wording (sec 5)

Chapter5/Figures/figure-calculation-formula&iterative-process-of-function.tex

@@ -11,7 +11,7 @@
     \node [anchor=west] (eq2) at (eq1.east) {$=$\ };
     \draw [-] ([xshift=0.3em]eq2.east) -- ([xshift=11.6em]eq2.east);
     \node [anchor=south west] (eq3) at ([xshift=1em]eq2.east) {$\sum_{k=1}^{K} c_{\mathbb{E}}(s_u|t_v;s^{[k]},t^{[k]})$};
-    \node [anchor=north west] (eq4) at (eq2.east) {$\sum_{s_u} \sum_{k=1}^{K} c_{\mathbb{E}}(s_u|t_v;s^{[k]},t^{[k]})$};
+    \node [anchor=north west] (eq4) at (eq2.east) {$\sum_{s'_u} \sum_{k=1}^{K} c_{\mathbb{E}}(s'_u|t_v;s^{[k]},t^{[k]})$};
 
    {
     \node [anchor=south] (label1) at ([yshift=-6em,xshift=3em]eq1.north west) {利用这个公式计算};

Chapter5/Figures/figure-em-algorithm-flow-chart.tex

@@ -14,7 +14,7 @@
 \node [anchor=north west] (line7) at ([yshift=-0.1em]line6.south west) {4: \quad \quad  \textbf{foreach} $k = 1$ to $K$ \textbf{do}};
 \node [anchor=north west] (line8) at ([yshift=-0.1em]line7.south west) {5: \quad \quad \quad \footnotesize{$c_{\mathbb{E}}(\seq{s}_u|\seq{t}_v;\seq{s}^{[k]},\seq{t}^{[k]}) = \sum\limits_{j=1}^{|\seq{s}^{[k]}|} \delta(s_j,s_u) \sum\limits_{i=0}^{|\seq{t}^{[k]}|} \delta(t_i,t_v) \cdot \frac{f(s_u|t_v)}{\sum_{i=0}^{l}f(s_u|t_i)}$}\normalsize{}};
 \node [anchor=north west] (line9) at ([yshift=-0.1em]line8.south west) {6: \quad \quad \textbf{foreach} $t_v$ appears at least one of $\{\seq{t}^{[1]},...,\seq{t}^{[K]}\}$ \textbf{do}};
-\node [anchor=north west] (line10) at ([yshift=-0.1em]line9.south west) {7: \quad \quad \quad $\lambda_{t_v}^{'} = \sum_{s_u} \sum_{k=1}^{K} c_{\mathbb{E}}(s_u|t_v;\seq{s}^{[k]},\seq{t}^{[k]})$};
+\node [anchor=north west] (line10) at ([yshift=-0.1em]line9.south west) {7: \quad \quad \quad $\lambda_{t_v}^{'} = \sum_{s'_u} \sum_{k=1}^{K} c_{\mathbb{E}}(s'_u|t_v;\seq{s}^{[k]},\seq{t}^{[k]})$};
 \node [anchor=north west] (line11) at ([yshift=-0.1em]line10.south west) {8: \quad \quad \quad \textbf{foreach} $s_u$ appears at least one of $\{\seq{s}^{[1]},...,\seq{s}^{[K]}\}$ \textbf{do}};
 \node [anchor=north west] (line12) at ([yshift=-0.1em]line11.south west) {9: \quad \quad \quad \quad $f(s_u|t_v) = \sum_{k=1}^{K} c_{\mathbb{E}}(s_u|t_v;\seq{s}^{[k]},\seq{t}^{[k]}) \cdot (\lambda_{t_v}^{'})^{-1}$};
 \node [anchor=north west] (line13) at ([yshift=-0.1em]line12.south west) {10: \textbf{return} $f(\cdot|\cdot)$};

Chapter5/chapter5.tex

@@ -330,7 +330,7 @@ $\seq{t}^{[2]}$ = So\; ,\; what\; is\; human\; \underline{translation}\; ?
 \label{eq:5-7}
 \end{eqnarray}
 
-\parinterval 公式\eqref{eq:5-7}相当于在函数$g(\cdot)$上做了归一化，这样等式右端的结果具有一些概率的属性，比如，$0 \le \frac{g(\seq{s},\seq{t})}{\sum_{\seq{t'}}g(\seq{s},\seq{t'})} \le 1$。具体来说，对于源语言句子$\seq{s}$，枚举其所有的翻译结果，并把所对应的函数$g(\cdot)$相加作为分母，而分子是某个翻译结果$\seq{t}$所对应的$g(\cdot)$的值。
+\parinterval 公式\eqref{eq:5-7}相当于在函数$g(\cdot)$上做了归一化，这样等式右端的结果具有一些概率的属性，比如，$0 \le \frac{g(\seq{s},\seq{t})}{\sum_{\seq{t'}}g(\seq{s},\seq{t'})} \le 1$。 具体来说，对于源语言句子$\seq{s}$，枚举其所有的翻译结果，并把所对应的函数$g(\cdot)$相加作为分母，而分子是某个翻译结果$\seq{t}$所对应的$g(\cdot)$的值。
 
 \parinterval 上述过程初步建立了句子级翻译模型，并没有直接求$\funp{P}(\seq{t}|\seq{s})$，而是把问题转化为对$g(\cdot)$的设计和计算上。但是，面临着两个新的问题：
 
@@ -1024,13 +1024,13 @@ f(s_u|t_v) &= &\lambda_{t_v}^{-1} \cdot \funp{P}(\seq{s}| \seq{t}) \cdot c_{\mat
 
 \parinterval 为了满足$f(\cdot|\cdot)$的概率归一化约束，易得$\lambda_{t_v}^{'}$为：
 \begin{eqnarray}
-\lambda_{t_v}^{'}&=&\sum\limits_{s_u} c_{\mathbb{E}}(s_u|t_v;\seq{s},\seq{t})
+\lambda_{t_v}^{'}&=&\sum\limits_{s'_u} c_{\mathbb{E}}(s'_u|t_v;\seq{s},\seq{t})
 \label{eq:5-43}
 \end{eqnarray}
 
 \parinterval 因此，$f(s_u|t_v)$的计算式可再一步变换成下式：
 \begin{eqnarray}
-f(s_u|t_v)&=&\frac{c_{\mathbb{E}}(s_u|t_v;\seq{s},\seq{t})}  { \sum\limits_{s_u} c_{\mathbb{E}}(s_u|t_v;\seq{s},\seq{t}) }
+f(s_u|t_v)&=&\frac{c_{\mathbb{E}}(s_u|t_v;\seq{s},\seq{t})}  { \sum\limits_{s'_u} c_{\mathbb{E}}(s'_u|t_v;\seq{s},\seq{t}) }
 \label{eq:5-44}
 \end{eqnarray}