Mutually independent hamiltonian cycles for the pancake graphs and the star graphs

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Contents lists available atScienceDirect

Discrete Mathematics

journal homepage:www.elsevier.com/locate/disc

Mutually independent hamiltonian cycles for the pancake graphs and

the star graphs

Cheng-Kuan Lin

a,∗

, Jimmy J.M. Tan

a

, Hua-Min Huang

b

, D. Frank Hsu

c

, Lih-Hsing Hsu

d

a_{Department of Computer Science, National Chiao Tung University, Hsinchu, 30010 Taiwan, ROC} b_{Department of Mathematics, National Central University, Chungli, 32001 Taiwan, ROC} c_{Department of Computer and Information Science, Fordham University, New York, NY 10023, USA}

d_{Department of Computer Science and Information Engineering, Providence University, Taichung, 43301 Taiwan, ROC}

a r t i c l e i n f o

Article history:

Received 19 September 2006 Accepted 8 December 2008 Available online 14 January 2009 Keywords:

Hamiltonian Pancake networks Star networks

a b s t r a c t

A hamiltonian cycle C of a graph G is an ordered sethu1,u2, . . . ,un(G),u1iof vertices such that ui 6= ujfor i 6= j and uiis adjacent to ui+1for every i ∈ {1,2, . . . ,n(G) −1}and

un(G)is adjacent to u1, where n(G)is the order of G. The vertex u1is the starting vertex and uiis the ith vertex of C . Two hamiltonian cycles C1 = hu1,u2, . . . ,un(G),u1iand

C2 = hv1, v2, . . . , vn(G), v1iof G are independent if u1 = v1and ui 6= vifor every i ∈ {2,3, . . . ,n(G)}. A set of hamiltonian cycles{C1,C2, . . . ,Ck}of G is mutually independent if its elements are pairwise independent. The mutually independent hamiltonicity IHC(G)of a graph 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.

In this paper, the mutually independent hamiltonicity is considered for two families of Cayley graphs, the n-dimensional pancake graphs Pnand the n-dimensional star graphs Sn. It is proven that IHC(P3) =1, IHC(Pn) =n−1 if n≥4, IHC(Sn) =n−2 if n∈ {3,4}and

IHC(Sn) =n−1 if n≥5.

In 1969, Lovász [32] asked whether every finite connected vertex transitive graph has a hamiltonian path, that is, a simple path that traverses every vertex exactly once. All known vertex transitive graphs have a hamiltonian path and moreover, only four vertex transitive graphs without a hamiltonian cycle are known. Since none of these four graph is a Cayley graph there is a folklore conjecture [9] that every Cayley graph with more than two vertices has a hamiltonian cycle. In the last decades this problem was extensively studied (see [2–5,7,12,19,33–36]) and for those Cayley graphs for which the existence of hamiltonian cycles is already proven, further properties related to this problem, such as edge-hamiltonicity, Hamilton-connectivity and Hamilton-laceability, are investigated (see [4,8]). In this paper we introduce one of such properties, the concept of mutually independent hamiltonian cycles which is related to the number of hamiltonian cycles in a given graph. In particular, mutually independent hamiltonian cycles of pancake graphs Pnand star graphs Sn(for definitions see Sections4

and5) are studied.

The paper is organized as follows. In Section2definitions and notations needed in the subsequent sections are introduced. In Section3applications of the mutually independent hamiltonicity concept are given. In Sections4and5the mutually independent hamiltonicity of pancake graphs Pnand star graphs Sn, respectively, is computed. And in the last section,

Section6, directions for further research on this topic are discussed.

∗_{Corresponding author.}

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2. Definitions

For definitions and notations not defined here see [6]. Let V be a finite set and E a subset of

{

(

u

, v) |

(

u

, v)

is an unordered pair of V

}

. Then G

=

(

V

,

E

)

is a graph with vertex set V and edge set E. The order of G, that is, the cardinality of the set V , is denoted by n

(

G

)

. For a subset S of V the graph G

[

S

]

induced by S is a graph with vertex set

V

(

G

[

S

]

) =

S and edge set E

(

G

[

S

]

) = {(

x

,

y

) | (

x

,

y

) ∈

E

(

G

)

and x

,

y

∈

S

}

. Two vertices u and

v

are adjacent if

(

u

, v)

is an edge of G. For a vertex u the set NG

(

u

) = {v | (

u

, v) ∈

E

}

is called the set of neighbors of u. The degree degG

(

u

)

of a vertex u in G, is

the cardinality of the set NG

(

u

)

. The minimum degree of G,

δ(

G

)

, is min

{

degG

(

x

) |

x

∈

V

}

. A graph G is k-regular if degG

(

u

) =

k

for every vertex u in G. The connectivity of G is the minimum number of vertices whose removal leaves the remaining graph disconnected or trivial. A path between vertices

v

0and

v

kis a sequence of vertices represented by

h

v

0

, v

1

, . . . , v

k

i

such that

there is no repeated vertex and

(v

i

, v

i+1

)

is an edge of G for every i

∈ {

0

. . .

k

−

1

}

. We use Q

(

i

)

to denote the ith vertex

v

iof

Q

= h

v

₁

, v

₂

, . . . , v

_k

i

. We also write the path

h

v

₀

, v

₁

, . . . , v

_k

i

as

h

v

₀

, . . . , v

_i

,

Q

, v

j

, . . . , v

k

i

, where Q is a path form

v

ito

v

j. A

path is a hamiltonian path if it contains all vertices of G. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two distinct vertices of G. A cycle is a sequence of vertices represented by

h

v

₀

, v

₁

, . . . , v

_k

, v

₀

i

such that

v

i

6=

v

j

for all i

6=

j,

(v

0

, v

k

)

is an edge of G, and

(v

i

, v

i+1

)

is an edge of G for every i

∈ {

0

, . . . ,

k

−

1

}

. A hamiltonian cycle of G is a

cycle that traverses every vertex of G. A graph is hamiltonian if it has a hamiltonian cycle.

A hamiltonian cycle C of graph G is described as

h

u1

,

u2

, . . . ,

un(G)

,

u1

i

to emphasize the order of vertices in C . Thus,

u1 is the starting vertex and ui is the ith vertex in C . Two hamiltonian cycles C1

= h

u1

,

u2

, . . . ,

un(G)

,

u1

i

and C2

=

h

v

₁

, v

₂

, . . . , v

_n₍_G₎

, v

₁

i

of G are independent if u1

=

v

1and ui

6=

v

i for every i

∈ {

2

, . . . ,

n

(

G

)}

. A set of hamiltonian

cycles

{

C1

,

C2

, . . . ,

Ck

}

of G are mutually independent if its elements are pairwise independent. The mutually independent

hamiltonicity IHC

(

G

)

of graph G the maximum integer k such that for any vertex u of G there exist k mutually independent hamiltonian cycles of G starting at u. Obviously, IHC

(

G

) ≤ δ(

G

)

if G is a hamiltonian graph.

