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Google recently announced that they have achieved "Quantum Supremacy": "that would be practically impossible for a classical machine."

Does this mean that they have definitely proved that BQP ≠ BPP ? And if that is the case, what are the implications for P ≠ NP ?

glS
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Alex Kinman
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    They have sampled from a random quantum circuit, which is suspected to be outside of $\mathrm{NP}$, based on theoretical work of others. They have not "definitely proved that $\mathrm{BQP}\ne\mathrm{BPP}$;" however, they have thrown down a gauntlet regarding the Extended Church-Turing Thesis – Mark Spinelli Oct 23 '19 at 16:05
  • @MarkS since this is not a decision problem, in what sense can one say the problem is P, NP, BQP etc.? – user1936752 Oct 23 '19 at 16:32
  • @user1936752 well, you could, for example, state the problem as a search problem, given a random quantum circuit on $n$ qubits of depth $m=O(poly:n)$, search for output strings that have an average cross-entropy fidelity of greater than (some reasonable number more than 0). See, for example, comment #13 on shtetl-optimized – Mark Spinelli Oct 23 '19 at 18:00
  • @MarkS so if I understand you correctly, a single solution to the problem is meaningless but a sufficient number of runs + imposing statistical guarantees on the set of solutions transform the problem into a decision problem? Sorry for hijacking the question a little. – user1936752 Oct 23 '19 at 18:03
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    @user1936752, well, a single solution - sample only one output $n$-bit string from a random quantum circuit of depth $m$, wherein the probability (amplitude squared) of the sampled string is greater than $1/2^n$ - is still likely a hard problem. See, for example, this question. You are right though - this is getting away from the OP. Can you ask another? Google's results challenge the hypothesis that it is physically impossible to build a scalable QC. – Mark Spinelli Oct 23 '19 at 18:18
  • @MarkS are you saying they have "violated" the extended Church-Turing thesis? But then IBM claims to be able to achieve the same result, but with a mega-super-computer and 2.5 days of compute time, so how does the Google result effect the thesis? – Alex Kinman Oct 24 '19 at 16:15
  • @AlexKinman Google reliably and repeatedly prepared highly-entangled states according to their own random quantum circuits, in a Hilbert space of dimension $2^{53}$, and sampled therefrom. I don't think IBM is even challenging this! They are only challenging whether $2^{53}$ is outside or barely inside the anthropomorphic realm of possibility. If $53$ is still barely inside the realm of possibility, then surely $57$ or $60$ would not be? Remember Deep Blue did not soundly beat Kasparov. The Wright Flyer's 1903 flight was less than a minute. – Mark Spinelli Oct 24 '19 at 17:02
  • @MarkS I'm still not following you: Per my understanding, whether a class of problems is Turing decidable or not is a binary, True or False proposition, and doesn't depend on problem size. We can't say class of problems X is Turing decidable for $|X| \leq N$ but undecidable for $|X| > N$ AFAIK. – Alex Kinman Oct 25 '19 at 00:26
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    @AlexKinman The Extended Church-Turing Thesis implies that all computational models have the same efficiency as those of a probabilistic Turing machine. A quantum computer (most likely) does not have the same efficiency as a probabilistic Turing machine. Google built a quantum computer, and showed that they performed a task in a manner orders of magnitude - indeed asymptotically faster - than a probabilistic Turing machine. There is no statement that Google performed a task that is formally undecidable - only that they asymptotically more efficient. – Mark Spinelli Oct 25 '19 at 01:59
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    @MarkS thanks. Now it makes sense. I was assuming that the Extended Church Turing thesis was the same as the Church–Turing–Deutsch principle, now I understand the difference. – Alex Kinman Oct 25 '19 at 03:13

2 Answers2

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Google's paper/results are kind of sideways to questions in computational complexity about the relation between $\mathrm{BPP}$ and $\mathrm{BQP}$ (and even further from questions about whether $\mathrm{P}\ne\mathrm{NP}$). It's more as if Google relies on the hypothesis that $\mathrm{BPP}\ne\mathrm{BQP}$ as evidence that their quantum computer performs a task many orders of magnitude faster than a classical computer could.

Google performed a sampling task on their quantum computer, that they have strong theoretical reasons to believe is not easily performed on a classical computer. If we say that these complexity classes live in some idealized platonic universe, then Google's results don't shed any light about the difficulty of proving whether or not they are equal to one another - because Google's paper assumes that they are not equal to one another.

