Protecting the Synchronization Vectors

Chapter III: Binary Codes Correcting 𝑘 Deletions/Insertions

3.4 Protecting the Synchronization Vectors

In this section we present a hash function with size 4𝑘log𝑛+𝑜(log𝑛) to protect the synchronization vector1𝑠 𝑦𝑛𝑐(c) from 𝑘 deletions in cand prove Lemma3.2.2.

We first prove Lemma3.2.1, which is decomposed to Proposition3.4.1and Propo- sition 3.4.2. In Proposition 3.4.1 we present an upper bound on the radius of the deletion ball for the synchronization vector. In Proposition3.4.2, we prove that the higher order parity check helps correct multiple deletions for sequences in which the there is a 0-run of length at least 3𝑘−1 between any two 1’s. Since1𝑠 𝑦𝑛𝑐(c)is such a sequence, we conclude that the higher order parity check helps recover1𝑠 𝑦𝑛𝑐(c).

After obtaining a bound on the difference between the higher order parity checks of two ambiguous sequence, we then apply Proposition3.4.2 on the synchronization vector1𝑠 𝑦𝑛𝑐(c)to prove Lemma3.2.1, which replaces the higher order parity checks in Proposition3.4.2by the higher parity checks modulo a numbers. After proving Lemma3.2.1, we use Lemma3.2.7to further compress the size of the higher order parity check that protects1𝑠 𝑦𝑛𝑐(c)and then prove Lemma3.2.2.

Proposition 3.4.1. Forc,c^′ ∈ {0,1}^𝑛, ifc^′∈ B𝑘(c), then1𝑠 𝑦𝑛𝑐(c^′) ∈ B₃𝑘(1𝑠 𝑦𝑛𝑐(c)). Proof. Since c^′ ∈ B𝑘(c), the sequences c^′ and c share a common subsequence after 𝑘 deletions in both. We now show that a single deletion in c causes at most two deletions and one insertion in its synchronization vector 1𝑠 𝑦𝑛𝑐(c). We first show that a deletion inccan destroy and generate at most 1 synchronization pattern.

This is because for any synchronization pattern that is destroyed or generated, there must be a deletion that occurs within the synchronization pattern. Hence any two destroyed or generated synchronization patterns cannot be caused by the same deletion. Therefore, we need to consider four cases in total. Let d^′ be the subsequence ofcafter a single deletion.

1. The deletion destroys a synchronization pattern (𝑐_𝑖+1, . . . , 𝑐_{𝑖+3𝑘+⌈log𝑘⌉+4}) for some𝑖and no synchronization pattern is generated. Then the sequence1𝑠 𝑦𝑛𝑐(d^′) can be obtained by deleting the 1 entry1𝑠 𝑦𝑛𝑐(c)_𝑖+3𝑘 in1𝑠 𝑦𝑛𝑐(c).

2. The deletion generates a new synchronization pattern(𝑐^′

𝑖^′+1, . . . , 𝑐^′

𝑖^′+3𝑘+⌈log𝑘⌉+4) for some𝑖^′ and destroys a synchronization pattern (𝑐_𝑖+1, . . . , 𝑐_{𝑖+3𝑘+⌈log𝑘⌉+4}). The sequence1𝑠 𝑦𝑛𝑐(d^′)can be obtained by deleting the 1 entry1𝑠 𝑦𝑛𝑐(c)_𝑖+3𝑘and the 0 entry1𝑠 𝑦𝑛𝑐(c)_{𝑖+3𝑘−1}in1𝑠 𝑦𝑛𝑐(c) and inserting a 1 entry at1𝑠 𝑦𝑛𝑐(c)𝑖^′+3𝑘. 3. The deletion generates a new synchronization pattern(𝑐^′

𝑖^′+1, . . . , 𝑐^′

𝑖^′+3𝑘+⌈log𝑘⌉+4) for some𝑖^′and no synchronization pattern is destroyed. Then the 1𝑠 𝑦𝑛𝑐(d^′) can be obtained by deleting two 0 entries 1𝑠 𝑦𝑛𝑐(c)𝑖^′+3𝑘 and 1𝑠 𝑦𝑛𝑐(c)𝑖^′+3𝑘+1

in1𝑠 𝑦𝑛𝑐(c) and inserting a 1 entry at1𝑠 𝑦𝑛𝑐(c)𝑖^′+3𝑘.

