完全圖的路徑分割

國 立 交 通 大 學

應用數學系

碩 士 論 文

完全圖的路徑分割

Decomposing the Complete Graph into Paths

研 究 生：張澍仁

指導教授：傅恆霖 教授

完全圖的路徑分割

Decomposing the Complete Graph into

Paths

研 究 生：張澍仁 Student：Shu-Ren Zhang

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

A Thesis

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 2006

Hsinchu, Taiwan, Republic of China

謝誌

首先，我想感謝我的指導老師傅恆霖教授。在撰寫論文這段期間，我碰

到了許多瓶頸，傅老師總是能提供我ㄧ些精闢的意見，使我少走了許多冤枉

路，也讓我接觸到各式各樣關於圖論的問題與想法。最重要的是，傅老師讓

我學到自己做研究的方法與態度。我想，這將是我碩士兩年中最大的收穫。

接著，我想感謝陳秋媛老師、翁志文老師、以及黃大原老師。大學時我

修的課程都偏向於純粹數學，其實當時數學讀的並不愉快，我似乎把以前對

數學的熱誠與愛好都遺忘在某個角落。組合數學則是我在研究所時才開始接

觸到的領域，感謝組合數學組的所有老師們，讓我重新燃起對數學的興趣。

最後，感謝黃明輝、郭志銘、陳宏賓、詹棨丰、羅元勳、張惠蘭等學長

姐在研究方面對我的幫助以及論文與投影片修改的建議。感謝研究所的各位

同學，筱凡、豪哥、老謝、教練、帥哥中、發誓、育慈、老闆、嘴砲吳、Re、

小強、大師兄、多啦 A 澍、妙妙、邱鈺傑、兆涵、黃皜文、威雄、敏筠、佩

純、陳子鴻。感謝你們，讓我研究所的生活過的這麼愉快、豐富，謝謝。

完全圖的路徑分割

研 究 生：張澍仁

指導老師：傅恆霖 教授

國 立 交 通 大 學

應 用 數 學 系

摘

要

已知當m可以整除完全圖的邊數時，在1≦m≦v-1的情況下一個v點的完全圖可以分 割成全部都是長度m的路徑。可是，任取一個正整數m滿足1≦m≦v-1，m並不一定 能夠整除v點的完全圖邊數。所以我們討論在這種情況下是否仍有類似的漂亮結果，即 當m不整除完全圖的邊數時，分割完全圖成為一些長度為m的路徑及一個長度為餘數的路 徑。在本論文中，我們證明：完全圖可以分割成為k個長度為m的路徑加上一個長度為r 的路徑，若且為若完全圖的邊數等於km+r且0≦r＜m≦v-1。

中 華 民 國 九 十 六 年 六 月

Decomposing the Complete Graph into Paths

Student: Shu-Ren Zhang

Advisor: Hung-Lin Fu

Department of Applied Mathematics National Chiao Tung University

Hsinchu, Taiwan 30050

Abstract

It is known that if m | v₂
, then the complete graph Kv can be decomposed

into paths of length m as long as 1 ≤ m ≤ v−1. But, for a given positive integer

1 ≤ m ≤ v −1, m may not be a factor of v₂
. Therefore, we are interested in the

case m - v2
. In these cases, we need a path of distinct length. Let Pt denote a

path with t edges. Then, it is proved in this thesis that the complete graph Kv

can be decomposed into k Pm’s and one Pr if and only if v₂
= km + r where

Contents

Abstract (in Chinese) i

Abstract (in English) ii

Contents iii

1 Introduction 1

2 Preliminaries 2

3 Known Results 5

3.1 Some results in cycle decomposition . . . 5 3.2 Some results in path decomposition . . . 6

4 Main Result 8

1

Introduction

Graph decomposition has been one of the most important topics in Graph Theory. Not only the study is close related to discover the structures of graphs, but also give rise another approach to construct combinatorial designs. It is well-known that the existence of a balanced incomplete block design (BIBD), (v, k, λ)-design, is equivalent to the decomposition of the multi-graph λKv into edge-disjoint complete graphs Kk.

As an analog, a λ-fold k-cycle system of order v is a decomposition of λKv into cycles

of length k and a λ-fold k-path system of order v is a decomposition of λKv into

paths of length k.

To construct a BIBD for each admissible triple v, k and λ is not an easy task, see [8, 12] for references. But, path systems have been obtained for all possible parameters v, k and λ, see [6, 13].

A bit of reflection, if Kv can be decomposed into paths of length k, then v ≥ k + 1

and k | v₂
. But, k | v₂
does not occur very often. In case that v₂
= qm + r, 0 < r < m, then decomposing Kv into paths of length m is not possible. Instead, we

try to decompose Kv into q paths each of length m and one path which is of length

2

Preliminaries

Note that the following definitions and notations can be referred to the book : INTRODUCTION TO GRAPH THEORY [14].

A graph G is a triple consisting of a vertex set V (G), an edge set E(G), and a relation that associates with each two vertices (not necessary distinct) called its endpoints.

A loop is an edge whose endpoints are equal. Multiple edges are edges having the same pair of endpoints. A simple graph is a graph having no loops or multiple edges. A complete graph Kv is a simple graph whose vertices are pairwise adjacent. A

complete graph with v vertices is denoted Kv.

A walk is a list v0, e1, v1, e2, v2, ..., ek, vk of vertices and edges such that, for 1 ≤

i ≤ k, the edge ei has endpoints vi−1and vi. If the edges e1, e2, ..., ek are distinct, then

this walk is called a trail. If, in addition, the vertices v0, v1, ..., vk are distinct, then

this walk is called a path. If this walk is a path and v0 = vk, then this walk is called

a cycle. A path with k vertices is denoted Pk and a cycle with k vertices is denoted

Ck. But, for convenience, we use Pk to denote a path with k edges throughout this

thesis.

The length of a walk, trail, path, or cycle is its number of edges. A walk or trail is closed if its endpoints are the same.

A factor of a graph G is a spanning subgraph of G. A k-factor is a spanning k-regular subgraph.

A matching of size k in a graph G is a set of k pairwise disjoint edges. The vertices belonging to the edges of a matching are saturated by the matching; the others are unsaturated. If a matching saturates every vertex of G, then it is a perfect matching or 1-factor. We shall denote 1-factor by I.

A Hamiltonian graph is a graph with a spanning cycle, also called a Hamiltonian cycle. A Hamiltonian path is a path containing all the vertices. An Eulerian circuit (or Eulerian trail) in a graph is a circuit (or trail) containing all the edges.

