BQP and the Polynomial Hierarchy

Introduction

A central task of quantum computing theory is to understand how $\mathsf{BQP}$—meaning Bounded-Error Quantum Polynomial-Time, the class of all problems feasible for a quantum computer—fits in with classical complexity classes. In their original 1993 paper defining $\mathsf{BQP}$, Bernstein and Vazirani showed that $\mathsf{BPP}\subseteq\mathsf{BQP}\subseteq\mathsf{P}^{\mathsf{\#P}}$.The upper bound was later improved to $\mathsf{BQP}\subseteq\mathsf{PP}$ by Adleman, DeMarrais, and Huang . Informally, this says that quantum computers are at least as fast as classical probabilistic computers and no more than exponentially faster (indeed, they can be simulated using an oracle for counting). Bernstein and Vazirani also gave evidence that $\mathsf{BPP}\neq\mathsf{BQP}$, by exhibiting an oracle problem called Recursive Fourier Sampling that requires $n^{\Omega\left(\log n\right)}$ queries on a classical computer but only $n$ queries on a quantum computer.For more about Recursive Fourier Sampling see Aaronson . The evidence for the power of quantum computers became dramatically stronger a year later, when Shor (building on work of Simon ) showed that Factoring and Discrete Logarithm are in $\mathsf{BQP}$. On the other hand, Bennett et al. gave oracle evidence that $\mathsf{NP}\not\subset\mathsf{BQP}$, and while no one regards such evidence as decisive, today it seems extremely unlikely that quantum computers can solve $\mathsf{NP}$-complete problems in polynomial time. A vast body of research, continuing to the present, has sought to map out the detailed boundary between those $\mathsf{NP}$ problems that are feasible for quantum computers and those that are not.

However, there is a complementary question that—despite being universally recognized as one of the “grand challenges” of the field—has had essentially zero progress over the last sixteen years:

Is $\mathsf{BQP}$ in $\mathsf{NP}$? More generally, is $\mathsf{BQP}$ contained anywhere in the polynomial hierarchy$\ \mathsf{PH}=\mathsf{NP}\cup\mathsf{NP}^{\mathsf{NP}}\cup\mathsf{NP}^{\mathsf{NP}^{\mathsf{NP}}}\cup\cdots$?

The “default” conjecture is presumably $\mathsf{BQP}\not\subset\mathsf{PH}$, since no one knows what a simulation of $\mathsf{BQP}$ in $\mathsf{PH}$ would look like. Before this work, however, there was no formal evidence for or against that conjecture. Almost all the problems for which we have quantum algorithms—including Factoring and Discrete Logarithm—are easily seen to be in $\mathsf{NP}\cap\mathsf{coNP}$.Here we exclude $\mathsf{BQP}$-complete problems such as approximating the Jones polynomial , which, by the very fact of being $\mathsf{BQP}$-complete, seem hard to interpret as “evidence” for $\mathsf{BQP}\not\subset\mathsf{PH}$. One notable exception is Recursive Fourier Sampling, the problem that Bernstein and Vazirani originally used to construct an oracle $A$ relative to which $\mathsf{BPP}^{A}\neq\mathsf{BQP}^{A}$. One can show, without too much difficulty, that Recursive Fourier Sampling yields oracles $A$ relative to which $\mathsf{BQP}^{A}\not\subset\mathsf{NP}^{A}$ and indeed $\mathsf{BQP}^{A}\not\subset\mathsf{MA}^{A}$. However, while it is reasonable to conjecture that Recursive Fourier Sampling (as an oracle problem) is not in $\mathsf{PH}$, it is open even to show that this problem (or any other $\mathsf{BQP}$ oracle problem) is not in $\mathsf{AM}$! Recall that $\mathsf{AM}=\mathsf{NP}$ under plausible derandomization assumptions . Thus, until we solve the problem of constructing an oracle $A$ such that $\mathsf{BQP}^{A}\not\subset\mathsf{AM}^{A}$, we cannot even claim to have oracle evidence (which is itself, of course, a weak form of evidence) that $\mathsf{BQP}\not\subset\mathsf{NP}$.

Before going further, we should clarify that there are two questions here: whether $\mathsf{BQP}\subseteq\mathsf{PH}$ and whether $\mathsf{P{}romiseBQP}\subseteq\mathsf{P{}romisePH}$. In the unrelativized world, it is entirely possible that quantum computers can solve promise problems outside the polynomial hierarchy, but that all languages in $\mathsf{BQP}$ are nevertheless in $\mathsf{PH}$. However, for the specific purpose of constructing an oracle $A$ such that $\mathsf{BQP}^{A}\not\subset\mathsf{PH}^{A}$, the two questions are equivalent, basically because one can always “offload” a promise into the construction of the oracle $A$.Here is a simple proof: let $\Pi=\left(\Pi_{\operatorname*{YES}},\Pi_{\operatorname*{NO}}\right)$ be a promise problem in $\mathsf{P{}romiseBQP}^{A}\setminus\mathsf{P{}romisePH}^{A}$, for some oracle $A$. Then clearly, every $\mathsf{P{}romisePH}^{A}$ machine $M$ fails to solve $\Pi$ on infinitely many inputs $x$ in $\Pi_{\operatorname*{YES}}\cup\Pi_{\operatorname*{NO}}$. This means that we can produce an infinite sequence of inputs $x_{1},x_{2},\ldots$ in $\Pi_{\operatorname*{YES}}\cup\Pi_{\operatorname*{NO}}$, whose lengths $n_{1},n_{2},\ldots$ are spaced arbitrarily far apart, such that every $\mathsf{P{}romisePH}^{A}$ machine $M$ fails to solve $\Pi$ on at least one $x_{i}$. Now let $B$ be an oracle that is identical to $A$, except that for each input length $n$, it reveals (i) whether $n=n_{i}$ for some $i$ and (ii) if so, what the corresponding $x_{i}$ is. Also, let $L$ be the unary language that contains $0^{n}$ if and only if (i) $n=n_{i}$ for some $i$ and (ii) $x_{i}\in\Pi_{\operatorname*{YES}}$. Then $L$ is in $\mathsf{BQP}^{B}$ but not $\mathsf{PH}^{B}$.

There are at least four reasons why the $\mathsf{BQP}$ versus $\mathsf{PH}$ question is so interesting. At a basic level, it is both theoretically and practically important to understand what classical resources are needed to simulate quantum physics. For example, when a quantum system evolves to a given state, is there always a short classical proof that it does so? Can one estimate quantum amplitudes using approximate counting (which would imply $\mathsf{BQP}\subseteq\mathsf{BPP}^{\mathsf{NP}}$)? If something like this were true, then while the exponential speedup of Shor’s factoring algorithm might stand, quantum computing would nevertheless seem much less different from classical computing than previously thought.

