Protecting 1 10 Indicator Vectors

Chapter II: Binary 2-Deletion/Insertion Correcting Codes

2.3 Protecting 1 10 Indicator Vectors

In this section, we prove Lemma2.2.2. The proof is based on Proposition2.1.3and Proposition2.2.3, which are proved at the end of this section. Sincec ∈ 𝐵₂(c^′) it follows that there exist integers𝑖₁, 𝑖₂, 𝑗₁, and 𝑗₂such that

c^del’−→^𝑖1 d^del’

𝑗₁

−→ e c^′del’−→^𝑖2 d^′del’

𝑗₂

−→ e,

and by Proposition 2.2.1it follows that there exist integers ℓ₁, ℓ₂, 𝑘₁, and 𝑘₂ such that

1₁₀(c) ^del’−→^ℓ11₁₀(d)^del’−→^𝑘1 1₁₀(e) 1₁₀(c^′) ^del’−→^ℓ21₁₀(d^′) ^del’−→^𝑘2 1₁₀(e).

Due to symmetry betweencandc^′, we distinguish between the following three cases.

In each case, the difference between the 𝑓 values ofcandc^′are given in terms of the function𝑔 (2.4). Further, the computation of these three differences, a somewhat tedious but straightforward task, is deferred to the appendix of this chapter.

Case (a). Ifℓ₁ ≤ℓ₂ < 𝑘₂ ≤ 𝑘₁(Fig.2.2), then

1₁₀(c)𝑡 =1₁₀(c^′)𝑡 if𝑡 < ℓ₁ orℓ₂ < 𝑡 < 𝑘₂ or𝑡 > 𝑘₁,

1₁₀(c)𝑡+1 =1₁₀(c^′)𝑡 ifℓ₁ ≤ 𝑡 ≤ ℓ₂−1, 1₁₀(c)𝑡 =1₁₀(c^′)_𝑡+1 if𝑘₂ ≤ 𝑡 ≤ 𝑘₁−1. Thus, for𝑒 ∈ {0,1,2},

(1₁₀(c) −1₁₀(c^′)) ·m^(𝑒)

=𝑔

m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))−

𝑔m^(𝑒)(𝑘₂,(1₁₀(c^′)^{(𝑘2, 𝑘1)},1₁₀(c)𝑘₁)). (2.13) Case (b). Ifℓ₁ ≤ℓ₂ < 𝑘₁ ≤ 𝑘₂(Fig.2.3), then

1₁₀(c)𝑡 =1₁₀(c^′)𝑡 if𝑡 < 𝑙₁ or𝑙₂ < 𝑡 < 𝑘₁. or𝑡 > 𝑘₂.

1₁₀(c)_𝑡+1=1₁₀(c^′)𝑡 ifℓ₁ ≤𝑡 ≤ ℓ₂−1 or𝑘₁≤ 𝑡 ≤ 𝑘₂−1. Thus, for𝑒 ∈ {0,1,2},

(1₁₀(c) −1₁₀(c^′)) ·m^(𝑒)

=𝑔

m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))+

𝑔m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂)). (2.14) Case (c). Ifℓ₁< 𝑘₁ ≤ ℓ₂< 𝑘₂(Fig.2.4), then

1₁₀(c)𝑡 =1₁₀(c^′)𝑡 if𝑡 < 𝑙₁ or𝑡 > 𝑘₂,

1₁₀(c)

1₁₀(c^′) ★

★ ★

= = ★ =

= = =

ℓ₁ ℓ₂ 𝑘₂ 𝑘₁

Figure 2.2: Case (a), whereℓ₁ ≤ ℓ₂ < 𝑘₂ ≤ 𝑘₁. The diagonal lines indicate equality between the respective entries of1₁₀(c)and1₁₀(c^′).

1₁₀(c)

1₁₀(c^′) ★

★

★

★

= = =

= = =

ℓ₁ ℓ₂ 𝑘₁ 𝑘₂

Figure 2.3: Case (b), whereℓ₁ ≤ ℓ₂< 𝑘₁ ≤ 𝑘₂. The diagonal lines indicate equality between the respective entries of1₁₀(c)and1₁₀(c^′).

1₁₀(c)𝑡+1 =1₁₀(c^′)𝑡 ifℓ₁ ≤𝑡 ≤ 𝑘₁−2 orℓ₂+1 ≤ 𝑡 ≤ 𝑘₂−1, 1₁₀(c)_𝑡+2 =1₁₀(c^′)𝑡 if𝑘₁−1 ≤ 𝑡 ≤ ℓ₂−1. Thus, for𝑒 ∈ {0,1,2},

(1₁₀(c) −1₁₀(c^′)) ·m^(𝑒)

=𝑔

m^(𝑒)(ℓ₁,(1₁₀(c)^{(ℓ1, 𝑘1−1)},1₁₀(c)^{(𝑘1+1,ℓ2+1)},1₁₀(c^′)ℓ₂))+

𝑔m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂)). (2.15)

1₁₀(c) 1₁₀(c^′)

★

★

★

= ★ =

= =

ℓ₁ 𝑘₁ ℓ₂ 𝑘₂

Figure 2.4: Case (c), whereℓ₁ < 𝑘₁ ≤ℓ₂< 𝑘₂. The diagonal lines indicate equality between the respective entries of1₁₀(c)and1₁₀(c^′).

