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Separations in query complexity using cheat sheets

Scott Aaronson
MIT
aaronson@csail.mit.edu
   Shalev Ben-David
MIT
shalev@mit.edu
   Robin Kothari
MIT
rkothari@mit.edu
Abstract

We show a power 2.52.5 separation between bounded-error randomized and quantum query complexity for a total Boolean function, refuting the widely believed conjecture that the best such separation could only be quadratic (from Grover’s algorithm).  We also present a total function with a power 44 separation between quantum query complexity and approximate polynomial degree, showing severe limitations on the power of the polynomial method.  Finally, we exhibit a total function with a quadratic gap between quantum query complexity and certificate complexity, which is optimal (up to log factors).  These separations are shown using a new, general technique that we call the cheat sheet technique.  The technique is based on a generic transformation that converts any (possibly partial) function into a new total function with desirable properties for showing separations.  The framework also allows many known separations, including some recent breakthrough results of Ambainis et al., to be shown in a unified manner.

1 Introduction

Query complexity (or decision tree complexity) is a model of computation that allows us to examine the strengths and weaknesses of resources such as access to randomness, nondeterminism, quantum computation and more.  As opposed to the Turing machine model where we can only conjecture that certain resources exponentially speed up computation, in this model such beliefs can be proved.  In particular, the query model is a natural setting to describe several well-known quantum algorithms, such as Grover’s algorithm [Gro96] and quantum walks [Amb07]; even Shor’s algorithm for integer factorization [Sho97] is based on a quantum query algorithm.

In the query model we measure a problem’s complexity, called its query complexity, by the minimum number of input bits that have to be read to solve the problem on any input.  For a function ff, we use D(f)D(f), R(f)R(f), and Q(f)Q(f) to denote the query complexity of computing ff with a deterministic, bounded-error randomized, and bounded-error quantum algorithm respectively.

For a total Boolean function f:{0,1}n{0,1}f:\{0,1\}^{n}\to\{0,1\}, we know that these measures are polynomially related: in one direction, we have Q(f)R(f)D(f)Q(f)\leq R(f)\leq D(f) as each model is more powerful than the next, and in the other direction, we have D(f)=O(R(f)3)D(f)=O(R(f)^{3}) [Nis91] and D(f)=O(Q(f))6D(f)=O(Q(f))^{6} [BBC+01].

Until recently, the best known separation between D(f)D(f) and R(f)R(f) was the Nand-tree function [SW86], which satisfies D(f)=Ω(R(f)1.3267)D(f)=\Omega(R(f)^{1.3267}) and even D(f)=Ω(R0(f)1.3267)D(f)=\Omega(R_{0}(f)^{1.3267}), where R0(f)R_{0}(f) denotes zero-error randomized query complexity.  The best known separation between D(f)D(f) or R(f)R(f) versus Q(f)Q(f) was only quadratic, achieved by the Or function: D(Or)R(Or)=Ω(n)D(\textsc{Or})\geq R(\textsc{Or})=\Omega(n) and Q(Or)=Θ(n)Q(\textsc{Or})=\Theta(\sqrt{n}) [Gro96, BBBV97].  No separation was known between R0(f)R_{0}(f) and R(f)R(f).  Furthermore, it was believed that all of these separations were optimal.

These beliefs were shattered by a recent breakthrough by Ambainis, Balodis, Belovs, Lee, Santha, and Smotrovs [ABB+15], who built upon the techniques of Göös, Pitassi, and Watson [GPW15], and showed the near-optimal separations D(f)=Ω~(R0(f)2)D(f)=\widetilde{\Omega}(R_{0}(f)^{2}) and R0(f)=Ω~(R(f)2)R_{0}(f)=\widetilde{\Omega}(R(f)^{2}).  Additionally, they presented new separations between quantum and classical models, exhibiting functions with D(f)=Ω~(Q(f)4)D(f)=\widetilde{\Omega}(Q(f)^{4}) and R0(f)=Ω~(Q(f)3)R_{0}(f)=\widetilde{\Omega}(Q(f)^{3}).  However, no super-quadratic separation was shown between a quantum and classical model when they are both allowed to err in the same way (e.g., bounded error, zero error, etc.)  For example, no super-quadratic separation was shown between Q(f)Q(f) and its classical analogue, R(f)R(f).

1.1 Results

Randomized versus quantum query complexity

We present the first super-quadratic separation between quantum and classical query complexity when both models are allowed bounded error.  This is a counterexample to a conjecture that had been widely believed about the nature of quantum speed-ups: that if the function is total, so that the inputs are not handpicked to orchestrate a quantum speed-up, then the best quantum advantage possible is the quadratic speed-up achieved by Grover’s algorithm.

Theorem 1.

There exists a total function ff such that R(f)=Ω~(Q(f)2.5)R(f)=\widetilde{\Omega}(Q(f)^{2.5}).

Theorem 1 can be boosted to a cubic separation, i.e., R(f)=Ω~(Q(f)3)R(f)=\widetilde{\Omega}(Q(f)^{3}), in a completely black-box manner if there exists a partial function (a function defined only on a subset of {0,1}n\{0,1\}^{n}) with Ω~(n)\widetilde{\Omega}(n) randomized query complexity but only poly(log(n))\operatorname{poly}(\log(n)) quantum query complexity (the best possible separation between these measures up to log factors).  It is conjectured that a recently studied partial function called kk-fold Forrelation achieves this separation [AA15].

Quantum query complexity versus polynomial degree

Nisan and Szegedy [NS95] introduced two new measures of a Boolean function ff called the degree and approximate degree, denoted deg(f)\deg(f) and deg~(f)\operatorname{\widetilde{\deg}}(f) respectively.  The (approximate) degree of a function ff is the minimum degree of a polynomial over the input variables that (approximately) equals f(x)f(x) at every input x{0,1}nx\in\{0,1\}^{n}.  They introduced these measures to lower bound classical query complexity, showing that deg(f)D(f)\deg(f)\leq D(f) and deg~(f)R(f)\operatorname{\widetilde{\deg}}(f)\leq R(f).  It turns out that these measures also lower bound the corresponding quantum measures, since deg(f)2QE(f)\deg(f)\leq 2\,Q_{E}(f) and deg~(f)2Q(f)\operatorname{\widetilde{\deg}}(f)\leq 2\,Q(f) [BBC+01], where QE(f)Q_{E}(f) denotes the exact quantum query complexity of ff.

Approximate degree has proved to be a fruitful lower bound technique for quantum query complexity, especially for problems like the collision and element distinctness problems [AS04], where it is known that the original quantum adversary method of Ambainis, another commonly used lower bound technique, cannot show nontrivial lower bounds.

For any lower bound technique, it is natural to ask whether there exist functions where the technique fails to prove a tight lower bound.  Answering this question, Ambainis showed that the approximate degree of a function can be asymptotically smaller than its quantum query complexity by exhibiting a function with Q(f)=Ω(deg~(f)1.3219)Q(f)=\Omega(\operatorname{\widetilde{\deg}}(f)^{1.3219}) [Amb03].  We dramatically strengthen this separation to obtain nearly a 4th4^{\mathrm{th}} power gap.

Theorem 2.

There exists a total function ff such that Q(f)deg~(f)4o(1)Q(f)\geq\operatorname{\widetilde{\deg}}(f)^{4-o(1)}.

Theorem 2 is optimal assuming the conjecture that D(f)=O(bs(f)2)D(f)=O(\operatorname{bs}(f)^{2}), where bs(f)\operatorname{bs}(f) is block sensitivity (to be defined in Section 2.2).

The cheat sheet technique

These separations are shown using a new technique for proving separations between query measures, which we call the cheat sheet technique.  Our technique is based on a generic transformation that takes any (partial or total) function and transforms it into a “cheat sheet version” of the function that has desirable properties for proving separations.

While the strategy is inspired by the recent breakthrough results [GPW15, ABB+15] (and bears some similarity to older works [LG06, AdW14]), it represents a more general approach: it provides a framework for proving separations and allows many separations to be shown in a unified manner.  Thus the task of proving separations is reduced to the simpler task of finding the right function to plug into this construction.  For example, it can be used to convert a partial function separation between two models into a weaker total function separation.

In this paper we demonstrate the power of the cheat sheet technique by using it to exhibit several new total function separations in query complexity.

Other separations

On the path to proving Theorem 2, we show several new separations.  First, we quadratically separate quantum query complexity from certificate complexity, which is essentially optimal since Q(f)C(f)2Q(f)\leq C(f)^{2}.

Theorem 3.

There exists a total function ff such that Q(f)=Ω~(C(f)2)Q(f)=\widetilde{\Omega}(C(f)^{2}).

Besides being a stepping stone to proving Theorem 2, the question of Q(f)Q(f) versus C(f)C(f) has been studied because it is known that the original adversary method of Ambainis (also known as the positive-weights adversary method) cannot prove a lower bound greater than C(f)C(f) [ŠS06, HLŠ07].

Plugging the function of Theorem 3 into the cheat sheet framework directly yields a function whose quantum query complexity is quadratically larger than its (exact) degree, improving the recent result of [GPW15], who exhibited a function with D(f)=Ω~(deg(f)2)D(f)=\widetilde{\Omega}(\deg(f)^{2}).

Theorem 4.

There exists a total function ff such that Q(f)=Ω~(deg(f)2)Q(f)=\widetilde{\Omega}(\deg(f)^{2}).

This theorem works in a very black-box way: any separation between a measure like Q(f)Q(f) or R(f)R(f) and C(f)C(f) can be plugged into the cheat sheet technique to obtain the same separation (up to log factors) between that measure and exact degree.  For example, since the And-Or function satisfies R(f)=Ω(C(f)2)R(f)=\Omega(C(f)^{2}), the cheat sheet version of And-Or satisfies R(f)=Ω~(deg(f)2)R(f)=\widetilde{\Omega}(\deg(f)^{2}).

This result also shows limitations on using deg(f)\deg(f) to lower bound QE(f)Q_{E}(f), since deg(f)\deg(f) can sometimes be quadratically smaller than QE(f)Q_{E}(f) and can even be quadratically smaller than Q(f)Q(f).

Summary of results

We summarize our new results in the table below.

Separation achieved Known relation Result
R(f)=Ω~(Q(f)2.5)R(f)=\widetilde{\Omega}(Q(f)^{2.5}) R(f)=O(Q(f)6)R(f)=O(Q(f)^{6}) Theorem 1
Q(f)deg~(f)4o(1)Q(f)\geq\operatorname{\widetilde{\deg}}(f)^{4-o(1)} Q(f)=O(deg~(f)6)Q(f)=O(\operatorname{\widetilde{\deg}}(f)^{6}) Theorem 2
Q(f)=Ω~(C(f)2)Q(f)=\widetilde{\Omega}(C(f)^{2}) Q(f)=O(C(f)2)Q(f)=O(C(f)^{2}) Theorem 3
Q(f)=Ω~(deg(f)2)Q(f)=\widetilde{\Omega}(\deg(f)^{2}) Q(f)=O(deg(f)3)Q(f)=O(\deg(f)^{3}) Theorem 4
Table 1: New separations shown in this paper

We are also able to use the cheat sheet technique to reprove many of the query separations of [GPW15, ABB+15, GJPW15].  Some of their results are subsumed by the results in Table 1.  We also prove R0(f)=Ω~(Q(f)3)R_{0}(f)=\widetilde{\Omega}(Q(f)^{3}) (Theorem 23) and R(f)=Ω~(QE(f)3/2)R(f)=\widetilde{\Omega}(Q_{E}(f)^{3/2}) (Theorem 24) in Section 6.4.  For more details, see Table 2 in which we summarize our results and the best known separations and relations between query measures.

It is also worth pointing out what we are unable to reproduce with our framework. The only separations from these papers that we do not reproduce are those that make essential use of “back pointers”.  Reproducing these separations in the cheat sheet framework is an interesting direction for future research.

Table 2: Best known separations between complexity measures
DD R0R_{0} RR CC RC\operatorname{RC} bs\operatorname{bs} QEQ_{E} deg\deg QQ deg~\operatorname{\widetilde{\deg}}
DD
2, 2
[ABB+15]
2*, 3
[ABB+15]
2, 2
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2, 3
[ABB+15]
2, 3
[GPW15]
4*, 6
[ABB+15]
4*, 6
[ABB+15]
R0R_{0}
1, 1
\oplus
2, 2
[ABB+15]
2, 2
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2, 3
[ABB+15]
2, 3
[GJPW15]
3, 6
[ABB+15]
4*, 6
[ABB+15]
RR
1, 1
\oplus
1, 1
\oplus
2, 2
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2*, 3
\wedge\circ\vee
1.5, 3
[ABB+15]
2, 3
[GJPW15]
2.5, 6
Th. 1
4*, 6
[ABB+15]
CC
1, 1
\oplus
1, 1
\oplus
1, 2
\oplus
2, 2
[GSS13]
2, 2
[GSS13]
1.1527, 3
[Amb13]
log36\log_{3}6, 3
[NW95]
2, 4
\wedge
2, 4
\wedge
RC\operatorname{RC}
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
1.5, 2
[GSS13]
1.1527, 2
[Amb13]
log36\log_{3}6, 2
[NW95]
2, 2
\wedge
2, 2
\wedge
bs\operatorname{bs}
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
1.1527, 2
[Amb13]
log36\log_{3}6, 2
[NW95]
2, 2
\wedge
2, 2
\wedge
QEQ_{E}
1, 1
\oplus
1.3267, 2
¯\bar{\wedge}-tree
1.3267, 3
¯\bar{\wedge}-tree
2, 2
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2, 3
Th. 4
2, 6
\wedge
4*, 6
Th. 2
deg\deg
1, 1
\oplus
1.3267, 2
¯\bar{\wedge}-tree
1.3267, 3
¯\bar{\wedge}-tree
2, 2
\wedge\circ\vee
2*, 3
\wedge\circ\vee
2*, 3
\wedge\circ\vee
1, 1
\oplus
2, 6
\wedge
2, 6
\wedge
QQ
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
2, 2
Th. 3
2*, 3
Th. 3
2*, 3
Th. 3
1, 1
\oplus
2, 3
Th. 4
4*, 6
Th. 2
deg~\operatorname{\widetilde{\deg}}
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus
7/6, 2
Ed\wedge\circ\textsc{Ed}
7/6, 3
Ed\wedge\circ\textsc{Ed}
7/6, 3
Ed\wedge\circ\textsc{Ed}
1, 1
\oplus
1, 1
\oplus
1, 1
\oplus

An entry a,ba,b in the row M1M_{1} and column M2M_{2} roughly111More precisely it means acrit(M1,M2)ba\leq\operatorname{crit}(M_{1},M_{2})\leq b, where we define crit(,)\operatorname{crit}(\cdot,\cdot) in Section 2. means M1(f)=O~(M2(f)b)M_{1}(f)=\widetilde{O}(M_{2}(f)^{b}) for all total ff and there exists a total ff with M1(f)=Ω~(M2(f)a)M_{1}(f)=\widetilde{\Omega}(M_{2}(f)^{a}).  Each cell contains a citation or a description of a separating function, where =Parity\oplus=\textsc{Parity}, =And\wedge=\textsc{And}, =And-Or\wedge\circ\vee=\textsc{And-Or}, ¯\bar{\wedge}-tree is the balanced Nand-tree function [SW86, San95], and Ed\wedge\circ\textsc{Ed} is the And function composed with Element Distinctness.222The Element Distinctness function accepts a list of nn numbers in [n][n] as input and asks if any two of them are equal.  The certificate complexity of the Andn\textsc{And}_{n} composed with element distinctness on nn bits is O~(n)\widetilde{O}(n), by composing certificates for the two functions.  However, its approximate degree is Ω~(n7/6)\widetilde{\Omega}(n^{7/6}), which follows from noting that its (negative) one-sided approximate degree is Ω(n2/3)\Omega(n^{2/3}) [BT15] and that composition with the Andn\textsc{And}_{n} function raises its approximate degree by a factor of n\sqrt{n} [BT15, She13].  (We thank Mark Bun for outlining this proof.) Entries followed by a star (e.g., 2*) correspond to separations that are optimal if D(f)=O(bs(f)2)D(f)=O(\operatorname{bs}(f)^{2}).  Separations colored red are new to this work. Separations colored gray are separations from recent papers that we reprove in this work.

1.2 Overview of techniques

Cheat sheet technique and randomized versus quantum query complexity

While we know large partial function separations between randomized and quantum query complexity, such as Simon’s problem [Sim97] or the Forrelation problem [AA15] that satisfies R(g)=Ω~(n)R(g)=\widetilde{\Omega}(\sqrt{n}) and Q(g)=1Q(g)=1, it is unclear how to make these functions total.  We could simply define gg to be 0 on inputs not in the domain, but then the quantum algorithm would need to be able to decide whether an input is in the domain.

To solve this problem, we compose the Forrelation problem gg with a total function h=And-Orh=\textsc{And-Or} on n2n^{2} bits, to obtain a partial function f=ghf=g\circ h on n3n^{3} bits that inherits the properties of both gg and hh.  Since And-Or has low certificate complexity, it is easier to certify that an input lies in the domain of ghg\circ h, since we only need to certify the outputs of the nn different hh gates.