Second, if $\mathsf{BQP}\not\subset\mathsf{PH}$, then many possibilities for new quantum algorithms might open up to us. One often hears the complaint that there are too few quantum algorithms, or that progress on quantum algorithms has slowed since the mid-1990s. In our opinion, the real issue here has nothing to do with quantum computing, and is simply that there are too few natural $\mathsf{NP}$-intermediate problems for which there plausibly could be quantum algorithms! In other words, instead of focussing on Graph Isomorphism and a small number of other $\mathsf{NP}$-intermediate problems, it might be fruitful to look for quantum algorithms solving completely different types of problems—problems that are not necessarily even in $\mathsf{PH}$. In this paper, we will see a new example of such a quantum algorithm, which solves a problem called Fourier Checking.

Third, it is natural to ask whether the $\mathsf{P}\overset{?}{=}\mathsf{BQP}$ question is related to that other fundamental question of complexity theory, $\mathsf{P}\overset{?}{=}\mathsf{NP}$. More concretely, is it possible that quantum computers could provide exponential speedups even if $\mathsf{P}=\mathsf{NP}$? If $\mathsf{BQP}\subseteq\mathsf{PH}$, then certainly the answer to that question is no (since $\mathsf{P}=\mathsf{NP}\Longrightarrow\mathsf{P}=\mathsf{PH}$). Therefore, if we want evidence that quantum computing could survive a collapse of $\mathsf{P}$ and $\mathsf{NP}$, we must also seek evidence that $\mathsf{BQP}\not\subset\mathsf{PH}$.

Fourth, a major challenge for quantum computing research is to get better evidence that quantum computers cannot solve $\mathsf{NP}$-complete problems in polynomial time. As an example, could we show that if $\mathsf{NP}\subseteq\mathsf{BQP}$, then the polynomial hierarchy collapses? At first glance, this seems like a wild hope; certainly we have no idea at present how to prove anything of the kind. However, notice that if $\mathsf{BQP}\subseteq\mathsf{AM}$, then the desired implication would follow immediately! For in that case,

where the last implication was shown by Boppana, Håstad, and Zachos . Similar remarks apply to the questions of whether $\mathsf{NP}\subseteq\mathsf{BQP}$ would imply $\mathsf{PH}\subseteq\mathsf{BQP}$, and whether the folklore result $\mathsf{NP}^{\mathsf{BPP}}\subseteq\mathsf{BPP}^{\mathsf{NP}}$ has the quantum analogue $\mathsf{NP}^{\mathsf{BQP}}\subseteq\mathsf{BQP}^{\mathsf{NP}}$. In each of these cases, we find that understanding some other issue in quantum complexity theory requires first coming to grips with whether $\mathsf{BQP}$ is contained in some level of the polynomial hierarchy.

2 Our Results

This paper presents the first formal evidence for the possibility that $\mathsf{BQP}\not\subset\mathsf{PH}$. Perhaps more importantly, it places the relativized $\mathsf{BQP}$ versus $\mathsf{PH}$ question at the frontier of (classical) circuit lower bounds. The heart of the problem, we will find, is to extend Braverman’s spectacular recent proof of the Linial-Nisan Conjecture, in ways that would reveal a great deal of information about small-depth circuits independent of the implications for quantum computing.

We have two main contributions. First, we achieve an oracle separation between $\mathsf{BQP}$ and $\mathsf{PH}$ for the case of relation problems. A relation problem is simply a problem where the desired output is an $n$-bit string (rather than a single bit), and any string from some nonempty set $S$ is acceptable. Relation problems arise often in theoretical computer science; one well-known example is finding a Nash equilibrium (shown to be $\mathsf{PPAD}$-complete by Daskalakis et al. ). Within quantum computing, there is considerable precedent for studying relation problems as a warmup to the harder case of decision problems. For example, in 2004 Bar-Yossef, Jayram, and Kerenidis gave a relation problem with quantum one-way communication complexity $O\left(\log n\right)$ and randomized one-way communication complexity $\Omega\left(\sqrt{n}\right)$. It took several more years for Gavinsky et al. to achieve the same separation for decision problems, and the proof was much more complicated. The same phenomenon has arisen many times in quantum communication complexity , though to our knowledge, this is the first time it has arisen in quantum query complexity.

There exists an oracle $A$ relative to which $\mathsf{FBQP}^{A}\not\subset\mathsf{FBPP}^{\mathsf{PH}^{A}}$, where $\mathsf{FBQP}$ and $\mathsf{FBPP}$ are the relation versions of $\mathsf{BQP}$ and $\mathsf{BPP}$ respectively.Confusingly, the $\mathsf{F}$ stands for “function”; we are simply following the standard naming convention for classes of relation problems ($\mathsf{FP}$, $\mathsf{FNP}$, etc).

Underlying Theorem 1 is a new lower bound against $\mathsf{AC}^{0}$ circuits (constant-depth circuits composed of AND, OR, and NOT gates). The close connection between $\mathsf{AC}^{0}$ and the polynomial hierarchy that we exploit is not new. In the early 1980s, Furst-Saxe-Sipser and Yao noticed that, if we have a $\mathsf{PH}$ machine $M$ that computes (say) the Parity of a $2^{n}$-bit oracle string, then by simply reinterpreting the existential quantifiers of $M$ as OR gates and the universal quantifiers as AND gates, we obtain an $\mathsf{AC}^{0}$ circuit of size $2^{\operatorname*{poly}\left(n\right)}$ solving the same problem. It follows that, if we can prove a $2^{\omega\left(\operatorname*{polylog}n\right)}$ lower bound on the size of $\mathsf{AC}^{0}$ circuits computing Parity, we can construct an oracle $A$ relative to which $\mathsf{\oplus P}^{A}\not\subset\mathsf{PH}^{A}$. The idea is the same for constructing an $A$ relative to which $\mathcal{C}^{A}\not\subset\mathsf{PH}^{A}$, where $\mathcal{C}$ is any complexity class.

Indeed, the relation between $\mathsf{PH}$ and $\mathsf{AC}^{0}$ is so direct that we get the following as a more-or-less immediate counterpart to Theorem 1:

In the unrelativized world (with no oracle), there exists a relation problem solvable in quantum logarithmic time but not in nonuniform $\mathsf{AC}^{0}$.

