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Author SHA1 Message Date
5e29891709 Merge pull request #7 from fraktc/patch-1
Avevi messo un indice di troppo :P
2026-01-19 02:44:05 +01:00
8a6e886fc9 Merge pull request #6 from fraktc/patch-2
Le virgole sono importanti
2026-01-19 02:43:34 +01:00
4e1020dbe0 Le virgole sono importanti
zao zao, mangia un po' di sushi anche x me :D
2026-01-18 23:32:00 +01:00
444404f1a8 Avevi messo un indice di troppo :P
zao
2026-01-18 23:27:17 +01:00
2 changed files with 5 additions and 5 deletions

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@ -395,7 +395,7 @@ Logistic regression has the following properties:
\bottomrule
\end{tabular}
\end{center}
A possible way to sample $\bar{x}^{(1)}$ is (w.r.t. the indexes of the examples in $x$) $[2, 3, 3, 2, 4, 6, 2, 4, 1, 9, 1]$.
A possible way to sample $\bar{x}^{(1)}$ is (w.r.t. the indexes of the examples in $x$) $[2, 3, 3, 2, 4, 6, 2, 4, 1, 9]$.
\end{example}
\end{description}

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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}