229 lines
7.9 KiB
TeX
229 lines
7.9 KiB
TeX
\chapter{Sequence alignment}
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\iffalse
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\begin{algorithm}
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\caption{Needleman-Wunsch Algorithm}
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\begin{algorithmic}[1]
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\Procedure{FillMatrix}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
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\Comment{$sub(a, b)$ is the substitution score, $del(a)$ and $ins(a)$ are the deletion and insertion penalty, in regard with the reference $S_{1}$ sequence}
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\State $M = $ Array($m+1$, $n+1$)
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\Comment{Initialize the matrix first column and first row}
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\State $P = $ Array($m$, $n$) \Comment{Store the direction of the cell we chose to build the next cell up on.}
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\State $M[0][0] = 0$
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\For {($i = 1$; $i < m+1$; $i++$)}
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\State $M[i][0] = M[i-1][0] + gap\_penalty$
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\EndFor
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\For {($j = 1$; $j < n+1$; $j++$)}
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\State $M[0][j] = M[0][j-1] + gap\_penalty$
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\EndFor
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\Comment{Fill the remaining matrix}
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\For {($i = 1$; $i < m+1$; $i++$)}
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\For {($j = 1$; $j < n+1$; $j++$)}
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\State $delete = M[i-1][j] + gap\_penalty$
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\State $insert = M[i][j-1] + gap\_penalty$
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\State $substitute = M[i-1][j-1] + sub(S_{1}[i-1], S_{2}[j-1])$
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\State $choice = \min \{delete, insert, substitute\}$
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\If {$substitute = choice$}
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\State $P[i-1][j-1] = '\nwarrow'$
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\ElsIf {$deletion = choice$}
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\State $P[i-1][j-1] = '\leftarrow'$
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\Else
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\State $P[i-1][j-1] = '\uparrow'$
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\EndIf
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\State $M[i][j] = choice$
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\EndFor
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\EndFor
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Needleman-Wunsch Algorithm (Backtrack)}
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\begin{algorithmic}[1]
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\Procedure{ShowAlignment}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
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\State $extend_{1} = ''$
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\State $extend_{2} = ''$
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\State $i = m$
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\State $j = n$
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\While{$i > 0$ and $j > 0$}
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\If {$P[i-1][j-1] = '\nwarrow'$}
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\State $extend_{1} = S_{1}[i-1] \circ extend_{1}$
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\State $extend_{2} = S_{2}[j-1] \circ extend_{2}$
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\State $i--$
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\State $j--$
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\ElsIf {$P[i-1][j-1] = '\uparrow'$}
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\State $extend_{1} = S_{1}[i-1] \circ extend_{1}$
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\State $extend_{2} = '-' \circ extend_{2}$
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\State $i--$
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\Else
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\State $extend_{1} = '-' \circ extend_{1}$
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\State $extend_{2} = S_{2}[j-1] \circ extend_{2}$
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\State $j--$
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\EndIf
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\EndWhile
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\While{$i > 0$}
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\State $extend_{1} = S_{1}[i-1] \circ extend_{1}$
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\State $extend_{2} = '-' \circ extend_{2}$
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\State $i--$
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\State \Call{Insert}{0, $alignment$,$tuple$}
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\EndWhile
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\While{$j > 0$}
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\State $extend_{1} = '-' \circ extend_{1}$
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\State $extend_{2} = S_{2}[j-1] \circ extend_{2}$
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\State $j--$
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\EndWhile
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\State \Call{print}{$extend_{1}$}
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\State \Call{print}{$extend_{2}$}
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\EndProcedure
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\State \Call{FillMatrix}{$S_{1}$, $S_{2}$}
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\State \Call{ShowAlignment}{$S_{1}$, $S_{2}$}
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\end{algorithmic}
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\end{algorithm}
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\fi
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\begin{algorithm}
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\caption{Needleman-Wunsch Algorithm, Build the matrix}
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\begin{algorithmic}[1]
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\Procedure{FillMatrix}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
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\State $M = $ Array($m+1$, $n+1$)
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\State $P = $ Array($m$, $n$)
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\Comment{Store the direction of the cell we chose to build the next cell up on.}
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\State $M[0][0] = 0$
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\For {($i = 1$; $i < m+1$; $i++$)}
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\State $M[i][0] = M[i-1][0] + gap\_penalty$
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\EndFor
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\For {($j = 1$; $j < n+1$; $j++$)}
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\State $M[0][j] = M[0][j-1] + gap\_penalty$
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\EndFor
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\For {($i = 1$; $i < m+1$; $i++$)}
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\For {($j = 1$; $j < n+1$; $j++$)}
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\State $delete = M[i-1][j] + gap\_penalty$
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\State $insert = M[i][j-1] + gap\_penalty$
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\State $substitute = M[i-1][j-1] + sub(S_{1}[i-1], S_{2}[j-1])$
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\State $M[i][j] = \min \{substitute, insert, delete\}$
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\EndFor
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\EndFor
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Needleman-Wunsch Algorithm, reconstruct the alignment}
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\begin{algorithmic}[1]
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\Procedure{BacktrackAlignment}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
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\State $alignment = LinkedList$
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\State $i = m$
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\State $j = n$
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\While{$i > 0$ and $j > 0$}