A separating set or vertex cut of a graph G is a set S ⊆ V (G) such that G − S has more than one component. The connectivity of G is the minimum size of a vertex set S such that G − S is disconnected or has only one vertex. A graph G is k-connected if its connectivity is at least k.

In this thesis, let v be an even integer and let Zv denote the vertex set of Kv. The

operations, addition and multiplication, are taken modulo v. We have listed some fundamental definitions of Graph Theory above. For our specific subject we also need the following notations :

1. Let di (difference i) be an edge set where 1 ≤ i ≤

v 2, then hx, yi ∈ di ⇔ i = min{| x − y |, v− | x − y |}. Note : | d1 | = | d2 | = ... = | dv₂−1 | = v, | dv 2 | = v 2.

2. Let A = ha0, a1, ..., asi be an ordered trail, then d(A) is the least number of

edges between two appearances of the same vertex along A (the length of the minimal cycle).

3. Let A = ha0, a1, ..., asi, B = hb0, b1, ..., bti be two ordered trails and as = b0,

then A + B = ha0, a1, ..., as, b1, b2, ..., bti.

4. Let A = ha0, a1, ..., asi and k ∈ Z,

then A + k = ha0 + k, a1+ k, ..., as+ ki (right shift k).

6. For x, y ∈ Zv, I(x, y) = hx, x + 1, x + 2, ..., y − 1, yi. 7. For v even, x ∈ Zv, a, b ∈ {1, 2, ..., v 2− 1} such that 1 ≤ a < b ≤ v 2 − 1,

S(v, x, a, b) = hx, x + a, x − 1, x + a + 1, x − 2, ..., yi, where the vertex y is chosen to make the length of the path (b − a + 1).

Note : The first edge is in da, the second edge is in da+1, and so on, and the

last edge is in db. Thus S(v, x, a, b) is a path containing exactly one edge of

da, da+1, ..., db respectively.

8. For v, x, a, b satisfying the conditions of the previous construction, L(v, x, a, b) = S(v, x, a, b) + hy, y + v

2i + S(v, x + v 2, a, b)

t_.

Note : This is a path of length 2(b − a) + 3 which uses two edges in differences a, a + 1, ..., b and one edge in difference v

2.

9. For v, x, a, b satisfying the conditions of the previous construction, Mv,a,b = {L(v, x, a, b) | 0 ≤ x ≤

v 2 − 1}. Note : This is a set of v

2 paths with length 2(b − a) + 3 which uses all the edges in da, da+1, ..., db and dv₂. Each vertex is the end vertex of exactly one path.

10. For v even, k ∈ {1, 2, ...,v 2 − 2},

C(v, k) = h0, k + 1, 1, k + 2, 2, k + 3, ..., k − 1, v − 1, k, 0i.

Note : This is a circuit of length 2v which contains all the edges in dk and dk+1.

11. For v even, a, b ∈ {1, 2, ...,v

2− 1} such that 1 ≤ a < b ≤ v

2− 1 and b − a is odd, P C(v, a, b) = C(v, a) + C(v, a + 2) + ... + C(v, b − 1).

3

Known Results

3.1

Some results in cycle decomposition

Theorem 3.1. [9] (1) For all odd integers n and all non-negative integer r satisfying 3r = n(n − 1)

2 there is a decomposition of Kn into r 3-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers r satisfying

3r = n(n − 2)

2 there is a decomposition of Kn− F into r 3-cycles which partitions the edges set of Kn− F .

We can establish the existence of cycle systems not only the 3-cycle system but also the m-cycle system for any m.

Theorem 3.2. [11] (1) For all odd integers n and all non-negative integer r and m satisfying mr = n(n − 1)

2 there is a decomposition of Kn into r m-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers

r and m satisfying mr = n(n − 2)

2 there is a decomposition of Kn− F into r m-cycles which partitions the edge set of Kn− F .

Theorem 3.3. [1] (1) For all odd integers n and all non-negative integers r, and s satisfying 3r + 5s = n(n − 1)

2 there is a decomposition of Kn into r 3-cycles and s 5-cycles which partitions the edge set of Kn. (2) For all even integers n and all

non-negative integers r, and s satisfying 3r + 5s = n(n − 2)

2 there is a decomposition of Kn− F into r 3-cycles and 5-cycles which partitions the edge set of Kn− F .

Theorem 3.4. [7] (1) For all odd integers n and all non-negative integer r, s and t satisfying 3r + 4s + 6t = n(n − 1)

2 there is a decomposition of Kn into r 3-cycles, s 4-cycles, and t 6-cycles which partition the edge set of Kn. (2) For all even integers

n and all non-negative integer r, s and t satisfying 3r + 4s + 6t = n(n − 2)

a decomposition of Kn− F into r 3-cycles, s 4-cycles, and t 6-cycles which partition

the edge set of Kn− F .

Theorem 3.5. [3] (1) For all odd integers n and all non-negative integer r and s satisfying 4r + 5s = n(n − 1)

2 there is a decomposition of Kn into r 4-cycles, s 5-cycles which partition the edge set of Kn. (2) For all even integers n and all

non-negative integer r and s satisfying 4r + 5s = n(n − 2)

2 there is a decomposition of Kn− F into r 4-cycles, s 5-cycles which partition the edge set of Kn− F .

The following useful contains three different lengths which are n, n − 1, n − 2.

Theorem 3.6. [7] Let S = {n − 2, n − 1, n}. If n is odd and a(n − 2) + b(n − 1) + cn = n(n − 1)

2 , then Kn = aCn−2+ bCn−1+ cCn. If n is even and a(n − 2) + b(n − 1) + cn = n(n − 2)

2 , then Kn− F = aCn−2+ bCn−1+ cCn.

Alspach Conjecture is also true if the cycles lengths mi are bounded by some linear

function of n and n is sufficiently large.

Theorem 3.7. [2] Assume n must be larger than N2 which is very large absolute

constants. If m1, ..., mt are integers with 3 ≤ mi ≤ b

n − 112 20 c and

Pt

i=1mi = n₂
(n

odd) or n₂
− n

2 (n even), then one can pack Kn (n odd) or Kn− I (n even) with cycles of lengths m1, ..., mt.

3.2

Some results in path decomposition

Theorem 3.8. [5] Let n be an even positive integer. Then Kn can be decomposed

into n

2 hamiltonian paths.

Theorem 3.9. [10] If n is odd and {ai : 1 ≤ i ≤ r} is a multiset of r positive integers

satisfying 1 ≤ ai ≤ n − 2 and Pr_i=1ai = n₂
. Then Kn can be decomposed into

Theorem 3.10. [13] Let m | λ n₂
, and m ≤ n − 1. Then λKn can be decomposed

into isomorphic paths of length m.

