††thanks: Non-local boxes

# Non-Locality distillation in tripartite NLBs

Talha Lateef<EMAIL_ADDRESS>
(2017-11-1)

###### Abstract

In quantum mechanics some spatially separated sub-systems behave as if they
are part of a single system, the superposition of states of this single system
cannot be written as products of states of individual sub-systems,we say that
the state of such system is entangled, such systems give rise to non-local
correlations between outcomes of measurements. The non-local correlations are
conditional probability distributions of some measurement outcomes given some
measurement settings and cannot be explained by shared informationBell (1964);
Clauser  _et al._ (1969).These correlations can be studied using a non-local
box(NLB) which can be viewed as a quantum system111we assume quantum
correlations.. A NLB is an abstract object which has number of
inputs(measurement settings) and number of outputs(outcomes), such NLBs can be
both quantum and super-quantumKhalfin and Csirel’son. (1985); Popescu and
Rohrlich (1994); Barrett  _et al._ (2005). The correlations are of use in
quantum information theory, the stronger the correlations the more useful they
are, hence we study protocols that have multiple weaker non-local systems,
application of these protocols to weaker systems may result in stronger non-
local correlations, we call such protocols non-locality distillation
protocolsForster  _et al._ (2009); Brunner and Skrzypczyk (2009); Allcock  _et
al._ (2009). In our work here we present non-locality distillation protocols
for tripartite NLBs specifically GHZ box and class 44,45 and 46 of no-
signalling polytope.

###### pacs:

Valid PACS appear here

##  I Introduction:

The no-signaling box under consideration has three parties, two measurement
settings, two measurement outcomes and is represented by the tuple $(3,2,2)$.
The measurement settings/inputs are $x,y,z\in\\{0,1\\}$ and for measurement
outcomes/outputs we use either $a,b,c\in\\{0,1\\}$ or the so called fourier
representation $a,b,c\in\\{1,-1\\}$. The three parties are Alice with input
$x$ and output $a$, Bob with input $y$ and output $b$ and Charlie with input
$z$ and output $c$. The correlations are in the form of conditional
probabilities $P(abc|xyz)$, these correlations satisfy three constraints of
positivity , normalization and no-signaling respectively:

$\displaystyle P(abc|xyz)\geq 0\text{\ \ \ \ for all $a,b,c,x,y,z$}$
$\displaystyle\sum_{a,b,c}P(abc|xyz)=1\text{\ \ \ \ for all $x,y,z$}$ and
$\displaystyle\sum_{a}P(abc|xyz)=\sum_{a}P(abc|x^{\prime}yz)\text{\ \ \ \ for
all $b,c,x,x^{\prime},y,z$}$
$\displaystyle\sum_{b}P(abc|xyz)=\sum_{b}P(abc|xy^{\prime}z)\text{\ \ \ \ for
all $a,c,x,y,y^{\prime},z$}$
$\displaystyle\sum_{c}P(abc|xyz)=\sum_{c}P(abc|xyz^{\prime})\text{\ \ \ \ for
all $a,b,x,y,z,z^{\prime}$}$

Positivity and normalization comes from the correlations being conditional
probabilities. No-signaling prevents faster than light communication between
parties. For the case of non-local tripartite boxes they also can not be
written in terms of shared information and correlations of individual parties
so they do not satisfy:

$P(abc|xyz)=\sum_{\lambda}p(\lambda)P(a|x,\lambda)P(b|y,\lambda)P(c|z,\lambda)$

where $p(\lambda)$ is probability distribution or shared information, we are
using a probability distribution because the variable(s) may be beyond our
control, from this it follows that $p(\lambda)\geq 0$ and
$\sum_{\lambda}p(\lambda)=1$. The constraints of positivity, normalization and
no-signaling is a description of a geometric object called polytope, it is
possible to change one description of polytope to another, application of
vertex enumeration algorithms to the constraints will result in all vertices
or boxes. There are total 53856 boxes with 64 being local and rest of them
non-local.These boxes are classified in 46 equivalent classesPironio  _et al._
(2011); Bancal  _et al._ (2013).

Non-locality distillation protocol combines multiple non-local boxes to
produce stronger non-locality using local operations without any classical
communication. The depth of the protocol is the number of non-local boxes that
are part of the protocol. In the case of depth $n$ protocol the three parties
Alice , Bob and Charlie input three bits $x$,$y$ and $z$ to the protocol, the
input to the first box is the same as the input to the protocol i.e. $x_{1}=x$
, $y_{1}=y$ and $z_{1}=z$, the output bits from the first box are $a_{1}$ ,
$b_{1}$ and $c_{1}$. The input to $n$th box are bits $x_{n}$ , $y_{n}$ and
$z_{n}$, which are result of each party using a local operation on their input
bit and output bit from the $n-1$th box i.e. $x_{n}=f_{n}(x_{n-1},a_{n-1})$ ,
$y_{n}=g_{n}(y_{n-1},b_{n-1})$ and $z_{n}=h_{n}(z_{n-1},c_{n-1})$, we get the
output $a_{n}$ , $b_{n}$ and $c_{n}$ from the $n$th box. The output of the
protocol is then define as $a=f_{n+1}(x,a_{1},...,a_{n})$ ,
$b=g_{n+1}(y,b_{1},...,b_{n})$ and $c=h_{n+1}(z,c_{1},...,c_{n})$.

