Longest common subsequence
Try to use includeonly to limit compilation time
This commit is contained in:
parent
fc0331e054
commit
18203b1e49
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@ -0,0 +1,41 @@
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sub createFolderStructure{
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system("bash ./createFolderStructure.sh");
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}
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createFolderStructure();
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$hash_calc_ignore_pattern{aux} =
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'^\\\\gdef\\\\minted@oldcachelist\{,'
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. '|^\s*default\.pygstyle,'
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. '|^\s*[[:xdigit:]]+\.pygtex';
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$pdflatex =
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'lualatex -shell-escape -file-line-error -interaction=nonstopmode -synctex=1 -output-directory=build %O '
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. '\'\PassOptionsToPackage{outputdir=build}{minted}\input{%S}\'';
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$aux_dir = 'build';
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$bibtex_use = 2;
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# Amend cleaned extensions
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$clean_ext .= " fdb_latexmk run.xml synctex.gz";
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# Make latexmk quiet
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$latexmk_silent = 1;
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# Makeglossaries
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add_cus_dep( 'acn', 'acr', 0, 'makeglossaries' );
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add_cus_dep( 'glo', 'gls', 0, 'makeglossaries' );
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$clean_ext .= " acr acn alg glo gls glg";
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sub makeglossaries {
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my ( $base_name, $path ) = fileparse( $_[0] );
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my @args = ( "-q", "-d", $path, $base_name );
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if ($silent) { unshift @args, "-q"; }
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return system "makeglossaries", "-d", $path, $base_name;
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}
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sub biber {
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my ( $base_name, $path ) = fileparse( $_[0] );
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my @args = ( "--output-directory", $path, $base_name );
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return system "biber", @args;
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}
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6
Makefile
6
Makefile
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@ -1,6 +0,0 @@
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options=-shell-escape -interaction=nonstopmode -file-line-error
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all: main.pdf
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%.pdf: %.tex
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lualatex $(options) $<
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@ -6,13 +6,12 @@
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\foreach \i in {0, ..., #2} {%
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\edef\FileName{content/chapters/#1/\i}%
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\IfFileExists{\FileName}{%
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\input{\FileName}%
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\include{\FileName}%
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}
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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{part2}{2}
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% \includechapters{part3}{1}
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@ -28,7 +28,7 @@ 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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\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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@ -248,13 +248,13 @@ 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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\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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\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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@ -4,7 +4,7 @@ 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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\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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@ -0,0 +1,113 @@
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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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\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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\part{Sequence alignment}
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\section{Simililarity between sequences}
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A function $d$ is a distance between two sequences $x$ and $y$ in an alphabet $\Sigma$ if
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\begin{itemize}
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\item $x, y \in \Sigma^{*}, d(x, x) = 0$
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\item $\forall x, y \in \Sigma^{*}$ $d(x,y) = d(y,x)$
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\item $\forall x, y, z \in \Sigma^{*}$ $d(x, z) \leq d(x, y) + d(x, z)$
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\end{itemize}
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Here we are interested by the distance that is able to represent the transformation of $x$ to $y$ using three types of basic operations:
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\begin{itemize}
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\item Substition
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\item Insertion
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\item Deletion
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\end{itemize}
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Example:
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\begin{itemize}
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\item $sub(a, b) = \begin{cases} 0 & \text{if} a = b \\ 1 &\text{otherwise} \end{cases}$.
