Constructions of Snake-in-the-Box Codes under $\ell_{\infty}$-metric for Rank Modulation

0.18 MB·

by  Xiang Wang

1
Constructions of Snake-in-the-Box Codes under the
ℓ
-Metric for Rank Modulation
∞
Xiang Wang and Fang-Wei Fu
Abstract
In the rank modulation scheme, Gray codes are very useful in the realization of flash memories. For a Gray code in this
scheme, two adjacent codewords areobtained by using one “push-to-the-top” operation. Moreover, snake-in-the-box codes under
the ℓ∞-metric are Gray codes, which can be capable of detecting one ℓ∞-error. In this paper, we give two constructions of ℓ∞-
6 snakes. On the one hand, inspired by Yehezkeally and Schwartz’s construction, we present a new construction of the ℓ∞-snake.
1 The length of this ℓ∞-snake is longer than the length of the ℓ∞-snake constructed by Yehezkeally and Schwartz. On the other
0 hand, we also give another construction of ℓ∞-snakes by using K-snakes and obtain the longer ℓ∞-snakes than the previously
2 known ones.

n
Index Terms
a
J Flash memory, rank modulation, Gray codes, snake-in-the-box codes, K-snakes, ℓ∞-snakes.
1
2
I. INTRODUCTION

] FLASH memory is a non-volatile storage medium that is both electrically programmable and erasable. It has been widely
T
used becauseofitsreliability,relativelowcost, andhighstoragedensity.Inflash memories,a blockwhichcontainsmany
I
. cells can maintain a block of charge levels to represent information. However, the flash memory has its inherent asymmetry
s
c betweencellprogramming(injectingcellswithcharge)andcellerasure(removingchargefromcells).Thatistosay,increasing
[
the charge level of a single cell (cell programming)is an easy operation, but decreasing the charge level of a single cell (cell

1 erasure)is a verydifficultprocess. In the currentflash memories’architecture,a single-cellerasureoperationincludescopying
v a large whole block which contains the single cell to a temporary location, erasing it, and then reprogramming all the cells
9 in the block. In the programming operation, some cells may be injected with extra charge. This will lead to overshooting
3
of charge. Then, overprogramming (overshooting of charge) is a severe problem because of some very difficult cell erasure
5
operations.Thus, in orderto avoidingoverprogramming,injectinga cell with chargebymanyiterationsmakesits chargelevel
5
0 gradually approach the desirable level in the cell programmingoperation. Due to charge leakage or reading disturbance, these
. may result in some errors in flash memory cells.
1
0 Therankmodulationschemehasbeenrecentlyproposedin[2]toovercometheseoutstandingproblems.Inthisscheme,one
6 permutation is induced by relative rankingsof the charge levels on a group of cells instead of using absolute values of charge
1 levels. Moreover, this permutation is used to represent information. Specifically, assume that c ,c ,...,c ∈ R represent the
1 2 n
v: chargelevelsofn∈Ncellsrespectively,thenthechargelevelsofthesecellsinduceonepermutationπ =[π(1),...,π(n)]∈Sn
i such that c > c > ··· > c , where S is the set of all the permutations over {1,2,...,n}. In order to avoid the
X π(1) π(2) π(n) n
overprogramming,the cell programming uses only “push-to-the-top” operations [2]. That is, a cell is programmed by raising
r
the charge level of this cell above those of all others in the block. Therefore, in the manner, the overprogramming is no
a
longer a problem. Moveover, when injection of some extra charge or leakage may not change their relative rankings, then
the permutation induced by their relative rankings will not change. Hence, this will not cause an encoding error. However,
when the relative rankings are changed because of injection of much extra charge or leakage in the cells, the permutation
induced by the relative rankings will be different from the desired permutation, i.e., this leads to an encoding error. To detect
and/or correct such errors, we require an appropriate distance measure in the permutations. There are several metrics on the
permutations such as the ℓ -metric [4], [14], the Ulam metric [16], and the Kendall’s τ-metric [3], [12], [17]. In this paper,
∞
we will consider only the ℓ -metric.
∞
Intherankmodulationscheme,Graycodesareimportantcodeswhichrepresentinformationinflashmemories.In[2],Jiang
et al. proposed the Gray codes by using “push-to-the-top”operations. For the Gray code, it was first presented in [11], where
itisa sequenceof distinctbinaryvectorsoffixedlengthandeveryadjacentpairdiffersina single coordinate.Inpractice,they
are widely used in many applications [7]. Moreover, a good survey on the Gray codes is given in [8]. Recently, Gray codes
for rank modulation have been studied in [3], [6], [9], [10]. In addition, a snake-in-the-box code is a Gray code in which the
distance of any two distinct codewordsin the code is at least 2. Thus, this code can detect a single error in one codeword. In
particular, the snake-in-the-boxcodes are usually studied in the context of binary codes in the Hamming scheme, e.g. [18]. In
this paper, we will only focus on the snake-in-the-box codes under the ℓ -metric.
∞
X. Wang and F.-W. Fu are with the Chern Institute of Mathematics and LPMC, Nankai University, Tianjin 300071, China (e-mail:
[email protected]; [email protected]).
2
In [6], Yehezkeally and Schwartz constructed directly a snake-in-the-box code of length ⌈n⌉!(⌊n⌋+(⌊n⌋−1)!) in S
2 2 2 n
under the ℓ -metric. In this paper, we will improve on this result. On the one hand, we will construct a snake of length
∞
⌈n⌉!(⌊n⌋+(⌊n⌋)!) in S . On the other hand, we will also construct the longer ℓ -snakes by using K-snakes.
2 2 2 n ∞
The rest of this paper is organized as follows. In Section II, we will give some basic definitions for the rank modulation
scheme and notations required in this paper. In Section III, we give directly two constructionsof ℓ -snakes in S . In Section
∞ n
IV,wepresentsomeexamplesofthesetwoconstructions.InSectionV,wecompareourresultswiththepreviousones.Section
VI concludes this paper.
II. PRELIMINARIES
In this section, we will use some definitions and notations mentioned in [1] and [15].