A protocol is said to be non-adaptive if all the inputs to the NLBs in the
protocol depends on the original input to the protocol i.e. $x_{n}=f_{n}(x)$ ,
$y_{n}=g_{n}(y)$ and $z_{n}=h_{n}(z)$ for all $n$ while in the case of
adaptive protocols inputs to NLBs also depend on outputs from the previous
NLBs in the protocol which can be seen in the original definition, hence non-
adaptive protocols are a subset of adaptive protocolsForster  _et al._ (2009);
Allcock  _et al._ (2009); Høyer and Rashid (2010).

Figure 1: Distillation protocol of depth n

##  II GHZ box:

This NLB is based on GHZ correlationsGreenberger  _et al._ (1990) and is
defined as:

$P_{abc|xyz=000}=\begin{cases}\frac{1}{4}&\text{if }a\oplus b\oplus c=0\\\
0&\text{if }a\oplus b\oplus c=1\\\ \end{cases}$

$P_{abc|xyz\neq 000}=\begin{cases}\frac{1}{4}&\text{if }a\oplus b\oplus c=1\\\
0&\text{if }a\oplus b\oplus c=0\\\ \end{cases}$

and in matrix form:

$P_{GHZ}=\left(\begin{array}[]{cccccccc}\frac{1}{4}&0&0&\frac{1}{4}&0&\frac{1}{4}&\frac{1}{4}&0\\\
0&\frac{1}{4}&\frac{1}{4}&0&\frac{1}{4}&0&0&\frac{1}{4}\\\
0&\frac{1}{4}&\frac{1}{4}&0&\frac{1}{4}&0&0&\frac{1}{4}\\\
0&\frac{1}{4}&\frac{1}{4}&0&\frac{1}{4}&0&0&\frac{1}{4}\end{array}\right)$

The inputs are limited to even parity i.e $xyz\in\\{000,011,101,110\\}$ and
outputs $a,b,c\in\\{0,1\\}$.

The GHZ box is not an extremal point of no-signaling polytope, however It is
possible to construct GHZ boxes using boxes of class 46 as given in Pironio
_et al._ (2011) by using:

$P_{GHZ}=\frac{1}{2}P_{46}+\frac{1}{2}P_{46}^{\prime}$

where $P_{46}$ and $P_{46}^{\prime}$ are both different boxes of class 46. The
GHZ box with even parity inputs can be constructed in similar way using boxes:

$\displaystyle P_{46}=\frac{1}{8}(1+abc(-1)^{xy+xz+yz})$ and $\displaystyle
P_{46}^{\prime}=\frac{1}{8}(1+abc(-1)^{x+y+z+xy+xz+yz})$ $\displaystyle\ \ \ \
\ \ \ \ \ \text{where }a,b,c\in\\{1,-1\\}$

or just using $P_{44}$ where $P_{44}$ is the extremal point of class 44 given
by:

$\displaystyle P_{44}=\frac{1}{8}(1+abc(-1)$ ${}^{x+y+z+xy+xz+yz+xyz})$
$\displaystyle\text{where }a,b,c$ $\displaystyle\in\\{1,-1\\}$

Figure 2: Noisy GHZ box

The local deterministic vertices of no-signalling polytope are grouped
together as local polytope $P_{L}$ and has 64 vertices which is expressed as:

$\displaystyle P_{L}^{\iota\kappa\mu\nu\sigma\tau}(abc|xyz)$
$\displaystyle=\begin{cases}1&if\>a=\iota x\oplus\kappa,\>b=\mu
y\oplus\nu,\>c=\sigma z\oplus\tau\\\ 0&otherwise\end{cases}$
$\displaystyle\text{where }\iota,\kappa,\mu,\nu,\sigma,\tau\in\\{0,1\\}$

The $P_{L}$ results in 53856 inequalities which has been classified in 46
different classesPitowsky and Svozil (2001); Sliwa (2003), the amount of non-
locality or box value for GHZ box is obtained using the bell inequalities of
class 2 :

$\displaystyle\normalsize{-2\leq\left\langle
A_{0}B_{0}C_{0}\right\rangle-\left\langle
A_{0}B_{1}C_{1}\right\rangle-\left\langle
A_{1}B_{0}C_{1}\right\rangle-\left\langle A_{1}B_{1}C_{0}\right\rangle\leq 2}$

(1)

where

$\left\langle
A_{x}B_{y}C_{z}\right\rangle=\sum_{a,b,c\in\\{0,1\\}}(-1)^{a\oplus b\oplus
c}P(abc|xyz)$

##  III GHZ distillation:

We now present the noisy form of a GHZ box with parameters $\epsilon\in[1,-1]$
and $\delta\in[1,-1]$

$P_{abc|xyz=000}=\begin{cases}\frac{1+\epsilon}{8}&\text{if }a\oplus b\oplus
c=0\\\ \frac{1-\epsilon}{8}&\text{if }a\oplus b\oplus c=1\\\ \end{cases}$

