1. Mutually independent Hamiltonian cycles on Cartesian product graphs Student: Kai-Siou Wu ( 吳凱修 ) Adviser: Justie Su-Tzu Juan 1National Chi Nan University

2. Outline Introduction Motivation & Contribution Main result – The lower bound of IHC(G 1  G 2 ) – IHC(C m  C n ) = 4 Conclusion & Future work 2National Chi Nan University

3. Outline Introduction Motivation & Contribution Main result – The lower bound of IHC(G 1  G 2 ) – IHC(C m  C n ) = 4 Conclusion & Future work 3National Chi Nan University

4. Introduction Hamiltonian cycle – In a graph G, a cycle is called Hamiltonian cycle if it contains all vertices of G exactly once. Independent – Two cycles C 1 =  u 0, u 1,..., u n–1, u 0  and C 2 =  v 0, v 1,..., v n–1, v 0  in G are independent if u 0 = v 0 and u i  v i for 1 ≤ i ≤ n – 1. National Chi Nan University4 0 4 32 1 G C 1 =  0, 1, 2, 3, 4, 0  C 2 =  0, 2, 3, 4, 1, 0 

5. Introduction Mutually independent – A set of Hamiltonian cycles {C 1, C 1, …, C k } of G are mutually independent if any two dierent Hamiltonian cycles of {C 1, C 1, …, C k } are independent. Mutually independent Hamiltonicity IHC(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. National Chi Nan University5 0 4 32 1 G C 1 =  0, 1, 2, 3, 4, 0  C 2 =  0, 2, 3, 4, 1, 0  C 3 =  0, 3, 4, 1, 2, 0  C 4 =  0, 4, 1, 2, 3, 0  IHC(G) = 4

6. Outline Introduction Motivation & Contribution Main result – The lower bound of IHC(G 1  G 2 ) – IHC(C m  C n ) = 4 Conclusion & Future work 6National Chi Nan University

7. Motivation & Contribution The lower bound of IHC(G 1  G 2 ) with different conditions – IHC(G 1  G 2 )  IHC(G 1 ) – IHC(G 1  G 2 )  IHC(G 1 ) + 2 Mutually independent Hamiltonianicity of C m  C n – IHC(C m  C n ) = 4, for m, n  3. National Chi Nan University7

8. Outline Introduction Motivation & Contribution Main result – The lower bound of IHC(G 1  G 2 ) – IHC(C m  C n ) = 4 Conclusion & Future work 8National Chi Nan University

9. The lower bound of IHC(G 1  G 2 ) By definition, we can get the trivial upper bound: IHC(G 1  G 2 )   (G 1  G 2 ) =  (G 1 ) +  (G 2 ). So, we want to discuss the lower bound of IHC(G 1  G 2 ). National Chi Nan University9

10. The lower bound of IHC(G 1  G 2 ) National Chi Nan University10 Theorem 1. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1 and n  2I 1, then IHC(G 1  G 2 )  I 1. Theorem 3. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 2, n is even and G 1 does not contain C 4, then IHC(G 1  G 2 )  I 1 + 2. Theorem 2. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 3, n is odd and m  n – 3, then IHC(G 1  G 2 )  I 1 + 2. Theorem 3. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 2, n is even and G 1 does not contain C 4, then IHC(G 1  G 2 )  I 1 + 2. Theorem 1. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1 and n  2I 1, then IHC(G 1  G 2 )  I 1. Theorem 2. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 3, n is odd and m  n – 3, then IHC(G 1  G 2 )  I 1 + 2. For any Hamiltonian graphs G 1 and G 2 : Let IHC(G 1 ) = I 1, |V(G 1 )| = m, |V(G 2 )| = n.

11. The lower bound of IHC(G 1  G 2 ) Let x = (u, v)  V(G 1  G 2 ), where u  V(G 1 ) and v  V(G 2 ). National Chi Nan University11 G1  G2G1  G2 (u 1, v 1 ) (u 2, v 1 ) (u 1, v 2 ) (u 2, v 2 ) (u 3, v 1 ) (u 3, v 2 ) … … G1G1 G2G2 H … H G1G1 G1  HG1  H G1G1 0 G1G1 G1G1 G1G1 G1G1 1 G1G1 n–2 G1G1 n–1

12. x 1,1 The lower bound of IHC(G 1  G 2 ) IHC(G 1 ) = I 1 HC 1 =  e, x 1,1, x 1,2, x 1,3, P 1,+, x 1,m–1, x 1,m–2, e 0  HC 2 =  e, x 2,1, x 2,2, x 2,3, P 2,+, x 2,m–1, x 2,m–2, e 0  HC National Chi Nan University12 G1G1 I1I1 e x 1,2 x 1,3 x 1, m –1 x 1, m –2 P 1, + e x 2, m –1 x 2, m –2 x 2,2 x 2,3 P 2, + x 2,1

