K 3 -design of K 2 n \ F with F a spanning odd forest

K3-design of K2n\Fwith Fa 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 K3of 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 problem Does 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 nor a 3-cycle system.

Problems and Conjectures It is natural to consider the following problem. Problem: For what subgraphsFof Kn, the graph Kn\F obtained by deleting E(F)fromKn 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

Conjectures • Conjecture 2. (???) If Fis a spanning odd forest of K2n and K2n\Fis 3-sufficient, then K2n\Fhas 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∪C5 when n = 9. Then Kn\Fhas 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.

Main Results In this talk, we study the Conjecture 2 for Fbeing 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 kis 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-designis 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 Fis a 3-regular caterpillar of length 3k+1. Then K6k+4\Fhas 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 X Y Z

k = 2 v v X Y Z

F= regular 3-caterpillar of length 3k • Lemma 7. Suppose that Fis a 3-regular caterpillar of length 3k. Then K6k+2\Fhas 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} withai+ bi = cior ai+ bi + ci= 3k for each isuch 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

k = 1 and D ={1,2,3} 1 2 v 3 6 5 4 7

k = 2 and D ={1,3,4} 1 2 7 8 9 v 10 3 5 6 4 11 13 12

k = 3 and D ={1,5,6} 15 14 1 2 4 5 11 12 13 v 3 16 17 18 19 7 8 9 10 6

k = 4 and D ={1,11,12} 11 24 6 7 8 9 10 1 2 4 5 12 v 3 21 22 20 23 25 19 17 18 15 16 14 13

F= regular 3-caterpillar • Theorem 9. Suppose that Fis a 3-regular caterpillar of length k. Then K2k+2\Fhas 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 Fkis a spanning complete 3-ary tree with level k of K2n. Then K2n\Fkhas 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\F3has 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 Fis a K1,q-factor of Kn, where q is odd. Then Kn\Fhas 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

K 3 -design of K 2 n \ F with F a spanning odd forest