Since R(And-Orn2)=Ω(n2)R(\textsc{And-Or}_{n^{2}})=\Omega(n^{2}), it can be shown that R(f)=R(gh)=Ω(n2.5)R(f)=R(g\circ h)=\Omega(n^{2.5}).  However, since Q(And-Orn2)=O(n)Q(\textsc{And-Or}_{n^{2}})=O(n), we have Q(f)=O(n)Q(f)=O(n).  Now ff is still a partial function, but there is a small certificate for the input being in the domain.  If we could ensure that only the quantum algorithm had access to this certificate, but the randomized algorithm did not, we would have our power 2.52.5 separation.

To achieve this, we hide a “cheat sheet” in the input that contains all the information the quantum algorithm needs.  We store it in a vast array of size n10n^{10} of potential cheat sheets, which cannot be searched quickly by brute force by any algorithm, classical or quantum.  Thus, we must provide the quantum algorithm the address of the correct cheat sheet.  But what information do we have that is known only to the quantum algorithm?  The value of ff on the input!  While one bit does not help much, we can use 10logn10\log n copies of ff acting on 10logn10\log n different inputs.  The outputs to these 10logn10\log n problems can index into an array of size n10n^{10}, and can be determined by the quantum algorithm using only O(nlogn)O(n\log n) queries.  At this index we will store the cheat sheet, which contains certificates for all 10logn10\log n instances of ff, convincing us that all the inputs lie in the domain.  Our cheat sheet function now evaluates to 11 if and only if all 10logn10\log n instances of ff lie in the domain, and the cell in the array pointed to by the outputs of these 10logn10\log n instances of ff contains a cheat sheet certifying that these inputs lie in the domain.

Quantum query complexity versus certificate complexity

Consider the k-sumk\textsc{-sum} problem, in which we are given nn numbers and have to decide if any kk of them sum to 0(modM)0\pmod{M} for some MM.  For large kk, the quantum query complexity of k-sumk\textsc{-sum} is nearly linear in nn [BŠ13].  While it is easy to certify 11-inputs by showing any kk elements that sum to 0, the 0-inputs are difficult to certify and hence the certificate complexity is also linear in nn.

Building on k-sumk\textsc{-sum}, we define a new function that we call Block k-sumk\textsc{-sum}, whose quantum query complexity and certificate complexity are both linear in nn.  However, Block k-sumk\textsc{-sum} has a curious property: for both 11-inputs and 0-inputs, the certificates themselves consist almost exclusively of input bits set to 11.  This means that if we compose this function with k-sumk\textsc{-sum}, the composed function on n2n^{2} bits has certificates of size O~(n)\widetilde{O}(n) because any certificate of Block k-sumk\textsc{-sum} consists almost entirely of 11s, which correspond to k-sumk\textsc{-sum} instances that output 11, which are easy to certify.  On the other hand, the quantum query complexity of the composed function, which we call BKK for Block k-sumk\textsc{-sum} of k-sumk\textsc{-sum}, is the product of the quantum query complexities of individual functions.  Thus the certificate complexity of BKK is O~(n)\widetilde{O}(n), but its quantum query complexity is Ω~(n2)\widetilde{\Omega}(n^{2}).

Quantum query complexity versus polynomial degree

We show that plugging any function that achieves Q(f)=Ω~(n2)Q(f)=\widetilde{\Omega}(n^{2}) and C(f)=O~(n)C(f)=\widetilde{O}(n) into the cheat sheet framework yields a function with Q(f)=Ω~(n2)Q(f)=\widetilde{\Omega}(n^{2}) and deg(f)=O~(n)\deg(f)=\widetilde{O}(n).  The quantum lower bound uses the hybrid method [BBBV97] and the recent strong direct product theorem for quantum query complexity [LR13].  The degree upper bound holds because for every potential cheat sheet location, there exists a degree O~(n)\widetilde{O}(n) polynomial that checks if this location is the one pointed to by the given input.  And since the input can point to at most one cheat sheet, the sum of these polynomials equals the function ff.

To achieve a fourth power separation between Q(f)Q(f) and deg~(f)\operatorname{\widetilde{\deg}}(f), we need to check whether a cheat sheet is valid using a polynomial of degree O~(C(f))\widetilde{O}(\sqrt{C(f)}).  Some certificates can be checked by Grover’s algorithm in time O~(C(f))\smash{\widetilde{O}}(\sqrt{C(f)}), which would yield an approximating polynomial of similar degree, but this is not true for all functions ff.  To remedy this, we construct a new function based on composing BKK with itself recursively logn\log n times, to obtain a new function we call RecBKK, and show that certificates for RecBKK can be checked quickly by a quantum algorithm.

2 Preliminaries

For any positive integer nn, let [n]:={1,,n}[n]:=\{1,\ldots,n\}.  We use f(n)=O~(g(n))f(n)=\widetilde{O}(g(n)) to mean there exists a constant kk such that f(n)=O(g(n)logkn)f(n)=O(g(n)\log^{k}n).  For example, f(n)=O~(1)f(n)=\widetilde{O}(1) means f(n)=O(logkn)f(n)=O(\log^{k}n) for some constant kk.  Similarly, f(n)=Ω~(g(n))f(n)=\widetilde{\Omega}(g(n)) denotes f(n)=Ω(g(n)/logkn)f(n)=\Omega(g(n)/\log^{k}n) for some constant kk.  Finally, f(n)=Θ~(g(n))f(n)=\widetilde{\Theta}(g(n)) means f(n)=O~(g(n))f(n)=\widetilde{O}(g(n)) and f(n)=Ω~(g(n))f(n)=\widetilde{\Omega}(g(n)).

2.1 Boolean functions

Let ff be function from a domain DΣnD\subseteq\Sigma^{n} to {0,1}\{0,1\}, where Σ\Sigma is some finite set.  Usually Σ={0,1}\Sigma=\{0,1\}, and we assume this for simplicity below, although most complexity measures are easily generalized to larger input alphabets.  A function ff is called a total function if D=ΣnD=\Sigma^{n}. Otherwise, we say ff is a partial function or a promise problem and refer to DD as the promised set or the promise.

\Treegghhz1z_{1}\edge\cdotszmz_{m}\edge\edge\cdots\edge\cdots\edge\cdotshhz(n1)m+1z_{(n-1)m+1}\edge\cdotsznmz_{nm}
Figure 1: Composed function ghg\circ h.

For any two (possibly partial) Boolean functions g:G{0,1}g:G\to\{0,1\}, where G{0,1}nG\subseteq\{0,1\}^{n}, and h:H{0,1}h:H\to\{0,1\}, where H{0,1}mH\subseteq\{0,1\}^{m}, we can define the composed function gh:D{0,1}g\circ h:D\to\{0,1\}, where D{0,1}nmD\subseteq\{0,1\}^{nm}, as follows.  For an input (z1,z2,,znm){0,1}nm(z_{1},z_{2},\ldots,z_{nm})\in\{0,1\}^{nm}, we define ghg\circ h as follows (as depicted in Figure 1):

gh(z):=g(h(z1,,zm),h(zm+1,,z2m),,h(z(n1)m+1,,znm)),g\circ h(z):=g(h(z_{1},\ldots,z_{m}),h(z_{m+1},\ldots,z_{2m}),\ldots,h(z_{(n-1)m+1},\ldots,z_{nm})), (1)

where the function ghg\circ h is only defined on zz where the inputs to the hh gates lie in the domain HH and the nn-bit input to gg, (h(z1,,zm),h(zm+1,,z2m),,h(z(n1)m+1,,znm))(h(z_{1},\ldots,z_{m}),h(z_{m+1},\ldots,z_{2m}),\ldots,h(z_{(n-1)m+1},\ldots,z_{nm})), lies in GG.

Some Boolean functions that we use in this paper are Andn\textsc{And}_{n} and Orn\textsc{Or}_{n}, the And and Or functions respectively on nn bits, and And-Orn2:{0,1}n2{0,1}\textsc{And-Or}_{n^{2}}:\{0,1\}^{n^{2}}\to\{0,1\}, defined as AndnOrn\textsc{And}_{n}\circ\textsc{Or}_{n}.

2.2 Complexity measures

Formal definitions of most measures introduced here may be found in the survey on query complexity by Buhrman and de Wolf [BdW02] and in the recent paper of Ambainis et al. [ABB+15].

In the model of query complexity, algorithms may only access the input by asking queries to an oracle that knows the input xDx\in D.  The queries are of the form “what is xix_{i}?” for some i[n]i\in[n] and the oracle responds with the value of xiΣx_{i}\in\Sigma.  The goal is to minimize the number of queries made to compute ff.

We use D(f)D(f) to denote the deterministic query complexity of computing ff, the minimum number of queries that have to be made by a deterministic algorithm that outputs f(x)f(x) on every input xx.  We use R(f)R(f) to denote bounded-error randomized query complexity, the minimum number of queries made by a randomized algorithm that outputs f(x)f(x) on input xx with probability at least 2/32/3.  We use R0(f)R_{0}(f) to denote zero-error randomized query complexity, the minimum number of queries made by a randomized algorithm that must output either the correct answer, f(x)f(x), on input xx or output *, indicating that it does not know the answer.  However, the probability (over the internal randomness of the algorithm) that it outputs * should be at most 1/21/2 on any xDx\in D.

Similarly, we study quantum analogues of these measures.  The quantum analogue of D(f)D(f) is exact quantum query complexity, denoted QE(f)Q_{E}(f), the minimum number of queries made by a quantum algorithm that outputs f(x)f(x) on every input xDx\in D with probability 1.  The quantum analogue of R(f)R(f) is bounded-error quantum query complexity, denoted Q(f)Q(f), where the algorithm is only required to be correct with probability at least 2/32/3.

We also study some measures related to the complexity of certifying the output.  For a function ff, let C1(f)C_{1}(f) denote the minimum number of bits of a 11-input, i.e., an input with f(x)=1f(x)=1, that have to be revealed to convince a verifier that f(x)=1f(x)=1.  Alternately, imagine the verifier is allowed to interact with an untrustworthy prover who claims that f(x)=1f(x)=1, and is allowed to provide an arbitrarily long witness for this.  After receiving the witness, the verifier makes at most C1(f)C_{1}(f) queries to the input and decides whether to accept that f(x)=1f(x)=1 or not.  We require that if f(x)=1f(x)=1, then there is some witness that makes the verifier accept and if f(x)=0f(x)=0 then no witness makes the verifier accept.  Similarly we define C0(f)C_{0}(f) as the analogous notion for 0-inputs.  Finally the certificate complexity of ff, or C(f)C(f), is defined as max{C0(f),C1(f)}\max\{C_{0}(f),C_{1}(f)\}.  Similarly, if the verifier is randomized and only needs to accept valid proofs and reject invalid proofs with probability greater than 2/32/3, we get the bounded-error randomized analogues RC0(f)\operatorname{RC}_{0}(f), RC1(f)\operatorname{RC}_{1}(f), and RC(f):=max{RC0(f),RC1(f)}\operatorname{RC}(f):=\max\{\operatorname{RC}_{0}(f),\operatorname{RC}_{1}(f)\} [Aar06].

We also study a combinatorial measure of a function called block sensitivity, denoted bs(f)\operatorname{bs}(f).  For a function ff and input xx, a block of bits B[n]B\subseteq[n] is said to be sensitive for xx if the input obtained by complementing all the bits xix_{i} for iBi\in B yields an input yDy\in D with f(x)f(y)f(x)\neq f(y).  The block sensitivity of ff on input xx is the maximum number of disjoint blocks sensitive for xx.  The block sensitivity of ff is the maximum block sensitivity over all inputs xDx\in D.

Lastly we study measures related to representing a function with a polynomial with real coefficients.  The (exact) degree of ff, denoted deg(f)\deg(f), is the minimum degree of a real polynomial pp over the variables x1,x2,,xnx_{1},x_{2},\ldots,x_{n}, such that p(x1,x2,,xn)=f(x)p(x_{1},x_{2},\ldots,x_{n})=f(x) for all xDx\in D.  Similarly, the approximate degree of ff, denoted deg~(f)\operatorname{\widetilde{\deg}}(f), is the minimum degree of a polynomial that approximates the value of f(x)f(x) on every input xDx\in D, i.e., for all xDx\in D the polynomial satisfies |p(x)f(x)|1/3|p(x)-f(x)|\leq 1/3.

All the measures introduced in this section lie between 0 and nn.  For most measures this follows because the algorithm may simply query all nn input variables and compute f(x)f(x).  Also, any Boolean function ff can be exactly represented by a real polynomial of degree at most nn.

2.3 Known relations and separations

We now summarize the best known relations between the complexity measures studied in this paper.  Figure 2 depicts all known relations of the type M1(f)=O(M2(f))M_{1}(f)=O(M_{2}(f)) for complexity measures M1(f)M_{1}(f) and M2(f)M_{2}(f).  An upward line from M1M_{1} to M2M_{2} in Figure 2 indicates M1(f)=O(M2(f))M_{1}(f)=O(M_{2}(f)) for all (partial or total) Boolean functions ff.  Most of these relations follow straightforwardly from definitions.  For example R0(f)D(f)R_{0}(f)\leq D(f) because any deterministic algorithm is also a zero-error randomized algorithm.  The relationships that do not follow from such observations are bs(f)=O(RC(f))\operatorname{bs}(f)=O(\operatorname{RC}(f)) [Aar06], deg(f)=O(QE(f))\deg(f)=O(Q_{E}(f)) [BBC+01], and deg~(f)=O(Q(f))\operatorname{\widetilde{\deg}}(f)=O(Q(f)) [BBC+01].

DDR0R_{0}QEQ_{E}CCRRRC\operatorname{RC}bs\operatorname{bs}deg\degQQdeg~\operatorname{\widetilde{\deg}}
Figure 2: Relations between complexity measures.

All other known relationships between these measures follow by combining the relationships in Figure 2 with the following relationships that hold for all total Boolean functions:

  • C(f)bs(f)2C(f)\leq\operatorname{bs}(f)^{2} [Nis91]

  • D(f)C(f)bs(f)D(f)\leq C(f)\operatorname{bs}(f) [BBC+01]

  • D(f)deg(f)3D(f)\leq\deg(f)^{3} [Mid04, Tal13]

  • RC(f)=O(deg~(f)2)\operatorname{RC}(f)=O(\operatorname{\widetilde{\deg}}(f)^{2}) [KT13]

  • R0(f)=O(R(f)2logR(f))R_{0}(f)=O(R(f)^{2}\log R(f)) [KT13]

We now turn to the best known separations between these measures.  To conveniently express these results, we use a notion called the critical exponent, as defined by [GSS13].  The the critical exponent for a measure M1M_{1} relative to a measure M2M_{2}, denoted crit(M1,M2)\operatorname{crit}(M_{1},M_{2}), is the infimum over all rr such that the relation M1(f)=O(M2(f)r)M_{1}(f)=O(M_{2}(f)^{r}) holds for all total functions ff. For example, because D(f)C(f)2D(f)\leq C(f)^{2} for all total ff, we have crit(D,C)2\operatorname{crit}(D,C)\leq 2.  Furthermore, since we also know that the And-Or function on k2k^{2} bits satisfies D(f)=k2D(f)=k^{2} and C(f)=kC(f)=k, we have crit(D,C)=2\operatorname{crit}(D,C)=2.  Note that the critical exponent between M1M_{1} and M2M_{2} is 22 even if we have M1(f)=O(M2(f)2logM2(f))M_{1}(f)=O(M_{2}(f)^{2}\log M_{2}(f)), or more generally if M1(f)M2(f)2+o(1)M_{1}(f)\leq M_{2}(f)^{2+o(1)}.

Table 2 lists the best known separations between these measures.  A cell in the table has the form a,ba,b, where aba\leq b are the best known lower and upper bounds on the critical exponent for the row measure relative to the column measure.  The cell also contains a citation for the function that provides the lower bound.  For example, in the cell corresponding to row DD and column deg\deg, we have the entry 2,32,3, which means 2crit(D,deg)32\leq\operatorname{crit}(D,\deg)\leq 3, and a function achieving the separation appears in [GPW15].

3 Randomized versus quantum query complexity

In this section we show a power 2.52.5 separation between bounded-error randomized query complexity, R(f)R(f), and bounded-error quantum query complexity, Q(f)Q(f).  In Section 3.1 we motivate the cheat sheet framework and provide a sketch of the separation.  In Section 3.2 we formally prove the separation.

3.1 Intuition

We begin with the best known separation between randomized and quantum query complexity for partial functions, which is currently Ω~(n)\widetilde{\Omega}(\sqrt{n}) versus 11 provided by the Forrelation problem [AA15].  (Note that Simon’s problem also provides a similar separation up to log factors [Sim97].)  Let gg be the Forrelation function on D{0,1}nD\subseteq\{0,1\}^{n} with R(g)=Ω~(n)R(g)=\widetilde{\Omega}(\sqrt{n}) and Q(g)=1Q(g)=1, although any function with these query complexities (up to log factors) would do.

