Embedding Hamiltonian paths in augmented cubes with a required vertex in a fixed position

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Computers and Mathematics with Applications

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

Embedding Hamiltonian paths in augmented cubes with a required

vertex in a fixed position

Chung-Meng Lee

a

, Yuan-Hsiang Teng

b,∗

, Jimmy J.M. Tan

c

, Lih-Hsing Hsu

a

a_{Department of Computer Science and Information Engineering, Providence University, Taichung County, 433, Taiwan, ROC} b_{Department of Computer Science and Information Engineering, Hungkuang University, Taichung County, 433, Taiwan, ROC} c_{Department of Computer Science, National Chiao Tung University, Hsinchu City, 300, Taiwan, ROC}

a r t i c l e i n f o Article history: Received 19 May 2008 Accepted 8 July 2009 Keywords: Hamiltonian Augmented cubes

a b s t r a c t

It is proved that there exists a path Pl(x,y)of length l if dAQn(x,y) ≤l≤2n−1 between

any two distinct vertices x and y of AQn. Obviously, we expect that such a path Pl(x,y)

can be further extended by including the vertices not in Pl(x,y)into a hamiltonian path

from x to a fixed vertex z or a hamiltonian cycle. In this paper, we prove that there exists a hamiltonian path R(x,y,z;l)from x to z such that dR(x,y,z;l)(x,y) =l for any three distinct

vertices x, y, and z of AQnwith n≥2 and for any dAQn(x,y) ≤l≤2n−1−dAQn(y,z).

Furthermore, there exists a hamiltonian cycle S(x,y;l)such that dS(x,y;l)(x,y) =l for any

two distinct vertices x and y and for any dAQn(x,y) ≤l≤2n−1.

1. Introduction

In this paper, a network is represented as a loopless undirected graph. For the graph definitions and notation, we follow [1]. Let G

=

(

V

,

E

)

be a graph if V is a finite set and E is a subset of

{

(

a

,

b

) | (

a

,

b

)

is an unordered pair of V

}

. We say that V is the vertex set and E is the edge set. Two vertices u and

v

are adjacent if

(

u

, v) ∈

E. We use NbdG

(

u

)

to

denote the set

{

v | (

u

, v) ∈

E

(

G

)}

. The degree of a vertex u in G, denoted by deg_G

(

u

)

, is

|

NbdG

(

u

)|

. We use

δ(

G

)

to denote

min

{

deg_G

(

u

) |

u

∈

V

(

G

)}

. A graph is k-regular if deg_G

(

u

) =

k for every vertex u in G. A path is a sequence of adjacent

vertices, written as

h

v

₀

, v

₁

, . . . , v

_m

i

, in which all the vertices

v

0

, v

1

, . . . , v

mare distinct except that possibly

v

0

=

v

m. We

also write the path

h

v

₀

,

P

, v

m

i

, where P

= h

v

0

, v

1

, . . . , v

m

i

. The length of a path P, denoted by l

(

P

)

, is the number of edges

in P. Let u and

v

be two vertices of G. The distance between u and

v

denoted by dG

(

u

, v)

is the length of the shortest path of G joining u and

v

. The diameter of a graph G, denoted by D

(

G

)

, is max

{

dG

(

u

, v) |

u

, v ∈

V

(

G

)}

. A cycle is a path with at least

three vertices such that the first vertex is the same as the last one. A hamiltonian cycle is a cycle of length V

(

G

)

. A hamiltonian

path is a path of length V

(

G

) −

1.

Interconnection networks play an important role in parallel computing/communication systems. The graph embedding problem is a central issue in evaluating a network. The graph embedding problem asked if the guest graph is a subgraph of a host graph, and an important benefit of the graph embeddings is that we can apply existing algorithm for guest graphs to host graphs. This problem has attracted numerous studies in recent years. Cycle networks and path networks are suitable for designing simple algorithms with low communication costs. The cycle embedding problem, which deals with all possible lengths of the cycles in a given graph, is investigated in a lot of interconnection networks [2–6]. The path embedding problem, which deals with all possible lengths of the paths between given two vertices in a given graph, is investigated in a lot of interconnection networks [5–12].

∗_{Corresponding author.}

E-mail address:[email protected](Y.-H. Teng).

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0 1 (a) AQ 1 00 01 10 11 (b) AQ 2 000 001 100 101 010 011 110 111 0000 0001 0100 0101 0010 0011 0110 0111 1000 1001 1100 1101 1010 1011 1110 1111 (c) AQ 3 (d) AQ4

Fig. 1. The augmented cubes AQ1, AQ2, AQ3and AQ4.

The hypercube Qn is one of the most popular interconnection networks for parallel computer/communication

system [13]. This is partly due to its attractive properties, such as regularity, recursive structure, vertex and edge symmetry, maximum connectivity, as well as effective routing and broadcasting algorithm. The augmented cube AQnis a variation of Qn, proposed by Choudum and Sunitha [14], and not only retains some favorable properties of Qnbut also processes some

embedding properties that Qndoes not [14–17,6]. For example, AQncontains cycles of all lengths from 3 to 2n, but Qncontains

only even cycles.

For the path embedding problem on the augmented cube, Ma et al. [6] proved that between any two distinct vertices

x and y of AQn, there exists a path Pl

(

x

,

y

)

of length l with dAQn

(

x

,

y

) ≤

l

≤

2

n

₋

_{1. Obviously, we expect that such a}

path Pl

(

x

,

y

)

can be further extended by including the vertices not in Pl

(

x

,

y

)

into a hamiltonian path from x to a fixed

vertex z or a hamiltonian cycle. For this reason, we prove that for any three distinct vertices x, y and z of AQn, and for any dAQn

(

x

,

y

) ≤

l

≤

2

n

₋

₁

₋

_d

AQn

(

y

,

z

)

there exists a hamiltonian path R

(

x

,

y

,

z

;

l

)

from x to z such that dR(x,y,z;l)

(

x

,

y

) =

l. As

a corollary, we prove that for any two distinct vertices x and y, and for any dAQn

(

x

,

y

) ≤

l

≤

2

n−1_{, there exists a hamiltonian}

cycle S

(

x

,

y

;

l

)

such that dS(x,y;l)

(

x

,

y

) =

l.

