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A Polynomial Time Algorithm for 3SAT

Robert Quigley

TL;DR

The paper proposes a polynomial-time algorithm for 3SAT based on two key techniques: clause Expansion (append any new term) and Implication (two clauses sharing a terminal of opposite polarity imply a shorter clause). By focusing on clauses of length at most 3 and systematically deriving implications, expansions, and reductions, the method aims to generate contradicting 1-terminal clauses if and only if the instance is unsatisfiable. The argument rests on a chain of Lemmas that connect long-implied clauses to shorter ones without enumerating clauses of length ≥4, ultimately reducing to a pair of contradicting 1-terminal clauses. If valid, this would imply $P = NP$, since 3SAT is NP-complete, causing all problems in NP to be solvable in polynomial time. However, the completeness and correctness of the construction hinge on intricate derivations and assumptions about derivability that are controversial in established complexity theory.

Abstract

It is shown that any two clauses in an instance of 3SAT sharing the same terminal which is positive in one clause and negated in the other can imply a new clause composed of the remaining terms from both clauses. Clauses can also imply other clauses as long as all the terms in the implying clauses exist in the implied clause. It is shown an instance of 3SAT is unsatisfiable if and only if it can derive contradicting 1-terminal clauses in exponential time. It is further shown that these contradicting clauses can be implied with the aforementioned techniques without processing clauses of length 4 or greater, reducing the computation to polynomial time. Therefore there is a polynomial time algorithm that will produce contradicting 1-terminal clauses if and only if the instance of 3SAT is unsatisfiable. Since such an algorithm exists and 3SAT is NP-Complete, P = NP.

A Polynomial Time Algorithm for 3SAT

TL;DR

The paper proposes a polynomial-time algorithm for 3SAT based on two key techniques: clause Expansion (append any new term) and Implication (two clauses sharing a terminal of opposite polarity imply a shorter clause). By focusing on clauses of length at most 3 and systematically deriving implications, expansions, and reductions, the method aims to generate contradicting 1-terminal clauses if and only if the instance is unsatisfiable. The argument rests on a chain of Lemmas that connect long-implied clauses to shorter ones without enumerating clauses of length ≥4, ultimately reducing to a pair of contradicting 1-terminal clauses. If valid, this would imply , since 3SAT is NP-complete, causing all problems in NP to be solvable in polynomial time. However, the completeness and correctness of the construction hinge on intricate derivations and assumptions about derivability that are controversial in established complexity theory.

Abstract

It is shown that any two clauses in an instance of 3SAT sharing the same terminal which is positive in one clause and negated in the other can imply a new clause composed of the remaining terms from both clauses. Clauses can also imply other clauses as long as all the terms in the implying clauses exist in the implied clause. It is shown an instance of 3SAT is unsatisfiable if and only if it can derive contradicting 1-terminal clauses in exponential time. It is further shown that these contradicting clauses can be implied with the aforementioned techniques without processing clauses of length 4 or greater, reducing the computation to polynomial time. Therefore there is a polynomial time algorithm that will produce contradicting 1-terminal clauses if and only if the instance of 3SAT is unsatisfiable. Since such an algorithm exists and 3SAT is NP-Complete, P = NP.
Paper Structure (9 sections, 19 theorems, 6 figures)

This paper contains 9 sections, 19 theorems, 6 figures.

Table of Contents

  1. Introduction
  2. Standard Definitions
  3. Algorithm-Specific Definitions
  4. Reformatting and Processing
  5. Lemmas
  6. Algorithm
  7. Time Complexity Analysis
  8. Proof of Correctness
  9. Conclusion

Key Result

Lemma 5.1

A clause can block an assignment.

Figures (6)

  • Figure 1: A graph illustrating the derivations described by the lemma
  • Figure 2: An illustration of the derivations described in the lemma
  • Figure 3: An illustration of the derivations by clauses $A$, $C_1$, and $E$
  • Figure 4: An illustration of the derivations described in the lemma
  • Figure 5: An illustration of the derivations described by the lemma
  • ...and 1 more figures

Theorems & Definitions (54)

  • Definition 2.1
  • Definition 2.2
  • Definition 2.3
  • Definition 2.4
  • Definition 2.5
  • Definition 2.6
  • Definition 2.7
  • Definition 2.8
  • Definition 3.1
  • Definition 3.2
  • ...and 44 more