One way to make gg total would be to define it to be 0 on the rest of the domain {0,1}nD\{0,1\}^{n}\setminus D.  But then it is unclear if the new function gg has low quantum query complexity, since the quantum algorithm would have to test whether the input lies in the promised set DD.  In general since partial function separations often require a stringent promise on the input, we may expect that it is necessary to examine most input bits to ascertain that xDx\in D.

Indeed, it is not even clear how to certify that xDx\in D for our function gg.  On the other hand, the domain certification problem is trivial for total functions.  So we can compose gg with a total function hh to obtain some of the desirable properties of both.  We can certify that an input to ghg\circ h lies in its domain by certifying the outputs of all instances of hh.  This means we want hh to have low certificate complexity, and since we will use ghg\circ h to separate randomized and quantum query complexity, we would also like hh to have Q(h)Q(h) smaller than R(h)R(h).  A function that fits the bill perfectly is the And-Orm2\textsc{And-Or}_{m^{2}} function that satisfies R(h)=Ω(m2)R(h)=\Omega(m^{2}), C(h)=mC(h)=m, and Q(h)=O(m)Q(h)=O(m).

We can now compute the various complexities of f:=ghf:=g\circ h.  First we have Q(f)=O~(m)Q(f)=\widetilde{O}(m), by composing algorithms for gg and hh.  There also exists a certificate of size O~(nm)\widetilde{O}(nm) that proves that the input satisfies the promise, since we can certify the outputs of the nn different hh gates using certificates of size C(h)=mC(h)=m.  Also, given query access to this certificate of size O~(nm)\widetilde{O}(nm), a quantum algorithm can check the certificate’s validity using Grover search in O~(nm)\widetilde{O}(\sqrt{nm}) queries.  Hence a quantum algorithm with access to a certificate can solve the problem with O~(m+nm)\widetilde{O}(m+\sqrt{nm}) queries.  On the other hand, it can be shown that R(f)=R(gh)=Ω(R(g)R(h))=Ω~(nm2)R(f)=R(g\circ h)=\Omega(R(g)R(h))=\widetilde{\Omega}(\sqrt{n}m^{2}).

Now if we set m=nm=n, then we see that Q(f)=O~(n)Q(f)=\widetilde{O}(n) and R(f)=Ω~(n2.5)R(f)=\widetilde{\Omega}(n^{2.5}), but ff is still a partial function.  However, ff has the desirable property that a quantum algorithm given query access to a certificate can decide whether the input satisfies the promise using O~(n)\widetilde{O}(n) queries.  But we cannot simply append the certificate to the input as we do not want the randomized algorithm to be able to use it.  It is, however, acceptable if the randomized algorithm finds the certificate after it spends R(f)R(f) queries, since that is the lower bound we want to prove.

What we would like is to hide the certificate somewhere in the input where only the quantum algorithm can find it.  What information do we have that is only known to the quantum algorithm and remains unknown to the randomized algorithm unless it spends R(f)R(f) queries?  Clearly the value of ff on the input has this property.

While one bit does not help much, we can obtain additional bits of this kind by using (say) 10logn10\log n copies of ff.  Now the answer string to these problems can address an array of size n10n^{10}, much larger than can be brute-force searched by a randomized algorithm in a reasonable amount of time.  Furthermore, this address can be found by a quantum algorithm using only O~(Q(f))\widetilde{O}(Q(f)) queries.  At this location we will store a cheat sheet: a collection of certificates for all 10logn10\log n instances of ff.

Now the new function, which we call fCSf_{\mathrm{CS}} (shown in Figure 3), evaluates to 11 if and only if the 10logn10\log n inputs are in the domain of ff, and the outputs of the 10logn10\log n copies of ff point to a valid cheat sheet, i.e., a set of valid certificates for the 10logn10\log n copies of ff.  This construction ensures that the quantum algorithm can find the cheat sheet using O~(Q(f))=O~(n)\widetilde{O}(Q(f))=\widetilde{O}(n) queries, and then verify it using O~(n)\widetilde{O}(n) queries, which gives Q(fCS)=O~(n)Q(f_{\mathrm{CS}})=\widetilde{O}(n). On the other hand, we intuitively expect that the randomized algorithm cannot even find the cheat sheet unless it computes ff by spending at least R(f)=Ω~(n2.5)R(f)=\widetilde{\Omega}(n^{2.5}) queries, which gives us the desired separation.

3.2 Implementation

In this section we prove Theorem 1, restated for convenience:

See 1

Let gg be the Forrelation function on D{0,1}nD\subseteq\{0,1\}^{n} with R(g)=Ω~(n)R(g)=\widetilde{\Omega}(\sqrt{n}) and Q(g)=O(1)Q(g)=O(1).  Let h:{0,1}m2{0,1}h:\{0,1\}^{m^{2}}\to\{0,1\} be the And-Orm2\textsc{And-Or}_{m^{2}} function, defined as AndmOrm\textsc{And}_{m}\circ\textsc{Or}_{m}. This function satisfies R(h)=Ω(m2)R(h)=\Omega(m^{2}), which can be shown using a variety of methods: it follows from the partition bound [JK10, Theorem 4] and also from our more general Theorem 5.  We also have C(h)=mC(h)=m, since one 11 from each Or gate is a valid 11-certificate and an Or gate with all zero inputs is a valid 0-certificate.  Lastly, Q(h)=O~(m)Q(h)=\widetilde{O}(m) follows from simply composing quantum algorithms for And and Or, and indeed Q(h)=O(m)Q(h)=O(m) [HMdW03].

Let f=ghf=g\circ h be a partial function on nm2nm^{2} bits.  We have Q(f)=O(m)Q(f)=O(m) by composition, since quantum algorithms can be composed in general without losing a log factor due to error reduction [Rei11, LMR+11].  There also exists a certificate of size O~(nm)\widetilde{O}(nm) that proves that the input satisfies the promise, since we can certify the outputs of the nn different hh gates using C(h)=mC(h)=m pointers to the relevant input bits.  Since each pointer uses O(logn)O(\log n) bits, the certificate is of size O~(nm)\widetilde{O}(nm).  Also note that a quantum algorithm with query access to this certificate can check its validity using Grover search in O~(nm)\widetilde{O}(\sqrt{nm}) queries.

Lastly, we claim that R(f)=R(gh)=Ω(R(g)R(h))=Ω~(nm2)R(f)=R(g\circ h)=\Omega(R(g)R(h))=\widetilde{\Omega}(\sqrt{n}m^{2}).  While a general composition theorem for bounded-error or zero-error randomized query complexity is unknown, we can show such a result when the inner function is the Or function.

Theorem 5.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D{0,1}nD\subseteq\{0,1\}^{n}. Then R(fOrm)=Ω(mR(f))R(f\circ\textsc{Or}_{m})=\Omega(mR(f)) and R0(fOrm)=Ω(mR0(f))R_{0}(f\circ\textsc{Or}_{m})=\Omega(mR_{0}(f)).  Therefore R(fAndmOrm)=Ω(m2R(f))R(f\circ\textsc{And}_{m}\circ\textsc{Or}_{m})=\Omega(m^{2}R(f)) and R0(fAndmOrm)=Ω(m2R0(f))R_{0}(f\circ\textsc{And}_{m}\circ\textsc{Or}_{m})=\Omega(m^{2}R_{0}(f)).

Proof.

We prove this for bounded-error algorithms; the argument for zero-error algorithms will be almost identical.  Let AA be the best randomized algorithm for fOrmf\circ\textsc{Or}_{m}.  We convert AA into an algorithm BB for ff, as follows.

Given input xx to ff, BB generates nn random inputs to Orm\textsc{Or}_{m}, denoted by Y1,Y2,,YnY_{1},Y_{2},\ldots,Y_{n}, in the following way.  First, each YiY_{i} is set to 0m0^{m}, the all-zero string.  Next, a random entry of YiY_{i} is chosen and replaced with a * symbol.  Finally, for each index ii, the * in YiY_{i} will be replaced by the bit xix_{i}.  This causes Orm(Yi)=xi\textsc{Or}_{m}(Y_{i})=x_{i} to be true for all ii.

To evaluate f(x)f(x), the algorithm BB can simply run AA on the string Y1Y2YnY_{1}Y_{2}\dots Y_{n}.  Since

fOrm(Y1Y2Yn)=f(Orm(Y1)Orm(Y2)Orm(Yn))=f(x1x2xn)=f(x),f\circ\textsc{Or}_{m}(Y_{1}Y_{2}\dots Y_{n})=f(\textsc{Or}_{m}(Y_{1})\textsc{Or}_{m}(Y_{2})\dots\textsc{Or}_{m}(Y_{n}))=f(x_{1}x_{2}\dots x_{n})=f(x), (2)

this algorithm evaluates f(x)f(x) with the same error as AA.  This process uses R(fOrm)R(f\circ\textsc{Or}_{m}) queries to the input YY.  However, not all of these queries require BB to query xx. In fact, BB only needs to query xx when algorithm AA queries a bit that was formerly a * in one of the YiY_{i}.  The expected number of queries BB makes is therefore the expected number of * entries found by the algorithm AA.  Using Markov’s inequality, we can turn BB into an R(f)R(f) algorithm, which uses a fixed number of queries to calculate f(x)f(x) with bounded error; it follows that the expected number of * entries found by AA is at least Ω(R(f))\Omega(R(f)).

We can now view AA as a randomized algorithm that finds Ω(R(f))\Omega(R(f)) star bits (in expectation) given input in Y1Y2YnY_{1}Y_{2}\dots Y_{n}.  Set Y=YiY=Y_{i} for a randomly chosen ii.  The number of queries AA makes to YY must be exactly R(fOrm)/nR(f\circ\textsc{Or}_{m})/n in expectation.  Let the probability that AA finds kk stars be pkp_{k}, for k=0,1,,nk=0,1,\dots,n.  Then

k=0nkpk=Ω(R(f)).\sum_{k=0}^{n}kp_{k}=\Omega(R(f)). (3)

For each kk, the probability that AA finds a star in YY given it found kk stars in total is k/nk/n.  The overall probability that AA finds a star in YY is therefore

k=0npk(k/n)=1nk=0nkpk=Ω(R(f)/n).\sum_{k=0}^{n}p_{k}(k/n)=\frac{1}{n}\sum_{k=0}^{n}kp_{k}=\Omega(R(f)/n). (4)

In other words, AA makes R(fOrm)/nR(f\circ\textsc{Or}_{m})/n expected queries to YY, and finds the * in YY with probability Ω(R(f)/n)\Omega(R(f)/n).  Now, finding the single * in an otherwise all-zero string is an unordered search problem; the chance of solving this problem after TT queries is at most T/mT/m.  In other words, any decision tree of height TT has probability only T/mT/m of solving the problem.  Since AA’s querying strategy on YY can be written as a probability distribution over decision trees with expected height R(fOrm)/nR(f\circ\textsc{Or}_{m})/n, it follows that the probability of AA finding the * is at most R(fOrm)/nmR(f\circ\textsc{Or}_{m})/nm.  Thus we have R(fOrm)/nm=Ω(R(f)/n)R(f\circ\textsc{Or}_{m})/nm=\Omega(R(f)/n), or R(fOrm)=Ω(mR(f))R(f\circ\textsc{Or}_{m})=\Omega(mR(f)).

If AA is a zero-error randomized algorithm instead of a bounded-error algorithm, the same argument follows verbatim, except that there’s no longer a need to use Markov’s inequality to conclude that the expected number of stars found by AA is at least Ω(R0(f))\Omega(R_{0}(f)).

The last part follows since we can show the same result for Andm\textsc{And}_{m}, using the fact that the query complexity of a function f(x1,,xn)f(x_{1},\ldots,x_{n}) equals that of f(x1¯,,xn¯)f(\overline{x_{1}},\ldots,\overline{x_{n}}) and the query complexity of ff and f¯\bar{f} is the same. ∎

a1a_{1}Forrelation\dotsn\wedge\dotsn\wedge\dotsn\vee\dotsn\vee\dotsn\vee\dotsn\vee\dotsna2a_{2}Forrelation\dotsn\wedge\dotsn\wedge\dotsn\vee\dotsn\vee\dotsn\vee\dotsn\vee\dotsna10logna_{10\log n}Forrelation\dotsn\wedge\dotsn\wedge\dotsn\vee\dotsn\vee\dotsn\vee\dotsn\vee\dotsn\dotsarray of size n10n^{10}
marked cell containing
pointers to certificates
proving the value of aa
Figure 3: The function ff that achieves a superquadratic separation between Q(f)=O~(n)Q(f)=\widetilde{O}(n) and R(f)=Ω~(n2.5)R(f)=\widetilde{\Omega}(n^{2.5}).

We now define the cheat sheet version of ff, which we call fCSf_{\mathrm{CS}}, depicted in Figure 3.  Let the input to fCSf_{\mathrm{CS}} consist of 10logn10\log n inputs to ff, each of size nm2nm^{2}, followed by n10n^{10} blocks of bits of size O~(mn)\widetilde{O}(mn) each, which we refer to as the array of cells or array of cheat sheets.  Each cell is large enough to hold certificates for all 10logn10\log n inputs to ff that certify for each input the output of ff evaluated on that input and that the promise holds.

Let us denote the input to fCSf_{\mathrm{CS}} as z=(x1,x2,,x10logn,Y1,Y2,,Yn10)z=(x^{1},x^{2},\ldots,x^{10\log n},Y_{1},Y_{2},\ldots,Y_{n^{10}}), where xix^{i} is an input to ff, and the YiY_{i} are the aforementioned cells of size O~(mn)\widetilde{O}(mn).  We define the value of fCS(z)f_{\mathrm{CS}}(z) to be 11 if and only if the following conditions hold:

  1. 1.

    For all ii, xix^{i} is in the domain of ff. If this condition is satisfied, let \ell be the positive integer corresponding to the binary string (f(x1),f(x2),,f(x10logn))(f(x^{1}),f(x^{2}),\ldots,f(x^{10\log n})).

  2. 2.

    YY_{\ell} certifies that all xix^{i} are in the domain of ff and that \ell equals the binary string formed by their output values, (f(x1),f(x2),,f(x10logn))(f(x^{1}),f(x^{2}),\ldots,f(x^{10\log n})).

We now upper bound the quantum query complexity of fCSf_{\mathrm{CS}}.  The quantum algorithm starts by assuming that the first condition of fCSf_{\mathrm{CS}} holds and simply computes ff on all 10logn10\log n inputs, which uses O~(Q(f))=O~(m)\widetilde{O}(Q(f))=\widetilde{O}(m) queries.  The answers to these inputs points to the cheat sheet YY_{\ell}, where \ell is the integer corresponding to the binary string (f(x1),f(x2),,f(x10logn))(f(x^{1}),f(x^{2}),\ldots,f(x^{10\log n})).  As discussed, verifying the cheat sheet of size O~(mn)\widetilde{O}(mn) requires only O~(mn)\widetilde{O}(\sqrt{mn}) queries by a recursive application of Grover search.  The algorithm outputs 11 if and only if the verification of YY_{\ell} succeeded. If the certificate is accepted by the algorithm, then both conditions of fCSf_{\mathrm{CS}} are satisfied and hence it is easy to see the algorithm is correct.  Hence Q(fCS)=O~(m+mn)Q(f_{\mathrm{CS}})=\widetilde{O}(m+\sqrt{mn}).

We also know that R(f)=Ω~(nm2)R(f)=\widetilde{\Omega}(\sqrt{n}m^{2}).  We now prove that this implies R(fCS)=Ω~(R(f))=Ω~(nm2)R(f_{\mathrm{CS}})=\widetilde{\Omega}(R(f))=\widetilde{\Omega}(\sqrt{n}m^{2}).  Proving this completes the proof of Theorem 1, since setting m=nm=n immediately yields Q(fCS)=O~(n)Q(f_{\mathrm{CS}})=\widetilde{O}(n) and R(fCS)=Ω~(n2.5)R(f_{\mathrm{CS}})=\widetilde{\Omega}(n^{2.5}).

To complete the proof, we show in general that R(fCS)=Ω~(R(f))R(f_{\mathrm{CS}})=\widetilde{\Omega}(R(f)).

Lemma 6.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n}, and let fCSf_{\mathrm{CS}} be the cheat sheet version of ff with c=10lognc=10\log n copies of ff.  Then R(fCS)=Ω(R(f)/c2)=Ω~(R(f))R(f_{\mathrm{CS}})=\Omega(R(f)/c^{2})=\widetilde{\Omega}(R(f)).

Proof.

Let AA be any bounded-error randomized algorithm for evaluating fCSf_{\mathrm{CS}}.  We will prove that AA makes Ω(R(f)/c2)\Omega(R(f)/c^{2}) queries.

We start by proving the following claim: Let x1,x2,,xcDom(f)x^{1},x^{2},\ldots,x^{c}\in\operatorname{Dom}(f) denote cc valid inputs to ff and let zz be an input to fCSf_{\mathrm{CS}} consisting of x1x2xcx^{1}x^{2}\cdots x^{c} with blank array—that is, it has 0 in all the entries of all cheat sheets.  Let us assume that the all zero string is not a valid cheat sheet for any input.  This can be enforced, for example, by requiring that all cheat sheets begin with the first bit equal to 11.  Let \ell be the binary number f(x1)f(x2)f(xc)f(x^{1})f(x^{2})\cdots f(x^{c}), and let hh_{\ell} be the th\ell^{\mathrm{th}} cheat sheet.  Then we claim that AA must query a bit of hh_{\ell} when run on zz with probability at least 1/31/3.

