1. Packing Graphs with 4-Cycles 學生 : 徐育鋒 指導教授 : 高金美教授 2013 組合新苗研討會 (2013.08.10 ~ 2013.08.11) 國立高雄師範大學

2. 1. Definition 2. Known Results 3. 4-Regular Graphs 4. Main Results 5. Future Works

3. Definition

4. 1. A graph G is an order pair (V, E), where V is a non-empty set called a vertex set and E is a set of two-element subsets of V called an edge set. 2. deg G (v) = the number of edges incident with a vertex v in G. 3. If all the vertices of a graph have the same degree r, then the graph is called r-regular. Definitions

5. V = {v 1, v 2, v 3, v 4, v 5, v 6 }. E = {v 1 v 2, v 1 v 3, v 1 v 5, v 1 v 6, v 2 v 3, v 2 v 4, v 2 v 6, v 3 v 4, v 3 v 5, v 4 v 5, v 4 v 6, v 5 v 6 }. The graph G is 4-regular. v1v1 v5v5 v4v4 v3v3 v2v2 v6v6 G:G: Definitions

6. 5. C n = (v 1,v 2,..., v n ) : n-cycle Definitions v1v1 C 5 = (v 1, v 2, v 3, v 4, v 5 ) v2v2 v3v3 v4v4 v5v5

7. 6. K n : the complete graph of order n. Definitions v1v1 K5K5 v2v2 v3v3 v4v4 v5v5

8. 7. K U,V : the complete bipartite graph with partite set U, V. If |U| = m, |V| = n, then K U,V can be denoted by K m,n. Definitions v2v2 U = {v 1, v 2, v 3 }, V = {v 4, v 5, v 6 } K U,V = K 3,3 v1v1 v3v3 v5v5 v4v4 v6v6

9. Let  = {H 1, H 2, , H s } be a set of subgraphs of G. If E(H 1 )  E(H 2 )    E(H s ) = E(G) and E(H i )  E(H j ) =  for i  j, then we call  is a decomposition (packing) of G. If H i is isomorphic to a subgraph H of G for each i = 1, 2, , s, then we say that G has an H decomposition (H system) or  is a H packing of G. If H i is isomorphic to a subgraph H of G for each i = 1, 2, , s–1, then we say that G can be packed with H and leave H s. That is, G – E(H s ) has an H decomposition. Definitions and leave H s. That is,  – {H s } is a H packing of G – E(H s ).

10. v4v4 v5v5 v6v6 v7v7 v8v8 v9v9 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v7v7 v8v8 v9v9 v1v1 v2v2 v3v3 G:G: G can be decomposed into H 1, H 2. H1:H1: H2:H2:  = {H 1, H 2 } is a packing of G.

11. v4v4 v5v5 v6v6 v7v7 v8v8 v9v9 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v7v7 v8v8 v9v9 v1v1 v2v2 v3v3  ‘= {(v 1, v 5, v 3, v 6 ), (v 1, v 2, v 5, v 4 ), (v 1, v 7, v 2, v 9 ), (v 2, v 3, v 7, v 6 ), (v 2, v 4, v 3, v 8 ), (v 1, v 3, v 9, v 8 )} is a 4-cycle packing of G. H1:H1: H2:H2: G has a 4-cycle system.

12. Alspach Conjecture : Let 3  m 1, m 2,..., m t  n such that m 1 + m 2 +... + m t = n(n–1)/2 for odd n (m 1 + m 2 +... + m t = n(n–2)/2 for even n). Then K n (K n – F) can be decomposed into cycles C 1, C 2,..., C t such that C i is a m i -cycle for i = 1, 2,..., t. D. Bryant, D. Horsley and W. Pettersson, Cycle decompositions V: Complete graphs into cycles of arbitrary lengths, arXiv:1204.3709v2 [math.CO], 2013. Cycle Decomposition

13. D. Sotteau, Decomposition of K m,n (K m,n *) into cycles (circuits) of length 2k, J. Combin. Theory B, 30 (1981) 75.81. Theorem 1: There exists a 2k-cycle decomposition of K m,n if and only if each vertex has even degree, mn is divisible by 2k, and m, n  k. Cycle Decomposition

14. Does there exist a 4-cycle system of K n – E(G) for any 4-regular subgraph G of K n ?

15. Known Results

16. A. Kotzig, On decomposition of the complete graph into 4k-gons, Mat.-Fyz. Cas., 15 (1965), 227-233. Theorem 2: There exists a 4-cycle system of K n if and only if n ≡ 1 (mod 8). Known Results

17. B. Alspach and S. Marshall, Even cycle decompositions of complete graphs minus a 1-factor, J. Combin. Des., 2 (1994), 441-458. Theorem 3: There exists a 4-cycle system on K n – F, where F is a 1-factor of K n, if and only if n ≡ 0 (mod 2). Known Results

18. H.-L. Fu and C. A. Rodger, Four-Cycle Systems with Two- Regular Leaves, Graphs and Comb., 17 (2001), 457-461. Theorem 4: Let F be a 2-regular subgraph of K n. There exists a 4-cycle system of K n – F if and only if n is odd and 4 divides the number of edges of K n – F. Known Results

19. C.-M. Fu, H.-L. Fu, C. A. Rodger and T. Smith, All graphs with Maximum degree three whose complements have 4-cycle Decompositions, Discrete Math., 308 (2008), 2901-2909. Theorem 5: Let G be a graph on n vertices, where n is even and  (G)  3. Then there exists a 4-cycle system of K n – E(G) if and only if (1) All vertices in G have odd degree, (2) 4 divides n(n–1)/2 – |E(G)|, and (3) G is not one of the two graphs of order 8 as follows. Known Results

20. Let G be a 4-regular subgraph of K n. Does there exist a 4-cycle system of K n – E(G)?

21. 4-Regular Graphs

22. Some 4-regular graphs

23. Question: Does there exist a 4-cycle system of K n – E(K 5 ) ? 1. n = 5, Yes! 2. n = 6, No! 3. n = 7, No! 4. n = ?, Yes!