Note that if 𝑓(c) = 𝑓(c^′), then 1₁₀(c) ·m^(𝑒) ≡ 1₁₀(c^′) ·m^(𝑒) mod𝑛_𝑒, where 𝑛₀ = 2𝑛, 𝑛₁ =𝑛²,and𝑛₂=𝑛³. Hence, from (2.13)–(2.15) we have that

𝑔m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))

−𝑔

m^(𝑒)(𝑘₂,(1₁₀(c^′)^{(𝑘2, 𝑘1)},1₁₀(c)𝑘₁)) ≡0 mod𝑛_𝑒,

𝑔m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂)) +𝑔

m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂)) ≡0 mod𝑛_𝑒, and 𝑔m^(𝑒)(ℓ₁,(1₁₀(c)^{(ℓ1, 𝑘1−1)},1₁₀(c)^{(𝑘1+1,ℓ2+1)},1₁₀(c^′)ℓ₂)) +𝑔

m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂)) ≡0 mod𝑛_𝑒, (2.16) for Case (a), Case (b), and Case (c), respectively.

In what follows, we show that these equalities also hold in their non modular version.

Specifically, we prove the following,

𝑔m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))

−𝑔

m^(𝑒)(𝑘₂,(1₁₀(c^′)^{(𝑘2, 𝑘1)},1₁₀(c)𝑘₁))=0, (2.17) 𝑔m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))

+𝑔

m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂))=0, and (2.18) 𝑔m^(𝑒)(ℓ₁,(1₁₀(c)^{(ℓ1, 𝑘1−1)},1₁₀(c)^{(𝑘1+1,ℓ2+1)},1₁₀(c^′)ℓ₂))

+𝑔

m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂))=0. (2.19) From (2.8), we have that

−m_{𝑟+𝑠−2}^(𝑒) ≤ 𝑔

m^(𝑒)(𝑟 ,x) ≤ m_{𝑟+𝑠−2}^(𝑒)

for anyx∈ {0,1}^𝑠 and any integer𝑟that satisfies𝑟+𝑠−2≤ 𝑛−1. Therefore,

−m^(𝑒)_ℓ

2 −m⁽_𝑘^𝑒)

1 ≤𝑔

m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))−

𝑔m^(𝑒)(𝑘₂,(1₁₀(c^′)^{(𝑘2, 𝑘1)},1₁₀(c)𝑘₁))

≤m_ℓ^(𝑒)

2 +m_𝑘^(𝑒)

1

,

−m^(𝑒)_ℓ

2 −m⁽_𝑘^𝑒)

2 ≤𝑔

m^(𝑒)(ℓ₁,(1₁₀(c)^(ℓ1,ℓ2),1₁₀(c^′)ℓ₂))+

𝑔m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂))

≤m_ℓ^(𝑒)

2 +m_𝑘^(𝑒)

2

, and

−m^(𝑒)_ℓ

2 −m⁽_𝑘^𝑒)

2 ≤𝑔

m^(𝑒)(ℓ₁,(1₁₀(c)^{(ℓ1, 𝑘1−1)},

1₁₀(c)^{(𝑘1+1,ℓ2+1)},1₁₀(c^′)ℓ₂)) +𝑔

m^(𝑒)(𝑘₁,(1₁₀(c)^{(𝑘1, 𝑘2)},1₁₀(c^′)𝑘₂))

≤m_ℓ^(𝑒)

2 +m_𝑘^(𝑒)

2

, (2.20)

for Case (a)–(c) respectively. Further, note that m⁽_ℓ^𝑒)

2 +m⁽_𝑘^𝑒)

1

< 𝑛_𝑒 andm⁽_ℓ^𝑒)

2 +m⁽_𝑘^𝑒)

2

< 𝑛_𝑒. (2.21)

Combining (2.16), (2.20), and (2.21), we conclude that if 𝑓(c) = 𝑓(c^′), then Eq. (2.17), (2.18), and (2.19) hold for Case (a), Case (b), and Case (c) respectively.

