Table Of ContentCoding for Racetrack Memories
Yeow Meng Chee,∗ Han Mao Kiah,∗ Alexander Vardy,† Van Khu Vu,∗ and Eitan Yaakobi‡
∗ Nanyang Technological University, Singapore
† University of California San Diego, La Jolla, CA92093, USA
‡ Technion — Israel Institute of Technology, Haifa, 32000 Israel
Emails:{ymchee,hmkiah,vankhu001}@ntu.edu.sg.edu, [email protected],
[email protected]
Abstract—Racetrack memory is a new technology which utilizes thus one of the domains is not read, which results with a dele-
magnetic domains along a nanoscopic wire in order to obtain ex- tion of the bit stored in this domain. In case the domains were
tremelyhighstoragedensity.Inracetrackmemory, eachmagnetic
not successfully shifted, then the same domain is read again
domain can store asinglebit of information, whichcan besensed
7 and we experience an insertion, however of the same bit. This
by a reading port (head). The memory has a tape-like structure
1 which supports a shift operation that moves the domains to be kind of insertion errors is also referred as repetition errors or
0
read sequentially by the head. In order to increase the memory’s sticky insertions in a sticky channel [2], [8].
2
speed,priorworkstudiedhowtominimizethelatencyoftheshift In this work we study codeswhich correctposition errorsin
n operation, whilethe no less important reliabilityof thisoperation
racetrackmemory.At first sight, this problemis not any differ-
a has received only a little attention.
entthanthewell-studiedproblemofdesigningcodescorrecting
J Inthisworkwedesigncodeswhichcombatshifterrorsinrace-
4 trackmemory,calledpositionerrors.Namely,shiftingthedomains deletions and insertions [1], [5]. However, we take another ap-
2 isnotanerror-freeoperationandthedomainsmaybeover-shifted proach to tackle the problem and leverage the special features
or are not shifted, which can be modeled as deletions and sticky of racetrack memory, where it is possible to use more than a
] insertions. While it is possible to use conventional deletion and single head in order to read the domains. Thus, each domain
T insertion-correcting codes, we tackle this problem with the spe-
is read more than once and the extra reads can be used in or-
I cial structure of racetrack memory, where the domains can be
s. read by multipleheads. Each head outputs a noisy version of the der to correct the position errors during the reading process.
c stored data and the multiple outputs are combined in order to Since every head reads all the bits, we can treat every head as
[ reconstruct the data. Under this paradigm, we will show that it a channel which returns a noisy version of the stored informa-
1 idselpeotisosnibslewittohcdo+rre1cth,ewaditshifatthmeohsetadassainreglweeblli-tseopfarreadtuedn.dSanimcyil,adr tion,andbasedonthesenoisyreadstheinformationisdecoded.
v This model falls under the general framework by Levenshtein
results are provided for burst of deletions, sticky insertions and
4 combinations of both deletions and sticky insertions. of the reconstructionproblem [6]. However,in our case, as op-
7
posed to the general one studied by Levenshtein, the position
8 I. INTRODUCTION errors are correlated and depend on the locations and distance
6
0 Racetrack memory, also known as domain wall memory, is between the different heads.
. anemergingnon-volatilememorywhichisbasedonspintronics In contrast to substitution errors, deletions/sticky insertions
1
technology. It attracts significant attention due to its promising behavedifferentially.Namely,tosuccessfullydecodeasubstitu-
0
7 ultra-high storage density, even comparing to other spintronics tion error,it is necessary to determine the location of the error.
