Computing 𝐹 𝐻

Chapter VI: Robust Indexing: Optimal Codes Correcting Deletion/Insertion

6.4 Computing 𝐹 𝐻

(3) Since s₂ is recovered, the strings {(𝑥_𝑖,1, . . . , 𝑥_{𝑖, 𝐿}′)}^𝑀

𝑖=1 = {a𝑖}^𝑀

𝑖=1 are known.

Sort{(𝑥_𝑖,1, . . . , 𝑥_{𝑖, 𝐿}′)}^𝑀

𝑖=1lexicographically in descending order. For each𝑖 ∈ [𝑀], find the unique𝜋(𝑖) ∈ [𝑀]such that𝑑_𝐻( (𝑥^′

𝜋(𝑖),1, . . . , 𝑥^′

𝜋(𝑖), 𝐿^′),

(𝑥_𝑖,1, . . . , 𝑥_{𝑖, 𝐿}′)) ≤ 𝑘 (note that𝑖₀ = 𝜋(1)). Similar to Step (1), we conclude that the stringx^′_𝜋(𝑖) is an erroneous copy ofx𝑖,𝑖 ∈ [𝑀], since the Hamming distance betweenx𝑗 andx𝑖 is at least 2𝑘 +1 for 𝑗 ≠𝑖. Hence, the identify of {(𝑥_𝑖,1, . . . , 𝑥_{𝑖, 𝐿}′)}^𝑀

𝑖=1are determined from{(𝑥^′

𝑖,1, . . . , 𝑥^′

𝑖, 𝐿^′)}^𝑀

𝑖=1.

(4) Since x^′_𝜋(𝑖) is an erroneous copy of x𝑖, 𝑖 ∈ [𝑀], it follows that the con- catenation s^′ = (x^′_𝜋(1), . . . ,x^′_𝜋(𝑀)) is an erroneous copy of (x₁, . . . ,x𝑀) = (m, 𝑅 𝑆_𝑘(m)), where m is defined in Step (4) in the encoding procedure.

Therefore,(x₁, . . . ,x𝑀)and thusd₂can be recovered from(x^′_𝜋(1), . . . ,x^′_𝜋(𝑀)) by using the Reed-Solomon decoder.

(5) Output(𝑑₁,d₂).

Therefore, the codeword{x𝑖}^𝑀

𝑖=1can be recovered. The redundancy of the code is 𝑟(C)=log

2^𝐿 𝑀

−log⌈(2^𝐿′ −𝑀 𝑄)^𝑀−1

(𝑀−1)! ⌉ − [𝑀(𝐿−𝐿^′) −4𝑘 𝐿^′−2𝑘⌈log𝑀 𝐿⌉]

(6.2)

(𝑎)

≤2𝑘log𝑀 𝐿+ (12𝑘+2)log𝑀+𝑂(𝑘³) +𝑂(𝑘log log𝑀 𝐿), (6.3) where(𝑎)will be proved in Appendix6.7. The complexity of the encoding/decoding is that of computing the function 𝐹^𝐻

𝑆 , which as will be discussed in Sec. 6.4, is 𝑝 𝑜𝑙 𝑦(𝑀 , 𝐿 , 𝑘).

consists of two steps. In the first step we map the integer 𝑑 ∈ [ ⌈⁽²

𝐿′

−𝑀 𝑄)^𝑀−1 (𝑀−1)! ⌉]

into 𝑀 − 1 integers 𝑞₁, . . . , 𝑞_𝑀 ∈ [2^𝐿′] such that 𝑞₁ = 2^𝐿′ and 𝑞_𝑖+1 ≤ 𝑞_𝑖 −𝑄 for 𝑖 ∈ [𝑀 − 1]. In the second step, we use 𝑞_𝑖 to generate a𝑖 sequentially for𝑖 ∈ [2, 𝑀]. The first step is given in the following lemma.