This claim is easy to prove.  Since the all zero string is an invalid cheat sheet, it follows that fCS(z)=0f_{\mathrm{CS}}(z)=0, so AA outputs 0 when run on zz with probability at least 2/32/3.  Now if we modify hh_{\ell} in zz to be the correct cheat sheet for the given input x1,x2,,xcx^{1},x^{2},\ldots,x^{c}, then we obtain a new input zz^{\prime} with fCS(z)=1f_{\mathrm{CS}}(z^{\prime})=1.  This means AA outputs 11 when run on zz^{\prime} with probability at least 2/32/3.  However, since these inputs only differ on hh_{\ell}, if AA does not query hh_{\ell} with high probability, it cannot distinguish input zz from input zz^{\prime}. Since AA is a valid algorithm for fCSf_{\mathrm{CS}}, the claim follows.

We now use the hybrid argument to prove the lemma.  For any input zz with blank array, let pz[0,1]2cp_{z}\in[0,1]^{2^{c}} be the vector such that for each i=[2c]i=[2^{c}], (pz)i(p_{z})_{i} is the probability that AA makes a query to the ithi^{\mathrm{th}} cheat sheet when given input zz.  Then the 11-norm of pzp_{z} is at most the running time of AA, which is at most R(f)R(f).  By the claim, if \ell is the relevant cheat sheet for zz, we have (pz)1/3(p_{z})_{\ell}\geq 1/3.  On the other hand, since pzp_{z} has 2cn10R(f)102^{c}\geq n^{10}\geq R(f)^{10} entries that sum to at most R(f)R(f), almost all of them have value less than, say, R(f)5R(f)^{-5}.

Next, consider hard distributions 𝒟0\mathcal{D}^{0} and 𝒟1\mathcal{D}^{1} over the 0- and 11-inputs to ff, respectively.  We pick these distributions such that distinguishing between them with probability at least 1/2+1/12c1/2+1/12c takes at least Ω(R(f)/c2)\Omega(R(f)/c^{2}) queries for any randomized algorithm.

For each i[2c]i\in[2^{c}], let qiq_{i} be the expectation over pzp_{z} when zz is made of cc inputs to ff generated from 𝒟i=𝒟i1×𝒟i2××𝒟ic\mathcal{D}^{i}=\mathcal{D}^{i_{1}}\times\mathcal{D}^{i_{2}}\times\dots\times\mathcal{D}^{i_{c}}, together with a blank array (here iji_{j} means the jthj^{\mathrm{th}} bit of ii when written in binary).  Then for all i[2c]i\in[2^{c}], we have (qi)i1/3(q_{i})_{i}\geq 1/3, and the 11-norm of qiq_{i} is at most R(f)R(f), so most entries of qiq_{i} are less than R(f)5R(f)^{-5}.  The entries of qiq_{i} can be interpreted as the probabilities of each cheat sheet being queried by the algorithm on an input sampled from 𝒟i\mathcal{D}^{i}.

Let k[2c]k\in[2^{c}] be such that (q0)k<1/6(q_{0})_{k}<1/6.  Let k0,k1,,kck_{0},k_{1},\dots,k_{c} be such that k0=0k_{0}=0, kc=kk_{c}=k, and consecutive kk’s differ by at most one bit (when represented as binary vectors of length cc).  Since (qk0)k<1/6(q_{k_{0}})_{k}<1/6 and (qkc)k1/3(q_{k_{c}})_{k}\geq 1/3, there must be some j[c]j\in[c] such that (qkj+1)k(qkj)k>1/6c(q_{k_{j+1}})_{k}-(q_{k_{j}})_{k}>1/6c and (qkj)k<1/3(q_{k_{j}})_{k}<1/3. Let a=(qkj+1)ka=(q_{k_{j+1}})_{k} and let b=(qkj)kb=(q_{k_{j}})_{k}.

We can now use this to distinguish the product distribution 𝒟kj+1\mathcal{D}^{k_{j+1}} from 𝒟kj\mathcal{D}^{k_{j}}, with probability of error at most 1/21/12c1/2-1/12c: simply run the algorithm AA, output 11 if it queried an entry of the kthk^{\mathrm{th}} cheat sheet, and otherwise output 11 with probability max(0,1ab2ab)\max(0,\frac{1-a-b}{2-a-b}).  If this maximum is 0, it means we have b>2/3b>2/3 and a<1/3a<1/3, so this algorithm has error at most 1/31/3.  Otherwise, it is not hard to check that the algorithm will determine if the input is from 𝒟kj+1\mathcal{D}^{k_{j+1}} with probability at most 1/2(ba)/2<1/21/12c1/2-(b-a)/2<1/2-1/12c.

Finally, note that since kj+1k_{j+1} and kjk_{j} differ in only one bit, distinguishing 𝒟kj+1\mathcal{D}^{k_{j+1}} from 𝒟kj\mathcal{D}^{k_{j}} allows us to distinguish between 𝒟0\mathcal{D}^{0} and 𝒟1\mathcal{D}^{1}.  This is because given any input from 𝒟0\mathcal{D}^{0} or 𝒟1\mathcal{D}^{1}, the algorithm can sample other inputs to construct an input from 𝒟kj+1\mathcal{D}^{k_{j+1}} or 𝒟kj\mathcal{D}^{k_{j}}, and then run the distinguishing algorithm.  It follows that the running time of AA is at least Ω(R(f)/c2)\Omega(R(f)/c^{2}), by the choice of the distributions 𝒟0\mathcal{D}^{0} and 𝒟1\mathcal{D}^{1}. ∎

4 Quantum query complexity versus certificate complexity and degree

We now show a nearly quadratic separation between Q(f)Q(f) and C(f)C(f), which yields a similar separation between Q(f)Q(f) and deg(f)\deg(f).  We will also use the functions introduced in this section as building blocks to obtain the nearly 4th4^{\mathrm{th}} power separation between Q(f)Q(f) and deg~(f)\operatorname{\widetilde{\deg}}(f) proved in Section 5.

4.1 Quadratic gap with certificate complexity

In this section we establish Theorem 3, restated for convenience:

See 3

Consider the k-sumk\textsc{-sum} problem, k-sum:[M]n{0,1}k\textsc{-sum}:[M]^{n}\to\{0,1\}, which asks if there are kk elements in the input string x1,x2,,xn[M]x_{1},x_{2},\ldots,x_{n}\in[M] that sum to 0(modM)0\pmod{M}.  Belovs and Špalek [BŠ13] showed that Q(k-sum)=Ω(nk/(k+1))Q(k\textsc{-sum})=\Omega(n^{k/(k+1)}) when the alphabet MM has size nkn^{k} and kk is constant.  In Appendix A, we show that their proof implies a bound of Q(k-sum)=Ω(nk/(k+1)/k)Q(k\textsc{-sum})=\Omega(n^{k/(k+1)}/\sqrt{k}) for super-constant kk.

Now the 11-certificate complexity of the k-sumk\textsc{-sum} problem is kk (assuming it costs one query to get an element of MM), since it suffices to provide the kk elements that sum to 0(modM)0\pmod{M}.  If we take k=lognk=\log n, we get Q(k-sum)=Ω(n/logn)Q(k\textsc{-sum})=\Omega(n/\sqrt{\log n}) and C1(k-sum)=O(logn)C_{1}(k\textsc{-sum})=O(\log n).  Although this function is not Boolean, turning it into a Boolean function will only incur an additional polylogarithmic loss.

While k-sumk\textsc{-sum} does not separate quantum query complexity from certificate complexity, its 11-certificate complexity is much smaller than its quantum query complexity.  Composing a function with small 0-certificates, such as the Andn\textsc{And}_{n} with k-sumk\textsc{-sum} already gives a function whose quantum query complexity is larger than its certificate complexity: in this case, we have certificate complexity O~(n)\widetilde{O}(n) and quantum query complexity Ω~(n3/2)\widetilde{\Omega}(n^{3/2}), which follows from the following general composition theorem [HLŠ07, Rei11, LMR+11, Kim12]:

Theorem 7 (Composition theorem for quantum query complexity).

Let f:D{0,1}f:D\to\{0,1\} and g:E{0,1}g:E\to\{0,1\} be partial functions where D{0,1}nD\subseteq\{0,1\}^{n} and E{0,1}mE\subseteq\{0,1\}^{m}.  Then Q(fg)=Θ(Q(f)Q(g))Q(f\circ g)=\Theta(Q(f)Q(g)).

To get an almost quadratic gap between Q(f)Q(f) and C(f)C(f), we use a variant of k-sumk\textsc{-sum} itself as the outer function instead of the And function.  From k-sumk\textsc{-sum}, we define a new Boolean function that we call Block k-sumk\textsc{-sum}, whose quantum query complexity is Θ~(n)\widetilde{\Theta}(n) and certificate complexity is also Θ~(n)\widetilde{\Theta}(n).  However, although its certificate complexity is linear, the certificates consist almost exclusively of input bits set to 11 and only O~(1)\widetilde{O}(1) input bits set to 0.  This means if we compose this function with k-sumk\textsc{-sum}, the composed function has certificates of size O~(n)\widetilde{O}(n), since the certificates of Block k-sumk\textsc{-sum} are essentially composed of 11s, which are easy to certify for k-sumk\textsc{-sum}.  We denote this composed function BKK:{0,1}n2{0,1}\textsc{BKK}:\{0,1\}^{n^{2}}\to\{0,1\}.  It satisfies C(BKK)=O~(n)C(\textsc{BKK})=\widetilde{O}(n) and Q(BKK)=Ω~(n2)Q(\textsc{BKK})=\widetilde{\Omega}(n^{2}), which yields the desired quadratic separation.

We now define the Block kk-sum problem.

Definition 8.

Let Block kk-sum be a total Boolean function on nn bits defined as follows.  We split the input into blocks of size 10klogn10k\log n each and say a block is balanced if it has an equal number of 0s and 11s.  Let the balanced blocks represent numbers in an alphabet MM of size Ω(nk)\Omega(n^{k}). The value of the function is 11 if and only if there are kk balanced blocks whose corresponding numbers sum to 0(modM)0\pmod{M} and all other blocks have at least as many 11s as 0s.

We then compose this function with k-sumk\textsc{-sum} to get the function BKK.

Definition 9.

Let BKKn2,k:{0,1}n2{0,1}\textsc{BKK}_{n^{2},k}:\{0,1\}^{n^{2}}\to\{0,1\} be the function Block kk-sum on nn bits composed with a Boolean version of k-sumk\textsc{-sum} on nn bits, and define BKKn2\textsc{BKK}_{n^{2}} to be BKKn2,k\textsc{BKK}_{n^{2},k} with k=lognk=\log n.

We are now ready to establish the various complexities of BKK.

Theorem 10.

For the total Boolean function BKKn2,k:{0,1}n2{0,1}\textsc{BKK}_{n^{2},k}:\{0,1\}^{n^{2}}\to\{0,1\}, we have

C(BKKn2,k)=O(k2nlogn)andQ(BKKn2,k)=Ω(n22/(k+1)k3log2n).C(\textsc{BKK}_{n^{2},k})=O(k^{2}n\log n)\quad\text{and}\quad Q(\textsc{BKK}_{n^{2},k})=\Omega\Bigl{(}\frac{n^{2-2/(k+1)}}{k^{3}\log^{2}n}\Bigr{)}. (5)
Proof.

We start by analyzing the certificates of Block kk-sum.  The key property we need is that every input of Block kk-sum has a certificate that uses very few 0s, but can use a large number of 11s.  To see this, note that we can certify a 11-input by showing the kk balanced blocks that sum to 0 which requires O(k2logn)O(k^{2}\log n) 0s, and all the 11s in every other block.  There are two kinds of 0 inputs to certify: A 0-input that has a block with more 0s than 11s can be certified by providing that block, which only uses O(klogn)O(k\log n) 0s.  A 0-input in which all blocks have at least as many 11s as 0s can be certified by providing all the 11s: this provides the number represented by each block if it were balanced (though it does not prove the block is actually balanced), which is enough to check that no kk of them sum to zero.  In conclusion, Block kk-sum can always be certified by providing O(n)O(n) 11s and O(k2logn)O(k^{2}\log n) 0s.

We now analyze the certificate complexity of BKKn2,k\textsc{BKK}_{n^{2},k}.  For each input, the outer Block kk-sum has a certificate using O(k2logn)O(k^{2}\log n) 0s and O(n)O(n) 11s.  The inner function, k-sumk\textsc{-sum}, has 11-certificates of size O(k2logn)O(k^{2}\log n) since there are kk numbers to exhibit and each uses klognk\log n bits when represented in binary, and has 0-certificates of size O(n)O(n).  Therefore, the composed function always has a certificate of size O(k2nlogn)O(k^{2}n\log n).  Hence C(BKKn2,k)=O(k2nlogn)C(\textsc{BKK}_{n^{2},k})=O(k^{2}n\log n).

The quantum query complexity of Block kk-sum is Ω(Q(k-sum)/klogn)\Omega(Q(k\textsc{-sum})/k\log n) by a reduction from k-sumk\textsc{-sum}.  Using the result in Appendix A, this is Ω(n11/(k+1)/k3/2logn)\Omega(n^{1-1/(k+1)}/k^{3/2}\log n).  Invoking the composition theorem for quantum query complexity (Theorem 7), we get Q(BKKn2,k)=Ω(n22/(k+1)k3log2n)Q(\textsc{BKK}_{n^{2},k})=\Omega\left(\frac{n^{2-2/(k+1)}}{k^{3}\log^{2}n}\right). ∎

Thus for the function BKK:{0,1}n2{0,1}\textsc{BKK}:\{0,1\}^{n^{2}}\to\{0,1\}, defined as BKK=BKKn2,logn\textsc{BKK}=\textsc{BKK}_{n^{2},\log n}, we have

C(BKK)=O(nlog3n)=O~(n)andQ(BKK)=Ω(n2log5n)=Ω~(n2).C(\textsc{BKK})=O(n\log^{3}n)=\widetilde{O}(n)\quad\text{and}\quad Q(\textsc{BKK})=\Omega\Bigl{(}\frac{n^{2}}{\log^{5}n}\Bigr{)}=\widetilde{\Omega}(n^{2}). (6)

This establishes Theorem 3, since BKK is a total Boolean function.

4.2 Quadratic gap with (exact) degree

We now show how to obtain a total function that nearly quadratically separates Q(f)Q(f) from deg(f)\deg(f) using any total function that achieves a similar separation between Q(f)Q(f) and C(f)C(f).  This proves Theorem 4:

See 4

Let f:{0,1}n{0,1}f:\{0,1\}^{n}\to\{0,1\} be a total function with Q(f)=Ω~(n)Q(f)=\widetilde{\Omega}(n) and C(f)=O~(n)C(f)=\widetilde{O}(\sqrt{n}), such as the BKK function introduced in the previous section.  Let fCSf_{\mathrm{CS}} denote the cheat sheet version of ff (as described in Section 3), created using 10logn10\log n copies of ff that point to a cheat sheet among n10n^{10} potential cheat sheets, where a valid cheat sheet contains certificates of size O~(n)\widetilde{O}(\sqrt{n}) for each of the 10logn10\log n inputs to ff and the binary string corresponding to their outputs equals the location of the cheat sheet.  In this case ff is a total function so the cheat sheets do not certify that the input satisfies the promise, but only the value of ff evaluated on the input.  We claim that the cheat sheet version of ff satisfies Q(fCS)=Ω~(n)Q(f_{\mathrm{CS}})=\widetilde{\Omega}(n) and deg(fCS)=O~(n)\deg(f_{\mathrm{CS}})=\widetilde{O}(\sqrt{n}).

Let us start with the degree upper bound, deg(fCS)=O~(n)\deg(f_{\mathrm{CS}})=\widetilde{O}(\sqrt{n}). Let [n10]\ell\in[n^{10}] be a potential location of the cheat sheet.  For any \ell, consider the problem of outputting 11 if and only if fCS(z)=1f_{\mathrm{CS}}(z)=1 and \ell is the location of the cheat sheet for the input zz.  Since \ell is known, this can be solved by a deterministic algorithm 𝒜\mathcal{A}_{\ell} that makes O~(n)\widetilde{O}(\sqrt{n}) queries, since it can simply check if the certificate stored at cell \ell in the array is valid: it can check all the 10logn10\log n certificates of size O~(n)\widetilde{O}(\sqrt{n}) for each of the 10logn10\log n instances of ff and then check if the outputs of ff evaluate to the location \ell.  Since polynomial degree is at most deterministic query complexity, we can construct a representing polynomial for 𝒜\mathcal{A}_{\ell} for any location \ell.  This is a polynomial pp_{\ell} of degree O~(n)\widetilde{O}(\sqrt{n}) such that p(z)=1p_{\ell}(z)=1 if and only if fCS(z)=1f_{\mathrm{CS}}(z)=1 and \ell is the position of the cheat sheet on input zz.  Now we can simply add all the polynomials pp_{\ell} together to obtain a new polynomial qq of the same degree.  We claim q(z)=fCS(z)q(z)=f_{\mathrm{CS}}(z) since if fCS(z)=0f_{\mathrm{CS}}(z)=0 then certainly all the polynomials p(z)=0p_{\ell}(z)=0 (since none of the cheat sheets is valid) and if fCS(z)=1f_{\mathrm{CS}}(z)=1 then q(z)=1q(z)=1 because exactly one of many p(z)p_{\ell}(z) will evaluate to 11, the one corresponding to the location of the cheat sheet for the input zz.  Note that the property used here is that in a 11-input to fCSf_{\mathrm{CS}}, exactly one location serves as the correct cheat sheet, i.e., the location of the cheat sheet for a 11-input is unique.