The mutually independent hamiltonicity of a graph can be interpreted as a Latin rectangle. A Latin square of order n is an

n

×

n array made from the integers 1 to n with the property that any integer occurs once in each row and column. If we delete

some rows from a Latin square, we will get a Latin rectangle. Let K5be the complete graph with vertex set

{

0

,

1

,

2

,

3

,

4

}

and

let C1

= h

0

,

1

,

2

,

3

,

4

,

0

i

, C2

= h

0

,

2

,

3

,

4

,

1

,

0

i

, C3

= h

0

,

3

,

4

,

1

,

2

,

0

i

, and C4

= h

0

,

4

,

1

,

2

,

3

,

0

i

. Obviously, C1, C2, C3, and

C4are mutually independent. Thus, IHC

(

K5

) =

4. We rewrite C1, C2, C3, and C4into the following Latin square:

1 2 3 4

2 3 4 1

3 4 1 2

4 1 2 3

In general, a Latin square of order n can be viewed as n mutually independent hamiltonian cycles with respect to the complete graph Kn+1.

Let H be a group and let S be the generating set of H such that S−1

₌

_{S. Then the Cayley graph Cayley}

₍

_S

_;

_H

₎

_{of the group H}

with respect to the generating set S is the graph with vertex set H and two vertex u and

v

are adjacent in Cayley

(

S

;

H

)

if and only if u−1

v ∈

S. Hamiltonian cycles in Cayley graphs naturally arise in computer science [25], in the study of word-hyperbolic groups and automatic groups [14], in changing-ringing [40], in creating Escher-like repeating patterns in hyperbolic plane [13], and in combinatorial designs [11].

3. Applications of the concept of mutually independent hamiltonian cycles

Mutually independent hamiltonicity of graphs can be applied to many areas. Consider the following scenario. In Christmas, we have a holiday of 10-days. A tour agency will organize a 10-day tour to Italy. Suppose that there will be a lot of people joining this tour. However, the maximum number of people stay in each local area is limited, say 100 people, for the sake of hotel contract. One trivial solution is on the First-Come-First-Serve basis. So only 100 people can attend this tour. (Note that we cannot schedule the tour in a pipelined manner because the holiday period is fixed.) Nonetheless, we observe that a tour is like a hamiltonian cycle based on a graph, in which a vertex is denoted as a hotel and any two vertices are joined with an edge if the associated two hotels can be traveled in a reasonable time. Therefore, we can organize several subgroups, that is, each subgroup has its own tour. In this way, we do not allow two subgroups stay in the same area during the same time period. In other words, any two different tours are indeed independent hamiltonian cycles. Suppose that there are 10 mutually independent hamiltonian cycles. Then we may allow 1000 people to visit Italy on Christmas vacation. For this reason, we would like to find the maximum number of mutually independent hamiltonian cycles. Such applications are useful for task scheduling and resource placement, which are also important for compiler optimization to exploit parallelism.

An interconnection network connects the processors of parallel computers. Its architecture can be represented as a graph in which the vertices correspond to processors and the edges correspond to connections. Hence, we use graphs and networks interchangeably. There are many mutually conflicting requirements in designing the topology for computer networks. The

n-cube is one of the most popular topologies [27]. The n-dimensional star network Snwas proposed in [1] as n attractive

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Fig. 1. The pancake graphs P2, P3, and P4.

has received considerable attention. Akers and Krishnameurthy [1] showed that the star graphs are vertex transitive and edge transitive. The diameter and fault diameters were computed in [1,26,37]. The hamiltonian and hamiltonian laceability of star graphs are studied in [16,17,21,23,31]. The spanning container of star graph is studied in [28].

Akers and Krishnameurthy [1] proposed another family of interesting interconnection networks, the n-dimensional pancake graph Pn. Hung et al. [22] studied the hamiltonian connectivity on the faulty pancake graphs. The embedding of

cycles and trees into the pancake graphs were discussed in [10,15,22,24]. The spanning container of pancake graph is studied in [28]. Gates and Papadimitriou [18] studied the diameter of the pancake graphs. Up to now, we do not know the exact value of the diameter of the pancake graphs [20].

4. The pancake graphs

Let n be a positive integer. We use

h

n

i

to denote the set

{

1

,

2

, . . . ,

n

}

. The n-dimensional pancake graph, Pn, is a graph with

the vertex set V

(

Pn

) = {

u1u2

. . .

un

|

ui

∈ h

n

i

and uj

6=

ukfor j

6=

k

}

. The adjacency is defined as follows: u1u2

. . .

ui

. . .

un

is adjacent to

v

1

v

2

. . . v

i

. . . v

nthrough an edge of dimension i with 2

≤

i

≤

n if

v

j

=

ui−j+1for all 1

≤

j

≤

i and

v

j

=

uj

for all i

<

j

≤

n. We will use boldface to denote a vertex of Pn. Hence, u1

,

u2

, . . . ,

undenote a sequence of vertices in Pn. In

particular, e denotes the vertex 12

. . .

n. The pancake graphs P2, P3, and P4are illustrated inFig. 1.

By definition, Pnis an

(

n

−

1

)

-regular graph with n

!

vertices. Akers and Krishnameurthy [1] showed that the connectivity

of Pnis

(

n

−

1

)

. Let u

=

u1u2

. . .

unbe an arbitrary vertex of Pn. We use

(

u

)

ito denote the ith component uiof u, and use P

{i}

n

to denote the ith subgraph of Pninduced by those vertices u with

(

u

)

n

=

i. Then Pncan be decomposed into n vertex disjoint

subgraphs Pn{i}, 1

≤

i

≤

n, and each P

{i}

n is isomorphic to Pn−1for all i, i

≤

n. Thus, the pancake graph can be constructed

recursively. Let H be any subset of

h

n

i

. We use PH

n to denote the subgraph of Pninduced by

∪

i∈HV

(

P

{i}

n

)

. By definition, there

is exactly one neighbor v of u such that u and v are adjacent through an i-dimensional edge with 2

≤

i

≤

n. We use

(

u

)

i_to

denote the unique i-neighbor of u. We have

((

u

)

i

₎

i

₌

_{u and}

₍

_u

₎

n

_∈

_P{(u)1}

n . For any two distinct elements i and j in

h

n

i

, we

use Eni,jto denote the set of edges between P

{i}

n and P

{j}

n .

Lemma 1. Let i and j be any two distinct elements in

h

n

i

with n

≥

3. Then

|

Eni,j

| =

(

n

−

2

)!

.

Lemma 2. Let u and v be any two distinct vertices of P_nwith d

(

u

,

v

) ≤

2. Then

(

u

)

₁

6=

(

v

)

₁.

Theorem 1 ([22]). Suppose that F is a subset of V

(

Pn

)

with

|

F

| ≤

n

−

4. Then Pn

−

F is hamiltonian connected.