$P_{abc|xyz\neq 000}=\begin{cases}\frac{1+\delta}{8}&\text{$if$ }a\oplus
b\oplus c=0\\\ \frac{1-\delta}{8}&\text{$if$ }a\oplus b\oplus c=1\\\
\end{cases}$

Figure 3: General depth 2 distillation protocol

and in matrix form:

${\scalebox{0.8}{\mbox{$\displaystyle
P_{GHZ(\epsilon,\delta)}$}}=\scalebox{0.8}{\mbox{$\displaystyle\frac{1}{8}$}}\left(\scalebox{0.8}{\mbox{$\displaystyle\begin{array}[]{cccccccc}1+\epsilon&1-\epsilon&1-\epsilon&1+\epsilon&1-\epsilon&1+\epsilon&1+\epsilon&1-\epsilon\\\
1+\delta&1-\delta&1-\delta&1+\delta&1-\delta&1+\delta&1+\delta&1-\delta\\\
1+\delta&1-\delta&1-\delta&1+\delta&1-\delta&1+\delta&1+\delta&1-\delta\\\
1+\delta&1-\delta&1-\delta&1+\delta&1-\delta&1+\delta&1+\delta&1-\delta\\\
\end{array}$}}\right)}$

using inequality (1) we get the single box value:

$V=\epsilon-3\delta$

To be able to achieve distillation we want to attain $V^{\prime}$ such that:

$V^{\prime}>V$

###### Lemma III.1

In a depth n GHZ setting there is no adaptive distillation protocol that is
more optimal than the non-adaptive distillation protocol.

  * Proof

Consider a general depth $n$ distillation protocol(see Fig. 1) with GHZ NLBs,
the inputs to the protocol $x$,$y$ and $z$ are restricted to
$\\{000,011,101,110\\}$ by the definition of distillation protocol and GHZ
box, now suppose that the protocol is non-adaptive i.e. all of the
$x_{k}=f_{k}(x_{k-1},a_{k-1})=x_{k-1}$ ,
$y_{k}=g_{k}(y_{k-1},b_{k-1})=y_{k-1}$ and
$z_{k}=h_{k}(z_{k-1},c_{k-1})=z_{k-1}$ (where $k\leq n$) are true, in this
case each box gets the same input as that of the protocol and hence every
input contribute towards $V^{\prime}$, we now suppose an adaptive protocol
i.e. atleast one of the $x_{k}=f_{k}(x_{k-1},a_{k-1})\neq x_{k-1}$ ,
$y_{k}=g_{k}(y_{k-1},b_{k-1})\neq y_{k-1}$ and
$z_{k}=h_{k}(z_{k-1},c_{k-1})\neq z_{k-1}$ is true, which results in the input
being changed to one of the $\\{001,010,100,111\\}$ and therefore will not be
accepted by the NLB and hence will not contribute towards $V^{\prime}$.

We now consider depth 2 non-adaptive protocols, the only protocols that
distill are parity protocols i.e the final output functions $f_{3}$,$g_{3}$
and $h_{3}$ are polynomials of degree at most $1$ over
$GF(2)$/$\mathbb{Z}_{2}$($\\{0,1\\}\,\oplus,\wedge)$) of outputs and or
input(protocol).

Since all distillation protocols are parity protocols we can generalize
protocols in a single protocol diagram (see Fig. 3) by introducing further
boolean variables $s_{a}$,$s_{b}$,$s_{c}$ and $t$. The value attain by this
protocol is then:

$V^{\prime}=(-1)^{1\oplus s_{a}\oplus s_{b}\oplus
s_{c}}\epsilon^{2}+(-1)^{s_{a}\oplus s_{b}\oplus s_{c}}3\delta^{2}$

The distillation plot is given in Fig. 4 the light gray curved part in the
plot highlights the region of distillation.

We now turn our attention to a specific non-adaptive depth $n$ distillation
protocol $NDP_{n}(GHZ)$(see Fig. 5) this is the tripartite version of Forster
et al.’sForster  _et al._ (2009) parity protocol with GHZ boxes instead of PR
boxes, just as we can represent an NLB in the matrix form, we can consider the
protocol as a new NLB with it’s inputs($x$,$y$,$z$) and outputs($a$,$b$,$c$)
without worrying about it’s internal working, this neat abstraction allows us
to write matrix form of the protocol as an NLB.

Figure 4: Distillation plot for depth 2 protocol with
$V^{\prime}=\epsilon^{2}-3\delta^{2}$

###### Lemma III.2

For the distillation protocol $NDP_{n}(GHZ)$ the matrix form is:

$\displaystyle
P_{NDP_{n}(GHZ)(\epsilon,\delta)}=\frac{1}{8}\left({\begin{array}[]{cccccccc}1+\epsilon^{n}&1-\epsilon^{n}&1-\epsilon^{n}&1+\epsilon^{n}&1-\epsilon^{n}&1+\epsilon^{n}&1+\epsilon^{n}&1-\epsilon^{n}\\\
1+\delta^{n}&1-\delta^{n}&1-\delta^{n}&1+\delta^{n}&1-\delta^{n}&1+\delta^{n}&1+\delta^{n}&1-\delta^{n}\\\
1+\delta^{n}&1-\delta^{n}&1-\delta^{n}&1+\delta^{n}&1-\delta^{n}&1+\delta^{n}&1+\delta^{n}&1-\delta^{n}\\\
1+\delta^{n}&1-\delta^{n}&1-\delta^{n}&1+\delta^{n}&1-\delta^{n}&1+\delta^{n}&1+\delta^{n}&1-\delta^{n}\\\
\end{array}}\right)\\\ $

with box value:

