to the following optimization program:

$$
\begin{align}
S_{\text{sum}}(W, k_1, k_2) &:= \underset{e_1, e_2, d_1, d_2}{\text{maximize}} \quad \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) d_1(i_1|y) \nonumber \\
&\quad + \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) d_2(i_2|y) \nonumber \\
&\text{subject to} \quad \sum_{x_1 \in \mathcal{X}_1} e_1(x_1|i_1) = 1, \forall i_1 \in [k_1] \nonumber \\
&\quad \sum_{x_2 \in \mathcal{X}_2} e_2(x_2|i_2) = 1, \forall i_2 \in [k_2] \nonumber \\
&\quad \sum_{j_1 \in [k_1]} d_1(j_1|y) = 1, \forall y \in \mathcal{Y} \nonumber \\
&\quad \sum_{j_2 \in [k_2]} d_2(j_2|y) = 1, \forall y \in \mathcal{Y} \nonumber \\
&\quad e_1(x_1|i_1), e_2(x_2|i_2), d_1(j_1|y), d_2(j_2|y) \ge 0 \tag{4.3}
\end{align}
$$

Note that we used independent decoders $d_1(j_1|y)$, $d_2(j_2|y)$ rather than a global $d(j_1j_2|y)$ here. This does not change the value of the optimization program. Indeed, since the program is convex, an optimal solution can be found on the extremal points of the search space. Thus, if we had used the variable $d(j_1j_2|y)$, we could always take it to be a function $d$ from $\mathcal{Y}$ to $[k_1] \times [k_2]$. Taking $d_1, d_2$ as the first and second coordinates of that function satisfies the equality $d(j_1j_2|y) = d_1(j_1|y)d_2(j_2|y)$, and therefore, the value of the program is the same in both cases. Note that it is also true for the program computing $S(W, k_1, k_2)$.

As for the usual (joint) success probability, we can define the sum-capacity region using Definition 2.12 by $C_{\text{sum}}(W) := C[S_{\text{sum}}](W)$. It turns out those two notions of success define the same capacity region:

**Proposition 4.2.** $C(W) = C_{\text{sum}}(W)$

*Proof.* Let us focus on error probabilities rather than success ones. Call them respectively $E(W, k_1, k_2) := 1 - S(W, k_1, k_2)$ and $E_{\text{sum}}(W, k_1, k_2) := 1 - S_{\text{sum}}(W, k_1, k_2)$. Let us fix a solution $e_1, d_1, e_2, d_2$ of the optimization program computing $S(W, k_1, k_2)$. Let us remark first that:

$$ \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) = k_1 k_2, $$

thus, the value of the maximum error for those encoders and decoders, which we call
$E(W, k_1, k_2, e_1, d_1, e_2, d_2)$, is:

$$ 
\begin{aligned}
& 1 - \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) d_1(i_1|y) d_2(i_2|y) \\
& = \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) \\
& - \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) d_1(i_1|y) d_2(i_2|y) \\
& = \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) [1 - d_1(i_1|y) d_2(i_2|y)].
\end{aligned} 
\quad (4.4) $$

Similarly, the value of the sum error for those encoder and decoders, which we call $E_{sum}(W, k_1, k_2, e_1, d_1, e_2, d_2)$, is:

$$ 
\begin{aligned}
& 1 - \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) \frac{d_1(i_1|y) + d_2(i_2|y)}{2} \\
& = \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) e_1(x_1|i_1) e_2(x_2|i_2) \left[ 1 - \frac{d_1(i_1|y) + d_2(i_2|y)}{2} \right].
\end{aligned} 
\quad (4.5) $$

However, for $x, y \in [0, 1]$, we have that:

$$ 1 - xy \geq \max(1-x, 1-y) \geq 1 - \frac{x+y}{2}, $$

and:

$$ 1 - xy \leq (1-x) + (1-y) = 2 \left( 1 - \frac{x+y}{2} \right). $$

This means that, for all $e_1, d_1, e_2, d_2$, we have:

$$ E_{sum}(W, k_1, k_2, e_1, d_1, e_2, d_2) \leq E(W, k_1, k_2, e_1, d_1, e_2, d_2), $$

and:

$$ E(W, k_1, k_2, e_1, d_1, e_2, d_2) \leq 2E_{sum}(W, k_1, k_2, e_1, d_1, e_2, d_2), $$

so, maximizing over all $e_1, d_1, e_2, d_2$, we get:

$$ E_{sum}(W, k_1, k_2) \leq E(W, k_1, k_2) \leq 2E_{sum}(W, k_1, k_2). $$

Thus, up to a multiplicative factor 2, the error is the same. In particular, when one of those errors tends to zero, the other one tends to zero as well. This implies that the capacity regions are the same. □

### 4.1.2 Non-Signaling Assisted Capacities

**Three-party non-signaling assistance** We now consider the case where the senders and the receiver are given non-signaling assistance. This resource, which is a theoretical but easier to manipulate generalization of quantum entanglement, can be characterized as
follows. A tripartite non-signaling box is described by a joint conditional probability distribution $P(x_1x_2(j_1j_2)|i_1i_2y)$ such that the marginal from any two parties is independent of the removed party's input, i.e., we have:

$$ 
\begin{align} 
\forall x_2, j_1, j_2, i_1, i_2, y, i'_1, \quad & \sum_{x_1} P(x_1x_2(j_1j_2)|i_1i_2y) = \sum_{x_1} P(x_1x_2(j_1j_2)|i'_1i'_2y), \nonumber \\ 
\forall x_1, j_1, j_2, i_1, i_2, y, i'_2, \quad & \sum_{x_2} P(x_1x_2(j_1j_2)|i_1i_2y) = \sum_{x_2} P(x_1x_2(j_1j_2)|i_1i'_2y), \tag{4.6} \\ 
\forall x_1, x_2, i_1, i_2, y, y', \quad & \sum_{j_1, j_2} P(x_1x_2(j_1j_2)|i_1i_2y) = \sum_{j_1, j_2} P(x_1x_2(j_1j_2)|i_1i_2y'). \nonumber 
\end{align} 
$$

This implies that one can consider for example $P(x_1x_2|i_1i_2)$ since it does not depend on $y$, or even $P(x_1|i_1)$ since it does not depend on $i_2, y$. Then, in our coding scenario, when the senders and the receiver are given non-signaling assistance, it means that they share a tripartite non-signaling box, the behavior of which is described by $P$. In this case, the expression $e_1(x_1|i_1)e_2(x_2|i_2)d(j_1j_2|y)$ in (4.1) is replaced by $P(x_1x_2(j_1j_2)|i_1i_2y)$, as depicted in Figure 4.2.

