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A Theory of Isolatability Scott Aaronson Andrew Drucker MIT

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Freeze-Dried Computation Motivating Question: How much useful computational work can one store in (say) an n-qubit quantum state, or a coin whose bias is an arbitrary real number? Potentially a huge amount! We give a new toolcalled isolatabilityfor ruling out the possibility of such extravagant encodings.

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Idea: Take some advice resource (such as a coin or a quantum state), and simulate it using a short classical string, together with a polynomial number of untrusted advice resources. In other words, all the relevant information in the advice resource gets packed into an ordinary string (which we say isolates the resource), and we're left with just a computational search problemof finding coins, quantum states, etc. that are consistent with the string.

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Part I: The Majority-Certificates Lemma Our basic tool Part II: Application to Quantum Advice BQP/qpoly QMA/poly Part III: Application to Advice Coins PSPACE/coin PSPACE/poly

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Part I: The Majority- Certificates Lemma

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that computes some Boolean function f:{0,1} n {0,1} belonging to a small set S (meaning, of size 2 poly(n) ). Someone wants to prove to us that f equals (say) the all-0 function, by having us check a polynomial number of outputs f(x 1 ),…,f(x m ). Intuition: Were given a black box f xf(x) This is trivially impossible! f0f0 f1f1 f2f2 f3f3 f4f4 f5f5 x1x1 010000 x2x2 001000 x3x3 000100 x4x4 000010 x5x5 000001 But what if we get 3 black boxes, and are allowed to simulate f=f 0 by taking the point-wise MAJORITY of their outputs?

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Majority-Certificates Lemma Lemma: Let S be a set of Boolean functions f:{0,1} n {0,1}, and let f * S. Then there exist m=O(n) certificates C 1,…,C m, each of size O(log|S|), such that (i)Some f i S is consistent with each C i, and (ii)If f 1 S is consistent with C 1, f 2 S is consistent with C 2, and so on, then MAJ(f 1,…,f m )=f *, where MAJ denotes pointwise majority. Definitions: A certificate is a partial Boolean function C:{0,1} n {0,1,*}. A Boolean function f:{0,1} n {0,1} is consistent with C, if f(x)=C(x) whenever C(x) {0,1}. The size of C is the number of inputs x such that C(x) {0,1}.

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Proof Idea By symmetry, we can assume f * is the all-0 function. Consider a two-player, zero-sum matrix game: Alice picks a certificate C of size k consistent with some f S Bob picks an input x {0,1} n Alice wins this game if f(x)=0 for all f S consistent with C. Crucial Claim: Alice has a mixed strategy that lets her win >90% of the time. The lemma follows from this claim! Just choose certificates C 1,…,C m independently from Alices winning distribution. Then by a Chernoff bound, almost certainly MAJ(f 1 (x),…,f m (x))=0 for all f 1,…,f m consistent with C 1,…,C m respectively and all inputs x {0,1} n. So clearly there exist C 1,…,C m with this property.

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Proof of Claim Use the Minimax Theorem! Given a distribution D over x, its enough to create a fixed certificate C such that Stage I: Choose x 1,…,x t independently from D, for some t=O(log|S|). Then with high probability, requiring f(x 1 )=…=f(x t )=0 kills off every f S such that Stage II: Repeatedly add a constraint f(x i )=b i that kills at least half the remaining functions. After log 2 |S| iterations, well have winnowed S down to just a single function f S.

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Part II: Application to Quantum Advice

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BQP/qpoly is the class of problems solvable in quantum polynomial time, with the help of polynomial-size quantum advice states Formally: a language L is in BQP/qpoly if there exists a polynomial time quantum algorithm A, as well as quantum advice states {| n } n on poly(n) qubits, such that for every input x of size n, A(x,| n ) decides whether or not x L with error probability at most 1/3 YQP (Yoda Quantum Polynomial-Time) is the same, except we also require that for every alleged advice state, A(x, ) outputs either the right answer or FAIL with probability at least 2/3 BQP YQP QMA BQP/qpoly

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Watrous 2000: For any fixed, finite black-box group G n and subgroup H n G n, deciding membership in H n is in BQP/qpoly The quantum advice state is just an equal superposition |H n over the elements of H n We dont know how to solve the same problem in BQP/poly A. 2004: BQP/qpoly PP/poly = PostBQP/poly Quantum advice can be simulated by classical advice, combined with postselection on unlikely measurement outcomes A. 2006: HeurBQP/qpoly = HeurYQP/poly Trusted quantum advice can be simulated on most inputs by trusted classical advice combined with untrusted quantum advice A.-Kuperberg 2007: There exists a quantum oracle separating BQP/qpoly from BQP/poly Q UANTUM ADVICE IS POWERFUL N O I T I SN T