$V^{\prime}_{NDP_{n}(GHZ)}=\epsilon^{n}-3\delta^{n}$

Let $V^{\prime}_{n}$ be the value for any depth $n$ non-adaptive GHZ protocol,
then $V^{\prime}_{n}\leq V^{\prime}_{NDP_{n}(GHZ)}$ i.e. $V^{\prime}_{n}$ is
bounded by $\boldsymbol{max_{1\leq k\leq
n}\left|\epsilon^{k}-3\delta^{k}\right|}$

  * Proof

We use Fourier transform for boolean function in our proofHøyer and Rashid
(2010). Let the $n$ bit tuples $(a_{i},b_{i},c_{i})$ be the output of the $n$
NLBs that Alice, Bob and Charlie obtain for inputs $x$,$y$ and $z$
respectively. The tuple $(a_{i},b_{i},c_{i})$ is drawn from
$\\{000,001,010,011,100,101,110,111\\}$ with respect to the distribution
$\mu=\frac{1}{8}\\{1+\epsilon,1-\epsilon,1-\epsilon,1+\epsilon,1-\epsilon,1+\epsilon,1+\epsilon,1-\epsilon\\}$,
where $\epsilon$ is the bias for the row corresponding to the inputs received
by the players($x=0,y=0,z=0$ in this case). For inputs $x$,$y$ and $z$ let
$\bar{A}$,$\bar{B}$,$\bar{C}\subseteq\\{0,1\\}^{n}$ be the set of strings for
which Alice, Bob and Charlie’s final output is $0$.

Given that Alice, Bob and Charlie input bits $x$,$y$ and $z$ into $n$ NLBs,
the probability that they receive bit strings $a$,$b$ and $c$ of length $n$ is
given by:

The probability $q_{(\bar{A},\bar{B},\bar{C})}(\epsilon)$ of obtaining output
$000$ is obtained by summing over all of the output bit strings $a$ in
$\bar{A}$, $b$ in $\bar{B}$ and $c$ in $\bar{C}$:

Representing the above expression in terms of $2^{n}$ characters of finite
group ($\mathbb{Z}_{2}^{n}$)

Using the homomorphism from additive group to multiplicative group

Let the functions $f$,$g$ and $h$ be +1 when $s$,$t$ and $u$ are in
$\bar{A}$,$\bar{B}$ and $\bar{C}$, respectively and -1 otherwise: we know that
fourier coefficient$\hat{f}_{S}=\left\langle
f,\chi_{S}\right\rangle=E_{x}\left[f(x).\chi_{S}(x)\right]=\frac{1}{2^{n}}\sum_{x\in\left\\{0,1\right\\}^{n}}f(x)\chi_{S}(x)$
and $\hat{f}_{0}=\left\langle f,1\right\rangle$ we obtain
$q_{(\bar{A},B,C)}$,$q_{(A,\bar{B},C)}$ and $q_{(A,B,\bar{C})}$ similarly by
flipping sign of $\hat{f}_{z}$,$\hat{g}_{z}$ and $\hat{h}_{z}$ The probability
of obtaining output when $A\oplus B\oplus C=0$ is then: The expected value for
the bias $\epsilon$ can now be expressed: The expected value for bias $\delta$
can be calculated similarly, using expected value of both biases in (1) we
obtain:

Figure 5: $NDP_{n}(GHZ)$ distillation protocol

Using Lemma III.1 and Lemma III.2 we conclude that non-adaptive parity
protocols are optimal for n copies of GHZ box.

##  IV Class 44,45 and 46:

We use the same representatives as in Barrett  _et al._ (2005); Pironio  _et
al._ (2011) and define each of class 44,45 and 46 perfect boxes as:

$\displaystyle P^{44}=\begin{cases}\frac{1}{4}&a\oplus b\oplus c=xyz\\\
0&otherwise\end{cases}$ $\displaystyle P^{45}=\begin{cases}\frac{1}{4}&a\oplus
b\oplus c=xy\oplus xz\\\ 0&otherwise\end{cases}$ $\displaystyle
P^{46}=\begin{cases}\frac{1}{4}&a\oplus b\oplus c=xy\oplus yz\oplus xz\\\
0&otherwise\end{cases}$

when we want to refer to any of the perfect boxes we use the notation:

$\displaystyle P^{N}=\begin{cases}\frac{1}{4}&a\oplus b\oplus c=f(x,y,z)\\\
0&otherwise\end{cases}$

where $N$ can be any of the perfect boxes and $f(x,y,z)$ is a boolean
polynomial of $x,y,z$ over $GF(2)$ for that box.

