1 / 42

Mutually independent Hamiltonian cycles on Cartesian product graphs

Mutually independent Hamiltonian cycles on Cartesian product graphs. Student : Kai- Siou Wu ( 吳凱修 ) Adviser: Justie Su-Tzu Juan . Outline. Introduction Motivation & Contribution Main result The lower bound of IHC( G 1  G 2 ) IHC( C m  C n ) = 4 Conclusion & Future work.

hope
Télécharger la présentation

Mutually independent Hamiltonian cycles on Cartesian product graphs

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. Mutually independent Hamiltonian cycles on Cartesianproduct graphs Student: Kai-SiouWu (吳凱修) Adviser: Justie Su-Tzu Juan National Chi Nan University

  2. Outline • Introduction • Motivation & Contribution • Main result • The lower bound of IHC(G1G2) • IHC(CmCn) = 4 • Conclusion & Future work National Chi Nan University

  3. Outline • Introduction • Motivation & Contribution • Main result • The lower bound of IHC(G1G2) • IHC(CmCn) = 4 • Conclusion & Future work National Chi Nan University

  4. Introduction G 0 • Hamiltonian cycle • In a graph G, a cycle is called Hamiltonian cycle if it contains all vertices of G exactly once. • Independent • Two cycles C1 = u0, u1, ..., un–1, u0 and C2 = v0, v1, ..., vn–1, v0 in G are independent if u0 = v0 and uivi for 1 ≤ i ≤ n – 1. 1 4 3 2 C1= 0, 1, 2, 3, 4, 0 C2 = 0, 2, 3, 4, 1, 0 National Chi Nan University

  5. Introduction G 0 • Mutually independent • A set of Hamiltonian cycles {C1, C1, …, Ck} of G are mutually independent if any two dierent Hamiltonian cycles of {C1, C1, …, Ck} are independent. • Mutually independent HamiltonicityIHC(G) = k • The mutually independent Hamiltonianicity of graph G, IHC(G), is the maximum integer k such that for any vertex u of G there exist k-mutually independent Hamiltonian cycles of G starting at u. 1 4 3 2 C1= 0, 1, 2, 3, 4, 0 C2 = 0, 2, 3, 4, 1, 0 C3 = 0, 3, 4, 1, 2, 0 C4= 0, 4, 1, 2, 3, 0 IHC(G) = 4 National Chi Nan University

  6. Outline • Introduction • Motivation & Contribution • Main result • The lower bound of IHC(G1G2) • IHC(CmCn) = 4 • Conclusion & Future work National Chi Nan University

  7. Motivation & Contribution • The lower bound of IHC(G1G2) with different conditions • IHC(G1G2)  IHC(G1) • IHC(G1G2)  IHC(G1) + 2 • Mutually independent Hamiltonianicity of CmCn • IHC(CmCn) = 4, for m, n  3. National Chi Nan University

  8. Outline • Introduction • Motivation & Contribution • Main result • The lower bound of IHC(G1G2) • IHC(CmCn) = 4 • Conclusion & Future work National Chi Nan University

  9. The lower bound of IHC(G1G2) • By definition, we can get the trivial upper bound: IHC(G1 G2)  (G1  G2) = (G1) + (G2). • So, we want to discuss the lower bound of IHC(G1G2). National Chi Nan University

  10. The lower bound of IHC(G1G2) • For any Hamiltonian graphs G1 and G2: • LetIHC(G1) = I1, |V(G1)| = m,|V(G2)| = n. • Theorem 1.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1 and n  2I1, then IHC(G1 G2)  I1. • Theorem 1.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1 and n  2I1, then IHC(G1 G2)  I1. • Theorem 2.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 3 , n is odd and m  n – 3, then IHC(G1 G2)  I1 + 2. • Theorem 2.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 3 , n is odd and m  n – 3, then IHC(G1 G2)  I1 + 2. • Theorem 3.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 2, n is even and G1 does not contain C4, then IHC(G1 G2)  I1 + 2. • Theorem 3.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 2, n is even and G1 does not contain C4, then IHC(G1 G2)  I1 + 2. National Chi Nan University

  11. The lower bound of IHC(G1G2) • Let x= (u, v) V(G1 G2), where uV(G1) and v V(G2). G1 H G1 G2 G1 G1 G1 G1 H H (u1, v2) (u1, v1) G1 0 1 n–2 n–1 • G1 • G1 • G1 • G1 (u2, v2) (u2, v1) G2 … (u3, v1) (u3, v2) … … National Chi Nan University