1 memory technologies such as STT-RAM [16]. However, for deletions/sticky insertions, the decoder can suc-
: A racetrack memory is composed of cells, also called do- cessfully decode the correct codeword without determining all
v
i mains, which are positioned on a tape-like stripe and are sepa- thelocationsofthe deletions/stickyinsertions,since itcouldbe
X
rated by domain walls. The magnetization of a domain is pro- any bitwhich belongsto the run where each deletion/stickyin-
r grammedtostoreasinglebitvalue,whichcanbereadbysens- sertion has occurred. Assume first that the heads are adjacent
a
ingitsmagnetizationdirection.Thereadingmechanismisoper- and on every cycle the domains are shifted by a single loca-
atedbyaread-onlyport,calledahead,togetherwithareference tion. Thus, if there are no position errors,the bitstored in each
domain. Since the head is fixed (i.e. cannot move), a shift op- domain is read twice. On the other hand, in the occurrence of
erationisrequiredin orderto readallthedomains.Shiftingthe position errors, the deletions/sticky insertions in the two heads
cells is accomplished by applying a shift current which moves are correlated. For example, if the ith bit is deleted in the first
the domain walls in one direction. Thus, shift operations move head then the (i+1)-st bit is deleted in the second head. In
all the domains one step either to the right or to the left. It is case these two deleted bits belong to the same run, then the
also possible to shift by more than a single step by applying noisy words from the two heads are identical and thus we did
a stronger current. When doing so, it is required to have more not benefit from the extra read by the additional head. On the
than a single head to read the domain walls [9]. hand,ifthe headsarewellseparatedandthereare nolongruns
There are several approaches to enhance the shift operation in the stored information, then the heads’ outputs will differ
in order to reduce its time and energy consumption [13], [15]. and under this setup we will show how it is possible to correct
Howeverthesemechanismssufferfromdegradedreliabilityand the position errors. Note that it is possible to correct a fixed
cannot ensure that domains are perfectly shifted so they are number of deletions and sticky insertions with a single head
aligned with the head. These errors, called position errors, can while the rate of the codes approaches 1 and the redundancy
bemodeledasdeletionsandstickyinsertions[16],whichisthe orderisΘ(log(n)) [1], [5].Hence,anycodeconstructionusing
motivation for this work. A deletion is the eventwhere the do- multiple heads should have rate approaching 1 and more than
mains are shifted by more than a single domain location and that,improvetheredundancyresultofΘ(log(n)).However,this
shouldbeaccomplishedwhileminimizingthedistancebetween comprised of magnetizable cells which can store a single bit.
the heads. The information is read back from the cells by sensing their
The rest of this paper is organized as follows. In Section II, magnetizationdirectionusingheadswhicharefixedintheirpo-
weformallydefinethemodelandproblemsstudiedinthepaper, sitions; see Fig. 1. Since the heads are fixed in their locations,
namelythereadingprocessinracetrackmemoryandcodescor-
recting deletions and sticky insertions using multiple heads. In
SectionIII,weconstructcodescorrectingasingledeletionusing
two heads with approximately0.36 redundancybits, by requir-
ing the distance between the heads to be at least ⌈log(n)⌉+1.
In Section IV, we extend this construction for codes correcting
Fig. 1: Racetrack memory with multiple heads
a burst of deletions where the length of the burst is either ex-
the memory cells move so they can all be read by the heads.
actlyboratmostb.AnotherextensionisgiveninSectionVfor
Thisshiftingoperationis performedbyapplyinga shiftcurrent
codes correcting multiple deletions. In this case our construc-
which moves all the cells on each cycle one or more steps in
tion can correct d deletions using d+1 heads with at most a
the same direction [9]. However, the shifting mechanism does
single bit of redundancy,by requiringthe distance between ad-
not work perfectly and may suffer from errors, called position
jacent heads to be at least d⌈log(n)⌉+d(d+1)/2+1. In the
errors. Thatis, cellsmaybeshiftedbymorethanasingleloca-
case thenumberofheadsmis strictlylessthand+1,we show
tion on each cycle or are not shifted. These position errors can
thatitispossibletocorrectm−1deletionswiththeheads,and
be modeledas deletionsandsticky insertions. Namely,a single
sothecodeshouldbeabletocorrecttheremainingd−(m−1)
deletion is the event where the cells are shifted by two loca-
deletions.Inthissection,wealsoreportonseveralmoreresults
tions instead of one and thus one of the bits is not read by the
we couldnotincludedue to the lack space. Someof the proofs
are omitted for the same reason. head. In case the cells were shifted by some b+1 > 2 loca-
tions, then b consecutive cells were not read and we say that a
II. PRELIMINARIES AND MODELDEFINITIONS deletionburstofsizebhasoccurred.Ontheotherhand,asticky
Let F denote the binary field. For a positive integer n, insertion is the event where the cells were not shifted and the
2
the set {1,2,...,n} is denoted by [n]. Let u = (u ,...,u ) same cell is read again and if this happens b > 1 times in a
1 n
and v = (v ,...,v ) be two vectors of length n and m, row, we say that a burst of b sticky insertions has occurred.