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\If {$M[i-1][j-1] = M[i][j] - sub(S_{1}[i-1], S_{2}[j-1])$}
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\State $tuple = \begin{pmatrix} S_{1}[i-1] \\ S_{2}[j-1] \end{pmatrix}$
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\State $i--$
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\State $j--$
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\ElsIf {$M[i-1][j-1] = M[i][j-1] - gap\_penalty$}
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\State $tuple = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix}$
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\State $i--$
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\Else
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\State $tuple = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix}$
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\State $j--$
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\EndIf
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\State \Call{Insert}{0, $alignment$,$tuple$}
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\EndWhile
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\While{$i > 0$}
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\State $tuple = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix}$
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\State $i--$
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\State \Call{Insert}{0, $alignment$,$tuple$}
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\EndWhile
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\While{$j > 0$}
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\State $tuple = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix}$
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\State $j--$
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\State \Call{Insert}{0, $alignment$,$tuple$}
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\EndWhile
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\EndProcedure
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\State \Call{FillMatrix}{$S_{1}$, $S_{2}$}
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\State \Call{BacktrackAlignment}{$S_{1}$, $S_{2}$}
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Backtrack a single alignment in a recursive way}
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\begin{algorithmic}[1]
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\State $S_{1}$: Array($m$), $S_{2}$: Array($n$), $M$: Array($m+1$, $n+1$),
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\Function{BacktrackRecurse}{$i$, $j$}
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\If {$i > 0$ and $j > 0$}
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\If {$M[i-1][j-1] = M[i][j] - sub(S_{1}[i-1], S_{2}[j-1])$}
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\State $z = $ \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i-1$, $j-1$}
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\State $z = \begin{pmatrix} S_{1}[i-1] \\ S_{2}[j-1] \end{pmatrix} \circ z$
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\ElsIf {$M[i-1][j] + gap\_penalty = M[i][j]$}
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\State $z = $ \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i-1$, $j$}
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\State $z = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix} \circ z$
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\Else
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\State $z = $ \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i$, $j-1$}
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\State $z = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix} \circ z$
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\EndIf
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\ElsIf {$i > 0$}
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\State $z = $ \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i-1$, $j$}
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\State $z = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix} \circ z$
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\ElsIf {$j > 0$}
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\State $z = $ \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i$, $j-1$}
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\State $z = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix} \circ z$
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\Else
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\State \Return []
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\EndIf
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\Else
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\State \Return $z$
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\EndIf
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\EndFunction
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\Function{Backtrack}{$S_{1}$: Array($m$), $S_{2}$: Array($n$), $M$: Array($m+1$, $n+1$)}
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\State \Return \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $m$, $n$}
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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{Backtrack all the optimum alignments in a recursive way}
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\begin{algorithmic}[1]
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\Procedure{BacktrackRecurse}{$S_{1}$: Array($m$), $S_{2}$: Array($n$), $M$: Array($m+1$, $n+1$), $i$, $j$}
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\If {$i > 0$ and $j > 0$}
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\If {$M[i-1][j-1] = M[i][j] - sub(S_{1}[i-1], S_{2}[j-1])$}
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\State $value = \begin{pmatrix} S_{1}[i-1] \\ S_{2}[j-1] \end{pmatrix}$
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\State $z' = value \circ z$
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\State \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i-1$, $j-1$, $z'$}
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\EndIf
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\If {$M[i-1][j] + gap\_penalty = M[i][j]$}
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\State $value = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix}$
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\State $z' = value \circ z$
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\State \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i-1$, $j$, $z'$}
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\EndIf
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\If {$M[i][j-1] + gap\_penalty = M[i][j]$}
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\State $value = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix}$
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\State $z' = value \circ z$
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\State \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i$, $j-1$, $z'$}
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\EndIf
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\ElsIf {$i > 0$}
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\State $value = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix}$
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\State $z' = value \circ z$
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\State \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i-1$, $j$, $z'$}
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\ElsIf {$j > 0$}
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\State $value = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix}$
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\State $z' = value \circ z$
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\State \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $i$, $j-1$, $z'$}
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\Else
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\State \Call{print}{$z$}
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\EndIf
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\EndProcedure
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\Procedure{Backtrack}{$S_{1}$: Array($m$), $S_{2}$: Array($n$), $M$: Array($m+1$, $n+1$)}
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\State \Return \Call{BacktrackRecurse}{$S_{1}$, $S_{2}$, $M$, $m$, $n$, []}
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\begin{figure}
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\centering
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\includegraphics{figures/part2/needle.pdf}
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\caption{Needleman-Wunsch global alignment matrix with an example of optimal path.}
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\end{figure}
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