For Case (a), Equation (2.17) and Proposition2.1.3imply that 1₁₀(c)ℓ₁ =. . .=1₁₀(c)ℓ₂ =1₁₀(c^′)ℓ₂

1₁₀(c^′)𝑘₂ =. . .=1₁₀(c^′)𝑘₁ =1₁₀(c)𝑘₁, which readily implies that

1₁₀(c^′)𝑡 =1₁₀(c)𝑡+1 =1₁₀(c)𝑡

forℓ₁ ≤𝑡 < ℓ₂and

1₁₀(c)𝑡 =1₁₀(c^′)_𝑡+1=1₁₀(c^′)𝑡

for 𝑘₂ ≤ 𝑡 < 𝑘₁. Together with 1₁₀(c)ℓ₂ = 1₁₀(c^′)ℓ₂ and 1₁₀(c^′)𝑘₁ = 1₁₀(c)𝑘₁, we have that1₁₀(c) =1₁₀(c^′).

For Case (b), Equation (2.18) and Proposition2.1.3implies that 1₁₀(c)ℓ₁ =. . .=1₁₀(c)ℓ₂ =1₁₀(c^′)ℓ₂

1₁₀(c^′)𝑘₁ =. . .=1₁₀(c^′)𝑘₂ =1₁₀(c)𝑘₂

and hence

1₁₀(c^′)𝑡 =1₁₀(c)𝑡+1 =1₁₀(c)𝑡

forℓ₁ ≤𝑡 < ℓ₂and 𝑘₁ ≤𝑡 < 𝑘₂, and thus1₁₀(c)=1₁₀(c^′).

Note that in Case (a) and (b) we used Proposition2.1.3, which implies that only two parity checks with weightsm⁽⁰⁾ andm⁽¹⁾ are needed. Case (c) is the most involved case and the only case when the parity check with weightm⁽²⁾ is needed. Hence, Proposition2.2.3is required. According to Equation (2.19) and Proposition2.2.3, we have either

1₁₀(c)ℓ₁ =. . .=1₁₀(c)𝑘₂ =1₁₀(c^′)ℓ₂ =1₁₀(c^′)𝑘₂ (2.22) or

1₁₀(c)ℓ₁ =. . .=1₁₀(c)𝑘₁−1=1₁₀(c)𝑘₁+1, 1₁₀(c)𝑖+1₁₀(c)𝑖+1=1 for𝑖 ∈ {𝑘₁, . . . , ℓ₂},

1₁₀(c^′)ℓ₂+1₁₀(c^′)𝑘₂ =1, and

1₁₀(c)ℓ₂+1 =. . .=1₁₀(c)𝑘₂ =1₁₀(c^′)𝑘₂. (2.23) If (2.22) is true, we can obtainc=c^′by following steps similar to those of Case (a) and Case (b). If (2.23) is true, we have

1₁₀(c^′)𝑡 =1₁₀(c)_𝑡+1 =1₁₀(c)𝑡

forℓ₁ ≤𝑡 ≤ 𝑘₁−2 andℓ₂+1≤ 𝑡 ≤ 𝑘₂−1. Furthermore, we have that 1₁₀(c^′)𝑡 =1₁₀(c)𝑡+2=1−1₁₀(c)𝑡+1=1₁₀(c)𝑡

for𝑘₁ ≤𝑡 ≤ ℓ₂−1. In addition, we have1₁₀(c^′)𝑘₁−1=1₁₀(c)𝑘₁+1=1₁₀(c)𝑘₁−1,1₁₀(c^′)ℓ₂ = 1−1₁₀(c^′)𝑘₂ =1−1₁₀(c)ℓ₂+1=1₁₀(c)ℓ₂ and1₁₀(c^′)𝑘₂ =1₁₀(c)𝑘₂. Hence, we have that1₁₀(c) = 1₁₀(c^′). This concludes the proof of Cases (a), (b), and (c), and thus the proof of Lemma2.2.2is completed.

We now prove Proposition2.1.3and Proposition2.2.3.

Proof. (of Proposition2.1.3) According to Eq. (2.12), if𝜆 =1, then Eq. (2.9) can be written as

𝑔m⁽⁰⁾(𝑟₁,x) −𝑔

m⁽⁰⁾(𝑟₂,y) =0, and 𝑔m⁽¹⁾(𝑟₁,x) −𝑔

m⁽¹⁾(𝑟₂,y) =0.

Therefore, it suffices to prove the claim for 𝜆 = −1. We distinguish between four cases according to the value of (𝑦₁, 𝑦_𝑠

2).

Case (a). (𝑦₁, 𝑦_𝑠

2) =(0,1). We have that 𝑔m^(𝑒)(𝑟₁,x) −𝑔

m^(𝑒)(𝑟₂,y)

=m𝑟^(𝑒)₁ 𝑥₁+

𝑠₁−1

∑︁

𝑡=2

(m_𝑡^(𝑒)_+𝑟

1−1−m_𝑡^(𝑒)_+𝑟

1−2)𝑥_𝑡−m⁽_𝑟^𝑒)

1+𝑠₁−2𝑥_𝑠

1− m𝑟^(𝑒)₂ 𝑦₁−

𝑠₂−1

∑︁

𝑡=2

(m^(𝑒)_𝑡+𝑟

2−1−m^(𝑒)_𝑡+𝑟

2−2)𝑦_𝑡 +m_𝑟^(𝑒)

2+𝑠₂−2𝑦_𝑠

2

≥ −m_𝑟^(𝑒)

1+𝑠₁−2−

𝑠₂−1

∑︁

𝑡=2

(m_𝑡^(𝑒)_+𝑟

2−1−m_𝑡^(𝑒)_+𝑟

2−2) + m_𝑟^(𝑒)

2+𝑠₂−2

=m𝑟⁽₂^𝑒) −m_𝑟^(𝑒)

1+𝑠₁−2 > 0, a contradiction.

