A New Shortest Path Routing Algorithm and Ernbedding Cycles of
Crossed Cube
Chien-Ping Changl, Ting-Yi Sung2 and Lih-Hsing Hsul
Department of Computer and Information Science
National Chiao Tung University
Hsinchu, Taiwan 30050, R.O.C.'
Institute of Information Science
Academia Sinica
Taipei, Taiwan 11529, R.0.C.2
Abstract
In order to evaluate the performance of a network
topology, we can consider the following measures:
shortest path routing complexity, vertex connectivity,
diameter and embedding of cycles. Efe presented a
shortest path routing algorithm of crossed cubes in
[l],which generated one shortest path for any pair
of vertices in O ( n 2 )time. In this paper, we define a
new distance measure which enables us to find more
shortest paths for any pair of vertices in O ( n ) time.
The vertex connectivity (simply abbreviated as
connectivity) of a network G = ( V , E ) ,denoted by
K ( G )or K , is the minimum number of vertices whose
removal leaves the remaining graph disconnected or
trivial. It follows from Menger's theorem that there
always exist K internally vertex-disjoint (abbreviated
as disjoznt) paths between any two vertices. Disjoint
paths between a pair of vertices contribute t o multipath communication between these two vertices and
provide alternative routes in the case of vertex or link
failures. Thus large conenctivity is preferred. It has
been shown in [1, 41 that K ( C Q ~=) K ( & , ) = n.
The problem of simulating one network by another
can be modeled as a graph emdedding problem. Embeddings of complete binary trees and cycles into
crossed cubes were pxesented in [3] and [6], respectively. In [6], the authors constructed one type of
cycles for an arbitrary length whereas we construct
various types of cycles in this paper.
The rest of this paper is organized as follows. Section 2 summarizes some known results on crossed
cubes and introduces aotation used in this paper. In
section 3 we give a nlzw shortest path routing algorithm which runs in O ( n ) time. Embedding of cycles
into crossed cubes by constructing various types of
cycles is presented in section 4. Finally, we make concluding remarks in section 5.
A n n-dzmensaonal crossed cube, CQn, zs a varzataon of hypercubes. In thzs paper, we gave a new shortest path routzng algorzthm based on a new dastance
measure defined herean. In comparzson wath Efe's algorzthm whzch generates one shortest path an O(n2)
tame, our algorathm can generate more shortest paths
zn O(n) tame. Furthermore, we show that CQn zs a
pancyclzc network and we construct varaous types of
cycles of an arbztrary length at least four
1
Introduction
Network topology is a crucial factor for interconnection networks since it determines the performance of
a network. Many interconnection network topologies
have been proposed in literature for connecting hundreds or thousands of processing elements. Network
topology is always represented by a graph where vertices represent processors and edges represent links between processors. Among these topologies, the binary
n-cube (or binary hypercube), denoted by Q n , IS one
of popular topologies. In a binary n-cube with N = 2n
vertices and
links, we say that the link from vertex
x = x n - l x n - - 2 . . ' 2 1 x 0 t o vertex y = yn-1yn-2
.ylyo
spans dimension a if and only if zt # yi and xJ = y j
for all j # i. We call this link the i-th link a t vertex
x or y. However, a binary n-cube does not make the
best use of its hardware in the following sense: given
N = 2" vertices and
links, it is possible to fashion networks with lower diameters than that of Q n .
One such topology is the crossed cube, which was first
proposed by Efe El]. An n-dimensional crossed cube,
denoted by CQn, has the same number of vertices and
links as Qn and is derived from changing the connection of some hypercube links. I t has a diameter of
[
9
1
an ,
improvement of approximately a factor of
2 , in a trade-off of reducing high degree of symmetry in Q n . Other properties of crossed cubes are also
studied in literature [1, 2 , 3, 4 , 61.
'
q
' '
1087-4089197 $10.00 0 1997 IEEE
2
Preliminaries; and Notation
Let G be a graph. We use V ( G ) and E ( G )
to denote the vertex set and the edge set of G,
125
respectively. Let x and y be two vertices. We
use d c ; ( z , y ) t o denote the shortest distance in G
between x and y. To define crossed cubes, we
first introduce the notation of "pair related " . Let
R = {(00,00),(10,10),(01,11),(11,01)}. Two binary
strings x = xlxo and y = y l y o are pair related if and
only if ( x ,y) E R.
