282 lines
9.1 KiB
TeX
282 lines
9.1 KiB
TeX
\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 $\mathcal{O}(m)$
|
|
|