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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.

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Mutually independent Hamiltonian cycles on Cartesian product graphs

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  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

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