††thanks: These authors contributed equally.††thanks: These authors
contributed equally.††thanks: These authors contributed equally.

# Exploring the Local Landscape in the Triangle Network

Elisa Bäumer  Institute for Theoretical Physics, ETH Zurich, 8093 Zürich,
Switzerland  Victor Gitton  Institute for Theoretical Physics, ETH Zurich,
8093 Zürich, Switzerland  Tamás Kriváchy  Atominstitut, Technische Universität
Wien, 1020 Vienna, Austria  Département de Physique Appliquée, Université de
Genève, CH-1211 Genève, Switzerland  Nicolas Gisin  Département de Physique
Appliquée, Université de Genève, CH-1211 Genève, Switzerland  Schaffhausen
Institute of Technology – SIT, Geneva, Switzerland  Renato Renner  Institute
for Theoretical Physics, ETH Zurich, 8093 Zürich, Switzerland

###### Abstract

Characterizing the set of distributions that can be realized in the triangle
network is a notoriously difficult problem. In this work, we investigate inner
approximations of the set of local (classical) distributions of the triangle
network. A quantum distribution that appears to be nonlocal is the Elegant
Joint Measurement (EJM) [Entropy. 2019; 21(3):325], which motivates us to
study distributions having the same symmetries as the EJM. We compare
analytical and neural-network-based inner approximations and find a remarkable
agreement between the two methods. Using neural network tools, we also
conjecture network Bell inequalities that give a trade-off between the levels
of correlation and symmetry that a local distribution may feature. Our results
considerably strengthen the conjecture that the EJM is nonlocal.

##  I Introduction

The problem of nonlocality, in general, requires to study classical (or
“hidden-variable”) causal models. While this study is already extensive in the
Bell scenario, featuring Alice and Bob sharing a common source, the case of
more general networks remains poorly understood, with many open questions [1].
From a mathematical and computational perspective, the presence of multiple
independent sources is the root of the difficulties associated to proving
network nonlocality. An interesting feature of network nonlocality is that, in
general, and contrarily to Bell nonlocality, it is not necessary to provide
different inputs to the parties to obtain a nonlocal behavior. Here, we
consider one of the simplest non-trivial networks, the triangle network, which
consists of Alice, Bob and Charlie sharing only independent bipartite sources
as depicted in Fig. 1. The first known example of quantum nonlocality in the
triangle network dates back to the Fritz distribution [2], which was shown to
admit an experimental implementation that was provably nonlocal [3]. This
example of triangle nonlocality is quite specific, since the Fritz
distribution is based on embedding a bipartite Bell test in the triangle
network. The other example of triangle nonlocality is the RGB4 family of
distributions introduced in [4] and later extended in [5, 6, 7, 8].

Figure 1: The triangle network features three observers (green circles),
connected by three independent bipartite sources (yellow ellipses): Alice has
access to the $\beta,\gamma$ sources, while Bob has access to $\gamma,\alpha$
and Charlie to $\alpha,\beta$.

Overall, compared to the Bell scenario, very few examples of triangle
nonlocality are known to date. In an attempt to find another example of
nonlocality in the triangle network, the so-called _Elegant Joint Measurement_
(EJM) was introduced in [9]. This measurement is a two-qubit measurement that
projects onto a partially entangled basis with a tetrahedral symmetry in the
Bloch sphere for the marginal basis states. The EJM can be used in different
networks: in [10], a network Bell inequality has been tailored to be violated
with the EJM applied in the bilocal scenario. This bilocal scenario can be
thought of as the triangle network with the source between Alice and Bob being
removed. The latter Bell inequality has been experimentally violated in [11].
When this measurement is used in the triangle network with the three sources
prepared in singlet states, the resulting outcome distribution is referred to
as the EJM distribution. This EJM distribution exhibits high correlations
between the three parties, as well as symmetry under both permutation of the
parties and joint permutation of the outcomes. The high level of symmetry of
the distribution coupled to the relatively low number of outputs per party,
namely, four, makes this distribution a prime candidate for the study of
triangle nonlocality. Indeed, the simplicity of the distribution suggests that
a simple proof of nonlocality should exist, although this remains an open
problem. This is for instance in analogy to the case of Werner states [12]:
their high level of symmetry is what enabled to show that there exist
entangled states that admit a Bell-type hidden variable model. The EJM
distribution has been initially conjectured to be nonlocal [9], and even
conjectured to be nonlocal under reasonable experimental noise [13]. The
interest for this distribution has been reinforced by a recent experimental
implementation of the EJM measurement in the triangle network [14]. Motivated
by the open question of the nonlocality of the EJM, we consider the following
problem: can we give a characterization of the local set in the triangle
network for those distributions that share the same symmetries as the EJM
distribution?

Our results are organized as follows. In Section II, we set the stage by
defining triangle (non)locality, and we describe in particular the
representation that we will use to depict a classical causal model in the
triangle network, i.e., a classical strategy for Alice, Bob and Charlie. We
discuss quantum strategies and in particular the EJM. We then introduce the
notion of _fully-symmetric_ distributions, which refers to the distributions
sharing the same symmetries as the EJM distribution. We then move on to
obtaining inner approximations of the set of fully-symmetric local
distributions. In Section III, we describe how we obtained a “large”
analytical inner approximation. We then describe in Section IV how a neural-
network tool [13] can be used to obtain another sort of inner approximation.
As we will see, the two methods agree surprisingly well, thus suggesting the
quality and completeness of either method. Moreover, aided with the neural
network approach, we conjecture network Bell inequalities which would hold for
any local distribution, even a non-symmetric one, and which are violated by
the EJM distribution. Our results suggest that the EJM distribution is
relatively far from the local set, which would make it a prime target for a
noise-robust proof of genuine triangle nonlocality. We conclude in Section V
with remaining open problems and future research directions.

##  II Setup: locality and symmetry

The setup we consider here is the triangle scenario, which is a network of
three observers, Alice, Bob and Charlie, and three independent sources
$\alpha$, $\beta$ and $\gamma$, as depicted in Fig. 1.

###  II.1 Classical setup

####  II.1.1 Local strategies

If the sources are classical, then Alice, Bob and Charlie would use a
classical strategy, i.e., a conditional distribution $p_{A}(a|\beta,\gamma)$
for Alice, $p_{B}(b|\gamma,\alpha)$ for Bob and $p_{C}(c|\alpha,\beta)$ for
Charlie. We say that a distribution $p(a,b,c)$ is _local_ or classical111A
more precise terminology would be to call such a distribution classically
local in the triangle network. This would be in contrast to _quantumly_ local
distributions in the triangle network, where “local” would simply refer to the
connectivity of the triangle network. However, following the standard
terminology of the field, we use the abbreviation “local distribution” in
place of “classically local distribution”. (in the triangle network) if there
exist $p_{A}$, $p_{B}$ and $p_{C}$ such that

$p(a,b,c)=\\\ \int_{[0,1]^{\times
3}}\mathrm{d}\alpha\,\mathrm{d}\beta\,\mathrm{d}\gamma\,p_{A}(a|\beta,\gamma)\,p_{B}(b|\gamma,\alpha)\,p_{C}(c|\alpha,\beta),$
(1)

where $\alpha,\beta,\gamma\in[0,1]$ represent the values distributed by each
source. Otherwise, $p(a,b,c)$ is said to be _nonlocal_. The set of local
distributions is non-convex, which makes its description quite complex. Note
that we assumed without loss of generality that the sources in this classical
setting are distributed uniformly over the interval $[0,1]$. We may also
assume without loss of generality that the local strategies determining the
output of each party are given by deterministic response functions
$a(\beta,\gamma)$, $b(\alpha,\gamma)$ and $c(\alpha,\beta)$, respectively. The
probabilistic response functions are then related to the deterministic
response functions by

$\displaystyle p_{A}(a|\beta,\gamma)$
$\displaystyle=\delta_{a,a(\beta,\gamma)},$ $\displaystyle
p_{B}(b|\gamma,\alpha)$ $\displaystyle=\delta_{b,b(\gamma,\alpha)},$
$\displaystyle p_{C}(c|\alpha,\beta)$
$\displaystyle=\delta_{c,c(\alpha,\beta)}.$

As every function depends on only two variables, we can nicely illustrate the
two independent parameters for each party as two dimensions and indicate the
corresponding outputs by different colors. This maps the response functions to
what we call “flags”, that might give a more intuitive understanding of the
different strategies. Let us illustrate that with the following example.

####  II.1.2 Example flags for a highly correlated distribution

Figure 2: Flags illustrating the response functions that the three parties
Alice, Bob and Charlie may use in the triangle network (Fig. 1). For instance,
Alice outputs her outcome $a$ given her two input values
$\beta,\gamma\in[0,1]$. The different values of the outcomes (such as $a$) are
indicated by different colors.

Figure 2 showcases output strategies of the three parties Alice, Bob and
Charlie, as functions of their respective local variables. Let us denote
“blue” by $1$, “pink” by $2$, “yellow” by $3$ and “green” by $4$. Then for
example Charlie’s strategy is given by

  * •

output “blue” $\equiv 1$ for $\alpha\in[0,0.5]$, $\beta\in[0,0.5]$,

  * •

output “pink” $\equiv 2$ for $\alpha\in[0,0.5]$, $\beta\in(0.5,1]$,

  * •

output “yellow” $\equiv 3$ for $\alpha\in(0.5,1]$, $\beta\in[0,0.5]$,

  * •

output “green” $\equiv 4$ for $\alpha\in(0.5,1]$, $\beta\in(0.5,1]$.

The strategies for Alice and Bob can be determined analogously. In this
example, the output probability distribution can be summarized as follows:

$\displaystyle p(a,b,c)$
$\displaystyle=\frac{1}{8}\big{(}[1,1,1]+[2,2,2]+[3,3,3]+[4,4,4]$
$\displaystyle\quad+[1,4,3]+[2,3,4]+[3,2,1]+[4,1,2]\big{)}.$ (2)

where we defined the deterministic distribution $[k,l,m]$ as
$[k,l,m](a,b,c)=\delta_{ak}\delta_{bl}\delta_{cm}$. Note that while here the
probability that all three parties output the same value is given by
$p(A=B=C)=\nicefrac{{1}}{{2}}$, indicating very strong correlations, the
distribution is not symmetric. In Section III we use this flag model to
investigate local strategies that yield fully symmetric probability
distributions.

###  II.2 Quantum setup

####  II.2.1 Quantum strategies

Let us move on to the quantum setting, where instead of outputting classical
values that follow a classical probability distribution, the sources
distribute entangled quantum systems. Note that in this setting, contrary to
the standard Bell nonlocality tests, the observers receive no setting, or
input, other than their respective two quantum systems. Quantum outcome
distributions include as a special case the local distributions of Eq. 1, but
can also create nonlocal distributions [2, 4]. However, in general, it is not
straightforward to create such nonlocal distributions or to demonstrate
nonlocality for a given distribution, i.e., to prove that the distribution
could not be reproduced by a classical local model.

In [4], nonlocal quantum distributions were found in the triangle network by
using quantum sources and joint measurements with entangled eigenstates. The
nonlocality of these distributions could not be traced back to the standard
violation of Bell inequalities. Thus, their nonlocality appears to be
fundamentally different, which is a major step toward characterizing true
quantum phenomena. However, no reasonable noise-robust proof of nonlocality
has yet been found, rendering an experimental implementation impossible. In
the next section, we present in more detail another entangled measurement
scheme, which is conjectured to be nonlocal with an appropriate noise-
robustness. In addition to high correlations, it features a very high level of
symmetry.