1 m
respectively. The concatenation of u and v is the vector We assume that there are several heads and each head reads
(u ,...,u ,v ,...,v ) of length n + m, which is denoted all the cells. In case there is only a single head, then the only
1 n 1 m
by u ◦ v. A subvector of a word u is a vector u[i ,i ] = approachtocorrectthepositionerrorsisbyusingacodewhich
1 2
(u ,u ,...,u ) ∈ Fn in which 1 6 i 6 i 6 n. The is capable of correcting deletions and sticky insertions. How-
i1 i1+1 i2 2 1 2
length of this subvector is 1 6 i − i + 1 6 n. In case ever,in case there are several heads, the cells are read multiple
2 1
i = i = i, we denote a subvector u[i,i] of length 1 by u[i] times byeach head andthus we study how thisinherentredun-
1 2
to specify the i-th element of vector u. dancy can be used to design better codes. The output of the
Let ℓ and m be two positive integers where ℓ 6 m. Then, headsdependontheir locations.Forexample,assume there are
a length-m vector v ∈ Fm which satisfies v = v for all three heads which are used to read the stored word u, where
2 i i+ℓ
1 6 i 6 m−ℓ is said to have period ℓ. For a vector u ∈ Fn, the distance between the first two heads is t and the distance
2 1
we denote by L(u,ℓ) the length of its longest subvectorwhich betweenthelasttwoheadsist .Ifadeletionoccursatposition
2
hasperiodℓ.NotethatbydefinitionL(u,ℓ)>ℓ,andforℓ=1, i in the first head then a deletion also occurs at position i+t
1
L(u,1) equals the length of the longest run in u. inthesecondheadandanotherdeletionatpositioni+t +t in
1 2
Example1. Let u = (u ,...,u ) = (0,0,1,1,0,1,0,1,1) ∈ the third head. Therefore, the output of the first, second, third
1 9
F92. Since the longest run in u is of length two, we have head is the vector u(δi),u(δi+t1),u(δi+t1+t2), respectively.
L(u,1) = 2. The subvector u[4,8] = (1,0,1,0,1) of u has The goal in this paper is to design codes correcting posi-
period2 since u =u =u =1 and u =u =0. Thisis the tion errors in the reading process. We say that a code is an
4 6 8 5 7
longest subvector of u of period 2, and hence L(u,2)=5. ✷ m-head b-position-error-correcting code if it can correct b po-
sitionerrorsusingmheads.Similarly,wealsodefinem-headb-
For a length-n word u ∈ Fn2 and i ∈ [n], we denote by deletion-correcting codes, m-head b-sticky-insertion-correcting
u(δi) the vector obtainedby u after deleting its ith bit, that is, codes, m-head b-burst-deletion-correcting codes, and m-head
u(δi) = (u1,...,ui−1,ui+1,...,un). For a set ∆ ⊆ {δi : i ∈ b-burst-sticky-insertion-correctingcodes.We notethattheloca-
[n]}, we denoteby u(∆) the vector of length n−|∆| obtained tionsoftheheadsisalsopartofthecodedesign.Sincethearea
fromuafterdeletingallthebitsspecifiedbythelocationsinthe forshiftingthe cellsisconstrained,theheadsshouldnotbetoo
set ∆. In case ∆ = {δi,...,δi+b−1} then we denote the vec- far apart and the distance between adjacent heads should thus
tor u(∆) by u(δ[i,b]) to specify a burst of b deletions starting be minimized. As always, the goal in designing these codes is
at the ith position. to minimize the redundancy of each code construction.
