Balanced Permutation Codes

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by  Ryan Gabrys

Balanced Permutation Codes
Ryan Gabrys∗† and Olgica Milenkovic∗
∗ECE Department, University of Illinois, Urbana-Champaign †Spawar Systems Center, Pacific
Abstract—Motivatedbychargebalancingconstraintsforrank choice of the parameters n and b>1. In contrast to the work
modulationschemes,weintroducethenotionofbalancedpermu- pertaining to permutation discrepancy, our work is not con-
tations and derive the capacity of balanced permutation codes.
cerned with finding the smallest discrepancy permutation but
We also describe simple interleaving methods for permutation
ratherwithconstructingcodesofrelativelysmalldiscrepancies
code constructions and show that they approach capacity.
and of size as large as possible. Furthermore, the discrepancy
I. INTRODUCTION constraint places a constraint on substrings of fixed length b,
We consider a new constraint on permutations that requires whereas the balancing constraint in (1) is more general as
movingaveragesofsymbolstobecloselyconcentratedaround balancing is required for all blocks of length b∈S, where S
6 the mean running average. This constraint and the resulting may contain more than one value. This code design approach
1 coding schemes aim to balance charges across cells in rank offers certain advantages in terms or built-in error detection
0 modulation systems for flash memories [5]. Balanced codes capabilities, and in addition, it covers the single blocklength b
2
may also potentially aid in detecting and correcting errors. balancing constraint by definition. To the best of the authors

n This can be accomplished by monitoring whether the balanc- knowledge, this coding problem has not been studied before
a
ing constraint is satisfied during the decoding process. Hence, in the literature.
J
theconstraintcomplementsexistingconstraintsimposedsoas One use of the results of [1] allows us to show that for
6
to contain cross-leakage between neighboring cells [2]1. The S = [2,n], C(n,S) is non-empty provided that D(b,n) ≥
2
permutationbalancingconstraintmayalsobeseenasablock- 4/(b(n+1)).Asanexample,itiseasytoseethatthesmallest

] by-block extension of classical bounded running digital sum discrepancy set of permutations for n=4 and b=2 equals:
T
codes [9], and some of the construction ideas pursued were
I (1,3,2,4), (1,4,2,3), (2,3,1,4), (2,4,1,3),
. inspired by the well known Knuth’s balancing algorithm [7].
s The constraint may be succinctly described as follows. (3,1,4,2), (3,2,4,1), (4,1,3,2), (4,2,3,1).
c
[ Letnbeafixedpositiveinteger,andlet[n]={1,2,...,n}.
Thediscrepancyofthepermutationsis1.Allowingforalarger
For any two non-negative integers a ≤ b, we use [a,b] to
1 discrepancyclearlyincreasesthesizeofthepermutationcode.
denotetheset{a,a+1,...,b−1,b}andweuseS todenote
v n For simplicity, our focus will be on two special cases for
7 the set of permutations of length n. We are concerned with
the set S, S = [1,n] and S = {2,4,...,2·s}, for some s,
8 studying codes C(n,S) ⊆ S , where S ⊆ [1,n], defined as 1 2
n suchthats< n ,andtheircorrespondingD(b,n)functions,
8 follows: π ∈ C(n,S) if and only if for every b ∈ S and (cid:100)n(cid:15)(cid:101)
6
1≤j ≤n−b+1, one has
0 4 16
1. n+1(1−D(b,n))≤ π(j)+π(j+1)+···+π(j+b−1) D1(b,n)= b, and D2(b,n)= bn(cid:15), (2)
2 b
0
(1) where (cid:15) ∈ (0,1). These two constraints were selected to
6
1 n+1 demonstrate that one can balance out the same set of per-
≤ (1+D(b,n)).
