Chapter V: Coding over Sets for DNA Storage: Substitution Errors
5.4 Bounds
In this section we use sphere packing arguments in order to establish an existence result of codes with low redundancy, and a lower bound on the redundancy of any๐- substitution code. The latter bound demonstrates the asymptotic optimality of the construction in Sec.5.5for๐ =1, up to constants. Our techniques rely on upper and lower bounds on the size of the ballB๐ (Definition5.2.1), which are given below.
However, since our measure for distance is not a metric, extra care is needed when applying sphere-packing arguments. We begin with the existential upper bound in Subsection5.4, continue to provide a lower bound for๐ =1 in Subsection5.4, and extend this bound to larger values of๐ in Subsection5.4.
Existential upper bound
In this subsection, let๐,๐, and๐ฟbe positive integers such that๐ โค ๐ ๐ฟand๐ โค 2๐ฟ. The subsequent series of lemmas will eventually lead to the following upper bound.
Theorem 5.4.1. There exists a ๐-substitution code C โ {0,1}๐ฟ
๐
such that๐(C) โค 2๐log(๐ ๐ฟ) +2.
We begin with a simple upper bound on the size of the ballB๐. Lemma 5.4.1. For every word๐ ={x๐}๐
๐=1 โ {0,1}
๐ฟ
๐
and every positive integer๐ โค ๐ ๐ฟ, we have that|B๐(๐) | โคร๐
โ=0 ๐ ๐ฟ
โ
.
Proof. Every word inB๐(๐) is obtained by flipping the bits inx๐ that are indexed by some ๐ฝ๐ โ [๐ฟ], for every ๐ โ [๐], where ร๐
๐=1|๐ฝ๐| โค ๐. Clearly, there are at mostร๐
โ=0 ๐ ๐ฟ
โ
ways to choose the index sets {๐ฝ๐}๐
๐=1. โก
For ๐ โ {0,1}
๐ฟ
โค๐
let R๐(๐) be the set of all words ๐ โ {0,1}
๐ฟ
๐
such that ๐ โ B๐(๐). That is, for a channel output ๐, the set R๐(๐) contains all potential codewords๐ whose transmission through the channel can result in๐, given that at most๐ substitutions occur. Further, for๐ โ {0,1}๐ฟ
๐
define theconfusable setof๐ asG๐(๐) โ โช๐โฒโB๐(๐)R๐(๐โฒ). It is readily seen that the words in the confusable set G๐(๐) of a word๐ cannot reside in the same ๐-substitution code as ๐, and therefore we have the following lemma.
Lemma 5.4.2. For every๐, ๐, and ๐ฟsuch that ๐ โค ๐ ๐ฟ and๐ โค 2๐ฟ there exists a๐-substitution codeCsuch that
|C | โฅ
$ 2๐ฟ
๐
๐ท
%
, where
๐ท โ max
๐โ({0,๐1}๐ฟ)
|G๐(๐) |.
Proof. Initialize a listP = {0,1}๐ฟ
๐
, and repeat the following process.
1. Choose๐ โ P.
2. RemoveG๐(๐)fromP.
Clearly, the resulting codeCis of the aforementioned size. It remains to show thatC corrects๐ substitutions, i.e., thatB๐(๐ถ) โฉ B๐(๐ถโฒ) =โ for every distinct๐ถ , ๐ถโฒ โ C.
Recall Definition5.2.1that B๐(๐ถ) can be obtained by substitution any ๐ symbols in strings of word๐ถ.
Assume for contradiction that there exist distinct๐ถ , ๐ถโฒ โ C and๐ โ {0โค,1}๐ฟ
๐
such that๐ โ B๐(๐ถ) โฉ B๐(๐ถโฒ), and w.l.o.g assume that๐ถ was chosen earlier than๐ถโฒin the above process. Since๐ โ B๐(๐ถ), it follows that R๐(๐) โ G๐(๐ถ). In addition, since๐ โ B๐(๐ถโฒ), it follows that๐ถโฒโ R๐(๐). Therefore, a contradiction is obtained, since๐ถโฒis in G๐(๐ถ), that was removed from the listPwhen๐ถwas chosen. โก Lemma 5.4.3. For an nonnegative integer ๐ โค ๐ and ๐ โ {๐โ๐0,1}๐ฟ
we have that|R๐(๐) | โค2(2๐ ๐ฟ)๐.
