Does BPP = P?

Question: does \(\mathrm{BPP} = \mathrm{P}\)? That is, can every decision problem solvable by a randomised polynomial-time algorithm with bounded two-sided error also be solved by a deterministic polynomial-time algorithm?

Randomised polynomial time sits between \(\mathrm{P}\) and the polynomial hierarchy (Sipser–Lautemann: \(\mathrm{BPP} \subseteq \Sigma_2^p\)), and every practical use of randomness in algorithms raises the question whether it is essential. Impagliazzo–Wigderson (1997) showed a sufficiently strong circuit lower bound for \(\mathrm{E}\) would derandomise everything (\(\mathrm{BPP} = \mathrm{P}\)), and Kabanets–Impagliazzo proved derandomisation is equivalent to lower bounds — so the question is stuck exactly where complexity theory is stuck. Unconditional derandomisation is known only in restricted models.