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# 数学代写|组合数学代写Combinatorial Mathematics代考|MATH4410 Tight designs in Johnson schemes

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## 数学代写|组合数学代写Combinatorial Mathematics代考|Existence and non-existence of tight designs

Let $V={1,2, \ldots, v}$, and let $X=\left(\begin{array}{l}V \ d\end{array}\right)$ be the set of $d$-element subsets of $V$. We assume $1 \leq t \leq d \leq \frac{v}{2}$. Let $\left(X,\left{R_i\right}_{0 \leq i \leq d}\right)$ be the Johnson scheme $J(v, d)$. Recall that a 2 -design $Y \subset X$ is tight if $\left.|Y|=m_0+m_1+\cdots+m_e=\sum_{i=0}^e\left(\begin{array}{l}v \ i\end{array}\right)-\left(\begin{array}{c}v \ i-1\end{array}\right)\right)=\left(\begin{array}{c}v \ e\end{array}\right)$ holds. In general, a 2e-design $Y$ satisfies the Fisher type inequality $|Y| \geq\left(\begin{array}{l}v \ v\end{array}\right)$. This inequality was first obtained by Petrenjuk [395] for the case $e=2$. Afterwards, Ray-Chaudhuri and Wilson announced that it holds for any $e$ [514]. For the proof, see Ray-Chaudhuri and Wilson [400]. Delsarte (1974) also gave a proof, and as was written in the book of Hiroshi Nagao [366], Noda and Bannai obtained the same result independently in 1972 . If there exists a tight $2 e$-design, by using Theorem 3.16, we can show the following theorem.
Theorem 3.27. If there exists a tight $2 e$-design in the Johnson scheme $J(v, d)$, then all $e$ zeros of the polynomial
$$\Psi_e(x)=\sum_{i=0}^e(-1)^{e-i} \frac{\left(\begin{array}{c} v-e \ i \end{array}\right)\left(\begin{array}{c} k-i \ e-i \end{array}\right)\left(\begin{array}{c} k-1-i \ e-i \end{array}\right)}{\left(\begin{array}{c} e \ i \end{array}\right)}\left(\begin{array}{c} x \ i \end{array}\right)$$
are positive integers. This polynomial $\Psi_e$ is called the Wilson polynomial or the RayChaudhuri-Wilson polynomial.

When $e=1, Y$ is a tight 2-design if and only if $b=v(b=|Y|)$ if and only if $Y$ is a symmetric 2-design (Chapter 1, Section 1.3, Definition 1.37). There exist quite a few symmetric 2-designs and their classification seems to be almost impossible.

When $e=2$, the non-trivial tight 4-designs are the Witt design 4- $(23,7,1)$ and its complementary design 4- $(23,16,52)$ only. (In the latter case, $d \leq \frac{v}{2}$ does not hold.) The classification was started by Noboru Ito $[245,246]$ and was almost completed by Enomoto, Ito, and Noda [179]. To be precise, the classification was completely solved by Bremner [111] and Stroeker [440] by determining a rational integral solution of the Diophantine equation $3 x^4-4 y^4-2 x^2+12 y^2-9=0$, which is related to an elliptic function. In the next subsection, we present the detailed proof by Noda.

## 数学代写|组合数学代写Combinatorial Mathematics代考|Classiffcation of tight 4-designs in Johnson schemes

Let $(V, \mathcal{B})$ be a tight 4-design in the Johnson scheme $J(v, k)$. Suppose $(V, \mathcal{B})$ is nontrivial. Assume $k \leq \frac{v}{2}$. Note that if $k>\frac{v}{2}$, the complementary design is also a tight 4-design. In this section, we prove the following theorem.

Theorem $3.32$ (Enomoto-Ito-Noda [179]). A non-trivial tight 4-design in the Johnson scheme is the 4- $(23,7,1)$ design or its complementary design 4-(23, 16, 52) only.

Remark 3.33. Noboru Ito $[245,246]$ started the proof of this theorem, and Enomoto, Ito, and Noda (1979) [179] almost completed the proof by correcting errors. In a part of the proof, a number theoretic result on the solution of a Diophantine equation was used (Bremner [111], Stroeker [440]). The proof given here is based on an unpublished note by Ryuzaburo Noda, which was written soon after [179]. We are grateful to Professor Noda, who permits us to use the contents of his note. It is similar to [179] that the problem is transformed into the Diophantine equation. However, compared to the proof combining three papers $[245,246,179]$, this proof is clearer and easier to read.
The proof of Theorem $3.32$ consists of steps $(\mathrm{A})-(\mathrm{K})$.
(A) Let $i, j$ be the cardinalities of intersections of two distinct blocks. Let $i<j$. Then $i, j$ are the roots of the following quadratic equation:
$$X^2-\left(\frac{2(k-1)(k-2)}{v-3}+1\right) X+\frac{k(k-1)^2(k-2)}{(v-2)(v-3)}=0 .$$
(B) We have
$$(v-2)(v-3) \mid 2 k(k-1)(k-2)$$
Proof. Since $b=\lambda_0=\left(\begin{array}{c}v \ 2\end{array}\right), \lambda_4=\frac{k(k-1)(k-2)(k-3)}{2(v-2)(v-3)}$ is an integer. Therefore, $2 \lambda_4=$ $\frac{k(k-1)(k-2)(k-3)}{(v-2)(v-3)}$ is an integer. On the other hand, by (A), $\frac{k(k-1)^2(k-2)}{(v-2)(v-3)}$ is an integer, and hence $\frac{k(k-1)^2(k-2)}{(v-2)(v-3)}-\frac{k(k-1)(k-2)(k-3)}{(v-2)(v-3)}=\frac{2 k(k-1)(k-2)}{(v-2)(v-3)}$ is also an integer.

## 数学代写|组合数学代写组合数学代考|紧密设计的存在与不存在

$$\Psi_e(x)=\sum_{i=0}^e(-1)^{e-i} \frac{\left(\begin{array}{c} v-e \ i \end{array}\right)\left(\begin{array}{c} k-i \ e-i \end{array}\right)\left(\begin{array}{c} k-1-i \ e-i \end{array}\right)}{\left(\begin{array}{c} e \ i \end{array}\right)}\left(\begin{array}{c} x \ i \end{array}\right)$$

## 数学代写|组合数学代写组合数学代考| Johnson方案中紧密的4-设计的分类

(A)设$i, j$为两个不同块的交集的基数。让$i<j$。那么$i, j$是下面的二次方程的根:
$$X^2-\left(\frac{2(k-1)(k-2)}{v-3}+1\right) X+\frac{k(k-1)^2(k-2)}{(v-2)(v-3)}=0 .$$
(B)我们有
$$(v-2)(v-3) \mid 2 k(k-1)(k-2)$$

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