Case (b). (𝑦₁, 𝑦_𝑠

2) = (1,0). From Proposition2.2.4and (2.9) we have𝑔

m^(𝑒)(𝑟₁,x) − 𝑔m^(𝑒)(𝑟₂,y) =0 for 𝑒 ∈ {0,1}, wherex ≜ 1−x andy ≜ 1−y. Since (𝑦

1, 𝑦

𝑠₂) = (0,1), from the previous case this leads to a contradiction.

Case (c). (𝑦₁, 𝑦_𝑠

2) =(1,1). Let 𝑆₁≜ {𝑗 :𝑦_𝑗−𝑟

2+1=1, 𝑟₂+1≤ 𝑗 ≤𝑟₂+𝑠₂−2}, and 𝑆^𝑐

1≜ {𝑗 :𝑦_𝑗−𝑟

2+1=0, 𝑟₂+1≤ 𝑗 ≤𝑟₂+𝑠₂−2}, and notice that

𝑔m⁽⁰⁾(𝑟₂,y)

=m𝑟⁽⁰⁾₂ −m_𝑟⁽⁰⁾

2+𝑠₂−2+

𝑠₂−1

∑︁

𝑗=2

(m⁽⁰⁾_𝑗+𝑟

2−1−m⁽⁰⁾_𝑗+𝑟

2−2)𝑦_𝑗

=−

𝑟₂+𝑠₂−2

∑︁

𝑗=𝑟₂+1

(m⁽_𝑗⁰⁾ −m⁽_𝑗⁰₋₁⁾ ) +

𝑠₂−1

∑︁

𝑗=2

(m⁽_𝑗⁰₊⁾_𝑟

2−1−m⁽_𝑗⁰₊⁾_𝑟

2−2)𝑦_𝑗

=−

𝑟₂+𝑠₂−2

∑︁

𝑗=𝑟₂+1

(m⁽⁰⁾_𝑗 −m⁽⁰⁾_𝑗−1) (1−𝑦_𝑗)

=− ∑︁

𝑗∈𝑆^𝑐

1

(m⁽_𝑗⁰⁾ −m⁽_𝑗⁰₋₁⁾ ) =−∑︁

𝑗∈𝑆^𝑐

1

1, and similarly, 𝑔m⁽¹⁾(𝑟₂,y)

=− ∑︁

𝑗∈𝑆^𝑐

1

(m⁽_𝑗¹⁾ −m⁽_𝑗¹₋₁⁾ ) =−∑︁

𝑗∈𝑆^𝑐

1

𝑗 . (2.24)

Now, on the one hand, if𝑥_𝑠

1 =0 we have 𝑔m⁽⁰⁾(𝑟₁,x) =m𝑟⁽⁰₁⁾𝑥₁+

𝑠₁−1

∑︁

𝑡=2

(m_𝑡⁽₊⁰⁾_𝑟

1−1−m_𝑡⁽₊⁰⁾_𝑟

1−2)𝑥_𝑡

≥0, (2.25)

and hence, (2.24) and (2.25) imply that𝑔

m⁽⁰⁾(𝑟₁,x) −𝑔

m⁽⁰⁾(𝑟₂,y) ≥0, and equality holds only when𝑔

m⁽⁰⁾(𝑟₁,x)and𝑔

m⁽⁰⁾(𝑟₂,y)are both 0, which by Proposition2.1.2 implies thatxandyare constant vectors. On the other hand, if𝑥_𝑠

1 =1 let𝑆₂ ={𝑡 : 𝑥_{max{𝑡−𝑟}

1+1,1} =0,1 ≤𝑡 ≤ 𝑟₁+𝑠₁−2}, and notice that 𝑔m⁽⁰⁾(𝑟₁,x)

=m𝑟⁽⁰⁾₁ 𝑥₁+

𝑠₁−1

∑︁

𝑡=2

(m⁽⁰⁾_𝑡+𝑟

1−1−m⁽⁰⁾_𝑡+𝑟

1−2)𝑥_𝑡−m_𝑟⁽⁰⁾

1+𝑠₁−2

=m𝑟⁽⁰⁾₁ (𝑥₁−1) +

𝑠₁−1

∑︁

𝑡=2

(m⁽⁰⁾_𝑡+𝑟

1−1−m_𝑡+𝑟⁽⁰⁾

1−2) (𝑥_𝑡−1)

