updates

Book/Chapter1/chapter1.tex

@@ -353,7 +353,7 @@ Candidate：the the the the
 \textrm{BP}=
 \begin{cases}
 1& c>r\\
-e^{(1-\frac{r}{c})}& c<r
+e^{(1-\frac{r}{c})}& c \le r
 \end{cases}
 \label{eq:brevity-penalty}
 \end{eqnarray}

Book/Chapter2/Figures/figure-Probability-density-function&Distribution-function.tex

@@ -13,7 +13,7 @@
 \draw [-latex,thick] (-0.5,-2.0) -- (-0.5,6);%纵轴
 \draw [-,dashed] (2,4) -- (2,-1.2);  %%图中纵轴虚线
 \draw [-,dashed] (-0.5,4) -- (2,4);  %%图中横轴虚线
-\node [anchor=north] at (-0.8,-1.2) {O};
+\node [anchor=north] at (-0.8,-1.2) {0};
 \node [anchor=north] at (2,-1.2) {$\mu$};
 \node [anchor=north] at (-1.4,4.75) {$\frac{1}{\sqrt{2\pi}\sigma}$};
 \node [anchor=north] at (-1.2,6.2) {\scriptsize{$f(x)$}};
@@ -30,7 +30,7 @@
 \draw [-,dashed] (3,1.5) -- (3,-1.2);  %%图中纵轴虚线
 \draw [-,dashed] (-0.5,1.5) -- (3,1.5);  %%图中横轴虚线
 \draw [-,dashed] (-0.5,4.2) -- (8.5,4.2);  %%图中横轴虚线
-\node [anchor=north] at (-0.8,-1.2) {O};
+\node [anchor=north] at (-0.8,-1.2) {0};
 \node [anchor=north] at (3,-1.2) {$\mu$};
 \node [anchor=north] at (-1.1,2.0) {0.5};
 \node [anchor=north] at (-0.8,4.7) {1};

Book/Chapter2/chapter2.tex

@@ -216,13 +216,8 @@
 
 \parinterval 首先介绍一下全概率公式：全概率公式（Law of total probability）是概率论中重要的公式，它可以将一个复杂事件发生的概率分解成不同情况的小事件发生概率的和。这里我们先介绍一个概念——划分。
 
-\parinterval 若集合S的一个划分事件为$B_1,…,B_n$是指它们满足
-%---------------------------------------------
-\begin{eqnarray}
-\bigcup_{i=1}^n B_i=S \textrm{且}B_iB_j=\varnothing , i,j=1,...,n,i\neq j
-\label{eq:2.2-8}
-\end{eqnarray}
-\parinterval 设$B_1,…,B_n$是S的一个划分，A为事件，则
+\parinterval 若集合S的一个划分事件为$B_1,…,B_n$是指它们满足$\bigcup_{i=1}^n B_i=S \textrm{且}B_iB_j=\varnothing , i,j=1,...,n,i\neq j$。设$B_1,…,B_n$是S的一个划分，A为事件，则
+
 \begin{eqnarray}
 \textrm{P}(A)=\sum_{k=1}^n \textrm{P}(A \mid B_k)\textrm{P}(B_k)
 \label{eq:2.2-9}
@@ -563,7 +558,7 @@
 \begin{figure}[htp]
 \centering
 \input{./Chapter2/Figures/figure-full-probability-word-segmentation-1}
-\caption{新投骰子结果}
+\caption{投掷一个很多面骰子的结果}
 \label{fig:2.3-8}
 \end{figure}
 %-------------------------------------------
@@ -586,7 +581,7 @@
 \centering
 \input{./Chapter2/Figures/figure-full-probability-word-segmentation-2}
 \setlength{\belowcaptionskip}{-0.2cm}
-\caption{换成汉字后结果}
+\caption{把数字换成汉字后的结果}
 \label{fig:2.3-9}
 \end{figure}
 %-------------------------------------------

Book/Chapter3/Chapter3.tex

@@ -173,7 +173,7 @@
 \vspace{-0.5em}
 \begin{eqnarray}
 \textrm{P}(x \leftrightarrow y; \mathbf{s},\mathbf{t}) & \equiv & \textrm{P}(x,y;\mathbf{s},\mathbf{t})   \nonumber \\
-                                                                             & =        & \frac{c(x,y;s,t)}{\sum_{x',y'} c(x',y';\mathbf{s},\mathbf{t})}
+                                                                             & =        & \frac{c(x,y;\mathbf{s},\mathbf{t})}{\sum_{x',y'} c(x',y';\mathbf{s},\mathbf{t})}
 \label{eqC3.1-new}
 \end{eqnarray}
 
