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Le virgole sono importanti

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zao zao, mangia un po' di sushi anche x me :D
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Francesca
2026-01-18 23:32:00 +01:00
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6
src/year2/natural-language-processing/sections/_language_models.tex
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@ -190,7 +190,7 @@
This is equivalent to computing the probability of the whole sentence, which expanded using the chain rule becomes:
\[
\begin{split}
\prob{w_1, \dots, w_{i-1} w_i} &= \prob{w_1} \prob{w_2 | w_1} \prob{w_3 | w_{1..2}} \dots \prob{w_n | w_{1..n-1}} \\
\prob{w_1, \dots, w_{i-1}, w_i} &= \prob{w_1} \prob{w_2 | w_1} \prob{w_3 | w_{1..2}} \dots \prob{w_n | w_{1..n-1}} \\
&= \prod_{i=1}^{n} \prob{w_i | w_{1..i-1}}
\end{split}
\]
@ -224,7 +224,7 @@
\begin{description}
\item[Estimating $\mathbf{N}$-gram probabilities]
Consider the bigram case, the probability that a token $w_i$ follows $w_{i-1}$ can be determined through counting:
\[ \prob{w_i | w_{i-1}} = \frac{\texttt{count}(w_{i-1} w_i)}{\texttt{count}(w_{i-1})} \]
\[ \prob{w_i | w_{i-1}} = \frac{\texttt{count}(w_{i-1}, w_i)}{\texttt{count}(w_{i-1})} \]
\end{description}
\begin{remark}
@ -378,4 +378,4 @@ Only for $n$-grams that occur enough times a representative probability can be e
c^* = \big( \texttt{count}(w_{i-1}w_i) + 1 \big) \frac{\texttt{count}(w_{i-1})}{\texttt{count}(w_{i-1}) + \vert V \vert}
\]
\end{example}
\end{description}
\end{description}