13. The lower bound of IHC(G 1  G 2 ) Theorem 1. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1 and n  2I 1, then IHC(G 1  G 2 )  I 1. Proof. We construct I 1 Hamiltonian cycles in G 1  G 2 starting at vertex e 0 first. Then prove these I 1 Hamiltonian cycles are mutually independent. National Chi Nan University13

14. The lower bound of IHC(G 1  G 2 ) National Chi Nan University14 H1H1 H2H2 G1G1 0 G1G1 1 G1G1 2 G1G1 3 G1G1 4 G1G1 5 G1G1 6

15. The lower bound of IHC(G 1  G 2 ) National Chi Nan University15 … … … … … H3H3 H j, 2  j  I 1 … G1G1 0 G1G1 1 G1G1 2 G1G1 3 G1G1 4 G1G1 5 G1G1 6 2j – 2 G1G1 G1G1 n–1

16. The lower bound of IHC(G 1  G 2 ) Theorem 1. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1 and n  2I 1, then IHC(G 1  G 2 )  I 1. Proof. For all 1  j < i  I 1, we prove that ith Hamiltonian cycle and jth Hamiltonian cycle are independent. Prove it by induction on i. For i = 2 is the base case. National Chi Nan University16

17. The lower bound of IHC(G 1  G 2 ) National Chi Nan University17 665432111– 0000+1–1112222+3–3334444+5–5556666+65432100 001222+ 23 3– 33 444+455– 55 666+ 100+ 00 H1H1 H2H2 n  4 m  7 x 1,1 0 x 2,1 0 x 1,m–1 0 x 2, m–1 0

18. The lower bound of IHC(G 1  G 2 ) Theorem 1. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1 and n  2I 1, then IHC(G 1  G 2 )  I 1. Proof. Suppose it is true when i = k – 1 for some 3  k  I 1. Now, consider i = k. We show these k Hamiltonian cycles are mutually independent by the following two cases: (1) H k v.s. H 1 ; (2) H k v.s. H j for j  {2, 3, …, k – 1}; National Chi Nan University18

19. The lower bound of IHC(G 1  G 2 ) National Chi Nan University19 G1  G2G1  G2 G1  HG1  H G1G1 0 G1G1 1 G1G1 n–2 G1G1 n–1

20. The lower bound of IHC(G 1  G 2 ) Theorem 2. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 3, n is odd and m  n – 3, then IHC(G 1  G 2 )  I 1 + 2. Proof. We construct I 1 + 2 Hamiltonian cycles in G 1  G 2 starting from vertex e 0, and prove these I 1 + 2 Hamiltonian cycles are mutually independent. The first I 1 Hamiltonian cycles H 1 ~ H I 1 are the same as those we constructed in Theorem 1. National Chi Nan University20

21. The lower bound of IHC(G 1  G 2 ) National Chi Nan University21 H I +1 1 H I +2 1 G1G1 0 G1G1 1 G1G1 2 G1G1 3 G1G1 4 G1G1 5 G1G1 6

22. (3) H k+1 v.s. H k+2. (2) H k+1 and H k+2 v.s. H j for j  {2, 3, …, I 1 }; The lower bound of IHC(G 1  G 2 ) Theorem 2. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 3, n is odd and m  n – 3, then IHC(G 1  G 2 )  I 1 + 2. Proof. Let I 1 = k. National Chi Nan University22 (1) H k+1 and H k+2 v.s. H 1 ; (3) H k+1 v.s. H k+2. (1) H k+1 and H k+2 v.s. H 1 ; (2) H k+1 and H k+2 v.s. H j for j  {2, 3, …, I 1 }; (1) H k+1 and H k+2 v.s. H 1 ; (2) H k+1 and H k+2 v.s. H j for j  {2, 3, …, I 1 }; (3) H k+1 v.s. H k+2.

