Appendix
A
Some Proofs and Useful Lemmas
Lemma 1 The graph P2 × Ck is Hamiltonian-connected when k ≥ 3 is odd.
Proof: Suppose V (P2 ) = {u0 , u1 }. To prove the lemma, we consider two super nodes, Ck0 and Ck1 ,
where Ck0 = {u0 } × Ck and Ck1 = {u1 } × Ck . For any two distinct vertices x and y in P2 × Ck ,
there are two cases.
Case 1: x and y are both in the same super node. Without loss of generality, we assume that
x, y ∈ Vk0 . If d(x, y) is odd (resp. even), a Hamiltonian x-y path can be constructed using
the routing strategy shown in Figure 1(a) (resp. Figure 1(b)).
Figure 1: Illustration of the proof of Lemma 1
Case 2: x and y are in different super nodes. Without loss of generality, we assume that x ∈ Vk0
and y ∈ Vk1 . If d(x, y) is odd (resp. even), a Hamiltonian x-y path can be constructed
Figure 2: Illustration of the proof of Lemma 2; the black vertices in (b) and (d) are excluded
from the u-v paths.
using the routing strategy shown in Figure 1(c) (resp. Figure 1(d)).
Lemma 2 The graph P2 × Ck is almost Hamiltonian-connected when k ≥ 3.
Proof: According to Lemma 1, we only need to prove this lemma when k is even. The proof is
similar to that presented in Lemma 1; the only difference is that the Hamiltonian u-v path is
constructed by using the routing strategy shown in Figure 2.
B
Application to Multiprocessor Systems
In this section, we apply our results to two multiprocessor systems: generalized hypercubes and
nearest neighbor mesh hypercubes.
Definition 1 (Generalized hypercubes GQmr ,mr−1 ,...,m1 ) First, we describe the mixed radix
number system. Let N be the total number of processors represented as N = mr · mr−1 · · · · · m1 ,
2
where mi > 1 is a positive integer for 1 ≤ i ≤ r. Then, each node (processor) X between 0 and
N − 1 can be expressed as an r-tuple (xr xr−1 . . . x1 ), where 0 ≤ xi ≤ mi − 1. We define the
weight wi for each xi so that
wi =
Then, X =
r
i=1
1,
i−1
if i = 1;
j=1 mj , otherwise.
(1)
xi wi . Each node X = (xr xr−1 . . . xi+1 xi xi−1 . . . x1 ) is connected to nodes
(xr xr−1 . . . xi+1 xi xi−1 . . . x1 ) for all 1 ≤ i ≤ r, where xi takes all integer values between 0 and
mi − 1, except xi itself. In other words, GQmr ,mr−1 ,...,m1 ∼
= Kmr × Kmr−1 × · · · × Km1 .
For example, let us consider N = 24 = 4 · 3 · 2. In this case, m1 = 2, m2 = 3, m3 = 4 and
w1 = 1, w2 = 2, w3 = 6. For a node X = (x3 x2 x1 ), where 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 2, and 0 ≤ x3 ≤ 3,
we have 0 ≤ X ≤ 23. Note that 010 = (000), 510 = (021), and 2310 = (321) in this mixed radix
system. The degree of a node in GQmr ,mr−1 ,...,m1 is l = ri=1 (mi − 1), and the number of edges
in GQmr ,mr−1 ,··· ,m1 is equal to
N
2
· l. Because GQmr ,mr−1 ,...,m1 = Kmr × GQmr−1 ,mr−2 ,...,m1 , there
exist mr super nodes GQjmr−1 ,mr−2 ,...,m1 for 0 ≤ j ≤ mr − 1 with the vertex sets Vmj r−1 ,mr−2 ,...,m1 =
{jxr−1 . . . x2 x1 | 0 ≤ xi ≤ mi − 1 for r − 1 ≤ i ≤ 1} that form a partition of V (GQmr ,mr−1 ,...,m1 ).
Figure 3(a) illustrates GQ4,3,2 .
