Add CDMO SAT resolution and DPLL

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\makenotesfront
\input{./sections/_constraint_programming.tex}
\input{./sections/_sat.tex}
\end{document}

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\chapter{SAT solvers}
\begin{description}
\item[Satifiability (SAT)] \marginnote{SAT}
Given a propositional formula $f$,
find an assignment of the variables to make the formula true.
If $f$ is satisfiable, we say that it is SAT. Otherwise, it is UNSAT.
\end{description}
\begin{theorem}(Cook-Levin) \marginnote{Cook-Levin theorem}
SAT is NP-complete.
\begin{descriptionlist}
\item[NP-complete]
A problem is NP-complete iff it is in NP and every NP problem can be reduced to it.
\begin{remark}
All problems in NP can be reduced to SAT.
\end{remark}
\end{descriptionlist}
\end{theorem}
\begin{description}
\item[Satisfiable formula] \marginnote{Satisfiable formula}
A formula $f$ is satisfiable iff there is an assignment of the variables that makes it true.
\item[Valid formula] \marginnote{Valid formula}
A formula $f$ is valid iff any assignment of the variables makes it true.
\begin{remark}
$f$ is valid iff $\lnot f$ is UNSAT.
\end{remark}
\item[SAT solver] \marginnote{SAT solver}
Given a problem, the following steps should be followed to solve it using a SAT solver:
\begin{enumerate}
\item Formalize the problem into a propositional formula.
\item Depending on the solver, it might be necessary to convert in into CNF.
\item Feed the formula to a SAT solver.
\item The result can be an assignment of the variables (SAT) or a failure (UNSAT).
\end{enumerate}
\end{description}
\section{Resolution}
\begin{description}
\item[Resolution] \marginnote{Resolution}
Basic approach for SAT.
Prove that a set of CNF clauses is UNSAT by deriving new clauses until a contradiction is reached.
Resolution rules to derive new clauses are:
\begin{descriptionlist}
\item[Resolution rule] \marginnote{Resolution rule}
\phantom{}
\begin{minipage}{0.4\linewidth}
\begin{prooftree}
\AxiomC{$p \vee V$}
\AxiomC{$\lnot p \vee W$}
\BinaryInfC{$V \vee W$}
\end{prooftree}
\end{minipage}
\item[Unit resolution] \marginnote{Unit resolution}
\begin{minipage}{0.4\linewidth}
\begin{prooftree}
\AxiomC{$p$}
\AxiomC{$\lnot p \vee W$}
\BinaryInfC{$W$}
\end{prooftree}
\end{minipage}
\begin{minipage}{0.4\linewidth}
\begin{prooftree}
\AxiomC{$\lnot p$}
\AxiomC{$p \vee V$}
\BinaryInfC{$V$}
\end{prooftree}
\end{minipage}
Note that this is equivalent to the rule of implication elimination:\\
\begin{minipage}{0.4\linewidth}
\begin{prooftree}
\AxiomC{$p$}
\AxiomC{$p \Rightarrow W$}
\BinaryInfC{$W$}
\end{prooftree}
\end{minipage}
\begin{minipage}{0.4\linewidth}
\begin{prooftree}
\AxiomC{$\lnot p$}
\AxiomC{$\lnot p \Rightarrow V$}
\BinaryInfC{$V$}
\end{prooftree}
\end{minipage}
\end{descriptionlist}
\end{description}
\begin{example}
Given the CNF formula:
\[ (p \vee q) \land (\lnot r \vee s) \land (\lnot q \vee r) \land (\lnot r \vee \lnot s) \land (\lnot p \vee r) \]
Resolution can be applied as follows:
\begin{itemize}
\item The starting clauses are:
$(p \vee q) \cdot (\lnot r \vee s) \cdot (\lnot q \vee r) \cdot (\lnot r \vee \lnot s) \cdot (\lnot p \vee r)$
\item From $(p \vee q)$ and $(\lnot q \vee r)$, one can derive $(p \vee r)$.
\item From $(\lnot p \vee r)$ and $(p \vee r)$, one can derive $r$.
\item From $(\lnot r \vee s)$ and $r$, one can derive $s$.
\item From $(\lnot r \vee \lnot s)$ and $s$, one can derive $\lnot r$.
\item From $r$ and $\lnot r$, one reaches $\bot$.
