Lecture 024 - Time Travel

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.

Markov Chain

The following transition matrix

\begin{bmatrix} 1-p & q\\ p & 1-q\\ \end{bmatrix}

has steady state

b = \begin{cases} 0 & \text{w.p. } \frac{q}{p+q}\\ 1 & \text{w.p. } \frac{p}{p+q}\\ \end{cases}

Quantum Time Travel

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


def SAT(b_0 qubit, C circuit):
  Flip n^2 coins
  if all head:
    b_1 = 0
    pick y in {0, 1}^n at random
    if C(y):
      b_1 = 1
      b_1 = b_0

When it is unsatisfiable, we have transition matrix

\begin{bmatrix} 1 & 2^{-n^2}\\ 0 & 1-2^{-n^2}\\ \end{bmatrix}

Therefore the steady state is:

\begin{cases} 0 & \text{w.p. 100\%}\\ 1 & \text{w.p. 0\%}\\ \end{cases}

When it is satisfiable, we have transition probability from 0 to 1:

\begin{align*} p \geq& Pr\{n^2 \text{ heads}\} Pr\{y \text{ is satisfying for } C\}\\ p =& 2^{-n} - 2^{-n^2}\\ \frac{p}{p+q} \geq& \frac{2^{-n} - 2^{-n^2}}{2^{-n}} \simeq 1 - 2^{-n^2}\\ \end{align*}

Therefore the steady state is:

\begin{cases} 1 & \text{w.p. } \simeq 1-2^{-n^2}\\ 0 & \text{w.p. } \simeq 2^{-n^2}\\ \end{cases}

So we solved SAT with one-sided error.

Other Complexity from Time Travel

With CTC:

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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