A simple algorithm to nd the steps of double-loop
networks
Robin Chi-Feng Chan, Chiuyuan Chen
∗, Zhi-Xin Hong
Department of Applied Mathematics, National Chiao Tung University, 1001 Ta-Hsueh Road, Hsinchu 300, Taiwan
Received 28 June 2000; received in revised form 20 March 2001; accepted 9 April 2001
Abstract
Double-loop networks have been widely studied as architecture for local area networks and it is well-known that the minimum distance diagram of a double-loop network yields an L-shape. Given an N, it is desirable to nd a double-loop network DL(N; s1; s2) with its diameter being
the minimum among all double-loop networks with N stations. Since the diameter can be easily computed from an L-shape, one method is to start with a desirable L-shape and then asks whether there exist s1 and s2 (also called the steps of the double-loop network) to realize it.
In this paper, we propose a simple and e6cient algorithm to nd s1 and s2, which is based on
the Smith normalization method of Aguil7o, Esqu7e and Fiol. ? 2002 Elsevier Science B.V. All rights reserved.
Keywords: Double-loop network; L-shape; Diameter; Algorithm
1. Introduction
A double-loop network DL(N; s1; s2) has N nodes 0; 1; : : : ; N − 1 and 2N links of
two types:
s1-links : i → i + s1 (mod N); i = 0; 1; : : : ; N − 1;
s2-links : i → i + s2 (mod N); i = 0; 1; : : : ; N − 1:
Double-loop networks have been widely studied as architecture for local area networks. For surveys about these networks, see [2,10,11,14].
Fiol et al. [8] proved that DL(N; s1; s2) is strongly connected if and only if
gcd(N; s1; s2) = 1. When DL(N; s1; s2) is strongly connected, then we can talk about
This research was partially supported by the National Science Council of the Republic of China under
the grant NSC89-2115-M009-026.
∗Corresponding author.
E-mail address: cychen@cc.nctu.edu.tw(C. Chen).
0166-218X/02/$ - see front matter ? 2002 Elsevier Science B.V. All rights reserved. PII: S0166-218X(01)00245-1
Fig. 1. Two examples of L-shapes.
Fig. 2. An L-shape with parameters.
a minimum distance diagram. This diagram gives a shortest path from node u to node v for any u; v. Since a double-loop network is node-symmetric, it su6ces to give a shortest path from node 0 to any other node. Let 0 occupy cell (0; 0). Then v occupies cell (i; j) if and only if ia+jb ≡ v (mod N) and i +j is the minimum among all (i; j) satisfying the congruence, where ≡ means congruent modulo N. Namely, a shortest path from 0 to v is through taking i s1-links and j s2-links (in any order). Note that in
a cell (i; j), i is the column index and j is the rowindex. A minimum distance diagram includes every node exactly once (in case of two shortest paths, the convention is to choose the cell with the smaller row index, i.e., the smaller j). Wong and Coppersmith [15] proved that the minimum distance diagram is always an L-shape (a rectangle is considered a degeneration). See Fig. 1 for two examples.
An L-shape is determined by four parameters l; h; p; n as shown in Fig. 2. These four parameters are the lengths of four of the six segments on the boundary of the L-shape. For example, DL(9; 4; 1) in Fig. 1 has l = 5, h = 3, p = 3, and n = 2. Let N = lh − pn. Fiol et al. [8,9] and Chen and Hwang [3] proved that there exists a DL(N; s1; s2)
realizing the L-shape(l; h; p; n) if and only if l ¿ n, h ¿ p, and gcd(l; h; p; n) = 1. The diameter d(N; s1; s2) of a double-loop network DL(N; s1; s2) is the largest
dis-tance between any pair of stations. It represents the maximum transmission delay be-tween two stations. Therefore, it is desirable to minimize the diameter. This is the problem discussed by many authors; see [1,5–7,9,12,15]. Let d(N) denote the best possible diameter of a double-loop network with N stations. Wong and Coppersmith [15] showed that d(N) ¿ √3N − 2.
Given an N, it is desirable to nd a double-loop network DL(N; s1; s2) with its
can be readily computed from the dimensions of its L-shape, one method is to start with a desirable L-shape and then asks whether there exist s1 and s2 to realize it.