<figure><img src="image_14.png" /><figcaption>Figure 4.2 – A non-signaling box $P$ replacing $e_1, e_2$ and $d$ in the coding problem for the MAC $W$.</figcaption></figure>

The cyclicity of Figure 4.2 is at first sight counter-intuitive. Note first that $P$ being a non-signaling box is completely independent of $W$: in particular, the variable $y$ does not need to follow any law in the definition of $P$ being a non-signaling box. Therefore, the remaining ambiguity is the apparent need to encode and decode at the same time. However, since $P$ is a non-signaling box, we do not need to do both at the same time. Indeed, $\forall y, P(x_1x_2|i_1i_2) = P(x_1x_2|i_1i_2y)$ by the non-signaling property of $P$. Thus, one can get the outputs $x_1, x_2$ on inputs $i_1, i_2$ without access to $y$, as that knowledge won't affect the laws of $x_1, x_2$. Then $y$ follows the law given by $W$ given those $x_1, x_2$. Finally, given access to $y$, the decoding process is described by:

$$ P((j_1j_2)|i_1i_2yx_1x_2) = \frac{P(x_1x_2(j_1j_2)|i_1i_2y)}{P(x_1x_2|i_1i_2y)} = \frac{P(x_1x_2(j_1j_2)|i_1i_2y)}{P(x_1x_2|i_1i_2)}, $$

so we recover globally $P((j_1j_2)|i_1i_2yx_1x_2) \times P(x_1x_2|i_1i_2) = P(x_1x_2(j_1j_2)|i_1i_2y)$ the prescribed conditional probability. The non-signaling condition ensures that it is possible
to consider the conditional probabilities of each party independently. This clarifies how one can effectively encode and then decode messages through a non-signaling box.

As in the unassisted case, we want to maximize over all non-signaling boxes $P$ the probability of successfully encoding and decoding the messages through $W$, i.e. the probability that $j_1 = i_1$ and $j_2 = i_2$, given that the input messages are taken uniformly in $[k_1]$ and $[k_2]$. We call this quantity $S^{\text{NS}}(W, k_1, k_2)$, which is characterized by the following optimization program:

$$
\begin{align}
S^{\text{NS}}(W, k_1, k_2) &:= \underset{P}{\text{maximize}} \quad \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P(x_1 x_2(i_1 i_2)|i_1 i_2 y) \nonumber \\
&\text{subject to} \quad \sum_{x_1} P(x_1 x_2(j_1 j_2)|i_1 i_2 y) = \sum_{x_1} P(x_1 x_2(j_1 j_2)|i_1' i_2 y) \nonumber \\
&\qquad \sum_{x_2} P(x_1 x_2(j_1 j_2)|i_1 i_2 y) = \sum_{x_2} P(x_1 x_2(j_1 j_2)|i_1 i_2' y) \nonumber \\
&\qquad \sum_{j_1, j_2} P(x_1 x_2(j_1 j_2)|i_1 i_2 y) = \sum_{j_1, j_2} P(x_1 x_2(j_1 j_2)|i_1 i_2 y') \nonumber \\
&\qquad \sum_{x_1, x_2, j_1, j_2} P(x_1 x_2(j_1 j_2)|i_1 i_2 y) = 1 \nonumber \\
&\qquad P(x_1 x_2(j_1 j_2)|i_1 i_2 y) \ge 0 \tag{4.7}
\end{align}
$$

Since it is given as a linear program, the complexity of computing $S^{\text{NS}}(W, k_1, k_2)$ is polynomial in the number of variables and constraints (see for instance Section 7.1 of [GM07]), which is a polynomial in $|\mathcal{X}_1|, |\mathcal{X}_2|, |\mathcal{Y}|, k_1$ and $k_2$. Also, as it is easy to check that a classical strategy is a particular case of a non-signaling assisted strategy, we have that $S^{\text{NS}}(W, k_1, k_2) \ge S(W, k_1, k_2)$.

As for the unassisted case, we define the non-signaling assisted capacity region using Definition 2.12 and $S^{\text{NS}}$ by $C^{\text{NS}}(W) := C[S^{\text{NS}}](W)$, and its zero-error equivalent is defined by $C_0^{\text{NS}}(W) := C_0[S^{\text{NS}}](W)$.