Lemma 6.4.1. There exists an invertible map 𝐹^𝐻

𝑄 : [ ⌈⁽²

𝐿′

−𝑀 𝑄)^𝑀−1

(𝑀−1)! ⌉] → ^[2𝐿

′] 𝑀

, computable in 𝑝 𝑜𝑙 𝑦(𝐿^′, 𝑀) time, that maps and integer𝑑 ∈ [ ⌈⁽²

𝐿′

−𝑀 𝑄)^𝑀−1

(𝑀−1)! ⌉] to an integer tuple(𝑞₁, . . . , 𝑞_𝑀) such that𝑞₁=2^𝐿′ and𝑞_𝑖 ≥ 𝑞_𝑖+1+𝑄 for𝑖 ∈ [𝑀−1]. Proof. Recall thecombinatorial numbering map 𝐹_𝑐𝑜𝑚 that maps an integer in the range [ ^𝑛

𝑚

] to a set of 𝑚 different and unordered integers in the range [𝑛] for integers 𝑛 and 𝑚 ≤ 𝑛. Since ⁽²

𝐿′

−𝑀 𝑄)^𝑀−1

(𝑀−1)! ≤ ^2𝐿

′−𝑀 𝑄+𝑀−1 𝑀−1

, we can map 𝑑 ∈ [ ⌈⁽²

𝐿′

−𝑀 𝑄)^𝑀−1

(𝑀−1)! ⌉] to𝑀 −1 integers 𝐹_𝑐𝑜𝑚(𝑑) ={𝑞^′

2, . . . , 𝑞^′

𝑀} such that 2^𝐿′ − 𝑀 𝑄+ 𝑀 − 1 ≥ 𝑞^′

2 > 𝑞^′

3 > . . . > 𝑞^′

𝑀. Let 𝑞₁ = 2^𝐿′, 𝑞_𝑖 = 𝑞^′

𝑖 + (𝑀 − 𝑖 +1) (𝑄 − 1) for𝑖 ∈ [2, 𝑀], and 𝐹^𝐻

𝑄(𝑑) = {𝑞₁, . . . , 𝑞_𝑀}. Then we have that𝑞₂ ≤ 2^𝐿′ −𝑄 and that𝑞_𝑖 ≥ 𝑞_𝑖+1+𝑄for𝑖 ∈ [2, 𝑀−1]. Since the map𝐹_𝑐𝑜𝑚is invertible and computed in 𝑝 𝑜𝑙 𝑦(𝐿^′, 𝑀) time, so is the map𝐹^𝐻

𝑄. □

We now turn to the second step. Given the integers 𝐹^𝐻

𝑄(𝑑) = (𝑞₁, . . . , 𝑞_𝑀), we generate the indexing bits{a𝑖 = (𝑥_𝑖,1, . . . , 𝑥_{𝑖, 𝐿}′)}^𝑀

𝑖=1 ∈ S^𝐻. First, we have thata₁= 1𝐿^′. The algorithm generates the indexing string a𝑖 sequentially for 𝑖 ∈ [2, 𝑀]. Each indexing stringa𝑖 is generated bit by bit in a recursive manner. We first give the following definition, on which the algorithm is based.

For a set of strings 𝐴 ⊂ {0,1}^𝐿′ and a stringa∈ {0,1}^ℓ of lengthℓ ∈ [𝐿^′]. Denote 𝑁_𝐻(a, 𝐴)= ∑︁

c:c∈𝐴

|{c^′: (𝑐^′

1, . . . , 𝑐^′

ℓ)=aand𝑑_𝐻(c^′,c) ≤2𝑘}|

as the sum of the number of sequences that have prefixaand have Hamming distance at most 2𝑘 fromcoverc ∈ 𝐴. The number 𝑁_𝐻(a, 𝐴) has the following properties that will be useful in our proof. The first property implies that

2^{𝐿′−ℓ}−𝑁_𝐻(a, 𝐴) =(2^{𝐿′−ℓ−1}− 𝑁_𝐻( (a,0), 𝐴)) + (2^{𝐿′−ℓ−1}−𝑁_𝐻( (a,1), 𝐴)), (6.4) which enables a recursion to generate each sequence a𝑖. The second property provides a way to compute𝑁_𝐻(a, 𝐴).