Similarly, we can contract those vertices in CQak+l
with the same prefix of length three into a vertex and
obtain a graph with eight vertices. Again, this eightvertex graph is isomorphic t o CQ3 as illustrated in
Figure 2(b). We can also obtain the following observations for any two vertices u , v in CQn with n odd
and n 2 3 [l]:
Definition 1 A n n-dimensional crossed cube CQn is
recursively constructed as follows: CQ1 is a complete graph with two vertices labeled b y 0 and 1, respectively. CQn consists of two identical ( n - 1)dimenszonal crossed cubes C Q i - , and CQh-l. The
vertex U = 0un-2...uo E V(CQ;-,) and the vertex
v = 1wn1~...vo E V(CQA-,) is an edge in CQn if
and only zf
( b l ) If p3(u) E p ~ ( v )then
,
U and v are in a subgraph
isomorphic t o CQn-3.
(i)
un-2
= vn-2 if n is even, and
(ii) f o r 0 5 i <
[+I,
(u~i+1uai,vai+1wai)
E R.
Examples of CQ3 and CQ4 are illustrated in Figure
1.
Throughout the paper, any vertex in CQn is represented by an n-dimensional binary vector, e.g.,
U
=
u n - 1 u n - z ~ . ~ u 0and v = vn-1wn-2 ' " W O .
For k < n , the k-prefix of U , p k ( u ) , is defined
as U n - l U n - z . . . U n - k . We can thus write U =
Pk(U)Un-k-1
' . "0. Let x be an 1-bit string with 1 5 n .
We use C Q n ( x ) t o denote the subgraph of CQn induced by the set of vertices with the prefix z . It is
shown in [3] that C Q n ( z )is isomorphic to CQn-jzl.
Let x and y be two distinct I-bit strings with I < n.
Let U and v be vertices of C Q n ( x ) and C Q n ( y ) ,respectively. If U and v are joined by an edge in CQn,
C Q n ( x ) and CQn(y) are called adjacent subgraphs of
CQn. In other words, C Q n ( x ) and CQn(y) are adjacent subgraphs of CQn if and only if ( x , y ) is an
edge in CQI and n - 1 is even. Let C Q n ( x ,y) denote
the subgraph of CQn induced by C Q n ( x )U C Q n ( y ) .
It is proved in [l] that C Q n ( x , y ) is isomorphic to
CQn-lsi+l if C Q n ( x ) and CQn(y) are adjacent subgraphs of CQn. In particular, let 1x1 = IyI = 2 and
(z,y) be an edge in CQ2. Obviously, CQ2k(x) and
CQzk(y) are isomorphic t o CQ2k-2. It follows from
above discussion that CQ,k(x) and CQak(y) are adjacent subgraphs and that C Q z k ( x ,y) is isomorphic
to CQZk-1. Thus, we can contract those vertices in
C Q 2 k having the same prefix of length two into a vertex and obtain a graph with four vertices. It is easy to
see that this four-vertex graph is isomorphic to CQz
as shown in Figure 2(a). For any two vertices u , u in
CQn with n even and n 2 4 , the following statements
can be easily observed [1]:
(b2) If ~ C Q ~ ( ~ ~ ( U ) , P=
~ (1,
Y then
) ) U and v are in a
subgraph isomorphic t o CQn-2.
(b3) If d C Q 3 ( P 3 ( U ) , P 3 ( U ) ) = 2! then a neighbor U' of U
and U are in a subgraph isomorphic to CQn-a.
Let ( U , v ) be an edge of CQn. When vertices U and
v have a leftmost differing bit at position d , we say
that v is the d-neighbor of U and t h a t the edge ( U , U )
is an edge of dzmension d or d i m ( u , v ) = d . We call
( U , v ) a dim-d edge. For example, let U = 10111, then
the 4-, 3-, 2-, 1- and 0-neighbors of U are given by
01101,11101,10001,10101 and 10110, respectively.
