = {y0, y1}. Clerly, y0x0y1o is a path of P. Let P0 = P \ {y0x0y1o} ∪ {x1x0y1o, y0x0ox1}, then P0 _{is a maximal P}

4-packing of Kn with leave L0. If n = 5, the leave L0 only contain edge {y0y1} and then |P0| = |P| + 1 = 3 = b n₂
/3c. If n ≥ 8, we can use the subgraph K3,2 and K3,3 (see case 1) to produce a maximal P4-packing with desired

size until the size of leave less than k. Case 3. n ≡ 1 ( mod 3).

Since n ≥ 5 (i.e. n ≥ 7 in these case), we have K3,3,1 is a subgraph of Kn. We have x2z0y0x0 and x1z0y2x2 are paths of P. Let P0 = P \ {x2z0y0x0} ∪ {x2x0y0z0, x0x1x2z0}, then P0 is a maximal P4-packing of Kn with leave L0. Note that x1z0y2x2 is still in P0 and there are three edges y0y1, y1y2 and y2y0 which

in E(K3,3,1) not be used. We will let the three edges be used. Consider P00 =

P0 _{\ {x}

1z0y2x2} ∪ {x1z0y2y0, y0y1y2x2}, then P00 is a maximal P4-packing of Kn

with leave L00. If n = 7, we are done. If n ≥ 10, we can produce a maximal

P4-packing with desired size by the subgraph K3,3,1 and K3,3 (see case 1) until the

size of leave less than k.

Lemma 3.4. Consider n = 4t + r ≥ 5, 0 ≤ r < 4.

1. S(n; P5) = {s | r + 1 ≤ s ≤ b n₂
/4c} = {2} if n = 5. 2. S(n; P5) = {s | 4t(t−1)₂ + tr ≤ s ≤ b n₂
/4c} if n ≥ 6.

Proof. When n = 5, we have a minimum P5-packing of K5 with size r + 1 =

b n₂
/4c} = {2} from Theorem 3.2.

If n ≥ 5, there is a minimum P5-packing P of Kn with leave L as defined in

the Theorem 3.2. Consider the four cases. Case 1. n ≡ 0 ( mod 4).

Take the subgraph K4,4 of Kn. Clearly, {yixiyi+1xi−1yi+2 | 0 ≤ i ≤ 3} ⊆ P is a packing of K4,4. Let P0 = P \ {y0x0y1x3y2} ∪ {x1x0y1x3y2, y0x0x3x2x1}, P00 = P0_\{y

{y3x3y0x2y1} ∪ {x3y3y0x2y1, x3y0y1y2y3}, then P0, P00 and P000 are maximal P5 -packing of Kn. If n = 8, there is no edge in L00 and then we have the packings with size from 4t(t−1)₂ + tr = 4 to b n₂
/4c = 7. If n ≥ 12, we can use the subgraph K4,4 to produce a maximal P5-packing with desired size until the size of leave less than k.

Case 2. n ≡ 3 ( mod 4).

Consider the subgraph K4,3 of Kn with the vertex classes X = {x0, x1, x2} and Y = {y0, y1, y2, o}. y0x0y1x2o, y1x1y2x0o are paths of K4,3. Let P0 = P \ {y0x0y1x2o} ∪ {oy0x0y1x2, y0y1y2ox2}, then P0 is a maximal P5-packing of Kn. Let P00 _{= P}0_{\ {y}

1x1y2x0o} ∪ {x0oy1x1y2, y0y2x0x1x2}, then P00 is a maximal P5-packing of Kn with leave L00. Note that there are only one edge which in E(K4,3) not be

used. If n = 7, we are done. If n ≥ 11, we can produce a maximal P5-packing with desired size by the subgraph K4,3 and K4,4 (see case 1) until the size of leave less

than k.

Case 3. n ≡ 2 ( mod 4).

K4,2 is the subgraph of Kn with vertex classes X = {x0, x1, o1, o2} and Y = {y0, y1}. Recall that P contains a path o1y0x0y1o2. Let P0 = P \ {o1y0x0y1o2} ∪ {o2o1y0x0y1, y1o2x0x1o1}, then P0 is a maximal P5-packing of Kn with leave L0. If n = 6, we are done since E(L0) = {x0o1, x1o2, y0y1}. Reuse the edge switching

of subgraph K4,2 and K4,4 (see case 1) when n ≥ 10, then we have the maximal

P5-packing with desired size until the size of leave less than k.