The relation problem that we use to separate $\mathsf{BQP}$ from $\mathsf{PH}$, and $\mathsf{BQLOGTIME}$ from $\mathsf{AC}^{0}$, is called Fourier Fishing. The problem can be informally stated as follows. We are given oracle access to $n$ Boolean functions $f_{1},\ldots,f_{n}:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$, which we think of as chosen uniformly at random. The task is to output $n$ strings, $z_{1},\ldots,z_{n}\in\left\{0,1\right\}^{n}$, such that the corresponding squared Fourier coefficients $\widehat{f}_{1}\left(z_{1}\right)^{2},\ldots,\widehat{f}_{n}\left(z_{n}\right)^{2}$ are “often much larger than average.” Notice that if $f_{i}$ is a random Boolean function, then each of its Fourier coefficients $\widehat{f}_{i}\left(z\right)$ follows a normal distribution—meaning that with overwhelming probability, a constant fraction of the Fourier coefficients will be a constant factor larger than the mean. Furthermore, it is straightforward to create a quantum algorithm that samples each $z$ with probability proportional to $\widehat{f}_{i}\left(z\right)^{2}$, so that larger Fourier coefficients are more likely to be sampled than smaller ones.

On the other hand, computing any specific $\widehat{f}_{i}\left(z\right)$ is easily seen to be equivalent to summing $2^{n}$ bits. By well-known lower bounds on the size of $\mathsf{AC}^{0}$ circuits computing the Majority function (see Håstad for example), it follows that, for any fixed $z$, computing $\widehat{f}_{i}\left(z\right)$ cannot be in $\mathsf{PH}$ as an oracle problem. Unfortunately, this does not directly imply any separation between $\mathsf{BQP}$ and $\mathsf{PH}$, since the quantum algorithm does not compute $\widehat{f}_{i}\left(z\right)$ either: it just samples a $z$ with probability proportional to $\widehat{f}_{i}\left(z\right)^{2}$. However, we will show that, if there exists a $\mathsf{BPP}^{\mathsf{PH}}$ machine $M$ that even approximately simulates the behavior of the quantum algorithm, then one can solve Majority by means of a nondeterministic reduction—which uses approximate counting to estimate $\Pr\left[M\text{ outputs }z\right]$, and adds a constant number of layers to the $\mathsf{AC}^{0}$ circuit. The central difficulty is that, if $M$ knew the specific $z$ for which we were interested in estimating $\widehat{f}_{i}\left(z\right)$, then it could choose adversarially never to output that $z$. To solve this, we will show that we can “smuggle” a Majority instance into the estimation of a random Fourier coefficient $\widehat{f}_{i}\left(z\right)$, in such a way that it is information-theoretically impossible for $M$ to determine which $z$ we care about.

Our second contribution is to define and study a new black-box decision problem, called Fourier Checking. Informally, in this problem we are given oracle access to two Boolean functions $f,g:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$, and are promised that either

$f$ and $g$ are both uniformly random, or

The problem is to decide whether (i) or (ii) is the case.

It is not hard to show that Fourier Checking is in $\mathsf{BQP}$: basically, one can prepare a uniform superposition over all $x\in\left\{0,1\right\}^{n}$, then query $f$, apply a quantum Fourier transform, query $g$, and check whether one has recovered something close to the uniform superposition. On the other hand, being forrelated seems like an extremely “global” property of $f$ and $g$: one that would not be apparent from querying any small number of $f\left(x\right)$ and $g\left(y\right)$ values, regardless of the outcomes of those queries. And thus, one might conjecture that Fourier Checking (as an oracle problem) is not in $\mathsf{PH}$.

In this paper, we adduce strong evidence for that conjecture. Specifically, we show that for every $k\leq 2^{n/4}$, the forrelated distribution over $\left\langle f,g\right\rangle$ pairs is $O\left(k^{2}/2^{n/2}\right)$-almost $k$-wise independent. By this we mean that, if one had $1/2$ prior probability that $f$ and $g$ were uniformly random, and $1/2$ prior probability that $f$ and $g$ were forrelated, then even conditioned on any $k$ values of $f$ and $g$, the posterior probability that $f$ and $g$ were forrelated would still be

We conjecture that this almost $k$-wise independence property is enough, by itself, to imply that an oracle problem is not in $\mathsf{PH}$. We call this the Generalized Linial-Nisan Conjecture.

Without the $\pm O\left(k^{2}/2^{n/2}\right)$ error term, our conjecture would be equivalentUp to unimportant variations in the parameters to a famous conjecture in circuit complexity made by Linial and Nisan in 1990. Their conjecture stated that polylogarithmic independence fools $\mathsf{AC}^{0}$: in other words, every probability distribution over $N$-bit strings that is uniform on every small subset of bits, is indistinguishable from the truly uniform distribution by $\mathsf{AC}^{0}$ circuits. When we began investigating this topic a year ago, even the original Linial-Nisan Conjecture was still open. Since then, Braverman (building on earlier work by Bazzi and Razborov ) has given a beautiful proof of that conjecture. In other words, to construct an oracle relative to which $\mathsf{BQP}\not\subset\mathsf{PH}$, it now suffices to generalize Braverman’s Theorem from $k$-wise independent distributions to almost $k$-wise independent ones. We believe that this is by far the most promising approach to the $\mathsf{BQP}$ versus $\mathsf{PH}$ problem.

Let us mention two further applications of Fourier Checking:

If the Generalized Linial-Nisan Conjecture holds, then just like with Fourier Fishing, we can “scale down by an exponential,” to obtain a promise problem that is in $\mathsf{BQLOGTIME}$ but not in $\mathsf{AC}^{0}$.

Without any assumptions, we can prove the new results that there exist oracles relative to which $\mathsf{BQP}\not\subset\mathsf{BPP}_{\mathsf{path}}$ and $\mathsf{BQP}\not\subset\mathsf{SZK}$. We can also reprove all previous oracle separations between $\mathsf{BQP}$ and classical complexity classes in a unified fashion.

Assuming the Generalized Linial-Nisan Conjecture, there exists an oracle $A$ relative to which $\mathsf{BQP}^{A}\not\subset\mathsf{PH}^{A}$, and there also exists a promise problem in $\mathsf{BQLOGTIME}\setminus\mathsf{AC}^{0}$. Unconditionally, there exists an oracle $A$ relative to which $\mathsf{BQP}^{A}\not\subset\mathsf{BPP}_{\mathsf{path}}^{A}$ and $\mathsf{BQP}^{A}\not\subset\mathsf{SZK}^{A}$.

