ding

[ker2014-2.git] / report / ass2-1.tex

\chapter{Probabilistic representation and reasoning (and burglars)}
\section{Formal description}
In our representation of the model we introduced a \textit{Noisy OR} to
represent the causal independence of \textit{Burglar} and \textit{Earthquake}
on \textit{Alarm}. The representation of the network is displayed in
Figure~\ref{bnetwork21}

\begin{figure}[H]
        \caption{Bayesian network alarmsystem}
        \label{bnetwork21}
        \centering
        \includegraphics[scale=0.5]{d1.eps}
\end{figure}

Days were chosen as unit to model the story. Calculation for the probability of
a\textit{Burglar} event happening at some day is then (assuming a gregorian
calendar and leap days):
$$\frac{1}{365 + 0.25 - 0.01 - 0.0025}=\frac{1}{365.2425}$$

The resultant probability distributions can be found in Table~\ref{probdist},
in order to avoid a unclear graph.

\begin{table}[H]
        \label{probdist}
        \begin{tabular}{|l|l|}
                \hline
                & Earthquake\\
                \hline
                T & $0.0027$\\
                F & $0.9973$\\
                \hline
        \end{tabular}
        %
        \begin{tabular}{|l|l|}
                \hline
                & Burglar\\
                \hline
                T & $0.0027$\\
                F & $0.9973$\\
                \hline
        \end{tabular}
        
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{$I_1$}\\
                Earthquake & T & F\\
                \hline
                T & $0.2$ & $0.8$\\
                F & $0$ & $1$\\
                \hline
        \end{tabular}
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{$I_2$}\\
                Burglar & T & F\\
                \hline
                T & $0.95$ & $0.05$\\
                F & $0$ & $1$\\
                \hline
        \end{tabular}
        \begin{tabular}{|ll|ll|}
                \hline
                && \multicolumn{2}{c|}{Alarm}\\
                $I_1$ & $I_2$ & T & F\\
                \hline
                T & T & $1$ & $0$\\
                T & F & $1$ & $0$\\
                F & T & $1$ & $0$\\
                F & F & $0$ & $1$\\
                \hline
        \end{tabular}
        
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{Watson}\\
                Alarm & T & F\\
                \hline
                T & $0.8$ & $0.2$\\
                F & $0.4$ & $0.6$\\
                \hline
        \end{tabular}
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{Gibbons}\\
                Alarm & T & F\\
                \hline
                T & $0.99$ & $0.01$\\
                F & $0.04$ & $0.96$\\
                \hline
        \end{tabular}
        \begin{tabular}{|l|ll|}
                \hline
                & \multicolumn{2}{c|}{Radio}\\
                Earthquake & T & F\\
                \hline
                T & $0.9998$ & $0.0002$\\
                F & $0.0002$ & $0.9998$\\
                \hline
        \end{tabular}
\end{table}

\section{Implementation}
This distribution results in the \textit{AILog} code in Listing~\ref{alarm.ail}

\begin{listing}[H]
        \label{alarm.ail}
        \caption{alarm.ail}
        \inputminted[linenos,fontsize=\footnotesize]{prolog}{./src/alarm.ail}
\end{listing}

\section{Queries}
Using the following queries the probabilities or as follows:\\
\begin{enumerate}[a)]
        \item $P(\text{Burglary})=
                0.002737757092501968$
        \item $P(\text{Burglary}|\text{Watson called})=
                0.005321803679438259$
        \item $P(\text{Burglary}|\text{Watson called}\wedge\text{Gibbons called})=
                0.11180941544755249$
        \item $P(\text{Burglary}|\text{Watson called}\wedge\text{Gibbons called}
                \wedge\text{Radio})=0.01179672476662423$
\end{enumerate}

\begin{listing}[H]
        \begin{minted}[fontsize=\footnotesize]{prolog}
ailog: predict burglar.
Answer: P(burglar|Obs)=0.002737757092501968.
  [ok,more,explanations,worlds,help]: ok.

ailog: observe watson.
Answer: P(watson|Obs)=0.4012587986186947.
  [ok,more,explanations,worlds,help]: ok.