Proof. Denote ๐ = {y1, . . . ,y๐โ๐} and let ๐ โ R๐(๐). Notice that by the definition of R๐(๐), there exists a ๐-substitution operation which turns๐ to ๐. Therefore, everyy๐in๐ is a result of a certain nonnegative number of substitutions in one or more strings in ๐. Hence, we denote by z11, . . . ,z๐1
1 the strings in ๐ that resulted in y1 after the ๐-substitution operation, we denote by z21, . . . ,z2๐
2 the strings which resulted in y2, and so on, up to z๐โ๐1 , . . . ,z๐๐โ๐
๐โ๐, which resulted in y๐โ๐. Therefore, since ๐ = โช๐โ๐
๐=1 {z1๐, . . . ,z๐๐
๐
}, it follows that there exists a
setL โ [๐] ร [๐ฟ], of size at most๐, such that
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ยซ z11
.. . z๐1
1
z21 .. . z๐๐โ๐โ1
๐โ๐โ1
z1๐โ๐ .. . z๐๐โ๐
๐โ๐
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ยซ y1
.. . y1 y2 .. . y๐โ๐โ1
y๐โ๐
.. . y๐โ๐
ยช
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ยฌ
(L)
, (5.1)
where (ยท)(L) is a matrix operator, which corresponds to flipping the bits that are indexed by L in the matrix on which it operates. In what follows, we bound the number of ways to chooseL, which will consequently provide a bound on|R๐(๐) |.
Note that the number of ways to choose L is summed over all possible tuples (๐1, . . . , ๐๐โ๐).
First, defineP={๐:๐๐ >1}, and denote๐ โ |P |. Therefore, sinceร๐โ๐
๐=1 ๐๐ =๐, it follows that
โ๏ธ
๐โP
๐๐ =
๐โ๐
โ๏ธ
๐=1
๐๐ โโ๏ธ
๐โP
๐๐
=๐ โ (๐ โ๐ โ๐)
=๐+๐. (5.2)
Second, notice that for every ๐ โ P, the set {z1๐, . . . ,z๐๐
๐
} contains ๐๐ different strings. Hence, since after the ๐-substitution operation they are all equal to y๐, it follows that at least๐๐โ1 of them must undergo at least one substitution. Clearly, there are ๐๐๐
๐โ1
=๐๐ different ways to choose who will these๐๐โ1 strings be, and additional ๐ฟ๐๐โ1different ways to determine the locations of the substitutions, and therefore๐๐ยท๐ฟ๐๐โ1ways to choose these๐๐โ1 substitutions.
Third, notice that
๐ โโ๏ธ
๐โP
(๐๐โ1) =๐ โโ๏ธ
๐โP
๐๐+๐
(5.2)
= ๐ โ๐ , (5.3)
and hence, there are at most ๐ โ๐ remaining positions to be chosen to L, after choosing the๐๐โ1 positions for every ๐ โ P as described above.
Now, letIbe the set of all tuples(๐1, . . . , ๐๐โ๐) of positive integers that sum to๐ (whose size is ๐๐โโ๐1โ1
by the famousstars and barstheorem). Let๐ :I โ Nbe a function which maps (๐1, . . . , ๐๐โ๐) โ I to the number of different๐ โ R๐(๐) for which there exist L โ [๐] ร [๐ฟ] of size at most ๐ such that (5.1) is satisfied.