Figure 3: (a) GQ4,3,2 = K4 × K3 × K2 and (b) NQ4,3,2 = C4 × C3 × C2
Definition 2 (Nearest neighbor mesh hypercubes ( NQmr ,mr−1 ,...,m1 )) The numbering system used to define the nodes of NQmr ,mr−1 ,...,m1 is the same as that used for GQmr ,mr−1 ,...,m1 .
NQmr ,mr−1 ,...,m1 is defined as follows: a node X = (xr xr−1 · · · xi+1 xi xi−1 · · · x1 ) is connected to
(xr xr−1 · · · xi+1 xi xi−1 · · · x1 ) for all 1 ≤ i ≤ r, where xi ≡ (xi ± 1) mod mi . In other words,
NQmr ,mr−1 ,...,m1 ∼
= Cmr × Cmr−1 × · · · × Cm1 .
3
Because NQmr ,mr−1 ,...,m1 = Cmr ×NQmr−1 ,mr−2 ,...,m1 , there exist mr super nodes NQmr−1 ,mr−2 ,...,m1
for 0 ≤ j ≤ mr − 1 with the vertex sets Vmj r−1 ,mr−2 ,...,m1 = {jxr−1 . . . x2 x1 | 0 ≤ xi ≤ mi − 1 for
r − 1 ≤ i ≤ 1} that form a partition of V (NQmr ,mr−1 ,...,m1 ). Figure 3(b) illustrates NQ4,3,2 . Note
that if r = 1, then GQm1 and NQm1 become Km1 and Cm1 respectively. Hereafter, we only con
sider the case where r ≥ 2. It is easy to see that GQmr ,mr−1 ,...,m1 is an l = ri=1 (mi − 1)-regular
r
graph and NQmr ,mr−1 ,...,m1 is an l =
i=1 min{2, (mi − 1)}-regular graph. Moreover, in both
GQmr ,mr−1 ,...,m1 and NQmr ,mr−1 ,...,m1 , the number of edges is equal to
N
2
· l.
The k-ary n-cube Qkn [4, 7], where k ≥ 2 and n ≥ 1 are positive integers, has N = k n vertices,
each of which has the form x = xn xn−1 . . . x1 , where xi ∈ {0, 1, . . . , k − 1} for 1 ≤ i ≤ n. Two
vertices x = xn xn−1 . . . x1 and y = yn yn−1 . . . y1 in Qkn are adjacent if and only if there exists an
integer j, where 1 ≤ j ≤ n, such that xj = yj ±1 (mod k) and xl = yl for all l ∈ {1, 2, ..., n}−{j}.
In fact, Qkn = Ck × Ck × · · · × Ck , where Ck is a cycle of k vertices. Note that each vertex in
n times
Qkn has degree 2n when k ≥ 3, and degree n when k = 2. Obviously, Qk1 is a cycle of length k,
Q2n is an n-dimensional hypercube, and Qk2 is a k × k wrap-around mesh. In particular, when
mr = mr−1 = · · · = m1 = k, NQmr ,mr−1 ,...,m1 is equal to Qkr . Figure 4 illustrates Q61 , Q52 , and Q33 .
Figure 4: (a) Q61 , (b) Q52 , and (c) Q33 .
We describe some useful properties.
Lemma 3 [3] The diam(GQmr ,mr−1 ,...,m1 ) = r and diam(NQmr ,mr−1 ,...,m1 ) =
r
mi
i=1 2 .
A graph G of order n is bipanconnected if there exists a path of length l joining any two
different vertices x and y with dG (x, y) ≤ l ≤ n − 1 and l − dG (x, y) is even. On the other hand,
G is bipancyclic if it contains cycles of every even length from its girth to n if n is even, or if it
contains cycles of every even length from its girth to n − 1 if n is odd [9]. In addition, G is said
4
to be edge-bipancyclic if every edge lies on a cycle of length l for every even l between its girth
to n.
Lemma 4 [10, 11] Let n ≥ 2 be a positive integer. Then, the following statements hold.
(1) Qkn is bipanconnected and edge-bipancyclic when k ≥ 3.
