# Better Non-Local Games from Hidden Matching

**ABSTRACT** We construct a non-locality game that can be won with certainty by a quantum strategy using log n shared EPR-pairs, while any classical strategy has winning probability at most 1/2+O(log n/sqrt{n}). This improves upon a recent result of Junge et al. in a number of ways. Comment: 11 pages, latex

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**ABSTRACT:**In this paper we obtain violations of general bipartite Bell inequalities of order $\frac{\sqrt{n}}{\log n}$ with $n$ inputs, $n$ outputs and $n$-dimensional Hilbert spaces. Moreover, we construct explicitly, up to a random choice of signs, all the elements involved in such violations: the coefficients of the Bell inequalities, POVMs measurements and quantum states. Analyzing this construction we find that, even though entanglement is necessary to obtain violation of Bell inequalities, the Entropy of entanglement of the underlying state is essentially irrelevant in obtaining large violation. We also indicate why the maximally entangled state is a rather poor candidate in producing large violations with arbitrary coefficients. However, we also show that for Bell inequalities with positive coefficients (in particular, games) the maximally entangled state achieves the largest violation up to a logarithmic factor.Communications in Mathematical Physics 07/2010; 306. · 1.97 Impact Factor

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arXiv:1007.2359v1 [quant-ph] 14 Jul 2010

Better Non-Local Games from Hidden Matching

Harry Buhrman∗

Giannicola Scarpa†

Ronald de Wolf‡

July 15, 2010

Abstract

We construct a non-locality game that can be won with certainty by a quantum strategy

using logn shared EPR-pairs, while any classical strategy has winning probability at most

1

2+ O

√n

. This improves upon a recent result of Junge et al. in a number of ways.

?

log n

?

1Introduction

One of the most striking features of quantum mechanics is the fact that entangled particles can

exhibit correlations which cannot be reproduced or explained by classical physics (i.e., by “local

hidden-variable theories”). This was first observed by Bell [Bel65] in response to Einstein-Podolsky-

Rosen’s challenge to the completeness of quantum mechanics [EPR35]. An appropriate experimen-

tal realization of such quantum correlations is the strongest proof we have that nature does not

behave in accordance with classical physics. Many such experiments have in fact already been

done starting with [AGR82]. All behave in accordance with quantum predictions, though so far

none has closed all conceivable “loopholes” that would allow some (usually very contrived) classical

explanation of the observed correlations.

Roughly speaking, the more entanglement the quantum experiment starts with, the further its

exhibited correlations can deviate from what is achievable classically. In this paper we study this

relation quantitatively. The setup is as follows [CHTW04]. Two spacelike separated parties, called

Alice and Bob, receive inputs x and y according to some fixed and known probability distribution π,

and are required to produce outputs a and b. There is a predicate V (ab|xy) specifying which output-

pairs a,b are considered “winning” on inputs x,y, while the others are “losing”. We will usually

write G = G(V,π) to denote such a game.

Quantum strategies for playing such a game start out with some fixed entangled state, say

with local dimension n; a typical example would be logn shared EPR-pairs

each input x, Alice has a set of measurement operators {Aa

?

state |ψ?, producing outputs a and b, respectively. The probability to output a,b is ?ψ|Aa

Note that no communication takes place between Alice and Bob. Assuming the predicate V (ab|xy)

∗CWI and University of Amsterdam, buhrman@cwi.nl. Supported by a Vici grant from NWO.

†CWI Amsterdam, g.scarpa@cwi.nl. Supported by a Vidi grant from NWO.

‡CWI Amsterdam, rdewolf@cwi.nl. Supported by a Vidi grant from NWO.

1

√2(|00? + |11?). For

x} (subject to the usual constraint

aAa

x= I, the n-dimensional identity) and for each y, Bob has measurement operators {Bb

(subject to?

y}

bBb

y= I). They apply the measurement corresponding to x and y to the entangled

x⊗Bb

y|ψ?.