Theorem 3.11. [4] If v is odd. Let m1, m2, ..., mt be t positive integers such that

1 ≤ mi ≤ n − 2,

Pt

i=1mi+ k(n − 1) = n₂
, and k ∈ {1, 2,

n − 1

2 }, then Kv can be decomposed into t + k paths P1, P2, ..., Pt+k such that the length of Pi is mi for

i = 1, 2, ..., t and the length of Pi _{is n − 1 for i > t.}

Theorem 3.12. [4] If v is odd. Let n − 1 ≥ m1 ≥ m2 ≥ ... ≥ mt ≥ 1 and

h ≤ mt ≤ n − h − 1 such that

Pt

i=1mi = n

2
, m1 = m2 = ... = mh = n − 1. Then

Kv can be decomposed into t paths P1, P2, ..., Pt such that the length of Pi is mi for

i = 1, 2, ..., t. Moreover, if there exists a h < t0 ≤ t such that h ≤ mt0 ≤ n − h − 1 or

h ≤Pt

i=t0m_i ≤ n − h − 1, then K_v can be decomposed into t paths P1, P2, ..., Pt such

that the length of Pi is mi for i = 1, 2, ..., t.

Theorem 3.13. [4] If v is odd. Let n − 1 ≥ m1 ≥ m2 ≥ ... ≥ mt ≥ 1, mt < h, and

mt−1− mt ≤ n − h − 1 such that

Pt

i=1mi = n

2
, m1 = m2 = ... = mh = n − 1. Then

Kv can be decomposed into t paths P1, P2, ..., Pt such that the length of Pi is mi for

i = 1, 2, ..., t.

Theorem 3.14. [4] If v is odd. Let n − 1 ≥ m1 ≥ m2 ≥ ... ≥ mt ≥ 1 and

n + h − 2 ≤ mt+ mt−1 ≤ 2n − h − 3 such that Pt_i=1mi = n₂
, m1 = m2 = ... =

mh = n − 1. Then Kv can be decomposed into t paths P1, P2, ..., Pt such that the

length of Pi _{is m}

i for i = 1, 2, ..., t. Moreover, if there exists a h < t0 ≤ t such

that n + h − 2 ≤ Pt

i=t0mi ≤ 2n − h − 3, then Kv can be decomposed into t paths

P1_{, P}2_{, ..., P}t _{such that the length of P}i _{is m}

4

Main Result

We shall prove the main theorem in what follows.

Theorem 4.1. Kv can be decomposed into k Pm’s and one Pr if and only if v₂
=

km + r where 0 ≤ r < m ≤ v − 1.

First of all, we obtain some lemmas below by using the preliminary definitions.

Lemma 4.2. The union of the i-th path of Mv,2,m₂ (the endpoints are i−1 and i−1+

v 2 ) and one of hi − 1, ii and hi − 1 + v

2, i + v

2i is a simple path of length m.

Proof. By the definition, the i-th path : hi − 1, i + 1, i − 2, i + 2, ..., yi + hy, y +v 2i + hy + v 2, ..., i + 2 + v 2, i − 2 + v 2, i + 1 + v 2, i − 1 + v

2i, where y is chosen to make the length of the path (m − 1).

Checking the segment hi − 1, i + 1, i − 2, i + 2, ..., yi which contains (m

2 − 1) edges. The subsequence of even indices which starts at the vertex i+1 is {i+1, i+2, i+3, ...}. The subsequence of odd indices which starts at the vertex i−1 is {i−1, i−2, i−3, ...}. Since m

2 − 1 < v

2− 1 and the length of these two subsequences are less than v

4. Thus the segment hi − 1, i + 1, i − 2, i + 2, ..., yi does not contain the vertex i.

Now, consider the segment hy +v

2, ..., i + 2 + v 2, i − 2 + v 2, i + 1 + v 2, i − 1 + v 2i t_.

Since the length is (m

2 − 1), the subsequence of even indices is an increasing sequence which starts at the vertex i + 1 +v

2 and the subsequence of odd indices is a decreasing sequence which starts at i − 1 + v

2. Then it is proved by the same way as above. Similarly, since the i-th path does not contain the vertex i + v

2, the union of the i-th path of Mv,2,m₂ (endpoints are i − 1 and i − 1 +

v

2 ) and one of hi − 1, ii and hi − 1 + v

2, i + v

2i is a path of length m.

Proof. Since C(v, k) = h0, k + 1, 1, k + 2, 2, k + 3, ..., k − 1, v − 1, k, 0i. The odd places of C(v, k) is the subsequence {0, 1, 2, ..., v − 1} and the even places of C(v, k) is the subsequence {k + 1, k + 2, ..., v − 1, 0, 1, ..., k − 1, k}. Thus, each vertex of Kv

appears twice in C(v, k); one in the odd places of C(v, k) and the other one in the even places of C(v, k). Let xeven (xodd) denote the vertex x ∈ Zv which appears in

the even (odd) places of C(v, k). If xodd appears before xeven, then the distance from

xodd to xeven is 2v − 2k − 1 ≥ 2k + 1. Else, the distance from xeven to xodd is 2k + 1.

This concludes the proof.

Lemma 4.4. d(P C(v, a, b)) = 2a + 1.

Proof. Looking at Lemma 4.3 and the structure of P C(v, a, b), it suffices to check whether the length of the cycle C which begins in C(v, k) and ends in C(v, k + 2) is larger than 2k + 1.

Let xe1 denote the vertex x ∈ Zv which appears in the even places of C(v, k) and

xo1 be the other one which appears in the odd places of C(v, k). Similarly, let xe2 and

xo2 denote the vertex x which appear in C(v, k +2). Let d(x, y) be the distance from x

to y. If the cycle C begins at xo1and ends at xo2, then d(xo1, xo2) = 2v ≥ 2k +1. If the

cycle C begins at xe1 and ends at xe2, then d(xe1, xe2) ≥ 2v − 4 ≥ 2k + 1. If the cycle

C begins at xo1 and ends at xe2, then d(xo1, xe2) = 2v − 2k − 5 ≥ 2k + 1. If the cycle

C begins at xe1 and ends at xo2, then d(xe1, xo2) = 2k + 1. Thus, d(P C(v, a, b)) =

min{d(C(v, a)), d(C(v, a + 2)), ..., d(C(v, b − 1))} = min{2a + 1, 2a + 5, ..., 2b − 1} = 2a + 1.