In Section2, we introduce the definition and some properties of the augmented cubes. In particular, we introduce another property, called 2RP, for augmented cubes. In Section3, we prove that any AQnsatisfies the 2RP-property if n

≥

2. Then we

apply the 2RP-property to prove the aforementioned properties in Section4.

2. Properties of augmented cubes

Assume that n

≥

1 is an integer. The graph of the n-dimensional augmented cube, denoted by AQn, has 2nvertices, each

labeled by an n-bit binary string V

(

AQn

) = {

u1u2

. . .

un

|

ui

∈ {

0

,

1

}}

. For n

=

1, AQ1is the graph K2with vertex set

{

0

,

1

}

.

For n

≥

2, AQncan be recursively constructed by two copies of AQn−1, denoted by AQn0−1and AQn1−1, and by adding 2nedges

between AQ_n0₋₁and AQ1

n−1as follows:

Let V

(

AQ_n0₋₁

) = {

0u2u3

. . .

un

|

ui

=

0 or 1 for 2

≤

i

≤

n

}

and V

(

AQn1−1

) = {

1

v

2

v

3

. . . v

n

|

v

i

=

0 or 1 for 2

≤

i

≤

n

}

. A

vertex u

=

0u2u3

. . .

unof AQn0−1is adjacent to a vertex v

=

1

v

2

v

3

. . . v

nof AQn1−1if and only if one of the following cases

holds.

(i) ui

=

v

i, for 2

≤

i

≤

n. In this case,

(

u

,

v

)

is called a hypercube edge. We set v

=

uh.

(ii) ui

= ¯

v

i, for 2

≤

i

≤

n. In this case,

(

u

,

v

)

is called a complement edge. We set v

=

uc.

The augmented cubes AQ1, AQ2, AQ3and AQ4are illustrated inFig. 1. It is proved in [14] that AQnis a vertex transitive,

(

2n

−

1

)

-regular, and

(

2n

−

1

)

-connected graph with 2nvertices for any positive integer n. Let i be any index with 1

≤

i

≤

n

and u

=

u1u2u3

. . .

un be a vertex of AQn. We use ui to denote the vertex v

=

v

1

v

2

v

3

. . . v

n such that uj

=

v

jwith

1

≤

j

6=

i

≤

n and ui

= ¯

v

i. Moreover, we use ui∗ to denote the vertex v

=

v

1

v

2

v

3

. . . v

nsuch that uj

=

v

i for j

<

i

and uj

= ¯

v

jfor i

≤

j

≤

n. Obviously, un

=

un∗, u1

=

uh, uc

=

u1∗, and NbdAQn

(

u

) = {

ui

|

1

≤

i

≤

n

} ∪ {

ui

∗

_|

₁

_≤

_i

_<

_n

_}

_.

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Proof. We prove this lemma by induction. Since AQ₂is isomorphic to the complete graph K4, the lemma holds for n

=

2.

Assume the lemma holds for 2

≤

k

<

n. Suppose that

{

u

,

v

} ⊂

V

(

AQi

n−1

)

for some i

∈ {

0

,

1

}

. By induction,

|

NbdAQn

(

u

) ∩

NbdAQn

(

v

)| ≥

2. Thus, consider the case that either v

=

u

h_{or v}

₌

_uc_{. Obviously,}

_{

_u2∗

_,

_uc

_{} ⊂}

_Nbd

AQn

(

u

) ∩

NbdAQn

(

v

)

if v

=

uh_{; and}

_{

_u2∗

_,

_uh

_{} ⊂}

_Nbd

AQn

(

u

) ∩

NbdAQn

(

v

)

if v

=

u

c_{. Then the statement holds.}

The following lemma can easily be obtained from the definition of AQn.

Lemma 2. Assume that n

≥

3. For any two different vertices u and v of AQn, there exists two other vertices x and y of AQnsuch that the subgraph of

{

u

,

v

,

x

,

y

}

containing a four cycle.

Lemma 3 ([16]). Let F be a subset of V

(

AQn

)

. Then there exists a hamiltonian path between any two vertices of V

(

AQn

) −

F if

|

F

| ≤

2n

−

4 for n

≥

4 and

|

F

| ≤

1 for n

=

3.

Lemma 4 ([14]). Let u and v be any two vertices in AQnwith n

≥

2. Suppose that both u and v are in AQni−1for i

=

0

,

1. Then dAQn

(

u

,

v

) =

dAQi

n−1

(

u

,

v

)

. Suppose that u is a vertex in AQ

i

n−1and v is a vertex in AQ 1−i

n−1. Then there exist two shortest paths P1 and P2of AQnjoining u to v such that

(

V

(

P1

) − {

v

}

) ⊂

V

(

AQni−1

)

and

(

V

(

P2

) − {

u

}

) ⊂

V

(

AQn1−−1i

)

.

WithLemma 4, we haveCorollary 1.

Corollary 1. Assume that n

≥

3. Let x and y be two vertices of AQnwith dAQn

(

x

,

y

) ≥

2. Then, there are two vertices p and q in

NbdAQn

(

x

)

with dAQn

(

p

,

y

) =

dAQn

(

q

,

y

) =

dAQn

(

x

,

y

) −

1.

Lemma 5 ([16]). Let

{

u

,

v

,

x

,

y

}

be any four distinct vertices of AQnwith n

≥

2. Then there exist two disjoint paths P1and P2 such that

(

1

)

P1is a path joining u and v,

(

2

)

P2is a path joining x and y, and

(

3

)

P1

∪

P2spans AQn.