As a candidate problem, Fourier Checking has at least five advantages over the Recursive Fourier Sampling problem of Bernstein and Vazirani . First, it is much simpler to define and reason about. Second, Fourier Checking has the almost $k$-wise independence property, which is not shared by Recursive Fourier Sampling, and which immediately connects the former to general questions about pseudorandomness against constant-depth circuits. Third, Fourier Checking can yield exponential separations between quantum and classical models, rather than just quasipolynomial ones. Fourth, one can hope to use Fourier Checking to give an oracle relative to which $\mathsf{BQP}$ is not in $\mathsf{PH}\left[n^{c}\right]$ (or $\mathsf{PH}$ with $n^{c}$ alternations) for any fixed $c$; by contrast, Recursive Fourier Sampling is in $\mathsf{PH}\left[\log n\right]$. Finally, it is at least conceivable that the quantum algorithm for Fourier Checking is good for something. We leave the challenge of finding an explicit computational problem that “instantiates” Fourier Checking, in the same way that Factoring and Discrete Logarithm instantiated Shor’s period-finding problem.

3 In Defense of Oracles

This paper is concerned with finding oracles relative to which $\mathsf{BQP}$ outperforms classical complexity classes. As such, it is open to the usual objections: “But don’t oracle results mislead us about the ‘real’ world? What about non-relativizing results like $\mathsf{IP}=\mathsf{PSPACE}$ ?”

In our view, it is most helpful to think of oracle separations, not as strange metamathematical claims, but as lower bounds in a concrete computational model that is natural and well-motivated in its own right. The model in question is query complexity, where the resource to be minimized is the number of accesses to a very long input string. When someone gives an oracle $A$ relative to which $\mathcal{C}^{A}\not\subset\mathcal{D}^{A}$, what they really mean is simply that they have found a problem that $\mathcal{C}$ machines can solve using superpolynomially fewer queries than $\mathcal{D}$ machines. In other words, $\mathcal{C}$ has has “cleared the first possible obstacle”—the query complexity obstacle—to having capabilities beyond those of $\mathcal{D}$. Of course, it could be (and sometimes is) that $\mathcal{C}\subseteq\mathcal{D}$ for other reasons, but if we do not even have a query complexity lower bound, then proving one is in some sense the obvious place to start.

Oracle separations have played a role in many of the central developments of both classical and quantum complexity theory. As mentioned earlier, proving query complexity lower bounds for $\mathsf{PH}$ machines is essentially equivalent to proving size lower bounds for $\mathsf{AC}^{0}$ circuits—and indeed, the pioneering $\mathsf{AC}^{0}$ lower bounds of the early 1980s were explicitly motivated by the goal of proving oracle separations for $\mathsf{PH}$.Yao’s paper was entitled “Separating the polynomial-time hierarchy by oracles”; the Furst-Saxe-Sipser paper was entitled “Parity, circuits, and the polynomial time hierarchy.” Within quantum computing, oracle results have played an even more decisive role: the first evidence for the power of quantum computers came from the oracle separations of Bernstein-Vazirani and Simon , and Shor’s algorithm contains an oracle algorithm (for the Period-Finding problem) at its core.

Having said all that, if for some reason one still feels averse to the language of oracles, then (as mentioned before) one is free to scale everything down by an exponential, and to reinterpret a relativized separation between $\mathsf{BQP}$ and $\mathsf{PH}$ as an unrelativized separation between $\mathsf{BQLOGTIME}$ and $\mathsf{AC}^{0}$.

Preliminaries

It will be convenient to consider Boolean functions of the form $f:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$. Throughout this paper, we let $N=2^{n}$; we will often view the truth table of a Boolean function as an “input” of size $N$. Given a Boolean function $f:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$, the Fourier transform of $f$ is defined as

We first define the Fourier Fishing problem, in both “distributional” and “promise” versions. In the distributional version, we are given oracle access to $n$ Boolean functions $f_{1},\ldots,f_{n}:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$, which are chosen uniformly and independently at random. The task is to output $n$ strings, $z_{1},\ldots,z_{n}\in\left\{0,1\right\}^{n}$, at least 75% of which satisfy $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 1$ and at least 25% of which satisfy $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 2$. (Note that these thresholds are not arbitrary, but were carefully chosen to produce a separation between the quantum and classical models!)

We now want a version of Fourier Fishing that removes the need to assume the $f_{i}$’s are uniformly random, replacing it with a worst-case promise on the $f_{i}$’s. Call an $n$-tuple $\left\langle f_{1},\ldots,f_{n}\right\rangle$ of Boolean functions good if

(We will show in Lemma 8 that the vast majority of $\left\langle f_{1},\ldots,f_{n}\right\rangle$ are good.) In Promise Fourier Fishing, we are given oracle access to Boolean functions $f_{1},\ldots,f_{n}:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$, which are promised to be good. The task, again, is to output strings $z_{1},\ldots,z_{n}\in\left\{0,1\right\}^{n}$, at least 75% of which satisfy $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 1$ and at least 25% of which satisfy $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 2$.

Next we define a decision problem called Fourier Checking. Here we are given oracle access to two Boolean functions $f,g:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$. We are promised that either

$\left\langle f,g\right\rangle$ was drawn from the uniform distribution $\mathcal{U}$, which sets every $f\left(x\right)$ and $g\left(y\right)$ by a fair, independent coin toss.

In other words, $f$ and $g$ individually are still uniformly random, but they are no longer independent: now $g$ is now extremely well correlated with the Fourier transform of $f$ (hence “forrelated”).

The problem is to accept if $\left\langle f,g\right\rangle$ was drawn from $\mathcal{F}$, and to reject if $\left\langle f,g\right\rangle$ was drawn from $\mathcal{U}$. Note that, since $\mathcal{F}$ and $\mathcal{U}$ overlap slightly, we can only hope to succeed with overwhelming probability over the choice of $\left\langle f,g\right\rangle$, not for every $\left\langle f,g\right\rangle$ pair.

We can also define a promise-problem version of Fourier Checking. In Promise Fourier Checking, we are promised that the quantity

is either at least $0.05$ or at most $0.01$. The problem is to accept in the former case and reject in the latter case.

2 Complexity Classes

See the Complexity Zoowww.complexityzoo.com for the definitions of standard complexity classes, such as $\mathsf{BQP}$, $\mathsf{AM}$, and $\mathsf{PH}$. When we write $\mathcal{C}^{\mathsf{PH}}$ (i.e., a complexity class $\mathcal{C}$ with an oracle for the polynomial hierarchy), we mean $\cup_{k\geq 1}\mathcal{C}^{\mathsf{\Sigma}_{k}^{\mathsf{P}}}$.