Lemma 4.5. When v is odd, v₂
= km + r and 0 ≤ r < m = v − 1. Kv can be

Proof. Let V (Kv) = {x∞, x1, x2, ..., xv−1}.

Let Ci _{= hx}

∞, xi, xv+i−2, xi+1, xv+i−3, xi+2, ..., xi+v−3₂ , xi+v−1₂ , x∞i (Indices take

mod-ulo v − 1). Then Kv can be decomposed into {Ci | 1 ≤ i ≤

v − 1

2 } and each C

i

is a hamiltonian cycle. Observe that hxi, xv+i−2i ∈ Ci for 1 ≤ i ≤

v − 1

2 . Thus, cutting these v − 1

2 edges from each C

i_{, we have} v − 1

2 hamiltonian path of length v − 1 = m. Now, the proof follows by combining the above v − 1

2 edges into the path hxv−1, x1, x2, ..., xv−1

2 i.

Proof of Theorem 4.1.

Since the necessary part is easy to see, it is left to prove the sufficiency. Note that when v is odd and 1 ≤ m ≤ v − 2, the condition is proved to be sufficient by [10]. Moreover, if v is odd and m = v − 1, the condition is proved to be sufficient by Lemma 4.5. Thus, we put the accent on the case : when v is even. The proof is split into four cases by taking Mv,a,b into consideration.

Case 1 : v − m ≡ 1 (mod 4)

Case 1.1 : m ≤ v − 5

Because M_v,1,m−1

2 covers all the edges in d1, d2, ..., d m−1 2 and d v 2 exactly once, P C(v,m + 1 2 , v

2− 1) covers all the edges in dm+12 , d m+3

2 , ..., d v

2−1 exactly once. Hence,

these two parts cover all the edges of Kv exactly once, i.e., E(Kv) = Mv,1,m−1₂ ∪

P C(v,m + 1 2 , v 2 − 1). By definition, M_v,1,m−1 2 is a set of v 2 paths of length 2( m − 1 2 − 1) + 3 = m. By Lemma 4.3, d(P C(v,m + 1 2 , v 2 − 1)) = 2( m + 1 2 ) + 1 = m + 2 > m. Then we can partition P C(v,m + 1 2 , v

2 − 1), starting from it’s beginning, into paths of length m, and the remainder is a path of length r.

Case 1.2 : m = v − 1

The proof follows by decomposing Kv into Hamiltonian paths.

Case 2 : v − m ≡ 3 (mod 4)

Case 2.1 : m ≤ v − 7

First, we claim that E(Kv) = Mv,2,m+1₂ ∪I(0, m)∪(I(m, 0)+P C(v,

m + 3 2 ,

v 2−1)). This is by the fact that M_v,2,m+1

2 contains all the edges in d2, d3, ..., d m+1

2 and d v 2,

{I(0, m) ∪ I(m, 0)} contains all the edges in d1, and P C(v,

m + 3 2 , v 2− 1) contains all the edges in dm+3 2 , ..., d v 2−1.

Next, we show that this construction provide a set of Pm’s and exactly one Pr. By

definition M_v,2,m+1

2 is a set of

v

2 Pm’s and I(0, m) is a Pm. Now the proof follows by claiming that d(I(m, 0)+P C(v,m + 1

2 , v 2−1)) > m. Since I(m, 0)+P C(v, m + 1 2 , v 2− 1) = hm, m + 1, m + 2, ..., v − 1, 0,m + 3 2 + 1, 1, m + 3 2 + 2, 2, m + 3 2 + 3, ...i. Therefore, I(m, 0) is increasing and P C(v,m + 1

2 , v

2 − 1) is alternately increasing. The first repeat vertex between I(m, 0) and P C(v,m + 1

2 , v

2−1) is m. So, if the first vertex m of P C(v,m + 1

2 , v

2−1) appears in the even (index) part, then the distance between these two vertices is (v − m) + (m − 4) > m. Otherwise, the distance is (v − m) + 2m > m. Thus d(I(m, 0) + P C(v,m + 1 2 , v 2− 1)) > m. Case 2.2 : m = v − 3 Since Mv,2,v₂−1 is a set of v

2 paths of length v − 3 = m, it covers all the edges of Kv exactly once except these edges in d1. Now, the cycle h0, 1, 2, ..., v − 1, 0i covers

all the edges in d1 and each segment of length less than v on the cycle is a path. This

Case 3 : v − m ≡ 2 (mod 4)

Case 3.1 : m ≤ v 2

Note that E(Kv) = (Mv,2,m 2 ∪ I(0, v 2)) ∪ (I( v 2, 0) + P C(v, m 2 + 1, v 2− 1)). Now, we prove that the construction is a set of Pm’s and exactly one Pr. By Lemma 4.2, The

union of the i-th path of Mv,2,m₂ (the endpoints are i − 1 and i − 1 +

v

2 ) and one of hi − 1, ii and hi − 1 +v

2, i + v

2i is a path of length m. Since I(0, v

2) = {hi, i − 1i | i = 1, 2, ...,v

2}, Mv,2,m2 ∪ I(0,

v

2) can be decomposed into paths of length m.

Similar to the proof of the claim in Case 2.1. We have to prove that the distance between repeating vertices is larger than m. Since the first repeat vertex between I(v 2, 0) and P C(v, m 2 + 1, v 2 − 1) is v 2, m ≤ v

2 and the length of I( v 2, 0) is

v 2. No matter the first repeat vertex v

2 belongs to the even part of P C(v, m

2 + 1, v

2 − 1) or not, the distance between these two vertices is larger than m. This concludes the proof of this subcase.

Case 3.2 : v

2 < m ≤ v − 6

Before the proof, we need some notations. Let f = min{[ v

2(v − m)], [

v(v − m − 2) 2(2m − v) ]} (Denote the integer part of x as [x]). Let Si = I(m, 0) + (i − 1)(m −

v 2), where 1 ≤ i ≤ f and SR = I(0, v 2 − f (v − m)) + (f − 1) v 2. Denote by Ti, 1 ≤ i ≤ f , f paths of length 2m − v each, cut along P C(v,m

2 + 1, v

2 − 1), and denote by TR, the final segment remaining of P C(v, m

2 + 1, v

2 − 1) after taking out T1, T2, ..., Tf. Finally, let Di = Si + Ti, 1 ≤ i ≤ f and DR = SR+ TR. Since the end of Si is

0 + (i − 1)(m − v

2) (mod v). And because l(Ti) = 2m − v is even and each Ti is gotten by cut along P C(v,m

2 + 1, v

2− 1). Thus, the beginning of Ti ≡ the end of Ti−1 ≡ the beginning of Ti−1+

2m − v

2 ≡ ... ≡ the beginning of Ti+ (i − 1)(m − v 2) ≡ 0 + (i − 1)(m − v

definition, (Mv,2,m₂ ∪ [d1\ (S1∪ S2∪ ... ∪ Sf∪ SR)]) ∪ D1∪ D2∪ ... ∪ Df∪ DRis obtained

from Mv,2,m₂ ∪d1∪P C(v,

m 2 +1,

v

2−1) which contains all the edges of Kv exactly once. Then we have E(Kv) = (Mv,2,m₂ ∪ [d1\ (S1, S2, ..., Sf, SR)]) ∪ D1∪ D2∪ ... ∪ Df∪ DR.