We let [n] , {1,2,...,n} and let π , [π(1),π(2),...,π(n)] be a permutation over [n]. And let S be the set of all the
n
permutationsover[n].Forσ,π ∈S ,theirmultiplicationπ◦σ isdenotedbythecompositionofσ onπ,i.e.,π◦σ(i)=σ(π(i)),
n
foralli∈[n]. Underthismultiplicationoperation,S is a noncommunicativegroup.Moreover,letπ−1 betheinverse element
n
of π, for π ∈S , and let A be the subgroup of all even permutations over [n].
n n
Assume that given a set S and a set of transformations T ⊂ {f|f : S → S}, a Gray code over S of size M, is a
sequence C = (c ,c ,...,c ) of M different elements from S, called codewords, in which for each i ∈ [M −1] there
0 1 M−1
exists some t˜ ∈ T such that c = t˜(c ). For convenience, we denote a transformation sequence of the Gray code C by
i i i i−1
T , i.e., T = (t˜ ,t˜,...,t˜ ). The Gray code is called complete if M = |S|, and cyclic if there exists t˜ ∈ T such that
C C 1 2 M−1 M
c =t˜ (c ). Thus, the transformation sequence T of the cyclic Gray code C is (t˜,t˜,...,t˜ ,t˜ ).
0 M M−1 C 1 2 M−1 M
Consider the Gray codes for rank modulation in flash memories, we have S =S and the set of transformationscomprises
n
ofallthe “push-to-the-top”operationsin S , definedbyT . Next,we denotebyt :S →S one“push-to-the-top”operation
n n i n n
on index i, for 2≤i≤n, that is,
t [a ,a ,....,a ,a ,a ,...,a ]=[a ,a ,a ,...,a ,a ,...,a ],
i 1 2 i−1 i i+1 n i 1 2 i−1 i+1 n
and a p-transition will be an abbreviation of a “push-to-the-top” operation. Therefore, T ={t ,t ,...,t }.
n 2 3 n
A sequence of p-transitionsis called a transition sequence. Given an initial permutation π in S and a transition sequence
0 n
(t ,t ,...,t ) with x(i)∈[n] for all i∈[l], we can obtain a sequence of permutationsπ ,π ,...,π in S , where π =
x(1) x(2) x(l) 0 1 l n i
t (π ) for all i∈[l]. When π =π and π 6=π for each pair 0≤i<j <l, the permutation sequence (π ,π ,...,π )
x(i) i−1 l 0 i j 0 1 l−1
is a cyclic Gray code by using the “push-to-the-top” operations, denoted by C . Moreover, the transition sequence T is
n Cn
(t ,t ,...,t ). For convenience, for a transition sequence T in T , we denote by f :{1,2,...,|T|}→{2,3,...,n} one
x(1) x(2) x(l) n
index function of the transition sequence. Hence, T =(t ,t ,...,t ).
f(1) f(2) f(|T|)
The ℓ -distance between two permutations π,σ ∈S , denoted by d (π,σ), is the maximal number of indices difference
∞ n ∞
between π and σ. A snake-in-the-boxcode C underthe ℓ -metric is a Gray code in which for each two distinctpermutations
∞
π,σ ∈C, we have that d (π,σ) ≥2. Moreover, a cyclic ℓ -snake is a cyclic Gray code. If not special specified, we call a
∞ ∞
cyclic ℓ -snake an ℓ -snake or a snake.
∞ ∞
Furthermore, we denote by an (n,M,ℓ )-snake an ℓ -snake of size M in S . And we let Cπ0 be an (n,M,ℓ )-snake,
where T is its transition sequence and π∞ is its first p∞ermutation. For simplicitny, we let Cπ0 ,TC(π ,π ,π ,...,π∞ ) and
T ,(tC ,t ,...,t ) such that π =0t (π ) for every i∈[M −1] and t (π TC)=π0. 1 2 M−1
C x(1) x(2) x(M) i x(i) i−1 x(M) M−1 0
In [2], Jiang et al. presented an n-length rank modulationGray code (n-RMGC) for flash memoriesin the rank modulation
scheme. For convenience, we denote a cyclic and complete n-RMGC by C , where T is its transition sequence. Hence,
Tn n
|C | = n!. And we define T , (t ,t ...,t ). In [14], Tamo and Schwartz discussed error-correcting and error-
Tn n in(1) in(2) in(n!)
detectingcodesin S underthe ℓ -metric,called as the limited-magnituderank-modulationcodes(LMRM codes). Moreover,
n ∞
in [14], they also proved that if C is an (n,M,ℓ )-snake, then M ≤ n! . Later Yehezkeally and Schwartz [6] constructed
∞ 2⌊n/2⌋
an (n,M,ℓ )-snake of size ⌈n⌉!(⌊n⌋+(⌊n⌋−1)!) for all n≥4.
∞ 2 2 2
Having the above definitions and notations, we will present two constructions of ℓ -snakes in the following section.
∞
III. MAIN RESULTS
According to the definition of the ℓ -distance, for any two permutations σ,π ∈ S , we have the following expression for
∞ n
d (σ,π) [6],
∞
d (σ,π)=max|σ(i)−π(i)|.
∞
i∈[n]
A. Construction of ℓ -snakes by using cyclic and complete RMGCs
∞
In this subsection, we give one construction of ℓ -snakes by using cyclic and complete RMGCs. In order to use the code
∞
constructions presented in [2], we will give the following lemma.
3
Lemma 1: [2, Theorems 7 and 8] For all n ≥ 3, there exists a cyclic and complete (n−1)-RMGC, denoted by C ,
Tn−1
where T , (t ,t ,...,t ). Moreover, for all n ≥ 4, the constructions in [2, Theorem 7] can yield a
n−1 in−1(1) in−1(2) in−1((n−1)!)
cyclic and complete n-RMGC, denoted by C , with its transition sequence T =(t ,t ...,t ), where
Tn n in(1) in(2) in(n!)
T =(t ,...,t ,t ,t ,...,t ,t ,...,t ,...,t ,t )
n n n ˆin−1(1) n n ˆin−1(2) n n ˆin−1((n−1)!)
n−1 n−1 n−1
and
| {z } | {z } | {z }
t =t for all j ∈[(n−1)!].
ˆin−1(j) n−in−1(j)+1
According to the above lemma, we can obtain some properties of this RMGC which we will later use.
Lemma 2: For any n ≥ 3, there exists a cyclic and complete n-RMGC, denoted by C , where its transition sequence
Tn
T =(t ,...,t ) such that
n in(1) in(n!)
t =t , t =t , and t =t
in(j) 2 in(k) n−1 in(l) n
for some j,k,l ∈[n!].
Proof: According to the construction of [2, Fig. 2], we have one transition sequence of a cyclic and complete 3-RMGC,
denoted by T , where T =(t ,t ,t ,t ,t ,t ). Hence, we have that
3 3 3 3 2 3 3 2
t =t and t =t =t . (1)
i3(1) 3 i3(3) i3(6) 2
By Lemma 1, when n = 4, we have a cyclic and complete 4-RMGC, denoted by C , with its transition sequence T =
T4 4
(t ,...,t ), where
i4(1) i4(4!)
