Two Spanning Disjoint Paths with Required Length in Augmented
Cubes
Chung-Meng Lee
Department of Computer Science
National Chiao Tung University
cmlee@pu.edu.tw
Yuan-Hsiang Teng
Department of Computer Science
National Chiao Tung University
gis93806@cis.nctu.edu.tw
Jimmy J. M. Tan
Department of Computer Science
National Chiao Tung University
jmtan@cs.nctu.edu.tw
Lih-Hsing Hsu
Department of Computer Science and Information Engineering
Providence University
lhhsu@pu.edu.tw
Abstract
In this article, we introduce 2RP-property in the aug-mented cubeAQn: Let{u, v, x, y} be any four distinct vertices of AQn. Let l1 and l2 be two integers with
l1≥ dAQn(u, v), l2≥ dAQn(x, y), and l1+ l2= 2n−2.
Then there exist two disjoint paths P1 andP2 such that (1)P1is a path joiningu and v with l(P1) = l1, (2)P2is a path joiningx and y with l(P2) = l2, and (3)P1∪ P2 spansAQnexcept some special conditions.
Keywords: hamiltonian, augmented cubes.
1 Introduction
Interconnection networks play an important role in par-allel computing/communication systems. The graph em-bedding problem is a central issue in evaluating a net-work. The graph embedding problem asked if the quest graph is a subgraph of a host graph, and an important ben-efit of the graph embeddings is that we can apply existing algorithm for guest graphs to host graphs. This problem has attracted a burst of studies in recent years. Cycle net-works and path netnet-works are suitable for designing simple algorithms with low communication costs. The cycle em-bedding problem, which deals with all possible lengths of the cycles in a given graph, is investigated in a lot of
inter-connection networks [6, 10, 12, 13]. The path embedding problem, which deals with all possible lengths of the paths between given two vertices in a given graph, is investi-gated in a lot of interconnection networks [3–5, 12–14].
In this article, a network is represented as a loopless undirected graph. For the graph definitions and nota-tion, we follow [1]. Let G = (V, E) be a graph if V is a finite set andE is a subset of {(a, b) | (a, b) is an unordered pair of V }. We say that V is the vertex set andE is the edge set. Two vertices u and v are
adja-cent if (u, v) ∈ E. We use N bdG(u) to denote the set
{v | (u, v) ∈ E(G)}. The degree of a vertex u in G,
denoted by degG(u), is |N bdG(u)|. We use δ(G) to de-note min{degG(u) | u ∈ V (G)}. A graph is k-regular if degG(u) = k for every vertex u in G. A path is a
se-quence of adjacent vertices, written asv0, v1, . . . , vm, in which all the verticesv0, v1, . . . , vmare distinct except that possiblyv0= vm. We also write the pathv0, P, vm, whereP = v0, v1, . . . , vm. The length of a path P , de-noted byl(P ), is the number of edges in P . Let u and v be two vertices ofG. The distance between u and v denoted
bydG(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.
The hypercubeQnis one of the most popular intercon-nection networks for parallel computer/comminication system [11]. This is partly due to its attractive prop-erties, such as regularity, recursive structure, vertex and edge symmetry, maximum connectivity, as well as effec-tive routing and broadcasting algorithm. The augmented cubeAQn is a variation ofQn, proposed by Choudum and Sunitha [2], and not only retains some favorable
prop-erties ofQnbut also processes some embedding proper-ties thatQndoes not [2, 7–9, 13]. For example,AQn con-tains cycles of all lengths from 3 to 2n, butQn contains only even cycles.
For the path embedding problem on the augmented cube, Ma et al. [13] proved that between any two dis-tinct verticesx and y of AQn, there exists a pathPl(x, y)
of length l with dAQn(x, y) ≤ l ≤ 2n − 1.
Obvi-ously, we expect that such a path Pl(x, y) can be
fur-ther extended by including the vertices not inPl(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 anydAQn(x, y) ≤ l ≤ 2n− 1 − dAQn(y, z) there
ex-ists 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 verticesx and y, and for any
dAQn(x, y) ≤ l ≤ 2n−1, there exists a hamiltonian cycle
S(x, y; l) such that dS(x,y;l)(x, y) = l.
In the following section, we introduce the definition and some properties of the augmented cubes. In Section 3, we introduce the 2RP-property for the augmented cube
AQn and prove that AQn satisfies the 2RP-property if
n ≥ 2. We make some remarks to illustrate that some
in-teresting properties of augmented cubes are consequences of 2RP-property in the final section.
2 Preliminaries
In this section, we introduce some properties of the augmented cubes. Assume thatn ≥ 1 is an integer. The
graph of then-dimensional augmented cube, denoted by
AQn, has 2n vertices, each labeled by an n-bit binary string V (AQn) = {u1u2. . . un | ui ∈ {0, 1}}. For n = 1, AQ1 is the graph K2 with vertex set {0, 1}. Forn ≥ 2, AQn can be recursively constructed by two
copies ofAQn−1, denoted byAQ0n−1andAQ1n−1, and by adding 2nedges betweenAQ0n−1andAQ1n−1as
fol-lows:
LetV (AQ0n−1) = {0u2u3. . . un | ui = 0 or 1 for
2≤ i ≤ n} and V (AQ1n−1) ={1v2v3. . . vn | vi = 0 or
1 for 2≤ i ≤ n}. A vertex u = 0u2u3. . . unofAQ0n−1 is adjacent to a vertexv = 1v2v3. . . vnofAQ1n−1if and only if one of the following cases holds.