@@ -220,11 +220,11 @@
 
 \qquad\qquad \; $\mathbf{s}^1$ = 机器\quad {\color{red}翻译}\; 就\; 是\; 用\; 计算机\; 来\; 进行\; {\color{red}翻译}
 
-\qquad\qquad\; $\mathbf{s}^1$ = machine\; {\color{red}translation}\; is\; just\; {\color{red}translation}\; by\; computer
+\qquad\qquad\; $\mathbf{s}^1$ = Machine\; {\color{red}translation}\; is\; just\; {\color{red}translation}\; by\; computer
 
 \qquad\qquad\; $\mathbf{s}^2$ = 那\quad 人工\quad {\color{red}翻译}\quad 呢\quad ?
 
-\qquad\qquad\; $\mathbf{t}^2$ = so\; what\; is\; human\; {\color{red}translation}\; ?
+\qquad\qquad\; $\mathbf{t}^2$ = So\; ,\; what\; is\; human\; {\color{red}translation}\; ?
 \label{example3-2}
 \end{example}
 
@@ -233,8 +233,8 @@
 \begin{eqnarray}
 {\textrm{P}(\textrm{``翻译''},\textrm{``translation''})} & = & {\frac{c(\textrm{``翻译''},\textrm{``translation''};\mathbf{s}^{1},\mathbf{t}^{1})+c(\textrm{``翻译''},\textrm{``translation''};\mathbf{s}^{2},\mathbf{t}^{2})}{\sum_{x',y'} c(x',y';\mathbf{s}^{1},\mathbf{t}^{1}) + \sum_{x',y'} c(x',y';\mathbf{s}^{2},\mathbf{t}^{2})}} \nonumber \\
                                                                             & = & \frac{4 + 1}{|\mathbf{s}^{1}| \times |\mathbf{t}^{1}| + |\mathbf{s}^{2}| \times |\mathbf{t}^{2}|} \nonumber \\
-                                                                            & = & \frac{4 + 1}{9 \times 7 + 5 \times 6} \nonumber \\
-                                                                            & = & \frac{5}{93}
+                                                                            & = & \frac{4 + 1}{9 \times 7 + 5 \times 7} \nonumber \\
+                                                                            & = & \frac{5}{98}
 \label{eqC3.6-new}
 \end{eqnarray}
 }
@@ -260,7 +260,7 @@
 \parinterval 计算句子级翻译概率并不简单。因为自然语言非常灵活，任何数据无法覆盖足够多的句子，因此我们也无法像公式\ref{eqC3.5-new}一样直接用简单计数的方式对句子的翻译概率进行估计。这里，我们采用一个退而求其次的方法：找到一个函数$g(\mathbf{s},\mathbf{t})\ge 0$来模拟翻译概率对译文可能性进行评价这种行为。我们假设：给定$\mathbf{s}$，翻译结果$\mathbf{t}$出现的可能性越大，$g(\mathbf{s},\mathbf{t})$的值越大；$\mathbf{t}$出现的可能性越小，$g(\mathbf{s},\mathbf{t})$的值越小。换句话说，$g(\mathbf{s},\mathbf{t})$的单调性和翻译概率呈正相关。如果存在这样的函数$g(\mathbf{s},\mathbf{t}
 )$，可以利用$g(\mathbf{s},\mathbf{t})$近似表示句子级翻译概率，如下：
 \begin{eqnarray}
-\textrm{P}(\mathbf{t}|\mathbf{s})  \approx  \frac{g(\mathbf{s},\mathbf{t})}{\sum_{\mathbf{t}'}g(\mathbf{s},\mathbf{t}')}
+\textrm{P}(\mathbf{t}|\mathbf{s})  \equiv  \frac{g(\mathbf{s},\mathbf{t})}{\sum_{\mathbf{t}'}g(\mathbf{s},\mathbf{t}')}
 \label{eqC3.7-new}
 \end{eqnarray}
 

