Add variation on Needleman - Wunsch algorithm
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@ -11,7 +11,7 @@
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}
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}
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\includechapters{part1}{3}
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\includechapters{part2}{2}
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\includechapters{part1}{4}
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\includechapters{part2}{4}
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% \includechapters{part3}{1}
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@ -0,0 +1,116 @@
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\chapter{Longest common subsequence}
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Let $S_{1} = \text{ATCTGAT}$ and $S_{2} = \text{TGCATA}$.
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In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
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\begin{algorithm}
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\caption{Construct a longest common subsequence matrix}
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\begin{algorithmic}[1]
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\Function{LCSQ\_Matrix}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
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\State $M \gets $ Array($m+1$, $n+1$)
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\For{($i = 0$; $i < n+1$; $i++$)}
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\For{$j = 0$; $j < m+1$; $j++$}
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\If {$i = 0$ or $j = 0$}
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\State $M[i][j] = 0$
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\Else
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\If {$S_{1}[i] = S_{2}[j]$}
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\State $match = M[i-1][j-1] + 1$
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\Else
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\State $match = M[i-1][j-1]$
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\EndIf
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\State $gap_{1} = M[i-1][j]$
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\State $gap_{2} = M[i][j-1]$
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\State $M[i][j] = \max \{ match, gap_{1}, gap_{2}\}$
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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 a longest common subsequence matrix keeping the path in memory}
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\begin{algorithmic}[1]
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\Function{LCSQ\_Matrix\_Path}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
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\State $M \gets $ Array($m+1$, $n+1$)
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\State $P \gets $ Array($m+1$, $n+1$)
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\For {($i = 0$; $i < n+1$, $i++$)}
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\State $M[i][0] \gets 0$
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\EndFor
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\For {($j = 0$; $j < m+1$; $j+$)}
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\State $M[0][j] \gets 0$
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\EndFor
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\For{($i = 1$; $i < n+1$; $i++$)}
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\For{($j = 1$; $j < m+1$; $j++$)}
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\If {$i = 1$ or $j = 0$}
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\State $M[i][j] = 0$
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\Else
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\If {$S_{1}[i-1] = S_{2}[j-1]$}
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\State $M[i][j] \gets M[i-1][j-1] + 1$
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\State $P[i][j] \gets '\nwarrow'$
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\ElsIf {$M[i][j-1] \geq M[i-1][j]$}
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\State $M[i][j] \gets M[i][j-1]$
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\State $P[i][j] \gets '\leftarrow'$
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\Else
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\State $M[i][j] \gets M[i-1][j]$
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\State $P[i][j] \gets '\downarrow'$
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\EndIf
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\EndFor
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\EndFor
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\State \Return $M, P$
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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 the longest common subsequence}
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\begin{algorithmic}[1]
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\Function{LCSQ}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
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\State $M, P \gets $ \Call{LCSQ\_Matrix}{$S_{1}$, $S_{2}$}
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\State $L \gets Array(M[n][m])$
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\State $k \gets 0$
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\State $i \gets n$
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\State $j \gets m$
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\While{$i > 0$ and $j > 0$}
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\If {$P[i][j] = '\nwarrow' $}
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\State $L[k] \gets S_{1}[i]$
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\State $i--$
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\State $j--$
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\State $k++$
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\ElsIf {$P[i][j] = '\leftarrow'$}
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\State $j--$
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\Else
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\State $i--$
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\EndIf
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\EndWhile
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\State \Return $L$
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\iffalse
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\begin{algorithm}
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\caption{Recursive reconstruction of the longest common subsequence}
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\begin{algorithmic}[1]
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\Procedure{LCSQ}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
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\State $M, P \gets $ \Call{LCSQ\_Matrix}{$S_{1}$, $S_{2}$}
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\State $i \gets n$
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\State $j \gets m$
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\State \Call{Aux}{$P$, $S_{1}$, $i$, $j$}
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\EndProcedure
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\Procedure{Aux}{$P$: Array($n+1$, $m+1$), $S_{1}$: Array($n$), $i$, $j$}
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\If {$P[i][j] = '\nwarrow' $}
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\State $l \gets S_{1}[i]$
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\State \Call{Aux}{$P$, $S_{1}$, $i-1$, $j-1$}
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\State \texttt{print}($l$)
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\ElsIf {$P[i][j] = '\leftarrow'$}
