The t-wise intersection and trellis of relative four-weight codes

Abstract

Based on the applications of codes with few weights, we define the so-called relative four-weight codes and present a method for constructing such codes by using the finite projective geometry method. Also, the t-wise intersection and the trellis of relative four-weight codes are determined.

Appendix

Key Lemmas for Theorem 4.3

In order to prove all the major classes in Theorem 4.3, we introduce the first key lemma which will be used in the cases 3, 4, 5, 6 and 7 in which m₁ < m₂, m₂ > m₃ and m₃ < m₄. Then, the generator matrix G of C can be written in the following form (G,G₃,G₄) = (G₁,G₂) in which G₁ consists of all points in PG(k − 1, 2) with each point repeating m₁ times and all the points in S₂ ∪ S₃ ∪ S₄, constitute the columns of G₂ with each point repeating m₂ − m₁ times. Columns of the generator matrix G₃ consist of all points of S₃ and each point repeats m₂ − m₃ times and columns of the generator matrix G₄ consist of all points of S₄ and each point repeats m₄ − m₃ times. Then, G₁ generates a k-dimensional constant- weight code $C^{\prime }$ with weight m₁2^k− 1 and length l₁ = m₁2^k − 1, G₂ generates a (k − k₁)-dimensional weight code $C^{\prime \prime }$ with weight $(m_{2}-m_{1})2^{k-k_{1}-1}$ and length $l_{2}=(m_{2}-m_{1})2^{k-k_{1}}-1$, G₃ generates a (k − k₂)-dimensional constant-weight code $C^{\prime \prime \prime }$ with weight $(m_{2}-m_{3})2^{k-k_{2}-1}$ and length $l_{3}=(m_{2}-m_{3})2^{k-k_{2}}-1$ and G₄ generates a (k − k₃)-dimensional constant-weight code $C^{\prime \prime \prime \prime }$ with weight $(m_{4}-m_{3})2^{k-k_{3}-1}$ and length $l_{4}=(m_{4}-m_{3})2^{k-k_{3}}-1$. Let c₁,⋯ ,c_t be a relative (t,t₁,t₂,t₃)(t₃ < t₂ < t₁ < t) subcode.

Denote [ c₁ c₂ ⋮ c_t ] = X_t×kG, [ c1′c2′ ⋮ ct′ ] = X_t×kG₁, [ c1″c2″ ⋮ ct″ ] = X_t×kG₂, [ c1″′c2″′ ⋮ ct″′ ] = X_t×kG₃, [ c1″″c2″″ ⋮ ct″″ ] = X_t×kG₄.

Then, we have for any i ∈{1, 2,⋯ ,t}, $(c_{i},c^{\prime \prime \prime }_{i},c^{\prime \prime \prime \prime }_{i})=(c^{\prime }_{i},c^{\prime \prime }_{i})$, where $c^{\prime }_{i}\in C^{\prime }$, $c^{\prime \prime }_{i}\in C^{\prime \prime }$, $c^{\prime \prime \prime }_{i}\in C^{\prime \prime \prime }$ and $c^{\prime \prime \prime \prime }_{i}\in C^{\prime \prime \prime \prime }.$ In addition that, $c^{\prime }_{1},\cdots ,c^{\prime }_{t}$ are linearly independent codewords, whereas $rank(c^{\prime \prime }_{1},\cdots , c^{\prime \prime }_{t}) =t-t_{1}$, $rank(c^{\prime \prime \prime }_{1},\cdots , c^{\prime \prime \prime }_{t})=t-t_{2}$ and $rank(c^{\prime \prime \prime \prime }_{1},\cdots , c^{\prime \prime \prime \prime }_{t})=t-t_{3}$ by Lemma 4.2. For satisfication, inter, inter₁, inter₂, inter₃ and inter₄ will be represented as follows. $inter=|\bigcap ^{t}_{i=1}\chi (c_{i})|$, $inter_{1}=|\bigcap ^{t}_{i=1}\chi (c^{\prime }_{i})|$, $inter_{2}=|\bigcap ^{t}_{i=1}\chi (c^{\prime \prime }_{i})|$, $inter_{3}=|\bigcap ^{t}_{i=1}\chi (c^{\prime \prime \prime }_{i})|$ and $inter_{4}=|\bigcap ^{t}_{i=1}\chi (c^{\prime \prime \prime \prime }_{i})|$. Based on the Lemma 4.1, we have inter = inter₁ + inter₂ − inter₃ − inter₄ with $inter_{1}=(\frac {1}{2})^{t-1}m_{1}2^{k-1}$, $0\leq inter_{2}\leq (\frac {1}{2})^{t-t_{1}-1}(m_{2}-m_{1})2^{k-k_{1}-1}$, $0\leq inter_{3}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{2}-m_{3})2^{k-k_{2}-1}$ and $0\leq inter_{4}\leq (\frac {1}{2})^{t-t_{3}-1}(m_{4}-m_{3})2^{k-k_{3}-1}$, then we have the following lemma.

