Add variation on Needleman - Wunsch algorithm
This commit is contained in:
parent
e945027027
commit
e64a1d711a
|
|
@ -11,7 +11,7 @@
|
|||
}
|
||||
}
|
||||
|
||||
\includechapters{part1}{3}
|
||||
\includechapters{part2}{2}
|
||||
\includechapters{part1}{4}
|
||||
\includechapters{part2}{4}
|
||||
|
||||
% \includechapters{part3}{1}
|
||||
|
|
|
|||
|
|
@ -0,0 +1,116 @@
|
|||
\chapter{Longest common subsequence}
|
||||
|
||||
Let $S_{1} = \text{ATCTGAT}$ and $S_{2} = \text{TGCATA}$.
|
||||
In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
|
||||
\begin{algorithm}
|
||||
\caption{Construct a longest common subsequence matrix}
|
||||
\begin{algorithmic}[1]
|
||||
\Function{LCSQ\_Matrix}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
|
||||
\State $M \gets $ Array($m+1$, $n+1$)
|
||||
\For{($i = 0$; $i < n+1$; $i++$)}
|
||||
\For{$j = 0$; $j < m+1$; $j++$}
|
||||
\If {$i = 0$ or $j = 0$}
|
||||
\State $M[i][j] = 0$
|
||||
\Else
|
||||
\If {$S_{1}[i] = S_{2}[j]$}
|
||||
\State $match = M[i-1][j-1] + 1$
|
||||
\Else
|
||||
\State $match = M[i-1][j-1]$
|
||||
\EndIf
|
||||
\State $gap_{1} = M[i-1][j]$
|
||||
\State $gap_{2} = M[i][j-1]$
|
||||
\State $M[i][j] = \max \{ match, gap_{1}, gap_{2}\}$
|
||||
\EndIf
|
||||
\EndFor
|
||||
\EndFor
|
||||
\State \Return $M$
|
||||
\EndFunction
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Construct a longest common subsequence matrix keeping the path in memory}
|
||||
\begin{algorithmic}[1]
|
||||
\Function{LCSQ\_Matrix\_Path}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
|
||||
\State $M \gets $ Array($m+1$, $n+1$)
|
||||
\State $P \gets $ Array($m+1$, $n+1$)
|
||||
\For {($i = 0$; $i < n+1$, $i++$)}
|
||||
\State $M[i][0] \gets 0$
|
||||
\EndFor
|
||||
\For {($j = 0$; $j < m+1$; $j+$)}
|
||||
\State $M[0][j] \gets 0$
|
||||
\EndFor
|
||||
\For{($i = 1$; $i < n+1$; $i++$)}
|
||||
\For{($j = 1$; $j < m+1$; $j++$)}
|
||||
\If {$i = 1$ or $j = 0$}
|
||||
\State $M[i][j] = 0$
|
||||
\Else
|
||||
\If {$S_{1}[i-1] = S_{2}[j-1]$}
|
||||
\State $M[i][j] \gets M[i-1][j-1] + 1$
|
||||
\State $P[i][j] \gets '\nwarrow'$
|
||||
\ElsIf {$M[i][j-1] \geq M[i-1][j]$}
|
||||
\State $M[i][j] \gets M[i][j-1]$
|
||||
\State $P[i][j] \gets '\leftarrow'$
|
||||
\Else
|
||||
\State $M[i][j] \gets M[i-1][j]$
|
||||
\State $P[i][j] \gets '\downarrow'$
|
||||
\EndIf
|
||||
\EndFor
|
||||
\EndFor
|
||||
\State \Return $M, P$
|
||||
\EndFunction
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Backtrack the longest common subsequence}
|
||||
\begin{algorithmic}[1]
|
||||
\Function{LCSQ}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
|
||||
\State $M, P \gets $ \Call{LCSQ\_Matrix}{$S_{1}$, $S_{2}$}
|
||||
\State $L \gets Array(M[n][m])$
|
||||
\State $k \gets 0$
|
||||
\State $i \gets n$
|
||||
\State $j \gets m$
|
||||
\While{$i > 0$ and $j > 0$}
|
||||
\If {$P[i][j] = '\nwarrow' $}
|
||||
\State $L[k] \gets S_{1}[i]$
|
||||
\State $i--$
|
||||
\State $j--$
|
||||