4. No synchronization pattern is generated or destroyed. Then1𝑠 𝑦𝑛𝑐(d^′)can be obtained by deleting a 0 entry1𝑠 𝑦𝑛𝑐(c)𝑗, where𝑗is the location of the deletion.

In summary, in each of the above cases, a single deletion in c causes at most two deletions and one insertion in 1𝑠 𝑦𝑛𝑐(c). Hence 𝑘 deletions in c and c^′ cause at

most 2𝑘deletions and𝑘insertions in1𝑠 𝑦𝑛𝑐(c)and1𝑠 𝑦𝑛𝑐(c^′)respectively. According to Lemma3.2.6, we have that1𝑠 𝑦𝑛𝑐(c^′) ∈ B₃𝑘(1𝑠 𝑦𝑛𝑐(c)) whenc^′ ∈ B𝑘(c). Hence,

Proposition3.4.1is proved. □

Let R𝑚 be the set of length 𝑛 sequences in which there is a 0 run of length at least 𝑚−1 between any two 1’s. Any two 1’s in a sequence c ∈ R𝑚 have index distance at least𝑚. The following lemma shows that the sequences inR₃𝑘 can be protected using higher order parity checks. Note that compared to the higher order parity checks 𝑓(c), the higher order parity checks in the following proposition do not have modulo operations.

Proposition 3.4.2. For sequencesc,c^′∈ R₃𝑘, ifc^′∈ B₃𝑘(c)andc·m^(𝑒) =c^′·m^(𝑒) for𝑒 ∈ [0,6𝑘], thenc=c^′.

Proof. We first compute the differencec·m^(𝑒) −c^′·m^(𝑒), 𝑒 ∈ [0,6𝑘]. Sincec^′ ∈ B₃𝑘(c), there exist two subsetsδ ={𝛿₁, . . . , 𝛿_3𝑘} ⊂ [1, 𝑛]andδ^′={𝛿^′

1, . . . , 𝛿^′

3𝑘} ⊂ [1, 𝑛] such that deleting bits with indices δ and δ^′ respectively from c and c^′ results in the same length 𝑛− 3𝑘 subsequence, i.e., (𝑐_𝑖 : 𝑖 ∉ δ) = (𝑐^′

𝑖 : 𝑖 ∉ δ^′).

Let 𝚫 = {𝑖 : 𝑐_𝑖 = 1} and 𝚫^′ = {𝑖 : 𝑐^′

𝑖 = 1} be the indices of 1 entries in c and c^′ respectively. Let 𝑆₁ = 𝚫 ∩δ be the indices of 1 entries that are deleted inc. Then𝑆^𝑐

1=𝚫∩ ( [1, 𝑛]\δ) denotes the indices of 1 entries that are not deleted.

Similarly, let 𝑆₂ = 𝚫^′∩δ^′ and 𝑆^𝑐

2 = 𝚫^′∩ ( [1, 𝑛]\δ^′) be the indices of 1 entries that are deleted and not in c^′ respectively. Let the elements in δ∪δ^′ be ordered by 1 ≤ 𝑝₁ ≤ 𝑝₂ ≤ . . .≤ 𝑝_6𝑘 ≤ 𝑛. Denote𝑝₀ =0 and 𝑝_6𝑘+1=𝑛. Then we have that

c·m^(𝑒) −c^′·m^(𝑒)

=∑︁

ℓ∈𝚫

m^(𝑒)_ℓ −∑︁

ℓ∈𝚫^′

m⁽_ℓ^𝑒)

=∑︁

ℓ∈𝚫

(

ℓ

∑︁

𝑖=1

𝑖^𝑒) −∑︁

ℓ∈𝚫^′

(

ℓ

∑︁

𝑖=1

𝑖^𝑒)

=

𝑛

∑︁

𝑖=1

( ∑︁

ℓ∈𝚫∩[𝑖,𝑛]

𝑖^𝑒) −

𝑛

∑︁

𝑖=1

( ∑︁

ℓ∈𝚫^′∩[𝑖,𝑛]

𝑖^𝑒)

=

𝑛

∑︁

𝑖=1

(|𝚫∩ [𝑖, 𝑛] | − |𝚫^′∩ [𝑖, 𝑛] |)𝑖^𝑒

=

𝑛

∑︁

𝑖=1

(|𝑆₁∩ [𝑖, 𝑛] | + |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆₂∩ [𝑖, 𝑛] |

