mutually independent hamiltonian cycles on cartesian product graphs

National Chi Nan University 1

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(G1 G2)– IHC(Cm Cn) = 4

• Conclusion & Future work

Outline

• Introduction• Motivation & Contribution • Main result– The lower bound of IHC(G1 G2)– IHC(Cm Cn) = 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 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.

C1 = 0, 1, 2, 3, 4, 0C2 = 0, 2, 3, 4, 1, 0

Introduction

• 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 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.

C1 = 0, 1, 2, 3, 4, 0C2 = 0, 2, 3, 4, 1, 0C3 = 0, 3, 4, 1, 2, 0C4 = 0, 4, 1, 2, 3, 0IHC(G) = 4

Outline

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.

Outline

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).

Theorem 1. For any Hamiltonian graphs G1 and G2, if m 4I1 – 1 and n 2I1, then IHC(G1 G2) I1.

Theorem 3. For any Hamiltonian graphs G1 and G2, if m 4I1 – 1, n 2I1 + 2, n is even and G1 does not contain C4, then IHC(G1 G2) I1 + 2.

Theorem 2. For any Hamiltonian graphs G1 and G2, if m 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, if m 4I1 – 1, n 2I1 + 2, n is even and G1 does not contain C4, then IHC(G1 G2) I1 + 2.

Theorem 2. For any Hamiltonian graphs G1 and G2, if m 4I1 – 1, n 2I1 + 3 , n is odd and m n – 3, then IHC(G1 G2) I1 + 2.

For any Hamiltonian graphs G1 and G2: Let IHC(G1) = I1, |V(G1)| = m, |V(G2)| = n.

• Let x = (u, v) V(G1 G2), where u V(G1) and v V(G2).

(u1, v1)

(u2, v1)

(u1, v2)

(u2, v2)

(u3, v1) (u3, v2) … …G1

G10 G1 G1 G1

G1n–2

G1n–1

• 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

x1,m–1

x1,m–2

ex2,m–1

x2,m–2 x2,2

x2,3 P2,+

Proof.We construct I1 Hamiltonian cycles in G1 G2 starting at vertex e0 first. Then prove these I1 Hamiltonian cycles are mutually independent.

G10 G1

1 G12 G1

3 G14 G1

… ……

… …

Hj, 2 j I1

G10 G1

1 G12 G1

3 G14 G1

2j – 2G1

2j–2 G1n–1

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.

For i = 2 is the base case.

66543211 1–

000 0+ 1– 111222 2+ 3– 333444 4+ 5– 555666 6+ 65432100

00122 2+ 23 3– 3344 4+ 45 5– 5566 6+ 10 0+ 00

n 4m 7

x1,m–10

x2, m–10

Proof.

Suppose it is true when i = k – 1 for some 3 k I1.

Now, consider i = k.We show these k Hamiltonian cycles are mutually independent by the following two cases: (1) Hk v.s. H1; (2) Hk v.s. Hj for j {2, 3, …, k – 1};

G10 G1

1 G1n–2

G1n–1

Theorem 2. For any Hamiltonian graphs G1 and G2, if m 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 ~ HI1

are the same as those we constructed in Theorem 1.

HI +11

HI +21

G10 G1

1 G12 G1

3 G14 G1

(3) Hk+1 v.s. Hk+2.(2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1};

Theorem 2. For any Hamiltonian graphs G1 and G2, if m 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;

(3) Hk+1 v.s. Hk+2.

(1) Hk+1 and Hk+2 v.s. H1; (1) Hk+1 and Hk+2 v.s. H1; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (1) Hk+1 and Hk+2 v.s. H1; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (3) Hk+1 v.s. Hk+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 ~ HI1

are the same as those we constructed in Theorem 1.

HI +11

HI +21

G10 G1

1 G12 G1

3 G14 G1

Theorem 3. For any Hamiltonian graphs G1 and G2, if m 4I1 – 1, n 2I1 + 2, n is even and G1 does not contain C4, then IHC(G1 G2) I1 + 2.Proof.Let I1 = k.

(3) Hk+1 v.s. Hk+2.(2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (1) Hk+1 and Hk+2 v.s. H1;

(3) Hk+1 v.s. Hk+2.