  12. The lower bound of IHC(G1G2) • IHC(G1) = I1 • HC1 = e, x1,1, x1,2, x1,3, P1,+, x1,m–1,x1,m–2, e0 • HC2= e, x2,1, x2,2,x2,3, P2,+, x2,m–1,x2,m–2, e0 • HC • e • e • x2,m–1 • x1,m–1 • x1,1 • x2,1 • x1,m–2 • x2,m–2 • x1,2 • x2,2 • G1 • I1 • x2,3 • x1,3 • P1,+ • P2,+ National Chi Nan University

  13. The lower bound of IHC(G1G2) Theorem 1.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1 and n  2I1, then IHC(G1 G2)  I1. Proof. We construct I1Hamiltonian cycles in G1 G2 starting at vertex e0 first. Then prove these I1Hamiltonian cycles are mutually independent. National Chi Nan University

  14. The lower bound of IHC(G1G2) H1 6 1 2 0 4 5 3 • G1 • G1 • G1 • G1 • G1 • G1 • G1 H2 National Chi Nan University

  15. The lower bound of IHC(G1G2) H3 2j–2 6 5 4 3 2 1 0 n–1 • G1 • G1 • G1 • G1 • G1 • G1 • G1 • G1 • G1 • Hj,2  j  I1 … … … … … … 2j – 2 National Chi Nan University

  16. The lower bound of IHC(G1G2) Theorem 1.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1 and n  2I1, then IHC(G1 G2)  I1. Proof. For all 1 j <iI1, we prove that ith Hamiltonian cycle and jth Hamiltonian cycle are independent. Prove it by induction on i. Fori= 2 is the base case. National Chi Nan University

  17. The lower bound of IHC(G1G2) x1,m–1 x2,1 x1,1 x2, m–1 0 0 0 0 0 0 0 0+ 1– 1 1 1 2 2 2 2+ 3– 3 3 3 4 4 4 4+ 5– 5 5 5 6 6 6 6+ 6 5 4 3 2 1 0 0 H1 2 3 0 0 1 2 2 2+ 3 3 5 5 3– 4 4 4+ 4 5 5– 6 6 6+ 6 6 5 4 3 2 1 1 1– 0 0 1 0 0+ H2 n  4 m  7 National Chi Nan University

  18. The lower bound of IHC(G1G2) Theorem 1.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1 and n  2I1, then IHC(G1 G2)  I1. Proof. Suppose it is true when i= k – 1 for some 3  k I1. Now, consider i= k. We show these kHamiltonian cycles are mutually independent by the followingtwocases: (1)Hkv.s. H1; (2) Hkv.s. Hj for j {2, 3, …, k– 1}; National Chi Nan University

  19. The lower bound of IHC(G1G2) G1 G2 G1 H 0 1 n–2 n–1 • G1 • G1 • G1 • G1 National Chi Nan University

  20. The lower bound of IHC(G1G2) Theorem 2.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 3, n is odd and m  n – 3, then IHC(G1 G2)  I1 + 2. Proof. We construct I1+ 2 Hamiltonian cycles in G1 G2 starting from vertex e0, and prove these I1+ 2 Hamiltonian cycles are mutually independent. The first I1 Hamiltonian cycles H1 ~ HI1are the same as those we constructed in Theorem 1. National Chi Nan University

  21. The lower bound of IHC(G1G2) HI +2 1 HI +1 1 6 0 2 1 4 5 3 • G1 • G1 • G1 • G1 • G1 • G1 • G1 National Chi Nan University

  22. The lower bound of IHC(G1G2) Theorem 2.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 3, n is odd and m  n – 3, then IHC(G1 G2)  I1 + 2. Proof. Let I1 = k. • (1) Hk+1 and Hk+2 v.s. H1; • (1)Hk+1 and Hk+2 v.s. H1; • (1)Hk+1 and Hk+2v.s. H1; • (1) Hk+1 and Hk+2 v.s. H1; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (3) Hk+1 v.s.Hk+2. • (3) Hk+1 v.s.Hk+2. • (3) Hk+1 v.s.Hk+2. National Chi Nan University

  23. The lower bound of IHC(G1G2) Theorem 3.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 2, n is even andG1 does not contain C4, then IHC(G1 G2)  I1 + 2. Proof. We construct I1+ 2 Hamiltonian cycles in G1 G2 starting from vertex e0, and prove these I1+ 2 Hamiltonian cycles are mutually independent. The first I1 Hamiltonian cycles H1 ~ HI1are the same as those we constructed in Theorem 1. National Chi Nan University