T =(t ,...,t ,t ,...,t ,...,t ,t ). (2)
4 4 4 ˆi3(1) 4 4 ˆi3(3!)
3 3
By Lemma 1 and (1), we have that
| {z } | {z }
t =t and t =t . (3)
ˆi3(1) 2 ˆi3(3) 3
Hence, by (2) and (3), we can obtain that
t =t , t =t , and t =t . (4)
i4(4) 2 i4(12) 3 i4(1) 4
ByLemma1,wecanalsogetacyclicandcomplete5-RMGC,denotedbyC ,withitstransitionsequenceT =(t ,...,t ),
T5 5 i5(1) i5(5!)
where
T =(t ,...,t ,t ,...,t ,...,t ,t )
5 5 5 ˆi4(1) 5 5 ˆi4(4!)
4 4
and
| {z } | {z }
t =t for all j ∈[4!]. (5)
ˆi4(j) 5−i4(j)+1
By (4) and (5), we have that
t =t =t =t
i5(20) ˆi4(4) 5−i4(4)+1 4
and
t =t =t =t .
i5(5) ˆi4(1) 5−i4(1)+1 2
Hence, we have that
t =t , t =t , and t =t .
i5(5) 2 i5(20) 4 i5(1) 5
Similarly, by induction, we can obtain that there exists a cyclic and complete n-RMGC, denoted by C , with its transition
Tn
sequence T =(t ,...,t ) such that
n in(1) in(n!)
t =t , t =t , and t =t
in(n) 2 in(n2−n) n−1 in(1) n
for all n≥3.
The following lemma gives one construction of a basic block which is useful for the construction of ℓ -snakes by using
∞
cyclic and complete RMGCs.
Lemma 3: For all n ≥ 6, let {a }k be a set of even integers of [n] and {b }l be a set of odd integers of [n], where
j j=1 j j=1
k =⌊n⌋ andl=⌈n⌉.Andletσ =[b ,a ,a ...,a ,a ,b ,b ,...,b ]beapermutationsuchthat|a −b |≥2.Then,thereexist
2 2 1 2 3 k 1 2 3 l 1 1
two noncyclic (n,k!+k,ℓ )-snakes. One noncyclic (n,k!+k,ℓ )-snake, denoted by Cσ,π1, is starting with σ and ending
∞ ∞ TC
with one permutation π , where
1
π =[a ,a ,...,a ,a ,a ,b ,b ,...,b ].
1 2 3 k−1 k 1 1 2 l
Another noncyclic (n,k!+k,ℓ )-snake, denoted by Cˆσ,π2, is starting with σ and ending with one permutation π , where
∞ TCˆ 2
π =[a ,a ,...,a ,a ,a ,b ,b ,...,b ].
2 2 3 k−1 1 k 1 2 l
4
Proof: We prove only the existence of Cσ,π1, since the proof of the existence of Cˆσ,π2 is similar. For convenience, let
Cσ,π1 ,(σ ,σ ,...,σ ) and T ,(t T,Ct ,...,t ). TCˆ
TC 0 1 k!+k−1 C α1(1) α1(2) α1(k!+k−1)
Now, by Lemma 1, there exists a cyclic and complete k-RMGC with its transition sequence T , where
k
T =(t ,t ,...,t ). (6)
k ik(1) ik(2) ik(k!)
By Lemma 2, since k ≥3, we have that
t =t and t =t for some s ,s ∈[k!]. (7)
ik(s1) k ik(s2) k−1 1 2
Hence, by (6) and (7), we can obtain two transition sequences, denoted by T1 and T2, where
k k
T1 =(t ,t ,...,t ,t ,t ,...,t )
k ik(s1+1) ik(s1+2) ik(k!) ik(1) ik(2) ik(s1)
and
T2 =(t ,t ,...,t ,t ,t ,...,t ).
k ik(s2+1) ik(s2+2) ik(k!) ik(1) ik(2) ik(s2)
For convenience, we let Tj , (t ,t ,...,t ) for j = 1,2. Applying some transition sequence Tj on one initial
k βj(1) βj(2) βj(k!) k
permutation πˆ, where πˆ ∈ S and πˆ = [c ,c ,...,c ], then we can obtain a cyclic and complete k-RMGC, denoted by Cπˆ ,
k 1 2 k Tj
with its last permutation π˜ for j =1,2. By the construction of Tj, when j =1, we have that k
j k
π˜ =[c ,c ,...,c ,c ,c ]. (8)
1 2 3 k−1 k 1
And when j =2, we have that
π˜ =[c ,c ,...,c ,c ,c ].
2 2 3 k−1 1 k
Next, we construct the transition sequence of Cσ,π1. We let σ ,σ, then σ =[b ,a ,...,a ,a ,b ,...,b ]. When 1≤j ≤
k−1, we let t =t . When j =k, we let t TC=t . If k0+1≤j ≤k!0+k−11,2we usekth1e tr2ansitioln sequence T1 to
α1(j) k α1(k) k+1 k
construct the p-transition t , and let t =t . Hence, we have that σ =t (σ ) for all 1≤j ≤k!+k−1.
α1(j) α1(j) β1(j−k) j α1(j) j−1
Finally, we will prove that for any 0 ≤ i < j ≤ k!+k−1, we have that d (σ ,σ ) ≥ 2. By the construction of t ,
∞ i j α1(j)
when 1≤j ≤k−2, we have that
σ =[a ,..,a ,b ,a ,...,a ,a ,b ,...,b ].
j k+1−j k 1 2 k−j 1 2 l
When j =k−1, we have that
σ =[a ,..,a ,b ,a ,b ,...,b ].
k−1 2 k 1 1 2 l
When j =k, we have that
σ =[a ,a ,...,a ,b ,b ,...,b ]. (9)
k 1 2 k 1 2 l
By (8) and (9), we can obtain that
π =σ =[a ,...,a ,a ,b ,...,b ]. (10)
1 k!+k−1 2 k 1 1 l
When 0≤i<j ≤k−1, we obtain easily that
d (σ ,σ )≥2. (11)
∞ i j
When 0≤i≤k−1 and k ≤j ≤k!+k−1, we have σ (k+1)=a and σ (k+1)=b , then
i 1 j 1
d (σ ,σ )≥|σ (k+1)−σ (k+1)|
∞ i j i j
=|a −b |
1 1
≥2. (12)
When k ≤i<j ≤k!+k−1, we know that the first k elements of σ and σ are differentpermutationsover {a }k . Since
i j j j=1
{a }k is a set of even integers, then
j j=1
d (σ ,σ )≥2. (13)
∞ i j
Hence, by (10) − (13), we can obtain a noncyclic (n,k! + k,ℓ )-snake Cσ,π1 starting with σ and ending with π =
∞ TC 1
[a ,a ,...,a ,a ,b ,b ,...,b ].