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\State \Call{Aux}{$P$, $S_{1}$, $i$, $j-1$}
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\Else
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\State \Call{Aux}{$P$, $S_{1}$, $i-1$, $j$}
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\EndIf
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\fi
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@ -55,6 +55,7 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
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\State $M[i][j] \gets M[i-1][j]$
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\State $P[i][j] \gets '\downarrow'$
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\EndIf
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\EndIf
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\EndFor
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\EndFor
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\State \Return $M, P$
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@ -88,7 +89,8 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\iffalse
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\begin{algorithm}
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\caption{Recursive reconstruction of the longest common subsequence}
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\begin{algorithmic}[1]
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\Procedure{LCSQ}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
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@ -110,4 +112,5 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
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\EndIf
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\end{algorithm}
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\fi
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@ -10,30 +10,37 @@
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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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\For {($i = 0$; $i < m+1$; $i++$)}
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\State $M[i][0] = i * del(S_{1}[i])$
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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 {($j = 0$; $j < n+1$; $j++$)}
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\State $M[0][j] = j * ins(S_{2}[j])$
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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 $delete = M[i-1][j] + del(S_{1}[i-1])$
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\State $insert = M[i][j-1] + ins(S_{2}[j-1])$
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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 $choice = \max \{delete, insert, substitute\}$
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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 {$insertion = choice$}
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\State $P[i-1][j-1] = '\uparrow'$
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\Else
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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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@ -47,18 +54,92 @@
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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 $extend_{2} =\quad '-' \circ extend_{2}$
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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 $extend_{1} =\quad '-' \circ extend_{1}$
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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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\State \Call{ShowAlignment}{$S_{1}$, $S_{2}$}
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\end{algorithmic}
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\end{algorithm}
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\begin{algorithm}
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\caption{Needleman-Wunsch Algorithm (Backtrack) }
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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, using proper notation (Backtrack)}
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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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39
tmp.tex
39
tmp.tex
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@ -11,6 +11,43 @@
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\input{definitions.tex}
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\begin{document}
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\begin{algorithm}
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\caption{Construct a longest common subsequence matrix keeping the path in memory}
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\begin{algorithmic}[1]
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\Function{LCSQ\_Matrix\_Path}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
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\State $M \gets $ Array($m+1$, $n+1$)
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\State $P \gets $ Array($m+1$, $n+1$)
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\For {($i = 0$; $i < n+1$, $i++$)}
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\State $M[i][0] \gets 0$
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\EndFor
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\For {($j = 0$; $j < m+1$; $j+$)}
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\State $M[0][j] \gets 0$
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\EndFor
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\For{($i = 1$; $i < n+1$; $i++$)}
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\For{($j = 1$; $j < m+1$; $j++$)}
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\If {$i = 1$ or $j = 0$}
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\State $M[i][j] = 0$
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\Else
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\If {$S_{1}[i-1] = S_{2}[j-1]$}
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\State $M[i][j] \gets M[i-1][j-1] + 1$
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\State $P[i][j] \gets '\nwarrow'$
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\ElsIf {$M[i][j-1] \geq M[i-1][j]$}
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\State $M[i][j] \gets M[i][j-1]$
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\State $P[i][j] \gets '\leftarrow'$
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\Else
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\State $M[i][j] \gets M[i-1][j]$
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\State $P[i][j] \gets '\downarrow'$
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\EndIf
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\EndIf
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\EndFor
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\EndFor
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\State \Return $M, P$
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\EndFunction
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\end{algorithmic}
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\end{algorithm}
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\iffalse
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\begin{algorithm}
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\caption{Backtrack the longest common subsequence}
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\begin{algorithmic}[1]
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@ -60,7 +97,7 @@
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\fi
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\end{document}
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\end{document}
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