  24. The lower bound of IHC(G1G2) HI +2 1 HI +1 1 0 5 1 3 4 2 • G1 • G1 • G1 • G1 • G1 • G1 National Chi Nan University

  25. The lower bound of IHC(G1G2) Theorem 3.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 2, n is even andG1 does not contain C4, then IHC(G1 G2)  I1 + 2. Proof. Let I1 = k. • (1) Hk+1 and Hk+2 v.s. H1; • (1)Hk+1 and Hk+2 v.s. H1; • (1)Hk+1 and Hk+2v.s. H1; • (1) Hk+1 and Hk+2 v.s. H1; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (2) Hk+1 andHk+2 v.s. Hj for j {2, 3, …, I1}; • (3) Hk+1 v.s.Hk+2. • (3) Hk+1 v.s.Hk+2. • (3) Hk+1 v.s.Hk+2. National Chi Nan University

  26. The lower bound of IHC(G1G2) Theorem 2.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 3 , n is odd and n – m  4,then IHC(G1 G2)  I1 + 2. Theorem 3.For any Hamiltonian graphs G1 and G2, ifm  4I1– 1,n  2I1 + 2, n is even andG1 does not contain C4, then IHC(G1 G2)  I1 + 2. Corollary 1. For m  7 is odd and n  6 is even, IHC(Cm Cn) = 4. National Chi Nan University

  27. Outline • Introduction • Motivation & Contribution • Main result • The lower bound of IHC(G1G2) • IHC(CmCn) = 4 • Conclusion & Future work National Chi Nan University

  28. IHC(CmCn) = 4 Corollary 1. For m  7 is odd and n  6 is even, IHC(Cm Cn) = 4. Theorem 4. For n3 is odd, IHC(Cn Cn) = 4. Theorem 5. For m, n3 and m, nare odd, IHC(Cm Cn) = 4. Theorem 6. For m  4 is even and n= 3, IHC(Cm Cn) = 4. Theorem 7. For m 4 is even and n= 5, IHC(Cm Cn) = 4. Theorem 8. For m= 4 and n 9 is odd, IHC(Cm Cn) = 4. Theorem 9. For n4 is even, IHC(Cn Cn) = 4. Theorem 10. For m 6 is even and n = 4, IHC(Cm Cn) = 4. Theorem 11. For m, n6, IHC(Cm Cn) = 4. National Chi Nan University

  29. IHC(CmCn) = 4 m n IHC(Cm Cn) = 4 Corollary 1. m  7is odd and n  6 is even Theorem 4. m = n 3 is odd Theorem 5. m, n 3 and m, nare odd Theorem 6. m 4 is even and n = 3 Theorem 7. m4 is even and n = 5 Theorem 8. m= 4 and n7 is odd Theorem 9. m = n 4 is even Theorem 10. m6 is even and n = 4 Theorem 11. m, n6 … National Chi Nan University

  30. Introduction C6  C5 (0, 0) (0, 1) (0, 2) (0, 3) (0, 4) C0 • Cm Cnfor m, n  3. • 4-regular • C0 • C4 (1, 0) (1, 1) (1, 2) (1, 3) (1, 4) C1 (2, 0) (2, 1) (2, 2) (2, 3) (2, 4) C2 (3, 0) (3, 1) (3, 2) (3, 3) (3, 4) C3 (4, 0) (4, 1) (4, 2) (4, 3) (4, 4) C4 (5, 0) (5, 1) (5, 2) (5, 3) (5, 4) C5 National Chi Nan University

  31. IHC(CmCn) = 4 Property 2. For any graph G, if there exist two paths P1, P2 and a subgraphCnG with Cn= x0, x1, ..., xn–1, x0. And for 0 jn – 1, v– u= a 0 andt1–t2 = b,P1(t1 + j) = xu+j, P2(t2+ j) = xv+j. If any one of the following conditionsholds, then these two paths do not meet the samevertex of Cnat the same time. (1) If a = 0 and b0. (2) If a0 and (b –aor n –a). Cn t1 xu G P1 xv t2 P2 National Chi Nan University

  32. IHC(CmCn) = 4 a = 1, (b –aor n –a) P1= x0, x1, ..., xn–1 P2= x1, x2, ...,xn–1, x0 n = 8 t1 – t2 = – 1 t1 – t2 = n– 1 t1 – t2 = 0 x2 x1 x3 t2 P2 x4 t1 C8 x0 P1 x7 x5 x6 National Chi Nan University

  33. IHC(CmCn) = 4 Property 3.For any graph G, if there exist two paths P1, P2 and a subgraphCnG with Cn= x0, x1, ..., xn–1, x0. And for 0 jn – 1, v– u= a 0and t1–t2 = b,P1(t1 + j) = xu–j , P2(t2+ j) = xv–j. If any one of the following conditionsholds, then these two paths do not meet the samevertex of Cnat the same time. (1) If a = 0 and b0. (2) If a0 and (baor a –n). National Chi Nan University

