# Fully Device-Independent Model on Quantum Networks

Ming-Xing Luo
Information Security and National Computing Grid Laboratory, Southwest
Jiaotong University, Chengdu 610031, China

###### Abstract

Bell inequality can provide a useful witness for device-independent
applications with quantum (or post-quantum) eavesdroppers. This feature holds
only for single entangled systems. Our goal is to explore device-independent
model for quantum networks. We firstly propose a Bell inequality to verify the
genuinely multipartite nonlocality of connected quantum networks including
cyclic networks and universal quantum computational resources for measurement-
based computation model. This is further used to construct new monogamy
relation in a fully device-independent model with multisource quantum
resources. It is finally applied for multiparty quantum key distribution,
blind quantum computation, and quantum secret sharing. The present model can
inspire various large-scale applications on quantum networks in a device-
independent manner.

The Bell theorem states that the statistics generated by local measurements on
a two-spin entanglement cannot be generated by any classical local model under
the locality and casualty assumptions. This provides an experimental method
for verifying the so-called nonlocality of entanglement 1 ; 2 . To date, Bell
theory has inspired many interesting applications in various areas, such as
quantum information processing 3 , quantum key distribution 4 ; 5 ; 6 ; 7 ; 8
and randomness amplification 9 ; 10 ; 11 .

Single entangled systems have experimental constraints in large-scale
applications because of the limited coherence time and transmission distance.
This inspires distributed settings in terms of quantum networks using various
independent entangled systems 12 ; 13 . Compared with single entangled
systems, it is great difficulty in characterizing multipartite correlations of
quantum networks because of the independent assumption of sources and local
joint measurement allowed for each party. Moreover, the participation of
multiple parties in a general network may provide new opportunities for
attackers. Thus, a natural problem is how to ensure secure information
processing on general quantum networks? The main difficulty is that these
multipartite quantum correlations can form non-convex semialgebraic sets 14 .
Several nonlinear Bell inequalities have recently been proposed for verifying
the non-multilocality of special quantum networks with the assumption of
source independence 15 ; 16 ; 16a ; 16b ; 17 ; luo ; 18 ; 18b ; 18c ; 18d ;
18e ; 18f . However, so far there is no way to verify the multipartite
nonlocality of general quantum networks, or no standard definition of
genuinely multipartite nonlocality for quantum networks.

In most of secure tasks, such as quantum key distribution (QKD) 5 , the
trustworthiness of quantum devices according to certain specifications should
be avoided in order to enable adversary (noise)-tolerant realizations 6 .
These so-called device-independent scenarios make only use of the statistics
of measurement outcomes 19 ; 20 ; 21 . Interestingly, the leaked information
in the case of quantum eavesdroppers (or the key rate of QKD) may be
characterized by the violation of specific inequality 6 ; 7 ; 8 , that is, the
higher the violation is, the lower the information leakage for the outcomes of
legal parties. Note that quantum devices may be correlated by an untrusted
producer (adversary or eavesdropper) in a device-independent model 6 . This
implies that the recent models 22 ; 23  are not device independent because of
the independent assumption of sources. The independence of sources in these
secure scenarios cannot be guaranteed by experiment improvements. Another
reason is that an untrusted party in network scenarios may correlate the
shared sources locally in secure applications. Hence, the question of how to
construct a device-independent model remains an open problem for general
quantum networks.

Our goal in this work is to propose a device-independent model for secure
information processing in general quantum networks against quantum (or post-
quantum) eavesdroppers. We first propose a new Bell inequality for verifying
the genuinely multipartite nonlocality of quantum networks in the biseparable
model 26 . Compared with a recent result for ring-shaped networks 24 , the
proposed inequality provides the first Bell test for verifying general cyclic
networks. This further implies a new feature for characterizing the leaked
information in device-independent tasks on quantum networks going beyond
single entangled systems 6 ; 7 ; 8 ; 19 ; 20 ; 21  or device-independent
models 22 ; 23 . It is then used to guarantee the security of various tasks,
such as multipartite quantum key distribution, blind quantum computation with
multiple servers, and quantum secret sharing using quantum networks. These
results can inspire interesting applications on large-scale quantum networks
in a device-independent manner.

Figure 1: Quantum networks.(a) An $n$-partite quantum network ${\cal N}_{q}$
consisting of $m$ entangled states $\otimes_{i=1}^{m}\rho_{i}$. (b) A directed
acyclic graph (DAG) 25  of ${\cal N}_{q}$ consisting of $m$ independent random
variables $\lambda_{1},\cdots,\lambda_{m}$. In device-independent
applications, a quantum (or post-quantum) eavesdropper can correlate shared
variables $\lambda_{i}$s into new variable $\lambda^{*}$.

Genuinely multipartite nonlocality of quantum networks. Consider an
$n$-partite quantum network ${\cal N}_{q}$, as shown in Fig.1(a), consisting
of $m$ entangled states $\otimes_{i=1}^{m}\rho_{i}$. ${\cal N}_{q}$ is
schematically represented by a directed acyclic graph (DAG) (Fig.1(b)) 25
with the same configuration as ${\cal N}_{q}$ using random variables
$\lambda_{1},\cdots,\lambda_{m}$. Different from the standard hidden variable
model 1 ; 2 , a quantum (or post-quantum) eavesdropper is able to correlate
all distributed variables $\lambda_{i}$s into a new variable $\lambda^{*}$
(Fig.1(b)). Unfortunately, this local model is insufficient for characterizing
device-independent applications including quantum secret sharing, where some
sharers may be eavesdropper. In this case, we consider the genuinely
multipartite nonlocality of quantum networks in the biseparable model 26 . Let
$I=\\{s_{1},\cdots,s_{i}\\}\subset\\{1,\cdots,n\\}$ and
$\overline{I}=\\{j|j\not\in I,1\leq j\leq n\\}$ be a bipartition of
$\\{1,\cdots,n\\}$. An $n$-partite state $\rho$ on Hilbert space
$\otimes_{i=1}^{n}{\cal H}_{i}$ is biseparable 26  if it has the following
decomposition

$\displaystyle\rho=\sum_{I}p_{I}\varrho_{I,\overline{I}}$ (1)

where $\\{p_{I}\\}$ is a probability distribution, $\varrho_{I,\overline{I}}$
denotes separable state on Hilbert space ${\cal H}_{I}\otimes{\cal
H}_{\overline{I}}$, ${\cal H}_{I}=\otimes_{j\in I}{\cal H}_{j}$ and ${\cal
H}_{\overline{I}}=\otimes_{j\in\overline{I}}{\cal H}_{j}$.

Let $M_{x_{k}(I)}$ denote dichotomic observable of the observer
$\textsf{A}_{k}$ for input index $x_{k}(I)$, where $x_{k}$ depends on the set
$I$. Denote $\textbf{M}_{I}=\prod_{k\in I}M_{x_{k}(I)}$ and
$\textbf{M}_{\overline{I}}=\prod_{k\in{}\overline{I}}M_{x_{k}(\overline{I})}$.
Our first result is to prove that any biseparable state satisfies the
following Bell inequality SI :

$\displaystyle\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}{\cal L}_{i}\leq
2^{n}\sqrt{2}-4\sqrt{2}+2$ (2)

where ${\cal L}_{i}$ denotes the summation of multipartite CHSH-type
quantities, that is, ${\cal L}_{i}=\sum_{|I|=i}CH_{I}$ with
$CH_{I}=(\textbf{M}_{I}+\hat{\textbf{M}}_{I})\textbf{M}_{\overline{I}}+(\textbf{M}_{I}-\hat{\textbf{M}}_{I})\hat{\textbf{M}}_{\overline{I}}$,
$\textbf{M}_{S}$ and $\hat{\textbf{M}}_{S}$ are observables depending on the
set $S=I$ or $\overline{I}$, and $\lfloor{}x\rfloor$ denotes the maximum
integer no more than $x$. The largest bound for quantum networks is
$2^{n}\sqrt{2}-2\sqrt{2}$ SI .

The inequality (2) can be regarded as Svetlichny-type inequality 26  with at
most eight inputs SI . The present inequality (2) may reduce to Svetlichny
inequality with specific settings 26 . It is useful for verifying the
genuinely multipartite nonlocality of connected quantum networks consisting of
EPR states 2  and GHZ states 30  or noisy quantum networks (including two
examples of triangle networks) Wrner ; SI . The present nonlocality is
stronger than the non-multilocality 15 ; 16 ; 16a ; 16b ; 17 ; luo ; 18 ; 18b
; 18c ; 18d ; 18e ; 18f  or the nonlocality 24 ; 30 ; 31 ; 32 .

Figure 2: (a) A device-independent model on an $n$-partite network ${\cal N}$
consisting of $m$ sources $\lambda_{1},\cdots,\lambda_{m}$. A quantum (post-
quantum) eavesdropper E holds some systems correlated with the shared sources.
$z_{i}$ and $e_{i}$ denote the respective input and outcome of eavesdropper
for recovering the output $a_{i}$. For a device-independent model, E may
correlate $\lambda_{i}$s into new source $\lambda^{*}$. (b) Consider a quantum
network ${\cal N}_{q}$, where $\textsf{A}_{i}$ shares a system $\rho_{i}$
consisting of some EPR states and GHZ states with $\textsf{A}_{j}$,
$i=1,\cdots,k$, and $j=k+1,\cdots,n$. $\textsf{E}_{i}$s are eavesdroppers
sharing random variables $\Lambda_{i}$ consisting of $\lambda_{j}$ with the
same network configuration as ${\cal N}_{q}$. $e_{i}$ is the output of
$\textsf{E}_{i}$ conditional on the inputs $z_{i}$ and $x_{i}$ and the
variables $\Lambda_{i}$. $\textsf{E}_{i}$s want to classically simulate
quantum correlations derived from local measurements on ${\cal N}_{q}$.

A device-independent model on quantum networks. For single entangled states,
the violation of a certain Bell inequality allows for device-independent
information processing against quantum (post-quantum) eavesdroppers 5 ; 6 ; 7
; 8 ; 9 ; 10 ; 11 ; 19 ; 20 ; 21 ; 26 . Our goal here is to propose a device-
independent model using quantum networks based on the inequality (2). Consider
a connected network ${\cal N}$ (Fig.2(a)). For a device-independent model,
eavesdropper may hold local systems correlated with some sources in order to
recover private information such as outcomes.

Denote the variation distance of two probability distributions $\\{p(x)\\}$
and $\\{q(x)\\}$ as: $D(p,q)=\frac{1}{2}\sum_{x}|p(x)-q(x)|$. The predictive
power of an eavesdropper to learn the outcomes of legitimate parties satisfies
SI :

$\displaystyle D_{e}\leq
2n-\frac{\varpi_{q}-2\sqrt{2}(\alpha-\lfloor{}\alpha\rfloor)}{2\lfloor{}\alpha\rfloor}$
(3)

where
$D_{e}=D(\prod_{i=1}^{n}P(e_{i}|a_{i};\textbf{x},z_{i}),\prod_{i=1}^{n}P(e_{i}|z_{i}))$,
$\alpha=\frac{1}{n}(2^{n}-1)$, $\textbf{x}=x_{1}\cdots{}x_{n}$,
$\varpi_{q}=\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}{\cal L}_{i}$, and ${\cal
L}_{i}$ is defined in the inequality (2) with the estimated quantum
correlations. The inequality (3) provides a monogamy relation for the
information leakage in a device-independent model on quantum networks.

If all inputs are achievable for eavesdroppers as shown in Fig.2(b), the
quantum correlations derived from $\mathcal{N}_{q}$ may be classically
simulated with shared randomness and finite classical communication SI . This
partially answers a recent conjecture 36 . It inspires special restraints on
the measurements to enhance the security for cryptographic applications.

Figure 3: Device-independent tripartite QKD. (a) Tripartite QKD based on a
triangle network. $a$, $b$, and $c$ ($x,y$, and $z$) denote the output (input)
of $\textsf{A}_{1}$, $\textsf{A}_{2}$ and $\textsf{A}_{3}$, respectively. (b)
Device-independent tripartite QKD. The output $e$ of eavesdropper E depends on
its input $E$ and these shared systems.

Device-independent multipartite quantum key distribution. Quantum entanglement
is useful for distributing random key for secure communication 5 ; 6 ; 7 ; 8 .
Our goal is to present a device-independent QKD with multiple parties using
quantum networks. For simplicity, consider a tripartite network consisting of
$\textsf{A}_{1}$, $\textsf{A}_{2}$ and $\textsf{A}_{3}$ (Fig.3(a)). Let $x,y$,
and $z$, respectively, denote the inputs of the three parties, and $a,b$, and
$c$ denote the corresponding outcomes. $\rho_{ABC}$ denotes the total system.
In a device-independent multipartite QKD (DIMQKD), the information available
to an eavesdropper Eve is represented by $E$ correlated with $\rho_{ABC}$
(Fig.3(b)). The total state is then denoted by $\rho_{ABCE}$ which should
satisfy $\textrm{tr}_{E}\rho_{ABCE}=\rho_{ABC}$. Here, suppose that Eve wants
to learn the output $a$ of $\textsf{A}_{1}$. For one copy resource, the
predictability for Eve is quantified by the guessing probability of $a$ as
$P_{gs}(a)=\max_{a}P(a|x)$, which is bounded by SI :

$\displaystyle P_{gs}(a)\leq\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi^{2}_{es}}$
(4)

where $\varpi_{es}$ denotes the estimate of the inequality (2) using the
practical quantum correlations in accordance with the practical quantum
network (Fig.3(a)). For $s$ copies of resources,the asymptotic secret-key rate
$R:=-\frac{\log_{2}P_{gs}({\bf a}|E)}{s}$ 20 ; 38 ; 52  is given by SI :

$\displaystyle
R\geq-\log_{2}(\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi_{es}^{2}})-H(a|b,c)$
(5)

where $P_{gs}({\bf a}|E)$ denotes the expected optimal guessing probability
conditional on the eavesdropper’s input $E$. The first term of Eq.(5)
represents the knowledge of Eve for the privacy amplification, and $H(a|b,c)$
quantifies the information needed for the error correction by
$\textsf{A}_{1}$. Similar evaluations may be conducted for other networks.

Figure 4: Quantum networks for measurement-based quantum computation.(a)
Honeycomb network consisting of singlets $e_{ij}$’s 40 . (b) Triangular
network consisting of GHZ states. (c) Square network consisting of four-
particle GHZ states. (d) Cyclic network consisting of 2-dimensional graph
states $|G_{i}\rangle$’s and 1-dimensional graph states $|L_{j}\rangle$’s.

Device-independent blind quantum computation. Measurement-based quantum
computation (MQC) provides a new computation model that makes use of simple
measurement on qubits prepared in a highly entangled state 40 . Some examples
are shown in Fig.4 by using specific systems 41 ; 42 . The inequality (2)
provides a useful way to verify these computational resources. Another way is
assisted with LOCC SI . MQC allows one party with limited computational power
to use other parties’ computational resources without revealing the real task,
inputs and outcomes 43 ; 44 ; 45 . So-called blind quantum computation (BQC)
provides a secure way to verify adversarial quantum processor utilizing the
statistics of local measurements 44 ; 47 . Inspired by a recent scheme 48 , a
two-server BQC scheme is shown in Fig.5(a) and DIBQC scheme is shown as
(Fig.5(b)) SI :

  * (i)

$\textsf{A}_{1}$ (verifier) secretly chooses independent strings
$\mathbf{x}=x_{1}\cdots{}x_{n}$ and $\mathbf{s}=s_{1}\cdots{}s_{n}$ with the
uniform probability distribution.

  * (ii)

$\textsf{A}_{2}$ and $\textsf{A}_{3}$ share some EPR states:
$\otimes_{j=1}^{m}|\phi_{x_{j},s_{j}}\rangle$, where
$|\phi_{x_{j},s_{j}}\rangle=\frac{1}{\sqrt{2}}(\mathbbm{1}\otimes\sigma_{x}^{s_{j}}\sigma_{z}^{x_{j}})(|00\rangle+|11\rangle)$
with Pauli operators $\sigma_{x}$ and $\sigma_{z}$.

  * (iii)

$\textsf{A}_{1}$ shares some EPR states $\otimes_{j}|\phi_{j}\rangle$ with
$\textsf{A}_{2}$ and $\textsf{A}_{3}$, respectively.

  * (iv)

$\textsf{A}_{1}$ measures the $j$-th entanglement shared with $\textsf{A}_{2}$
in the basis
$\\{\frac{1}{\sqrt{2}}(|0\rangle\pm{}e^{-i\theta_{j}^{\prime}}|1\rangle)\\}$
after a local qubit operation $U=|0\rangle\langle
0|+e^{-2i\theta_{j}^{\prime}}|1\rangle\langle 1|$, in which
$\theta_{j}^{\prime}=(-1)^{s_{j}}\theta_{j}+x_{j}\pi$ and $\theta_{j}$ is
randomly chosen from $\\{\frac{k\pi}{4}|k=0,1,\cdots,7\\}$. $\textsf{A}_{2}$
measures the $j$-th entanglement shared with $\textsf{A}_{1}$ in the basis
randomly chosen from
$\\{\\{\frac{1}{\sqrt{2}}(|0\rangle\pm{}e^{-i\theta_{j}^{\prime}}|1\rangle)\\},\forall
j\\}$. They detect eavesdroppers with public discussions, and then distil a
random key $k_{1}\cdots{}k_{t}$.

  * (v)

$\textsf{A}_{2}$ measures the local qubit of the $j$-th Bell state shared with
$\textsf{A}_{3}$ in the basis
$\\{\frac{1}{\sqrt{2}}(|0\rangle\pm{}e^{-i\theta_{j}^{\prime}}|1\rangle)\\}$,
and then sends the outcome $b_{j}\in\\{0,1\\}$ in accord with the one-time pad
with $k_{1}\cdots{}k_{t}$.

  * (vi)

$\textsf{A}_{1}$ and $\textsf{A}_{3}$ start the modified single-server
protocol 43  with the following encoding:
$\hat{\theta}_{j}\mapsto\hat{\theta}^{\prime}_{j}+b_{j}\pi$, $j=1,2,\cdots,m$.
Here, the modified scheme means that all of $\textsf{A}_{1}$’s classical
information are encoded into the one-time pad using the key which is shared by
$\textsf{A}_{1}$ and $\textsf{A}_{3}$ 44 .