− |𝑆^𝑐

2∩ [𝑖, 𝑛] |)𝑖^𝑒

=

6𝑘

∑︁

𝑗=0 𝑝_𝑗+1

∑︁

𝑖=𝑝𝑗+1

(|𝑆₁∩ [𝑖, 𝑛] | − |𝑆₂∩ [𝑖, 𝑛] | + |𝑆^𝑐

1∩ [𝑖, 𝑛] |

− |𝑆^𝑐

2∩ [𝑖, 𝑛] |)𝑖^𝑒

(𝑎)=

6𝑘

∑︁

𝑗=0 𝑝_𝑗+1

∑︁

𝑖=𝑝_𝑗+1

(|𝑆₁∩ [𝑝_𝑗+1, 𝑛] |

− |𝑆₂∩ [𝑝_𝑗+1, 𝑛] | + |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] |)𝑖^𝑒, (3.5) where (𝑎) holds since by definition of 𝑝_𝑗, there is no deleted 1 entry in inter- val (𝑝_𝑗, 𝑝_𝑗+1) ={𝑝_𝑗 +1, . . . , 𝑝_𝑗+1−1}, 𝑗 ∈ [0,6𝑘]. In the following we show

Statement 1: −1 ≤ |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] | ≤ 1 for𝑖 ∈ [1, 𝑛].

Statement 2:For each interval(𝑝_𝑗, 𝑝_𝑗+1] ={𝑝_𝑗+1, . . . , 𝑝_𝑗+1}, 𝑗 =0, . . . ,6𝑘, we have either |𝑆^𝑐

1 ∩ [𝑖, 𝑛] | − |𝑆^𝑐

2 ∩ [𝑖, 𝑛] | ≤ 0 for all 𝑖 ∈ (𝑝_𝑗, 𝑝_𝑗+1] or

|𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] | ≥0 for all𝑖 ∈ (𝑝_𝑗, 𝑝_𝑗+1].

We first proveStatement 1. Note that deleting bits with indicesδincand deleting bits with indicesδ^′ inc^′result in the same subsequence. Hence, for every𝑖 ∈ 𝑆^𝑐

1, there is a unique corresponding index𝑖^′ ∈ 𝑆^𝑐

2such that the two 1 entries 𝑐_𝑖 and𝑐^′

𝑖^′

end in the same location after deletions, i.e.,𝑖− |δ∩ [1, 𝑖−1] | =𝑖^′− |δ^′∩ [1, 𝑖^′−1] |. This implies that|𝑖^′−𝑖| ≤ 3𝑘. Fix integers𝑖and𝑖^′. Then by definition of𝑖and𝑖^′, for every𝑥 ∈𝑆^𝑐

1∩ [𝑖+1, 𝑛], there is a unique corresponding𝑦 ∈𝑆^𝑐

2∩ [𝑖^′+1, 𝑛]such that the two 1 entries𝑐_𝑥 and𝑐^′

𝑦 end in the same location after deletions. Therefore, we have that|𝑆^𝑐

1∩ [𝑖+1, 𝑛] | = |𝑆^𝑐

2∩ [𝑖^′+1, 𝑛] |, and thus that|𝑆^𝑐

1∩ [𝑖, 𝑛] | =|𝑆^𝑐

2∩ [𝑖^′, 𝑛] |. If𝑖^′ ≥𝑖, then

|𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] |

=|𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖^′, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑖^′−1] |

=− |𝑆^𝑐

2∩ [𝑖, 𝑖^′−1] |

(𝑎)

≥ − |𝑆^𝑐

2∩ [𝑖, 𝑖+3𝑘−1] |

(𝑏)

≥ −1,

where (𝑎) follows from the fact that 𝑖^′ ≤ 𝑖 +3𝑘 and (𝑏) follows from the fact thatc,c^′∈ R₃𝑘. Also we have that|𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] | =−|𝑆^𝑐

2∩ [𝑖, 𝑖^′−1] | ≤ 0.