We refer toLemma 5as 2P-property of the augmented cube. This property is used for many applications of the augmented cubes [15,16]. Obviously, l

(

P1

) ≥

dAQn

(

u

,

v

)

and l

(

P2

) ≥

dAQn

(

x

,

y

)

, and l

(

P1

) +

l

(

P2

) =

2n

−

2. We expect that l

(

P1

)

, hence, l

(

P2

)

can be an arbitrarily integer with the above constraint. However, such expectation is almost true. Let us consider AQ3. Suppose that u

=

001, v

=

110, x

=

101, and y

=

010. Thus, dAQ3

(

u

,

v

) =

1 and dAQ3

(

x

,

y

) =

1. We can find

P1and P2with l

(

P1

) ∈ {

1

,

3

,

5

}

. Note that

{

x

,

y

} =

NbdAQ3

(

u

) ∩

NbdAQ3

(

v

)

. We cannot find P1 with l

(

P1

) =

2. Again,

{

u

,

v

} =

NbdAQ3

(

x

) ∩

NbdAQ3

(

y

)

. We cannot find P2with l

(

P2

) =

2. Hence, we cannot find P1with l

(

P1

) =

4. Similarly, we

consider AQ4. Suppose that u

=

0000, v

=

1001, x

=

0001 and y

=

1000. Thus, dAQ4

(

u

,

v

) =

2 and dAQ4

(

x

,

y

) =

2. We

can find P1and P2with l

(

P1

) ∈ {

3

,

4

, . . . ,

11

}

. Note that

{

x

,

y

} =

NbdAQ4

(

u

) ∩

NbdAQ4

(

v

)

(

P1

) =

2.

Again,

{

u

,

v

} =

NbdAQ4

(

x

) ∩

NbdAQ4

(

y

)

(

P2

) =

2.

Now, we propose the 2RP-property of AQnwith n

≥

2: Let

{

u

,

v

,

x

,

y

}

be any four distinct vertices of AQn. Let l1and l2be two integers with l1

≥

dAQn

(

u

,

v

)

, l2

≥

dAQn

(

x

,

y

)

, and l1

+

l2

=

2n

−

2. Then there exist two disjoint paths P1and P2such that

(1) P1is a path joining u and v with l

(

P1

) =

l1, (2) P2is a path joining x and y with l

(

P2

) =

l2, and (3) P1

∪

P2spans AQnexcept for the following cases: (a) l1

=

2 with dAQn

(

u

,

v

) =

1 such that

{

x

,

y

} =

NbdAQn

(

u

)∩

NbdAQn

(

v

)

; (b) l2

=

2 with dAQn

(

x

,

y

) =

1

such that

{

u

,

v

} =

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

; (c) l1

=

2 with dAQn

(

u

,

v

) =

2 such that

{

x

,

y

} =

NbdAQn

(

u

) ∩

NbdAQn

(

v

)

; and (d)

l2

=

2 with dAQn

(

x

,

y

) =

2 such that

{

u

,

v

} =

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. 3. The 2RP-property of augmented cubes

Theorem 1. Assume that n is a positive integer with n

≥

2. Then AQnsatisfies 2RP-property.

Proof. We prove this theorem by induction. By brute force, we check the theorem holds for n

=

2

,

3

,

4. Assume the theorem holds for any AQkwith 4

≤

k

<

n. Without loss of generality, we can assume that l1

≥

l2. Thus, l2

≤

2n−1

−

1. By the

symmetric property of AQn, we can assume that at least one of u and v, say u, is in V

(

AQn0−1

)

. Thus, we have the following

cases:

Case 1: v

∈

V

(

AQ_n0₋₁

)

and

{

x

,

y

} ⊂

V

(

AQ_n1₋₁

)

.

Subcase 1.1: dAQn

(

x

,

y

) ≤

l2

≤

2

n−1

₋

_{3 except that (1) l}

2

=

2n−1

−

4 and (2) l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

)∩

NbdAQn

(

y

)

. ByLemma 3, there exists a hamiltonian path R of AQ

0

n−1joining u to v. Since l

(

R

) =

2n−1

−

1,

we can write R as

h

u

,

R1

,

p

,

q

,

R2

,

v

i

for some vertices p and q such that

{

ph

,

qh

} ∩ {

x

,

y

} = ∅

. By induction, there exist two

disjoint paths S1and S2such that (1) S1is a path joining phto qhwith l

(

S1

) =

2n−1

−

l2

−

2, (2) S2is a path joining x to y

with l

(

S2

) =

l2, and (3) S1

∪

S2spans AQn1−1. We set P1as

h

u

,

R1

,

p

,

ph

,

S1

,

qh

,

q

,

R2

,

v

i

and set P2as S2. Obviously, P1and P2

are the required paths.

Subcase 1.2: l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Obviously, there exists a path P2of length 2

in AQn

− {

u

,

v

}

joining x to y. ByLemma 3, there exists a hamiltonian path P1of AQn

−

V

(

P2

)

joining u to v. Obviously, P1

(4)

Subcase 1.3: l2

=

2n−1

−

4. Obviously, there exists a vertex p in V

(

AQn1−1

) − {

x

,

y

,

u

h

_,

_vh

_}

_{, a vertex q in Nbd}

AQn1−1

(

p

) − {

x

,

y

}

,

and a vertex r in Nbd_AQ1

n−1

(

q

) − {

x

,

y

,

p

}

. Suppose that rh

6∈ {

u

,

v

}

. By induction, there exist two disjoint paths Q1and Q2

such that (1) Q1is a path joining u to ph, (2) Q2is a path joining rhto v, and (3) Q1

∪

Q2spans AQn0−1. ByLemma 3, there exists

a hamiltonian path P2of AQn1−1

− {

p

,

q

,

r

}

joining x to y. We set P1as

h

u

,

Q1

,

ph

,

p

,

q

,

r

,

rh

,

Q2

,

v

i

. Suppose that rh

∈ {

u

,

v

}

.

Without loss of generality, we assume that rh

=

v. ByLemma 3, there exists a hamiltonian path R of AQ_n0₋₁

− {

v

}

joining u to ph_{. We set P}

1as

h

u

,

R

,

ph

,

p

,

q

,

r

,

rh

=

v

i

. Obviously, P1and P2are the required paths.

Subcase 1.4: l2

=

2n−1

−

2. Obviously, there exist a vertex p

∈

V

(

AQn1−1

) − {

x

,

y

,

uh

,

uc

,

vh

,

vc

}

. ByLemma 5, there exists

two disjoint paths Q1and Q2such that (1) Q1is a path joining u and ph, (2) Q2is a path joining pcand v, and (3) Q1

∪

Q2spans AQ0

n−1. ByLemma 3, there exists a hamiltonian path P2of AQn0−1

− {

p

}

h

u

,

Q1

,

ph

,

p

,

pc

,

Q2

,

v

i

.

Obviously, P1and P2are the required paths.

Subcase 1.5: l2

=

2n−1

−

1. ByLemma 3, there exists a hamiltonian path P1of AQn0−1joining u and v and there exists a

hamiltonian path P2of AQn1−1joining x to y. Obviously, P1and P2are the required paths.