: 2 mutations for almost all choices of b simultaneously, although
v
not with the same balancing function D(b,n) - the balancing
i Here, D(b,n) takes values in the interval [0,1] and is allowed
X function is weaker for smaller b. For the first balancing
to be a function of b and n. We refer to the constraint in
r (1) as a balancing constraint, as it requires any b-consecutive constraint where D(b,n) = D1(b,n) and S = S1, the
a
resulting permutation code has capacity one. On the other
sum of elements in a permutation to stay close to the mean
hand, it is impossible to have the same permutation code
b-consecutive sum, which equals b(n+1)/2.
have a discrepancy uniformly scaling with n−(cid:15) for all choices
A constraint related to the balancing constraint was studied
of b ∈ [1,n]. Some divisibility properties on b have to be
under the name of low-discrepancy permutation constraint
imposed. At the same time, the code rate is strictly less than
in a handful of papers [1]. The authors of [1] studied the
one, equal to 1−(cid:15).
problem of finding a smallest discrepancy permutation. In
Codes C(n,S) with codewords satisfying (1) and discrep-
particular, they defined the discrepancy of a permutation
π ∈S according to ancy D1 and D2 are referred to as a D1- and D2-balanced
n
(cid:12) (cid:12) permutation codes. Let
(cid:12) b (cid:12)
(cid:12)(cid:88) n+1(cid:12) log |C(n,S)|
disc(π,b)=1≤mi≤anx−b (cid:12)(cid:12)(cid:12)j=1 πi+j −b 2 (cid:12)(cid:12)(cid:12), Ri =nl→im∞sup log n! , (3)
and focused their attention on computing disc(n,b) = fori=1,2,denotethecapacityofaDi-balancedpermutation
minπ∈Sn disc(π,b). They showed that disc(n,b)≤2 for any code expressed in bits2. Our main results demonstrate that
1Theseconstraintsoftengounderthenametwo-neighborconstraints,and 2Hereandthroughoutthepaperweassumethatalllogarithmsareevaluated
willbediscussedinmoredetailattheendofthepaper basetwo.
when S = S , R = 1. For S = S , we get that R = implies that len(O(j−1)) (cid:54)= 0. Similarly, 1 (cid:80)j−1π((cid:96)) ≥
1 1 2 2 2 j−1 (cid:96)=1
1−(cid:15). We also provide a sampling of results for D1− and n+1, implies that len(O(j−1))(cid:54)=0.
D −balanced codes that satisfy the two-neighbor constraint 2 1
int2roduced in [2]. Proof: Suppose that 1 (cid:80)j−1π((cid:96)) < n+1 and that on
j−1 (cid:96)=1 2
the contrary, len(O(j−1)) = 0. Let P represent the set of
II. THECAPACITYOFD1-BALANCEDCODES symbols in the seque2nce (π(1),...,π(j−1)). The set P, and
In this section, we present a construction for D1-balanced the sets of elements contained in O(j−1) and O(j−1) form a
1 2
permutation codes that achieve rate one. For simplicity of partition of [n]. Clearly, the average symbol value of the set
exposition, we suppose that n is an even number. [n] equals n+1. Hence, if len(O(j−1))=0, then
2 2
The idea behind the code construction is to partition the set
 
[n] into two subsets, P = [n/2] and P = [n]\[n/2]. The
1 2 n+1 1 (cid:88) (cid:88) n+1
symbolsinthesetP arearrangedaccordingtoapermutation =  y+ z< ,
1 2 n  2
γ1 ∈ Sn, and the resulting sequence is denoted by O1. y∈P z∈O(j−1)
2 1
Similarly, the symbols in the set P2 are arranged according which is a contradiction. The case where 1 (cid:80)j−1π((cid:96)) ≥
to a permutation γ2 ∈ Sn2, and the resulting sequence is n+1 may be handled similarly. j−1 (cid:96)=1
denoted by O . We form permutations π ∈ C(n,S ), which 2
2 1 We have the following claim.
we subsequently prove to be D -balanced as follows.
1
We initialize the construction by choosing the first element Claim 1. Let π ∈ C(n,S1) be constructed according to the
of π to be the first element of O ; we consequently remove previously described running sums A. For any integer 1 ≤
1
that element from O . The second element of π is set to the j ≤n,
1
first element of O2 and this element is subsequently removed n+1 (cid:88)j n+1
from O . Suppose next that j−1, j >1, symbols of π have j· −(n+1)≤ π((cid:96))≤j· +(n+1).