Figure 5: DIBQC with two servers.(a) A two-server BQC scheme. $\textsf{A}_{1}$
delegates a computational task to two servers $\textsf{A}_{2}$ and
$\textsf{A}_{3}$ who share some EPR states of
$\otimes_{j}|\phi_{x_{j},s_{j}}\rangle$. $y_{j}$ and $z_{j}$ are bit encodings
of the respective $\theta_{j}^{\prime}$ and $\theta_{j}^{\prime}+b_{j}\pi$.
(b) DIBQC with two servers. $\textsf{A}_{1}$ and $\textsf{A}_{2}$ (or
$\textsf{A}_{3}$) share a random key $k_{j}$ (or $\hat{k}_{j}$) using the
shared EPR states (QKD with purple lines). The measurement outcomes
$y_{j},z_{j},b_{j}$, and $c_{j}$ are then encrypted with the one-time pad. An
eavesdropper E can correlate its system with the shared sources.

The main difference between two protocols in Fig.5 is the transmission of
classical information. $\textsf{A}_{1}$ uses the statistics to verify the
shared random key with $\textsf{A}_{2}$ or $\textsf{A}_{3}$. Hence, the
present DIBQC is sensitive to cheating 45 ; 48 . $\textsf{A}_{3}$ cannot
obtain $\textsf{A}_{2}$’s outputs from $\textsf{A}_{1}$’s inputs 47  because
two servers are forbidden to communicate in the present model. Similar to
single-server protocol 45 , both servers cannot learn the hidden computational
task. From Eq.(4) the guessing probability about $\textsf{A}_{1}$’s outcome is
given by $P_{gs}({\bf
a}|E)\leq(\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi_{es}^{2}}+m^{-1/4}_{es})^{m}$,
where $m_{es}$ denotes the number of distilled key. One may further estimate
the guessing probability for $\textsf{A}_{1}$’s computational task 44 ; 45 .
Here, $\textsf{A}_{1}$ is able to perform quantum measurement while EPR states
should be prepared and distributed by the servers 45 ; 48 .

Device-independent quantum secret sharing. In an $(n,k)$ secret sharing
scheme, a secret $x$ is split into $n$ parts $x_{1},x_{2},\cdots,x_{n}$ such
that $x$ is recoverable from any $k$ parts 49 ; 50 . Inspired by recent scheme
with GHZ state MBNC , our goal is to propose a device-independent quantum
secret sharing (DIQSS) using quantum networks based on the inequality (2). For
simplicity, a $(3,2)$ DIQSS is constructed using the triangular network ${\cal
N}_{q}$ in Fig.3(a). For a secret bit $s$, the goal is to distribute the split
parts $s_{1},s_{2}$, and $s_{3}$ to $\textsf{A}_{1}$, $\textsf{A}_{2}$ and
$\textsf{A}_{3}$, respectively, where $s_{i}\oplus{}s_{j}=s$ with $1\leq
i\not=j\leq 3$. In Bell model (Fig.3), supposes that $a_{1}=s_{1}$,
$a_{2}=s_{2}$ and $b_{2}=s_{3}$ for specified inputs $x_{i}$, $y_{i}$, and
$z_{i}$. This implies that any two parties can cooperate to recover the secret
$s$, but one party fails. The violation of Mermin inequality 32  can rule out
the outer eavesdropper but not inner untrustful party. The main reason is that
a local correlating operation of receiver is equivalent to a measurement
dependent hidden variable model MBNC . Suppose that $\textsf{A}_{i}$ is
eavesdropper who recovers the output $a_{1}$ of $\textsf{A}_{1}$ with
$i\not=1$. The guessing probability $P(e=a_{1}|\textbf{x},u)$ by eavesdroppers
can be reshaped using the conditional probability
$P(\textbf{a},e|\textbf{x},u)$ into a linear optimization as

$\displaystyle\\!\\!\\!\\!\\!\\!\max_{P(\textbf{a},e|\textbf{x},u)}$
$\displaystyle\\!\\!\\!\\!\\!\\!P(e=a_{1}|\textbf{x},u)$ (6) s.t.
$\displaystyle\\!\\!\\!\\!\sum_{\textbf{a},e}P(\textbf{a},e|\textbf{x},u)=1,\forall\textbf{x},u$
(11) $\displaystyle\\!\\!\\!\\!\\!\\!P(\textbf{a},e|\textbf{x},u)\geq
0,\forall\textbf{a},\textbf{x},e,u$
$\displaystyle\\!\\!\\!\\!\\!\\!\sum_{w}(P(\textbf{a},e|\textbf{x},u)-P(\textbf{a},e|\textbf{x}^{\zeta},u))=0,\forall\zeta$
$\displaystyle\\!\\!\\!\\!\\!\\!P(\textbf{a}|\textbf{x})=\sum_{u,e}p(u)P(\textbf{a},e|\textbf{x},u)$
$\displaystyle\\!\\!\\!\\!\\!\\!\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}{\cal
L}_{i}(P(\textbf{a}|\textbf{x}))>2^{n}\sqrt{2}-4\sqrt{2}+2$

where $\textbf{a}=a_{1}\cdots a_{n}$, $\zeta{}$ is a subset of
$\\{a_{1},\cdots,a_{n}\\}$, and $\textbf{x}^{\zeta{}}=x_{1}^{\zeta}{}\cdots
x_{n}^{\zeta}{}$ satisfies $x_{i}^{\zeta}{}=x_{i}$ with
$a_{i}\not\in{}\zeta{}$. Eq.(11) is non-signalling condition for any subset
$\zeta{}$ with respect to different inputs of the corresponding parties.
Eqs.(11) and (11) are key to estimate the guessing probability of
$P(e=a_{1}|\textbf{x},u)$, where the projected distribution
$P(\textbf{a}|\textbf{x})$ is the $n$-variable probability derived from local
measurements on ${\cal N}_{q}$. The linear optimization problem is
algorithmically solvable using the semi-definite program 54  and NPA hiearchy
51 ; MBNC ; SI . Eve’s guess probability $p(e=c|x_{1},x_{2},x_{3})$ in quantum
settings satisfies $p(e=c|x_{1},x_{2},x_{3})\leq 3/4$ for $\varpi\geq 8.25$
and $p(e=c|x_{1},x_{2},x_{3})=1/2$ for $\varpi=6\sqrt{2}$. This provides a
useful method for characterizing secure quantum secret sharing. Similar
optimization problems can be formed for parties being eavesdroppers.

### Conclusion

The multipartite correlations generated by a $k$-independent network form a
star-convex set 18 . There is a convex correlation set with relaxed input
assumption arising from the inequality (2). Compared with previous nonlinear
Bell inequalities 15 ; 16 ; 17 ; 18 , the inequality (2) enables a standard
Bell test without assumptions of source independence for verifying general
quantum networks, including acyclic networks 18 , cyclic networks 24 , and
universal computational resources for measurement-based quantum computation 41
; 42 ; 43 . The main drawback is the lower visibility for noisy networks SI .
It may be applicable when both the number of parties and the number of
entangled sources are not large enough. This remains an interesting problem
for further improvement.

The independence of sources cannot be guaranteed in network scenarios because
any untrusted party may correlate the shared independent sources into a new
one. This is essentially different from the independence assumption of
measurement settings 55 , which is resolvable both in theory 56  and
experimental improvements 57 ; 58 . The present inequality (2) allows remote
legitimate parties to share a common random key in a fully device-independent
manner. Another example is device-independent BQC 45 ; 47 . The last example
is sharing of a classical secret 51 . The formal security proofs of these
applications and other candidates should be of interest in quantum information
processing using practical quantum resources.

## Acknowledgements

We thank the helps of Ronald de Wolf, Carlos Palazuelos, Luming Duan, Yaoyun
Shi, M. Orgun, J. Pieprzyk, Yuan Su, Huiming Li, Xiubo Chen, Yixian Yang. This
work was supported by the National Natural Science Foundation of China
(No.61772437), Sichuan Youth Science and Technique Foundation (No.2017JQ0048),
Fundamental Research Funds for the Central Universities (No.2018GF07), and EU
ICT COST CryptoAction (No.IC1306).

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##  Appendix A Proof of the inequality (1)

Consider an $n$-partite quantum network ${\cal N}_{q}$ (as shown in Fig.1(a)
in the main text) consisting of $n$ parties
$\textsf{A}_{1},\cdots,\textsf{A}_{n}$. Suppose that the total state of ${\cal
N}_{q}$ is in a biseparable state $\rho$ on Hilbert space
$\otimes_{i=1}^{n}{\cal H}_{i}$ 4 , that is, it can be decomposed into

$\displaystyle\rho=\sum_{I}p_{I}\varrho_{I,\overline{I}}$ (12)

where $I=\\{s_{1},\cdots,s_{k}\\}\subset\\{1,\cdots,n\\}$ and
$\overline{I}=\\{j|j\not\in I,1\leq j\leq n\\}$ are bipartition of
$\\{1,\cdots,n\\}$, $\\{p_{I}\\}$ is a probability distribution,
$\varrho_{I,\overline{I}}$ denotes separable state on Hilbert space ${\cal
H}_{I}\otimes{\cal H}_{\overline{I}}$, ${\cal H}_{I}=\otimes_{j\in I}{\cal
H}_{j}$ and ${\cal H}_{\overline{I}}=\otimes_{j\in\overline{I}}{\cal H}_{j}$.
Here, the summation is over all the bipartitions of $\\{1,\cdots,n\\}$.

Denote $M_{x_{i}}^{a_{i}}$ as positive-operator-value measurement (POVM for
short) performed by $\textsf{A}_{i}$, where $M^{a_{i}}_{x_{i}}$ satisfies
$\sum_{x_{i}=0,1}M^{a_{i}}_{x_{i}}=\mathbbm{1}$ for $a_{i}\in\\{-1,1\\}$,
$i=1,\cdots,n$. These measurement operators satisfy the commutativity
condition
$[M^{a_{i}}_{x_{i}},M^{a_{j}}_{x_{j}}]=M^{a_{i}}_{x_{i}}M^{a_{j}}_{x_{j}}-M^{a_{j}}_{x_{j}}M^{a_{i}}_{x_{i}}=0$
for any $i$ and $j$ with $i\not=j$. The binary output $a_{i}\in\\{-1,1\\}$ is
conditional on the input $x_{i}\in\\{0,1\\}$. All parties perform local
measurements on the state $\rho$. From Born rule, the output statistics is
characterized by the joint probability conditional on inputs, that is

$\displaystyle P(\textbf{a}|\textbf{x})={\rm
tr}(M_{x_{1}}^{a_{1}}\otimes\cdots{}\otimes{}M_{x_{n}}^{a_{n}}\rho)$ (13)

where $\textbf{x}=x_{1}\cdots{}x_{n}$, and $\textbf{a}=a_{1}\cdots{}a_{n}$.
For each input $x_{i}$, define $M_{x_{i}}$ as a dichotomic observable given by
$M_{x_{i}}=M_{x_{i}}^{a_{i}=0}-M_{x_{i}}^{a_{i}=1}$, $i=1,\cdots,n$. With the
distribution of $P(\textbf{a}|\textbf{x})$, an $n$-partite correlation in
terms of $\rho$ is defined by

$\displaystyle\langle M_{x_{1}}\cdots{}M_{x_{n}}\rangle$ $\displaystyle=$
$\displaystyle{\rm tr}(M_{x_{1}}\otimes\cdots{}\otimes{}M_{x_{n}}\rho)$ (14)
$\displaystyle=$
$\displaystyle\sum_{\textbf{a}}(-1)^{\sum_{i=1}^{n}(a_{i}+3)/2}P(\textbf{a}|\textbf{x})$

In what follows, each party makes use of multiple observables. Especially, for
each bipartition $I$ and $\overline{I}$, define $M_{x_{j}(I)}$ as dichotomic
observable of $\textsf{A}_{j}$ for the input $x_{j}$ depending on $I$, that
is, the input $x_{j}$ depends on $I$. For simplicity, denote

$\displaystyle\textbf{M}_{I}=\prod_{j\in
I}M_{x_{j}(I)},\textbf{M}_{\overline{I}}=\prod_{j\in\overline{I}}M_{x_{j}(\overline{I})}$
(15)

Since the left side of the inequality (2) is linear for all the measurement
observables $M_{x_{j}}$. All the biseparable states consist a convex set of
$\\{\sum_{I,\overline{I}}p_{I,\overline{I}}\varrho_{I}\otimes\varrho_{\overline{I}}\\}$.
So, the maximal bound of $\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}{\cal L}_{i}$
is achievable at the vertex of
$\\{\sum_{I,\overline{I}}p_{I,\overline{I}}\varrho_{I}\otimes\varrho_{\overline{I}}\\}$,
that is, product states. Hence, it is sufficient to prove the inequality (2)
by using the product state $\rho=\varrho_{I}\otimes\varrho_{\overline{I}}$.

From the inequality (2) $CH_{I}$ is defined by

$\displaystyle
CH_{I}=(\textbf{M}_{I}+\hat{\textbf{M}}_{I})\textbf{M}_{\overline{I}}+(\textbf{M}_{I}-\hat{\textbf{M}}_{I})\hat{\textbf{M}}_{\overline{I}}$
(16)

which is multipartite CHSH-type operator 31  or Mermin operator 40 , where
$|\textbf{M}_{I}|,|\hat{\textbf{M}}_{I}|,|\textbf{M}_{\overline{I}}|,|\hat{\textbf{M}}_{\overline{I}}|\leq
1$.

Now, for each bipartition $I_{0}$ and $\overline{I}_{0}$ (which may be
different from the bipartition $I$ and $\overline{I}$) there are two subcases
for any product state $\rho=\varrho_{I}\otimes\varrho_{\overline{I}}$.

  * (i)

The state $\rho$ is separable in terms of the bipartition $I_{0}$ and
$\overline{I}_{0}$. In this case, we can regard all the parties
$\textsf{A}_{j}$ with $j\in I$ as one party, while all the parties
$\textsf{A}_{t}$ with $t\in\overline{I}$ as the other. The quantity of
$CH_{I}$ in Eq.(16) is then regarded as a bipartite CHSH-type quantity 31 . It
follows that

$\displaystyle CH_{I_{0}}\leq 2$ (17)

which holds for any product state
$\rho=\varrho_{I}\otimes\varrho_{\overline{I}}$.

  * (ii)

The state $\rho$ is entangled in terms of the bipartition $I_{0}$ and
$\overline{I}_{0}$. In this case, there are two parties $\textsf{A}_{s}$ and
$\textsf{A}_{t}$ with $s,t\in I$, or $\overline{I}$ such that they are
entangled in the bipartition $I_{0}$ and $\overline{I}_{0}$, and $s\in
I_{0},t\in\overline{I}_{0}$. It follows that

$\displaystyle CH_{I_{0}}\leq 2\sqrt{2}$ (18)

Here, the upper bound $2\sqrt{2}$ may be achievable. Take
$\varrho_{I}=|\phi\rangle_{12}\langle\phi|$ and
$\varrho_{\overline{I}}=|\phi\rangle_{34}\langle\phi|$ as an example, where
$|\phi\rangle$ is an EPR state. $I_{0}=\\{1,3\\}$ and
$\overline{I}_{0}=\\{2,4\\}$. $\textsf{A}_{1}$ and $\textsf{A}_{2}$ shares one
EPR state. The maximal bound in the inequality (18) can be followed from the
CHSH test 31 .

Moreover, for each product state of
$\rho=\varrho_{I}\otimes\varrho_{\overline{I}}$ with the given bipartition $I$
and $\overline{I}$, there is at least one term $CH_{I}$ in the summation of
$\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor}{\cal L}_{i}$ satisfies that
$CH_{I}\leq 2$. Combined with the linearity of ${\cal L}_{i}$, it follows that

$\displaystyle\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor}{\cal L}_{i}\leq
2(N-1)\sqrt{2}+2$ (19)

for any product state $\rho$, where $N$ denotes the total number of $CH_{I}$s.

In what follows, we need to evaluate $N$. For an even integer $n$, define
$m_{i}$ as the total number of $CH_{I}$ for a given $i$ with
$i\leq\frac{n}{2}$. From the standard combination theory Graph  we obtain

$\displaystyle m_{i}$ $\displaystyle=$ $\displaystyle C(n,i),1\leq
i<\frac{n}{2},$ $\displaystyle m_{j}$ $\displaystyle=$
$\displaystyle\frac{1}{2}C(n,j),j=\frac{n}{2}$ (20)

where $C(n,i)$ denotes the combination number of choosing $i$ balls from $n$
different balls without choosing order. From Eq.(15), it follows that

$\displaystyle N$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n/2}m_{i}$ (21)
$\displaystyle=$ $\displaystyle\frac{1}{2}\sum_{i=1}^{n-1}C(n,i)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\sum_{i=0}^{n}C(n,i)-1$ (22)
$\displaystyle=$ $\displaystyle 2^{n-1}-1$ (23)

where Eq.(21) is obtained by using the equalities: $C(n,i)=C(n,n-i)$ with
$i=1,\cdots,\frac{n}{2}-1$. Eq.(22) is from the equalities: $C(n,0)=C(n,n)=1$.
Eq.(23) is from the equality: $\sum_{i=0}^{n}C(n,i)=2^{n}$.

For an odd integer $n$, similar to Eq.(20), using the standard combination
theory we obtain

$\displaystyle m_{i}=C(n,i),1\leq i\leq\lfloor{}\frac{n}{2}\rfloor$ (24)

where $\lfloor{}x{}\rfloor$ denotes the maximal integer which is no more than
$x$. Similar to Eqs.(20)-(23) it is easy to prove that

$\displaystyle N=$ $\displaystyle\sum_{i=1}^{\frac{n}{2}}m_{i}=2^{n-1}-1$ (25)

From Eqs.(19) and (23), and the linearity of ${\cal L}_{i}$, it yields that

$\displaystyle\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor}{\cal L}_{i}\leq
2^{n}\sqrt{2}-4\sqrt{2}+2$ (26)

For a general quantum network, $CH_{I}\leq 2\sqrt{2}$ for each bipartition $I$
and $\overline{I}$. From Eqs.(19), (23) and (25), it follows that

$\displaystyle\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor}{\cal
L}_{i}\leq(2^{n}-2)\sqrt{2}$ (27)

The upper bound is achievable for some quantum networks, which will be proved
in the next section. This completes the proof.