To calculate box value we select the following inequality of class 41:

$\scalebox{0.85}{\mbox{$\displaystyle-\left\langle
A_{0}\right\rangle-\left\langle B_{0}\right\rangle-\left\langle
C_{0}\right\rangle+\left\langle A_{0}B_{1}\right\rangle+\left\langle
A_{1}B_{0}\right\rangle-\left\langle A_{1}B_{1}\right\rangle+\left\langle
A_{0}C_{1}\right\rangle$}}\\\
\scalebox{0.85}{\mbox{$\displaystyle+\left\langle
A_{1}C_{0}\right\rangle-\left\langle A_{1}C_{1}\right\rangle+\left\langle
B_{0}C_{0}\right\rangle+3\left\langle
A_{0}B_{0}C_{0}\right\rangle+\left\langle
A_{0}B_{0}C_{1}\right\rangle+\left\langle A_{0}B_{1}C_{0}\right\rangle$}}\\\
\scalebox{0.85}{\mbox{$\displaystyle+2\left\langle
A_{0}B_{1}C_{1}\right\rangle+4\left\langle
A_{1}B_{0}C_{0}\right\rangle-\left\langle
A_{1}B_{0}C_{1}\right\rangle-\left\langle
A_{1}B_{1}C_{0}\right\rangle-2\left\langle A_{1}B_{1}C_{1}\right\rangle\leq
7$}}$ (2)

calculating $V$ for each of the perfect boxes using (2), we get the same value
of $11$.

$P^{c}$ is the tripartite correlated box with even parity and is defined as:

$\displaystyle P^{c}=\begin{cases}\frac{1}{4}&a\oplus b\oplus c=0\\\
0&otherwise\end{cases}$

we are restricting ourselves to correlated noise model hence we define noisy
box as:

$\displaystyle P_{\delta}^{N}=$ $\displaystyle\left[\delta
P^{N}+(1-\delta)P^{c}\right]$ where $\delta\in[0,1]$

Using (2) we get the box value $V$ for $P_{\delta}^{N}$:

$V=4\delta+7$

Finally we present couple of notations that we use in next sections:

$\displaystyle P_{(1-\delta)}^{N}=$ $\displaystyle\left[(1-\delta)P^{N}+\delta
P^{c}\right]$ where $\delta\in[0,1]$ $\displaystyle
P^{f(x_{1},x_{2},.....,x_{n})}\rightarrow a\oplus b\oplus
c=f(x_{1},x_{2},.....,x_{n})$ where $f(x_{1},x_{2},.....,x_{n})$ is polynomial
of inputs.

##  V $P_{\delta}^{N}$ distillation:

###  V.1 Common protocols

In this section we are going to present two distillation protocols of depth
$2$ that are common to these three classes:

Figure 6: Protocol 1

####  V.1.1 Protocol 1

This protocol(see Fig. 6) is the similar to the non-adaptive protocol of depth
2 which has been presented previously for GHZ box, we now give proof of it’s
distillability:

###### Lemma V.1

Protocol 1(see Fig. 6) distills for two copies of $P_{\delta}^{N}$ with the
same amount of $V^{\prime}$ for each class

  * Proof

We start with two copies of $P_{\delta}^{N}$:

$\displaystyle P_{\delta}^{N}P_{\delta}^{N}$ $\displaystyle=\left[\delta
P^{N}+(1-\delta)P^{c}\right]\left[\delta P^{N}+(1-\delta)P^{c}\right]$
$\displaystyle=\delta^{2}P^{N}P^{N}+\delta(1-\delta)\left(P^{N}P^{c}+P^{c}P^{N}\right)+(1-\delta)^{2}P^{c}P^{c}$

Hence we need each of the four relations above to get to the final box:
$P^{N}P^{N}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=f(x_{1},y_{1},z_{1})=f(x,y,z)$, for box 2 we have $a_{2}\oplus
b_{2}\oplus c_{2}=f(x_{2},y_{2},z_{2})=f(x_{1},y_{1},z_{1})=f(x,y,z)$,
equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus
1\oplus c_{1}\oplus c_{2}=0$, $a\oplus b\oplus c=0$, hence $P^{c}$

$P^{N}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=f(x_{1},y_{1},z_{1})=f(x,y,z)$, $0=f(x,y,z)\oplus a_{1}\oplus
b_{1}\oplus c_{1}$, for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$,
equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus
1\oplus c_{1}\oplus c_{2}=f(x,y,z)$, $a\oplus b\oplus c=f(x,y,z)$, hence
$P^{N}$

$P^{c}P^{N}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=f(x_{2},y_{2},z_{2})=f(x_{1},y_{1},z_{1})=f(x,y,z)$, $0=f(x,y,z)\oplus
a_{2}\oplus b_{2}\oplus c_{2}$ equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=f(x,y,z)$,
$a\oplus b\oplus c=f(x,y,z)$, hence $P^{N}$

$P^{c}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$,
$a\oplus b\oplus c=0$, hence $P^{c}$

The final box is given by:

$\displaystyle P_{\delta}^{N}P_{\delta}^{N}$
$\displaystyle=\left(\delta(2-2\delta)\right)P^{N}+\left(1-\delta\left(2-2\delta\right)\right)P^{c}$

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=-8\delta^{2}+8\delta+7$

The protocol distills($V^{\prime}>V$) in the region of
$0<\delta<\dfrac{1}{2}$, the grey curve in fig. 11 highlights the protocol.

