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diff --git a/assignments/assignment2/assignment2.tex b/assignments/assignment2/assignment2.tex index 0b30bb2..a7b36e4 100644 --- a/assignments/assignment2/assignment2.tex +++ b/assignments/assignment2/assignment2.tex @@ -34,6 +34,47 @@ \maketitle \section{Chocolate Bar Machine} +\begin{enumerate} + \item + L* creates large hypothesis model from the beginning. + This means that it works faster with inefficient equivalence oracles. + On the other hand, Rivest-Schapire and TTT create small models which can perform better with a fast equivalence oracles, + because these algorithms spend less time on creating a well-based hypothesis. + + \item + We used L* with the following counterexamples: + + \begin{itemize} + \item 10ct 10ct 5ct snickers + \item 10ct 10ct 10ct 10ct 10ct snickers snickers + \item 10ct 10ct 10ct snickers 10ct twix + \item 10ct 10ct 10ct 5ct snickers twix + \end{itemize} + + The hypotheses are given in \cref{fig:lstar-run} (p.~\pageref{fig:lstar-run}). + + \begin{figure}[p] + \includegraphics[width=.5\textwidth]{LStar_hypothesis0} + \includegraphics[width=.5\textwidth]{LStar_hypothesis1} + \includegraphics[width=.5\textwidth]{LStar_hypothesis2} + \includegraphics[width=.5\textwidth]{LStar_hypothesis3} + \caption{Learning the chocolate bar machine with L* (top to bottom, left to right).\label{fig:lstar-run}} + \end{figure} + + \item + States are uniquely identified by the amount injected in the machine. + The machine does not accept more than 40ct, and the granularity is 5ct. + Hence, there are $\frac{40}{5}+1=9$ states. + The learned model also contains states, therefore they must be equivalent. + + \item + + \item + + \item + + \item +\end{enumerate} \section{Bounded Retransmission Protocol} |