Remark S1. The present inequality (2) may be trivial if all observables of
$\textbf{M}_{I},\hat{\textbf{M}}_{I},\textbf{M}_{\overline{I}},\hat{\textbf{M}}_{\overline{I}}$
are different for each bipartition $I$ and $\overline{I}$, where each party
has $2$ inputs and two outputs for each bipartition $I$ and $\overline{I}$. In
this case, the present inequality (2) is a summation of different CHSH
inequalities 31 , where each party has $2N$ inputs and two outputs. However,
it is not trivial when each party has less than $2N$ inputs, or only a few
inputs. This is the main concern in what follows. Generally, the number of
inputs will depend on the involved state or quantum network. We can prove that
the total number of inputs is no more than $8$. This will be followed from the
proof for quantum networks in the next sections.

Example S1. Consider a triangle network ${\cal N}_{q}$ 13  shared by three
parties Alice, Bob and Charlie. We get the following inequality:

$\displaystyle
A_{1}B_{1}C_{1}+A_{2}B_{1}C_{1}+A_{1}B_{2}C_{2}-A_{2}B_{2}C_{2}$
$\displaystyle+A_{3}B_{3}C_{1}+A_{3}B_{4}C_{1}+A_{4}B_{3}C_{2}-A_{4}B_{4}C_{2}$
$\displaystyle+A_{3}B_{5}C_{3}+A_{3}B_{5}C_{4}+A_{4}B_{6}C_{3}-A_{4}B_{6}C_{4}\leq
4\sqrt{2}+2$ (28)

which holds for all biseparable states. $A_{i},B_{j}$, and $C_{k}$ are
dichotomic observable of $\textsf{A}_{1}$, $\textsf{A}_{2}$, and
$\textsf{A}_{3}$, respectively, $i,k=1,\cdots,4;j=1,\cdots,6$. It means that
Alice and Charlie have 4 inputs while Bob has 6 inputs.

Example S2. Different from Example 1, we can get another inequality with less
inputs as:

$\displaystyle
A_{1}B_{1}C_{1}+A_{2}B_{1}C_{1}+A_{1}B_{2}C_{2}-A_{2}B_{2}C_{2}$
$\displaystyle+A_{3}B_{3}C_{1}+A_{3}B_{4}C_{1}+A_{4}B_{3}C_{2}-A_{4}B_{4}C_{2}$
$\displaystyle+A_{3}B_{1}C_{3}+A_{3}B_{1}C_{4}+A_{4}B_{2}C_{3}-A_{4}B_{2}C_{4}\leq
4\sqrt{2}+2$ (29)

which holds for all biseparable states. It means that each party has 4 inputs.

##  Appendix B Genuinely multipartite nonlocality of quantum networks with
generalized EPR states

Let ${\cal N}_{q}$ be an $n$-partite connected quantum network shared by
$\textsf{A}_{1},\cdots,\textsf{A}_{n}$. Suppose that ${\cal N}_{q}$ consists
of generalized EPR states $|\phi_{1}\rangle,\cdots,|\phi_{m}\rangle$, where
$|\phi_{i}\rangle$ is given by

$\displaystyle|\phi_{i}\rangle=\cos_{i}|00\rangle+\sin\theta_{i}|11\rangle$
(30)

$\theta_{i}\in(0,\frac{\pi}{4})$. Here, the connectedness of quantum network
means that for each pair of $\textsf{A}_{i},\textsf{A}_{j}$, there is a chain-
type subnetwork ${\cal N}_{i\to
j}=\\{\textsf{A}_{i},\textsf{A}_{s_{1}},\cdots,\textsf{A}_{s_{k}},\textsf{A}_{j}\\}$,
where two adjacent parties share entangled states. The main goal is to prove
the genuinely multipartite nonlocality of ${\cal N}_{q}$ with even $n$ in this
section. The case of odd $n$ will be proved in the next section.

Result S1. There are local observables for each party such that the output
statistics violates the inequality (2) with an even integer $n\geq 4$, that
is, ${\cal N}_{q}$ is genuinely multipartite nonlocal in the biseparable
model.

Lemma S1. For an even $n\geq 4$, ${\cal N}_{q}$ can be locally transformed
into a new network ${\cal N}_{q}^{\prime}$ without classical communication
such that all parties have odd particles, where ${\cal N}_{q}^{\prime}$
consists of generalized EPR states and $4$-particle GHZ states.

Proof of Lemma S1. By representing an EPR state as one edge with two nodes,
${\cal N}_{q}$ can be schematically represented by a connected graph ${\cal
G}$ in which each party owns some nodes of edges. Here, a connected graph
means that for any two nodes, there is a path connecting them. For each pair
of two parties who share EPR states are connected by edges. The result is
easily followed for an acyclic graph which has no cycle, that is, a tree graph
${\cal G}$ Graph . In fact, note that the total number of particles is even.
The total number of parties who have odd number of particles is even. It means
that the total number of parties who have even number of particles is even.
Consider one pair $\textsf{A}_{i}$ and $\textsf{A}_{j}$ who has even number of
particles. Since ${\cal G}$ is connected, there is a connected subgraph ${\cal
G}_{i\to{}j}=\\{\textsf{A}_{i},\textsf{A}_{s_{1}},\cdots,\textsf{A}_{s_{k}},\textsf{A}_{j}\\}$,
where $\textsf{A}_{s_{1}},\cdots,\textsf{A}_{s_{k}}$ have odd number of
particles. In this case, each adjacent pair of ${\cal G}_{i\to{}j}$ can
transform one shared EPR state $|\phi_{i}\rangle|0\rangle|0\rangle$ into a
four-particle GHZ state
$|\hat{\phi}_{i}\rangle=\cos\theta_{i}|0000\rangle+\sin\theta_{i}|1111\rangle$
by using local controlled-not operation and one axillary particle in the state
$|0\rangle$ for each party. After these local operations being performed, all
the parties in ${\cal G}_{i\to{}j}$ have odd number of particles. This
procedure can be iteratively completed for all parties who have even number of
particles.

In what follows, we propose Algorithm 1 to locally transform a cyclic graph.
Here, a cyclic graph ${\cal G}$ means that it contains at least one cycle.

Input: A connected graph ${\cal G}$ with $n$ vertexes, where $n$ is even
integer

Output: A connected graph ${\cal G}^{\prime}$ with $n$ vertexes, where each
vertex has odd number of nodes

  * S0

(Initialization) Each cyclic graph ${\cal G}$ is decomposed into a set of
cyclic subgraphs $\\{{\cal G}_{1},\cdots,{\cal G}_{k}\\}$, where ${\cal
G}_{i}$ denotes a subgraph with one cycle in which any two adjacent parties
share some EPR states. This can be efficiently completed 111For $i$ from $1$
to $n$, find new cycle ${\cal G}_{j}$ started from the vertex $\textsf{A}_{i}$
such that ${\cal G}_{j}\in\\{{\cal G}_{1},\cdots,{\cal G}_{j-1}\\}$.
Otherwise, there is no new cycle. For each $j$, there are at least one new
vertex which will be added in ${\cal G}_{j}$ going beyond ${\cal
G}_{1},\cdots,{\cal G}_{j-1}$. This procedure will stop if all the vertex will
be included in some cycle. The total time complexity is $O(n^{2})$, where each
cycle has at most $n$ vertexes.. Denote $\textsf{A}_{i}$s as the vertexes (or
parties in the network) who have even number of nodes (or particles in the
network). Denote $\textsf{B}_{j}$s as the vertexes (or parties in the network)
who have odd number of nodes (or particles in the network), where the
subindexes $i,j$ of $\textsf{A}_{i}$s and $\textsf{B}_{j}$s are different for
convenience.

  * S1

If all vertexes have odd number of nodes, then output the graph ${\cal G}$;

  * S2

There are vertexes which have even number of nodes. For convenience, assume
that in each subgraph ${\cal G}_{i}$ there are vertexes which have even number
of nodes. Otherwise, delete the subgraphs without vertexes which have even
number of nodes.

    * For $\ell$ from $1$ to $k$, consider ${\cal G}_{\ell}$ as follows.

      * 1)

For two adjacent vertexes $\textsf{A}_{i}$ and $\textsf{A}_{j}$ in ${\cal
G}_{\ell}$, if both of them have even number of nodes in ${\cal G}_{\ell}$,
they can transform one connected edge into one hyper edge with four nodes by
adding one new node for each vertex Graph . This can be completed by
transforming one shared EPR state $|\phi_{i}\rangle|0\rangle_{i}|0\rangle_{j}$
into one four-particle GHZ state
$|\hat{\phi}_{i}\rangle=\cos\theta_{i}|0000\rangle+\sin\theta_{i}|1111\rangle$
by using local controlled-not operations and one axillary particle in
networks. After these operations being performed, both vertexes have odd
number of nodes. They are then relabeled as $\textsf{B}_{i}$ and
$\textsf{B}_{j}$.

      * 2)

For two adjacent vertexes $\textsf{A}_{i}$ and $\textsf{A}_{j}$ in ${\cal
G}_{\ell}$, if one of them, $\textsf{A}_{i}$ for example, has even number of
nodes in ${\cal G}_{\ell}$, there exist other vertexes
$\textsf{A}_{s_{1}},\cdots$, $\textsf{A}_{s_{t}}$ which own even number of
nodes. Choose one vertex $\textsf{A}_{s_{1}}$ such that there are other
vertexes $\textsf{B}_{j_{1}},\cdots,\textsf{B}_{j_{\ell}}$ satisfying that
$\textsf{A}_{i},\textsf{B}_{j_{1}},\cdots,\textsf{B}_{j_{\ell}},\textsf{A}_{s_{1}}$
consist of a chain-shaped connected subgraph. Now, each pair of adjacent
vertexes can transform one edge into one hyper edge with four nodes by adding
one new node for each vertex. After the local operations being performed,
these vertexes have odd number of nodes, where one node is added for
$\textsf{A}_{i}$ and $\textsf{A}_{s_{1}}$, while two nodes are added for
$\textsf{B}_{j_{1}},\cdots,\textsf{B}_{j_{\ell}}$. $\textsf{A}_{i}$ and
$\textsf{A}_{s_{1}}$ are then relabeled as $\textsf{B}_{i}$ and
$\textsf{B}_{s_{1}}$ respectively. These two steps can be iteratively
performed for all vertexes in ${\cal G}_{\ell}$.

Algorithm 1 Transforming ${\cal N}_{q}$ with local unitary operations and
axillary particles in the state $|0\rangle$

Figure 6: Schematically transforming a network using local operations and
auxiliary particles without classical communication. (a) Cyclic network with
four parties. Four parties $\textsf{A}_{1},\cdots,\textsf{A}_{4}$ share four
EPR states. (b) Cyclic network with two cycles. Four parties
$\textsf{A}_{1},\cdots,\textsf{A}_{4}$ share five EPR states.

For ${\cal G}_{\ell}$, each vertex has only added even number of nodes, even
if local transformations may change the number of nodes for these vertexes
contained in ${\cal G}_{j}$ for $j<\ell$. In this case, all the vertexes
$\textsf{B}_{s}\in{\cal G}_{j}$ have odd number of nodes after local
transformations being performed by all vertexes in ${\cal G}_{\ell}$. It means
that one does not need to consider all the subgraphs ${\cal G}_{j}$ with
$j<\ell$ after ${\cal G}_{\ell}$ has been changed. This is key to stop
Algorithm 1.

We take use of this algorithm to prove Lemma S1. Note that $n$ is even and
each edge has two nodes. Moreover, two nodes will be added in each time. For
each round, there are two vertexes $\textsf{A}_{i}$ and $\textsf{A}_{j}$ being
relabeled as $\textsf{B}_{i}$ and $\textsf{B}_{j}$ while other vertexes
$\textsf{B}_{s}$ are unchanged. By induction on the number $n$ of vertexes,
all the vertexes $\textsf{A}_{j}$ can be relabeled as $\textsf{B}_{j}$ because
the network ${\cal N}_{q}$ is connected. This can be proved by contradiction.
In fact, assume that one vertex has even number of particles while $n-1$
vertexes have odd number of nodes. There are odd nodes because $n-1$ is odd
integer. This contradicts with the fact that there is even number of vertexes
from two nodes of each edge. This completes the proof of Lemma S1. $\Box$

Example S3. Two networks are shown in Fig.S6. In Fig.S6(a), there are four
parties who share four EPR states. In step S2, $\textsf{A}_{1}$ and
$\textsf{A}_{2}$ can locally change an EPR state into four-particle GHZ state
with local operations and axillary particles. Similar result holds for
$\textsf{A}_{3}$ and $\textsf{A}_{4}$. After these local operations being
performed, they obtain a new network consisting of two EPR states and two
four-particle GHZ states, where each party has three particles. In Fig.S6(b),
there are four parties who share five EPR states. Different from the network
shown in Fig.S6(a), there are two cycles, that is, ${\cal
G}_{1}=\\{\textsf{A}_{1},\textsf{A}_{2},\textsf{A}_{3}\\}$ and ${\cal
G}_{2}=\\{\textsf{A}_{1},\textsf{A}_{3},\textsf{A}_{4}\\}$. In step S2,
consider ${\cal G}_{1}$ firstly. $\textsf{A}_{1}$ and $\textsf{A}_{2}$ can
change an EPR state into four-particle GHZ state with local operations and
axillary particles. And then, consider ${\cal G}_{2}$. $\textsf{A}_{1}$ has
four particles by adding the particles involved in cycle ${\cal G}_{1}$. So,
it is sufficient to consider the EPR state shared by $\textsf{A}_{1}$.
Specifically, $\textsf{A}_{1}$ and $\textsf{A}_{4}$ transform the shared EPR
state into four-particle GHZ state with local operations and axillary
particles. After these local operations, they obtain a new network consisting
of three EPR states and two four-particle GHZ states, where each party has odd
number of particles.

Proof of Result S1. Let $M_{x_{i}}$ be dichotomic observable of
$\textsf{A}_{i}$ for the input $x_{i}\in\\{0,1\\}$. In what follows, it is
sufficient to prove that there are local observable for each party on ${\cal
N}_{q}$ such that the quantum correlations generated by local measurements
violate the inequality (2).

From Lemma S1, after local transformations being performed, each party
$\textsf{A}_{j}$ is relabeled as $\textsf{B}_{j}$ who has $n_{j}$ number of
particles. From Eq.(30), the total state of ${\cal N}_{q}$ is given by

$\displaystyle\rho=\otimes_{i=1}^{m}|\phi_{i}\rangle\langle\phi_{i}|$ (31)

where $|\phi_{i}\rangle$ is EPR state or 4-particle GHZ state. Define
$\varpi_{q}$ as the quantum violation given by

$\displaystyle\varpi_{q}=\sum_{i=1}^{\lfloor{\frac{n}{2}}\rfloor}{\cal L}_{i}$
(32)

where ${\cal L}_{i}$s are defined in the inequality (2) associated with the
state $\rho$ in Eq.(31). It is sufficient to evaluate $\varpi_{q}$ by
maximizing all the multipartite quantities $CH_{I}$ in the inequality (2).

For each set $I=\\{s_{1},\cdots,s_{k}\\}$, consider $CH_{I}$ firstly. Note
that ${\cal N}_{q}$ is connected. For each pair of parties $\textsf{B}_{i}$
and $\textsf{B}_{j}$ with $i\in I$ and $j\not\in I$, there is one chain-shaped
subnetwork ${\cal N}_{i\to j}\subset{\cal N}^{\prime}_{q}$ connecting them.
There is one longest chain-shaped subnetwork in these subnetworks $\\{{\cal
N}_{s_{t}\to\ell}\\}$, that is, ${\cal N}_{s_{t}\to\ell}\subset{\cal
N}^{\prime}_{q}$ connecting the parties $\textsf{B}_{s_{t}}$ and
$\textsf{B}_{\ell}$, where $s_{t}\in{}I$ and $\ell\not\in{}I$. For simplicity,
define ${\cal N}_{s_{t}\to\ell}$ as

$\displaystyle{\cal
N}_{s_{t}\to\ell}=\\{\textsf{B}_{s_{t}},\textsf{B}_{l_{1}},\cdots,\textsf{B}_{l_{v}},\textsf{B}_{\ell}\\}$
(33)

where $l_{1},\cdots,l_{v}\not\in I$.

