3e0f9b8a
Francisco Coelho
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%
% Local commands
%
\newcommand{\todo}[1]{{\color{orange}TODO #1}}
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%
% Identificação deste documento
%
\title{Zugzwang}
\subtitle{Stochastic Adventures in Inductive Logic}
\author{Francisco Coelho}
\institute[\texttt{fc@uevora.pt}]{
Departamento de Informática, Universidade de Évora\\
High Performance Computing Chair\\
NOVA-LINCS
}
\begin{document}
%
\begin{frame}[plain]
\titlepage
\end{frame}
\section{Introduction}
\begin{frame}{Notation and Assumptions}
% --------------------------------
\begin{itemize}
% --------------------------------
\item $\co{x} = 1 - x$.
% --------------------------------
\item \textbf{Probabilistic Atomic Choice (PAC):} $x :: a$ defines $a \lor \neg a$ and probabilities $\pr{a} = x, \pr{\neg a} = \co{x}$.
% --------------------------------
\item $\delta a$ denotes $a \lor \neg a$ and $\delta\! \set{x :: a, a \in \fml{A}} = \set{\delta a, a \in \fml{A}}$ for a set of atoms $\fml{A}$.
% --------------------------------
\item \textbf{Closed World Assumption:} $\naf p \models \neg p$.
% --------------------------------
% \item Probabilistic choices and sub-goals are independent.
% --------------------------------
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{General Setting}
% --------------------------------
\begin{itemize}
% --------------------------------
\item \textbf{Atoms} $\fml{A}$,
$\overline{\fml{A}} = \cset{\neg a}{a \in \fml{A}}$, and \textbf{literals} $\fml{L} = \fml{A} \cup \co{\fml{A}}$.
% --------------------------------
\item \textbf{Samples} $z \in \fml{Z} \iff z \subseteq \fml{L}$.
% --------------------------------
\item \textbf{Events} or \textit{consistent samples} $\fml{E}$ :
$$\fml{E} = \cset{z \in \fml{Z} }{ \forall a \in \fml{A}~\envert{\set{a,\neg a} \cap z} \leq 1}.$$
% --------------------------------
\item \textit{PASP Problem} or \textbf{Specification:} $P = C \land F \land R$ where
% --------------------------------
\begin{itemize}
% --------------------------------
\item $C = C_P = \cset{x_i :: a_i }{ i \in 1:n \land a_i \in \fml{A}}$ \textit{pacs}.
% --------------------------------
\item $F = F_P$ \textit{facts}.
% --------------------------------
\item $R = R_P$ \textit{rules}.
% --------------------------------
\item $\fml{A}_P, \fml{Z}_P$ and $\fml{E}_P$: \textit{atoms}, \textit{samples} and \textit{events} of $P$.
\end{itemize}
% --------------------------------
\item \textbf{Stable Models} of $P$, $\fml{S} = \fml{S}_P$, are the stable models of $\delta P = \delta C + F + R$.
% --------------------------------
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{Distribution Semantics}
% --------------------------------
\begin{itemize}
% --------------------------------
\item \textbf{Total Choices:} $\Theta = \Theta_C = \Theta_P$ elements are $\theta = \cset{t_c}{c \in C}$ where $c=x::a$ and $t_c$ is $a$ or $\neg a$.
% --------------------------------
%\item For $s\in\fml{S}$ let $\theta_s \subseteq s$ (unique \textit{total choice})
%\item Define $\fml{S}_\theta = \cset{s \in \fml{S}}{\theta \subset s}$.
% --------------------------------
% --------------------------------
\item \textbf{Total Choice Probability:}
\begin{equation}
\pr{\theta} = \prod_{a \in \theta}x \prod_{\neg a \in \theta}\co{x}.\label{eq:prob.tc}
\end{equation}
% --------------------------------
\end{itemize}
% --------------------------------
This is the \emph{distribution semantic} as set by Sato.
\end{frame}
% ================================================================
\begin{frame}
% --------------------------------
\begin{block}{Problem Statement}
How to \textit{extend} probability from total choices to stable models, events and samples?
\end{block}
% --------------------------------
\begin{quotation}
There's a problem right at extending to stable models.
\end{quotation}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{The Disjunction Case}
% --------------------------------
\begin{exampleblock}{Disjuntion Example}
The specification
% --------------------------------
$$
\begin{aligned}
0.3 :: a &, \cr
b \lor c &\larr a .
\end{aligned}
$$
% --------------------------------
has three stable models,
% --------------------------------
$$
\begin{aligned}
s_1 &= \set{\neg a}, & s_2 &= \set{a, b}, & s_3 &= \set{a, c}.
\end{aligned}
$$
\end{exampleblock}
% --------------------------------
\begin{itemize}
% --------------------------------
\item\label{prop:unique.ext.tcsm}\textit{Any stable model contains exactly one total choice.~$\blacksquare$}
% --------------------------------
\item $\pr{\set{\neg a}} = 0.7$ is straightforward.