23. The lower bound of IHC(G 1  G 2 ) Theorem 3. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 2, n is even and G 1 does not contain C 4, then IHC(G 1  G 2 )  I 1 + 2. Proof. We construct I 1 + 2 Hamiltonian cycles in G 1  G 2 starting from vertex e 0, and prove these I 1 + 2 Hamiltonian cycles are mutually independent. The first I 1 Hamiltonian cycles H 1 ~ H I 1 are the same as those we constructed in Theorem 1. National Chi Nan University23

24. The lower bound of IHC(G 1  G 2 ) National Chi Nan University24 H I +1 1 H I +2 1 G1G1 0 G1G1 1 G1G1 2 G1G1 3 G1G1 4 G1G1 5

25. The lower bound of IHC(G 1  G 2 ) Theorem 3. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 2, n is even and G 1 does not contain C 4, then IHC(G 1  G 2 )  I 1 + 2. Proof. Let I 1 = k. National Chi Nan University25 (3) H k+1 v.s. H k+2. (2) H k+1 and H k+2 v.s. H j for j  {2, 3, …, I 1 }; (1) H k+1 and H k+2 v.s. H 1 ; (3) H k+1 v.s. H k+2. (1) H k+1 and H k+2 v.s. H 1 ; (2) H k+1 and H k+2 v.s. H j for j  {2, 3, …, I 1 }; (1) H k+1 and H k+2 v.s. H 1 ; (2) H k+1 and H k+2 v.s. H j for j  {2, 3, …, I 1 }; (3) H k+1 v.s. H k+2.

26. The lower bound of IHC(G 1  G 2 ) Theorem 2. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 3, n is odd and n – m  4, then IHC(G 1  G 2 )  I 1 + 2. Theorem 3. For any Hamiltonian graphs G 1 and G 2, if m  4I 1 – 1, n  2I 1 + 2, n is even and G 1 does not contain C 4, then IHC(G 1  G 2 )  I 1 + 2. Corollary 1. For m  7 is odd and n  6 is even, IHC(C m  C n ) = 4. National Chi Nan University26

27. Outline Introduction Motivation & Contribution Main result – The lower bound of IHC(G 1  G 2 ) – IHC(C m  C n ) = 4 Conclusion & Future work 27National Chi Nan University

28. IHC(C m  C n ) = 4 Corollary 1. For m  7 is odd and n  6 is even, IHC(C m  C n ) = 4. Theorem 4. For n  3 is odd, IHC(C n  C n ) = 4. Theorem 5. For m, n  3 and m, n are odd, IHC(C m  C n ) = 4. Theorem 6. For m  4 is even and n = 3, IHC(C m  C n ) = 4. Theorem 7. For m  4 is even and n = 5, IHC(C m  C n ) = 4. Theorem 8. For m = 4 and n  9 is odd, IHC(C m  C n ) = 4. Theorem 9. For n  4 is even, IHC(C n  C n ) = 4. Theorem 10. For m  6 is even and n = 4, IHC(C m  C n ) = 4. Theorem 11. For m, n  6, IHC(C m  C n ) = 4. National Chi Nan University28

29. IHC(C m  C n ) = 4 National Chi Nan University29 3456789101112… 3… 4… 5… 6… 7… 8… 9… 10… 11… 12… …………………………… m n … IHC(C m  C n ) = 4 Corollary 1. m  7 is odd and n  6 is even Theorem 4. m = n  3 is odd Theorem 5. m, n  3 and m, n are 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

30. Introduction C m  C n for m, n  3. – 4-regular –C0–C0 –C4–C4 30 (0, 0)(0, 1)(0, 2)(0, 3)(0, 4) (1, 0)(1, 1)(1, 2)(1, 3)(1, 4) (2, 0)(2, 1)(2, 2)(2, 3)(2, 4) (3, 0)(3, 1)(3, 2)(3, 3)(3, 4) (4, 0)(4, 1)(4, 2)(4, 3)(4, 4) C 6  C 5 (5, 0)(5, 1)(5, 2)(5, 3)(5, 4) C0C0 C1C1 C2C2 C3C3 C4C4 C5C5 National Chi Nan University

31. IHC(C m  C n ) = 4 Property 2. For any graph G, if there exist two paths P 1, P 2 and a subgraph C n  G with C n =  x 0, x 1,..., x n–1, x 0 . And for 0  j  n – 1, v – u = a  0 and t 1 – t 2 = b, P 1 (t 1 + j) = x u+j, P 2 (t 2 + j) = x v+j. If any one of the following conditions holds, then these two paths do not meet the same vertex of C n at the same time. (1) If a = 0 and b  0. (2) If a  0 and (b  – a or n – a). 31 G CnCn P1P1 P2P2 xuxu xvxv t1t1 t2t2 National Chi Nan University

32. IHC(C m  C n ) = 4 P 1 =  x 0, x 1,..., x n–1  P 2 =  x 1, x 2,..., x n–1, x 0  32 a = 1, (b  – a or n – a) C8C8 P1P1 P2P2 x0x0 t1t1 t2t2 x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 n = 8 t 1 – t 2 = – 1 t 1 – t 2 = n – 1 t 1 – t 2 = 0 National Chi Nan University