Theorem 2. Let

{

a1

,

a2

, . . . ,

ar

}

be a subset of

h

n

i

for some positive integer r

∈ h

n

i

with n

≥

5. Assume that u and v are two

distinct vertices of Pnwith u

∈

P

{a1}

n and v

∈

P

{ar}

n . Then there is a hamiltonian path

h

u

=

x1

,

H1

,

y1

,

x2

,

H2

,

y2

, . . . ,

xr

,

Hr

,

yr

=

v

i

of

∪

r i=1P

{ai}

n joining u to v such that x1

=

u, yr

=

v, and Hiis a hamiltonian path of P

{ai}

n joining xito yifor every i, 1

≤

i

≤

r.

Proof. We set x1as u and yras v. We know that P {ai}

n is isomorphic to Pn−1for every i

∈ h

r

i

. ByTheorem 1, this statement

holds for r

=

1. Thus, we assume that r

≥

2. ByLemma 1,

|

Eai,ai+1

n

| =

(

n

−

2

)! ≥

6 for every i

∈ h

r

−

1

i

. We choose

(

y_i

,

xi+1

) ∈

E

ai,ai+1

n for every i

∈ h

r

−

1

i

with y1

6=

x1and xr

6=

yr. ByTheorem 1, there is a hamiltonian path Hiof P

{ai}

n

joining xito yifor every i

∈ h

r

i

. Then

h

u

=

x1

,

H1

,

y1

,

x2

,

H2

,

y2

, . . . ,

xr

,

Hr

,

yr

=

v

i

is the desired path. SeeFig. 2for

illustration on Pn.

Lemma 3. Let k

∈ h

n

i

with n

≥

4, and let x be a vertex of Pn. There is a hamiltonian path P of Pn

− {

x

}

joining the vertex

(

x

)

n

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Fig. 2. Illustration forTheorem 2on Pn.

Proof. Suppose that n

=

4. Since P4is vertex transitive, we may assume that x

=

1234. The required paths of P4

− {

1234

}

are listed below:

k=1 h4321,3421,2431,4231,1324,3124,2134,4312,1342,2143,4132,2314,3214,4123,2143,3412,1432,2341,3241,1423,2413,4213,1243i k=2 h4321,3421,2431,4231,1324,3124,2134,4312,1342,3142,2413,4213,1243,2143,3412,1432,4132,2314,3214,4123,1423,3241,2341i k=3 h4321,3421,2431,4231,1324,3124,2134,4312,1342,3142,4132,2314,3214,4123,1423,2413,4213,1342,2143,3412,1432,2341,3241i k=4 h4321,3421,2431,1342,3142,4132,2314,3214,4123,2143,1243,4213,2413,1423,3241,2341,1432,3412,4312,2134,3124,1324,4231i

WithTheorem 1, we can find the required hamiltonian path in Pnfor every n, n

≥

5.

Lemma 4. Let a and b be any two distinct elements in

h

n

i

with n

≥

4, and let x be a vertex of Pn. There is a hamiltonian path P

of Pn

− {

x

}

joining a vertex u with

(

u

)

1

=

a to a vertex v with

(

v

)

1

=

b.

Proof. Suppose that n

=

4. Since P4is vertex transitive, we may assume that x

=

1234. Without loss of generality, we may

assume that a

<

b. The required paths of P4

− {

1234

}

are listed below: a=1 and b=2 h1423,4123,3214,2314,1324,3124,4213,2413,3142,4132,1432,3412,2143,1243,3421,4321,2341,3241,4231,2431,1342,4312,2134i a=1 and b=3 h1423,4123,2143,1243,4213,2413,3142,1342,2431,3421,4321,2341,3241,4231,1324,3124,2134,4312,3412,1432,4132,2314,3214i a=1 and b=4 h1423,2413,3142,1342,2431,3421,4321,2341,3241,4231,1324,2314,3214,4123,2143,1243,4213,3124,2134,4312,3412,1432,4132i a=2 and b=3 h2134,4312,1342,3142,2413,4213,1243,2143,3412,1432,4132,2314,3214,4123,1423,3241,2341,4321,3421,2431,4231,1324,3124i a=2 and b=4 h2134,3124,1324,2314,3214,4123,2143,1243,4213,2413,1423,3241,4231,2431,3421,4321,2341,1432,3412,4312,1342,3142,4132i a=3 and b=4 h3214,4123,2143,1243,4213,3124,2134,4312,3412,1432,2341,4321,3421,2431,1342,3142,2413,1423,3241,4231,1324,2314,4132i

WithTheorem 1, we can find the required hamiltonian path on Pnfor every n, n

≥

5.

h

n

i

with n

≥

4. Assume that x and y are two adjacent vertices of Pn. There

is a hamiltonian path P of Pn

− {

x

,

y

}

(

u

)

1

=

(

v

)

1

=

b.

Proof. Since Pnis vertex transitive, we may assume that x

=

e and y

=

(

e

)

ifor some i

∈ {

2

,

3

, . . . ,

n

}

. Without loss of

generality, we assume that a

<

b. Thus, a

6=

n and b

6=

1. We prove this statement by induction on n. For n

=

4, the required paths of P4

− {

1234

, (

1234

)

i

}

are listed below:

y=2134 a=1 and b=2 h1432,2413,3142,4132,1432,3412,4312,1342,2431,3421,4321,2341,3241,4231,1324,3124,4213,1243,2143,4123,3214,2314i a=1 and b=3 h1432,4123,2143,1243,3421,4321,2341,3241,4231,2431,1342,4312,3412,1432,4132,3142,2413,4213,3124,1324,2314,3214i a=1 and b=4 h1432,4123,3214,2314,1324,3124,4213,2413,3142,4132,1432,2341,3241,4231,2431,1342,4312,3412,2143,1243,3421,4321i a=2 and b=3 h2314,3214,4123,2143,1243,4213,3124,1324,4231,2431,1342,4312,3412,1432,4132,3142,2413,1423,3241,2341,4321,3421i a=2 and b=4 h2314,3214,4123,2143,3412,4312,1342,2431,3421,1243,4213,3124,1324,4231,3241,1423,2413,3142,4132,1432,2341,4321i a=3 and b=4 h3214,4123,2143,1243,3421,2431,4231,3241,1423,2413,4213,3124,1324,2314,4132,3142,1342,4312,3412,1432,2341,4321i y=3214 a=1 and b=2 h1423,4123,2143,1243,3421,4321,2341,3241,4231,2431,1342,3142,2413,4213,3124,1324,2314,4132,1432,3412,4312,2134i a=1 and b=3 h1423,4123,2143,1243,4213,2413,3142,1342,2431,3421,4321,2341,3241,4231,1324,2314,4132,1432,3412,4312,2134,3124i a=1 and b=4 h1423,4123,2143,1243,3421,2431,1342,3142,2413,4213,3124,2134,4312,3412,1432,4132,2314,1324,4231,3241,2341,4321i a=2 and b=3 h2134,4312,1342,2431,4231,3241,1423,4123,2143,3412,1432,2341,4321,3421,1243,4213,2413,3142,4132,2314,1324,3124i a=2 and b=4 h2134,3124,4213,2413,3142,1342,4312,3412,1432,4132,2314,1324,4231,2431,3421,1243,2143,4123,1423,3241,2341,4321i a=3 and b=4 h3124,2134,4312,1342,3142,2413,4213,1243,3421,2431,4231,1324,2314,4132,1432,3412,2143,4123,1423,3241,2341,4321i

