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

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.

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]

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”

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”

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

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.

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

“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]

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

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?

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 2-state Markov Chain and get a sample from its stationary distribution.

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

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

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.

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 + 1 CTC bit = “PostBPP” = BPPpath

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 = 1. [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]

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

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.

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, …)

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]

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. 

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?

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?

= 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).

Thanks! Any questions?