2
2 2
beenselected.Todeterminethenextsymbolinπ,wecompute (cid:96)=1
the accumulated average sum A(j−1) = 1 (cid:80)j−1π((cid:96)). If Proof: The proof proceeds by induction on j. The result
j−1 (cid:96)=1 holds for j = 1, and this establishes the base case. Suppose
A(j−1)< n+1, we set the j-th element of π to be equal to
2 next that the result holds for all j <J and consider the case
the first element of O and remove the element from O . If
2 2
j =J ≤n. Clearly,
A(j)≥ n+1, then we set the j-th element of π to be equal to
2
the first element of O and remove this element from O . J−1 J j−1
1 1 (cid:88) (cid:88) (cid:88)
π((cid:96))+π(J)≤ π((cid:96))≤ π((cid:96))+π(J).
The procedure is illustrated via the example below.
(cid:96)=1 (cid:96)=1 (cid:96)=1
Example 1. Let n = 12. In this setting, we have P =
1 If (cid:80)J−1π((cid:96)) < (J −1)· n+1, then according to Lemma 1,
{1,2,3,4,5,6}, P = {7,8,9,10,11,12}. By choosing γ = (cid:96)=1 2
2 1 we have n+1 < π(J) ≤ n. Furthermore, using the inductive
(3,4,1,2,5,6) and γ = (6,5,4,3,2,1), we arrive at O = 2
2 1 hypothesis along with (cid:80)J−1π((cid:96))<(J−1)·n+1 shows that
(3,4,1,2,5,6) and O2 =(12,11,10,9,8,7). (cid:96)=1 2
For the given choice of γ ,γ , we initialize π as J
1 2 n+1 (cid:88) n+1
π = (3,12,...) and obtain O = (4,1,2,5,6), O = J · −(n+1)≤ π((cid:96))≤J · +(n+1),
1 2 2 2
(11,10,9,8,7). Next, we evaluate A(2)= 1(3+12)=7.5≥ (cid:96)=1
2 which established the validity of the claim for the case that
6.5,andsubsequentlyselectthefirstelementfromO toobtain
1 (cid:80)J−1π((cid:96)) < (J −1)· n+1. The case where (cid:80)J−1π((cid:96)) ≥
π = (3,12,4,...). We then compute A(3) = 6.33 and arrive (cid:96)=1 2 (cid:96)=1
(J −1)· n+1 can be handled similarly.
at π = (3,12,4,11,...). Continuing until all elements are 2
From Claim 1, we can prove that the code C(n,S ) satis-
used up, we obtain (3,12,4,11,1,10,2,9,8,5,7,6). 1
fies (1).
Itisstraightforwardtoseethatifπ isconstructedaccording
Lemma 2. For any 2≤b≤n and j ∈[n−b+1],
tothepreviousprocedureusingtwopermutationsγ ,γ ,while
1 2
π(cid:48) is constructed from γ1(cid:48),γ2(cid:48) where (γ1,γ2) (cid:54)= (γ1(cid:48),γ2(cid:48)), then b· n+1 −2(n+1)≤π(j)+π(j+1)
π (cid:54)=π(cid:48).Therefore,thecardinalityofC(n,S )⊆S equalsthe 2
1 n
n+1
number of possible choices for the permutations γ1,γ2, i.e., +...+π(j+b−1)≤b· +2(n+1).
|C(n,S )|=(cid:0)n!(cid:1)2, which implies that R =1. 2
1 2 1 Proof: Clearly,
NotethatthesequencesO ,O areupdatedaftereachstep,
1 2
i.e., after each extension of the permutation π. For notational π(j)+π(j+1)+...+π(j+b−1)=
convenience, we let O(j) denote the sequence O after j π(1)+π(2)+...+π(j+b−1)−
1 1
elements have been added to the permutation π; similarly,
(π(1)+π(2)+...+π(j−1)).
we let O(j) denote the sequence O after j elements have
2 2
Using the result of Claim 1, we obtain
been added to the permutation π. For ease of notation, we
use len(O) to denote the length of the sequence O. The π(j)+π(j+1)+...+π(j+b−1)
next lemma establishes an important property of the encoding n+1
procedure described in Example 1. ≤(j+b−1)· +(n+1)−
2
(cid:18) (cid:19)
n+1 n+1
Lemma 1. For any π ∈C(n,S ) constructed using two given
1 (j−1)· −(n+1) =b· +2(n+1).