Hence when𝑖^′ ≤ 𝑖, we have that−1 ≤ |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] | ≤ 0. Similarly,

when𝑖^′ < 𝑖, we have that

|𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] |

=|𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖^′, 𝑛] | + |𝑆^𝑐

2∩ [𝑖^′, 𝑖−1] |

=|𝑆^𝑐

2∩ [𝑖^′, 𝑖−1] |

≤|𝑆^𝑐

2∩ [𝑖^′, 𝑖^′+3𝑘−1] |

≤1, and that |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2 ∩ [𝑖, 𝑛] | = |𝑆^𝑐

2∩ [𝑖^′, 𝑖 − 1] | ≥ 0. Therefore, we have that 0 ≤ |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] | ≤1 when𝑖^′< 𝑖. ThusStatement 1is proved.

We now prove Statement 2 by contradiction. Suppose on the contrary, there exist𝑖₁, 𝑖₂∈ (𝑝_𝑗, 𝑝_𝑗+1] such that𝑖₁< 𝑖₂and

(|𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] |) (|𝑆^𝑐

1∩ [𝑖₂, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₂, 𝑛] |)

< 0 FromStatement 1we have that|𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] | ∈ [−1,1]and that|𝑆^𝑐

1∩ [𝑖₂, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₂, 𝑛] | ∈ [−1,1]. Hence by symmetry it can be assumed that |𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] | =−1 and|𝑆^𝑐

1∩ [𝑖₂, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₂, 𝑛] | =1. As shown in proof of Statement 1, for every element𝑖 ∈ 𝑆^𝑐

1, there is a corresponding element𝑖^′ ∈𝑆^𝑐

2such that the two 1 entries𝑐_𝑖and𝑐^′

𝑖^′ end in the same location after deletions. Hence, for 𝑦=min𝑖∈𝑆^𝑐

2∩[𝑖₁,𝑛]𝑖, there exists an integer𝑥 ∈𝑆^𝑐

1such that the two 1 entries𝑐_𝑥and𝑐^′

𝑦

are in the same location after deletions, i.e.,𝑥− |δ∩ [1, 𝑥−1] | =𝑦− |δ^′∩ [1, 𝑦−1] |. Since|𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] | =−1, we have that𝑥 ∈ 𝑆^𝑐

1∩ [1, 𝑖₁−1]. Otherwise, we have that 𝑥 ∈ 𝑆^𝑐

1∩ [𝑖₁, 𝑛] and for every integer 𝑖^′ ∈ 𝑆^𝑐

2∩ (𝑦, 𝑛], there exists an integer 𝑖 ∈ 𝑆^𝑐

1∩ (𝑥 , 𝑛] such that 𝑐_𝑖 and 𝑐_𝑖′ end up in the same location after deletions. This implies that |𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] | ≥ 0, contradicting the fact that|𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] | =−1. Therefore,

𝑖₁− |δ∩ [1, 𝑖₁−1] | >𝑖₁−1− |δ∩ [1, 𝑖₁−1] |

≥𝑥− |δ∩ [1, 𝑥−1] |

=𝑦− |δ^′∩ [1, 𝑦−1] |

≥𝑖₁− |δ^′∩ [1, 𝑖₁−1] |, which implies that

|δ∩ [1, 𝑖₁−1] | < |δ^′∩ [1, 𝑖₁−1] |. (3.6)

Similarly, from|𝑆^𝑐

1∩ [𝑖₂, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₂, 𝑛] | =1 we have that

|δ∩ [1, 𝑖₂−1] | > |δ^′∩ [1, 𝑖₂−1] |. (3.7) Eq. (3.6) and Eq. (3.7) implies that

|δ∩ [1, 𝑖₂−1] | − |δ∩ [1, 𝑖₁−1] |

≥|δ^′∩ [1, 𝑖₂−1] | +1− |δ^′∩ [1, 𝑖₁−1] | +1

≥2. (3.8)

However, since𝑖₁, 𝑖₂ ∈ (𝑝_𝑗, 𝑝_𝑗+1] and no deletion occurs in the interval(𝑝_𝑗, 𝑝_𝑗+1], we have that |δ ∩ [1, 𝑖₁] | = |δ∩ [1, 𝑖₂−1] | and |δ^′∩ [1, 𝑖₁] | = |δ^′∩ [1, 𝑖₂−1] |, which implies that

|δ∩ [1, 𝑖₂−1] | − |δ∩ [1, 𝑖₁−1] |

≤|δ∩ [1, 𝑖₂−1] | − |δ∩ [1, 𝑖₁] | +1

=1,

contradicting Eq. (3.8). Hence there do not exist different integers𝑖₁, 𝑖₂ ∈ (𝑝_𝑗, 𝑝_𝑗+1] such that

(|𝑆^𝑐

1∩ [𝑖₁, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₁, 𝑛] |) (|𝑆^𝑐

1∩ [𝑖₂, 𝑛] | − |𝑆^𝑐

2∩ [𝑖₂, 𝑛] |)

< 0. HenceStatement 2is proved.