Example2. Let u = (0,0,1,1,0,1,0,1,1) ∈ F9, then
2
u(δ4) = (0,0,1,0,1,0,1,1). For ∆ = {δ4,δ7,δ9} then III. TWO-HEADSINGLE-DELETION-CORRECTING CODES
u(∆)=(0,0,1,0,1,1), and u(δ )=(0,0,0,1,1). ✷
[3,4] In this section we study how to construct two-head single-
In this work, we assume that the information stored in the deletion-correctingcodes. Our main result states thatif the dis-
racetrack memory is represented by a word u. The memory is tancebetweenthetwoheadsisatleast⌈log(n)⌉+1cells,where
n isthelengthofthecodewords,thensuchcodesexistwithre- t, the rates of these codes is strictly less than 1. Since we can
dundancy of roughly 0.36 bits. achieve codes with rate approaching 1 by simply using a sin-
Construction1.Forallt6n,letC (n,1,t)beacodeoflength gle head and a single-deletion-correcting code of redundancy
1
nsuchthatthelengthofthelongestrunofeverycodewordisat at most log(n+1) [5], we are interested only in codes with
mostt.Thatis,C (n,1,t)={c∈Fn|L(c,1)6t}. rate approaching 1 and will optimize their redundancy. Thus,
1 2
we followa similar approachto the onetakenin [12] forcodes
The following theorem proves the correctness of this con- correcting a burst of deletions and let t be a function of the
struction. code length n. In particular, by choosing t = ⌈log(n)⌉+1, it
was observedin [12], using the derivationsfrom [10] and [11],
Theorem2.ThecodeC1(n,1,t)isatwo-headsingle-deletion- that the redundancy of the code C1(n,1,⌈log(n)⌉ + 1) is
correctingcodewhentheheadsarepositionedtlocationsapart. approximately 0.36, and for t = ⌈log(n)⌉+2 efficient encod-
ing and decoding algorithms were recently found for these
Proof:Letc=(c ,...,c )∈C (n,1,t)beastoredcode-
1 n 1 codes using a single bit of redundancy [7]. We conclude this
word of length n and assume that a single deletion occurred at
discussion with the following corollary.
position i. Then, the outputs from the two heads are:
Corollary3.Thereexistsatwo-headsingle-deletion-correcting
Head 1: c(δi)=(c1,...,ci−1,ci+1,...,cn), codewhentheheadsarepositionedt = ⌈log(n)⌉+1locations
Head 2: c(δi+t)=(c1,...,ci+t−1,ci+t+1,...,cn). apartwithredundancyofapproximatelylog(e)/4≈0.36bits.
Consider the first i+t−1 bits in these two sequences: IV. CODES CORRECTING A BURST OFDELETIONS
Head 1:c(δi)[1,i+t−1]=(c1,...,ci−1,ci+1,ci+2,...,ci+t), In this section we study the setup where the domains are
over-shifted by more than a single location, so a burst of dele-
Head 2:c(δi+t)[1,i+t−1]=(c1,...,ci−1,ci,ci+1,...,ci+t−1).
tions occurs in each head. We will focus on two cases: the
We claim that c(δi)[1,i+t−1]6=c(δi+t)[1,i+t−1]. Other- length of the burst is exactly b or at most b.
wise, we will get that
A. Two-Head b-Burst-Deletion-Correcting Codes
ci =ci+1 =···=ci+t−1 =ci+t,
Here we investigatecodescorrectinga burstof exactlyb ad-
which implies that there is a run of length t+1 in c in con-
jacent deletionsusing two heads. Suppose we use two heads at
tradiction to the construction of the code C (n,1,t). Let j be
1 1 distance t to correct a burst of size b in the stored codeword
the leftmost index that differs between c(δ )[1,i+t−1] and
i c=(c ,...,c ).Recallthatfori∈[n]andb∈[n−i],thevec-
c(δ )[1,i+t−1].Suchanindexexistssincec(δ )[1,i+t−1]6= 1 n
i+t i torobtainedfromc afterdeletingthesubvectorc[i,i+b−1]=
c(δ )[1,i+t−1] and so j 6i+t−1. Furthermore,j >i
sincie+tthe first i−1 bits in th1e outputs from two heads ar1e the (ci,...,ci+b−1)is c(δ[i,b]). Therefore,we knowthatif theout-
put from the first head is c(δ ) for some i and b, then the
same as in the stored codeword. Note that j can be differ- [i,b]
1 outputfrom the second head is c(δ ), where the heads are
ent than i in case the ith bit which was deleted is in a middle [i+t,b]
located t positions apart. The following is the construction of
of a run and so the first occurrence where c(δ )[1,i+t−1]
i such codes.