Since this quantity is at most the number of ways to choose a suitableL, the above arguments demonstrate that
๐(๐1, . . . , ๐๐โ๐) โค ๐ ๐ฟ
๐ โ๐ ร
๐โP
๐๐๐ฟ๐๐โ1. Then, we have
|R๐(๐) | โคโ๏ธ
I
๐(๐1, . . . , ๐๐โ๐)
โค โ๏ธ
I
๐ ๐ฟ ๐โ๐
ร
๐โP
๐๐๐ฟ๐๐โ1
โค โ๏ธ
I
(๐ ๐ฟ)๐โ๐๐ฟ
ร
๐(๐๐โ1) ร
๐โP
๐๐
(5.3)
โค โ๏ธ
I
(๐ ๐ฟ)๐โ๐๐ฟ๐ ร
๐โP
๐๐. (5.4)
Since the geometric mean of positive numbers is always less than the arithmetic one, we have
ร
๐โP๐๐ 1/๐
โค 1
๐
ร
๐โP๐๐, and hence, (5.4) โค โ๏ธ
I
(๐ ๐ฟ)๐โ๐๐ฟ๐ ร
๐๐๐
๐
๐
(5.2)
โค โ๏ธ
I
(๐ ๐ฟ)๐โ๐๐ฟ๐( (๐ +๐)/๐)๐
โค
๐ โ1 ๐โ๐ โ1
(๐ ๐ฟ)๐โ๐๐ฟ๐( (๐+๐)/๐)๐
โค ๐
๐
(๐ ๐ฟ)๐โ๐๐ฟ๐( (๐ +๐)/๐)๐
โค (๐ ๐ฟ)๐โ๐(๐ ๐ฟ)๐( (๐+๐)/๐)๐
โค (๐ ๐ฟ)๐( (๐ +๐)/๐)๐
(๐)
โค (๐ ๐ฟ)๐2๐
โค (2๐ ๐ฟ)๐, (5.5)
where(๐)will be proved in Appendix5.9. โก
Proof. (of Theorem 5.4.1) It follows from Lemma5.4.1, Lemma 5.4.3, and from the definition of๐ทthat
๐ท โค max
๐โ({0,๐1}๐ฟ)
|B๐(๐) | ยท max
๐โ({0,๐1}๐ฟ)
|R๐(๐) |
โค
๐
โ๏ธ
โ=0
๐ ๐ฟ โ
!
(2๐ ๐ฟ)๐.
Therefore, the codeCthat is constructed in Lemma5.4.2satisfies ๐(C) โค log
2๐ฟ ๐
โlog|C |
โค log ๐
โ๏ธ
โ=0
๐ ๐ฟ โ
!
(2๐ ๐ฟ)๐
!
=log
๐
โ๏ธ
โ=0
๐ ๐ฟ โ
!
+๐(log(๐ ๐ฟ) +1)
โค log
๐ ๐ ๐ฟ
๐
+๐(log(๐ ๐ฟ) +1)
โค
log(๐) โlog(๐!) +log(๐ ๐ฟ๐) +๐(log(๐ ๐ฟ) +1)
โค 2๐log(๐ ๐ฟ) +2. โก
Lower bound for a single substitution code
Notice that the bound in Lemma5.4.1is tight, e.g., in cases where๐๐ป(x๐,x๐) โฅ2๐+1 for all distinct๐, ๐ โ [๐]. This might occur only if ๐ is less than the maximum size of a ๐-substitution correcting vector-code, i.e., when๐ โค 2๐ฟ/(ร๐
๐=0 ๐ฟ ๐
) [79, Sec. 4.2]. When the minimum Hamming distance between the strings in a codeword is not large enough, different substitution errors might result in identical words, and the size of the ball is smaller than the given upper bound. This is illustrated in the following example.
Example 5.4.1. For ๐ฟ = 3 and ๐ = 2, consider the word ๐ = {010,011}. By flipping either the two underlined symbols, or the two bold symbols, the word๐โฒ= {010,110} is obtained. Hence, different substitution operations might result in identical words. As a result, the size of the ball
B2(๐) ={{010,011},{110,011},{000,011},{011}, {010,111},{010,001},{010},{100,011},
{111,011},{110,111},{110,001},{110,010}, {001,011},{000,111},{000,001},{000,010}, {010,101}}
is17, which is smaller than the upper bound of theB2(๐)ball size ๐ ๐ฟ0 + ๐ ๐ฟ
1
+
๐ ๐ฟ 2
=22.
