1 / 35

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

K 3 -design of K 2 n 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.

jory
Download Presentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 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

  2. OUTLINE • Introduction • Known Results • Main Results

  3. 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.

  4. 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?

  5. 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.

  6. 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.

  7. 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

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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

  13. 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

  14. 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).

  15. 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.

  16. 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

  17. k = 2 X Y Z

  18. k = 2 v v X Y Z

  19. 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.

  20. 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

  21. 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}.

  22. The specific STS(6k+1) Theorem 8. The set {1,2,3,…,3k} can have a Skolem Triple System with a triple D, where

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

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

  25. 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

  26. 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

  27. 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.

  28. 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

  29. Fk= complete 3-ary tree with level k F1 F2 F3

  30. 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.

  31. 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

  32. 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

  33. 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).

  34. 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,…)

  35. Thanks for your attention

More Related