Longest common subsequence

Try to use includeonly to limit compilation time

.latexmkrc

@@ -0,0 +1,41 @@
+sub createFolderStructure{
+   system("bash ./createFolderStructure.sh");
+}
+
+createFolderStructure();
+
+$hash_calc_ignore_pattern{aux} =
+'^\\\\gdef\\\\minted@oldcachelist\{,'
+  . '|^\s*default\.pygstyle,'
+  . '|^\s*[[:xdigit:]]+\.pygtex';
+
+$pdflatex =
+'lualatex -shell-escape -file-line-error -interaction=nonstopmode -synctex=1 -output-directory=build %O '
+  . '\'\PassOptionsToPackage{outputdir=build}{minted}\input{%S}\'';
+$aux_dir    = 'build';
+$bibtex_use = 2;
+
+# Amend cleaned extensions
+$clean_ext .= " fdb_latexmk run.xml synctex.gz";
+
+# Make latexmk quiet
+$latexmk_silent = 1;
+
+# Makeglossaries
+add_cus_dep( 'acn', 'acr', 0, 'makeglossaries' );
+add_cus_dep( 'glo', 'gls', 0, 'makeglossaries' );
+$clean_ext .= " acr acn alg glo gls glg";
+
+sub makeglossaries {
+    my ( $base_name, $path ) = fileparse( $_[0] );
+    my @args = ( "-q", "-d", $path, $base_name );
+    if ($silent) { unshift @args, "-q"; }
+    return system "makeglossaries", "-d", $path, $base_name;
+}
+
+sub biber {
+    my ( $base_name, $path ) = fileparse( $_[0] );
+    my @args = ( "--output-directory", $path, $base_name );
+    return system "biber", @args;
+}
+

Makefile

@@ -1,6 +0,0 @@
-options=-shell-escape -interaction=nonstopmode -file-line-error
-
-all: main.pdf
-
-%.pdf: %.tex
-	lualatex $(options) $<

content/chapters/include.tex

@@ -6,13 +6,12 @@
 	\foreach \i in {0, ..., #2} {%
 			\edef\FileName{content/chapters/#1/\i}%
 			\IfFileExists{\FileName}{%
-				\input{\FileName}%
+				\include{\FileName}%
 			}
 		}
 }
 
-\includechapters{part1}{2}
+\includechapters{part1}{3}
-
+\includechapters{part2}{2}
-% \includechapters{part2}{2}
 
 % \includechapters{part3}{1}

content/chapters/part1/2.tex

@@ -28,7 +28,7 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
 
 
 \begin{table}
-    \includegraphics{figures/part1/comparison_matrix_repetitions.pdf}
+    \includegraphics{./figures/part1/comparison_matrix_repetitions.pdf}
     \caption{Comparison matrix for $seq = $``ACGUUACGUUGUU"}
 \end{table}
 
@@ -51,8 +51,8 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
         \State \Return $M$
         \EndFunction
     \end{algorithmic}
-
 \end{algorithm}
+
 \begin{algorithm}
     \caption{Construct the top half of a comparison matrix}
     \begin{algorithmic}[1]
@@ -248,18 +248,18 @@ The suffix language of $S$ is $\{S, ACTACT, CTACT, TACT, ACT, CT, T\}$.
 
 \begin{figure}
     \centering
-    \includegraphics{figures/part1/minimal_suffix_automaton_exercise.pdf}
+    \includegraphics{./figures/part1/minimal_suffix_automaton_exercise.pdf}
     \caption{Suffix automaton for $S = $ AACTACT}
 \end{figure}
 
 \begin{figure}
     \centering
-    \includegraphics{figures/part1/minimal_suffix_automaton_exercise_bis.pdf}
+    \includegraphics{./figures/part1/minimal_suffix_automaton_exercise_bis.pdf}
     \caption{Suffix automaton for $S = $ TCATCATT}
 \end{figure}
 
-    \begin{algorithm}
+\begin{algorithm}
-        \caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$}
+    \caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$, built from the automaton}
     \begin{algorithmic}[1]
         \Function{CheckMotifInSuffixAutomaton}{$W$: Array($m$), $A$: $\langle S, s_{0}, T, \Sigma,f \rangle$}
         \Returns{Boolean valued to \True{} if the motif is in the sequence}
@@ -276,6 +276,6 @@ The suffix language of $S$ is $\{S, ACTACT, CTACT, TACT, ACT, CT, T\}$.
         \EndIf
         \EndFunction
     \end{algorithmic}
-    \end{algorithm}
+\end{algorithm}
-    The complexity of the pattern matching algorithm is $\mathcal{O}(n + m)$, because building the automaton is $\mathcal{O}(m)$
+The complexity of the pattern matching algorithm is $\mathcal{O}(n + m)$, because building the automaton is $\mathcal{O}(m)$
 