(i) ui= vi, for 2≤ i ≤ n. In this case, (u, v) is called a
hypercube edge. We setv = uh.
(ii) ui = ¯vi, for 2≤ i ≤ n. In this case, (u, v) is called a complement edge. We setv = uc.
The augmented cubesAQ1,AQ2,AQ3 andAQ4 are illustrated in Figure 1. It is proved in [2] thatAQnis a vertex transitive, (2n−1)-regular, and (2n−1)-connected
graph with 2nvertices for any positive integern. Let i be any index with 1 ≤ i ≤ n and u = u1u2u3. . . un be a vertex ofAQn. We use ui to denote the vertexv =
v1v2v3. . . vn such thatuj = vjwith 1≤ j = i ≤ n and
ui = ¯vi. Moreover, we useui∗ to denote the vertexv =
v1v2v3. . . vnsuch thatuj = viforj < i and uj = ¯vjfor i ≤ j ≤ n. Obviously, un =un∗,u1 =uh,uc =u1∗, andN bdAQn(u) = {ui | 1 ≤ i ≤ n} ∪ {ui∗ | 1 ≤ i <
n}.
Lemma 1. Assume that n ≥ 2. Then |N bdAQn(u) ∩
N bdAQn(v)| ≥ 2 if (u, v) ∈ E(G).
Proof. We prove this lemma by induction. SinceAQ2 is isomorphic to the complete graphK4, the lemma holds forn = 2. Assume the lemma holds for 2 ≤ k < n. Sup-pose that{u, v} ⊂ V (AQin−1) for some i ∈ {0, 1}. By induction,|NbdAQn(u) ∩ NbdAQn(v)| ≥ 2. Thus,
con-sider the case that eitherv = uhorv = uc. Obviously,
{u2∗, uc} ⊂ Nbd AQn(u) ∩ NbdAQn(v) if v = uh; and 0 1 (a) AQ1 00 01 10 11 (b) AQ2 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) AQ3 (d) AQ4
Figure 1: The augmented cubes AQ1, AQ2, AQ3 and
AQ4.
{u2∗, uh} ⊂ NbdAQ
n(u)∩NbdAQn(v) if v = uc. Then
the statement holds.
The following lemma can easily be obtained from the definition ofAQn.
Lemma 2. Assume thatn ≥ 3. For any two different
verticesu and v of AQn, there exists two other vertices
x and y of AQn such that the subgraph of{u, v, x, y} containing a four cycle.
Lemma 3. [8] Let F be a subset of V (AQn). Then there exists a hamiltonian path between any two vertices ofV (AQn)− F if |F | ≤ 2n − 4 for n ≥ 4 and |F | ≤ 1 forn = 3.
Lemma 4. [2] Letu and v be any two vertices in AQn
withn ≥ 2. Suppose that both u and v are in AQin−1 fori = 0, 1. Then dAQn(u, v) = dAQin−1(u, v). Suppose
thatu is a vertex in AQin−1andv is a vertex in AQ1−in−1.
Then there exist two shortest pathsP1 and P2 ofAQn
joiningu to v such that (V (P1)− {v}) ⊂ V (AQin−1)
and (V (P2)− {u}) ⊂ V (AQ1−in−1).
With Lemma 4, we have the following corollary.
Corollary 5. Assume thatn ≥ 3. Let x and y be two
two verticesp and q in NbdAQn(x) with dAQn(p, y) =
dAQn(q, y) = dAQn(x, y) − 1.
Lemma 6. [8] Let{u, v, x, y} be any four distinct
ver-tices ofAQn withn ≥ 2. Then there exist two disjoint
pathsP1andP2such that (1)P1is a path joiningu and
v, (2) P2is a path joiningx and y, and (3) P1∪ P2spans
AQn.
We refer to Lemma 6 as 2P-property of the augmented cube. This property is used for many applications of the augmented cubes [7, 8]. Obviously,l(P1)≥ dAQn(u, v)
andl(P2) ≥ dAQn(x, y), and l(P1) + l(P2) = 2n− 2.
We expect thatl(P1), hence, l(P2) can be an arbitrarily
integer with the above constraint. However, such expec-tation is almost true. Let us considerAQ3. Suppose that u = 001, v = 110, x = 101, and y = 010. Thus,
dAQ3(u, v) = 1 and dAQ3(x, y) = 1. We can find
P1 andP2 withl(P1) ∈ {1, 3, 5}. Note that {x, y} =
N bdAQ3(u) ∩ NbdAQ3(v). We can not find P1 with
l(P1) = 2. Again,{u, v} = NbdAQ3(x) ∩ NbdAQ3(y).
We can not findP2withl(P2) = 2. Hence, we cannot find
P1withl(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
findP1 andP2 withl(P1) ∈ {3, 4, . . . , 11}. Note that
{x, y} = NbdAQ4(u) ∩ NbdAQ4(v). We can not find
P1 with l(P1) = 2. Again, {u, v} = NbdAQ4(x) ∩
N bdAQ4(y). We can not find P2withl(P2) = 2.
