1. Ryan O’Donnell includes joint work with A. C. Cem Say for Manuel Blum, on his Magic77 th birthday

2. Apparently it’s possible!

3. This was first proven by Kurt Gödel in 1949. Apparently, it’s not inconsistent with the laws of General Relativity.

4. “Yes, but it’s logically impossible because of the academic great- grandfather paradox.” Hawking & Ellis:

5. 2003: learns complexity theory uses accumulated knowledge to time-travel 1986: learns complexity theory 1992: learns complexity theory severely distracts him from TCS

6. “Grandfather paradox” in a nutshell

7. time Closed Timelike Curve (CTC) space The state of the universe at the two yellow points is, by definition, the same.

8. Closed Timelike Curve (CTC) 8:5 9

9. Closed Timelike Curve (CTC) 9:0 0 1

10. Closed Timelike Curve (CTC) 9:0 1 You can do some stuff now, but whatever you put in at 5:00 will get sent back in time to 9:00.

11. Closed Timelike Curve (CTC) 9:0 2

12. Closed Timelike Curve (CTC) 9:3 6

13. Closed Timelike Curve (CTC) 3:2 5 Grandfather paradox ≡ you apply a NOT gate

14. Closed Timelike Curve (CTC) 4:5 9 0

15. 5:0 0 Closed Timelike Curve (CTC) Now what? ;)

16. Assume you apply a NOT gate. If the bit is 0 at time 9:00 then it is 1 at time 5:00. If the bit is 1 at time 9:00 then it is 0 at time 5:00. Contradiction, time travel is impossible, QED. Wrong! The laws of physics are stochastic. No reason for the bit’s state to be deterministic. It could be “50% chance of 0, 50% chance of 1.” No contradiction! David Deutsch

17. Assume you apply a NOT gate. If the bit is 0 at time 9:00 then it is 1 at time 5:00. If the bit is 1 at time 9:00 then it is 0 at time 5:00. Contradiction, time travel is impossible, QED. Wrong! The laws of physics are stochastic. No reason for the bit’s state to be deterministic. It could be “50% chance of 0, 50% chance of 1.” No contradiction! David Deutsch

18. Let’s say your microwave (CTC) can fit 1 bit. 8:5 9

19. Let’s say your microwave (CTC) can fit 1 bit. In general, if you open it and see a 0, you might put in a 1 with probability p and put in a 0 with probability 1−p. And if you open it and see a 1, you might put in a 0 with probability q and put in a 1 with probability 1−q.

20. 01 p q 1−p 1−q 8:5 9

21. 01 p q 1−p 1−q Is there a (probabilistic) state for the bit that is self-consistent? Yes, the stationary distribution!

22. The Deutschian model of w-bit CTCs (defined in [Deu91,Bac04,Aar05,AW09]) You can decide on any w-input, w-output randomized circuit C. This implicitly defines a 2 w -state Markov chain. When you move it into the CTC, Nature automagically sets the bits to the chain’s stationary distribution.

23. The Deutschian model of w-bit CTCs (defined in [Deu91,Bac04,Aar05,AW09]) Note 1: Actually, Deutsch allowed qubits. M.C. becomes a quantum channel. But let’s stick with bits for now. Note 2: If the chain isn’t irreducible, its stationary distribution is not unique. This proves to be annoying, but in a boring way. Allow me to ignore it.

24. As always, I ask: What computational power is conferred by access to a w-bit CTC? [Deu91,Bru03,Bac04,Aar05,SY12]: Given a 1-bit CTC, we can solve SAT in randomized polynomial time.

25. The proof is very cute. I won’t give it. I’ll just flash it.

26. NP ⊆ BPP + 1 CTC bit On input ϕ, a formula with n variables… Construct C ϕ implementing this 2-state chain: 01 11−2 –n 2 2 –n 2 On input state i ∈ {0,1}… With probability 2 – n 2, output state J = 0. Else pick x ∈ {0,1} n randomly. If x satisfies ϕ, output state J = 1. Else output state J = i.

27. ϕ is unsatisfiable: 01 2 –n 2 Stationary Distribution: 100% on 0. 0 ϕ is satisfiable: 01 2 –n 2 Stationary Distribution: 99.99% on 1. ≳ 2 – n Thus you solve SAT whp by looking at the bit! NP ⊆ BPP + 1 CTC bit

28. coNP ⊆ BPP + 1 CTC bit (because it’s clearly closed under complementation) What is the exact power of BPP + 1 CTC bit? [SY12]:It’s BPP path, a cool complexity class you’ve probably never heard of. I could tell you, but… Long story short, it’s probably P NP ||

29. More results [AW09]: P + poly(n) CTC bits = BQP + poly(n) CTC qubits = PSPACE [SY12]: BQP + 1 CTC bit = PP [OS14]: BPP + O(log n) CTC bits = BPP path [OS15]: BQP + O(log n) CTC qubits = PP

30. Summary 1. If you’re going to send a (“realistic”-length) message composed of qubits back in time, you can compress your message to 1 bit. 2. If you are boring and don’t like time travel, everything is equivalent to understanding the computational complexity of finding the stationary distribution of an implicitly given Markov Chain / quantum channel.

31. 8:5 9