3
Shortest path routing
The ith double bit of vertex U is defined as a 2-bit
string u2i+1u2i for 0 5 i 51?J- 1, and as simply a
single bit u2i for i =I;]and R odd. Bit 1 is called the
most significant bit between U and v if pn-l-l(u) =
p n - / - 1 ( u ) and U { # V I . Let i*
called the most
significant double bit. We define a function p on u , v
as follows:
=if],
pj(u,U ) = 0 for all j
2 i* + 1,
Subsequently, we recursively define p j ( u , v ) for j
i* - 1 as follows:
0 i f ~ 2 ~ + 1 and
~ 2 jv2j+lv2j are
P j ( U , U)
=
distance-preserving pair related,
1 otherwise,
We say that u2j+lu2j and ~ 2 ~ + 1 v zfor
j , j 5 i* - 1, are
distance-preserving pair related (abbreviated as d.-p.
pair related) if one of the following conditions holds:
( a l ) If p a ( " ) = p 2 ( v ) , then U and v are in a subgraph
isomorphic to CQn-2.
(a2) If ~ C Q ~ ( ~ Z ( U ) , ~ =
~ ( 1,
V then
) )
subgraph isomorphic t o CQn-l
U
and v are in a
(as) If&Q,(PZ(U),pz(v)) = 2 , then a neighboi U' of
and v are in a subgraph isomorphic to CQ,a-l
U
("Zj+l?L2.7,
I26
5
VZj+lVZj)
E ((00,OO),
(10,lO))
We write
-
d.-p.
u2j+lu2j
Vzj+lVzj
if
uzj+luzj
Initialization: Calculate p ( u , v ) .
Define Q i = Qz = { j I p j ( u ,w) # 0 } ,
T = { j I p j ( u , v ) # 0 , j < i*, and either
and
+
d.-p.
are d.-p. pair related, and u 2 j + l u z j
We define the p a i r related d i s t a n c e
between U and v , denoted by p(u, v ) , as
w2j+lw2j
u 2 j + 1 ~ 2 j otherwise.
For example, let U = 01101101100001 and v =
01010011100101 be vertices in CQ14. We recursively
find pi as follows: p s ( u ,v ) = 0 , p5(u,v ) = 2, p4(u,v ) =
1,P 3 ( U , U) = 0,P Z ( U , v ) = 0,P l ( U , U) = 1, po(u, v ) = 0 ,
and p(u, v) = 4.
Now consider the relationship between ~ C Q( ,U , v )
and p ( u , v ) . Let P be a shortest path from U to v.
If uzi*+!uzi* = i&*+1V2?*,it requires two steps in P .
Otherwise, by the definition of i” it requires a single
Step 2: If Q1 = 0
then output shortest paths and go t o Step 4.
Step 3: I f either
uzi+lUzi -’
vzi+lvzi holds for some i E Q1
then choose such smallest i;
else choose i = mux j I j E Q1}.
Call One-St’ep-Route i, Q 1 ) and go t o Step 2.
-+
1(
after identifying
2 with those a t v using
p,(u, v ) steps, we can reach a vertex such that
bits a t positions 2 j 1 and 2 j identical t o vzj+1 and
v2j+1v2jl
Step 4: If Q2 = 0
then output shortest paths and STOP
+
Step 5: Choose 1 = m u x { j I j E Q 2 ) and
m = m a z { j I j E Q z and j < I } .
d.-p.
If uzj+lu2j
v2j+1v2jl a t least one step is
required in P to change the 2 j 1)-th and the 2j-th
bits t o v2j+l and v z j . With t e above observation, we
have the following remark.
Remark:
+
h
I f p l ( u , v ) = 1 and
-
d.-p.
U~m+1~2m
v2m+1Wzm
2 p ( u >U ) .
Proof: We prove this lemma by induction on p ( u , v ) .