1

Page 2

takes value +1 on winning outputs and value −1 on losing outputs, the advantage (i.e., difference

between winning and losing probabilities) of this quantum strategy can be succinctly expressed as:

ωq(G,{Aa

x},{Bb

y},ψ) =

E

x,y,a,b[V (ab|xy)] =

?

x,y,a,b

π(x,y)?ψ|Aa

x⊗ Bb

y|ψ?V (ab|xy).

The quantum value of the game, when restricted to entangled states with local dimension n, is

defined as

ωq,n(G) =max

{Aa

This is a number between −1 and 1; a value of 1 indicates that the strategy wins with certainty, a

value of −1 that it always fails (for instance if V (ab|xy) = −1 for all a,b,x,y).

This quantum value should be contrasted with the best value that can be obtained by classi-

cal strategies. In a classical strategy the shared entangled state is replaced by a shared random

variable R, sometimes called the “hidden variable”. Its distribution is independent of the inputs

and its value r is seen by both Alice and Bob, who can use this to coordinate their behaviour. A

classical strategy is described by two functions A : x,r ?→ a and B : y,r ?→ b which are used by

Alice and Bob, respectively, to determine their output as a function of their input and of the shared

random variable. The value of such a classical strategy for game G is

x},{Ba

y},ψ∈Cn×nωq(G,{Aa

x},{Bb

y},ψ).

ωc(G,A,B,R) =

?

x,y,r

Pr[R = r]π(x,y)V (A(x,r)B(y,r)|xy)

and the classical value of the game is what can be achieved by the best classical strategy:

ωc(G) = max

A,B,Rωc(G,A,B,R).

It turns out that this value can be achieved by deterministic strategies, so we could drop R from

this definition.

It is well-known that there are games where ωq,n(G) is substantially higher than ωc(G); the

CHSH game [CHSH69] is a famous example of this with n = 2 (Alice and Bob share one EPR-

pair). In fact, there are games G where the ratio ωq,n(G)/ωc(G) is unbounded when the local

dimension n of the entangled state grows. We are interested in the maximal value this ratio can

take as a function of n. Our starting point is a recent paper by Junge et al. [JPP+09] who studied

the same question using tools from Operator Space theory. On the one hand, they proved that the

ratio cannot be larger than O(n); on the other hand they proved the existence of a game where the

ratio is Ω(√n/log2n).

Our main result in this paper is a simple game where the local dimension is n and the ratio

between the optimal quantum and classical values is Ω(√n/logn): there is a quantum strategy that

achieves the maximal value 1 using logn EPR-pairs as its entangled state |ψ?, while no classical

strategy can have an advantage better than O(logn/√n). We also give a classical strategy achieving

advantage of Ω(1/√n), so our bounds are nearly optimal.

Our game is a variant of the “Hidden Matching” problem. This was introduced in the context

of quantum communication complexity by Bar-Yossef et al. [BJK08], and other variants of it were

subsequently studied in [GKRW09, GKK+08, Gav08, Gav09]. A precise definition will be given

below. The main mathematical tool we use in our analysis is the so-called “KKL inequality” from

Fourier analysis of Boolean functions (see [O’D08, Wol08] for surveys of this area). This inequality

2

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was used before to analyze another variant of Hidden Matching in [GKK+08], though their analysis

is different and more complicated.

Our result has several advantages over the one of Junge et al. [JPP+09]. First, our game is

simple and explicit, while they only give a non-explicit existence proof (via a probabilistic argument

based on Gaussian matrices). Clearly, explicitness is necessary for experimental realization. Second,

our quantum-classical separation is slightly stronger, Ω(√n/log n) instead of Ω(√n/log2n). The

stronger the separation, the more resistant it is to noise (for instance, if noise can change numerator

and denominator of a large ratio by some small ε, the ratio won’t be significantly affected). Third,

the number of inputs for Alice and Bob is smaller: in their game there are roughly 2nlog2npossible

inputs, while in our case there are 2npossible inputs x for Alice and roughly 2nlognpossible inputs

y for Bob (in fact, the later could easily be reduced to much less than 2n). The fewer possible inputs

and measurement settings there are, the easier it should be to experimentally realize a quantum

strategy for the game.