Now we continue to prove Proposition3.4.2. Denote 𝑠_𝑖 ≜ |𝑆₁∩ [𝑖, 𝑛] | − |𝑆₂∩ [𝑖, 𝑛] | + |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] |. (3.9) Note that no deletion occurs in the interval(𝑝_𝑗, 𝑝_𝑗+1], it follows that

|𝑆₁∩ [𝑖, 𝑛] | − |𝑆₂∩ [𝑖, 𝑛] |

=|𝑆₁∩ [𝑝_𝑗+1, 𝑛] | − |𝑆₂∩ [𝑝_𝑗+1, 𝑛] | (3.10) for 𝑖 ∈ (𝑝_𝑗, 𝑝_𝑗+1]. Combining (3.10) with Statement 1 and Statement 2, we conclude that for each interval(𝑝_𝑗, 𝑝_𝑗+1], 𝑗 ∈ {0, . . . ,6𝑘}, either 𝑠_𝑖 ≥ 0 for all𝑖 ∈ (𝑝_𝑗, 𝑝_𝑗+1] or𝑠_𝑖 ≤ 0 for all𝑖 ∈ (𝑝_𝑗, 𝑝_𝑗+1]. Letx =(𝑥₀, . . . , 𝑥_6𝑘) ∈ {−1,1}^6𝑘+1be a vector defined by

𝑥_𝑖 =











−1, if𝑠_𝑗 < 0 for some 𝑗 ∈ (𝑝_𝑖, 𝑝_𝑖+1] 1, else.

.

Then from Eq. (3.5) and Eq. (3.9), the differencec·m^(𝑒) −c^′·m^(𝑒) is given by c·m^(𝑒)−c^′·m^(𝑒) =

6𝑘

∑︁

𝑗=0

(

𝑝_𝑗+1

∑︁

𝑖=𝑝𝑗+1

|𝑠_𝑖|𝑖^𝑒)𝑥_𝑗. (3.11)

Let 𝐴 be a 6𝑘 +1×6𝑘 +1 matrix with entries defined by 𝐴_{𝑒, 𝑗} =Í^𝑝𝑗

𝑖=𝑝𝑗−1+1|𝑠_𝑖|𝑖^𝑒−1 for𝑒, 𝑗 ∈ [1,6𝑘 +1]. If c·m^(𝑒) =c^′·m^(𝑒) for𝑒 ∈ [0,6𝑘], we have the following linear equation

𝐴x=















 Í^𝑝1

𝑖=𝑝₀+1|𝑠_𝑖|𝑖⁰ . . . Í^𝑝6𝑘+1

𝑖=𝑝_6𝑘+1|𝑠_𝑖|𝑖⁰ ..

.

...

.. . Í^𝑝1

𝑖=𝑝₀+1|𝑠_𝑖|𝑖^6𝑘 . . . Í^𝑝6𝑘+1

𝑖=𝑝_6𝑘+1|𝑠_𝑖|𝑖^6𝑘































 𝑥₀

.. . 𝑥_6𝑘

















=0, (3.12)

with a solution𝑥_𝑖 ∈ {−1,1}for𝑖 ∈ [0,6𝑘]. We show that this is impossible unless𝐴 is a zero matrix. Suppose on the contrary that 𝐴is nonzero, let 𝑗₁ < . . . < 𝑗_𝑄 be the indices of all nonzero columns of 𝐴. Let 𝐴^∗ be a submatrix of 𝐴, obtained by choosing the intersection of the first𝑄 rows and columns with indices 𝑗₁, . . . , 𝑗_𝑄. Then taking the first𝑄linear equations from the equation set (3.13) and noting that the nonzero columns in 𝐴are the 𝑗₁, . . . , 𝑗_𝑄-th columns, we have that

𝐴^∗x^′

=

















 Í^𝑝𝑗₁

𝑖=𝑝_𝑗

1−1+1|𝑠_𝑖|𝑖⁰ . . . Í^𝑝𝑗𝑄

𝑖=𝑝_𝑗

𝑄−1+1|𝑠_𝑖|𝑖⁰ ..