By convention, we define p ( u > v )= 0 if and only if
d c ~ , ( u , v )= 0. We first show that the lemma is
true for p ( u , v ) = 1. Let 1 be the most significant
bit, and let i” =
Since p ( u , v ) = 1, it follows that
uzi+lu2i
-
-
~ c Q , ( u ~=
v ) 1 and v is a neighbor of U . Suppose
that the lemma is true for p ( u , v ) = d c g , ( u , v ) < m
and m 2 2. Now let p ( u , v ) = m. There is a neighbor U’ of U such that p ( u ’ , v ) < m. By the induction hypothesis, we have p(u’,v ) = ~ c Q , ( u ‘U]) < m
and p ( u , u ’ ) = ~ C Q , ( U , U / ) = 1. Since m = p ( u , v ) 5
~ c Q , ( uU>) 5 d c g n ( u ,U ’ ) S & Q , ( U ’ )
v ) 5 m , it follows
that p(u,v ) = ~ C Q , ( U , , U )= m. Hence the lemma fol0
lows.
Based on the above lemma, we present a shortest
path routing algorithm which is different from the one
proposed in [l].
Algorithm: Shortest Path-Routing
Input: Vertices U and v
Output: Shortest paths from
U
i
One-StepRoute(j, Q)
If p j ( u , v ) = 2
then route to U’, i.e., the 2j-th or ( 2 j 1)-th
neighbor of U ,
U) = Pj ( U , U) - 1;
else route t o U ’ , the ( 2 j 1)-th neighbor of U
+
d.-p.
v2i+1wzi for all 0 5 i <
i”. Since p i * ( u , v )= 1 is odd, u ~ i + l u z id.-p. v z i + ~ v z i
implies ( u z i + l u z i , v z i + l v z i ) E R. Hence p ( u , v ) =
p p ( U , v) = 1 and
v~m+1~2m
then (call One-Step-Route m , Q 2 ) or
call One-Step-Route 1 , Q 2 ) ;
else c.all One-Step-Route(1, Q z ) .
Go t o Step 4.
Lemma 1 p ( u , v ) = dcQ,(u, v )
I;].
-
d.-p.
uZm+lGZm
then call One-Step-Route(m, Q 2 ) ;
else
if p l ( u , w) = 1 and
p ( u ,v ) is a lower bound for ~ c Q = (v),
u,
d C Q , ( u , U)
-
d.-p.
U~i+lu~i
v z i + l v ~ or
i
d.-p.
d.-p.
+
or
VZj+lV2j).
Step 1: I f T = 0
then go to Step 2 ;
else find a ,J’ E T and call
One-Step-Route(j, Q 1 ) .
j =O
v2j.
VZjtlVZj
-
d.-p.
UZj+l.llZj
edge in P.If uzj+luzj
bits a t positions p 2 2 j
-
d.-p.
UZjtlUZj
-
+
d.-p.
U2j+luzj,
vZj+lvZj
if
and the 2j-th
d.-p.
neighbor of U if uzjtli&
v2j+1v2j,
set Q 1Q - { j } ,
Pj(.,.)
= P j ( U , V ) - 1.
= U’.
-
Set U
end
To illustrate this algorithm, we still use U =
01101101100001 and v = 01010011100101 in CQ14
as example. At initial stage, we have ps(u,w) = 2,
P ~ ( u , ~ J )1, P ~ ( u , v =
) 1, Q i = Q 2 = { 5 , 4 , 1 )
and T = 8. Sincse T = 8 and we cannot find
d.-p.
any i E Q1 satisfying Uzi+1u2i
W Z i + l W 2 i or
d.-p.
uzi+1UZi
v2i+lwzi, in Step 3 we choose i = 5,
route to 01000111100011 or 01110111100011 and reduce p s ( ~v ,) by 1. Now, the set of Q 1 is still {5,4,1}.
Repeating Step 3, we find i = 1 in this iteration and
1
-
to v .
127
N
specified in Step 1 and on the smallest i E
route to 01000111100101 or 01110111100101. It reduces Q 1 to {5,4}. Subsequently, i = 4 and i = 5
are found in the following iterations of Step 3, which
render Q 1 to be empty. In summary, performing Step
1 to Step 3 we can attain the following two shortest
paths:
PI = ( 01101101100001,
0100011 1100101,
01010011100101 ).
P2 = (
P4 = ( 01101101100001,
01110001100001,
01010011100101 ) .