(5)

y=4321 a=1 and b=2 h1423,4123,3214,2314,4132,3142,2413,4213,3124,1324,4231,3241,2341,1432,3412,2143,1243,3421,2431,1342,4312,2134i a=1 and b=3 h1423,4123,2143,3412,1432,2341,3241,4231,1324,3124,2134,4312,1342,2431,2431,1243,4213,2413,3142,4132,2314,3214i a=1 and b=4 h1423,2413,4213,3124,2134,4312,3412,2143,1243,3421,2431,1342,3142,4132,1432,2341,3241,4231,1324,2314,3214,4123i a=2 and b=3 h2134,4312,1342,3142,4132,2314,3214,4123,2143,3412,1432,2341,3241,1423,2413,4213,1243,3421,2431,4231,1324,3124i a=2 and b=4 h2134,3124,4213,2413,1423,3241,2341,1432,4132,3142,1342,4312,3412,2143,1243,3421,2431,4231,1324,2314,3214,4123i a=3 and b=4 h3214,2314,1324,4231,3241,2341,1432,4132,3142,1342,2431,3421,1243,2143,3412,4312,2134,3124,4213,2413,1423,4123i

Suppose that this statement holds for Pkfor every k, 4

≤

k

<

n. We have the following cases:

Case 1. y

=

(

e

)

i_{for some i}

₆₌

_{1 and i}

₆₌

_{n, that is, y}

_∈

_P{n}

n . Let c be an element in

h

n

−

1

i − {

a

}

. By induction, there is

a hamiltonian path R of Pn{n}

− {

e

, (

e

)

i

}

(

u

)

1

=

a to a vertex z with

(

z

)

1

=

c. We choose a vertex

v in Pnhn−1i−{c}with

(

v

)

1

=

b. ByTheorem 2, there is a hamiltonian path H of P hn−1i

n joining the vertex

(

z

)

n to v. Then

h

u

,

R

,

z

, (

z

)

n

_,

_H

_,

_v

_i

_{is the desired path.}

Case 2. y

=

(

e

)

n_{, that is, y}

_∈

_P{1}

n . Let c be an element in

h

n

−

1

i − {

1

,

a

}

, and let d be an element in

h

n

−

1

i − {

1

,

b

,

c

}

. By

Lemma 4, there is a hamiltonian path R of Pn{n}

− {

e

}

(

u

)

1

=

a to a vertex w with

(

w

)

1

=

c. Again,

there is a hamiltonian path H of Pn{1}

− {

(

e

)

n

}

joining a vertex z with

(

z

)

1

=

d to a vertex v with

(

v

)

1

=

b. ByTheorem 2,

there is a hamiltonian path Q of Pnhn−1i−{1}joining the vertex

(

w

)

nto the vertex

(

z

)

n. Then

h

u

,

R

,

w

, (

w

)

n

,

Q

, (

z

)

n

,

z

,

H

,

v

i

is

the desired path.

h

n

i

with n

≥

4. Let x be any vertex of Pn. Assume that x1and x2are two

distinct neighbors of x. There is a hamiltonian path P of Pn

− {

x

,

x1

,

x2

}

(

u

)

1

=

(

v

)

1

=

b.

Proof. Since Pnis vertex transitive, we may assume that x

=

e. Moreover, we assume that x1

=

(

e

)

iand x2

=

(

e

)

jfor some

{

i

,

j

} ⊂ h

n

i − {

1

}

with i

<

j. Without loss of generality, we assume that a

<

b. Thus, a

6=

n and b

6=

1. We prove this lemma by induction on n. For n

=

4, the required paths of P4

− {

1234

, (

1234

)

i

, (

1234

)

j

}

are listed below:

x1=2134 and x2=3214 a=1 and b=2 h1423,4123,2143,1243,3421,4321,2341,3241,4231,2431,1342,4312,3412,1432,4132,3142,2413,4213,3124,1324,2314i a=1 and b=3 h1423,4123,2143,1243,3421,4321,2341,3241,4231,2431,1342,4312,3412,1432,4132,2314,1324,3124,4213,2413,3142i a=1 and b=4 h1423,4123,2143,1243,3421,4321,2341,3241,4231,2431,1342,3142,2413,4213,3124,1324,2314,4132,1432,3412,4312i a=2 and b=3 h2143,4123,1423,3241,4231,2431,1342,4312,3412,1432,2341,4321,3421,1243,4213,2413,3142,4132,2314,1324,3124i a=2 and b=4 h2143,4123,1423,2413,3142,1342,4312,3412,1432,4132,2314,1324,3124,4213,1243,3421,2431,4231,3241,2341,4321i a=3 and b=4 h3124,4213,2413,3142,1342,4312,3412,1432,4132,2314,1324,4231,2431,3421,1243,2143,4123,1423,3241,2341,4321i x1=2134 and x2=4321 a=1 and b=2 h1423,2413,3142,4132,1432,2341,3241,4231,1324,3124,4213,1243,3421,2431,1342,4312,3412,2143,4123,3214,2314i a=1 and b=3 h1423,4123,2143,1243,3421,2431,1342,4312,3412,1432,2341,3241,4231,1342,3124,4213,2413,3142,4132,2314,3214i a=1 and b=4 h1423,4123,3214,2314,1324,3124,4213,2413,3142,1342,4312,3412,2143,1243,3421,2431,4231,3241,2341,1432,4132i a=2 and b=3 h2314,3214,4123,2143,3412,4312,1342,3142,4132,1432,2341,3241,1423,2413,4213,1243,3421,2431,4231,1324,3124i a=2 and b=4 h2314,3214,4123,2143,1243,3421,2431,4231,1324,3124,4213,2413,1423,3241,2341,1432,3412,4312,1342,3142,4132i a=3 and b=4 h3214,2314,4132,3142,1342,4312,3412,1432,2341,3241,1423,2413,4213,3124,1324,4231,2431,3421,1243,2143,4123i