permutations γ ,γ , and any j ≤ n, 1 (cid:80)j−1π((cid:96)) < n+1, 2 2
1 2 j−1 (cid:96)=1 2
Using the same approach, one may show that 1, the element following 4 in O , and another arrow pointing
1
at 12, the first element in O .
n+1 2
π(j)+π(j+1)+...+π(j+b−1)≥b· −2(n+1), In the third step, since ∆(2) ≥ 0 and π(2)(3) < n+1,
2 2
no transposition is performed, and we simply move the left-
and this completes the proof.
most pointer to the next element in O so that π(3) =
These results lead to the following theorem. 1
(3,12,4,11,1,10,2,9,5,8,6,7).Notethatafterthreestepsof
Theorem1. ThecapacityoftheD -constraintequalsR =1. ↑ ↑
1 1 encodings,thepositionsofthreeelementsinπarepermanently
Proof: As a consequence of Lemma 2, we know that the fixed. We terminate after n elements in π have been fixed, in
code construction from Example 1 satisfies the D -constraint which case we obtain π =(3,12,4,11,1,10,2,9,8,5,7,6).
1
and given that there are (n)! choices for each of the two
permutations γ and γ , it f2ollows that R =1. III. THECAPACITYOFD2-BALANCEDCODES
1 2 1
Notethatanaiveimplementationoftheencodingprocedure We now consider the case where D(b,n) scales inversely
requiresmaintainingthesequencesO ,O androughlyO(n2) both with b and (cid:100)n(cid:15)(cid:101), (cid:15) ∈ (0,1) and where S =
1 2 2
operationsforre-computingtheaveragevaluesofthesymbols {2,4,...,2·s}. The reason behind this choice of problem
n times. Clearly, significantly less complex implementations parametersisthatwecannotsimultaneouslysatisfyastringent
are possible. discrepancy constraint with S =[n]. Such a constraint would
One approach would be to divide the input information require that for all j ∈[n], one has
sequence into two blocks of the same size or sizes that
n+1 n+1
differ by one, and than use the two parts to “encode” for −8((cid:98)n1−(cid:15)+n−(cid:15)(cid:99))≤π(j)≤ +8((cid:98)n1−(cid:15)+n−(cid:15)(cid:99)),
2 2
the permutations γ1 and γ2. Here, encoding may refer to whichisclearlyimpossible.Asimilarproblemisencountered
generating a permutation at a given position in the lexico- when S contains two consecutively valued symbols.
graphical order of permutations, and efficient, straightforward Thus, we limit our attention to the case where S contains
algorithmsforthisandmoregeneralencodingsareknown[3], elements which are multiples of two. The proof follows by
[4], [6]. This approach would remove the storage require- noting the D -constraint requires that for every i ∈ [n], π(i)
2
ment for the permutation γ1 and γ2, and subsequently only is close in value to π(i−2).
require transposing adjacent symbols in the permutations.
Lemma 3. The capacity R = 0 for s > 32(n1−(cid:15)+1)+1,
The procedure, which we next illustrate with an example, 2
and R ≤1−(cid:15) for s<32(n1−(cid:15)+1)+1.
may be seen as an extension of Knuth’s balancing principle, 2
where complementation used for binary strings is replaced by Throughout the remainder of this section, we write N =
transpositions in permutations3. (cid:100)n(cid:15)(cid:101) to ease notational burden and for simplicity assume that
N|n and that N is divisible by four. We focus our attention
Example 2. Suppose once more that n = 12,
on deriving a lower bound on R by constructing a balanced
P = {1,2,3,4,5,6}, P = {7,8,9,10,11,12}, 2
1 2
code C(n,S ), with S ={2,4,...,2·(n −1)} and of rate
γ1 = (3,4,1,2,5,6), and γ2 = (6,5,4,3,2,1), so that 2 2 N
O1W=e (fo3r,m4,1a,n2a,5u,x6il)iaaryndpeOr2m=uta(t1io2n,1π1(,11)0∈,9S,8n,7b)y.interleav- logn→∞ log|lCo(gnn,!S2)| =1−(cid:15).