=−∑︁

𝑡∈𝑆₂

1, and similarly, 𝑔m⁽¹⁾(𝑟₁,x) =−∑︁

𝑡∈𝑆₂

𝑡 . (2.26)

Inserting (2.24) and (2.26) into (2.9), we have

−∑︁

𝑡∈𝑆₂

1+ ∑︁

𝑗∈𝑆^𝑐

1

1=0,

−∑︁

𝑡∈𝑆₂

𝑡+ ∑︁

𝑗∈𝑆^𝑐

1

𝑗 =0. This implies that the sets 𝑆^𝑐

1 and 𝑆₂ have the same cardinality and the same sum of elements. However, the maximum element in 𝑆₂ is smaller than the minimum element in𝑆^𝑐

1. Therefore,𝑆^𝑐

1and𝑆₂are empty, which implies thatxis the 0 vector andyis the all 1’s vector.

Case (d). (𝑦₁, 𝑦_𝑠

2) = (0,0). From Proposition 2.2.4 and Eq. (2.9) we have 𝑔m^(𝑒)(𝑟₁,x) −𝑔

m^(𝑒)(𝑟₂,y) = 0 for 𝑒 ∈ {0,1}, where x ≜ 1− x and y ≜ 1− y.

Since (𝑦

1, 𝑦

𝑠₂) = (1,1), from the previous casex and y are constant vectors, and

thus so arexandy. □

Proof. (of Proposition2.2.3) We distinguish between four cases according to the value of(𝑥_𝑠

1+𝑠₂+1, 𝑦_𝑠

2+𝑠₃+1).

Case (a). (𝑥_𝑠

1+𝑠₂+1, 𝑦_𝑠

2+𝑠₃+1) = (0,0). Similar to (2.25), we have that𝑔

m⁽⁰⁾(𝑟₁,x) + 𝑔m⁽⁰⁾(𝑟₂,y) ≥ 0, where equality holds only ifxandyare constant 0 vectors.

Case (b). (𝑥_𝑠

1+𝑠₂+1, 𝑦_𝑠

2+𝑠₃+1) = (1,1). From Proposition 2.2.4 and Eq. (2.10) we have that𝑔

m⁽⁰⁾(𝑟₁,x)+𝑔

m⁽⁰⁾(𝑟₂,y) =0. On the other hand, since (𝑥_𝑠

1+𝑠₂+1, 𝑦

𝑠₂+𝑠₃+1) = (0,0) , it follows that 𝑔

m⁽⁰⁾(𝑟₁,x) +𝑔

m⁽⁰⁾(𝑟₂,y) ≥ 0 where equality holds whenx andyare constant 1 vectors.

Case (c). (𝑥_𝑠

1+𝑠₂+1, 𝑦_𝑠

2+𝑠₃+1) = (0,1). On the one hand, for𝑦₁=0 we have 𝑔m⁽⁰⁾(𝑟₁,x) +𝑔

m⁽⁰⁾(𝑟₂,y)

=m𝑟⁽⁰⁾₁ 𝑥₁+

𝑠₁+1

∑︁

𝑡=2

(m_𝑡⁽⁰⁾_+𝑟

1−1−m_𝑡⁽⁰⁾_+𝑟

1−2)𝑥_𝑡+

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m𝑡⁽⁰⁾+𝑟₁ −m_𝑡+𝑟⁽⁰⁾

1−1)𝑥_𝑡+1 +

𝑠₂

∑︁

𝑡=2

(m_𝑡⁽₊⁰_𝑟⁾

2−1−m⁽_𝑡+⁰_𝑟⁾

2−2)𝑦_𝑡+

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m⁽⁰⁾_𝑡+𝑟

2−1−m_𝑡⁽⁰⁾_+𝑟

2−2)𝑦_𝑡−m⁽⁰⁾_𝑟

2+𝑠₂+𝑠₃−1

=m𝑟⁽⁰⁾₁ 𝑥₁+

𝑠₁+1

∑︁

𝑡=2

(m_𝑡+𝑟⁽⁰⁾

1−1−m_𝑡+𝑟⁽⁰⁾

1−2)𝑥_𝑡+

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m𝑡⁽⁰⁾+𝑟₁ −m_𝑡+𝑟⁽⁰⁾

1−1) (𝑥_𝑡+𝑥_𝑡+1) +

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m⁽⁰⁾_𝑡+𝑟

2−1−m⁽⁰⁾_𝑡+𝑟

2−2)𝑦_𝑡−m⁽⁰⁾

𝑟₂+𝑠₂+𝑠₃−1

≤m𝑟⁽⁰₁⁾+

𝑠₁+1

∑︁

𝑡=2

(m_𝑡⁽₊⁰⁾_𝑟

1−1−m_𝑡⁽₊⁰⁾_𝑟

1−2)+

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m_𝑡+𝑟⁽⁰⁾

1 −m_𝑡⁽₊⁰⁾_𝑟

1−1) +

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m⁽⁰⁾_𝑡+𝑟

2−1−m⁽⁰⁾_𝑡+𝑟

2−2) −m_𝑟⁽⁰⁾

2+𝑠₂+𝑠₃−1 =0, where equality equality holds when

𝑥_𝑡 =1 for𝑡 ∈ {1, . . . , 𝑠₁+1},

𝑥_𝑡+𝑥_𝑡+1 =1 for𝑡 ∈ {𝑠₁+1, . . . , 𝑠₁+𝑠₂−1}, and 𝑦_𝑡 =1 for𝑡 ∈ {𝑠₂+1, . . . , 𝑠₂+𝑠₃},