**Independent non-signaling assistance** One can also consider the case where non-signaling assistance is shared independently between the first sender and the receiver as well as between the second encoder and the receiver, which we call independent non-signaling assistance. The precise scenario is depicted in Figure 4.3:
<figure><img src="image_15.png" /><figcaption>Figure 4.3 – Non-signaling boxes $P^1, P^2$ replacing $e_1, d_1$ and $e_2, d_2$ in the coding problem for the MAC $W$.</figcaption></figure>

This leads to the following definition of the success probability $S^{\text{NSSR}}(W, k_1, k_2)$:

$$
\begin{array}{ll}
\underset{P^1,P^2}{\text{maximize}} & \displaystyle \frac{1}{k_1 k_2} \sum_{i_1,i_2,x_1,x_2,y} W(y|x_1x_2) P^1(x_1i_1|i_1y) P^2(x_2i_2|i_2y) \\
\text{subject to} & \displaystyle \sum_{x_1} P^1(x_1j_1|i_1y) = \sum_{x_1} P^1(x_1j_1|i'_1y) \\
& \displaystyle \sum_{j_1} P^1(x_1j_1|i_1y) = \sum_{j_1} P^1(x_1j_1|i_1y') \\
& \displaystyle \sum_{x_1,j_1} P^1(x_1j_1|i_1y) = 1 \\
& \displaystyle \sum_{x_2} P^2(x_2j_2|i_2y) = \sum_{x_2} P^2(x_2j_2|i'_2y) \\
& \displaystyle \sum_{j_2} P^2(x_2j_2|i_2y) = \sum_{j_2} P^2(x_2j_2|i_2y') \\
& \displaystyle \sum_{x_2,j_2} P^2(x_2j_2|i_2y) = 1 \\
& P^1(x_1j_1|i_1y), P^2(x_2j_2|i_2y) \ge 0
\end{array}
\qquad (4.8)
$$
As before, one can also consider the sum-success probability $S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2)$:

$$
\begin{align}
\text{maximize} \quad & \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P^1(x_1 i_1 | i_1 y) \sum_{j_2} P^2(x_2 j_2 | i_2 y) \nonumber \\
& + \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P^2(x_2 i_2 | i_2 y) \sum_{j_1} P^1(x_1 j_1 | i_1 y) \nonumber \\
\text{subject to} \quad & \sum_{x_1} P^1(x_1 j_1 | i_1 y) = \sum_{x_1} P^1(x_1 j_1 | i'_1 y) \nonumber \\
& \sum_{j_1} P^1(x_1 j_1 | i_1 y) = \sum_{j_1} P^1(x_1 j_1 | i_1 y') \nonumber \\
& \sum_{x_1, j_1} P^1(x_1 j_1 | i_1 y) = 1 \tag{4.9} \\
& \sum_{x_2} P^2(x_2 j_2 | i_2 y) = \sum_{x_2} P^2(x_2 j_2 | i'_2 y) \nonumber \\
& \sum_{j_2} P^2(x_2 j_2 | i_2 y) = \sum_{j_2} P^2(x_2 j_2 | i_2 y') \nonumber \\
& \sum_{x_2, j_2} P^2(x_2 j_2 | i_2 y) = 1 \nonumber \\
& P^1(x_1 j_1 | i_1 y), P^2(x_2 j_2 | i_2 y) \ge 0 \nonumber
\end{align}
$$

We define the independent non-signaling assisted capacity (resp. sum-capacity) region using Definition 2.12 by $C^{\text{NSSR}}(W) := C[\mathbb{S}^{\text{NSSR}}](W)$ (resp. $C^{\text{NSSR}}_{\text{sum}}(W) := C[\mathbb{S}^{\text{NSSR}}_{\text{sum}}](W)$). It turns out those two notions of success define the same capacity region:

**Proposition 4.3.** $C^{\text{NSSR}}(W) = C^{\text{NSSR}}_{\text{sum}}(W)$

*Proof.* Given non-signaling boxes $P^1, P^2$, the maximum success probability of encoding and decoding correctly with those is given by:

$$S^{\text{NSSR}}(W, k_1, k_2, P^1, P^2) := \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P^1(x_1 i_1 | i_1 y) P^2(x_2 i_2 | i_2 y).$$

This should be compared to the sum success probability of encoding and decoding correctly with those, which we call $S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2, P^1, P^2)$ and is equal to:

$$
\frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P^1(x_1 i_1 | i_1 y) \sum_{j_2} P^2(x_2 j_2 | i_2 y) \\
+ \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P^2(x_2 i_2 | i_2 y) \sum_{j_1} P^1(x_1 j_1 | i_1 y). \tag{4.10}
$$

Similarly to what was done in Proposition 4.2, we focus on error probabilities rather than success probabilities. We have that:

$$\frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \sum_{j_1, j_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) = 1,$$
so we get that $E^{\text{NSSR}}(W, k_1, k_2, P^1, P^2)$ is equal to:

$$ 
\begin{aligned}
& \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \sum_{j_1, j_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) \\
& - \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) P^1(x_1 i_1 | i_1 y) P^2(x_2 i_2 | i_2 y) \\
& = \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \sum_{(j_1, j_2) \neq (i_1, i_2)} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y).
\end{aligned} 
\quad (4.11) $$

On the other hand, since:

$$ 
\begin{aligned}
& \sum_{j_1, j_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) - P^1(x_1 i_1 | i_1 y) \sum_{j_2} P^2(x_2 j_2 | i_2 y) \\
& = \sum_{j_1 \neq i_1, j_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y),
\end{aligned} 
\quad (4.12) $$

and symmetrically, we get that $E_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2, P^1, P^2)$ is equal to:

$$ 
\begin{aligned}
& \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \sum_{j_1 \neq i_1, j_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) \\
& + \frac{1}{2k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \sum_{j_1, j_2 \neq i_1} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) \\
& = \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \sum_{j_1 \neq i_1, j_2 \neq i_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) \\
& + \frac{1}{k_1 k_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \frac{\sum_{(j_1, j_2) \in S} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y)}{2},
\end{aligned} 
\quad (4.13) $$

with $S := \{(j_1, i_2) : j_1 \in [k_1] - \{i_1\}\} \cup \{(i_1, j_2) : j_2 \in [k_2] - \{i_2\}\}$. However, we have that:

$$ 
\begin{aligned}
& \sum_{j_1 \neq i_1, j_2 \neq i_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) + \frac{\sum_{(j_1, j_2) \in S} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y)}{2} \\
& \le \sum_{j_1 \neq i_1, j_2 \neq i_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) + \sum_{(j_1, j_2) \in S} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) \\
& = \sum_{(j_1, j_2) \neq (i_1, i_2)} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) \\
& \le 2 \left( \sum_{j_1 \neq i_1, j_2 \neq i_2} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y) + \frac{\sum_{(j_1, j_2) \in S} P^1(x_1 j_1 | i_1 y) P^2(x_2 j_2 | i_2 y)}{2} \right).
\end{aligned} 
\quad (4.14) $$

