GENIOMHE
/
sequence-algorithms
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Compare commits

base: GENIOMHE:18203b1e4998940c569f3f90116ed7a19e607f73
Branches Tags
GENIOMHE:main
..
compare: GENIOMHE:5f909eb9550a946cc391ddc37425240e3987df7e
Branches Tags
GENIOMHE:main

No commits in common. "18203b1e4998940c569f3f90116ed7a19e607f73" and "5f909eb9550a946cc391ddc37425240e3987df7e" have entirely different histories.
18203b1e49 ... 5f909eb955

21 changed files with 136 additions and 921 deletions
Show Stats Expand all files Collapse all files

41
.latexmkrc
View File

@ -1,41 +0,0 @@
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;
}

6
Makefile
View File

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

7
content/chapters/include.tex
View File

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

2
content/chapters/part1/0.tex
View File

@ -58,6 +58,6 @@
\EndIf
\EndFor
\State \Return $c$
\EndFunction
\EndFunction
\end{algorithmic}
\end{algorithm}

226
content/chapters/part1/2.tex
View File

@ -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}
@ -40,32 +40,32 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\Function{ComparisonMatrix}{$S$: Array($n$)}
\State $M \gets $ Array($n$, $n$)
\For{($i = 0$; $i < n$; $i++$)}
\For{$j = 0$; $j < n$; $j++$}
\If {$S[i] = S[j]$}
\State $M[i][j] = 1$
\Else
\State $M[i][j] = 0$
\EndIf
\EndFor
\For{$j = 0$; $j < n$; $j++$}
\If {$S[i] = S[j]$}
\State $M[i][j] = 1$
\Else
\State $M[i][j] = 0$
\EndIf
\EndFor
\EndFor
\State \Return $M$
\EndFunction
\end{algorithmic}
\end{algorithm}
\end{algorithm}
\begin{algorithm}
\caption{Construct the top half of a comparison matrix}
\begin{algorithmic}[1]
\Function{ComparisonMatrix}{$S$: Array($n$)}
\State $M \gets$ Array($n$,$n$)
\For{($i = 0$; $i < n$; $i++$)}
\For{j=i; j < n; j++}
\If {S[i] = S[j]}
\State M[i][j] = 1
\Else
\State M[i][j] = 0
\EndIf
\EndFor
\For{j=i; j < n; j++}
\If {S[i] = S[j]}
\State M[i][j] = 1
\Else
\State M[i][j] = 0
\EndIf
\EndFor
\EndFor
\State \Return M
\EndFunction
@ -83,18 +83,18 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\State $pos = \{\}$
\State $visited = \{\}$
\For {($i_{start} = 0$; $i_{start} < n$; $i_{start}++$)}
\For {($j_{start} = i_{start}+1$; $j_{start} < n$; $j_{start}++$)}
\If{$M[i_{start}][j_{start}] = 1$ and $(i_{start}, j_{start}) \notin visited$}
\State $i = i_{start}$
\State $j = j_{start}$
\While {$M[i][j] = 1$}
\State $i++$
\State $j++$
\State $visited = visited \cup \{(i, j)\}$
\EndWhile
\State $pos = pos \cup \{(i_{start}, i), (j_{start},j)\}$
\EndIf
\EndFor
\For {($j_{start} = i_{start}+1$; $j_{start} < n$; $j_{start}++$)}
\If{$M[i_{start}][j_{start}] = 1$ and $(i_{start}, j_{start}) \notin visited$}
\State $i = i_{start}$
\State $j = j_{start}$
\While {$M[i][j] = 1$}
\State $i++$
\State $j++$
\State $visited = visited \cup \{(i, j)\}$
\EndWhile
\State $pos = pos \cup \{(i_{start}, i), (j_{start},j)\}$
\EndIf
\EndFor
\EndFor
\EndFunction
\end{algorithmic}
@ -109,22 +109,22 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\State $M$ = \Call{ComparisonMatrix}{S}
\State $pos = \{\}$
\For {($diag = 1$; $diag < n$; $diag++$)}
\State $j = diag$
\State $i = 0$
\While {$i < n$ and $j < n$}
\If {$M[i][j] = 1$}
\State $i_{start} = i$
\State $j_{start} = j$
\While {$i < n$ and $j < n$ and $M[i][j] = 1$}
\State i++
\State j++
\EndWhile
\State $pos = pos \cup \{((i_{start},i-1),(j_{start},j-1))\}$
\EndIf
\State $i++$
\State $j++$
\State
\EndWhile
\State $j = diag$
\State $i = 0$
\While {$i < n$ and $j < n$}
\If {$M[i][j] = 1$}
\State $i_{start} = i$
\State $j_{start} = j$
\While {$i < n$ and $j < n$ and $M[i][j] = 1$}
\State i++
\State j++
\EndWhile
\State $pos = pos \cup \{((i_{start},i-1),(j_{start},j-1))\}$
\EndIf
\State $i++$
\State $j++$
\State
\EndWhile
\EndFor
\EndFunction
\end{algorithmic}