####  II.2.2 The Elegant Joint Measurement

The Elegant Joint Measurement (EJM) was first introduced in 2017 and describes
a measurement of two qubits projected onto a basis of partially entangled
states with a tetrahedral symmetry (see [9] for more details). When applied to
the triangle scenario, the setting considered in [9] starts with all three
parties sharing pairwise the maximally entangled singlet state
$|\Psi^{-}\rangle=\frac{1}{\sqrt{2}}(|01\rangle-|10\rangle)$ and then
performing the EJM onto their two respective qubits. Each party obtains an
output $a,b,c\in\\{1,2,3,4\\}$, respectively, and as the resulting probability
distribution is highly symmetric, it can be fully described by only three
cases: all outcomes are equal, exactly two outcomes are equal, or all outcomes
are different. The distribution can be described as follows: for all
$a,b,c\in\\{1,2,3,4\\}$,

$p(a,b,c)=\left\\{\begin{aligned} \displaystyle\frac{25}{256}&\hskip
10.0pt\text{if }a=b=c,\\\ \displaystyle\frac{5}{256}&\hskip 10.0pt\text{if
}a\neq b\neq c\neq a,\\\ \displaystyle\frac{1}{256}&\hskip
10.0pt\text{else.}\end{aligned}\right.$

Although this distribution has strong correlations, i.e., a large probability
${p(A=B=C)}$, one could also find classical models with even higher
correlations (see the example in Section II.1.2). It is however conjectured
that this specific distribution is nonlocal due to its additional high degree
of symmetry [15], i.e., that the distribution cannot be written as Eq. 1. In
addition, unlike previous nonlocal quantum correlations, the EJM’s nonlocality
is also conjectured to be noise-robust [13]. This served as another motivation
to investigate fully symmetric distributions in the triangle network.

###  II.3 Fully symmetric distributions

We now focus on the case of four outcomes per party, since this is the case
for the EJM distribution of interest. Let us first define what we mean by
fully symmetric distributions: a distribution $p$ if fully symmetric if it is
symmetric under permutation of the parties as well as under joint permutation
of the outcomes (note that distributions which are symmetric under joint
permutations of the outcomes were referred to as output-permutation-invariant
(OPI) distributions in [16]). This implies that a fully symmetric distribution
$p$ can be characterized by only three values,

$\displaystyle s_{111}$ $\displaystyle=p(A=B=C)=4\,p(1,1,1),$ $\displaystyle
s_{112}$ $\displaystyle=p(A=B\neq C)+p(A=C\neq B)+p(B=C\neq A)$
$\displaystyle=36\,p(1,1,2),$ $\displaystyle s_{123}$ $\displaystyle=p(A\neq
B\neq C\neq A)=24\,p(1,2,3).$ (3)

In fact, the normalization $s_{111}+s_{112}+s_{123}=1$ implies that only two
values could be used. Additionally, let us define the three extremal fully
symmetric distributions, $p_{111}$, $p_{112}$ and $p_{123}$ as follows:

$\displaystyle p_{111}$ $\displaystyle=\frac{1}{4}\sum_{k}[k,k,k],$
$\displaystyle p_{112}$ $\displaystyle=\frac{1}{36}\sum_{k\neq
l}[k,k,l]+[k,l,k]+[l,k,k],$ $\displaystyle p_{123}$
$\displaystyle=\frac{1}{24}\sum_{k\neq l\neq m\neq k}[k,l,m].$

With the above definitions, we have that any fully symmetric distribution $p$
can be written as

$\displaystyle p=s_{111}p_{111}+s_{112}p_{112}+s_{123}p_{123},$

or in other words, for all $a,b,c\in\\{1,2,3,4\\}$,

$p(a,b,c)=s_{111}p_{111}(a,b,c)+s_{112}p_{112}(a,b,c)\\\
+s_{123}p_{123}(a,b,c).$

For instance, the EJM (Section II.2.2) is characterized by

$\displaystyle(s_{111},s_{112},s_{123})=\left(\frac{25}{64},\frac{9}{64},\frac{30}{64}\right).$

The Finner inequality [17] states that both local and quantum strategies in
the triangle network lead to distributions satisfying

$\displaystyle p(a,b,c)\leq\sqrt{p(A=a)p(B=b)p(C=c)}.$ (4)

In the special case of fully symmetric distributions, the marginals are
maximally mixed, since by symmetry we must have $p(A=a)=p(A=\sigma(a))$ for
all $\sigma\in S_{4}$, and similarly for the other marginals. Hence, Eq. 4
simplifies to

$\displaystyle p(a,b,c)\leq\frac{1}{8},$

which implies nothing for $s_{112}$ and $s_{123}$ (since it only states that
$s_{112}\leq\nicefrac{{36}}{{8}}$ and $s_{123}\leq\nicefrac{{24}}{{8}}$), but
it implies that

$\displaystyle s_{111}\leq\frac{1}{2}$ (5)

for all local and quantum distributions [18]. Note that the Finner inequality
may not be valid in physical theories that merely satisfy the non-signaling
principle, as investigated in [16].

##  III Analytical construction of local models

In this section, we describe analytical local models leading to fully
symmetric distributions. In the context of trying to find a local model for
the EJM distribution, a similar kind of parametrized local model was proposed
in [9]. The resulting local distributions are fully symmetric: the results of
[9] prove that for all $t\in[0,1]$, the fully symmetric distribution described
by

$(s_{111},s_{112},s_{123})=\left(\frac{52+9t}{256},\frac{180+9t}{256},\frac{24-18t}{256}\right)$

is local. This family of distributions can be seen as a special case of the
more general constructions that we describe in the following.

###  III.1 Description of the constructions

To obtain an inner approximation of the set of local distributions within the
symmetric subspace, we construct analytical local models that give rise to a
fully symmetric distribution $p(A,B,C)$. We do so using the flag model
introduced in Section II.1. The first step in doing so was to devise a method
for generating flags that yield an outcome-symmetric distribution. This method
is described in Sections A.2 and A.3. It essentially relies on ensuring that
outcome permutations can be “cancelled” by a suitable permutation of the
values of the three classical sources $\alpha,\beta,\gamma$. The flags that we
construct typically come with a few analytical parameters, such as $q$ and
$\nu$ in Fig. 3. We could then compute the output distribution of such flags
in terms of those parameters, and enforce party symmetry by imposing suitable
relations between those parameters. The explicit constructions are described
in Sections A.4, A.5 and A.6. As an example, we depict in Fig. 3 flags that
yield a highly correlated and fully symmetric distribution (upon imposing the
appropriate relation on $q$ and $\nu$, see Section A.4).

Figure 3:  Flags illustrating response functions $a(\beta,\gamma)$,
$b(\alpha,\gamma)$ and $c(\alpha,\beta)$ that yield the maximal three-party
correlation $p(A=B=C)=\nicefrac{{1}}{{4}}$ that a fully-symmetric local
distribution can achieve within the analytical inner approximation that we
considered. The flags are parameterized by $q\in[0,\nicefrac{{1}}{{2}}]$ and
$\nu\in[0,1]$, and they give rise to the distribution of Eq. 6 with $r=\eta=1$
if $\nu\in[0,\nicefrac{{1}}{{3}}]$ and $q=\nicefrac{{\nu}}{{(1-\nu)}}$. See
Section A.4 for more details on this construction.

The most general family of flags that we found, described in Section A.5,
comes with three parameters: $r,\eta\in[0,1]$ and
$\nu\in[0,\nicefrac{{1}}{{2}}]$, satisfying the relation

$0\leq\frac{1-r}{3}+\frac{\nu}{1-\nu}\frac{4\eta-1}{3}\leq\frac{1}{2}.$

They yield the following family of fully symmetric distributions:

$\displaystyle s_{111}$
$\displaystyle=\frac{1}{4}\big{(}(1-\nu)r+\eta\nu\big{)}$ $\displaystyle
s_{112}$ $\displaystyle=\frac{3}{4}\big{(}(1-\nu)(1-r)+3\eta\nu\big{)}$
$\displaystyle s_{123}$
$\displaystyle=\frac{1}{4}\big{(}1+(1-\nu)2r+(3-10\eta)\nu\big{)}.$ (6)

In a different construction, presented in Section A.6, we allowed Alice, Bob
and Charlie to further anti-correlate their strategies, achieving
distributions outside of the family of distributions described above. This
results in a line that can be described by

$\displaystyle(s_{111},s_{112},s_{123})=\Big{(}\frac{r}{48},\frac{4-r}{16},\frac{18+r}{24}\Big{)},$
(7)

for $r\in[0,1]$. As described in Section A.7, we can extend this line of
feasible distributions into a more general family of distributions by applying
a generic “decorrelation” procedure in which the players and the sources
modify their behaviors with a certain probability: this yields the little two-
dimensional “spike” at the bottom right of Fig. 4.

###  III.2 Visualizing the local set

To visualize the set of fully symmetric local models, we can plot all
combinations $(s_{111},s_{112},s_{123})$ in a triangle, where the three
corners correspond to the three extremal points $s_{111}=1$ (top), $s_{112}=1$
(bottom left) and $s_{123}=1$ (bottom right), respectively. All other points
are convex combinations and lie inside of the triangle, or on an edge if one
of the coefficient $(s_{111},s_{112},s_{123})$ is $0$. In Table 1, we describe
some segments of local distributions in the fully symmetric subspace that can
be obtained as special cases of Eq. 6: some of these visibly lie on the
boundary of the local region implied by Eq. 6. The local distributions of Eqs.
6 and 7 are plotted in Fig. 4.

Fixed values | Line $(s_{111},s_{112},s_{123})$ | Color in
---|---|---
From | To | Fig. 4
$\eta=1,\ r=1$ | $\displaystyle\left(\frac{1}{4},\frac{3}{4},0\right)$ | $\displaystyle\left(\frac{1}{4},0,\frac{3}{4}\right)$ | purple
$\eta=1,r=\displaystyle\frac{7\nu-1}{2(1-\nu)}$ | $\displaystyle\left(\frac{1}{28},\frac{27}{28},0\right)$ | $\displaystyle\left(\frac{1}{4},\frac{3}{4},0\right)$ | red
$\eta=1,\ r=0$ | $\displaystyle\left(0,\frac{3}{4},\frac{1}{4}\right)$ | $\displaystyle\left(\frac{1}{28},\frac{27}{28},0\right)$ | grey
$\eta=0,\displaystyle r=\frac{1-2\nu}{1-\nu}$ | $\displaystyle\left(0,\frac{3}{8},\frac{5}{8}\right)$ | $\displaystyle\left(\frac{1}{4},0,\frac{3}{4}\right)$ | dark green
$\eta=0,\ r=0$ | $\displaystyle\left(0,\frac{3}{4},\frac{1}{4}\right)$ | $\displaystyle\left(0,\frac{3}{8},\frac{5}{8}\right)$ | light blue
$\nu=0$ | $\displaystyle\left(0,\frac{3}{4},\frac{1}{4}\right)$ | $\displaystyle\left(\frac{1}{4},0,\frac{3}{4}\right)$ | dark blue
Table 1: We fix some of the parameters of Eq. 6 to obtain a few key lines,
which are also represented in Fig. 4. Figure 4: The set of fully symmetric
local distributions given by Eq. 6. The respective lines are indicated
explicitly in Table 1. Additionally, we plot the construction described around
Eq. 7: it is indicated by the $(*)$ marker. We also indicate the Finner
inequality, see Eq. 5. The fact that the EJM distribution $p_{\textup{EJM}}$
(see Section II.2.2) is quite far from all the local distributions we found
supports the conjecture that it is nonlocal.

From the construction in Section A.5 we can see that the cases with $\eta=1$,
which are at the top left of the set, describe the strategies where Alice and
Bob use their common source to maximally correlate, while for the cases with
$\eta=0$, which are on the bottom right of the set, they use it to maximally
anti-correlate. Note that the case $\eta=1$, $r=1$, which corresponds to a
horizontal line at $s_{111}=\nicefrac{{1}}{{4}}$, corresponds to the maximally
correlated flags constructed in Section A.4 and shown in Fig. 3. The case of
$\nu=0$, dividing the cases with $\eta=1$ and $\eta=0$, describes the
strategies where Alice and Bob do not use their common source at all. The
resulting distributions could be obtained already in the bilocal network,
which corresponds to the triangle network but without the source between Alice
and Bob. These include the maximally mixed distribution, where
$p(a,b,c)=\nicefrac{{1}}{{64}}$, yielding
$(s_{111},s_{112},s_{123})=(\nicefrac{{1}}{{16}},\nicefrac{{9}}{{16}},\nicefrac{{6}}{{16}})$,
which can be seen as the most trivial symmetric strategy, as it can be reached
by every party just randomly outputting each of the four outcomes with equal
probability.

##  IV Neural Network Approaches

Numerical searches of local models in the triangle network are difficult, as
the problem is non-convex and one often ends up in local minima. However,
modeling local response functions with artificial neural networks has been
shown to be a relatively reliable heuristic, reproducing benchmark results, as
well as providing new conjectures, which have since been partially proven [13,
6]. In general, a feed-forward artificial neural network is a numeric model
for a multivariate, multidimensional function. It can be trained, i.e., its
parameters can be fit, in order to minimize an objective function that depends
on the neural network’s outputs in a differentiable manner.