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New Result: BQP/qpoly = YQP/poly Trusted quantum advice is equivalent in power to trusted classical advice combined with untrusted quantum advice. (Quantum states never need to be trusted) Let ρ be any quantum state on n qubits. Then for all m,ε, there exists a 2-local Hamiltonian H=H 1 + +H L on poly(n,m,1/ε) qubits, such that any ground state |φ of H can be used to simulate ρ (with error ε) on all quantum circuits of size at most m. In other words, there exists an efficient mapping CC such that for all circuits C of size m, P HYSICS I MPLICATION

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What Does It Mean? Preparing quantum advice states is no harder than preparing ground states of local Hamiltonians This explains a once-mysterious relationship between quantum proofs and quantum advice: efficient preparability of ground states would imply both QMA=QCMA and BQP/qpoly=BQP/poly Quantum Karp-Lipton Theorem: NP-complete problems are not efficiently solvable using quantum advice, unless some uniform complexity classes collapse unexpectedly QCMA/qpoly QMA/poly: classical proofs and quantum advice can be simulated with quantum proofs and classical advice

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BQP YQPQCMABQP/poly BQP/qpoly =YQP/poly QCMA/polyQMA QCMA/qpoly QMA/polyPP PP/polyQMA/qpoly PSPACE/poly A.06 This work

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Majority- Certificates Lemma Real Majority- Certificates Lemma Circuit Learning (Bshouty et al.) Minimax Theorem Safe Winnowing Lemma Holevos Theorem Random Access Code Lower Bound (Ambainis et al.) BQP/qpoly=YQP/poly HeurBQP/qpoly=HeurYQP/poly (A.06) Quantum advice no harder than ground state preparation Fat-Shattering Bound (A.06) Covering Lemma (Alon et al.) Learning of p- Concept Classes (Bartlett & Long) L OCAL H AMILTONIANS is QMA-complete (Kitaev) Cook-Levin Theorem QMA=QMA+ (Aharonov & Regev) Used as lemma Generalizes

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Lifting the Majority-Certificates Lemma Boolean Majority-CertificatesBQP/qpoly=YQP/poly Proof Set S of Boolean functionsSet S of p(n)-qubit mixed states True function f * STrue advice state | n Other functions f 1,…,f m Other states 1,…, m Certificate C i to isolate f i Measurement E i to isolate I New DifficultySolution The class of p(n)-qubit quantum states is infinitely large! And even if we discretize it, its still doubly-exponentially large Result of A.06 on learnability of quantum states (building on Ambainis et al. 1999) Instead of Boolean functions f:{0,1} n {0,1}, now we have real functions f :{0,1} n [0,1] representing the expectation values Learning theory has tools to deal with this: fat-shattering dimension, -covers… (Alon et al. 1997) How do we verify a quantum witness without destroying it? QMA=QMA+ (Aharonov & Regev 2003) What if a certificate asks us to verify Tr(E )a, but Tr(E ) is right at the knife-edge? Safe Winnowing Lemma

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Majority-Certificates Lemma, Real Case Lemma: Let S be a set of functions f:{0,1}[0,1], let f S, and let ε>0. Then we can find m=O(n/ε²) functions f 1,…,f m S, sets X 1,…,X m {0,1} each of size and for which the following holds. All functions g 1,…,g m S that satisfy for all i [m] also satisfy where

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Theorem: BQP/qpoly = YQP/poly. Proof Sketch: YQP/poly BQP/qpoly is immediate. For the other direction, let L BQP/qpoly. Let M be a quantum algorithm that decides L using advice state | n. Define Let S = {f : }. Then S has fat-shattering dimension at most poly(n), by A.06. So we can apply the real version of the Majority-Certificates Lemma to S. This yields certificates C 1,…,C m (for some m=poly(n)), such that any states 1,…, m consistent with C 1,…,C m respectively satisfy for all x {0,1} n (regardless of entanglement). To check the C i s, we use the QMA+ super-verifier of Aharonov & Regev.

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Quantum Karp-Lipton Theorem Our quantum analogue: If NP BQP/qpoly, then coNP NP QMA PromiseQMA. Karp-Lipton 1982: If NP P/poly, then coNP NP = NP NP. Proof Idea: A coNP NP statement has the form x y R(x,y). By the hypothesis and BQP/qpoly = YQP/poly, there exists an advice string s, such that any quantum state consistent with s lets us solve NP problems (and some such is consistent). In QMA PromiseQMA, first guess an s thats consistent with some state. Then use the oracle to search for an x and such that, if is consistent with s, then R(x,Q(x, )) holds, where Q is a quantum algorithm that searches for a y such that R(x,y).

