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Francisco Coelho
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text/summaries/00_PASP_credal.md
text/summaries/00_pasp.md
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+# Probabilistic Answer Set Programming
+
+## Non-stratified programs
+
+> Minimal example of **non-stratified program**.
+
+The following annotated LP, with clauses $c_1, c_2, c_3$ respectively, is non-stratified (because has a cycle with negated arcs) but no head is disjunctive:
+```prolog
+0.3::a.                 % c1
+s :- not w, not a.      % c2
+w :- not s.             % c3
+```
+
+This program has three stable models:
+$$
+\begin{aligned}
+m_1 &= \set{ a, w } \cr
+m_2 &= \set{ \neg a, s } \cr
+m_3 &= \set{ \neg a, w }
+\end{aligned}
+$$
+
+The probabilistic clause `0.3::a.` defines a **total choice**
+$$
+\Theta = \set{
+    \theta_1 = \set{ a }, 
+    \theta_2 = \set{ \neg a }
+}
+$$
+such that
+$$
+\begin{aligned}
+P(\Theta = \set{ a }) &= 0.3\cr
+P(\Theta = \set{ \neg a }) &= 0.7 \cr
+\end{aligned}
+$$
+
+> While it is natural to extend $P( m_1 ) = 0.3$ from $P(\theta_1) = 0.3$ there is no clear way to assign $P(m_2), P(m_3)$ since both models result from the total choice $\theta_2$.
+
+
+
+Under the **CWA**, $\sim\!\!q \models \neg q$, so $c_2, c_3$ induce probabilities:
+
+$$
+\begin{aligned}
+p_a &= P(a | \Theta) \cr
+p_s &= P(s | \Theta) &= (1 - p_w)(1 - p_a) \cr
+p_w &= P(w | \Theta) &= (1 - p_w)
+\end{aligned}
+$$
+from which results
+$$
+\begin{equation}
+p_s = p_s(1 - p_a).
+\end{equation}
+$$
+
+So, if $\Theta = \theta_1 = \set{ a }$ (one stable model):
+
+- We have $p_a = P(a | \Theta = \set{ a }) = 1$.
+- Equation (1) becomes $p_s = 0$.
+- From $p_w = 1 - p_s$ we get $P(w | \Theta) = 1$.
+
+and if $\Theta = \theta_2 = \set{ \neg a }$ (two stable models):
+
+- We have $p_a = P(a | \Theta = \set{ \neg a }) = 0$.
+- Equation (1) becomes $p_s = p_s$; Since we know nothing about $p_s$, we let $p_s =  \alpha \in \left[0, 1\right]$.
+- We still have the relation $p_w = 1 - p_s$ so $p_w = 1 - \alpha$.
+
+We can now define the **marginals** for $s, w$:
+$$
+\begin{aligned}
+P(s) &=\sum_\theta P(s|\theta)P(\theta)= 0.7\alpha            \cr
+P(w) &=\sum_\theta P(s|\theta)P(\theta)= 0.3 + 0.7(1 - \alpha) \cr 
+\alpha &\in\left[ 0, 1 \right]
+\end{aligned}
+$$
+
+> The parameter $\alpha$ not only **expresses insufficient information** to sharply define $p_s$ but also **relates** $p_s$ and $p_w$.
+
+## Disjunctive heads
+
+> Minimal example of **disjunctive heads** program.
+
+Consider this LP
+
+```prolog
+0.3::a.
+b ; c :- a.
+```
+
+with three stable models:
+$$
+\begin{aligned}
+m_1 &= \set{ \neg a } \cr
+m_2 &= \set{ a, b } \cr
+m_3 &= \set{ a, c }
+\end{aligned}
+$$
+
+Again, $P(m_1) = 0.3$ is quite natural but there are no clear assignments for $P(m_2), P(m_3)$.
+
+The total choices here are
+$$
+\Theta = \set{
+    \theta_1 = \set{ a }
+    \theta_2 = \set{ \neg a }
+}
+$$
+such that
+$$
+\begin{aligned}
+P(\Theta = \set{ a }) &= 0.3\cr
+P(\Theta = \set{ \neg a }) &= 0.7 \cr
+\end{aligned}
+$$
+and the LP induces
+$$
+P(b \vee c | \Theta) = P(a | \Theta).