However, in some cases it is possible to bound the size ofB๐ from below by using tools from Fourier analysis of Boolean functions. In the following it is assumed that that๐ =1. A word๐ โ {0,1}
๐ฟ
๐
corresponds to a Boolean function ๐๐ :{ยฑ1}๐ฟ โ {ยฑ1} as follows. For x โ {0,1}๐ฟ let x โ {ยฑ1}๐ฟ be the vector which is obtained fromxbe replacing every 0 by 1 and every 1 byโ1. Then, we define ๐๐(x) = โ1 if x โ ๐, and 1 otherwise. Considering the set {ยฑ1}๐ฟ as the hypercube graph3, theboundary๐ ๐๐ of ๐๐ is the set of all edges{x1,x2} โ {ยฑ1}๐ฟ
2
in this graph such that ๐๐(x1) โ ๐๐(x2).
Lemma 5.4.4. For every word๐ โ {0,1}๐ฟ
๐
we have that |B1(๐) | โฅ |๐ ๐๐| +1.
Proof. Let๐ = {x1, . . . ,x๐} be a word. Every edge๐ ={x๐,xโฒ} on the boundary of ๐๐corresponds to a substitution operation inx๐from๐that results in a word๐๐ = {x1, . . . ,x๐โ1,xโฒ,x๐+1, . . . ,x๐} โ (B1(๐)\{๐}) โฉ {0,1}๐ฟ
๐
. To show that every edge ๐ on the boundary corresponds to a unique word ๐๐ in B1(๐), assume for contradiction that๐๐ =๐๐โฒ for two distinct edges๐ = {x1,x2} and๐โฒ = {y1,y2}, where x1,y1 โ ๐ andx2,y2 โ ๐. Since both ๐๐ and ๐๐โฒ contain precisely one element which is not in๐, and are missing one element which is in๐, it follows that x1 = y1 and x2 = y2, a contradiction. Therefore, there exists an injective mapping between the boundary of ๐๐ andB1(๐)\{๐}, and the claim follows. โก Notice that the bound in Lemma5.4.4 is tight, e.g., in cases where the minimum Hamming distance between the strings of๐is at least 2. This implies the tightness of the bound which is given below in these cases. Having established the connection betweenB1(๐)and the boundary of ๐๐, the following Fourier analytic claim will aid in proving a lower bound. Let thetotal influenceof ๐๐be๐ผ(๐๐) โ ร๐ฟ
๐=1Prx(๐๐(x) โ ๐๐(xโ๐)), wherexโ๐is obtained fromxby changing the sign of the๐-th entry, andx is chosen uniformly at random.
3The nodes of the hypercube graph of dimension๐ฟare identified by{ยฑ1}๐ฟ, and every two nodes are connected if and only if the Hamming distance between them is 1.
Lemma 5.4.5. [72, Theorem 2.39] For every function ๐ : {ยฑ1}๐ฟ โ R, we have that๐ผ(๐) โฅ2๐ผlog(1/๐ผ), where๐ผ=๐ผ(๐) โ min{Prx(๐(x) =1),Prx(๐(x)=โ1)}, andx โ {ยฑ1}๐ฟ is chosen uniformly at random.
Lemma 5.4.6. Let ๐ฟ and ๐ be integers such that ๐ โค 2(1โ๐)๐ฟ and ๐ฟ > 1
๐ for some0 < ๐ <1. For every word๐ โ {0,1}
๐ฟ
๐
we have that |๐ ๐๐| โฅ ๐ ๐ ๐ฟ.
Proof. Since๐ โค 2(1โ๐)๐ฟ and๐ผ= ๐ผ(๐๐) =min{(2๐ฟ โ๐)/2๐ฟ, ๐/2๐ฟ}, it follows that๐ผ= ๐/2๐ฟwhenever๐ฟ > 1
๐, which holds for every non-constant๐ฟ. In addition, since Prx(๐๐(x) โ ๐๐(xโ๐))equals the fraction of dimension๐edges that lie on the boundary of ๐๐ ([OโDonnell]), Lemma5.4.4implies that
๐ผ(๐๐) = |๐ ๐๐| 2๐ฟโ1
.
Therefore, since ๐ โค 2(1โ๐)๐ฟ and from Lemma 5.4.5 we have that |๐ ๐๐| = 2๐ฟโ1๐ผ(๐๐) โฅ2๐ฟ๐ผlog(1/๐ผ) = ๐log(2๐ฟ/๐) โฅ ๐ ๐ ๐ฟ. โก Corollary 5.4.1. For integers ๐ฟ and ๐ and a constant 0 < ๐ < 1such that ๐ โค 2(1โ๐)๐ฟ and ๐ฟ > 1
๐, any 1-substitution code C โ {0,1}๐ฟ
๐
satisfies that ๐(C) โฅ log(๐ ๐ฟ) โlog 1/๐.
Proof. According to Lemma 5.4.4 and Lemma 5.4.6, every codeword of every C excludes at least ๐ ๐ ๐ฟ other words from belonging to C. Hence, we have that
|C | โค 2
๐ฟ
๐
/๐ ๐ ๐ฟ, and by the definition of redundancy, it follows that ๐(C)=log
2๐ฟ ๐
โlog(|C |)
โฅ log(๐ ๐ ๐ฟ)
=log(๐ ๐ฟ) โlog 1/๐ . โก
Remark 5.4.1. A similar lower bound was presented in [62], where it was shown that for a code correcting๐ loss of strings,๐substitution errors in each of๐กstrings, the redundancy is lower bounded by
๐ ๐ฟ+๐กlog๐+๐ก ๐log๐ฟโlog(๐ !๐ก!๐!๐ก) +๐(1),
when๐ =2๐ฝ ๐ฟ for some0< ๐ฝ < 1. Taking๐ =0and๐ =๐ก =1results in the lower boundlog(๐ ๐ฟ) +๐(1), which is order-wise the same as, and stronger by a constant log 1/๐ than the lower boundlog(๐ ๐ฟ) โlog 1/๐ in Corollary5.4.1. Compared to the bound in [62], the bound in Corollary5.4.1does not require๐to be exponential in๐ฟ, and thus applies to broader parameter range for๐ and๐ฟ.
Lower bound for more than one substitution
Similar techniques to the ones in Subsection5.4can be used to obtain a lower bound for larger values of๐. Specifically, we have the following theorem.
Theorem 5.4.2. For integers๐ฟ,๐,๐, and positive constants๐ , ๐ <1such that๐ โค 2(1โ๐)๐ฟ, ๐ฟ > 1
๐, and ๐ โค ๐๐
โ
๐, a ๐-substitution code C โ {0,1}๐ฟ
๐
satisfies that๐(C) โฅ ๐(log(๐ ๐ฟ) โ2 log(๐)) โ๐(1).
To prove this theorem, it is shown that certain special๐-subsets of๐ ๐๐correspond to words inB๐(๐), and by bounding the number of these special subsets from below, the lower bound is attained. A subset of ๐ boundary edges is called special, if it does not contain two edges that intersect on a node (i.e., a string) in๐. Formally, a subset S โ ๐ ๐๐ is special if |S | = ๐, and for every {x1,y1},{x2,y2} โ S with ๐๐(x1) = ๐๐(x2) =โ1 and ๐๐(y1) = ๐๐(y2) = 1 we have thatx1 โ x2. We begin by showing how special sets are relevant to proving Theorem5.4.2.
Lemma 5.4.7. For every word๐ โ {0,1}๐ฟ
๐
we have that
|B๐(๐) | โฅ |{S โ๐ ๐๐|S is special}|
๐๐
.
Proof. It is shown that every special set corresponds to a word in B๐(๐), and at most ๐๐ different special sets can correspond to the same word (namely, there exists a mapping from the family of special sets to B๐(๐), which is at most ๐๐ to 1). Let S = {{x๐,y๐}}๐
๐=1 be special, where ๐๐(x๐) = โ1 and ๐๐(y๐) = 1 for every ๐ โ [๐]. Let ๐S โ {0,1}
๐ฟ
โค๐
be obtained from ๐ by removing the x๐โs and adding the y๐โs, i.e., ๐S โ (๐ \ {x๐}๐
๐=1) โช {y๐}๐
๐=1 for some๐ โค ๐; notice that there are exactly ๐ distinctx๐โs but at most ๐ distinct y๐โs, since S is special, and therefore we assume w.l.o.g thaty1, . . . ,y๐ are the distincty๐โs. It is readily verified that๐S โ B๐(๐), since๐S can be obtained from๐ by performing๐ substitution operations in ๐, each of which corresponds to an edge in S. Moreover, every S corresponds to a unique surjective function ๐S : [๐] โ [๐] such that ๐S(๐) = ๐ if there exists ๐ โค ๐ such that {x๐,y๐} โ S, and hence at most ๐๐ โค ๐๐ different special setsScan correspond to the same word inB๐(๐). โก We now turn to prove a lower bound on the number of special sets.
