Let $v$ be an even, make a token for each element of $V$ as follows $P_{+}=\left\{0,1,\cdots ,\frac{v}{2}-1\right\}$, $P_{-}=\left\{\frac{v}{2},\frac{v}{2}+1,\cdots ,v-1\right\}$, $p_{+}=\left|P_{+}\right|$, $p_{-}=\left|P_{-}\right|$ then $v=2p_{+}=2p_{-}$, $r_{i,j}$ denotes the element of Latin rectangle, $i$ is the row indicator $i\in P_{+}$ and $j$ is the column indicator $j\in P_{-}$. If the element of a Latin rectangle is equal to its row or column indicator, such a point is called a fixed point.

The Latin rectangle constructed in this paper cannot contain fixed points, if they contain fixed points that the elements in the corresponding set have the same contradict the definition of the set, two Latin rectangles without fixed points if they satisfy the definition of orthogonal Latin rectangle, the orthogonal Latin rectangle formed also do not contain fixed points. Latin rectangle without fixed points can form a Steiner system with parameters $\left(2,3,v\right)$ and also form an antichain. Their collection forms an antichain code, which results in a combinatorial neural code with special parameters.

The following shows the construction of an antichain based on orthogonal Latin rectangle with parameters $t=2,k=4$.

If the first row of the Latin rectangle is all $P_{+}$ or $P_{-}$, then after the cycle will inevitably appear in an element of row will be equal to its row index or column index, so the first row needs to contain both $P_{+}$ and $P_{-}$, in order to facilitate the construction of the Latin rectangle is now divided into two sub-rectangle, $\Re$ represents the Latin rectangle, ${\Re }_{p_{+}}$ means that all the element of the positive sub-Latin rectangle are all from the $P_{+}$, ${\Re }_{p_{-}}$ means that all the negative sub-Latin rectangle are all from the $P_{-}$, and its rows and columns to do a substitution. It is easy to find that if ${\Re }_{p_{+}}$ and ${\Re }_{p_{-}}$ do not have fixed points it means that $\Re$ does not have fixed points either, if $r_{i,j}=k\in {\Re }_{p_{+}}$ then $i\in P_{+}$, $j\in P_{-}$, $\mathrm{mod}{p}_{+}$ for any element, and if $r_{i,j}=k\in {\Re }_{p_{-}}$ then $i\in P_{-}$, $j\in P_{+}$, $\mathrm{mod}2p_{+}$ for any element.

The row index and column index of Latin moments and the corresponding elements in the above can form a ternary array. For example, $r_{i,j}=k$ can form a ternary array of $\left\{i,j,k\right\}$. It is not hard to see that knowing two of these three elements tells us whether $r_{i,j}$ is an element in a positive sub-Latin rectangle or a negative sub-Latin rectangle. For the convenience of representation, the row index $i$ and the corresponding element $k$ form a pair $\left\{i,k\right\}$, and the set formed by such a pair of positive sub-Latin rectangles is called the set of positive sub-Latin rectangles, and the set formed by such a pair of negative sub-Latin rectangles is called the set of negative sub-Latin rectangles.

In order for the set to which the Latin rectangle are pairs to satisfy the definition of set and t-intersectivity, the following three conditions are satisfied on top of the above:

Condition1 (Update rule): If $r_{i,j}=k$ then $r_{i-1,j+1}=k-1,i\in P_{+},j\in P_{-}$.

Condition2 (Symmetry condition): If $r_{i,j}=h\in {\Re }_{p_{+}}$ then $r_{h,j}=i\in {\Re }_{p_{+}}$; if $r_{i,j}=h\in {\Re }_{p_{-}}$ then $r_{h,j}=i\in {\Re }_{p_{-}}$ (Note the row and column substitution).

Condition3 (Regularity condition): Different pairs $\left\{i,h\right\}$ and $\left\{i^{\prime },h^{\prime }\right\}$ are belonging to positive sub-Latin rectangle negative sub-Latin rectangle then there is $i-i^{\prime }\ne h-h^{\prime }$.

Note: Consider condition 1, e.g. $r_{2,0}=5\to r_{1,1}=4$ and it can be found that the elements of the Latin rectangle are completely determined by the first column, so that a good choice of the first column constructs the Latin square; Condition 2, e.g. $r_{0,8}=7$ and $r_{7,8}=0$ form the same triad $\left(0,7,8\right)$. The symmetry condition groups the elements of the Latin rectangle into pairs that result in identical elements in which exactly half of them are identical.