3 The 2RP-property of the
augmented cubes
In this section, we introduce the 2RP-property for the augmented cubeAQn and prove that AQn satisfies the 2RP-property if n ≥ 2. First, we propose the 2RP-property of AQn withn ≥ 2: Let {u, v, x, y} be any
four distinct vertices of AQn. Letl1 and l2 be two
in-tegers with l1 ≥ dAQn(u, v), l2 ≥ dAQn(x, y), and
l1 + l2 = 2n − 2. Then there exist two disjoint paths
P1 and P2 such that (1)P1 is a path joiningu and v
withl(P1) = l1, (2)P2 is a path joiningx and y with
l(P2) = l2, and (3)P1∪ P2 spansAQn except for the
following cases: (a)l1 = 2 with dAQn(u, v) = 1 such
that{x, y} = NbdAQn(u) ∩ NbdAQn(v); (b) l2 = 2 withdAQn(x, y) = 1 such that {u, v} = NbdAQn(x) ∩
N bdAQn(y); (c) l1 = 2 with dAQn(u, v) = 2 such that
{x, y} = NbdAQn(u) ∩ NbdAQn(v); and (d) l2 = 2
withdAQn(x, y) = 2 such that {u, v} = NbdAQn(x) ∩
N bdAQn(y).
Theorem 7. Assume thatn 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 forn = 2, 3, 4. As-sume the theorem holds for anyAQk with 4 ≤ k < n. Without loss of generality, we can assume thatl1 ≥ l2. Thus,l2≤ 2n−1−1. By the symmetric property of AQn, we can assume that at least one ofu and v, say u, is in
V (AQ0n−1). Thus, we have the following cases: Case 1:v ∈ V (AQ0n−1) and{x, y} ⊂ V (AQ1n−1).
Subcase 1.1: 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} = NbdAQn(x) ∩ NbdAQn(y). See
Fig-ure 2(a) for an illustration. By Lemma 3, there exists a hamiltonian pathR of AQ0n−1 joiningu to v. Since
l(R) = 2n−1− 1, we can write R as u, R1, p, q, R2, v for some verticesp and q such that {ph, qh} ∩ {x, y} =
∅. By induction, there exist two disjoint paths S1 and
S2 such that (1) S1 is a path joining ph to qh with
l(S1) = 2n−1− l2− 2, (2) S2is a path joining x to y withl(S2) = l2, and (3)S1∪ S2spansAQ1n−1. We set
P1 as u, R1, p, ph, S1, qh, q, R2, v and set P2 as S2. Obviously,P1andP2are the required paths.
u v x y p ph qh R1 R2 S2 q S1 u v x y p r ph Q1 Q2 P2 q rh
(a)
(b)
Figure 2: Subcase 1.1 and Subcase 1.3.
Subcase 1.2: l2 = 2 if dAQn(x, y) = 1 or 2 with
{u, v} = NbdAQn(x) ∩ NbdAQn(y). Obviously, there
exists a pathP2 of length 2 inAQn − {u, v} joining x toy. By Lemma 3, there exists a hamiltonian path P1of
AQn− V (P2) joiningu to v. Obviously, P1andP2are the required paths.
Subcase 1.3:l2= 2n−1− 4. See Figure 2(b) for an
illus-tration. Obviously, there exists a vertexp in V (AQ1n−1)−
{x, y, uh, vh}, a vertex q in Nbd AQ1
n−1(p) − {x, y},
and a vertexr in NbdAQ1
n−1(q) − {x, y, p}. Suppose
that rh ∈ {u, v}. By induction, there exist two dis-/
joint pathsQ1 andQ2 such that (1)Q1 is a path join-ing u to ph, (2) Q2 is a path joiningrh to v, and (3)
Q1 ∪ Q2 spans AQ0n−1. By Lemma 3, there exists a
hamiltonian pathP2 ofAQ1n−1− {p, q, r} joining x to y. We set P1asu, Q1, ph, p, q, r, rh, Q2, v. Suppose thatrh ∈ {u, v}. Without loss of generality, we assume that rh = v. By Lemma 3, there exists a hamiltonian
pathR of AQ0n−1− {v} joining u to ph. We setP1 as
u, R, ph, p, q, r, rh =v. Obviously, P
1andP2are the required paths.
Subcase 1.4:l2= 2n−1− 2. Obviously, there exist a
ver-texp ∈ V (AQ1n−1)− {x, y, uh, uc, vh, vc}. By Lemma
6, there exists two disjoint pathsQ1andQ2such that (1)
Q1is a path joiningu and ph, (2)Q2is a path joiningpc andv, and (3) Q1∪Q2spansAQ0n−1. By Lemma 3, there
exists a hamiltonian pathP2ofAQ0n−1− {p} joining x toy. We set P1asu, Q1, ph, p, pc, Q2, v. Obviously,
P1andP2are the required paths.
Subcase 1.5:l2= 2n−1− 1. By Lemma 3, there exists a
hamiltonian pathP1ofAQ0n−1joiningu and v and there exists a hamiltonian path P2 of AQ1n−1 joiningx to y. Obviously,P1andP2are the required paths.
Case 2:v ∈ V (AQ0n−1) and exactly one ofx and y is in
V (AQ0n−1). Without loss of generality, we assume that
x ∈ V (AQ0
n−1).
Subcase 2.1: l2 = 1. Obviously, dAQn(x, y) = 1. We
setP2asx, y. By Lemma 3, there exists a hamiltonian pathP1ofAQn− {x, y} joining u to v. Obviously, P1 andP2are the required paths.
Subcase 2.2: l2 = 2 if dAQn(x, y) = 1 or 2 with
{u, v} = NbdAQn(x) ∩ NbdAQn(y). The proof is the
same to Subcase 1.2.
Suppose thatdAQn(x, y) = 1. There exists a vertex p
inN bdAQ0
n−1(x) − {u, v}. By Lemma 3, there exists a
hamiltonian pathP1ofAQn− {x, y, p, ph} joining u to v. We set P2asx, p, ph, y. Obviously, P1andP2are the required paths.