In what follows, define observables for each party as

$\displaystyle
M_{x_{s_{t}}(I)}=\cos\theta\sigma_{z}^{\otimes{}n_{s_{t}}}+\sin\theta\sigma_{x}^{\otimes{}n_{s_{t}}},$
(34)
$\displaystyle\hat{M}_{x_{s_{t}}(I)}=\cos\theta\sigma_{z}^{\otimes{}n_{s_{t}}}-\sin\theta\sigma_{x}^{\otimes{}n_{s_{t}}},$
(35) $\displaystyle
M_{x_{j}(I)}=\sigma_{z}^{\otimes{}n_{j}},\hat{M}_{x_{j}(I)}=\sigma_{x}^{\otimes{}n_{j}},\forall
j\not=s_{t}$ (36)

where $n_{j}$ denotes the number of particles owned by $\textsf{B}_{j}$, and
$X^{\otimes k}$ denotes the tensor of $k$ copies of $X$. It is easy to show
that
$M_{s_{1}}M_{s_{1}}^{\dagger}=\hat{M}_{s_{1}}\hat{M}_{s_{1}}^{\dagger}=\mathbbm{1}$
because $n_{1}$ is odd integer. Here, $x_{j}(I)$ means that the input $x_{j}$
depends on the set $I$. With these observables, from Eqs.(15), (34)-(36) it
follows that

$\displaystyle\langle\textbf{M}_{I}\textbf{M}_{\overline{I}}\rangle$
$\displaystyle=$ $\displaystyle{\rm
tr}((\cos\theta\sigma_{z}^{\otimes{}n_{s_{t}}}+\sin\theta\sigma_{x}^{\otimes{}n_{s_{t}}})(\otimes_{j\not=s_{t}}\sigma_{z}^{\otimes{}n_{j}})\rho)$
(37) $\displaystyle=$ $\displaystyle\cos\theta{\rm
tr}((\sigma_{z}^{\otimes{}n_{s_{t}}}\otimes_{j\not=s_{t}}\sigma_{z}^{\otimes{}n_{j}})(\rho_{s_{t}}\otimes_{j\not=s_{t}}\rho_{j}))$
$\displaystyle=$ $\displaystyle\cos\theta$ (38)

In Eq.(37), $\rho_{j}$ denotes the state owned by $\textsf{B}_{j}$, for
$j\in\\{1,\cdots,n\\}$. Eq.(37) follows from Lemma S1. Note that ${\rm
tr}(\sigma_{x}\otimes\sigma_{z}|\phi_{i}\rangle\langle\phi_{i}|)={\rm
tr}(\sigma_{z}\otimes\sigma_{x}|\phi_{i}\rangle\langle\phi_{i}|)=0$ for any
generalized EPR state $|\phi_{i}\rangle$. Moreover, ${\rm tr}(M_{1}\otimes
M_{2}\otimes M_{3}\otimes M_{4}|\phi_{i}\rangle\langle\phi_{i}|)=0$ for any
generalized 4-particle GHZ state $|\phi_{i}\rangle$ when
$M_{i}\in\\{\sigma_{z},\sigma_{x}\\}$ and $M_{i}$s are different. With these
equalities, it follows from Eq.(31) that ${\rm
tr}((\sigma_{x}^{\otimes{}n_{s_{t}}})(\otimes_{j\not=s_{t}}\sigma_{z}^{\otimes{}n_{j}})(\rho_{s_{t}}\otimes_{j\not=s_{t}}\rho_{j}))$.
Moreover, we get Eq.(38) from the equalities of ${\rm
tr}(\sigma_{z}\otimes\sigma_{z}|\phi_{i}\rangle\langle\phi_{i}|)=1$ for any
generalized EPR state $|\phi_{i}\rangle$, and ${\rm tr}(\sigma_{z}^{\otimes
4}|\varphi_{i}\rangle\langle\varphi_{i}|)=1$ for any generalized 4-particle
GHZ state $|\varphi_{i}\rangle$.

Similarly, we get that

$\displaystyle\langle\hat{\textbf{M}}_{I}\textbf{M}_{\overline{I}}\rangle$
$\displaystyle=$ $\displaystyle\cos\theta$ (39)
$\displaystyle\langle\textbf{M}_{I}\hat{\textbf{M}}_{\overline{I}}\rangle$
$\displaystyle=$
$\displaystyle\langle\hat{\textbf{M}}_{I}\hat{\textbf{M}}_{\overline{I}}\rangle=\sin\theta\prod_{j=1}^{m}\sin
2\theta_{j}$ (40)

From Eqs.(16) and (38)-(40), it follows that

$\displaystyle CH_{I}$ $\displaystyle=$ $\displaystyle{\rm
tr}(((\textbf{M}_{I}+\hat{\textbf{M}}_{I})\textbf{M}_{\overline{I}}+(\textbf{M}_{I}-\hat{\textbf{M}}_{I})\hat{\textbf{M}}_{\overline{I}})\rho)$
$\displaystyle=$ $\displaystyle 2\cos\theta+2\sin\theta\prod_{j=1}^{m}\sin
2\theta_{j}$ $\displaystyle=$ $\displaystyle 2\sqrt{1+\prod_{j=1}^{m}(\sin
2\theta_{j})^{2}}$ (42)

when $\theta$ satisfies $\cos\theta=1/\sqrt{1+\prod_{j=1}^{m}(\sin
2\theta_{j})^{2}}$.

Generally, for each set $I\subset\\{1,\cdots,n\\}$, from the assumption in the
inequality (2), each party $\textsf{B}_{j}$ can choose proper observables
depending on the input $x_{j}(I)$ such that the joint statistics satisfies the
inequality (42). The main reason is that one may choose different input
$x_{j}(I)$ depending on $I$. This implies that the inequality (42) holds for
all the multipartite quantities $CH_{I}$s simultaneously. Hence, there exist
observables for each observer such that the quantum correlations satisfy

$\displaystyle\varpi_{q}$ $\displaystyle=$ $\displaystyle\sum_{i}{\cal L}_{i}$
(43) $\displaystyle=$ $\displaystyle 2N(1+\prod_{j=1}^{m}(\sin
2\theta_{j})^{2})^{1/2}$ $\displaystyle>$ $\displaystyle
2^{n}\sqrt{2}-4\sqrt{2}+2$

if all $\theta_{j}$s satisfy the following inequality:

$\displaystyle\prod_{j=1}^{m}\sin
2\theta_{j}>(N^{2}-2(2-\sqrt{2})N+3-2\sqrt{2})^{1/2}$ (44)

where $N=2^{n-1}-1$. This completes the proof for even $n$. $\Box$

##  Appendix C Genuinely multipartite nonlocality of general connected quantum
networks

Let ${\cal N}_{q}$ be an $n$-partite connected quantum network consisted of
$\textsf{A}_{1},\cdots,\textsf{A}_{n}$. Suppose that ${\cal N}_{q}$ consists
of generalized GHZ states $|\phi_{1}\rangle,\cdots,|\phi_{m}\rangle$, where
$|\phi_{i}\rangle$ is given by

$\displaystyle|\phi_{i}\rangle=\cos_{i}|0\rangle^{\otimes
m_{i}}+\sin\theta_{i}|1\rangle^{\otimes m_{i}}$ (45)

$\theta_{i}\in(0,\frac{\pi}{4})$. The main goal in this section is to prove
the genuinely multipartite nonlocality of ${\cal N}_{q}$ by using the
inequality (2).

Result S2. There are local observables for each party such that the output
statistics violates the inequality (2), that is, ${\cal N}_{q}$ is genuinely
multipartite nonlocal in the biseparable model.

Proof of Result S2. Note that $m_{i}$ can be assumed to be even integer for
each $i$. Otherwise, one party can change it into new GHZ state with even
number of particles by using local operation and axillary particle. With this
assumption, we firstly prove that ${\cal N}_{q}$ is genuinely $n$-partite
nonlocal for even $n$ with $n\geq 3$. Similar to local transformations defined
in Algorithm 1, ${\cal N}_{q}$ can be locally transformed into new network
${\cal N}^{\prime}_{q}$ without classical communication such that all parties
have odd number of particles, where ${\cal N}^{\prime}_{q}$ consists of
generalized EPR states and GHZ states with even number of particles. The proof
is similar to Algorithm 1. The rest proof of the genuinely multipartite
nonlocality is similar to the procedure from Eq.(32) to Eq.(43).

In what follows, we prove the result for odd $n$ with $n\geq 3$. Similar to
local transformations defined in Algorithm 1, ${\cal N}_{q}$ can be locally
transformed into a new network ${\cal N}^{\prime}_{q}$ without classical
communication such that there are $n-1$ observers who have odd number of
particles, where ${\cal N}^{\prime}_{q}$ consists of generalized EPR states
and GHZ states with even number of particles. The proof is similar to
Algorithm 1 by considering a connected subnetwork consisting of $n-1$ parties.

There are three subcases to prove the genuinely multipartite nonlocality of
${\cal N}^{\prime}_{q}$.

  * S1

Consider the subnetwork consisting of $n-1$ parties
$\textsf{A}_{1},\cdots,\textsf{A}_{n-1}$. These parties can locally transform
the shared network into another subnetwork using local unitary operations
without the help of classical communication such that they have odd number of
particles. After these local transformations, these parties are relabeled by
$\textsf{B}_{1},\cdots,\textsf{B}_{n-1}$. The rest proof is similar to
Eqs.(31)-(43) when $\textsf{A}_{n}$ has odd number of particles. Otherwise,
there are two kinds of multipartite CHSH-type quantities $CH_{I}$ which will
be distinguished from their inputs.

  * S2

Consider $CH_{I}$ with $n\not\in{}I$. Similar to Eq.(33), there exists one
longest chain-shaped network ${\cal
N}_{s_{t}\to\ell}=\\{\textsf{B}_{s_{t}},\textsf{B}_{j_{1}},\cdots,\textsf{B}_{j_{t}},\textsf{B}_{\ell}\\}$,
where $j_{1},\cdots,j_{t},\ell\not\in I$. For the party $\textsf{A}_{n}$,
define its local observable as

$\displaystyle M_{x_{n}(I)}=\sigma_{z}^{\otimes{}h_{n}},$
$\displaystyle\hat{M}_{x_{n}(I)}=\sigma_{x}^{\otimes{}h_{n}}$ (46)

where $h_{n}$ denotes the number of particles owned by the party
$\textsf{B}_{n}$. With this definition, the inequality (35) holds for each
$I\subset\\{1,\cdots,n\\}$ satisfying $n\not\in{}I$. From the assumption in
the inequality (2) in the main text, i.e., $t_{I},\hat{t}_{I}$ are different
for any $I$, the party $\textsf{A}_{n}$ can choose proper local observable for
these post-selective terms $CH_{I}$ simultaneously. Hence, the inequality (42)
holds for all $I$s satisfying $n\not\in{}I$.

  * S3

Consider $CH_{I}$ with $n\in{}I$. In this case, from the assumption of the
inequality (2), for each $I$ $\textsf{A}_{n}$ and $\textsf{A}_{i}$, who share
at least one GHZ state can transform locally the shared states, such that
$\textsf{A}_{n}$ owns odd number of particles after local operations. The
observables of $\textsf{A}_{n}$ are similar to these given in Eqs.(32) and
(34). And then, there exist another set of observables for all parties such
that the inequality (42) holds for $CH_{I}$. Note that all $CH_{I}$s can be
distinguished by all parties using proper inputs. This means that the
inequality (42) holds for all the terms $CH_{I}$. Consequently, the inequality
(42) holds for each odd $n$.

This completes the proof. $\Box$

##  Appendix D Evaluating the inputs of the inequality (2)

In this section, we show that there are eight inputs for each party involved
in the inequality (2) for verifying general networks. The main idea is from
the proof in Appendixes B and C.

Consider an $n$-partite connected quantum network ${\cal N}_{q}$ consisting of
EPR states and GHZ states. For an even $n\geq 3$, from Lemma S1 ${\cal N}_{q}$
can be locally transformed into new network ${\cal N}^{\prime}_{q}$ satisfying
that each party has odd number of particles, where ${\cal N}^{\prime}_{q}$
consists of generalized EPR states and GHZ states with even number of
particles.

Similar to Eq.(33), for each $I\subset\\{1,\cdots,n\\}$, there is one largest
chain-type subnetwork ${\cal
N}_{s_{t}\to\ell}=\\{\textsf{B}_{s_{t}},\textsf{B}_{l_{1}},\cdots,\textsf{B}_{l_{v}},\textsf{B}_{\ell}\\}$
connecting two parties $\textsf{B}_{s_{t}}$ and $\textsf{B}_{\ell}$ with
$s_{t}\in I$ and $l_{1},\cdots,l_{v},\ell\not\in I$. The quantum network
${\cal N}^{\prime}_{q}$ can be verified by using the inequality (2) with four
inputs. In fact, for each party $\textsf{B}_{j}$, there are two subcases.

  * (i)

$j=s_{t}$. In this case, there are two input indexes $x_{j}(I)\in\\{1,2\\}$
associated with the observables shown in Eqs.(34) and (35).

  * (ii)

$j\not=s_{t}$. In this case, there are two input indexes
$x_{j}(I)\in\\{3,4\\}$ associated with the observables shown in Eq.(36).

Note that $s_{t}$ depends the longest chain-type subnetwork ${\cal
N}_{s_{t}\to\ell}$, which may be relaxed as a general chain-type subnetwork
without requirement of the length. With this relaxation, one can assume that
$s_{t}=\min\\{s_{j}|s_{j}\in I\\}$ and $\ell=\min\\{j|j\not\in I\\}$. Denote

$\displaystyle\textbf{M}_{I}=M_{x_{s_{t}}=1}\prod_{j\in
I,j\not=s_{t}}M_{x_{j}=3},$
$\displaystyle\hat{\textbf{M}}_{I}=M_{x_{s_{t}}=2}\prod_{j\in
I,j\not=s_{t}}M_{x_{j}=4},$
$\displaystyle\textbf{M}_{\overline{I}}=\prod_{j\in\overline{I}}M_{x_{j}=3},$
$\displaystyle\hat{\textbf{M}}_{\overline{I}}=\prod_{j\in\overline{I}}M_{x_{j}=4}$
(47)

With these definitions, the inequality (2) can be rewritten into

$\displaystyle\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}{\cal L}_{i}\leq
2^{n}\sqrt{2}-4\sqrt{2}+2$ (48)

where ${\cal L}_{i}=\sum_{|I|=i}CH_{I}$ with
$CH_{I}=(\textbf{M}_{I}+\hat{\textbf{M}}_{I})\textbf{M}_{\overline{I}}+(\textbf{M}_{I}-\hat{\textbf{M}}_{I})\hat{\textbf{M}}_{\overline{I}}$.
Similar to Appendixes B and C, we can prove that Results S1 and S2 holds for
any $n$-partite connected quantum network consisted of generalized EPR states
and GHZ states.

For an odd $n$, from the proof in Appendix C, the quantum network of ${\cal
N}_{q}$ can be locally transformed into two new networks ${\cal
N}^{\prime\prime}_{q}$ and ${\cal N}^{\prime\prime\prime}_{q}$ satisfying that
each party in ${\cal N}^{\prime\prime}_{q}$ or ${\cal
N}^{\prime\prime\prime}_{q}$ has odd number of particles. For the network of
${\cal N}^{\prime\prime}_{q}$, the genuinely multipartite nonlocality can be
verified by the inequality (48) with four inputs. Similar result holds for
${\cal N}^{\prime\prime\prime}_{q}$. Note that for each party
$\textsf{B}_{j}$, there are four inputs as shown in Eq.(47) if local states of
$\textsf{B}_{j}$ are same as each other in both networks ${\cal
N}^{\prime\prime}_{q}$ and ${\cal N}^{\prime\prime\prime}_{q}$. Otherwise,
there are at most eight inputs associated with two sets of observables as
shown in Eq.(47). For some special network, it may be reduced to six inputs,
where there are two different observables for verifying ${\cal
N}^{\prime\prime\prime}_{q}$ and ${\cal N}^{\prime\prime\prime}_{q}$ (see the
following example). Note that the number of inputs for each party depends on
the transformed networks ${\cal N}^{\prime\prime}_{q}$ or ${\cal
N}^{\prime\prime\prime}_{q}$, which is from the network configuration of
${\cal N}_{q}$. We cannot get the explicit form of the inequality (48) in this
case.

Continuing Example S1. Consider the triangle network consisting of three EPR
states by using the inequality (28). Assume that two parties of Alice, Bob and
Charlie share one generalized EPR state:
$|\phi_{i}\rangle_{2i-1,2i}=\cos\theta_{i}|00\rangle+\sin\theta_{i}|11\rangle$
with $\theta_{i}\in(0,\pi)$ and $i=1,2$, and $3$. The total system is denoted
as $|\phi_{1}\rangle_{12}|\phi_{2}\rangle_{34}|\phi_{3}\rangle_{56}$. Alice
owns the particles 1 and 6. Bob owns the particles 2 and 3 while Charlie owns
the particles 4 and 5. Here, $|{\phi}_{1}\rangle$ and $|{\phi}_{3}\rangle$
will be changed into four-particle GHZ states with local operations and
auxiliary qubits. Define local observables as

$\displaystyle A_{1}=B_{3}=C_{3}=\cos\theta\sigma_{z}^{\otimes
3}+\sin\theta\sigma_{x}^{\otimes 3},$ $\displaystyle
A_{2}=B_{4}=C_{4}=\cos\theta\sigma_{z}^{\otimes
3}-\sin\theta\sigma_{x}^{\otimes 3},$ $\displaystyle
A_{3}=B_{1}=\sigma_{z}^{\otimes 3},A_{4}=B_{2}=\sigma_{x}^{\otimes 3},$
$\displaystyle B_{5}=\sigma_{z}^{\otimes 4},B_{6}=\sigma_{x}^{\otimes 4},$
$\displaystyle
C_{1}=\sigma_{z}\otimes\sigma_{z},C_{2}=\sigma_{x}\otimes\sigma_{x},$ (49)

From a forward evaluation we get

$\displaystyle\varpi_{q}$ $\displaystyle:=$
$\displaystyle\langle{}A_{1}B_{1}C_{1}\rangle+\langle{}A_{2}B_{1}C_{1}\rangle+\langle{}A_{1}B_{2}C_{2}\rangle-\langle{}A_{2}B_{2}C_{2}\rangle$
(50)
$\displaystyle+\langle{}A_{3}B_{3}C_{1}\rangle+\langle{}A_{3}B_{4}C_{1}\rangle+\langle{}A_{4}B_{3}C_{2}\rangle-\langle{}A_{4}B_{4}C_{2}\rangle$
$\displaystyle+\langle{}A_{3}B_{5}C_{3}\rangle+\langle{}A_{3}B_{5}C_{4}\rangle+\langle{}A_{4}B_{6}C_{3}\rangle-\langle{}A_{4}B_{6}C_{4}\rangle$
$\displaystyle=$ $\displaystyle 6\cos\theta+6\sin\theta\sin 2\theta_{1}\sin
2\theta_{2}\sin 2\theta_{3}$ $\displaystyle>$ $\displaystyle 4\sqrt{2}+2$

which violates the inequality (28) when $\theta_{i}$s satisfy

$\displaystyle\sin 2\theta_{1}\sin 2\theta_{2}\sin 2\theta_{3}>0.7928$ (51)

Moreover, consider the noisy network ${\cal N}_{q}$ given by

$\displaystyle\rho=\rho_{1}\otimes\rho_{2}\otimes\rho_{3}$ (52)

where $\rho_{i}$ denotes Werner state Wrner  given by
$\rho_{i}=v|\phi_{i}\rangle\langle\phi_{i}|+\frac{1-v}{4}\mathbbm{1}$ with
$\theta_{i}=\frac{\pi}{4}$, and $v\in(0,1]$. The visibility is given by
$v=0.966$. This is going beyond a recent result without the robustness 23 .

