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# 数学代写|组合数学代写Combinatorial Mathematics代考|MATH069 Distance-regular graphs revisited

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## 数学代写|组合数学代写Combinatorial Mathematics代考|Distance-regular graphs revisited

simple graph (Chapter 1, Section 1.1), where $X$ is the vertex set and $R$ is the edge set. For $x, y \in X$, let $\partial(x, y)$ be the length of a shortest path connecting $x$ and $y$. If there is no such path, then we set $\partial(x, y)=\infty$. We let $d=\operatorname{Max}{\partial(x, y) \mid x, y \in X}$ denote the maximum of the distances between vertices in $X$, and call $d$ the diameter of $\Gamma$. In what follows, we assume that the diameter satisfies $d<\infty$, that is, $\Gamma$ is a connected graph. Let $S^X$ be the symmetric group on $X$. An element $\sigma \in S^X$ is said to be an automorphism of $\Gamma$ if $\left(x^\sigma, y^\sigma\right) \in R$ for any edge $(x, y) \in R$. The set of all automorphisms of $\Gamma$ will be denoted by Aut( $\Gamma$ ). The set $\operatorname{Aut}(\Gamma)$ forms a subgroup of $S^X$, and is called the automorphism group of the graph $\Gamma$. It acts on $X \times X$ by $(x, y)^\sigma=\left(x^\sigma, y^\sigma\right)(\sigma \in \operatorname{Aut}(\Gamma))$. For $0 \leq i \leq d$, we set
$$R_i={(x, y) \in X \times X \mid \partial(x, y)=i}$$
Each $R_i$ is invariant under the action of $\operatorname{Aut}(\Gamma)$. Namely, Aut $(\Gamma)$ acts on each $R_i$. We say that $\Gamma$ is a distance-transitive graph if the action of Aut( $\Gamma$ ) on $R_i$ is transitive. That is, we call $\Gamma$ a distance-transitive graph if there exists $\sigma \in \operatorname{Aut}(\Gamma)$ such that $x^{\prime}=x^\sigma, y^{\prime}=y^\sigma$ for any $x, y, x^{\prime}, y^{\prime} \in X$ satisfying $\partial(x, y)=\partial\left(x^{\prime}, y^{\prime}\right)$.

We now forget about the group action, and return again to a finite connected simple graph $\Gamma=(X, R)$. Recall that $d$ denotes the diameter of $\Gamma$. For $x, y \in X$ and $i, j \in{0,1, \ldots, d}$, let
$$p_{i, j}(x, y)=\left|\left{z \in X \mid(x, z) \in R_i,(z, y) \in R_j\right}\right| .$$
If $\Gamma$ is a distance-transitive graph, then for any $i, j, k \in{0,1, \ldots, d}$ there exists a constant $p_{i, j}^k$ such that
$$p_{i, j}^k=p_{i, j}(x, y) \quad\left((x, y) \in R_k\right)$$