@ -138,24 +138,24 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\State $M$ = \Call{ComparisonMatrix}{S}
\State $pos = \{\}$
\For {($diag = 1$; $diag < n$; $diag++$)}
\State $j = diag$
\State $i = 0$
\State $l = 0$
\While {$i < n$ and $j < n$}
\If {$M[i][j] = 1$}
\State $l++$
\Else
\If {$l > 0$}
\State $pos = pos \cup \{(i-l,j-l,l)\}$
\State $l = 0$
\EndIf
\EndIf
\State $i++$
\State $j++$
\EndWhile
\If {$l > 0$}
\State $pos = pos \cup \{((i-l,j-l,l))\}$
\EndIf
\State $j = diag$
\State $i = 0$
\State $l = 0$
\While {$i < n$ and $j < n$}
\If {$M[i][j] = 1$}
\State $l++$
\Else
\If {$l > 0$}
\State $pos = pos \cup \{(i-l,j-l,l)\}$
\State $l = 0$
\EndIf
\EndIf
\State $i++$
\State $j++$
\EndWhile
\If {$l > 0$}
\State $pos = pos \cup \{((i-l,j-l,l))\}$
\EndIf
\EndFor
\State \Return $pos$
\EndFunction
@ -171,18 +171,18 @@ Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\State $M$ = \Call{ComparisonMatrix}{S}
\State $pos = \{\}$
\For {$i_{start} = 0$; $i_{start} < n$; $i_{start}++$}
\For {$j_{start} = i_{start}+1$; $j_{start} < n$; $j_{start}++$}
\If{$M[i_{start}][j_{start}] = 1$}
\State $i = i_{start}$
\State $j = j_{start}$
\While {$M[i][j] = 1$}
\State $M[i][j] = 0$ \Comment{Ensure that the segment is not explored again}
\State $i++$
\State $j++$
\EndWhile
\State $pos = pos \cup \{((i_{start}, i-1), (j_{start},j-1))\}$
\EndIf
\EndFor
\For {$j_{start} = i_{start}+1$; $j_{start} < n$; $j_{start}++$}
\If{$M[i_{start}][j_{start}] = 1$}
\State $i = i_{start}$
\State $j = j_{start}$
\While {$M[i][j] = 1$}
\State $M[i][j] = 0$ \Comment{Ensure that the segment is not explored again}
\State $i++$
\State $j++$
\EndWhile
\State $pos = pos \cup \{((i_{start}, i-1), (j_{start},j-1))\}$
\EndIf
\EndFor
\EndFor
\EndFunction
\end{algorithmic}
@ -215,23 +215,23 @@ An automaton is a tuple $\langle S, s_{0}, T, \Sigma,f\rangle$
\caption{Check wether a word belong to a language for which we have an automaton}
\begin{algorithmic}[1]
\Function{WordInLanguage}{$W$: Array($n$), $A$: $\langle S, s_{0}, T, \Sigma,f \rangle$}
\Returns{A Boolean valued to \True{} if the word is recognized by the language automaton}
\State $s \gets s_{0}$
\State $i \gets 0$
\While {$i < n$}
\State $a \gets W[i]$
\If {$\exists f(s, a)$}
\State $s \gets f(s, a)$
\Else
\State \Return \False
\EndIf
\State i++
\EndWhile
\If {$s \in T$}
\State \Return \True
\Else
\State \Return \False
\EndIf
\Returns{A Boolean valued to \True{} if the word is recognized by the language automaton}
\State $s \gets s_{0}$
\State $i \gets 0$
\While {$i < n$}
\State $a \gets W[i]$
\If {$\exists f(s, a)$}
\State $s \gets f(s, a)$
\Else
\State \Return \False
\EndIf
\State i++
\EndWhile
\If {$s \in T$}
\State \Return \True
\Else
\State \Return \False
\EndIf
\EndFunction
\end{algorithmic}
\end{algorithm}
@ -248,34 +248,34 @@ 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}
\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}
\State $s \gets s_{0}$
\State $i \gets 0$
\While {$i < m$ and $\exists f(s, W[i])$}
\State $s \gets f(s, W[i])$
\State $i++$
\EndWhile
\If {$i=n$}
\State \Return \True
\Else
\State \Return \False
\EndIf
\EndFunction
\end{algorithmic}
\end{algorithm}
The complexity of the pattern matching algorithm is $\mathcal{O}(n + m)$, because building the automaton is $\mathcal{O}(m)$
\begin{algorithm}
\caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$}
\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}
\State $s \gets s_{0}$
\State $i \gets 0$
\While {$i < m$ and $\exists f(s, W[i])$}
\State $s \gets f(s, W[i])$
\State $i++$
\EndWhile
\If {$i=n$}
\State \Return \True
\Else
\State \Return \False
\EndIf
\EndFunction
\end{algorithmic}
\end{algorithm}
The complexity of the pattern matching algorithm is $\mathcal{O}(n + m)$, because building the automaton is $\mathcal{O}(m)$

84
content/chapters/part1/3.tex
View File

@ -1,84 +0,0 @@
\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}

113
content/chapters/part1/4.tex
View File

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

24
content/chapters/part2/0.tex
View File

@ -1,24 +0,0 @@
\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}

BIN
figures/part1/motif_search_automaton.pdf (Stored with Git LFS)
View File

Binary file not shown.

42
figures/part1/motif_search_automaton.tex
View File

@ -1,42 +0,0 @@
\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}

BIN
figures/part1/motif_search_automaton_bis.pdf (Stored with Git LFS)
View File

Binary file not shown.

47
figures/part1/motif_search_automaton_bis.tex
View File

@ -1,47 +0,0 @@
\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}

176
figures/part2/*
View File

@ -1,176 +0,0 @@
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()

181
figures/part2/lcsq.lua
View File

@ -1,181 +0,0 @@
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
}

BIN
figures/part2/lcsq.pdf (Stored with Git LFS)
View File

Binary file not shown.

18
figures/part2/lcsq.tex
View File

@ -1,18 +0,0 @@
\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}

8
folder-structure.sh
View File

@ -1,8 +0,0 @@
#!/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

BIN
main.pdf (Stored with Git LFS)
View File

Binary file not shown.

13
main.tex
View File

@ -12,8 +12,7 @@
fontsize=10pt,
fleqn,
oneside
]{scrbook}
]{scrbook}
\usepackage{mus}
@ -40,11 +39,11 @@
\hypersetup{
pdftitle={
Course - Sequence algorithms
Course - Sequence algorithms
},
pdfauthor={
Samuel Ortion
},
Samuel Ortion
},
pdfsubject={},
pdfkeywords={},
pdfcreator={LaTeX}
@ -65,7 +64,6 @@
\definecolor{clementine}{HTML}{dfa000}
\colorlet{primary}{clementine}
% \includeonly{content/chapters/part1/1}
\makeindex%
\makeglossary%
\begin{document}
@ -79,7 +77,10 @@
\newpage
% \input{content ./introduction}
\input{content/chapters/include}
% \input{content/conclusion}
\end{document}

BIN
tmp.pdf (Stored with Git LFS)
View File

Binary file not shown.

50
tmp.tex
View File

@ -11,56 +11,6 @@
\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}