content/chapters/part1/3.tex

@@ -0,0 +1,84 @@
+\chapter{Automata for motif search}
+
+Let $M$ be a motif $M = $ ACAT.
+
+\begin{figure}
+    \centering
+    \includegraphics{./figures/part1/motif_search_automaton.pdf}
+    \caption{Motif search automaton for $M = $ ACAT}
+\end{figure}
+
+The alphabet of motif is the same as the alphabet of the sequence.
+The search automaton is complete.
+If the there exists a letter $c$ in the sequence that is not
+in the motif alphabet, we can make a virtual transition from 
+each state to the initial state whenever we encounter an unknown letter.
+
+\begin{algorithm}
+    \caption{Search a motif in a sequence with an automaton}
+    \begin{algorithmic}[1]
+        \Function{SearchMotif}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
+            \Returns{A set of positions where the motif has been found}
+            \State $s \gets s_0$
+            \State $i \gets 0$
+            \State $pos \gets \{\}$
+            \While {$i < n$} % $\exists f(s, S[i])$  We assume $S$ and $P$ are formed on the same alphabet, so we could remove the second check, as $A$ is complete
+                \If {$s \in T$}
+                    \State $pos \gets pos \cup \{ i - m \}$
+                \EndIf
+                \State $s \gets f(s, S[i])$
+                \State $i++$
+            \EndWhile
+            \State \Return $pos$
+        \EndFunction
+    \end{algorithmic}
+\end{algorithm}
+
+\begin{algorithm}
+    \caption{Check if the a motif automaton recognizes only the prefix of size $m-1$ of a motif $P$ of size $m$ }
+    \begin{algorithmic}[1]
+        \Function{SearchMotifLastPrefix}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
+            \Returns{A set of positions where the motif has been found}
+            \State $s \gets s_0$
+            \State $i \gets 0$
+            \State $T_{new} \gets \{\}$
+            \For {$s \in S$}
+                \For {$a \in \Sigma$}
+                    \For {$t \in T$}
+                        \If {$\exists f(s, a)$ and $f(s, a) = t$}
+                            \State $T_{new} \gets T_{new} \cup s$
+                        \EndIf
+                    \EndFor
+                \EndFor
+            \EndFor
+            \While {$i < n$}
+                \If {$s \in T_{new}$}
+                    \State \Return \True
+                \EndIf
+                \State $s \gets f(s, S[i])$
+                \State $i++$
+            \EndWhile
+            \State \Return \False 
+        \EndFunction
+    \end{algorithmic}
+\end{algorithm}
+
+\begin{algorithm}
+    \caption{Check if the a motif automaton recognizes only the prefix of size $m-1$ of a motif $P$ of size $m$, knowing the sequence of the motif}
+    \begin{algorithmic}[1]
+        \Function{SearchMotifLastPrefix}{$S$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma, f \rangle$, $P$: Array($m$)}
+        \Returns{A set of positions where the motif has been found}
+        \State $s \gets s_0$
+        \State $i \gets 0$
+        \While {$i < n$ and $f(s, P[m-1]) \notin T$}
+        \State $s \gets f(s, S[i])$
+        \State $i++$
+        \EndWhile
+        \If{$f(s, P[m-1]) \in T$} 
+        \State \Return \True
+        \Else
+        \State \Return \False 
+        \EndIf
+        \EndFunction
+    \end{algorithmic}
+\end{algorithm}

content/chapters/part1/4.tex

@@ -0,0 +1,113 @@
+\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}
+
+\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}

content/chapters/part2/0.tex

@@ -0,0 +1,24 @@
+\part{Sequence alignment}
+
+\section{Simililarity between sequences}
+
+A function $d$ is a distance between two sequences $x$ and $y$ in an alphabet $\Sigma$ if
+\begin{itemize}
+  \item $x, y \in \Sigma^{*}, d(x, x) = 0$
+  \item $\forall x, y \in \Sigma^{*}$ $d(x,y) = d(y,x)$
+  \item $\forall x, y, z \in \Sigma^{*}$ $d(x, z) \leq d(x, y) + d(x, z)$
+\end{itemize}
+
+Here we are interested by the distance that is able to represent the transformation of $x$ to $y$ using three types of basic operations:
+\begin{itemize}
+  \item Substition
+  \item Insertion
+  \item Deletion
+\end{itemize}
+
+Example:
+\begin{itemize}
+\item  $sub(a, b) = \begin{cases} 0 & \text{if} a = b \\ 1 &\text{otherwise} \end{cases}$.
+\item $del(a) = 1$
+\item $ins(a) = 1$
+\end{itemize}