\end{itemize}
\end{example}
\section{DPLL}
\begin{description}
\item[DPLL] \marginnote{DPLL}
Algorithm based on unit resolution.
Given a set of clauses, DPLL is applied as follows:
\begin{enumerate}
\item Apply unit resolution as long as possible
(i.e. for every clause containing a single literal $l$, remove $\lnot l$ from all clauses.
Clauses containing $l$ can also be removed as they are redundant).
\item Choose a variable $p$.
\item Recursively apply DPLL by assuming $p$ in one call and $\lnot p$ in another.
\end{enumerate}
\begin{theorem}
DPLL is complete.
\end{theorem}
\begin{remark}
DPLL efficiency strongly depends on the variables.
\end{remark}
\end{description}
\begin{example}
Given a CNF formula with clauses:
\[ (\lnot p \vee \lnot s) \cdot (\lnot p \vee \lnot r) \cdot (\lnot q \vee \lnot t) \cdot (p \vee r) \cdot (p \vee s) \cdot (r \vee t) \cdot (\lnot s \vee t) \cdot (q \vee s) \]
There are no unit clauses, so we choose the variable $p$ and consider the cases where $p$ and where $\lnot p$.
By assuming $p$, unit resolution can be applied as follows:
\begin{itemize}
\item $p$ with $(\lnot p \vee \lnot s)$ derives $\lnot s$.
\item $p$ with $(\lnot p \vee \lnot r)$ derives $\lnot r$.
\item $\lnot s$ with $(q \vee s)$ derives $q$.
\item $\lnot r$ with $(r \vee t)$ derives $t$.
\item $q$ with $(\lnot q \vee \lnot t)$ derives $\not t$.
\item $t$ and $\lnot t$ bring to $\bot$. This branch is UNSAT.
\end{itemize}
By assuming $\lnot p$, unit resolution can be applied as follows:
\begin{itemize}
\item $\lnot p$ with $(p \vee r)$ derives $r$.
\item $\lnot p$ with $(p \vee s)$ derives $s$.
\item $s$ with $(\lnot s \vee t)$ derives $t$.
\item $t$ with $(\lnot q \vee \lnot t)$ derives $\lnot q$.
\end{itemize}
We reached a full assignment $\langle \lnot p, r, s, t, \lnot q \rangle$. The formula is SAT.
\end{example}
\begin{description}
\item[DPLL implementation] \marginnote{DPLL algorithm}
An efficient implementation of DPLL uses a list $M$ of the literals that have been decided or derived.
\begin{description}
\item[Notations] \phantom{}
\begin{itemize}[leftmargin=*]
\item A literal $l$ holds in $M$ ($M \models l$) iff $l$ is in $M$.
\item $M$ contradicts a clause $C$ ($M \models \lnot C$) iff for every literal $l$ in $C$, $M \models \lnot l$.
\item A literal $l$ is undefined iff $M \cancel{\models} l$ and $M \cancel{\models} \lnot l$.
\item A decision literal (i.e. literal assumed to hold) is denoted as $l^\mathcal{D}$.
\end{itemize}
\end{description}
The algorithm follows four rules:
\begin{descriptionlist}
\item[\texttt{UnitPropagate}]
If the CNF contains a clause $(A \vee l)$ such that $M \models \lnot A$ and $l$ is an undefined literal,
then there is a transition:
\[ M \rightarrow Ml \]
\item[\texttt{Decide}]
When \texttt{UnitPropagate} it not possible, an undefined literal $l$ is assumed to hold (it becomes a decision literal):
\[ M \rightarrow Ml^\mathcal{D} \]
\item[\texttt{Backtrack}]
When a branch is UNSAT, the algorithm backtracks to the nearest decision literal and negates it.
Given a clause $C$ of the CNF, if $Ml^\mathcal{D}N \models \lnot C$ and $N$ does not contain other decision literals,
then there is a transition:
\[ Ml^\mathcal{D}N \rightarrow M \lnot l \]
\item[\texttt{Fail}]
Given a clause $C$ of the CNF, if $M \models \lnot C$ and $M$ contains a definitive assignment of all the variables (i.e. no decision literals),
then there is a transition:
\[ M \rightarrow \text{fail} \]
\end{descriptionlist}
In the end, if the formula is SAT, $M$ contains an assignment of the variables.
\end{description}