Aguil7o, Esqu7e and Fiol [1,7] proposed the Smith normalization method to nd s1 and
s2 for a given L-shape, but no explicit algorithm was given in their paper. In [3], Chen
and Hwang proposed a simple method, based on the sieve method in number theory, to nd s1 and s2 for a given L-shape.
In this paper, we propose a simple and e6cient algorithm to nd s1 and s2 for a
given L-shape. Our algorithm is based on the Smith normalization method of Aguil7o, Esqu7e and Fiol [1,7], but unlike their method, our algorithm does not require any matrix operation. Our algorithm takes at most O((log N)2) time and if gcd(l; n) = 1 or
gcd(l; p) = 1 or gcd(h; p) = 1 or gcd(h; n) = 1, then our algorithm could nd the steps of a double-loop network in only O(log N) time.
2. Preliminary
It is well-known that
Lemma 1. If a and b are integers; not both zero; then there exist integers and such that a + b = gcd(a; b).
We nowprove that
Lemma 2. If ; a; ; b are integers; not all zero; such that a+b = 1; then gcd(a; ) = 1. Proof. Assume that a + b = 1 and gcd(a; ) = k. Then k|a and k|. Thus k|a + b = 1. So k = 1.
Theorem 3. If a and b are integers; not both zero; then there exist integers x and y such that xa + yb = gcd(a; b) and (y; gcd(a; b)) = 1.
Proof. Set r = gcd(a; b) for easy writing. By Lemma 1, there exist integers and such that a + b = r. If gcd(; r) = 1, then we are done. In the following, assume that gcd(; r) = k ¿ 1. Suppose k = ps1
1ps22· · · psmm, where p1¡ p2¡ · · · ¡ pmare the prime
factors of k and suppose r = pr1
1pr22· · · prmmprm+1m+1prm+2m+2· · · prnn, where p1¡ p2¡ · · · ¡ pn
are the prime factors of r. Since k|r, we have ri¿ si for all i, 1 6 i 6 m. Let
r= prm+1
m+1prm+2m+2· · · prnn; a= a=r; and b= b=r:
Note that gcd(r; ) = 1; otherwise, we will have gcd(; r) ¿ k. Since gcd(r; ) = 1 and k|, we have gcd(r; k) = 1. Since a + b = r, we have a+ b= 1. By Lemma 2, we have gcd(a; ) = 1. Since gcd(a; ) = 1 and k|, we have gcd(a; k) = 1. Since k| and gcd(r; k) = 1 and gcd(a; k) = 1, we have gcd( −ra; k) = 1. Since gcd(r; ) = 1 and r|ra, we have gcd( − ra; r) = 1. Since gcd( − ra; k) = 1 and
gcd( − ra; r) = 1 and every prime factor of r is either a prime factor of k or a prime factor of r, we have gcd( − ra; r) = 1. Consider
x = + rb and y = − ra:
Then xa + yb = ( + rb)a + ( − ra)b = r and gcd(y; r) = gcd( − ra; r) = 1. We have this theorem.
The proof of Theorem 3 leads to the following algorithm for nding x and y in Theorem 3.
ALGORITHM-MODIFIED-EUCLIDEAN
Input: Integers a and b, not both zero, and r = gcd(a; b).
Output: Integers x and y such that xa + yb = r and gcd(y; r) = 1. 1. Find integers and such that a + b = r.
2. If gcd(; r) = 1, then let x = , y = , return x; y and stop this algorithm. 3. Let k = gcd(; r), r= r, and d = k. 4. WHILE (d ¿ 1) DO BEGIN r= r=d; d = gcd(r; k); END
5. Let a= a=r, b= b=r, x = + rb and y = − ra. Return x, y.
We give an example to showhowStep 4 is executed. Suppose before Step 4 is executed, d = k = 23 × 32× 74 and r= r = 24 × 38× 715× 11 × 23. After the rst
iteration of the while-loop, r= 2 × 36× 711× 11 × 23 and d = 2 × 32 × 74. After
the second iteration, r= 34× 77× 11 × 23 and d = 32× 74. After the third iteration,
r= 32×73×11×23 and d = 32×73. After the fourth iteration, r= 11×23 and d = 1. Since d = 1, we stop the iteration.