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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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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] = {}
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for j=0,n2 do
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matrix[i][j] = 0
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end
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end
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-- Fill the rest of the matrix
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local match, delete, insert
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for i=1,n1 do
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for j=1,n2 do
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if string.sub(seq1, i, i) == string.sub(seq2, j, j) then
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match = matrix[i-1][j-1] + match_score
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else
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match = matrix[i-1][j-1]
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end
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gap1 = matrix[i-1][j] + gap_penalty
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gap2 = matrix[i][j-1] + gap_penalty
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matrix[i][j] = math.max(match, gap1, gap2)
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end
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end
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return matrix
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end
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local function has_value (tab, val)
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for index, value in ipairs(tab) do
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if value == val then
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return true
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end
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end
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return false
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end
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function repr_matrix(matrix)
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repr = ""
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for i=1,#matrix do
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for j=1,#matrix do
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repr = repr .. matrix[i][j] .. " "
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end
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repr = repr .. "\n"
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end
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return repr
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end
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function draw_lcsq_matrix_graph(seq1, seq2)
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local matrix = lcsq_matrix(seq1, seq2)
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local tikz_code = ""
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function coordinate(i, j)
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return i .. "_" .. j
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end
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local steps = {
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{-1, 0},
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{-1, -1},
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{0, -1}
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}
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local n1 = string.len(seq1)
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local n2 = string.len(seq2)
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local path = {}
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local i = n1
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local j = n2
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while i >= 0 and j >= 0 do
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path[#path+1] = coordinate(i, j)
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local min = matrix[i][j]
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local min_step = steps[1]
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for index, step in ipairs(steps) do
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local k = i + step[1]
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local l = j + step[2]
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if k >= 0 and l >= 0 and matrix[k][l] <= min then
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min_step = step
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min = matrix[k][l]
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end
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end
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i = i + min_step[1]
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j = j + min_step[2]
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print(i, j)
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end
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-- Draw the matrix as tikz node with matrix value
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for i=0,n1 do
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for j=0,n2 do
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local options = ""
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if has_value(path, coordinate(i, j)) then
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options = "[fill=gray, draw, minimum size=1]"
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end
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tikz_code = tikz_code .. "\\node" .. options .. " (" .. coordinate(i, j) .. ") at (" .. i .. ", " .. -j .. ")" .. " {" .. matrix[i][j] .. "};"
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end
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end
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-- Add nucleotide labels
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for i=1,n1 do
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local nt = string.sub(seq1, i, i)
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tikz_code = tikz_code .. "\\node at (".. i .. "," .. 1 .. ")" .. "{$" .. nt .."$};"
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end
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for i=1,n2 do
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local nt = string.sub(seq2, i, i)
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tikz_code = tikz_code .. "\\node at (" .. -1 .. ", " .. -i .. ")" .. "{$ ".. nt .."$};"
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end
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-- For seq2
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for i=0,n1 do
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for j=0,n2 do
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local min = math.huge
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for index, step in ipairs(steps) do
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local k = i + step[1]
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local l = j + step[2]
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if k >= 0 and l >= 0 and matrix[k][l] < min then
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min = matrix[k][l]
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end
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end
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local highlighted = false
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for index, step in ipairs(steps) do
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local k = i + step[1]
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local l = j + step[2]
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if k >= 0 and l >= 0 and matrix[k][l] == min then
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tikz_code = tikz_code .. "\\draw[->] (" .. coordinate(i, j) .. ")" .. " -- " .. "(" .. coordinate (k, l) .. ");"
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end
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end
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end
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end