We will consider not only decision problems, but also relation problems (also called function problems). In a relation problem, the output is not a single bit but a $\operatorname*{poly}\left(n\right)$-bit string $y$. There could be many valid $y$’s for a given instance, and the algorithm’s task is to output any one of them.

The definitions of $\mathsf{FP}$ and $\mathsf{FNP}$ (the relation versions of $\mathsf{P}$ and $\mathsf{NP}$) are standard. We now define $\mathsf{FBPP}$ and $\mathsf{FBQP}$, the relation versions of $\mathsf{BPP}$ and $\mathsf{BQP}$.

$\mathsf{FBPP}$ is the class of relations $R\subseteq\left\{0,1\right\}^{\ast}\times\left\{0,1\right\}^{\ast}$ for which there exists a probabilistic polynomial-time algorithm $A$ that, given any input $x\in\left\{0,1\right\}^{n}$, produces an output $y$ such that

where the probability is over $A$’s internal randomness. (In particular, this implies that for every $x$, there exists at least one $y$ such that $\left(x,y\right)\in R$.) $\mathsf{FBQP}$ is defined the same way, except that $A$ is a quantum algorithm rather than a classical one.

An important point about $\mathsf{FBPP}$ and $\mathsf{FBQP}$ is that, as far as we know, these classes do not admit amplification. In other words, the value of an algorithm’s success probability might actually matter, not just the fact that the probability is bounded above $1/2$. This is why we adopt the convention that an algorithm “succeeds” if it outputs $\left(x,y\right)\in R$ with probability $1-o\left(1\right)$. In practice, we will give oracle problems for which the $\mathsf{FBQP}$ algorithm succeeds with probability $1-1/\exp\left(n\right)$, while any $\mathsf{FBPP}^{\mathsf{PH}}$ algorithm succeeds with probability at most (say) $0.99$. How far the constant in this separation can be improved is an open problem.

Another important point is that, while $\mathsf{BPP}^{\mathsf{PH}}=\mathsf{P}^{\mathsf{PH}}$ (which follows from $\mathsf{BPP}\subseteq\mathsf{\Sigma}_{\mathsf{2}}^{\mathsf{P}}$), the class $\mathsf{FBPP}^{\mathsf{PH}}$ is strictly larger than $\mathsf{FP}^{\mathsf{PH}}$. To see this, consider the relation

where we are given $n$, and asked to output any string of Kolmogorov complexity at least $n$. Clearly this problem is in $\mathsf{FBPP}$: just output a random $2n$-bit string. On the other hand, just as obviously the problem is not in $\mathsf{FP}^{\mathsf{PH}}$. This is why we need to construct an oracle $A$ such that $\mathsf{FBQP}^{A}\not\subset\mathsf{FBPP}^{\mathsf{PH}^{A}}$: because constructing an oracle $A$ such that $\mathsf{FBQP}^{A}\not\subset\mathsf{FP}^{\mathsf{PH}^{A}}$ is trivial and not even related to quantum computing.

We now discuss some “low-level” complexity classes. $\mathsf{AC}^{0}$ is the class of problems solvable by a nonuniform family of AND/OR/NOT circuits, with depth $O\left(1\right)$, size $\operatorname*{poly}\left(n\right)$, and unbounded fanin. When we say “$\mathsf{AC}^{0}$ circuit,” we mean a constant-depth circuit of AND/OR/NOT gates, not necessarily of polynomial size. Any such circuit can be made into a formula (i.e., a circuit of fanout $1$) with only a polynomial increase in size. The circuit has depth $d$ if it consists of $d$ alternating layers of AND and OR gates (without loss of generality, the NOT gates can all be pushed to the bottom, and we do not count them towards the depth). For example, a DNF (Disjunctive Normal Form) formula is just an $\mathsf{AC}^{0}$ circuit of depth $2$.

We will also be interested in quantum logarithmic time, which can be defined naturally as follows:

$\mathsf{BQLOGTIME}$ is the class of languages $L\subseteq\left\{0,1\right\}^{\ast}$ that are decidable, with bounded probability of error, by a $\mathsf{LOGTIME}$-uniform family of quantum circuits $\left\{C_{n}\right\}_{n}$ such that each $C_{n}$ has $O\left(\log n\right)$ gates, and can include gates that make random-access queries to the input string $x=x_{1}\ldots x_{n}$ (i.e., that map $\left|i\right\rangle\left|z\right\rangle$ to $\left|i\right\rangle\left|z\oplus x_{i}\right\rangle$ for every $i\in\left[n\right]$).

One other complexity class that arises in this paper, which is less well known than it should be, is $\mathsf{BPP}_{\mathsf{path}}$. Loosely speaking, $\mathsf{BPP}_{\mathsf{path}}$ can be defined as the class of problems that are solvable in probabilistic polynomial time, given the ability to “postselect” (that is, discard all runs of the computation that do not produce a desired result, even if such runs are the overwhelming majority). Formally:

$\mathsf{BPP}_{\mathsf{path}}$ is the class of languages $L\subseteq\left\{0,1\right\}^{\ast}$ for which there exists a $\mathsf{BPP}$ machine $M$, which can either “succeed” or “fail” and conditioned on succeeding either “accept” or “reject,” such that for all inputs $x$:

$\Pr\left[M\left(x\right)\text{ succeeds}\right]>0$.

$x\in L\Longrightarrow\Pr\left[M\left(x\right)\text{ accepts }|~{}M\left(x\right)\text{ succeeds}\right]\geq\frac{2}{3}$.

$x\notin L\Longrightarrow\Pr\left[M\left(x\right)\text{ accepts }|~{}M\left(x\right)\text{ succeeds}\right]\leq\frac{1}{3}$.