Theorem 4. ALGORITHM-MODIFIED-EUCLIDEAN is correct and it takes at most O((log N)2) time; where N = max{a; b}.
Proof. Note that Steps 1, 2, 3, and 5 of ALGORITHM-MODIFIED-EUCLIDEAN are translated directly from the proof of Theorem 3, so they are correct. Steps 1 and 2 take O(log N) time; Steps 3 and 5 take O(1) time. It remains to consider Step 4. Let k = ps1
1ps22· · · pmsm and r = pr11pr22· · · prmmprm+1m+1prm+2m+2· · · prnn be dened as in the proof of
Theorem 3. Note that ri¿ si for all i, 1 6 i 6 m. In the proof of Theorem 3, we need
r= prm+1
m+1prm+2m+2· · · prnn. The purpose of Step 4 is to derive r= pm+1rm+1prm+2m+2· · · prnn.
Before Step 4 is executed, r= r = pr1
1pr22· · · prmmpm+1rm+1prm+2m+2· · · prnn. We then use a
while-loop to remove pr1
1pr22· · · prmm from r. Before an iteration of the while-loop, if
r
i, then after the iteration, the power of pi is decreased by si or ri, whichever is
smaller. At the end of Step 4, d = gcd(r; k) = 1; that is, r= prm+1
m+1prm+2m+2· · · prnn. Set
= max{r1=s1; r2=s2; : : : ; rm=sm} for easy writing. Step 4 iterates times. Thus
Step 4 takes O( log N) time. Since = O(log N), Step 4 takes at most O((log N)2)
time.
The above arguments showthat ALGORITHM-MODIFIED-EUCLIDEAN is correct and it takes at most O((log N)2) time.
3. The Smith normalization method
Let L(l; h; p; n) be an L-shape such that l ¿ n, h ¿ p, and gcd(l; h; p; n) = 1. Aguil7o and Fiol [1], and also Esqu7e et al. [7] proposed the following method of computing s1
and s2 such that DL(N; s1; s2) realizes L. They considered the integral matrix
M =
l −p
−n h
and computed the Smith normal form of M, S(M) = 1 0 0 N :
Then S(M) = LMR, where L and R are two nonsingular unimodular (determinant
±1) integral matrices. They proved that if
L = ;
then s1= (mod N) and s2= (mod N) in DL(N; s1; s2). No algorithm on computing
the Smith normal form was actually given in their paper except a reference to [13]. In [13], the reader was referred to three theorems (Theorem II.1, Theorem II.2, and Theorem II.9) for learning howto compute the Smith normal form.
The following is a brief description of what the three theorems in [13] say. Let 1
and 2 be two integers, no both zero, and let = gcd(1; 2). Theorem II.1 says that
there exists an integral matrix
1 2
with rst row [; 2] and determinant ; note that the elements and may be
determined by the Euclidean algorithm. Theorem II.2 uses Theorem II.1 to showthat the (1,1) element of a matrix may be replaced by the greatest common divisor of the rst column of the matrix. Theorem II.9 uses Theorem II.2 to derive the Smith normal form. To make the readers easy to understand the Smith normalization method, we nowgive an explicit algorithm for it.
THE-SMITH-NORMALIZATION-METHOD
Input: l; h; p; n of an L-shape L, where l ¿ n, h ¿ p, and gcd(l; h; p; n) = 1. Output: s1 and s2 such that DL(N; s1; s2) realizes the L-shape L(l; h; p; n).
1. Let M = l −p −n h ; M0= M, i = 0; j = 0; k = 0.
2. Repeat sub-steps 2.1–2.2 until the (1,1) element of Mj divides both the (2,1)
ele-ment and the (1,2) eleele-ment of Mj.
2.1 If the (1,1) element of Mj does not divide the (2,1) element of Mj, then let
i = i + 1, j = j + 1, and nd a nonsingular unimodular integral matrix Li such
that the (1,1) element of Mj= LiMj−1 is the greatest common divisor of the
rst column of Mj−1.