.

.. .

.. . Í^𝑝𝑗₁

𝑖=𝑝𝑗

1−1+1|𝑠_𝑖|𝑖^𝑄−1 . . . Í^𝑝𝑗𝑄

𝑖=𝑝𝑗

𝑄−1+1|𝑠_𝑖|𝑖^𝑄−1

































 𝑥_𝑗

1

.. . 𝑥_𝑗

𝑄

















=0. (3.13)

The determinant of 𝐴^∗is given by det(𝐴^∗)

=det

©

­

­

­

­

« Í^𝑝𝑗1

𝑖=𝑝𝑗

1−1+1|𝑠_𝑖|𝑖⁰ . . . Í^𝑝𝑗𝑄

𝑖=𝑝𝑗

𝑄−1+1|𝑠_𝑖|𝑖⁰ ..

.

...

.. . Í^𝑝𝑗₁

𝑖=𝑝_𝑗

1−1+1|𝑠_𝑖|𝑖^𝑄−1 . . . Í^𝑝𝑗𝑄

𝑖=𝑝_𝑗

𝑄−1+1|𝑠_𝑖|𝑖^𝑄−1 ª

®

®

®

®

¬

(𝑎)= ∑︁

𝑖₁∈(𝑝𝑗 1−1, 𝑝𝑗

1],..., 𝑖_𝑄∈(𝑝_𝑗

𝑄−1, 𝑝_𝑗

𝑄]

det

©

­

­

­

«

|𝑠_𝑖

1|𝑖⁰

1 . . . |𝑠_𝑖

𝑄|𝑖⁰

𝑄

.. .

... .. .

|𝑠_𝑖

1|𝑖

𝑄−1

1 . . . |𝑠_𝑖

𝑄|𝑖

𝑄−1 𝑄

ª

®

®

®

¬

(𝑏)= ∑︁

𝑖₁∈(𝑝𝑗 1−1, 𝑝𝑗

1],..., 𝑖_𝑄∈(𝑝_𝑗

𝑄−1, 𝑝_𝑗

𝑄]

" 𝑄

Ö

𝑞=1

|𝑠_𝑖

𝑞|det

©

­

­

­

« 𝑖⁰

1 . . . 𝑖⁰

𝑄

.. .

... .. . 𝑖^𝑄−1

1 . . . 𝑖^𝑄−1

𝑄

ª

®

®

®

¬

#

(𝑐)= ∑︁

𝑖₁∈(𝑝𝑗 1−1, 𝑝𝑗

1],..., 𝑖𝑄∈(𝑝𝑗

𝑄−1, 𝑝𝑗 𝑄]

[

𝑄

Ö

𝑞=1

|𝑠_𝑖

𝑞| Ö

1≤𝑚 <ℓ≤𝑄

(𝑖_ℓ −𝑖_𝑚)], (3.14)

where equality(𝑎)follows from the multi-linearity of the determinant det( [v₁ . . . 𝑎v𝑖+𝑏v . . . v𝑞])

=𝑎det( [v₁ . . . v𝑖 . . . v𝑞])

+𝑏det( [v₁ . . . v𝑖−1v v𝑖+1 . . . v𝑞])

for any integers 𝑞 and 𝑖 and 𝑞-dimensional vectors v₁, . . . ,v𝑞, v. Equality (𝑏) follows from the linearity of the determinant

det( [v₁ . . . 𝑎v𝑖 . . . v𝑞]) =𝑎det( [v₁ . . . v𝑖 . . . v𝑞])

for any integers𝑞and𝑖 and𝑞-dimensional vectorsv₁, . . . ,v𝑞. Equality(𝑐)follows from the determinant of Vandermonde matrix

det

©

­

­

­

« 𝑖⁰

1 . . . 𝑖⁰

𝑞

.. .