This implies that:

$$ E_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2, P^1, P^2) \le E^{\text{NSSR}}(W, k_1, k_2, P^1, P^2) \le 2E_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2, P^1, P^2), $$

and by maximizing over all $P^1$ and $P^2$:

$$ E_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2) \le E^{\text{NSSR}}(W, k_1, k_2) \le 2E_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2). $$

Thus, as before, the capacity regions are the same. □
## 4.2 Hardness of Approximation for Multiple-Access Channel Coding

Since MACs are more general than point-to-point channels (by defining $W(y|x_1x_2) := \hat{W}(y|x_1)$ for $\hat{W}$ a point-to-point channel and looking only at its first input), computing $S(W, k_1, k_2)$ is NP-hard, and it is even NP-hard to approximate within a better ratio than $1 - e^{-1}$, as a consequence of the hardness result for point-to-point channels from [BF18].

On the other hand, it has been shown in [LALS20] that it is NP-hard to decide whether a rate pair belongs to the classical capacity region of a MAC.

The goal of this section is to show that, under some reasonable hardness assumptions, the multiple-access channel coding problem cannot be approximated in polynomial time within a $\Omega(1)$ factor. We will achieve this through a simple reduction from the bipartite version of the densest k-subgraph problem [FKP01], where one aims at finding a subgraph of k vertices with the highest number of edges. In the bipartite version, given a bipartite graph $G = (A \sqcup B, E)$ and two parameters $k_1$ and $k_2$, the goal is to find subsets $S \subseteq A$ and $T \subseteq B$ such that $|S| = k_1, |T| = k_2$ with the maximum number of edges contained in $S \cup T$, i.e. $|(S \times T) \cap E|$. With a simple reduction from the maximum edge biclique problem [Pee03], the bipartite densest subgraph problem is NP-complete.

A well-known result by Alon et al. [AAM$^{+}$11] shows that, under the following complexity hypothesis on random clauses with k literals (k-SAT formulas) introduced by Feige in [Fei02], there is no polynomial-time $\Omega(1)$-approximation algorithm for the bipartite densest subgraph problem; see [AAM$^{+}$11] for further details.

**Hypothesis 4.1** ([Fei02]). For some constant $c > 0$, for every $k$, for $\Delta$ a sufficiently large constant independent of $n$, there is no polynomial-time algorithm that on most random k-SAT formulas with $n$ variables and $m = \Delta n$ clauses outputs **typical**, but never outputs **typical** on k-SAT formulas with at least $m2^{-c\sqrt{k}}$ satisfiable clauses.

We will consider the following reformulation of the multiple-access channel coding problem:

**Proposition 4.4.** Given a MAC $W$ and integers $k_1, k_2$, the multiple-access channel coding problem entails maximizing:

$$f_W(S, T) := \frac{1}{|S||T|} \sum_{y \in \mathcal{Y}} \max_{x_1 \in S, x_2 \in T} W(y|x_1x_2),$$

over all sets $S \subseteq \mathcal{X}_1, T \subseteq \mathcal{X}_2$ with $|S| = k_1, |T| = k_2$.

*Proof.* As the multiple-access channel coding problem is a convex program, its optimal solution can be found on integer solutions, i.e. when $e_1, e_2$ and $d$ are functions. Note that this restriction does not affect the hardness of the problem as a deterministic code with a better or equal value can be retrieved easily from any code.

Therefore, given encoders $e_1, e_2$, the objective value becomes:

$$\max_d \frac{1}{k_1 k_2} \sum_{i_1, i_2, y} W(y|e_1(i_1)e_2(i_2)) \mathbb{1}_{d(y)=(i_1, i_2)} = \frac{1}{k_1 k_2} \sum_y \max_{i_1, i_2} W(y|e_1(i_1)e_2(i_2)),$$
by taking $d(y) \in \argmax_{i_1, i_2}\{W(y|e_1(i_1)e_2(i_2))\}$. Finally, optimizing over functions $e_1, e_2$ is the same as optimizing over sets $S, T$ of size $k_1, k_2$ (corresponding to their images). Thus, we have that the multiple-access channel coding can be reformulated as:

$$ \max_{S \subseteq \mathcal{X}_1, T \subseteq \mathcal{X}_2: |S|=k_1, |T|=k_2} \frac{1}{|S||T|} \sum_{y \in \mathcal{Y}} \max_{x_1 \in S, x_2 \in T} W(y|x_1 x_2). \quad \square $$

The previously mentioned hardness of approximation for the bipartite densest subgraph problem from [AAM$^+$11] will imply the following hardness of approximation for the multiple-access channel coding problem:

**Theorem 4.5.** Under Hypothesis 4.1, there is no polynomial-time $\Omega(1)$-approximation algorithm for the multiple-access channel coding problem.

*Proof.* Given a bipartite graph $G = (A \cup B, E)$, we define its associated MAC $W_G$ on $\mathcal{X}_1 := A, \mathcal{X}_2 := B$ and $\mathcal{Y} := A \times B$ in the following way:

$$ W((a', b')|ab) = \begin{cases} 1_{a=a'} 1_{b=b'} & \text{if } (a, b) \in E, \\ \frac{1}{|A||B|} & \text{otherwise.} \end{cases} \quad (4.15) $$

It is indeed a MAC as $\sum_{a', b'} W((a', b')|ab) = \sum_{a', b'} 1_{a=a'} 1_{b=b'} = 1$ if $(a, b) \in E$ and $\sum_{a', b'} W((a', b')|ab) = \sum_{a', b'} \frac{1}{|A||B|} = 1$ otherwise.