Lemma 6.4.2. 1. For any sequence a ∈ {0,1}^ℓ of length ℓ ∈ [𝐿^′ − 1] and set𝐴 ⊂ {0,1}^𝐿′, we have

𝑁_𝐻(a, 𝐴) =𝑁_𝐻( (a,0), 𝐴) +𝑁_𝐻( (a,1), 𝐴), (6.5)

where (a,0)or (a,1)is the concatenation ofaand a0or1bit respectively.

2. For anya ∈ {0,1}^ℓ and𝐴 ⊂ {0,1}^𝐿′, we have

𝑁_𝐻(a, 𝐴) = ∑︁

c:c∈𝐴

2𝑘−𝑑𝐻(a,(𝑐₁,...,𝑐ℓ))

∑︁

𝑖=0

𝐿^′−ℓ 𝑖

. (6.6)

Proof. Note that for any sequence c, the ℓ + 1-th bit of any sequence c^′ satisfy- ing(𝑐^′

1, . . . , 𝑐^′

ℓ) =ais either 0 or 1. Hence

|{c^′:(𝑐^′

1, . . . , 𝑐^′

ℓ) =aand𝑑_𝐻(c^′,c) ≤ 2𝑘}|

=|{c^′:(𝑐^′

1, . . . , 𝑐^′

ℓ+1) =(a,0)and𝑑_𝐻(c^′,c) ≤2𝑘}|

+ |{c^′:(𝑐^′

1, . . . , 𝑐^′

ℓ+1) = (a,1)and𝑑_𝐻(c^′,c) ≤2𝑘}|,

which implies Eq. (6.5). Moreover, for any sequencec∈ {0,1}^𝐿′, we have that

|{c^′: (𝑐^′

1, . . . , 𝑐^′

ℓ)=aand𝑑_𝐻(c^′,c) ≤2𝑘}| =

2𝑘−𝑑_𝐻(a,(𝑐₁,...,𝑐_ℓ))

∑︁

𝑖=0

𝐿^′−ℓ 𝑖

. Hence the number 𝑁_𝐻(a, 𝐴)can be computed by Eq. (6.6). □ Next, we present the algorithm that takes𝐹^𝐻

𝑄 (𝑑) = (𝑞₁, . . . , 𝑞_𝑀) as input and out- putsa𝑖 such that{a₁, . . . ,a𝑀} ∈ S^𝐻 and that the decimal presentation decimal(a𝑖) ofa𝑖, 𝑖 ∈ [𝑀] satisfies

decimal(a𝑖) =𝑞_𝑖−1+ ∑︁

ℓ:𝑎_{𝑖 ,ℓ}=1and^{ℓ∈[𝐿′]}

𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ−1,0),{a𝑗}^𝑖−1

𝑗=1). (6.7) We then show that the sequencesa𝑖, 𝑖 ∈ [𝑀]satisfying (6.7) are decodable, i.e., we can recover the tuple(𝑞₁, . . . , 𝑞_𝑀)from{a₁, . . . ,a𝑀}.

Encoding:

for𝑖 ∈ [𝑀], do 𝑞 =𝑞_𝑖.

forℓ ∈ [𝐿^′], do

if 2^{𝐿′−ℓ} −𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ−1,0),{a𝑗}^𝑖−1

𝑗=1) ≥ 𝑞, then𝑎_𝑖,ℓ =0.

else

𝑞 =𝑞− (2^{𝐿′−ℓ} −𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ−1,0),{a𝑗}^𝑖−1

𝑗=1)), 𝑎_𝑖,ℓ =1.

end if end for end for

return{a₁, . . . ,a𝑀}.