(2) Qkn is almost Hamiltonian-connected when k ≥ 2 is even.
Lemma 5 [8] Let n ≥ 2 be a positive integer. Then, the following statements hold.
(1) Qkn is k2 n-panconnected when k ≥ 3 is odd.
(2) Qkn is k-pancyclic when k ≥ 3 is odd.
(3) Qkn is both bipanconnected and edge-bipancyclic when k ≥ 2 is even.
Lemma 6 [5] If graphs G1 and G2 are traceable, then G1 × G2 is not Hamiltonian if and only
if both have an odd order and neither has an odd cycle.
Lemma 7 If graphs G1 and G2 are Hamiltonian, then G1 × G2 is also Hamiltonian.
Corollary 1 For m, n ≥ 2, the graph Km × Kn is Hamiltonian.
Corollary 2 The following two statements hold:
(1) Let G1 and G2 be two connected graphs, but G1 ×G2 is not isomorphic to K1 or K2 . If G1 and
G2 are edge-Hamiltonian, or one is isomorphic to K1 or K2 and the other is edge-Hamiltonian,
then G1 × G2 is also edge-Hamiltonian.
(2) For any positive integers m and n with mn ≥ 3, the graph Km × Kn is edge-Hamiltonian.
Lemma 8 Let G1 and G2 be two graphs with orders m ≥ 3 and n ≥ 3 respectively. If both
graphs are edge-Hamiltonian, then G1 × G2 is also edge-Hamiltonian.
Lemma 9 Let G1 and G2 be two graphs with orders m and n respectively, where m ≥ 3 or
n ≥ 3. If G1 and G2 are Hamiltonian-connected, then G1 × G2 is also Hamiltonian-connected.
Corollary 3 If m = 2 or n = 2, the graph Km × Kn is Hamiltonian-connected.
Lemma 10 Given two graphs, G1 and G2 , if G1 is k1 -panconnected and G2 is k2 -panconnected
with |V (Gi )| > ki + 1, where ki ≥ 1, for i = 1, 2, then G1 × G2 is (k1 + k2 )-panconnected.
5
B.1
Properties derived for nearest neighbor mesh hypercubes
Theorem 1 If r ≥ 2 and mi ≥ 3 for 1 ≤ i ≤ r, N Qmr ,mr−1 ,...,m1 is almost Hamiltonianconnected.
Corollary 4 If N Qmr ,mr−1 ,...,m1 ∈ N Qodd , it is Hamiltonian-connected.
Proof: This can be proved by the method used in Theorem 1.
Next, we investigate the panconnectivity of nearest neighbor mesh hypercubes.
Proposition 1 For all a, b ∈ N , either a ≤
a+b
2
or b ≤
a+b
.
2
Moreover, if a = b, either
− 1 or b ≤ a+b
− 1.
a ≤ a+b
2
2
Lemma 11 Let m, n ≥ 3 be odd integers; and let k1 = m − y1 + y2 and k2 = n − y2 + y1 , where
0 ≤ y1 ≤ m2 and 1 ≤ y2 ≤ n2 . Then, the following statements hold: (1) either k1 or k2 is
less than or equal to the lowest odd number that is greater than or equal to
m+n
2
− 1; and (2) if
k1 and k2 are both even, either k1 or k2 is less than or equal to the lowest even number that is
greater than or equal to
Proof: Let d =
m+n
2
m+n
2
− 1.
− 1. Then,
k1 +k2
2
=
m+n
2
= d + 1; hence, either k1 ≤
We first show that Statement (1) holds. If d is even, d + 1 =
m+n
2
m+n
2
m+n
2
or k2 ≤
m+n
.
2
is odd, which implies that
is the lowest odd number that is greater than d. However, if d is odd, then
m+n
2
is even,
which implies that m = n because m and n are both odd. By Proposition 1, we have either
k1 ≤
m+n
2
− 1 = d or k2 ≤
m+n
2
− 1 = d.