####  V.1.2 Protocol 2

Figure 7: Protocol 2

We now turn our attention to another common protocol(see fig. 7), protocol 2
is a depth 2 adaptive protocol with final box and expression given below:

###### Lemma V.2

Protocol 2(see fig. 7) distills for two copies of $P_{\delta}^{N}$ with the
different amount of $V^{\prime}$ for each class

  * Proof

For each class we derive a different final box and expression:
Class 44:

$P^{44}P^{44}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}z_{1}=xyz$, $0=xyz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for
box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=x_{2}y_{2}z_{2}=x_{1}y_{1}z_{1}b_{1}=xyzb_{1}$, $0=xyzb_{1}\oplus
a_{2}\oplus b_{2}\oplus c_{2}$ equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xyz\oplus
xyzb_{1}$, $a\oplus b\oplus c=xyz\left(1\oplus b_{1}\right)$, hence
$\frac{1}{2}\left(P^{44}+P^{c}\right)$

$P^{44}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}z_{1}=xyz$, $0=xyz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for
box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus
c_{2}=xyz$, $a\oplus b\oplus c=xyz$, hence $P^{44}$

$P^{c}P^{44}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=x_{2}y_{2}z_{2}=x_{1}y_{1}z_{1}b_{1}=xyzb_{1}$, $0=xyzb_{1}\oplus
a_{2}\oplus b_{2}\oplus c_{2}$ equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xyzb_{1}$,
$a\oplus b\oplus c=xyzb_{1}$, hence $\frac{1}{2}\left(P^{c}+P^{44}\right)$

$P^{c}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$,
$a\oplus b\oplus c=0$, hence $P^{c}$

The final box is given by:

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=-4\delta^{2}+6\delta+7$

Class 45:

$P^{45}P^{45}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus x_{1}z_{1}=xy\oplus xz$, $0=xy\oplus xz\oplus
a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=x_{2}y_{2}\oplus x_{2}z_{2}=x_{1}y_{1}b_{1}\oplus
x_{1}z_{1}=xyb_{1}\oplus xz$, $0=xyb_{1}\oplus xz\oplus a_{2}\oplus
b_{2}\oplus c_{2}$, equating we get: $1\oplus a_{1}\oplus a_{2}\oplus
b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xy\oplus xyb_{1}\oplus
xz\oplus xz$, $a\oplus b\oplus c=xy\left(1\oplus b_{1}\right)$, hence
$\frac{1}{2}\left(P^{xy}+P^{c}\right)$

$P^{45}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus x_{1}z_{1}=xy\oplus xz$, $0=xy\oplus xz\oplus
a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=0$, equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus
b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xy\oplus xz$, $a\oplus b\oplus
c=xy\oplus xz$, hence $P^{45}$

$P^{c}P^{45}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}\oplus
x_{2}z_{2}=x_{1}y_{1}b_{1}\oplus x_{1}z_{1}=xyb_{1}\oplus xz$,
$0=xyb_{1}\oplus xz\oplus a_{2}\oplus b_{2}\oplus c_{2}$ equating we get:
$1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus
c_{2}=xyb_{1}\oplus xz$, $a\oplus b\oplus c=xyb_{1}\oplus xz$, hence
$\frac{1}{2}\left(P^{xz}+P^{45}\right)$

$P^{c}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating $a_{1}\oplus
b_{1}\oplus c_{1}=a_{2}\oplus b_{2}\oplus c_{2}$, $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$, $a\oplus
b\oplus c=0$, hence $P^{c}$

The final box is given by:

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=-6\delta^{2}+9\delta+7$

Class 46:

$P^{46}P^{46}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus y_{1}z_{1}\oplus x_{1}z_{1}=xy\oplus yz\oplus xz$,
$0=xy\oplus yz\oplus xz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we
have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}\oplus y_{2}z_{2}\oplus
x_{2}z_{2}=x_{1}y_{1}b_{1}\oplus y_{1}z_{1}b_{1}\oplus
x_{1}z_{1}=xyb_{1}\oplus yzb_{1}\oplus xz$, $0=xyb_{1}\oplus yzb_{1}\oplus
xz\oplus a_{2}\oplus b_{2}\oplus c_{2}$, equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xyb_{1}\oplus
yzb_{1}\oplus xz\oplus xy\oplus yz\oplus xz$, $a\oplus b\oplus c=xyb_{1}\oplus
xy\oplus yzb_{1}\oplus yz$, $a\oplus b\oplus c=\left(1\oplus
b_{1}\right)\left(xy\oplus yz\right)$,hence $\frac{1}{2}\left(P^{xy\oplus
yz}+P^{c}\right)$

$P^{46}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus y_{1}z_{1}\oplus x_{1}z_{1}=xy\oplus yz\oplus xz$,
$0=xy\oplus yz\oplus xz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we
have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xy\oplus
yz\oplus xz$, $a\oplus b\oplus c=xy\oplus yz\oplus xz$, hence $P^{46}$