The generation of a𝑖, 𝑖 ∈ [𝑀] in the encoding procedure can be intuitively char- acterized as walking on a complete binary tree of 𝐿^′+1 layers. The walk starts at layer 1, i.e., the root of the binary tree, and ends at layer 𝐿^′+1 at one of the leaf nodes. At each step, it goes to one of its two child nodes, which represent the bits 0 and 1 respectively. Each string a𝑖, 𝑖 ∈ [𝑀] is represented by the path of a walk. For each patha𝑖 = (𝑎_𝑖,1, . . . , 𝑎_{𝑖, 𝐿}′) and each layer ℓ ∈ [𝐿^′], assign the weight𝑤(𝑎_𝑖,ℓ) =2^{𝐿′−ℓ} −𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ),{a𝑗}^𝑖−1

𝑗=1)to node𝑎_𝑖,ℓ in theℓ-th layer, and the weight𝑤(𝑎¯_𝑖,ℓ) =2^{𝐿′−ℓ}−𝑁_𝐻( (𝑎_𝑖,1, . . . ,1−𝑎_𝑖,ℓ),{a𝑗}^𝑖−1

𝑗=1)to the brother node of node𝑎_𝑖,ℓ, i.e., the node that shares the same parent node with𝑎_𝑖,ℓ. From Eq. (6.5) we have that 𝑤(𝑎_𝑖,ℓ) = 𝑤(𝑎_𝑖,ℓ+1) +𝑤(𝑎¯_𝑖,ℓ+1) forℓ ∈ [𝐿^′−1]. Moreover, we have that 0 < 𝑞 ≤ 𝑤(𝑎_𝑖,ℓ) after theℓ-th inner for loop in the𝑖-th outer for loop. This is formalized in the following lemma, which can be used to prove that Eq. (6.7) holds and that{a₁, . . . ,a𝑀} ∈ S^𝐻.

Lemma 6.4.3. After theℓ-th,ℓ ∈ [𝐿^′], inner for loop in the𝑖-th,𝑖 ∈ [𝑀], outer for loop in the encoding procedure, we have that

0< 𝑞 ≤2^{𝐿′−ℓ}−𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ),{a𝑗}^𝑖_𝑗⁻¹₌₁). (6.8) At the end of the𝑖-th outer for loop, we have that𝑞 =1.

Proof. We prove Eq. (6.8) by induction onℓ. Forℓ =1, according to Lemma6.4.1, we have 0 < 𝑞 = 𝑞_𝑖 ≤ 2^𝐿′ − (𝑖 −1)𝑄 at the beginning of the 𝑖-th outer for loop.

If 𝑎_𝑖,1 = 0, then according to the if condition in the encoding procedure, we have that 0 < 𝑞 ≤ 2^{𝐿′−ℓ} − 𝑁_𝐻(0,{a𝑗}^𝑖−1

𝑗=1) for ℓ = 1, which proves (6.8). Otherwise if𝑎_𝑖,1=1, we have

0 < 𝑞=𝑞_𝑖− (2^{𝐿′−ℓ}−𝑁_𝐻(0,{a𝑗}^𝑖−1

𝑗=1))

≤ 2^𝐿′ − (𝑖−1)𝑄− (2^{𝐿′−ℓ}−𝑁_𝐻(0,{a𝑗}^𝑖−1

𝑗=1))

(𝑎)

=(2^𝐿′−1−𝑁_𝐻(1,{a𝑗}^𝑖−1_𝑗=1)),

where(𝑎)holds since by definition of𝑁_𝐻(a, 𝐴), we have that 𝑁_𝐻(0,{a𝑗}^𝑖_𝑗⁻¹₌₁) +𝑁_𝐻(1,{a𝑗}^𝑖_𝑗⁻¹₌₁) =

𝑖−1

∑︁

𝑗=1

|{c:𝑑_𝐻(c,a𝑗) ≤ 2𝑘}|

=

𝑖−1

∑︁

𝑗=1

𝑄

= (𝑖−1)𝑄 .