Next, we show that Statement (2) holds. If d is odd, then
k1 +k2
2
k1 +k2
2
is even, which implies that
is the lowest even number that is greater than d. However, if d is even, then
k1 +k2
2
is odd,
which implies that k1 = k2 because k1 and k2 are both even. Therefore, by Proposition 1, we
have either k1 ≤
k1 +k2
2
− 1 = d or k2 ≤
k1 +k2
2
− 1 = d.
Lemma 12 If m, n ≥ 3 are both odd integers, then N Qm,n = Cm × Cn is ( m2 + n2 )panconnected.
Proof: We attempt to construct an x-y path of any length l for m2 + n2 ≤ l ≤ mn − 1 for two
arbitrary vertices x and y in NQm,n . By Lemma 3, we have d(x, y) ≤ m2 + n2 =
6
m+n
2
− 1.
Without loss of generality, we assume that x = 00 and y = y1 y2 , where 0 ≤ y1 ≤
1 ≤ y2 ≤
n−1
.
2
m−1
2
and
We have the following cases.
Case 1: y1 = 0 and y2 is even.
Case 1.1: l is even. An x-y Hamiltonian path can be generated, and a path of every even
length can be obtained from it by subtracting 2 edges successively, until the length of
d(x, y) is obtained, using the routing strategy shown in Figure 5(a).
Figure 5: Illustration of the proof of Case 1 in Lemma 12
Case 1.2: l is odd. Two different x-y paths of length mn − 2 can be constructed using
two different routing strategies (see Figures 5(b) and 5(c)). Then, a path of every odd
length can be obtained from the two paths by subtracting 2 edges successively, until
the length of m − y1 + y2 and the length n − y2 + y1 are obtained, using the routing
strategies shown in Figures 5(b) and 5(c) respectively. Moreover, by Lemma 11(1), a
7
desired path of every odd length can be obtained from one of the above two routing
strategies.
Case 2: y1 = 0 and y2 is odd.
Case 2.1: l is odd. An x-y path of length mn − 2 can be generated, and a path of every
odd length can be obtained from it by subtracting 2 edges successively, until the length
of d(x, y) is obtained, using the routing strategy shown in Figure 6(a).
Figure 6: Illustration of the proofs of Cases 2 and 3 in Lemma 12
Case 2.2: l is even. Two Hamiltonian x-y paths can be constructed by using two different
routing strategies (see Figures 6(b) and 6(c)). Then, a path of every even length can
be obtained from the two paths by subtracting 2 edges successively, until the length
of m − y1 + y2 and n − y2 + y1 are obtained, using the routing strategies shown in
Figures 6(b) and 6(c) respectively. Moreover, by Lemma 11(2), the desired path of
8
every even length can be obtained from one of the above two routing strategies.
Case 3: y1 and y2 are both even.
Case 3.1: l is even. By using the routing strategy of a Hamiltonian x-y path illustrated
in Figure 6(d), the desired paths can be constructed using a method similar to that
used in Case 1.1.
Case 3.2: l is odd. By using two different routing strategies to construct two x-y paths of
length mn − 2, as shown in Figures 6(e) and 6(f) respectively, the desired paths can
be constructed by using a method similar to that employed in Case 1.2.
Case 4: y1 and y2 are odd.
Case 4.1: l is even. By using the routing strategy in Figure 7(a) to construct a Hamiltonian x-y path, the desired paths can be constructed using a method similar to that
adopted in Case 1.1.
Case 4.2: l is odd. By using two different routing strategies to construct two x-y paths of
length mn − 2, as shown in Figures 7(b) and 7(c) respectively, the desired paths can
be constructed by using a method similar to that employed in Case 1.2.
Case 5: y1 is even and y2 is odd.
Case 5.1: l is odd. By using the routing strategy in Figure 7(d) to construct an x-y path
of length mn − 2, the desired paths can be constructed by using a method similar to
that used in Case 2.1.
Case 5.2: l is even. By using two different routing strategies to construct two different
x-y paths of length mn − 2, as shown in Figures 7(e) and 7(f) respectively, the desired
paths can be constructed using a method similar to that used in Case 2.2.
Taken together, the above five cases complete the proof.
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