It can be found that the Latin rectangles as long as the construction of the first column in accordance with the update rule can be constructed throughout the Latin rectangle, but the Latin rectangle also to meet the inside of all the elements are not duplicated, so the construction of the first column to meet the first column cannot be duplicated meta, but also note that the peer cannot be duplicated meta, and the condition of the 2 is to meet the requirements of this. For example: if the pairs $\left\{6,2\right\}$ and $\left\{4,0\right\}$ are part of the set of positive sub-Latin rectangle, $r_{6,0}=2\to r_{5,3}=1\to r_{4,4}=0$ and $r_{4,0}=0$, the presence of two 0 in the same rows contradicts the definition of a Latin rectangle. Thus, the regularity condition ensures that there are no same elements in each row.

Construct 1: $p_{+}=p_{-}$ is a multiple of 4 if $v=4m$.

The positive sub-Latin rectangle set of the first Latin rectangle ${\Re }_1$ has elements $\left\{i,-i-1\right\}$, where $i\in \left[\frac{p_{+}}{4}-1\right]$; the negative sub-Latin rectangle set has elements $\left\{i,-i-1\right\}$, where $i\in \left[\frac{p_{+}}{4},\frac{p_{+}}{4}+1,\cdots ,\frac{p_{+}}{2}-1\right]$;

The positive sub-Latin rectangle of the second Latin rectangle ${\Re }_2$ has elements $\left\{i,-i-1\right\}$, where $i\in \left[\frac{p_{+}}{4},\frac{p_{+}}{4}+1,\cdots ,\frac{p_{+}}{2}-1\right]$; the negative sub-Latin rectangle has element $\left\{i,-i-1\right\}$, where $i\in \left[\frac{p_{+}}{4}-1\right]$.

Orthogonal Latin rectangle $\left(x_1,x_2\right)\in v\times v$ are composed and $\left(i,j\right)\in p_{+}\times p_{-}$, ${\Re }_1\left(i,j\right)=x_1$, ${\Re }_2\left(i,j\right)=x_2$.

Lemma 3: The Latin rectangle in Construction 1 has no fixed points, nor does the orthogonal Latin rectangle contain fixed points.

Proof: it is only necessary to prove that one Latin rectangle has no fixed point, and the second Latin rectangle is proved using the same steps.

1) There is no such thing as an element appearing twice in the first column, and the subsequent columns are sequentially minus one from the first, so there is also no such thing as an element existing twice, i.e., there are no identical elements in the same column.

2) Let different points $\left\{i,j\right\},\left\{i^{\prime },j^{\prime }\right\}$ such that $i-i^{\prime }\ne j-j^{\prime }$, by construction 1 $j=-i-1$, $j^{\prime }=-i^{\prime }-1$ thus $i-i^{\prime }=-i-1-\left(-i^{\prime }-1\right)=-\left(i-i^{\prime }\right)$, i.e. $i-i^{\prime }=0$ or $i-i^{\prime }=\frac{p_{+}}{2}$, obviously $i-i^{\prime }=0$ needs to be rounded off since $i$ is not the same as $i^{\prime }$, so $i-i^{\prime }=\frac{p_{+}}{2}$. From Construction 1 it follows that if $i,i^{\prime }\in P_{+}$, $\max \left(i-i^{\prime }\right)=\frac{p_{+}}{2}-2,\min \left(i-i^{\prime }\right)=1$.

If $i,i^{\prime }\in P_{-}$,

$\max i-i^{\prime }=p_{+}-2,\min \left(i-i^{\prime }\right)=\frac{p_{+}}{2}-2$

Neither satisfies $i-i^{\prime }=\frac{p_{+}}{2}$, so only one of $i$ and $i\text{<msup> <mo> ′ </mo> </msup>}$belongs to $P_{-}$ and one to $P_{+}$, which contradicts that both have to belong to positive sub-Latin rectangle or both have to belong to negative sub-Latin rectangle, i.e. there is no $r_{i,j}=r_{i^{\prime },j^{\prime }}$, and so the peers don’t have the same element.

3) Let $r_{i,j}=i$ $\to$ $r_{i+1,j-1}=i+1$ $\underset{\cdots }{\to }$ $r_{i+j,0}=i+j$, contradictory to Construct 1. For $r_{i,j}=j$ the case is the same as for $r_{i,j}=i$, i.e. it is proved that there is no fixed point.

The two Latin rectangles have no fixed points, i.e., neither do their mutually orthogonal Latin rectangle. #

Example 1: Let $v=16,p_{+}=p_{-}=8$, constructed Latin rectangles ${\Re }_1$ and ${\Re }_2$ ( Figure 3 and Figure 4 ).

Constructed orthogonal Latin rectangles ( Figure 5 ).

Rounding off the Latin rectangle after repeating 3 elements ( Figure 6 ).