The claim that Q(fCS)=Ω~(n)Q(f_{\mathrm{CS}})=\widetilde{\Omega}(n) seems intuitive since Q(f)=Ω~(n)Q(f)=\widetilde{\Omega}(n) and the cheat sheet version of ff cannot be easier than ff itself.  This intuitive claim is true and we show below that Q(fCS)=Ω(Q(f))Q(f_{\mathrm{CS}})=\Omega(Q(f)) in general, which completes the proof of Theorem 4.

To prove this general result for quantum query complexity, we will need the following strong direct product theorem due to Lee and Roland [LR13].

Theorem 11.

Let ff be a (partial) function with Q1/4(f)T{Q}_{1/4}\left(f\right)\geq T.  Then any TT-query quantum algorithm that outputs the value of ff evaluated on cc independent instances has success probability at most O((3/4)c/2)O(\left(3/4\right)^{c/2}).

We now prove the lower bound on the quantum query complexity of cheat sheet functions.

Lemma 12.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n}, and let fCSf_{\mathrm{CS}} be a cheat sheet version of ff with c=10lognc=10\log n copies of ff.  Then Q(fCS)=Ω(Q(f))Q(f_{\mathrm{CS}})=\Omega(Q(f)).

Proof.

By Theorem 11, we know that given c=10lognc=10\log n instances of ff, any quantum algorithm that makes fewer than T=Q1/4(f)T={Q}_{1/4}(f) queries will correctly decide all the instances with probability at most O((3/4)c/2)O(\left(3/4\right)^{c/2}).

Now suppose by way of contradiction that Q1/4(fCS)T{Q}_{1/4}\left(f_{\mathrm{CS}}\right)\leq T.  Then we will show how to decide cc copies of ff with probability Ω(1/T2)\Omega\left(1/T^{2}\right) by making TT quantum queries.  Since

1T21n212c/5(34)c/2,\frac{1}{T^{2}}\geq\frac{1}{n^{2}}\geq\frac{1}{2^{c/5}}\gg\left(\frac{3}{4}\right)^{c/2}, (7)

this is enough to contradict Theorem 11.

Let QQ be a quantum algorithm that decides fCSf_{\mathrm{CS}} using at most TT queries.  Then consider running QQ on an input z=(x1,,xc,Y1,Y2,,Y2c)z=(x^{1},\ldots,x^{c},Y_{1},Y_{2},\ldots,Y_{2^{c}}), where the xix^{i} are in the domain of ff and whose cheat-sheet array (Y1,Y2,,Y2c)(Y_{1},Y_{2},\ldots,Y_{2^{c}}) has been completely zeroed out.  We assume again that the all-zero cheat sheet is invalid for any input.  This can always be enforced by requiring a valid cheat sheet to have the first bit set to 1.

For all y[2c]y\in\left[2^{c}\right] and t[T]t\in\left[T\right], define my,tm_{y,t}, or the query magnitude on yy at query tt, to be the probability that QQ would be found querying some bit in cell YyY_{y}, were we to measure in the standard basis immediately before the ttht^{\mathrm{th}} query.

For an input zz where we have zeroed out the cheat sheet array, clearly fCS(z)=0f_{\mathrm{CS}}(z)=0 and hence QQ outputs 0 on this input with high probability.  On the other hand, if we let =f(x1)f(xc)[2c]\ell=f\left(x^{1}\right)\cdots f\left(x^{c}\right)\in[2^{c}], by modifying only the th\ell^{\mathrm{th}} cell in the array, YY_{\ell}, we could produce an input zz^{\prime} such that fCS(z)=1f_{\mathrm{CS}}\left(z^{\prime}\right)=1.  From these facts, together with the standard BBBV hybrid argument [BBBV97], it follows that

m,1++m,T=Ω(1).\sqrt{m_{\ell,1}}+\cdots+\sqrt{m_{\ell,T}}=\Omega\left(1\right). (8)

So by the Cauchy–Schwarz inequality, we have

m,1++m,T=Ω(1T).m_{\ell,1}+\cdots+m_{\ell,T}=\Omega\left(\frac{1}{T}\right). (9)

This implies that, if we simply choose a t[T]t\in\left[T\right] uniformly at random, run QQ until the ttht^{\mathrm{th}} query with the cheat-sheet array zeroed out, and then measure in the standard basis, we will observe =f(x1)f(xc)\ell=f\left(x^{1}\right)\cdots f\left(x^{c}\right) with probability Ω(1/T2)\Omega\left(1/T^{2}\right).  This completes the proof. ∎

5 Quantum query complexity versus approximate degree

We now show how to obtain a nearly quartic separation between quantum query complexity and approximate degree from a function that quadratically separates Q(f)Q(f) from C(f)C(f).  As in Section 3, we first motivate the construction in Section 5.1 and then formally give the separation in Section 5.2.  The main result of this section is Theorem 2:

See 2

5.1 Intuition

To obtain the quartic separation with approximate degree, we could try the same approach as in the previous section for exact degree.  Using notation from Section 4.2, consider the algorithm 𝒜\mathcal{A}_{\ell} that makes O~(n)\widetilde{O}(\sqrt{n}) queries.  Instead of using a polynomial to exactly represent 𝒜\mathcal{A}_{\ell}, we could try to construct more efficient approximating polynomials.  If there were a quantum algorithm that checked the certificate quadratically faster than the deterministic 𝒜\mathcal{A}_{\ell}, then we would be done.  This is because such a quantum algorithm would yield a polynomial of degree O~(n1/4)\widetilde{O}(n^{1/4}) that approximates 𝒜\mathcal{A}_{\ell}.  Then we could sum up these polynomials as before, with some error reduction to ensure that the error in each polynomial is small enough not to affect the sum of the polynomials.  This error reduction only adds an overhead of O(logn)O(\log n), which gives us a polynomial of degree O~(n1/4)\widetilde{O}(n^{1/4}) that approximates fCSf_{\mathrm{CS}}.

However, it is unclear if there is a quantum algorithm to check a certificate of size O~(n)\widetilde{O}(\sqrt{n}) quickly.  Reading the certificate itself could require Ω~(n)\widetilde{\Omega}(\sqrt{n}) queries.  All we know is that the certificate can be checked using O~(n)\widetilde{O}(\sqrt{n}) queries classically, but this does not imply a quadratic quantum speedup.  To obtain a quadratic speedup the certificates need some structure that can, for example, be exploited with Grover’s algorithm.  But the certificates in this case are simply 0-inputs of k-sumk\textsc{-sum} of size O~(n)\widetilde{O}(\sqrt{n}), and it is unclear how to quantumly check if an input is a 0-input (even with query access to any certificate) any quicker than querying the entire input.  What we would like is for the certificate to be checkable using O~(n1/4)\widetilde{O}({n}^{1/4}) queries to the certificate and the input.  So we construct a new function that has this property by modifying the BKK function used in Section 4.2.

Let ff be BKK:{0,1}n2{0,1}\textsc{BKK}:\{0,1\}^{n^{2}}\to\{0,1\} for some large, but constant nn.  We choose nn to be large enough that Adv±(f)n1.99\operatorname{Adv^{\pm}}(f)\geq n^{1.99} and C(f)n1.01C(f)\leq n^{1.01}.  Such an nn exists because asymptotically C(BKK)=O~(n)C(\textsc{BKK})=\widetilde{O}(n) and Q(BKK)=Ω~(n2)Q(\textsc{BKK})=\widetilde{\Omega}(n^{2}) (and Q(BKK)=Θ(Adv±(BKK))Q(\textsc{BKK})=\Theta(\operatorname{Adv^{\pm}}(\textsc{BKK})) [LMR+11]).  Composing this function with itself dd times gives us a new function g:{0,1}N{0,1}g:\{0,1\}^{N}\to\{0,1\}, where N=n2dN=n^{2d}.  This function has C(g)C(f)dn1.01dN0.51C(g)\leq C(f)^{d}\leq n^{1.01d}\leq N^{0.51}.  Since the general adversary bound satisfies a perfect composition theorem [LMR+11], we have Q(g)=Ω(Adv±(g))=Ω(Adv±(f)d)=Ω(n1.99d)=Ω(N0.99)Q(g)=\Omega(\operatorname{Adv^{\pm}}(g))=\Omega(\operatorname{Adv^{\pm}}(f)^{d})=\Omega(n^{1.99d})=\Omega(N^{0.99}).

If we view the function gg as a tree of depth dd and fanin n2n^{2}, it has NN leaves but only N0.51N^{0.51} of these leaves are part of a certificate.  Consider the subtree of gg that consists of all nodes that are ancestors of these N0.51N^{0.51} certificate leaves.  This subtree has size O(N0.51)O(N^{0.51}) and the set of values of all nodes in this subtree is a certificate for gg.  Furthermore, this certificate can be checked by a quantum algorithm in only O(N0.26)O(N^{0.26}) queries, since each level of the tree consists of at most N0.51N^{0.51} instances of ff acting on a constant number of bits.  Checking a constant-sized ff costs only O(1)O(1) queries, and searching over all N0.51N^{0.51} instances for an invalid certificate costs O(N0.26)O(N^{0.26}) queries by Grover’s algorithm.  This can be done for each level, resulting in a quantum algorithm with query complexity O~(N0.26)\widetilde{O}(N^{0.26}).

Thus we have constructed a function gg for which Q(g)Q(g) is close to NN, but the complexity of quantumly checking a certificate (given query access to it) is close to N1/4N^{1/4}.  Plugging this into the cheat sheet construction yields a nearly 4th4^{\mathrm{th}} power separation between quantum query complexity and approximate degree.

5.2 Implementation

Recall the function BKK:{0,1}n2{0,1}\textsc{BKK}:\{0,1\}^{n^{2}}\to\{0,1\} introduced in Theorem 10.  We introduce a function we call RecBKK which consists of recursively composing BKK with itself dd times; that is, we replace each bit in BKK with a new copy of BKK, then replace each bit in the new copies with yet more copies of BKK, and so on.  The resulting function will look like a tree of height dd where each vertex has n2n^{2} children and will have total input size N=n2dN=n^{2d}.  We will choose d=(4/25)logn/loglognd=(4/25)\log n/\log\log n to optimize our construction.  The resulting function RecBKK has the following properties.

Theorem 13.

There exists a total function RecBKK:{0,1}N{0,1}\textsc{RecBKK}:\{0,1\}^{N}\to\{0,1\} such that given query access to a certificate of size N1/2+o(1)N^{1/2+o(1)}, a quantum algorithm can check the validity of the certificate using at most N1/4+o(1)N^{1/4+o(1)} queries.  Furthermore, Q(RecBKK)=N1o(1)Q(\textsc{RecBKK})=N^{1-o(1)}.

Proof.

Our function RecBKK is defined as the dd-fold composition of BKK:{0,1}n2{0,1}\textsc{BKK}:\{0,1\}^{n^{2}}\to\{0,1\} with itself.  This yields a function on N=n2dN=n^{2d} bits and we set d=(4/25)logn/loglognd=(4/25)\log n/\log\log n.

With this choice of dd, we can show more precisely that the function RecBKK has a certificate that can be checked by a quantum algorithm using O(N1/4LN[1/2,1])=N1/4+o(1)O(N^{1/4}L_{N}[1/2,1])=N^{1/4+o(1)} queries, where LN[a,c]=exp((c+o(1))logaNloglog1aN)L_{N}[a,c]=\exp((c+o(1))\log^{a}N\log\log^{1-a}N), and Q(RecBKK)=Ω(N/LN[1/2,1])=N1o(1)Q(\textsc{RecBKK})=\Omega({N}/{L_{N}[1/2,1]})=N^{1-o(1)}.

With this choice of dd, the parameters are now related as follows: Since N=n2dN=n^{2d}, we have logN=2dlogn=(8/25)log2n/loglogn\log N=2d\log n=(8/25)\log^{2}n/\log\log n.  This gives logn=(5/4+o(1))logNloglogN\log n=(5/4+o(1))\sqrt{\log N\log\log N}, so we get n=LN[1/2,5/4]n=L_{N}[1/2,5/4].  Additionally, since (logn)d=exp(dloglogn)=exp((4/25)logn)=n4/25(\log n)^{d}=\exp(d\log\log n)=\exp((4/25)\log n)=n^{4/25}, it follows that (logn)d=LN[1/2,1/5](\log n)^{d}=L_{N}[1/2,1/5].

We start with the quantum lower bound for Q(RecBKK)Q(\textsc{RecBKK}).  By Theorem 10 and the optimality of the general adversary bound [LMR+11], we have

Adv±(BKK)=Ω(Q(BKK))=Ω(n2log5n)=n2(logn)5+o(1).\operatorname{Adv^{\pm}}(\textsc{BKK})=\Omega(Q(\textsc{BKK}))=\Omega\left(\frac{n^{2}}{\log^{5}n}\right)=\frac{n^{2}}{(\log n)^{5+o(1)}}. (10)

By using the fact that Adv±(fd)=Adv±(f)d\operatorname{Adv^{\pm}}(f^{d})=\operatorname{Adv^{\pm}}(f)^{d} for Boolean functions ff [LMR+11], we get

Q(RecBKK)=Ω(Adv±(BKK)d)=n2d(logn)(5+o(1))d=NLN[1/2,1/5]5+o(1)=Ω(N/LN[1/2,1]).Q(\textsc{RecBKK})=\Omega(\operatorname{Adv^{\pm}}(\textsc{BKK})^{d})=\frac{n^{2d}}{(\log n)^{(5+o(1))d}}=\frac{N}{L_{N}[1/2,1/5]^{5+o(1)}}=\Omega({N}/{L_{N}[1/2,1]}). (11)

Now we show the upper bound on quantumly checking a certificate.  First note that every non-leaf node in the RecBKK tree corresponds to a BKK instance.  For each such node, there is therefore a set of C(BKK)C(\textsc{BKK}) children that constitute a certificate for that BKK instance.  We can therefore certify the RecBKK instance by starting from the top of the tree, listing out C(BKK)C(\textsc{BKK}) children that constitute a certificate, then listing out C(BKK)C(\textsc{BKK}) children for each of those, and so on.  In each layer ii of the tree, we thus have C(BKK)iC(\textsc{BKK})^{i} nodes that belong to a certificate.

We require our quantumly checkable certificate to provide, for each non-leaf node that belongs to a certificate, pointers to C(BKK)C(\textsc{BKK}) of the node’s children that constitute a certificate for that BKK instance, starting with the root of the tree.  A quantum algorithm can then use Grover search to search for a bad certificate.  More precisely, the algorithm checks the certificate for the root to see if the C(BKK)C(\textsc{BKK}) children of the root pointed to and their claimed values do indeed form a certificate.  It then checks if all the claimed values in the first level are correct assuming the claimed values of nodes in the second level and so on.  The total number of certificates to check is

1+C(BKK)+C(BKK)2++C(BKK)d12C(BKK)d1,1+C(\textsc{BKK})+C(\textsc{BKK})^{2}+\dots+C(\textsc{BKK})^{d-1}\leq 2C(\textsc{BKK})^{d-1}, (12)

where each certificate has size C(BKK)C(\textsc{BKK}).  Therefore, the quantum algorithm will make

O~(C(BKK)(d1)/2C(BKK))=O~(C(BKK)(d+1)/2)\widetilde{O}(C(\textsc{BKK})^{(d-1)/2}C(\textsc{BKK}))=\widetilde{O}(C(\textsc{BKK})^{(d+1)/2}) (13)

queries, where we have logarithmic factors due to the pointers being encoded in binary and error reduction.  From Theorem 10, we have C(BKK)=O(nlog3n)=n(logn)3+o(1)C(\textsc{BKK})=O(n\log^{3}n)=n(\log n)^{3+o(1)}, so the search takes

O~(n(d+1)/2(logn)(3+o(1))(d+1)/2)=O~(N1/4n1/2(logn)(3/2+o(1))d)=O~(N1/4LN[1/2,37/40]).\widetilde{O}(n^{(d+1)/2}(\log n)^{(3+o(1))(d+1)/2})=\widetilde{O}(N^{1/4}n^{1/2}(\log n)^{(3/2+o(1))d})=\widetilde{O}(N^{1/4}L_{N}[1/2,37/40]). (14)

quantum queries, which is is O(N1/4LN[1/2,1])O(N^{1/4}L_{N}[1/2,1]). ∎

Using this function RecBKK we can now establish Theorem 2.  This proof is very similar to the separation between quantum query complexity and exact degree in Section 4.2.