... .. . 𝑖^𝑞−1

1 . . . 𝑖^𝑞−1_𝑞 ª

®

®

®

¬

= Ö

1≤𝑚 <ℓ≤𝑞

(𝑖_ℓ −𝑖_𝑚)

for any integers 𝑞, 𝑖₁, . . . , 𝑖_𝑞. . The determinant 𝑑𝑒𝑡(𝐴^∗) is positive since𝑖_ℓ > 𝑖_𝑚 for ℓ > 𝑚. and for 𝑖₁ ∈ (𝑝_𝑗

1−1, 𝑝_𝑗

1], . . . , 𝑖_𝑄 ∈ (𝑝_𝑗

𝑄−1, 𝑝_𝑗

𝑄]. Note that all the columns of 𝐴^∗ are nonzero. Therefore, the linear equation 𝐴^∗x^′ = 0 does not have nonzero solutions, contradicting the fact thatx^′ = (𝑥_𝑗

1, . . . , 𝑥_𝑗

𝑄) ∈ {−1,1}^𝑄. Hence 𝐴is a zero matrix, meaning that

|𝑆₁∩ [𝑖, 𝑛] | − |𝑆₂∩ [𝑖, 𝑛] | + |𝑆^𝑐

1∩ [𝑖, 𝑛] | − |𝑆^𝑐

2∩ [𝑖, 𝑛] |

=|𝚫∩ [𝑖, 𝑛] | − |𝚫^′∩ [𝑖, 𝑛] | =0

for𝑖 ∈ {1, . . . , 𝑛}. This implies𝚫 =𝚫^′and thusc=c^′. Hence Proposition3.4.2is

proved. □

Proof of Lemma3.2.1

We are now ready to prove Lemma 3.2.1, which states that 1𝑠 𝑦𝑛𝑐(c) = 1𝑠 𝑦𝑛𝑐(c^′) for sequences c and c^′ ∈ B𝑘(c) satisfying 𝑓(1𝑠 𝑦𝑛𝑐(c)) = 𝑓(1𝑠 𝑦𝑛𝑐(c)). From

Proposition 3.4.1 we have that 1𝑠 𝑦𝑛𝑐(c^′) ∈ B₃𝑘(1𝑠 𝑦𝑛𝑐(c)). Then, it is not hard to see that (1𝑠 𝑦𝑛𝑐(c^′)𝑖, . . . ,1𝑠 𝑦𝑛𝑐(c^′)𝑛) ∈ B₃𝑘( (1𝑠 𝑦𝑛𝑐(c)𝑖, . . . ,1𝑠 𝑦𝑛𝑐(c)𝑛)). This im- plies that ||𝚫 ∩ [𝑖, 𝑛] | − |𝚫^′ ∩ [𝑖, 𝑛] || ≤ 3𝑘, where 𝚫 = {𝑖 : 1𝑠 𝑦𝑛𝑐(c)𝑖 = 1}

and𝚫^′={𝑖:1𝑠 𝑦𝑛𝑐(c^′)𝑖 =1}. According to the forth line in Eq. (3.5), we have that

|1𝑠 𝑦𝑛𝑐(c) ·m^(𝑒)−1𝑠 𝑦𝑛𝑐(c^′) ·m^(𝑒)|

=|

𝑛

∑︁

𝑖=1

(|𝚫∩ [𝑖, 𝑛] | − |𝚫^′∩ [𝑖, 𝑛] |)𝑖^𝑒|,

≤

𝑛

∑︁

𝑖=1

3𝑘 𝑖^𝑒

<3𝑘 𝑛^𝑒+1. (3.15)

If 𝑓(1𝑠 𝑦𝑛𝑐(c)) = 𝑓(1𝑠 𝑦𝑛𝑐(c^′)), then

1𝑠 𝑦𝑛𝑐(c) ·m^(𝑒) ≡ 1𝑠 𝑦𝑛𝑐(c^′) ·m^(𝑒) mod 3𝑘 𝑛^𝑒+1 (3.16) for𝑒 ∈ [0,6𝑘]. Equations (3.16) and (3.15) imply that1𝑠 𝑦𝑛𝑐(c) ·m^(𝑒) =1𝑠 𝑦𝑛𝑐(c^′) · m^(𝑒)for𝑒 ∈ [0,6𝑘]. Since1𝑠 𝑦𝑛𝑐(c^′) ∈ B₃𝑘(1𝑠 𝑦𝑛𝑐(c))and1𝑠 𝑦𝑛𝑐(c),1𝑠 𝑦𝑛𝑐(c^′) ∈ R₃𝑘, from Proposition3.4.2we conclude that1𝑠 𝑦𝑛𝑐(c) =1𝑠 𝑦𝑛𝑐(c^′). Hence Lemma3.2.1 is proved.