(6)

x1=3214 and x2=4321 a=1 and b=2 h1423,4123,2143,1243,3421,2431,1342,4312,3412,1432,2341,3241,4231,1324,2314,4132,3142,2413,4213,3124,2134i a=1 and b=3 h1423,4123,2143,3412,1432,2341,3241,4231,1324,2314,4132,3142,2413,4213,1243,3421,2431,1342,4312,2134,3124i a=1 and b=4 h1423,2413,4213,3124,2134,4312,3412,1432,2341,3241,4231,1324,2314,4132,3142,1342,2431,3421,1243,2143,4123i a=2 and b=3 h2134,4312,3412,1432,2341,3241,4231,1324,2314,4132,3142,1341,2431,3421,1243,2143,4123,1423,2413,4213,3124i a=2 and b=4 h2134,3124,4213,2413,3142,1342,4312,3412,1432,2341,3241,1423,4123,2143,1243,3421,2431,4231,1324,2314,4132i a=3 and b=4 h3124,2134,4312,3412,1432,2341,3241,4231,1324,2314,4132,3142,1342,2431,3421,1243,2143,4123,1423,2413,4213i

Suppose that this statement holds for Pkfor every k, 4

≤

k

<

n. We have the following cases:

Case 1. j

6=

n, that is, x1

∈

P {n}

n and x2

∈

P {n}

n . Let c

∈ h

n

−

1

i − {

1

,

a

}

. By induction, there is a hamiltonian path R of

Pn{n}

− {

e

,

x1

,

x2

}

(

u

)

1

=

(

z

)

1

=

c. We choose a vertex v in P {1}

n with

(

v

)

1

=

b.

ByTheorem 2, there is a hamiltonian path H of Pnhn−1ijoining the vertex

(

z

)

nto v. We set P

= h

u

,

R

,

z

, (

z

)

n

,

H

,

v

i

. Then P is

the desired path.

Case 2. j

=

n, that is, x1

∈

P {n}

n and x2

∈

P {1}

n . Let c

∈ h

n

−

1

i − {

1

,

a

}

and d

∈ h

n

−

1

i − {

1

,

b

,

c

}

. ByLemma 5, there is

a hamiltonian path R of Pn{n}

− {

e

,

x1

}

(

u

)

1

=

(

z

)

1

=

c. ByLemma 4, there is a

hamiltonian path H of Pn{1}

− {

x2

}

joining a vertex w with

(

w

)

1

=

d to a vertex v with

(

v

)

1

=

b. ByTheorem 2, there is a

hamiltonian Q of Pnhn−1i−{1}joining the vertex

(

z

)

nto the vertex

(

w

)

n. We set P

= h

u

,

R

,

z

, (

z

)

n

,

Q

, (

w

)

n

,

w

,

H

,

v

i

. Then P is

the desired path.

Our main result for the pancake graph Pnis stated in the following theorem.

Theorem 3. IHC

(

P3

) =

1 and IHC

(

Pn

) =

n

−

1 if n

≥

4.

Proof. It is easy to see that P3is isomorphic to a cycle with six vertices. Thus, IHC

(

P3

) =

1. Since Pnis

(

n

−

1

)

-regular

graph, it is clear that IHC

(

Pn

) ≤

n

−

1. Since Pnis vertex transitive, we only need to show that there exist

(

n

−

1

)

mutually

independent hamiltonian cycles of Pnstarting form the vertex e. For n

=

4, we prove that IHC

(

P4

) ≥

3 by listing the required

hamiltonian cycles as follows: C1= h1234,2134,4312,3412,2143,1243,4213,3124,1324,4231,3241,2341,1432,4132,2314,3214,4123,1423,2413,3142,1342,2431,3421,4321,1234i C2= h1234,3214,2314,1324,3124,4213,2413,1423,4123,2143,1243,3421,4321,2341,3241,4231,2431,1342,3142,4132,1432,3412,4312,2134,1234i C3= h1234,4321,2341,1432,4132,2314,1324,4231,3241,1423,2413,3142,1342,2431,3421,1243,4213,3124,2134,4312,3412,2143,4123,3214,1234i

Suppose that n

≥

5. Let B be the

(

n

−

1

) ×

n matrix with bi,j

=

i

+

j

−

1 if i

+

j

−

1

≤

n

,

i

+

j

−

n

+

1 if n

≥

i

+

j

.

More precisely, B

=







1 2 3 4

· · ·

n

−

1 n 2 3 4 5

· · ·

n 1

... ... ... ... ... ...

...

n

−

1 n 1 2

· · ·

n

−

3 n

−

2







.

It is not hard to see that bi,1bi,2

. . .

bi,nforms a permutation of

{

1

,

2

, . . . ,

n

}

for every i with 1

≤

i

≤

n

−

1. Moreover,

bi,j

6=

bi0,jfor any 1

≤

i

<

i0

≤

n

−

1 and 1

≤

j

≤

n. In other words, B forms a Latin rectangle with entries in

{

1

,

2

, . . . ,

n

}

.

For every k

∈ h

n

−

1

i

, we construct Ckas follows:

(

1

)

k

=

1. ByLemma 3, there is a hamiltonian path H1of P {b1,n}

n

−{

e

}

joining a vertex x with x

6=

(

e

)

n−1and

(

x

)

1

=

n

−

1 to

the vertex

(

e

)

n−1_{. By}_{Theorem 2, there is a hamiltonian path H}

2of

∪

nt=−11P {b1,t}

(

e

)

nto the vertex

(

x

)

nwith

H2

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

P {b1,j}

n for every i

∈ h

(

n

−

1

)!i

and for every j

∈ h

n

−

1

i

. We set C1

= h

e

, (

e

)

n

,

H2

, (

x

)

n

,

x

,

H1

, (

e

)

n−1

,

e

i

.

(

2

)

k

=

2. ByLemma 5, there is a hamiltonian path Q1of P {b2,n−1}

n

− {

e

, (

e

)

2

}

joining a vertex y with

(

y

)

1

=

n

−

1 to a

vertex z with

(

z

)

1

=

1. ByTheorem 2, there is a hamiltonian Q2of

∪

nt=−12P {b2,t}

((

e

)

2

)

nto the vertex

(

y

)

n

such that Q2

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

P {b2,j}

n for every i

∈ h

(

n

−

1

)!i

and for every j

∈ h

n

−

2

i

. ByTheorem 1, there is a hamiltonian

path Q3of P {b2,n}

(

z

)

nto the vertex

(

e

)

n. We set C2

= h

e

, (

e

)

2

, ((

e

)

2

)

n

,

Q2

, (

y

)

n

,

y

,

Q1

,

z

, (

z

)

n

,

Q3

, (

e

)

n

,

e

i

.