$P^{c}P^{46}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}\oplus
y_{2}z_{2}\oplus x_{2}z_{2}=x_{1}y_{1}b_{1}\oplus y_{1}z_{1}b_{1}\oplus
x_{1}z_{1}=xyb_{1}\oplus yzb_{1}\oplus xz$, $0=xyb_{1}\oplus yzb_{1}\oplus
xz\oplus a_{2}\oplus b_{2}\oplus c_{2}$, equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xyb_{1}\oplus
yzb_{1}\oplus xz$, $a\oplus b\oplus c=b_{1}\left(xy\oplus yz\right)\oplus xz$,
hence $\frac{1}{2}\left(P^{xz}+P^{46}\right)$

$P^{c}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating $a_{1}\oplus
b_{1}\oplus c_{1}=a_{2}\oplus b_{2}\oplus c_{2}$, $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$, $a\oplus
b\oplus c=0$, hence $P^{c}$

The final box is given by:

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=-10\delta^{2}+9\delta+7$

For class 45 protocol 2 distills in the region of $0<\delta<\dfrac{5}{6}$ and
both for class 44 and class 46 in the region of $0<\delta<\dfrac{1}{2}$, white
curves in fig. 11 refer to protocol 2.

###  V.2 Unique protocols

We present three different protocols that atleast distill for each of their
respective classes:

####  V.2.1 Protocol 3

Figure 8: Protocol 3

###### Lemma V.3

Protocol 3(see fig. 8) distills for two copies of $P_{\delta}^{44}$

  * Proof

We start with two copies of $P_{\delta}^{44}$:

$P^{44}P^{44}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}z_{1}=xyz$, $0=xyz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for
box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}z_{2}=x_{1}(1\oplus
y_{1})z_{1}=xz\oplus xyz$, $0=xz\oplus xyz\oplus a_{2}\oplus b_{2}\oplus
c_{2}$ equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus
b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xyz\oplus xz\oplus xyz$, $a\oplus
b\oplus c=xz$, hence $P^{xz}$

$P^{44}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}z_{1}=xyz$, $0=xyz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for
box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus
c_{2}=xyz$, $a\oplus b\oplus c=xyz$, hence $P^{44}$

$P^{c}P^{44}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}z_{2}=x_{1}(1\oplus
y_{1})z_{1}=xz\oplus xyz$, $0=xz\oplus xyz\oplus a_{2}\oplus b_{2}\oplus
c_{2}$ equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus
b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xz\oplus xyz$, $a\oplus b\oplus
c=xz\oplus xyz$, hence $P^{xz\oplus xyz}$

$P^{c}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$,
$a\oplus b\oplus c=0$, hence $P^{c}$

The final box is given by:

$\displaystyle P_{\delta}^{44}P_{\delta}^{44}$
$\displaystyle=\delta\left(\delta P^{xz}+(1-\delta)P^{xz\oplus
xyz}\right)+(1-\delta)P_{\delta}^{44}$

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=6\delta+7$

The protocol distills in the region of $0<\delta\leqslant 1$.

####  V.2.2 Protocol 4

Figure 9: Protocol 4

###### Lemma V.4

Protocol 4(see fig. 9) distills for two copies of $P_{\delta}^{45}$

  * Proof

We start with two copies of $P_{\delta}^{45}$:

$P^{45}P^{45}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus x_{1}z_{1}=xy\oplus xz$, $0=xy\oplus xz\oplus
a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=x_{2}y_{2}\oplus x_{2}z_{2}=0\oplus x_{1}z_{1}=xz$, $0=xz\oplus
a_{2}\oplus b_{2}\oplus c_{2}$, equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xy\oplus
xz\oplus xz$, $a\oplus b\oplus c=xy$, hence $P^{xy}$

$P^{45}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus x_{1}z_{1}=xy\oplus xz$, $0=xy\oplus xz\oplus
a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we have $a_{2}\oplus b_{2}\oplus
c_{2}=0$, equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus
b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xy\oplus xz$, $a\oplus b\oplus
c=xy\oplus xz$, hence $P^{45}$

$P^{c}P^{45}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}\oplus
x_{2}z_{2}=0\oplus x_{1}z_{1}=xz$, $0=xz\oplus a_{2}\oplus b_{2}\oplus c_{2}$
equating we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus
1\oplus c_{1}\oplus c_{2}=xz$, $a\oplus b\oplus c=xz$, hence $P^{xz}$

$P^{c}P^{c}\rightarrow$ for box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating $a_{1}\oplus
b_{1}\oplus c_{1}=a_{2}\oplus b_{2}\oplus c_{2}$, $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$, $a\oplus
b\oplus c=0$, hence $P^{c}$

The final box is given by:

$\displaystyle P_{\delta}^{45}P_{\delta}^{45}$
$\displaystyle=\delta\left(\delta
P^{xy}+(1-\delta)P^{xz}\right)+(1-\delta)P_{\delta}^{45}$

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=-4\delta^{2}+10\delta+7$

The region of distillation for protocol 4 is $0<\delta\leqslant 1$.