% --------------------------------
\item But, no \textit{informed} choice for $x\in\intcc{0,1}$ in
$$
\begin{aligned}
\pr{\set{a, b}} &= 0.3 x, \cr
\pr{\set{a, c}} &= 0.3 \co{x}.
\end{aligned}
$$
% --------------------------------
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{Lack of Information \& Parametrization}
% --------------------------------
\begin{itemize}
% --------------------------------
\item The specification \textit{lacks information} to set $x\in\intcc{0,1}$ in
$$
\begin{aligned}
\pr{\set{a, b}} &= 0.3 x, \cr
\pr{\set{a, c}} &= 0.3 \co{x}.
\end{aligned}
$$
\item A \textit{random variable} captures this uncertainty, \alert{assuming} that the stable models are statistically independent:
$$
\begin{aligned}
\pr{\set{\neg a} \given X = x } &= 0.7, \cr
\pr{\set{a, b} \given X = x } &= 0.3 x, \cr
\pr{\set{a, c} \given X = x } &= 0.3 \co{x}.
\end{aligned}
$$
\item Other uncertainties may lead to further conditions:
$$
\pr{s \given X_1 = x_1, \ldots, X_n = x_n }.
$$
% --------------------------------
\end{itemize}
Reducing \textbf{uncertainty}, \textit{e.g.} setting $X = 0.21$, must result from \textbf{external} sources, since the specification lacks information for further assertions.
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{Independence of Stable Models}
% --------------------------------
\begin{itemize}
\item[Q:] Why are the stable models assumed statistically independent?
% --------------------------------
\item[A:] Because dependence can be \textit{explicitly} modelled.
% --------------------------------
\item So, it is assumed \textit{intention} of the \textit{modeller} to not explicit express such dependences.
% --------------------------------
\item \textbf{For example:} \todo{Some key examples}.
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}%{Main Research Question}
% --------------------------------
A \textit{random variable} captures this uncertainty:
$$
\begin{aligned}
\pr{\set{\neg a} \given X = x } &= 0.7, \cr
\pr{\set{a, b} \given X = x } &= 0.3 x, \cr
\pr{\set{a, c} \given X = x } &= 0.3 \co{x}.
\end{aligned}
$$
% --------------------------------
\begin{block}{Main Research Question}
Can \textit{all} specification uncertainties be neatly expressed as that example?
\end{block}
% --------------------------------
\begin{itemize}
% --------------------------------
\item Follow ASP syntax; for each case, what are the uncertainty scenarios?
% --------------------------------
\item The disjunction example illustrates one such scenario.
% --------------------------------
\item \textit{Neat} means a function $d: \fml{S} \to \intcc{0, 1}$ such that
$$
\sum_{s\in\fml{S}_\theta} d\at{s} = 1
$$
for each $\theta \in \Theta$.
% --------------------------------
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{Leap into Inductive Programming}
% --------------------------------
Given a method that produces a distribution of samples, $p$, from a specification, $P$ and:
% --------------------------------
\begin{itemize}
% --------------------------------
\item $Z$, a dataset (of samples).
% --------------------------------
\item $e$, the respective empirical distribution.
% --------------------------------
\item $D$, some probability divergence, \textit{e.g.} Kullback-Leibler.
% --------------------------------
\end{itemize}
% --------------------------------
\begin{block}{Specification Performance \& Inductive Programming}
% --------------------------------
\begin{itemize}
% --------------------------------
\item $D\at{P} = D\at{e, p}$ is a \textbf{performance} measure of $P$.
% --------------------------------
\item Predictor performance measures, such as accuracy, are common in \textit{Machine Learning} tasks.
% --------------------------------
\item For \textit{Inductive Programming} this performance can be used, \textit{e.g.} as fitness, by algorithms searching for \textbf{optimal specifications of a dataset}.
% --------------------------------
\end{itemize}
% --------------------------------
\end{block}
% --------------------------------
\end{frame}
% ================================================================
\section{Extending Probability to Samples}
% ================================================================
\begin{frame}{Resolution Path}
Prior to \textit{conciliation} with data:
\begin{enumerate}
\item \alert{Hopefully}, \textit{conditional parameters} extend probability from total choices to \textit{standard models}.
\item \textbf{How} to extend it to \textit{events}?
\begin{itemize}
\item $\pr{x} = 0$ for $x$ \textit{excluded} by the specification, including \textit{inconsistent} samples.
\item $\pr{x}$ depends on the $s \in \fml{S}$ that contain/are contained in $x$.