Book/mt-book-xelatex.idx

@@ -41,12 +41,12 @@
 \indexentry{Chapter2.4.2|hyperpage}{63}
 \indexentry{Chapter2.4.2.1|hyperpage}{64}
 \indexentry{Chapter2.4.2.2|hyperpage}{65}
-\indexentry{Chapter2.4.2.3|hyperpage}{67}
+\indexentry{Chapter2.4.2.3|hyperpage}{66}
 \indexentry{Chapter2.5|hyperpage}{68}
-\indexentry{Chapter2.5.1|hyperpage}{69}
+\indexentry{Chapter2.5.1|hyperpage}{68}
 \indexentry{Chapter2.5.2|hyperpage}{70}
-\indexentry{Chapter2.5.3|hyperpage}{74}
-\indexentry{Chapter2.6|hyperpage}{76}
+\indexentry{Chapter2.5.3|hyperpage}{75}
+\indexentry{Chapter2.6|hyperpage}{77}
 \indexentry{Chapter3.1|hyperpage}{81}
 \indexentry{Chapter3.2|hyperpage}{83}
 \indexentry{Chapter3.2.1|hyperpage}{83}

Book/mt-book-xelatex.ptc

@@ -95,17 +95,17 @@
 \defcounter {refsection}{0}\relax 
 \contentsline {subsubsection}{古德-图灵估计法}{65}{section*.49}
 \defcounter {refsection}{0}\relax 
-\contentsline {subsubsection}{Kneser-Ney平滑方法}{67}{section*.51}
+\contentsline {subsubsection}{Kneser-Ney平滑方法}{66}{section*.51}
 \defcounter {refsection}{0}\relax 
 \contentsline {section}{\numberline {2.5}句法分析（短语结构）}{68}{section.2.5}
 \defcounter {refsection}{0}\relax 
-\contentsline {subsection}{\numberline {2.5.1}句子的句法树表示}{69}{subsection.2.5.1}
+\contentsline {subsection}{\numberline {2.5.1}句子的句法树表示}{68}{subsection.2.5.1}
 \defcounter {refsection}{0}\relax 
 \contentsline {subsection}{\numberline {2.5.2}上下文无关文法}{70}{subsection.2.5.2}
 \defcounter {refsection}{0}\relax 
-\contentsline {subsection}{\numberline {2.5.3}规则和推导的概率}{74}{subsection.2.5.3}
+\contentsline {subsection}{\numberline {2.5.3}规则和推导的概率}{75}{subsection.2.5.3}
 \defcounter {refsection}{0}\relax 
-\contentsline {section}{\numberline {2.6}小结及深入阅读}{76}{section.2.6}
+\contentsline {section}{\numberline {2.6}小结及深入阅读}{77}{section.2.6}
 \defcounter {refsection}{0}\relax 
 \contentsline {part}{\@mypartnumtocformat {II}{统计机器翻译}}{79}{part.2}
 \ttl@stoptoc {default@1}

Section03-Word-Based-Models/section03.tex

@@ -926,7 +926,7 @@
 \node [anchor=north west] (t1) at ([yshift=0.4em]s1.south west) {$t_1=$ Machine translation is just translation by computer};
 
 \node [anchor=north west] (s2) at (t1.south west) {$s_2=$ 那 人工 翻译 呢 ?};
-\node [anchor=north west] (t2) at ([yshift=0.4em]s2.south west) {$t_2=$ so , what is human translation ?};
+\node [anchor=north west] (t2) at ([yshift=0.4em]s2.south west) {$t_2=$ So , what is human translation ?};
 
 \end{tikzpicture}
 \end{flushleft}
@@ -937,7 +937,7 @@
 \begin{eqnarray}
 &   & \textrm{P}(\textrm{'翻译'},\textrm{'translation'}) \nonumber \\
 & = & \frac{c(\textrm{'翻译'},\textrm{'translation'};s^{[1]},t^{[1]})+c(\textrm{'翻译'},\textrm{'translation'};s^{[2]},t^{[2]})}{\sum_{x',y'} c(x',y';s^{[1]},t^{[1]}) + \sum_{x',y'} c(x',y';s^{[2]},t^{[2]})} \nonumber \\
-\visible<3->{& = & \frac{4 + 1}{|s^{[1]}| \times |t^{[1]}| + |s^{[2]}| \times |t^{[2]}|} = \frac{4 + 1}{9 \times 7 + 5 \times 7} = \frac{5}{102}} \nonumber
+\visible<3->{& = & \frac{4 + 1}{|s^{[1]}| \times |t^{[1]}| + |s^{[2]}| \times |t^{[2]}|} = \frac{4 + 1}{9 \times 7 + 5 \times 7} = \frac{5}{98}} \nonumber
 \end{eqnarray}
 }