-
01000111100011,
01010011100011,
01110111100011,
01010011100011,
Proof:
Let U be the source and w be the destination. Observing from Step 2 and Step 3, we note
that One-Step-Route is performed at those double bits
with p i ( u , w) # 0. Furthermore, One-Step-Route is
performed IQ11 times if p p ( u , w ) = 1 and I Q 1
1
times i f p p ( u , w) = 2, i.e., exactly p ( u , w) = d c y , [ u , v )
times. Let u c to denote a current vertex reached by
the algorithm and QE and QS be subsets of Q 1 and
Q 2 , respectively, obtained by the algorithm. We note
that One-Step-Route is performed at the i-th dou-
+
-
-
d.-p.
d.-p.
or % ! i + l U Z i
v2i+lv2i.
Therefore, at the end of p ( u , w ) steps of
performing One-Step-Route the i-th double bit becomes identical to that of U . We further note another fact. Let i' be the smallest index in QE sat-
-
d.-p.
TJZi+1~2i
$ ,d.;P.
Wzi+iv2i.
W z i + l v 2 i or ~ ; i + l 2%
isfying U ; i + l ~ ; i
Then after performing ONE-STEP-ROUTE(i',
and reaching
-
U',
we obtain that zl&+luii
d.-p.
-
d.-p.
-I j
d.-p.
-
d.-p.
= 1 and U;,+~U;,
w 2 m + l v 2 m l i.e.,
(U;m+lU;m,V2m+lu2m)
E {(00,11), (101 Ol), (111 l o ) ,
(01,00)}, there is exactly one common neighbor
U i I U ; m + l E ; m O f u % 1 u ; m + 1 U ; m and W w 2 m + l v 2 m . Thus
we can perform ONE-STEP-ROUTE(m, Q ; ) to the
2m-th neighbor, say U ' , of U'. Since the 1-th double bit does not change and the m can be found as
m a x { j 1 j E Q 2 and j < I } in Step 5, we obtain that
pl(u', w) = 1 and pm(u', U ) = 1 is satisfied.
If
p i ( u C , w)
d.-p.
If ~ I ( u ' , T =
J ) 1 and ii;m+lu$m
%m+lTJ2m
i.e., (ugm+lu;,, T J Z ~ + I W ~ ) E ((00,l o ) , (1O,OO),
(11,l l ) , (Ol,Ol)}, there are two common neighbors
U;Iu;,+lu;m and U ~ ~ ~ ~ , +of~ UU~ , U~ ; ,, + ~ U ; , and
w21w2,+1wZm.
Thus we can perform One-Step-Route
at the (21)-th or (2m 1)-th neighbors of U ' . We can
use the same arguments as above when we perform
ONE-STEP-ROUTE m, Q;). Therefore, we consider
the case performing NE-STEP-ROUTE(Z, QZ). After performing ONE-STEP-ROUTE(I, Q ; ) and reach-
-
2
+
d.-p.
or U ~ ; + l y i
W Z ~ + I T J Z ~let
,
2' = m m { j
E Q 1 ) as found in Step 3 . It follows
that we can choose 'i = i* first and then continue
the procedure ONE-STEP-ROUTE i , Q l ) in the increasing order of i E Q 1 until Q 1 = . Otherwise, we
can always perform One-Step-Route on any t E T as
wzi+luzi
-
d.-p.
w ~ i t 1 ~ 2ori u ~ i + l U ~ i
-
&E)
v2it1wZi
or uLitliiii
w ~ i + l w 2 i is satisfied for all i < 'i and
i E QE.
If initially T = 8 and no i E Q 1 satisfies
?2i+lii2i
-
d.-p.
m i + l w z i can
always be found, until Q1 = 0.
Let 1 be m m { j I j E Q2} with its initial
value i* and let m be m a z { j I j E Q Z and
j < 1 } as defined in Step 5. If p p ( u , w ) = 2,
we shall perform ONE-STEP-ROUTE at i* double
bit to reduce p p ( U , w), however, after performing
ONE-STEP-ROUTE(i* ,Q4), the Q; does not change
and p i * ( u , w ) is changed to one with p m ( u , v ) = 1.
Thus, we shall focus our attenation on p l ( u , w ) =
1 and p m ( y , w ) = 1. If p l ( u , w ) = 1, it follows that either u 2 1 + 1 u z I = G21+1w21 or u 2 1 + 1 u z I =
v a r + i h and CQn(un-iu,-z.. . u z ~ + ~ u z ~ and
+z)
CQn(wn-1v,-2..