Suppose thatdAQn(x, y) = 2. By Lemma 4, there
exists a path x, p, y from x to y such that p ∈
V (AQ1n−1). By Lemma 1, there exists a vertex q ∈ N bdAQ1
n−1(p) ∩ NbdAQ1n−1(y). By Lemma 3, there
ex-ists a hamiltonian pathP1ofAQn− {x, y, p, q} joining
u to v. We set P2asx, p, q, y. Obviously, P1andP2 are the required paths.
Suppose thatdAQn(x, y) = 3. By Lemma 4, there
exists a pathP2fromx to y such that (V (P2)− {x}) ⊂
V (AQ1n−1). By Lemma 3, there exists a hamiltonian path
P1ofAQn− V (P2) joiningu to v. Obviously, P1and
P2are the required paths.
Subcase 2.4: 4≤ l2≤ 2n−1− 2 except that l2= 2n−1− 3.
Suppose thatdAQn(x, y) = 1 or 2. We first claim that
there exists a vertexp in NbdAQn(x) ∩ NbdAQn(y).
As-sume thatdAQn(x, y) = 1. Obviously, either y = xhor y = xc. We setp = xcify = xh; and we setp = xhif
y = xc. Assume thatdAQ
n(x, y) = 2. See Figure 3 for
an illustration. By Lemma 4, there exists a pathx, p, y fromx to y such that p ∈ V (AQ1n−1). Obviously, p
satisfies our claim. By Lemma 3, there exists a hamil-tonian path R of AQ0n−1− {x} joining u to v. Since
l(R) = 2n−1− 3, we can write R as u, R1, s, t, R2, v such that{sh, th} ∩{p, y} = ∅. By induction, there exist
two disjoint pathsS1andS2such that (1)S1is a path join-ingshtothwithl(S1) = 2n−1− 1 − l2, (2)S2is a path joiningp to y with l(S2) = l2− 1, and (3) S1∪ S2spans
AQ1n−1. We setP1asu, R1, s, sh, S1, th, t, R2, v and
P2asx, p, S2, y. Obviously, P1andP2are the required
paths. x y u v R2 R1 p s t sh th S 1 S2 Figure 3: Subcase 2.4.
Suppose thatdAQn(x, y) ≥ 3. By Lemma 4, there
exists a vertexp in V (AQ1n−1) such that dAQn(p, y) =
dAQn(x, y) − 1. By Lemma 3, there exists a hamiltonian
pathR of AQ0n−1− {x} joining u to v. We can write R as u, R1, s, t, R2, v such that {sh, th} ∩ {p, y} = ∅. By induction, there exist two disjoint pathsS1 and S2 such that (1)S1is a path joiningsh tothwithl(S1) = 2n−1 − 1 − l2, (2) S2 is a path joining p to y with
l(S2) = l2− 1, and (3) S1∪ S2spansAQ1n−1. We setP1 asu, R1, s, sh, S1, th, t, R2, v and P2 asx, p, S2, y.
Obviously,P1andP2are the required paths.
Subcase 2.5:l2= 2n−1− 3 or l2= 2n−1− 1. Let k = 3
ifl2= 2n−1− 3 and k = 1 if l2= 2n−1− 1. There
ex-ists a vertexp in NbdAQ0
n−1(x)−{u, v, y
n}. By Lemma
3, there exists a hamiltonian pathR of AQ0n−1− {x, p} joiningu to v. We can write R as u, R1, s, t, R2, v such that{s, t} ∩ {p, yn} = ∅. By induction, there exist two disjoint pathsS1 andS2such that (1)S1is a path join-ingsn totnwithl(S1) = k, (2) S2is a path joiningpn toy with l(S2) = 2n−1− k − 2, and (3) S1∪ S2spans
AQ1n−1. We setP1asu, R1, s, sn, S1, tn, t, R2, v and
re-quired paths.
Case 3:{v, x, y} ⊂ V (Q0n−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} = NbdAQn(x) ∩ NbdAQn(y). The proof is the
same as Subcase 1.2.
Subcase 3.3: dAQn(x, y) ≤ l2 ≤ 2n−2− 1. See Figure
4(a) for an illustration. By induction, there exist two dis-joint pathsR1andR2such that (1)R1is a path joiningu tov with l(R1) = 2n−1− l2− 2, (2) R2is a path joining x to y with l(R2) = l2, (3)R1∪ R2spansAQ0n−1. We
can writeR1asu, R3, p, q, R4, v. By Lemma 3, there exists a hamiltonian pathS of AQ1n−1joiningphtoqh. We setP1asu, R3, p, ph, S, qh, q, R4, v and P2asR2. Obviously,P1andP2are the required paths.
Subcase 3.4: 2n−2+ 1 ≤ l2 ≤ 2n−1− 1 except that
l2 = 2n−2+ 2. See Figure 4(b) for an illustration. By induction, there exist two disjoint pathsR1andR2 such that (1)R1is a path joiningu to v with l(R1) = 2n−2−1,
(2)R2 is a path joiningx to y with l(R2) = 2n−2− 1,
and (3) R1 ∪ R2 spans AQ0n−1. We can write R1 as
u, R3, p, q, R4, v and write R2 asx, R5, s, t, R6, y. By induction, there exist two disjoint paths S1 and S2 such that (1)S1is a path joiningphtoqh withl(S1) =
2n−1− l2+ 2n−2− 2, (2) S2is a path joiningshtoth withl(S2) = l2− 2n−2, and (3)S1∪ S2spansAQ1n−1. We setP1 as u, R3, p, ph, S1, qh, q, R4, v and P2 as
x, R5, s, sh, S2, th, t, R6, y. Obviously, P1andP2are the required paths.
