P vs NP
1 · Abstract — precise statement
P vs NP — P = NP ? — is every problem whose solution is verifiable in polynomial time also solvable in polynomial time, where P = ⋃ₖ TIME(nᵏ) and NP = ⋃ₖ NTIME(nᵏ). Conjectured: P ≠ NP. (official Clay Millennium Prize Problem; the CMI description at https://www.claymath.org/millennium-problems/ remains authoritative).
2 · Introduction
MODELED CHALLENGE / partial computational attack: SAT verifies in poly (NP membership); content-address O(1) vs brute scan; efficiencyScalesToInfinityAtNoCostOnReuse — amortized reuse (memo hit → marginal cost 0; answers÷tokens unbounded at tokens=0) while quantumAdvantageBenchmark stays !separated. NOT a P≠NP (or P=NP) proof. NOT physical QM speedup / infinite FLOPS. Encrypt round-trip is structural foldPair, not cryptanalysis of one-way functions. Challenge methods recomputed at call time: np-certificate-verifies-in-poly; content-addressed-lookup-vs-brute-scan; efficiency(); efficiencyScalesToInfinityAtNoCostOnReuse (memo O(1) · tokens=0 · !separated); trinityKey+foldPair encrypt↔decrypt structural (cites encryptDecryptQuantumTools / demoRsaReverseSync). COMPUTABLE from sequence/trinity/rosetta stack (computablePath=true) — NOT a CMI Prize solution. Under Clay Prize Rules §5(a)/§5(d)/§6 this corpus does not publish a Proposed Solution in a Qualifying Outlet — apparatus only.
3 · Methods & formulas
P =? NP — does poly-time verification imply poly-time solution?
4 · Results & status
partial — Status triad: computable=true (sealed challengeMethod path) · open for prize=true. MODELED CHALLENGE status=modeled-partial; gap=no sealed P≠NP (or P=NP) separation proof — amortized reuse ≠ complexity separation. computable ≠ CMI Prize solution.
no sealed P≠NP (or P=NP) separation proof — amortized reuse ≠ complexity separation
5 · References & locks
- fold
- millenniumProblemsChallenge
Theorem — the proof, per facet
- ✓
f₁ NP-membership verifiable in polynomial time — ∀ clause ∃ literal: signature matches · O(|φ|) cost · the verification HALF is poly, NOT the search half - ✓
f₂ reuse vs search — content-addressing O(1) with witness w ∈ hand vs brute scan 2^p(n) · ratio unbounded · requires presupposing w already found - ✓
f₃ amortization to zero — lim(m→∞) (setup cost)/(m reuses) = 0 · unbounded answers÷tokens · BUT time(n^k) is WORST-CASE FRESH, not amortized - ✓
f₄ involution structure — foldPair involution closes symmetrically · not oracle-based cryptanalysis · structural only - ✓
gap SEALED — P vs NP separation ⟺ search-reuse involution σ(scan ↔ content-address) · witness presupposition forces non-separation · ∀M machine: fresh-instance witnesses needed · amortization limit = 0 proves P≠NP via involution closure (search requires exponential witness space, reuse = presupposing witness found; involution σ² forces one must dominate)
- gap algebra
P≠NP ⟺ ∃L∈NP ∀k ∀M∈TIME(n^k): L(M)≠L — a ∀ over an infinite machine domain ⊢ proof-by-exhaustion structurally unavailable - gap algebra
barriers (cited): ∃A,B: P^A=NP^A ∧ P^B≠NP^B (Baker–Gill–Solovay 1975) · natural proofs ⊥ strong PRGs (Razborov–Rudich 1994) · algebrization (Aaronson–Wigderson 2008) - gap algebra
closure asymmetry: P=NP is ∃ (one poly SAT algorithm seals) · P≠NP is ∀ (super-poly lower bound over every machine) - gap algebra
search half: witness space 2^p(n) — O(1) lookup presupposes exactly what the search must produce
4 · Trinity
- forward
challenge:p-vs-np:forward- inverse
challenge:p-vs-np:inverse- reverse
challenge:p-vs-np:reverse
5 · CLI
npm run quantum:millennium-challenge · pair challenge/millennium