###  IV.1 Minimizing distance to target distributions

For local models in networks, one can model each of the response functions in
Eq. 1 with neural networks, i.e., the neural network for Alice would take as
inputs some $\beta_{i}$ and $\gamma_{i}$ values and output (a normalized, not
necessarily deterministic)
$p_{A}^{\text{NN}}(a|\beta_{i},\gamma_{i})\in\mathbb{R}^{4}$, and similarly
for Bob and Charlie (this architecture is often referred to as LHV-Net). An
example of response functions that LHV-Net “thought about” is displayed in
Fig. 5. Sampling over $M\gg 1$ triples
$(\alpha_{i},\beta_{i},\gamma_{i})\in[0,1]^{\times 3}$, one can then
numerically calculate a Monte Carlo estimate of Eq. 1,

$p_{\text{NN}}=\frac{1}{M}\sum_{i=1}^{M}p_{A}^{\text{NN}}(a|\beta_{i},\gamma_{i})p_{B}^{\text{NN}}(b|\gamma_{i},\alpha_{i})p_{C}^{\text{NN}}(c|\alpha_{i},\beta_{i}).$

Crucially, the neural networks of Alice, Bob, and Charlie only have access to
the respective hidden variables allowed by the triangle structure, thus any
distribution given by LHV-Net is by definition local. The three neural
networks are then jointly optimized by minimizing the objective function
$||p_{\text{target}}-p_{\text{NN}}||^{2}$. In our case, we use a multilayer
perceptron of depth 4 and width 30 with rectified linear activation functions
for each party, with an Adadelta optimizer, and stochastic gradient descent
used for fine-tuning the weights. We designated 260 target distributions in
the symmetric subspace, and trained the neural network 10 times independently
for each of them. Finally, we kept the best resulting model for each point.

Figure 5: Illutration made from some of the response functions that the neural
network came up with. To obtain this figure, we turned the probabilistic
reponse functions into deterministic ones by picking the most likely outcome
for each input values. See Section D.3 for more details about these specific
flags.

The results are displayed in Fig. 6. It is important to note that the neural
network’s output distributions are not forced to be symmetric, i.e., they are
not necessarily actually within the fully symmetric subspace. However, those
that are close in 2-norm to their respective (fully symmetric) targets are
naturally close to being symmetric. Hence, the resulting dark blue region
portrayed in Fig. 6 gives a good indication of a region that is local in the
fully symmetric subspace. Further plots of scans of the symmetric subspace for
$N$ outcomes with $N=3,5,6$ can be found in Appendix B.

Figure 6: Results from the neural network trained to find local strategies
yielding a distribution close to each point inside of the fully symmetric
subspace. The color indicates the distance to the actual distribution, which
should be very small (i.e., dark blue) in order to consider a local strategy
as “detected”. The orange line delimits the local region described in Fig. 4
and is drawn to emphasize the good agreement between the analytical local
models we found and what the neural network perceives as local. We also
indicate the Finner inequality, see Eq. 5, as well as the $l=1$ conjectured
Bell inequality of Eq. 9.

###  IV.2 Finding inequalities

In general causal scenarios, Bell-type inequalities are those which are
satisfied by correlations from local models, and hopefully violated by some
quantum correlations. Obtaining such inequalities for the triangle network has
proven difficult, with previous attempts not finding any genuine quantum
violations or resulting in difficult-to-interpret inequalities [19, 20, 21].

Regarding the symmetric subspace, our previous analytic and numeric findings
give strong evidence that within the symmetric subspace the $s_{111}$ of local
models is limited by some value $s_{111}^{*}$, with
$s_{111}^{*}\approx\nicefrac{{1}}{{4}}$, allowing for a simple inequality that
would rule out the Elegant distribution. However, outside this subspace, local
models can reach high (in fact, maximal) $s_{111}$ values. Formalizing this
intuition (that it is difficult for local models to simultaneously have large
$s_{111}$ and be symmetric) as an inequality for generic distributions is a
priori difficult.

By changing the objective function of LHV-Net, we can test different ansätze
for Bell-type inequalities for the triangle network. We do this by penalizing
asymmetry with a penalty term on each of the three types of probabilities
appearing in symmetric distributions (111, 112, 123). Specifically, we sum up
the joint deviation from the mean for each of these types of probabilities,

$\displaystyle\Delta_{l}$
$\displaystyle=\sum_{X\in\\{111,112,123\\}}\Delta_{l,X},$
$\displaystyle\Delta_{l,X}$
$\displaystyle=\sum_{\\{a,b,c\\}\in\mathcal{I}_{X}}|M_{X}-p(a,b,c)|^{l},$
$\displaystyle M_{X}$
$\displaystyle=\frac{1}{|\mathcal{I}_{X}|}\sum_{\\{a,b,c\\}\in\mathcal{I}_{X}}p(a,b,c),$

where $\mathcal{I}_{X}$ is the index set of $X$-type outcomes, and
$l\in\\{1,2\\}$. In particular $\mathcal{I}_{111}$ will contain 4 elements,
$\mathcal{I}_{112}$ 36, and $\mathcal{I}_{123}$ 24 elements.

Consider now the following function:

$\displaystyle f_{w}(p)=w\cdot s_{111}(p)-(1-w)\Delta_{l}(p).$

By maximizing this quantity for LHV models, we can see whether it can
outperform $f_{w}(p_{\textup{EJM}})$ for any value of $w$. Intuitively, we are
trying to maximize $s_{111}$ (with weight $w$) and minimize the penalty
$\Delta_{l}$ (with weight $1-w$). We define the gap

$\displaystyle\delta_{w}:=f_{w}(p_{\textup{EJM}})-\max_{p\in\mathcal{L}}f_{w}(p),$

which, if positive, defines an inequality

$\displaystyle f_{w}(p)\leq f_{w}(p_{\textup{EJM}})-\delta_{w},$ (8)

which all local distributions must obey.

Finding the exact value of $\delta_{w}$ is difficult, as one must optimize
over local models of the form of Eq. 1. However, using LHV-Net, one can obtain
an estimate of $\delta_{w}$, by setting the objective function to be
$-f_{w}(p)$. For each $w$ value, we train the neural network from scratch to
try to violate that given inequality and plot the resulting $\delta_{w}$
values in Appendix D, for both absolute value and square penalty ($l=1,2$). We
find the largest $\delta_{w}$ for $l=1$ ($l=2$) at $w^{*}\approx 0.678$
($w^{*}\approx 0.161$) with $\delta_{w^{*}}\approx 0.069$
($\delta_{w^{*}}\approx 0.012$), getting that approximately

$\displaystyle s_{111}(p)-0.475\,\Delta_{l=1}(p)$ $\displaystyle\leq 0.289,$
(9) $\displaystyle s_{111}(p)-5.211\,\Delta_{l=2}(p)$ $\displaystyle\leq
0.316,$ (10)

should both hold for all local models. Recently, the inequality with $l=2$ has
been violated for a range of $w$ values by experimentally obtained data [14].
In the symmetric subspace, the penalty term vanishes, which means that Eq. 9
is the most constraining of the two inequalities. This is the inequality that
we plotted in Fig. 6.

Interestingly, when maximizing these inequalities, the neural network seems to
find something better than the $s_{111}=\nicefrac{{1}}{{4}}$ strategy that was
found analytically. This apparent out-performance of the $\nicefrac{{1}}{{4}}$
bound also appears in Fig. 6, most prominently at values where $p(112)\approx
p(123)$. In Appendix D we include details about the local strategy that the
neural network found with $s_{111}\approx 0.289$ ($\Delta_{l=1}\approx 0.009$,
$\Delta_{l=2}\approx 2.3\cdot 10^{-6}$), and its discretized, deterministic
approximation, which is displayed in Fig. 5, with $s_{111}\approx 0.294$
($\Delta_{l=1}\approx 0.014$, $\Delta_{l=2}\approx 4.7\cdot 10^{-6}$). It is
currently an open question whether there exists an exactly symmetric local
distribution with $s_{111}>\nicefrac{{1}}{{4}}$ (e.g. $s_{111}=0.289$ as
implied by the inequality), or whether these distributions only exist very
close to the symmetric subspace. Note that in the case of 3 outcomes per
party, it is possible to find a fully symmetric local distribution with
$s_{111}>\nicefrac{{1}}{{3}}$ as shown in Section C.1.

Finally, one could try a variety of different penalty functions for
constructing inequalities. However, several simple ones that do not work are
using only $\Delta_{l}=\Delta_{l,111}$ or when imposing symmetry only at the
level of the single-party marginals. Both of these types of penalty functions
have the local distribution in Section II.1.2 as an example of why such
penalty functions would not work: this local distribution would get zero
penalty, however, its $s_{111}$ is larger than that of the EJM distribution.

##  V Conclusion

The high level of symmetry in the correlations obtained from performing the
Elegant Joint Measurement [9] in the triangle network, which are conjectured
to lead to noise-robust nonlocal quantum correlations, inspired us to
investigate fully symmetric distributions in this setting, i.e., distributions
that are symmetric under permutation of the parties and under joint
permutations of the outcomes.

We analytically constructed classical local model and applied neural network
techniques to substantiate our findings. The agreement between the two methods
is best witnessed in Fig. 6. Both methods are fundamentally inner
approximations, i.e., they can only certify that a given distribution _is_
local. However, the good agreement between the two methods suggests that, in
this case, these methods are essentially able to find a local model for a
distribution if the distribution is local. Of course, the exact location of
the boundary of the local set is still hard to pinpoint, but in any case, this
boundary seems fairly far away from the Elegant Joint Measurement
distribution. Moreover, we formalized the trade-off that local models face
between being highly correlated and highly symmetric via the conjectured Bell
inequalities (see Section IV.2).

This led to the conjecture that local models yielding a fully symmetric
outcome distribution with four outcomes per party have a maximal correlation
very close to $p(A=B=C)=\nicefrac{{1}}{{4}}$. While the $\nicefrac{{1}}{{4}}$
maximum value is what we found with our analytical construction, the neural
network approach supports that indeed the correlation cannot be much higher,
but also indicates that the true value might be slightly above
$\nicefrac{{1}}{{4}}$.

######  Open Problem.

Does there exist a distribution $p(A,B,C)$ that is local in the triangle
network and fully symmetric such that $p(A=B=C)>\nicefrac{{1}}{{4}}$ ?

However, as the EJM is well above that bound, even a slightly higher upper
bound would still imply that it leads to noise-robust nonlocal quantum
correlations. For $N\geq 3$ outcomes per party, we can construct strategies
with $p(A=B=C)=\nicefrac{{1}}{{N}}$, and let neural networks substantiate that
the local upper bound must be close to $\nicefrac{{1}}{{N}}$ (see Appendix B).
However, in the $N=3$ outcome case, we found an analytical local model that
gives rise to $s_{111}>\nicefrac{{1}}{{3}}$ (see Section C.1), but were not
able to generalize this to $N\geq 4$ outcomes.

It still remains open to find a proof that the Elegant Joint Measurement is
nonlocal in the triangle network. Ideally, it would be even better to obtain a
more general bound on the maximal correlations for local models like our
suggested conjecture, or an even more general network Bell inequality that can
be used to identify nonlocality irrespectively of symmetry. In parallel, it
would be useful to develop similar inner approximation tools for the set of
quantum distributions in the triangle network. Such an exploration could in
particular provide additional examples of quantum nonlocality on top of the
few examples that are known today.

##  VI Data availibility

A data appendix is available at Ref. [22]. The following are included.

  * •

The ($w,\delta_{w}$) pairs that were found by LHV-Net for Eq. 8 (both for
$l=1$ and $l=2$).

  * •

The almost symmetric distribution found by LHV-Net that has
$s_{111}>\nicefrac{{1}}{{4}}$, as well as the corresponding (discretized,
deterministic) flags that generate it.

  * •

Data used in the LHV-Net maps of the symmetric subspace for $N=3,4,5,6$ (Figs.
6 and 14).

  * •

Accompanying scripts to load and evaluate this data.

  * •

A Wolfram Mathematica script that displays the different analytical flags that
we used in Section III, and computes the associated outcome distributions.

  * •

A Python script that displays the inner approximation of Fig. 4.