(

3

)

k

∈ h

n

−

1

i − {

1

,

2

}

. ByLemma 6, there is a hamiltonian path Rk

1of P {bk,n−k+1}

n

− {

e

, (

e

)

k−1

, (

e

)

k

}

joining a vertex

wk with

(

wk

)

1

=

n

−

1 to a vertex vk with

(

vk

)

1

=

1. ByTheorem 2, there is a hamiltonian path Rk2of

∪

n−k t=1P

{bk,t}

(7)

Fig. 3. Illustration forTheorem 3on P5. joining the vertex

((

e

)

k

₎

n_{to the vertex}

₍

_w

k

)

nsuch that Rk2

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

P {bk,j}

n for every i

∈ h

(

n

−

1

)!i

and

for every j

∈ h

n

−

k

i

. Again, there is a hamiltonian path R₃k of

∪

n_t₌_n₋_k₊₂Pn{bk,t}joining the vertex

(

vk

)

n to the vertex

((

e

)

k−1

₎

n_{such that R}k

3

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

P

{bk,n−k+j+1}

n for every i

∈ h

(

n

−

1

)!i

and for every j

∈ h

k

−

1

i

. We set

Ck

= h

e

, (

e

)

k

, ((

e

)

k

)

n

,

Rk2

, (

wk

)

n

,

wk

,

Rk1

,

vk

, (

vk

)

n

,

Rk3

, ((

e

)

k

−1

₎

n

_{, (}

_e

₎

k−1

_,

_e

_i

_.

Then

{

C1

,

C2

, . . . ,

Cn−1

}

forms a set of

(

n

−

1

)

mutually independent hamiltonian cycles of Pnstarting from the vertex e.

Example. We illustrate the proof ofTheorem 3with n

=

5 as follows: We set B

=







1 2 3 4 5 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3





 .

Then we construct

{

C1

,

C2

,

C3

,

C4

}

as follows:

(

1

)

k

=

1. ByLemma 3, there is a hamiltonian path H1of P {b1,5}

5

− {

e

}

joining a vertex x with x

6=

(

e

)

4_and

₍

_x

₎

1

=

4 to

the vertex

(

e

)

4_{. By}_{Theorem 2, there is a hamiltonian path H}

2of

∪

4t=1P {b1,t}

5 joining the vertex

(

e

)

5to the vertex

(

x

)

5with

H2

(

i

+

24

(

j

−

1

)) ∈

P {b1,j}

5 for every i

∈ h

24

i

and for every j

∈ h

4

i

. We set C1

= h

e

, (

e

)

5

,

H2

, (

x

)

5

,

x

,

H1

, (

e

)

4

,

e

i

.

(

2

)

k

=

2. ByLemma 5, there is a hamiltonian path Q1of P {b2,4}

5

− {

e

, (

e

)

2

_}

_{joining a vertex y with}

₍

_y

₎

1

=

4 to a vertex z

with

(

z

)

1

=

∪

3t=1P {b2,t}

((

e

)

2

)

5to the vertex

(

y

)

5such that

Q2

(

i

+

24

(

j

−

1

)) ∈

P {b2,j}

5 for every i

∈ h

24

i

and for every j

∈ h

3

i

. ByTheorem 1, there is a hamiltonian path Q3of P {b2,5}

5

joining the vertex

(

z

)

5_{to the vertex}

₍

_e

₎

5_{. We set C}

2

= h

e

, (

e

)

2

, ((

e

)

2

)

5

,

Q2

, (

y

)

5

,

y

,

Q1

,

z

, (

z

)

5

,

Q3

, (

e

)

5

,

e

i

.

(

3

)

k

∈ {

3

,

4

}

. ByLemma 6, there is a hamiltonian path Rk₁of P₅{bk,6−k}

−{

e

, (

e

)

k−1

_{, (}

_e

₎

k

_}

_{joining a vertex w}

kwith

(

wk

)

1

=

4

to a vertex vkwith

(

vk

)

1

=

∪

5−k t=1P

{bk,t}

((

e

)

k

)

5to the

vertex

(

w_k

)

5_{such that R}k

2

(

i

+

24

(

j

−

1

)) ∈

P {bk,j}

5 for every i

∈ h

24

i

and for every j

∈ h

5

−

k

i

. Again, there is a hamiltonian

path Rk₃of

∪

5_t₌₇₋_kP₅{bk,t}joining the vertex

(

vk

)

5to the vertex

((

e

)

k−1

)

5such that Rk3

(

i

+

24

(

j

−

1

)) ∈

P {bk,6−k+j}

5 for every

i

∈ h

24

i

and for every j

∈ h

k

−

1

i

. We set Ck

= h

e

, (

e

)

k

, ((

e

)

k

)

5

,

Rk2

, (

wk

)

5

,

wk

,

Rk1

,

vk

, (

vk

)

5

,

Rk3

, ((

e

)

k

−1

₎

5

_{, (}

_e

₎

k−1

_,

_e

_i

_.

Then

{

C1

,

C2

,

C3

,

C4

}

forms a set of 4 mutually independent hamiltonian cycles of P5starting from the vertex e. SeeFig. 3

for illustration. 5. The star graphs

Let n be a positive integer. The n-dimensional star graph, Sn, is a graph with the vertex set V

(

Sn

) = {

u1

. . .

un

|

ui

∈ h

n

i

and uj

6=

ukfor j

6=

k

}

. The adjacency is defined as follows: u1

. . .

ui

. . .

unis adjacent to

v

1

. . . v

i

. . . v

nthrough an edge of

dimension i with 2

≤

i

≤

n if

v

j

=

ujfor every j

∈ h

n

i − {

1

,

i

}

,

v

1

=

ui, and

v

i

=

u1. The star graphs S2, S3, and S4are

illustrated inFig. 4. In [1], it showed that the connectivity of Snis

(

n

−

1

)

. We use boldface to denote vertices in Sn. Hence,

u₁

,

u₂

, . . . ,

u_ndenotes a sequence of vertices in Sn.

By definition, Snis an

(

n

−

1

)

-regular graph with n

!

vertices. We use e to denote the vertex 12

. . .

n. It is known that Sn

is a bipartite graph with one partite set containing the vertices corresponding to odd permutations and the other partite set containing those vertices correspond to even permutations. We use white vertices to represent those even permutation vertices and we use black vertices to represent those odd permutation vertices. Let u

=

u1u2

. . .

unbe an arbitrary vertex

of the star graph Sn. We say that uiis the ith coordinate of u,

(

u

)

i, for 1

≤

i

≤

n. For 1

≤

i

≤

n, let S

{i}

n be the subgraph of

Sninduced by those vertices u with

(

u

)

n

=

i. Then Sncan be decomposed into n subgraph S

{i}

n , 1

≤

i

≤

n, and each S

{i}

n is

(8)

Fig. 4. The star graphs S2, S3, and S4. the subgraph of Sninduced by

∪

i∈IV

(

S

{i}

n

)

. For any two distinct elements i and j in

h

n

i

, we use E i,j

n to denote the set of edges

between Sn{i}and S

{j}

n . By the definition of Sn, there is exactly one neighbor v of u such that u and v are adjacent through an

i-dimensional edge with 2

≤

i

≤

n. For this reason, we use

(

u

)

i_{to denote the unique i-neighbor of u. We have}

₍₍

_u

₎

i

₎

i

₌

_u

and

(

u

)

n

_∈

_S{(u)1}

n .