2.2 If the (1,1) element of Mj does not divide the (1,2) element of Mj, then let
j = j + 1, k = k + 1, and nd a nonsingular unimodular integral matrix Rk such
that the (1,1) element of Mj= Mj−1Rk is the greatest common divisor of the
rst rowof Mj−1.
3. If the (2,1) element of Mj is not zero, then let i = i+1, j = j+1, and nd a
nonsin-gular unimodular integral matrix Li to make the (2,1) element of Mj= LiMj−1
zero.
4. If the (1,2) element of Mj is not zero, then let j = j + 1, k = k + 1, and nd a
non-singular unimodular integral matrix Rk to make the (1,2) element of Mj= Mj−1Rk
zero.
5. If the (1,1) element of Mj does not divide the (2,2) element of Mj, then add
column 2 of Mj to column 1 of Mj and go to Step 2.
6. Now Mj is the Smith normal form of M, i.e.,
Mj= Li· · · L2L1MR1R2· · · Rk= S(M) = 1 0 0 N : Let L = Li· · · L2L1. If L = ;
then let s1= (mod N) and let s2= (mod N). Return s1, s2.
Since [1,7] did not provide the time complexity analysis of the Smith normalization method, we now analyze its time complexity. Its time complexity is dominated by Step 2. Each execution of Step 2 takes O(log N) time. Since each execution of Step 2.1 and Step 2.2 makes the (1,1) element of Mj contains less prime factors than before,
Step 2 is executed at most O(log N) times. Therefore the Smith normalization method takes at most O((log N)2) time.
4. The sieve method
Let L(l; h; p; n) be an L-shape such that l ¿ n, h ¿ p, and gcd(l; h; p; n) = 1. Chen and Hwang [3] (see also [11]) proposed the following method, based on the sieve method in number theory, to nd s1 and s2.
For k = 0; 1; : : : ; dene ak= kn + h;
bk= kl + p:
Let Fk denote the set of prime factors of gcd(ak; bk) and F denote the set of prime
factors of N. They proved that there exists a k such that f ∈ Fk for all f ∈ F; then
s1= ak (mod N) and s2= bk (mod N) realize L. Note that if f ∈ F appears in Fk for
some k and kf is the smallest such k, then f appears in every fth k after kf.
For example, suppose N = 2×3×5×7×11×59, and L(l; h; p; n) = L(22×107; 22×3×
5×7; 2×3×5×7; 32×23). Then F 0= {2; 3; 5; 7}, F1= {11}, and F = {2; 3; 5; 7; 11; 59}. Thus 2 ∈ F appears in a0; b0; a2; b2; a4; b4; a6; b6; a8; b8; etc., 3 ∈ F appears in a0; b0; a3; b3; a6; b6; a9; b9; etc., 5 ∈ F appears in a0; b0; a5; b5; a10; b10; a15; b15; etc., 7 ∈ F appears in a0; b0; a7; b7; a14; b14; a21; b21; etc., 11 ∈ F appears in a1; b1; a12; b12; a23; b23; a34; b34; etc.
The rst pair ak; bk that is not crossed out by the sieve method is a11; b11. Thus
s1= a11 (mod N) and s2= b11 (mod N) realize L.
The sieve method is simple and easy to implement. Note that [3,11] did not give the time complexity analysis of the sieve method. Although we are also unable to give an exact time complexity analysis for the method, we give an upper bound for it. Let "(N) denote the number of prime factors of N and let Pi denote the ith prime,
i.e., P1= 2; P2= 3, etc. Since smaller primes cross out more pairs (ak; bk) than larger
primes can cross out, the sieve method would take the longest time when N contains the smallest "(N) primes. In this case, the nal k is bounded above by P"(N). Since
checking if f ∈ Fk for all f ∈ F is equivalent to check if gcd(gcd(ak; bk); N) = 1
(which could be checked by using the Euclidean algorithm twice), the sieve method takes at most O(P"(N)log N) time.
5. Our algorithm
Given an L-shape, we propose the following algorithm to nd s1 and s2.
ALGORITHM-COMPUTING-STEPS
Output: s1 and s2 such that DL(N; s1; s2) realizes L.