Lemma A.1

Assume q = 2 and $w_{3}>\max \limits \{w_{1},w_{2},w_{4}\}$ and let D =< c₁,⋯ ,c_t > be a relative (t,t₁,t₂,t₃)(t₃ < t₂ < t₁ < t) subcode of C with inter₄≠ 0, then $inter_{3}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}$.

Proof

Write [ c₁ c₂ ⋮ c_t ] = X_t×kG, then similar to the proof of the Lemma 4.2, there exists an invertible matrix Y_t×t such that

$$ \begin{array}{@{}rcl@{}} Y_{t\times t}X_{t\times k}&=& \begin{bmatrix} X^{\prime}_{t_{1}\times k_{1}} & 0_{t_{1}\times (k_{2}-k_{1})} & 0_{t_{1}\times (k_{3}-k_{2})} & 0_{t_{1}\times (k-k_{3})} \\ X^{\prime\prime}_{(t_{2}-t_{1})\times k_{1}} & X^{\prime\prime}_{(t_{2}-t_{1})\times (k_{2}-k_{1})} & 0_{(t_{2}-t_{1})\times (k_{3}-k_{2})} & 0_{(t_{2}-t_{1})\times (k-k_{3})} \\ X^{\prime\prime}_{(t_{3}-t_{2})\times k_{1}} & X^{\prime\prime}_{(t_{3}-t_{2})\times (k_{2}-k_{1})} & X^{\prime\prime}_{(t_{3}-t_{2})\times (k_{3}-k_{2})} & 0_{(t_{3}-t_{2})\times (k-k_{3})}\\ X^{\prime\prime\prime}_{(t-t_{3})\times k_{1}} & X^{\prime\prime\prime}_{(t-t_{3})\times (k_{2}-k_{1})} & X^{\prime\prime\prime}_{(t-t_{3})\times (k_{3}-k_{2})} & X^{\prime\prime\prime}_{(t-t_{3})\times (k-k_{3})} \end{bmatrix}, \end{array} $$

with $rank(X^{\prime }_{t_{{1}\times k_{1}}})=t_{1}$, $rank(X^{\prime \prime }_{(t_{2}-t_{1})\times (k_{2}-k_{1})})=t_{2}-t_{1}$, $rank(X^{\prime \prime }_{(t_{3}-t_{2})\times (k_{3}-k_{2})})=t_{3}-t_{2}$ and $rank(X^{\prime \prime \prime }_{(t-t_{3})\times (k-k_{3})})=t-t_{3}$.

Thus,

$$ Y_{t\times t}\left( \begin{array}{c|c} c^{\prime\prime\prime}_{1} & c^{\prime\prime\prime\prime}_{1}\\ \vdots&\vdots\\ c^{\prime\prime\prime}_{t} & c^{\prime\prime\prime\prime}_{t} \end{array}\right) =\left( \begin{array}{c|c} 0 & 0\\ \vdots&\vdots\\ 0 & 0\\ \overline{c}^{\prime\prime\prime}_{t_{2}+1} & 0\\ \vdots&\vdots\\ \overline{c}^{\prime\prime\prime}_{t_{3}} & 0\\ \overline{c}^{\prime\prime\prime}_{t_{3}+1} & \overline{c}^{\prime\prime\prime\prime}_{t_{3}+1}\\ \vdots&\vdots\\ \overline{c}^{\prime\prime\prime}_{t} & \overline{c}^{\prime\prime\prime\prime}_{t} \end{array}\right), $$