##  VII Acknowledgements

This work was supported by the Swiss National Science Foundation (NCCR
SwissMAP, as well as project No. 200021_188541 and P1GEP2_199676 (TK)) and by
the Air Force Office of Scientific Research via grant FA9550-19-1-0202. TK was
additionally funded by the European Research Council (Consolidator grant
’Cocoquest’ 101043705) and the Austrian Federal Ministry of Education via the
Austrian Research Promotion Agency–FFG (flagship project FO999897481, funded
by EU program NextGenerationEU). Most of the neural network computations were
performed at University of Geneva on “Baobab” HPC cluster.

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##  Appendix A General flag constructions

In this section, we describe the general four-outcome flag models that
underlie Section III.

###  A.1 Two-party marginals of the Elegant Joint Measurement

We start by noting that the Elegant Joint Measurement has local two-party
marginals in the following sense. In Fig. 7, we provide an explicit strategy
such that all the two-party marginals are equal to those of the Elegant Joint
Measurement, i.e.,

$\displaystyle p(A=k,B=l)$
$\displaystyle=p(A=k,C=l)=p(B=k,C=l)=\left\\{\begin{aligned}
\displaystyle\frac{7}{64}&\hskip 10.0pt\text{if }k=l,\\\
\displaystyle\frac{3}{64}&\hskip 10.0pt\text{if $k\neq
l$}.\end{aligned}\right.$

However, the full distribution $p(A,B,C)$ obtained from this strategy is not
equal to that of the Elegant Joint Measurement. In particular, the three-party
correlations are quite small, with

$\displaystyle p(A=B=C=k)=\frac{1}{16}.$ (11)

Additionally, we note that while the resulting three-party-distribution is
symmetric under cyclic permutations of the parties, it is not symmetric under
permutation of the outcomes.

Figure 7: Flags illustrating response functions $a(\beta,\gamma)$,
$b(\alpha,\gamma)$ and $c(\alpha,\beta)$ that yield the same two-party
marginals as the EJM, but not the right three-party probabilities.

###  A.2 Defining outcome-symmetric local models

As we would like to characterize the set of fully symmetric distributions with
four outcomes in the triangle network that admit a local model, we start by
analytically constructing such local models. A straightforward but potentially
limiting way to construct fully symmetric flags is to allow each party only
strategies that are symmetric under permutation of the outputs. Let us now
describe such strategies formally.

Figure 8: We divide the range of each source into four equal partitions which
leads to a $4\times 4$ grid of $16$ strategies per party that we label by
$q_{A}^{jk}$, $q_{B}^{ki}$ and $q_{C}^{ij}$, respectively.

Let

$\displaystyle q_{A}^{jk}:\\{1,2,3,4\\}\times[0,1]^{\times 2}\to[0,1]$

be Alice’s “sub-response-functions”, for $j,k\in\\{1,2,3,4\\}$. Let
$I_{j}=[(j-1)/4,j/4[$ for $j\in\\{1,2,3,4\\}$. The full response function of
Alice is defined as a piecewise function: for all $\beta,\gamma\in[0,1]$, with
$j,k$ be such that $\beta\in I_{j}$ and $\gamma\in I_{k}$,

$\displaystyle p_{A}(a|\beta,\gamma)=q_{A}^{jk}(a|4\beta-j+1,4\gamma-k+1).$

We now require the following: for all $\sigma\in S_{4}$ (the group of
permutations of $\\{1,2,3,4\\}$),

$\displaystyle
q_{A}^{\sigma(j)\sigma(k)}(\sigma(a)|\beta,\gamma)=q_{A}^{jk}(a|\beta,\gamma).$
(12)

We repeat this construction for Bob’s and Charlie’s response functions, with
associated sub-response-functions $q_{B}^{kl}$ and $q_{C}^{li}$ satisfying a
constraint analogous to Eq. 12. A straightforward calculation shows that the
output distribution of such a local model is

$\displaystyle p(a,b,c)=\frac{1}{4^{3}}\sum_{i,j,k=1}^{4}p^{ijk}(a,b,c),$

where

$\displaystyle p^{ijk}(a,b,c)=\int_{[0,1]^{\times
3}}\mathrm{d}\alpha\,\mathrm{d}\beta\,\mathrm{d}\gamma\,q_{A}^{jk}(a|\beta,\gamma)q_{B}^{ki}(b|\gamma,\alpha)q_{C}^{ij}(c|\alpha,\beta).$

The fact that $p(\sigma(a),\sigma(b),\sigma(c))=p(a,b,c)$, for all
$a,b,c\in\\{1,2,3,4\\}$ and $\sigma\in S_{4}$, follows directly from the
observation that Eq. 12 implies

$\displaystyle
p^{ijk}(\sigma(a),\sigma(b),\sigma(c))=p^{\sigma^{-1}(i)\sigma^{-1}(j)\sigma^{-1}(k)}(a,b,c).$

Thus, this construction yields distributions that are symmetric under
permutation of the outputs, but not yet necessarily under permutation of the
parties. We give more details on how to construct such families of response
functions in Section A.3.

###  A.3 Generating outcome-symmetric local models

In this section, we describe how to generate flags that satisfy the
constraints described in Section A.2. Let $q_{A}^{11}(a|\beta,\gamma)$ be an
arbitrary sub-response-function that satisfies the constraint

$\displaystyle
q_{A}^{11}(2|\beta,\gamma)=q_{A}^{11}(3|\beta,\gamma)=q_{A}^{11}(4|\beta,\gamma).$
(13)

Let $\tau_{j}\in S_{4}$ be the transposition $1\leftrightarrow j$ (note that
$\tau_{j}^{-1}=\tau_{j}$ and $\tau_{1}$ is the identity). Then, define
$q_{A}^{jj}$ for $j\in\\{2,3,4\\}$ as follows:

$\displaystyle
q_{A}^{jj}(a|\beta,\gamma)=q_{A}^{11}(\tau_{j}(a)|\beta,\gamma).$ (14)

Notice that Eq. 14 also holds trivially for $j=1$. Let
$q_{A}^{12}(a|\beta,\gamma)$ be an arbitrary sub-response-function that
satisfies the constraint

$\displaystyle q_{A}^{12}(3|\beta,\gamma)=q_{A}^{12}(4|\beta,\gamma).$ (15)

For all $j,k\in\\{1,2,3,4\\}$, $j\neq k$, define $\pi_{jk}\in S_{4}$ to be the
permutation such that $\pi_{jk}(1)=j$, $\pi_{jk}(2)=k$ (and complete it
arbitrarily — Eq. 16 will still be well-defined thanks to Eq. 15). Define
$q_{A}^{jk}$ for $j\neq k$, $(j,k)\neq(1,2)$ as follows:

$\displaystyle
q_{A}^{jk}(a|\beta,\gamma)=q_{A}^{12}(\pi_{jk}^{-1}(a)|\beta,\gamma).$ (16)

Notice that Eq. 16 also holds trivially for $(j,k)=(1,2)$.

######  Lemma 1.

Any such family $\\{q_{A}^{jk}\\}_{j,k=1}^{4}$ satisfies the constraint of Eq.
12.

###### Proof.

The case of $j=k$. We have to show in this case that

$\displaystyle
q_{A}^{jj}(\sigma(a)|\beta,\gamma)=q_{A}^{\sigma^{-1}(j)\sigma^{-1}(j)}(a|\beta,\gamma).$
(17)

Using Eq. 14, this simplifies to

$\displaystyle
q_{A}^{11}(\tau_{j}\circ\sigma(a)|\beta,\gamma)=q_{A}^{11}(\tau_{\sigma^{-1}(j)}(a)|\beta,\gamma).$
(18)

Thanks to Eq. 13, this follows if either

$\displaystyle(\tau_{j}\circ\sigma(a)=1\textup{ and
}\tau_{\sigma^{-1}(j)}(a)=1)$ (19) or
$\displaystyle(\tau_{j}\circ\sigma(a)\neq 1\textup{ and
}\tau_{\sigma^{-1}(j)}(a)\neq 1).$ (20)

Let $P$ be the proposition $\tau_{j}\circ\sigma(a)=1$ and $Q$ the proposition
$\tau_{\sigma^{-1}(j)}(a)=1$. We want to show $(P\land Q)\lor(\lnot
P\land\lnot Q)$. This is trivially true if $P\Leftrightarrow Q$. This is
exactly what happens: we have that $P$ is equivalent to $\sigma(a)=j$, while
$Q$ is equivalent to $a=\sigma^{-1}(j)$, which is clearly equivalent to $P$.

The case of $j\neq k$. We have to show in this case that

$\displaystyle
q_{A}^{jk}(\sigma(a)|\beta,\gamma)=q_{A}^{\sigma^{-1}(j)\sigma^{-1}(k)}(a|\beta,\gamma).$
(21)

Using Eq. 16, this simplifies to

$\displaystyle
q_{A}^{12}(\pi^{-1}_{jk}\circ\sigma(a)|\beta,\gamma)=q_{A}^{12}(\pi^{-1}_{\sigma^{-1}(j)\sigma^{-1}(k)}(a)|\beta,\gamma).$
(22)

Thanks to Eq. 15, this follows if either

$\displaystyle(\pi^{-1}_{jk}\circ\sigma(a)=1\textup{ and
}\pi^{-1}_{\sigma^{-1}(j)\sigma^{-1}(k)}(a)=1)$ (23) or
$\displaystyle(\pi^{-1}_{jk}\circ\sigma(a)=2\textup{ and
}\pi^{-1}_{\sigma^{-1}(j)\sigma^{-1}(k)}(a)=2)$ (24) or
$\displaystyle(\pi^{-1}_{jk}\circ\sigma(a)\notin\\{1,2\\}\textup{ and
}\pi^{-1}_{\sigma^{-1}(j)\sigma^{-1}(k)}(a)\notin\\{1,2\\}).$ (25)

Let $P_{x}$ be the proposition $\pi^{-1}_{jk}\circ\sigma(a)=x$ and $Q_{x}$ be
the proposition $\pi^{-1}_{\sigma^{-1}(j)\sigma^{-1}(k)}(a)=x$. We want to
show

$\displaystyle(P_{1}\land Q_{1})\lor(P_{2}\land
Q_{2})\lor\Big{(}\lnot(P_{1}\lor P_{2})\land\lnot(Q_{1}\lor Q_{2})\Big{)}.$
(26)

Notice that $P_{1}\Leftrightarrow Q_{1}$: indeed, $P_{1}$ is equivalent to
$\sigma(a)=\pi_{jk}(1)=j$, i.e., $\sigma(a)=j$, while $Q_{1}$ is equivalent to
$a=\pi_{\sigma^{-1}(j)\sigma^{-1}(k)}(1)=\sigma^{-1}(j)$, i.e.,
$a=\sigma^{-1}(j)$. Similarly, $P_{2}\Leftrightarrow Q_{2}$. Thus, Eq. 26
simplifies to

$\displaystyle P_{1}\lor P_{2}\lor\lnot(P_{1}\lor P_{2}),$ (27)

which is trivially true. ∎

###  A.4 Maximizing the correlations

In this section, we describe flags that satisfy the symmetry constraints of
Section A.2 and that maximize the correlations $p(A=B=C)$. Indeed, the
combination of high symmetry and strong correlations seems to be one of the
essential characteristics implying nonlocality. We came up with the following
parametrized construction, yielding the flags depicted in Fig. 3 and
reproduced in Fig. 9 for easier reference.