Case 4. n ≡ 1 ( mod 4).

In the case, K4,4,1 is a subgraph of Kn. Recall that {xi−1z0yixiyi+1 | 0 ≤ i ≤ 3} ⊆ P.Let P0 _{= P \ {x} 3z0y0x0y1} ∪ {x3z0y0x0x1, x1x2x3x0y1}, P00 _{= P}0_{\ {x} 2z0y3x3y0} ∪ {x2z0y3y0x3, y0y1y2y3x3} and, P000 = P00\ {x0z0y1x1y2, x1z0y2x2y3} ∪ {z0y1x1y2y0, z0y2x2y3y1, x2x0z0x1x3}, then P0,P00 and P000 all are the maximal P5-packing of Kn. Clearly, there is no edge in the leave of P000. If n = 9, we are done. If n ≥ 13, we can produce a maximal

P5-packing with desired size by the subgraph K4,4,1 and K4,4 (see case 1) until the size of leave less than k.

Lemma 3.5. Consider n = 5t + r ≥ 6, 0 ≤ r < 5.

1. S(n; P6) = {s | r + 1 ≤ s ≤ b n₂
/5c} if n = 6, 7 or 8. 2. S(n; P6) = {s | 5t(t−1)₂ + tr ≤ s ≤ b n₂
/5c} if n ≥ 9. Proof. First, consider n = 6, 7 or 8. In fact

{r + 1 ≤ s ≤ bn 2  /5c} =          {2, 3}, if n = 6 {3, 4}, if n = 7 {4, 5}, if n = 8.

There are a minimum P6-packing of K6 with size 2, a minimum P6-packing of

K7 with size 3 and a minimum P6-packing of K8 with size 4 from Theorem 3.2.

Besides, we have a maximum P6-packing of K6 with size 3, a maximum P6-packing of K7 with size 4 and a maximum P6-packing of K8 with size 5 from Corollary 2.2.

Final, consider n ≥ 9. We have a minimum P6-packing P of Kn with leave L

as defined in Theorem 3.2. There are five cases of n.

Case 1. n ≡ 0 ( mod 5).

Consider the subgraph K5,5. We have {yixiyi+1xi−1yi+2xi−2 | 0 ≤ i ≤ 4} ⊆ P from Lemma 2.5 and Theorem 3.1. Switch the edges as following:

• Replace {y0x0y1x4y2x3} by {x1x0y1x4y2x3, y0x0x4x3x2x1}. • Replace {y1x1y2x0y3x4} by {x3x1y2x0y3x4, y1x1x4x2x0x3}. • Replace {y3x3y4x2y0x1} by {y3x3y4x2y0y2, y2y4y1y3y0x1}. • Replace {y4x4y0x3y1x2} by {x4y4y0x3y1x2, x4y0y1y2y3y4}.

We can produce some P6-packings of Kn and the new packings are all maximal. If

n ≥ 10, the step 3 can operate until the leave contains no edge by use subgraph K5,5.

Case 2. n ≡ 4 ( mod 5).

Take the subgraph K5,4 of Kn and then {yixiyi+1xi−1yi+2o | 0 ≤ i ≤ 3} ⊆ P from the remark of Open problem 3 and Theorem 3.1. Switch the edges as following:

• Replace {y1x1y2x0y3o} by {x3x1y2x0y3o, y1x1ox2x0x3}.

• Replace {y2x2y3x1y0o, y3x3y0x2y1o} by {y1x2y3x1y0o, y2y1y3x3y0x2, oy1y0y3y2 x2}.

We produce some P6-packings of Kn and the new packings are all maximal. If

n = 9, it’s finished. If n ≥ 14, the step 3 can operate until the leave contains only

one edge by use subgraph K5,4 and K5,5 (see case 1). Case 3. n ≡ 3 ( mod 5).