and c(δ )[1,i+t−1] differ is only at the end of this run.
i+t
We conclude that c[1,j1] = c(δi+t)[1,j1] and c[j1 +1,n] = Construction4.LetC2(n,b,t)beacodeoflengthnsuchthat
c(δ )[j ,n−1].Hence,theoriginalcodewordccanberecovered thelengthofthelongestsubvectorwhichhasperiodbofevery
i 1
byconcatenatingthefirstj1 bitsfromc(δi+t)andthelastn−j1 codewordc∈C2(n,b,t)isatmostt.Thatis,C2(n,b,t)={c∈
bitsfromc(δi).Thatis,c=c(δi+t)[1,j1]◦c(δi)[j1,n−1].This Fn2 |L(c,b)6t}.
proof also provides a simple decoding algorithm for the code
The proof that this construction can correct a burst of deletion
C (n,1,t).
1 of length b follows similar ideas from the proof of Theorem 2.
The next example demonstrates this code construction and
its decoder.
Example3. Letn=9,t=3 andc=(0,0,1,1,0,1,0,1,1)be Theorem5.ThecodeC (n,b,t)isatwo-headb-burst-deletion-
astoredcodewordinC1(n,1,t).Letusassumethattheoutputs correctingcodewhenthe2headsarepositionedtlocationsapart.
from two heads are:
Next we turn to evaluate the size of the code C (n,b,t). In
Head 1: c(δ )=(0,0,1,0,1,0,1,1), 2
3 particular, as done in the previoussection, we will find a value
Head 2: c(δ6)=(0,0,1,1,0,0,1,1). oftforwhichtheredundancyofthecodewillbeapproximately
0.36 bits. Let us start with the following definition.
Hence, j = 4 is the leftmost index that differs between two
1
vectors and thus the stored codeword is decoded according to
c=c(δ )[1,4]◦c(δ )[4,8]=(0,0,1,1,0,1,0,1,1). ✷ Definition6.Letu=(u1,...,um)∈Fm2 bealength-mbinary
6 3 vector.Forb > m,theb-period check vectorofuisthevector
By a suitable mapping described in Section IV, the code pb(u)=(u1+u1+b,...,um−b+um)∈Fm2 −boflengthm−b.
C (n,1,t) can be transformed into a code that satisfies the The following lemma can be readily verified.
1
(0,t−1) Run Length Limited (RLL) constraint [3]. While effi-
cient encoding and decoding algorithms are known for codes Lemma7.Aworducontainsasubvectoroflengthtwithperiod
which satisfy the (0,t−1) RLL constraint for fixed value of bifandonlyifp (u)containsarunoft−bzeroes.
b
For a vector u, we denote by L0(u) the length of the longest V. CODES CORRECTING MULTIPLEDELETIONS
run of zeroes in u. For example L (0110100010)=3. For all
0 In this section we move to the more challenging task of
n and t6n, we define the code R(n,t) to be
correcting multiple deletions and construct m-head d-deletion-
R(n,t)={c∈Fn |L (u)6t}. correcting codes. For simplification, we first consider the case
2 0
d = 2 and show that the code C (n,6 2,t ), which can cor-
Using Lemma 7, we can construct a bijection between 3 1
rect a burst of at most two deletions by using two heads, is a
C (n,b,t) and the set Fb ×R(n − b,t −b) for n > b+ 1.
2 2 three-head double-deletion-correcting code, when the distance
Specifically, we define the following maps.
between every adjacent heads is at least t = 2(t − 1). We
• Φ : C (n,b,t) → Fb ×R(n−b,t−b), where Φ(u) = 1
2 2 will then use this result as a building block for a more general
(u[1,b],p (u)).
b claim on codes which can correct d deletions using m heads.