Now, let x be the end vertex of Si. Then Si = hx + m, x + m + 1, ..., xi and

Ti = hx, x+k, x+1, x+k+1, x+2, ..., x+m− v 2i. By the property of P C(v, m 2+1, v 2−1), we have m 2 < k < v

2. Thus, the vertices in the even places of Ti belong to the open interval (x + m

2, x + m). And because the vertices in the odd places of Ti is {x, x + 1, ..., x + m − v

2}, Si = {x + m, x + m + 1, ..., x}. The vertex x is the only common vertex of Si and Ti. Hence, for each i, Di = Si + Ti is a path of length

l(Si) + l(Ti) = (v − m) + (2m − v) = m. Since Si = ( I(v 2− (i)(v − m), v 2 − (i − 1)(v − m)) i : even I(v − (i)(v − m), v − (i − 1)(v − m)) i : odd ; and

SR =    I(v 2, v − (f )(v − m)) f : even I(0,v 2 − (f )(v − m)) f : odd ,

S1, S2, ..., Sf, SRare all distinct and that exactly one of hx, x+1i and hx+

v 2, x+

v 2+1i belongs to [d1\ (S1∪ S2∪ ... ∪ Sf∪ SR)]. Then we can obtain

v

2 Pm’s by Lemma 4.2. Finally, if f = [ v

2(v − m)], then l(SR) < l(Si) = v − m. Therefore, by the same argument as above, we obtain d(DR) > d(Di) = m.

On the other hand, if f = [l(P C(v,

m 2 + 1, v 2 − 1)) 2m − v ] = [ v(v − m − 2) 2(2m − v) ] < [ v 2(v − m))]. This implies (v − m − 2)(v − m) < (2m − v). Because v − m ≡ 2 (mod 4) which implies that l(P C(v,m 2 + 1, v 2− 1)) = v(v − m − 2) 2 = ( v − m − 2 2 )(2m − v) + (v − m − 2)(v − m), f = (v − m − 2) 2 and l(TR) = (v − m − 2)(v − m) := q is even. Since (v − m − 2)(v − m) < (2m − v) < 2v, TR is contained in the last circuit

C(v,v 2 − 2) of P C(v, m 2 + 1, v 2 − 1) and TR = hv − q 2, v − q 2 − 1, ..., v − 1, v 2 − 2, 0i. And f = (v − m − 2) 2 implies that SR = I( v 2, v − q 2). Hence DR= SR+ TR is also a

path. This concludes the proof of this subcase.

Case 3.3 : m = v − 2

The proof follows by the fact that E(Kv) = (Mv,2,v₂−1∪ I(0,

v

2)) ∪ I( v 2, 0).

Case 4 : v − m ≡ 0 (mod 4)

Note that if tm < v, then the proof follows by decomposing Kv into paths of

length tm and a path of length less than tm (may be zero). Therefore, it suffices to consider the cases v > m ≥ v

2. Case 4.1 : v 2 ≤ m < 3v 4 , m ≤ v − 8 and v ≡ 0 (mod 4) or v 2 ≤ m < 3v 4 − 1 2, m ≤ v − 8 and v ≡ 2 (mod 4)

First, we need some notations. Let L0(v, x, 2,m 2) = [S(v, x, 2, m 2)] t_{+ hx, x +}v 2+ 1, x + v 2i + S(v, x + v 2, 2, m 2), M = {L0(v, x, 2,m 2) | 0 ≤ x ≤ v 2− 1}, A = I(0,v 2) and B = h0, v 2, 1, v 2 + 1, 2, v 2 + 2, 3, ..., v − 1, v 2i. Let A be obtained from A by replacing the last (m −v

2) edges with the last 2(m − v 2) edges of B and B be obtained from B by replacing the last 2(m − v

2) edges with the last (m − v

2) edges of A. Let B = D + E, where l(D) = 3v 2 − 2m, l(E) = m. Then, we have E(Kv) = M ∪ A ∪ E ∪ (P C(v, m 2 + 1, v 2− 2) + D). So, it is sufficient to prove that M ∪ A ∪ E ∪ (P C(v,m

2 + 1, v

2 − 2) + D) is a set of Pm’s and one Pr by the following steps :

(1) M is a set of v

2 paths of length m. (2) A is a path of length m.

(3) E is a path of length m. (4) d(P C(v,m

2 + 1, v

2 − 2) + D) = m + 3. Step (1) : By definition, M is a set of v

2 paths of length m. Step (2) : Because A = h0, 1, 2, ..., v −m,3v 2 −m, v −m+1, 3v 2 −m+1, ..., v −1, v 2i. Thus, A is a path if and only if there is no repeat vertex of the following vertex sets : {0, 1, 2, ..., v − m}, {v − m + 1, v − m + 2, ...,v 2} and { 3v 2 − m, 3v 2 − m + 1, ..., v − 1}. Since v 2 < 3v

2 − m ⇔ m < v, the proof follows. Step (3) : Notice that if v ≡ 0 (mod 4), then

C = h3v 4 − m, 5v 4 − m, 3v 4 − m + 1, 5v 4 − m + 1, ..., 3v 2 − m − 1, v − m, v − m + 1, v − m + 2, ..., v 2i.

Thus, E is a path if and only if there is no repeat vertex of the following vertex sets : {3v 4 − m, 3v 4 − m + 1, ..., v − m}, {v − m + 1, v − m + 2, ..., v 2} and { 5v 4 − m, 5v 4 − m + 1, ...,3v 2 − m − 1}. Since v 2 < 5v 4 − m ⇔ m < 3v

4 , we have the claim. On the other hand, if v ≡ 2 (mod 4), then

C = h5v 4 − m − 1 2, 3v 4 − m + 1 2, 5v 4 − m + 1 2, 3v 4 − m + 3 2, ..., 3v 2 − m − 1, v − m, v − m + 1, v − m + 2, ..., v 2i.