figures/part1/motif_search_automaton.tex

@@ -0,0 +1,42 @@
+\documentclass[tikz]{standalone}
+
+\usepackage{tikz}
+
+\begin{document}
+\usetikzlibrary{automata,positioning}
+
+\iffalse
+  Let $M = $ ACACT be a motif.
+  $\Sigma = \{ACT\}$
+  \fi
+\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid, auto]
+% TODO
+  \node[state, initial] (q_0) {$q_{0}$};
+  \node[state,right=of q_0] (q_1) {$q_{1}$};
+  \node[state,right=of q_1] (q_2) {$q_{2}$};
+  \node[state,right=of q_2] (q_3) {$q_{3}$};
+   \node[state,right=of q_3] (q_4) {$q_{4}$};
+   \node[state,right=of q_4,accepting] (q_5) {$q_{5}$};
+
+  % M itself
+  \path[->] (q_0) edge node {A} (q_1)
+  (q_1) edge node {C} (q_2)
+  (q_2) edge node {A} (q_3)
+  (q_3) edge node {C} (q_4)
+  (q_4) edge node {T} (q_5);
+
+
+  % Make it complete so that it recognize all motif in the sequence
+  \path[->] (q_0) edge[loop above] node {C, T} (q_1);
+  \path[->] (q_1) edge[loop above] node {A} (q_1);
+  \path[->] (q_1) edge[bend left=30] node {T} (q_0);
+  \path[->] (q_2) edge[bend left=30] node {C, T} (q_0);
+  \path[->] (q_3) edge[bend right=30] node[above] {A} (q_1);
+  \path[->] (q_3) edge[bend left=40] node {T} (q_0);
+  \path[->] (q_4) edge[bend right=50] node[above] {A} (q_3);
+  \path[->] (q_4) edge[bend left=40] node {C} (q_0);
+  \path[->] (q_5) edge[bend right=40] node[above] {A} (q_1);
+  \path[->] (q_5) edge[bend left=50] node {C,T} (q_0);
+\end{tikzpicture}
+
+\end{document}

figures/part1/motif_search_automaton_bis.tex

@@ -0,0 +1,47 @@
+\documentclass[tikz]{standalone}
+
+\usepackage{tikz}
+
+\begin{document}
+\usetikzlibrary{automata,positioning}
+
+\iffalse
+  Let $M = $ CCATCAT be a motif.
+  $\Sigma = \{A,C,T\}$
+  \fi
+\begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid, auto]
+  \node[state, initial] (q_0) {$q_{0}$};
+  \node[state,right=of q_0] (q_1) {$q_{1}$};
+  \node[state,right=of q_1] (q_2) {$q_{2}$};
+  \node[state,right=of q_2] (q_3) {$q_{3}$};
+   \node[state,right=of q_3] (q_4) {$q_{4}$};
+   \node[state,right=of q_4] (q_5) {$q_{5}$};
+   \node[state,right=of q_5] (q_6) {$q_{6}$};
+    \node[state,right=of q_6,accepting] (q_7) {$q_{7}$};
+
+  % M itself
+  \path[->] (q_0) edge node {C} (q_1)
+  (q_1) edge node {C} (q_2)
+  (q_2) edge node {A} (q_3)
+  (q_3) edge node {T} (q_4)
+  (q_4) edge node {C} (q_5)
+  (q_5) edge node {A} (q_6)
+  (q_6) edge node {T} (q_7)
+  ;
+
+  \path[->] (q_0) edge [loop above] node {A,T} (q_0);
+  \path[->] (q_1) edge [bend right=40] node {A, T} (q_0);
+  \path[->] (q_2) edge [bend right=40] node {C} (q_1);
+  \path[->] (q_2) edge [bend right=40] node {T} (q_0);
+  \path[->] (q_3) edge [bend right=40] node {A} (q_0);
+  \path[->] (q_3) edge [bend right=40] node {C} (q_1);
+  \path[->] (q_4) edge [bend right=40] node {A, T} (q_0);
+  \path[->] (q_5) edge [bend left=40] node[above] {C} (q_2);
+  \path[->] (q_5) edge [bend left=40] node[above] {T} (q_0);
+  \path[->] (q_6) edge [bend left=40] node[above] {A} (q_0);
+  \path[->] (q_6) edge [bend right=40] node {C} (q_1);
+  \path[->] (q_7) edge [bend right=40] node {A, C, T} (q_0);
+  
+\end{tikzpicture}
+
+\end{document}

figures/part2/*

@@ -0,0 +1,176 @@
+function lcsq_matrix(seq1, seq2)
+    local gap_penalty = 0
+    local match_score = 1
+    local n1 = string.len(seq1)
+    local n2 = string.len(seq2)
+    -- Create a n1 x n2 matrix
+    local matrix = {}
+    for i=0,n1 do
+        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]
+            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=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()

figures/part2/lcsq.lua

@@ -0,0 +1,181 @@
+function lcsq_matrix(seq1, seq2)
+    local gap_penalty = 0
+    local match_score = 1
+    local n1 = string.len(seq1)
+    local n2 = string.len(seq2)
+    -- Create a n1 x n2 matrix
+    local matrix = {}
+    for i=0,n1 do
+        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]
+            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
+}

figures/part2/lcsq.tex

@@ -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}

folder-structure.sh

@@ -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

main.tex

@@ -12,7 +12,8 @@
 	fontsize=10pt,
 	fleqn,
 	oneside
-]{scrbook}
+	]{scrbook}
+
 	
 \usepackage{mus}
 
@@ -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}

tmp.tex

@@ -11,6 +11,56 @@
 \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}
 
 \end{document}