Commit bcd1e724 by Martin Jonas

### More polishing

parent a6c04467
 ... ... @@ -312,13 +312,13 @@ combinations of the represented Boolean functions using the recursive procedure \texttt{Apply}~\cite{Bry86}. The main idea of symbolic \sat solvers is to convert a \cnf formula to the corresponding \robdd and check the root of the resulting \bdd. If the resulting \bdd is equivalent to the \bdd for the constant $0$ function, the formula is unsatisfiable, and it is satisfiable otherwise. However, in order to keep the size of the \bdd small, it is necessary to existentially quantify the variables during the computation. This technique is known as \emph{early quantification}~\cite{HKB96}. A simplified symbolic \sat algorithm can be found for example in the survey of \sat solving by Darwiche and Pipatsriswat~\cite{DP09}. equivalent to the \bdd for the function that is $0$ everywhere, the formula is unsatisfiable, and it is satisfiable otherwise. However, in order to keep the size of the \bdd small, it is necessary to existentially quantify the variables during the computation. This technique is known as \emph{early quantification}~\cite{HKB96}. A simplified symbolic \sat algorithm can be found for example in the survey of \sat solving by Darwiche and Pipatsriswat~\cite{DP09}. Look-ahead based algorithm, in contrast to the \cdcl, are employing expensive heuristics to guide the \dpll search to a satisfying ... ...
 ... ... @@ -32,8 +32,8 @@ that at least one of the last 4 bits of the variable $a$ has to be $1$, i.e. the value of $a$ is not divisible by $16$. After conjoining this \bdd to the \bdd for $a=16 \cdot b \,+\, 16 \cdot c$, the formula is decided unsatisfiable, as the resulting \bdd represents the constant $0$ function. On the other hand, if one considers only quantifier instantiations by the subterm of the input formula, function that is $0$ everywhere. On the other hand, if one considers only quantifier instantiations by the subterm of the input formula, exponentially many quantifier instances have to be added to the formula to show its unsatisfiability, as there is no subterm of $\varphi$ that can be instantiated as $x$ to yield an unsatisfiable ... ...
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