2 3 k 1 1 2 l
Similarly, we can construct another noncyclic (n,k!+k,ℓ )-snake Cˆσ,π2. Let T , (t ,t ,...,t ) and
Cˆσ,π2 , (σˆ ,σˆ ,...,σˆ ). Analogously, when 1 ≤ j ≤ k∞−1, we letTtCˆ = t C.ˆWhenαj2(1=) kα,2w(2e) let tα2(k!+=k−t1) . If
TCˆ 0 1 k!+k−1 α2(j) k α2(k) k+1
k+1≤j ≤k!+k−1,weusethetransitionsequenceT2 toconstructthetransitiont ,andlett =t .Moreover,
k α2(j) α2(j) β2(j−k)
we let σˆ =σ. Then, we have that σˆ =t (σˆ ) for all 1≤j ≤k!+k−1. As the above discussion, we can also obtain
0 j α2(j) j−1
another noncyclic (n,k!+k,ℓ )-snake Cˆσ,π2 starting with σ and ending with π =[a ,a ,...,a ,a ,a ,b ,b ,...,b ].
∞ TCˆ 2 2 3 k−1 1 k 1 2 l
5
In the following, by Lemma 3, we will give the construction of an (n,M,ℓ )-snake of size M = ⌈n⌉!(⌊n⌋+⌊n⌋!).
∞ 2 2 2
Suppose p , ⌈n⌉ and q , ⌊n⌋, then [n] has p odd elements and q even ones. We let E be the set of even integers of [n],
2 2 n
and let O be the set of odd integers of [n]. Consider n≥6, we let σ be the first permutation of the ℓ -snake, where
n 0 ∞
σ =[1,4,...,2q−2,2,2q,3,5...,2p−1]. (14)
0
Firstly, we construct one transition sequence, denoted by T = {t ,t ,...,t }. By Lemma 1, we take a cyclic and
i(1) i(2) i(M)
complete p-RMGC by using the following transition sequence
T =(t ,t ,...,t ). (15)
p ip(1) ip(2) ip(p!)
Moreover, by Lemma 3, we can obtain two noncyclic (n,M ,ℓ )-snakes of size M =q!+q, denoted by Cσ,π1 and Cˆσ,π2,
respectively. And Cσ,π1 is given by the following transitionqseq∞uence q TC TCˆ
TC
T =(t ,t ,...,t ) (16)
C α1(1) α1(2) α1(Mq−1)
starting with σ = [b ,a ,...,a ,a ,a ,b ,...,b ] and ending with π = [a ,...,a ,a ,a ,b ,b ,...,b ], where σ,π ∈
1 2 q−1 q 1 2 p 1 2 q−1 q 1 1 2 p 1
S , E = {a |j = 1,...,q}, O = {b |j = 1,...,p}, and |a −b | ≥ 2. Similarly, Cˆσ,π2 is determined by the following
trannsitinon sequjence n j 1 1 TCˆ
T =(t ,t ,...,t ) (17)
Cˆ α2(1) α2(2) α2(Mq−1)
starting with σ = [b ,a ,...,a ,a ,a ,b ,...,b ] and ending with π = [a ,...,a ,a ,a ,b ,b ,...,b ], where σ,π ∈
1 2 q−1 q 1 2 p 2 2 q−1 1 q 1 2 p 2
S , E ={a |j =1,...,q}, O ={b |j =1,...,p}, and |a −b |≥2.
n n j n j 1 1
By (14)−(17), we construct the transition sequence T = (t ,t ,...,t ). According to the construction of σ , we
i(1) i(2) i(M) 0
have that a =2q and a =2.
1 q
We consider 0≤l ≤p!−1. When |σ (i (l+1)+q)−2q|6=1, if σ (q+1)=2q, then we let
l·(q!+q) p l·(q!+q)
t =t , (18)
i(j) α1(j−l·(q!+q))
otherwise we let
t =t (19)
i(j) α2(j−l·(q!+q))
for all l·(q!+q)+1≤j ≤(l+1)·(q!+q)−1. When |σ (i (l+1)+q)−2q|=1, if σ (q+1)=2, then we let
l·(q!+q) p l·(q!+q)
t =t , (20)
i(j) α1(j−l·(q!+q))
otherwise we let
t =t (21)
i(j) α2(j−l·(q!+q))
for all l·(q!+q)+1≤j ≤(l+1)·(q!+q)−1.
And we let
t =t , for all 1≤j ≤p!. (22)
i(j·(q!+q)) ip(j)+q
Hence,T andσ canyieldonepermutationsequence,definedbyC =(σ ,σ ,...,σ ).By(18)−(22)anditsfirstpermutation
0 T 0 1 M
σ , we have that σ = t (σ ) for all 1 ≤ j ≤ M. In the following, we will obtain that σ = σ . Moreover, we let
0 j i(j) j−1 0 M
Cσ0 ,(σ ,σ ,...,σ ). Then we will prove that Cσ0 is an ℓ -snake in the following theorem.
T 0 1 M−1 T ∞
Theorem 4: For all n≥6, there exist an (n,M,ℓ )-snake of size M =⌈n⌉!(⌊n⌋+⌊n⌋!).
∞ 2 2 2
Proof: According to the construction of Cσ0, we consider 0 ≤ l ≤ p!−1. If |σ (i (l +1)+q)−2q| 6= 1 and
T l·(q!+q) p
σ (q+1)=2q or 2, by Lemma 3 and (14), we have that
l·(q!+q)
σ =[σ (i (l+1)+q),4,...,2q−2,2,2q,ˆb ,...,ˆb ], (23)
(l+1)·(q!+q) l·(q!+q) p 2 p
where {σ (i (l+1)+q),ˆb ,...,ˆb }=O . Similarly,if |σ (i (l+1)+q)−2q|=1 andσ (q+1)=2q or 2,
l·(q!+q) p 2 p n l·(q!+q) p l·(q!+q)
by Lemma 3 and (14), we can obtain that
σ =[σ (i (l+1)+q),4,...,2q−2,2q,2,ˆb ,...,ˆb ]. (24)
(l+1)·(q!+q) l·(q!+q) p 2 p
When |σ (i (l+1)+q)−2q|=1, since |2q−2|≥4, then we have
l·(q!+q) p
|σ (i (l+1)+q)−2|≥2. (25)
l·(q!+q) p
By (23)−(25), then we have |σ (1)−σ (q +1)| ≥ 2 for all 0 ≤ l ≤ p!−1. Since σ (1) = 1 and
(l+1)·(q!+q) (l+1)·(q!+q) 0
σ (q+1) = 2q, then |σ (1)−σ (q+1)| ≥ 2. Thus, for all 0 ≤ l ≤ p!−1, σ satisfies the condition of Lemma 3.