Subcase 3.5:l2 = 2n−2or 2n−2+ 2. Let k = 0 if l2= 2n−2andk = 2 if l2= 2n−2+2. By induction, there exist
two disjoint pathsR1andR2 such that (1)R1 is a path joiningu to v with l(R1) = 2n−2− k, (2) R2is a path joiningx to y with l(R2) = 2n−2+k−2, and (3) R1∪R2
spansAQ0n−1. We can writeR1 as u, R3, p, q, R4, v and writeR2asx, R5, s, t, R6, y. By Lemma 3, there exists a hamiltonian pathS of AQ1n−1− {sn, tn} joining pn to qn. We set P
1 as u, R3, p, pn, S, qn, q, R4, v andP2asx, R5, s, sn, tn, t, R6, y. Obviously, P1and
P2are the required paths.
x y u v p q ph qh x y u v p q ph qh s t sh th R 3 R 4 S R2 R3 R4 R 5 R6 S 1 S 2
(a)
(b)
Figure 4: Subcase 3.3 and Subcase 3.4.
Case 4:{x, v, y} ⊂ V (AQ1n−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} = NbdAQn(x) ∩ NbdAQn(y). See Figure 5(a)
for an illustration. Obviously, there exists a vertexp in
N bdAQ1
disjoint pathsS1andS2such that (1)S1is a path joining p to v with l(S1) = l1− 2n−1, (2)S2is a path joiningx toy with l(S2) = l2, and (3)S1∪ S2spansAQ1n−1. By Lemma 3, there exists a hamiltonian pathR of AQ0n−1 joiningu and ph. We setP1 asu, R, ph, p, S1, v and we setP2 asS2. Obviously,P1 andP2are the required paths. u v x y p ph R S1 S2 u v x y p ph s R 1 R 2 S 2 2 r S21 rh sh S 1
(a)
(b)
Figure 5: Subcase 4.1 and Subcase 4.5.
Subcase 4.2: l2 = 2 if dAQn(x, y) = 1 or 2 with
{u, v} = NbdAQn(x) ∩ NbdAQn(y). The proof is the
same to Subcase 1.2.
Subcase 4.3: l2 = 2n−1− 4. Obviously, there exists a
vertexp in NbdAQ1
n−1(v) − {x, y}, and there exists a
vertexq in NbdAQ1
n−1(p) − {x, y, v, u
h}. By Lemma 3,
there exists a hamiltonian pathR of AQ0n−1joiningu to qh, and there exists a hamiltonian pathP
2 ofAQ1n−1− {v, p, q} joining x to y. We set P1asu, R, qh, q, p, v. Obviously,P1andP2are the required paths.
Subcase 4.4: l2 = 2n−1− 2. Let v be an element in
{vh, vc} − {u}. By Lemma 3, there exists a
hamilto-nian pathR of AQ0n−1 joiningu to v, and there exists a hamiltonian pathP2of AQ1n−1− {v} joining x to y. We setP1asu, R, v, v. Obviously, P1andP2are the required paths.
Subcase 4.5: l2 = 2n−1 − 1. See Figure 5(b) for an illustration. Obviously, there exists a vertex p in
N bdAQ1
n−1(v) − {x, y}. By induction, there exist two
disjoint pathsS1andS2such that (1)S1is a path joining p to v with l(S1) = 1, (2) S2is a path joiningx to y with
l(S2) = 2n−1− 3, and (3) S1∪ S2spansAQ1n−1.
Ob-viously, we can writeS2as x, S21, r, s, S22, y for some vertexr and s such that u /∈ {rh, sh}. Again by induc-tion, there exist two disjoint pathsR1 andR2 such that (1)R1is a path joiningu to phwithl(R1) = 2n−1− 3,
(2)R2is a path joiningrhtoshwithl(R2) = 1, and (3)
R1∪ R2spansAQ0n−1. We setP1 asu, R1, ph, p, v and setP2 as x, S12, r, rh, sh, s, S22, y. Obviously, P1 andP2are the required paths.
Case 5: v ∈ V (AQ1n−1) and|{x, y} ∩ V (AQ0n−1)| = 1. Without loss of generality, we assume that x ∈
V (AQ0n−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} = NbdAQn(x) ∩ NbdAQn(y). The proof is the
same to Subcase 1.2.
Suppose thatdAQn(x, y) = 1. Obviously, there exists
a vertex p in NbdAQ0
n−1(x) − {u, v
h}. We set P
2 as
x, p, ph, y. By Lemma 3, there exists a hamiltonian
pathP1ofAQn− V (P2) joiningu to v. Obviously, P1
andP2are the required paths.
Suppose thatdAQn(x, y) = 2. Assume that {u, v} =
N bdAQn(x)∩NbdAQn(y). Thus, we have either v = xh
orv = xc. Moreover,u = xα, andy = vα for some
α ∈ {i | 2 ≤ i ≤ n} ∪ {i∗ | 2 ≤ i ≤ n − 1}. We set P2asx, xh∗, (xh∗)α, ((xh)α) = y in the case of v =
xh. Otherwise, we setP
2as x, xh, (xh)α, ((xh∗)α) = y. By Lemma 3, there exists a hamiltonian path P1 of AQn − V (P2) joining u to v. Obviously, P1 and
P2 are the required paths. Now, assume that {u, v} =
N bdAQn(x) ∩ NbdAQn(y). By Lemma 1, there exists
a vertex p in (NbdAQn(x) ∩ NbdAQn(y)) − {u, v}.