Let ff be the total function RecBKK:{0,1}n{0,1}\textsc{RecBKK}:\{0,1\}^{n}\to\{0,1\} with Q(f)=n1o(1)Q(f)=n^{1-o(1)} and C(f)=n1/2+o(1)C(f)=n^{1/2+o(1)}, and more importantly given query access to this certificate, a quantum algorithm can check its validity using n1/4+o(1)n^{1/4+o(1)} queries.  Let fCSf_{\mathrm{CS}} denote the cheat sheet version of ff created using 10logn10\log n copies of ff as before.  From Lemma 12 we know that Q(fCS)=Ω(Q(f))Q(f_{\mathrm{CS}})=\Omega(Q(f)) and hence Q(fCS)=n1o(1)Q(f_{\mathrm{CS}})=n^{1-o(1)}.  We now show deg~(fCS)=n1/4+o(1)\operatorname{\widetilde{\deg}}(f_{\mathrm{CS}})=n^{1/4+o(1)}.

Let [n10]\ell\in[n^{10}] be a potential location of the cheat sheet.  For any \ell, consider the problem of outputting 11 if and only if fCS(z)=1f_{\mathrm{CS}}(z)=1 and \ell is the location of the cheat sheet for the input zz.  For any fixed \ell, this can be solved by a quantum algorithm 𝒬\mathcal{Q}_{\ell} that makes n1/4+o(1)n^{1/4+o(1)} queries as shown in Theorem 13, since it can simply check if the certificate stored at cell \ell in the array is valid: it can check all the 10logn10\log n certificates for each of the 10logn10\log n instances of ff and then check if the outputs of ff evaluate to the location \ell.

Since approximate polynomial degree is at most quantum query complexity, we can construct a representing polynomial for 𝒬\mathcal{Q}_{\ell} for any location \ell.  This is a polynomial pp_{\ell} of degree n1/4+o(1)n^{1/4+o(1)} such that p(z)=1p_{\ell}(z)=1 if and only if fCS(z)=1f_{\mathrm{CS}}(z)=1 and \ell is the position of the cheat sheet on input zz.  Now we can simply add all the polynomials pp_{\ell} together to obtain a new polynomial qq of the same degree, except that we first reduce the error in each polynomial to below 1/n101/n^{10} so that the total error is bounded.  (This can be done, for example, using the amplification polynomial construction of [BNRdW07, Lemma 1].)  Now q(z)=fCS(z)q(z)=f_{\mathrm{CS}}(z) as argued in Section 4.2.

6 Cheat sheet functions

In this section we define the cheat sheet framework more generally.  Once the framework is set up, proofs based on the framework are short and conveniently separate out facts about the framework from results about the separation under consideration.  To demonstrate this, we reprove some known separations in Section 6.4.

6.1 Certifying functions and cheat sheets

Intuitively, a certifying function for a function f:D{0,1}f:D\to\{0,1\}, where D[M]nD\subseteq[M]^{n}, is a function ϕ\phi that takes in an input x[M]nx\in[M]^{n}, a claimed value y1y_{1} for f(x)f(x), and a proof (y2,y3,,yk)(y_{2},y_{3},\ldots,y_{k}) that we indeed have xDx\in D and f(x)=y1f(x)=y_{1}.  The value of the certifying function will be 1 only when these conditions hold.  We would also like the certifying function to depend nontrivially on the certificate yy, i.e., for every xDx\in D there should be some yy that makes the function output 11, and some yy that makes it output 0.  For convenience of analysis, we enforce that the certificate y=0ky=0^{k} is invalid for all xx.  Any nontrivial certifying function can be made to have this property by requiring that (say) the second bit of yy, y2y_{2}, is 11 for all valid certificates.

Definition 14 (Certifying function).

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n} for some integer M2M\geq 2.  We say a total function ϕ:[M]n×{0,1}k{0,1}\phi:[M]^{n}\times\{0,1\}^{k}\to\{0,1\}, where kk is any positive integer, is a certifying function for ff if the following conditions are satisfied:

  1. 1.

    xD\forall x\notin D, ϕ(x,y)=0\phi(x,y)=0 (invalid inputs should not have certificates)

  2. 2.

    x[M]n\forall x\in[M]^{n}, y{0,1}k\forall y\in\{0,1\}^{k}, if y1f(x)y_{1}\neq f(x) then ϕ(x,y)=0\phi(x,y)=0 (certificate asserts y1=f(x)y_{1}=f(x))

  3. 3.

    xD\forall x\in D, y{0,1}k\exists y\in\{0,1\}^{k} such that ϕ(x,y)=1\phi(x,y)=1 (valid inputs should have certificates)

  4. 4.

    x[M]n\forall x\in[M]^{n}, ϕ(x,0k)=0\phi(x,0^{k})=0 (nontriviality condition)

Typically, a certifying function will be defined so that its value is only 11 if yy includes pointers to a certificate in xx and if ff is a partial function, we will also want yy to include a proof that xx satisfies the promise of ff.  Thus the query complexity of ϕ\phi may be smaller than that of ff.

We now define the cheat sheet version of a function ff with certifying function ϕ\phi.  In the separations in the previous sections we used 10logn10\log n copies of the function ff to create the address of the cheat sheet.  For generality we now use c=10logD(f)c=\lceil 10\log D(f)\rceil instead so that the construction only adds logarithmic factors in D(f)D(f), as opposed to logarithmic factors in nn, which may be larger than D(f)D(f).  This does not make a difference in our applications, however.

Definition 15 (Cheat sheet version of ff).

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n} for some integer M2M\geq 2, and let c=10logD(f)c=\lceil 10\log D(f)\rceil. Let ϕ:[M]n×{0,1}k{0,1}\phi:[M]^{n}\times\{0,1\}^{k}\to\{0,1\}, for some positive integer kk, be a certifying function for ff.  Then the cheat sheet version of ff with respect to ϕ\phi, denoted fCS(ϕ)f_{\mathrm{CS(\phi)}}, is a total function

fCS(ϕ):([M]n)c×(({0,1}k)c)2c{0,1}f_{\mathrm{CS(\phi)}}:([M]^{n})^{c}\times((\{0,1\}^{k})^{c})^{2^{c}}\to\{0,1\} (15)

acting on an input (X,Y1,Y2,,Y2c)(X,Y_{1},Y_{2},\ldots,Y_{2^{c}}), where X([M]n)cX\in([M]^{n})^{c} and Yi({0,1}k)cY_{i}\in(\{0,1\}^{k})^{c}.  Also, let X([M]n)cX\in([M]^{n})^{c} be written as X=(x1,x2,,xc)X=(x^{1},x^{2},\ldots,x^{c}), where xi[M]nx^{i}\in[M]^{n}.  Then we define the value of fCS(ϕ)(X,Y1,Y2,Y2c)f_{\mathrm{CS(\phi)}}(X,Y_{1},Y_{2},\ldots Y_{2^{c}}) to be 11 if and only if conditions (1) and (2) hold.

  1. (1)

    For all i[c]i\in[c], xiDx^{i}\in D.

If condition (1) is satisfied, let [2c]\ell\in[2^{c}] be the positive integer corresponding to the binary string (f(x1),f(x2),,f(xc))(f(x^{1}),f(x^{2}),\ldots,f(x^{c})) and let Y=(y1,y2,yc)Y_{\ell}=(y^{1},y^{2},\ldots y^{c}) where each yi{0,1}ky^{i}\in\{0,1\}^{k}.

  1. (2)

    For all i[c]i\in[c], xiDx^{i}\in D and ϕ(xi,yi)=1\phi(x^{i},y^{i})=1.

Intuitively, the input to fCS(ϕ)f_{\mathrm{CS(\phi)}} is interpreted as cc inputs to ff, which we call (x1,x2,,xc)(x^{1},x^{2},\ldots,x^{c}), followed by 2c2^{c} strings (Y1,Y2,Y2c)(Y_{1},Y_{2},\ldots Y_{2^{c}}) of length ckck called cheat sheets.  The value of the function will be 0 if any of the cc inputs to ff do not satisfy the promise of ff.  If they do satisfy the promise, let [2c]\ell\in[2^{c}] be the number encoded by the binary string f(x1)f(x2)f(xc)f(x^{1})f(x^{2})\dots f(x^{c}), where 0c0^{c} and 1c1^{c} encode the numbers 11 and 2c2^{c} respectively.  If the th\ell^{\mathrm{th}} cheat sheet in the input, YY_{\ell}, is written as Y=(y1,y2,,yc)Y_{\ell}=(y^{1},y^{2},\ldots,y^{c}), where yi{0,1}ky^{i}\in\{0,1\}^{k} for all ii, then the value of fCS(ϕ)f_{\mathrm{CS(\phi)}} is 11 if ϕ(xi,yi)=1\phi(x^{i},y^{i})=1 for all i[c]i\in[c], and 0 otherwise.

In other words, the value of fCS(ϕ)f_{\mathrm{CS(\phi)}} is 11 if the first part of its input consists of cc valid inputs to ff that together encode a location of a cheat sheet in the second part of the input, and this cheat sheet in turn contains pointers that certify the values of the cc inputs to ff.  The idea is that the only way to find the cheat sheet is to solve the cc copies of ff, so that fCS(ϕ)f_{\mathrm{CS(\phi)}} has query complexity at least that of ff, but once the cheat sheet has been found one can verify the value of fCS(ϕ)f_{\mathrm{CS(\phi)}} using only the query complexity of ϕ\phi, which may be much smaller.

We now characterize the query complexity of fCS(ϕ)f_{\mathrm{CS(\phi)}} in terms of the query complexities of ff and ϕ\phi in various models.  The following result summarizes our results for the query complexity of cheat sheet functions.

Theorem 16 (Query complexity of cheat sheet functions).

Let f:D{0,1}f:D\to\{0,1\}, where D{0,1}nD\subseteq\{0,1\}^{n}, be a partial function, and let ϕ\phi be a certifying function for ff.  Then the following relations hold.

  • D(fCS(ϕ))=Θ~(D(f)+D(ϕ))D(f_{\mathrm{CS(\phi)}})=\widetilde{\Theta}(D(f)+D(\phi))

  • R(fCS(ϕ))=Θ~(R(f)+R(ϕ))R(f_{\mathrm{CS(\phi)}})=\widetilde{\Theta}(R(f)+R(\phi))

  • Q(fCS(ϕ))=Θ~(Q(f)+Q(ϕ))Q(f_{\mathrm{CS(\phi)}})=\widetilde{\Theta}(Q(f)+Q(\phi))

  • deg(fCS(ϕ))=Θ~(deg(ϕ))\deg(f_{\mathrm{CS(\phi)}})=\widetilde{\Theta}(\deg(\phi))

  • deg~(fCS(ϕ))=Θ~(deg~(ϕ))\operatorname{\widetilde{\deg}}(f_{\mathrm{CS(\phi)}})=\widetilde{\Theta}(\operatorname{\widetilde{\deg}}(\phi))

  • R0(fCS(ϕ))=Ω~(R0(f)+R0(ϕ))R_{0}(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(R_{0}(f)+R_{0}(\phi)) (but the corresponding upper bound relation is false)

  • QE(fCS(ϕ))=O~(QE(f)+QE(ϕ))Q_{E}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(Q_{E}(f)+Q_{E}(\phi)) (we conjecture that the corresponding lower bound holds)

The algorithmic measures, DD, RR, and QQ behave as expected since Theorem 16 asserts that the straightforward way to compute fCS(ϕ)f_{\mathrm{CS(\phi)}} in these models, which is to first compute all cc copies of ff and then check if they point to a valid cheat sheet, is essentially optimal.  We conjecture that QEQ_{E}, the quantum analogue of DD, should behave similarly to DD or QQ, but we can only prove the upper bound (see Lemma 17). In fact, if the lower bound for QEQ_{E} can be proved, then we would get a near cubic separation between Q(f)Q(f) and QE(f)Q_{E}(f).

For the zero-error class R0R_{0}, while we can show that the lower bound holds, the upper bound relation is false (see Theorem 28 for a counterexample).  We believe the zero-error class Q0Q_{0} behaves similarly: we conjecture the analogous lower bound holds for Q0Q_{0}, but are unable to prove it.  For Q0Q_{0}, the corresponding upper bound is false, for reasons similar to those for R0R_{0}.

Notably, one-sided error measures do not behave well for cheat sheet functions at all.  In Theorem 27, we demonstrate ff and ϕ\phi such that R1(fCS(ϕ))R1(f)R_{1}(f_{\mathrm{CS(\phi)}})\ll R_{1}(f), showing that the lower bound fails.  The upper bound need not hold either.

Lastly, note that deg\deg and deg~\operatorname{\widetilde{\deg}} behave fundamentally differently on cheat sheet functions than the algorithmic measures.  The (approximate) degree of the function ff does not even play a role in determining the (approximate) degree of fCS(ϕ)f_{\mathrm{CS(\phi)}}.

Theorem 16 is proved in the next two sections.  In Appendix B we present some examples where the naïvely expected relations do not hold.

6.2 Upper bounds on the complexity of cheat sheet functions

We start by showing that the usual algorithmic models can compute fCS(ϕ)f_{\mathrm{CS(\phi)}} in roughly the number of queries they require for ff and ϕ\phi.

Lemma 17.

Let f:D{0,1}f:D\to\{0,1\}, where D[M]nD\subseteq[M]^{n}, be a partial function, and ϕ\phi be a certifying function for ff.  Then the following upper bounds hold.

  • D(fCS(ϕ))=O~(D(f)+D(ϕ))D(f_{\mathrm{CS(\phi)}})=\widetilde{O}(D(f)+D(\phi))

  • R(fCS(ϕ))=O~(R(f)+R(ϕ))R(f_{\mathrm{CS(\phi)}})=\widetilde{O}(R(f)+R(\phi))

  • Q(fCS(ϕ))=O~(Q(f)+Q(ϕ))Q(f_{\mathrm{CS(\phi)}})=\widetilde{O}(Q(f)+Q(\phi))

  • QE(fCS(ϕ))=O~(QE(f)+QE(ϕ))Q_{E}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(Q_{E}(f)+Q_{E}(\phi))

Proof.

The algorithm to evaluate fCS(ϕ)f_{\mathrm{CS(\phi)}} is the following: First, evaluate ff on the cc inputs to ff contained in zz, assuming they satisfy the promise of ff.  If one of these inputs is discovered to contradict the promise of ff, then output 0 and halt.  This takes cD(f)cD(f) queries for a deterministic algorithm (and cQE(f)cQ_{E}(f) queries for an exact quantum algorithm), O(R(f)clogc)O(R(f)c\log c) for a bounded-error randomized algorithm, and O(Q(f)clogc)O(Q(f)c\log c) for a bounded-error quantum algorithm, where the factor of O(logc)O(\log c) comes from reducing the error enough to ensure that all cc functions are computed correctly with high probability.  From the cc outputs we get a binary number \ell.  We can then go to the appropriate cheat sheet, and evaluate ϕ\phi on all the (xi,yi)(x^{i},y^{i}) pairs and output 11 if they are all 11.  This takes cD(ϕ)cD(\phi) queries for a deterministic algorithm (and cQE(ϕ)cQ_{E}(\phi) queries for an exact quantum algorithm), O(R(ϕ)clogc)O(R(\phi)c\log c) queries for a randomized algorithm, and O(Q(ϕ)clogc)O(Q(\phi)c\log c) queries for a quantum algorithm. Since ϕ(x,y)=0\phi(x,y)=0 when xx does not satisfy the promise of ff, this algorithm is correct.  ∎

Before proving the upper bounds for the degree measures, we prove a technical lemma that shows that log(D(f))\log(D(f)) and log(C(ϕ))\log(C(\phi)) or log(deg~(ϕ))\log(\operatorname{\widetilde{\deg}}(\phi)) are the same up to constants.  We prove this to show that the log factors that appear in our construction really can be neglected compared to the measures we consider.

Lemma 18.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n}, and let ϕ\phi be a certifying function for ff.  Then D(f)C(1)(ϕ)2D(f)\leq C^{(1)}(\phi)^{2} and C(f)C(1)(ϕ)C(f)\leq C^{(1)}(\phi).  As a consequence, we have D(f)D(ϕ)2D(f)\leq D(\phi)^{2} and C(f)C(ϕ)C(f)\leq C(\phi).

Proof.

Note that every 11-certificate for ϕ(x,y)\phi(x,y) proves that f(x)=y1f(x)=y_{1}.  In particular, such a 11-certificate must reveal a certificate for ff in xx.  This already shows that C(f)C(1)(ϕ)C(f)\leq C^{(1)}(\phi).

Moreover, every 11-certificate of ϕ(x,y)\phi(x,y) where y1=0y_{1}=0 must conflict with every 11-certificate of ϕ(x,y)\phi(x,y) where y1=1y_{1}=1 on some bit in xx.  This is because otherwise, there would be some xx in the promise of ff that can be certified to be both a 0- and a 11-input for ff.