\end{itemize}
\end{enumerate}
\alert{Consider probabilities \textbf{conditional} on the total choice!}
\end{frame}
% ================================================================
\begin{frame}{Bounds of Events}
% --------------------------------
\begin{itemize}
% --------------------------------
\item For $x\in\fml{E}$:
% --------------------------------
\begin{itemize}
% --------------------------------
\item \textbf{Lower Models:} $\lset{x} = \cset{s\in \fml{S} }{ s \subseteq x}$.
% --------------------------------
\item \textbf{Upper Models:} $\uset{x} = \cset{s\in \fml{S} }{ x \subseteq s}$.
% --------------------------------
\end{itemize}
% --------------------------------
\item\label{prop:lucases} \textbf{Proposition.} Exactly \textit{one} of the following cases takes place:
% --------------------------------
\begin{enumerate}
% --------------------------------
\item\label{prop:lucases.a} $\lset{x} = \set{x} = \uset{x}$ and $x$ is a stable model. Then:
\begin{equation}
\pr{x \given C = \theta_x} = d\at{x}.
\end{equation}
% --------------------------------
\item\label{prop:lucases.b} $\lset{x} \neq \emptyset \land \uset{x} = \emptyset$. Then:
\begin{equation}
\pr{x \given C = \theta_s, s \in \lset{x}} = \prod_{s\in\lset{x}} d\at{s}.
\end{equation}
% --------------------------------
\item\label{prop:lucases.c} $\lset{x} = \emptyset \land \uset{x} \neq \emptyset$. Then:
\begin{equation}
\pr{x \given C = \theta_s, s \in \uset{x}} = \sum_{s\in\uset{x}} d\at{s}.
\end{equation}
% --------------------------------
\item\label{prop:lucases.d} $\lset{x} = \emptyset = \uset{x}$. Then:
\begin{equation}
\pr{x} = 0.
\end{equation}
% --------------------------------
\end{enumerate}
because stable models are \textit{minimal}.
% --------------------------------
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{Conditional on Total Choices}
% --------------------------------
\begin{itemize}
% --------------------------------
\item A stable model is entailed by an atomic choice plus the facts and rules of the specification.
\item We express that entailment as a \textit{conditional}. For example:
$$\pr{\set{a,b} \given X = x} = \pr{b \given X = x, \Theta = a}\pr{\theta = a}$$
\item And now $\pr{b \given X = x, \Theta = a} = x$, since $X$ is a proxy for the stable models of the total choice $\theta = a$, we can further.
% --------------------------------
\end{itemize}
% --------------------------------
\end{frame}
% ================================================================
\begin{frame}{Disjunction Example | The Events Lattice}
\begin{center}
\begin{tikzpicture}
% --------------------------------
%\draw [help lines, color=gray!20] grid (11,7);
% --------------------------------
\node at (7, 7) {$\pr{\Theta=a} = 0.3$};
\node at (7, 6) {$x = \pr{S = ab \given \Theta}$};
\node at (7, 5) {$\co{x} = \pr{S \not= ab \given \Theta}$};
\node at (7, 7.5) {$\pr{E = abc \given \Theta} = \pr{S = ab, S = ac \given \Theta }$};
% --------------------------------
% \node [rrect] (sub) at (2, 7) {sub};
% --------------------------------
% \node [ fill=gray!10] (sup) at (3, 7) {sup};
% --------------------------------
% \node (ind) at (4, 7) {ind};
% --------------------------------
%
% --------------------------------
\node [ sub,
pin=45:\textcolor{violet}{$1$} ]
(E) at (5.5,0) {$\emptyset$};
% --------------------------------
%
% --------------------------------
\node [ sub,
pin=0:\textcolor{blue!75}{$1$} ]
(a) at (1.5,1.5) {$a$};
% --------------------------------
\node [ sub,
pin=315:\textcolor{blue!50}{$x$}]
(b) at (0,1.5) {$b$};
% --------------------------------
\node [ sub,
pin=315:\textcolor{blue!50}{$\co{x}$}]
(c) at (4.5,1.5) {$c$};
% --------------------------------
% \node [ sm,
% pin=270:\textcolor{teal}{$1$}]
% (A) at (8.5,1.5) {$\co{a}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (B) at (9.5, 1.5) {$\co{b}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (C) at (10.5, 1.5) {$\co{c}$};
% --------------------------------
%
% --------------------------------
\node [ sm,
pin=0:\textcolor{teal}{$x$}]
(ab) at (0,4) {$ab$};
% --------------------------------
\node [ sm,
pin=0:\textcolor{teal}{$\co{x}$}]
(ac) at (3,4) {$ac$};
% --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (aB) at (1,4) {$a\co{b}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (aC) at (2,4) {$a\co{c}$};