What Google's paper does do, is provide evidence that the hypothesis that "a probabilistic Turing machine can efficiently simulate any realistic model of computation" is incorrect. They have prepared and maintained coherence of a state of their choosing in a Hilbert space of dimension $2^{53}$. As Aaronson argues, is akin to the Wright Flyer providing evidence that "heavier-than-air human-controlled powered flight is impossible" is incorrect.

Mark Spinelli
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    @Mark S, How have Google's paper managed to provide evidence against the hypothesis that "a probabilistic Turing machine can efficiently simulate any realistic model of computation"? Pl. share relevant logic or parts of paper which show such evidence. – Ashish Oct 24 '19 at 11:47
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    @Ashish It's an interesting question about whether or how one can conclude that random circuit sampling helps invalidate the Extended Church-Turing Thesis (ECT) - I claim it does, but that seems separate from the OP's, which appears to be more along the lines of whether or how random circuit sampling helps validate that $\mathrm{BQP}\ne\mathrm{BPP}$ - I claim it doesn't. However, if you were to formally ask here a well-phrased question about random circuit sampling and the ECT, it might be well-received. – Mark Spinelli Oct 24 '19 at 13:38
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    @Ashish I don't know if I understand your question properly. let me try, since Random Circuit Sampling (RCS) is #P-hard (see here: https://arxiv.org/pdf/1803.04402.pdf) So, it is hard classically, and there is an efficient algorithm to compute it in quantum. RCS can be computed with few quibts, so it is very good candidate problem to put it in 'quantum supremacy' experiment. Assume that all complexity classes are hold, then by Google experiment, we know that ECT is no longer a true statement; since probabilistic TM cannot simulate RCS in best classical computer (this is a counterexample.) – YOUSEFY Oct 29 '19 at 10:09
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Paraphrasing some tweets on the matter earlier, the result is rather underwhelming because it plays on a discrepancy between what they mean by quantum supremacy (QS) and what people tend to think QS means.

What I find most people think QS is supposed to mean, and what I assumed it meant until a month or so ago, was that there exists a computable problem (in the CTT sense of computation) and an actual quantum computer, such that, at some scales, the problem is tractable on the quantum computer but intractable on all classical computers.

The problem the Google QC folks have demonstrated is not computation in the CTT sense. It is a physical process of sampling that involves computations as part of the process, and as with any physical process, it can be simulated approximately by computation. They have good reason to believe (proof? I'm not sure but it should reasonably be assumed true by default anyway) that computation to similate the process is going to be intractably slow. This is not surprising at all. It's a fundamental consequence of quantum mechanics that lots of physical processes will have that property.

That's not to say it's entirely uninteresting. There are likely useful applications of the sampling problem they implemented, and as I understand it, it provides examples of large classes of physical systems which are not amenable to efficient computational simulation. But it has nothing to do with whether or how soon a QC will be able to compete with a (classical, CTT) computer solving computable problems.

