Closed Timelike Curves (CTC): you could have two points in spacetime (t_0, x_0, y_0, z_0), (t_1, x_1, y_1, z_1) such that the state of particles is the same.
The following transition matrix
has steady state
Theorem: with 1 time-traveling bit, we can efficiently solve CIRCUIT-SAT (and therefore any NP-complete problems and coNP-complete probems)
NP, coNP \subseteq BPP + 1 C.T.C. bit
Algorithm:
def SAT(b_0 qubit, C circuit):
Flip n^2 coins
if all head:
b_1 = 0
else:
pick y in {0, 1}^n at random
if C(y):
b_1 = 1
else:
b_1 = b_0
When it is unsatisfiable, we have transition matrix
Therefore the steady state is:
When it is satisfiable, we have transition probability from 0 to 1:
Therefore the steady state is:
So we solved SAT with one-sided error.
With CTC:
BPP + k CTC = BPP-path = BPP + CTC for any constant k (O'D-Say '14)
BQP + 1 CTC = PP (Aaronson '08)
PP != BPP-path unless polynomial hierarchy (PH) collapses
P or BPP or BQP or PSPACE + poly(n) CTC = PSPACE
Above result assume you can send bits back in poly(n) time. If you assume you can send bits back in O(1) time, then all decidable language can be decided in O(1) expected time.
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