Lemma 5.4.8. Let ๐ฟ and ๐ be integers such that ๐ โค 2(1โ๐)๐ฟ and ๐ฟ > 1
๐ for some0 < ๐ < 1. If there exists a positive constant๐ < 1 such that ๐ โค ๐ยท๐
โ ๐, then there are at least(1โ๐2) |๐ ๐๐|
๐
special setsS โ ๐ ๐๐.
Proof. Clearly, the number of ways to choose a๐-subset of๐ ๐๐which isnotspecial, i.e., contains ๐ distinct edges of ๐ ๐๐ but at least two of those are adjacent to the samex โ๐, is at most
๐ ยท ๐ฟ
2
ยท
|๐ ๐๐| ๐ โ2
=๐ ยท ๐ฟ
2
ยท ๐(๐โ1)
(|๐ ๐๐| โ๐+2) (|๐ ๐๐| โ๐+1) ยท
|๐ ๐๐| ๐
. Observe that the multiplier of |๐ ๐๐๐|
in the above expression can be bounded as follows.
๐ ยท ๐ฟ
2
ยท ๐(๐โ1)
(|๐ ๐๐| โ๐+2) (|๐ ๐๐| โ๐+1)
โค๐ ยท ๐ฟ
2
ยท ๐(๐โ1)
(๐ ๐ ๐ฟโ๐+2) (๐ ๐ ๐ฟโ๐ +1)
โค๐ ยท๐ฟ2ยท ๐2 ๐2๐2๐ฟ2
,
where the former inequality follows since|๐ ๐๐| โฅ ๐ ๐ ๐ฟby Lemma5.4.6; the latter inequality follows since ๐ โค ๐๐
โ
๐ implies that๐ ๐ ๐ฟ โ๐ +2 โฅ ๐ ๐ ๐ฟโ๐ +1 โฅ
โ1
2 ยท๐ ๐ ๐ฟ whenever ๐
โ
โ 2
2โ1 โค ๐ฟ
โ
๐, which holds for every non-constant ๐ and ๐ฟ. Therefore, since
๐ ยท๐ฟ2ยท ๐2 ๐2๐2๐ฟ2
= ๐2 ๐2๐
โค ๐2, it follows that these are at least(1โ๐2) ยท |๐ ๐๐|
๐
special subsets in๐ ๐๐. โก Lemma 5.4.7 and Lemma 5.4.8 readily imply that |B๐(๐) | โฅ (1โ๐2)
๐๐ ๐ ๐ ๐ฟ
๐
for every๐ โ {0,1}
๐ฟ
๐
, from which we can prove Theorem5.4.2.
Proof. (of Theorem 5.4.2) Clearly, no two ๐-balls around codewords in C can intersect, and therefore we must have|C | โค 2
๐ฟ
๐
/min๐โC|B๐(๐) |. Therefore, ๐(C) =log
2๐ฟ ๐
โlog|C |
โฅ log
(1โ๐2) ๐๐
๐ ๐ ๐ฟ ๐
=log ๐ ๐ ๐ฟ
๐
โ๐log(๐) โ๐(1)
โฅ log
(๐ ๐ ๐ฟโ๐+1)๐ ๐๐
โ ๐log(๐) โ๐(1)
โฅ logยฉ
ยญ
ยญ
ยซ โ1
2๐ ๐ ๐ฟ ๐
๐๐ ยช
ยฎ
ยฎ
ยฌ
โ ๐log(๐) โ๐(1)
=๐
log(โ๐
2) +log(๐ ๐ฟ))
โ2๐log(๐) โ๐(1)
โฅ ๐log(๐ ๐ฟ) โ2๐log(๐) โ๐(๐). โก