We claim the following linear relation between the objective functions of the bipartite densest subgraph problem on $G$ and the multiple-access channel coding problem on $W_G$:

**Lemma 4.6.** For $G = (A \cup B, E)$, $S \subseteq A$ and $T \subseteq B$:

$$ \frac{f_{W_G}(S, T) - \frac{1}{|S||T|}}{1 - \frac{1}{|A||B|}} = \frac{|(S \times T) \cap E|}{|S||T|}. $$

*Proof of Lemma 4.6.* We have that:

$$ \begin{aligned} f_{W_G}(S, T) &= \frac{1}{|S||T|} \sum_{a', b'} \max_{a \in S, b \in T} W((a', b')|ab) \\ &= \frac{1}{|S||T|} \left( \sum_{(a', b') \in (S \times T) \cap E} \max_{a \in S, b \in T} W((a', b')|ab) + \sum_{(a', b') \notin (S \times T) \cap E} \max_{a \in S, b \in T} W((a', b')|ab) \right) \\ &= \frac{1}{|S||T|} \left( \sum_{(a', b') \in (S \times T) \cap E} 1 + \sum_{(a', b') \notin (S \times T) \cap E} \frac{1}{|A||B|} \right) \\ &= \frac{1}{|S||T|} \left( |(S \times T) \cap E| + \frac{|A||B| - |(S \times T) \cap E|}{|A||B|} \right) \\ &= \frac{1}{|S||T|} + \frac{|(S \times T) \cap E|}{|S||T|} \left( 1 - \frac{1}{|A||B|} \right), \end{aligned} \quad (4.16) $$

which implies the claimed equality. $\square$
Let us now assume that a polynomial-time constant approximation of the multiple-access channel coding problem exists. Let us call $\lambda \in (0, 1)$ the ratio of that algorithm. Then, for any $G = (A \sqcup B, E)$, $k_1, k_2$, we can apply our algorithm to $W_G$, $k_1, k_2$ and get $S_{approx}$, $T_{approx}$ of size $k_1, k_2$ such that:

$$f_{W_G}(S_{approx}, T_{approx}) \geq \lambda \cdot \max_{S,T:|S|=k_1,|T|=k_2} f_{W_G}(S,T).$$

But, thanks to Lemma 4.6, we have that:

$$\frac{\max_{S,T:|S|=k_1,|T|=k_2} f_{W_G}(S,T) - \frac{1}{k_1 k_2}}{1 - \frac{1}{|A||B|}} = \frac{\max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E|}{k_1 k_2}.$$

So in all, we have:

$$\begin{align}
\frac{|(S_{approx} \times T_{approx}) \cap E|}{k_1 k_2} &= \frac{f_{W_G}(S_{approx}, T_{approx}) - \frac{1}{k_1 k_2}}{1 - \frac{1}{|A||B|}} \nonumber \\
&\geq \frac{\lambda \cdot \max_{S,T:|S|=k_1,|T|=k_2} f_{W_G}(S,T) - \frac{1}{k_1 k_2}}{1 - \frac{1}{|A||B|}} \tag{4.17} \\
&= \lambda \cdot \frac{\max_{S,T:|S|=k_1,|T|=k_2} f_{W_G}(S,T) - \frac{1}{k_1 k_2}}{1 - \frac{1}{|A||B|}} - \frac{\frac{1-\lambda}{k_1 k_2}}{1 - \frac{1}{|A||B|}} \nonumber \\
&= \lambda \cdot \frac{\max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E|}{k_1 k_2} - \frac{\frac{1-\lambda}{k_1 k_2}}{1 - \frac{1}{|A||B|}}, \nonumber
\end{align}$$

and thus:

$$|(S_{approx} \times T_{approx}) \cap E| \geq \lambda \cdot \max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E| - \frac{1-\lambda}{1 - \frac{1}{|A||B|}}.$$

Let us consider instances of the bipartite densest subgraph problem with $|E| \geq \min(k_1, k_2)$, which implies directly that $\max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E| \geq \min(k_1, k_2)$ by picking $\min(k_1, k_2)$ edges and taking their endpoints to construct $S$ and $T$. As $|A||B| \geq 2$, we have that $\frac{1-\lambda}{1-\frac{1}{|A||B|}} \leq 2$. Then:

$$\begin{align}
|(S_{approx} \times T_{approx}) \cap E| &\geq \lambda \cdot \max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E| - 2 \nonumber \\
&\geq (1-c)\lambda \cdot \max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E| + c\lambda \min(k_1, k_2) - 2 \quad \text{for any } c \in (0, 1) \nonumber \\
&= \left(1 - \frac{2}{\lambda \min(k_1, k_2)}\right) \lambda \cdot \max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E|, \tag{4.18}
\end{align}$$

by choosing $c = \frac{2}{\lambda \min(k_1, k_2)}$. Finally, for instances such that $k_1, k_2 \geq \frac{4}{\lambda}$, we have that:

$$|(S_{approx} \times T_{approx}) \cap E| \geq \frac{\lambda}{2} \cdot \max_{S,T:|S|=k_1,|T|=k_2} |(S \times T) \cap E|.$$

Therefore, if a polynomial-time approximation of the multiple-access channel coding problem of constant ratio $\lambda$ exists, then we have a polynomial-time $\frac{\lambda}{2}$-approximation of
the bipartite densest subgraph problem for instances satisfying $|E| \ge \min(k_1, k_2)$ and $k_1, k_2 \ge \frac{4}{\lambda}$. Note that the bipartite densest subgraph problem can be solved in polynomial time for instances which do not satisfy these constraints (as $\frac{1}{\lambda}$ is constant). Thus, we have in all a constant polynomial-time approximation for the bipartite densest subgraph problem, which is in contradiction with Hypothesis 4.1 as proved in [AAM$^{+}$11]. Hence, under Hypothesis 4.1, there is no polynomial-time $\Omega(1)$-approximation algorithm for the multiple-access channel coding problem. $\square$