The orthogonal Latin rectangle forms six different quaternions:

{0, 7, 8, 15}, {1, 5, 9, 14}, {0, 3, 10, 13}, {1, 6, 8, 13}, {1, 4, 9, 12}, {2, 5, 8, 11}.

These seven different quaternion feet do not contain an inclusion relation, i.e., they form a set that also forms an antichain:

{0, 7, 8, 15}, {1, 5, 9, 14}, {0, 3, 10, 13}, {1, 6, 8, 13}, {1, 4, 9, 12}, {2, 5, 8, 11},

Based on the relationship between antichain codes and combinatorial neural codes, this results in a combinatorial neural code with parameters

$\left(n=16,K=6,{\delta }_r=3r+3\right)$

and a weight of 4, a character in a binary code

(10000001100000001) (0100010001000010) (1001000000100100) (0100001010000100) (0100100001001000) (0010010010010000).

Consider below the case where $\frac{p_{+}}{2}$ is odd and $v=4m$.

Construct 2: $v=4m$, $\frac{p_{+}}{2}$ is odd.

The elements of the set of positive sub-Latin rectangle of the first Latin rectangle ${\Re }_1$ are $\left\{i,-i-1\right\}$ at $i\in \left[\frac{P_{+}-2}{4}-1\right]$, at $i\in \left\{\frac{P_{+}-2}{4},\frac{P_{+}-2}{4}+1,\cdots ,\frac{P_{+}}{2}-2\right\}$ for $\left\{i,-i-2\right\}$; the element of the set of negative sub-Latin rectangle is $\left\{\frac{P_{+}}{2}-1,-\frac{P_{+}-2}{4}-1\right\}$ at $i=\frac{P_{+}}{2}-1$.

The elements of the set of positive sub-Latin rectangle of the second Latin matrix ${\Re }_2$ are $\left\{\frac{P_{+}}{2}-1,-\frac{P_{+}-2}{4}-1\right\}$ at $i=\frac{P_{+}}{2}-1$; the elements of the set of negative sub-Latin rectangle are $\left\{i,-i-1\right\}$ at $i\in \left[\frac{P_{+}-2}{4}-1\right]$ and at $i\in \left\{\frac{P_{+}-2}{4},\frac{P_{+}-2}{4}+1,\cdots ,\frac{P_{+}}{2}-2\right\}$ for $\left\{i,-i-2\right\}$.

The orthogonal Latin rectangle $\left(x_1,x_2\right)\in v\times v$ are composed and $\left(i,j\right)\in p_{+}\times p_{-}$, ${\Re }_1\left(i,j\right)=x_1$, ${\Re }_2\left(i,j\right)=x_2$.

Lemma 4: The Latin rectangle in Construction 2 has no fixed points, nor does the orthogonal Latin rectangle contain fixed points.

Proof: It is only necessary to prove that one Latin rectangle has no fixed point, and the second Latin rectangle is proved using the same steps.

1) There is no such thing as an element appearing twice in the first column, and the subsequent columns are in order minus one from the first, so there can be no such thing as an element appearing twice, i.e., there are no identical elements in the same column.

2) Let $\left\{i,j\right\}$ and $\left\{i^{\prime },j^{\prime }\right\}$ be two different elements belonging to the set of positive Latin rectangle, and let $i^{\prime }-i=j^{\prime }-j$.

a) When $j=-i-1$, $j^{\prime }=-i^{\prime }-1$,

$i-i^{\prime }=j-j^{\prime }=-i-1-\left(-i^{\prime }-1\right)=\left(i-i^{\prime }\right)$,

We get $i-i^{\prime }=\frac{p_{+}}{2}$, by construction 2 if

$i,i^{\prime }\in \left[\frac{P_{+}-2}{4}-1\right]\to \max \left(i-i^{\prime }\right)=\frac{p_{+}}{2}-3,\min \left(i-i^{\prime }\right)=1$

so $i-i^{\prime }\in \left\{1,2,\cdots ,\frac{p_{+}}{2}-3\right\}$, a contradiction.

b) When $j=-i-1$, $j^{\prime }=-i^{\prime }-2$,

$i-i^{\prime }=j-j^{\prime }=-i-1-\left(-i^{\prime }-2\right)=-\left(i-i^{\prime }\right)+1\to 2\left(i-i^{\prime }\right)=1$,

Obviously not available, since $P_{+}$ is even, and in the case of $\mathrm{mod}{P}_{+}$, $P_{+}+1$ being odd cannot equal $2\left(i-i^{\prime }\right)$, a contradiction.