(A.1)

in which $rank(\overline {c}^{\prime \prime \prime }_{t_{2}+1},\cdots , \overline {c}^{\prime \prime \prime }_{t})=t-t_{2}$ and $rank(\overline {c}^{\prime \prime \prime \prime }_{t_{3}+1},\cdots , \overline {c}^{\prime \prime \prime \prime }_{t})=t-t_{3}$. We have $rank\left (\begin {array}{c|c} c^{\prime \prime \prime }_{1} & c^{\prime \prime \prime \prime }_{1}\\ {\vdots } &\vdots \\ c^{\prime \prime \prime }_{t} & c^{\prime \prime \prime \prime }_{t} \end {array}\right )=t-t_{2}$. Without loss of generality, let $(c^{\prime \prime \prime }_{t_{2}+1},\cdots ,c^{\prime \prime \prime }_{t})$ be a maximal linearly independent set of $(c^{\prime \prime \prime }_{{1}},\cdots ,c^{\prime \prime \prime }_{t})$. Then, the last (t − t₂) rows of the matrix $\left (\begin {array}{c|c} c^{\prime \prime \prime }_{1} & c^{\prime \prime \prime \prime }_{1}\\ \vdots &\vdots \\ c^{\prime \prime \prime }_{t} & c^{\prime \prime \prime \prime }_{t} \end {array}\right )$, which is $\left (\begin {array}{c|c} c^{\prime \prime \prime }_{t_{2}+1} & c^{\prime \prime \prime \prime }_{t_{2}+1}\\ {\vdots } &\vdots \\ c^{\prime \prime \prime }_{t} & c^{\prime \prime \prime \prime }_{t} \end {array}\right )$, is a maximal linearly independent set of its all rows. So, there exists a matrix $\begin {pmatrix} p_{1\times (t_{2}+1)}& {\cdots } &p_{1\times t}\\ {\vdots } & {\ddots } &\vdots \\ p_{t_{2}\times (t_{2}+1)}& \cdots & p_{t_{2}\times t}\\ \end {pmatrix}$such that $\left (\begin {array}{c|c} c^{\prime \prime \prime }_{1} & c^{\prime \prime \prime \prime }_{1}\\ {\vdots } &\vdots \\ c^{\prime \prime \prime }_{t} & c^{\prime \prime \prime \prime }_{t} \end {array}\right )=\begin {pmatrix} p_{1\times (t_{2}+1)}& {\cdots } &p_{1\times t}\\ {\vdots } & {\ddots } &\vdots \\ p_{t_{2}\times (t_{2}+1)}& {\cdots } & p_{t_{2}\times t}\\ \end {pmatrix} \left (\begin {array}{c|c} c^{\prime \prime \prime }_{t_{2}+1} & c^{\prime \prime \prime \prime }_{t_{2}+1}\\ {\vdots } &\vdots \\ c^{\prime \prime \prime }_{t} & c^{\prime \prime \prime \prime }_{t} \end {array}\right )$, that is

$$ \begin{pmatrix} c^{\prime\prime\prime}_{1}\\ \vdots\\ c^{\prime\prime\prime}_{t_{2}}\\ \end{pmatrix}=\begin{pmatrix} p_{1\times (t_{2}+1)}& {\cdots} &p_{1\times t}\\ {\vdots} & {\ddots} &\vdots\\ p_{t_{2}\times (t_{2}+1)}& {\cdots} & p_{t_{2}\times t}\\ \end{pmatrix} \left( \begin{array}{c} c^{\prime\prime\prime}_{t_{2}+1} \\ \vdots\\ c^{\prime\prime\prime}_{t} \end{array}\right) $$

(A.2)

and

$$ \begin{pmatrix} c^{\prime\prime\prime\prime}_{1}\\ \vdots\\ c^{\prime\prime\prime\prime}_{t_{2}}\\ \end{pmatrix}=\begin{pmatrix} p_{1\times (t_{2}+1)}& {\cdots} &p_{1\times t}\\ {\vdots} & {\ddots} &\vdots\\ p_{t_{2}\times (t_{2}+1)}& {\cdots} & p_{t_{2}\times t}\\ \end{pmatrix} \left( \begin{array}{c} c^{\prime\prime\prime\prime}_{t_{2}+1} \\ \vdots\\ c^{\prime\prime\prime\prime}_{t} \end{array}\right). $$

(A.3)

