A deterministic algorithm solves 3-SAT in polynomial time, suggesting P equals NP.
We present a deterministic polynomial-time algorithm that solves any instance of 3-SAT, implying P = NP. The algorithm traverses the implicit binary decision tree Tₙ of depth n induced by the Boolean variable space \0,1\ⁿ. At each of the n levels, instead of searching the subtree exhaustively, it fixes the current variable and verifies whether the resulting partial assignment is consistent with the formula — a check that costs $O(m)$ per level, where m is the number of clauses. The total complexity is O(n · m), which is polynomial in the input size. The key insight is that NP is defined by the existence of a polynomial-time verifier; we show that this verifier, applied to partial assignments at each level of Tₙ, is sufficient to navigate the tree to a solution without exhaustive search.
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Kaoru Aguilera Katayama (2026) studied this question.
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