$\mathsf{BPP}_{\mathsf{path}}$ was defined by Han, Hemaspaandra, and Thierauf , who also showed that $\mathsf{MA}\subseteq\mathsf{BPP}_{\mathsf{path}}$ and $\mathsf{P}_{||}^{\mathsf{NP}}\subseteq\mathsf{BPP}_{\mathsf{path}}\subseteq\mathsf{BPP}_{||}^{\mathsf{NP}}$. Using Fourier Checking, we will construct an oracle $A$ relative to which $\mathsf{BQP}^{A}\not\subset\mathsf{BPP}_{\mathsf{path}}^{A}$. This result might not sound amazing, but (i) it is new, (ii) it does not follow from the “standard” quantum algorithms, such as those of Simon and Shor , and (iii) it supersedes almost all previous oracle results placing $\mathsf{BQP}$ outside classical complexity classes.The one exception is the result of Green and Pruim that there exists an $A$ relative to which $\mathsf{BQP}^{A}\not\subset\mathsf{P}^{\mathsf{NP}^{A}}$, but that can also be easily reproduced using Fourier Checking. As another illustration of the versatility of Fourier Checking, we use it to give an $A$ such that $\mathsf{BQP}^{A}\not\subset\mathsf{SZK}^{A}$, where $\mathsf{SZK}$ is Statistical Zero Knowledge. The opposite direction—an $A$ such that $\mathsf{SZK}^{A}\not\subset\mathsf{BQP}^{A}$—was shown by Aaronson in 2002.

Quantum Algorithms

In this section, we show that Fourier Fishing and Fourier Checking both admit simple quantum algorithms.

Here is a quantum algorithm, FF-ALG, that solves Fourier Fishing with overwhelming probability in $O\left(n^{2}\right)$ time and $n$ quantum queries (one to each $f_{i}$). For $i:=1$ to $n$, first prepare the state

then apply Hadamard gates to all $n$ qubits, then measure in the computational basis and output the result as $z_{i}$.

Intuitively, FF-ALG samples the Fourier coefficients of each $f_{i}$ under a distribution that is skewed towards larger coefficients; the algorithm’s behavior is illustrated pictorially in Figure 1. We now give a formal analysis. Recall the definition of a “good” tuple $\left\langle f_{1},\ldots,f_{n}\right\rangle$ from Section 2.1. Assuming $\left\langle f_{1},\ldots,f_{n}\right\rangle$ is good, it is easy to analyze FF-ALG’s success probability.

Assuming $\left\langle f_{1},\ldots,f_{n}\right\rangle$ is good, FF-ALG succeeds with probability $1-1/\exp\left(n\right)$.

Proof. Let $\left\langle z_{1},\ldots,z_{n}\right\rangle$ be the algorithm’s output. For each $i$, let $X_{i}$ be the event that $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 1$ and let $Y_{i}$ be the event that $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 2$. Also let $p_{i}:=\Pr\left[X_{i}\right]$ and $q_{i}:=\Pr\left[Y_{i}\right]$, where the probability is over FF-ALG’s internal (quantum) randomness. Then clearly

By a Chernoff/Hoeffding bound, it follows that

Hence FF-ALG succeeds with $1-1/\exp\left(n\right)$ probability by the union bound.

$\left\langle f_{1},\ldots,f_{n}\right\rangle$ is good with probability $1-1/\exp\left(n\right)$, if the $f_{i}$’s are chosen uniformly at random.

Proof. Choose $f:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$ uniformly at random. Then for each $z$, the Fourier coefficient $\widehat{f}\left(z\right)$ follows a normal distribution, with mean and variance $1$. So in the limit of large $N$,

Since the $f_{i}$’s are chosen independently of one another, it follows by a Chernoff bound that

with probability $1-1/\exp\left(n\right)$ over the choice of $\left\langle f_{1},\ldots,f_{n}\right\rangle$.

Combining Lemmas 7 and 8, we find that FF-ALG succeeds with probability $1-1/\exp\left(n\right)$, where the probability is over both $\left\langle f_{1},\ldots,f_{n}\right\rangle$ and FF-ALG’s internal randomness.

2 Quantum Algorithm for Fourier Checking

We now turn to Fourier Checking, the problem of deciding whether two Boolean functions $f,g$ are independent or forrelated. Here is a quantum algorithm, FC-ALG, that solves Fourier Checking with constant error probability using $O\left(1\right)$ queries. First prepare a uniform superposition over all $x\in\left\{0,1\right\}^{n}$. Then query $f$ in superposition, to create the state

Then apply Hadamard gates to all $n$ qubits, to create the state

Then query $g$ in superposition, to create the state

Then apply Hadamard gates to all $n$ qubits again, to create the state

Finally, measure in the computational basis, and “accept” if and only if the outcome $\left|0\right\rangle^{\otimes n}$ is observed. If needed, repeat the whole algorithm $O\left(1\right)$ times to boost the success probability.

It is clear that the probability of observing $\left|0\right\rangle^{\otimes n}$ (in a single run of FC-ALG) equals

Recall that Promise Fourier Checking was the problem of deciding whether $p\left(f,g\right)\geq 0.05$ or $p\left(f,g\right)\leq 0.01$, promised that one of these is the case. Thus, we immediately get a quantum algorithm to solve Promise Fourier Checking, with constant error probability, using $O\left(1\right)$ queries to $f$ and $g$.

For the distributional version of Fourier Checking, we also need the following theorem.

If $\left\langle f,g\right\rangle$ is drawn from the uniform distribution $\mathcal{U}$, then

If $\left\langle f,g\right\rangle$ is drawn from the forrelated distribution $\mathcal{F}$, then

Proof. The first part follows immediately by symmetry (i.e., the fact that all $N=2^{n}$ measurement outcomes of the quantum algorithm are equally likely).

or the squared inner product between the vectors $\operatorname*{flat}\left(w\right)$ and $H\operatorname*{flat}\left(Hw\right)$. Note that $w^{T}\cdot HHw=w^{T}w=1$. So the whole problem is to understand the “discretization error” incurred in replacing $w^{T}$ by $\operatorname*{flat}\left(w\right)^{T}$ and $HHw$ by $H\operatorname*{flat}\left(Hw\right)$. By the triangle inequality, the angle between $\operatorname*{flat}\left(w\right)$ and $H\operatorname*{flat}\left(Hw\right)$ is at most the angle between $\operatorname*{flat}\left(w\right)$ and $w$, plus the angle between $w$ and $H\operatorname*{flat}\left(Hw\right)$. In other words:

Recall that each $v_{x}$ is an independent real Gaussian with mean and variance $1$, meaning that each $\left|v_{x}\right|$ is an independent nonnegative random variable with expectation $\sqrt{2/\pi}$. So by standard tail bounds, for all constants $\varepsilon>0$ we have

Since $H$ is unitary, the same analysis applies to $w^{T}H\operatorname*{flat}\left(Hw\right)$. Therefore, for all constants $\varepsilon>0$, with $1-1/\exp\left(N\right)$ probability we have

Therefore, with $1-1/\exp\left(N\right)$ probability over $\left\langle f,g\right\rangle$ drawn from $\mathcal{F}$,

in which case $p\left(f,g\right)\geq\left(\cos 1.3\right)^{2}\approx 0.0\allowbreak 72$.