Based on (A.3) and $rank(c^{\prime \prime \prime \prime }_{1}, \cdots , c^{\prime \prime \prime \prime }_{t_{1}})=t-t_{3}$, without loss of generality, we assume $(c^{\prime \prime \prime \prime }_{t_{3}+1},\cdots , c^{\prime \prime \prime \prime }_{t})$ to be a maximal linearly independent set of $(c^{\prime \prime \prime \prime }_{1},\cdots ,c^{\prime \prime \prime \prime }_{t}).$ Then, there exists a matrix $\begin {pmatrix} r_{(t_{2}+1)\times (t_{3}+1)}& {\cdots } &r_{(t_{2}+1)\times t}\\ {\vdots } & {\ddots } &\vdots \\ r_{t_{3}\times (t_{3}+1)}& {\cdots } & r_{t_{3}\times t}\\ \end {pmatrix}$, such that

$$ \begin{pmatrix} c^{\prime\prime\prime\prime}_{t_{2}+1}\\ \vdots\\ c^{\prime\prime\prime\prime}_{t_{3}}\\ \end{pmatrix} =\begin{pmatrix} r_{(t_{2}+1)\times (t_{3}+1)}& {\cdots} &r_{(t_{2}+1)\times t}\\ {\vdots} & {\ddots} &\vdots\\ r_{t_{3}\times (t_{3}+1)}& {\cdots} & r_{t_{3}\times t}\\ \end{pmatrix}\begin{pmatrix} c^{\prime\prime\prime\prime}_{t_{3}+1}\\ \vdots\\ c^{\prime\prime\prime\prime}_{t}\\ \end{pmatrix}. $$

(A.4)

Substitute the (A.4) in (A.3), we get

$$ \begin{pmatrix} c^{\prime\prime\prime\prime}_{1}\\ \vdots\\ c^{\prime\prime\prime\prime}_{t_{2}}\\ \end{pmatrix}= \begin{pmatrix} p_{1\times (t_{2}+1)}& {\cdots} &p_{1\times t}\\ {\vdots} & {\ddots} &\vdots\\ p_{t_{2}\times (t_{2}+1)}& {\cdots} & p_{t_{2}\times t}\\ \end{pmatrix} \begin{pmatrix} r_{(t_{2}+1)\times (t_{3}+1)}& {\cdots} &r_{(t_{2}+1)\times t}\\ {\vdots} & {\ddots} &\vdots\\ r_{t_{3}\times (t_{3}+1)}& {\cdots} & r_{t_{3}\times t}\\ 1& {\cdots} &0\\ {\vdots} & {\ddots} &\vdots\\ 0& {\cdots} & 1\\ \end{pmatrix} \begin{pmatrix} c^{\prime\prime\prime\prime}_{t_{3}+1}\\ \vdots\\ c^{\prime\prime\prime\prime}_{t}\\ \end{pmatrix}. $$

(A.5)

Since inter₄≠ 0, there must be a co-ordinate position j₀ ∈{1, 2,⋯ ,l₄} such that $j_{0}\in \chi (c^{\prime \prime \prime \prime }_{i})$, ∀1 ≤ i ≤ t.

Then, (A.4) implies that $\begin {pmatrix} r_{(t_{2}+1)\times (t_{3}+1)}& {\cdots } &r_{(t_{2}+1)\times t}\\ {\vdots } & {\ddots } &\vdots \\ r_{t_{3}\times (t_{3}+1)}& {\cdots } & r_{t_{3}\times t}\\ \end {pmatrix}\begin {pmatrix} 1\\ \vdots \\ 1\\ \end {pmatrix}=\begin {pmatrix} 1\\ \vdots \\ 1\\ \end {pmatrix}$ and (A.5) implies $\begin {pmatrix} p_{1\times (t_{2}+1)}& {\cdots } &p_{1\times t}\\ {\vdots } & {\ddots } &\vdots \\ p_{t_{2}\times (t_{2}+1)}& {\cdots } & p_{t_{2}\times t}\\ \end {pmatrix} \begin {pmatrix} r_{(t_{2}+1)\times (t_{3}+1)}& {\cdots } &r_{(t_{2}+1)\times t}\\ {\vdots } & {\ddots } &\vdots \\ r_{t_{3}\times (t_{3}+1)}& {\cdots } & r_{t_{3}\times t}\\ 1& {\cdots } &0\\ {\vdots } & {\ddots } &\vdots \\ 0& {\cdots } & 1\\ \end {pmatrix} \begin {pmatrix} 1\\ \vdots \\ 1\\ \end {pmatrix}=\begin {pmatrix} 1\\ \vdots \\ 1\\ \end {pmatrix}$.