% --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$1$}]
% (Ab) at (4,4) {$\co{a}b$};
% % --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$1$}]
% (Ac) at (5,4) {$\co{a}c$};
% % --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$1$}]
% (AB) at (6,4) {$\co{a}\co{b}$};
% % --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$1$}]
% (AC) at (7,4) {$\co{a}\co{c}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (bc) at (10,4) {$bc$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (bC) at (11,4) {$b\co{c}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (Bc) at (9.5,3.5) {$\co{b}c$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (BC) at (10.5,3.5) {$\co{b}\co{c}$};
% --------------------------------
%
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!75}{$x\co{x}$}]
(abc) at (1.5,6)
{$abc$};
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!50}{$x$}]
(abC) at (0,6) {$ab\co{c}$};
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!50}{$\co{x}$}]
(aBc) at (3,6) {$a\co{b}c$};
% --------------------------------
% \node [ ind,
% pin=90:\textcolor{purple}{$0$}]
% (aBC) at (5,6) {$a\co{b}\co{c}$};
% --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$1$}]
% (Abc) at (7,6) {$\co{a}bc$};
% % --------------------------------
% \node [ sup,
% pin=270:\textcolor{blue!50}{$1$}]
% (AbC) at (8,6) {$\co{a}b\co{c}$};
% % --------------------------------
% \node [ sup,
% pin=270:\textcolor{blue!50}{$1$}]
% (ABc) at (9,6) {$\co{a}\co{b}c$};
% % --------------------------------
% \node [ sup,
% pin=270:\textcolor{blue!50}{$1$}]
% (ABC) at (10,6) {$\co{a}\co{b}\co{c}$};
% --------------------------------
%
% --------------------------------
\draw [->] (ab) to [out=270,in=180] (E);
\draw [->] (ab) to [out=270,in=90] (a);
\draw [->] (ab) to [out=270,in=90] (b);
\draw [->] (ab) to [out=90,in=270] (abc);
\draw [->] (ab) to [out=90,in=270] (abC);
%
\draw [->] (ac) to [out=270,in=180] (E);
\draw [->] (ac) to [out=270,in=90] (a);
\draw [->] (ac) to [out=270,in=90] (c);
\draw [->] (ac) to [out=90,in=270] (abc);
\draw [->] (ac) to [out=90,in=270] (aBc);
%
% \draw [->] (A) to [out=270,in=0] (E);
% %
% \draw [->] (A) to [out=90,in=270] (Abc);
% \draw [->] (A) to [out=90,in=270] (AbC);
% \draw [->] (A) to [out=90,in=270] (ABc);
% \draw [->] (A) to [out=90,in=270] (ABC);
% %
% \draw [->] (A) to [out=90,in=270] (Ab);
% \draw [->] (A) to [out=90,in=270] (Ac);
% \draw [->] (A) to [out=90,in=270] (AB);
% \draw [->] (A) to [out=90,in=270] (AC);
\end{tikzpicture}
\end{center}
\end{frame}
% ================================================================
\begin{frame}{Disjunction Example | The Events Lattice}
\begin{center}
\begin{tikzpicture}
% --------------------------------
%\draw [help lines, color=gray!20] grid (11,7);
% --------------------------------
\node [sm] (sm) at (5, 7) {$\pr{\Theta=\set{\co{a}}} = \co{0.3}$};
% --------------------------------
% \node [rrect] (sub) at (2, 7) {sub};
% --------------------------------
% \node [ fill=gray!10] (sup) at (3, 7) {sup};
% --------------------------------
% \node (ind) at (4, 7) {ind};
% --------------------------------
%
% --------------------------------
\node [ sub,
pin=45:\textcolor{violet}{$1$} ]
(E) at (5.5,0) {$\emptyset$};
% --------------------------------
%
% --------------------------------
% \node [ sub,
% pin=270:\textcolor{blue!75}{$1$} ]
% (a) at (1.5,1.5) {$a$};
% % --------------------------------
% \node [ sub,
% pin=270:\textcolor{blue!50}{$x$}]
% (b) at (0,1.5) {$b$};
% % --------------------------------
% \node [ sub,
% pin=270:\textcolor{blue!50}{$\co{x}$}]
(c) at (4.5,1.5) {$c$};
% --------------------------------
\node [ sm,
pin=45:\textcolor{teal}{$1$}]
(A) at (8.5,1.5) {$\co{a}$};
% --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (B) at (9.5, 1.5) {$\co{b}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (C) at (10.5, 1.5) {$\co{c}$};
% --------------------------------
%
% --------------------------------
% \node [ sm,
% pin=90:\textcolor{teal}{$x$}]
% (ab) at (0,4) {$ab$};
% % --------------------------------
% \node [ sm,
% pin=90:\textcolor{teal}{$\co{x}$}]
% (ac) at (3,4) {$ac$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (aB) at (1,4) {$a\co{b}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (aC) at (2,4) {$a\co{c}$};
% --------------------------------
\node [ sup,
pin=135:\textcolor{blue!50}{$1$}]
