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e64a1d711a
...
18203b1e49
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@ -1,5 +1,5 @@
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sub createFolderStructure{
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system("bash ./folder-structure.sh");
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system("bash ./createFolderStructure.sh");
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}
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createFolderStructure();
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@ -11,7 +11,7 @@
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}
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}
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\includechapters{part1}{4}
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\includechapters{part2}{4}
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\includechapters{part1}{3}
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\includechapters{part2}{2}
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% \includechapters{part3}{1}
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@ -1,116 +0,0 @@
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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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@ -55,7 +55,6 @@ 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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@ -89,7 +88,6 @@ 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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\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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@ -113,4 +111,3 @@ In this case the longest common subsequence of $S_{1}$ and $S_{2}$ is $TCTA$.
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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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@ -22,16 +22,3 @@ Example:
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\item $del(a) = 1$
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\item $ins(a) = 1$
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\end{itemize}
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Let $X = x_{0} x_{1} \ldots x_{m-1}$, $Y = y_{0} y_{1} \ldots y_{n-1} $
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An alignment is noted as $z = \begin{pmatrix} \bar{x}_{0} \\ \bar{y}_{0} \end{pmatrix} \ldots \begin{pmatrix} \bar{x}_{p-1} \\ \bar{y}_{p-1} \end{pmatrix}$ of size $p$. $n \leq p \leq n + m$
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$\bar{x}_{i} = x_{j}$ or $\bar{x}_{i} = \varepsilon$ for $0 \leq i \leq p-1$ and $0 \leq j \leq m - 1$
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$\bar{y}_{i} = y_{j}$ or $\bar{y}_{i} = \varepsilon$ for $0 \leq i \leq p-1$ and $0 \leq j \leq n - 1$
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$X' = \bar{x}_{0} \bar{x}_{1} \ldots \bar{x}_{i} \ldots \bar{x}_{p-1}$
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$Y' = \bar{y}_{0} \bar{y}_{1} \ldots \bar{y}_{i} \ldots \bar{y}_{p-1}$
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for $0 \leq i \leq p-1$, $\nexists i$, such that $\bar{x}_{i} = \bar{y}_{i} = \varepsilon$
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@ -1,145 +0,0 @@
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\chapter{Section alignment}
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\section{Needleman - Wunsch algorithm}
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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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\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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@ -0,0 +1,176 @@
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function lcsq_matrix(seq1, seq2)
|
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local gap_penalty = 0
|
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local match_score = 1
|
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local n1 = string.len(seq1)
|
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local n2 = string.len(seq2)
|
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-- Create a n1 x n2 matrix
|
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local matrix = {}
|
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for i=0,n1 do
|
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matrix[i] = {}
|
||||
for j=0,n2 do
|
||||
matrix[i][j] = 0
|
||||
end
|
||||
end
|
||||
-- Fill the rest of the matrix
|
||||
local match, delete, insert
|
||||
for i=1,n1 do
|
||||
for j=1,n2 do
|
||||
if string.sub(seq1, i, i) == string.sub(seq2, j, j) then
|
||||
match = matrix[i-1][j-1] + match_score
|
||||
else
|
||||
match = matrix[i-1][j-1]
|
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end
|
||||
gap1 = matrix[i-1][j] + gap_penalty
|