Thus,

$$ \begin{pmatrix} p_{1\times (t_{2}+1)}& {\cdots} &p_{1\times t}\\ {\vdots} & {\ddots} &\vdots\\ p_{t_{2}\times (t_{2}+1)}& {\cdots} & p_{t_{2}\times t}\\ \end{pmatrix} \begin{pmatrix} 1\\ \vdots\\ 1\\ \end{pmatrix}=\begin{pmatrix} 1\\ \vdots\\ 1\\ \end{pmatrix}. $$

(A.6)

We set $\cap ^{t}_{i=t_{2}+1}\chi (c^{\prime \prime \prime }_{i})=\{j_{1},j_{2},\cdots ,j_{r}\}$ and let $\begin {pmatrix} 1& {\cdots } &1\\ {\vdots } & {\ddots } &\vdots \\ 1& {\cdots } & 1\\ \end {pmatrix}$bethe matrix which consists of the $\mathrm {j}_{1}^{\text {th}},~\mathrm {j}_{2}^{\text {th}},~\cdots ,~\mathrm {j}_{r}^{\text {th}}$ columns of the matrix $\left (\begin {array}{c} c^{\prime \prime \prime }_{t_{2}+1} \\ \vdots \\ c^{\prime \prime \prime }_{t} \end {array}\right )$. From (A.2) and (A.6), we have $\begin {pmatrix}\label {f} p_{1\times (t_{2}+1)}& {\cdots } &p_{1\times t}\\ {\vdots } & {\ddots } &\vdots \\ p_{t_{2}\times (t_{2}+1)}& {\cdots } & p_{t_{2}\times t}\\ \end {pmatrix} \begin {pmatrix} 1& {\cdots } &1\\ {\vdots } & {\ddots } &\vdots \\ 1& {\cdots } & 1\\ \end {pmatrix}=\begin {pmatrix} 1& {\cdots } &1\\ {\vdots } & {\ddots } &\vdots \\ 1& {\cdots } & 1\\ \end {pmatrix}$, which gives $\bigcap ^{t}_{i=1}\chi (c_{i}^{\prime \prime \prime })=\bigcap ^{t}_{i=t_{2}+1}\chi (c_{i}^{\prime \prime \prime })$. When $inter_{3}=|\bigcap ^{t}_{i=t_{2}+1}\chi (c_{i}^{\prime \prime \prime })|$, we will get $inter_{3}=(\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}.$

Since $(c^{\prime \prime }_{t_{2}+1},\cdots ,c^{\prime \prime }_{t})$ are t − t₂ linearly independent codewords of constant-weight $c^{\prime \prime \prime }$ with weight $(m_{3}-m_{2})2^{k-k_{2}-1}$, using Lemma 4.1, it will follow that $|\bigcap ^{t}_{i=t_{2}+1}\chi (c^{\prime \prime \prime }_{i})|=(\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}.$ Thus, $inter_{3}=(\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}.$□