\State $k++$
|
||||
\ElsIf {$P[i][j] = '\leftarrow'$}
|
||||
\State $j--$
|
||||
\Else
|
||||
\State $i--$
|
||||
\EndIf
|
||||
\EndWhile
|
||||
\State \Return $L$
|
||||
\EndFunction
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\iffalse
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Recursive reconstruction of the longest common subsequence}
|
||||
\begin{algorithmic}[1]
|
||||
\Procedure{LCSQ}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
|
||||
\State $M, P \gets $ \Call{LCSQ\_Matrix}{$S_{1}$, $S_{2}$}
|
||||
\State $i \gets n$
|
||||
\State $j \gets m$
|
||||
\State \Call{Aux}{$P$, $S_{1}$, $i$, $j$}
|
||||
\EndProcedure
|
||||
|
||||
\Procedure{Aux}{$P$: Array($n+1$, $m+1$), $S_{1}$: Array($n$), $i$, $j$}
|
||||
\If {$P[i][j] = '\nwarrow' $}
|
||||
\State $l \gets S_{1}[i]$
|
||||
\State \Call{Aux}{$P$, $S_{1}$, $i-1$, $j-1$}
|
||||
\State \texttt{print}($l$)
|
||||
\ElsIf {$P[i][j] = '\leftarrow'$}
|
||||
\State \Call{Aux}{$P$, $S_{1}$, $i$, $j-1$}
|
||||
\Else
|
||||
\State \Call{Aux}{$P$, $S_{1}$, $i-1$, $j$}
|
||||
\EndIf
|
||||
\EndProcedure
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
\fi
|
||||
|
|
@ -55,6 +55,7 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
|
|||
\State $M[i][j] \gets M[i-1][j]$
|
||||
\State $P[i][j] \gets '\downarrow'$
|
||||
\EndIf
|
||||
\EndIf
|
||||
\EndFor
|
||||
\EndFor
|
||||
\State \Return $M, P$
|
||||
|
|
@ -88,6 +89,7 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
|
|||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\iffalse
|
||||
\begin{algorithm}
|
||||
\caption{Recursive reconstruction of the longest common subsequence}
|
||||
\begin{algorithmic}[1]
|
||||
|
|
@ -111,3 +113,4 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
|
|||
\EndProcedure
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
\fi
|
||||
|
|
|
|||
|
|
@ -10,30 +10,37 @@
|
|||
\State $M = $ Array($m+1$, $n+1$)
|
||||
\Comment{Initialize the matrix first column and first row}
|
||||
\State $P = $ Array($m$, $n$) \Comment{Store the direction of the cell we chose to build the next cell up on.}
|
||||
\For {($i = 0$; $i < m+1$; $i++$)}
|
||||
\State $M[i][0] = i * del(S_{1}[i])$
|
||||
\State $M[0][0] = 0$
|
||||
\For {($i = 1$; $i < m+1$; $i++$)}
|
||||
\State $M[i][0] = M[i-1][0] + gap\_penalty$
|
||||
\EndFor
|
||||
\For {($j = 0$; $j < n+1$; $j++$)}
|
||||
\State $M[0][j] = j * ins(S_{2}[j])$
|
||||
\For {($j = 1$; $j < n+1$; $j++$)}
|
||||
\State $M[0][j] = M[0][j-1] + gap\_penalty$
|
||||
\EndFor
|
||||
\Comment{Fill the remaining matrix}
|
||||
\For {($i = 1$; $i < m+1$; $i++$)}
|
||||
\For {($j = 1$; $j < n+1$; $j++$)}
|
||||
\State $delete = M[i-1][j] + del(S_{1}[i-1])$
|
||||
\State $insert = M[i][j-1] + ins(S_{2}[j-1])$
|
||||
\State $delete = M[i-1][j] + gap\_penalty$
|
||||
\State $insert = M[i][j-1] + gap\_penalty$
|
||||
\State $substitute = M[i-1][j-1] + sub(S_{1}[i-1], S_{2}[j-1])$
|
||||
\State $choice = \max \{delete, insert, substitute\}$