The organization of the paper is as follows. While our focus is non-locality, it will actually

be useful to first study the original version of the Hidden Matching problem in the context of

protocols where communication from Alice to Bob is allowed. In Section 2 we prove a tight bound

of 1/2 + Θ(c/√n) on the maximal success probability Alice and Bob can obtain with c bits of

classical communication.1Section 3 then ports those results from the communication setting to

the non-locality setting, establishing the results mentioned above.

2 The Hidden Matching problem

2.1Problem definition and quantum protocol

In this section we describe the original Hidden Matching communication problem and an efficient

quantum protocol for it, both from [BJK08].

Definition 1 (Hidden Matching (HM)). Let n be a power of 2 and Mn the set of all perfect

matchings on the set [n] = {1,...,n} (a perfect matching is a partition of [n] into n/2 disjoint

pairs (i,j), also called “edges”). Alice is given x ∈ {0,1}nand Bob is given M ∈ Mn, distributed

according to the uniform distribution U. We allow 1-way communication from Alice to Bob, and

Bob outputs an edge (i,j) ∈ M and v ∈ {0,1}. They win if v = xi⊕ xj.

Theorem 1. There exists a quantum protocol for HM with logn qubits of 1-way communication,

such that always v = xi⊕ xj.

Proof. The protocol is the following:

1. Alice sends Bob the state |ψ? =

2. Bob measures |ψ? in the basis B = {1

measurement is a state

1

√n

?n

i=1(−1)xi|i?.

√2(|i? ± |j?) | (i,j) ∈ M}. If the outcome of the

√2(|i?+|j?) then Bob outputs (i,j) and v = 0. If the outcome of the

1

√2(|i? − |j?), Bob outputs (i,j) and v = 1.

1

measurement is a state

1This bound includes the main lower bound of [BJK08] as a special case: namely, the communication c needs to

be Ω(√n) bits if we want to have success probability at least 2/3. The proof of [BJK08] is based on information

theory rather than Fourier analysis.

3

Page 4

For each (i,j) ∈ M the probability to get outcome

1

√2(|i? + |j?) is

|?ψ|1

√2(|i? + |j?)?|2=

?

2/n

0

if xi⊕ xj= 0

if xi⊕ xj= 1

A similar argument holds for

1

√2(|i? − |j?). Hence Bob’s output is always correct.

2.2 Bound on classical communication protocols for HM

Here we show that classical protocols for Hidden Matching with little communication cannot have

a good success probability.

Theorem 2. Every classical deterministic protocol for HM with c bits of 1-way communication

from Alice to Bob where Bob outputs (i,j),v, has

Pr

U[v = xi⊕ xj] ≤1

2+ O

?

c

√n

?

.

The intuition behind the proof is the following. If the communication c is small, the set Xmof

inputs x for which Alice sends message m, will typically be large, meaning Bob has little knowledge

of most of the bits of x. By the KKL inequality, this implies that for most of the?n

outputs, but that freedom is limited to the n/2 (i,j)-pairs in his matching M, and it turns out

that on average he won’t be able to guess any of those parities well.

2

?(i,j)-pairs,

Bob cannot guess the parity xi⊕ xj well. Of course, Bob has some freedom in which (i,j) he

Proof. Fix a classical deterministic protocol. For each m ∈ {0,1}c, let Xm⊆ {0,1}nbe the set

of Alice’s inputs for which she sends message m. These sets Xmtogether partition Alice’s input

space {0,1}n. Define pm=|Xm|

2cmessages m. Define ε such that PrU[v = xi⊕ xj] =1

Bob received m] =1

2+ εm. Then ε =?

Bob received m]. We have qm(i,j) ≤

and for fixed i ?= j we have PrM[(i,j) ∈ M] = 1/(n − 1) (each j is equally likely to be paired up

with i). Also define βm

given message m, is to output the value of xi⊕xjthat occurs most often among the x ∈ Xm. The

fraction of x ∈ Xmwhere xi⊕xj= 0 is 1/2+βm

guessing xi⊕ xjis 1/2 + |βm

2+|βm

2

x∈Xm

M∈Mn

2n . Note that?

mpm= 1, so p is a probability distribution over the

2+ ε, and εmsuch that PrU[v = xi⊕ xj|

mpmεm.