1. Find r1= gcd(l; −n).
2. Find integers 1 and 1 such that 1l + 1(−n) = r1.
3. Find r2= gcd(r1; −1p + 1h).
4. Find integers 2 and 2 such that 2r1+ 2(−1p + 1h) = r2 and gcd(2; r2) = 1.
5. s1= 2n − 2h (mod N) and s2= 2l − 2p (mod N).
For example, let l = 5, h = 3, p = 3, and n = 2. Then our algorithm derives r1= 1,
1= 1, 1= 2, r2= 1, 2=−2, and 2= 1. Thus s1=−7 (mod 9) and s2=−13 (mod 9),
i.e., s1= 2 and s2= 5. It can be veried that DL(9; 2; 5) realizes
L-shape(5,3,3,2). We nowprove that
Theorem 5. ALGORITHM-COMPUTING-STEPS is correct and it takes at most O((log N)2) time.
Proof. Note that N = lh − pn. Let M = l −p −n h :
Consider column 1 of M: it contains l and −n. After Step 1 is performed, we have r1= gcd(l; −n) and 1l + 1(−n) = r1. Let L1= 1 1 n r1 l r1 : and let M1= L1M. Then
M1= 1 1 n r1 l r1 l −p −n h = r1 −1p + 1h 0 N r1 :
Consider row1 of M1: it contains r1 and −1p + 1h. After Step 2 is performed,
we have r2= gcd(r1; −1p+1h), 2r1+2(−1p+1h) = r2, and gcd(2; r2) = 1. Let
R1= 2 −(−1rp+2 1h) 2 rr12 : and let M2= M1R1. Then
M2= r1 −1p + 1h 0 N r1 2 −(−1rp+2 1h) 2 rr12 = r2 0 N2 r1 N r2 :
Consider column 1 of M2: it contains r2 and N2=r1. Let r3= gcd(r2; N2=r1). Note
that in Step 2 we choose gcd(2; r2) = 1. Thus
r3= gcd r2;Nr 2 1 = gcd r2;Nr 1 = gcd r1; −1p + 1h;Nr 1 :
We claim that r3= 1. Suppose this is not true and r3¿ 1. Then every entry of M1 is
a multiple of r3. Since M1= L1M, we have
M = L−1 1 M1=det(L1 1) l r1 −1 −n r1 1 r1 −1p + 1h 0 Nr 1 : That is, M = 1 det(L1) l r1 −1 −n r1 1 r3 r1 r3 −1p+1h r3 0 N r1r3 : Since r3= gcd(r1; −1p + 1h;rN1), r1 r3 −1p+1h r3 0 N r1r3
is integral. Since det(L1) = ± 1, every entry of M must be a multiple of r3. Then
gcd(l; h; p; n) ¿ r3¿ 1; this contradicts with the assumption that gcd(l; h; p; n) = 1.
Therefore r3= 1.
Since r3= gcd(r2; N2=r1) and r3= 1, by Lemma 1, there exist integers 3 and 3
such that 3r2+ 3(N2=r1) = 1. Let
L2= 3 3 −N2 r1 r2
and let M3= L2M2. Then
M3= 3 3 −N2 r1 r2 r2 0 N2 r1 N r2 = 1 3N r2 0 N : Let R2= 1 −3N r2 0 1 : and let M4= M3R2. Then
M4= 1 3N r2 0 N 1 −3N r2 0 1 = 1 0 0 N = S(M):
From the above, L2L1MR1R2= S(M). Moreover, L1, L2, R1 and R2 are
uni-modular integral matrices. Let L = L1L2. Then
L = 3 3 −N2 r1 r2 1 1 n r1 l r1 = 31+r11n 31+r31l −N21+r2n r1 −N21+r2l r1 :
Using the facts that N = lh−pn and 1l+1(−n) = r1 and 2r1+2(−1p+1h) = r2,
we have (−N21+ r2n)=r1= 2n − 2h and (−N21+ r2l)=r1= 2l − 2p. Thus if
It is clear that Steps 1, 2, and 3 can be done in O(log N) time by using the Euclidean algorithm. Step 4 can be done in O((log N)2) time by using
ALGORITHM-MODIFIED-EUCLIDEAN. Step 5 can be done in O(1) time. Thus ALGORITHM-COMPUTING-STEPS takes at most O((log N)2).