Case 2: v

∈

V

(

AQ_n0₋₁

)

and exactly one of x and y is in V

(

AQ_n0₋₁

)

. Without loss of generality, we assume that x

∈

V

(

AQ_n0₋₁

)

.

Subcase 2.1: l2

=

1. Obviously, dAQn

(

x

,

y

) =

1. We set P2as

h

x

,

y

i

. ByLemma 3, there exists a hamiltonian path P1of

AQn

− {

x

,

y

}

joining u to v. Obviously, P1and P2are the required paths.

Subcase 2.2: l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. The proof is the same to Subcase 1.2.

Subcase 2.3: l2

=

3. Suppose that dAQn

(

x

,

y

) =

1. There exists a vertex p in NbdAQ_n0₋₁

(

x

) − {

u

,

v

}

. ByLemma 3, there exists

a hamiltonian path P1of AQn

− {

x

,

y

,

p

,

ph

}

joining u to v. We set P2as

h

x

,

p

,

ph

,

y

i

. Obviously, P1and P2are the required

paths.

Suppose that d_AQ_n

(

x

,

y

) =

2. ByLemma 4, there exists a path

h

x

,

p

,

y

i

from x to y such that p

∈

V

(

AQ1

n−1

)

. ByLemma 1,

there exists a vertex q

∈

Nbd_AQ1

n−1

(

p

) ∩

NbdAQ 1

n−1

(

y

)

. ByLemma 3, there exists a hamiltonian path P1of AQn

− {

x

,

y

,

p

,

q

}

h

x

,

p

,

q

,

y

i

Suppose that dAQn

(

x

,

y

) =

3. ByLemma 4, there exists a path P2from x to y such that

(

V

(

P2

) − {

x

}

) ⊂

V

(

AQn1−1

)

. By Lemma 3, there exists a hamiltonian path P1of AQn

−

V

(

P2

)

joining u to v. Obviously, P1and P2are the required paths. Subcase 2.4: 4

≤

l2

≤

2n−1

−

2 except that l2

=

2n−1

−

3.

Suppose that dAQn

(

x

,

y

) =

1 or 2. We first claim that there exists a vertex p in NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Assume that

dAQn

(

x

,

y

) =

1. Obviously, either y

=

x

h_{or y}

₌

_xc_{. We set p}

₌

_xc_{if y}

₌

_xh_{; and we set p}

₌

_xh_{if y}

₌

_xc_{. Assume that} dAQn

(

x

,

y

) =

2. ByLemma 4, there exists a path

h

x

,

p

,

y

i

from x to y such that p

∈

V

(

AQ

1

n−1

)

. Obviously, p satisfies our

claim. ByLemma 3, there exists a hamiltonian path R of AQ_n0₋₁

− {

x

}

joining u to v. Since l

(

R

) =

2n−1

₋

_{3, we can write R as}

h

u

,

R1

,

s

,

t

,

R2

,

v

i

such that

{

sh

,

th

} ∩ {

p

,

y

} = ∅

. By induction, there exist two disjoint paths S1and S2such that (1) S1is a

path joining shto thwith l

(

S1

) =

2n−1

−

1

−

l2, (2) S2is a path joining p to y with l

(

S2

) =

l2

−

1, and (3) S1

∪

S2spans AQn1−1.

We set P1as

h

u

,

R1

,

s

,

sh

,

S1

,

th

,

t

,

R2

,

v

i

and P2as

h

x

,

p

,

S2

,

y

i

Suppose that dAQn

(

x

,

y

) ≥

3. ByLemma 4, there exists a vertex p in V

(

AQ

1

n−1

)

such that dAQn

(

p

,

y

) =

dAQn

(

x

,

y

) −

1.

ByLemma 3, there exists a hamiltonian path R of AQ0

n−1

− {

x

}

joining u to v. We can write R as

h

u

,

R1

,

s

,

t

,

R2

,

v

i

such

that

{

sh

_,

_th

_{} ∩ {}

_p

_,

_y

_{} = ∅}

_{. By induction, there exist two disjoint paths S}

1and S2such that (1) S1is a path joining shto th

with l

(

S1

) =

2n−1

−

1

−

l2, (2) S2is a path joining p to y with l

(

S2

) =

l2

−

1, and (3) S1

∪

h

u

,

R1

,

s

,

sh

,

S1

,

th

,

t

,

R2

,

v

i

and P2as

h

x

,

p

,

S2

,

y

i

Subcase 2.5: l2

=

2n−1

−

3 or l2

=

2n−1

−

1. Let k

=

3 if l2

=

2n−1

−

3 and k

=

1 if l2

=

2n−1

−

1. There exists a vertex p

in Nbd_AQ0

n−1

(

x

) − {

u

,

v

,

y

n

_}

_{. By}_{Lemma 3}_{, there exists a hamiltonian path R of AQ}0

n−1

− {

x

,

p

}

joining u to v. We can write R

as

h

u

,

R1

,

s

,

t

,

R2

,

v

i

such that

{

s

,

t

} ∩ {

p

,

yn

} = ∅

. By induction, there exist two disjoint paths S1and S2such that (1) S1is a

path joining sn_{to t}n_{with l}

₍

_S

1

) =

k, (2) S2is a path joining pnto y with l

(

S2

) =

2n−1

−

k

−

2, and (3) S1

∪

S2spans AQn1−1.

We set P1as

h

u

,

R1

,

s

,

sn

,

S1

,

tn

,

t

,

R2

,

v

i

and P2as

h

x

,

p

,

pn

,

S2

,

y

i

. Obviously, P1and P2are the required paths. Case 3:

{

v

,

x

,

y

} ⊂

V

(

Q_n0₋₁

)

.

Subcase 3.1: l2

=

1. The proof is the same as Subcase 2.1.

Subcase 3.2: l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. The proof is the same as Subcase 1.2.