# 4.3 Properties of Non-Signaling Assisted Codes

## 4.3.1 Symmetrization

One can prove an equivalent formulation of the linear program computing $S^{\text{NS}}(W, k_1, k_2)$ with a number of variables and constraints polynomial in only $|\mathcal{X}_1|, |\mathcal{X}_2|$ and $|\mathcal{Y}|$ and independent of $k_1$ and $k_2$:

**Proposition 4.7.** For a MAC $W: \mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}$ and $k_1, k_2 \in \mathbb{N}^*$, we have:

$$
\begin{align}
S^{\text{NS}}(W, k_1, k_2) = & \underset{r,r^1,r^2,p}{\text{maximize}} \quad \frac{1}{k_1 k_2} \sum_{x_1,x_2,y} W(y|x_1 x_2) r_{x_1,x_2,y} \nonumber \\
& \text{subject to} \quad \sum_{x_1,x_2} r_{x_1,x_2,y} = 1 \nonumber \\
& \qquad \sum_{x_1} r_{x_1,x_2,y}^1 = k_1 \sum_{x_1} r_{x_1,x_2,y} \nonumber \\
& \qquad \sum_{x_2} r_{x_1,x_2,y}^2 = k_2 \sum_{x_2} r_{x_1,x_2,y} \tag{4.19} \\
& \qquad \sum_{x_1} p_{x_1,x_2} = k_1 \sum_{x_1} r_{x_1,x_2,y}^2 \nonumber \\
& \qquad \sum_{x_2} p_{x_1,x_2} = k_2 \sum_{x_2} r_{x_1,x_2,y}^1 \nonumber \\
& \qquad 0 \le r_{x_1,x_2,y} \le r_{x_1,x_2,y}^1, r_{x_1,x_2,y}^2 \le p_{x_1,x_2} \nonumber \\
& \qquad p_{x_1,x_2} - r_{x_1,x_2,y}^1 - r_{x_1,x_2,y}^2 + r_{x_1,x_2,y} \ge 0 \nonumber
\end{align}
$$

*Proof.* One can check that given a solution of the original program, the following choice of variables is a valid solution of the second program achieving the same objective value:

$$
\begin{align}
r_{x_1,x_2,y} &:= \sum_{i_1,i_2} P(x_1 x_2 (i_1 i_2) | i_1 i_2 y), & r_{x_1,x_2,y}^1 &:= \sum_{j_1,i_1,i_2} P(x_1 x_2 (j_1 i_2) | i_1 i_2 y), \nonumber \\
r_{x_1,x_2,y}^2 &:= \sum_{j_2,i_1,i_2} P(x_1 x_2 (i_1 j_2) | i_1 i_2 y), & p_{x_1,x_2} &:= \sum_{j_1,j_2,i_1,i_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y). \tag{4.20}
\end{align}
$$

Note that $p_{x_1,x_2}$ is well-defined since $\sum_{j_1,j_2,i_1,i_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y)$ is independent of $y$ as $P$ is a non-signaling box.

For the other direction, given those variables, a non-signaling probability distribution $P(x_1 x_2 (j_1 j_2) | i_1 i_2 y)$ achieving the same objective value is given by, for $j_1 \neq i_1$ and
$j_2 \neq i_2:$

$$
\begin{align}
P(x_1 x_2 (i_1 i_2) | i_1 i_2 y) &:= \frac{r_{x_1, x_2, y}}{k_1 k_2}, \nonumber \\
P(x_1 x_2 (j_1 i_2) | i_1 i_2 y) &:= \frac{r_{x_1, x_2, y}^1 - r_{x_1, x_2, y}}{k_1 k_2 (k_1 - 1)}, \nonumber \\
P(x_1 x_2 (i_1 j_2) | i_1 i_2 y) &:= \frac{r_{x_1, x_2, y}^2 - r_{x_1, x_2, y}}{k_1 k_2 (k_2 - 1)}, \tag{4.21} \\
P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) &:= \frac{p_{x_1, x_2} - r_{x_1, x_2, y}^1 - r_{x_1, x_2, y}^2 + r_{x_1, x_2, y}}{k_1 k_2 (k_1 - 1) (k_2 - 1)}. \nonumber
\end{align}
$$

<figure><img src="image_16.png" /></figure>

This symmetrization can also be done for the program computing $S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2)$:

**Proposition 4.8.**

$$
\begin{align}
S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2) &= \underset{r^1, r^2, p^1, p^2}{\text{maximize}} \quad \frac{1}{2k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) (p_{x_2}^2 r_{x_1, y}^1 + p_{x_1}^1 r_{x_2, y}^2) \nonumber \\
&= \frac{1}{2k_1} \sum_{x_1, y} W_{p^2, k_2}^1(y|x_1) r_{x_1, y}^1 + \frac{1}{2k_2} \sum_{x_2, y} W_{p^1, k_1}^2(y|x_2) r_{x_2, y}^2 \nonumber \\
&\quad \text{with } W_{p^2, k_2}^1(y|x_1) := \frac{1}{k_2} \sum_{x_2} W(y|x_1 x_2) p_{x_2}^2 \nonumber \\
&\quad \text{and } W_{p^1, k_1}^2(y|x_2) := \frac{1}{k_1} \sum_{x_1} W(y|x_1 x_2) p_{x_1}^1 \nonumber \\
&\quad \text{subject to } \sum_{x_1} r_{x_1, y}^1 = 1, \sum_{x_2} r_{x_2, y}^2 = 1 \nonumber \\
&\qquad \sum_{x_1} p_{x_1}^1 = k_1, \sum_{x_2} p_{x_2}^2 = k_2 \nonumber \\
&\qquad 0 \le r_{x_1, y}^1 \le p_{x_1}^1, 0 \le r_{x_2, y}^2 \le p_{x_2}^2 \tag{4.22}
\end{align}
$$