(Ab) at (4,4) {$\co{a}b$};
% --------------------------------
\node [ sup,
pin=135:\textcolor{blue!50}{$1$}]
(Ac) at (5,4) {$\co{a}c$};
% --------------------------------
\node [ sup,
pin=135:\textcolor{blue!50}{$1$}]
(AB) at (6,4) {$\co{a}\co{b}$};
% --------------------------------
\node [ sup,
pin=135:\textcolor{blue!50}{$1$}]
(AC) at (7,4) {$\co{a}\co{c}$};
% --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (bc) at (10,4) {$bc$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (bC) at (11,4) {$b\co{c}$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (Bc) at (9.5,3.5) {$\co{b}c$};
% % --------------------------------
% \node [ ind,
% pin=270:\textcolor{purple}{$0$}]
% (BC) at (10.5,3.5) {$\co{b}\co{c}$};
% --------------------------------
%
% --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!75}{$x\co{x}$}]
% (abc) at (1.5,6)
% {$abc$};
% % --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$x$}]
% (abC) at (0,6) {$ab\co{c}$};
% % --------------------------------
% \node [ sup,
% pin=90:\textcolor{blue!50}{$\co{x}$}]
% (aBc) at (3,6) {$a\co{b}c$};
% % --------------------------------
% \node [ ind,
% pin=90:\textcolor{purple}{$0$}]
% (aBC) at (5,6) {$a\co{b}\co{c}$};
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!50}{$1$}]
(Abc) at (7,6) {$\co{a}bc$};
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!50}{$1$}]
(AbC) at (8,6) {$\co{a}b\co{c}$};
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!50}{$1$}]
(ABc) at (9,6) {$\co{a}\co{b}c$};
% --------------------------------
\node [ sup,
pin=45:\textcolor{blue!50}{$1$}]
(ABC) at (10,6) {$\co{a}\co{b}\co{c}$};
% --------------------------------
%
% --------------------------------
% \draw [->] (ab) to [out=270,in=180] (E);
% \draw [->] (ab) to [out=270,in=90] (a);
% \draw [->] (ab) to [out=270,in=90] (b);
% \draw [->] (ab) to [out=90,in=270] (abc);
% \draw [->] (ab) to [out=90,in=270] (abC);
% %
% \draw [->] (ac) to [out=270,in=180] (E);
% \draw [->] (ac) to [out=270,in=90] (a);
% \draw [->] (ac) to [out=270,in=90] (c);
% \draw [->] (ac) to [out=90,in=270] (abc);
% \draw [->] (ac) to [out=90,in=270] (aBc);
%
\draw [->] (A) to [out=270,in=0] (E);
%
\draw [->] (A) to [out=90,in=270] (Abc);
\draw [->] (A) to [out=90,in=270] (AbC);
\draw [->] (A) to [out=90,in=270] (ABc);
\draw [->] (A) to [out=90,in=270] (ABC);
%
\draw [->] (A) to [out=90,in=270] (Ab);
\draw [->] (A) to [out=90,in=270] (Ac);
\draw [->] (A) to [out=90,in=270] (AB);
\draw [->] (A) to [out=90,in=270] (AC);
\end{tikzpicture}
\end{center}
\end{frame}
% ================================================================
\begin{frame}
\begin{itemize}
\item Consider the ASP program $P = C \land F \land R$ with total choices $\Theta $ and stable models $\fml{S}$.
\item Let $d : \fml{S} \to \intcc{0,1}$ such that $\sum_{s\in\fml{S}_\theta} d\at{s} = 1$ for each $\theta \in \Theta$.
\end{itemize}
\end{frame}
% ================================================================
\begin{frame}
For each $z\in\fml{Z}$ only one of the following cases takes place
\begin{enumerate}
\item $z$ is inconsistent. Then \textbf{define}
\begin{equation}
w_d\at{x} = 0.\label{def:w.inconsistent}
\end{equation}
%
\item $z$ is an event and $\lset{z} = \set{z} = \uset{z}$. Then $z$ is a stable model and \textbf{define}
\begin{equation}
w_d\at{z} = w\at{z} = d\at{z} \pr{\theta_z}.\label{eq:prob.sm}
\end{equation}
%
\item $z$ is an event and $\lset{z} \neq \emptyset \land \uset{x} = \emptyset$. Then \textbf{define}
\begin{equation}
w_d\at{z} = \sum_{s \in \lset{z}} w_d\at{s}.\label{def:w.disj}
\end{equation}
%
\item $z$ is an event and $\lset{z} = \emptyset \land \uset{z} \neq \emptyset$. Then \textbf{define}
\begin{equation}
w_d\at{z} = \prod_{s \in \uset{z}} w_d\at{s}.\label{def:w.conj}
\end{equation}
%
\item $z$ is an event and $\lset{z} = \emptyset \land \uset{z} = \emptyset$. Then \textbf{define}
\begin{equation}
w_d\at{z} = 0.\label{def:w.empty}
\end{equation}
\end{enumerate}
\end{frame}
% ================================================================
\begin{frame}
\begin{enumerate}
%
\item The last point defines a ``weight'' function on the samples that depends not only on the total choices and stable models of a PASP but also on a certain function $d$ that must respect some conditions. To simplify the notation we use the subscript in $w_d$ only when necessary.