ing γ1 and γ2, which in the above case leads to π(1) = We partition the set [n] into N subsets of equal size n,
(3,12,4,11,1,10,2,9,5,8,6,7). We maintain two pointers, N
comprising consecutive integers, subsequently denoted by P ,
↑ ↑ 1
eachrequiringlog(n)bits.Theinitialpositionsofthepointers P ,..., P 4. For each i ∈ [N], we order the elements of
2 N
are at the location of the second element of O and the first P arbitrarily and denote the resulting sequence by O . Since
1 i i
elementofO2,aswewouldliketotestiftransposingthesetwo thereare Nn!waystoarrangeeachsetPi,|C(n,S2)|≥(cid:0)Nn!(cid:1)N,
elements will reduce the running sum. We also initialize the and hence
discrepancyto∆(1)=π(1)(1)−(n+1)/2=3−6.5=−3.5, log(cid:0)n!(cid:1)N
lim N =1−(cid:15),
and store ∆(1) = −3.5, which requires O(log(n)) bits of n→∞ logn!
overhead.
which implies the lower bound.
In the second step of encoding, since ∆(1) < 0 and
Figure 1 illustrates the encoding process. Similar to what
π(1)(2)> n+1, the pointer at 12 is moved up to the position
2 we did before, we incrementally build a permutation π ∈
of the next element in O which is 11 as shown below π(2) =
2 C(n,S ). Each cell in the figure involves appending two
(3,12,4,11,1,10,2,9,5,8,6,7). The updated discrepancy is 2
↑ ↑ elements to π, chosen from one of two possible sets. For
computed according to ∆(2)=∆(1)+π(2)(2)−6.5=2. instance, based on Figure 1, visiting the first cell requires
Note thatif ∆(1)≥0 and π(1)(2)> n+1, thenthe element appending either a) one element from O and one element
2 2
pointed at by the second arrow would have been deleted from from O or b) appending one element from O and one
N 1
π(1) and reinserted back into the permutation at the position element from O to π. Note that since each of our sets
N−1
ofthefirstarrowusingasingleadjacenttransposition.Atthis has size n, each cell in Figure 1 will be visited 2· n times.
N N
point, π(2) would have had one arrow pointing at the element Wenextexplainthisoutlinedencodingprocessinmoredetail.
3Thereappearstobenonaturalextensionofthenotionofcomplementation 4Clearly,forvaluesofN notsatisfyingthegivendivisibilityproperties,the
in a binary string for permutations, as “reflecting” values of a prefix of a sets Pi,i=1,...,N may have different cardinalities. This small technical
permutationaroundn+1maynotresultinapermutation. detaildoesnotchangethevalidityoftheargumentnortheclaimedresult.
Thefirst2symbolsofπ areselectedasfollows.Setthefirst steps, we compute A(8) = 16.5, A(10) = 16.4,
element of π to be the first element in O and then remove A(12) = 16.67, and A(14) = 16.28. Based
1
this element from O . Set the second element of π to be the on these values, the permutation π is augmented
1
first element of O and remove the chosen element from recursively as π = (2,25,8,32,3,26,7,29,4,28,...),
N−1
O . This selection process is captured by the first cell in π = (2,25,8,32,3,26,7,29,4,28,6,30,...),
N−1
Figure 1, indicating that the first element of the permutation π = (2,25,8,32,3,26,7,29,4,28,6,30,1,27,...),
is taken from O or O and the second element is from O π = (2,25,8,32,3,26,7,29,4,28,6,30,1,27,5,31,...).
1 2 N
or O . At this point, we move onto the second cell in Figure 1.
N−1
We next compute A(2) = 1(π(1)+π(2)) and if For clarity, we now write π = (π(1),π(2)), where
2
len(O ) (cid:54)= 0 or len(O ) (cid:54)= 0, we revisit the first cell. π(1) = (2,25,8,32,3,26,7,29,4,28,6,30,1,27,5,31).
1 2
If A(2) < n+1, we append the first element of the set In the next iteration of the algorithm, we compute
2
O followed by the first element of the set O to π and A(16) = 16.5 and augment π(2) to π(2) = (11,17).