c) When $j=-i-2$, $j^{\prime }=-i^{\prime }-1$,

$i-i^{\prime }=j-j^{\prime }=-i-2-\left(-i^{\prime }-1\right)=-\left(i-i^{\prime }\right)-1$,

the same obviously cannot be obtained.

d) When $j=-i-2$, $j^{\prime }=-i^{\prime }-2$,

$i-i^{\prime }=j-j^{\prime }=-i-1-\left(-i^{\prime }-1\right)=-\left(i-i^{\prime }\right)$,

We get $i-i^{\prime }=\frac{p_{+}}{2}$, by construction 2 it follows that if

$i,i^{\prime }\in \left\{\frac{P_{+}-2}{4},\frac{P_{+}-2}{4}+1,\cdots ,\frac{P_{+}}{2}-2\right\}$, $i-i^{\prime }\in \left\{\frac{p_{+}}{2}+3,\cdots ,p_{+}-2\right\}$,

a contradiction, that satisfies regularity for positive sub-Latin rectangle, discussed below for negative sub-Latin rectangle:

$i-i^{\prime }=\frac{P_{+}}{2}-1-\left(-\frac{P_{+}-2}{4}-1\right)=\frac{3p_{+}-2}{4}$ $j-j^{\prime }=-\frac{P_{+}-2}{4}-1-\left(\frac{p_{+}}{2}-1\right)=\frac{2-3p_{+}}{4}$

so $i-i^{\prime }\ne j-j^{\prime }$, satisfying regularity.

3) Let $r_{i,j}=i\to r_{i+1,j-1}=i+1\underset{\cdots }{\to }{r}_{i+j,0}=i+j$, contradiction with Construct 2. For $r_{i,j}=j$ the case is the same as for $r_{i,j}=i$, i.e. it is proved that there is no fixed point.

There is no fixed point for two Latin rectangle, i.e. there is no fixed point for their mutually orthogonal Latin rectangle. #

Example 2: Let $v=20,\frac{p_{+}}{2}=5$, constructed Latin rectangles ${\Re }_1$ and ${\Re }_2$ ( Figure 7 and Figure 8 ).

Constructed orthogonal Latin rectangles ( Figure 9 )

Rounding off the Latin rectangle after repeating 3 elements. ( Figure 10 )

Orthogonal Latin rectangle forms an array of 12 distinct quaternions:

{0, 9, 10, 19}, {0, 7, 11, 18}, {0, 4, 12, 16}, {0, 3, 14, 17}, {1, 8, 10, 17}, {1, 5, 11, 15}, {2, 6, 10, 14}, {3, 5, 10, 12}, {2, 4, 11, 13}, {6, 7, 12, 13}, {0, 2, 16, 18}, {1, 9, 14, 16}.

These 12 different quaternions arrays do not have an inclusion relationship therefore the set formed constitutes an antichain:

{{0, 9, 10, 19}, {0, 7, 11, 18}, {0, 4, 12, 16}, {0, 3, 14, 17}, {1, 8, 10, 17}, {1, 5, 11, 15}, {2, 6, 10, 14}, {3, 5, 10, 12}, {2, 4, 11, 13}, {6, 7, 12, 13}, {0, 2, 16, 18}, {1, 9, 14, 16}}.

This results in a parameter of $\left(n=20,k=12,{\delta }_r=3r+3\right)$ and a combined neural code with a weight of 4. Its corresponding codeword is:

(1000000001100000001) (10000001000100000010) (100010000000100001000) (10010000000000100100) (01000000101000000100) (01000100000100010000) (00100010001000100000) (00010100001001000000) (00101000000101000000) (00000011000011000000) (10100000000000001010) (01000000010000101000).

Consider the case $v=4m+1$ below.

Construction 3: One column on top of construction 12 constitutes $\left[2m\right]\times \left[2m+1\right]$ Latin rectangle, the first Latin rectangle adds a column $r_{i,2m+1}=\frac{p_{+}}{2}+i\in p_{+}$, and the second Latin rectangle adds a column $r_{i,2m+1}=\frac{p_{+}}{2}-i\in p_{-}$.

Lemma 5: Construction 3 is an orthogonal Latin rectangle without fixed points.

Proof: the first Latin rectangle obviously occurs at most once in the same column of elements, $r_{0,0},r_{0,1},\cdots ,r_{0,2m+1}$ the elements belonging to $p_{+}$ in this row are all odd, and the one element added is even. For row $i$, if $i$ is even, $-i-1$ is odd, and $\frac{p_{+}}{2}+i$ is even; if $i$ is odd, $-i-1$ is even, and $\frac{p_{+}}{2}+i$ is odd; therefore the same element cannot occur in the same row. The same proof for the second Latin rectangle, i.e., that there are no fixed points for either Latin rectangle, and no fixed points for the orthogonal Latin rectangle.