||||
gap2 = matrix[i][j-1] + gap_penalty
|
||||
matrix[i][j] = math.max(match, gap1, gap2)
|
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end
|
||||
end
|
||||
return matrix
|
||||
end
|
||||
|
||||
local function has_value (tab, val)
|
||||
for index, value in ipairs(tab) do
|
||||
if value == val then
|
||||
return true
|
||||
end
|
||||
end
|
||||
|
||||
return false
|
||||
end
|
||||
|
||||
function repr_matrix(matrix)
|
||||
repr = ""
|
||||
for i=1,#matrix do
|
||||
for j=1,#matrix do
|
||||
repr = repr .. matrix[i][j] .. " "
|
||||
end
|
||||
repr = repr .. "\n"
|
||||
end
|
||||
return repr
|
||||
end
|
||||
|
||||
|
||||
function draw_lcsq_matrix_graph(seq1, seq2)
|
||||
local matrix = lcsq_matrix(seq1, seq2)
|
||||
local tikz_code = ""
|
||||
function coordinate(i, j)
|
||||
return i .. "_" .. j
|
||||
end
|
||||
local steps = {
|
||||
{-1, 0},
|
||||
{-1, -1},
|
||||
{0, -1}
|
||||
}
|
||||
|
||||
local n1 = string.len(seq1)
|
||||
local n2 = string.len(seq2)
|
||||
local path = {}
|
||||
local i = n1
|
||||
local j = n2
|
||||
while i >= 0 and j >= 0 do
|
||||
path[#path+1] = coordinate(i, j)
|
||||
local min = matrix[i][j]
|
||||
local min_step = steps[1]
|
||||
for index, step in ipairs(steps) do
|
||||
local k = i + step[1]
|
||||
local l = j + step[2]
|
||||
if k >= 0 and l >= 0 and matrix[k][l] <= min then
|
||||
min_step = step
|
||||
min = matrix[k][l]
|
||||
end
|
||||
end
|
||||
i = i + min_step[1]
|
||||
j = j + min_step[2]
|
||||
print(i, j)
|
||||
end
|
||||
-- Draw the matrix as tikz node with matrix value
|
||||
for i=0,n1 do
|
||||
for j=0,n2 do
|
||||
local options = ""
|
||||
if has_value(path, coordinate(i, j)) then
|
||||
|
||||
options = "[fill=gray, draw, minimum size=1]"
|
||||
end
|
||||
tikz_code = tikz_code .. "\\node" .. options .. " (" .. coordinate(i, j) .. ") at (" .. i .. ", " .. -j .. ")" .. " {" .. matrix[i][j] .. "};"
|
||||
end
|
||||
end
|
||||
-- Add nucleotide labels
|
||||
for i=1,n1 do
|
||||
local nt = string.sub(seq1, i, i)
|
||||
tikz_code = tikz_code .. "\\node at (".. i .. "," .. 1 .. ")" .. "{$" .. nt .."$};"
|
||||
end
|
||||
for i=1,n2 do
|
||||
local nt = string.sub(seq2, i, i)
|
||||
tikz_code = tikz_code .. "\\node at (" .. -1 .. ", " .. -i .. ")" .. "{$ ".. nt .."$};"
|
||||
end
|
||||
-- For seq2
|
||||
for i=0,n1 do
|
||||
for j=0,n2 do
|
||||
local min = math.huge
|
||||
for index, step in ipairs(steps) do
|
||||
local k = i + step[1]
|
||||
local l = j + step[2]
|
||||
if k >= 0 and l >= 0 and matrix[k][l] < min then
|
||||
min = matrix[k][l]
|
||||
end
|
||||
end
|
||||
local highlighted = false
|
||||
for index, step in ipairs(steps) do
|
||||
local k = i + step[1]
|
||||
local l = j + step[2]
|
||||
if k >= 0 and l >= 0 and matrix[k][l] == min then
|
||||
tikz_code = tikz_code .. "\\draw[->] (" .. coordinate(i, j) .. ")" .. " -- " .. "(" .. coordinate (k, l) .. ");"
|
||||
end
|
||||
end
|
||||
end
|
||||
end
|
||||
return tikz_code
|
||||
end
|
||||
|
||||
function draw_lcsq_matrix(seq1, seq2)
|
||||
-- print(string.format(" Path: %s -> %s", seq1, seq2))
|
||||
local matrix = lcsq_matrix(seq1, seq2)
|
||||
local n1 = string.len(seq1)
|
||||
local n2 = string.len(seq2)
|
||||
-- Draw the matrix as tikz nodes
|
||||
for i=0,n1-1 do
|
||||
for j=0,n2-1 do
|
||||
print(string.format("\\node[draw, minimum width=1cm, minimum height=1cm] at (%d, -%d) {};", i, j, matrix[i][j]))
|
||||
end
|
||||
end
|
||||
-- Draw the sequence labels
|
||||
for i=1,n1 do
|
||||
print(string.format("\\node at (%d, -%d) {%s};", i-1, -1, string.sub(seq1, i, i)))
|
||||
end
|
||||
for i=1,n2 do
|
||||
print(string.format("\\node at (%d, -%d) {%s};", -1, i-1, string.sub(seq2, i, i)))
|
||||
end
|
||||
-- Add a path from the bottom right corner to the top left corner, following the minimum of the three possible moves at each step
|
||||
local i, j, value, previous_value
|
||||
i = n1-1
|
||||
j = n2-1
|
||||
print(string.format("\\draw[-,line width=2, gray] (%d, -%d) --", i, j))
|
||||
while i > 0 and j > 0 do
|
||||
value = math.min(matrix[i-1][j-1], table[i-1][j], table[i][j-1])
|
||||
if value == matrix[i-1][j-1] then
|
||||
i = i - 1
|
||||
j = j - 1
|
||||
elseif value == matrix[i-1][j] then
|
||||
i = i - 1
|
||||
else
|
||||
j = j - 1
|
||||
end
|
||||
print(string.format(" (%d, -%d) -- ", i, j))
|
||||
end
|
||||
print(string.format("(0, 0) -- (-1, 1);", i, j))
|
||||
end
|
||||
|
||||
function main()
|
||||
local seq1 = "ATCTGAT"
|
||||
local seq2 = "TGCATA"
|
||||
|
||||
local matrix = lcsq_matrix(seq1, seq2)
|
||||
print(repr_matrix(matrix))
|
||||
end
|
||||
|
||||
main()
|
||||
39
tmp.tex
39
tmp.tex
|
|
@ -11,43 +11,6 @@
|
|||
\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]
|
||||
|
|
@ -97,7 +60,7 @@
|
|||
\EndProcedure
|
||||
\end{algorithmic}
|
||||
\end{algorithm}
|
||||
\fi
|
||||
|
||||
\end{document}
|
||||
|
||||
\end{document}
|
||||
|
|
|
|||
Loading…
Reference in New Issue