. w 2 m + 3 ~ 2 ~ + 2 ) are adjacent
subgraphs. Thus we can contract those vertices in CQ,,
U Z , + ~ U ~ into
~
a verices in CQn with prewZm+1vZm
into a vertex. Since
t possiblities at double bits I
and m of U and w with p l ( u , w ) = 1, there are eight
of 2those satisfy pair revertices, and u 2 1 - 1 . . ' u ~ ~ +
lated definition. It is easy to see that these eight
vertices graph is isomorphic to CQ3. For ease of
exposition, we can represent u,-1 . . U Z m + l U Z m and
Wn-1"
. ~ 2 m + l ~ by
m U Z l U 2 m + l U 2 m and w 2 l w z m + l v 2 m
if U Z I + ~ U Z I= w?r+iVzr, and by U Z I + I U Z ~ + I U Z and
~,
~ . 1 + 1 ~ m + 1 w ~ifm u z i + l u z i = G z + l w z i , respectively.
Without loss of generality, we may assume u 2 1 + 1 u 2 1 =
w 2 l + l G z l . Since pl(u,w) = 1 and pm(u,TJ) = 1, we have
u21u2m+lu2m
and w21w2m+lw2m are nonadjacent vertices of C Q 3 . Since the diameter of CQ3 is two, there
is at least one vertex which is a common neighbor of
U 2 I U 2 m + l U 2 m and w 2 1 V Z m + l w 2 m .
uzi+1uzi
01000111100011,
01000001101111,
Theorem 1 The Shortest-Path-Routing algorithm
correctly finds shortest paths from the source to the
destination.
uZi+li&
N
v z i + l v 2 i as specified in Step
212i+lwzi or u2i+li&
3. In subsequent iterations of repeating Step 3, it follows from the above fact that an i E Q 1 satisfying
The routings of P 3 and P4 are t o change the 5th, 4th,
5th and 1st double bits sequentially.
ble bit if
d.-p.
d.-p.
Performing Step 4 to Step 5 we can attain two shortest
paths P3 and P4 as follows:
01101101100001,
01000001100001,
01010011100101 ) .
(actu-
ally arbitrary i E Q1 allowed) satisfying U 2 i + 1 u 2 i
01101101100001,
01110111100011,
01110111100101,
01110001101111,
01010011100101 ) .
P3 = (
Q 1
6
0,
ing
128
-
d.-p.
U',
we obtain that f i ~ m + l u ~ m
w2m+lU2m
is
satisfied. Thus, in the next iteration 1 and m can be
found from Q; as in step 5 satisfying p ~ ( u 'w)
, = 1
and p m ( u ' , w ) = 1 and we can continuously perform
One-Step-Route .
+
P1,2 = ( U O O , u10,
w10,wll,wo1,W O O ) ,
(Uoo,u1o,ull,uo1,
W l l , wo1,W O O ) ,
P1,4 = ( U O O , uo1, u l l , u l O , w10,wll,WO1,WOO),
P1,3 =
d.-p.
If PI("',")
= 1, csm+1u;m
+
d.-p.
wZm+lw2m,
and u;,+16;,
~ m + 1 ~ 2 mi.e.,
,
(U;,+iu;,,
w ~ m + ~ ~Em {(OO,O1),
)
(10,11), (11,00), (01,lo)},
there is exactly one common neighbor 6zzu;m+lu;m
of u ~ , u $ m + l u $ mand T J ~ z w ~ , + ~ v ~Thus
, . we can perform ONE-STEP-ROUTE(l,Q;). After we routed t o
the (21)-th neighbor U' of U ' , the m-th double bits of
-
-
d.-p.
and w differ at the (2m)-th bit, i.e., ~ ' 2 ~ + 1 ~ ' 2 m
7 ~ 2 ~ + 1 2 1 threrfore,
2~,
in the next iteration 1 and m can
be found from Q; as in step 5 satisfying pl(u', w) = 1
and pm(u', w) = 1. Similarly, we have the same statements for U21+1uZm+lu2m, W ~ I + I W Z ~ +InI subse~Z~,
quent iterations of repeating Step 5, it follows that
we can always perform One-Step-Route, until Qz = @.