For each m and (i,j) define the probability distribution qm(i,j) = PrM∈Mn[Bob outputs (i,j) |

1

n−1, because we assume Bob always outputs an edge in M

ij= Ex∈Xm[(−1)xi· (−1)xj]. The best Bob can do when guessing xi⊕ xj

ij/2, hence Bob’s optimal success probability when

ij|/2. This implies, for fixed m,

E

(i,j)∼qm

?1

ij|

?

≥ Pr[v = xi⊕ xj] =1

2+ εm.

As explained in [Wol08, section 4.1], it follows from the KKL inequality [KKL88] that

?

i,j:i?=j

(βm

ij)2≤ O

?

log

1

pm

?2

.(1)

4

Page 5

This allows us to upper bound εm:

2εm≤

E

(i,j)∼qm

[|βm

ij|] =

?

i,j

qm(i,j)|βm

ij|

(∗)

≤

??

i,j

qm(i,j)2·

??

i,j

|βm

ij|2

(∗∗)

≤

1

√n − 1· O

?

log

1

pm

?

,

where (∗) is Cauchy-Schwarz and (∗∗) follows from?

i,jqm(i,j)2≤ maxi,jqm(i,j) ·?

i,jqm(i,j) ≤

1

n−1and Eq. (1). Now we can bound ε:

ε =

?

m

pmεm≤

?

m

pmO(log(1/pm))

√n − 1

=

1

√n − 1

?

m

pmO(log(1/pm)) =

1

√n − 1O(H(p)) = O

?

c

√n

?

where H denotes the entropy function, and H(p) ≤ c since the distribution p is on 2celements.

2.3 Classical communication protocol for HM

Here we design a classical communication protocol that achieves the above upper bound on the

success probability.

Theorem 3. For every positive integer c ≤√n, there exists a classical protocol for HM with c bits

of 1-way communication from Alice to Bob, such that

Pr

U[v = xi⊕ xj] =1

2+ Ω

?

c

√n

?

.

Proof. We will first handle the case c ≥ 2, dealing with c = 1 at the end of the proof. Assume for

simplicity√n is divisible by c. The protocol is as follows:

1. Alice considers the first√n bits of x, and divides them into c consecutive blocks B1...Bcof

k =√n/c bits each. She sends Bob a c-bit string m = m1...mcwhere mi= Majority(Bi)

(the value that occurs more often in Bi). Let B(i) denote the index of the block containing xi.

With slight abuse of notation, we will also use B(i) for the corresponding block itself (i.e., a

k-bit string), and for the set of k indices of the bits in this block.

2. Let E denote the event that there is an edge (i,j) ∈ M satisfying i,j ∈ [√n],B(i) ?= B(j). If

E occurs then Bob chooses uniformly at random one of the edges (i,j) satisfying the above

condition, and outputs (i,j) and mB(i)⊕mB(j). If E does not occur then he outputs a random

(i,j) ∈ M and a random bit (in which case they win with probability 1/2).

We first show that PrM[E] ≥ 1/10. The probability (over a uniformly random matching M) that

none of the i ∈ [√n] is paired up with a j ∈ [√n], is at most

√n

?

On the other hand, if some i ∈ [√n] is paired up with another j ∈ [√n], then since this j will be

uniformly distributed over [√n]\{i}, the probability that j lands in the same k-bit block as i (i.e.,

that B(i) = B(j)) is at most (k − 1)/(√n − 1) ≤ 1/c ≤ 1/2. Hence PrM[¬E] ≤ 1/e + 1/2 < 9/10.

i=1

n −√n − i + 1

n − i + 1

≤

?

1 −

1

√n

?√n

≤ 1/e.