Hence the claim holds forℓ =1. Suppose Eq. (6.8) holds forℓ =𝑚. Forℓ =𝑚+1, if𝑎_𝑖,𝑚+1=0, then from Step (3), we have 0< 𝑞 ≤ 2^{𝐿′−𝑚−1}−𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,𝑚,0), {a𝑗}^𝑖−1

𝑗=1). Otherwise if 𝑎_𝑖,𝑚+1=1, we have that

0 < 𝑞=𝑞_𝑖− (2^{𝐿′−𝑚−1}− 𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ,0),{a𝑗}^𝑖−1

𝑗=1))

≤ 2^{𝐿′−𝑚} −𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,𝑚),{a𝑗}^𝑖−_𝑗=1¹)

− (2^{𝐿′−𝑚−1}−𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,𝑚,0),{a𝑗}^𝑖_𝑗⁻¹₌₁))

(𝑏)= (2^{𝐿′−𝑚−1}−𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,𝑚,1),{a𝑗}^𝑖_𝑗⁻¹₌₁)),

where(𝑏)follows from Eq. (6.5). Therefore, Eq. (6.8) holds forℓ =𝑚+1 and thus holds forℓ ∈ [𝐿^′]. Hence at the end of Step (2) we have that

0< 𝑞 ≤ 2^{𝐿′−𝐿′}−𝑁_𝐻(a𝑖,{a𝑗}^𝑖−1_𝑗=1) ≤1. (6.9)

Hence𝑞equals 1 at the end of Step (2). □

We now show that the strings {a₁, . . . ,a𝑀} generated in the encoding procedure belong toS𝐻. By Lemma6.4.3, we have

𝑞 =2^{𝐿′−𝐿′} −𝑁_𝐻(a𝑖,{a𝑗}^𝑖−1

𝑗=1) =1,

at the end of each round of Step (2) in the encoding procedure. This implies that𝑁_𝐻(a𝑖,{a𝑗}^𝑖−1

𝑗=1) =0 and thus𝑑_𝐻(a𝑖,a𝑗) ≥2𝑘+1 for𝑖 ∈ [2, 𝑀]and 𝑗 ∈ [𝑖−1]. Moreover, since𝑞₁=2^𝐿′, we have thata₁=1𝐿^′. Therefore,{a𝑖}^𝑀

𝑖=1∈ S𝐻. Next, we use Lemma6.4.3to show that the strings{a𝑖}^𝑀

𝑖=1satisfy Eq. (6.7).

Lemma 6.4.4. The output{a𝑖}^𝑀

𝑖=1of the encoding algorithm satisfies Eq.(6.7).

Proof. Note that in each inner for loop, the number 𝑞 is subtracted by 2^{𝐿′−ℓ} − 𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ−1,0),{a𝑗}^𝑖−1

𝑗=1) only when 𝑎_𝑖,ℓ = 1 and ℓ ∈ [𝐿^′]. Since the number𝑞equals𝑞_𝑖 at the beginning of each outer for loop, and from Lemma6.4.3 equals 1 at the end of each outer for loop, hence we have that

𝑞_𝑖− ∑︁

ℓ:𝑎𝑖 ,ℓ=1and^{ℓ∈[𝐿′]}

(2^{𝐿′−ℓ}−𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ−1,0),{a𝑗}^𝑖_𝑗⁻¹₌₁)) =1,

which implies (6.7). □

Remark 6.4.1. By definition of𝑁_𝐻(a, 𝐴), we have the following alternative char- acterization ofdecimal(a𝑖),𝑖 ∈ [𝑀].

decimal(a𝑖)=𝑞_𝑖−1+

𝑖−1

∑︁

𝑗=1

|{c:𝑑𝑒 𝑐𝑖 𝑚 𝑎𝑙(c) < 𝑑𝑒 𝑐𝑖 𝑚 𝑎𝑙(a𝑖)and𝑑_𝐻(c,a𝑗) ≤2𝑘}|, (6.10) which is 𝑞_𝑖 −1 plus the sum of number of strings that are lexicographically less thana𝑖and have Hamming distance at most2𝑘 froma𝑗 over 𝑗 < 𝑖.

Lemma6.4.4immediately implies a decoding algorithm that transforms{a𝑖}^𝑀

𝑖=1back to(𝑞₁, . . . , 𝑞_𝑀).

