### P versus NP

Frank Vega
2020 Zenodo
$P$ versus $NP$ is considered as one of the most important open problems in computer science. This consists in knowing the answer of the following question: Is $P$ equal to $NP$? It was essentially mentioned in 1955 from a letter written by John Nash to the United States National Security Agency. However, a precise statement of the $P$ versus $NP$ problem was introduced independently by Stephen Cook and Leonid Levin. Since that date, all efforts to find a proof for this problem have failed. To
more » ... ttack the $P$ versus $NP$ question the concept of $\textit{NP-completeness}$ has been very useful. A well-known $\textit{NP-complete}$ problem is $\textit{MAX-3SAT}$. Given a natural number $K$ and a Boolean formula $\phi$ in $3CNF$, $\textit{MAX-3SAT}$ consists in deciding whether there is a truth assignment in $\phi$ with at most $K$ clauses unsatisfied. In this paper, we consider the problem of computing the sum of the weighted densities of states of a Boolean formula in $3CNF$. Given a Boolean formula $\phi$, the density of states $n(E)$ counts the number of truth assignments that leave exactly $E$ clauses unsatisfied in $\phi$. The weighted density of states $m(E)$ is equal to $E \times n(E)$. The sum of the weighted densities of states of a Boolean formula in $3CNF$ with $m$ clauses is equal to $\sum_{E = 0}^{m} m(E)$. We prove that we can calculate the sum of the weighted densities of states in polynomial time. The lowest value of $E$ with a non-zero density (i.e. $min_{E}\{E|n(E) > 0\}$) is the solution of the corresponding $\textit{MAX-3SAT}$ problem. Diophantine equations of the form $\sum_{i = 0}^{r} a_{i} \times x_{i} = c$ are solvable in polynomial time for arbitrary values of $r$. We can apply this Diophantine equation such that $c$ is the value of the sum of the weighted densities of states from a Boolean formula $\phi$ in $3CNF$, $a_{i} \times x_{i}$ is the value of each term $E \times n(E)$, where $n(E)$ corresponds to the unknown value of $x_{i}$. In this way, we are able to find the lowest value of \$E [...]