Consider below $v=4m+2,p_{+}=p_{-}=2m+1$.

Construct 4: $v=4m+2,p_{+}=p_{-}=2m+1$, construct orthogonal Latin rectangle of $\left[2m+1\right]\times \left[2m+1\right]$.

The element of the set of positive sub-Latin rectangle of the first Latin moment ${\Re }_1$ is $\left\{i,-i-1\right\}$ at $i\in \left[\frac{P_{+}-1}{2}-1\right]$ the elements of the set of negative sub-Latin rectangle are $\left\{\frac{P_{+}-1}{2},\frac{P_{+}-1}{2}-1\right\}$ at $i=\frac{P_{+}-1}{2}$

The elements of the set of positive sub-Latin rectangle of the second Latin rectangle ${\Re }_2$ are $\left\{\frac{P_{+}-1}{2},\frac{P_{+}-1}{2}-1\right\}$ for $i=\frac{P_{+}-1}{2}$; the elements of the set of negative sub-Latin rectangle are $\left\{i,-i-1\right\}$ for $i\in \left[\frac{P_{+}-1}{2}-1\right]$.

Lemma 6: Construction 4 is a Latin rectangle without fixed points, and orthogonal Latin rectangle does not contain fixed points.

Proof: it is sufficient to prove that one Latin rectangle has no fixed point, and the second Latin rectangle is proved using the same steps.

1) There is no such thing as an element appearing twice in the first column, and the subsequent columns are sequentially minus one from the first column, so there can be no such thing as an element appearing twice, i.e., there are no identical elements in the same column.

2) Let $\left\{i,j\right\}$ and $\left\{i^{\prime },j^{\prime }\right\}$ be two different elements belonging to the set of positive Latin rectangle, and let $i^{\prime }-i=j^{\prime }-j$, By the construction 4 know

$i^{\prime }-i=j^{\prime }-j=-i^{\prime }-1-\left(-i-1\right)=-\left(i^{\prime }-i\right)$,

that is, $i^{\prime }-i=0$ or $2\left(i^{\prime }-i\right)=p_{+}$ obviously cannot be satisfied, a contradiction; and belongs to the negative sub-Latin rectangle of the set of a column of a row of only one element therefore the peer will not be duplicated, that is, to satisfy the regularity.

3) Let $r_{i,j}=i\to r_{i+1,j-1}=i+1\underset{\cdots }{\to }{r}_{i+j,0}=i+j$, Contradiction with Construct 4. For $r_{i,j}=j$ the case is the same as for $r_{i,j}=i$, it is proved that there is no fixed point.

The two Latin rectangles have no fixed points, i.e., neither do their mutually orthogonal Latin rectangle.

Example 3: $v=10,p_{+}=p_{-}=5$, rounding off the Latin rectangle after repeating 3 elements. ( Figure 11 )

The orthogonal Latin rectangle forms 3 distinct quaternions:

{0, 4, 9, 5}, {0, 2, 6, 8}, {1, 3, 7, 5}.

These 3 different quaternions do not have an inclusion relationship therefore the set formed constitutes an antichain:

{{0, 4, 9, 5}, {0, 2, 6, 8}, {1, 3, 7, 5}}

This results in a parameter of $\left(n=10,k=3,{\delta }_r=3r+3\right)$ and the combined neural code with weight 4 has the corresponding codeword:

(1001100001) (1010001010) (0101010100)

Example 4: $v=14,p_{+}=p_{-}=7$, the orthogonal Latin rectangle forms 3 distinct quaternion arrays ( Figure 12 ):

The orthogonal Latin rectangle forms 6 distinct quaternions:

{0, 7, 6, 13}, {0, 4, 8, 12}, {0, 2, 11, 9}, {1, 5, 7, 11}, {1, 3, 8, 10}, {2, 4, 7, 9}.

These six different quaternions do not have an inclusion relationship therefore the set formed constitutes an antichain:

{{0, 7, 6, 13}, {0, 4, 8, 12}, {0, 2, 11, 9}, {1, 5, 7, 11}, {1, 3, 8, 10}, {2, 4, 7, 9}}.

This results in a parameter of $\left(n=14,k=6,{\delta }_r=3r+3\right)$ and the combined neural code with weight 4 has the corresponding codeword:

(10100000010100) (01000101000100) (01010000101000) (00101001010000) (10000011000001) (10001000100010).

Consider the case $v=4m+3$ below.