Continuing Example S2. Assume that two parties of Alice, Bob and Charlie share
a generalized EPR state:
$|\phi_{i}\rangle_{2i-1,2i}=\cos\theta_{i}|00\rangle+\sin\theta_{i}|11\rangle$
with $\theta_{i}\in(0,\pi)$ and $i=1,2$, and $3$. The total system is denoted
as $|\phi_{1}\rangle_{12}|\phi_{2}\rangle_{34}|\phi_{3}\rangle_{56}$. Alice
owns the particles 1 and 6. Bob owns the particles 2 and 3 while Charlie owns
the particles 4 and 5. To verify the genuinely tripartite nonlocality by using
the inequality (29), Alice and Bob will change $|\phi_{1}\rangle_{12}$ into
four-particle GHZ state
$|\phi_{1}\rangle_{12ab}=\cos\theta_{1}|0000\rangle+\sin\theta_{1}|1111\rangle$
with local two-particle unitary operations for the inputs $3$ and $4$. Define
observables of three observers as follows:

$\displaystyle
A_{1}=\cos\theta\sigma_{z}\otimes\sigma_{z}+\sin\theta\sigma_{x}\otimes\sigma_{z},$
$\displaystyle
A_{2}=\cos\theta\sigma_{z}\otimes\sigma_{z}-\sin\theta\sigma_{x}\otimes\sigma_{z},$
$\displaystyle A_{3}=\sigma_{z}^{\otimes 3},A_{4}=\sigma_{x}^{\otimes 3},$
$\displaystyle
B_{1}=\sigma_{z}\otimes\sigma_{z},B_{2}=\sigma_{x}\otimes\sigma_{x},$
$\displaystyle B_{3}=\cos\theta\sigma_{z}^{\otimes
3}+\sin\theta\sigma_{x}^{\otimes 3},$ $\displaystyle
B_{4}=\cos\theta\sigma_{z}^{\otimes 3}-\sin\theta\sigma_{x}^{\otimes 3},$
$\displaystyle
C_{1}=\sigma_{z}\otimes\sigma_{z},C_{2}=\sigma_{x}\otimes\sigma_{z},$
$\displaystyle
C_{3}=\cos\theta\sigma_{z}\otimes\sigma_{z}+\sin\theta\sigma_{x}\otimes\sigma_{z},$
$\displaystyle
C_{4}=\cos\theta\sigma_{z}\otimes\sigma_{z}-\sin\theta\sigma_{x}\otimes\sigma_{z},$
(53)

From a forward evaluation we get

$\displaystyle\varpi_{q}$ $\displaystyle:=$
$\displaystyle\langle{}A_{1}B_{1}C_{1}\rangle+\langle{}A_{2}B_{1}C_{1}\rangle+\langle{}A_{1}B_{2}C_{2}\rangle$
(54)
$\displaystyle-\langle{}A_{2}B_{2}C_{2}\rangle+\langle{}A_{3}B_{3}C_{1}\rangle+\langle{}A_{3}B_{4}C_{1}\rangle$
$\displaystyle+\langle{}A_{4}B_{3}C_{2}\rangle-\langle{}A_{4}B_{4}C_{2}\rangle+\langle{}A_{3}B_{1}C_{3}\rangle$
$\displaystyle+\langle{}A_{3}B_{1}C_{4}\rangle+\langle{}A_{4}B_{2}C_{3}\rangle-\langle{}A_{4}B_{2}C_{4}\rangle$
$\displaystyle=$ $\displaystyle 6\cos\theta+2\sin\theta(\sin 2\theta_{1}\sin
2\theta_{2}$ $\displaystyle+\sin 2\theta_{2}\sin 2\theta_{3}+\sin
2\theta_{1}\sin 2\theta_{3})$ $\displaystyle>$ $\displaystyle 4\sqrt{2}+2$

which violates the inequality (29), where $\theta_{i}$s satisfy

$\displaystyle\sin 2\theta_{1}\sin 2\theta_{2}+\sin 2\theta_{2}\sin
2\theta_{3}+\sin 2\theta_{1}\sin 2\theta_{3}>2\times{}2^{1/4}\approx 2.3784$
(55)

For special case of $\theta_{1}=\theta_{2}=\theta_{3}$, it follows that
$\theta>\frac{1}{2}\arcsin((\frac{32}{81})^{1/8})\approx 0.1748\pi$, where
$\theta$ satisfies $\cos\theta=\frac{1}{\sqrt{1+(\sin 2\theta)^{4}}}\approx
0.59$. If $\theta_{1}=\theta_{2}=\frac{\pi}{4}$, it follows that
$\theta_{3}>0.121\pi$. The same visibility can be followed from the
inequalities (28) and (29).

##  Appendix E Robustness of the inequality (2)

The goal in this section is to prove the result for noisy networks. Consider
an $n$-partite connected network ${\cal N}_{q}$ consisting of noisy states
$\rho_{i},\varrho_{j}$, where $\rho_{i}$ and $\varrho_{i}$ are Werner states
Wrner  given respectively by

$\displaystyle\rho_{i}=v_{i}|\phi_{i}\rangle\langle\phi_{i}|+\frac{1-v_{i}}{4}\mathbbm{1}_{4}$
$\displaystyle\varrho_{i}=w_{i}|\psi_{i}\rangle\langle\psi_{i}|+\frac{1-w_{i}}{2^{k_{i}}}\mathbbm{1}_{2^{k_{i}}}$
(56)

where $|\phi_{i}\rangle$ is EPR state, $|\psi_{j}\rangle$ is GHZ state with
$k_{j}$ particles, $\mathbbm{1}_{j}$ is the identity operator with the rank
$j$, $v_{i},w_{j}\in(0,1]$, $i=1,\cdots,m_{1};j=1,\cdots,m_{2}$.

Result S3. The network ${\cal N}_{q}$ is genuinely $n$-partite nonlocal if
$v_{i},w_{j}$ satisfy the following inequality:

$\displaystyle\prod_{i=1}^{m_{1}}\prod_{j=1}^{m_{2}}v_{i}w_{j}>1-\frac{2-\sqrt{2}}{2N}$
(57)

with $N=2^{n-1}-1$.

Proof of Result S3. The proof is similar to these stated in Appendixes B and
C. Note that all observables are dichotomic. We get
$CH_{I}^{\prime}=\prod_{i=1}^{m_{1}}\prod_{j=1}^{m_{2}}v_{i}w_{j}CH_{I}$,
where $CH_{I}$ is defined in the inequality (2) for the network ${\cal N}_{q}$
without noise while $CH^{\prime}_{\vec{s}_{i}}$ denotes the same quantity
associated with noisy network shown in Eq.(56). From Eqs.(31)-(42) it implies
that

$\displaystyle\varpi_{q}$ $\displaystyle=$ $\displaystyle\sum_{i}{\cal L}_{i}$
(58) $\displaystyle\geq$ $\displaystyle
2N\sqrt{2}\prod_{i=1}^{m_{1}}\prod_{j=1}^{m_{2}}v_{i}w_{j}$ $\displaystyle>$
$\displaystyle 2^{n}\sqrt{2}-4\sqrt{2}+2$

when $v_{i},w_{j}$ satisfy the inequality of
$\prod_{i=1}^{m_{1}}\prod_{j=1}^{m_{2}}v_{i}w_{j}>1-\frac{2-\sqrt{2}}{2N}$.
Numeric simulations are shown in Fig.S7 for the visibility of $n$-partite
cyclic networks. $\Box$

Figure 7: Visibility of Werner states. Here, all $v_{i}$s and $w_{j}$s are
equal to each other. $n$ denotes the number of parties while $m=m_{1}+m_{2}$
denotes the number of entangled states.

##  Appendix F Proof of the inequality (3)

In this section, we prove the inequality (3) for connected quantum networks.
Consider a general network ${\cal N}$ (Fig.2) consisting of $n$ parties
$\textsf{A}_{1},\cdots,\textsf{A}_{n}$, who share $m$ sources
$\lambda_{1},\cdots,\lambda_{m}$. An eavesdropper may gain information related
to the outcome of $\textsf{A}_{i}$ by measuring the correlated system with
input $z_{i}$ and outcome $e_{i}$. Suppose that all the inputs of
$\textbf{x}=x_{1}\cdots{}x_{n}$ are accessible to the eavesdropper. Note that
in secure applications all the legal parties will not send out their
measurement outcomes before they begin to detect the eavesdropper. This means
that a smart eavesdropper will recover one outcome $a_{i}$ by using the prior
knowledge of $a_{i},\textbf{x},z_{i}$, i.e.,
$P(e_{i}|a_{i},\textbf{x},z_{i})$. Hence, we consider the variation distance
of two probability distributions
$\\{\prod_{i}P(e_{i}|a_{i},\textbf{x},z_{i})\\}$ and
$\\{\prod_{i}P(e_{i}|\textbf{x},z_{i})\\}$. From the non-signalling condition
we get that

$\displaystyle P(e_{i}|\textbf{x},z_{i})=P(e_{i}|z_{i}),$ $\displaystyle
P(e_{i}|a_{i},\textbf{x},z_{i})=P(e_{i}|a_{i},x_{i},z_{i})$ (59)

Denote
$D_{e}:=D(\prod_{i=1}^{n}P(e_{i}|a_{i};\textbf{x},z_{i}),\prod_{j=1}^{n}p(e_{j}|z_{j}))$
for convenience. From Eq.(59), the left side of the inequality (3) can be
rewritten into

$\displaystyle D_{e}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\sum_{\textbf{e}}p(e_{1}|a_{1};\textbf{x},z_{1})|\prod_{i=2}^{n}p(e_{i}|a_{i};\textbf{x},z_{i})-\prod_{i=2}^{n}p(e_{i}|z_{i})|$
(60)
$\displaystyle+\frac{1}{2}\sum_{\textbf{e}}|p(e_{1}|a_{1};\textbf{x},z_{1})-p(e_{1}|z_{1})|\prod_{i=2}^{n}p(e_{i}|z_{i})$

Here, the inequality (60) is followed from the triangle inequality
$|x-y|\leq|x-z|+|z-y|$. Moreover, we have

$\displaystyle D_{e}$ $\displaystyle\leq$ $\displaystyle
D(p(e_{1}|a_{1};{\textbf{x}},z_{1}),p(e_{1}|z_{1}))+D(\prod_{i=2}^{n}p(e_{i}|a_{i};{\textbf{x}},z_{i}),\prod_{i=2}^{n}p(e_{i}|z_{i}))$
(61) $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{n}D(p(e_{i}|a_{i};{\textbf{x}},z_{i}),p(e_{i}|z_{i}))$
(62) $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}D(p(e_{i}|a_{i};x_{i},x_{j},z_{i}),p(e_{i}|z_{i}))$
(63) $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{n}I_{2}(P(a_{i},a_{j}|x_{i},x_{j})$ (64)

Eq.(61) is from the normalization conditions of
$\sum_{e_{1}}p(e_{1}|z_{1})=1$, and
$\sum_{e_{j}}p(e_{j}|a_{j};{\textbf{x}},z_{j})=1$, $j=2,\cdots,n$. The
inequality (62) is obtained by a similar process of the inequality (60) with
$n-1$ iterations. Eq.(63) is followed from Eq.(59), where $\textsf{A}_{i}$ and
$\textsf{A}_{j}$ shares at least one hidden variable. The inequality (64) is
obtained by using the inequality of
$D(p(e_{i}|a_{i};x_{i},x_{j},z_{i}),p(e_{i}|z_{i}))\leq
I_{2}(P(a_{i},a_{j}|x_{i},x_{j}))$ with the chained Bell inequality 19 ; 51 ,
where
$I_{2}(P(a_{i},a_{j}|x_{i},x_{j}))=P(a_{i}=a_{j}|x_{i}=0;x_{j}=1)+\sum_{|x_{i}-x_{j}|=1}P(a_{i}\not=a_{j}|x_{i};x_{j})$,
$x_{i}$ and $a_{i}$ are the input and outcome of $\textsf{A}_{i}$, and $x_{j}$
and $a_{j}$ are the input and outcome of $\textsf{A}_{j}$.

Consider a quantum network ${\cal N}_{q}$ on which all parties have multiple
inputs and binary outcomes. The quantum violation $\varpi_{es}$ of practical
quantum correlations is shown as follows

$\displaystyle\varpi_{es}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}\sum_{I}[\langle(\textbf{M}_{I}+\hat{\textbf{M}}_{I})\textbf{M}_{\overline{I}}\rangle+\langle(\textbf{M}_{I}-\hat{\textbf{M}}_{I})\hat{\textbf{M}}_{\overline{I}}\rangle]$
(65) $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}\sum_{I}[|\langle(\textbf{M}_{I}+\hat{\textbf{M}}_{I})\textbf{M}_{\overline{I}}\rangle|+|\langle(\textbf{M}_{I}-\hat{\textbf{M}}_{I})\hat{\textbf{M}}_{\overline{I}}\rangle|]$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}\sum_{j\in
I}[|\langle(M_{x_{j}(I)}+\hat{M}_{x_{j}(I)}){M}_{x_{k_{j}}(\overline{I})}\rangle|+|\langle(M_{x_{j}(I)}-\hat{M}_{x_{j}(I)})\hat{{M}}_{x_{k_{j}}(\overline{I})}|]$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}\sum_{I}CH_{M_{j}{M}_{k_{j}}}$
(66)

Here, the inequality (65) follows from the inequality of $\langle
A\rangle\leq|\langle A\rangle|$. The inequality (65) is from the assumptions
of separable observables ($M_{x_{i}}\leq 1$) except for
$M_{x_{k_{j}}(\overline{I})}$ and $\hat{M}_{x_{k_{j}}(\overline{I})}$ given in
Eq.(36). $M_{x_{k_{j}}}$ and $\hat{M}_{x_{k_{j}}}$ denote two observables of
$\textsf{A}_{x_{k_{j}}}$ with $k_{j}\not\in{}I$ associated with $j\in I$.
$CH_{M_{j}{M}_{k_{j}}}$ in Eq.(66) is defined by

$\displaystyle
CH_{M_{j}{M}_{k_{j}}}=\langle(M_{x_{j}(I)}+\hat{M}_{x_{j}(I)}){M}_{x_{k_{j}}(\overline{I})}\rangle+\langle(M_{x_{j}(I)}-\hat{M}_{x_{j}(I)})\hat{{M}}_{x_{k_{j}}(\overline{I})}\rangle$

where we have
$\langle(M_{x_{j}(I)}+\hat{M}_{x_{j}(I)}){M}_{x_{k_{j}}(\overline{I})}\rangle\geq
0$ and
$\langle(M_{x_{j}(I)}-\hat{M}_{x_{j}(I)})\hat{{M}}_{x_{k_{j}}(\overline{I})}\rangle\geq
0$ in applications (Appendixes B and C).

From the evaluations in Eqs.(42) and (44), we obtain that

$\displaystyle\varpi_{es}\leq\lfloor{}\frac{N}{n}\rfloor\sum_{j=1}^{n}CH_{M_{j}{M}_{k_{j}}}+2\sqrt{2}(\frac{N}{n}-\lfloor{}\frac{N}{n}\rfloor)$
(67)

where $N=2^{n}-1$. From $\langle XY\rangle=2p(X=Y)-1$ it follows that
$I_{2}(P(a_{i},a_{j}|x_{i},x_{j}))=2-\frac{1}{2}CH_{X_{i}X_{j}}$. Using the
inequalities (64) and (67), $D_{e}$ is evaluated as follows

$\displaystyle D_{e}\leq$ $\displaystyle
2n-\frac{1}{2}\sum_{j=1}^{n}CH_{M_{j}M_{k_{j}}}$ $\displaystyle\leq$
$\displaystyle
2n-\frac{\varpi_{es}-2\sqrt{2}(n_{0}-\lfloor{}n_{0}\rfloor)}{2\lfloor{}n_{0}\rfloor}$
(68)

with $n_{0}=\frac{N}{n}$. This completes the proof.

##  Appendix G Classical simulation of quantum correlations with finite
classical communication and shared randomness

There are two assumptions for guaranteeing the security against quantum or
post-quantum eavesdroppers. One is that an eavesdropper may access all inputs,
and the other is that an eavesdropper (not a party) cannot communicate with
other parties. Otherwise, the bipartite quantum correlations can be precisely
simulated with shared classical random variables 34 ; 36 . A similar result
holds for multisource quantum networks. In fact, suppose that a general
network ${\cal N}_{q}$ in Fig.2(b) is $k$-independent, that is, there are $k$
parties $\textsf{A}_{1},\cdots,\textsf{A}_{k}$ (for simplicity) who do not
share any entangled state with each other. ${\cal N}_{q}$ is schematically
represented by a directed acyclic graph (DAG) (Fig.1(b)) with the same
configuration as ${\cal N}_{q}$ using random variables
$\lambda_{1},\cdots,\lambda_{m}$. $(\Omega_{i},\Sigma_{i},\mu_{i})$ denotes
the measure space of $\lambda_{i}$ with $i=1,\cdots,m$. Each observer
$\textsf{A}_{i}$ obtains states from some random sources
$\Lambda_{i}=\\{\lambda_{j_{1}},\cdots,\lambda_{j_{\ell_{i}}}\\}$. The outcome
$a_{i}$ of $\textsf{A}_{i}$ depends on the shared variables $\Lambda_{i}$ and
the type of local measurement $x_{i}$. The joint conditional distributions of
the outcomes conditional on the inputs are given

$\displaystyle
P_{c}(\textbf{a}|\textbf{x})=\int_{\times_{i=1}^{m}\Omega_{i}}\prod_{j=1}^{n}P_{c}(a_{j}|x_{j},\Lambda_{j})\prod_{i=1}^{m}d\mu_{i}(\lambda_{i})$
(69)

Define a classical correlation as

$\displaystyle
E_{c}(\vec{\textbf{x}})=\sum_{\Xi_{1}}P_{c}(\textbf{a}_{k}|\vec{\textbf{x}}_{k})-\sum_{\Xi_{2}}P_{c}(\textbf{a}_{k}|\vec{\textbf{x}}_{k})$
(70)

where $P_{c}(\textbf{a}_{k}|\vec{\textbf{x}}_{k})$ denote the marginal
probability of $\textbf{a}_{k}$ conditional on $\vec{\textbf{x}}_{k}$, and
$\Xi_{1}$ denotes all combinations with even number of output $-1$ while
$\Xi_{2}$ denotes all combinations with odd number of output $-1$,
$\textbf{a}=a_{1}\cdots a_{n}$, $\textbf{x}=x_{1}\cdots x_{n}$,
$\textbf{a}_{k}=a_{1}\cdots a_{k}$, and
$\vec{\textbf{x}}_{k}=\vec{x}_{1}\cdots{}\vec{x}_{k}$.