We can then use the following strategy to compute f(x)f(x).  On input xDom(f)x\in\operatorname{Dom}(f), pick a 11-certificate of some ϕ(x,y)\phi(x,y) with y1=0y_{1}=0, and query all of its entries in xx.  This reveals at least one bit of every 11-certificate of ϕ\phi that has y1=1y_{1}=1.  We then find another 11-certificate of some ϕ(x,y)\phi(x,y) where y1=0y_{1}=0, consistent with the inputs revealed in xx so far.  We once again reveal all of its entries in xx. After C(1)(ϕ)C^{(1)}(\phi) iterations of this process, we have either eliminated all 11-certificates of ϕ\phi with y1=0y_{1}=0, or else all 11-certificates of ϕ\phi with y1=1y_{1}=1.  This tells us the value of f(x)f(x), because since xDom(f)x\in\operatorname{Dom}(f), there must be some yy such that ϕ(x,y)=1\phi(x,y)=1, and hence there must be some 11-certificate of ϕ\phi consistent with xx.  The total number of queries used was at most C(1)(ϕ)2C^{(1)}(\phi)^{2}. ∎

Next, we show the upper bounds on the degree measures of fCS(ϕ)f_{\mathrm{CS(\phi)}} and show that they only depend on the degree of ϕ\phi and not on any measure of the function ff.

Lemma 19.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D{0,1}nD\subseteq\{0,1\}^{n}, and ϕ\phi be a certifying function for ff.  Then deg(fCS(ϕ))=O~(deg(ϕ))\deg(f_{\mathrm{CS(\phi)}})=\widetilde{O}(\deg(\phi)) and deg~(fCS(ϕ))=O~(deg~(ϕ))\operatorname{\widetilde{\deg}}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(\operatorname{\widetilde{\deg}}(\phi)).

Proof.

We start by showing this for deg(fCS(ϕ))\deg(f_{\mathrm{CS(\phi)}}).  For each [2c]\ell\in[2^{c}], we construct a polynomial which is 11 if fCS(ϕ)(z)=1f_{\mathrm{CS(\phi)}}(z)=1 and the cheat sheet of zz is the th\ell^{\mathrm{th}} cheat sheet, and 0 otherwise.  We can do this by simply multiplying together the polynomials of ϕ(xi,yi)\phi(x^{i},y^{i}) for all i[c]i\in[c], where xix^{i} are the inputs to ff in zz and yiy^{i} are the entries found in the th\ell^{\mathrm{th}} cheat sheet of zz.

The resulting polynomial, pp_{\ell}, has degree cdeg(ϕ)c\deg(\phi), is 11 on input zz if zz is a 11-input for fCS(ϕ)f_{\mathrm{CS(\phi)}} and \ell is the location of the cheat sheet for input zz, and is 0 on all other inputs.  To get a polynomial for fCS(ϕ)f_{\mathrm{CS(\phi)}}, we simply sum up the polynomials pp_{\ell} for all [2c]\ell\in[2^{c}].  The resulting polynomial is equal to fCS(ϕ)(z)f_{\mathrm{CS(\phi)}}(z) on all inputs zz, and has degree cdeg(ϕ)c\deg(\phi).

The same trick works for deg~\operatorname{\widetilde{\deg}}, except in that case we cannot simply construct the And of cc polynomials by multiplying them together: we must first perform error reduction on the polynomials and map the output range [0,1/3][2/3,1][0,1/3]\cup[2/3,1] to [0,1/3c][11/3c,1][0,1/3c]\cup[1-1/3c,1].  It is possible to do this amplification while increasing the degree of the polynomials by at most O(logc)O(\log c) using, for example, the amplification polynomial construction of [BNRdW07, Lemma 1].  Therefore we have deg~(fCS(ϕ))=O(cdeg~(ϕ)logc)=O~(deg~(ϕ))\operatorname{\widetilde{\deg}}(f_{\mathrm{CS(\phi)}})=O(c\operatorname{\widetilde{\deg}}(\phi)\log c)=\widetilde{O}(\operatorname{\widetilde{\deg}}(\phi)).

Finally, note that cc is at most logarithmic in deg~(ϕ)\operatorname{\widetilde{\deg}}(\phi), since c=O(logD(f))=O(logC(1)(ϕ))=O(logC(ϕ))=O(logdeg~(ϕ))c=O(\log D(f))=O(\log C^{(1)}(\phi))=O(\log C(\phi))=O(\log\operatorname{\widetilde{\deg}}(\phi)), where the last equality follows from [BdW02] and from the fact that ϕ\phi is a total function. ∎

6.3 Lower bounds on the complexity of cheat sheet functions

Lemma 20.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D{0,1}nD\subseteq\{0,1\}^{n}, and let ϕ\phi be a certifying function for ff.  Then the complexity of fCS(ϕ)f_{\mathrm{CS(\phi)}} is at least that of ϕ\phi with respect to any complexity measure that does not increase upon learning input bits, which includes all reasonable complexity models such as QQ, DD, RR, R0R_{0}, R1R_{1}, Q1Q_{1}, Q0Q_{0}, QEQ_{E}, CC, RCRC, bsbs, deg\deg, and deg~\operatorname{\widetilde{\deg}}.

Proof.

Consider restricting fCS(ϕ)f_{\mathrm{CS(\phi)}} to a promise that guarantees that all cheat sheets of the input are identical.  An input zz to this promise problem evaluates to 11 if and only if the ϕ\phi inputs from the first cheat sheet all evaluate to 11.  We add the additional promise that this is true for all but the last input to ϕ\phi.  The value of an input zz to this function then becomes exactly ϕ\phi evaluated on a certain portion of the input.  The complexity measures mentioned do not increase when the function is restricted to a promise.  Thus the complexity of fCS(ϕ)f_{\mathrm{CS(\phi)}} under these measures is at least that of ϕ\phi. ∎

Lemma 21.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n}, and let ϕ\phi be a certifying function for ff.  Then D(fCS(ϕ))=Ω(D(f))D(f_{\mathrm{CS(\phi)}})=\Omega(D(f)).

Proof.

We describe an adversary strategy that can be used to force a deterministic algorithm to make Ω(D(f))\Omega(D(f)) queries.  Let x1,x2,,xcx^{1},x^{2},\dots,x^{c} represent the cc inputs to ff.  Whenever the algorithm queries a bit not in x1,x2,,xcx^{1},x^{2},\dots,x^{c}, the adversary returns 0.  Whenever the algorithm queries a bit in xix^{i}, the adversary uses the adversarial strategy for ff on that input; that is, it returns answers that are consistent with the promise of ff, but that force any deterministic algorithm to use D(f)D(f) queries to discern f(xi)f(x^{i}).

Now if a deterministic algorithm uses fewer than D(f)D(f) queries on an input to fCS(ϕ)f_{\mathrm{CS(\phi)}}, then there must be many cheat sheets in the input that it did not query at all.  In addition, such an algorithm cannot determine the value of f(xi)f(x^{i}) for any ii.  It follows that this algorithm could not have discerned the value of fCS(ϕ)f_{\mathrm{CS(\phi)}} on the given input, and thus D(fCS(ϕ))D(f)D(f_{\mathrm{CS(\phi)}})\geq D(f). ∎

Lemma 22.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n}, and let ϕ\phi be a certifying function for ff.  Then R0(fCS(ϕ))=Ω(R0(f))R_{0}(f_{\mathrm{CS(\phi)}})=\Omega(R_{0}(f)).

Proof.

Let AA be a zero-error randomized algorithm for fCS(ϕ)f_{\mathrm{CS(\phi)}}.  Then AA must always find a certificate for fCS(ϕ)f_{\mathrm{CS(\phi)}}.  Consider running AA on an input zz consisting of cc 0-inputs to ff, together with an all-zeros array.  Certifying such an input requires presenting 0-certificates for some of the cc inputs to ff, and presenting at least one bit from each cheat sheet whose index has not been disproven by the 0-certificates of the ff inputs.

Now, since there are at least D(f)10D(f)^{10} cheat sheets, only a small fraction can be provided in a certificate of size at most O(R0(f))O(R_{0}(f)).  It follows that the algorithm AA must find certificates to at least half of the cc 0-inputs to ff with high probability.

We can now turn AA into a new algorithm BB that finds a certificate in a given 0-input to ff sampled from the hard distribution 𝒟0\mathcal{D}^{0} over the 0-inputs of ff.  Given such an input xx, the algorithm BB will generate c1c-1 additional inputs from 𝒟0\mathcal{D}^{0}, shuffle them together, and add an all-zero array. BB will then feed this input to AA.  We know that with high probability, AA will find a certificate for at least half the inputs to ff, so it must find a certificate for xx with probability close to 1/21/2 (say, probability at least 1/31/3).  If it does not return a certificate, BB can repeat this process.  The expected running time of BB is at most 33 times that of AA.

Similarly, if we feed AA an input consisting of cc 11-inputs to ff together with an all-zero array, we can use a similar argument to construct an algorithm CC for finding 11-certificates in inputs sampled from 𝒟1\mathcal{D}^{1} (the hard distribution over 11-inputs for ff).  We can then alternate running steps of BB and steps of CC to get an algorithm that finds a certificate in an input sampled from either 𝒟0\mathcal{D}^{0} or 𝒟1\mathcal{D}^{1}, which uses an expected number of queries that is at most 66 times that of AA.

This last algorithm evaluates ff on inputs sampled from the hard distribution for ff, and therefore must have expected running time at least R0(f)R_{0}(f).  It follows that the expected running time of AA is at least R0(f)/6R_{0}(f)/6. ∎

The lower bounds for bounded-error randomized query complexity and bounded-error quantum query complexity are exactly as in Lemma 6 and Lemma 12.

6.4 Proofs of known results using cheat sheets

Theorem 23 ([ABB+15]).

There exists a total function ff such that R0(f)=Ω~(Q(f)3)R_{0}(f)=\widetilde{\Omega}(Q(f)^{3}).

Proof.

This proof is similar to that of Theorem 1, except with a different function gg.  Let g:D{0,1}g:D\to\{0,1\}, where D{0,1}nD\in\{0,1\}^{n}, be a partial function that satisfies Q(g)=1Q(g)=1 and R0(g)=Ω(n)R_{0}(g)=\Omega(n), such as the Deutsch–Jozsa problem [DJ92, CEMM98].  R0(g)=Ω(n)R_{0}(g)=\Omega(n) follows from the fact that even certifying that the input is all zeros or all ones requires a certificate of linear size.

Let h:{0,1}m{0,1}h:\{0,1\}^{m}\to\{0,1\} be the And-Or function on mm bits.  Let f=ghf=g\circ h and let ϕ\phi be a certifying function that requires the obvious certificate for ff (as used in the proof of Theorem 1).  Then we have Q(ϕ)=O(n+nm1/4)Q(\phi)=O(n+\sqrt{n}m^{1/4}) and Q(f)=Q(gh)=O(m)Q(f)=Q(g\circ h)=O(\sqrt{m}).  From Theorem 5 we have that R0(f)=R0(gh)=Ω~(nm)R_{0}(f)=R_{0}(g\circ h)=\widetilde{\Omega}(nm).  Using Theorem 16, we have Q(fCS(ϕ))=O~(Q(f)+Q(ϕ))=O~(m+n+nm1/4)Q(f_{\mathrm{CS(\phi)}})=\widetilde{O}(Q(f)+Q(\phi))=\widetilde{O}(\sqrt{m}+n+\sqrt{n}m^{1/4}) and R(fCS(ϕ))=Ω~(R0(f))=Ω~(nm)R(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(R_{0}(f))=\widetilde{\Omega}(nm).  Plugging in m=n2m=n^{2}, we get Q(fCS(ϕ))=O~(n)Q(f_{\mathrm{CS(\phi)}})=\widetilde{O}(n) and R(fCS(ϕ))=Ω~(n3)R(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(n^{3}). ∎

Theorem 24 ([ABB+15]).

There exists a total function ff such that R(f)=Ω~(QE(f)3/2)R(f)=\widetilde{\Omega}(Q_{E}(f)^{3/2}) .

Proof.

Let g:D{0,1}g:D\to\{0,1\}, where D={x{0,1}n:|x|=0 or |x|=1}D=\{x\in\{0,1\}^{n}:|x|=0\textrm{ or }|x|=1\}, be the partial function that evaluates to 0 if the input is the all-zeros string and evaluates to 11 if the input string has Hamming weight 11.  This function satisfies R(g)=Ω(n)R(g)=\Omega(n), since the block sensitivity of the 0-input is linear and QE(g)=O(n)Q_{E}(g)=O(\sqrt{n}), since Grover’s algorithm can be made exact in this case [BHMT02].

Let h:{0,1}m{0,1}h:\{0,1\}^{m}\to\{0,1\} be the And function on mm bits, which satisfies R(h)=Ω(m)R(h)=\Omega(m) and QE(h)=O(m)Q_{E}(h)=O(m).  Let f=ghf=g\circ h, and let ϕ\phi be a certifying function for ff.  The certificate complexity of this function is O(n)O(n) in the no case (by showing one 0-input to each of the nn And gates), and is O(m)O(m) in the yes case (by showing all the 11 inputs to an And gates).  We choose the certifying function ϕ\phi that requires this certificate for ff and hence QE(ϕ)D(ϕ)=O~(n+m)Q_{E}(\phi)\leq D(\phi)=\widetilde{O}(n+m).

Now by composing quantum algorithms for gg and hh we have QE(f)=O(mn)Q_{E}(f)=O(m\sqrt{n}), and by Theorem 5 we have R(f)=Ω(mn)R(f)=\Omega(mn).  Hence using Theorem 16, we have QE(fCS(ϕ))=O~(QE(f)+QE(ϕ))=O~(n+m+mn)Q_{E}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(Q_{E}(f)+Q_{E}(\phi))=\widetilde{O}(n+m+m\sqrt{n}), and R(fCS(ϕ))=Ω~(mn)R(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(mn).  Choosing m=nm=\sqrt{n} gives QE(fCS(ϕ))=O~(n)Q_{E}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(n) and R(fCS(ϕ))=Ω~(n3/2)R(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(n^{3/2}). ∎

Acknowledgements

We thank Ansis Rosmanis for comments on Appendix A and Mark Bun for discussions on approximate degree lower bounds. We thank Hardik Bansal for spotting an error in an earlier version of the proof of Theorem 5. We also thank Iordanis Kerenidis, Frédéric Magniez, and Ashwin Nayak for catching a bug in an earlier version of the proof of Lemma 6.

This work was partially funded by the ARO grant Contract Number W911NF-12-1-0486, as well as by an Alan T. Waterman Award from the National Science Foundation, under grant no. 1249349. This preprint is MIT-CTP #4730.

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Appendix A Quantum query complexity of k-sumk\textsc{-sum}

Belovs and Špalek [BŠ13] proved a lower bound of Ω(nk/(k+1))\Omega(n^{k/(k+1)}) on the quantum query complexity of k-sumk\textsc{-sum} for constant kk, as long as the alphabet is large enough.  However, in their result, Belovs and Špalek do not keep track of the dependence on kk.  We trace their proof to determine this dependence, ultimately proving the following theorem.

Theorem 25.

For the function k-sum:[q]n{0,1}k\textsc{-sum}:[q]^{n}\to\{0,1\}, if |q|2(nk)|q|\geq 2{\binom{n}{k}}, we have

Q(k-sum)=Ω(nk/(k+1)/k),Q(k\textsc{-sum})=\Omega(n^{k/(k+1)}/\sqrt{k}), (16)

where the constant in the Ω\Omega is independent of kk.

Proof.

We proceed by tracing Belovs and Špalek’s proof, which can be found in section 33 of [BŠ13], to extract an unsimplified lower bound.  We then tune some parameters to determine the best dependence on kk.

The proof in [BŠ13] proceeds by the generalized adversary method.  Belovs and Špalek construct an adversary matrix Γ\Gamma, which depends on parameters αm\alpha_{m} for m=0,1,,nkm=0,1,\dots,n-k.  This construction can be found in Section 3.13.1 of their paper.  They then show a lower bound on Γ\|\Gamma\| and an upper bound on ΓΔ1\|\Gamma\circ\Delta_{1}\| in terms of the parameters αm\alpha_{m} (the upper bounds on ΓΔi\|\Gamma\circ\Delta_{i}\| for i1i\neq 1 follow by symmetry).  Finally, they pick the values for the parameters that give the best lower bound.

A lower bound on Γ\|\Gamma\| without Ω\Omega notation, which therefore does not ignore factors of kk, can be found in Section 3.63.6 of their paper.  In that section, they show

Γα0(nk)(11q(nk)),\|\Gamma\|\geq\alpha_{0}\sqrt{{\binom{n}{k}}\left(1-\frac{1}{q}{\binom{n}{k}}\right)}, (17)

where qq is the alphabet size.  When q43(nk)q\geq\frac{4}{3}{\binom{n}{k}}, this gives

Γα02(nk).\|\Gamma\|\geq\frac{\alpha_{0}}{2}\sqrt{{\binom{n}{k}}}. (18)

For the upper bound on ΓΔ1\|\Gamma\circ\Delta_{1}\|, it turns out to be more convenient to switch to a different matrix, Γ~\tilde{\Gamma}, which satisfies ΓΔ1Γ~Δ1\|\Gamma\circ\Delta_{1}\|\leq\|\tilde{\Gamma}\circ\Delta_{1}\|.  This is explained in the beginning of Section 33 of their paper and in Section 3.13.1.  To upper bound Γ~Δ1\|\tilde{\Gamma}\circ\Delta_{1}\|, in Section 3.33.3 the authors switch to yet another matrix, Γ~1\tilde{\Gamma}_{1}, with the property that Γ~1Δ1=Γ~Δ1\tilde{\Gamma}_{1}\circ\Delta_{1}=\tilde{\Gamma}\circ\Delta_{1} and that Γ~1Δ12Γ~1\|\tilde{\Gamma}_{1}\circ\Delta_{1}\|\leq 2\|\tilde{\Gamma}_{1}\|.  This gives ΓΔ12Γ~1\|\Gamma\circ\Delta_{1}\|\leq 2\|\tilde{\Gamma}_{1}\|, so it suffices to upper bound Γ~1\|\tilde{\Gamma}_{1}\|.

