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**One time-travelling bit is as good as logarithmically many**

Ryan O’Donnell A. C. Cem Say Carnegie Mellon University Boğaziçi University

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**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.

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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]

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**x ∈ {0,1}n w-input (w+1)-output randomized circuit**

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

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**Cx defines a 2w-state Markov Chain**

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

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

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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.

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

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**“Time travel”-free summary**

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

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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

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**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?

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**Prior Work: an example fact**

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

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**Stationary Distribution:**

NP ⊆ BPP + 1 CTC bit On input ϕ, a formula with n variables… Construct Cϕ implementing this 2-state chain: On input state i ∈ {0,1}… With probability 2–n2, output state J = 0. Else… Else output state J = i. Stationary Distribution: 100% on 0. 1 1−2–n2 2–n2

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**Stationary Distribution:**

NP ⊆ BPP + 1 CTC bit On input ϕ, a formula with n variables… Construct Cϕ implementing this 2-state chain: On input state i ∈ {0,1}… With probability 2–n2, 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. 1 1 1−2–n2 2–n2

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**One sample from stationary dist. decides SAT whp.**

NP ⊆ BPP + 1 CTC bit Stationary Distribution: 100% on 0. 1 ϕ is unsatisfiable: 2–n2 ≳ 2–n Stationary Distribution: 99.99% on 1. 1 ϕ is satisfiable: 2–n2 One sample from stationary dist. decides SAT whp.

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**1 CTC bit ≡ Aaronson’s postselection.**

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

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**What is BPPpath ?? 1. [HHT93]: L ∈ BPPpath ⇔ ∃ NTM M such that**

x∈L ⇒ #accM(x) ≥ 2#rejM(x) x∉L ⇒ #rejM(x) ≥ 2#accM(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 = [BGM03] noticed CBPP = BPPpath (independently so did [Aar05]). 3. [YS13]: Equivalently, PostBPP = “RestartingBPP”. 4. [OS15]: BPPpath = P||ApproxCounting = P||ApproxCounting[2] 5. [SU05]: Standard derand. assumption ⇒ BPPpath = P||NP = PNP[log]

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**Prior Work Summary BPP + 1 CTC bit = BPPpath = P||NP**

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

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**This Work Theorem [O–Say ’14]: BPP + w CTC bits = BPP + 1 CTC bit**

= BPPpath 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.

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**3-Slide Proof Suffices / ~equivalent to show: ₹ output J∈[n]**

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 BPPpath. (Remark: When C is a black box, it’s a problem studied by Lovász–Winkler, Propp–Wilson, …)

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**Markov Chain Tree Theorem:**

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 Tr in G with root r ∏ arcs (i,j) in Tr PrM[i→j] ∑ rooted arborescences T in G ∏ arcs (i,j) in T PrM[i→j]

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**3-Slide Proof ∑ ∏ PrM[i→j] ∑ ∏ PrM[i→j]**

arborescences Tr in G with root r ∏ arcs (i,j) in Tr PrM[i→j] ∑ rooted arborescences T in G ∏ arcs (i,j) in T PrM[i→j] If you stare at it for a while, it’s kind of obvious that you can sample from this distribution in PostBPP = BPPpath.

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**of implicitly given w-qubit quantum channel?**

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?

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**of implicitly given w-qubit quantum channel?**

An open problem closed 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?

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**= BQP + 1 CTC classical bit**

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).

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Thanks! Any questions?

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