Pt1420 Unit 1 Math Paper

We work with a Boolean Language whose basic symbols are
$\vee,\wedge,\neg,\Rightarrow$ endowed with a finite set of atoms $\mathcal{A}$. The set of literals, $\mathcal{L}$, is the set
$\mathcal{A}\cup\{\neg p|p\in\mathcal{A}\}$. A pair formed by a literal together with its negation is called a conjugated pair.

Here we give the (classical) definition of valuation over a Boolean formula. That is the definition we will use henceforth.

\begin{definition}\label{BOOLFOR:def}
Given a finite set of {\em atoms}, $\mathcal{A}$, $\psi$ belongs to the set of Boolean formulas $\mathtt{F}$ if
\begin{enumerate}
\item $\psi\in\mathcal{A}$;
\item $\psi=\chi\wedge\phi$ if both $\chi$ and $\phi$ belong to $\mathtt{F}$;
\item $\psi=\chi\vee\phi$ if both $\chi$

\end{enumerate}
\end{definition}

\begin{definition}\label{VALUA:def}
A valuation over a set of atoms $P$ is a mapping $v$ from $P$ to the set $\{True, False\}$ and, as usual, we extend the definition of $v$ to $v'$ as,
\begin{enumerate}
\item If $\psi\in\mathcal{A}$, $v'(\psi)=v(\psi)$;
\item If $\psi=\chi\wedge\phi$, $v'(\psi)=True$ if

The obvious, high consuming, expsize long search is a direct attempt to test all the $\mathbf{m}^{\mathbf{n}}$ trials.

We refer to Example \ref{CARD:ex} for a card
$\mathbf{3}\times\mathbf{2}$, a satisfiable one. Analogously to deciding Boolean formulas (satisfiable, unsatisfiable), we can, of course, solve cards and decide them all, but the game, likewise, is how fast can we solve a card.

A card has $\mathbf{m}\times{\mathbf{n}}$ entries The number of disjunction in each entry is bounded by the maximum number of distinct disjunctions of at most two literals one can write given a finite number of atoms. In Section \ref{LASTIDEA:sec}, we discuss one effective, polynomially bounded, strategy. In Section \ref{BC:sec}, we discuss the complexity of the solution we gave.
Here, we built the tools in order solve polynomially, in time and space, a card.
The starting point is to define the cylindrical digraph associated with a given card. The vertices of a cylindrical digraph are associated with the disjunction of each entry. If $cl\equiv p\vee q$ is a disjunction of some entry, then both $\neg p\Rightarrow q$ and $\neg q\Rightarrow