From Finner’s inequality [17] (see Eq. 4), we know that in order to maximize
the volume where $A=B=C$, we should aim at having unicolored “cubes” as large
as possible. Assuming the grid for the symmetrized strategy as in Fig. 8, we
can at most have such cubes of side length $\nicefrac{{1}}{{4}}$. Thus, it
seems that the best strategy would be for each party to output the outcome $k$
in the sub-response-function $q_{A}^{kk}$, $q_{B}^{kk}$ and $q_{C}^{kk}$,
respectively. Next, we need to maximize the correlation in the off-diagonal
sub-response-functions $q_{A}^{jk}$, $q_{B}^{ki}$ and $q_{C}^{ij}$,
respectively. We start by considering Alice and Bob and allow them to take a
fraction $\nu\in[0,\nicefrac{{1}}{{3}}]$ (this upper bound on $\nu$ will be
explained later) of each interval $I_{k}=[(k-1)/4,j/4[$ of their $\gamma$
input in which they perfectly correlate by just outputting the color $k$. To
correlate with Charlie, they output $j$ and $i$ in their remaining parts of
the sub-response-functions $q_{A}^{jk}$ and $q_{B}^{ki}$, respectively. Note
that Alice’s and Bob’s strategies are the same up to a reflection and uniform
among all $\alpha\in I_{i}$ and $\beta\in I_{j}$. It remains to define
Charlie’s sub-response-functions $q_{C}^{ij}$ for $i\neq j$, which we can
describe, thanks to the symmetries in Alice’s and Bob’s strategies, with just
one variable $q\in[0,\nicefrac{{1}}{{2}}]$, such that

$\int_{[0,1]^{\times
2}}\mathrm{d}\alpha\,\mathrm{d}\beta\,q_{C}^{ij}(i|\alpha,\beta)=\int_{[0,1]^{\times
2}}\mathrm{d}\alpha\,\mathrm{d}\beta\,q_{C}^{ij}(j|\alpha,\beta)=q,$

and for all $k\notin\\{i,j\\}$,

$\int_{[0,1]^{\times
2}}\mathrm{d}\alpha\,\mathrm{d}\beta\,q_{C}^{ij}(k|\alpha,\beta)=\frac{1}{2}-q.$

This yields the strategy given by the flags in Fig. 9.

Figure 9: Flags illustrating response functions $a(\beta,\gamma)$,
$b(\alpha,\gamma)$ and $c(\alpha,\beta)$ that yield the (conjectured to be)
maximal three-party correlation $p(A=B=C)=\nicefrac{{1}}{{4}}$ as a function
of $\nu\in[0,\nicefrac{{1}}{{3}}]$ and $q=\nicefrac{{\nu}}{{(1-\nu)}}$.

Since the flags satisfy the constraints of Section A.2, it is already
guaranteed that the resulting distribution is symmetric under joint
permutation of the outcomes. Note that in Fig. 9, the constraint of Eq. 15 is
not strictly satisfied: we should have that the yellow and green components of
$q_{C}^{12}$ are mixed together. However, the flags that we drew in Fig. 9 are
_equivalent_ (as far as the resulting distribution goes) with flags where the
yellow and green regions would be mixed together — the latter being harder to
clearly depict.

Finally, we need to relate $q$ to $\nu$ such that the distribution is
symmetric under permutation of the parties. The flags as drawn in Fig. 9
result in the following distribution: for all $k,l,m\in\\{1,2,3,4\\}$, $k\neq
l\neq m\neq k$,

$\displaystyle p(k,k,k)$ $\displaystyle=\frac{1}{16},$ $\displaystyle
p(k,k,l)$ $\displaystyle=\frac{\nu}{16},$ $\displaystyle p(k,l,k)$
$\displaystyle=p(l,k,k)=(1-\nu)\frac{q}{16},$ $\displaystyle p(k,l,m)$
$\displaystyle=(1-\nu)\frac{1}{16}\Big{(}\frac{1}{2}-q\Big{)}.$

We must thus require $p(k,k,l)=p(k,l,k)$ to achieve full symmetry, which
implies $q=\nicefrac{{\nu}}{{(1-\nu)}}$. Noting that $q\in[0,\frac{1}{2}]$, we
must thus have $\nu\in[0,\frac{1}{3}]$. Thus, all distributions resulting from
our construction above can be characterized by

$\displaystyle\left(s_{111},s_{112},s_{123}\right)$
$\displaystyle=\left(\frac{1}{4},\frac{9\nu}{4},\frac{3-9\nu}{4}\right).$ (28)

Building also on the intuition we gained while constructing different local
models, this leads us to the conjecture that any local model in the triangle
scenario that yields a fully symmetric distribution with four outcomes per
party can reach a maximal correlation that is very close to
$s_{111}=p(A=B=C)=\nicefrac{{1}}{{4}}$. Finding the exact maximum value and
proving such an upper bound remains an open question. A direct implication of
this conjecture would be the nonlocality with noise-robustness of the
distribution obtained from the Elegant Joint Measurement. Note that while here
we focus on the scenario with four outcomes, we can directly extend our
construction to strategies with $N$ outcomes that yield
$s_{111}^{(N)}=\nicefrac{{1}}{{N}}$ for $N\geq 3$ (see Appendix B) and state a
generalized conjecture.

###  A.5 More general flags

For a more general characterization of the local distributions in the
symmetric subspace, we should consider also strategies with smaller
correlations or even anti-correlations. Thus, we allow Alice to choose a
fraction $\eta$ out of her fraction $\nu$ in which she previously correlated
to Bob, in which she still correlates to Bob, while in the $(1-\eta)\nu$
remaining fraction she completely anti-correlates to Bob. The flag
corresponding to her more general strategy is illustrated in Fig. 10 (note
that Bob’s strategy stays the same as before). In a similar way, Charlie could
decide to anti-correlate in his strategies $q_{C}^{ii}$ (i.e. on the diagonal
of his flag) by choosing output $i$ only with probability $r\in[0,1]$ and else
sample uniformly from the other outputs, as also illustrated in Fig. 10.
Again, we note that the flag of Charlie depicted in Fig. 10 does not strictly
satisfy the constraint of Eq. 13, but is equivalent to one where the purple,
yellow and green colors are mixed together in the $q_{C}^{11}$ sub-response-
function.

Figure 10: Alice’s and Charlie’s more general flags, which allow for more
anti-correlations. Alice is changing her strategies in the fraction $\nu$,
while Charlie is only replacing his strategies $q_{C}^{ii}$, i.e., those that
lie on the diagonal of his grid.

Again, our construction already guarantees that the resulting distribution is
symmetric under permutation of the outcomes, but we still need to ensure that
it is also symmetric under permutation of the parties. For that we need to
relate $q$ to $r$, $\nu$ and $\eta$. Let us determine the following
probabilities: for all $k,l,m\in\\{1,2,3,4\\}$, with $k\neq l\neq m\neq k$,

$\displaystyle p(k,k,k)$
$\displaystyle=\frac{1}{16}\left[(1-\nu)r+\nu\eta\right],$ $\displaystyle
p(k,k,l)$
$\displaystyle=\frac{1}{16}\left[(1-\nu)\frac{1-r}{3}+\nu\eta\right],$
$\displaystyle p(k,l,k)$
$\displaystyle=p(l,k,k)=\frac{1}{16}\left[(1-\nu)q+\nu\frac{1-\eta}{3}\right],$
$\displaystyle p(k,l,m)$
$\displaystyle=\frac{1}{16}\left[(1-\nu)(\frac{1}{2}-q)+\nu\frac{1-\eta}{3}\right].$

To achieve symmetry under permutation of the outcomes we require
$p(k,k,l)=p(k,l,k)$ which implies

$q=\frac{1-r}{3}+\frac{\nu}{1-\nu}\frac{4\eta-1}{3}.$

Recall that $q$ is subject to the constraint $q\in[0,\nicefrac{{1}}{{2}}]$,
which implies a constraint on $r,\nu,\eta$. This yields the distribution in
Eq. 6.

###  A.6 Even more anti-correlated flags

To further explore the set of anti-correlated strategies, a new approach was
based on the idea to minimize the probabilities $p(A=B)$, $p(A=C)$ and
$p(B=C)$ by completely anti-correlating all off-diagonal elements of the
flags. Using again the construction of Section A.2, this can be done when each
party outputs $k$ in their substrategies $q_{A}^{jk}$, $q_{B}^{jk}$ and
$q_{C}^{jk}$, respectively, for $j\neq k$. Then we only need to determine the
“diagonal” strategies $q_{A}^{kk}$, $q_{B}^{kk}$ and $q_{C}^{kk}$.
Uncorrelating Alice and Bob by returning all outputs $i\neq k$ in strategies
$q_{A}^{kk}$ and $q_{B}^{kk}$ with equal probability independently of $\gamma$
allows to parameterize Charlie’s strategy in the way depicted in Fig. 11. Note
that Charlie has only one parameter $r$ to tune the correlation with Alice and
Bob, as everything else results from the symmetry constraints.

Figure 11: Even more anti-correlated flags that yield the distribution in Eq.
7.

This leads to the distribution of Eq. 7. To obtain the full “spike” at the
bottom right of the local distributions shown in Fig. 4, one should
furthermore take into the decorrelation currents presented in Section A.7.

###  A.7 Decorrelation currents

Given a distribution which is guaranteed to be local, but whose local model is
unknown, it is occasionally possible to deduce that other distributions are
also local. Such deductions can be obtained by modifying the original unknown
local model: the sources and the parties may randomly deviate from their
original behavior. The deviations from the original behavior must be chosen in
such a way that the new distribution which is also local can be written
entirely in terms of the original output distribution and not in terms of its
unknown original local model. If the deviations from the original model are
parametrized by $\varepsilon$, such that for $\varepsilon=0$, there is no
deviation, and for $\varepsilon=1$, the distribution is “more mixed” then the
original distribution, then we have what we call a _decorrelation current_ ,
which flows continuously from the original local distribution to other local
distributions which are more mixed.

We now give an explicit example of such a decorrelation current. This is the
current that we use to extend the line of local distributions described in Eq.
7 to the two-dimensional local region labeled $(*)$ in Fig. 4.

######  Lemma 2.

Suppose that the fully symmetric distribution $p^{(0)}$ described by
$(s_{111}^{(0)},s_{112}^{(0)},s_{123}^{(0)})$ is local. Then, for all
$\varepsilon\in[0,1]$, for all $l\in[0,1]$, the fully symmetric distribution
$p^{(\varepsilon,l)}$ described by

$\displaystyle s_{111}^{(\varepsilon,l)}$
$\displaystyle=(1-\varepsilon)^{3}s_{111}^{(0)}+3\left(\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\right)\frac{1-l}{4}+\frac{1}{64}\varepsilon^{3}$
$\displaystyle s_{112}^{(\varepsilon,l)}$
$\displaystyle=(1-\varepsilon)^{3}s_{112}^{(0)}+3\left(\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\right)\frac{3-l}{4}+\frac{9}{64}\varepsilon^{3}$
$\displaystyle s_{123}^{(\varepsilon,l)}$
$\displaystyle=(1-\varepsilon)^{3}s_{123}^{(0)}+3\left(\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\right)\frac{l}{2}+\frac{6}{64}\varepsilon^{3}$

is also local.

###### Proof.

Since the distribution $p^{(0)}$ is local, it admits a local model of the form

$p^{(0)}(a,b,c)=\int_{[0,1]^{\times
3}}\mathrm{d}\alpha\,\mathrm{d}\beta\,\mathrm{d}\gamma\,p_{A}^{(0)}(a|\beta,\gamma)p_{B}^{(0)}(b|\gamma,\alpha)p_{C}^{(0)}(c|\alpha,\beta).$

We now define a local model for the distribution $p^{(\varepsilon,l)}$. We let
the three sources distribute tuples of the form

$\displaystyle\textup{$\alpha$ source: }(\alpha,x,b_{BC},c_{BC}),$
$\displaystyle\textup{$\beta$ source: }(\beta,y,a_{AC},c_{AC}),$
$\displaystyle\textup{$\gamma$ source: }(\gamma,z,a_{AB},b_{AB}),$

where $\alpha,\beta,\gamma\in[0,1]$, $x,y,z\in\\{0,1\\}$, and
$a_{AB},a_{AC},b_{AB},b_{BC},c_{AC},c_{BC}\in\\{1,2,3,4\\}$. This is
summarized in Fig. 12.

Figure 12: The tuples sent out by the three sources.