Combining Theorem 9 with Markov’s inequality, we immediately get the following:

The Classical Complexity of Fourier Fishing

In Section 3.1, we gave a quantum algorithm for Fourier Fishing that made only one query to each $f_{i}$. By contrast, it is not hard to show that any classical algorithm for Fourier Fishing requires exponentially many queries to the $f_{i}$’s. In this section, we prove a much stronger result: that Fourier Fishing is not in $\mathsf{FBPP}^{\mathsf{PH}}$. This result does not rely on any unproved conjectures.

Our starting point will be the following $\mathsf{AC}^{0}$ lower bound, which can be found in the book of Håstad for example.

Any depth-$d$ circuit that accepts all $n$-bit strings of Hamming weight $n/2+1$, and rejects all strings of Hamming weight $n/2$, has size $\exp\left(\Omega\left(n^{1/\left(d-1\right)}\right)\right)$.

We now give a corollary of Theorem 11, which (though simple) seems to be new, and might be of independent interest. Consider the following problem, which we call $\varepsilon$-Bias Detection. We are given a string $y=y_{1}\ldots y_{m}\in\left\{0,1\right\}^{m}$, and are promised that each bit $y_{i}$ is $1$ with independent probability $p$. The task is to decide whether $p=1/2$ or $p=1/2+\varepsilon$.

Let $\mathcal{U}\left[\varepsilon\right]$ be the distribution over $\left\{0,1\right\}^{m}$ where each bit is $1$ with independent probability $1/2+\varepsilon$. Then any depth-$d$ circuit $C$ such that

has size $\exp\left(\Omega\left(1/\varepsilon^{1/\left(d+2\right)}\right)\right)$.

Proof. Suppose such a distinguishing circuit $C$ exists, with depth $d$ and size $S$, for some $\varepsilon>0$ (the parameter $m$ is actually irrelevant). Let $n=1/\varepsilon$, and assume for simplicity that $n$ is an integer. Using $C$, we will construct a new circuit $C^{\prime}$ with depth $d^{\prime}=d+3$ and size $S^{\prime}=O\left(nS\right)+\operatorname*{poly}\left(n\right)$, which accepts all strings $x\in\left\{0,1\right\}^{n}$ of Hamming weight $n/2+1$, and rejects all strings of Hamming weight $n/2$. By Theorem 11, this will imply that the original circuit $C$ must have had size

So fix an input $x\in\left\{0,1\right\}^{n}$, and suppose we choose $m$ bits $x_{i_{1}},\ldots,x_{i_{m}}$ from $x$, with each index $i_{j}$ chosen uniformly at random with replacement. Call the resulting $m$-bit string $y$. Observe that if $x$ had Hamming weight $n/2$, then $y$ will be distributed according to $\mathcal{U}\left[0\right]$, while if $x$ had Hamming weight $n/2+1$, then $y$ will be distributed according to $\mathcal{U}\left[\varepsilon\right]$. So by assumption,

for some constants $\alpha$ and \delta\neq 0\(we can assume $\delta>0$ without loss of generality).

Now suppose we repeat the above experiment $T=kn$ times, for some constant $k=k\left(\alpha,\delta\right)$. That is, we create $T$ strings $y_{1},\ldots,y_{T}$ by choosing random bits of $x$, so that each $y_{i}$ is distributed independently according to either $\mathcal{U}\left[0\right]$ or $\mathcal{U}\left[\varepsilon\right]$. We then apply $C$ to each $y_{i}$. Let

be the number of $C$ invocations that accept. Then by a Chernoff bound, if $\left|x\right|=n/2$ then

By taking $k$ large enough, we can make both of these probabilities less than $2^{-n}$. By the union bound, this implies that there must exist a way to choose $y_{1},\ldots,y_{T}$ so that

for every $x$ with $\left|x\right|\in\left\{n/2,n/2+1\right\}$ simultaneously. In forming the circuit $C^{\prime}$, we simply hardwire that choice.

The last step is to decide whether $Z\leq\alpha T+\frac{\delta}{3}T$ or $Z\geq\alpha T+\frac{2\delta}{3}T$. This can be done using an $\mathsf{AC}^{0}$ circuit for the Approximate Majority problem (see Viola for example), which has depth $3$ and size $\operatorname*{poly}\left(T\right)$. The end result is a circuit $C^{\prime}$ to distinguish $\left|x\right|=n/2$ from $\left|x\right|=n/2+1$, which has depth $d+3$ and size $TS+\operatorname*{poly}\left(T\right)=O\left(nS\right)+\operatorname*{poly}\left(n\right)$.

2 Secretly Biased Fourier Coefficients

In this section, we prove two lemmas indicating that one can slightly bias one of the Fourier coefficients of a random Boolean function $f:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$, and yet still have $f$ be information-theoretically indistinguishable from a random Boolean function (so that, in particular, an adversary has no way of knowing which Fourier coefficient was biased). These lemmas will play a key role in our reduction from $\varepsilon$-Bias Detection to Fourier Fishing.

Fix a string $s\in\left\{0,1\right\}^{n}$. Let $\mathcal{A}\left[s\right]$ be the probability distribution over functions $f:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$ where each $f\left(x\right)$ is $1$ with independent probability $\frac{1}{2}+\left(-1\right)^{s\cdot x}\frac{1}{2\sqrt{N}}$, and let $\mathcal{B}\left[s\right]$ be the distribution where each $f\left(x\right)$ is $1$ with independent probability $\frac{1}{2}-\left(-1\right)^{s\cdot x}\frac{1}{2\sqrt{N}}$. Then let $\mathcal{D}\left[s\right]=\frac{1}{2}\left(\mathcal{A}\left[s\right]+\mathcal{B}\left[s\right]\right)$ (that is, an equal mixture of $\mathcal{A}\left[s\right]$ and $\mathcal{B}\left[s\right]$).

Suppose Alice chooses $s\in\left\{0,1\right\}^{n}$ uniformly at random, then draws $f$ according to $\mathcal{D}\left[s\right]$. She keeps $s$ secret, but sends the truth table of $f$ to Bob. After examining $f$, Bob outputs a string $z$ such that $\left|\widehat{f}\left(z\right)\right|\geq\beta$. Then

where the probability is over all runs of the protocol.

