feat: Add more algorithm on automata

This commit is contained in:
Samuel Ortion 2024-03-15 11:40:26 +01:00
parent 5fb762e73f
commit fc0331e054
7 changed files with 94 additions and 8 deletions

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@ -11,7 +11,7 @@
}
}
\includechapters{part1}{2}
\includechapters{part1}{3}
% \includechapters{part2}{2}

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@ -58,6 +58,6 @@
\EndIf
\EndFor
\State \Return $c$
\EndFunction
\EndFunction
\end{algorithmic}
\end{algorithm}

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@ -51,8 +51,8 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\State \Return $M$
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Construct the top half of a comparison matrix}
\begin{algorithmic}[1]
@ -259,7 +259,7 @@ The suffix language of $S$ is $\{S, ACTACT, CTACT, TACT, ACT, CT, T\}$.
\end{figure}
\begin{algorithm}
\caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$}
\caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$, built from the automaton}
\begin{algorithmic}[1]
\Function{CheckMotifInSuffixAutomaton}{$W$: Array($m$), $A$: $\langle S, s_{0}, T, \Sigma,f \rangle$}
\Returns{Boolean valued to \True{} if the motif is in the sequence}

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@ -0,0 +1,84 @@
\chapter{Automata for motif search}
Let $M$ be a motif $M = $ ACAT.
\begin{figure}
\centering
\includegraphics{figures/part1/motif_search_automaton.pdf}
\caption{Motif search automaton for $M = $ ACAT}
\end{figure}
The alphabet of motif is the same as the alphabet of the sequence.
The search automaton is complete.
If the there exists a letter $c$ in the sequence that is not
in the motif alphabet, we can make a virtual transition from
each state to the initial state whenever we encounter an unknown letter.
\begin{algorithm}
\caption{Search a motif in a sequence with an automaton}
\begin{algorithmic}[1]
\Function{SearchMotif}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
\Returns{A set of positions where the motif has been found}
\State $s \gets s_0$
\State $i \gets 0$
\State $pos \gets \{\}$
\While {$i < n$} % $\exists f(s, S[i])$ We assume $S$ and $P$ are formed on the same alphabet, so we could remove the second check, as $A$ is complete
\If {$s \in T$}
\State $pos \gets pos \cup \{ i - m \}$
\EndIf
\State $s \gets f(s, S[i])$
\State $i++$
\EndWhile
\State \Return $pos$
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Check if the a motif automaton recognizes only the prefix of size $m-1$ of a motif $P$ of size $m$ }
\begin{algorithmic}[1]
\Function{SearchMotifLastPrefix}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
\Returns{A set of positions where the motif has been found}
\State $s \gets s_0$
\State $i \gets 0$
\State $T_{new} \gets \{\}$
\For {$s \in S$}
\For {$a \in \Sigma$}
\For {$t \in T$}
\If {$\exists f(s, a)$ and $f(s, a) = t$}
\State $T_{new} \gets T_{new} \cup s$
\EndIf
\EndFor
\EndFor
\EndFor
\While {$i < n$}
\If {$s \in T_{new}$}
\State \Return \True
\EndIf
\State $s \gets f(s, S[i])$
\State $i++$
\EndWhile
\State \Return \False
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Check if the a motif automaton recognizes only the prefix of size $m-1$ of a motif $P$ of size $m$, knowing the sequence of the motif}
\begin{algorithmic}[1]
\Function{SearchMotifLastPrefix}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
\Returns{A set of positions where the motif has been found}
\State $s \gets s_0$
\State $i \gets 0$
\While {$i < n$ and $f(s, P[m-1]) \notin T$}
\State $s \gets f(s, S[i])$
\State $i++$
\EndWhile
\If{$f(s, P[m-1]) \in T$}
\State \Return \True
\Else
\State \Return \False
\EndIf
\EndFunction
\end{algorithmic}
\end{algorithm}

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main.pdf (Stored with Git LFS)

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tmp.pdf (Stored with Git LFS)

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@ -13,4 +13,6 @@
\begin{document}
\end{document}
\end{document}