443 lines
14 KiB
TeX
443 lines
14 KiB
TeX
\chapter{Motif}
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\section{Searching a substring in a string}
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\begin{algorithm}
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\caption{Brute-force search of a motif in a sequence}
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\begin{algorithmic}[1]
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\Function{FindMotif}{$S$: Array($n$), $M$: Array($m$)}
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\Returns{a list of position}
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\State $pos \gets \{\}$
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\State $i \gets 0$
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\While {$i < n - m + 1$}
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\State $j \gets 0$
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\While {$j < m$ and $S[i+j] = M[j]$}
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\State $j++$
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\EndWhile
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\If {$j = m$}
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\State $pos \gets pos \cup \{i\}$
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\EndIf
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\State $i++$
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\EndWhile
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\State \Return $pos$
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\EndFunction
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\end{algorithmic}
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\label{alg:naive-motif-matching}
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\end{algorithm}
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\begin{algorithm}
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\caption{Knuth-Morris-Pratt algorithm}
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\begin{algorithmic}[1]
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\Function{KMP\_Search}{$S$: Array($n$), $M$: Array($m$)}
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\Returns{Integer}
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\State $table \gets$ \Call{KMP\_Table}{$M$}
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\State $c \gets 0$ \Comment{Count the number of matches}
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\State $i \gets 0$
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\State $j \gets 0$
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\While {$i < n$}
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\If{$S[i] = M[i]$}
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\State $i \gets i + 1$
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\State $j \gets j + 1$
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\EndIf
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\If {$j = m$}
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\State $c \gets c + 1$
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\State $j \gets table[j-1]$
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\ElsIf {$j < n$ and $M[j] \neq S[i]$}
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\If {$j \neq 0$}
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\State $j \gets table[j-1]$
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\Else
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\State $i \gets i + 1$
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\EndIf
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\EndIf`
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\EndWhile
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\State \Return $c$
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\EndFunction
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\Function{KMP\_Table}{M: Array(m)}
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\State \textbf{Returns} Array(m)
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\State $previous \gets 0$
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\State $table \gets $ array of zeros of size m
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\For {$i = 0$; $i < m$; $i++$}
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\If {$M[i] = M[previous]$}
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\State $previous \gets previous + 1$
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\State $table[i] \gets previous$
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\State $i \gets i + 1$
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\Else
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\If {$previous = 0$}
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\State $previous \gets table[previous - 1]$
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\Else
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\State $table[i] \gets 0$
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\State $i \gets 1$
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\EndIf
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\EndIf
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\EndFor
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\section{Using matrices to search motifs}
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Let $S_{1}$ and $S_{2}$ be two sequences.
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$S_{1} = $ ACGUUCC
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$S_{2} = $ GUU
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\begin{table}
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\centering
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\begin{tabular}{c|ccccccc}
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& A & C & G & U & U & C & C \\
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\hline
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G & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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U & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
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U & 0 & 0 & 0 & 1 & 1 & 0 & 0
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\end{tabular}
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\caption{Comparison matrix}
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\end{table}
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Let $n = |S_{1}|$, $m = |S_{2}|$
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The complexity of this algorithm is $\mathcal{O}(n \cdot m)$ to build the matrix, and it requires also to find the diagonals and thus it is a bit less efficient than the \autoref{alg:naive-motif-matching}.
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To find repetitions, we can use a comparison matrix with a single sequence against itself. A repetition would appear as a diagonal of ones, not on the main diagonal.