Thus, E is a path if and only if there is no repeat vertex of the following vertex sets : {3v 4 − m + 1 2, 3v 4 − m + 3 2, ..., v − m}, {v − m + 1, v − m + 2, ..., v 2} and {5v 4 − m − 1 2, 5v 4 − m + 1 2, ..., 3v 2 − m − 1}. Since v 2 < 5v 4 − m − 1 2 ⇔ m < 3v 4 − 1 2, the proof follows.

Step (4) : Because the length of E is larger then (m − v

2). Thus, D is contained in B. Then D is a segment of the first (3v

2 − 2m) edges of C(v, v

2 − 1). By Lemma 4.4, we are done.

Case 4.2 : 3v 4 − 1 2 ≤ m < 3v 4 , m ≤ v − 8 and v ≡ 2 (mod 4)

These conditions implies that m = 3v

4 −

1

2. We shall use the same notations for L0, M, A, B, A, B as in Case 4.1. Let B = F + G, where l(F ) = m, l(G) = 1. Then, we have E(Kv) = M ∪ A ∪ F ∪ (P C(v,

m 2 + 1,

v

2 − 2) ∪ G).

By applying the idea of the proof in Case 4.1, we only have to check F is a Pm

and P C(v,m 2 + 1,

v

2 − 2) ∪ G can be decomposed into Pm’s and exactly one Pr. For m = 3v 4 − 1 2, we know that B = h0, v 2, 1, v 2 + 1, 2, ..., 3v 2 − m − 1, v − m, v − m + 1, v − m + 2, ...,v 2 − 1, v

2i of length m + 1. Because the length of the path h0,v

2, 1, v

2 + 1, 2, ..., 3v

2 − m − 1, v − mi is even. Which implies that the number of vertices in F is 3v

2 − m = m + 1 = l(F ) + 1. Thus, F is a path of length m. Let C0(v, k) = hv 2− 1, v 2 − 1 + k + 1, v 2, v 2− 1 + k + 2, v 2+ 1, ..., v 2− 1 + k, v 2− 1i, where 1 ≤ k ≤ v 2 − 2. Then P C(v, a, b) = C(v, a) + C(v, a + 2) + ... + C(v, b − 1) = C0(v, a) + C0(v, a + 2) + ... + C0(v, b − 1). By Lemma 4.4, we have d(C0(v, a) + C0(v, a + 2) + ... + C0(v, b − 1)) = 2a + 1. Because the end vertex of C0(v,v

2− 3) is the beginning vertex of G. Thus P C(v, m 2 + 1, v 2 − 2) ∪ G = C 0_(v,m 2 + 1) + C 0_(v,m 2 + 3) + ... + C0(v,v 2−3)+h v 2−1, v

2i. Since the distance between the last vertex (the second vertex) v

2 of C

0_(v,v

2− 3) and the end vertex v 2− 1 of C 0_(v,v 2− 3) is 2(v − 4 − v 2) = v − 8 ≥ m. Thus, we can cut C0(v,m

2 + 1) + C 0_(v,m 2 + 3) + ... + C 0_(v,v 2− 3) + h v 2− 1, v 2i, starting from its end vertex, into paths of length m, and the remainder is a path of length r. This concludes the proof of this subcase.

Case 4.3 : v

2 ≤ m < 3v

4 and m = v − 4

Since 8 ≤ v < 16, there are only four cases left to prove.

i v = 8, m = 4

ii v = 10, m = 6

Let V (Kv) = {0, 1, 2, 3, ..., 8, 9}. Then, the decomposition is

h1, 0, 2, 9, 4, 7, 5i ∪ h2, 1, 3, 0, 5, 8, 6i ∪ h3, 2, 4, 1, 6, 9, 7i ∪ h4, 3, 5, 2, 7, 0, 8i∪ h5, 4, 6, 3, 8, 1, 9i ∪ h0, 4, 8, 9, 3, 7, 1i ∪ h1, 5, 9, 0, 6, 7, 8i ∪ h8, 2, 6, 5i.

iii v = 12, m = 8

Let V (Kv) = {0, 1, 2, 3, ..., 10, 11}. Then, the decomposition is

h0, 2, 11, 3, 9, 5, 8, 6, 7i ∪ h2, 1, 3, 0, 4, 10, 6, 9, 7i ∪ h3, 2, 4, 1, 5, 11, 7, 10, 8i∪ h4, 3, 5, 2, 6, 0, 8, 11, 9i ∪ h5, 4, 6, 3, 7, 1, 9, 0, 10i ∪ h6, 5, 7, 4, 8, 2, 10, 1, 11i∪ h7, 0, 1, 6, 11, 10, 3, 8, 9i ∪ h11, 0, 5, 10, 9, 2, 7, 8, 1i ∪ h11, 4, 9i.

iv v = 14, m = 10

Let V (Kv) = {0, 1, 2, 3, ..., 12, 13}. Then, the decomposition is

h1, 0, 2, 13, 3, 12, 5, 10, 6, 9, 7i ∪ h2, 1, 3, 0, 4, 13, 6, 11, 7, 10, 8i∪ h3, 2, 4, 1, 5, 0, 7, 12, 8, 11, 9i ∪ h4, 3, 5, 2, 6, 1, 8, 13, 9, 12, 10i∪ h5, 4, 6, 3, 7, 2, 9, 0, 10, 13, 11i ∪ h5, 7, 4, 8, 3, 10, 1, 11, 0, 12, 13i∪ h7, 6, 8, 5, 9, 4, 11, 2, 12, 1, 13i ∪ h0, 6, 5, 11, 12, 4, 10, 9, 8, 7, 13i∪ h5, 13, 0, 8, 2, 10, 11, 3, 9, 1, 7i ∪ h6, 12i. Case 4.4 : m ≥ 3v 4 , m = v − 4 and v ≡ 0 (mod 4) Let A = hv 2 + 4, 5, v 2+ 6, 7, v 2+ 8, ..., v 2− 1, 0i + I(0, 4) + h4, v 2+ 5, 6, v 2 + 7, ..., v 2i, B = hv 2, 1, v 2 + 2, 3, v 2 + 4i + I( v 2 + 4, 0) + h0, v 2 + 1, 2, v 2+ 3, 4i and M = Mv,2,m₂ ∪ I(4, v

2 + 4). Then, we have E(Kv) = M ∪ A ∪ B.

It suffices to check A is a path of length m and B is a path of length v 2 + 4.