Construct 5: One column on top of construct 4 constitutes the Latin rectangle of $\left[2m+1\right]\times \left[2m+2\right]$,

the first Latin rectangle adds a column $r_{i,2m+2}=\frac{p_{+}-1}{2}+i\in p_{-}$,

the second Latin rectangle adds a column $r_{i,2m+2}=\frac{p_{+}-1}{2}+i\in p_{+}$.

Lemma 7: Construction 5 is an orthogonal Latin rectangle without fixed points.

Proof: two proofs of similarity of Latin rectangle. The following only proves the first one. Obviously the same column element occurs at most once, $r_{0,0},r_{0,1},\cdots ,r_{0,2m+1}$ is only one element belonging to $p_{-}$ in this row and it is odd, the added one is an element belonging to $p_{-}$ and it is even. For row $i$, if $i$ is even, $-i-1$ is odd and $\frac{p_{+}-1}{2}+i$ is even; if $i$ is odd, $-i-1$ is even and $\frac{p_{+}-1}{2}+i$ is odd, so there can be no identical element in the row.

$F$ is a family of sets, If the set in $F$ can be split into $F_1,F_2,\cdots ,F_n$ satisfying that $\left(V,F_i\right)$ is a $\left(t^{\prime },k,v\right)$ system $1\le i\le n$, then $\left(V,F\right)$ is $t^{\prime }$-partitionable. $\left(t,k,v\right)$ In layman’s terms, this means that each set of size $k$ in the family of sets formed by $v$ elements has at most $t-1$ elements that intersect every two sets.

Let $\left(V_1,F^1\right)$ be $t_1$-partitionable, with partition classes $F_1^1,F_2^1,\cdots ,F_n^1$, then $\left(V_2,F^2\right)$ is $t_2$-partitionable, with partition classes $F_1^2,F_2^2,\cdots ,F_n^2$, assuming that $V_1$ and $V_2$ do not intersect, then a new system can be formed, blocks are

$\left\{A\cup B:A\in F_i^1,D\in F_i^2,1\le i\le n\right\}$

Lemma 8 [16] [19] : The above $V_1$ and $V_2$ form a new system with parameters

$\left(t=\max \left(k_2+t_1,k_1+t_2\right),k_1+k_2,v_1+v_2\right)$. (5)

Proof: the new system contains $v=v_1+v_2$ points, each block has $k=k_1+k_2$ points, and the block

$\left\{A\cup B:A\in F_i^1,B\in F_i^2,1\le i\le n\right\}$.

Let $A_1,A_2\in F_i^1$, $B_1,B_2\in F_i^2$ then

$t=\max \left(\left(A_1\cup B_1\right)\cap \left(A_2\cup B_2\right)\right)$.

And because the value of $\max \left(\left(A_1\cup B_1\right)\cap \left(A_2\cup B_2\right)\right)$ is only in two cases: the first one where $B_1$ is the same as $B_2$,

$A_1\cap A_2=t_1-1$;

the second where $A_1$ is the same as $A_2$,

$B_1\cap B_2=t_2-1$, with $\left|B_i\right|=k_2,\left|A_i\right|=k_1$,

Since,

$\max \left(\left(A_1\cup B_1\right)\cap \left(A_2\cup B_2\right)\right)=\max \left(k_2+t_1,k_1+t_2\right)-1$,

So,

$t=\max \left(\left(A_1\cup B_1\right)\cap \left(A_2\cup B_2\right)\right)$,

That is, the parameters of the system constituting the new

$\left(t=\max \left(k_2+t_1,k_1+t_2\right),k_1+k_2,v_1+v_2\right)$.

Construct 6: Based on 1-partitionable and 0-partitionable constructs $\left(2,4,v\right)$.

The 1-factorization is $\left(2,4,2m\right)$ system 1-partitionable with $2m-1$ classes, the 1-factorization can be expressed as $\left(1,2,2m\right)$ and $2m-1$ points can form $\left(1,1,2m-1\right)$ with one point as a block of districts, which is 0-partitionable. By Lemma 6 a new system parameter can be formed as $\left(2,3,4m-1\right)$. Obviously the number of block of $\left(2,3,4m-1\right)$ is $m\left(2m-1\right)=2m^2-m$, and $2m^2-m$ points different from the previous ones are added to each block of $\left(2,3,4m-1\right)$ respectively to form a new system parameter as $\left(2,4,2m^2+3m-1\right)$.

$\left(2,4,2m^2+3m-1\right)$ forms $2m^2-m$ quaternion, each of which has at most one intersecting element and no containment relation, so the set family formed by these quaternion is an antichain, i.e., it can be constituted as a 4-weighted combinatorial neural code with the parameters $\left(n=2m^2+3m-1,k=2m^2-m,{\delta }_r=3r+3\right)$.