Proof. By Yao’s principle, we can assume without loss of generality that Bob’s strategy is deterministic. For each $z$, let $\mathcal{F}\left[z\right]$ be the set of all $f$’s that cause Bob to output $z$. Then the first step is to lower-bound $\Pr_{\mathcal{D}\left[z\right]}\left[f\right]$, for some fixed $z$ and $f\in\mathcal{F}\left[z\right]$. Let $N_{f}\left[z\right]$ be the number of inputs $x\in\left\{0,1\right\}^{n}$ such that $f\left(x\right)=\left(-1\right)^{z\cdot x}$. It is not hard to see that $N_{f}\left[z\right]=\frac{N}{2}+\frac{\sqrt{N}\widehat{f}\left(z\right)}{2}$. So

Here the second-to-last line takes the limit as $N\rightarrow\infty$, while the last line follows from the assumption $\left|\widehat{f}\left(z\right)\right|\geq\beta$, together with the fact that $e^{y}+e^{-y}$ increases monotonically away from $y=0$.

Now let $\mathcal{D}=\operatorname*{E}_{s}\left[\mathcal{D}\left[s\right]\right]$ (that is, an equal mixture of all the $\mathcal{D}\left[s\right]$’s). We claim that $\mathcal{D}$ is extremely close in variation distance to $\mathcal{U}$, the uniform distribution over all Boolean functions $f:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$.

$\left\|\mathcal{D}-\mathcal{U}\right\|\leq\frac{e-1}{2\sqrt{2eN}}$.

Proof. By a calculation from Lemma 13, for all $f$ and $s$ we have

Clearly $\operatorname*{E}_{f}\left[\Pr_{\mathcal{D}}\left[f\right]\right]=1/2^{N}$. Our goal is to upper-bound the variance $\operatorname*{Var}_{f}\left[\Pr_{\mathcal{D}}\left[f\right]\right]$, which measures the distance from $\mathcal{D}$ to the uniform distribution. In the limit of large $N$, we have

An immediate corollary of Lemma 14 is that, if a Fourier Fishing algorithm succeeds with probability $p$ on $\left\langle f_{1},\ldots,f_{n}\right\rangle$ drawn from $\mathcal{U}^{n}$, then it also succeeds with probability at least

on $\left\langle f_{1},\ldots,f_{n}\right\rangle$ drawn from $\mathcal{D}^{n}$.

3 Putting It All Together

Using the results of Sections 4.1 and 4.2, we are now ready to prove a lower bound on the constant-depth circuit complexity of Fourier Fishing.

Any depth-$d$ circuit that solves the Fourier Fishing problem, with probability at least $0.99$ over $f_{1},\ldots,f_{n}$ chosen uniformly at random, has size $\exp\left(\Omega\left(N^{1/\left(2d+8\right)}\right)\right)$.

Proof. Let $C$ be a circuit of depth $d$ and size $s$. Let $G$ be the set of all $\left\langle f_{1},\ldots,f_{n}\right\rangle$ on which $C$ succeeds: that is, for which it outputs $z_{1},\ldots,z_{n}$, at least 75% of which satisfy $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 1$ and at least 25% of which satisfy $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq 2$. Suppose

Using the above fact, we will convert $C$ into a new circuit $C^{\prime}$ that solves the $\varepsilon$-Bias Detection problem of Corollary 12, with $\varepsilon:=\frac{1}{2\sqrt{N}}$. This $C^{\prime}$ will have depth $d^{\prime}=d+2$ and size $S^{\prime}=O\left(NS\right)$. By Corollary 12, this will imply that $C$ itself must have had size

Let $M=N^{2}n$, and let $R=r_{1}\ldots r_{M}\in\left\{0,1\right\}^{M}$ be a string of bits where each $r_{j}$ is $1$ with independent probability $p$. We want to decide whether $p=1/2$ or $p=1/2+\varepsilon$—that is, whether $R$ was drawn from $\mathcal{U}\left[0\right]$ or $\mathcal{U}\left[\varepsilon\right]$. We can do this as follows. First, choose strings $s_{1},\ldots,s_{n}\in\left\{0,1\right\}^{n}$, bits $b_{1},\ldots,b_{n}\in\left\{0,1\right\}$, and an integer $k\in\left[n\right]$ uniformly at random. Next, define Boolean functions $f_{1},\ldots,f_{n}:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$ using the first $Nn$ bits of $R$, like so:

Finally, feed $\left\langle f_{1},\ldots,f_{n}\right\rangle$ as input to $C$, and consider $z_{k}$, the $k^{th}$ output of $C$ (discarding the other $n-1$ outputs). We are interested in $\Pr\left[z_{k}=s_{k}\right]$, where the probability is over $R$, $s_{1},\ldots,s_{n}$, $b_{1},\ldots,b_{n}$, and $k$.

If $p=1/2$, notice that $f_{1},\ldots,f_{n}$ are independent and uniformly random regardless of $s_{1},\ldots,s_{n}$. So $C$ gets no information about $s_{k}$, and $\Pr\left[z_{k}=s_{k}\right]=1/N$.

On the other hand, if $p=1/2+\varepsilon$, then each $f_{i}$ is drawn independently from the distribution $\mathcal{D}\left[s\right]$ studied in Lemma 13. So by the Lemma, for every $i\in\left[n\right]$, if $\left|\widehat{f}_{i}\left(z_{i}\right)\right|\geq\beta$ then

So assuming $C$ succeeds (that is, $\left\langle f_{1},\ldots,f_{n}\right\rangle\in G$), we have

So for a random $\left\langle f_{1},\ldots,f_{n}\right\rangle$ drawn according to $\mathcal{D}^{n}$,

Notice that this is bounded above $1/N$ by a multiplicative constant.

Now let us repeat the above experiment $N$ times. That is, for all $j:=1$ to $N$, we generate Boolean functions $f_{j1},\ldots,f_{jn}:\left\{0,1\right\}^{n}\rightarrow\left\{-1,1\right\}$ by the same probabilistic procedure as before, but each time using a new $Nn$-bit substring of $R_{j}$ of $R$, as well as new $s$, $b$, and $k$ values (denoted $s_{j1},\ldots,s_{jn}$, $b_{j1},\ldots,b_{jn}$, and $k_{j}$). We then apply $C$ to each $n$-tuple $\left\langle f_{j1},\ldots,f_{jn}\right\rangle$. Let $z_{j}$ be the $k_{j}^{th}$ string that $C$ outputs when run on $\left\langle f_{j1},\ldots,f_{jn}\right\rangle$. Then by the above, for each $j\in\left[N\right]$ we have