%
\item At first, it may seem counter-intuitive that $w\at{\emptyset} = \sum_{s\in\fml{S}} w\at{s}$ is the largest ``weight'' in the lattice. But $\emptyset$, as an event, sets zero restrictions on the ``compatible'' stable models. The ``complement'' of $\bot = \emptyset$ is the \emph{maximal inconsistent} sample $\top = \fml{A} \cup \cset{\neg a }{ a \in \fml{A}}$.
%
\item \textbf{We haven't yet defined a probability measure.} To do so we must define a set of samples $\Omega$, a set of events $F\subseteq \pset{\Omega}$ and a function $P:F\to\intcc{0,1}$ such that:
\begin{enumerate}
\item $\pr{E} \in \intcc{0, 1}$ for any $E \in F$.
\item $\pr{\Omega} = 1$.
\item if $E_1 \cap E_2 = \emptyset$ then $\pr{E_1 \cup E_2} = \pr{E_1} + \pr{E_2}$.
\end{enumerate}
%
\item In the following, assume that the stable models are iid.
%
\item Let the sample space $\Omega = \fml{Z}$ and the event space $F = \pset{\Omega}$. Define $Z = \sum_{\zeta\in\fml{Z}} w\at{\zeta}$ and
\begin{equation}
\pr{z} = \frac{w\at{z}}{Z}, z \in \Omega \label{eq:def.prob}
\end{equation}
and
\begin{equation}
\pr{E} = \sum_{x\in E} \pr{x}, E \subseteq \Omega. \label{eq:def.prob.event}
\end{equation}
Now:
\begin{enumerate}
\item $P(E) \in \intcc{0,1}$ results directly from the definitions of $P$ and $w$.
\item $\pr{\Omega} = 1$ also results directly from the definitions.
\item Consider two disjunct events $A, B \subset \Omega \land A \cap B = \emptyset$. Then
$$
\begin{aligned}
\pr{A \cup B} &= \sum_{x \in A \cup B} \pr{x} \cr
&= \sum_{x \in A} \pr{x} + \sum_{x \in B} \pr{x} - \sum_{x \in A \cap B} \pr{x} \cr
&= \sum_{x \in A} \pr{x} + \sum_{x \in B} \pr{x} &\text{because}~A\cap B = \emptyset \cr
&= \pr{A} + \pr{B}.
\end{aligned}
$$
\item So $\del{\Omega = \fml{Z}, F = \pset{\Omega}, P}$ is a probability space. {$\blacksquare$}
\end{enumerate}
\end{enumerate}
\end{frame}
% ================================================================
\section{Cases \& Examples}
% ================================================================
\subsection{Programs with disjunctive heads}
% ================================================================
\begin{frame}
Consider the program:
$$
\begin{aligned}
c_1 &= a \lor \neg a, \cr
c_2 &= b \lor c \larr a.
\end{aligned}
$$
This program has two total choices,
$$
\begin{aligned}
\theta_1&= \set{ \neg a }, \cr
\theta_2&= \set{ a }.
\end{aligned}
$$
and three stable models,
$$
\begin{aligned}
s_1 &= \set{ \neg a }, \cr
s_2 &= \set{ a, b }, \cr
s_3 &= \set{ a, c }.
\end{aligned}
$$
\end{frame}
% ================================================================
\begin{frame}
Suppose that we add an annotation $x :: a$, which entails $\co{x} :: \neg a$. This is enough to get $w\at{s_1} = \co{x}$ but, on the absence of further information, no fixed probability can be assigned to either model $s_2, s_3$ except that the respective sum must be $x$. So, expressing our lack of knowledge using a parameter $d \in \intcc{0, 1}$ we get:
$$
\begin{cases}
w\at{s_1 } = &\co{x}\cr
w\at{s_2 } = &dx\cr
w\at{s_3} = &\co{d}x.
\end{cases}
$$
\end{frame}
% ================================================================
\begin{frame}
In this diagram:
\begin{itemize}
\item Negations are represented as \emph{e.g.} $\co{a}$ instead of $\neg a$; Stable models are denoted by shaded nodes as \tikz{\node[fill=gray!50] {$ab$}}.
\item Events in $\lset{x}$ are \emph{e.g.} \tikz{\node[ circle] {$a$}} and those in $\uset{x}$ are \emph{e.g.} \tikz{\node[draw] {$\co{a}b$}}. The remaining are simply denoted by \emph{e.g.} \tikz{\node {$a\co{b}$}}.