####  V.2.3 Protocol 5

Figure 10: Protocol 5

###### Lemma V.5

Protocol 5(see fig. 10) distills for two copies of $P_{\delta}^{46}$

  * Proof

We start with two copies of $P_{\delta}^{46}$:

$P^{46}P^{46}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus y_{1}z_{1}\oplus x_{1}z_{1}=xy\oplus yz\oplus xz$,
$0=xy\oplus yz\oplus xz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we
have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}\oplus y_{2}z_{2}\oplus
x_{2}z_{2}=0\oplus 0\oplus(1\oplus x_{1})z_{1}=(1\oplus x)z=z\oplus xz$,
$0=z\oplus xz\oplus a_{2}\oplus b_{2}\oplus c_{2}$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus
c_{2}=xy\oplus yz\oplus xz\oplus z\oplus xz$, $a\oplus b\oplus c=xy\oplus
yz\oplus z$, $a\oplus b\oplus c=xy\oplus yz\oplus z$, hence $P^{xy\oplus
yz\oplus z}$

$P^{46}P^{c}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus
c_{1}=x_{1}y_{1}\oplus y_{1}z_{1}\oplus x_{1}z_{1}=xy\oplus yz\oplus
xz$,$0=xy\oplus yz\oplus xz\oplus a_{1}\oplus b_{1}\oplus c_{1}$, for box 2 we
have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus a_{1}\oplus
a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=xy\oplus
yz\oplus xz$, $a\oplus b\oplus c=xy\oplus yz\oplus xz$, hence $P^{46}$

$P^{c}P^{46}\rightarrow$ For box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=x_{2}y_{2}\oplus
y_{2}z_{2}\oplus x_{2}z_{2}=0\oplus 0\oplus(1\oplus x_{1})z_{1}=(1\oplus
x)z=z\oplus xz$, $0=z\oplus xz\oplus a_{2}\oplus b_{2}\oplus c_{2}$ equating
we get: $1\oplus a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus
c_{1}\oplus c_{2}=z\oplus xz$, $a\oplus b\oplus c=z\oplus xz$, hence
$P^{z\oplus xz}$

$P^{c}P^{c}\rightarrow$ for box 1 we have $a_{1}\oplus b_{1}\oplus c_{1}=0$,
for box 2 we have $a_{2}\oplus b_{2}\oplus c_{2}=0$, equating we get: $1\oplus
a_{1}\oplus a_{2}\oplus b_{1}\oplus b_{2}\oplus 1\oplus c_{1}\oplus c_{2}=0$,
$a\oplus b\oplus c=0$, hence $P^{c}$

The final box is given by:

$\displaystyle P_{\delta}^{46}P_{\delta}^{46}$
$\displaystyle=\delta\left(\delta P^{xy\oplus yz\oplus z}+(1-\delta)P^{z\oplus
xz}\right)+(1-\delta)P_{\delta}^{46}$

Applying (2) to the final box we get:

$\displaystyle V^{\prime}=8\delta^{2}-2\delta+7$

The region of distillation for protocol 5 is $\dfrac{3}{4}<\delta\leqslant 1$.

From fig. 11 we can conclude that protocol 2 as compared to protocol 1
performs much better for class 45, for class 46 protocol 2 performs slightly
better than protocol 1, however for class 44 protocol 1 performs better than
protocol 2, each of the unique protocols represented by black curves in fig.
11 perform better than common protocols for their respective classes and
distillation regions, among unique protocols protocol 4 for class 45 performs
the best.

Figure 11: Plot for $P_{\delta}^{N}$ distillation protocols

##  VI Final comments:

We presented here a limited study of distillation in tripartite NLBs, in
comparison to the bipartite($(2,2,2)$) case the tripartite($(3,2,2)$) case is
much more complicated with 53856 extremal points classified into 46
classesPironio  _et al._ (2011), because of this complexity we selected a
limited part of tripartite no-signalling polytope specifically a
representative of each class 44,45,46 and GHZ box, the reason for this
selection is based on a common criteria that each of these genuine tripartite
boxes in their definition have no one and two party expectations. As is the
case with tripartite extremal points the distillation is also complicated and
does lead to numerous distillation protocols, for GHZ box we selected a pair
of class 2 inequalities while for $P_{\delta}^{N}$ we selected a class 41
inequality out of 53856 inequalitiesSliwa (2003), this inequality has a
limited no-signalling violation for $P^{N}$ which provides a common box value
for $P^{N}$, while GHZ box is distilled with only single protocol,
$P_{\delta}^{N}$ has much more distillation protocols, for $P_{\delta}^{N}$ we
selected two common protocols, protocol 2 is selected for large no-signalling
violation of $P_{\delta}^{45}$, with the same selection criteria we selected
unique protocols, this has led to the conjecture that common protocols cannot
outperform unique protocols, clearly in our setting as far as no-signalling
violation is concerned class 45 has outperformed other classes. In future we
intend to study distillation in other classes however such a task will require
a suitable representation of boxes in terms of boolean polynomials which while
is not impossible but still is much more complicated, of course such a
exercise will lead to numerous protocols, which perhaps can be classified
according to some criteria.

###### Acknowledgements.

  * •

I am thankful to Jibran Rashid for his initial guidance.

  * •

I thank Cezary $\acute{\textnormal{S}}$liwa for his discussion on bell
inequalities.

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