Thus we can find shortest paths from u to w , correctly.
U'
0
Step 1 takes O(1) time.
-
If in Step 3 either
-
d.-p.
d.-p.
U~i+1~2i
W ~ ~ + I W or
Z ~ ~2i+lU2i
7)Zi+lV2i, we
d.-p.
prefrom One-Step-Route. Otherwise,
w 2 i + l w z i when we have performed One-Step-Route on
j-th double bit t o U' with j E Q1 and j > i. Thus we
examine each double bit, p i ( u , v ) # 0, at most twice
in right to left order. Steps 2 and 3 are executed at
most 2p(u, w) times; Steps 4 and 5 are executed exactly
p(u, w) times and Steps 4 and 5 are O(1). Hence, the
time complexity of the algorithm is O ( n ) . The algorithm proposed in [l] can generate one shortest path
in O(n2) time, while ours can generate more shortest
paths in O ( n ) time. Furthermore, our algorithm can
be modified to generate all shortest paths in O(n2)
time.
-
Theorem 2 The diameter of C Q n is
Proof:
Let U , w be two distinct vertices. Since
double bits, it follows that & @ % ( U , w) =
there are
p ( u , w) 5
In particular, we choose U * = 00.. '0
and w* = 11 . . . 1 such that p ( u * , w * ) =
Hence
the theorem follows.
0
If]
[q].
4
[TI.
Embedding of cycles
We present embedding of cycles into C Q n that our
embedding of cycles is different from the one proposed
in [6]. Let U , w be binary strings of length n - 2 satisfying (,U, w) E E(CQn-2). In order to construct large
cycles in C Q n , we define two families of przmztzwe paths
as follows:
Type 1: Four primitive paths from uOO to WOO given
by
P1,1
= ( U O O , uo1, u l l , wo1,WOO),
as shown in Figure 3(a)
Type 2: Four primitive paths from u10 to WOO given
by
= (u10, u l l , u O l , uoo, W O O ) ,
= (U10, w10,wll,u01,uOO,
WOO),
P2,3 = ( U 1 0 , u l l , V o 1 , v l l , U o 1 , U o o , Woo),
P 2 , 4 = ( U 1 0 , W l O , W l l , W O l , U 1 1 , U 0 1 , u00,w00),
P2,l
P2,2
as shown in Figure 3(b)
Theorem 3 CQn zs a pancyclac network for n 2 2 ,
a e , CQn contaans C'I for all 4 5 1 5 2n as subgraphs
Proof: The theorem can be easily proved for n = 2 , 3
by using type 1 primitive paths For n = 4, since CQ3
is a subgraph of CQLl,it suffices to construct cycles C1
for 9 5 1 5 16 as follows
c
9=
(0000,0001,0011,0010,0110,1110,1111,
0101,0100,0000),
ClO = (0000,0001,0011,0010,0110,1110,1111,
1101,1100,0100,0000),
C11 = (0000,0001,0011,0010,0110,0111,0101,
1111,1101,1100,0100,0000},
ClZ = (0000,0001,0011,0010,0110,1110,1111,
0101,0111,1101,1100,0100,0000),
C13 = (0000,000~,00~~,0010,0110,0111,01O1,
0100,1100, Pl', 1000, OOOO),
C14 = (0000,0001,0011,0010,0110,0111,0101,
0100,1100,Pi,1000, OOOO),
c
1
5 = (0000,0001,0011,0010,0110,0111,0101,
0100,1100, Pi, 1000, OOOO),
C16 = (0000,0001,0011,0010,0110,0111,0101,
0100,1100, Pi, 1000,0000)
We next prove the theorem for n 2 5 by induction.
Since CQn-l is a subgraph of C Q n , it suffices t o construct cycles of length 2n-1
1 1 2n Let s be an
arbitrary integer satisfying 2n-3
1 5 s 5 2"-'
By
the induction hypothesis, there is a cycle C, in CQn-2
given by
cs=z (72, ,uS,u1)
+ < <
+
When s is even, construct C, based on C, as follows:
c,= (
u100, U101, U111,u110,U210, U211,U201,
~'00,
~ ~ 0~ 0~ ,0~ 1~, 1~ 1~ ,1~ 0~ , 1 0, ,
P', d 0 0 ,UlOO),
U ~ - ~ o o , u ~ - ~ o o ,
129
References
where P’ is a type-1 primitive path from u ‘ - ~ O O t o
usOO. It follows that i = 4(s - 2) IP’I 1 and to be
precise, i = 4s - 3 , 4 s - 2 , 4 s - 1 , 4 s .
When s is odd, we use C, to construct Cj similar
t o C; as follows:
+
c, = (
+
K. Efe, “The crossed cube architecture for parallel
computing,” IEEE Trans. Parallel and Distributed
Systems, vol. 3, no. 5, pp. 513-524, Sept. 1992.
uloo, ulO1, ulll,u110,u210, u211, u201,
u200, u300, u301, u311,u310,. . . ,
us-210, u s - ~ l o P”,
, usOO, u100),
K. Efe,“A variation on the hypercube with lower
diameter,” IEEE Trans. Comp., vol. 40, no. 11,
pp. 1, 312-1, 316, Nov. 1991.
P. Kulasinghe and S. Bettayeb, “Embedding binary
trees into crossed cubes”, IEEE Trans. Computers,
vol. 44, no. 7, pp. 923-929, July 1995.
where P” is a type-2 primitive path from u “ - ~ ~ Oto
~ ~ 0 0It .follows that j can be 4s - 3 , 4 s - 2 , 4 s 1 , 4 s . Therefore, we have attained cycles Ci such that
4(2”-3 + 1) - 3 5 1 5 42n-2, i.e. 2“-’
1 5 1 5 2”.
0
Hence the theorem follows.
Based on the above proof idea, we can easily construct a cycle of arbitrary length. We here illustrate
an example of constructing C19 in CQg. First, we find
a cycle of length s =
= 5 in CQ3, which is given
by
= (000,001,011,101,100,000).
Since s is odd and 19 = 4s - 1, we can use Cg and a
P 2 , 3 from 10110 to 10000 to construct a C19 as follows:
+
Priyalal D. Kulasinghe, “Connectivity of the
crossed cube,” Information Processing Letters, vol.
61, pp. 221-226, Feb. 1997.
F. T. Leighton, Introduction to Parallel Algorithms
and Architectures: Arrays, Trees, Hypercubes, San
Mateo, Calif.: Mrogan Kaufmann, 1992.
cg
c
1
9=
(
S. Q. Zheng and Shahram Latifi, “Optimal
simulation of linear multiprocessor architectures
on multiply-twisted cube using generalized Gray
code,” IEEE Trans. Parallel and Distributed Systems, vol. 7 , no. 6, pp. 612-619, June. 1996.
00000,00001,00011,00010,00110,00111,
00101,00100,01100,01101,01111,01110,
10110,10111,10001,10011,10101,10100,
10000,00000).
Embedding cycles presented in [6] can construct
one type of cycles for an arbitrary length, while ours
can construct various types of cycles for an arbitrary
length since we can start with any vertex and make
slight modification of type 1 and type 2 primitive
paths then follows the above method. For example,
t o construct another type Cig in CQ5 we first find
another type C
A given by
c; = (011,010,000,100,101,011).
A and a Pi,3, where Pi,3is another
We can use the C
type 2 primitive path, t o construct another type Ci9
as follows:
c:, = (
5
01111,01110,01100,01101,01011,01010,
01000,01001,00011,00010,00000,00001,
10011,10001,10000,10010,10110,10100,
10101,01111).
Concluding Remarks
In this paper, we introduced a new notion called
pair related distance and used it to evaluate the distance of distinct vertices. Based on pair related distance, we have presented a new shortest path routing algorithm in O ( n ) time. We also proved that the
crossed cube is a pancyclic network. Hence, crossed
cube is a good topology for interconnection networks.
130
Fig. 1. CQ,and CQ,
p*,3
(a)
Fig. 3. Primitive paths of type 1 and type 2
131
zzz3izz
vll
v10
U11
U10