glS
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R.. GitHub STOP HELPING ICE
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    I wouldn't say it's "underwhelming". What they mean by quantum supremacy is exactly what everyone in the community means with the term, and pretty much the kind of result that people (people working on the field I mean) expected from them. Also, they do solve a "computational problem", only it's a "sampling problem", which is a type of problem unfamiliar to many. Also, there are computational complexity results underlying the claim of hardness of solving this problem classically – glS Oct 24 '19 at 09:35
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    It is not computation in CTT sense. – R.. GitHub STOP HELPING ICE Oct 24 '19 at 10:26
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    I guess by "CTT" you mean "Church-Turing thesis". What do you mean exactly with "computation in the CTT" sense? That it's not a decision problem? – glS Oct 24 '19 at 11:37
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    @R.. - Your answer seems to challenge whether or not sampling from a random quantum circuit qualifies as computation. Although this may be asked and answered as a separate question, it's not clear whether your answer addresses the OP's question: "Does this mean that they have definitely proved that BQP ≠ BPP ? And if that is the case, what are the implications for P ≠ NP ?" – Mark Spinelli Oct 24 '19 at 12:15
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    @glS: Decision problems, computable functions, etc. are equivalent, so whichever form you like. Sampling is not even a function, much less a computable one. It's a physical process outside the scope of computing/functions. – R.. GitHub STOP HELPING ICE Oct 24 '19 at 15:00
  • @MarkS: My understanding of BPP (maybe incorrect?) is that it's defined in terms of a probabilistic solution to a (deterministic) comptable decision problem, not probabilistic simulations of physical problems. The sampling problem here is the latter not the former, so I don't think the result says anything about the relationship between BQP and BPP, much less between P and NP. – R.. GitHub STOP HELPING ICE Oct 24 '19 at 15:03
  • @R.. You are stating some positions that, if framed as a separate question along the lines of "Why is it argued that sampling from a random quantum circuit invalidates the Extended Church-Turing Thesis, when it's not clear that sampling is even a function, much less a computable function?" can probably be answered. I'd ask you again to consider formalizing your comments as such, and asking (here, or e.g. TCS, etc.) – Mark Spinelli Oct 24 '19 at 15:36
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    @R.. but results about sampling problems do tell you about what you call "CTT" complexity classes. For example, if you can classically solve the sampling problem of simulating a boson sampling device, then it has been proven (up to some reasonable cc assumptions) that the polynomial hierarchy collapses at the third level. Similar results hold for simulating random circuits, see e.g. Bouland et al. – glS Oct 24 '19 at 15:37
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    @glS: I don't see how the abstract supports what you're saying. – R.. GitHub STOP HELPING ICE Oct 30 '19 at 04:20
  • @glS If you get some time, could you elaborate on how exactly the random circuit sampling problem is a computation in the Church Turing sense, in a new (self-answered) Q&A thread? I believe that's a common confusion for many people. – Sanchayan Dutta Oct 30 '19 at 20:43
  • I've asked for clarification regarding this here. – Sanchayan Dutta Oct 30 '19 at 21:39
  • @SanchayanDutta: If people are going to object to my claim that it's "not computation in the CTT sense" by arguing about what the CTT means, should I just clarify this answer to state that there is no [partial] recursive function yielding the result of random circuit sampling (or any encoding of it) as output? That is the key point that seems to be getting lost. – R.. GitHub STOP HELPING ICE Oct 30 '19 at 23:24
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    It could be okay to say that serious supremacy requires deterministic output. However, Google's experiment is much closer to CS theory than just that physical systems can be hard to simulate. A system may be hard to simulate when it has too much state (like water flow) or when we don't know its dynamic (like neutron stars). But the dynamics of the Sycamore chip is totally nailed down, and its state is only 53 qubits, which is a lot like just 53 bits. Instead, Sycamore is serious evidence of a different regime of computational complexity. – Greg Kuperberg Oct 31 '19 at 00:49
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    @R.. Yes, you could do that and claim that the Google supremacy experiment doesn't satisfy that stringent criteria (we all agree on that). I suppose nothing short of a classically intractable experimental verification of Shor's algorithm would! The point here is that even the hardcore CS theorists like Aaronson and Kuperberg are not really using your particular version of the deterministic CTT (for good reasons). – Sanchayan Dutta Oct 31 '19 at 04:03
  • @GregKuperberg: I don't have any objection to the output of the QC being non-deterministic; the output of any physically instantiated computer is (neutrino-induced bitflips, etc.). As long as it can get the right result with probability higher than the wrong result, bounded away from 50% independent of the particular input, you can repeat the process to get arbitrarily high probability of correct result... – R.. GitHub STOP HELPING ICE Oct 31 '19 at 12:33
  • ... Rather, what I object to is that the probabilistic result is not the result of a computation in the sense I mean computation, and thereby doesn't tell us anything about the ability of a QC to solve currently-intractable computational (in the sense I mean it, again) problems. – R.. GitHub STOP HELPING ICE Oct 31 '19 at 12:50
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    @SanchayanDutta: That's basically my view, yes. Not necessarily Shor's algorithm, but some classically intractable (deterministic) computational problem. – R.. GitHub STOP HELPING ICE Oct 31 '19 at 12:50
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    @R.. - Yes, I get it. You don't mind slightly uncertain output, but you want it to meet the standard of BQP, that the true answer is deterministic and the computer's output is probably correct. In your interpretation of ECT, which is not crazy, quantum supremacy doesn't violate it. But other interpretations of ECT aren't crazy either. By any standard, quantum supremacy is evidence against ECT, even if it isn't a death blow. – Greg Kuperberg Oct 31 '19 at 15:11
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    @SanchayanDutta - Look, in all fairness, Scott and I already disbelieved ECT even before the quantum supremacy experiment. I would also concede that it's not by itself a death blow to ECT. I see it as a brutal punch to the jaw to a boxer who I already thought was the weaker contender in the ring. I do think that ECT is doomed, but for now still standing. :-) – Greg Kuperberg Oct 31 '19 at 15:19