33. IHC(C m  C n ) = 4 Property 3. For any graph G, if there exist two paths P 1, P 2 and a subgraph C n  G with C n =  x 0, x 1,..., x n–1, x 0 . And for 0  j  n – 1, v – u = a  0 and t 1 – t 2 = b, P 1 (t 1 + j) = x u–j, P 2 (t 2 + j) = x v–j. If any one of the following conditions holds, then these two paths do not meet the same vertex of C n at the same time. (1) If a = 0 and b  0. (2) If a  0 and (b  a or a – n). 33National Chi Nan University

34. IHC(C m  C n ) = 4 Theorem 8. For n  7 is odd, IHC(C 4  C n ) = 4. Proof. Since C 4  C n is 4-regular, IHC(C 4  C n )  4. Consider n  9 and without loss of generality we assume that four Hamiltonian cycles starting at e = (0, 0). National Chi Nan University34

35. IHC(C m  C n ) = 4 National Chi Nan University35 123 161514 171819 323130 45 1312 2021 2928 67 1110 2223 2726 8 9 24 25 36 35 34 33 01 2 3 45 67 8 0 1 2 3 01 2 3 45 67 8 0 1 2 3 13419 23332 345 363518 2021 3130 67 1716 2223 2928 89 1514 24 27 10 13 25 26 11 12 13635 11109 141516 252423 3433 87 1718 2221 3231 65 1929 2030 3 4 28 27 2 12 13 26 11112 362625 352728 2109 1314 2423 2930 87 1516 2221 3132 65 17 20 33 4 18 19 34 3 C 4  C 9 123 161514 171819 323130 45 1312 2021 2928 67 1110 2223 2726 8 9 24 25 36 35 34 33 13419 23332 345 363518 2021 3130 67 1716 2223 2928 89 1514 24 27 10 13 25 26 11 12 13635 11109 141516 252423 3433 87 1718 2221 3231 65 1929 2030 3 4 28 27 2 12 13 26 11112 362625 352728 2109 1314 2423 2930 87 1516 2221 3132 65 17 20 33 4 18 19 34 3

36. IHC(C m  C n ) = 4 National Chi Nan University36 For m = 4, n  9 2+1– 0+1–2+3–32100 0001–2+ 2230– 03–0+1–2+110 033 0 0+3–01 H3H3 H4H4 H1H1 H2H2 (0, 1) (0, 2) (0, n–1) (0, n–2) Property 3 Property 2

37. IHC(C m  C n ) = 4 Theorem 8. For n  7 is odd, IHC(C 4  C n ) = 4. Proof. (1) IHC(C 4  C n )  4. (2) C 4  C n is 4-regular, IHC(C 4  C n )  4. Hence, IHC(C 4  C n ) = 4 can be concluded. National Chi Nan University37

38. Outline Introduction Motivation & Contribution Main result – The lower bound of IHC(G 1  G 2 ) – IHC(C m  C n ) = 4 Conclusion & Future work 38National Chi Nan University

39. Conclusion In this thesis, we discuss the lower bound of IHC(G 1  G 2 ) and get Theorems 1, 2 and 3. For any Hamiltonian graphs G 1 and G 2, let IHC(G 1 ) = I 1, |V(G 1 )| = m, |V(G 2 )| = n. Theorem 1: If m  4I 1 – 1 and n  2I 1, then IHC(G 1  G 2 )  I 1. Theorem 2: If m  4I 1 – 1, n  2I 1 + 3, n is odd and m  n – 3, then IHC(G 1  G 2 )  I 1 + 2. Theorem 3: If m  4I 1 – 1, n  2I 1 + 2, n is even and G 1 does not contain C 4, then IHC(G 1  G 2 )  I 1 + 2. National Chi Nan University39

40. Conclusion We also study on IHC(C m  C n ) and get the optimal results as IHC(C m  C n ) = 4 for m, n  3 in Theorems 4, 5, …, and 11. Corollary 2. For any graphs G 1, G 2, if G 1 and G 2 are Hamiltonian and |V(G 1 )|, |V(G 2 )|  3, then IHC(G 1  G 2 )  4. National Chi Nan University40 2 1 l Corollary 3. For graph G = C k  C k  …  C k, l  2, if k 1 ·k 2  15, k 3  11 and k 3, k 4, …, k l are odd. For all 3  i  l, k i  4(i – 1) + 3 and k 1 ·k 2 · … · k i–1  k i – 3. Then IHC(G) = 2l.

41. Future work The exact value of IHC(G 1  G 2 ) Other interesting graphs National Chi Nan University41

42. Thanks for listening. Q & A National Chi Nan University42