Without loss of generality, we may assume that p is in AQ0n−1. By Lemma 1, there exists a vertex q in
(N bdAQ0
n−1(p) ∩ NbdAQ0n−1(x)) − {u}. By Lemma 3,
there exists a hamiltonian pathP1ofAQn− {x, q, p, y} joiningu to v. We set P2asx, q, p, y. Obviously, P1 andP2are the required paths.
Suppose thatdAQn(x, y) = 3. By Lemma 4, there are
two shortest pathsR1andR2ofAQnjoiningx to y such thatR1can be written asx, r1, r2, y with {r1, r2} ⊂
V (AQ0n−1) and R2can be written asx, s1, s2, y with
{s1, s2} ⊂ V (AQ1n−1). Suppose that u = r2 or v = s1. Without loss of generality, we assume that u = r2. By Corollary 5, there exists a vertex t ∈
N bdAQ0
n−1(x) ∩ NbdAQ0n−1(r2)− {u}. We set P2 as
x, t, r2, y. By Lemma 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 = r2 and v = s1. By Corollary 5, there exists a vertexp in NbdAQ0
n−1(x) ∩ NbdAQ0n−1(u). Obviously,
dAQn(p, y) = 2. By Lemma 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 = v. We set
P2 asx, p, q, y. By Lemma 3, there exists a hamilto-nian pathP1ofAQn− V (P2) joiningu to v. Obviously,
P1andP2are the required paths.
Subcase 5.4: 4≤ l2≤ 2n−1− 1 with dAQn(x, y) = 1.
Suppose that l2 = 4. See Figure 6(a) for an illustration. Obviously, there exists a vertex p in
N bdAQ0
n−1(x) − {u, v
h}. By Lemma 1, there exists
a vertexq in (NbdAQ0
n−1(x) ∩ NbdAQ0n−1(p)) − {u}.
By Lemma 3, there exists a hamiltonian path P1 of
AQn − {x, y, p, ph, q} joining u to v. We set P2 as
x, q, p, ph, y. Obviously, P
1 andP2 are the required paths.
Suppose that 5 ≤ l2 ≤ 2n−1− 1 except that l2 = 2n−1− 2. See Figure 6(b) for an illustration. Obviously,
there exist a vertexp in NbdAQ0
n−1(x)−{u, v
h, yh} and
a vertexs in NbdAQ0
n−1(u) − {x, p, v
h, yh}. By
induc-tion, there exist two disjoint pathsR1andR2such that (1)
R1is a path joiningu to s with l(R1) = 2n−1−2−l2, (2)
R2is a path joiningp to x with l(R2) = l2− 2, and (3)
R1∪R2spansAQ0n−1. By Lemma 3, there exists a hamil-tonian pathS of AQ1n−1− {y, ph} joining shtov. We setP1asu, R1, s, sh, S, v and P2asx, R2, p, ph, y. Obviously,P1andP2are the required paths.
Suppose thatl2 = 2n−1− 2. See Figure 6(c) for an illustration. Lets and p be two vertices in V (AQ0n−1)−
{u, x, vh, yh}. By induction, there exist two disjoint
pathsR1andR2such that (1)R1is a path joiningu to s withl(R1) = 2n−2, (2)R2is a path joiningp to x with
l(R2) = 2n−2− 2, (3) R1∪ R2spansAQ0n−1. Similarly,
there exist two disjoint pathsS1 andS2such that (1)S1 is a path joiningshtov with l(S1) = 2n−2− 1, (2) S2 is a path joiningphtoy with l(S2) = 2n−2− 1, and (3)
x y u v ph p s sh R1 R 2 S x y q u v ph p x y u v ph p s sh R1 R 2 S2 S1
(a)
(b)
(c)
Figure 6: Subcase 5.4.S1∪ S2spansAQ1n−1. We setP1asu, R1, s, sh, S1, v andP2asx, R2, p, ph, S2, y. Obviously, P1andP2are the required paths.
Subcase 5.5: 4≤ l2 ≤ 2n−1− 1 except l2 = 2n−1− 3
withdAQn(x, y) ≥ 2.
Suppose that dAQn(x, y) = 2 with {u, v} =
N bdAQn(x) ∩ NbdAQn(y). See Figure 7(a) for an
illus-tration. Thus, we have eitherv = xhorv = xc. More-over,u = xαandy = (xh)αfor someα ∈ {i | 2 ≤ i ≤
n} ∪ {i∗ | 2 ≤ i ≤ n − 1}. Obviously, there exists a
ver-text in NbdAQ1
n−1(v) − {x
h, y, xc, uh}. By induction,
there exist two disjoint pathsR1andR2such that (1)R1 is a path joiningt to v with l(R1) = 2n−1−1−l2, (2)R2 is a path joiningxc toy with l(R2) = l2− 1 in the case ofv = xh; otherwiseR2is a path joiningxhtoy with
l(R2) = l2−1, and (3) R1∪R2spansAQ1n−1. By Lemma
3, there exists a hamiltonian pathS of AQ0n−1−{x} join-ingthtou. We set P1 asu, S, th, t, R1, v and P2 as
x, xc, R2, y in the case of v = xh; otherwise, we setP2
asx, xh, R2, y. Obviously, P1andP2are the required paths.