The source distributions are defined as follows:

$\displaystyle p(\alpha,x,b_{BC},c_{BC})$
$\displaystyle=r(x)q(b_{BC},c_{BC}),$ $\displaystyle p(\beta,y,a_{AC},c_{AC})$
$\displaystyle=r(y)q(a_{AC},c_{AC}),$ $\displaystyle
p(\gamma,z,a_{AB},b_{AB})$ $\displaystyle=r(z)q(a_{AB},b_{AB}),$

i.e., $\alpha,\beta,\gamma$ are still uniformly distributed in $[0,1]$,
$x,y,z$ are distributed according to $r$, and the pair of outcomes $a,b$ is
distributed as $q(a,b)$. We let $r(0)=1-\varepsilon$ and $r(1)=\varepsilon$.
We define the distribution $q$ as:

$q(a,b)=\left\\{\begin{aligned} \frac{1-l}{4}&\textup{ if }a=b,\\\
\frac{l}{12}&\textup{ if }a\neq b.\end{aligned}\right.$ (29)

We now define the response functions of the parties. The idea is the
following: Alice receives two bits, $y$ and $z$. If both of those bits are
$0$, Alice follows the original strategy looking at $\beta$ and $\gamma$. If
only one of those bits is $1$, Alice outputs the $a$ that is sent along that
bit. If both of those bits are $1$, then Alice chooses uniformly at random
whether to output $a_{AB}$ or $a_{AC}$. Bob and Charlie follow a similar
strategy. This results in the following response functions:

$\displaystyle p_{A}(a|\beta,y,a_{AC},c_{AC},\gamma,z,a_{AB},b_{AB})$
$\displaystyle=\delta_{y,0}\delta_{z,0}p_{A}^{(0)}(a|\beta,\gamma)+\delta_{y,1}\delta_{z,0}\delta_{a,a_{AC}}+\delta_{y,0}\delta_{z,1}\delta_{a,a_{AB}}+\delta_{y,1}\delta_{z,1}\frac{1}{2}(\delta_{a,a_{AB}}+\delta_{a,a_{AC}}),$
$\displaystyle p_{B}(b|\gamma,z,a_{AB},b_{AB},\alpha,x,b_{BC},c_{BC})$
$\displaystyle=\delta_{x,0}\delta_{z,0}p_{B}^{(0)}(b|\gamma,\alpha)+\delta_{x,1}\delta_{z,0}\delta_{b,b_{BC}}+\delta_{x,0}\delta_{z,1}\delta_{b,b_{AB}}+\delta_{x,1}\delta_{z,1}\frac{1}{2}(\delta_{b,b_{AB}}+\delta_{b,b_{BC}}),$
$\displaystyle p_{C}(c|\alpha,x,b_{BC},c_{BC},\beta,y,a_{AC},c_{AC})$
$\displaystyle=\delta_{x,0}\delta_{y,0}p_{C}^{(0)}(c|\alpha,\beta)+\delta_{x,1}\delta_{y,0}\delta_{c,c_{BC}}+\delta_{x,0}\delta_{y,1}\delta_{c,c_{AC}}+\delta_{x,1}\delta_{y,1}\frac{1}{2}(\delta_{c,c_{AC}}+\delta_{c,c_{BC}}).$

We used the Kronecker delta $\delta_{a,b}$ which is $1$ if $a=b$ and 0 else.
We can now compute the output distribution of such a local model, averaging
over the different cases for the bits $x,y,z$. One may of course proceed
analytically, but looking at individual terms makes the calculation easier. We
summarize this information in Table 2.

Case $(x,y,z)$ | Analogous cases | Probability of case | Description | Resulting distribution
---|---|---|---|---
$(0,0,0)$ | $\emptyset$ | $(1-\varepsilon)^{3}$ |  The parties simply follow the original local model. | $p^{(0)}(a,b,c)$
$(1,0,0)$ | $\begin{aligned} \\{&(0,1,0),\\\ &(0,0,1)\\}\end{aligned}$ | $\varepsilon(1-\varepsilon)^{2}$ |  Bob and Charlie output $b_{BC}$ and $c_{BC}$, distributed according to $q$. Alice follows her original strategy but neither Bob nor Charlie are looking at $\beta$ and $\gamma$. Her output is thus distributed according to the marginal $p^{(0)}(a)$. | $q(b,c)p^{(0)}(a)$
$(1,1,0)$ | $\begin{aligned} \\{&(1,0,1),\\\ &(0,1,1)\\}\end{aligned}$ | $\varepsilon^{2}(1-\varepsilon)$ |  Alice and Bob output $a_{AC}$ and $b_{BC}$, respectively. Charlie outputs $c_{AC}$ or $c_{BC}$ with probability $\nicefrac{{1}}{{2}}$. | $\frac{1}{2}\big{(}q(a,c)q(b)+q(b,c)q(a)\big{)}$
$(1,1,1)$ | $\emptyset$ | $\varepsilon^{3}$ |  Alice chooses to output $a_{AB}$ or $a_{AC}$ with probability $\nicefrac{{1}}{{2}}$, and similarly for Bob and Charlie. | $\begin{gathered}\textstyle\frac{1}{4}\big{(}q(a,b)q(c)+q(a,c)q(b)\\\ +\,q(b,c)q(a)+q(a)q(b)q(c)\big{)}\end{gathered}$
Table 2: The different cases for the bits $x,y,z$ and the resulting output
distributions.

Thus, the output distribution is

$\displaystyle p^{(\varepsilon,l)}(a,b,c)=(1-\varepsilon)^{3}p^{(0)}(a,b,c)$
$\displaystyle+\varepsilon(1-\varepsilon)^{2}\left(q(a,b)p^{(0)}(c)+q(a,c)p^{(0)}(b)+q(b,c)p^{(0)}(a)\right)$
$\displaystyle+\varepsilon^{2}(1-\varepsilon)\Big{(}q(a,b)q(c)+q(a,c)q(b)+q(b,c)q(a)\Big{)}$
$\displaystyle+\frac{1}{4}\varepsilon^{3}\Big{(}q(a,b)q(c)+q(a,c)q(b)+q(b,c)q(a)+q(a)q(b)q(c)\Big{)}.$

We now recall the definition of $q$, see Eq. 29. It implies that the marginal
$q(a)$ is maximally mixed, i.e., $q(a)=\nicefrac{{1}}{{4}}$. Furthermore,
since $p^{(0)}$ was assumed to be fully symmetric, its single party marginals
such as $p^{(0)}(a)$ are also maximally mixed. Thus, we obtain

$\displaystyle p^{(\varepsilon,l)}(a,b,c)$
$\displaystyle=(1-\varepsilon)^{3}p^{(0)}(a,b,c)+\Big{(}\varepsilon(1-\varepsilon)^{2}+\varepsilon^{2}(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\Big{)}\frac{q(a,b)+q(a,c)+q(b,c)}{4}+\frac{1}{256}\varepsilon^{3}$
$\displaystyle=(1-\varepsilon)^{3}p^{(0)}(a,b,c)+\Big{(}\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\Big{)}\frac{q(a,b)+q(a,c)+q(b,c)}{4}+\frac{1}{256}\varepsilon^{3}.$

Inserting the definition of $q$, we find that

$\displaystyle p^{(\varepsilon,l)}(1,1,1)$
$\displaystyle=(1-\varepsilon)^{3}p^{(0)}(1,1,1)+3\left(\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\right)\frac{1-l}{16}+\frac{1}{256}\varepsilon^{3},$
$\displaystyle p^{(\varepsilon,l)}(1,1,2)$
$\displaystyle=(1-\varepsilon)^{3}p^{(0)}(1,1,2)+\left(\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\right)\frac{3-l}{48}+\frac{1}{256}\varepsilon^{3},$
$\displaystyle p^{(\varepsilon,l)}(1,2,3)$
$\displaystyle=(1-\varepsilon)^{3}p^{(0)}(1,2,3)+3\left(\varepsilon(1-\varepsilon)+\frac{1}{4}\varepsilon^{3}\right)\frac{l}{48}+\frac{1}{256}\varepsilon^{3}.$

Using Section II.3, we obtain the claim. ∎

Figure 13: The $N$ outcome flags yielding the local distributions of Eq. 31.
The color blue labels the outcome 0, the color red labels the outcome 1, the
color yellow labels the outcome 2, the outcomes 3 to $N-1$ are left implicit,
and the color green labels the outcome $N$.

##  Appendix B $N$ outcomes per party

We now turn to some investigations of fully symmetric distributions with
$N\geq 3$ outcomes. Analogously to the four outcome case, such fully symmetric
distributions can be characterized by the following three numbers:

$\displaystyle s_{111}=Np(1,1,1),\qquad s_{112}=3N(N-1)p(1,1,2),\qquad
s_{123}=N(N-1)(N-2)p(1,2,3),$

such that $s_{111}+s_{112}+s_{123}=1$. The corresponding extremal
distributions are defined as

$\displaystyle p_{111}^{(N)}$
$\displaystyle=\frac{1}{N}\sum_{k=1}^{N}[k,k,k],$ $\displaystyle
p_{112}^{(N)}$
$\displaystyle=\frac{1}{3N(N-1)}\sum_{\begin{subarray}{c}k,l=1\\\ k\neq
l\end{subarray}}^{N}[k,k,l]+[k,l,k]+[l,k,k],$ $\displaystyle p_{123}^{(N)}$
$\displaystyle=\frac{1}{N(N-1)(N-2)}\sum_{\begin{subarray}{c}k,l,m=1\\\ k\neq
l\neq m\neq k\end{subarray}}^{N}[k,l,m].$

Notice that in this case, the Finner inequality of Eq. 4 simplifies to

$\displaystyle p(a,b,c)\leq\frac{1}{N^{3/2}},$

since again the marginals of fully symmetric distributions are uniform. This
implies that for all fully symmetric distributions that have either a local or
quantum model,

$\displaystyle s_{111}\leq\frac{1}{\sqrt{N}}.$ (30)

Analogously to the construction for $N=4$, we can generalize our construction
of strategies that yield strongly correlated probabilities to $N$ outcomes.
Similar to the division illustrated in Fig. 8, each strategy is divided now in
$N$ by $N$ substrategies that are invariant under permutation of the outcomes
as in Eq. 12. The substrategies need to be adapted to more outcomes, yielding
the general construction as depicted in Fig. 13, resulting in the following
distribution: for all $k,l,m\in\\{1,..,N\\}$, $k\neq l\neq m\neq k$,

$\displaystyle p(k,k,k)$ $\displaystyle=\frac{1}{N^{2}},$ $\displaystyle
p(k,k,l)$ $\displaystyle=\frac{\nu}{N^{2}},$ $\displaystyle p(k,l,k)$
$\displaystyle=p(l,k,k)=\frac{q}{N^{2}}(1-\nu),$ $\displaystyle p(k,l,m)$
$\displaystyle=\frac{(1-\nu)(1-2q)}{N^{2}(N-2)}.$

Again, for symmetry under permutation of the parties we require
$p(k,k,l)=p(k,l,k)$, which implies $q=\frac{\nu}{1-\nu}$. Noting that
$q\in[0,\frac{1}{2}]$, we must thus have $\nu\in[0,\frac{1}{3}]$.

Thus, the resulting distributions can be characterized for any $N\geq 3$ by

$\displaystyle\left(s_{111},s_{112},s_{123}\right)$
$\displaystyle=\left(\frac{1}{N},\frac{3\nu(N-1)}{N},\frac{(1-3\nu)(N-1)}{N}\right),\qquad\nu\in\left[0,\frac{1}{3}\right].$
(31)

We have then also applied LHV-net to characterize the set of local
distributions and obtained results shown in Fig. 14. Again, these results
support the conjecture that classically, the highest $s_{111}$ with $N$
outcomes is close to $\nicefrac{{1}}{{N}}$.

(a) $3$ outcomes per party.

(b) $4$ outcomes per party.

(c) $5$ outcomes per party.

(d) $6$ outcomes per party.

$\left\lVert
p_{\text{target}}-p_{\text{NN}}\right\lVert_{2}$$\scriptstyle\Err$

Figure 14:  Distance of symmetric distributions to the local set according to
LHV-Net ($N=3,4,5,6$ for the (a,b,c,d) subplots, respectively, with
$s_{111}<\nicefrac{{1}}{{2}}$ target distributions considered only). For each
point, LHV-Net independently minimized the distance to the local set. Color
represents Euclidean distance of the numerically found closest (not
necessarily symmetric) distribution; values are artificially cut off at $0.1$
in order to provide a collective color scale. Additionally, the Finner
inequality and $\nicefrac{{1}}{{N}}$ line is depicted in each map, with the
latter being conjectured to be close to the true upper bound. Additional
points: $p_{\text{unif}}$ is the uniform distribution, $(\star)$ marks the
counter-example of a classical distribution that has
$s_{111}>\nicefrac{{1}}{{N}}$ for $N=3$ (see Section C.1); $(\dagger)$ marks
the $s_{112}=0$, $s_{111}=\nicefrac{{1}}{{3}}$ distribution, which is proven
to be the only symmetric distribution with $s_{112}=0$ for $N=3$ that is
local, see Section C.2; $(?)$ marks the local distribution with
$s_{111}>\nicefrac{{1}}{{4}}$ for $N=4$, which is not entirely symmetric, but
is close (see Section D.3).