*Proof.* One can check that given a solution of the original program, the following choice of variables is a valid solution of the second program achieving the same objective value:

$$
\begin{align}
r_{x_1, y}^1 &:= \sum_{i_1} P^1(x_1 i_1 | i_1 y), \quad p_{x_1}^1 := \sum_{j_1, i_1} P^1(x_1 j_1 | i_1 y), \nonumber \\
r_{x_2, y}^2 &:= \sum_{i_2} P^2(x_2 i_2 | i_2 y), \quad p_{x_2}^2 := \sum_{j_2, i_2} P^2(x_2 j_2 | i_2 y). \tag{4.23}
\end{align}
$$

Note that $p_{x_1}^1$ and $p_{x_2}^2$ are well-defined since $\sum_{j_1, i_1} P^1(x_1 j_1 | i_1 y)$ and $\sum_{j_2, i_2} P^2(x_2 j_2 | i_2 y)$ are independent of $y$ as $P^1$ and $P^2$ are non-signaling boxes.

For the other direction, given those variables, non-signaling probability distributions $P^1(x_1 j_1 | i_1 y)$ and $P^2(x_2 j_2 | i_2 y)$ achieving the same objective value are given by, for $j_1 \neq i_1$
and $j_2 \neq i_2$:

$$
\begin{align}
P^1(x_1 i_1 | i_1 y) &:= \frac{r_{x_1, y}^1}{k_1}, \nonumber \\
P^1(x_1 j_1 | i_1 y) &:= \frac{p_{x_1, y}^1 - r_{x_1, y}^1}{k_1(k_1 - 1)}, \tag{4.24} \\
P^2(x_2 i_2 | i_2 y) &:= \frac{r_{x_2, y}^2}{k_2}, \nonumber \\
P^2(x_2 j_2 | i_2 y) &:= \frac{p_{x_2, y}^2 - r_{x_2, y}^2}{k_2(k_2 - 1)}. \nonumber
\end{align}
$$

<figure><img src="image_17.png" /></figure>

4.3.2 Properties of S<sup>NS</sup>(W, k<sub>1</sub>, k<sub>2</sub>), C<sup>NS</sup>(W) and C<sub>0</sub><sup>NS</sup>(W)

Proposition 4.9. For a MAC W : $\mathcal{X}_1 \times \mathcal{X}_2 \rightarrow \mathcal{Y}$ and $k_1, k_2 \in \mathbb{N}^*$, we have:

1. $\frac{1}{k_1 k_2} \le S^{\text{NS}}(W, k_1, k_2) \le 1$.

2. $S^{\text{NS}}(W, k_1, k_2) \le \min \left( \frac{|\mathcal{X}_1|}{k_1}, \frac{|\mathcal{X}_2|}{k_2}, \frac{|\mathcal{Y}|}{k_1 k_2} \right)$.

3. If $k'_1 \le k_1$ and $k'_2 \le k_2$, then $S^{\text{NS}}(W, k'_1, k'_2) \ge S^{\text{NS}}(W, k_1, k_2)$.

4. For $W' : \mathcal{X}'_1 \times \mathcal{X}'_2 \to \mathcal{Y}'$ and $k_1, k_2 \in \mathbb{N}^*$, we have $S^{\text{NS}}(W \otimes W', k_1 k'_1, k_2 k'_2) \ge S^{\text{NS}}(W, k_1, k_2) \cdot S^{\text{NS}}(W', k'_1, k'_2)$. In particular, for any positive integer $n$, we have $S^{\text{NS}}(W^{\otimes n}, k_1^n, k_2^n) \ge [S^{\text{NS}}(W, k_1, k_2)]^n$ and $S^{\text{NS}}(W \otimes W', k_1, k_2) \ge S^{\text{NS}}(W, k_1, k_2)$.

Proof. 1. We will start by proving that $S^{\text{NS}}(W, k_1, k_2) \ge \frac{1}{k_1 k_2}$. Take $p_{x_1,x_2} := \frac{k_1 k_2}{|\mathcal{X}_1||\mathcal{X}_2|}$, $r_{x_1,x_2,y}^1 := \frac{p_{x_1,x_2}}{k_2}$, $r_{x_1,x_2,y}^2 := \frac{p_{x_1,x_2}}{k_1}$ and $r_{x_1,x_2,y} := \frac{p_{x_1,x_2}}{k_1 k_2} = \frac{1}{|\mathcal{X}_1||\mathcal{X}_2|}$. One can easily check that it is indeed a valid solution of the linear program computing $S^{\text{NS}}(W, k_1, k_2)$. Thus we have:

$$
\begin{align}
S^{\text{NS}}(W, k_1, k_2) &\ge \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) r_{x_1, x_2, y} \nonumber \\
&= \frac{1}{k_1 k_2} \sum_{x_1, x_2} \frac{1}{|\mathcal{X}_1||\mathcal{X}_2|} \sum_y W(y|x_1 x_2) \tag{4.25} \\
&= \frac{1}{k_1 k_2} \sum_{x_1, x_2} \frac{1}{|\mathcal{X}_1||\mathcal{X}_2|} = \frac{1}{k_1 k_2}. \nonumber
\end{align}
$$