and hence (2.11) holds. On the other hand, when𝑦₁=1, let 𝑆₁={𝑡 :𝑥_{max{𝑡−𝑟}

1+1,1} =1,1≤ 𝑡 ≤ 𝑠₁+𝑟₁}, 𝑆₂={𝑡 :𝑥_𝑡−𝑟

1 +𝑥_𝑡−𝑟

1+1=0, 𝑟₂+1 ≤𝑡 ≤ 𝑟₂+𝑠₂−1}, 𝑆₃={𝑡 :𝑦_𝑡−𝑟

2+1 =0, 𝑟₂+𝑠₂≤ 𝑡 ≤𝑟₂+𝑠₂+𝑠₃−1}, and notice that

𝑔m⁽⁰⁾(𝑟₁,x) +𝑔

m⁽⁰⁾(𝑟₂,y)

=m𝑟⁽⁰⁾₁ 𝑥₁+

𝑠₁+1

∑︁

𝑡=2

(m_𝑡+𝑟⁽⁰⁾

1−1−m_𝑡+𝑟⁽⁰⁾

1−2)𝑥_𝑡+

m⁽⁰⁾𝑠₁+𝑟₁+

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m𝑡⁽⁰⁾+𝑟₁−m⁽⁰⁾_𝑡+𝑟

1−1) (𝑥_𝑡+𝑥_𝑡+1)+

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m_𝑡⁽₊⁰_𝑟⁾

2−1−m_𝑡⁽₊⁰_𝑟⁾

1−2)𝑦_𝑡−m⁽_𝑟⁰⁾

2+𝑠₂+𝑠₃−1

= ∑︁

𝑡∈𝑆₁

(m⁽⁰⁾𝑡 −m_𝑡⁽⁰⁾₋₁) −∑︁

𝑡∈𝑆₂

(m⁽⁰⁾𝑡 −m_𝑡⁽⁰⁾₋₁)−

∑︁

𝑡∈𝑆₃

(m⁽⁰⁾𝑡 −m_𝑡−1⁽⁰⁾)

=∑︁

𝑡∈𝑆₁

1−∑︁

𝑡∈𝑆₂

1−∑︁

𝑡∈𝑆₃

1. (2.27)

Similarly, we have that

𝑔m⁽¹⁾(𝑟₁,x) +𝑔

m⁽¹⁾(𝑟₂,y) =∑︁

𝑡∈𝑆₁

𝑡−∑︁

𝑡∈𝑆₂

𝑡−∑︁

𝑡∈𝑆₃

𝑡 . (2.28)

Equations (2.10), (2.27), and (2.28) imply that the cardinality of𝑆₁equals the sum of cardinalities of𝑆₂and𝑆₃, and in addition, the sum of elements of𝑆₁equals the sum of elements of𝑆₂and𝑆₃. Note that the minimum element of𝑆₂∪𝑆₃is larger than the maximum element of𝑆₁. This is impossible, unless𝑆₁, 𝑆₂, and𝑆₃are empty, which implies that𝑥_𝑡 =0 for𝑡 ∈ {1, . . . , 𝑠₁+1},𝑥_𝑡+𝑥_𝑡+1=1 for𝑡 ∈ {𝑠₁+1, . . . , 𝑠₁+𝑠₂−1}, and𝑦_𝑡 =1 for𝑡 ∈ {𝑠₂+1, . . . , 𝑠₂+𝑠₃}, and hence (2.11) holds.

Case (d). (𝑥_𝑠

1+𝑠₂+1, 𝑦_𝑠

2+𝑠₃+1) = (1,0). On the one hand, for𝑦₁=0, let 𝑆₁={𝑡 :𝑥_{max{𝑡−𝑟}

1+1,1} =0,1≤ 𝑡 ≤ 𝑠₁+𝑟₁}, 𝑆₂={𝑡 :𝑥_𝑡−𝑟

1 +𝑥_𝑡−𝑟

1+1=0, 𝑟₂+1 ≤𝑡 ≤ 𝑟₂+𝑠₂−1}, 𝑆₃={𝑡 :𝑦_𝑡−𝑟

2+1 =1, 𝑟₂+𝑠₂≤ 𝑡 ≤𝑟₂+𝑠₂+𝑠₃−1}. We have

𝑔m⁽⁰⁾(𝑟₁,x) +𝑔

m⁽⁰⁾(𝑟₂,y)