Subcase 3.3: dAQn

(

x

,

y

) ≤

l2

≤

2

n−2

₋

_{1. By induction, there exist two disjoint paths R}

1and R2such that (1) R1is a path

joining u to v with l

(

R₁

) =

2n−1

₋

_l

2

−

2, (2) R2is a path joining x to y with l

(

R2

) =

l2, (3) R1

∪

R2spans AQn0−1. We

can write R1as

h

u

,

R3

,

p

,

q

,

R4

,

v

i

. ByLemma 3, there exists a hamiltonian path S of AQn1−1joining p

h_{to q}h_{. We set P}

1as

h

u

,

R3

,

p

,

ph

,

S

,

qh

,

q

,

R4

,

v

i

and P2as R2. Obviously, P1and P2are the required paths.

Subcase 3.4: 2n−2

₊

₁

_≤

_l

2

≤

2n−1

−

1 except that l2

=

2n−2

+

2. By induction, there exist two disjoint paths R1and R2such

that (1) R1is a path joining u to v with l

(

R1

) =

2n−2

−

1, (2) R2is a path joining x to y with l

(

R2

) =

2n−2

−

1, and (3) R1

∪

R2

spans AQ_n0₋₁. We can write R1as

h

u

,

R3

,

p

,

q

,

R4

,

v

i

and write R2as

h

x

,

R5

,

s

,

t

,

R6

,

y

i

. By induction, there exist two disjoint

(5)

l

(

S2

) =

l2

−

2n−2, and (3) S1

∪

h

u

,

R3

,

p

,

ph

,

S1

,

qh

,

q

,

R4

,

v

i

and P2as

h

x

,

R5

,

s

,

sh

,

S2

,

th

,

t

,

R6

,

y

i

.

Subcase 3.5: l2

=

2n−2or 2n−2

+

2. Let k

=

0 if l2

=

2n−2and k

=

2 if l2

=

2n−2

+

2. By induction, there exist two

disjoint paths R1and R2such that (1) R1is a path joining u to v with l

(

R1

) =

2n−2

−

k, (2) R2is a path joining x to y with l

(

R2

) =

2n−2

+

k

−

2, and (3) R1

∪

R2spans AQn0−1. We can write R1as

h

u

,

R3

,

p

,

q

,

R4

,

v

i

and write R2as

h

x

,

R5

,

s

,

t

,

R6

,

y

i

.

ByLemma 3, there exists a hamiltonian path S of AQ1

n−1

− {

sn

,

tn

}

joining pnto qn. We set P1as

h

u

,

R3

,

p

,

pn

,

S

,

qn

,

q

,

R4

,

v

i

and P2as

h

x

,

R5

,

s

,

sn

,

tn

,

t

,

R6

,

y

i

. Obviously, P1and P2are the required paths. Case 4:

{

x

,

v

,

y

} ⊂

V

(

AQ1

n−1

)

.

Subcase 4.1: dAQn

(

x

,

y

) ≤

l2

≤

2n−1

−

3 except that (1) l2

=

2n−1

−

4 and (2) l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Obviously, there exists a vertex p in NbdAQn1−1

(

v

) − {

x

,

y

,

u

h

_}

_{. By induction, there exist}

two disjoint paths S1and S2such that (1) S1is a path joining p to v with l

(

S1

) =

l1

−

2n−1, (2) S2is a path joining x to y with l

(

S2

) =

l2, and (3) S1

∪

S2spans AQn1−1. ByLemma 3, there exists a hamiltonian path R of AQn0−1joining u and ph. We set P1

as

h

u

,

R

,

ph

_,

_p

_,

_S

1

,

v

i

and we set P2as S2. Obviously, P1and P2are the required paths.

Subcase 4.2: l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

Subcase 4.3: l2

=

2n−1

−

4. Obviously, there exists a vertex p in NbdAQ_n1₋₁

(

v

) − {

x

,

y

}

, and there exists a vertex q in Nbd_AQ1

n−1

(

p

) − {

x

,

y

,

v

,

u

h

_}

_{. By}_{Lemma 3}_{, there exists a hamiltonian path R of AQ}0

n−1joining u to qh, and there exists a

hamiltonian path P2of AQn1−1

− {

v

,

p

,

q

}

h

u

,

R

,

qh

,

q

,

p

,

v

i

paths.

Subcase 4.4: l2

=

2n−1

−

2. Let v0be an element in

{

vh

,

vc

} − {

u

}

. ByLemma 3, there exists a hamiltonian path R of AQn0−1

joining u to v0, and there exists a hamiltonian path P2of AQn1−1

− {

v

}

h

u

,

R

,

v0

,

v

i

. Obviously, P1

and P2are the required paths.

Subcase 4.5: l2

=

2n−1

−

1. Obviously, there exists a vertex p in NbdAQn1−1

(

v

) − {

x

,

y

}

. By induction, there exist two disjoint

paths S1and S2such that (1) S1is a path joining p to v with l

(

S1

) =

1, (2) S2is a path joining x to y with l

(

S2

) =

2n−1

−

3,

and (3) S1

∪

S2spans AQn1−1. Obviously, we can write S2as

h

x

,

S21

,

r

,

s

,

S22

,

y

i

for some vertex r and s such that u

6∈ {

rh

,

sh

}

.

Again by induction, there exist two disjoint paths R1and R2such that (1) R1is a path joining u to phwith l

(

R1

) =

2n−1

−

3,

(2) R2is a path joining rhto shwith l

(

R2

) =

1, and (3) R1

∪

R2spans AQn0−1. We set P1as

h

u

,

R1

,

ph

,

p

,

v

i

and set P2as

h

x

,

S₂1

,

r

,

rh

,

sh

,

s

,

S₂2

,

y

i

Case 5: v

∈

V

(

AQ_n1₋₁

)

and

|{

x

,

y

} ∩

V

(

AQ_n0₋₁

)| =

1. Without loss of generality, we assume that x

∈

V

(

AQ_n0₋₁

)

.

Subcase 5.1: l2

=

1. The proof is the same to Subcase 2.1.

Subcase 5.2: l2

=

2 if dAQn

(

x

,

y

) =

1 or 2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

Subcase 5.3: l2

=

(

x

,

y

) =

1. Obviously, there exists a vertex p in NbdAQn0−1

(

x

) − {

u

,

v

h

_}

_{. We set P}

2as

h

x

,

p

,

ph

_,

_y

_i

_{. By}_{Lemma 3}_{, there exists a hamiltonian path P}

1of AQn

−

V

(

P2

)

joining u to v. Obviously, P1and P2are the

required paths.