\item The edges connect stable models with related events. Up arrow indicate links to $\uset{s}$ and down arrows to $\lset{s}$.
\item The \emph{weight propagation} sets:
$$
\begin{aligned}
w\at{abc} &= w\at{ab} w\at{ac} = x^2d\co{d}, \cr
w\at{\co{a}\cdot\cdot} &= w\at{\neg a} = \co{x}, \cr
w\at{a} &= w\at{ab} + w\at{ac} = x(d + \co{d}) = x, \cr
w\at{b} &= w\at{ab} = dx, \cr
w\at{c} &= w\at{ac} = \co{d}x, \cr
w\at{\emptyset} &= w\at{ab} + w\at{ac} + w\at{\neg a} = dx + \co{d}x + \co{x} = 1, \cr
w\at{a\co{b}} &= 0.
\end{aligned}
$$
\item The total weight is
$$
\begin{aligned}
Z &= w\at{abc} + 8 w\at{\co{a}b}\cr
&+ w\at{ab} + w\at{ac} + w\at{\co{a}}\cr
&+ w\at{a}+ w\at{b}+ w\at{c}\cr
&+ w\at{\emptyset}\cr
%
&= - x^{2} d^{2} + x^{2} d + 2 x d - 7 x + 10
\end{aligned}
$$
\item Now, if $x$ has an annotation to \emph{e.g.} $0.3$ we get
$$
Z = - 0.09 d^{2} + 0.69 d + 7.9
$$
\item Now some statistics are possible. For example we get
$$
\pr{abc \mid x = 0.3} = \frac{0.09 d \left(d - 1\right)}{0.09 d^{2} - 0.69 d - 7.9}
$$.
\item This expression can be plotted for $d\in\intcc{0,1}$
\begin{center}
\includegraphics[height=15em]{Pabc_alpha03.pdf}
\end{center}
\item If a data set $E$ entails \emph{e.g.} $\pr{abc \mid E} = 0.0015$ we can numerically solve
$$
\begin{aligned}
\pr{abc \mid x = 0.3} &= \pr{abc \mid E} \cr
\iff\cr
\frac{0.09 d \del{d - 1}}{0.09 d^{2} - 0.69 d - 7.9} &= 0.0015
\end{aligned}
$$
which has two solutions, $d \approx 0.15861$ or $d \approx 0.83138$.
\end{itemize}
\end{frame}
% ================================================================
\subsection{Non-stratified programs}
% ================================================================
\begin{frame}
The following LP is non-stratified, because has a cycle with negated arcs:
$$
\begin{aligned}
c_1 &= a\lor \neg a,\cr
c_2 &= b \larr \naf c \land \naf a, \cr
c_3 &= c \larr \naf b.
\end{aligned}
$$
This program has three stable models
$$
\begin{aligned}
s_1 &= \set{ a, c }, \cr
s_2 &= \set{ \neg a, b }, \cr
s_3 &= \set{ \neg a, c }.
\end{aligned}
$$
\end{frame}
\begin{frame}
The disjunctive clause $a\lor\neg a$ defines a set of \textbf{total choices}
$$
\Theta = \set{
\theta_1 = \set{ a },
\theta_2 = \set{ \neg a }
}.
$$
\end{frame}
% ================================================================
\begin{frame}
Looking into probabilistic events of the program and/or its models, we define $x = \pr{\Theta = \theta_1}\in\intcc{0, 1}$ and $\pr{\Theta = \theta_2} = \co{x}$.
Since $s_1$ is the only stable model that results from $\Theta = \theta_1$, it is natural to extend $\pr{ s_1 } = \pr{\Theta = \theta_1} = x$. However, there is no clear way to assign $\pr{s_2}, \pr{s_3}$ since \emph{both models result from the single total choice} $\Theta = \theta_2$. Clearly,
$$\pr{s_2 \mid \Theta} + \pr{s_3 \mid \Theta} =
\begin{cases}
0 & \text{if}~\Theta = \theta_1\cr
1 & \text{if}~\Theta = \theta_2
\end{cases}
$$
but further assumptions are not supported \emph{a priori}. So let's \textbf{parameterize} the equation above,
$$
\begin{cases}
\pr{s_2 \mid \Theta = \theta_2} = &\beta \in \intcc{0, 1} \cr
\pr{s_3 \mid \Theta = \theta_2} = &\co{\beta},
\end{cases}
$$
in order to explicit our knowledge, or lack of, with numeric values and relations.
\end{frame}
% ================================================================
\begin{frame}
Now we are able to define the \textbf{joint distribution} of the boolean random variables $A,B,C$:
$$
\begin{array}{cc|l}
A, B, C& P & \text{Obs.}\cr
\hline
a, \neg b, c & x & s_1, \Theta=\theta_1\cr
\neg a, b, \neg c & \co{x}\beta & s_2, \Theta=\theta_2\cr
\neg a, \neg b, c & \co{x}\co{\beta} & s_3, \Theta=\theta_2\cr
\ast & 0&\text{not stable models}
\end{array}
$$
where $x, \beta\in\intcc{0,1}$.