In Section 3.43.4, the authors express Γ~12\|\tilde{\Gamma}_{1}\|^{2} as the norm of a sum of matrices, with the sum ranging over a set 𝒮\mathcal{S}.  They split 𝒮=𝒮1𝒮2\mathcal{S}=\mathcal{S}_{1}\cup\mathcal{S}_{2}; by the triangle inequality, it then suffices to separately upper bound the norm of the sum over 𝒮1\mathcal{S}_{1} and the norm of the sum over 𝒮2\mathcal{S}_{2}.  The former is upper bounded by maxmαm2(m+k1k1)\max_{m}\alpha_{m}^{2}{m+k-1\choose k-1}, and the latter by k(n1k)maxm(αmαm+1)2k{n-1\choose k}\max_{m}(\alpha_{m}-\alpha_{m+1})^{2}.  We therefore conclude that

ΓΔ124maxmαm2(m+k1k1)+4k(n1k)maxm(αmαm+1)2.\|\Gamma\circ\Delta_{1}\|^{2}\leq 4\max_{m}\alpha_{m}^{2}{m+k-1\choose k-1}+4k{n-1\choose k}\max_{m}(\alpha_{m}-\alpha_{m+1})^{2}. (19)

It remains to pick values for αm\alpha_{m} to optimize the resulting adversary bound.  There are no restrictions on the αm\alpha_{m} parameters.  To maximize the adversary bound, we want α0\alpha_{0} to be large, consecutive α\alpha’s to be close (to minimize maxm(αmαm+1)2\max_{m}(\alpha_{m}-\alpha_{m+1})^{2}), and αm\alpha_{m} for large mm to be small (to minimize maxmαm2(m+k1k1)\max_{m}\alpha_{m}^{2}{m+k-1\choose k-1}).  We choose to make the αm\alpha_{m} parameters decrease to 0 in an arithmetic sequence, with αmαm+1=c>0\alpha_{m}-\alpha_{m+1}=c>0 for all mα0/cm\leq\alpha_{0}/c, and αm=0\alpha_{m}=0 for all mα0/cm\geq\alpha_{0}/c.  Let β=α0/c\beta=\alpha_{0}/c.  Then αm=α0cm\alpha_{m}=\alpha_{0}-cm for mβm\leq\beta.

We can write the final adversary bound as follows:

Adv±(k-sum)2=Ω(α02(nk)maxmβ(α0cm)2(m+k1k1)+k(n1k)c2)\operatorname{Adv^{\pm}}(k\textsc{-sum})^{2}=\Omega\left(\frac{\alpha_{0}^{2}{\binom{n}{k}}}{\max_{m\leq\beta}(\alpha_{0}-cm)^{2}{m+k-1\choose k-1}+k{n-1\choose k}c^{2}}\right) (20)
=Ω(min{(nk)maxmβ(1m/β)2(m+k1k1),(nk)k(n1k)β2})=\Omega\left(\min\left\{\frac{{\binom{n}{k}}}{\max_{m\leq\beta}(1-m/\beta)^{2}{m+k-1\choose k-1}},\frac{{\binom{n}{k}}}{k{n-1\choose k}}\beta^{2}\right\}\right) (21)

The second term in the minimization simplifies to nk(nk)β2\frac{n}{k(n-k)}\beta^{2}, which is Ω(β2/k)\Omega(\beta^{2}/k). To deal with the first term, we use the well-known inequalities (n/k)k(nk)(en/k)k(n/k)^{k}\leq{\binom{n}{k}}\leq(en/k)^{k}, which hold for all nn and kk.  This gives

Adv±(k-sum)2=Ω(min{nk(k1)maxmβ(1m/β)2(e(m+k1))k1(k1k)k,β2k})\operatorname{Adv^{\pm}}(k\textsc{-sum})^{2}=\Omega\left(\min\left\{\frac{n^{k}}{(k-1)\max_{m\leq\beta}(1-m/\beta)^{2}(e(m+k-1))^{k-1}}\left(\frac{k-1}{k}\right)^{k},\frac{\beta^{2}}{k}\right\}\right) (22)
=Ω(min{nkkmaxmβ(1m/β)2(e(m+k))k1,β2k}).=\Omega\left(\min\left\{\frac{n^{k}}{k\max_{m\leq\beta}(1-m/\beta)^{2}(e(m+k))^{k-1}},\frac{\beta^{2}}{k}\right\}\right). (23)

We can solve the maximization in the denominator using calculus.  The unique maximum occurs at m=β(2k+1)(βk)m=\beta-\left(\frac{2}{k+1}\right)(\beta-k).  Substituting this in and simplifying, we get

Adv±(k-sum)2=Ω(min{kβ(1k/β)2(ne(β+3))k,β2k})=Ω(min{kβ(n3β)k,β2k}),\operatorname{Adv^{\pm}}(k\textsc{-sum})^{2}=\Omega\left(\min\left\{\frac{k\beta}{(1-k/\beta)^{2}}\left(\frac{n}{e(\beta+3)}\right)^{k},\frac{\beta^{2}}{k}\right\}\right)=\Omega\left(\min\left\{k\beta\left(\frac{n}{3\beta}\right)^{k},\frac{\beta^{2}}{k}\right\}\right), (24)

where for the last equality we assumed β2k\beta\geq 2k and β3e/(3e)\beta\geq 3e/(3-e).

Finally, we set β=(1/3)nk/(k+1)\beta=(1/3)n^{k/(k+1)}.  This gives

Adv±(k-sum)2=Ω(min{knk/(k+1)(nnk/(k+1))k,n2k/(k+1)k})=Ω(n2k/(k+1)/k),\operatorname{Adv^{\pm}}(k\textsc{-sum})^{2}=\Omega\left(\min\left\{kn^{k/(k+1)}\left(\frac{n}{n^{k/(k+1)}}\right)^{k},\frac{n^{2k/(k+1)}}{k}\right\}\right)=\Omega(n^{2k/(k+1)}/k), (25)

so

Q(k-sum)=Ω(Adv±(k-sum))=Ω(nk/(k+1)/k).Q(k\textsc{-sum})=\Omega(\operatorname{Adv^{\pm}}(k\textsc{-sum}))=\Omega\left(n^{k/(k+1)}\middle/\sqrt{k}\right). (26)

Note that we assumed β2k\beta\geq 2k.  Since β=nk/(k+1)/3\beta=n^{k/(k+1)}/3, we need nk/(k+1)6kn^{k/(k+1)}\geq 6k, or n(6k)1+1/kn\geq(6k)^{1+1/k}.  Since (6k)1/k=exp(ln(6k)/k)=1+Θ(log(k)/k)(6k)^{1/k}=\exp(\ln(6k)/k)=1+\Theta(\log(k)/k), it suffices to have n6k+ω(logk)n\geq 6k+\omega(\log k).  In particular, this bound works as long as we have kn/10k\leq n/10.  When kn/10k\geq n/10, we can directly prove a lower bound of Ω(n)\Omega(\sqrt{n}) by a reduction from Grover search; this means there are no restrictions on the size of nn and kk in this result. ∎

Appendix B Measures that behave curiously with cheat sheets

In this appendix we show that R1R_{1} and Q1Q_{1} can behave strangely on cheat sheet functions, potentially decreasing from R1(f)R_{1}(f) to R(f)R(f) and from Q1(f)Q_{1}(f) to Q(f)Q(f).

Theorem 26.

Let f:D{0,1}f:D\to\{0,1\} be a partial function, where D[M]nD\subseteq[M]^{n}, and let ϕ\phi be a certifying function for ff.  Then

R1(fCS(ϕ))=O~(R(f)+R0(ϕ))andQ1(fCS(ϕ))=O~(Q(f)+Q0(ϕ)).R_{1}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(R(f)+R_{0}(\phi))\qquad\text{and}\qquad Q_{1}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(Q(f)+Q_{0}(\phi)). (27)
Proof.

We show a randomized (and quantum) algorithm for fCS(ϕ)f_{\mathrm{CS(\phi)}} that uses the required number of queries and finds a 11-certificate with constant probability.  Such an algorithm could be made to have one-sided error by outputting 11 only if it finds a 11-certificate.

The algorithm works as follows.  First, use the randomized (resp. quantum) algorithm on the cc inputs to ff (with some amplification), to determine the number \ell of the correct cheat sheet with high probability (assuming the promises of ff all hold).  Next, go to the th\ell^{\mathrm{th}} cheat sheet, and use a zero-error algorithm to evaluate ϕ(xi,yi)\phi(x_{i},y_{i}) for i=1,2,,ci=1,2,\dots,c, where xix_{i} are the inputs to ff and yiy_{i} are the cheat sheet strings.  This finds certificates for each input to ϕ\phi.

Now, if the value of the function is 11 on the given input, then with constant probability, all the cc certificates found should be 11-certificates for ϕ\phi.  These 11-certificates certify the value of the inputs to ff.  Therefore, taken together, they constitute a valid 11-certificate for fCS(ϕ)f_{\mathrm{CS(\phi)}}.  It follows that this algorithm uses O~(R(f)+R0(ϕ))\widetilde{O}(R(f)+R_{0}(\phi)) randomized queries and finds a 11-certificate with constant probability.  The result for quantum algorithms follows similarly. ∎

We now show that it is possible for R1(fCS(ϕ))R_{1}(f_{\mathrm{CS(\phi)}}) to be much smaller than R1(f)R_{1}(f), unlike some of the other measures studied.  Intuitively, this is because 11-inputs of fCS(ϕ)f_{\mathrm{CS(\phi)}} contains a cheat sheet with a certificate and hence even if the algorithm is not sure of its answer before finding the cheat sheet, the cheat sheet may convince the algorithm of its answer.

Theorem 27.

There is a total function f:{0,1}n{0,1}f:\{0,1\}^{n}\to\{0,1\} and a certifying function ϕ\phi for ff such that R1(fCS(ϕ))=O~(R1(f))R_{1}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(\sqrt{R_{1}(f)}).

Proof.

We can use Theorem 26 to construct an explicit function ff and a certifying function ϕ\phi for ff such that R1(fCS(ϕ))R_{1}(f_{\mathrm{CS(\phi)}}) is smaller than R1(f)R_{1}(f).  The construction is as follows.  In [ABB+15], a Boolean function was constructed that has R0=Ω~(n2)R_{0}=\widetilde{\Omega}(n^{2}) and R1=O~(n)R_{1}=\widetilde{O}(n).  This function has certificate complexity roughly equal to nn.  Now, by taking the Xor of this function with an additional bit, we get a function ff for which R1R_{1} is roughly n2n^{2} but RR and CC are still roughly nn.

We define ϕ\phi to be a certifying function that simply checks if yy provides pointers to a certificate for ff.  Then R0(ϕ)=O~(n)R_{0}(\phi)=\widetilde{O}(n). It follows that R1(fCS(ϕ))=O~(n)R_{1}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(n), while R1(f)=Ω~(n2)R_{1}(f)=\widetilde{\Omega}(n^{2}). ∎

Lastly we show that even zero-error randomized query complexity behaves curiously with cheat sheets.  We might have expected that the obvious upper bound R0(fCS(ϕ))=O~(R0(f)+R0(ϕ))R_{0}(f_{\mathrm{CS(\phi)}})=\widetilde{O}(R_{0}(f)+R_{0}(\phi)) holds, but in fact it does not.  The reason is subtle and relates to the behavior of the zero-error algorithm on inputs outside of the domain DD.  For a partial function f:D{0,1}f:D\to\{0,1\}, a zero-error algorithm’s behavior is not constrained on inputs outside the domain DD.  The algorithm could, for example, run forever on such inputs.  The obvious algorithm, which simply runs the zero-error algorithm for ff on all cc inputs to ff now fails to output any answer on 0-inputs to fCS(ϕ)f_{\mathrm{CS(\phi)}} in which the inputs to ff do not lie in DD.  We exploit this observation to prove the following counterexample.

Theorem 28.

There is a partial function f:D{0,1}f:D\to\{0,1\}, where D{0,1}nD\subseteq\{0,1\}^{n}, and a certifying function ϕ\phi for ff such that R0(fCS(ϕ))=Ω~(R0(f)3/2+R0(ϕ)3/2)R_{0}(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(R_{0}(f)^{3/2}+R_{0}(\phi)^{3/2}).

Proof.

Let g:{0,1}4m{0,1}g:\{0,1\}^{4m}\to\{0,1\} be the partial function defined as follows.  On input (x,y)(x,y) with x,y{0,1}2mx,y\in\{0,1\}^{2m}, define g(x,y)=0g(x,y)=0 if the Hamming weight of xx is mm and the Hamming weight of yy is 0, and define g(x,y)=1g(x,y)=1 if the Hamming weight of yy is mm and the Hamming weight of xx is 0.  The promise of gg is that one of these two conditions holds for every input (x,y)(x,y).

It is easy to see that R0(g)=O(1)R_{0}(g)=O(1).  Let ff be the composition of gg with an mm by mm And-Or, and let n=4m3n=4m^{3}.  Then R0(f)=O(m2)=O(n2/3)R_{0}(f)=O(m^{2})=O(n^{2/3}).  Let ϕ\phi be a certifying function for ff that takes an input to ff and an additional string, and asserts that the latter provides pointers to a certificate for each And-Or instance in the input to ff.  This assertion can be verified (or refuted) in O~(m2)\widetilde{O}(m^{2}) deterministic queries, so R0(ϕ)D(ϕ)=O~(m2)R_{0}(\phi)\leq D(\phi)=\widetilde{O}(m^{2}).

We now show that R0(fCS(ϕ))=Ω~(m3)R_{0}(f_{\mathrm{CS(\phi)}})=\widetilde{\Omega}(m^{3}).  Consider an input consisting of cc inputs to ff, each of which has all its And-Or instances evaluate to 0.  Note that these inputs to ff do not satisfy the promise of ff.  We attach a blank cheat sheet array after these cc inputs.  By definition of fCS(ϕ)f_{\mathrm{CS(\phi)}}, the resulting input is therefore a 0-input to fCS(ϕ)f_{\mathrm{CS(\phi)}}.  However, any 0-certificate of it (of reasonable size) must “partially certify” at least half of the cc inputs to ff; that is, for at least half the inputs to ff, it must either prove the input is not a 0-input or prove it is not a 11-input.  The only way to do this is to display (c/2)m(c/2)m 0-certificates for And-Or instances.

This means we have an algorithm that takes in 4cm4cm zero-inputs to And-Or, and finds a certificate for at least cm/8cm/8 of them using R0(fCS(ϕ))R_{0}(f_{\mathrm{CS(\phi)}}) expected queries.  It follows that given one input from the hard distribution over 0-inputs to ff, we can find a certificate for it by generating 4cm14cm-1 additional inputs from this distribution, mixing them together, and running the R0(fCS(ϕ))R_{0}(f_{\mathrm{CS(\phi)}}) algorithm.  This finds a certificate with probability at least 1/81/8, and uses R0(fCS(ϕ))/cmR_{0}(f_{\mathrm{CS(\phi)}})/cm expected queries.  By repeating the algorithm if necessary, we find a certificate using 8R0(fCS(ϕ))/cm8R_{0}(f_{\mathrm{CS(\phi)}})/cm expected queries.

We construct a function ff^{\prime} is the same as ff except with Not-And-Or instead of And-Or.  Repeating the argument provides a randomized algorithm that finds a 11-certificate for And-Or using 8R0(fCS(ϕ)/cm8R_{0}(f^{\prime}_{CS(\phi^{\prime})}/cm expected queries.  By running both algorithms in parallel, we get

R0(And-Or)8R0(fCS(ϕ))/cm+8R0(fCS(ϕ))/cm.R_{0}(\textsc{And-Or})\leq 8R_{0}(f_{\mathrm{CS(\phi)}})/cm+8R_{0}(f^{\prime}_{\mathrm{CS}(\phi^{\prime})})/cm. (28)

Since R0(And-Or)=Ω(m2)R_{0}(\textsc{And-Or})=\Omega(m^{2}), it follows that either R0(fCS(ϕ))=Ω(m3)R_{0}(f_{\mathrm{CS(\phi)}})=\Omega(m^{3}) or R0(fCS(ϕ))=Ω(m3)R_{0}(f^{\prime}_{\mathrm{CS}(\phi^{\prime})})=\Omega(m^{3}).  The desired result follows. ∎