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One time-travelling bit is as good as logarithmically many Ryan O’DonnellA. C. Cem Say Carnegie Mellon UniversityBoğaziçi University.

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Presentation on theme: "One time-travelling bit is as good as logarithmically many Ryan O’DonnellA. C. Cem Say Carnegie Mellon UniversityBoğaziçi University."— Presentation transcript:

1 One time-travelling bit is as good as logarithmically many Ryan O’DonnellA. C. Cem Say Carnegie Mellon UniversityBoğaziçi University

2 Time Travel Apparently it’s possible! Apparently it’s not inconsistent with the laws of General Relativity. This was first proven by Kurt Gödel in 1949.

3 Time Travel Apparently it’s not inconsistent with the laws of General Relativity. David Deutsch: If it exists, could we use it for computational speedups? You need to define the model carefully! [Deu91,Bac04,Aar05,AW09]

4 Closed Timelike Curve (CTC) x ∈ {0,1} n poly(n)-time algorithm w bits “wide” CxCx w-input (w+1)-output randomized circuit output bit w “CTC bits”

5 Closed Timelike Curve (CTC) x ∈ {0,1} n poly(n)-time algorithm w bits “wide” w-input (w+1)-output randomized circuit CxCx C x defines a 2 w -state Markov Chain output bit

6 The Deutschian Model The CTC (time-travelling) bits are automagically set to the stationary distribution of the Markov Chain defined by circuit C x.

7 Remark 1: We say “the” stationary distribution, though some annoying Markov Chains (“reducible” ones) don’t have a unique stationary distribution. Dealing with this annoyance is a major but boring technical component of paper. So let’s ignore it for the talk.

8 Remark 2: Deutsch et al. allowed quantum circuits and time-travelling qubits. Markov ChainQuantum Channel We only worry about classical, randomized computation.

9 “Time travel”-free summary Given access to a CTC of width w… On input x, algorithm writes a randomized circuit C x implementing transitions of 2 w -state Markov Chain. Then it freely gets one sample from the chain’s stationary distribution to help it decide if x ∈ L. CxCx input i ∈ [2 w ] output J ∈ [2 w ]

10 Prior Work Aaronson & Watrous (2009) showed that w = poly(n) CTC bits are super-powerful: “Everything collapses to PSPACE.” BPP + poly(n) CTC bits = PSPACE = BQP + poly(n) CTC qubits

11 Prior Work Poly(n) time-travelling bits: how unrealistic! Need to find a new wormhole whenever input length gets larger. [AW09] main open problem: Computational power of “narrow” CTCs? E.g.: What is BPP + 1 CTC bits? + 2 CTC bits? + 3 CTC bits?

12 Prior Work: an example fact [Bru03,Bac04,Aar05,AW09,SY12]: I.e.: You can solve SAT if you can set up a 2-state Markov Chain and get a sample from its stationary distribution. NP ⊆ BPP + 1 CTC bit

13 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… Else output state J = i. Stationary Distribution: 100% on 0.

14 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. Stationary Distribution: 100% on 0. NP ⊆ BPP + 1 CTC bit

15 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 One sample from stationary dist. decides SAT whp. NP ⊆ BPP + 1 CTC bit

16 coNP ⊆ BPP + 1 CTC bit Theorem [Say–Yakaryılmaz ’12]: 1 CTC bit ≡ Aaronson’s postselection. Hence: BQP + 1 CTC bit = “PostBQP” = PP [Aar05] & BPP + 1 CTC bit = “PostBPP” = BPP path

17 What is BPP path ?? 1. [HHT93]: L ∈ BPP path ⇔ ∃ NTM M such that x ∈ L ⇒ #acc M (x) ≥ 2#rej M (x) x ∉ L ⇒ #rej M (x) ≥ 2#acc M (x) 2. [AFFKK01]: Paper about stock market. (Unaware of [HHT93].) Defined postselection, wrote CBPP (instead of PostBPP) for class BPP with 2-bit outputs R, S: answer is R, conditioned on S = 1. [BGM03] noticed CBPP = BPP path (independently so did [Aar05]). 3. [YS13]: Equivalently, PostBPP = “RestartingBPP”. 4. [OS15]: BPP path = P || ApproxCounting = P || ApproxCounting[2] 5. [SU05]: Standard derand. assumption ⇒ BPP path = P || NP = P NP[log]

18 Prior Work Summary BPP + 1 CTC bit = BPP path = P || NP (probably) BPP + poly(n) CTC bits = PSPACE BPP + 2 CTC bits = ??? BPP + 3 CTC bits = ??? Maybe a hierarchy, like P NP[1] ⊊ P NP[2] ⊊ P NP[3] ⊊ …?

19 This Work Theorem [O–Say ’14]: BPP + w CTC bits = BPP + 1 CTC bit = BPP path for all constant w, even up to w = O(log n). A sample from the stationary distribution of a (circuit-defined) 2-state Markov chain is as useful as one from a poly(n)-state chain.

20 3-Slide Proof Suffices / ~equivalent to show: C input i ∈ [n] output J ∈ [n] Given randomized circuit C of poly(n) size implementing transitions of n-state Markov chain M, you can sample from M’s stationary distribution in BPP path. (Remark: When C is a black box, it’s a problem studied by Lovász–Winkler, Propp–Wilson, …)

21 3-Slide Proof What is M’s stationary distribution? Markov Chain Tree Theorem: (“The most rediscovered result in probability theory” —Aldous) Let G be underlying digraph for M. Then stationary probability on state r is… ∑ arborescences T r in G with root r ∏ arcs (i,j) in T r Pr M [i→j] ∑ rooted arborescences T in G ∏ arcs (i,j) in T Pr M [i→j]

22 3-Slide Proof ∑ arborescences T r in G with root r ∏ arcs (i,j) in T r Pr M [i→j] ∑ rooted arborescences T in G ∏ arcs (i,j) in T Pr M [i→j] If you stare at it for a while, it’s kind of obvious that you can sample from this distribution in PostBPP = BPP path.

23 An open problem What is the power of 1 CTC qubit? Or 2, 3, 4, … CTC qubits? Equivalently, what is the complexity of approximating the stationary density matrix of implicitly given w-qubit quantum channel?

24 An open problem What is the power of 1 CTC qubit? Or 2, 3, 4, … CTC qubits? Equivalently, what is the complexity of approximating the stationary density matrix of implicitly given w-qubit quantum channel? closed

25 An open problem closed Theorem [O–Say ’15]: BQP + w CTC qubits = BQP + 1 CTC classical bit = PostBQP = PP for all constant w, even up to w = O(log n).

26 Thanks! Any questions?


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