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

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

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

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 University G C 1 =  0, 1, 2, 3, 4, 0  C 2 =  0, 2, 3, 4, 1, 0 

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 University 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

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

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

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

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

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 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 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 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 For any Hamiltonian graphs G 1 and G 2 : Let IHC(G 1 ) = I 1, |V(G 1 )| = m, |V(G 2 )| = n.

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

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

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

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

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

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

The lower bound of IHC(G 1  G 2 ) National Chi Nan University – – – – – – H1H1 H2H2 n  4 m  7 x 1,1 0 x 2,1 0 x 1,m–1 0 x 2, m–1 0

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

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

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 Proof. We construct I Hamiltonian cycles in G 1  G 2 starting from vertex e 0, and prove these I 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

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

(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 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.

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 Proof. We construct I Hamiltonian cycles in G 1  G 2 starting from vertex e 0, and prove these I 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

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

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 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.

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 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 Corollary 1. For m  7 is odd and n  6 is even, IHC(C m  C n ) = 4. National Chi Nan University26

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

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

IHC(C m  C n ) = 4 National Chi Nan University … 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

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

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

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

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

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

IHC(C m  C n ) = 4 National Chi Nan University C 4  C

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

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

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

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 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 National Chi Nan University39

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 University 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.

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

Thanks for listening. Q & A National Chi Nan University42