Open
Unresolved assertion in the cited scope ↗
Observed Oct 10, 2026
Conjecture · version 2
3SAT cannot be decided in deterministic time 2^{o(n)} times a polynomial in the input length, where n is the number of variables.
A stronger quantitative hypothesis than P ≠ NP.
Open
Observed Oct 10, 2026
In this problem · publication v12 · Selected context
Different statement · related variant · Open
This statement contributes context in its stated scope; it does not settle the canonical P versus NP question.
ETH rules out even subexponential 3SAT algorithms in the stated model.
P ≠ NP does not by itself establish exponential runtime lower bounds.
Open · version 2
3SAT cannot be decided in deterministic time 2^{o(n)} times a polynomial in the input length, where n is the number of variables.
Scope: A stronger quantitative hypothesis than P ≠ NP.
Open · version 3
There exists a language in NP that is not decidable by any deterministic polynomial-time algorithm.
Scope: The conjectural negative answer to the same canonical question.
Relationship statement · Proved · version 2
If ETH holds, then P ≠ NP.
Scope: A sufficient stronger hardness hypothesis; P ≠ NP does not establish ETH.
Relationship not independently reviewed.
Connection in P versus NP →Relationship not independently reviewed. This concerns its use in this problem.
In this problem · publication v14
Different statement · related variant · Open
This statement contributes context in its stated scope; it does not settle the canonical P versus NP question.
ETH rules out even subexponential 3SAT algorithms in the stated model.
P ≠ NP does not by itself establish exponential runtime lower bounds.
Open · version 2
3SAT cannot be decided in deterministic time 2^{o(n)} times a polynomial in the input length, where n is the number of variables.
Scope: A stronger quantitative hypothesis than P ≠ NP.
Open · version 3
There exists a language in NP that is not decidable by any deterministic polynomial-time algorithm.
Scope: The conjectural negative answer to the same canonical question.
Relationship statement · Proved · version 2
If ETH holds, then P ≠ NP.
Scope: A sufficient stronger hardness hypothesis; P ≠ NP does not establish ETH.
Relationship not independently reviewed.
Connection in P versus NP →Relationship not independently reviewed. This concerns its use in this problem.
3SAT cannot be decided in deterministic time 2^{o(n)} times a polynomial in the input length, where n is the number of variables.
paraphrase · Only metadata and original OpenMath paraphrases are published; proof PDFs remain at their source.
If ETH holds, then P ≠ NP.
paraphrase · Only metadata and original OpenMath paraphrases are published; proof PDFs remain at their source.
Russell Impagliazzo · Ramamohan Paturi · Published · Publication date Mar 2001
Published author exposition or original paper.
Source revision: JCSS-2001. Observed Oct 10, 2026.
AI involvement is not reported by this cited source; this is not a claim of its absence.
Impagliazzo–Paturi · On the complexity of k-SAT — proof text: JCSS-2001. External source text. Its mathematical claims and independent OpenMath review are assessed separately.
There exists a language in NP that is not decidable by any deterministic polynomial-time algorithm.
paraphrase · Only metadata and original OpenMath paraphrases are published; proof PDFs remain at their source.
Stephen Cook · Published
Published author exposition or original paper.
Source revision: official-description-sha256-018f3d473d16. Observed Oct 10, 2026.
AI involvement is not reported by this cited source; this is not a claim of its absence.
Statement versions and problem publication versions are separate. A new publication can update context or sources while retaining the same statement.
Assessments retained in these publication records. Earlier statement versions retain their own source histories.
Open
Observed Oct 10, 2026 · Source passage →