Lemma 7. Let i and j be any two distinct elements in

h

n

i

with n

≥

3. Then

|

Eni,j

| =

(

n

−

2

)!

. Moreover, there are

(

n

−

2

)!/

2

edges joining black vertices of Sn{i}to white vertices of S

{j}

n .

Lemma 8. Let u and v be two distinct vertices of Snwith d

(

u

,

v

) ≤

2. Then

(

u

)

1

6=

(

v

)

1.

Theorem 4 ([21]). Let n

≥

4. Suppose that u is a white vertex of Snand v is a black vertex of Sn. Then there is a hamiltonian

path of Snjoining u to v.

Theorem 5. Let

{

a1

,

a2

, . . . ,

ar

}

be a subset of

h

n

i

for some r

∈ h

n

i

with n

≥

5. Assume that u is a white vertex in S

{a1}

n and v

is a black vertex in S{ar}

n . Then there is a hamiltonian path

h

u

=

x1

,

H1

,

y1

,

x2

,

H2

,

y2

, . . . ,

xr

,

Hr

,

yr

=

v

i

of

∪

ri=1S {ai}

n joining u

to v such that x1

=

u, yr

=

v, and Hiis a hamiltonian path of S

{ai}

n joining xito yifor every i, 1

≤

i

≤

r.

Proof. We set x1as u and yras v. ByTheorem 4, this theorem holds on r

=

1. Suppose that r

≥

2. ByLemma 7, there are

(

n

−

2

)!/

2

≥

3 edges joining black vertices of S{ai}

n to white vertices of S

{ai+1}

n for every i

∈ h

r

−

1

i

. We can choose an edge

(

y_i

,

xi+1

) ∈

E

ai,ai+1

n with yibeing a black vertex and xi+1being a white vertex for every i

∈ h

r

−

1

i

. ByTheorem 4, there is a

hamiltonian path Hiof S

{ai}

n joining xito yifor every i

∈ h

r

i

. Then the path

h

u

=

x1

,

H1

,

y1

,

x2

,

H2

,

y2

, . . . ,

xr

,

Hr

,

yr

=

v

i

is

the desired path.

Theorem 6 ([21]). Let w be a black vertex of Snwith n

≥

4. Assume that u and v are two distinct white vertices of Sn

− {

w

}

.

Then there is a hamiltonian path H of Sn

− {

w

}

joining u to v.

Lemma 9 ([30]). Let i be any element in

h

n

i

with n

≥

4. Assume that r and s are two adjacent vertices of Snand u is a white

vertex of Sn

− {

r

,

s

}

. Then there is a hamiltonian path of Sn

− {

r

,

s

}

joining u to some black vertex v with

(

v

)

1

=

i.

h

n

i

with n

≥

4. Assume that x is a white vertex of Sn, and assume that

x1and x2are two distinct neighbors of x. Then there is a hamiltonian path P of Sn

− {

x

,

x1

,

x2

}

joining a white vertex u with

(

u

)

1

=

a to a white vertex v with

(

v

)

1

=

b.

Proof. Since Snis vertex transitive and edge transitive, we may assume that x

=

e, x1

=

(

e

)

2, and x2

=

(

e

)

3. Without loss

of generality, we may also assume that a

<

b. We have a

6=

n and b

6=

1. We prove this statement by induction on n. For

n

=

4, the required paths of S4

− {

1234

,

2134

,

3214

}

are listed below:

a=1 and b=2 h1324,3142,4132,1432,3412,4312,2314,1324,3124,4123,2143,1243,4213,2413,1423,3421,4321,2341,3241,4231,2431i a=1 and b=3 h1423,2413,4213,1243,2143,4123,3124,1324,2314,4312,3412,1432,4132,3142,1342,2341,4321,3421,2431,4231,3241i a=1 and b=4 h1324,3142,4132,1432,3412,4312,2314,1324,3124,4123,2143,1243,4213,2413,1423,3421,2431,4231,3241,2341,4321i a=2 and b=3 h2314,1324,3124,4123,2143,1243,4213,2413,1423,3421,4321,2341,3241,4231,2431,1432,4132,3142,1342,4312,3412i a=2 and b=4 h2314,1324,3124,4123,2143,1243,4213,2413,1423,3421,4321,2341,3241,4231,2431,1432,3412,4312,1342,3142,4132i a=3 and b=4 h3124,1324,2314,4312,3412,1432,4132,3142,1342,2341,4321,3421,2431,4231,3241,1243,2143,4123,1423,2413,4213i

(9)

Suppose that this statement holds for Skfor every k, 4

≤

k

≤

n

−

1. Let c be any element in

h

n

−

1

i − {

1

,

a

}

. By induction,

there is a hamiltonian path H of Sn{n}

− {

e

, (

e

)

2

, (

e

)

3

}

joining a white vertex u with

(

u

)

1

=

a to a white vertex z with

(

z

)

1

=

c.

We choose a white vertex v in Sn{1}with

(

v

)

1

=

b. ByTheorem 5, there is a hamiltonian path R of S hn−1i

n joining the black

vertex

(

z

)

n_{to v. Then}

_h

_u

_,

_H

_,

_z

_{, (}

_z

₎

n

_,

_R

_,

_v

_i

_{is the desired path of S}

n

− {

e

, (

e

)

2

, (

e

)

3

}

.

The following theorem is our main result for the star graph Sn.

Theorem 7. IHC

(

S3

) =

1, IHC

(

S4

) =

2, and IHC

(

Sn

) =

n

−

1 if n

≥

5.

Proof. It is easy to see that S3is isomorphic to a cycle with six vertices. Thus, IHC

(

S3

) =

1. Using a computer, we have

IHC

(

S4

) =

2 by brute force checking. Thus, we assume that n

≥

5. We know that Sn is

(

n

−

1

)

-regular graph. Hence,

IHC

(

Sn

) ≤

n

−

1. Since Snis vertex transitive, we only need to show that there are

(

n

−

1

)

mutually independent hamiltonian

cycles of Snstarting from e. Let B be the

(

n

−

1

) ×

n matrix with

bi,j

=

i

+

j

−

1 if i

+

j

−

1

≤

n

,

i

+

j

−

n

+

1 if n

<

i

+

j

−

1

.

We construct

{

C1

,

C2

, . . . ,

Cn−1

}

as follows:

(

1

)

k

=

1. We choose a black vertex x in Sn{b1,n}

− {

(

e

)

n−1

}

with

(

x

)

1

=

n

−

1. ByTheorem 6, there is a hamiltonian path

H1of S {b1,n}

n

− {

e

}

joining x to the black vertex

(

e

)

n−1. ByTheorem 5, there is a hamiltonian path H2of

∪

n −1

t=1S {b1,t}

n joining

the black vertex

(

e

)

nto the white vertex

(

x

)

nwith H2

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

S {b1,j}

n for every i

∈ h

(

n

−

1

)!i

and for every

j

∈ h

n

−

1

i

. We set C1

= h

e

, (

e

)

n

,

H2

, (

x

)

n

,

x

,

H1

, (

e

)

n−1

,

e

i

.