+$$
+
+Since the disjunctive expands as
+$$
+\begin{equation}
+P(b \vee c | \Theta) = P(b | \Theta) + P( c | \Theta)  - P(b \wedge c | \Theta)
+\end{equation}
+$$
+and we know that $P(b \vee c | \Theta) = P(a | \Theta)$ we need two independent parameters, for example
+$$
+\begin{aligned}
+P(b | \Theta) &= \beta \cr
+P(c | \Theta) &= \gamma \cr
+\end{aligned}
+$$
+where
+$$
+\begin{aligned}
+    \alpha & \in \left[0, 0.3\right] \cr
+    \beta & \in \left[0, \alpha\right]
+\end{aligned}
+$$
+
+This example also calls for reconsidering the CWA since it entails that **we should assume that $b$ and $c$ are conditionally independent given $a$.**
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+# Probabilistic ASP
+
+## Weighted Approach
+
+1. **Total Choices.** $N(C = x) = \prod_{a \in x} w_a \prod_{\neg a \in x} (1 - w_a)$.
+2. **Stable Models.** $N(S = x | C = c) = \alpha_{x,c}$,
+    where the set of parameters $\alpha_{x,c}$ is such that:
+    $$
+    \begin{cases}
+        \alpha_{x,c} \geq 0, & \forall c, x\cr
+        \alpha_{x,c} = 0, & \forall x \not\supseteq c \cr
+        \sum_{x} \alpha_{x,c} = 1, & \forall c.
+    \end{cases}
+    $$
+3. **Worlds.** $N(W = x)$
+   1. If $x$ is a _total choice_: $
+    N(W = x) = \prod_{a \in x} w_a \prod_{\neg a \in x} (1 - w_a).
+    $$
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		1	+# Probabilistic Answer Set Programming
		2	+
		3	+## Non-stratified programs
		4	+
		5	+> Minimal example of non-stratified program.
		6	+
		7	+The following annotated LP, with clauses $c_1, c_2, c_3$ respectively, is non-stratified (because has a cycle with negated arcs) but no head is disjunctive:
		8	+```prolog
		9	+0.3::a. % c1
		10	+s :- not w, not a. % c2
		11	+w :- not s. % c3
		12	+```
		13	+
		14	+This program has three stable models:
		15	+$$
		16	+\begin{aligned}
		17	+m_1 &= \set{ a, w } \cr
		18	+m_2 &= \set{ \neg a, s } \cr
		19	+m_3 &= \set{ \neg a, w }
		20	+\end{aligned}
		21	+$$
		22	+
		23	+The probabilistic clause `0.3::a.` defines a total choice
		24	+$$
		25	+\Theta = \set{
		26	+ \theta_1 = \set{ a },
		27	+ \theta_2 = \set{ \neg a }
		28	+}
		29	+$$
		30	+such that
		31	+$$
		32	+\begin{aligned}
		33	+P(\Theta = \set{ a }) &= 0.3\cr
		34	+P(\Theta = \set{ \neg a }) &= 0.7 \cr
		35	+\end{aligned}
		36	+$$
		37	+
		38	+> While it is natural to extend $P( m_1 ) = 0.3$ from $P(\theta_1) = 0.3$ there is no clear way to assign $P(m_2), P(m_3)$ since both models result from the total choice $\theta_2$.
		39	+
		40	+
		41	+
		42	+Under the CWA, $\sim\!\!q \models \neg q$, so $c_2, c_3$ induce probabilities:
		43	+
		44	+$$
		45	+\begin{aligned}
		46	+p_a &= P(a \| \Theta) \cr
		47	+p_s &= P(s \| \Theta) &= (1 - p_w)(1 - p_a) \cr
		48	+p_w &= P(w \| \Theta) &= (1 - p_w)
		49	+\end{aligned}
		50	+$$
		51	+from which results
		52	+$$
		53	+\begin{equation}
		54	+p_s = p_s(1 - p_a).
		55	+\end{equation}
		56	+$$
		57	+
		58	+So, if $\Theta = \theta_1 = \set{ a }$ (one stable model):
		59	+
		60	+- We have $p_a = P(a \| \Theta = \set{ a }) = 1$.
		61	+- Equation (1) becomes $p_s = 0$.
		62	+- From $p_w = 1 - p_s$ we get $P(w \| \Theta) = 1$.
		63	+
		64	+and if $\Theta = \theta_2 = \set{ \neg a }$ (two stable models):
		65	+
		66	+- We have $p_a = P(a \| \Theta = \set{ \neg a }) = 0$.