We introduce the second key lemma in the cases 8, 9 and 10 in which m₁ < m₂, m₂ < m₃ and m₃ > m₄ holds. Their generator matrix G of the code C can be written as (G,G₃,G₄) = (G₁,G₂) in which G₁ consists of all points in PG(k − 1, 2) with each point repeating m₁ times and all the points in S₂ ∪ S₃ ∪ S₄ constitute the columns of G₂ with each point repeating m₂ − m₁ times. Columns of the generator matrix G₃ consist of all points of S₃ and each point repeats m₃ − m₂ times and columns of the generator matrix G₄ consist of all points of S₄ and each point repeats m₃ − m₄ times. Then, G₁ generates a k-dimensional constant-weight code $C^{\prime }$ with weight m₁2^k− 1 and length l₁ = m₁(2^k − 1), G₂ generates a (k − k₁)-dimensional weight code $C^{\prime \prime }$ with weight $(m_{2}-m_{1})2^{k-k_{1}-1}$ and length $l_{2}=(m_{2}-m_{1})(2^{k-k_{1}}-1)$, G₃ generates a (k − k₂)-dimensional weight code $C^{\prime \prime \prime }$ with weight $(m_{3}-m_{2})2^{k-k_{2}-1}$ and length $l_{3}=(m_{3}-m_{2})(2^{k-k_{2}}-1)$ and G₄ generates a (k − k₃)-dimensional weight code $C^{\prime \prime \prime \prime }$ with weight $(m_{3}-m_{4})2^{k-k_{3}-1}$ and length $l_{4}=(m_{3}-m_{4})(2^{k-k_{3}}-1)$. Assume that c₁,⋯ ,c_t with the matrix form $\begin {pmatrix} c_{1}\\ \vdots \\ c_{t}\\ \end {pmatrix}=X_{t\times k}G$, are any t linearly independent codewords in C. Obviously, for any i ∈{1, 2,⋯ ,t}, we have $(c_{i},c_{i}^{\prime \prime \prime },c_{i}^{\prime \prime \prime \prime })=(c_{i}^{\prime },c_{i}^{\prime \prime })$, where $c_{i}^{\prime }\in C^{\prime }$, $c_{i}^{\prime \prime }\in C^{\prime \prime }$, $c_{i}^{\prime \prime \prime }\in C^{\prime \prime \prime }$ and $c_{i}^{\prime \prime \prime \prime }\in C^{\prime \prime \prime \prime }$. Additionally $rank(c_{1}^{\prime },\cdots ,c_{t}^{\prime })=t$, based on Lemma 4.2, we have $rank(c_{1}^{\prime \prime },\cdots ,c_{t}^{\prime \prime })=t-t_{1}$, $rank(c_{1}^{\prime \prime \prime },\cdots ,c_{t}^{\prime \prime \prime })=t-t_{2}$ and $rank(c_{1}^{\prime \prime \prime \prime },\cdots ,c_{t}^{\prime \prime \prime \prime })=t-t_{3}$. Furthermore, inter = inter₁ + inter₂ − inter₃ − inter₄ with $inter_{1}=(\frac {1}{2})^{t-1}m_{1}2^{k-1}$, $0\leq inter_{2}\leq (\frac {1}{2})^{t-t_{1}-1}(m_{2}-m_{1})2^{k-k_{1}-1}$, $0\leq inter_{3}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}$ and $0\leq inter_{4}\leq (\frac {1}{2})^{t-t_{3}-1}(m_{3}-m_{4})2^{k-k_{3}-1}$ by Lemma 4.1.

We introduce the third key lemma in cases 11, 12, 13, 14 and 15 in which we deduce that w₁ is greater than w₂, w₂ is less than w₃ and w₃ is greater than w₄ yields m₁ > m₂, m₂ < m₃ and m₃ > m₄. Then, the generator matrix G of C can be written in the following form (G,G₁) = (G₂,G₃,G₄) in which the block matrix G₁ consists of all points in S₂ ∪ S₃ ∪ S₄ and constitute the columns of G₁ with each point repeating m₁ − m₂ times. G₂ consists of all points in PG(k − 1, 2) and each point appears m₁ times. All points in S₃ constitute columns of G₃ and each point occurs m₃ − m₂ times and all the points in S₄ constitute columns of G₄ and each point occurs m₃ − m₄ times. Thus, G₁ generates a (k − k₁)-dimensional constant-weight code $C^{\prime }$ with weight $(m_{1}-m_{2})2^{k-k_{1}-1}$ and length $l_{1}=(m_{1}-m_{2})(2^{k-k_{1}}-1)$. G₂ generates a k- dimensional constant-weight code $C^{\prime \prime }$ with weight m₁2^k− 1 and length l₂ = m₁(2^k − 1). G₃ generates a (k − k₂)-dimensional constant-weight code $C^{\prime \prime \prime }$ with weight $(m_{3}-m_{2})2^{k-k_{2}-1}$ and length $l_{3}=(m_{3}-m_{2})(2^{k-k_{2}}-1)$ and G₄ generates (k − k₃)-dimensional constant-weight $(m_{3}-m_{4})2^{k-k_{3}-1}$ and length $l_{4}=(m_{3}-m_{4})(2^{k-k_{3}}-1).$ Assume that, c₁,⋯ ,c_t are any t linearly independent codewords with the matrix form $\begin {pmatrix} c_{1}\\ \vdots \\ c_{t}\\ \end {pmatrix}=X_{t\times k}G$, $\begin {pmatrix} c_{1}^{\prime }\\ \vdots \\ c_{t}^{\prime }\\ \end {pmatrix}=X_{t\times k}G_{1}$, $\begin {pmatrix} c_{1}^{\prime \prime }\\ \vdots \\ c_{t}^{\prime \prime }\\ \end {pmatrix}=X_{t\times k}G_{2}$, $\begin {pmatrix} c_{1}^{\prime \prime \prime }\\ \vdots \\ c_{t}^{\prime \prime \prime }\\ \end {pmatrix}=X_{t\times k}G_{3}$ and $\begin {pmatrix} c_{1}^{\prime \prime \prime \prime }\\ \vdots \\ c_{t}^{\prime \prime \prime \prime }\\ \end {pmatrix}=X_{t\times k}G_{4}$. Obviously, for any i ∈{1, 2,⋯ ,t}, we have $(c_{i},c_{i}^{\prime })=(c_{i}^{\prime \prime },c_{i}^{\prime \prime \prime },c_{i}^{\prime \prime \prime \prime })$, where $c_{i}^{\prime }\in C^{\prime }$, $c_{i}^{\prime \prime }\in C^{\prime \prime }$, $c_{i}^{\prime \prime \prime }\in C^{\prime \prime \prime }$ and $c_{i}^{\prime \prime \prime \prime }\in C^{\prime \prime \prime \prime }$. Additionally, $rank(c_{1}^{\prime \prime },\cdots ,c_{t}^{\prime \prime })=t$. From Lemma 4.2, we have $rank(c_{1}^{\prime },\cdots ,c_{t}^{\prime })=(t-t_{1})$, $rank(c_{1}^{\prime \prime },\cdots ,c_{t}^{\prime \prime })=t-t_{2}$, $rank(c_{1}^{\prime \prime \prime },\cdots ,c_{t}^{\prime \prime \prime })=t-t_{3}$ and $rank(c_{1}^{\prime \prime \prime \prime },\cdots ,c_{t}^{\prime \prime \prime \prime })=t-t_{3}.$ Therefore, we have inter = inter₂ + inter₃ + inter₄ − inter₁ with $inter_{2}=(\frac {1}{2})^{t-1}m_{1}2^{k-1}$, $0\leq inter_{1}\leq (\frac {1}{2})^{t-t_{1}-1}(m_{1}-m_{2})2^{k-k_{1}-1}$, $0\leq inter_{3}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}$ and $0\leq inter_{4}\leq (\frac {1}{2})^{t-t_{3}-1}(m_{3}-m_{4})2^{k-k_{3}-1}$ by Lemma 4.1.