Consider special applications with a joint system in the state $\rho$, such as
multipartite QKD. All the independent parties of
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$ will finally perform qubit measurements
after the others’ local measurements and classical communication. The joint
conditional probability
$P(\textbf{a}|\vec{\textbf{x}}_{k},x_{k+1}\cdots{}x_{n})$ is defined

$\displaystyle
P_{q}(\textbf{a}|\vec{\textbf{x}}_{k},x_{k+1}\cdots{}x_{n})={\rm
tr}(\vec{x}_{1}\otimes\cdots\vec{x}_{n}\rho)$ (71)

where $\vec{x}_{i}$ is a qubit measurement vector under the Pauli basis
$(\sigma_{x},\sigma_{y},\sigma_{z})$. Similar to Eq.(70), define a quantum
correlation as

$\displaystyle
E_{q}(\vec{\textbf{x}})=\sum_{\Xi_{1}}P_{q}(\textbf{a}_{k}|\vec{\textbf{x}}_{k})-\sum_{\Xi_{2}}P_{q}(\textbf{a}_{k}|\vec{\textbf{x}}_{k})$
(72)

Consider a protocol completed in a device-independent manner as shown in
Fig.3(b). Suppose that $\textsf{E}_{i}$s share variables $\Lambda_{i}$
consisting of independent random variables $\lambda_{j}$s with the same
network configuration as ${\cal N}_{q}$, that is, all the parties
$\textsf{A}_{i}$ are replaced with $\textsf{E}_{i}$ in the hidden variable
model of ${\cal N}_{q}$. The goal in this section is to simulate the quantum
correlation $E_{q}(\vec{\textbf{x}})$ in Eq.(72) by using the classical
correlation $E_{c}(\vec{\textbf{x}})$ in Eq.(70).

Result S4. Suppose that eavesdroppers $\textsf{E}_{i}$s can access all
measurement inputs $\vec{\textit{x}}$. $\textsf{E}_{i}$s can generate a
classical correlation $E_{c}(\vec{\textbf{x}})$ such that

$\displaystyle E_{c}(\vec{\textbf{x}})=E_{q}(\vec{\textbf{x}})$ (73)

by using shared randomness and finite classical communication, where the
classical correlations are defined by
$E_{c}(\vec{\textbf{x}})=\sum_{\Xi_{1}}P_{c}(\textbf{e}_{k}|\vec{\textbf{x}}_{k})-\sum_{\Xi_{2}}P_{c}(\textbf{e}_{k}|\vec{\textbf{x}}_{k})$,
$P_{c}(\textbf{e}_{k}|\vec{\textbf{x}}_{k})$ denotes the joint distribution of
eavesdroppers $\textsf{E}_{1},\cdots,\textsf{E}_{k}$ conditional on the inputs
$\vec{\textbf{x}}_{k}$, and their own inputs
$\textbf{e}_{k}=e_{1}\cdots{}e_{k}$.

Remark S2. The classical simulation requires only finite classical
communication for $\textsf{E}_{i}$s (see Fig.2(b)). Here, eavesdroppers cannot
simulate the joint distribution $P({\bf a}|{\bf x})$ or the exact outputs
${\bf a}$ for each set of inputs ${\bf x}$. That is, the present simulation in
Eq.(73) provides only potential information leakage of correlations based on
generalized entanglement swapping going beyond chain-shaped networks without
correlating assumptions 36 . However, it does not imply the information
leakage of measurement outputs in device-independent applications.

Figure 8: Schematically long-distance entanglement swapping network. (a) Long-
distance entanglement swapping network consisting of $k$ independent sources
$\rho_{i}=|\phi_{i}\rangle\langle\phi_{i}|$ with generalized EPR states
$|\phi_{i}\rangle=\alpha_{i}|01\rangle-\beta_{i}|10\rangle$. All the middle
observers $\textsf{A}_{2},\cdots,\textsf{A}_{k}$ can help $\textsf{A}_{1}$ and
$\textsf{A}_{k+1}$ to build single entangled state with LOCC. (b) Classical
simulation of correlations generated by a long-distance entanglement swapping
experiment. Eavesdroppers share $n$ random variables
$\lambda_{1},\cdots,\lambda_{k}$. In simulation experiment, eavesdroppers can
access to all measurement inputs and finite classical communication. The red
dashed arrows represent to exchange classical messages.

Proof of Result S4. Consider a general quantum network ${\cal N}_{q}$
(Fig.2(a) shown in the main text), which consists of all entangled pure
states. Suppose that there are $k$ independent observers
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$, who do not share entangled states. Our
goal here is to show classical simulation of the multipartite correlations by
eavesdroppers who can access all inputs and commutate with each other. A
classical simulation with the same network configuration is shown in Fig.2(b)
in the main text. One example of Fig.2 is the long-distance entanglement
swapping network given in Fig.S8(a). In experiment, all the middle observers
$\textsf{A}_{2},\cdots,\textsf{A}_{k}$ perform Bell measurements
$\sigma_{z}\otimes\sigma_{z},\sigma_{x}\otimes\sigma_{x}$ and send the
outcomes to $\textsf{A}_{1}$ and $\textsf{A}_{k+1}$ who can recover a single
entangled state with local unitary operations. To complete the simulation
shown in Fig.S8(b), suppose that $\textsf{A}_{1}$ and $\textsf{A}_{k+1}$
perform the qubit projective measurements represented by respective Bloch
vector $\vec{x}_{1},\vec{x}_{k+1}$ in terms of the Pauli basis
$(\sigma_{x},\sigma_{y},\sigma_{z})$. Suppose that $\textsf{A}_{1}$ and
$\textsf{A}_{k+1}$ obtain binary outcomes $a_{1},a_{k+1}=\pm 1$ respectively.
The correlations of $\textsf{A}_{1}$ and $\textsf{A}_{k+1}$ exhibit the
following form:

$\displaystyle E_{q}(\vec{x}_{1},\vec{x}_{k+1}):=$
$\displaystyle\sum_{\Xi_{1}}P_{q}(\textbf{a}|\vec{\textbf{x}})-\sum_{\Xi_{2}}P_{q}(\textbf{a}|\vec{\textbf{x}})$
(74)

where $P_{q}(\textbf{a}|\vec{\textbf{x}})$ is defined in Eq.(72). An open
problem 3  is to determine the classical simulation capability of
$E_{q}(\vec{x}_{1},\vec{x}_{k+1})$ for $n\geq 3$ with independent sources and
finite communication.

Generally, the proof of Eq.(73) is divided into three cases. One is to prove
classical simulation of long-distance entanglement swapping shown in Fig.S8(a)
with EPR states for $k\geq 1$. The second is to prove classical simulation of
generalized entanglement swapping for any acyclic networks consisting of EPR
states. The last one is to prove the result for any acyclic networks
consisting of generalized EPR states.

Case 1. Long-distance entanglement swapping on chain-shaped networks

For $k=1$, the network shown in Fig.S8(a) reduces to standard Bell network 34
. For $k=2$, it is a standard entanglement swapping network that can be
simulated classically 36 . In what follows, we prove the result for $k>2$. The
proof is completed by induction $k$.

We firstly prove the result for $k=3$, as shown in Fig.S9. $\textsf{A}_{i}$
and $\textsf{A}_{i+1}$ share one or two EPR states shown in Fig.S9(a).
$\textsf{A}_{1}$ and $\textsf{A}_{4}$ obtain the respective measurement input
$\vec{x}_{1}=(\cos\theta_{a},0,\sin\theta_{a})$ and
$\vec{x}_{4}=(\cos\theta_{b},0,\sin\theta_{b})$. The correlation given in
Eq.(74) is rewritten into

$\displaystyle{E}_{q}(\vec{x}_{1},\vec{x}_{4})$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{q}(\textbf{a}|\vec{\textbf{x}})-\sum_{\Xi_{2}}P_{q}(\textbf{a}|\vec{\textbf{x}})$
(75) $\displaystyle=$ $\displaystyle\cos(\theta_{a}-\theta_{b})$

Figure 9: Classical simulation of quantum correlations from entanglement
swapping. (a) Entanglement swapping with $4$ observers
$\textsf{A}_{1},\cdots,\textsf{A}_{4}$, who share EPR states
$\rho_{1}\otimes\rho_{2}\otimes\rho_{3}$. (b) The classical simulation of
quantum correlations generated by entanglement swapping by eavesdroppers
$\textsf{E}_{1},\cdots,\textsf{E}_{4}$ who share random variables
$\lambda_{1},\lambda_{2},\lambda_{3}$. The classical communication is
represented by red dashed arrows. The classical simulation is completed by two
simulations S1 and S2 without assumptions of correlating sources.

The classical simulation is shown in Fig.S9(b), where $\textsf{E}_{i}$ and
$\textsf{E}_{i+1}$ share one or two random variables $\lambda_{i}$s with the
same network configuration as shown in Fig.S9(a). The finite classical
communication is allowed for $\textsf{E}_{i}$s who know the inputs of
$\textsf{A}_{1}$ and $\textsf{A}_{4}$. The simulation protocol consists of two
separable simulations S1 and S2 as follows:

  * S1

$\textsf{E}_{i}$s set up the first classical simulation according to finite
classical communication.

    * $\bullet$

$\textsf{E}_{1},\textsf{E}_{2}$, and $\textsf{E}_{3}$ obtain a classical
correlation as follows

$\displaystyle{E}_{c}(\vec{x}^{\prime}_{1},\vec{y}_{1})$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}|\vec{x}^{\prime}_{1},\vec{y}_{1})-\sum_{\Xi_{2}}P_{c}(\textbf{e}|\vec{x}^{\prime}_{1},\vec{y}_{1})$
(76) $\displaystyle=$ $\displaystyle 2\cos\theta_{a}$

from a tripartite classical simulation protocol 36  with independent sources
and finite communication. Here, the inputs are assumed to be
$\vec{x}^{\prime}_{1}=(2\cos\theta_{a},0,0)$ and $\vec{y}_{1}=(1,0,0)$.

    * $\bullet$

$\textsf{E}_{3}$ and $\textsf{E}_{4}$ obtain a classical correlation as
follows

$\displaystyle{E}_{c}(\vec{y}_{1},\vec{x}^{\prime}_{4})$ $\displaystyle=$
$\displaystyle
P_{c}(e_{3}=e_{4}|\vec{y}_{1},\vec{x}^{\prime}_{4})-P_{c}(e_{3}\not=e_{4}|\vec{y}_{1},\vec{x}^{\prime}_{4})$
(77) $\displaystyle=$ $\displaystyle\cos\theta_{b}$

from a bipartite classical simulation protocol 34 , where
$\vec{x}^{\prime}_{4}=(\cos\theta_{b},0,0)$. Note that two steps are
independent. It follows that

$\displaystyle{E}_{c}(\vec{x}^{\prime}_{1},\vec{x}^{\prime}_{4})$
$\displaystyle=$ $\displaystyle
E_{c}(\vec{x}^{\prime}_{1},\vec{y}_{1})E_{c}(\vec{y}_{1},\vec{x}^{\prime}_{4})$
(78) $\displaystyle=$ $\displaystyle 2\cos\theta_{a}\cos\theta_{b},$

which can be easily followed from the separable Bell measurement of
$\textsf{A}_{3}$, i.e., $\langle
M_{x_{1}}M_{x_{2}}M_{x_{3}}M_{x_{4}}\rangle=\langle{}M_{x_{1}}M_{x_{2}}M_{x_{3};1}\rangle\langle{}M_{x_{3};2}M_{x_{4}}\rangle$
with separable measurement $M_{x_{3};1}\otimes M_{x_{3};2}$ of
$\textsf{A}_{3}$, where $M_{x_{1}}$ is observable with dichotomic outcomes.
Equivalently, one can prove the result from the conditional independence of
full correlations
$P_{q}(\textbf{e}|\vec{x})=P_{q}(\textbf{e}_{1}|\vec{x}_{1})P_{q}(\textbf{e}_{2}|\vec{x}_{2})$,
where $P_{q}(\textbf{e}_{1}|\vec{x}_{1})$ and
$P_{q}(\textbf{e}_{2}|\vec{x}_{2})$ denote the respective correlation obtained
from the tripartite network (consisting of $\textsf{E}_{1},\textsf{E}_{2}$,
and $\textsf{E}_{3}$) and the bipartite network (consisting of
$\textsf{E}_{3}$ and $\textsf{E}_{4}$). Here, $\textsf{E}_{3}$ in Fig.S9(b) or
$\textsf{A}_{3}$ in Fig.S9(a) performs the separable measurement.

  * S2

$\textsf{E}_{i}$s set up the second classical simulation according to finite
classical communication.

    * $\bullet$

$\textsf{E}_{1},\textsf{E}_{2}$, and $\textsf{E}_{3}$ obtain a classical
correlation as follows

$\displaystyle{E}_{c}(\vec{x}^{\prime\prime}_{1},\vec{y}_{1})$
$\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}|\vec{x}^{\prime\prime}_{1},\vec{y}_{1})-\sum_{\Xi_{2}}P_{c}(\textbf{e}|\vec{x}^{\prime\prime}_{1},\vec{y}_{1})$
(79) $\displaystyle=$ $\displaystyle 2\sin\theta_{a}$

from a tripartite classical simulation protocol 36  by using independent
sources and finite communication, where their inputs are assumed to be
$\vec{x}^{\prime\prime}_{1}=(0,0,2\sin\theta_{a})$ and $\vec{y}_{1}=(0,0,1)$.

    * $\bullet$

$\textsf{E}_{3}$ and $\textsf{E}_{4}$ obtain a classical correlation as
follows

$\displaystyle{E}_{c}(\vec{y}_{1},\vec{x}^{\prime\prime}_{4})$
$\displaystyle=$ $\displaystyle
P_{c}(e_{3}=e_{4}|\vec{y}_{1},\vec{x}^{\prime\prime}_{4})-P_{c}(e_{3}\not=e_{4}|\vec{y}_{1},\vec{x}^{\prime\prime}_{4})$
(80) $\displaystyle=$ $\displaystyle\sin\theta_{b}$

from a bipartite classical simulation protocol 34  without correlating
sources, where $\vec{x}^{\prime\prime}_{4}=(0,0,\sin\theta_{b})$. These two
steps are independent. It follows that

$\displaystyle{E}_{c}(\vec{x}^{\prime\prime}_{1},\vec{x}^{\prime\prime}_{4})$
$\displaystyle=$
$\displaystyle{E}_{c}(\vec{x}^{\prime\prime}_{1},\vec{y}_{1}){E}_{c}(\vec{y}_{1},\vec{x}^{\prime\prime}_{4})$
(81) $\displaystyle=$ $\displaystyle 2\sin\theta_{a}\sin\theta_{b}$

From Eqs.(78) and (81), it follows that

$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}|\vec{\textbf{x}})-\sum_{\Xi_{2}}P_{c}(\textbf{e}|\vec{\textbf{x}})$
$\displaystyle:=$
$\displaystyle\frac{1}{2}({E}_{c}(\vec{x}^{\prime}_{1},\vec{x}^{\prime}_{4})+{E}_{c}(\vec{x}^{\prime\prime}_{1},\vec{x}^{\prime\prime}_{4}))$
(82) $\displaystyle=$ $\displaystyle\cos(\theta_{a}-\theta_{b})$
$\displaystyle=$ $\displaystyle{E}_{q}(\vec{x}_{1},\vec{x}_{4})$

from Eq.(75), where S1 and S2 are two independent simulations which are
recombined into one simulation with equal probability. No correlating
operation is required for any sources. In this protocol, total classical
communication is finite because each simulation uses finite communication.

Now, by induction we can prove the result for any $k\geq 3$, where the
classical simulations are divided into two separable chained subnetworks
consisting of $\textsf{E}_{1},\cdots,\textsf{E}_{k}$, and $\textsf{E}_{k}$ and
$\textsf{E}_{k+1}$, respectively. Similar result holds for generalized
measurement in terms of the Pauli basis 34 ; 36  when all measurement inputs
are accessible to eavesdroppers without assumption of correlating sources.

Figure 10: Classical simulation of quantum correlations from generalized
entanglement swapping. (a) A generalized entanglement swapping on any acyclic
connected quantum network ${\cal N}_{q}$. The resources consist of generalized
EPR states $\otimes_{i}|\phi_{i}\rangle$ with
$|\phi_{i}\rangle=\alpha_{i}|01\rangle+\beta_{i}|10\rangle$.
$\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$ perform the multiple-particle Bell
measurement on local systems and send outcomes to other observers
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$, who can recover a multipartite
generalized GHZ state by proper local unitary operations. (b) The classical
simulation of generalized entanglement swapping by non-signalling
eavesdroppers $\textsf{E}_{1},\cdots,\textsf{E}_{n}$ who share variables
$\lambda_{i}$s. The red dashed arrows represent exchanging classical messages.
The classical simulation network is decomposed into $k$ chain-shaped
subnetworks ${\cal N}_{1},\cdots,{\cal N}_{k}$. Finite classical communication
is allowed for all eavesdroppers.