0 0 0 l·(q!+q)
Hence, by the construction of Cσ0 and Lemma 3, for all 0≤l≤p!−1, 0≤j <k ≤q!+q−1, we have that
T
d (σ ,σ )≥2.
∞ l(q!+q)+j l(q!+q)+k
6
Furthermore, for l,˜l ∈ [p!] and l < ˜l, since the code generated by its transition sequence T = (t ,t ,...,t )
p ip(1) ip(2) ip(p!)
is a cyclic and complete p-RMGC code, we are assured that for all 0 ≤ j,˜j ≤ q!+q−1, the last p−1 elements of both
σ and σ are all odd and represent two distinct permutations. Hence, we have that
(l−1)(q!+q)+j (˜l−1)(q!+q)+˜j
d (σ ,σ )≥2.
∞ (l−1)(q!+q)+j (˜l−1)(q!+q)+˜j
Finally,wewillprovethatt (σ )=σ .SincethecodegeneratedbythetransitionsequenceT =(t ,t ,
i(p!(q!+q)) p!(q!+q)−1 0 p ip(1) ip(2)
...,t ) is a cyclic and complete p-RMGC code, and by the construction of σ , we have σ (i (p!−1)+q)=3.
ip(p!) 0 (p!−1)(q!+q) p
Thus, we have that |2q−3|≥2 for all n≥6. Hence, we can obtain that
σ =[3,4,...,2q−2,2,2q,5,...,2i (p!)−3,2i (p!)−1,1,2i (p!)+1,...,2p−1]. (26)
(p!−1)(q!+q) p p p
Since σ (i (p!)+q)=1andσ (q+1)=2q,thenwe havethat|σ (i (p!)+q)−2q|6=1. Then,
(p!−1)(q!+q) p (p!−1)(q!+q) (p!−1)(q!+q) p
by (23) and (26), we can obtain that
σ =[1,4,...,2q−2,2,2q,3,5,...,2p−1]=σ .
p!(q!+q) 0
Therefore, C is an (n,M,ℓ )-snake of size M =⌈n⌉!(⌊n⌋+⌊n⌋!).
n,TCn ∞ 2 2 2
B. Construction of ℓ -snakes by using K-snakes
∞
In this subsection, we will construct ℓ -snakes by using K-snakes. In order to complete the construction, we need some
∞
notations and lemmas of K-snakes.
For a K-snake C over S , it is a Gray code over S . Furthermore, for any two distinct codewords σ,π ∈C, we have that
n n
d (σ,π)≥2. Here, d (σ,π) is the K-distance between σ and π in [3], where
K K
d (σ,π)=|{(i,j):σ−1(i)<σ−1(j)∧π−1(i)>π−1(j)}|.
K
Moreover,the Kendall’sτ-metricis rightinvariant[19], thatis, foreverythreepermutationsσ,π,ρ∈S , we haved (σ,π)=
n K
d (σ◦ρ,π◦ρ). We denote by an (n,M,K)-snake a K-snake of size M in S .
K n
In [6], Yehezkeally and Schwartz constructed a (2n+1,M ,K)-snake of length M =(2n+1)(2n−1)M in
2n+1 2n+1 2n−1
A , from a (2n−1,M ,K)-snake in A . Later Horovitz and Etzion [1] improved on this result by constructing a
2n+1 2n−1 2n−1
(2n+1,M ,K)-snake oflengthM =((2n+1)2n−1)M in A , froma (2n−1,M ,K)-snake in A .
2n+1 2n+1 2n−1 2n+1 2n−1 2n−1
Theauthorsin[1]alsopresentedadirectconstructionaimingatobtainingasnakeinA ofsize (2n+1)!−2n+1.Recently,
2n+1 2
Zhang and Ge [13] gave a rigorous proof for the Horovitz-Etzionconstruction of a snake in A of size (2n+1)! −2n+1.
2n+1 2
In order to use the K-snake construction in [13], we will give the following lemma.
Lemma 5: [13, Theorem 4] The construction in [13] can yield a (2n+1,M ,K)-snake in A of size M =
2n+1 2n+1 2n+1
(2n+1)! −2n+1 with the transition sequence including t for all n≥2.
2 2n+1
Furthermore, we require the following lemmas for constructing ℓ -snakes by using K-snakes.
∞
Lemma 6: Suppose{a }n ,n≥2,isasetofintegersofthesameparity.Letσ =[σ (1),..,σ (n),σ (n+1),b ,...,b ]∈
j j=1 i i i i n+2 m
S fori=1,2,whereσ 6=σ ,{σ (j)}n+1 ={a }n ∪{x}fori=1,2,andtheparityofxdiffersfromthatoftheelements
m 1 2 i j=1 j j=1
of {a }n . If σ and σ are both odd permutations or even permutations, then d (σ ,σ )≥2.
j j=1 1 2 ∞ 1 2
Proof: Since σ 6= σ , then d (σ ,σ ) ≥ 1. Suppose d (σ ,σ ) < 2, we have that d (σ ,σ ) = 1. Without loss
1 2 ∞ 1 2 ∞ 1 2 ∞ 1 2
of generality, we let σ = [a ,a ,..,a ,x,b ,..,b ], |a − x| = 1, and |a − x| = 1, where j ,j ∈ [n]. Since
1 1 2 n n+2 m j1 j2 1 2
{a }n are the same parity and d (σ ,σ ) = 1, then σ = [a ,...,a ,x,a ,...,a ,a ,b ,...,b ] or σ =
j j=1 ∞ 1 2 2 1 j1−1 j1+1 n j1 n+2 m 2
[a ,...,a ,x,a ,...,a ,a ,b ,...,b ]. Thus, σ can be obtained from σ using one transposition of a and x
1 j2−1 j2+1 n j2 n+2 m 2 1 ji
for i = 1 or 2. Hence, the parity of σ differs from the parity of σ , which causes a contradiction. Then, we have that
1 2
d (σ ,σ )≥2.