(1) Hk+1 and Hk+2 v.s. H1; (1) Hk+1 and Hk+2 v.s. H1; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (1) Hk+1 and Hk+2 v.s. H1; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (2) Hk+1 and Hk+2 v.s. Hj for j {2, 3, …, I1}; (3) Hk+1 v.s. Hk+2.

Theorem 2. For any Hamiltonian graphs G1 and G2, if m 4I1 – 1, n 2I1 + 3 , n is odd and n – m 4, then IHC(G1 G2) I1 + 2.

Corollary 1. For m 7 is odd and n 6 is even, IHC(Cm Cn) = 4.

Outline

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, n are 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.

IHC(Cm Cn) = 43 4 5 6 7 8 9 10 11 12 …

3 …4 …5 …6 …7 …8 …9 …10 …11 …12 …… … … … … … … … … … …

IHC(Cm Cn) = 4Corollary 1. m 7 is odd and n 6 is evenTheorem 4. m = n 3 is oddTheorem 5. m, n 3 and m, n are oddTheorem 6. m 4 is even and n = 3Theorem 7. m 4 is even and n = 5Theorem 8. m = 4 and n 7 is oddTheorem 9. m = n 4 is evenTheorem 10. m 6 is even and n = 4Theorem 11. m, n 6

Introduction

• Cm Cn for m, n 3.– 4-regular– C0

– C4

(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)

(5, 0) (5, 1) (5, 2) (5, 3) (5, 4)

National Chi Nan University

IHC(Cm Cn) = 4

Property 2. For any graph G, if there exist two paths P1, P2 and a subgraph Cn G with Cn = x0, x1, ..., xn–1, x0. And for 0 j n – 1, v – u = a 0 and t1 – t2 = b, P1(t1 + j) = xu+j , P2(t2 + j) = xv+j. If any one of the following conditions holds, then these two paths do not meet the same vertex of Cn at the same time. (1) If a = 0 and b 0. (2) If a 0 and (b – a or n – a).

IHC(Cm Cn) = 4

P1 = x0, x1, ..., xn–1 P2 = x1, x2, ..., xn–1, x0

a = 1, (b – a or n – a)

n = 8t1 – t2 = – 1t1 – t2 = n – 1t1 – t2 = 0

IHC(Cm Cn) = 4

Property 3. For any graph G, if there exist two paths P1, P2 and a subgraph Cn G with Cn = x0, x1, ..., xn–1, x0. And for 0 j n – 1, v – u = a 0 and t1 – t2 = b, P1(t1 + j) = xu–j , P2(t2 + j) = xv–j. If any one of the following conditions holds, then these two paths do not meet the same vertex of Cn at the same time. (1) If a = 0 and b 0. (2) If a 0 and (b a or a – n).

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).

IHC(Cm Cn) = 4

16 15 14

17 18 19

32 31 30

0 1 2 3 4 5 6 7 8

1 34 19

2 33 32

36 35 18

1 36 35

11 10 9

14 15 16

25 24 23

1 11 12

36 26 25

35 27 28

2 10 9

16 15 14

17 18 19

32 31 30

1 34 19

2 33 32

36 35 18

1 36 35

11 10 9

14 15 16

25 24 23

1 11 12

36 26 25

35 27 28

2 10 9

IHC(Cm Cn) = 4For m = 4, n 9

2+ 1–

0+ 1– 2+ 3– 32100

000 1– 2+ 223 0–

0 3– 0+ 1– 2+ 110

03300+3–01

(0, 1)

(0, 2)

(0, n–1)

(0, n–2)

Property 3

Property 2

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.

Outline

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, let IHC(G1) = I1, |V(G1)| = m, |V(G2)| = n. Theorem 1: If m 4I1 – 1 and n 2I1, then IHC(G1 G2) I1.

Theorem 2: If m 4I1 – 1, n 2I1 + 3, n is odd and m n – 3, then IHC(G1 G2) I1 + 2.Theorem 3: If m 4I1 – 1, n 2I1 + 2, n is even and G1 does not contain C4, then IHC(G1 G2) I1 + 2.

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.

21 lCorollary 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.

Future work

• The exact value of IHC(G1 G2)

• Other interesting graphs

Thanks for listening.Q & A

mutually independent hamiltonian cycles on cartesian product graphs

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