K3-design of K2n\F with F a spanning odd forest

黃 國 卿Department of Financial and Computational Mathematics

Providence University

A joint work with Professor Hung-Lin Fu

OUTLINE Introduction Known Results Main Results

Introduction A K3-design of a graph G is a collection of K3 of G

such that the union of K3’s partitions the edge set of G.

The necessary conditions for a graph to have a K3-design is 3-sufficient, that is, (1) each vertex is of even degree and(2) the number of edges is divisible by 3.

A graph being 3-sufficient may not have a K3-design. For example, G = K6,6.

Introduction The motivation of K3-design problem comes

from the so-called Kirkman problem.

Kirkman problemDoes the complete graph Kn have a K3-

design?

Kirkman Problem Theorem 1. (Kirkman)

The complete graph Kn has a K3-design if and only if it is 3-sufficient, that is, n ≡ 1,3(mod 6).

Note that a K3-design of Kn is also called the Steiner Triple System (STS(n)) of order n or a 3-cycle system.

Problems and Conjectures

It is natural to consider the following problem.

Problem:For what subgraphs F of Kn, the graph Kn\F

obtained by deleting E(F) from Kn can have a K3-design.

Conjectures Conjecture 1. (Nash-William, 1970)

If G is 3-sufficient with , then G has a K3-design.

Theorem 2. (Gustavsson, 1991)The Conjecture 1 holds if

|)(|43)( GVG

.|)(|)101()( 24 GVG

Conjectures Conjecture 2. (???)

If F is a spanning odd forest of K2n and K2n\F is 3-sufficient, then K2n\F has a K3-design.

An odd forest is a forest that each vertex is of odd degree.

Known Results F = disjoint union of complete graphs

Theorem 3: (Hanani)The balanced complete multipartite

graph has a K3-design if and only if it is 3-sufficient.

Known Results F = 2-factor, i.e., 2-regular spanning subgraph

Theorem 4: (Colbourn and Rosa, 1986)Suppose that F is a 2-factor of Kn except

F = C4 C∪ 5 when n = 9. Then Kn\F has a K3-design if and only if it is 3-sufficient.

Known Results Theorem 5: (Hanani, 1975)

Kn\H has a K3-design, where H and n are shown as the following table.

F: 1-factor, F1: spanning odd forest with n/2 +1 edges, and C4 is a cycle of length 4.

n(mod 6) 0 1 2 3 4 5

H F F F1 C4

Main Results

In this talk, we study the Conjecture 2 for F being the following graphs.

F = regular 3-caterpillar F = complete 3-ary tree with level k F = K1,q-factor, where q is odd

F = regular 3-caterpillar A caterpillar is a tree that it remains a path Pk after

deleting all of the leaves. In this case, Pk is called the skeleton and the value k is called the length.

A 3-regular caterpillar of length k is a caterpillar that each non-leaf vertex is of degree 3 and its skeleton is Pk.

A 3-regular caterpillar of length 5

F = regular 3-caterpillar of length k

The necessary condition for K2k+2\F having a K3-design is that K2k+2\F is 3-sufficient, i.e., k ≡ 0,1(mod 3).

F = regular 3-caterpillar of length 3k+1

Lemma 6.Suppose that F is a 3-regular

caterpillar of length 3k+1. Then K6k+4\F has a K3-design if and only if it is 3-sufficient.

Sketch proof of Lemma 6 Construct a specific Steiner Triple System

of order 6k+3. Choose some K3’s such that the union of

these K3’s contains F － v, where v is the end vertex of the skeleton of F.

Add the vertex v and construct the desired K3-design

k = 2

5])[( YG

X

Y

Z

k = 2

X

Y

Z

v v

F = regular 3-caterpillar of length 3k

Lemma 7.Suppose that F is a 3-regular

caterpillar of length 3k. Then K6k+2\F has a K3-design if and only if it is 3-sufficient.

Sketch proof of Lemma 7 Construct a specific STS(6k+1) by Skolem

Triple System of order k Choose some K3’s such that the union of

these K3’s contains F － v, where v is the end vertex of the skeleton of F.

Add the vertex v and construct the desired K3-design

Skolem Triple System of order k

A Skolem Triple System of order k is a collection of triples {ai, bi, ci} with ai + bi = ci

or ai + bi + ci = 3k for each i such that the triples partition the set {1,2,3,…,3k}.

The specific STS(6k+1)

Theorem 8.The set {1,2,3,…,3k} can have a

Skolem Triple System with a triple D, where

)4(mod3}2

33,2

13,1{

)4(mod2}12

3,2

3,1{

)4(mod1,0}3,13,1{

kifkk

kifkk

kifkk

D

k = 1 and D ={1,2,3}

1

7

6

54

32 v

k = 2 and D ={1,3,4}

1 7

65 4

32 v

131211

1098

k = 3 and D ={1,5,6}

1

7 6

5432 v131211

1098 16

15

191817

14

k = 4 and D ={1,11,12}

1 765432 v

13

12111098

1615 19181714 20 21 22 23

24

25

F = regular 3-caterpillar

Theorem 9.Suppose that F is a 3-regular

caterpillar of length k. Then K2k+2\F has a K3-design if and only if it is 3-sufficient.

Fk = complete 3-ary tree with level k

A complete 3-ary tree with level k, denoted by Fk, is a rooted tree such that (1) the root has 3 children(2) each parent except the root has 2 children(3) the distance from the root to a leaf is k

Fk = complete 3-ary tree with level k

F1 F2

F3

Fk = complete 3-ary tree with level k

Theorem 10.Suppose that Fk is a spanning

complete 3-ary tree with level k of K2n. Then K2n\Fk has a K3-design if and only if it is 3-sufficient.

Sketch proof of Theorem 10 Recursive construction A (2n － 3)-regular graph of order 2n is 1-

factorable, i.e., the graph obtained from K2n by deleting a 2-factor is 1-factorable.

For n 3, there exists an idempotent latin ≧square of order n

K22\F3 has a K3-design

K10\F2 has a K3-design

One K3-factor 3 1-factors

One K3-factor 3 1-factors

Idempotent latin square of order 6

F = K1,q-factor, where q is odd

Theorem 11.Suppose that F is a K1,q-factor of Kn,

where q is odd. Then Kn\F has a K3-design if and only if it is 3-sufficient except possibly the case that q ≡ 5(mod 6) and n = 6(q+1).

Problem review Conjecture 2.

If F is a spanning odd forest of K2n and K2n\F is 3-sufficient, then K2n\F has a K3-design. Δ(F) = 3 Useful tools(embedding,…)

Thanks for your attention