Case 2. Generalized entanglement swapping on any connected acyclic quantum
networks

Consider a connected acyclic quantum network ${\cal N}_{q}$ consisting of EPR
states $|\phi_{i}\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$, as shown
in Fig.S10(a). One goal of ${\cal N}_{q}$ is that all parties
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$ who have no prior-shared entanglement
create a multipartite GHZ state with the help of other parties
$\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$. In experiment, each party
$\textsf{A}_{k+j}$ performs a multi-particle Bell measurement on the local
system and sends outcome for other parties $\textsf{A}_{i}$s, who can recover
a $k$-particle GHZ state $|GHZ\rangle=\frac{1}{\sqrt{2}}(|0\rangle^{\otimes
k}+|1\rangle^{\otimes k})$ by performing proper local unitary operations.
Suppose that all parties $\textsf{A}_{i}$s finally obtain binary outcomes
$a_{i}\in\\{-1,1\\}$ after performing local qubit measurements. Here,
$\textsf{A}_{i}$s perform the single qubit measurement with measurement input
$\vec{x}_{i}=(\cos\theta_{i},0,\sin\theta_{i})$ on Bloch sphere. Their
outcomes have correlations as follows:

$\displaystyle E_{q}(\vec{\textbf{x}})$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{q}(\textbf{a}|\vec{\textbf{x}})-\sum_{\Xi_{2}}P_{q}(\textbf{a}|\vec{\textbf{x}})$
(83) $\displaystyle=$
$\displaystyle\prod_{i=1}^{k}\cos\theta_{i}+\prod_{i=1}^{k}\sin\theta_{i}$

where $\vec{\textbf{x}}=\vec{x}_{1}\cdots\vec{x}_{k}$ are measurement vectors
under the Pauli basis of single qubit system.

In what follows, we show that these correlations can be simulated classically,
as shown in Fig.S10(b) with even $k$. Similar proof holds for odd $k$. The
classical simulation is completed by powerful eavesdroppers $\textsf{E}_{i}$s
who can access all the measurement inputs with fully independent variables
$\lambda_{j}$s. The proof is completed by the following separable steps:

  * S1

Eavesdroppers divide the simulation network shown in Fig.S10(b) into $n$
chain-shaped subnetworks ${\cal N}_{1},\cdots,{\cal N}_{k}$ with the help of
classical communication, where each subnetwork ${\cal N}_{i}$ contains
$\textsf{E}_{i}$ and $\textsf{E}_{k+j}$, and all subnetworks are connected and
${\cal N}_{i}\cap{\cal N}_{j}\not=\varnothing$.

  * S2

For the subnetwork ${\cal N}_{1}$, eavesdroppers set up the first simulation
according to finite classical communication. They obtain classical correlation
as follows

$\displaystyle{E}_{c}^{1}$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}_{1}|\vec{\textbf{x}}_{1})-\sum_{\Xi_{2}}P_{c}(\textbf{e}_{1}|\vec{\textbf{x}}_{1})$
(84) $\displaystyle=$ $\displaystyle 2\cos\theta_{1}$

from the multipartite classical simulation protocol given in Case 1, where
$\vec{\textbf{x}}_{1}$ and $\textbf{e}_{1}$ denote the respective inputs and
outcomes of all eavesdroppers included in the subnetwork ${\cal N}_{1}$, the
input of the eavesdropper $\textsf{E}_{1}$ is given by
$\vec{x}^{\prime}_{1}=(2\cos\theta_{1},0,0)$ while the inputs of other
eavesdroppers are given by $(1,0,0)$. Note that this simulation procedure does
not require to correlate variables $\lambda_{i}$s.

  * S3

By repeating S2 for all subnetworks ${\cal N}_{i}$, eavesdroppers can obtain
classical correlation as follows

$\displaystyle{E}_{c}^{i}$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}_{i}|\vec{\textbf{x}}_{i})-\sum_{\Xi_{2}}P_{c}(\textbf{e}_{i}|\vec{\textbf{x}}_{i})$
(85) $\displaystyle=$ $\displaystyle\cos\theta_{i}$

for $i=2,\cdots,n$, where $\vec{\textbf{x}}_{i}$ and $\textbf{e}_{i}$ denote
the respective inputs and outcomes of eavesdroppers included in the subnetwork
${\cal N}_{i}$, the input of the eavesdropper $\textsf{E}_{i}$ is given by
$\vec{x}^{\prime}_{i}=(\cos\theta_{i},0,0)$ while the inputs of all the other
eavesdroppers are given by $(1,0,0)$ in the subnetwork ${\cal N}_{i}$. Note
that these steps are independent. It implies that

$\displaystyle E_{c}(\vec{\textbf{x}}^{\prime})$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}|\vec{\textbf{x}}^{\prime})-\sum_{\Xi_{2}}P_{c}(\textbf{e}|\vec{\textbf{x}}^{\prime})$
(86) $\displaystyle=$ $\displaystyle\prod_{i=1}^{n}{E}_{c}^{i}$
$\displaystyle=$ $\displaystyle 2\prod_{i=1}^{k}\cos\theta_{i}$

which can be followed from the Bell observables of $\textsf{A}_{k+j}$s shown
in Fig.S10(b), where $\vec{\textbf{x}}^{\prime}$ and e denote the respective
inputs and outputs of eavesdroppers included in ${\cal N}_{q}$.

  * S4

Eavesdroppers set up the second classical simulation with the help of finite
classical communication. They obtain multipartite correlations

$\displaystyle\hat{E}_{c}^{i}$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}^{\prime}_{i}|\vec{\textbf{x}}^{\prime\prime}_{i})-\sum_{\Xi_{2}}P_{c}(\textbf{e}^{\prime}_{i}|\vec{\textbf{x}}^{\prime\prime}_{i})$
(87) $\displaystyle=$ $\displaystyle\sin\theta_{i}$

for $i\geq 2$ and $2\sin\theta_{1}$ for $i=1$ from the classical simulation
protocol given in Case 1, where the input of $\textsf{E}_{i}$ is given by
$\vec{x}^{\prime\prime}_{i}=(0,0,\sin\theta_{i})$ for $i\geq 2$ or
$\vec{x}^{\prime\prime}_{1}=(0,0,2\sin\theta_{1})$ while the inputs of other
eavesdroppers are given by $(0,0,1)$ in the subnetwork ${\cal N}_{i}$,
$i=1,\cdots,n$. Note that these steps are independent. It follows that

$\displaystyle\hat{E}_{c}(\vec{\textbf{x}}^{\prime\prime})$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{k}\hat{E}_{c}^{i}$ (88) $\displaystyle=$
$\displaystyle 2\prod_{i=1}^{k}\sin\theta_{i}$

Finally, from Eqs.(86) and (88) it implies that

$\displaystyle\sum_{\Xi_{1}}P_{c}(\textbf{e}|\vec{\textbf{x}})-\sum_{\Xi_{2}}P_{c}(\textbf{e}|\vec{\textbf{x}})$
$\displaystyle:=$
$\displaystyle\frac{1}{2}(\hat{E}_{c}(\vec{\textbf{x}}^{\prime})+\hat{E}_{c}(\vec{\textbf{x}}^{\prime\prime}))$
(89) $\displaystyle=$
$\displaystyle\prod_{i=1}^{k}\cos\theta_{i}+\prod_{i=1}^{k}\sin\theta_{i}$
$\displaystyle=$ $\displaystyle E_{q}(\vec{\textbf{x}})$

from Eq.(83), where two independent simulations are recombined into one
simulation with equal probability. Similar result holds for generalized
measurement in terms of the Pauli basis 34 ; 36 .

Figure 11: Decomposing simulation networks. (a) An acyclic network consisting
of five random variables. (b) A network with one cycle consisting of 9 random
variables. (c) A network with two cycles consisting of 11 random variables.
Each variable is represented by two dots (or one dot and one circle) connected
by one line. All the empty circles denote independent observers who want to
create an entanglement in terms of the quantum network. The shadowed node
denotes the root of the equivalent tree. The right side of each subnetwork
consists of several decomposed simulation networks with several chain-shaped
subnetworks, where each subnetwork is represented by one colored path. The
dotted lines denote the reused random variables. The grey figures are useless
for generalized entanglement swapping experiment. The proof holds for special
cyclic networks shown in Fig.S5(b) and Fig.S5(c), where all the independent
agents are not included in any cycle.

Case 3. Entanglement swapping on acyclic networks consisting of generalized
EPR states

Consider an $n$-partite connected acyclic quantum network ${\cal N}_{q}$
consisting of generalized EPR states
$|\psi_{i}\rangle=\alpha_{i}|00\rangle+\sqrt{1-\alpha_{i}^{2}}|11\rangle$, as
shown in Fig.S10(a). One goal of ${\cal N}_{q}$ is to help all parties
$\textsf{A}_{i}$s to build a generalized multipartite GHZ state assisted by
local measurements of other parties $\textsf{A}_{k+j}$, $i=1,\cdots,k$;
$j=1,\cdots,n-k$. In experiment, each party $\textsf{A}_{k+j}$ performs a
multi-particle Bell measurement on the local systems and sends outcome to
$\textsf{A}_{i}$s who can recover a generalized $k$-partite GHZ state
$|GHZ\rangle=\gamma|0\rangle^{\otimes k}+\sqrt{1-\gamma^{2}}|1\rangle^{\otimes
k}$ with success probability $p$ by performing proper local unitary
operations, where $p$ and $\gamma$ depend on $\alpha_{i}$s and $\beta_{i}$s.
Note that the probability distribution $p$ can be generated by one party in a
classical simulation experiment. So, it is sufficient to prove the result for
one generalized GHZ state $|GHZ\rangle$. Suppose that $\textsf{A}_{i}$s obtain
binary outcomes $a_{i}\in\\{-1,1\\}$. In the classical simulation,
$\textsf{A}_{i}$s finally perform the single qubit measurement with
measurement input $\vec{x}_{i}=(\cos\theta_{i},0,\sin\theta_{i})$. Their
outcomes for even $k$ exhibit nonlocal correlations of the following form:

$\displaystyle E_{q}(\vec{\textbf{x}})$ $\displaystyle=$
$\displaystyle\sum_{\Xi_{1}}P_{q}(\textbf{a}|\vec{\textbf{x}})-\sum_{\Xi_{2}}P_{q}(\textbf{a}|\vec{\textbf{x}})$
(90) $\displaystyle=$ $\displaystyle
2\gamma\sqrt{1-\gamma^{2}}\prod_{i=1}^{k}\cos\theta_{i}+\prod_{i=1}^{k}\sin\theta_{i}$

Consider a classical network ${\cal N}$ consisting of $m$ variables
$\lambda_{i}$s with the similar network configuration as ${\cal N}_{q}$.
Suppose that ${\cal N}$ is shared by classical eavesdroppers
$\textsf{E}_{i}$s. For classical simulation, eavesdroppers can perform a
probabilistic simulation with an input $(2\gamma\sqrt{1-\gamma^{2}},0,0)$.
From a similar procedure of Case 2, it is easy to prove that
$E(\vec{\textbf{x}})$ given in Eq.(90) can be simulated classically using
several independent simulations by eavesdroppers who can access to all
measurement inputs.

Before closing the proof, we show that S1 in Case 2 can be completed with
finite classical communication, where other steps require finite classical
communication from Case 2. Note that ${\cal N}$ is acyclic Graph . There is
one node that is connected to each agent $\textsf{E}_{i}$ with only one path.
This path is contained in the subnetwork ${\cal N}_{i}$. In theory, an acyclic
graph is equivalent to a tree where the root is the desired center node, and
some edges are allowed to be in different paths, i.e., one random variable can
be used more than one time in classical simulation. Some examples are shown in
Fig.S11(a). Generally, a similar result holds for special cyclic networks
shown in Fig.S11(b) and Fig.S11(c), where all the independent agents are not
included in any cycle. This completes the proof. $\Box$

##  Appendix H Proof of the inequality (4)

From the inequality (2), the expect violation of quantum correlations is
represented by

$\displaystyle\varpi_{es}$ $\displaystyle=$
$\displaystyle\sum_{a_{i},b_{i},c_{i}\atop{x_{i},y_{i},z_{i}\in{S}_{in}}}[(-1)^{a_{1}+b_{1}+c_{1}+x_{1}y_{1}z_{1}}P(a_{1}b_{1}c_{1}|x_{1}y_{1}z_{1})$
(91)
$\displaystyle+(-1)^{a_{2}+c_{2}+b_{2}+x_{2}y_{2}z_{2}}P(a_{2}b_{2}c_{2}|x_{2}y_{2}z_{2})+(-1)^{a_{3}+b_{3}+c_{3}+x_{3}y_{3}z_{3}}P(a_{3}b_{3}c_{3}|x_{3}y_{3}z_{3})]$

where
$P(a_{i}b_{i}c_{i}|x_{i}y_{i}z_{i})=\textrm{tr}[(A^{a_{i}}_{x_{i}}B^{b_{i}}_{y_{i}}C^{c_{i}}_{z_{i}})\rho]$,
$A^{a_{i}}_{x_{i}}$, $B^{b_{i}}_{y_{i}}$ and $C^{c_{i}}_{z_{i}}$ denote the
POVM operators of $\textsf{A}_{1}$, $\textsf{A}_{2}$ and $\textsf{A}_{3}$,
respectively, $\rho$ denotes the total state of the triangle network (Fig.3),
and ${S}_{in}$ denotes the input set satisfying the conditions $x_{1}=y_{1}$,
$x_{2}=z_{2}$ and $y_{3}=z_{3}$. From the symmetry of inputs $x_{i}$, $y_{i}$,
and $z_{i}$, it follows that

$\displaystyle\varpi_{es}=3\\!\\!\\!\\!\\!\\!\sum_{a_{1},b_{1},c_{1}\atop{x_{1},y_{1},z_{1}\in{S}_{in}}}\\!\\!\\!\\!\\!\\!(-1)^{a_{1}+b_{1}+c_{1}+x_{1}y_{1}z_{1}}P(a_{1}b_{1}c_{1}|x_{1}y_{1}z_{1})$
(92)

In what follows, we only consider the case of $a_{1}=0$ and $x_{1}=0$ to prove
the following inequality

$\displaystyle
P(a_{1}|x_{1})\leq\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi^{2}_{es}}$ (93)

which is inspired by bipartite QKD scheme 20 ; 52 . Note that the right side
of the inequality (4) is concave function of $\varpi_{es}$. Denote
$W=\sum_{a_{1},b_{1},c_{1}\atop{x_{1},y_{1},z_{1}\in{\cal{}S}_{in}}}(-1)^{a_{1}+b_{1}+c_{1}+x_{1}y_{1}z_{1}}A^{a_{1}}_{x_{1}}B^{b_{1}}_{y_{1}}C^{c_{1}}_{z_{1}}$
as tripartite quantum operator. Inspired by the semidefinite programming
relaxation of the bipartite scheme 20 ; 52 , from Eq.(92) the inequality (93)
is equivalent to the following operator inequality:

$\displaystyle
A^{0}_{0}\leq\frac{1}{2}+\frac{6}{\sqrt{72-\varpi^{2}}}-\frac{\varpi}{12\sqrt{72-\varpi^{2}}}W$
(94)

This operator inequality is then rewritten into

$\displaystyle\frac{1}{2}+\frac{1}{2}A_{0}\leq\frac{1}{2}+\frac{6}{\sqrt{72-\varpi^{2}}}-\frac{\varpi}{12\sqrt{72-\varpi^{2}}}W$
(95)

where $A_{0}=A^{0}_{0}-A^{1}_{0}$ is Hermitian operator.

Now, it is sufficient to prove the operator inequality (95) for the guessing
probability in Eq.(93). Define four operators $L_{i}$s as follows 20 ; 51 :

$\displaystyle L_{1}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{\gamma}}{2}A_{1}+\frac{\varpi\sqrt{\gamma}}{24}(B_{0}C_{0}-B_{1}C_{1})+\frac{\varpi}{48\sqrt{\gamma}}(A_{1}B_{0}C_{0}+A_{1}B_{1}C_{1}),$
(96) $\displaystyle L_{2}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{\gamma}}{8}-\frac{\varpi}{24\sqrt{\gamma}}A_{0}+\frac{1}{8\sqrt{\gamma}}(B_{0}C_{0}+B_{1}C_{1})$
(97)
$\displaystyle-\frac{\sqrt{\gamma}}{2}(A_{0}B_{0}C_{0}+A_{0}B_{1}C_{1})+\frac{(36-\varpi^{2})\sqrt{\gamma}}{144}(A_{1}B_{0}C_{0}-A_{1}B_{1}C_{1})$
$\displaystyle L_{3}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{\gamma}}{4}(A_{0}B_{0}C_{0}-A_{0}B_{1}C_{1})-\frac{1}{8\sqrt{\gamma}}(B_{0}C_{0}-B_{1}C_{1})$
(98)
$\displaystyle-\frac{(36-\varpi^{2})\sqrt{\gamma}}{144}(A_{1}B_{0}C_{0}+A_{1}B_{1}C_{1})$

$\displaystyle L_{4}$ $\displaystyle=$
$\displaystyle\frac{1}{4\sqrt{\gamma}}-\frac{(36-\varpi^{2})\sqrt{\gamma}}{72}A_{0}-\frac{\sqrt{\gamma}\varpi}{24}(B_{0}C_{0}+B_{1}C_{1})$
(99)
$\displaystyle-\frac{\varpi}{48\sqrt{\gamma}}(A_{1}B_{0}C_{0}-A_{1}B_{1}C_{1})$

where $\gamma=\frac{3}{\sqrt{72-\varpi^{2}}}$. It is from a forward evaluation
to check that

$\displaystyle\sum_{i=1}^{4}L^{\dagger}_{i}L_{i}$ $\displaystyle=$
$\displaystyle\frac{6}{\sqrt{72-\varpi^{2}}}-\frac{1}{2}A_{0}+\frac{\varpi}{12\sqrt{72-\varpi^{2}}}W$
(100) $\displaystyle\geq$ $\displaystyle 0$ (101)

for $\varpi<6\sqrt{2}$. Eq.(100) makes use of the equalities
$A_{x}^{2}=B_{y}^{2}=C_{z}^{2}=\mathbbm{1}$, and
$[A_{x},B_{y}]=[A_{x},C_{z}]=[B_{y},C_{z}]=0$ (commute relations). The
inequality (101) follows from the fact that the right side of Eq.(100) is
positive semidefinite, i.e., $\sum_{i=1}^{4}L^{\dagger}_{i}L_{i}\geq 0$.
Finally, the inequality (95) is obtained from the concavity of its right side
for $\varpi$ with $\varpi\in[4\sqrt{2}+2,6\sqrt{2}]$.