##  Appendix C 3 outcomes per party

###  C.1 An example 3-outcome local distribution with
$s_{111}>\nicefrac{{1}}{{3}}$

While the results mentioned earlier in this work seem to indicate that the
upper bound for $s_{111}$ is close to $\nicefrac{{1}}{{N}}$ given $N$ outcomes
per party, we were able to find a counter example for the case of $N=3$, where
we can create classical correlations that yield $s_{111}>\nicefrac{{1}}{{3}}$.
The corresponding flags are in Fig. 15, which yield the distribution

$\displaystyle\left(s_{111},s_{112},s_{123}\right)$
$\displaystyle=\left(\frac{7}{18},\frac{7}{18},\frac{2}{9}\right).$ (32)

Figure 15: The 3 outcome flags yielding the fully symmetric distribution of
Eq. 32.

For more than three outcomes, the generalization of this strategy is not
straightforward, since the increased number of combinations for 112-type and
123-type outcomes increases the difficulty of satisfying the symmetry
constraints.

###  C.2 (Almost) unique local strategy for 3 outcomes under the constraint
$s_{112}=0$

The Elegant Joint Measurement distribution is particularly fascinating, as it
displays a high probability of having $(1,1,1)$-type events, while maintaining
a very low probability of having $(1,1,2)$ events. This motivates us to study
examples of local distributions where $(1,1,2)$-type events are scarce, or do
not appear at all.

We show that for the triangle network with 3 outcomes for each party and under
the symmetry constraints (described in Appendix B), if one considers
distributions with $s_{112}=0$, then $s_{111}=\nicefrac{{1}}{{3}}$ is not only
an upper bound for $s_{111}$, it is also the only possible value for
$s_{111}$. Let us call this distribution $p_{\dagger}$, which is portrayed in
the symmetric subspace in Fig. 14(a). Moreover, we show that the local
strategy to achieve it is essentially unique, and after reordering the hidden
variable values, one flag can be made to have a 3$\times$3 Latin-square
structure, where each row and column contains only one of each color, while
the other two can be 3$\times$3, 3$\times$2 or 3$\times$1 “Latin squares”.
Technically, a Latin square is an $n\times n$ array, colored with $n$ colors
such that each row and column contains exactly one of each color. Hence, when
having $n\times m$ grids ($m<n$) we work with generalizations (or cropped
versions) of Latin squares.

We will work in the discrete local hidden variable picture with deterministic
outcomes labeled by colors (red, green, blue, or
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$).
We can do this without loss of generality, by using maximally 24 symbols for
each local hidden variable [23]. With a slight abuse of notation, we will
label the set of all symbols of source $\alpha$ as $\alpha$, and similarly for
$\beta,\gamma$.

####  C.2.1 Qualitative response functions

We will first show that the response functions must take the Latin square
structure. The steps are as follow.

Step 1. Choosing an RRR. First, start with a triple of symbols
$(\alpha_{R0},\beta_{R0},\gamma_{R0})$, where
$\alpha_{R0}\in\alpha,\beta_{R0}\in\beta$, and $\gamma_{R0}\in\gamma$, such
that the outcome is
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
(formally,
$a(\beta_{R0},\gamma_{R0})={\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
$b(\gamma_{R0},\alpha_{R0})={\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
and
$c(\alpha_{R0},\beta_{R0})={\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$)
. These
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
responses are portrayed in Fig. 16(a).

Step 2. Rearranging $\gamma$ symbols. Next, notice that in Alice’s flag, for
any $\gamma^{\prime}\neq\gamma_{R0}$, the color of Alice’s response
$a(\beta_{R0},\gamma^{\prime})$ uniquely determines Bob’s response
$b(\gamma^{\prime},\alpha_{R0})$, according to the following table.

$a(\beta_{R0},\gamma^{\prime})$ |  | $b(\gamma^{\prime},\alpha_{R0})$
---|---|---
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$ | $\implies$ | ${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$ | $\implies$ | ${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$ | $\implies$ | ${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$

This is true due to the strict $s_{112}=0$ condition, and since for all these
cases Charlie’s response is
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
i.e.
$c(\alpha_{R0},\beta_{R0})={\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$.
In some sense, this can be seen as a “color inversion around
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$”,
induced by Charlie’s
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
which leaves
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
invariant, but flips
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
and
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$.

So let us now rearrange the symbols of $\gamma$, such that on Alice’s flag, in
the column defined by $\beta_{R0}$, we first have
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
responses below Alice’s first
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
rectangle, and then
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
responses, and finally all the
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
responses, as shown in Fig. 16(b). Due to the table above, this immediately
implies that Bob’s first row will be ordered as
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
as shown in Fig. 16(b).

Step 3. Rearranging $\beta$ symbols. We can apply the same procedure to the
$\beta$ axis, grouping Alice’s responses in her first $\gamma_{R0}$ row in the
order
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$.
This implies, due to Bob’s
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
in his $\alpha_{R0}$ row, that Charlie’s first $\alpha_{R0}$ row must be in
the order
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$.
This is illustrated in Fig. 16(c).

Step 4. Filling the rest of Alice’s flag. The rest of Alice’s flag is uniquely
determined by what we have already established on Bob and Charlie’s flags, and
the $s_{112}=0$ condition, i.e. by the table used in Step 2, and similar
versions for
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
and
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$.
The filling is shown in Fig. 16(d).

Step 5. Bob and Charlie’s rows (rearranging $\alpha$ symbols). Before moving
on, note that even with Bob and Charlie having only a single row, we have all
(1,1,1)-type and (1,2,3)-type outcome events appearing, namely
(${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$).
In fact, as we later show, if one stretches the first row of Bob and Charlie
to cover their whole flags ($\alpha_{R0}=1)$, one can already obtain
$p_{\dagger}$.

Again, we may rearrange $\alpha$ such that Bob has the order of
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
in his first column (as shown in Fig. 16(e) and Fig. 16(f)). Any
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
below Bob’s first
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
will generate the same row as before on both Bob and Charlie’s flag, whereas
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
below this first
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
will force Charlie to have a
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
row, and Bob to have a
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
row (Fig. 16(e).

In a similar manner, if there is
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
anywhere in Bob’s first column, then Charlie’s row becomes
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
and Bob’s is forced to be
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
as shown in Fig. 16(f).

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (a)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (b)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (c)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (d)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (e)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (f)

Figure 16: (a) Step 1: Move an
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
event into one corner. (b) Step 2: Rearrange $\gamma$, such that Alice has the
order of
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
in her first column. Due to the $s_{112}=0$ condition, this implies what Bob’s
first row must look like. (c) Step 3: Rearrange $\beta$, such that Alice has
the order of
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
in her first row. This implies what Charlie’s first row must look like. (d)
Step 4: The rest of Alice’s flag is uniquely determined by the $s_{112}=0$
condition. (e) Step 5: Rearrange $\alpha$ s.t. the
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
events come below the
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
events in Bob’s first column. A single
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
event uniquely determines both Bob and Charlie’s second row. (Note that in the
figure, we have not drawn any
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
event below the original one, however, this would just imply a wider first row
for Bob and Charlie, with the same coloring.) (f) In a similar manner, a
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
event in Bob’s flag below Bob’s first (top left)
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
rectangle determines both Bob and Charlie’s final row. At this point we have
exhausted all possibilities for coloring the flags.

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (a)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (b)

CharlieBobAlice$\alpha$${\scriptstyle 0}$${\scriptstyle
1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$$\beta$${\scriptstyle
0}$${\scriptstyle 1}$$\gamma$${\scriptstyle 0}$${\scriptstyle 1}$ (c)

Figure 17: All possible flags that reproduce $p_{\dagger}$ for 3 outcomes per
party, (up to permutations of colors, parties and local hidden variable
values). Note that any $\\{p(\alpha_{i})\\}_{i=1}^{3}$ values are valid as
long as they sum to one, as as illustrated in the examples (a-c), where (a)
$p(\alpha_{2})=p(\alpha_{3})=0$, (b) $p(\alpha_{3}=0)$, and (c)
$p(\alpha_{i})>0,\forall i$. Recall that $p(\alpha_{1})$ is the height of the
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
row of Charlie, $p(\alpha_{2})$ is the height of the
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}},{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
row of Charlie, and $p(\alpha_{3})$ is the height of the
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}},{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}},{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$
row of Charlie.

####  C.2.2 Cardinalities

Before moving on to fixing the length of the rectangle sides within the flags,
let us stop for a note on the cardinalities of the local hidden variables. For
3 outcomes per party, there are three 111-type events (e.g.,
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$),
and six 123-type events (e.g.,
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$).
That means that in the whole cube (as parametrized by $\alpha,\beta,\gamma$),
there must be at least 9 distinct types of sub-cuboids
(${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$,
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}{\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}{\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$).
In order to generate these cuboids, the product of the three local hidden
variable cardinalities must be at least 9, e.g.,
$|\alpha|=1,|\beta|=3,|\gamma|=3$ is already sufficient (if it can satisfy all
the desired constraints, which it can as we have seen in Fig. 17(a)). Smaller
cardinalities, e.g., $(1,1,9)$ or $(1,2,5)$ are insufficient because each
party (Alice, Bob and Charlie) must output at least 3 distinct colors. The
above reasoning implies that for any distribution where $s_{111}>0$ and
$s_{123}>0$, at least one party’s flag is at least a 3x3 grid (with no two
columns or two rows being the same in the flag). Hence the construction of
Alice’s flag in the previous sections was not merely a demonstration, but it
was necessary that she had these many different colors in her first row and
column.

Also note that, as we have shown, due to the limited possibilities when
enforcing $s_{112}=0$, (3,3,3) is effectively the largest LHV cardinality
triple. Anything larger can be rearranged to take this 3 by 3 Latin square
form.

####  C.2.3 Lengths of the rectangle sides

We start from an effectively $3\times 3\times 3$ grid, with the elements of
$\\{\alpha_{i}\\}_{i=1}^{3},\\{\beta_{i}\\}_{i=1}^{3}$ and
$\\{\gamma_{i}\\}_{i=1}^{3}$ denoting the symbols of the respective
$\alpha,\beta$ or $\gamma$ random variables.

Let us now consider the lengths of the rectangle sides, i.e. the probabilities
of $\alpha_{i},\beta_{j},\gamma_{k}$ appearing. First, let Bob and Charlie
only have stripes, i.e., $p(\alpha_{2})=p(\alpha_{3})=0$, as in Fig. 17(a).
Then from the fact that the marginals must all be $\nicefrac{{1}}{{3}}$ on Bob
and Charlie’s flags, we immediately get that $\forall
i:\,p(\beta_{i})=p(\gamma_{i})=\nicefrac{{1}}{{3}}$. Extending this strategy
to the case that $p(\alpha_{2})>0$ and $p(\alpha_{3})\geq 0$, we see that next
to these $p(\beta_{i}),p(\gamma_{i})$ values, one can have any
$p(\alpha_{1}),p(\alpha_{2}),p(\alpha_{3})\geq 0$ such that
$\sum_{i}p(\alpha_{i})=1$. However, can $p(\beta_{i})$ and $p(\gamma_{i})$
take on different values than $\nicefrac{{1}}{{3}}$? No, as this is already
guaranteed by the symmetry constraints for the marginals on Alice’s flag. If
any of the rectangles is larger, then the other rectangles will be smaller and
will not be able to satisfy the constraint that
$p(A={\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}})=p(A={\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}})=p(A={\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}})=\nicefrac{{1}}{{3}}$.
An optimization in the Mathematica software confirms this intuition. More
precisely, setting the constraints that the marginals must be $1/3$ for a
given flag and positivity and normalization of the hidden variables’
probabilities, we get that at least one of its sources (local hidden
variables) must have equal values (e.g. $\beta_{1}=\beta_{2}=\beta_{3})$. This
is true for each of the three flags, and since there are three of them, at
least two hidden variables must have this property. As a consequence, at least
one of the flags must be composed of a 3$\times$3 grid of equal sided squares.

In summary, any local distribution with 3 outcomes, for which $s_{112}=0$ and
for which the symmetry constraints are satisfied (as described in Appendix B),
must have $s_{111}=1/3$. The only type of strategy is the Latin square
strategy, with $p(\beta_{i})=p(\gamma_{i})=1/3$ for $i=1,2,3$, and for any
$p(\alpha_{1}),p(\alpha_{2}),p(\alpha_{3})\geq 0$ s.t.
$\sum_{i}p(\alpha_{i})=1$. These strategies are depicted in Fig. 17.