  34. IHC(CmCn) = 4 Theorem 8. For n7 is odd, IHC(C4 Cn) = 4. Proof. Since C4  Cn is 4-regular, IHC(C4  Cn) 4. Consider n 9 and without loss of generality we assume that four Hamiltonian cycles starting at e = (0, 0). National Chi Nan University

  35. IHC(CmCn) = 4 C4C9 0 1 36 35 34 33 1 36 35 34 33 3 1 2 3 4 5 32 31 3 1 2 3 4 5 8 32 31 2 6 7 8 2 6 7 36 36 1 11 10 9 8 7 11 10 9 8 7 4 16 15 14 13 12 6 5 4 16 15 14 13 12 9 6 5 12 11 10 9 12 11 10 35 35 2 14 15 16 17 18 14 15 16 17 18 28 17 18 19 20 21 19 29 28 17 18 19 20 21 24 19 29 13 22 23 24 13 22 23 34 34 3 25 24 23 22 21 25 24 23 22 21 27 32 31 30 29 28 20 30 27 32 31 30 29 28 25 20 30 26 27 26 25 26 27 26 33 33 8 4 5 0 1 2 8 3 6 7 4 5 0 1 2 3 6 7 0 1 34 19 20 21 1 34 19 20 21 24 22 23 24 22 23 25 25 1 11 12 13 14 1 11 12 13 14 17 15 16 17 15 16 18 18 1 2 33 32 31 30 2 33 32 31 30 27 29 28 27 29 28 26 26 36 26 25 24 23 36 26 25 24 23 20 22 21 20 22 21 19 19 2 3 4 5 6 7 3 4 5 6 7 10 8 9 10 8 9 11 11 35 27 28 29 30 35 27 28 29 30 33 31 32 33 31 32 34 34 3 36 35 18 17 16 36 35 18 17 16 13 15 14 13 15 14 12 12 2 10 9 8 7 2 10 9 8 7 4 6 5 4 6 5 3 3 National Chi Nan University

  36. IHC(CmCn) = 4 (0, 1)(0, 2) Property 2 For m= 4, n 9 Property 3 (0, n–1)(0, n–2) 0+ 1– 2+ 3– 3 2 1 0 0 H1 3– 2 2 3 0– 0 0 0 1– 2+ H2 0 3– 0+ 1– 2+ 1 1 0 H3 0 1 2+ 3– 0+ 1– 0 3 3 0 H4 National Chi Nan University

  37. IHC(CmCn) = 4 Theorem 8. For n 7 is odd, IHC(C4 Cn) = 4. Proof. (1) IHC(C4  Cn) 4. (2)C4  Cn is 4-regular, IHC(C4  Cn)  4. Hence, IHC(C4  Cn) = 4 can be concluded. National Chi Nan University

  38. Outline • Introduction • Motivation & Contribution • Main result • The lower bound of IHC(G1G2) • IHC(CmCn) = 4 • Conclusion & Future work National Chi Nan University

  39. Conclusion In this thesis, we discuss the lower bound of IHC(G1G2) and get Theorems 1, 2 and 3. For any Hamiltonian graphs G1 and G2, letIHC(G1) = I1,|V(G1)| = m,|V(G2)| = n. Theorem 1: Ifm  4I1– 1 and n  2I1, then IHC(G1 G2)  I1. Theorem 2:Ifm  4I1– 1,n  2I1 + 3, n is odd and m  n – 3, then IHC(G1 G2)  I1 + 2.Theorem 3: Ifm  4I1– 1,n  2I1 + 2, n is even andG1 does not contain C4, then IHC(G1 G2)  I1 + 2. National Chi Nan University

  40. Conclusion We also study on IHC(CmCn) and get the optimal results as IHC(CmCn) = 4 for m, n3 in Theorems 4, 5, …, and 11. Corollary 2. For any graphs G1, G2, if G1 and G2 are Hamiltonian and |V(G1)|, |V(G2)|  3, then IHC(G1G2)  4. • Corollary 3. For graph G = Ck Ck  …  Ck, l 2, if k1·k2 15, k3 11 and k3, k4, …, kl are odd. For all 3 il, ki4(i– 1) + 3 and k1·k2· … · ki–1 ki – 3. Then IHC(G) = 2l. 1 l 2 National Chi Nan University

  41. Future work • The exact value of IHC(G1G2) • Other interesting graphs National Chi Nan University

  42. Thanks for listening. Q & A National Chi Nan University

More Related