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Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
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\begin{table}
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\includegraphics{./figures/part1/comparison_matrix_repetitions.pdf}
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\caption{Comparison matrix for $seq = $``ACGUUACGUUGUU"}
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\end{table}
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\begin{algorithm}
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\caption{Construct a comparison matrix}
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\begin{algorithmic}[1]
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\Function{ComparisonMatrix}{$S$: Array($n$)}
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\State $M \gets $ Array($n$, $n$)
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\For{($i = 0$; $i < n$; $i++$)}
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\For{$j = 0$; $j < n$; $j++$}
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\If {$S[i] = S[j]$}
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\State $M[i][j] = 1$
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\Else
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\State $M[i][j] = 0$
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\EndIf
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\EndFor
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\EndFor
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\State \Return $M$
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Construct the top half of a comparison matrix}
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\begin{algorithmic}[1]
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\Function{ComparisonMatrix}{$S$: Array($n$)}
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\State $M \gets$ Array($n$,$n$)
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\For{($i = 0$; $i < n$; $i++$)}
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\For{j=i; j < n; j++}
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\If {S[i] = S[j]}
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\State M[i][j] = 1
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\Else
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\State M[i][j] = 0
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\EndIf
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\EndFor
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\EndFor
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\State \Return M
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Find repetitions (with a set of visited segments)}
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\begin{algorithmic}[1]
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\Function{FindRepetions}{$S$: Array($n$)}
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\Returns{A list of start and end positions for repeated sequences}
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\State $M = $ \Call{ComparisonMatrix}{S}
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\State $pos = \{\}$
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\State $visited = \{\}$
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\For {($i_{start} = 0$; $i_{start} < n$; $i_{start}++$)}
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\For {($j_{start} = i_{start}+1$; $j_{start} < n$; $j_{start}++$)}
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\If{$M[i_{start}][j_{start}] = 1$ and $(i_{start}, j_{start}) \notin visited$}
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\State $i = i_{start}$
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\State $j = j_{start}$
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\While {$M[i][j] = 1$}
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\State $i++$
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\State $j++$
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\State $visited = visited \cup \{(i, j)\}$
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\EndWhile
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\State $pos = pos \cup \{(i_{start}, i), (j_{start},j)\}$
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\EndIf
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\EndFor
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\EndFor
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Find repetitions with an exploration of diagonals}
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\begin{algorithmic}[1]
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\Function{FindRepetions}{$S$: Array($n$)}
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\Returns{A list of start and end positions for repeted sequences}
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\State $M$ = \Call{ComparisonMatrix}{S}
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\State $pos = \{\}$
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\For {($diag = 1$; $diag < n$; $diag++$)}
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\State $j = diag$
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\State $i = 0$
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\While {$i < n$ and $j < n$}
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\If {$M[i][j] = 1$}
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\State $i_{start} = i$
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\State $j_{start} = j$
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\While {$i < n$ and $j < n$ and $M[i][j] = 1$}
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\State i++
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\State j++
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\EndWhile
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\State $pos = pos \cup \{((i_{start},i-1),(j_{start},j-1))\}$
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\EndIf
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\State $i++$
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\State $j++$
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\State
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\EndWhile
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\EndFor
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Find repetitions with an exploration of diagonals, without nested while}
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\begin{algorithmic}[1]
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\Function{FindRepetions}{$S$: Array($n$)}
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\Returns{A list of start positions for repeted sequences and match length}
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\State $M$ = \Call{ComparisonMatrix}{S}
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\State $pos = \{\}$
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\For {($diag = 1$; $diag < n$; $diag++$)}
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\State $j = diag$
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\State $i = 0$
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\State $l = 0$
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\While {$i < n$ and $j < n$}
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\If {$M[i][j] = 1$}
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\State $l++$
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\Else
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\If {$l > 0$}
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\State $pos = pos \cup \{(i-l,j-l,l)\}$
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\State $l = 0$
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\EndIf
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\EndIf
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\State $i++$
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\State $j++$
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\EndWhile
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\If {$l > 0$}
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\State $pos = pos \cup \{((i-l,j-l,l))\}$
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\EndIf
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\EndFor
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\State \Return $pos$
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Find repetitions}
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\begin{algorithmic}[1]
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\Function{FindRepetions}{$S$: Array($n$)}
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\Returns{A list of start and end positions for repeted sequences}
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\State $M$ = \Call{ComparisonMatrix}{S}
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\State $pos = \{\}$
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\For {$i_{start} = 0$; $i_{start} < n$; $i_{start}++$}
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\For {$j_{start} = i_{start}+1$; $j_{start} < n$; $j_{start}++$}
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\If{$M[i_{start}][j_{start}] = 1$}
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\State $i = i_{start}$
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\State $j = j_{start}$
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\While {$M[i][j] = 1$}
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\State $M[i][j] = 0$ \Comment{Ensure that the segment is not explored again}
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\State $i++$
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\State $j++$
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\EndWhile
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\State $pos = pos \cup \{((i_{start}, i-1), (j_{start},j-1))\}$
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\EndIf
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\EndFor
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\EndFor
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\section{Automata}
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An automaton is a tuple $\langle S, s_{0}, T, \Sigma,f\rangle$
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\begin{itemize}
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\item $S$ the set of states
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\item $s_{0}$ the initial state
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\item $T$ the set of terminal states
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\item $\Sigma$ the alphabet
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\item $f$ the transition function $f: (s_{1}, c) \to s_{2}$
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\end{itemize}
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\paragraph{Example} Given the language $L$ on the alphabet $\Sigma = \{A, C, T\}$, $L = \{A^{*}, CTT, CA^{*}\}$
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\begin{definition}[Deterministic automaton]
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An automaton is deterministic, if for each couple $(p, a) \in S \times \Sigma$ it exists at most a state $q$ such as $f(p, a) = q$
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\end{definition}
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\begin{definition}[Complete automaton]
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An automaton is complete, if for each couple $(p, a) \in S \times \Sigma$ it exists at least a state $q$ such as $f(p, a) = q$.