We apply the fourth key lemma in the cases 16, 17 and 18 in which m₁ > m₂, m₂ < m₃ and m₃ < m₄. Then, the generator matrix G can be written in the following form (G,G₁,G₄) = (G₂,G₃) in which the block matrix G₁ includes all the points in S₂ ∪ S₃ ∪ S₄, which constitute the columns of G₁ with each point repeating m₁ − m₂ times. Columns of the generator matrix G₂ consist of all points in PG(k − 1, 2) and each point appears m₁ times and all points in S₃ constitute columns of G₃ and each point occurs m₃ − m₂ times, columns of the generator matrix G₄ consist of all points of S₄ and each point repeats m₄ − m₃ times. Then, G₁ generates a (k − k₁)-dimensional constant-weight code $C^{\prime }$ with weight $(m_{1}-m_{2})2^{k-k_{1}-1}$ and length $l_{1}=(m_{1}-m_{2})(2^{k-k_{1}}-1)$. G₂ generates a k-dimensional constant-weight code $C^{\prime \prime }$ with weight m₁2^k− 1 and length l₂ = m₁(2^k − 1), G₃ generates (k − k₃)-dimensional constant-weight code $C^{\prime \prime \prime }$ with weight $(m_{3}-m_{2})2^{k-k_{2}-1}$ and length $l_{3}=(m_{3}-m_{2})(2^{k-k_{2}}-1)$ and G₄ generates (k − k₄)-dimensional constant-weight $(m_{4}-m_{3})2^{k-k_{3}-1}$ and length $l_{4}=(m_{4}-m_{3})(2^{k-k_{3}}-1).$ Using the same procedure as above, we get the intersection inter = inter₂ + inter₃ − inter₁ − inter₄ with $inter_{2}=(\frac {1}{2})^{t-1}m_{1}2^{k-1}$, $0\leq inter_{1}\leq (\frac {1}{2})^{t-t_{1}-1}(m_{1}-m_{2})2^{k-k_{1}-1}$, $0\leq inter_{3}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{3}-m_{2})2^{k-k_{2}-1}$ and $0\leq inter_{4}\leq (\frac {1}{2})^{t-t_{3}-1}(m_{4}-m_{3})2^{k-k_{3}-1}$.