∞ 1 2
Lemma 7: Suppose C is an (n,M ,K)-snake in A with its first permutation π and one transition sequence T =
n n n 0 Cn
(t ,t ,...,t ). For any σ ∈S , by applying the transition sequence T on the permutation σ , then we can obtain
ˆi(1) ˆi(2) ˆi(Mn) 0 n Cn 0
another (n,M ,K)-snake, denoted by Cˆ =(σ ,σ ,...,σ ), where σ =t (σ ) for all j ∈[M −1]. Moreover, the
n n 0 1 Mn−1 j ˆi(j) j−1 n
parities of all the permutations of Cˆ are same.
n
Proof: According to the [5, Lemma 3], we have that Cˆ is an (n,M ,K)-snake. Suppose C = (π ,π ,...,π ).
n n n 0 1 Mn−1
According to the [5, Lemma 2], we can obtain that
σ ◦σ −1 =π ◦π −1 for all j ∈[M −1]. (27)
j 0 j 0 n
Since the Kendall’s τ-metric is right invariant, and by (27), then for any two distinct permutationsσ ,σ ∈Cˆ , we have that
j k n
d (σ ,σ )=d (σ ◦(σ −1◦π ),σ ◦(σ −1◦π ))
K j k K j 0 0 k 0 0
=d (σ ◦σ −1,σ ◦σ −1)
K j 0 k 0
=d (π ◦π −1,π ◦π −1)
K j 0 k 0
=d (π ,π ). (28)
K j k
7
Furthermore, since C ⊂A , then for two distinct permutations π ,π , we can obtain that
n n j k
d (π ,π )=2s (29)
K j k
where s is a positive integer. By (28) and (29), we can obtain that the parities of all the permutations of Cˆ are same.
n
The following lemma gives the construction of a basic block which is useful for the construction of ℓ -snakes by using
∞
K-snakes.
Lemma 8: Let {a }k be a set of integers of the same parity, and let {b }l be also a set of integers of the same parity
j j=1 j j=1
such that {a }k ∪{b }l = [n]. And we let σ ,[a ,b ,b ,b ,...,b ,a ,a ...,a ]. Suppose we have an (l+1,M ,K)-
j j=1 j j=1 1 1 2 3 l 2 3 k l+1
snake in A with one transition sequence T = (t ,t ,...,t ) such that t = t . Then, there exists a
l+1 K,l+1 ˆi(1) ˆi(2) ˆi(Ml+1) ˆi(Ml+1) l+1
noncyclic (n,M ,ℓ )-snake starting with σ and ending with the permutation π =[b ,b ,..,b ,a ,a ,...,a ].
l+1 ∞ 1 2 l 1 2 k
Proof:WeletCσ,π betheclaimednoncyclicℓ -snake,whereCσ,π =(σ ,σ ,...,σ )andTˆ =(t ,t ,...,
Tˆl+1 ∞ Tˆl+1 0 1 Ml+1−1 l+1 α(1) α(2)
t ).
α(Ml+1−1)
Firstly, we denote by σ ,σ. Next, we construct the transition sequence Tˆ . We let
0 l+1
t =t for all j ∈[M −1]. (30)
α(j) ˆi(j) l+1
By (30) and its first permutation σ , we have that
0
σ =[σ (1),...,σ (l+1),a ,a ,...,a ]
j j j 2 3 k
for all j ∈[M −1]. By (30) and Lemma 7, due to the (l+1,M ,K)-snake in A , we have that Cσ,π is a noncyclic
Gray code andl+t1he parities of all the permutations of Cσ,π are saml+e.1Since t =l+1t , then we havTˆel+π1 = σ =
Tˆl+1 ˆi(Ml+1) l+1 Ml+1−1
[b ,b ,...,b ,a ,a ,...,a ].
1 2 l 1 2 k
Finally, for any two distinct permutations σ ,σ ∈ Cσ,π , since their parities are same and σ = [σ (1),...,σ (l +
j1 j2 Tˆl+1 ji ji ji
1),a ,a ,...,a ], for i=1 or 2, by Lemma 6, we have that
2 3 k
d (σ ,σ )≥2.
∞ j1 j2
Hence, we can obtain that Cσ,π is a noncyclic (n,M ,ℓ )-snake starting with σ and ending with the permutation π =
Tˆl+1 l+1 ∞
[b ,b ,..,b ,a ,a ,...,a ].
1 2 l 1 2 k
When n=4k+1, k ≥1, then [n] has 2k even elements and 2k+1 odd ones. In the following, by Lemma 8, we will give
one construction of an (n,M,ℓ )-snake by using some K-snakes. Firstly, we denote by σ an initial permutation, where
∞ 0
σ =[1,2,4,...,4k,3,5...,4k+1].
0
And we construct a transition sequence, denoted by T =(t ,t ,...,t ). By the transition sequence T and the initial
C i(1) i(2) i(M) C
permutationσ , we can get a permutationsequence,denotedby Cσ0 =(σ ,σ ,...,σ ). Given a (2k+1,M ,K)-snake
0 TC 0 1 M−1 2k+1
in A with one transition sequence (t ,t ,...,t ) and t = t , by Lemma 8, we take a noncyclic
2k+1 α(1) α(2) α(M2k+1) α(M2k+1) 2k+1
(n,M ,ℓ )-snake by using the following transition sequence
2k+1 ∞
Tˆ =(t ,t ,...,t ). (31)
2k+1 α(1) α(2) α(M2k+1−1)
Moreover, by Lemma 1, we can obtain a cyclic and complete (2k+1)-RMGC by using the following transition sequence
T =(t ,t ,...,t ). (32)
2k+1 i2k+1(1) i2k+1(2) i2k+1((2k+1)!)
By (31)−(32), we construct the transition sequence T =(t ,t ,...,t ) such that M =M (2k+1)!. We let
C i(1) i(2) i(M) 2k+1
t =t (33)
i(j) α(j−l·M2k+1)
for all l·M +1≤j ≤(l+1)·M −1 and 0≤l≤(2k+1)!−1, and we let
2k+1 2k+1
t =t (34)
i(j·M2k+1) i2k+1(j)+2k
for all 1≤j ≤(2k+1)!.
Secondly, by (33)− (34) and the initial permutation σ , we obtain the permutation sequence σ = t (σ ) for all
0 j i(j) j−1
1≤j ≤M (2k+1)!−1.
2k+1
Finally, in the following theorem, we will prove that Cσ0 is a (4k+1,M (2k+1)!,ℓ )-snake.
TC 2k+1 ∞
Similarly, when n= 4k+3,k ≥1, then [n] has 2k+1 even elements and 2k+2 odd ones. In the following, by Lemma
8, we will give another construction of an (n,Mˆ,ℓ )-snake by using some K-snakes. Firstly, we denote by σˆ an initial
∞ 0
permutation, where
σˆ =[2,1,3,5,...,4k+3,4,6...,4k+2]. (35)
0
8
And we construct another transition sequence, denoted by T =(t ,t ,...,t ). By the transition sequence T and the
Cˆ ˆi(1) ˆi(2) ˆi(Mˆ) Cˆ
initial permutation σˆ , we can get a permutation sequence, denoted by Cˆσˆ0 =(σˆ ,σˆ ,...,σˆ ).
