The Equivalence of Sampling and Searching

1. The Equivalence of Sampling and Searching Scott Aaronson MIT

2. In complexity theory, we love at least four types of problems Given an input x{0,1}n… Languages / Decision Problems. Decide if xL or xL Promise Problems. Decide if xYES or xNO Search Problems. Output an element of a (nonempty) set Ax{0,1}m, with probability 1-, in poly(n,1/) time Sampling Problems. Sample from a probability distribution Dx over m-bit strings, with error  in variation distance, in poly(n,1/) time

3. Suppose we want to know whether quantum computers are stronger than classical computers (To pick a random example of a complexity question) Then which formal question do we “really” mean to ask? BPP vs. BQP? PromiseBPP vs. PromiseBQP? FBPP vs. FBQP? SampBPP vs. SampBQP?

4. Easy Implications SampBPP=SampBQP FBPP=FBQP  PromiseBPP=PromiseBQP  BPP=BQP Crucial question: Can these implications be reversed? We show that at least one of them can: FBPP=FBQP SampBPP=SampBQP

5. Application to Linear Optics [A.-Arkhipov, STOC’11] study a rudimentary type of quantum computer based entirely on linear optics: identical, non-interacting photons passing through a network of beamsplitters Our model doesn’t seem to be universal for quantum computing (or even classical computing)—but it can solve sampling problems that we give evidence are hard classically Using today’s result, we automatically also get a search problem solvable with linear optics that ought to be hard classically

6. But the QC stuff is just one application of a much more general result… Informal Statement: Let S={Dx}x be any sampling problem. Then there exists a search problem RS={Ax}x that’s equivalent to S, in the following sense: For any “reasonable” complexity class C (BPP, BQP, BPPSPACE, etc.), RSFC  SSampC

7. Intuition Suppose our sampling problem is to sample uniformly from a set A{0,1}n First stab at an “equivalent” search problem: output any element of A That clearly doesn’t work—finding an A element could be much easier than sampling a random element! Better idea: output an element yA whose Kolmogorov complexity K(y) is close to log2|A|

8. Clearly, if we can sample a random yA, then with high probability K(y)log2|A| But conversely, if a randomized machine M outputs a y with K(y)log2|A|, it can only do so by sampling y almost-uniformly from A. For otherwise, M would yield a succinct description of y, contrary to assumption! Technical part: Generalize to nonuniform distributions Requires notion of a universal randomness test from algorithmic information theory

9. Comments If we just wanted a search problem at least as hard as S, that would be easy: Kolmogorov complexity only comes in because we need RS to be equivalent to S Our “reduction” from sampling to search is non-black-box: it requires the assumption that we have a Turing machine to solve RS! Our result provides an extremely natural application of Kolmogorov complexity to “standard” complexity: one that doesn’t just amount to a counting argument

10. Kolmogorov Review K(y | x): Prefix-free Kolmogorov complexity of y, conditioned on x Kolmogorentropy Lemma: Let D={py} be a distribution, and let y be in its support. Then where K(D) is the length of the shortest program to sample from D. Same holds if we replace K(y) by K(y|x) and K(D) by K(D|x).

11. Constructing the Search Problem We’re given a sampling problem S={Dx}x, where on input x{0,1}n, >0, the goal is to sample an m-bit string from a distribution C that’s -close to D=Dx, in poly(n,1/) time. Let Then the search problem RS is this: on input x{0,1}n, >0, output an N-tuple Y=y1,…,yNAx, with probability 1-, in poly(n,1/) time

12. Equivalence Proof Lemma: Let C be any distribution over {0,1}m such that |C-Dx|. Then In other words, any algorithm that solves the sampling problem also solves the search problem w.h.p. Proof: Counting argument.

13. Lemma: Given a probabilistic Turing machine B, suppose Let C be the distribution over m-bit strings obtained by running B(x,), then picking one its N outputs y1,…,yN randomly. Then there exists a constant QB such that In other words: if B solves the search problem w.h.p., then it also solves the sampling problem Proof Sketch: Use Kolmogorentropy Lemma to show B(x,)’s output distribution has small KL-divergence from DN. Similar to Parallel Repetition Theorem, this implies C has small KL-divergence from D. By Pinsker’s Inequality, this implies |C-Dx| is small.

14. Wrapping Up Theorem: Let O be any oracle that, given x, 01/, and a random string r, outputs a sample from a distribution C such that |C-Dx|. Then RSFBPPO. Let B be any probabilistic Turing machine that, given x,01/, outputs a YAx, with probability 1-. Then SSampBPPB.

15. Application to Quantum Complexity • Suppose FBPP=FBQP. • Let SSampBQP. • Then RSFBQP [RSS reduction] • RSFBPP [by hypothesis] • SSampBPP. [SRS reduction] • Therefore SampBPP=SampBQP.

16. Open Problems Can we show there’s no black-box equivalence between search and sampling problems? (I.e., that our use of Kolmogorov complexity was necessary?) The converse direction: Given a search problem, can we construct an equivalent sampling problem? • What if we want the search problem to be checkable? • Can redo proof with space-bounded Kolmogorov complexity to put search problem in PSPACE, but seems hard to do better More equivalence theorems—ideally, involving decision and promise problems?