• Ψ:Fb×R(n−b,t−b)→C (n,b,t),whereΨ(v,w)=u
2 2 While we don’t design new code constructions, a key point
and
in the construction is finding the required minimum distance
v , if i6b,
i
u = between two adjacent heads for its success.
i
(ui−b+wi−b, otherwise. Westartbypresentingourresultfortheconstructionofthree-
In the context of error-correcting codes for tandem duplica- head double-deletion-correctingcodes.
tions[4], Jain et al. demonstratedLemma 7 and the fact thatΦ
and Ψ are bijections when t = 2b−1. It is straightforward to Theorem14. ThecodeC3(n,6 2,t1) isathree-headdouble-
extendtheprooffort>b.Hence,wehavethefollowinglemma deletion-correctingcode when the distance between adjacent
that is useful in evaluating the size of the code C2(n,b,t). headsisatleastt=2(t1−1).
Lemma8.Foralln,b,t,|C2(n,b,t)|=2b·|R(n−b,t−b)|. Proof:Let c=(c1,...,cn)∈C3(n,62,t1) be the stored
codeword and t = 2(t −1) be the distance between adjacent
The size of the code R(n,t) can be calculated using the re- 1
heads.Letusassumethatthetwodeletionsoccurredinthefirst
sults from Section III and by applying Lemma 8 for b = 1 to
head are in positions i ,i , where i <i . Hence the deletions
getthatforallnandt6n,|R(n,t)|=|C (n+1,1,t+1)|/2. 1 2 1 2
1 in thesecondheadarein positionsi +t,i +t andin thethird
1 2
We can now conclude with the following corollary.
head they are in positions i +2t,i +2t. That is, the outputs
1 2
Corollary9.Foralln,b,t, from the three heads are:
|C (n−b+1,1,t−b+1)|
|C2(n,b,t)|=2b· 1 2 . Head 1: c(δi1,δi2)
According to Corollaries 3 and 9 we conclude the following. =(c1,...,ci1−1,ci1+1,...,ci2−1,ci2+1,...,cn),
Corollary10. There exists a two-head b-burst-deletion- Head 2: c(δi1+t,δi2+t)
correctingcodewhentheheadsarepositionedt =⌈log(n)⌉+b =(c1,...,ci1+t−1,ci1+t+1,...,ci2+t−1,ci2+t+1,...,cn),
locationsapartwithredundancyofapproximatelylog(e)/4 ≈
Head 3: c(δ ,δ )
0.36bits. i1+2t i2+2t
B. Correcting a Burst of Length at Most b =(c1,...,ci1+2t−1,ci1+2t+1,...,ci2+2t−1,ci2+2t+1,...,cn).
Thegoalofthissectionistodesignacodecorrectingaburst We prove that it is possible to correct the two deletions by
of at most b deletions using two heads. We follow the same explicitly showing how to decode them. This will be done in
ideas presented thus far and use the following construction. three steps.
Construction11.LetC (n,6b,t)beacodeoflengthnwhich 1) First, use the first two heads to correctthe first deletionin
3
istheintersectionofthecodesC (n,ℓ,t)for16ℓ6b.Thatis, the first head.
2
2) Then, use the second and third heads to correct the first
C (n,6b,t)=∩b C (n,ℓ,t)
3 ℓ=1 2 deletion in the second head.
={c∈Fn|L(c,ℓ)6t, forallℓ6b}.
2 3) Atthispoint,thefirstandsecondheadshaveonlya single
Theorem12.ThecodeC (n,6b,t)cancorrectuptobconsec- deletion and thus we proceed to correct this deletion as
3
was done in Theorem 2.
utivedeletionsusingtwoheadsatdistancet.