Because v ≡ 0 (mod 4), the cycle h0,v 2 + 1, 2, v 2+ 3, 4, v 2 + 5, 6, ..., v − 1, v 2, 1,v 2 + 2, 3, v 2+ 4, 5, v 2 + 6, 7, ..., v − 2, v 2 − 1, 0i

contains all the edges in dv

2−1 and each vertex of Kv appears exactly once. Since

A = hv 2 + 4, 5, v 2 + 6, 7, v 2 + 8, ..., v 2 − 1, 0i + I(0, 4) + h4, v 2 + 5, 6, v 2 + 7, ..., v 2i is obtained from the union of two segments of this cycle and a path I(0, 4), moreover, the vertices in {1, 2, 3} appear in B (thus the vertices in {1, 2, 3} will not appear in A). So, A is a path of length (v − 8) + 4 = v − 4 = m. By a similar idea as above, we prove that B is a Pv

2+4. This concludes the proof of this subcase.

Case 4.5 : m ≥ 3v 4 , m = v − 4 and v ≡ 2 (mod 4) Let A = h4,v 2 + 5, 6, v 2 + 7, 8, ..., 0i + I(0, 3) + h3, v 2 + 4, 5, v 2+ 6, ..., v 2i, B = hv 2, 1, v 2 + 2, 3i + I(3, 4) + h4, v 2+ 3, 2, v 2 + 1, 0i + [I( v 2+ 4, 0)] t _and M = Mv,2,m₂ ∪ I(4, v 2 + 4). Then, E(Kv) = M ∪ A ∪ B.

Now, it suffices to check that A is a path of length m and B is a path of length v

2 + 4. Because v ≡ 2 (mod 4), the following cycles

h0,v 2 + 1, 2, v 2 + 3, 4, v 2+ 5, 6, ..., v − 4, v 2− 3, v − 2, v 2 − 1, 0i and h1,v 2 + 2, 3, v 2 + 4, 5, v 2+ 6, 7, ..., v − 3, v 2− 2, v − 1, v 2, 1i

contain all the edges in dv

2−1 and each vertex of Kv appears exactly once. Since

A = h4,v 2+ 5, 6, v 2+ 7, 8, ..., 0i + I(0, 3) + h3, v 2+ 4, 5, v 2+ 6, ..., v

2i is obtained from the union of two segments of these cycles and a path I(0, 3), and the vertices in {1, 2} appear in B (thus the vertices in {1, 2} will not appear in A), A is a path of length (v − 7) + 3 = v − 4 = m. By the same idea as above, we also prove that B is a Pv

2+4.

Case 4.6 : 3v

4 ≤ m ≤ v − 8 and v ≡ 0 (mod 4)

First, we need some notations. Let A = h3v 2 − m, v − m + 1, 3v 2 − m + 2, v − m + 3, ..., v 2 − 1, 0i + I(0, v − m)+ hv − m, 3v 2 − m + 1, v − m + 2, 3v 2 − m + 3, ..., v − 1, v 2i, B = hv 2, 1, v 2+ 2, 3, v 2 + 4, 5, ..., v − m − 1, 3v 2 − mi + I( 3v 2 − m, 0)+ h0,v 2 + 1, 2, v 2+ 3, 4, ..., 3v 2 − m − 1, v − mi, T = P C(v,m 2 + 1, v

2 − 2) and K be the last 2(2m − 3v

2 ) edges of T . Let B be obtained from B by replacing the last (2m − 3v

2 ) edges of the segment I(3v

2 − m, 0) with K. Also, let f = min{[ v

2(v − m) − 2], [

l(T \ K)

2m − v ]} (Denote the integer part of x as [x]), Si = I(m, 0) + (i − 1)(m −

v 2), where 1 ≤ i ≤ f and SR = I(2(v − m), v 2 − f (v − m)) + (f − 1) v

2. Finally, denote by Ti, 1 ≤ i ≤ f , the collection of f paths of length 2m − v each, cut along T \ K, and denote by TR, the

final segment remaining of T \K after taking away T1, T2, ..., Tf. Now, let Di = Si+Ti,

1 ≤ i ≤ f and DR = SR+ TRand M = Mv,2,m₂ ∪ {d1\ [(I(0, v − m) ∪ I(

3v 2 − m,

5v 2 − 2m) ∪ S1 ∪ S2∪ ... ∪ sf ∪ SR]}. Similar to the proof of Case 3.2, we know that Di

and DR are well defined. And since {I(0, v − m) ∩ I(

3v 2 − m, 5v 2 − 2m) ∩ S1∩ S2∩ ... ∩ Sf ∩ SR} = φ, M is a set of v

2 paths of length m. Then, by routine checking, E(Kv) = A ∪ B ∪ M ∪ D1∪ D2∪ ... ∪ Df ∪ DR.

We prove that Kv can be decomposed into a set of Pm’s and one Prby the following

steps :

(1) A is a path of length m. (2) B is a path of length m. (3) Di is a path of length m.

(4) d(DR) > m.

of (3) and (4) are similar to that of Case 3.2. Step (1) : Since v ≡ 0 (mod 4), the cycle

h0,v 2 + 1, 2, v 2+ 3, 4, v 2 + 5, 6, ..., v − 1, v 2, 1,v 2 + 2, 3, v 2+ 4, 5, v 2 + 6, 7, ..., v − 2, v 2 − 1, 0i

contains all the edges in dv

2−1 and each vertex of Kv appears exactly once. Since

A = h3v 2 − m, v − m + 1, 3v 2 − m + 2, v − m + 3, ..., v 2− 1, 0i + I(0, v − m) + hv − m, 3v 2 − m + 1, v − m + 2,3v 2 − m + 3, ..., v − 1, v

2i which is obtained from the union of two segments of this cycle and a path I(0, v − m) and the vertices in {1, 2, 3, ..., v − m − 1} appear in B (thus the vertices in {1, 2, 3, ..., v − m − 1} will not appear in A), A is a path of length (m − v

2) + (v − m) + (m − v

2) = m.

Step (2) : Similar to the proof of Step (1), B is a path of length (v − m) + (m − v 2) + (v − m) = 3v 2 − m. Since K = h 5v 2 − 2m, 5v

2 − 2m + 1, ..., 0i, the vertex set of K is {5v 2 − 2m, 5v 2 − 2m + 1, ..., 0} ∪ {2(v − m) − 2, 2(v − m) − 1, ..., v 2− 3} and each vertex of {2(v − m) − 2, 2(v − m) − 1, ...,v

2 − 3} will not appear in B. Hence B is a path of length (3v 2 − m) − (2m − 3v 2 ) + 2(2m − 3v 2 ) = m.