The following theorem will be used in the follow-up discussions. Theorem 6 (Chen and Hwang [4]). Suppose s
1; s2realize L-shape(l; h; n; p). Let x and
y be integers such that s
2x − s1y = 1. Then s1= nx − hy (mod N) and s2= lx −
py (mod N) realize L-shape(l; h; p; n); moreover; s1; s2 can be derived from s1; s2 in
O(log N) time.
Theorem 7. If gcd(l; n) = 1 or gcd(l; p) = 1 or gcd(h; p) = 1 or gcd(h; n) = 1; then we could use ALGORITHM-COMPUTING-STEPS to 9nd the steps s1 and s2 of
L-shape(l; h; p; n) in only O(log N) time. Proof. There are four cases:
Case 1: gcd(l; n) = 1. Then, clearly, r1= gcd(l; −n) = 1. Hence r2= 1 and Step 4 of
COMPUTING-STEPS takes only O(log N) time. Thus ALGORITHM-COMPUTING-STEPS nds the steps s1 and s2 in only O(log N) time.
Case 2: gcd(l; p) = 1. By an argument similar to that in Case 1, we could use ALGORITHM-COMPUTING-STEPS to nd the steps s
1 and s2 of L-shape(l; h; n; p)
in only O(log N) time. Then, by Theorem 6, s1; s2 could be derived from s1; s2 in
O(log N) time.
Case 3: gcd(h; p) = 1. By an argument similar to that in Case 1, we could use ALGORITHM-COMPUTING-STEPS to nd the steps s
1 and s2 of L-shape(h; l; n; p)
in only O(log N) time. Since L-shape(h; l; n; p) is the Mipping of L-shape(l; h; p; n), s1= s2 and s2= s1.
Case 4: gcd(h; n) = 1. Again, by an argument similar to that in Case 1, we could use ALGORITHM-COMPUTING-STEPS to nd the steps s
1 and s2 of L-shape(h; l; p; n)
in only O(log N) time. Then, by Theorem 6, the steps s
1; s2 of L-shape(h; l; n; p) could
be derived from s
1; s2 in O(log N) time. Since L-shape(h; l; n; p) is the Mipping of
L-shape(l; h; p; n), s1= s2 and s2= s1.
We nowcompare the three existing algorithms for computing the steps of double-loop networks: the Smith normalization method [1,7], the sieve method [3,11], and our algo-rithm. Both the Smith normalization method and our algorithm take at most O((log N)2)
time. In the Smith normalization method, one needs to nd nonsingular unimodular in-tegral matrices Li; : : : ; L2; L1; R1; R2; : : : ; Rk such that
Li· · · L2L1MR1R2· · · Rk= S(M):
Our algorithm is based on the Smith normalization method, but our algorithm does not require any matrix operation; moreover, as could be seen from the proof of Theorem
5, we prove that there exist nonsingular unimodular integral matrices L2; L1; R1; R2
such that
L2L1MR1R2= S(M):
Therefore, our algorithm greatly simplies the computation of the Smith normalization method.
Both the sieve method and our algorithm are very simple and easy to implement. The sieve method shows that the steps of a double-loop network are of the form
s1= kn + h (mod N); s2= kl + p (mod N);
and our algorithm shows that the steps of a double-loop network are of the form s1= 2n − 2h (mod N); s2= 2l − 2p (mod N):
The sieve method takes at most O(P"(N)log N) time. However, we are unable to predict
the value of P"(N) and therefore unable to tell which algorithm is more e6cient.
It is open whether the steps of a double-loop network can be found in O(log N) time. Note that Cheng and Hwang [5] gave an O(log N) time algorithm to com-pute the L-shape of a double-loop network DL(N; s1; s2). It is also open whether
we can nd integers x and y such that xa + yb = gcd(a; b) and (y; gcd(a; b)) = 1 in only O(log N) time, where a and b are integers, not both zero. If this is true, then ALGORITHM-COMPUTING-STEPS would take only O(log N) time and the steps of a double-loop network can be nd in O(log N) time.
Acknowledgements
We thank Prof. Frank K. Hwang for many helpful comments. We also thank the referees for many constructive comments that greatly improve the presentation of this paper.
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