Next, we apply the fifth key lemma in the cases 19, 20 and 21 in which m₁ > m₂, m₂ > m₃ and m₃ < m₄. Then, the generator matrix G of C can be written in the following form (G,G₁,G₂) = (G₃,G₄) in which the block matrix G₁ consists of all points in S₂ ∪ S₃ ∪ S₄, constitute the columns of G₁ with each point repeating m₁ − m₂ times. Columns of the generator matrix G₂ consist of all points of S₃ and each point repeats m₂ − m₃ times and columns of the generator matrix G₄ consist of all points of S₄ and each point repeats m₂ − m₃ times. G₃ consists of all points in PG(k − 1, 2) and each point appears m₁ times and all the points in S₄, constitute columns of G₄ and each point occurs m₃ − m₄ times. Then, G₁ generates a (k − k₁)-dimensional constant-weight code $C^{\prime }$ with weight $(m_{1}-m_{2})2^{k-k_{1}-1}$ and length $l_{1}=(m_{1}-m_{2})(2^{k-k_{1}}-1)$. G₂ generates (k − k₂)-dimensional constant-weight code $C^{\prime \prime }$ with weight $(m_{2}-m_{3})2^{k-k_{2}-1}$ and length $l_{2}=(m_{2}-m_{3})(2^{k-k_{2}}-1)$, G₃ generates a k-dimensional constant-weight code $C^{\prime \prime \prime }$ with weight m₁2^k− 1 and length l₃ = m₁(2^k − 1) and G₄ generates (k − k₃)-dimensional constant-weight $(m_{4}-m_{3})2^{k-k_{3}-1}$ and length $l_{4}=(m_{4}-m_{3})(2^{k-k_{3}}-1)$. Similar to this, using the above procedure, we have inter = inter₃ + inter₄ − inter₁ − inter₂ with $inter_{3}=(\frac {1}{2})^{t-1}m_{1}2^{k-1}$, $0\leq inter_{1}\leq (\frac {1}{2})^{t-t_{1}-1}(m_{1}-m_{2})2^{k-k_{1}-1}$, $0\leq inter_{2}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{2}-m_{3})2^{k-k_{2}-1}$ and $0\leq inter_{4}\leq (\frac {1}{2})^{t-t_{3}-1}(m_{4}-m_{3})2^{k-k_{3}-1}$ by Lemma 4.1.

Again, we use the sixth key lemma in the cases 22, 23 and 24 in which m₁ < m₂, m₂ > m₃ and m₃ > m₄. Then, the generator matrix G of C can be written in the following form (G,G₂,G₃,G₄) = (G₁) in which the block matrix G₁ consists of all points in PG(k − 1, 2) and each point appears m₁ times. Columns of the generator matrix G₂ consist of all points of S₂ ∪ S₃ ∪ S₄ and each point repeats m₂ − m₁ times. All points in S₃ constitute columns of G₃ and each point occurs m₂ − m₃ times , G₄ consists of all points in S₄ and each point appears m₃ − m₄ times. Thus, G₁ generates a k-dimensional constant-weight code $C^{\prime }$ with weight m₁2^k− 1 and length l₁ = m₁(2^k − 1), G₂ generates (k − k₁)-dimensional constant-weight code $C^{\prime \prime }$ with weight $(m_{2}-m_{1})2^{k-k_{1}-1}$ and length $l_{2}=(m_{2}-m_{1})(2^{k-k_{1}}-1)$, G₃ generates (k − k₂)-dimensional constant-weight code $C^{\prime \prime \prime }$ with weight $(m_{2}-m_{3})(2^{k-k_{2}-1})$ and length $l_{3}=(m_{2}-m_{3})2^{k-k_{2}}-1$ and G₄ generates a k-dimensional constant-weight code $C^{\prime \prime \prime \prime }$ with weight $(m_{3}-m_{4})2^{k-k_{3}-1}$ and length $l_{4}=(m_{3}-m_{4})(2^{k-k_{3}}-1).$ Likewise, using the same procedure as above, we have inter = inter₁ − inter₂ − inter₃ − inter₄ with $inter_{1}=(\frac {1}{2})^{t-1}m_{1}2^{k-1}$, $0\leq inter_{2}\leq (\frac {1}{2})^{t-t_{1}-1}(m_{2}-m_{1})2^{k-k_{1}-1}$, $0\leq inter_{3}\leq (\frac {1}{2})^{t-t_{2}-1}(m_{2}-m_{3})2^{k-k_{2}-1}$ and $0\leq inter_{4}\leq (\frac {1}{2})^{t-t_{3}-1}(m_{3}-m_{4})2^{k-k_{3}-1}$ by Lemma 4.1.