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by doormatt
310 days ago
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Just to clarify, how exactly does proving exponential lower bounds for Algphys translate into a proof that no polynomial time algorithm exists for NP-complete problems in the standard Turing model? Isn’t it possible that your "hard" instances could be solvable in polynomial time by some algorithm that doesn’t rely on geometric modeling or Hamiltonian dynamics? How do you justify that every polynomial-time Turing machine algorithm can be modeled as a trajectory in your Hamiltonian system? |
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1. Algphys is shown to be equivalent to P, meaning any polynomial-time Turing algorithm can be modeled in Algphys. The paper constructs "frustrated" 3-SAT instances requiring exponential time in Algphys due to high combinatorial complexity and spectral properties (e.g., Hessian eigenvalues growing as ~ 2^n). Since Algphys = P, this implies no polynomial-time Turing algorithm can solve NP-complete problems.
2. The equivalence of Algphys and P means any polynomial-time algorithm, regardless of approach, can be modeled in Algphys. The exponential lower bound for these instances in Algphys applies to all polynomial-time Turing algorithms, suggesting these "hard" instances are inherently exponential, no matter the method.
3. The paper establishes P ~ Algphys by mapping Turing machine states to points on a symplectic manifold, with the cost function H encoding computation steps. The Hamiltonian dynamics (γ̇(t) = J∇H(γ(t))) simulate the algorithm’s execution path, ensuring every polynomial-time algorithm corresponds to a trajectory in Algphys.