2025-04-24 19:00:14 +08:00
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分布式计算和集中式计算是完全等价的。将分布式算法中的本地观测
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$$
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b_i(k) = \sum_j a_{ij} x_j(k)
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$$
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代入瑞利商公式
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$$
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y(k) = \frac{\sum_i x_i(k) b_i(k)}{\sum_i x_i(k)^2}
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$$
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即可得到集中式计算的Rayleigh商形式
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$$
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\frac{x(k)^T (A x(k))}{x(k)^T x(k)},
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$$
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这与特征值估计的幂迭代公式
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$$
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\lambda_{\max} \approx \frac{x^T A x}{x^T x}
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$$
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完全一致。
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2025-04-23 11:37:26 +08:00
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---
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2025-04-24 19:00:14 +08:00
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### 代数示例
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考虑一个简单的2×2矩阵
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$$
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A = \begin{pmatrix}2 & 1\\1 & 2\end{pmatrix}, \quad x = \begin{pmatrix}2\\1\end{pmatrix}.
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$$
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1. **集中式计算**
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$$
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\lambda \approx \frac{x^T A x}{x^T x}
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= \frac{\begin{pmatrix}2 & 1\end{pmatrix} \begin{pmatrix}5\\4\end{pmatrix}}{2^2 + 1^2}
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= \frac{10 + 4}{5}
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= \frac{14}{5} = 2.8.
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$$
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2. **分布式计算**
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各节点分别计算本地观测值
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$$
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b_1 = (2 \cdot 2 + 1 \cdot 1) = 5, \quad b_2 = (1 \cdot 2 + 2 \cdot 1) = 4,
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$$
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然后通过全网共识计算
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$$
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y = \frac{2 \cdot 5 + 1 \cdot 4}{2^2 + 1^2}
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= \frac{14}{5} = 2.8.
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$$
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两种方法得到的结果完全相同。
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