Suppose that dAQn

(

x

,

y

) =

2. Assume that

{

u

,

v

} =

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Thus, we have either v

=

x

h_{or v}

₌

_xc_.

Moreover, u

=

xα, and y

=

vαfor some

α ∈ {

i

|

2

≤

i

≤

n

} ∪ {

i

∗ |

2

≤

i

≤

n

−

1

}

. We set P2as

h

x

,

xh∗

, (

xh∗

)

α

, ((

xh

)

α

) =

y

i

in the case of v

=

xh_{. Otherwise, we set P}

2as

h

x

,

xh

, (

xh

)

α

, ((

xh∗

)

α

) =

y

i

. ByLemma 3, there exists a hamiltonian path P1of AQn

−

V

(

P2

)

joining u to v. Obviously, P1and P2are the required paths. Now, assume that

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

.

ByLemma 1, there exists a vertex p in

(

NbdAQn

(

x

) ∩

NbdAQn

(

y

)) − {

u

,

v

}

. Without loss of generality, we may assume that p is

in AQ_n0₋₁. ByLemma 1, there exists a vertex q in

(

Nbd_AQ0

n−1

(

p

) ∩

NbdAQn0−1

(

x

)) − {

u

}

. ByLemma 3, there exists a hamiltonian

path P1of AQn

− {

x

,

q

,

p

,

y

}

h

x

,

q

,

p

,

y

i

Suppose that dAQn

(

x

,

y

) =

3. ByLemma 4, there are two shortest paths R1and R2of AQnjoining x to y such that R1

can be written as

h

x

,

r1

,

r2

,

y

i

with

{

r1

,

r2

} ⊂

V

(

AQn0−1

)

and R2can be written as

h

x

,

s1

,

s2

,

y

i

with

{

s1

,

s2

} ⊂

V

(

AQn1−1

)

.

Suppose that u

6=

r2 or v

6=

s1. Without loss of generality, we assume that u

6=

r2. By Corollary 1, there exists a

vertex t

∈

Nbd_AQ0

n−1

(

x

) ∩

NbdAQn0−1

(

r2

) − {

u

}

. We set P2as

h

x

,

t

,

r2

,

y

i

. ByLemma 3, there exists a hamiltonian path P1

of AQn

−

V

(

P2

)

joining u to v. Obviously, P1 and P2 are the required paths. Thus, we consider u

=

r2and v

=

s1. By Corollary 1, there exists a vertex p in Nbd_AQ0

n−1

(

x

) ∩

NbdAQ 0

n−1

(

u

)

. Obviously, dAQn

(

p

,

y

) =

2. ByLemma 4, there exists a

vertex q in V

(

AQ1

n−1

) ∩

NbdAQn

(

p

) ∩

NbdAQn

(

y

)

. Since dAQn

(

q

,

y

) =

1 and dAQn

(

v

,

y

) =

2, q

6=

v. We set P2as

h

x

,

p

,

q

,

y

i

. By

Lemma 3, there exists a hamiltonian path P1of AQn

−

V

(

P2

)

joining u to v. Obviously, P1and P2are the required paths. Subcase 5.4: 4

≤

l2

≤

2n−1

−

1 with dAQn

(

x

,

y

) =

1. Suppose that l2

=

4. Obviously, there exists a vertex p in

Nbd_AQ0

n−1

(

x

) − {

u

,

v

h

_}

_{. By}_{Lemma 1}_{, there exists a vertex q in}

₍

_Nbd

AQ_n0₋₁

(

x

) ∩

NbdAQ_n0₋₁

(

p

)) − {

u

}

. ByLemma 3, there exists a

hamiltonian path P1of AQn

− {

x

,

y

,

p

,

ph

,

q

}

h

x

,

q

,

p

,

ph

,

y

i

(6)

Suppose that 5

≤

l2

≤

2n−1

−

1 except that l2

=

2n−1

−

2. Obviously, there exist a vertex p in NbdAQ_n0−1

(

x

) − {

u

,

vh

_,

_yh

_}

and a vertex s in Nbd_AQ0

n−1

(

u

) − {

x

,

p

,

v

h

_,

_yh

_}

_{. By induction, there exist two disjoint paths R}

joining u to s with l

(

R1

) =

2n−1

−

2

−

l2, (2) R2is a path joining p to x with l

(

R2

) =

l2

−

2, and (3) R1

∪

R2spans AQn0−1.

ByLemma 3, there exists a hamiltonian path S of AQ1

n−1

− {

y

,

ph

}

joining shto v. We set P1as

h

u

,

R1

,

s

,

sh

,

S

,

v

i

and P2as

h

x

,

R2

,

p

,

ph

,

y

i

Suppose that l2

=

2n−1

−

2. Let s and p be two vertices in V

(

AQn0−1

) − {

u

,

x

,

vh

,

yh

}

. By induction, there exist two

disjoint paths R1and R2such that (1) R1is a path joining u to s with l

(

R1

) =

2n−2, (2) R2is a path joining p to x with l

(

R2

) =

2n−2

−

2, (3) R1

∪

R2spans AQn0−1. Similarly, there exist two disjoint paths S1and S2such that (1) S1is a path joining

sh_{to v with l}

₍

_S

1

) =

2n−2

−

1, (2) S2is a path joining phto y with l

(

S2

) =

2n−2

−

1, and (3) S1

∪

h

u

,

R1

,

s

,

sh

,

S1

,

v

i

and P2as

h

x

,

R2

,

p

,

ph

,

S2

,

y

i

Subcase 5.5: 4

≤

l2

≤

2n−1

−

1 except l2

=

2n−1

−

3 with dAQn

(

x

,

y

) ≥

(

x

,

y

) =

2 with

{

u

,

v

} =

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Thus, we have either v

=

x

h_{or v}

₌

_xc_{. Moreover, u}

₌

_xα_{and y}

₌

₍

_xh

₎

α_{for some}

α ∈ {

i

|

2

≤

i

≤

n

} ∪ {

i

∗ |

2

≤

i

≤

n

−

1

}

. Obviously, there exists a vertex t in Nbd_AQ1

n−1

(

v

) − {

x

h

_,

_y

_,

_xc

_,

_uh

_}

_{. By induction,}

there exist two disjoint paths R1and R2such that (1) R1is a path joining t to v with l

(

R1

) =

2n−1

−

1

−

l2, (2) R2is a

path joining xcto y with l

(

R2

) =

l2

−

1 in the case of v

=

xh; otherwise R2is a path joining xhto y with l

(

R2

) =

l2

−

1,

and (3) R1

∪

R2spans AQn1−1. ByLemma 3, there exists a hamiltonian path S of AQn0−1

− {

x

}

joining thto u. We set P1as

h

u

,

S

,

th

,

t

,

R1

,

v

i

and P2as

h

x

,

xc

,

R2

,

y

i

in the case of v

=

xh; otherwise, we set P2as

h

x

,

xh

,

R2

,

y

i

. Obviously, P1and P2are

the required paths.

