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@@ -110,34 +110,34 @@ $$
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-[推导]:由公式(10.3)可得,
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+[推导]:由公式(10.3)可得
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$$
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b_{ij}=-\frac{1}{2}(dist^2_{ij}-b_{ii}-b_{jj})
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$$
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-由公式(10.6)和(10.9)可得,
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+由公式(10.6)和(10.9)可得
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$$
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\begin{aligned}
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tr(\boldsymbol B)&=\frac{1}{2m}\sum^m_{i=1}\sum^m_{j=1}dist^2_{ij}\\
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&=\frac{m}{2}dist^2_{\cdot}
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\end{aligned}
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$$
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-由公式(10.4)和(10.8)可得,
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+由公式(10.4)和(10.8)可得
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$$
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\begin{aligned}
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b_{jj}&=\frac{1}{m}\sum^m_{i=1}dist^2_{ij}-\frac{1}{m}tr(\boldsymbol B)\\
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&=dist^2_{\cdot j}-\frac{1}{2}dist^2_{\cdot}
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\end{aligned}
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$$
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-由公式(10.5)和(10.7)可得,
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+由公式(10.5)和(10.7)可得
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$$
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\begin{aligned}
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b_{ii}&=\frac{1}{m}\sum^m_{j=1}dist^2_{ij}-\frac{1}{m}tr(\boldsymbol B)\\
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&=dist^2_{i\cdot}-\frac{1}{2}dist^2_{\cdot}
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\end{aligned}
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$$
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-综合可得,
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+综合可得
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$$
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\begin{aligned}
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b_{ij}&=-\frac{1}{2}(dist^2_{ij}-b_{ii}-b_{jj})\\
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