(

2

)

k

=

2. We choose a white vertex y in S{nb2,n−1}

− {

e

, (

e

)

2

}

with

(

y

)

1

=

n

−

1. ByLemma 9, there is a hamiltonian path

Q1of S {b2,j}

n

− {

e

, (

e

)

2

}

joining y to a black vertex z with

(

z

)

1

=

∪

n −2

t=1S {b2,t}

n

joining the white vertex

((

e

)

2

₎

n_{to the black vertex}

₍

_y

₎

n_{such that Q}

2

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

S {b2,j}

n for every i

∈ h

(

n

−

1

)!i

and

for every j

∈ h

n

−

2

i

. Again, there is a hamiltonian path Q3of S {b2,n}

n joining the white vertex

(

z

)

nto the black vertex

(

e

)

n.

We set C2

= h

e

, (

e

)

2

, ((

e

)

2

)

n

,

Q2

, (

y

)

n

,

y

,

Q1

,

z

, (

z

)

n

,

Q3

, (

e

)

n

,

e

i

.

(

3

)

3

≤

k

≤

n

−

1. ByLemma 10, there is a hamiltonian path R₁kof Sn{bk,n−k+1}

− {

e

, (

e

)

k−1

, (

e

)

k

}

joining a white vertex wk

with

(

wk

)

1

=

n

−

1 to a white vertex vkwith

(

vk

)

1

=

∪

n−k t=1S

{bk,t}

n joining

the white vertex

((

e

)

k

₎

n_{to the black vertex}

₍

_w

k

)

nsuch that Rk2

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

S {bk,j}

n for every i

∈ h

(

n

−

1

)!i

and

for every j

∈ h

n

−

k

−

1

i

. Again, there is a hamiltonian path Rk

3of

∪

nt=n−k+2S {bk,t}

n joining the black vertex

(

vk

)

nto the black

vertex

((

e

)

k−1

₎

n_{such that R}k

3

(

i

+

(

j

−

1

)(

n

−

1

)!) ∈

S

{bk,n−k+j+1}

n for every i

∈ h

(

n

−

1

)!i

and for every j

∈ h

k

−

1

i

. We set

Ck

= h

e

, (

e

)

k

, ((

e

)

k

)

n

,

Rk2

, (

wk

)

n

,

wk

,

Rk1

,

vk

, (

vk

)

n

,

Rk3

, ((

e

)

k

−1

₎

n

_{, (}

_e

₎

k−1

_,

_e

_i

_.

Then

{

C1

,

C2

, . . . ,

Cn−1

}

forms a set of

(

n

−

1

)

mutually independent hamiltonian cycles of Snstarting form the vertex e.

6. Discussion

In this paper, we discuss the mutually independent hamiltonian cycles for the pancake graphs and the star graphs. The concept of mutually independent hamiltonian cycle can be viewed as a generalization of Latin rectangles. Perhaps one of the most interesting topics in Latin square is orthogonal Latin square. Two Latin squares of order n are orthogonal if the

n-squared pairs formed by juxtaposing the two arrays are all distinct. Similarly, two Latin rectangles of order n

×

m are orthogonal if the n

×

m pairs formed by juxtaposing the two arrays are all distinct. With this in mind, let G be a hamiltonian

graph and C1and C2be two sets of mutually independent hamiltonian cycles of G from a given vertex x. We say C1and C2are

orthogonal if their corresponding Latin rectangles are orthogonal. For example, we know that IHC

(

P4

) =

3. The following

Latin rectangle represents three mutually independent hamiltonian cycles beginning at 1234.

2134, 4312, 1342, 2431, 3421, 1243, 4213, 3124, 1324, 4231, 3241, 1423, 2413, 3142, 4132, 2314, 3214, 4123, 2143, 3412, 1432, 2341, 4321 3214, 2314, 4132, 1432, 3412, 4312, 1342, 3142, 2413, 4213, 1243, 2143, 4123, 1423, 3241, 2341, 4321, 3421, 2431, 4231, 1324, 3124, 2134 4321, 2341, 1432, 3412, 2143, 4123, 1423, 3241, 4231, 1324, 3124, 2134, 4312, 1342, 2431, 3421, 1243, 4213, 2413, 3142, 4132, 2314, 3214

Yet, the following Latin rectangle also represents three mutually independent hamiltonian cycles beginning at 1234.

2134, 3124, 4213, 1243, 2143, 4123, 1423, 2413, 3142, 4132, 1432, 3412, 4312, 1342, 2431, 3421, 4321, 2341, 3241, 4231, 1324, 2314, 3214 3214, 2314, 4132, 3142, 2413, 4213, 1243, 3421, 2431, 1342, 4312, 2134, 3124, 1324, 4231, 3241, 1423, 4123, 2143, 3412, 1432, 2341, 4321 4321, 3421, 1243, 2143, 3412, 4312, 1342, 2431, 4231, 1324, 2314, 3214, 4123, 1423, 3241, 2341, 1432, 4132, 3142, 2413, 4213, 3124, 2134

We can check that these two Latin rectangles are orthogonal. Thus, we have two sets of three mutually independent hamiltonian cycles that are orthogonal. With this example in mind, we can consider the following problem. Let G be any

(10)

hamiltonian graph. We can define MOMH

(

G

)

as the largest integer k such that there exist k sets of mutually independent hamiltonian cycle of G beginning from any vertex x such that each set contains exactly IHC

(

G

)

hamiltonian cycles and any two different sets are orthogonal. It would be interesting to study the value of MOMH

(

G

)

for some hamiltonian graphs G.

We can also discuss mutually independent hamiltonian paths for some graphs. Let P1

= h

v

1

, v

2

, . . . , v

n

i

and P2

=

h

u1

,

u2

, . . . ,

un

i

be two hamiltonian paths of a graph G. We say that P1and P2are independent if u1

=

v

1, un

=

v

n, and

ui

6=

v

ifor 1

<

i

<

n. We say a set of hamiltonian paths

{

P1

,

P2

, . . . ,

Ps

}

of G between two distinct vertices are mutually

independent if any two distinct paths in the set are independent. There are some study on mutually independent hamiltonian paths [29,39].

Recently, people are interested in a mathematical puzzle, called Sudoku [38]. Sudoku can be viewed as a 9

×

9 Latin square with some constraints. There are several variations of Sudoku have been introduced. Mutually independent hamiltonian cycles can also be considered as a variation of Sudoku.

Acknowledgements

The second author’s research was partially supported by the Aiming for the Top University and Elite Research Center Development Plan and also his work was supported in part by the National Science Council of the Republic of China under Contract NSC 96-2221-E-137-MY3.

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