Suppose that dAQn

(

x

,

y

) =

2 with

{

u

,

v

} 6=

NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Then, there exists a vertex p in

(

NbdAQn

(

x

) ∩

NbdAQn

(

y

)) − {

u

,

v

}

. Without loss of generality, we may assume that p

∈

V

(

AQ

1

n−1

)

. Obviously, there exists a vertex t in Nbd_AQ1

n−1

(

v

) − {

y

,

p

,

u

h

_,

_xh

_}

1and R2such that (1) R1is a path joining t to v

with l

(

R1

) =

2n−1

−

1

−

l2, (2) R2is a path joining p to y with l

(

R2

) =

l2

−

1, and (3) R1

∪

R2spans AQn1−1. ByLemma 3, there

exists a hamiltonian path S of AQ0

n−1

− {

x

}

joining thto u. We set P1as

h

u

,

S

,

th

,

t

,

R1

,

v

i

and P2as

h

x

,

p

,

R2

,

y

i

. Obviously, P1and P2are the required paths.

Suppose that dAQn

(

x

,

y

) =

k

≥

3. By Lemma 4, there are two shortest paths S1 and S2 of AQn joining x to y

such that S1 can be written as

h

x

=

r0

,

r1

,

r2

, . . . ,

rk−1

,

y

i

with

(

V

(

S1

) − {

y

}

) ⊂

V

(

AQn0−1

)

and S2can be written as

h

x

,

s1

,

s2

, . . . ,

sk−1

,

y

i

with

(

V

(

S2

) − {

x

}

) ⊂

V

(

AQn1−1

)

. Suppose that u

6=

rk−1. We set p

=

rk−1. Again, there exists a

vertex s in Nbd_AQ0

n−1

(

u

) − {

x

,

p

,

y

h

_,

_vh

_}

joining u to s with l

(

R1

) =

2n−1

−

1

−

l2, (2) R2is a path joining p to x with l

(

R2

) =

l2

−

1, and (3) R1

∪

R2spans AQn0−1. By Lemma 3, there exists a hamiltonian path S of AQ1

n−1

−{

y

}

h

u

,

R1

,

s

,

sh

,

S

,

v

i

and P2as

h

x

,

R2

,

p

,

y

i

.

Now we assume that rk−1

=

u and s1

=

v. Since dAQn

(

rk−2

,

y

) =

2, byLemma 4, there exists a vertex p

∈

NbdAQn

(

rk−2

)

in V

(

AQ1

n−1

)

such that dAQn

(

p

,

y

) =

1. Suppose that l2

=

4 with dAQn

(

x

,

y

) =

3. Thus,

h

x

,

r1

,

p

,

y

i

is a shortest path joining

x and y. ByLemma 1, there exists a vertex q

∈

Nbd_AQ1

n−1

(

p

) ∩

NbdAQ 1

n−1

(

y

) − {

v

}

. ByLemma 3, there exists a hamiltonian path P1of AQn

− {

x

,

r1

,

p

,

q

,

y

}

h

x

,

r1

,

p

,

q

,

y

i

. Obviously, P1 and P2are the required paths.

Suppose that l2

=

4 with dAQn

(

x

,

y

) =

4. Thus, P2

= h

x

,

r1

,

r2

,

p

,

y

i

is a shortest path joining x and y. ByLemma 3,

there exists a hamiltonian path P1of AQn

− {

x

,

r1

,

r2

,

p

,

y

}

joining u to v. Obviously, P1 and P2 are the required paths.

Suppose that 5

≤

l2

≤

2n−2with dAQn

(

x

,

y

) ≥

3. Obviously, there exists a vertex s in NbdAQn0−1

(

u

) − {

x

,

rk−2

,

y

h

_,

_vh

_}

_{. By}

induction, there exist two disjoint paths R1and R2such that (1) R1is a path joining u to s with l

(

R1

) =

2n−1

−

l2, (2) R2

is a path joining rk−2to x with l

(

R2

) =

l2

−

2, and (3) R1

∪

R2spans AQn0−1. ByLemma 3, there exists a hamiltonian path S of AQ_n1₋₁

− {

p

,

y

}

h

u

,

R1

,

s

,

sh

,

S

,

v

i

and P2as

h

x

,

R2

,

rk−2

,

p

,

y

i

. Obviously, P1and P2are the

required paths. Suppose that 2n−2

₊

₁

_≤

_l

2

<

2n−1

−

1 except 2n−1

−

3 with dAQn

(

x

,

y

) ≥

3. Obviously, there exists a vertex s in Nbd_AQ0

n−1

(

u

) − {

x

,

rk−2

,

y

h

_,

_vh

_}

1and R2such that (1) R1is a path joining

u to s with l

(

R1

) =

2n−2

+

1, (2) R2is a path joining rk−2to x with l

(

R2

) =

2n−2

−

3, and (3) R1

∪

R2spans AQn0−1. Again by

induction, there exist two disjoint paths S1and S2such that (1) S1is a path joining shto v with l

(

S1

) =

2n−1

−

l2

+

2n−2

−

4,

(2) S2is a path joining p to y with l

(

S2

) =

l2

−

2n−2

+

2, and (3) S1

∪

h

u

,

R1

,

s

,

sh

,

S1

,

v

i

and P2

as

h

x

,

R2

,

rk−2

,

p

,

S2

,

y

i

Subcase 5.6: l2

=

2n−1

−

3 or l2

=

2n−1

−

1 with dAQn

(

x

,

y

) ≥

2. Let t

=

0 if l2

=

2n−1

−

3 and t

=

1 if l2

=

2n−1

−

1.