2 N
remove these elements from their respective sets. Otherwise, In the next four steps we compute A(18) = 16.22,
if A(2)≥ n+1 we append the first element from O followed A(20) = 16.5, A(22) = 16.36, A(24) = 16.5,
2 1
by the first element from O to π and remove these A(26) = 16.42, A(28) = 16.5, and A(30) = 16.3. The
N−1
two elements from their respective sets. We then consider corresponding updates in π(2) result in (11,17,16,22,...),
A(4) = 1(π(1)+π(2)+π(3)+π(4)), and if len(O ) (cid:54)= 0 (11,17,16,22,10,20,...), (11,17,16,22,10,20,13,23,...),
4 1
or len(O ) (cid:54)= 0, we revisit the first cell. If A(4) < n+1, (11,17,16,22,10,20,13,23,12,19,...),
2 2
we append the first element from O followed by the first (11,17,16,22,10,20,13,23,12,19,14,21,...),
2
element from O to π and remove these elements from their (11,17,16,22,10,20,13,23,12,19,14,21,9,18,...)
N
respective sets. Otherwise, we append the first element from (11,17,16,22,10,20,13,23,12,19,14,21,9,18,15,24).
O followedbythefirstelementfromO toπ andremove
1 N−1
Next, we prove that the chosen balancing constraint is
these elements from their respective sets. This process is
satisfied by the previously outlined encoding procedure. The
continued until we have added 4· n elements to π so that
N firstresultinthisdirectionisageneralizationofClaim1from
len(O ) = len(O ) = len(O ) = len(O ) = 0. Notice
1 2 N N−1
the previous section. It follows directly from the encoding
thatsincetwoelementsareappendedatonce,wehavevisited
procedureillustratedinExample3,anditsproofishenceforth
the first cell 2· n times.
N omitted.
Next,weagaincomputetherunningaverageoftheelements
fixed(orappended)toπ thusfar.Iftherunningaverageisless Claim 2. Let π ∈ C(n,S ), where C(n,S ) is constructed
2 2
than n+1,thenwechoosethenexttwoelementsofπ tobethe according to the described encoding algorithm. For any even
2
first elements of the sets O4,ON−2, which we then remove integer j ≤n, it holds that
fromtheirrespectivesets.Otherwise,thenexttwoelementsof
j
π are chosen from O3,ON−3, added, and removed from their n+1 2n (cid:88) n+1 2n
j· − ≤ π((cid:96))≤j· + .
respective sets. Afterwards, we repeatedly add elements from 2 N 2 N
the sets O ,O ,O ,O until len(O )=len(O )= (cid:96)=1
4 N−2 3 N−3 4 N−2
len(O )=len(O )=0. This process is continued until π Suppose now that j−i+1∈S . Then, j−i+1<2· n,
3 N−3 2 N
has length n. and so if (c−1)· 4n <i≤c· 4n, then
N N
4n
j+1≤(c+1)· . (4)
N
Thus,ifthei-thelementinapermutationisencodedaccording
to cell index c in Figure 1 and j−i+1∈S , the (j+1)-st
i 2
elementinthatpermutationisencodedaccordingtocellindex
Fig.1. EncodingforaD2-BalancedCode c orc +1.Supposethatj−i+1∈S .Asaresultofthenext
i i 2
claim,weknowthatanyelementencodedaccordingtocellc
We illustrate the procedure with an example. i
has a symbol value close to an element encoded according to
Example 3. Letn=32,N =8,andd=2,andsupposethat cell c .
j
O =(2,3,4,1), O =(8,7,6,5), O =(11,10,12,9),
1 2 3 Claim 3. Let π ∈ C(n,S ). Suppose that i,j are such that
O =(16,13,14,15), O =(17,20,19,18), 2
4 5 j−i+1∈S . Then,
O =(22,23,21,24), O =(25,26,28,27), and 2
6 7
O =(32,29,30,31). 4n 4n
8
2|(j−i+1), (c−1)· <i<j ≤(c+1)·
We set the first two elements of π to be the first element N N
from O followed by the first element from O so that π =
1 7 for some positive integer c, and
(2,25,...) and A(2) = 2+25 = 13.5. Since A(2) < n+1, we
2 2
extend π to π =(2,25,8,32,...). 4n 4n
π(i)− ≤π(j+1)≤π(i)+ .
Now,wecomputeA(4)=16.75whichimpliesthatπshould N N
be extended to π =(2,25,8,32,3,26,...).
The next lemma establishes that our balancing criteria is
Next, we find A(6) = 16 and arrive at
satisfied.