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\end{definition}
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\begin{algorithm}
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\caption{Check wether a word belong to a language for which we have an automaton}
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\begin{algorithmic}[1]
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\Function{WordInLanguage}{$W$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma,f \rangle$}
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\Returns{A Boolean valued to \True{} if the word is recognized by the language automaton}
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\State $s \gets s_{0}$
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\State $i \gets 0$
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\While {$i < n$}
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\State $a \gets W[i]$
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\If {$\exists f(s, a)$}
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\State $s \gets f(s, a)$
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\Else
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\State \Return \False
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\EndIf
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\State i++
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\EndWhile
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\If {$s \in T$}
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\State \Return \True
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\Else
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\State \Return \False
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\EndIf
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\subsection{Suffix Automaton}
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Let $S = $ AACTACT
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A suffix automata recognize all suffix of a given sequence.
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The suffix language of $S$ is $\{S, ACTACT, CTACT, TACT, ACT, CT, T\}$.
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\begin{figure}
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\centering
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\includegraphics{./figures/part1/minimal_suffix_automaton_exercise.pdf}
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\caption{Suffix automaton for $S = $ AACTACT}
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\end{figure}
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\begin{figure}
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\centering
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\includegraphics{./figures/part1/minimal_suffix_automaton_exercise_bis.pdf}
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\caption{Suffix automaton for $S = $ TCATCATT}
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\end{figure}
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\begin{algorithm}
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\caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$, built from the automaton}
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\begin{algorithmic}[1]
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\Function{CheckMotifInSuffixAutomaton}{$W$: Array($m$), $A$: $\langle S, s_{0}, T, \Sigma,f \rangle$}
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\Returns{Boolean valued to \True{} if the motif is in the sequence}
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\State $s \gets s_{0}$
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\State $i \gets 0$
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\While {$i < m$ and $\exists f(s, W[i])$}
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\State $s \gets f(s, W[i])$
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\State $i++$
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\EndWhile
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\If {$i=n$}
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\State \Return \True
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\Else
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\State \Return \False
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\EndIf
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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The complexity of the pattern matching algorithm is $\mathcal{O}(n + m)$, because building the automaton is $\mathcal{O}(m)$
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\subsection{Automata for motif search}
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Let $M$ be a motif $M = $ ACAT.
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\begin{figure}
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\centering
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\includegraphics{./figures/part1/motif_search_automaton.pdf}
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\caption{Motif search automaton for $M = $ ACAT}
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\end{figure}
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The alphabet of motif is the same as the alphabet of the sequence.
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The search automaton is complete.
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If the there exists a letter $c$ in the sequence that is not
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in the motif alphabet, we can make a virtual transition from
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each state to the initial state whenever we encounter an unknown letter.
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\begin{algorithm}
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\caption{Search a motif in a sequence with an automaton}
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\begin{algorithmic}[1]
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\Function{SearchMotif}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
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\Returns{A set of positions where the motif has been found}
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\State $s \gets s_0$
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\State $i \gets 0$
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\State $pos \gets \{\}$
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\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
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\If {$s \in T$}
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\State $pos \gets pos \cup \{ i - m \}$
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\EndIf
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\State $s \gets f(s, S[i])$
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\State $i++$
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\EndWhile
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\State \Return $pos$
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Check if the a motif automaton recognizes only the prefix of size $m-1$ of a motif $P$ of size $m$ }
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\begin{algorithmic}[1]
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\Function{SearchMotifLastPrefix}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
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\Returns{A set of positions where the motif has been found}
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\State $s \gets s_0$
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\State $i \gets 0$
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\State $T_{new} \gets \{\}$
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\For {$s \in S$}
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\For {$a \in \Sigma$}
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\For {$t \in T$}
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\If {$\exists f(s, a)$ and $f(s, a) = t$}
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\State $T_{new} \gets T_{new} \cup s$
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\EndIf
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\EndFor
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\EndFor
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\EndFor
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\While {$i < n$}
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\If {$s \in T_{new}$}
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\State \Return \True
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\EndIf
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\State $s \gets f(s, S[i])$
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\State $i++$
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\EndWhile
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\State \Return \False
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\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}
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\begin{algorithmic}[1]
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\Function{SearchMotifLastPrefix}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
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\Returns{A set of positions where the motif has been found}
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\State $s \gets s_0$
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\State $i \gets 0$
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\While {$i < n$ and $f(s, P[m-1]) \notin T$}
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\State $s \gets f(s, S[i])$
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\State $i++$
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\EndWhile
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\If{$f(s, P[m-1]) \in T$}
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\State \Return \True
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\Else
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\State \Return \False
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\EndIf
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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