K5,5,3 is a subgraph of Kn in the case. We have y0x0y1x4y2x3, x1y2x0y3x4z0, y3x3y4x2y0x1 and z0y4x4y0x3y1 are the paths in P. Switch the edges as following:

• Replace {y0x0y1x4y2x3} by {x1x0y1x4y2x3, y0x0x4x3x2x1}. • Replace {x1y2x0y3x4z0} by {x1x3x0y3x4x2, x2x0y2x1x4z0}. • Replace {y3x3y4x2y0x1} by {y3x3y4x2y0y2, y2y4y1y3y0x1}. • Replace {z0y4x4y0x3y1} by {y3y4x4y0x3y1, z0y4y0y1y2y3}.

We produce some P6-packings of Kn and the new packings are all maximal. Hence,

it’s trivial for n = 13. Consider the subgraph K5,5,3 and K5,5 (see case 1) in step 1 if n ≥ 18, then the step 3 can operate until the leave contains only three edges.

Case 4. n ≡ 2 ( mod 5).

If n = 12, we only consider the subgraph K5,5,2 in step 1. We have a packing

{yi+1z1xi−2z0yixi | 0 ≤ i ≤ 4} of K5,5,2 from Lemma 2.7. Switch the edges as following:

• Replace {y1z1x3z0y0x0} by {x1x2x3z0y0x0, y1z1x3x4x0x1}. • Replace {y2z1x4z0y1x1} by {x1x2x4z0y1x1, y2z1x4x1x3x0}. • Replace {y4z1x1z0y3x3} by {y4z1x1x0y3x1, y1y4y2y0y3x3}. • Replace {y0z1x2z0y4x4} by {y0z1x2z0y4y3, x4y4y0y1y2y3}.

We produce some P6-packings of Kn and the new packings are all maximal. Since

the edge number of the leave of the last new packing is one, we are done. If

subgraph by the same way, then we have packing with desired size.

Case 5. n ≡ 1 ( mod 5).

Similarly to the other cases, K5,5,1 is a subgraph of Kn. From Lemma 2.7 and

Theorem 3.1, {xi−2z0yixiyi+1xi−1 | 0 ≤ i ≤ 4 | 0 ≤ i ≤ 3} ⊆ P. Switch the edges as following:

• Replace {x3z0y0x0y1x4} by {z0y0x0y1x4x3, x4x0x1x2x3z0}. • Replace {x4z0y1x1y2x0} by {y1x1y2x0x2x4, x0x3x1x4z0y1}. • Replace {x1z0y3x3y4x2} by {x1z0y3x3y4y1, y1y3y0y2y4x2}. • Replace {x2z0y4x4y0x3} by {x2z0y4x4y0y1, y1y2y3y4y0x3}.

We produce some P6-packings of Kn and the new packings are all maximal. If

n = 11, we are done since the leave of the last new packing contains no edge.

Consider the subgraph K5,5,1 and K5,5 (see case 1) if n ≥ 16, then we have packing with desired size by edge switching.

Lemma 3.6. Consider n = 6t + r ≥ 7, 0 ≤ r < 6.

1. S(n; P7) = {s | r + 1 ≤ s ≤ b n₂
/6c} if n = 7, 8. 2. S(n; P7) = {s | 6t(t−1)₂ + tr ≤ s ≤ b n₂
/6c} if n ≥ 9. Proof. First, consider n = 7, 8. Note that,

{r + 1 ≤ s ≤ bn 2  /6c} =    {2, 3}, if n = 7 {3, 4}, if n = 8.

Since the minimum number of the set is the size of minimum P7-packing of Kn

and the maximum number of the set is the size of maximum P7-packing of Kn, we

are done from Theorem 3.2 and Corollary 2.2.

We have a minimum P7-packing P of Knwith leave L as defined in Theorem 3.2

if n ≥ 9. Consider the edge switching of Kn by the following six cases of n. Case 1. n ≡ 0 ( mod 6).

Consider K6,6 ⊆ Kn, {yixiyi+1xi−1yi+2xi−2yi+3 | 0 ≤ i ≤ 4} ⊆ P from Lemma 2.5 and Theorem 3.1. Switch the edges as following:

• Replace {y0x0y1x5y2x4y3} by {x1x0y1x5y2x4y3, y0x0x5x4x3x2x1}. • Replace {y1x1y2x0y3x5y4} by {x4x1y2x0y3x5y4, y1x1x3x0x4x2x5}. • Replace {y2x2y3x1y4x0y5, y3x3y4x2y5x1y0} by {x3y3x1y4x0y5x2, y5x1x5x3y4x2x0, x1y0y2x2y3y5x1}. • Replace {y4x4y5x3y0x2y1} by {y1y4x4y5x3y0x2, x2y1y3y0y4y2y5}. • Replace {y5x5y0x4y1x3y2} by {x5y5y0x4y1x3y2, x5y0y1y2y3y4y5}.