Since the last two steps are very similar to the first one, we
In this case we will not able to providean exactapproxima-
only discuss the first step.
tionfortheredundancyofthecodeC (n,6b,t)asinprevious
3 Inordertoprovethefirststep,weshowthatc(δ ,δ )[1,i +
cases. However, we will find a value of t for which the redun- i1 i2 1
t−1] 6=c(δ ,δ )[1,i +t−1]. Assume in the contrary,
dancy of the code is at most a single bit. For this purpose, we i1+t i2+t 1
then we distinguish between the following two cases:
follow similar ideas to the ones presented by Schoeny et al. in
• Case 1: If i −i >t +1 then the two subvectors
[12] when studying the redundancy of the so-called universal 2 1 1
RLL constraint. This result is stated in the next theorem.
c(δi1,δi2)[1,i1+t1−1]=(c1,...,ci1−1,ci1+1,...,ci1+t1),
Theorem13.Foralln,b,t, t−b c(δi1+t,δi2+t)[1,i1+t1−1]=(c1,...,ci1−1,ci1,...,ci1+t1−1)
1
|C (n,6b,t)|>2n 1−n· .
3 (cid:18)2(cid:19) ! areidentical,sothesubvector(ci1,ci1+1,...,ci1+t1−1,ci1+t1)
forms a run of length t +1, in contradiction to the con-
Inparticular,fort=⌈log(n)⌉+b+1theredundancyofthecode 1
C (n,6b,t)isatmostasinglebit. struction of the code C3(n,62,t1).
3
• Case 2: If i −i 6 t then i + t = i +2t − 2 > d-deletion-correctingcodewherethedistancebetweenadjacent
2 1 1 1 1 1
i +t −2. Therefore,the first i +t −3 bits in the first headsist>dt −d(d+1)/2+1.Inparticularunderthissetup:
2 1 2 1 1
two heads, which are subvectors 1) Thereexistsa(d+1)-headd-deletion-correctingcodewith
atmostasinglebitofredundancy
c(δ ,δ )[1,i +t −3]
i1 i2 2 1 2) Thereexistsad-headd-deletion-correctingcodewithre-
=(c1,...,ci1−1,ci1+1,...,ci2−1,ci2+1,...,ci2+t1−1), dundancyatmost⌈log(n+1)⌉+1.
c(δ ,δ )[1,i +t −3]
i1+t i2+t 2 1
Lastly, we reporton our results for the other cases solved in
=(c1,...,ci1−1,ci1,...,ci2−2,ci2−1,...,ci2+t1−3), this work, which we could not include their details.
are identical, which implies that (ci2−1,ci2,...,ci2+t1−1) Theorem17.
isa subvectorof lengtht +1 withperiod2, againin con-
tradiction to the construc1tion of the code C (n,62,t ). 1) ThecodeC1(n,1,t)cancorrectdburstsofstickyinsertions
3 1
eachoflengthatmostt−1usingd+1headswhilethedis-
Letj betheleftmostindexthatc(δ ,δ )andc(δ ,δ )
1 i1 i2 i1+t i2+t tancebetweenadjacentheadsisatleastt.Specifically,for
differ. Similarly to the argument in the proof of Theorem
t = ⌈log(n)⌉+1,theredundancyofthecodeisapproxi-
2, such an index exists and we can obtain the vector
mately0.36bits.
c(δ ) by concatenating the first j bits in c(δ ,δ )
i2 1 i1+t i2+t 2) Thecode C (n,1,t) isatwo-headsingle-position-error-
and the last n − 1 − j bits in c(δ ,δ ), i.e., c(δ ) = 1
1 i1 i2 i2 correctingcodewhenthedistancebetweentwoheadsisat
c(δ ,δ )[1,j ]◦c(δ ,δ )[j ,n−2].
i1+t i2+t 1 i1 i2 1 leastt.
The next example demonstratesthe decodingprocedure pre-
3) ThecodeC (n,62,t )isathree-headtwo-position-error-
sented in Theorem 14. 3 1
correctingcodewhenthedistancebetweenadjacentheads
Example4. Let n=11,t =3,t=4 and the stored codeword isatleastt=3t −2.
1 1
is c = (0,0,1,1,0,1,1,0,1,1,1) ∈ C (11,6 2,3). Assume
2
that the outputs from the three heads are:
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3 1