Suppose that dAQn(x, y) = 2 with {u, v} =
N bdAQn(x) ∩ NbdAQn(y). See Figure 7(b) for an
illus-tration. Then, there exists a vertexp in (NbdAQn(x) ∩
N bdAQn(y)) − {u, v}. Without loss of generality, we
may assume thatp ∈ V (AQ1n−1). Obviously, there exists
a vertext in NbdAQ1
n−1(v) − {y, p, uh, xh}. By
induc-tion, there exist two disjoint pathsR1andR2such that (1)
R1is a path joiningt to v with l(R1) = 2n−1− 1 − l2, (2)R2is a path joiningp to y with l(R2) = l2− 1, and (3) R1 ∪ R2 spansAQ1n−1. By Lemma 3, there exists a hamiltonian pathS of AQ0n−1− {x} joining th tou. We setP1asu, S, th, t, R1, v and P2asx, p, R2, y. Obviously,P1andP2are the required paths.
x y u v xc t th R 1 R2 S x y u p v t th R1 R 2 x y u p=r k-1 v s sh R1 R2 x y u=r k-1 v s sh R1 R 2 rk-2 p S1 S2 (a) (b) (c) (d) S S Figure 7: Subcase 5.5.
Lemma 4, there are two shortest paths S1 and
S2 of AQn joining x to y such that S1 can be written as x = r0, r1, r2, . . . , rk−1, y with
(V (S1) − {y}) ⊂ V (AQ0n−1) and S2 can be writ-ten as x, s1, s2, . . . , sk−1, y with (V (S2)− {x}) ⊂
V (AQ1n−1). Suppose that u = rk−1. See Figure 7(c) for an illustration. We setp = rk−1. Again, there exists a vertexs in NbdAQ0
n−1(u) − {x, p, y
h, vh}. By
induc-tion, there exist two disjoint pathsR1andR2such that (1)
R1is a path joiningu to s with l(R1) = 2n−1− 1 − l2, (2)R2is a path joiningp to x with l(R2) = l2− 1, and (3) R1 ∪ R2 spansAQ0n−1. By Lemma 3, there exists a hamiltonian pathS of AQ1n−1− {y} joining shtov. We setP1asu, R1, s, sh, S, v and P2asx, R2, p, y. Obviously,P1andP2are the required paths.
Now we assume thatrk−1 =u and s1=v. See
Fig-ure 7(d) for an illustration. SincedAQn(rk−2, y) = 2,
by Lemma 4, there exists a vertexp ∈ NbdAQn(rk−2)
inV (AQ1n−1) such that dAQn(p, y) = 1. Suppose that
l2 = 4 with dAQn(x, y) = 3. Thus, x, r1, p, y is a
shortest path joiningx and y. By Lemma 1, there ex-ists a vertexq ∈ NbdAQ1
n−1(p) ∩ NbdAQ1n−1(y) − {v}.
By Lemma 3, there exists a hamiltonian path P1 of
AQn − {x, r1, p, q, y} joining u to v. We set P2 as
x, r1, p, q, y. Obviously, P1 andP2 are the required paths. Suppose thatl2 = 4 with dAQn(x, y) = 4. Thus,
P2 = x, r1, r2, p, y is a shortest path joining x and
y. By Lemma 3, there exists a hamiltonian path P1 of
AQn−{x, r1, r2, p, y} joining u to v. Obviously, P1and
P2 are the required paths. Suppose that 5 ≤ l2 ≤ 2n−2 withdAQn(x, y) ≥ 3. Obviously, there exists a vertex
s in NbdAQ0
n−1(u) − {x, rk−2, y
h, vh}. By induction,
there exist two disjoint pathsR1andR2such that (1)R1 is a path joiningu to s with l(R1) = 2n−1− l2, (2)R2 is a path joiningrk−2tox with l(R2) = l2− 2, and (3)
R1∪R2spansAQ0n−1. By Lemma 3, there exists a
hamil-tonian pathS of AQ1n−1− {p, y} joining shtov. We set P1 as u, R1, s, sh, S, v and P2 as x, R2, rk−2, p, y. Obviously, P1 andP2 are the required paths. Suppose that 2n−2+ 1 ≤ l2 < 2n−1− 1 except 2n−1− 3 with dAQn(x, y) ≥ 3. Obviously, there exists a vertex s in
N bdAQ0
n−1(u) − {x, rk−2, yh, vh}. By induction, there
exist two disjoint pathsR1andR2such that (1)R1 is a path joiningu to s with l(R1) = 2n−2+ 1, (2) R2 is a path joiningrk−2 tox with l(R2) = 2n−2− 3, and (3)
R1∪ R2 spansAQ0n−1. Again by induction, there exist
two disjoint paths S1 andS2 such that (1)S1 is a path joiningshtov with l(S1) = 2n−1−l2+ 2n−2−4, (2) S2 is a path joiningp to y with l(S2) = l2−2n−2+2, and (3)
S1∪ S2spansAQ1n−1. We setP1asu, R1, s, sh, S1, v andP2asx, R2, rk−2, p, S2, y. Obviously, P1andP2 are the required paths.
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
ifl2 = 2n−1− 1. Obviously, there exist two vertices s
andp in AQ0n−1− {u, x, vn, yn}. By induction, there exist two disjoint pathsR1andR2such that (1)R1 is a path joiningu to s with l(R1) = 2n−2− t, (2) R2 is a path joiningp to x with l(R2) = 2n−2+ t − 2, and
(3) R1∪ R2 spans AQ0n−1. Similarly, there exist two disjoint pathsS1andS2such that (1)S1is a path joining sn tov with l(S1) = 2n−2− t, (2) S2is a path joining
pn toy with l(S
2) = 2n−2+ t − 2, and (3) S1∪ S2 spansAQ1n−1. We setP1asu, R1, s, sn, S1, v and P2 as x, R2, p, pn, S2, y. Obviously, P1 andP2 are the required paths.