=m𝑟⁽⁰⁾₁ 𝑥₁+

𝑠₁+1

∑︁

𝑡=2

(m⁽⁰⁾_𝑡+𝑟

1−1−m⁽⁰⁾_𝑡+𝑟

1−2)𝑥_𝑡+

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m⁽⁰⁾_𝑡+𝑟1−m⁽⁰⁾_𝑡+𝑟

1−1) (𝑥_𝑡+𝑥_𝑡+1)−

m_𝑟⁽⁰⁾

1+𝑠₁+𝑠₂−1+

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m⁽_𝑡+⁰_𝑟⁾

2−1−m_𝑡⁽₊⁰_𝑟⁾

1−2)𝑦_𝑡

=−m𝑟⁽⁰⁾₁ (1−𝑥₁) −

𝑠₁+1

∑︁

𝑡=2

(m_𝑡+𝑟⁽⁰⁾

1−1−m_𝑡+𝑟⁽⁰⁾

1−2) (1−𝑥_𝑡)−

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m⁽⁰⁾_𝑡+𝑟1−m⁽⁰⁾_𝑡+𝑟

1−1) (1−𝑥_𝑡−𝑥_𝑡+1)+

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m_𝑡⁽₊⁰⁾_𝑟

2−1−m_𝑡⁽₊⁰⁾_𝑟

1−2)𝑦_𝑡

=−∑︁

𝑡∈𝑆₁

(m𝑡⁽⁰⁾ −m_𝑡−⁽⁰⁾₁) −∑︁

𝑡∈𝑆₂

(m⁽⁰⁾𝑡 −m_𝑡−⁽⁰⁾₁)+

∑︁

𝑡∈𝑆₃

(m_𝑡⁽⁰⁾−m⁽⁰⁾_𝑡−1)

=−∑︁

𝑡∈𝑆₁

1−∑︁

𝑡∈𝑆₂

1+∑︁

𝑡∈𝑆₃

1=0.

Then similar to the previous case, we obtain sets with identical cardinalities and sum of elements, and yet the smallest element in one is greater than the largest element in the others. Therefore, it follows that 𝑆₁, 𝑆₂, and 𝑆₃ are empty. Then we have 𝑥_𝑡 = 1 for𝑡 ∈ {1, . . . , 𝑠₁+1}, 𝑥_𝑡 +𝑥_𝑡+1 = 1 for𝑡 ∈ {𝑠₁+1, . . . , 𝑠₁+𝑠₂−1}, and 𝑦_𝑡 =0 for𝑡 ∈ {𝑠₂+1, . . . , 𝑠₂+𝑠₃}, and hence (2.11) holds.

On the other hand, for𝑦₁=1, let 𝑆₁={𝑡 :𝑥_{max{𝑡−𝑟}

1+1,1} =1,1≤ 𝑡 ≤ 𝑠₁+𝑟₁}, 𝑆₂={𝑡 :𝑥_𝑡−𝑟

1 +𝑥_𝑡−𝑟

1+1=0, 𝑟₂+1 ≤𝑡 ≤ 𝑟₂+𝑠₂−1}, 𝑆₃={𝑡 :𝑦_𝑡−𝑠

2+1=1, 𝑟₂+𝑠₂ ≤ 𝑡 ≤ 𝑟₂+𝑠₂+𝑠₃−1}. We have

𝑔m⁽⁰⁾(𝑟₁,x) +𝑔

m⁽⁰⁾(𝑟₂,y)