Step (3) : Let x be the end vertex of Si. Let Si = hx + m, x + m + 1, ..., xi

and Ti = hx, x + k, x + 1, x + k + 1, x + 2, ..., x + m −

v

2i. Then by the property of P C(v,m 2 + 1, v 2− 1), we have m 2 < k < v

2. Thus, the vertices in the even places of Ti belong to the open interval (x +m

2, x + m), since the vertices in the odd places of Ti is {x, x + 1, ..., x + m − v

2} and Si = {x + m, x + m + 1, ..., x}. Moreover, the vertex x is the only common vertex of Si and Ti. Hence, for each i, Di = Si + Ti is a path

of length l(Si) + l(Ti) = (v − m) + (2m − v) = m.

Step (4) : First, if f = [ v

2(v − m)] − 2, then l(SR) < l(Si) = v − m. Therefore, by the same argument as above, we obtain d(DR) > d(Di) = m.

On the other hand, f = [l(T \ K) 2m − v ] < [

v

2(v − m))] − 2. This implies that (v − m − 2)(v − m) < (2m − v). Thus, l(T \K) = v(v 2− m 2 − 2) − 4m + 3v = (v − m − 6) 2 (2m − v) + (v − m − 2)(v − m) and f = (v − m − 6) 2 , l(TR) = (v − m − 2)(v − m) := q < (2m − v) < 2v which is even. Since l(K) = 4m − 3v is even and l(K) + l(TR) <

(4m − 3v) + (2m − v) = 6m − 4v < 2v. TR is contained in the last circuit C(v,

v 2− 3) of P C(v,m 2 + 1, v 2− 2)\K and TR = h( 5v 2 − 2m) − q 2, 2(v − m) − 2 − q 2, ..., 5v 2 − 2mi. Since f = (v − m − 6)

2 implies that SR = I(2(v − m), ( 5v 2 − 2m) − q 2), DR is a path. Case 4.7 : 3v 4 ≤ m ≤ v − 8 and v ≡ 2 (mod 4)

We start with some new notations. Let A = hv − m,3v 2 − m + 1, v − m + 2, 3v 2 − m + 3, ..., v 2 − 1, 0i + I(0, v − m − 1)+ hv − m − 1,3v 2 − m, v − m + 2, 3v 2 − m + 2, ..., v − 1, v 2i and B = hv 2, 1, v 2+ 2, 3, ..., 3v 2 − m − 2, v − m − 1i + I(v − m − 1, v − m)+ hv − m,3v 2 − m − 1, v − m − 2, 3v 2 − m − 3, ..., v 2 + 1, 0i + [I( 3v 2 − m, 0)] t_.

Let B be obtained from B by replacing the first (2m − 3v

2 ) edges of the segment [I(3v

2 − m, 0)]

t _{with the last 2(2m −}3v

2 ) segment of P C(v, m

2 + 1, v

2− 2). Finally, let M, Di, Df and DR be defined as in Case 4.6. Then, we have E(Kv) = A ∪ B ∪ M ∪

D1∪ D2 ∪ ... ∪ Df ∪ DR.

Since v ≡ 2 (mod 4), the following two cycles

h0,v 2 + 1, 2, v 2 + 3, 4, v 2+ 5, 6, ..., v − 4, v 2− 3, v − 2, v 2 − 1, 0i and h1,v 2 + 2, 3, v 2 + 4, 5, v 2+ 6, 7, ..., v − 3, v 2− 2, v − 1, v 2, 1i

contain all the edges in dv

2−1 and each vertex of Kv appears in these two cycles exactly

once. Moreover, since A = hv − m,3v

2 − m + 1, v − m + 2, 3v 2 − m + 3, ..., v 2− 1, 0i + I(0, v − m − 1) + hv − m − 1,3v 2 − m, v − m + 2, 3v 2 − m + 2, ..., v − 1, v 2i is obtained from

the union of two segments of these cycles and a path I(0, v − m − 1) and the vertices in {1, 2, ..., v − m − 2} appear in B (thus the vertices in {1, 2, ..., v − m − 2} will not appear in A), A is a path of length (m −v

2− 1) + (v − 1 − m) + (m − v

2+ 1) = m. By the same way as above, we can prove that B is a Pv

2+4. Thus, by a similar argument

5

Conclusion

In this thesis, we have generalized the idea of decomposing Kv into paths of length

m to a maximum packing of Kv with paths of length m and the leave is also a path.

But, our long-term project is to settle the following problem.

Problem 5.1. Let v and t be positive integers such that t ≥ v

2. Let m1, m2, ..., mt be t positive integers less than v such that Pt

i=1mi = v

2
. Prove that Kv can be

decomposed into t paths P1_{, P}2_{, ..., P}t _{such that the length of P}i _{is m}

i for i = 1, 2, ..., t.

So far, partial results have been obtained especially when v is odd. But, for the case when v is even, not much is know.

References

[1] P. Adams, D. E. Bryant and A. Khodkar, (3,5)-cycles decompositions. J. Combin. Designs, 6 (1998), 91-110.

[2] P. N. Balister. On the Alspach conjecture. Combin., Probability and com-puting, 10 (2001), 95-125.

[3] D. E. Bryant, A. Khodkar and H. L. Fu, (m,n)-cycles systems. J. Statist, Planning and Inference, 74 (1998), 365-370.

[4] P. K. Chuang, Decomposing Complete Graph into Paths with Prescribed Lengths, M. Sc. Thesis, National Chiao Tung University, 2003.

[5] F. Harary, Graph Theory, Addison-Wesley, Reading MA, 1972.

[6] K. Heinrich, Path-decompositions, Le matematiche, XLVII (1992), 241-258.

[7] K. Heinrich, P. Horak, A. Rosa, On Alspach’s conjecture, Discrete Math., 77 (1989), 97-121.

[8] D. R. Hughes, F. C. Piper, Design theory, Cambr., Cambridge University Press, (1985).

[9] Rev. T. P. Kirkman, On the problem in combinations, Cambr. and Dublin Math. J., 2 (1847), 191-204.

[10] K. W. P. Ng, On Path decompositions of Complete Graphs, M. Sc. Thesis, Simon Fraser University, 1985.

[11] M. Sajna, Cycle decompositions III : complete graphs and fixed length cycles, J. Combin. Designs, 10 (2002), 27-78.

[12] D. R. Stinson, Combinatorial Designs : constructions and analysis, New York, Springer, 2004.

[13] M. Tarsi, Decomposition of a Complete Multigraph into Simple Paths: Nonbalanced Handcuffed Designs, J. Combin. Theory, A 34 (1983), 60-70.

[14] D. B. West, Introduction to Graph Theory 2nd, New Jersey, Prentice-Hall, 2001.