Decoding:

(1) Order the strings{a𝑖}^𝑀

𝑖=1such thata₁ > a₂ > . . . >a𝑀. (2) For𝑖 ∈ [𝑀],

𝑞_𝑖 =decimal(a𝑖) +1+ ∑︁

ℓ:𝑎_{𝑖 ,ℓ}=1and^{ℓ∈[𝐿′]}

𝑁_𝐻( (𝑎_𝑖,1, . . . , 𝑎_𝑖,ℓ−1,0),{a𝑗}^𝑖−1

𝑗=1). (6.11) To show that the decoding is correct, we prove that the stringa𝑖,𝑖 ∈ [𝑀] generated in the encoding procedure satisfies

a₁ > a₂> . . . >a𝑀. (6.12) Then we conclude that the string a𝑖 obtained by ordering{a𝑖}^𝑀

𝑖=1in Step (1) in the decoding procedure satisfies Eq. (6.7). Hence we have Eq. (6.23) and thus 𝑞_𝑖, 𝑖 ∈ [𝑀]can be recovered. Suppose on the contrary, there exista𝑖₁ > a𝑖₂for some𝑖₁> 𝑖₂.

Letℓ^∗be the most significant bit wherea𝑖₁ anda𝑖₂ differ, i.e.,(𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ^∗−1) = (𝑎_𝑖

2,1, . . . , 𝑎_𝑖

2,ℓ^∗−1) and𝑎_𝑖

1,ℓ^∗ = 1 and𝑎_𝑖

2,ℓ^∗ =0. Then according to the if statement in the encoding procedure, we have that

𝑞_𝑖

1− ∑︁

ℓ:𝑎𝑖

1,ℓ=1and^{ℓ∈[ℓ∗]}

(2^{𝐿′−ℓ}−𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖1−1

𝑗=1)) > 0 and 𝑞_𝑖

2− ∑︁

ℓ:𝑎_𝑖

1,ℓ=1and^{ℓ∈[ℓ∗]}

(2^{𝐿′−ℓ}−𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖2−1

𝑗=1)) ≤0, which implies that

𝑞_𝑖

2−𝑞_𝑖

1 <

∑︁

ℓ:𝑎𝑖

1,ℓ=1and^{ℓ∈[ℓ∗]}

(2^{𝐿′−ℓ} −𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖2−1

𝑗=1))

− ∑︁

ℓ:𝑎𝑖

1,ℓ=1and^{ℓ∈[ℓ∗]}

(2^{𝐿′−ℓ} −𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖1−1

𝑗=1))

= ∑︁

ℓ:𝑎𝑖

1,ℓ=1and^{ℓ∈[ℓ∗]}

(𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖1−1

𝑗=1)

−𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖2−1

𝑗=1))

= ∑︁

ℓ:𝑎𝑖

1,ℓ=1and^{ℓ∈[ℓ∗]}

𝑁_𝐻( (𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ−1,0),{a𝑗}^𝑖_𝑗=𝑖¹⁻¹

2)

(𝑎)

≤

𝑖₁−1

∑︁

𝑗=𝑖₂

|c: 𝑑_𝐻(c,a𝑗) ≤2𝑘|

=(𝑖₁−𝑖₂)𝑄 , (6.13)

where (𝑎) follows from the definition of 𝑁_𝐻(a, 𝐴) and the fact that the strings which have(𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ₁−1,0)and(𝑎_𝑖

1,1, . . . , 𝑎_𝑖

1,ℓ₂−1,0)as prefixes, respectively, where𝑎_𝑖

1,ℓ₁=1, 𝑎_𝑖

1,ℓ₂=1 andℓ₁ ≠ ℓ₂, are different. Eq. (6.13) contradicts to the fact that the integers(𝑞₁, . . . , 𝑞_𝑀) =𝐹^𝐻

𝑄 (𝑑)satisfy𝑞_𝑖−𝑞_𝑖+1 > 𝑄for𝑖 ∈ [𝑀−1], which implies𝑞_𝑖

1−𝑞_𝑖

2 ≥ (𝑖₁−𝑖₂)𝑄.

Since the calculation of𝑁_𝐻(a, 𝐴)has polynomial complexity, the complexity of the encoding/decoding procedure is polynomial in 𝑀and 𝐿^′.