5

Page 6

Next we show PrU[mB(i)⊕ mB(j)= xi⊕ xj | E] =

without mentioning this further. It will be convenient to use ±1-valued bits instead of 0/1-valued

bits, because then the parity of two bits corresponds to their product. Let X ∈ {±1}n, uniformly

distributed, be the random variable for Alice’s input. Let I,J be the random variables for Bob’s

output edge, then I,J are uniformly distributed over distinct blocks B(I),B(J) respectively. Let

MB(I)∈ {±1} be the majority value of the block B(I) ∈ {±1}k, and similarly for MB(J). Note

that MB(I)has the same sign as the sum of the entries of the block B(I).

Since I is uniformly distributed over B(I), the bit MB(I)(which Bob knows) has some positive

correlation with the bit XI(which Bob would like to know):

1

2+ Ω

?

c

√n

?

. Below we condition on E

E

X,I[MB(I)XI] = E

X,I

1

k

?

ℓ∈B(I)

MB(I)Xℓ

= E

X,I

1

k

??????

?

ℓ∈B(I)

Xℓ

??????

= Ω

?

1

√k

?

,

where the last step follows from the binomial distribution: the sum of k uniform random coin flips

is at least√k away from its expectation with at least constant probability (roughly 5%). The same

lower bound holds for EX,J[MB(J)XJ].

Bob uses MB(I)MB(J)to guess the parity XIXJ. Because I and J are each uniformly distributed

over distinct blocks, the random variables MB(I)XIand MB(J)XJare independent. Hence we have

the following positive correlation in the guess of the parity:

E

X,I,J[(MB(I)MB(J))(XIXJ)] = E

X,I[MB(I)XI] · E

X,J[MB(J)XJ] = Ω

?

1

√k

?

· Ω

?

1

√k

?

= Ω

?1

k

?

.

This says that, conditioned on the event E, the players win with probability1

E does not hold, the winning probability is exactly 1/2. Since PrM[E] ≥ 1/10, the claimed lower

bound on the overall winning probability follows.

To handle the case c = 1, note that if c = 2 it suffices if Alice sends m1⊕m2instead of m1,m2.

This gives a 1-bit protocol with winning probability 1/2 + Ω(1/√n).

2+ Ω

?

c

√n

?

. In case

3The Non-Local Hidden Matching problem

3.1Problem definition and quantum protocol

We now port the problem and results from Section 2 to the non-local setting.

Definition 2 (Non-Local Hidden Matching (HMnl)). Let n be a power of 2 and Mnthe set of all

perfect matchings on the set [n]. Alice is given x ∈ {0,1}nand Bob is given M ∈ Mn, distributed

according to the uniform distribution U. Alice’s output is a string a ∈ {0,1}log nand Bob’s output

is an edge (i,j) ∈ M and b ∈ {0,1}log n. They win the game if the following condition is true

(a ⊕ b)(i ⊕ j) = xi⊕ xj.

Next we show ωq,n(HMnl) = 1, while in the next section we show ωc(HMnl) = O(logn/√n).

Together this gives our main result: ωq,n(HMnl)/ωc(HMnl) = Ω(√n/logn).

(2)

Theorem 4. There exists a quantum protocol for HMnl with a shared entangled state of logn

EPR-pairs, such that condition (2) is always satisfied.

6

Page 7

Proof. The protocol is as follows. Alice and Bob share a state |ψ? =

1

√n

?

i∈{0,1}log n|i?|i?.

1. Alice performs a phase-flip to her part of |ψ? according to her input x. The state becomes

|ψ′? =

1

√n

?

i∈{0,1}log n(−1)xi|i?|i?.

2. Bob performs a projective measurement with projectors Pij= |i??i| + |j??j|, with (i,j) ∈ M.

The state collapses to |ψ′′? =

3. Both players apply Hadamard transforms H⊗logn, so they get

1

√2[(−1)xi|i?|i? + (−1)xj|j?|j?] for some (i,j) ∈ M.

H⊗2logn|ψ′′? =

1

√2n

?

a,b∈{0,1}log n

?

(−1)xi+(a·i)+(b·i)+ (−1)xj+(a·j)+(b·j)?

|a?|b?,

where x · y is the bitwise inner product of x and y modulo 2. In this last state, only a,b satisfying

condition (2) have nonzero amplitude, hence Alice and Bob win the game with certainty.