0 TCˆ 0 1 Mˆ−1
Givena(2k+3,M ,K)-snakeinA with onetransitionsequence(t ,t ,...,t )andt =t ,
2k+3 2k+3 β(1) β(2) β(M2k+3) β(M2k+3) 2k+3
by Lemma 8, we take a noncyclic (4k+3,M ,ℓ )-snake by using the following transition sequence
2k+3 ∞
Tˆ =(t ,t ,...,t ). (36)
2k+3 β(1) β(2) β(M2k+3−1)
Moreover, by Lemma 1, we can obtain a cyclic and complete (2k+1)-RMGC by using the following transition sequence
T =(t ,t ,...,t ). (37)
2k+1 i2k+1(1) i2k+1(2) i2k+1((2k+1)!)
By (36)−(37), we construct the transition sequence T =(t ,t ,...,t ) such that Mˆ =M (2k+1)!. We let
Cˆ ˆi(1) ˆi(2) ˆi(Mˆ) 2k+3
t =t (38)
ˆi(j) β(j−l·M2k+3)
for all l·M +1≤j ≤(l+1)·M −1 and 0≤l≤(2k+1)!−1, and we let
2k+3 2k+3
t =t (39)
ˆi(j·M2k+3) i2k+1(j)+2k+2
for all 1≤j ≤(2k+1)!.
Secondly, by (38)−(39) and its first permutation σˆ , we obtain the permutationsequence σˆ =t (σˆ ) for all 1≤j ≤
0 j ˆi(j) j−1
M (2k+1)!−1.
2k+3
Finally, in the following theorem, we will also prove that Cˆσˆ0 is a (4k+3,M (2k+1)!,ℓ )-snake.
TCˆ 2k+3 ∞
Theorem 9: When n = 4k+1 and k ≥ 1, given a (2k+1,M ,K)-snake in A , there exists an (n,M,ℓ )-snake
2k+1 2k+1 ∞
of size M = M ·(2k+1)!. Moreover, when n = 4k+3 and k ≥ 1, given a (2k+3,M ,K)-snake in A , there
2k+1 2k+3 2k+3
exists an (n,Mˆ,ℓ )-snake of size Mˆ =M ·(2k+1)!.
∞ 2k+3
Proof: When n = 4k+1, we will prove that the above Cσ0 is an ℓ -snake. Since σ = [1,2,4,...,4k,3,5,...,4k+1],
TC ∞ 0
by the construction of this ℓ -snake, we have that for all 0 ≤ l ≤ (2k+1)!−1, σ satisfies the condition of Lemma
∞ l·M2k+1
8. Hence, by the construction of Cσ0 and Lemma 8, for all 0≤l ≤(2k+1)!−1 and 0≤j <k ≤M −1, we have
TC 2k+1
d (σ ,σ )≥2.
∞ l·M2k+1+j l·M2k+1+k
Furthermore,forl,˜l∈[(2k+1)!]andl <˜l,sincethecodegeneratedbythetransitionsequenceT =(t ,t ,...,
2k+1 i2k+1(1) i2k+1(2)
t ) is a cyclic and complete (2k+1)-RMGC code, we are assured that for all 0≤j,˜j ≤M −1, the last 2k
i2k+1((2k+1)!) 2k+1
elements of both σ and σ are all odd and represent two distinct permutations. Hence, we have that
(l−1)M2k+1+j (˜l−1)M2k+1+˜j
d (σ ,σ )≥2.
∞ (l−1)M2k+1+j (˜l−1)M2k+1+˜j
Finally, we note that t (σ )=σ , since the code generated by the transition sequence T is
cyclic. Therefore, Cσ0 isi(aMn2k(+n1,(M2k+,ℓ1)!))-sMna2kk+e1o(2fks+i1z)e!−M1 =M0 (2k+1)!. 2k+1
TC ∞ 2k+1
Similarly, when n=4k+3, according to the construction of Cˆσˆ0, we can obtain that Cˆσˆ0 is a (4k+3,Mˆ,ℓ )-snake of
TCˆ TCˆ ∞
size Mˆ =M (2k+1)!.
2k+3
Corollary 10: If n = 4k+1 and k ≥ 2, there exists an (n,M,ℓ )-snake of size M = (2k+1)! −2k+1 ·(2k+1)!.
∞ 2
Moreover, if n=4k−1 and k ≥2, there also exists an (n,Mˆ,ℓ )-snake of size Mˆ = (2k+1)! −2k+1 ·(2k−1)!
∞ 2(cid:0) (cid:1)
Proof: According to Theorem 9 and Lemma 5, we can prove this corollary.
(cid:0) (cid:1)
IV. EXAMPLESOFℓ∞-SNAKES
A. One example of ℓ -snakes by using cyclic and complete RMGCs
∞
In this subsection, we give an example of ℓ -snakes which is constructed by using cyclic and complete RMGCs. Consider
∞
n=6, then we have that p=q =3. By Lemma3, firstly we will constructtwo kindsof noncyclicℓ -snakeswhich are basic
∞
building blocks for ℓ -snakes. Secondly we will give a cyclic ℓ -snake.
∞ ∞
Now, we will start this example with an initial permutation, denoted by σ . By (14), we have that
0
σ =[1,4,2,6,3,5].
0
Inordertoconstructtheblocks,weneedonetransitionsequenceofacyclicandcomplete3-RMGC,i.e,T =(t ,t ,t ,t ,t ,t ).
3 3 3 2 3 3 2
By Lemma 3, we can obtain two transition sequences T and T , where
C Cˆ
T =(t ,t ,t ,t ,t ,t ,t ,t )
C 3 3 4 2 3 3 2 3
and
T =(t ,t ,t ,t ,t ,t ,t ,t ).
Cˆ 3 3 4 3 3 2 3 3
9
Next, we will give two noncyclic (6,3!+3,ℓ )-snakes by the two transition sequences and σ . Then, one noncyclic (6,3!+
∞ 0
3,ℓ )-snake is constructed by T and σ , which is depicted by Figure 1 as follows
∞ C 0
[1,4,2,6,3,5]
↓ t3
[2,1,4,6,3,5]
↓ t3
[4,2,1,6,3,5]
↓ t4
[6,4,2,1,3,5]
↓ t2
[4,6,2,1,3,5]
↓ t3
[2,4,6,1,3,5]
↓ t3
[6,2,4,1,3,5]
↓ t2
[2,6,4,1,3,5]
↓ t3
[4,2,6,1,3,5]
Fig.1. Anoncyclic(6,3!+3,ℓ∞)-snakeconstructedbyTC andσ0.