ailog: predict burglar.
Answer: P(burglar|Obs)=[0.005321803679438259,0.005321953115441623].
  [ok,more,explanations,worlds,help]: ok.

ailog: observe gibbons.
Answer: P(gibbons|Obs)=[0.04596053565368094,0.045962328885721306].
  [ok,more,explanations,worlds,help]: ok.

ailog: predict burglar.
Answer: P(burglar|Obs)=[0.11180941544755249,0.1118516494624678].
  [ok,more,explanations,worlds,help]: ok.

ailog: observe radio.
Answer: P(radio|Obs)=[0.02582105837443645,0.025915745316785182].
  [ok,more,explanations,worlds,help]: ok.

ailog: predict burglar.
Answer: P(burglar|Obs)=[0.01179672476662423,0.015584580594335082].
  [ok,more,explanations,worlds,help]: ok.
        \end{minted}
\end{listing}

\section{Comparison with manual calculation}
Querying the \textit{Alarm} variable gives the following answer
\begin{minted}{prolog}
        ailog: predict alarm.
        Answer: P(alarm|Obs)=0.0031469965467367292.
                            
          [ok,more,explanations,worlds,help]: ok.
\end{minted}

Using formula: $P(i_1|C_1)+P(i_2|C_2)(1-P(i_1|C_1))$ we can calculate the
probability of the \textit{Alarm} variable using variable elimination. This
results in the following answer: 
$$0.2*0.0027+0.95*0.0027*(1-0.2*0.0027)=0.00314699654673673$$
TODOOOOOOOOOOO

\newpage
\section{Burglary problem with extended information}
$P(burglary)\cdot\left(
        P(\text{first house is holmes'})+
        P(\text{second house is holmes'})+
        P(\text{third house is holmes'})\right)=\\
0.5102041\cdot\left(
        \frac{1}{10000}+
        \frac{9999}{10000}\cdot\frac{1}{9999}+
        \frac{9999}{10000}\cdot\frac{9998}{9999}\cdot\frac{1}{9998}\right)=
\frac{3}{19600}\approx0.000153$

\section{Bayesian networks}
A bayesian network representation of the burglary problem with a multitude of
houses and burglars is possible but would be very big and tedious because all
the constraints about the burglars must be incorporated in the network.
The network would look something like in figere~\ref{bnnetworkhouses}
accompanied with the probability distributions below.

\begin{tabular}{|l|l|}
        \hline
        Joe &\\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}
\begin{tabular}{|l|l|}
        \hline
        William &\\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}
\begin{tabular}{|l|l|}
        \hline
        Jack & \\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}
\begin{tabular}{|l|l|}
        \hline
        Averall & \\
        \hline
        T & $\nicefrac{5}{7}$\\
        F & $\nicefrac{2}{7}$\\
        \hline
\end{tabular}

\begin{tabular}{|llll|ll|}
        \hline
        & & & & Burglary &\\
        Joe & William & Jack & Averall & T & F\\
        \hline
        F& F& F& F & $0$ & $1$\\
        F& F& F& T & $0$ & $1$\\
        F& F& T& F & $0$ & $1$\\
        F& F& T& T & $0$ & $1$\\
        F& T& F& F & $0$ & $1$\\
        F& T& F& T & $0$ & $1$\\
        F& T& T& F & $0$ & $1$\\
        F& T& T& T & $0$ & $1$\\
        T& F& F& F & $0$ & $1$\\
        T& F& F& T & $0$ & $1$\\
        T& F& T& F & $1$ & $0$\\
        T& F& T& T & $0$ & $1$\\
        T& T& F& F & $1$ & $0$\\
        T& T& F& T & $0$ & $1$\\
        T& T& T& F & $1$ & $0$\\
        T& T& T& T & $1$ & $0$\\
        \hline
\end{tabular}
\begin{tabular}{|lll|}
        \hline
        & Holmes &\\
        Burglary & T & F\\
        \hline
        T & $0.000153$ & $0.999847$\\
        F & $0$ & $1$\\
        \hline
\end{tabular}

\begin{figure}[H]
        \caption{Bayesian network of burglars and houses}
        \label{bnnetworkhouses}
        \centering
        \includegraphics[scale=0.5]{d2.eps}
\end{figure}