All the new packings created after the edge switching are maximal since L contains

no P7 and then we have the packing with the desired size when n ≥ 12.

Case 2. n ≡ 5 ( mod 6).

We have a subgraph K6,5 with vertex classes X = {x0, x1, x2, x3, x4} and Y = {y0, y1, y2, y3, y4, o} in the case. {yixiyi+1xi−1yi+2xi−2o | 0 ≤ i ≤ 4} ⊆ P. Switch the edges as following:

• Replace {y0x0y1x4y2x3o} by {oy0x0y1x4y2x3, x3oy4y3y2y1y0}. • Replace {y1x1y2x0y3x4o} by {y4y1x1y2x0y3x4, x4oy2y4y0y3y1}. • Replace {y3x3y4x2y0x1o} by {x1oy3x3y4x2y0, y2y0x1x4x2x3x0}. • Replace {y4x4y0x3y1x2o} by {x0x4y0x3y1x2x1, y4x4x3x1x0x2o}.

If n = 11, we can have the maximal P7-packing with the desired size. Consider the subgraph K6,5 and K6,6 (see case 1) if n ≥ 17, then we are done.

Case 3. n ≡ 4 ( mod 6).

K6,4 is a subgraph of Kn. Recall that the vertex classes of K6,4 are X =

{x0, x1, x2, x3, o1, o2} and Y = {y0, y1, y2, y3}, and {o1yixiyi+1xi−1yi+2o2 | 0 ≤ i ≤ 3} ⊆ P. Switch the edges as following:

• Replace {o1y0x0y1x3y2o2} by {o2o1y0x0y1x3y2, y2o2x0x1x2x3o1}. • Replace {o1y1x1y2x0y3o2} by {x2o1y1x1y2x0y3, o1x1x3x0x2o2y3}. • Replace {o1y2x2y3x1y0o2} by {x0o1y2x2y3y0y1, y1y2y3x1y0o2x3}.

If n = 10, we are done. If n ≥ 16, we can produce a maximal P7-packing with

desired size by consider the subgraph K6,4 and K6,6 (see case 1). Case 4. n ≡ 3 ( mod 6).

Consider the subgraph K6,3 if n = 9. According the construction of P, we

have a subset {x0y0x1y2x5y1x3, x1y1x2y0x3y2x4, x2y2x0y1x4y0x5} of P and it is is a packing of K6,3. Switch the edges as following:

• Replace {x0y0x1y2x5y1x3} by {x0x1y2x5y1x3x2, x3x4x5x0y0x1x2}. • Replace {x1y1x2y0x3y2x4} by {x0x4x1y1x2y0y2, y0x3y2x4x2x5x1}. • Replace {x2y2x0y1x4y0x5} by {y2x0y1x4y0x5x3, x1x3x0x2y2y1y0}.

After the edge switching, we have the maximal P7-packing with the desired size.

If n ≥ 15, the step 3 can operate by use the edge switching of subgraph K6,3 and

K6,6 (see case 1).

Case 5. n ≡ 2 ( mod 6).

In step 1, we take the subgraph K6,6,2 and K6,6 (see case 1) if n ≥ 14. We

have {yi+2z1xi−2z0yixiyi+1 | 0 ≤ i ≤ 5} ∪ {y2x4y3x5y4x0y5, y4x2y5x3y0x4y2} ⊆ P is

a packing of K6,6,2 from Lemma 2.7 and Theorem 3.1. In step 3, switch the edges

as following: • Replace {y2z1x4z0y0x0y1} by {y2z1x4z0y0x0x1, x1x2x3x4x5x0y1}. • Replace {y3z1x5z0y1x1y2} by {x0x3x5z0y1x1y2, y3z1x5x2x4x1x3}. • Replace {y2x4y3x5y4x0y5, y4x2y5x3y0x4y1} by {x2y4x0x4y3x5x1, x0x2y5x3y0y2x4, x0y5y1x4y0y4x5}. • Replace {y0z1x2z0y4x4y5} by {z1x2z0y4x4y5y3, y5y2y4y1y3y0z1}. • Replace {y1z1x3z0y5x5y0} by {y0y1z1x3z0y5x5, x5y0y5y4y3y2y1}.