3.2Bound on classical protocols for HMnl

Theorem 5. Every classical protocol for HMnlhas

Pr

U[(a ⊕ b)(i ⊕ j) = xi⊕ xj] ≤1

2+ O

?logn

√n

?

.

Proof. A protocol that wins HMnlwith success probability 1/2+ε can be turned into a protocol for

HM with logn bits of communication and the same probability to win: the players play HMnl, with

Alice producing a and Bob producing i,j,b; Alice then sends a to Bob, who outputs i,j,(a⊕b)(i⊕j).

This requires c = logn bits of communication (the length of a), so Theorem 2 gives the bound.

3.3Classical protocol for HMnl

Here we show that our upper bound of1

strategies for HMnlis nearly optimal:

2+ O

?

logn

√n

?

on the best success probability of classical

Theorem 6. There exists a classical protocol for HMnlsuch that

Pr

U[(a ⊕ b)(i ⊕ j) = xi⊕ xj] =1

2+ Ω

?

1

√n

?

.

Proof. For any positive integer c ≤√n, Alice and Bob can “simulate” the communication protocol

of Theorem 3 using shared randomness, as follows. They share a uniformly random string r ∈

{0,1}c. Alice knows which message m she would have sent to Bob in the communication protocol.

If r = m (which happens with probability 1/2c) then Alice outputs a = 0logn, otherwise she outputs

a uniformly random a ∈ {0,1}log n. Bob treats r as the communication Alice would have sent him

in the original protocol, and computes i,j,v accordingly. He outputs i,j, and a string b satisfying

b(i⊕j) = v. Note that if m = r then a = 0lognand hence (a⊕b)(i⊕j) = v, in which case Alice and

Bob win HMnlwith probability1

√n). On the other hand, if m ?= r then a⊕ b is a uniformly

2+ Ω(c

7

Page 8

random string and i ⊕ j ?= 0logn, so (a ⊕ b)(i ⊕ j) is a uniformly random bit. Hence the overall

winning probability is:

Pr

U[(a ⊕ b)(i ⊕ j) = xi⊕ xj] =1

Taking c = 1 gives the claimed result.

2c·

?1

2+ Ω

?

c

√n

??

+

?

1 −1

2c

?

·1

2=12+ Ω

?

c

2c√n

?

.

3.4 An alternative classical protocol based on the Grothendieck inequality

In this section we describe a classical protocol that works for arbitrary input distributions π, instead

of just uniform. This protocol will be based on the famous Grothendieck inequality. Let HMnl(π)

denote the Non-Local Hidden Matching game with probability distribution π on the input.

Theorem 7. For every input distribution π there exists a classical protocol for HMnl(π) such that

Pr

π[(a ⊕ b)(i ⊕ j) = xi⊕ xj] =1

2+ Ω

?

1

√nlogn

?

.

Proof. Alice and Bob start with a shared uniformly random i ∈ [n] and r ∈ [logn]. There will be a

unique j such that (i,j) ∈ M. Bob outputs that (i,j). Now we need to explain how they compute

a,b ∈ {0,1}log nsuch that (a ⊕ b)(i ⊕ j) = xi⊕ xjwith advantage Ω(1/√nlogn).

We start by defining the following systems of unit vectors {vx},{vy} ⊆ Rn, and matrix N ∈

Rn×n. Here x and y range over {0,1}nand Mn, respectively, while i ∈ [n] is as above.

1

√n

k∈{0,1}log n

vy= |j?, for (i,j) ∈ y

Nxy= π(x,y) · (−1)xi⊕xj, for (i,j) ∈ y.

We have:

vx=

?

(−1)xi⊕xk|k?

?

It follows from the Grothendieck inequality that there is constant2KG such that the following

holds: there exist classical strategies A : x ?→ {0,1} and B : y ?→ {0,1} such that

1

KG

{vx},{vy}

x,y

Nxy?vx|vy? =

?

x,y

Nxy

1

√n

?

k

(−1)xi⊕xk?k|j? =

1

√n

?

x,y

Nxy(−1)xi⊕xj=

1

√n

?

x,y

π(x,y) =

1

√n.