π = (2,25,8,32,3,26,7,29,...). In the next three
Lemma 4. For any i,j ∈[n] i<j, b=j−i+1∈S2, and From[2],atrivialupperboundonR1((cid:100)n(cid:15)k(cid:101))is 1+2(cid:15)k.Using
π ∈C(n,S ), we have similar ideas as in Section II and the construction from [2],
2
onecanderiveamatchinglowerbound.Similarlytowhatwas
n+1 (n+1)
b· −8 ≤π(i)+π(i+1)+...+π(j) proposed in [2], in our setting the k-constraint is imposed by
2 N
selecting two elements at a time from a set with elements that
n+1 (n+1)
≤b· +8 . differ by at most k. These ideas are illustrated by the next
2 N
example, while details of the proof are deferred to the full
Proof: Recall from Equation (4) and Claim 3 that
version of the paper.
4n 4n
(c−1)· <i<j ≤(c+1)· Example 4. Let n = 24 and k = 4. We begin by parti-
N N
tioning the set [24] into six sets of size four each, where
for some integer c. Since j−i+1∈S2, j−i+1 is an even P1 = {1,2,3,4}, P2 = {5,6,7,8}, P3 = {9,10,11,12},
integer. Thus, one of the values i,j is even. Suppose for now and P = {13,14,15,16}, P = {17,18,19,20}, P =
4 5 6
that j is even. Then using Claim 2, we have {21,22,23,24}. We choose an ordering for each of these sets
to obtain
j i−1
(cid:88) (cid:88)
π(i)+π(i+1)+...+π(j)= π((cid:96))− π((cid:96))
O =(3,4,1,2), O =(8,7,6,5), O =(9,10,12,11),
1 2 3
(cid:96)=1 (cid:96)=1
n+1 4n O4 =(13,16,14,15),O5 =(20,19,18,17),O6 =(21,22,23,24).
≤(j−i+1)· + .
2 N We select the first two elements from O , remove these
1
Otherwise, if j is odd, we may write elements from O , and arrive at π = (3,4,...). Next, we
1
computetherunningaverageofπasA(2)= 3+4 =3.5.Since
j+1 i 2
(cid:88) (cid:88) 3.5 < n+1 = 12.5, we choose one of the sets O ,O ,O to
π(i)+π(i+1)+...+π(j)= π((cid:96))− π((cid:96))+ 2 4 5 6
select two additional elements from. Suppose we pick O so
(cid:96)=1 (cid:96)=1 4
n+1 4n 4n that π = (3,4,13,16,... and so that the symbols 13,16 are
π(i)−π(j+1)≤(j−i+1)· + + ,
subsequently removed from O . Then, since the symbol aver-
2 N N 4
age A(4) = 9 < 12.5, we choose elements from O ,O ,O .
where the inequality follows from Claims 2 and 3. The 4 5 6
Suppose we pick O so that π = (3,4,13,16,20,19,...).
inequality in the other direction for the case of j even or j 5
Then 20,19 are removed from O . In this case, A(6) ≥ 12.5
odd can be proved using similar arguments. 5
and so we pick the next two elements from one of the sets
As a consequence of Lemmas 3 and 4, the following
O ,O ,O . Suppose, we choose the set O so that π =
theorem holds. 1 2 3 2
(3,4,13,16,20,19,8,7 and so A(8) = 11.25. Suppose O
6
Theorem 2. For (cid:15)∈(0,1), R =1−(cid:15). is chosen next and so π = (3,4,13,16,20,19,8,7,21,22.
2
Suppose we continue the same procedure and choose from
IV. THEBALANCEDTWO-NEIGHBORCONSTRAINT
the following sets (in order): O ,O ,O ,O ,O ,O ,O . The
1 5 6 2 3 4 3
We now turn our attention to a short treatment regard- result is the permutation π =(3,4,13,16,20, 19,8,7,21,22,
ing combined balanced and constrained codes [8]. We will 1,2,18,17,23,24,6 ,5,9,10,14,15,12,11)∈S .
24
focus on the two-neighbor symmetric constraint coding as
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crepancyConsecutivek-sums,”J.ofComb.Theory,Vol.100,no.2,pp.
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Theorem 3. For (cid:15)k ∈(0,1), R1((cid:100)n(cid:15)k(cid:101))= 1+2(cid:15)k.