##  Appendix I Proof of the inequality (5)

The length of the real secret $k$ is lower bounded by $H_{min}(A|E)-s_{p}$ (up
to terms of order $\sqrt{s}$), where $H_{min}(A|E)$ denotes the min-entropy 20
; 39  of $A$ depending on Eve’s information, $s_{p}\approx sH(A|B)$ is used
for error correcting 20 ; 39 , $H(A|B)=-\sum_{a,b}p(a,b)\log_{2}p(a|b)$
denotes the conditional entropy depending on the average probability with
specific outcomes $a,b$ being observed. Hence, it is essentially to get the
asymptotic secret-key rate $R$ in Eq.(5) by evaluating $H_{min}(A|E)$ 20 .
After the local measurement of $\textsf{A}_{1}$, the joint state of
$\textsf{A}_{1}$ and Eve can be represented by $\rho_{AE}=\sum_{\bf a}P({\bf
a}|{\bf x}_{r})|{\bf a}\rangle\langle{\bf a}|\otimes\rho_{E}^{\bf a}$, where
$\rho_{E}^{\bf a}$ denotes the reduced state owned by Eve conditional on
$\textsf{A}_{1}$’s output ${\bf a}$. Let $H_{min}({\bf a}|E)$ be the min-
entropy 20 ; 39  of ${\bf a}$ depending on Eve’s information related to
$\rho_{AE}$.

Now, we estimate the guessing probability of eavesdropper Eve about the raw
key ${\bf a}=a_{1}\cdots{}a_{s}$ using $s$ pairs of the triangular network
(Fig.3). Let $P({\bf a}|{\bf x}_{r},ez)$ denote the probability of ${\bf a}$
conditional on Eve’s information. The expected optimal guessing probability
for Eve is shown as follows

$\displaystyle P_{gs}({\bf a}|E)=\max_{z}\sum_{e}P(e|z)\max_{\bf a}P({\bf
a}|{\bf x}_{r},ez)$ (102)

Denote $\rho_{ABC|ez}$ as the initial state when Eve measures $z$ and obtains
the outcome $e$, where $\rho_{ABC|ez}$ satisfies
$\rho_{ABC}=\sum_{e}P(e|z)\rho_{ABC|ez}$. Denote ${\bf A}^{\bf a}_{{\bf
x}_{r}}=\prod_{i=1}^{s}A^{a_{i}}_{x_{r}}$ as POVMs for all outputs ${\bf a}$
conditional on $\textsf{A}_{1}$’s inputs ${\bf x}_{r}$. We get that

$\displaystyle
P(\textbf{a}|\textbf{x}_{r},ez)=\textrm{tr}[\rho_{ABC|ez}^{\otimes{}s}{\bf
A}^{\textbf{a}}_{\textbf{x}_{r}}]$ (103)

where ${\bf x}_{r}$ denote all parties’ inputs with $x_{r}$ for each
experiment. By using the semidefinite programming relaxation 20 ; 52 , it
follows that

$\displaystyle{\bf
A}^{\textbf{a}}_{\textbf{x}_{r}}\leq\textbf{A}(\varpi)=\prod_{i=1}^{s}[\alpha(\varpi)\mathbbm{1}+\beta(\varpi)O_{i}]$
(104)

where
$\alpha(\varpi)=\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi^{2}}+\frac{\varpi^{2}}{12\sqrt{72-\varpi^{2}}}$,
$\beta(\varpi)=-\frac{\varpi}{12\sqrt{72-\varpi^{2}}}$ and
$O_{i}=\sum_{abc,xyz}g^{abc}_{xyz}A^{a_{i}}_{x_{i}}B^{b_{i}}_{y_{i}}C^{c_{i}}_{z_{i}}$.
From the inequality (104) we obtain that

$\displaystyle P_{gs}(\textbf{a}|E)$ $\displaystyle=$
$\displaystyle\max_{z}\sum_{e}P(e|z)\max_{\textbf{a}}\rm{tr}[\rho_{ABC|ez}^{\otimes{}s}{\bf
A}^{\textbf{a}}_{\textbf{x}_{r}}]$ (105) $\displaystyle\leq$
$\displaystyle\max_{z}\sum_{e}P(e|z)\min_{\varpi}\textrm{tr}[\rho_{ABC|ez}^{\otimes{}s}\textbf{A}(\varpi)]$
$\displaystyle\leq$
$\displaystyle\min_{\varpi}\textrm{tr}[\rho_{ABC}^{\otimes{}s}\textbf{A}(\varpi)]$
$\displaystyle\leq$
$\displaystyle\min_{\varpi}(\alpha(\varpi)+\beta(\varpi)\varpi_{es}+s^{-1/4}_{es})^{s}$
$\displaystyle\leq$
$\displaystyle(\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi_{es}^{2}}+s^{-1/4}_{es})^{s}$
(106)

The inequality (105) follows from the result 20 ; 52  by combining the systems
B and C, where $s_{es}$ denotes the number of the binary key distilled from
the raw key. The inequality (106) follows from the inequality

$\displaystyle\min_{\varpi}\\{\alpha(\varpi)+\beta(\varpi)\varpi_{es}\\}\leq\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi_{es}^{2}}$
(107)

using the concavity.

Denote $R:=\frac{H_{min}({\bf a}|E)}{s}=-\frac{\log_{2}P_{gs}({\bf a}|E)}{s}$
as the asymptotic secret-key rate of the tripartite QKD scheme 20 ; 39 . It
follows from the inequality (106) that

$\displaystyle R$ $\displaystyle=$ $\displaystyle-\frac{\log_{2}P_{gs}({\bf
a}|E)}{s}$ (108) $\displaystyle\geq$
$\displaystyle-\log_{2}(\frac{1}{2}+\frac{1}{12}\sqrt{72-\varpi_{es}^{2}})-H(a|b,c)$

where $P_{gs}({\bf a}|E)$ denotes the expected optimal guessing probability
conditional on the eavesdropper’s input $E$.

##  Appendix J Verifying general quantum networks with LOCC

In this section, we verify general quantum networks with LOCC. Take the
network shown in Fig.S12 as an example. For Fig.S12(a), each pair of
$\textsf{A}_{1}$ and $\textsf{A}_{2}$, or $\textsf{A}_{2}$ and
$\textsf{A}_{3}$ shares one bipartite entangled pure state $|\psi_{i}\rangle$.
Note that any two parties, $\textsf{A}_{2}$ and $\textsf{A}_{3}$ for example,
can recover a new bipartite entangled state $|\phi_{2}\rangle$ when the other
party $\textsf{A}_{1}$ performs a two-particle joint measurement and sends out
outcomes. Now, $\textsf{A}_{2}$ and $\textsf{A}_{3}$ share one generalized
bipartite entangled state $|\phi_{i}\rangle$. It means that any two parties
can rebuild a bipartite entangled state assisted by LOCC. Moreover, consider
the network as shown in Fig.S12(b), where each pair of $\textsf{A}_{1}$ and
$\textsf{A}_{2}$, $\textsf{A}_{2}$ and $\textsf{A}_{3}$, and $\textsf{A}_{3}$
and $\textsf{A}_{1}$ shares one bipartite entangled pure state
$|\psi_{i}\rangle$. Any two adjacent parties ($\textsf{A}_{2}$ and
$\textsf{A}_{3}$ for example) can recover a new bipartite entangled state
$|\psi_{4}\rangle$ when the other ($\textsf{A}_{1}$ in example) performs a
two-particle joint measurement and sends out outcomes. $\textsf{A}_{2}$ and
$\textsf{A}_{3}$ share two generalized bipartite entangled states
$|\psi_{2}\rangle|\psi_{4}\rangle$. So, any two parties rebuild a bipartite
entangled state assisted by LOCC 28 . This fact is useful for detecting the
genuinely multipartite nonlocality of all the multipartite entangled pure
states using tailed CHSH inequalities. Interestingly, it is also applicable
for network scenarios. Here, the detecting experiment consists of two steps.
One is to construct one bipartite entangled state for each pair with LOCC. The
other is to verify a generalized bipartite network.

Figure 12: Entanglement swapping network. (a) An entanglement swapping network
consisting of three parties. (b) A triangle network consisting of three
parties. $|\psi_{i}\rangle$ are bipartite entangled pure states.

Result S5. The genuinely multipartite nonlocality of any connected quantum
networks consisting of genuinely entangled multipartite pure states can be
verified with the help of LOCC.

Proof of Result S5. The proof is divided into three steps inspired by a recent
result for single entangled system 28 . The first is to prove that any
genuinely multipartite entangled pure states can be useful for entanglement
swapping. The second is to prove any two observers in a connected network
consisting of genuinely multipartite entangled pure states can recover a new
bipartite entangled pure state with the help of others’ LOCC. The last is to
verify that any two parties with shared $k$ bipartite entangled pure states
using the maximal violation of some bipartite Bell inequality.

Lemma S2. Consider a $4$-partite network consisting of Alice, Bob, Charlie and
Tom, where Alice, Bob and Tom share one genuinely entangled pure state
$|\psi_{1}\rangle$, and Bob and Charlie and Tom share the other genuinely
entangled pure state $|\psi_{2}\rangle$. Alice and Charlie can create a new
bipartite entangled pure state with the help of LOCC of Bob and Tom.

Compared with the standard tripartite entanglement swapping with two EPR
states 8s , there are other parties (Tom, one party or multiple parties) which
can be viewed as controllers. The proof is completed by two steps. One is to
obtain two bipartite entangled pure states for Alice and Bob (Bob and Charlie)
with the help of LOCC of Tom. The other is to use tripartite entanglement
swapping 8s . Similarly, we have

Lemma S3. Consider a $k$-partite connected network consisting of
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$, where $\textsf{A}_{i}$ shares at least
one genuinely multipartite entangled pure state $|\psi_{i}\rangle$ with other
parties. Any two parties who do not share any entangled state can create a new
bipartite entangled pure state with the help of LOCC of other parties.

Now, we continue to prove the result. Consider an $n$-partite connected
network ${\cal N}$ consisting of $\otimes_{i}|\psi_{i}\rangle$, where
$|\psi_{i}\rangle$ is genuinely multipartite entangled pure state on Hilbert
space $\otimes_{j=1}^{s_{i}}\mathcal{H}_{j}$. The proof is similar to its for
single-source network 7s. For any pair of parties $\textsf{A}_{i_{1}}$ and
$\textsf{A}_{i_{2}}$, two cases will be discussed. One is that
$\textsf{A}_{i_{1}}$ and $\textsf{A}_{i_{2}}$ share one entangled pure state
9s . The other is that $\textsf{A}_{i_{1}}$ and $\textsf{A}_{i_{2}}$ do not
share any entangled pure state.

Step 1. ${\cal N}_{q}$ has $n$-partite genuinely nonlocality if there exists a
covering set such that each pair in it can create entanglement in all branches
of LOCC protocol for other parties.

By using the tilted-CHSH family of Bell inequalities 6s ; 7s; 10s , there is a
linear Bell inequality defined by

$\displaystyle B_{i_{1}i_{2}}=\sum_{{\bf s}}\sum_{\bf a,x}B^{({\bf s})}_{{\bf
x};{\bf a}}\leq C$ (109)

when at least one pair of $\textsf{A}_{i_{1}}$ and $\textsf{A}_{i_{2}}$ cannot
create one bipartite entangled pure state. Here, $B_{i_{1}i_{2}}$ is used to
verify the final entangled pure state for $\textsf{A}_{i_{1}}$ and
$\textsf{A}_{i_{2}}$, and $B^{({\bf s})}_{{\bf x};{\bf a}}$ depends on the
specific branch ${\bf s}$ and inputs ${\bf a}$ and outputs ${\bf x}$ of all
involved parties 6s ; 7s; 10s . Otherwise, the achievable quantum upper bound
is $C^{\prime}>C$. Here, we choose $C,C^{\prime}$ to be independent of
$i_{1}i_{2}$ 6s ; 7s. Define

$\displaystyle{\cal B}=\sum_{(i_{1},i_{2})}B_{i_{1}i_{2}}$ (110)

where the summation is taken over all pairs in a covering set.

Assume that the total state of ${\cal N}_{q}$ is a biseparable state in
Eq.(12). From the definition of LOCC, the joint system shared by
$\textsf{A}_{i_{1}}$ and $\textsf{A}_{i_{2}}$ which are in different
partitions remain in a separable state in each branch of the protocol 7s. So
the maximum of $B_{i_{1}i_{2}}$ is at most $C$ if $i_{1}\in I$ and
$i_{2}\in\overline{I}$. It implies that $\max{\cal
B}=(k-1)C^{\prime}+C<kC^{\prime}$ for any biseparable state $\rho_{bs}$ while
$\max{\cal B}=kC^{\prime}$ for a fully entangled ${\cal N}_{q}$.

Step 2. ${\cal N}_{q}$ consisting of generalized EPR states and genuinely
entangled GHZ states satisfies the assumption in Step 1.

Consider an $n$-partite connected network ${\cal N}_{q}$ consisting of
generalized EPR states $\otimes_{i=1}^{m_{1}}|\phi_{i}\rangle$ and genuinely
entangled GHZ states $\otimes_{j=1}^{m_{2}}|\psi_{j}\rangle$. Since ${\cal
N}_{q}$ is connected,
$|\Omega\rangle:=\otimes_{i=1}^{m_{1}}\otimes_{j=1}^{m_{2}}|\phi_{i}\rangle|\psi_{j}\rangle$
can be viewed as an $n$-partite entangled pure state with local dimension $d$,
where $d$ is the maximal dimension of all local systems. Otherwise, there
exist at least two parties who cannot share an entangled pure state with LOCC.
From Lemmas 1 and 2, it means that two parties are not connected, i.e., the
network is disconnected. A simple fact of these networks is as follows: for
each $\textsf{A}_{i}$, there are projections on local systems such that the
remained network of the other $n-1$ parties also consists of EPR states and
GHZ states with LOCC. Here, the measurement outcomes are sent to $n-2$
parties. After these operations, any pair of parties can share at least one
generalized EPR state with LOCC of other $n-2$ parties.

Step 3. Bell test for two parties with shared bipartite entangled pure states
$\otimes_{i}|\phi_{i}\rangle$.

For a connected network, any pair of parties $\textsf{A}_{i},\textsf{A}_{j}$
can create $k_{ij}$ bipartite entangled pure states
$\otimes_{s=1}^{k_{ij}}|\phi_{s}\rangle$, where
$|\phi_{s}\rangle=\cos\theta_{s}|00\rangle+\sin\theta_{s}|11\rangle$. We can
rewrite the total state into

$\displaystyle\otimes_{s=1}^{k_{ij}}|\phi_{s}\rangle=\sum_{t=0}^{2^{k}-1}\alpha_{t}|tt\rangle$
(111)

where $\sum_{t=0}^{2^{k_{ij}}-1}\alpha_{t}|tt\rangle=:|G\rangle$ is a
bipartite state on Hilbert space
$\mathbb{C}^{2^{k_{ij}}}\times\mathbb{C}^{2^{k_{ij}}}$, and $\alpha_{t}$’s are
coefficients depending on $\theta_{s}$’s. There is a tailored Bell inequality
which will be maximally violated by this state. One example is the tailed CHSH
6s  for verifying the nonlocality of $|G\rangle$ for some $\theta_{i}\not=0$.
Another method is using a linear Bell inequality with $3$ and 4 inputs and
$2^{k_{ij}}$ outputs 10s .

Step 4. ${\cal N}_{q}$ consisting of all genuinely entangled pure states
satisfies the assumption in Step 1.

For each pair of $\textsf{A}_{i}$ and $\textsf{A}_{j}$, there exists a
generalized chain-shaped subnetwork consisting of
$\textsf{A}_{i},\textsf{A}_{s_{1}},\cdots,\textsf{A}_{s_{\ell}},\textsf{A}_{j}$.
For each entangled pure state $\rho_{1\cdots{}m}$, there exist local
measurements on the $m$-th particle such that the resultant is entangled pure
state $\varrho_{1\cdots{}m-1}$ 7s. By using these local operations for all the
other parties (iteratively for each multipartite state), they can recover a
standard chain-shaped subnetwork consisting of all $d$-dimensional bipartite
entangled pure states $|\Phi_{j}\rangle$. Similar to Lemma S3,
$\textsf{A}_{i}$ and $\textsf{A}_{j}$ can recover at least one bipartite
entangled pure state using the local $d$-dimensional rotations after
$\textsf{A}_{s_{1}},\cdots,\textsf{A}_{s_{\ell}}$ perform local joint
measurement under the generalized Bell basis:
$\\{|B_{st}\rangle=\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}\exp(\frac{2isj\pi}{d})|j\rangle|d-j\rangle\\}$.
Finally, the nonlocality of the resultant can be verified by $\textsf{A}_{i}$
and $\textsf{A}_{j}$ using the Bell inequality with multiple setting 10s .
$\square$

Result S6. Consider a connected $n$-partite network ${\cal N}_{q}$ consisting
of EPR and GHZ states with white noise. ${\cal N}_{q}$ has genuinely
$n$-partite nonlocality with the help of LOCC if each noisy state can be
verified with a Bell inequality.

Proof. Firstly, consider the entanglement swapping network consisting of
Alice, Bob and Charlie, who share two Werner states Wrner :

$\displaystyle\rho_{i}=v_{i}|\phi\rangle\langle\phi|+\frac{1-v_{i}}{4}\mathbbm{1}_{4},i=1,2$
(112)

where $|\phi\rangle$ is EPR state, $\mathbbm{1}_{4}$ is the identity operator
with the rank $4$, and $v_{i}\in(0,1]$. In what follows, we consider the
resultant of Alice and Charlie with the help of LOCC of Bob. Specially, Bob
performs a two-qubit joint measurement under Bell basis:
$\\{\frac{1}{\sqrt{2}}(|00\rangle\pm|11\rangle),\frac{1}{\sqrt{2}}(|01\rangle\pm|10\rangle)\\}$,
and sends the outcome to Alice or Charlie, who can then recover a bipartite
state as follows:

$\displaystyle\rho=$ $\displaystyle
v_{1}v_{2}|\phi\rangle\langle\phi|+\frac{v_{1}+v_{2}-2v_{1}v_{2}}{2}(|00\rangle\langle
00|+|11\rangle\langle 11|)+\frac{(1-v_{1})(1-v_{2})}{4}\mathbbm{1}_{4}$ (113)

The fidelity of $\rho$ with respect to EPR state $|\phi\rangle$ is given by

$\displaystyle F(\rho,|\phi\rangle\langle\phi|)=$ $\displaystyle{\rm