Theorem 1: Construct 6 forms a combinatorial neural code with parameter

$\left(n=2m^2+3m-1,k=2m^2-m,{\delta }_r=3r+3\right)$ (6)

Group Divisible Design (GD design) [18] : with $X$ as a given set of positive integers, $K$ and $M$ as a given set of positive integers, let $D=\left(V,G,A\right)$ be a finite associative structure, where $V$ is a v-element set, $G$ constitutes a division of $V$, the elements of $\text{A}$ are called the block, and the elements of $G$ are called the group, if the following conditions are satisfied:

Then $D$ is said to be a Group Divisible Design or GD design, denoted $GD\left(K,X,M;V\right)$.

Theorem 2: $GD\left(4,1,3;v\right)$, $v$ is a multiple of 12,

$\left|B\left(4,1;v+1\right)\right|=\frac{v}{3}+\frac{v\left(v-3\right)}{12}$ (7)

Example 7: Let $v=\left\{1,2,\cdots ,12\right\}$, the groups in $GD\left(4,1,3;v\right)$ are {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}; blocks: {1, 4, 7, 10}, {1, 5, 8, 11}, {1, 6, 9, 12}, {2, 4, 8, 12}, {2, 5, 9, 10}, {2, 6, 7, 11}, {3, 4, 9, 11}, {3, 5, 7, 12}, {3, 6, 8, 10}. Adding a point 13 to each of these groups results in 4 new blocks: {1, 2, 3, 13}, {4, 5, 6, 13}, {7, 8, 9, 13}, {10, 11, 12, 13}. Together with the above, the 9 blocks form a $B\left(4,1;13\right)$, $\left|B\left(4,1;13\right)\right|=13$. The above quaternion array forms a combinatorial neural code with parameters $\left(n=13,K=13,{\delta }_r=3r+3\right)$ with the following structure:

(1001001001000) (1000100100100) (1000010010010) (0101000100010) (0100100011000) (0100011000100) (0010101000010) (0011000010100) (1110000000001) (0010010101000) (0001110000001) (0000001110001) (0000000001111)

Transversal design [18] : i) The set $V$ has $kv$ elements called points; ii) Partition of $V$ into sets $\left\{G_i:i\in \left[k\right]\right\}$, each $G_i$ contains $v$ points called groups; iii) A set $B$ of $k$-subsets is called a block if it satisfies the following two conditions: a. Each group and each block intersect by exactly 1 point; b. $t$-subsets of $V$ appear in only one block or two and more points are in one group but not all points.

Theorem 3: $TD\left(t,k,v\right)$, $v=km$, $m,k$ is even, then

$\left|\left(k-1,k,v\right)=m^{k-1}+2^{\frac{k}{2}}\times \left(\begin{array}{c}m\\ 2\end{array}\right)\times {\left(\frac{m}{2}\right)}^{\frac{k-2}{2}}\right|$ (8)

Proof: $v=km$ divide the $v$ points into $k$ groups, each group has $m$ points, will be $v$ points with two-dimensional label $\left[k\right]\times \left[m\right]$, based on the transversal design there are already $m^{k-1}$ blocks, because it is determined that one is selected for each group, and the last is determined by determining the $k-1$ in front of it. Below add new block which is formed by

$\left\{a_1,a_2,\cdots ,a_{k-1},a_k\right\}$,

where $\left\{a_1,a_2\right\}\in {\Phi }_i^{g_1}$, $\left\{a_3,a_4\right\}\in {\Phi }_i^{g_2}$, $\cdots$, $\left\{a_{k-1},a_k\right\}\in {\Phi }_i^{g_{\frac{k}{2}}}$, and

$g_1\in \left[0,1\right],g_2\in \left[2,3\right],\cdots ,g_{\frac{k}{2}}\in \left[k-2,k-1\right],{\Phi }^{\left(g\right)}=\left\{{\Phi }_1^{\left(g\right)},{\Phi }_2^{\left(g\right)},\cdots ,{\Phi }_{m-1}^{\left(g\right)}\right\}$

is a 1-factor decomposition of $m$ elements in each group. There are a total of $k$ groups for $g\in \left[k-1\right]$ because $g$ consists of $\left\{g_1,g_2,\cdots ,g_{\frac{k}{2}}\right\}$, and each $g_i$ has two choices, i.e., there are $2^{\frac{k}{2}}$ choices for $g$. If $\left\{g_1,g_2,\cdots ,g_{\frac{k}{2}}\right\}$ is fixed, $\left\{a_1,a_2\right\}$ are related so there are $\left(\begin{array}{c}m\\ 2\end{array}\right)$ choices, and $\left\{a_3,a_4\right\}$ have $\frac{m}{2}$ choices, and similarly $\left\{a_{k-1},a_k\right\}$ also have $\frac{m}{2}$ choices, then $\left\{a_1,a_2,\cdots ,a_{k-1},a_k\right\}$ has a total of $2^{\frac{k}{2}}\times \left(\begin{array}{c}m\\ 2\end{array}\right)\times {\left(\frac{m}{2}\right)}^{\frac{k-2}{2}}$ choices. So