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Part III: Application to Advice Coins

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Erik Demaine (motivated by a computational genetics problem): Suppose a PSPACE machine can flip a coin with Bernoulli probability p an unlimited number of times. Can it extract an exponential amount of information (or even more) about p? Me: Im sure whatever the answer is, its obvious... Didnt seem too likely there could be superpowerful Advice Coins

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Indeed, Hellman & Cover proved the following in 1970... Suppose a finite automaton M is trying to decide whether a coin has p=½ or p=½+. Then even if it can flip the coin an unlimited number of times, M needs (1/ ) states to succeed with probability (say) 2/3. This result seems to imply PSPACE/coin PSPACE/poly, since we could take the first poly(n) bits of p as the advice. But it breaks down if p is close to 0 or 1!

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Furthermore, quantum mechanics nullifies the Hellman-Cover Theorem! Theorem: For any >0, its possible to distinguish a coin with p=½ from a coin with p=½+ using a single qubit of memory, with error probability independent of. Keep flipping the coin. Whenever the coin lands heads, rotate /100 radians counterclockwise. Whenever it lands tails, rotate /100 radians clockwise. Halt with probability ~ 2 /100 at each time step Expected difference in final angle after halting, in p=½ vs. p=½+ cases: 1 radian Standard deviation in angle:

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Theorem: Despite these obstacles, BQPSPACE/coin = PSPACE/coin = PSPACE/poly. Were interested in a fixed-point of p : a mixed state p such that Proof: Suffices to show BQPSPACE/coin PSPACE/poly. Let 0 = quantum operation applied to our memory qubits whenever coin lands heads, 1 = operation applied when it lands tails Then induced operation at each time step:

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Fixed-Points of Superoperators Already studied by [A.-Watrous 2008], in the context of quantum computing with closed timelike curves 0.000000000000000110101111101 Our result there: BQP CTC = P CTC = PSPACE Quantum computers with CTCs have exactly the same power as classical computers with CTCs, namely PSPACE (or: CTCs make time and space equivalent as computational resources)

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Key Lemma: Let a(p) be the probability that an S-state quantum finite automaton accepts, if each input bit is 1 with independent probability p. Then a(p) is a degree-S 2 rational function of p. Proof Idea: We can write a(p) as Furthermore, each entry of p can be written as a degree-S 2 rational function of p, by using Cramers Rule on S 2 S 2 matrices.

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Now, by calculus, a degree-S 2 rational function can cross the line y=½ at most 2S 2 times… p a x (p) Hence a(p) is itself a degree-S 2 rational function of p, except possibly at a finite number of singularities: A further continuity argument rules out the singularities, except possibly at p=0 and p=1.

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p a x (p) Given a BQPSPACE/coin machine M, let a x (p) be its acceptance probability on input x {0,1} n and a coin with Bernoulli probability p. Challenge: How can we describe p well enough to compute a x (p) for every x, using only poly(n) bits?

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Finishing the Argument (Sketch) Weve upper-bounded how many distinct Boolean functions f:{0,1} n {0,1} can be expressed, as we vary p from 0 to 1. So by the Majority-Certificates Lemma, we can simulate PSPACE/coin using a PSPACE/poly machine, combined with poly(n) untrusted advice coins. We then get a computational search problem: finding coin biases p 1,…,p m that are consistent with the /poly advice string. We can solve that problem in PSPACE, using NC algorithms for root-finding developed by Neff and Pan

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When Can An O(1)-State Finite Automaton Detect an Change to the Bias of a Coin?

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Open Problems Quantum finite automata: are their limiting acceptance probabilities continuous functions of p, for p (0,1)? Find other applications of isolatability Circuit complexity? Communication complexity? Learning theory? Quantum information? Optimality of the Majority-Certificates Lemma? Prove a classical oracle separation between BQP/poly and BQP/qpoly=YQP/poly

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Promised Application to Physics Furthermore, in their reduction, the witness is a history state So given any language L BQP/qpoly=YQP/poly, we can use the Kitaev et al. reduction to get a local Hamiltonian H whose unique ground state is |. We can then use | to recover the YQP witness |, and thereby decide L By Kitaev et al., we know L OCAL H AMILTONIANS is QMA-complete. Measuring this state yields the original QMA witness | 1 with (1/poly(n)) probability. Hence | 1 can be recovered from

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