24. Question: Does there exist a 4-cycle system of K n – E(K 5 ) ? n  5 is odd and 4 | n(n – 1) / 2 – 10 ⇒ 4 | (n 2 – n – 20) / 2 ⇒ 8 | (n – 4)(n – 5) ⇒ n  5 (mod 8).

25. Question: Does there exist a 4-cycle system of K n – E(K 5 ) ? Answer: n  5 (mod 8), Yes! Let n = 8k + 5. K 8k+5 – E(K 5 ) = K 8k+1  K 4, 8k.... K 8k+1 K 4, 8k

26. Lemma 6: There exists a 4-cycle system of K n – E(K 5 ) if and only if n  5 (mod 8).

27. Question: Does there exist a 4-cycle system of K n – E(G) ? n  6 is odd and 4 | n(n – 1) / 2 – 12 ⇒ 4 | n(n – 1) / 2 ⇒ 8 | n(n – 1) ⇒ n  1 (mod 8). G:G:

28. Question: Does there exist a 4-cycle system of K 9 – E(G) ? G:G: K 9 – E(G) :

29. Question: Does there exist a 4-cycle system of K 9 – E(G) ? Answer: Yes !

30. Question: Does there exist a 4-cycle system of K n – E(G) ? Answer: n  1 (mod 8), Yes ! Let n = 8k + 1. K n – E(G) = (K 9 – E(G))  K 8k–8  K 8k–8,9 = (K 9 – E(G))  K 8k–7  K 8k–8,8 G:G: K 8k–8 K n – E(G) K 9 – E(G) G

31. Lemma 7: There exists a 4-cycle system of K n – E(G) if and only if n  1 (mod 8). G:G:

32. Question: Does there exist a 4-cycle system of K n – E(G)? n  6 is odd and 4 | n(n – 1) / 2 – 12 ⇒ 4 | n(n – 1) / 2 ⇒ 8 | n(n – 1) ⇒ n  1 (mod 8). G:G:

33. Question: Does there exist a 4-cycle system of K 9 – E(G)? Answer: No! G:G:K 9 – E(G) :

34. Lemma 8: There exists a 4-cycle system of K n – E(G) if and only if n  1 (mod 8) and n  17. G:G:

35. t|Q(t)| 51 61 72 86 916 1059 11265 121544 1310778 Q(t) = {G | G is any connected 4-regular graph with t vertices}.

36. Definition 9: Let G be a 4-regular graph of order t. If there exists S  V(G), |S| = s and a graph H where V(H) = N 1 (S) and E(H)  E(G) = ∅ such that (G – S)  H is 4-regular, then we call the graph G is s-reducible. s-reducible

37. S = { ∞ 1 } V(H) = N 1 (S) = {v 1, v 2, v 3, v 4 } E(H) = {v 1 v 4, v 2 v 3 } ∞1∞1 v3v3 v1v1 v2v2 v4v4 v5v5 v6v6 G:G: G – S: v3v3 v1v1 v2v2 v4v4 v5v5 v6v6 v3v3 v1v1 v2v2 v4v4 v5v5 v6v6 (G – S)  H: G is 1-reducible.

38. Theorem 10: Let t  8 and G be a 4-regular graph of order t. If G contains a component with at least 6 vertices, then G is 3-reducible. 3-reducible

39. Theorem 11: Let G be a 4-regular of order t. If there exists a 4-cycle system of K n – E(G), then (1) n ≣ 1 (mod 8), for t is even and (2) n ≣ 5 (mod 8), for t is odd. Sufficient Condition

40. G is 3-reducible 4-regular graph of order t. K n – E(G) = [K n–4 – E((G – S)  H)]  R. (G – S)  H K n–4 – E((G – S)  H) S K n – E(G) HConstruction n ≣ 1 (mod 8), t is even. n ≣ 5 (mod 8), t is odd. n ≣ 1 (mod 8), t is even. n ≣ 5 (mod 8), t is odd.

41. Main Results

42. Main Theorem: Let G be any 4-regular graph with t vertices. There exists a 4-cycle system of K n – E(G), if n is odd, 4 | n(n – 1)/2 – 2t, and (1) G is a vertex-disjoint union of t/5 copies of K 5. (2) n  (4t – 5)/3. (3) n > 9 for the following two graphs.

43. Future Work

44. Question 1. Let G be a 4-regular graph of order t. Does there exist a 4-cycle system of K n – E(G) for t  n < (4t – 5)/3? Question 2. Let G be a 4-regular graph of order t and t ≣ 5 (mod 8). Is G 5-reducible? Question 3. Let G be a 4-regular spanning subgraph of K n. Does there exist a 4-cycle system of K n – E(G) for n ≡ 5 (mod 8)?

45. Thanks for your patient.