Hopefully such an explicit strategy can serve as a basis point for
understanding the nonlocality of distributions such as the Elegant Joint
Measurement distribution, which has a large $s_{111}$ value and a small
(though non-zero) $s_{112}$ value. In this aspect, note that the
generalization of these arguments to, e.g., 4 outputs per party is not so
straightforward, as having a
${\color[rgb]{0.65234375,0.00390625,0.40234375}\definecolor[named]{pgfstrokecolor}{rgb}{0.65234375,0.00390625,0.40234375}\mathrm{R}}$
and
${\color[rgb]{0.00390625,0.453125,0.33203125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.453125,0.33203125}\mathrm{G}}$
on two flags implies that the third can be
${\color[rgb]{0.9609375,0.60546875,0.14453125}\definecolor[named]{pgfstrokecolor}{rgb}{0.9609375,0.60546875,0.14453125}\mathrm{Y}}$
or
${\color[rgb]{0.00390625,0.015625,0.45703125}\definecolor[named]{pgfstrokecolor}{rgb}{0.00390625,0.015625,0.45703125}\mathrm{B}}$,
i.e., it is not fixed uniquely. Similar strategies as these, however, can be
constructed, though they are not as unique, as can be seen in the main text in
Fig. 3.

##  Appendix D Finding inequality parameters with LHV-Net, and perhaps a
$s_{111}>\nicefrac{{1}}{{4}}$ local distribution?

LHV-Net, the neural network which parametrizes local models obeying the
triangle structure, can be used to maximize the left hand side of the
inequality of Eq. 8, namely of

$\displaystyle f_{w}(p)\leq f_{w}(p_{\textup{EJM}})-\delta_{w},$ (33)

where $f_{w}(p)=w\cdot s_{111}(p)-(1-w)\Delta_{l}(p)$ is a function that is
large for large $s_{111}$ and for symmetric distributions ($\Delta_{l}$
quantifies asymmetry). A strictly positive $\delta_{w}$ indicates that the EJM
distribution outperforms local models for this inequality. For several
different values of $w$ we see whether LHV-Net can outperform the value of the
EJM distribution. We plot $\max_{\\{p\text{ by LHV-
Net}\\}}f_{w}(p)-f_{w}(p_{\textup{EJM}})$ in Fig. 18. We can see that for a
range of $w$ values LHV-Net can not go over the value 0, hence for these $w$
values we can conjecture inequalities. The corresponding gaps, $\delta_{w}$,
can be read off from the vertical distance between the blue line and LHV-Net’s
result for these $w$’s. There is no unique best inequality in general,
however, in the maintext we chose those $w$ values where $\delta_{w}$ was
largest, namely $w^{*}=0.678$, $\delta_{w^{*}}=0.069$ for $l=1$, and
$w^{*}=0.16$, $\delta_{w^{*}}=0.012$ for $l=2$,

###  D.1 Absolute value penalty inequalities ($l=1$)

When examining the plot for $l=1$, we find that there seem to be two natural
regimes: 1) where the asymmetry is strongly punished (small $w$ values): in
this regime fully symmetric local models with high $s_{111}$ seem to perform
best; and 2) where asymmetry is not punished strongly (large $w$ values), for
which the all-111 strategy performs best, where all parties always output 1
(or blue). The performance of these two extremal strategies are also depicted
in Fig. 18. Notice that LHV-Net seems to find a distribution with
$s_{111}>\nicefrac{{1}}{{4}}$. This can be seen from the slope of the LHV-
Net’s results as a function of $w$. From the distributions found by LHV-Net we
extract that $s_{111}\approx 0.289$ could maybe be the maximal $s_{111}$
within the symmetric subspace.

A strategy found by the neural network for $s_{111}\approx 0.294$
($\Delta_{l=1}\approx 0.0136,\Delta_{l=2}\approx 4.716\cdot 10^{-6}$) can be
found in Fig. 19. It is peculiar that there are some approximate symmetries
also in this response function (Alice and Bob’s symmetry under exchange of
blue and red and of green and yellow, if the $\gamma$ axis is swapped, as well
as the same color change when Charlie flips her strategy about the diagonal
axis). Though not exact, such symmetries hint at some deeper structure that is
not yet well understood.

Finally, note that any known local distribution can be used to obtain an upper
bound on the $\delta_{w}$ value, by rearranging Eq. 8 as

$\delta_{w}\leq\Delta_{l}+w\left(\frac{100}{256}-s_{111}-\Delta_{l}\right)$
(34)

For example for $l=1$, if we evaluate this for the deterministically all-1
distribution ($p(abc)=[1,1,1]$), we find that

$\delta_{w}\leq\frac{3}{2}-\frac{135}{64}w\qquad(l=1),$ (35)

which immediately implies that only $w<0.711$ are candidate $w$ values, as
$\delta_{w}>0$ is required. Moreover, for $w=0.678$ this expression gives us
$\delta_{w}\leq 0.069$, in correspondence with the results in the maintext.
Based on the numerics (Fig. 18) we believe that for the $l=1$ case this bound
(and the corresponding bound for the largest $s_{111}$ symmetric distribution)
will ultimately give the appropriate value of $\delta_{w}$. Writing this bound
for the $s_{111}=0.25$ distribution, we get

$\delta_{w}\leq\frac{9}{64}w\qquad(l=1),$ (36)

which is perhaps not tight, as there might be a symmetric distribution with
$s_{111}\approx 0.289$ which would give an even tighter bound. In fact,
evaluating the (not precisely symmetric) model found by LHV-Net gives

$\delta_{w}\leq 0.013627+0.082912\,w\qquad(l=1),$ (37)

giving $\delta_{w}\leq 0.0698$ for $w=0.678$.

###  D.2 Squared penalty inequalities ($l=2$)

For the $l=2$ penalty function, there seems to be an additional intermediate
regime, where neither the all-111 nor the fully symmetric $s_{111}\approx
0.25$ distribution are optimal. We find that here LHV-Net actually recovers
the distribution of Section II.1.2, which we name “squares” distribution due
to the four large squares appearing in the response functions.

We evaluate several local distributions to get the following bounds on
$\delta_{w}$:

$\displaystyle\delta_{w}$ $\displaystyle\leq\frac{9}{64}w$
$\displaystyle\qquad(l=2;s_{111}=0.25),$ (38) $\displaystyle\delta_{w}$
$\displaystyle\leq 4.7163\cdot 10^{-6}+0.0965\,w$
$\displaystyle\qquad(l=2;\text{LHV-Net, }s_{111}=0.294,\Delta_{2}=4.7163\cdot
10^{-6}),$ (39) $\displaystyle\delta_{w}$
$\displaystyle\leq\frac{5}{96}-\frac{124}{768}w$
$\displaystyle\qquad(l=2;\text{``squares'' strategy,
\lx@cref{creftype~refnum}{eq:high111local}})$ (40)

In particular these bounds imply that only $w<\frac{10}{31}\approx 0.3226$ are
viable $w$ values, and that $\delta_{w}<0.015$ for $w^{*}\approx 0.16$, which
is of course a bit farther from the recovered $\delta_{w}^{*}\approx 0.012$,
as can be seen from the inset in Fig. 18 (right). Moreover, notice that the
$\delta_{w}$ bound of the fully symmetric $s_{111}=0.25$ distribution and the
intermediate “squares” distribution Section II.1.2 meet at $w\approx 0.1724$,
close to the numerically extracted optimal value of $w^{*}\approx 0.16$,
hinting that the transition region between these two distributions is highly
relevant for finding an inequality for $l=2$.

Figure 18: LHV-Net trying to maximize $f_{w}(p)-f_{w}(p_{\textup{EJM}})$ (see
Eq. 8) as a function of $w$, for $l=1$ (left) and $l=2$ (right). Additionally
we plot the EJM distribution (grey, at $y=0$), deterministically outputting-1
distribution (i.e. $p(abc)=[1,1,1]$, sparsely dashed violet, highly
asymmetric, but high $s_{111}$), a symmetric distribution with
$s_{111}=\frac{1}{4}$ (dashed violet, current best analytic strategy that is
fully symmetric), and the distribution found by LHV-Net that is almost
symmetric and has $s_{111}\approx 0.29$ (green dot-dashed). Moreover, for
$l=2$ we also depict the values of the distribution in Section II.1.2 (called
“Squares”, dotted violet). Candidate $w$ parameters are those where the LHV-
Net’s results are below the EJM’s value, i.e. below zero. These correspond to
a positive $\delta_{w}$. The best $\delta_{w}$ estimates are signaled by
vertical lines and are for $l=1$ ($l=2$) at $w^{*}\approx 0.678$
($w^{*}\approx 0.16$) with $\delta_{w^{*}}\approx 0.069$
($\delta_{w^{*}}\approx 0.012$).

###  D.3 $s_{111}>\nicefrac{{1}}{{4}}$ local distribution?

When maximizing the inequalities for $l=1$ for small $w$ values (i.e. in the
regime symmetry is crucial), LHV-Net finds a distribution which is almost
symmetric, but has $s_{111}>\nicefrac{{1}}{{4}}$. Below are some
characteristics of the distributions, as well as for a deterministic
approximation of the LHV-Net’s distribution (whose flags can be seen in Fig.
19(left)), and a reference distribution which is analytically symmetric and
has $s_{111}=\nicefrac{{1}}{{4}}$. The full LHV-Net and its deterministic
approximation distributions can be found in the data appendix [22]. Moreover,
we plot the distributions in Fig. 19.

|  $s_{111}=\nicefrac{{1}}{{4}}$ distr. | LHV-Net mean ($\pm$ std. dev.) | LHV-Net range | det. approx. mean ($\pm$ std. dev.) | det. approx. range
---|---|---|---|---|---
$p(111)$ | 0.0625 | $0.0723\pm 0.5\%$ | $[0.0719,0.0727]$ | $0.0735\pm 0.2\%$ | $[0.0733,0.0738]$
$p(112)$ | 0.0125 | $0.0118\pm 1.5\%$ | $[0.0114,0.0123]$ | $0.0116\pm 2.4\%$ | $[0.0112,0.0124]$
$p(123)$ | 0.0125 | $0.0119\pm 1.2\%$ | $[0.0115,0.0122]$ | $0.0120\pm 2.3\%$ | $[0.0116,0.0128]$

Though the distribution is not exactly symmetric, several pieces of numeric
evidence point towards the existence of a distribution which has
$s_{111}>\nicefrac{{1}}{{4}}$ and is fully symmetric, summarized in the
following list.

  * •

The nonlocality of the Elegant distribution has previously been conjectured be
robustness to noise at the source (up to $20\%$ noise) and at the detectors
(up to $14\%$ noise) [13]. The extracted noise robustness values indicate that
the noisy elegant distribution is local for $s_{111}=0.289$ and
$s_{111}=0.286$ for the two noise models, respectively.

  * •

For the 3 outcome-per-party scenario ($N=3$), there is an analytic example of
a distribution for which $s_{111}=7/18>1/3$ (see Section C.1).

  * •

The maps of the symmetric subspace as found by LHV-Net indicate a bump in the
local set around the area of $p(112)\approx p(123)$, where $s_{111}>1/N$ is
possible. The bump is visible for the examined cases of $N=3,4,5,6$ (see Fig.
14).

  * •

When maximizing the inequality for $l=1$ in Fig. 18 (left), the slope of the
maximum values are above the line that corresponds to a $s_{111}=0.25$ maximum
value (and instead line up with a hypothesis of $s_{111}\approx 0.29$ as a
maximum value).

Though many of these evidences rely on the numerics of LHV-Net, they have been
obtained in different manners. Moreover, the explicit construction of a
distribution with $s_{111}>1/3$ for $N=3$ is an analytic result obtained
independently from numerics. Together, these lead us to believe that the
structure of the local set in the symmetric subspace is more surprising than
it seems at first sight, and that it can not be perfectly characterized by a
simple bound on $s_{111}$.

Figure 19: (Left) Flags for the deterministic approximation of the strategy
found by LHV-Net, which has $s_{111}\approx 0.29$. (Right) Distribution found
by LHV-Net and its deterministic approximation, compared to the Elegant
distribution and an analytically symmetric $s_{111}=0.25$ distribution with
$p(112)=p(123)$. Exact data for these flags and distributions can be found in
the data appendix [22].