Obviously, there exist two vertices s and p in AQ0

n−1

− {

u

,

x

,

vn

,

yn

}

. By induction, there exist two disjoint paths R1and R2

such that (1) R1is a path joining u to s with l

(

R1

) =

2n−2

−

t, (2) R2is a path joining p to x with l

(

R2

) =

2n−2

+

t

−

2,

and (3) R1

∪

R2spans AQn0−1. Similarly, there exist two disjoint paths S1and S2such that (1) S1is a path joining snto v with l

(

S1

) =

2n−2

−

t, (2) S2is a path joining pnto y with l

(

S2

) =

2n−2

+

t

−

2, and (3) S1

∪

h

u

,

R1

,

s

,

sn

,

S1

,

v

i

and P2as

h

x

,

R2

,

p

,

pn

,

S2

,

y

i

(7)

4. The applications of the 2RP-property

Theorem 2. Assume that n is a positive integer with n

≥

2. For any three distinct vertices x, y and z of AQn and for any dAQn

(

x

,

y

) ≤

l

≤

2

n

₋

₁

₋

_d

AQn

(

y

,

z

)

, there exists a hamiltonian path R

(

x

,

y

,

z

;

l

)

from x to z such that dR(x,y,z;l)

(

x

,

y

) =

l. Proof. Obviously, the theorem holds for n

=

2. Thus, we consider that n

≥

3. We have the following cases:

Case 1: dAQn

(

x

,

y

) =

1 and dAQn

(

y

,

z

) =

1. ByLemma 1, there exists a vertex w in

(

NbdAQn

(

y

) ∩

NbdAQn

(

z

)) − {

x

}

. Similarly,

there exists a vertex p in

(

NbdAQn

(

x

) ∩

NbdAQn

(

y

)) − {

z

}

. Suppose that l

=

2. ByTheorem 1, there exist two disjoint paths S1

and S2such that (1) S1is a path joining x to p with l

(

S1

) =

1, (2) S2is a path joining y to z with l

(

S2

) =

2n

−

3, and (3) S1

∪

S2

spans AQn. We set R as

h

x

,

p

,

y

,

S2

,

z

i

. Obviously, R forms a hamiltonian path from x to z such that dR

(

x

,

y

) =

l. Suppose that l

=

2n

−

3. ByTheorem 1, there exist two disjoint paths Q1and Q2such that (1) Q1is a path joining x to y with l

(

Q1

) =

2n

−

3,

(2) Q2is a path joining w to z with l

(

Q2

) =

1, and (3) Q1

∪

Q2spans AQn. We set R as

h

x

,

Q1

,

y

,

w

,

z

i

. Obviously, R forms a

hamiltonian path from x to z such that dR

(

x

,

y

) =

l. Suppose that 1

≤

l

≤

2n

−

2 with l

6∈ {

2

,

2n

−

3

}

. ByTheorem 1, there

exist two disjoint paths P1and P2such that (1) P1is a path joining x to y with l

(

P1

) =

l, (2) P2is a path joining w to z with l

(

P2

) =

2n

−

2

−

l, and (3) P1

∪

P2spans AQn. We set R as

h

x

,

P1

,

y

,

w

,

P2

,

z

i

. Obviously, R forms a hamiltonian path from x

to z such that dR

(

x

,

y

) =

l.

Case 2: dAQn

(

x

,

y

) =

1 and dAQn

(

y

,

z

) 6=

1. ByLemma 1, there exists a vertex p in NbdAQn

(

x

) ∩

NbdAQn

(

y

)

. Suppose that l

=

2.

ByTheorem 1, there exist two disjoint paths S1and S2such that (1) S1is a path joining x to p with l

(

S1

) =

1, (2) S2is a path

joining y to z with l

(

S2

) =

2n

−

3, and (3) S1

∪

S2spans AQn. We set R as

h

x

,

p

,

y

,

S2

,

z

i

. Obviously, R forms a hamiltonian

path from x to z such that dR

(

x

,

y

) =

l. Suppose that 1

≤

l

≤

2n

−

1

−

dAQn

(

y

,

z

)

with l

6=

2. ByCorollary 1, there exists a

vertex w in NbdAQn

(

y

) − {

x

}

such that dAQn

(

w

,

z

) =

dAQn

(

y

,

z

) −

1. ByTheorem 1, there exist two disjoint paths P1and P2

such that (1) P1is a path joining x to y with l

(

P1

) =

l, (2) S2is a path joining w to z with l

(

P2

) =

2n

−

2

−

l, and (3) P1

∪

P2

spans AQn. We set R as

h

x

,

P1

,

y

,

w

,

P2

,

z

i

. Obviously, R forms a hamiltonian path from x to z such that dR

(

x

,

y

) =

l. Case 3: dAQn

(

x

,

y

) 6=

1 and dAQn

(

y

,

z

) =

1. This case is similar as Case 2 by interchanging the roles of x and z.

Case 4: dAQn

(

x

,

y

) 6=

1 and dAQn

(

y

,

z

) 6=

1. Let l be any integer with dAQn

(

x

,

y

) ≤

l

≤

2

n

₋

₁

₋

_d

AQn

(

y

,

z

)

. Let w be a vertex

in NbdAQn

(

y

)

. ByTheorem 1, there exist two disjoint paths S1and S2such that (1) S1is a path joining x to y with l

(

S1

) =

l,

(2) S2is a path joining w to z with l

(

S2

) =

2n

−

2

−

l, and (3) S1

∪

S2spans AQn. We set R as

h

x

,

S1

,

y

,

w

,

S2

,

z

i

. Obviously, R

forms a hamiltonian path from x to z such that dR

(

x

,

y

) =

l.

The theorem is proved.

Corollary 2. Assume that n is a positive integer with n

≥

2. For any two distinct vertices x and y and for any dAQn

(

x

,

y

) ≤

l

≤

2n−1_{, there exists a hamiltonian cycle S}

₍

_x

_,

_y

_;

_l

₎

_{such that d}

S(x,y;l)

(

x

,

y

) =

l.

Proof. Let z be a vertex in NbdAQn

(

x

) − {

y

}

. By Theorem 2, there exists a hamiltonian path R joining x to z such that

dR(x,y,z;l)

(

x

,

y

) =

l. We set S as

h

x

,

R

,

z

,

x

i

. Obviously, S forms the required hamiltonian cycle. References

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