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return tikz_code
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end
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function draw_lcsq_matrix(seq1, seq2)
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-- print(string.format(" Path: %s -> %s", seq1, seq2))
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local matrix = lcsq_matrix(seq1, seq2)
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local n1 = string.len(seq1)
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local n2 = string.len(seq2)
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-- Draw the matrix as tikz nodes
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for i=0,n1-1 do
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for j=0,n2-1 do
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print(string.format("\\node[draw, minimum width=1cm, minimum height=1cm] at (%d, -%d) {};", i, j, matrix[i][j]))
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end
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end
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-- Draw the sequence labels
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for i=1,n1 do
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print(string.format("\\node at (%d, -%d) {%s};", i-1, -1, string.sub(seq1, i, i)))
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end
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for i=1,n2 do
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print(string.format("\\node at (%d, -%d) {%s};", -1, i-1, string.sub(seq2, i, i)))
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end
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-- Add a path from the bottom right corner to the top left corner, following the minimum of the three possible moves at each step
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local i, j, value, previous_value
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i = n1-1
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j = n2-1
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print(string.format("\\draw[-,line width=2, gray] (%d, -%d) --", i, j))
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while i > 0 and j > 0 do
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value = math.min(matrix[i-1][j-1], table[i-1][j], table[i][j-1])
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if value == matrix[i-1][j-1] then
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i = i - 1
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j = j - 1
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elseif value == matrix[i-1][j] then
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i = i - 1
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else
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j = j - 1
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end
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print(string.format(" (%d, -%d) -- ", i, j))
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end
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print(string.format("(0, 0) -- (-1, 1);", i, j))
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end
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function main()
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local seq1 = "ATCTGAT"
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local seq2 = "TGCATA"
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local matrix = lcsq_matrix(seq1, seq2)
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print(repr_matrix(matrix))
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end
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main()
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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] = {}
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for j=0,n2 do
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matrix[i][j] = 0
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end
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end
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-- Fill the rest of the matrix
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local match, delete, insert
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for i=1,n1 do
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for j=1,n2 do
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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]
|
||||
end
|
||||
gap1 = matrix[i-1][j] + gap_penalty
|
||||
gap2 = matrix[i][j-1] + gap_penalty
|
||||
matrix[i][j] = math.max(match, gap1, gap2)
|
||||
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=0,#matrix do
|
||||
for j=0,#matrix[i] do
|
||||
repr = repr .. matrix[i][j] .. " "
|
||||
end
|
||||
repr = repr .. "\n"
|
||||
end
|
||||
return repr
|
||||
end
|
||||
|
||||
|
||||
function draw_lcsq_matrix_graph(seq1, seq2, matrix)
|
||||
local tikz_code = ""
|
||||
function coordinate(i, j)
|
||||
return i .. "_" .. j
|
||||
end
|
||||
local steps = {
|
||||
{-1, -1},
|
||||
{0, -1},
|
||||
{-1, 0},
|
||||
}
|
||||
|
||||
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 max = matrix[i][j]
|
||||
local max_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] > max then
|
||||
max_step = step
|
||||
max = matrix[k][l]
|
||||
end
|
||||
end
|
||||
i = i + max_step[1]
|
||||
j = j + max_step[2]
|
||||
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 max = 0
|
||||
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] > max then
|
||||
max = 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] == max 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)
|
||||
local repr = ""
|
||||
-- Draw the matrix as tikz nodes
|
||||
for i=0,n1-1 do
|
||||
for j=0,n2-1 do
|
||||
repr = repr .. " " .. 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
|
||||
repr = repr .. " " .. string.format("\\node at (%d, -%d) {%s};", i-1, -1, string.sub(seq1, i, i))
|
||||
end
|
||||
for i=1,n2 do
|
||||
repr = repr .. " " .. 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
|
||||
repr = repr .. " " 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], matrix[i-1][j], matrix[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
|
||||
repr = repr .. " " .. string.format(" (%d, -%d) -- ", i, j)
|
||||
end
|
||||
repr = repr .. " " .. string.format("(0, 0) -- (-1, 1);", i, j)
|
||||
return repr
|
||||
end
|
||||
|
||||
function main()
|
||||
local seq1 = "ATCTGAT"
|
||||
local seq2 = "TGCATA"
|
||||
local matrix = lcsq_matrix(seq1, seq2)
|
||||
print(draw_lcsq_matrix_graph(seq1, seq2, matrix))
|
||||
end
|
||||
|
||||
-- main()
|
||||
|
||||
return {
|
||||
lcsq_matrix=lcsq_matrix,
|
||||
draw_lcsq_matrix_graph=draw_lcsq_matrix_graph,
|
||||
draw_lcsq_matrix=draw_lcsq_matrix
|
||||
}
|
||||
Binary file not shown.
|
|
@ -0,0 +1,18 @@
|
|||
\documentclass[tikz]{standalone}
|
||||
|
||||
\usepackage{tikz}
|
||||
\usepackage{luatextra}
|
||||
\begin{document}
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{luacode}
|
||||
lcsq = require('lcsq')
|
||||
seq2 = "ATCTGAT"
|
||||
seq1 = "TGCATA"
|
||||
matrix = lcsq.lcsq_matrix(seq1, seq2)
|
||||
tikz_code = lcsq.draw_lcsq_matrix_graph(seq1, seq2, matrix)
|
||||
tex.print(tikz_code)
|
||||
\end{luacode}
|
||||
\end{tikzpicture}
|
||||
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,8 @@
|
|||
#!/bin/sh
|
||||
|
||||
find ./content -type d > folder_list.txt
|
||||
|
||||
mkdir -p build
|
||||
cd build
|
||||
cat ../folder_list.txt | xargs mkdir -p
|
||||
rm ../folder_list.txt
|
||||
5
main.tex
5
main.tex
|
|
@ -14,6 +14,7 @@
|
|||
oneside
|
||||
]{scrbook}
|
||||
|
||||
|
||||
\usepackage{mus}
|
||||
|
||||
\RequirePackage{algorithm}
|
||||
|
|
@ -64,6 +65,7 @@
|
|||
\definecolor{clementine}{HTML}{dfa000}
|
||||
\colorlet{primary}{clementine}
|
||||
|
||||
% \includeonly{content/chapters/part1/1}
|
||||
\makeindex%
|
||||
\makeglossary%
|
||||
\begin{document}
|
||||
|
|
@ -77,10 +79,7 @@
|
|||
\newpage
|
||||
|
||||
% \input{content ./introduction}
|
||||
|
||||
\input{content/chapters/include}
|
||||
|
||||
|
||||
% \input{content/conclusion}
|
||||
|
||||
\end{document}
|
||||
|
|
|
|||
48
tmp.tex
48
tmp.tex
|
|
@ -11,7 +11,55 @@
|
|||
\input{definitions.tex}
|
||||
|
||||
\begin{document}
|
||||
\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}
|
||||
|
||||
\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}
|
||||
|
||||
\end{document}
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue