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Improve typography and formatting

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Samuel Ortion 2024-03-12 14:11:33 +01:00
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.gitignore vendored
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build/
.bak*
.auctex-auto
## Core latex/pdflatex auxiliary files:

9
content/chapters/part1/0.tex
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\caption{Count the occurrences of an element in an array}
\begin{algorithmic}[1]
\Function{Search}{$A$: Array($n$), $E$: element} \State $c \gets 0$
\For{($i = 0$; $i < n$; $i++$)} \If {A[i] $ = $ E} \State $c \gets c + 1$ \EndIf
\EndFor \State \Return $c$ \EndFunction
\For{($i = 0$; $i < n$; $i++$)}
\If {$A[i] = E$}
\State $c \gets c + 1$
\EndIf
\EndFor
\State \Return $c$
\EndFunction
\end{algorithmic}
\end{algorithm}

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content/chapters/part1/1.tex
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\begin{algorithm}
\caption{Brute-force search of a motif in a sequence}
\begin{algorithmic}[1]
\Function{FindMotif}{S: Array(n), M: Array(m)} {
\Returns{list of position}
\Function{FindMotif}{$S$: Array($n$), $M$: Array($m$)}
\Returns{a list of position}
\State $pos \gets \{\}$
\State $i \gets 0$
\While {$i < n - m + 1$} {
\While {$i < n - m + 1$}
\State $j \gets 0$
\While {$j < m$ and S[i+j] $ = $ M[j]} {
\While {$j < m$ and $S[i+j] = M[j]$}
\State $j++$
}
\EndWhile
\If {$j = m$}
\State $pos \gets pos \cup \{i\}$
\EndIf
\State $i++$
}
\EndWhile
\State \Return $pos$
}
\EndFunction
\end{algorithmic}
\label{alg:naive-motif-matching}
@ -29,9 +26,9 @@
\begin{algorithm}
\caption{Knuth-Morris-Pratt algorithm}
\begin{algorithmic}[1]
\Function{KMP\_Search}{S: Array(n), M: Array(m)}
\Function{KMP\_Search}{$S$: Array($n$), $M$: Array($m$)}
\Returns{Integer}
\State $table \gets \Call{KMP\_Table}{M}$
\State $table \gets$ \Call{KMP\_Table}{$M$}
\State $c \gets 0$ \Comment{Count the number of matches}
\State $i \gets 0$
\State $j \gets 0$

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content/chapters/part1/2.tex Executable file → Normal file
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\chapter{Matrices}
Let $S_{1}$ and $S_{2}$ be two sequences.
$S_{1} = $ ACGUUCC
$S_{2} = $ GUU
\begin{table}
\centering
\begin{tabular}{c|ccccccc}
& A & C & G & U & U & C & C \\
\hline
G & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
U & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
U & 0 & 0 & 0 & 1 & 1 & 0 & 0
\end{tabular}
\caption{Comparison matrix}
\end{table}
Let $n = |S_{1}|$, $m = |S_{2}|$
The complexity of this algorithm is $\mathcal{O}(n \cdot m)$ to build the matrix, and it requires also to find the diagonals and thus it is a bit less efficient than the \autoref{alg:naive-motif-matching}.
To find repetitions, we can use a comparison matrix with a single sequence against itself. A repetition would appear as a diagonal of ones, not on the main diagonal.
Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\begin{table}
\includegraphics{figures/part1/comparison_matrix_repetitions.pdf}
\caption{Comparison matrix for $seq = $``ACGUUACGUUGUU"}
\end{table}
\begin{algorithm}
\caption{Construct a comparison matrix}
\begin{algorithmic}[1]
\Function{ComparisonMatrix}{$S$: Array($n$)}
\State $M \gets $ Array($n$, $n$)
\For{($i = 0$; $i < n$; $i++$)}
\For{$j = 0$; $j < n$; $j++$}
\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}
\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
\EndFor
\State \Return M
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions (with a set of visited segments)}
\begin{algorithmic}[1]
\Function{FindRepetions}{$S$: Array($n$)}
\Returns{A list of start and end positions for repeated sequences}
\State $M = $ \Call{ComparisonMatrix}{S}
\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
\EndFor
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions with an exploration of diagonals}
\begin{algorithmic}[1]
\Function{FindRepetions}{$S$: Array($n$)}
\Returns{A list of start and end positions for repeted sequences}
\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
\EndFor
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions with an exploration of diagonals, without nested while}
\begin{algorithmic}[1]
\Function{FindRepetions}{$S$: Array($n$)}
\Returns{A list of start positions for repeted sequences and match length}
\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
\EndFor
\State \Return $pos$
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions}
\begin{algorithmic}[1]
\Function{FindRepetions}{$S$: Array($n$)}
\Returns{A list of start and end positions for repeted sequences}
\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
\EndFor
\EndFunction
\end{algorithmic}
\end{algorithm}
\section{Automata}
An automaton is a tuple $\langle S, s_{0}, T, \Sigma,f\rangle$
\begin{itemize}
\item $S$ the set of states
\item $s_{0}$ the initial state
\item $T$ the set of terminal states
\item $\Sigma$ the alphabet
\item $f$ the transition function $f: (s_{1}, c) \to s_{2}$
\end{itemize}
\paragraph{Example} Given the language $L$ on the alphabet $\Sigma = \{A, C, T\}$, $L = \{A^{*}, CTT, CA^{*}\}$
\begin{definition}[Deterministic automaton]
An automaton is deterministic, if for each couple $(p, a) \in S \times \Sigma$ it exists at most a state $q$ such as $f(p, q) = q$
\end{definition}
\begin{definition}[Complete automaton]
An automaton is complete, if for each couple $(p, a) \in S \times \Sigma$ it exists at least a state $q$ such as $f(p, q) = q$.
\end{definition}
\begin{algorithm}
\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
\EndFunction
\end{algorithmic}
\end{algorithm}
\section{Suffix Automaton}
Let $S = $ AACTACT
A suffix automata recognize all suffix of a given sequence.
The suffix language of $S$ is $\{S, ACTACT, CTACT, TACT, ACT, CT, T\}$.
\begin{figure}
\centering
\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}
\caption{Suffix automaton for $S = $ TCATCATT}
\end{figure}
\begin{algorithm}
\caption{Check if a sequences matches a motif, from a suffix automaton $\mathcal{O}(m)$}
\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 $\mathcal{O}(m)$

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content/chapters/part1/3.tex
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\chapter{Matrices}
Let $S_{1}$ and $S_{2}$ be two sequences.
$S_{1} = $ ACGUUCC
$S_{2} = $ GUU
\begin{table}
\centering
\begin{tabular}{c|ccccccc}
& A & C & G & U & U & C & C \\
\hline
G & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
U & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
U & 0 & 0 & 0 & 1 & 1 & 0 & 0
\end{tabular}
\caption{Comparison matrix}
\end{table}
Let $n = |S_{1}|$, $m = |S_{2}|$
The complexity of this algorithm is $\mathcal{O}(n \cdot m)$ to build the matrix, and it requires also to find the diagonals and thus it is a bit less efficient than the \autoref{alg:naive-motif-matching}.
To find repetitions, we can use a comparison matrix with a single sequence against itself. A repetition would appear as a diagonal of ones, not on the main diagonal.
Let $S = $ ACGUUACGUU. Let's write the comparison matrix.
\begin{table}
\includegraphics{figures/part1/comparison_matrix_repetitions.pdf}
\caption{Comparison matrix for seq="ACGUUACGUUGUU"}
\end{table}
\begin{algorithm}
\caption{Construct a comparison matrix}
\begin{algorithmic}[1]
\Function{ComparisonMatrix}{S: Array(n)}
\State $M \gets Array(n, n)$
\For{$i = 0$; $i < n$; $i++$}
\For{$j=0$; $j < n$; $j++$}
\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}
\begin{algorithm}
\caption{Construct the top half of 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
\EndFor
\State \Return M
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions (with the set of visited segments)}
\begin{algorithmic}[1]
\Function{FindRepetions}{S: Array(n)}
\Returns{A list of start and end positions for repeted sequences}
\State M = \Call{ComparisonMatrix}{S}
\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
\EndFor
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions with an exploration of diagonals}
\begin{algorithmic}[1]
\Function{FindRepetions}{S: Array(n)}
\Returns{A list of start and end positions for repeted sequences}
\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
\EndFor
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions with an exploration of diagonals, without nested while}
\begin{algorithmic}[1]
\Function{FindRepetions}{S: Array(n)}
\Returns{A list of start positions for repeted sequences and match length}
\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
\EndFor
\State \Return pos
\EndFunction
\end{algorithmic}
\end{algorithm}
\begin{algorithm}
\caption{Find repetitions}
\begin{algorithmic}[1]
\Function{FindRepetions}{S: Array(n)}
\Returns{A list of start and end positions for repeted sequences}
\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
\EndFor
\EndFunction
\end{algorithmic}
\end{algorithm}
\section{Automata}
An automaton is a tuple $\langle S, s_{0}, T, \Sigma,f\langle$
\begin{itemize}
\item $S$ the set of states
\item $s_{0}$ the initial state
\item $T$ the set of terminal states
\item $\Sigma$ the alphabet
\item $f$ the transition function $f: (s_{1}, c) \to s_{2}$
\end{itemize}
\paragraph{Example} Given the language $L$ on the alphabet $\Sigma = \{A, C, T\}$, $L = \{A^{*}, CTT, CA^{*}\}$
\begin{definition}[Deterministic automaton]
An automaton is deterministic, if for each couple $(p, a) \in S \times \Sigma$ it exists at most a state $q$ such as $f(p, q) = q$
\end{definition}
\begin{definition}[Complete automaton]
An automaton is complete, if for each couple $(p, a) \in S \times \Sigma$ it exists at least a state $q$ such as $f(p, q) = q$.
\end{definition}
\begin{algorithm}
\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
\EndFunction
\end{algorithmic}
\end{algorithm}
\section{Suffix Automaton}
Let $S = $ AACTACT
A suffix automata recognize all suffix of a given sequence.
The suffix language of $S$ is $\{S, ACTACT, CTACT, TACT, ACT, CT, T\}$.
\begin{figure}
\centering
\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}
\caption{Suffix automaton for $S = $ TCATCATT}
\end{figure}
\end{definition}
\begin{algorithm}
\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 $\mathcal{O}(m)$
\end{algorithm}

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localSettings.yaml
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defaultIndent: " "
specialBeginEnd:
If:
begin: '\\If'
middle: '\\ElsIf'
end: '\\EndIf'
lookForThis: 1
While:
begin: '\\While'
end: '\\EndWhile'
lookForThis: 1
For:
begin: '\\For'
end: '\\EndFor'
specialBeforeCommand: 1

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main.pdf (Stored with Git LFS)
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tmp.tex
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\begin{document}
\end{document}