=m𝑟⁽⁰₁⁾𝑥₁+

𝑠₁+1

∑︁

𝑡=2

(m_𝑡⁽₊⁰⁾_𝑟

1−1−m_𝑡⁽₊⁰⁾_𝑟

1−2)𝑥_𝑡+m⁽𝑠⁰₁⁾+𝑟₁+

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m_𝑡+𝑟⁽⁰⁾

1 −m_𝑡⁽₊⁰⁾_𝑟

1−1) (𝑥_𝑡+𝑥_𝑡+1)−

m_𝑟⁽⁰⁾

1+𝑠₁+𝑠₂−1+

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m_𝑡+𝑟⁽⁰⁾

2−1−m_𝑡+𝑟⁽⁰⁾

1−2)𝑦_𝑡

=m𝑟⁽⁰⁾₁ 𝑥₁+

𝑠₁+1

∑︁

𝑡=2

(m_𝑡+𝑟⁽⁰⁾

1−1−m_𝑡+𝑟⁽⁰⁾

1−2)𝑥_𝑡−

𝑠₁+𝑠₂−1

∑︁

𝑡=𝑠₁+1

(m𝑡⁽⁰⁾+𝑟₁ −m_𝑡+𝑟⁽⁰⁾

1−1) (1−𝑥_𝑡−𝑥_𝑡+1)+

𝑠₂+𝑠₃

∑︁

𝑡=𝑠₂+1

(m⁽_𝑡+⁰_𝑟⁾

2−1−m_𝑡⁽₊⁰_𝑟⁾

1−2)𝑦_𝑡

=∑︁

𝑡∈𝑆₁

(m𝑡⁽⁰⁾ −m_𝑡⁽⁰⁾₋₁) −∑︁

𝑡∈𝑆₂

(m⁽⁰⁾𝑡 −m_𝑡⁽⁰⁾₋₁)+

∑︁

𝑡∈𝑆₃

(m𝑡⁽⁰⁾ −m_𝑡−1⁽⁰⁾)

=∑︁

𝑡∈𝑆₁

1−∑︁

𝑡∈𝑆₂

1+∑︁

𝑡∈𝑆₃

1=0. (2.29)

Similarly, we have

𝑔m⁽¹⁾(𝑟₁,x) +𝑔

m⁽¹⁾(𝑟₂,y) =∑︁

𝑡∈𝑆₁

𝑡−∑︁

𝑡∈𝑆₂

𝑡+∑︁

𝑡∈𝑆₃

𝑡 𝑔m⁽²⁾(𝑟₁,x) +𝑔

m⁽²⁾(𝑟₂,y) =∑︁

𝑡∈𝑆₁

𝑡²−∑︁

𝑡∈𝑆₂

𝑡²+∑︁

𝑡∈𝑆₃

𝑡². (2.30) According to (2.29) and (2.30), the following linear equation

𝐴x=0, (2.31)

where

𝐴=













 Í

𝑡∈𝑆₁1 Í

𝑡∈𝑆₂1 Í

𝑡∈𝑆₃1 Í

𝑡∈𝑆₁𝑡 Í

𝑡∈𝑆₂𝑡 Í

𝑡∈𝑆₃𝑡 Í

𝑡∈𝑆₁𝑡² Í

𝑡∈𝑆₂𝑡² Í

𝑡∈𝑆₃𝑡²













 , x=













 𝑥₁ 𝑥₂ 𝑥₃













 ,

has a nonzero solution (𝑥₁, 𝑥₂, 𝑥₃) = (1,−1,1)^⊤. We show that this is impossible unless 𝐴 =0. Suppose on the other hand, 𝐴 ≠ 0. If all columns of 𝐴are not zero columns, then according to the multi-linearity of the determinant,

det(𝐴) = ∑︁

𝑖∈𝑆₁, 𝑗∈𝑆₂, 𝑘∈𝑆₃

det©

­

­

«

1 1 1

𝑖 𝑗 𝑘

𝑖² 𝑗² 𝑘² ª

®

®

¬

= ∑︁

𝑖∈𝑆₁, 𝑗∈𝑆₂, 𝑘∈𝑆₃

(𝑗−𝑖) (𝑘−𝑖) (𝑘 − 𝑗) (2.32) is strictly positive since max𝑖∈𝑆₁𝑖 < min𝑗∈𝑆₂ 𝑗 < min𝑘∈𝑆₃𝑘. Hence the equation cannot have nonzero solutions. It is also obvious that the equation (2.31) cannot have solution(𝑥₁, 𝑥₂, 𝑥₃) = (1,−1,1)^⊤when only one column𝐴is non-zero. For the

case when 𝐴contains two non-zero columns, e.g., the first column and the second column, then we have that

𝐴^′x=

"

Í

𝑡∈𝑆₁1 Í

𝑡∈𝑆₂1 Í

𝑡∈𝑆₁𝑡 Í

𝑡∈𝑆₂𝑡

# "

𝑥₁ 𝑥₂

#

=0. (2.33)

Again, similar to Eq. (2.32), we have det(𝐴^′)= ∑︁

𝑖∈𝑆₁, 𝑗∈𝑆₂, 𝑘∈𝑆₃

det 1 1 𝑖 𝑗

!

= ∑︁

𝑖∈𝑆₁, 𝑗∈𝑆₂

(𝑗−𝑖) > 0, (2.34)

which implies that the equation Eq. (2.33) cannot have nonzero solutions. Thus, Eq. (2.31) cannot have solution (𝑥₁, 𝑥₂, 𝑥₃) = (1,−1,1)^⊤ unless 𝐴 = 0, which implies that 𝑆₁, 𝑆₂, and 𝑆₃ are empty. Therefore, 𝑥_𝑡 = 0 for 𝑡 ∈ {1, . . . , 𝑠₁+1}, 𝑥_𝑡+𝑥_𝑡+1=1 for𝑡 ∈ {𝑠₁+1, . . . , 𝑠₁+𝑠₂−1}, and𝑦_𝑡 =0 for𝑡 ∈ {𝑠₂+1, . . . , 𝑠₂+𝑠₃},

which implies (2.11). □