Another noncyclic (6,3!+3,ℓ )-snake is constructed by T and σ , which is depicted by Figure 2 as follows
∞ Cˆ 0
[1,4,2,6,3,5]
↓ t3
[2,1,4,6,3,5]
↓ t3
[4,2,1,6,3,5]
↓ t4
[6,4,2,1,3,5]
↓ t3
[2,6,4,1,3,5]
↓ t3
[4,2,6,1,3,5]
↓ t2
[2,4,6,1,3,5]
↓ t3
[6,2,4,1,3,5]
↓ t3
[4,6,2,1,3,5]
Fig.2. Anoncyclic(6,3!+3,ℓ∞)-snakeconstructedbyTCˆ andσ0.
By the two kinds of basic noncyclic ℓ -snakes, we have a cyclic (6,3!(3!+3),ℓ )-snake described in Figure 3 as follows
∞ ∞
σ0 [1,4,2,6,3,5]
⇓(2)
σ8 [4,6,2,1,3,5]
↓ t6=t3+3
σ9 [5,4,6,2,1,3]
⇓(2)
σ17 [4,2,6,5,1,3]
↓ t6=t3+3
σ18 [3,4,2,6,5,1]
⇓(2)
σ26 [4,6,2,3,5,1]
↓ t5=t2+3
σ27 [5,4,6,2,3,1]
⇓(2)
σ35 [4,2,6,5,3,1]
↓ t6=t3+3
σ36 [1,4,2,6,5,3]
⇓(1)
σ44 [4,2,6,1,5,3]
↓ t6=t3+3
σ45 [3,4,2,6,1,5]
⇓(1)
σ53 [4,2,6,3,1,5]
↓ t5=t2+3
σ54 [1,4,2,6,3,5]
Fig.3. A(6,54,ℓ∞)-snakeobtainedbyusingacyclicandcomplete3-RMGC.
In Figure 3, “⇓ (1)” stands for an omitted transition sequence T = (t ,t ,t ,t ,t ,t ,t ,t ). While “⇓ (2)” stands for
C 3 3 4 2 3 3 2 3
another omitted transition sequence T = (t ,t ,t ,t ,t ,t ,t ,t ). Hence, when n = 6, by using one cyclic and complete
Cˆ 3 3 4 3 3 2 3 3
3-RMGC, we can construct a cyclic ℓ -snake of size 54.
∞
10
B. One example of ℓ -snakes by using K-snakes
∞
In this subsection, we present an example of ℓ -snakes constructed by using K-snakes. Consider n = 7, then we have
∞
4 odd elements and 3 even ones in {1,2,3,4,5,6,7}. Moreover, Horovitz and Etzion [1] gave a (5,57,K)-snake in A
5
with one transition sequence, denoted by T = (Tˆ,Tˆ,Tˆ), where Tˆ is a partial transition sequence of T and Tˆ =
K,5 K,5
(t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ).
3 3 5 3 3 5 3 5 5 3 3 5 3 3 5 3 5 5 5
By Theorem 9, given this (5,M ,K)-snake of size 57 in [1], we can obtain a (7,M,ℓ )-snake of size M =342.
5 ∞
Firstly, we will start this example with one initial permutation, denoted by σˆ . By (35), we have that
0
σˆ =[2,1,3,5,7,4,6].
0
By Lemma 8 and T , we can construct one transition sequence, denoted by Tˆ , where
K,5 K,5
Tˆ =(Tˆ,Tˆ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ,t ).
K,5 3 3 5 3 3 5 3 5 5 3 3 5 3 3 5 3 5 5
Hence, we will give one noncyclic (7,57,ℓ )-snake by the transition sequence Tˆ and σˆ . Then, the noncyclic (7,57,ℓ )-
∞ K,5 0 ∞
snake determined by Tˆ and σˆ is depicted by Figure 4 as follows
K,5 0
231731532732132537127157137257217237512513517523521527531
123173153273213253712715713725721723751251351752352152753
312317315327321325371271571372572172375125135175235215275
555222771555777112533322255133355511237772223317773331127
777555227111555771255533322511133355123337772231117773312
444444444444444444444444444444444444444444444444444444444
666666666666666666666666666666666666666666666666666666666
Fig.4. Anoncyclic(7,57,ℓ∞)-snakeconstructedbyTˆK,5andσˆ0.
Here,everycolumninFigure4representsonepermutationover{1,2,3,4,5,6,7}.Moreover,wehaveacyclicandcomplete
3-RMGC with the transition sequence T = (t ,t ,t ,t ,t ,t ). By Figure 4 and T , we can obtain a (7,342,ℓ )-snake
3 3 3 2 3 3 2 3 ∞
depicted by Figure 5 as follows
σ0 [2,1,3,5,7,4,6]
⇓
σ56 [1,3,5,7,2,4,6]
↓ t7=t3+4
σ57 [6,1,3,5,7,2,4]
⇓
σ113 [1,3,5,7,6,2,4]
↓ t7=t3+4
σ114 [4,1,3,5,7,6,2]
⇓
σ170 [1,3,5,7,4,6,2]
↓ t6=t2+4
σ171 [6,1,3,5,7,4,2]
⇓
σ227 [1,3,5,7,6,4,2]
↓ t7=t3+4
σ228 [2,1,3,5,7,6,4]
⇓
σ284 [1,3,5,7,2,6,4]
↓ t7=t3+4
σ285 [4,1,3,5,7,2,6]
⇓
σ341 [1,3,5,7,4,2,6]
↓ t6=t2+4
σ342 [2,1,3,5,7,4,6]
Fig.5. A(7,342,ℓ∞)-snakeconstructedbyusingaK-snakeinA5.
In Figure 5, “⇓” stands for an omitted transition sequence Tˆ . Therefore, when n=7, by using K-snakes in A , we can
K,5 n
obtain a cyclic (7,342,ℓ )-snake.
∞
V. COMPARISON
Yehezkeally and Schwartz [6] presented one construction of an (n,M ,ℓ )-snake of size
n,0 ∞
n n n
M =⌈ ⌉!(⌊ ⌋+(⌊ ⌋−1)!) for all n≥4, (40)
n,0
2 2 2
which is far less than the upper bound n! on the length of ℓ -snakes in S .
2⌊n2⌋ ∞ n