$\left|\left(k-1,k,v\right)\right|=m^{k-1}+2^{\frac{k}{2}}\times \left(\begin{array}{c}m\\ 2\end{array}\right)\times {\left(\frac{m}{2}\right)}^{\frac{k-2}{2}}$

$TD\left(t,k,v\right)$ constructs a family of sets of set size $k$ and the sets in the family intersect at most $t-1$ identically, which do not have an inclusion relation, thus constituting a combinatorial neural code. $TD\left(k-1,k,v\right)$ constitutes a combinatorial neural code with parameters

$\left(n=v,K=m^{k-1}+2^{\frac{k}{2}}\times \left(\begin{array}{c}m\\ 2\end{array}\right)\times {\left(\frac{m}{2}\right)}^{\frac{k-2}{2}},{\delta }_r=2r+2\right)$

Since there are four choices for it follows that the total number of added blocks is 48.

Example 8: Suppose that is a transversal design whose with points of the form $Z_4\times \left\{0,1,2,3\right\}$, and blocks

$\left\{\left\{\left(i_0,0\right),\left(i_1,1\right),\left(i_2,2\right),\left(i_3,3\right)\right\}:i_0,i_1,i_2,i_3\in Z_4,i_3=i_0+i_1+i_2\right\}$

Since there are four choices for each of $i_0,i_1,i_2$ it follows that we have $4^3=64$ blocks.

The factorization of interest reads as

$\begin{array}{c}{\Phi }^g=\left\{{\Phi }_1^{\left(g\right)},{\Phi }_2^{\left(g\right)},{\Phi }_3^{\left(g\right)}\right\}\\ =\{\left\{\left\{\left(0,g\right),\left(1,g\right)\right\},\left\{\left(2,g\right),\left(3,g\right)\right\}\right\},\\ \left\{\left\{\left(0,g\right),\left(2,g\right)\right\},\left\{\left(1,g\right),\left(3,g\right)\right\}\right\},\\ \left\{\left\{\left(0,g\right),\left(3,g\right)\right\},\left\{\left(1,g\right),\left(2,g\right)\right\}\right\}\}\end{array}$

Where $g\in \left\{0,1,2,3\right\}$, for a $g_1\in \left\{0,1\right\}$ and a $g_2\in \left\{2,3\right\}$, we obtain 12 blocks:

$\begin{array}{l}\{\left\{\left(0,g_1\right),\left(1,g_1\right),\left(0,g_2\right),\left(1,g_2\right)\right\},\left\{\left(0,g_1\right),\left(1,g_1\right),\left(2,g_2\right),\left(3,g_2\right)\right\},\\ \left\{\left(2,g_1\right),\left(3,g_1\right),\left(0,g_2\right),\left(1,g_2\right)\right\},\left\{\left(2,g_1\right),\left(3,g_1\right),\left(2,g_2\right),\left(3,g_2\right)\right\},\\ \left\{\left(0,g_1\right),\left(2,g_1\right),\left(0,g_2\right),\left(2,g_2\right)\right\},\left\{\left(0,g_1\right),\left(2,g_1\right),\left(1,g_2\right),\left(3,g_2\right)\right\},\\ \left\{\left(1,g_1\right),\left(3,g_1\right),\left(0,g_2\right),\left(2,g_2\right)\right\},\left\{\left(1,g_1\right),\left(3,g_1\right),\left(1,g_2\right),\left(3,g_2\right)\right\},\\ \left\{\left(0,g_1\right),\left(3,g_1\right),\left(0,g_2\right),\left(3,g_2\right)\right\},\left\{\left(0,g_1\right),\left(3,g_1\right),\left(1,g_2\right),\left(2,g_2\right)\right\},\\ \left\{\left(1,g_1\right),\left(2,g_1\right),\left(0,g_2\right),\left(3,g_2\right)\right\},\left\{\left(1,g_1\right),\left(2,g_1\right),\left(1,g_2\right),\left(2,g_2\right)\right\}\}\end{array}$

Since there are four choices for $g_1,g_2$ it follows that the total number of added blocks is 48. There are 112 blocks in total, i.e., they constitute a combinatorial neural code with parameters $\left(n=16,K=112,{\delta }_r=2r+2\right)$.