\end{frame}
% ================================================================
\section{Conclusions}
% ================================================================
\begin{frame}
\begin{itemize}
\item We can use the basics of probability theory and logic programming to assign explicit \emph{parameterized} probabilities to the (stable) models of a program.
\item In the covered cases it was possible to define a (parameterized) \emph{family of joint distributions}.
\item How far this approach can cover all the cases on logic programs is (still) an issue \emph{under investigation}.
\item However, it is non-restrictive since \emph{no unusual assumptions are made}.
\end{itemize}
\end{frame}
% ================================================================
\section*{ASP \& related definitions}
% ================================================================
\begin{frame}
\begin{itemize}
\item An \deft{atom} is $r(t_1, \ldots t_n)$ where
\begin{itemize}
\item $r$ is a $n$-ary predicate symbol and each $t_i$ is a constant or a variable.
\item A \deft{ground atom} has no variables; A \deft{literal} is either an atom $a$ or a negated atom $\neg a$.
\end{itemize}
\item An \deft{ASP Program} is a set of \deft{rules} such as $h_1 \vee \cdots \vee h_m \leftarrow b_1 \wedge \cdots \wedge b_n$.
\begin{itemize}
\item The \deft{head} of this rule is $h_1 \vee \cdots \vee h_m$, the \deft{body} is $b_1 \wedge \cdots \wedge b_n$ and each $b_i$ is a \deft{subgoal}.
\item Each $h_i$ is a literal, each subgoal $b_j$ is a literal or a literal preceded by $\naf\;$ and $m + n > 0$.
\item A \deft{propositional program} has no variables.
\item A \deft{non-disjunctive rule} has $m \leq 1$; A \deft{normal rule} has $m = 1$; A \deft{constraint} has $m = 0$; A \deft{fact} is a normal rule with $n = 0$.
\end{itemize}
\item The \deft{Herbrand base} of a program is the set of ground literals that result from combining all the predicates and constants of the program.
\begin{itemize}
\item An \deft{event} is a consistent subset (\emph{i.e.} doesn't contain $\set{a, \neg a}$) of the Herbrand base.
\item Given an event $I$, a ground literal $a$ is \deft{true}, $I \models a$, if $a \in I$; otherwise the literal is \deft{false}.
\item A ground subgoal, $\naf b$, where $b$ is a ground literal, is \deft{true}, $I \models \naf b$, if $b \not\in I$; otherwise, if $b \in I$, it is \deft{false}.
\item A ground rule $r = h_1 \vee \cdots \vee h_m \leftarrow b_1 \wedge \cdots \wedge b_n$ is \deft{satisfied} by the event $I$, \emph{i.e.} $I \models r$, iff
$$
\forall j \exists i~I \models b_j \implies I \models h_i.
$$
\item A \deft{model} of a program is an event that satisfies all its rules. Denote $\fml{M}_P$ the set of all models of $P$.
\end{itemize}
\item The \deft{dependency graph} of a program is a digraph where:
\begin{itemize}
\item Each grounded atom is a node.
\item For each grounded rule there are edges from the atoms in the body to the atoms in the head.
\item A \deft{negative edge} results from an atom with $\naf\;$; Otherwise it is a \deft{positive edge}.
\item An \deft{acyclic program} has an acyclic dependency graph; A \deft{normal program} has only normal rules; A \deft{definite program} is a normal program that doesn't contains $\neg$ neither $\naf\;$.
\item In the dependency graph of a \deft{stratified program} no cycle contains a negative edge.
\item \textbf{A stratified program has a single minimal model} that assigns either true or false to each atom.
\end{itemize}
\item Every \emph{definite program} has a unique minimal model: its \deft{semantic}.
\item Programs with negation may have no unique minimal model.
\item Given a program $P$ and an event $I$, their \deft{reduct}, $P^I$, is the propositional program that results from
\begin{enumerate}
\item Removing all the rules with $\naf b$ in the body where $b \in I$.
\item Removing all the $\naf b$ subgoals from the remaining rules.
\end{enumerate}
\item A \deft{stable model} (or \deft{answer set}) of the program $P$ is an event $I$ that is the minimal model of the reduct $P^I$.
\item Denote $\fml{S}_P$ the set of all stable models of program $P$. The \deft{semantics} (or \deft{answer sets}) of a program $P$ is the set $\fml{S}_P$.
\begin{itemize}
\item Some programs, such as $a \leftarrow \naf a$, have no stable models.
\item A stable model is an event closed under the rules of the program.
\end{itemize}
\end{itemize}
\end{frame}
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\end{document}
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