|
||||
\State $choice = \min \{delete, insert, substitute\}$
|
||||
\If {$substitute = choice$}
|
||||
\State $P[i-1][j-1] = '\nwarrow'$
|
||||
\ElsIf {$insertion = choice$}
|
||||
\State $P[i-1][j-1] = '\uparrow'$
|
||||
\Else
|
||||
\ElsIf {$deletion = choice$}
|
||||
\State $P[i-1][j-1] = '\leftarrow'$
|
||||
\Else
|
||||
\State $P[i-1][j-1] = '\uparrow'$
|
||||
\EndIf
|
||||
\State $M[i][j] = choice$
|
||||
\EndFor
|
||||
\EndFor
|
||||
\EndProcedure
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Needleman-Wunsch Algorithm (Backtrack)}
|
||||
\begin{algorithmic}[1]
|
||||
\Procedure{ShowAlignment}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
|
||||
\State $extend_{1} = ''$
|
||||
\State $extend_{2} = ''$
|
||||
|
|
@ -47,18 +54,92 @@
|
|||
\State $j--$
|
||||
\ElsIf {$P[i-1][j-1] = '\uparrow'$}
|
||||
\State $extend_{1} = S_{1}[i-1] \circ extend_{1}$
|
||||
\State $extend_{2} =\quad '-' \circ extend_{2}$
|
||||
\State $extend_{2} = '-' \circ extend_{2}$
|
||||
\State $i--$
|
||||
\Else
|
||||
\State $extend_{1} =\quad '-' \circ extend_{1}$
|
||||
\State $extend_{1} = '-' \circ extend_{1}$
|
||||
\State $extend_{2} = S_{2}[j-1] \circ extend_{2}$
|
||||
\State $j--$
|
||||
\EndIf
|
||||
\EndWhile
|
||||
\While{$i > 0$}
|
||||
\State $extend_{1} = S_{1}[i-1] \circ extend_{1}$
|
||||
\State $extend_{2} = '-' \circ extend_{2}$
|
||||
\State $i--$
|
||||
\State \Call{Insert}{0, $alignment$,$tuple$}
|
||||
\EndWhile
|
||||
\While{$j > 0$}
|
||||
\State $extend_{1} = '-' \circ extend_{1}$
|
||||
\State $extend_{2} = S_{2}[j-1] \circ extend_{2}$
|
||||
\State $j--$
|
||||
\EndWhile
|
||||
\State \Call{print}{$extend_{1}$}
|
||||
\State \Call{print}{$extend_{2}$}
|
||||
\EndProcedure
|
||||
\State \Call{FillMatrix}{$S_{1}$, $S_{2}$}
|
||||
\State \Call ShowAlignment($S_{1}$, $S_{2}$)
|
||||
\State \Call{ShowAlignment}{$S_{1}$, $S_{2}$}
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Needleman-Wunsch Algorithm (Backtrack) }
|
||||
\begin{algorithmic}[1]
|
||||
\Procedure{FillMatrix}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
|
||||
\State $M = $ Array($m+1$, $n+1$)
|
||||
\State $P = $ Array($m$, $n$)
|
||||
\Comment{Store the direction of the cell we chose to build the next cell up on.}
|
||||
\State $M[0][0] = 0$
|
||||
\For {($i = 1$; $i < m+1$; $i++$)}
|
||||
\State $M[i][0] = M[i-1][0] + gap\_penalty$
|
||||
\EndFor
|
||||
\For {($j = 1$; $j < n+1$; $j++$)}
|
||||
\State $M[0][j] = M[0][j-1] + gap\_penalty$
|
||||
\EndFor
|
||||
\For {($i = 1$; $i < m+1$; $i++$)}
|
||||
\For {($j = 1$; $j < n+1$; $j++$)}
|
||||
\State $delete = M[i-1][j] + gap\_penalty$
|
||||
\State $insert = M[i][j-1] + gap\_penalty$
|
||||
\State $substitute = M[i-1][j-1] + sub(S_{1}[i-1], S_{2}[j-1])$
|
||||
\State $M[i][j] = \min \{substitute, insert, delete\}$
|
||||
\EndFor
|
||||
\EndFor
|
||||
\EndProcedure
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Needleman-Wunsch Algorithm, using proper notation (Backtrack)}