		67	+- Equation (1) becomes $p_s = p_s$; Since we know nothing about $p_s$, we let $p_s = \alpha \in \left[0, 1\right]$.
		68	+- We still have the relation $p_w = 1 - p_s$ so $p_w = 1 - \alpha$.
		69	+
		70	+We can now define the marginals for $s, w$:
		71	+$$
		72	+\begin{aligned}
		73	+P(s) &=\sum_\theta P(s\|\theta)P(\theta)= 0.7\alpha \cr
		74	+P(w) &=\sum_\theta P(s\|\theta)P(\theta)= 0.3 + 0.7(1 - \alpha) \cr
		75	+\alpha &\in\left[ 0, 1 \right]
		76	+\end{aligned}
		77	+$$
		78	+
		79	+> The parameter $\alpha$ not only expresses insufficient information to sharply define $p_s$ but also relates $p_s$ and $p_w$.
		80	+
		81	+## Disjunctive heads
		82	+
		83	+> Minimal example of disjunctive heads program.
		84	+
		85	+Consider this LP
		86	+
		87	+```prolog
		88	+0.3::a.
		89	+b ; c :- a.
		90	+```
		91	+
		92	+with three stable models:
		93	+$$
		94	+\begin{aligned}
		95	+m_1 &= \set{ \neg a } \cr
		96	+m_2 &= \set{ a, b } \cr
		97	+m_3 &= \set{ a, c }
		98	+\end{aligned}
		99	+$$
		100	+
		101	+Again, $P(m_1) = 0.3$ is quite natural but there are no clear assignments for $P(m_2), P(m_3)$.
		102	+
		103	+The total choices here are
		104	+$$
		105	+\Theta = \set{
		106	+ \theta_1 = \set{ a }
		107	+ \theta_2 = \set{ \neg a }
		108	+}
		109	+$$
		110	+such that
		111	+$$
		112	+\begin{aligned}
		113	+P(\Theta = \set{ a }) &= 0.3\cr
		114	+P(\Theta = \set{ \neg a }) &= 0.7 \cr
		115	+\end{aligned}
		116	+$$
		117	+and the LP induces
		118	+$$
		119	+P(b \vee c \| \Theta) = P(a \| \Theta).
		120	+$$
		121	+
		122	+Since the disjunctive expands as
		123	+$$
		124	+\begin{equation}
		125	+P(b \vee c \| \Theta) = P(b \| \Theta) + P( c \| \Theta) - P(b \wedge c \| \Theta)
		126	+\end{equation}
		127	+$$
		128	+and we know that $P(b \vee c \| \Theta) = P(a \| \Theta)$ we need two independent parameters, for example
		129	+$$
		130	+\begin{aligned}
		131	+P(b \| \Theta) &= \beta \cr
		132	+P(c \| \Theta) &= \gamma \cr
		133	+\end{aligned}
		134	+$$
		135	+where
		136	+$$
		137	+\begin{aligned}
		138	+ \alpha & \in \left[0, 0.3\right] \cr
		139	+ \beta & \in \left[0, \alpha\right]
		140	+\end{aligned}
		141	+$$
		142	+
		143	+This example also calls for reconsidering the CWA since it entails that we should assume that $b$ and $c$ are conditionally independent given $a$.
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		1	+# Probabilistic ASP
		2	+
		3	+## Weighted Approach
		4	+
		5	+1. Total Choices. $N(C = x) = \prod_{a \in x} w_a \prod_{\neg a \in x} (1 - w_a)$.
		6	+2. Stable Models. $N(S = x \| C = c) = \alpha_{x,c}$,
		7	+ where the set of parameters $\alpha_{x,c}$ is such that:
		8	+ $$
		9	+ \begin{cases}
		10	+ \alpha_{x,c} \geq 0, & \forall c, x\cr
		11	+ \alpha_{x,c} = 0, & \forall x \not\supseteq c \cr
		12	+ \sum_{x} \alpha_{x,c} = 1, & \forall c.
		13	+ \end{cases}
		14	+ $$
		15	+3. Worlds. $N(W = x)$
		16	+ 1. If $x$ is a _total choice_: $
		17	+ N(W = x) = \prod_{a \in x} w_a \prod_{\neg a \in x} (1 - w_a).
		18	+ $$
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