Thus, Theorem 7 is proved.
4 Concluding remarks
Now, we make some remarks to illustrate that some
in-teresting properties of augmented cubes are consequences of Theorem 7.
Remark 1. The hamiltonian connected property of
aug-mented cubes, proved in [8], states that there exists a hamiltonian path ofAQn joining any two different ver-ticesu and y. Now, we prove that AQn is hamiltonian connected by Theorem 7. Obviously,AQnis hamiltonian connected forn = 1. Since n ≥ 2, we can choose a pair of adjacent verticesv and x such that {v, x}∩{u, y} = ∅. By Theorem 7, there are two disjoint paths P1 andP2 such that (1)P1 is a path joiningu to v, (2) P2is a path joiningx to y, and (3) P1∪ P2spansAQn. Obviously,
u, P1, v, x, P2, y forms a hamiltonian path joining u to y. Thus, AQnis hamiltonian connected.
Remark 2. The panconnected property ofAQn, proved in [13], stated that between any two different verticesx andy of AQnthere exists a pathPl(x, y) of length l for
anydAQn(x, y) ≤ l ≤ 2n− 1. Now, we prove that AQn
is panconnected by Theorem 7. Obviously,AQn is pan-connected forn = 1, 2. Now, we consider that n ≥ 3.
Suppose that l = 2n − 1. By Remark 1, AQn is hamiltonian connected. Obviously, the hamiltonian path ofAQnjoiningx and y is of length 2n− 1. Suppose that
l = 2n − 2. Let u be a vertex in NbdAQn(y) − {x}.
By Lemma 1, there exists a vertexv in (NbdAQn(u) ∩
N bdAQn(y)) − {x}. By Theorem 7, there exist two
dis-joint pathsP1 andP2 such that (1)P1 is a path joining x to u with l(P1) = 2n− 3, (2) P2 is a path joiningy tov with l(P2) = 1, and (3) P1∪ P2 spansAQn. Ob-viously,x, P1, u, y is a path of length 2n− 2 joining x
toy. Suppose that l = 2n− 3. We can find two adja-cent verticesu and v such that {u, v} ∩ {x, y} = ∅. By Theorem 7, there exist two disjoint pathsP1andP2such that (1)P1is a path joiningx to y with l(P1) = 2n− 3,
P1∪ P2 spans AQn. Obviously,P1 is a path of length
2n− 3 joining x to y. Suppose that l ≤ 2n − 4. By
Lemma 2, there exist two vertices u and v such that
dAQn(u, v) = 2, {x, y} = NbdAQn(u) ∩ NbdAQn(v),
and{u, v} = NbdAQn(x) ∩ NbdAQn(y). By Theorem
7, there exist two disjoint pathsP1andP2 such that (1)
P1is a path joiningx to y with l(P1) = l, (2) P2is a path joiningu to v with l(P2) = 2n− 2 − l, and (3) P1∪ P2
spansAQn. Obviously,P1is a path of lengthl joining x
toy. Thus, AQnis panconnected.
Remark 3. The edge-pancyclic property property of
AQn stated that for any edge e = (x, y) and for any
3 ≤ l ≤ 2n, there exists a cycle of lengthl containing
e if n ≥ 2. We prove that AQnis edge-pancyclic by The-orem 7. Obviously,AQn is edge-pancyclic forn = 2.
Thus, we consider thatn ≥ 3.
Suppose thatl = 3. By Lemma 1, there exists u ∈ N bdAQn(x)∩NbdAQn(y). Obviously, x, y, u, x forms
a cycle of length three containinge. Now, we consider thatl = 2n andl = 2n− 1. By Lemma 1, there exists
v ∈ (NbdAQn(u) ∩ NbdAQn(y)) − {x}. By Theorem 7,
there exist two disjoint pathsP1andP2such that (1)P1is a path joiningx to u with l(P1) = 2n− 3, (2) P2is a path joiningv to y with l(P2) = 1, and (3) P1∪P2spansAQn. Obviously,x, P1, u, v, y, x forms a cycle of length 2n containinge and x, P1, u, y, x forms a cycle of length
2n− 1 containing e. Suppose l = 2n− 2. By Theorem
7, there exist two disjoint pathsQ1andQ2such that (1)
Q1is a path joiningx to y with l(Q1) = 2n− 3, (2) Q2 is a path joiningu to v with l(Q2) = 1, and (3) Q1∪
Q2 spansAQn. Obviously,x, Q1, y, x forms a cycle of length 2n− 2 containing e. Suppose that 4 ≤ l ≤
2n− 3. By Lemma 2, there exists two vertices p and q of
AQnsuch thatdAQn(p, q) = 2, {x, y} = NbdAQn(p)∩
N bdAQn(q), and {p, q} = NbdAQn(x) ∩ NbdAQn(y).
By Theorem 7, there exist two disjoint pathsR1 andR2 such that (1)R1is a path joiningx to y with l(R1) = l−1,
(2)R2is a path joiningu to v with l(R2) = 2n− l − 1,
and (3) R1 ∪ R2 spans AQn. Obviously,x, R1, y, x forms a cycle of lengthl containing e.
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