## 数学代写|组合数学代写Combinatorial Mathematics代考|Q-polynomial schemes revisited

As a preparation for Section $6.4$, where P- and Q-polynomial schemes will be discussed, we review the definition of Q-polynomial schemes briefly following the manner of the previous subsection, where P-polynomial schemes are discussed. Especially, the definition in terms of Terwilliger algebras will play an important role later. Let $\mathfrak{X}=\left(X,\left{R_i\right}_{0 \leq i \leq d}\right)$ be a symmetric association scheme and let $E_0, E_1, \ldots, E_d$ be the primitive idempotents of the Bose-Mesner algebra $\mathfrak{A}$ of $\mathfrak{X}$. By $\left(4^{\prime \prime}\right)$ in Chapter 2 , Section 2.3, we have
$$E_i \circ E_j=\frac{1}{|X|} \sum_{k=0}^d q_{i, j}^k E_k$$
with respect to the Hadamard product of $\mathfrak{A}$. Note that Krein numbers $q_{i, j}^k$ are nonnegative real (Theorem 2.26, Corollary 2.37). Then the following conditions (i) , (ii) $^{\prime}$, and (iii) ${ }^{\prime}$ on $q_{i, j}^k$ are equivalent, and a symmetric association scheme satisfying these conditions is called a Q-polynomial scheme. For the proof of the equivalence, the proof in the previous subsection is valid if we replace the ordinary matrix product by the Hadamard product and the intersection number $p_{i, j}^k$ by the Krein number $q_{i, j}^k$. We have: (i) ${ }^{\prime}$
$$q_{1, j}^k \begin{cases}=0, & \text { if } 1<|k-j|, \ \neq 0, & \text { if } 1=|k-j|\end{cases}$$ (ii) ${ }^{\prime} \quad E_i$ is a polynomial of degree $i$ in $E_1$ with respect to the Hadamard product ( $0 \leq$ $i \leq d)$ $\left(\right.$ (iii) $^{\prime}$ $$q_{i, j}^k \begin{cases}=0, & \text { if } i<|k-j| \text { or } i>k+j, \ \neq 0, & \text { if } i=|k-j| \text { or } i=k+j\end{cases}$$
If we set $b_i^=q_{1, i+1}^i, a_i^=q_{1, i}^i, c_i^=q_{1, i-1}^i$, then (i) ${ }^{\prime}$ is rewritten as (i) $$E_1 \circ E_j=\frac{1}{|X|}\left(b_{j-1}^ E_{j-1}+a_j^* E_j+c_{j+1}^* E_{j+1}\right) \quad(0 \leq j \leq d),$$
and $b_{i-1}^* c_i^* \neq 0(1 \leq i \leq d)$, where $b_{-1}^$ is indeterminate and $c_{d+1}^=1, E_{-1}=0, E_{d+1}=0$. Moreover, if we set $m=b_0^=q_{1,1}^0$, we have $$m=a_0^+b_0^=c_i^+a_i^+b_i^=c_d^+a_d^ \quad(1 \leq i \leq d-1)$$
(Proposition 2.24).

## 数学代写|组合数学代写Combinatorial Mathematics代考|Distance-regular graphs revisited

$$R_i=(x, y) \in X \times X \mid \partial(x, y)=i$$

\left 缺少或无法识别的分隔符

$$p_{i, j}^k=p_{i, j}(x, y) \quad\left((x, y) \in R_k\right)$$

## 数学代号|组合数学代与Combinatorial Mathematics代考|Q-polynomial schemes revisited

\left 缺少或无法识别的分隔符 是一个对称关联方宣，让 $E_0, E_1, \ldots, E_d$ 是 Bose-Mesner 代数的原始昌等元 A的 $\mathfrak{X}$. 经过 $\left(4^{\prime \prime}\right)$ 在第 2 章 $2.3$ 节中，我们有
$$E_i \circ E_j=\frac{1}{|X|} \sum_{k=0}^d q_{i, j}^k E_k$$

$$q_{1, j}^k{=0, \quad \text { if } 1<|k-j|, \neq 0, \quad \text { if } 1=|k-j|$$ (二) ${ }^{\prime} \quad E_i$ 是次数的多项式 $i$ 在 $E_1$ 关于 Hadamard 产品 $(0 \leq i \leq d)\left((\text { H })^{\prime}\right.$ $q_{i, j}^k{=0, \quad$ if $i<|k-j|$ or $i>k+j, \neq 0, \quad$ if $i=|k-j|$ or $i=k+j$

$c_{d+1} \overline{\bar{d}}i 1, E{-1}=0, E_{d+1}=0$. 此外，如果我们设置 $m=b_0^{=} q_{1,1}^0$ ， 我们有
$$m=a_0^{+} b_0^{=} c_i^{+} a_i^{+} b_i^{-} c_d^{+} a_d \quad(1 \leq i \leq d-1)$$
(提宲 2.24)。

## MATLAB代写

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