?

x,y

Nx,y(−1)A(x)⊕B(y)≥

max

?

x,y

Nxy?vx|vy? ≥

1

KG√n.

(3)

This implies that (for every i, and for j defined by (i,j) ∈ y), we have Prπ[A(x)⊕B(y) = xi⊕xj] ≥

1

2+

1

2KG√n. It remains to define the output strings a and b, which we do as follows:

• Since Alice knows i and x, she knows the bit A(x). She outputs a = A(x)er, where er ∈

{0,1}log nis a string whose only 1-bit sits at position r.

2This constant (which is independent of the matrix N and the vectors) is called the Grothendieck constant. Its

exact value is unknown, but [Kri79, Dav84, Ree91] show that 1.68 ? KG ? 1.78. The relationship between non-local

games and the Grothendieck inequality has been studied extensively in [BBT09].

8

Page 9

• Since Bob knows i and y, he knows the bit B(y). Since r is uniformly random and i ?= j,

with probability at least 1/log n the strings i and j differ at position r. If this is the case

then Bob outputs b = B(x)er. This ensures (a ⊕ b)(i ⊕ j) = A(x) ⊕ B(y), so then Alice and

Bob win with probability at least 1/2 + 1/(2KG√n). If i and j do not differ at position r,

then Bob outputs a uniformly random b ∈ {0,1}log n, and they win with probability 1/2.

The overall winning probability is at least 1/2 + 1/(2KG√nlogn).

If π induces a uniform marginal distribution on M, then i and j in the above proof differ at

position r with probability 1/2 instead of 1/log n. This gives an alternative proof of Theorem 6.

3.5 Quantum/classical ratio as a function of the number of possible outputs

We can also study the ratio between the quantum and classical values as a function of the number of

possible outputs, rather than the local dimension of the entangled state. Let ωq(G) = supnωq,n(G)

be the quantum value with unlimited (but finite) entanglement.

G with k possible outputs for Alice (i.e., values for a) and ℓ possible outputs for Bob, we have

ωq(G)/ωc(G) = O(kℓ) [DKLR08, JPP+09].

As presented above, our game HMnlhas n possible outputs for Alice and roughly n3for Bob.

However, it can easily be modified to have only n possible outputs for Bob, by restricting Mnto

matchings M = {(i,j)} where i ≤ n/2 and j > n/2, and where there is a bijection between (i,j)

and i ⊕ j. Now Alice behaves as before, and we just require Bob to output i ⊕ j (which he can do

in logn − 1 bits since the most significant bit is always 1), and the bit w = b(i ⊕ j). The number

of possible outputs for Bob is now 2logn−1·2 = n, and the original relation (a⊕b)(i ⊕j) = xi⊕ xj

is equivalent to a(i ⊕ j) = xi⊕ xj⊕ w.

For this modified version of HMnl, we also have ωq(G)/ωc(G) = Ω(√n/logn). First, it is easy

to see that the proof of Theorem 2 works as before, the only change being that now qa(i,j) ≤ 2/n

instead of ≤ 1/(n−1), because we are choosing matchings from a subset of Mn. Second, the proof

of Theorem 1 also works when Bob’s output is modified as above, so we have the claimed bound.

It is known that for a game

4Conclusion and future work

We presented a simple non-local game where the ratio between the quantum value (with n-

dimensional entanglement) and the classical value scales as roughly√n. On the other hand, Junge

et al. [JPP+09] showed that this ratio is O(n) for all possible games. It is an interesting open

problem to close this quadratic gap: can we find a non-local game where the quantum/classical

ratio is Ω(n) or find a better upper bound on all games? A second open problem is closing the

gap on the ratio between the quantum and classical values as a function of the number of possible

outputs. We have still more than a fourth-power gap with our lower bound of Ω(√n/log n).

Acknowledgements

We thank Jop Bri¨ et for useful discussions in the early stages of this research.

9

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