Proof of Lemma3.2.2

Based on Lemma 3.2.1, we now show Lemma 3.2.2. Specifically, we show that there exists a function 𝑝 : {0,1}^𝑛 → [1,2^{2𝑘log𝑛+𝑜(log𝑛)}] such that 1𝑠 𝑦𝑛𝑐(c) = 1𝑠 𝑦𝑛𝑐(c^′)for sequencescandc^′∈ B𝑘(c)satisfying(𝑓(1𝑠 𝑦𝑛𝑐(c)) mod 𝑝(c), 𝑝(c))= (𝑓(1𝑠 𝑦𝑛𝑐(c^′)) mod 𝑝(c^′), 𝑝(c^′)).

Lemma 3.2.1 implies that 𝑓(1𝑠 𝑦𝑛𝑐(c)) ≠ 𝑓(1𝑠 𝑦𝑛𝑐(c^′)) for c^′ ∈ B𝑘(c)\{c}, if 1𝑠 𝑦𝑛𝑐(c) ≠ 1𝑠 𝑦𝑛𝑐(c^′). Hence |𝑓(1𝑠 𝑦𝑛𝑐(c)) − 𝑓(1𝑠 𝑦𝑛𝑐(c^′)) | ≠ 0 for c^′ ∈ B𝑘(c)\{c}, where 𝑓(1𝑠 𝑦𝑛𝑐(c)) and 𝑓(1𝑠𝑦𝑛𝑐(c^′)) denote the integer presentation of their vec- tor form. The integers are in the range [0,(3𝑘)^6𝑘+1𝑛^{(3𝑘+1) (6𝑘+1)} −1]. According to Lemma 3.2.7, the number of divisors of |𝑓(1𝑠 𝑦𝑛𝑐(c)) − 𝑓(1𝑠 𝑦𝑛𝑐(c^′)) | is upper bounded by

2^{2[(3𝑘+1) (6𝑘+1)ln𝑛+(6𝑘+1)ln 3𝑘]/ln( (3𝑘+1) (6𝑘+1)ln𝑛+(6𝑘+1)ln 3𝑘)}

=2^{𝑜(log𝑛)},

where the equality holds since𝑘 =𝑜(√︁

log log𝑛). For any sequencec ∈ {0,1}^𝑛, let

P (c) ={𝑝 : 𝑝divides|𝑓(1𝑠 𝑦𝑛𝑐(c^′)) − 𝑓(1𝑠 𝑦𝑛𝑐(c)) | for somec^′∈ B𝑘(c)\{c}such that1𝑠 𝑦𝑛𝑐(c)≠ 1𝑠 𝑦𝑛𝑐(c^′)}

be the set of all divisors of the numbers {|𝑓(1𝑠 𝑦𝑛𝑐(c^′)) − 𝑓(1𝑠 𝑦𝑛𝑐(c)) | : c^′ ∈ B𝑘(c)\{c} and1𝑠 𝑦𝑛𝑐(c)≠ 1𝑠 𝑦𝑛𝑐(c^′)}. Since |B𝑘(c) | ≤ ^𝑛

𝑘

2

2^𝑘 ≤ 2𝑛^2𝑘, we have that

|P (c) | ≤2𝑛^2𝑘2^{𝑜(log𝑛)}

=2^{2𝑘log𝑛+𝑜(log𝑛)}.

Therefore, there exists a number 𝑝(c) ∈ [1,2^{2𝑘log𝑛+𝑜(log𝑛)}] such that 𝑝(c) does not divide|𝑓(1𝑠 𝑦𝑛𝑐(c^′)) − 𝑓(1𝑠 𝑦𝑛𝑐(c)) |for allc^′∈ B𝑘(c)\{c} satisfying1𝑠 𝑦𝑛𝑐(c) ≠ 1𝑠 𝑦𝑛𝑐(c^′). Hence, if 𝑓(1𝑠 𝑦𝑛𝑐(c^′)) ≡ 𝑓(1𝑠 𝑦𝑛𝑐(c))mod 𝑝(c)andc^′ ∈ B𝑘(c), we have that 𝑝(c) divides |𝑓(1𝑠 𝑦𝑛𝑐(c^′)) − 𝑓(1𝑠 𝑦𝑛𝑐(c)) |, and thus that 1𝑠 𝑦𝑛𝑐(c^′) = 1𝑠 𝑦𝑛𝑐(c).

This completes the proof of Lemma3.2.2.