The P7-packings still maximal after the edge switching, we can produce the packing with the desired size.

Case 6. n ≡ 1 ( mod 6).

We will consider K6,6,1 in the last case. Recall that, K6,6,1 has a packing

• Replace {x4z0y0x0y1x5y2} by {x3x4z0y0x0x5y2, x3x2x1x0y1x5x4}. • Replace {x5z0y1x1y2x0y3} by {x3x5z0y1x1y2x0, y3x0x3x1x4x2x5}. • Replace {x0z0y2x2y3x1y4, x1z0y3x3y4x2y5} by {x0x2y3x1y4y0y2, x5x1z0y3x3y4x2, y1y5x2y2z0x0x4}. • Replace {x2z0y4x4y5x3y0} by {x2z0y4x3y5y3y0, y3y1y4y2y5x3y0}. • Replace {x3z0y5x5y0x4y1} by {x3z0y5x5y0y1y2, y2y3y4y5y0x4y1}.

We have some maximal P7-packings of Kn with different size. It is take only one

subgraph K6,6,1 if n = 13, and take subgraph K6,6,1 and K6,6 (see case 1) both in step 1. We are done.

From Lemma 3.3 to Lemma 3.6, we have the theorem.

Theorem 3.7. Suppose n = tk + r ≥ k + 1, 0 ≤ r < k. If k ≤ 6, then

1. S(n; Pk+1) = {s | r + 1 ≤ s ≤ b n₂
/kc}, if t = 1 and either 1 ≤ r < dk₂e or both k and r are odd.

2. S(n; Pk+1) = {s | tk(t−1)

2 + tr ≤ s ≤ b

n

References

[1] A.E. Brouwer, Optimal packing of K4’s into a Kn, J. Combin. Theory Ser. A 26 (1979) 278 - 297.

[2] B¨ela Bollob¨as, Graduate Texts in Mathematics: Modern Graph Theory,

Pub-lished by Springer (1998).

[3] D. Bryant, Packing paths in complete graphs, J. Combin. Theory Ser. B 100

(2010) 206 - 215.

[4] Douglas B.West, Introduction to Graph Theory, Published by Prentice Hall

(2001).

[5] Y. Caro and R. Yuster, Packing graphs: The packing problem solved, Elect.

J. of Combin. 4 (1997), #R1.

[6] C.C. Chen, H.L. Fu and K.C Huang, (P3∪ P2)-packing of graphs.

[7] H.E. Dudeney, Amusements in Mathematics, Nelson, Edinburgh, 1917, reprinted by Dover Publications, New York, 1959.

[8] H.L. Fu, K.C. Huang and C.L Shiue, Maximal configurations of stars, JCMCC 8 (1990) 51 - 59.

[9] P. Hell, A. Rosa, Graph decompositions, handcuffed prisoners and balanced P -designs, Discrete Math. 2 (1972) 229 - 252.

[10] S.H.Y. Hung, N.S. Mendelsohn, Handcuffed designs, Aequationes Math. 11 (1974) 256 - 266.

[11] J.F. Lawless, Further results concerning the existence of handcuffed designs, Aequationes Math. 11 (1974) 97 - 106.

[12] Y. Roditty, Packing and covering of the complete graph with a graph G of

four vertices or less, J. Combin. Theory Ser. A 34 (1983) 231 - 243.

[13] Y. Roditty, Packing and covering of the complete graph IV, the trees of order

[14] A. Rosa and S.Znam, Packing pentagons into complete graphs: how clumsy

can you get, Discrete Math. 128 (1994), 305 - 316.

[15] M. Tarsi, Decomposition of a complete multigraph into simple paths:

Non-balanced handcuffed designs, J. Combin. Theory Ser. A 34 (1983) 60 - 70.

[16] M. Truszczynski, Note on the decomposition of λKm,n (λKm,n∗ ) into paths,

Discrete Math. 55 (1985), 89 - 96.

[17] J. Schonheim, On maximal systems of k-tuples, Studia Sci. Math. Hunger.

(1996), 363 - 368.

[18] J. Schonheim and A. Bialostocki, Packing and covering of the complete graph