|
||||
\begin{algorithmic}[1]
|
||||
\Procedure{BacktrackAlignment}{$S_{1}$: Array($m$), $S_{2}$: Array($n$)}
|
||||
\State $alignment = LinkedList$
|
||||
\State $i = m$
|
||||
\State $j = n$
|
||||
\While{$i > 0$ and $j > 0$}
|
||||
\If {$M[i-1][j-1] = M[i][j] - sub(S_{1}[i-1], S_{2}[j-1])$}
|
||||
\State $tuple = \begin{pmatrix} S_{1}[i-1] \\ S_{2}[j-1] \end{pmatrix}$
|
||||
\State $i--$
|
||||
\State $j--$
|
||||
\ElsIf {$M[i-1][j-1] = M[i][j-1] - gap\_penalty$}
|
||||
\State $tuple = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix}$
|
||||
\State $i--$
|
||||
\Else
|
||||
\State $tuple = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix}$
|
||||
\State $j--$
|
||||
\EndIf
|
||||
\State \Call{Insert}{0, $alignment$,$tuple$}
|
||||
\EndWhile
|
||||
\While{$i > 0$}
|
||||
\State $tuple = \begin{pmatrix} S_{1}[i-1] \\ \varepsilon \end{pmatrix}$
|
||||
\State $i--$
|
||||
\State \Call{Insert}{0, $alignment$,$tuple$}
|
||||
\EndWhile
|
||||
\While{$j > 0$}
|
||||
\State $tuple = \begin{pmatrix} \varepsilon \\ S_{2}[j-1] \end{pmatrix}$
|
||||
\State $j--$
|
||||
\State \Call{Insert}{0, $alignment$,$tuple$}
|
||||
\EndWhile
|
||||
\EndProcedure
|
||||
\State \Call{FillMatrix}{$S_{1}$, $S_{2}$}
|
||||
\State \Call{BacktrackAlignment}{$S_{1}$, $S_{2}$}
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
|
|
|||
39
tmp.tex
39
tmp.tex
|
|
@ -11,6 +11,43 @@
|
|||
\input{definitions.tex}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\begin{algorithm}
|
||||
\caption{Construct a longest common subsequence matrix keeping the path in memory}
|
||||
\begin{algorithmic}[1]
|
||||
\Function{LCSQ\_Matrix\_Path}{$S_{1}$: Array($n$), $S_{2}$: Array($m$)}
|
||||
\State $M \gets $ Array($m+1$, $n+1$)
|
||||
\State $P \gets $ Array($m+1$, $n+1$)
|
||||
\For {($i = 0$; $i < n+1$, $i++$)}
|
||||
\State $M[i][0] \gets 0$
|
||||
\EndFor
|
||||
\For {($j = 0$; $j < m+1$; $j+$)}
|
||||
\State $M[0][j] \gets 0$
|
||||
\EndFor
|
||||
\For{($i = 1$; $i < n+1$; $i++$)}
|
||||
\For{($j = 1$; $j < m+1$; $j++$)}
|
||||
\If {$i = 1$ or $j = 0$}
|
||||
\State $M[i][j] = 0$
|
||||
\Else
|
||||
\If {$S_{1}[i-1] = S_{2}[j-1]$}
|
||||
\State $M[i][j] \gets M[i-1][j-1] + 1$
|
||||
\State $P[i][j] \gets '\nwarrow'$
|
||||
\ElsIf {$M[i][j-1] \geq M[i-1][j]$}
|
||||
\State $M[i][j] \gets M[i][j-1]$
|
||||
\State $P[i][j] \gets '\leftarrow'$
|
||||
\Else
|
||||
\State $M[i][j] \gets M[i-1][j]$
|
||||
\State $P[i][j] \gets '\downarrow'$
|
||||
\EndIf
|
||||
\EndIf
|
||||
\EndFor
|
||||
\EndFor
|
||||
\State \Return $M, P$
|
||||
\EndFunction
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\iffalse
|
||||
\begin{algorithm}
|
||||
\caption{Backtrack the longest common subsequence}
|
||||
\begin{algorithmic}[1]
|
||||
|
|
@ -60,7 +97,7 @@
|
|||
\EndProcedure
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
|
||||
\fi
|
||||
\end{document}
|
||||
|
||||
\end{document}
|
||||
|
|
|
|||
Loading…
Reference in New Issue