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# P versus NP

Version 1
: Received: 1 August 2019 / Approved: 5 August 2019 / Online: 5 August 2019 (03:31:23 CEST)

Version 2 : Received: 26 November 2019 / Approved: 29 November 2019 / Online: 29 November 2019 (07:28:14 CET)

Version 3 : Received: 4 April 2020 / Approved: 6 April 2020 / Online: 6 April 2020 (12:57:55 CEST)

Version 4 : Received: 15 April 2020 / Approved: 16 April 2020 / Online: 16 April 2020 (10:17:03 CEST)

Version 5 : Received: 18 September 2020 / Approved: 19 September 2020 / Online: 19 September 2020 (09:51:02 CEST)

Version 2 : Received: 26 November 2019 / Approved: 29 November 2019 / Online: 29 November 2019 (07:28:14 CET)

Version 3 : Received: 4 April 2020 / Approved: 6 April 2020 / Online: 6 April 2020 (12:57:55 CEST)

Version 4 : Received: 15 April 2020 / Approved: 16 April 2020 / Online: 16 April 2020 (10:17:03 CEST)

Version 5 : Received: 18 September 2020 / Approved: 19 September 2020 / Online: 19 September 2020 (09:51:02 CEST)

How to cite:
Vega, F. P versus NP. *Preprints* **2019**, 2019080037 (doi: 10.20944/preprints201908.0037.v5).
Vega, F. P versus NP. Preprints 2019, 2019080037 (doi: 10.20944/preprints201908.0037.v5).

## Abstract

$P$ 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 $NP$? The 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. Another major complexity class is $\textit{P-Sel}$. $\textit{P-Sel}$ is the class of decision problems for which there is a polynomial time algorithm (called a selector) with the following property: Whenever it's given two instances, a $``yes"$ and a $``no"$ instance, the algorithm can always decide which is the $``yes"$ instance. It is known that if $NP$ is contained in $\textit{P-Sel}$, then $P = NP$. We consider the problem of computing the sum of the weighted densities of states of a Boolean formula in $3CNF$. Given a Boolean formula $\phi$ with $m$ clauses, the density of states $n(E)$ for some integer $0 \leq E \leq m$ 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. Given two Boolean formulas $\phi_{1}$ and $\phi_{2}$ in $3CNF$ with $n$ variables and $m$ clauses, the combinatorial optimization problem $\textit{SELECTOR-3SAT}$ consists in selecting the formula which is satisfiable, where every clause from $\phi_{1}$ and $\phi_{2}$ can be unsatisfied for some truth assignment. We assume that the formula that is satisfiable has the minimum sum of the weighted densities of states. In this way, we solve $\textit{SELECTOR-3SAT}$ with an exact polynomial time algorithm. We claim this problem is a selector of $3SAT$ and thus, $P = NP$.

## Subject Areas

complexity classes; combinatorial optimization; polynomial time; reduction; logarithmic space; one-way

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Commenter: Frank Vega

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