Furthermore, in order to show that it is at most 1, let us consider an optimal solution of $S^{\text{NS}}(W, k_1, k_2)$. We have:

$$
\begin{align}
S^{\text{NS}}(W, k_1, k_2) &= \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) r_{x_1, x_2, y} \le \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) p_{x_1, x_2} \nonumber \\
&= \frac{1}{k_1 k_2} \sum_{x_1, x_2} p_{x_1, x_2} \sum_{y} W(y|x_1 x_2) = \frac{1}{k_1 k_2} \sum_{x_1, x_2} p_{x_1, x_2} = 1, \tag{4.26}
\end{align}
$$

$$
\text{since } \sum_{x_1,x_2} p_{x_1,x_2} = k_1 \sum_{x_1,x_2} r_{x_1,x_2,y}^2 = k_1 k_2 \sum_{x_1,x_2} r_{x_1,x_2,y} = k_1 k_2.
$$
2. First let us show that $S^{\text{NS}}(W, k_1, k_2) \le \frac{|\mathcal{X}_1|}{k_1}$ (the case $S^{\text{NS}}(W, k_1, k_2) \le \frac{|\mathcal{X}_2|}{k_2}$ is symmetric):

$$
\begin{align}
S^{\text{NS}}(W, k_1, k_2) &= \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) r_{x_1, x_2, y} \nonumber \\
&\le \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) r_{x_1, x_2, y}^2 \nonumber \\
&\le \frac{1}{k_1 k_2} \sum_{x_2, y} \left( \sum_{x'_1} W(y|x'_1 x_2) \right) \cdot \left( \sum_{x_1} r_{x_1, x_2, y}^2 \right) \tag{4.27} \\
&= \frac{1}{k_1 k_2} \sum_{x_2, y} \left( \sum_{x'_1} W(y|x'_1 x_2) \right) \cdot \left( \frac{1}{k_1} \sum_{x_1} p_{x_1, x_2} \right) \nonumber \\
&= \frac{1}{k_1^2 k_2} \sum_{x_1, x_2} p_{x_1, x_2} \sum_{x'_1} \left( \sum_y W(y|x'_1 x_2) \right) \nonumber \\
&= \frac{|\mathcal{X}_1|}{k_1^2 k_2} \sum_{x_1, x_2} p_{x_1, x_2} = \frac{|\mathcal{X}_1|}{k_1} . \nonumber
\end{align}
$$

Let us show now that $S^{\text{NS}}(W, k_1, k_2) \le \frac{|\mathcal{Y}|}{k_1 k_2}$:

$$
\begin{align}
S^{\text{NS}}(W, k_1, k_2) &= \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) r_{x_1, x_2, y} \nonumber \\
&\le \frac{1}{k_1 k_2} \sum_y \left( \max_{x_1, x_2} W(y|x_1 x_2) \right) \sum_{x_1, x_2} r_{x_1, x_2, y} \tag{4.28} \\
&\le \frac{1}{k_1 k_2} \sum_y \sum_{x_1, x_2} r_{x_1, x_2, y} = \frac{|\mathcal{Y}|}{k_1 k_2}. \nonumber
\end{align}
$$

3. Let us assume that $k'_1 \le k_1$ and that $k'_2 = k_2$, since this latter case will follow by symmetry. Consider an optimal solution of $S^{\text{NS}}(W, k_1, k_2) = \frac{1}{k_1} \sum_{i_1 \in [k_1]} f(i_1)$ with:

$$
f(i_1) := \frac{1}{k_2} \sum_{x_1, x_2, y, i_2} W(y|x_1 x_2) P(x_1 x_2(i_1 i_2)|i_1 i_2 y),
$$

and $P$ non-signaling. Let us consider $S \in \underset{S' \subseteq [k_1]: |S'|=k'_1}{\operatorname{argmax}} \sum_{i_1 \in S'} f(i_1)$. Then, by construction, we have that $\frac{1}{k'_1} \sum_{i_1 \in S} f(i_1) \ge \frac{1}{k_1} \sum_{i_1 \in [k_1]} f(i_1) = S^{\text{NS}}(W, k_1, k_2)$, since we have taken the average of the $k'_1$ largest values of the sum.

Let us define the strategy $P'$ on the smallest set $\mathcal{X}_1 \times \mathcal{X}_2 \times (S \times [k_2]) \times S \times [k_2] \times \mathcal{Y}$:

$$
\begin{align}
P'(x_1 x_2 (j_1 j_2) | i_1 i_2 y) &:= P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + C(x_1 x_2 j_2 | i_1 i_2 y), \nonumber \\
\text{with } C(x_1 x_2 j_2 | i_1 i_2 y) &:= \frac{1}{k'_1} \sum_{j'_1 \in [k_1] - S} P(x_1 x_2 (j'_1 j_2) | i_1 i_2 y). \tag{4.29}
\end{align}
$$
$P'$ is a correct conditional probability distribution. Indeed, it is nonnegative by construction, and we have that $\sum_{x_1,x_2,j_1 \in S, j_2} P'(x_1 x_2 (j_1 j_2) | i_1 i_2 y)$ is equal to:

$$ 
\begin{align} 
& \sum_{x_1,x_2,j_1 \in S, j_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \sum_{x_1,x_2,j_1 \in S, j_2} C(x_1 x_2 j_2 | i_1 i_2 y) \nonumber \\
&= \sum_{x_1,x_2 j_2} \sum_{j_1 \in S} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \sum_{x_1,x_2,j_2} \sum_{j_1 \in S} \frac{1}{k_1'} \sum_{j_1' \in [k_1] - S} P(x_1 x_2 (j_1' j_2) | i_1 i_2 y) \nonumber \\
&= \sum_{x_1,x_2,j_2} \sum_{j_1 \in S} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \sum_{x_1,x_2,j_2} \sum_{j_1' \in [k_1] - S} P(x_1 x_2 (j_1' j_2) | i_1 i_2 y) \nonumber \\
&= \sum_{x_1,x_2,j_1,j_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) = 1. \tag{4.30}
\end{align} 
$$

Let us show that $P'$ is non-signaling:

a) First with $x_1$:

$$ 
\begin{align} 
\sum_{x_1} P'(x_1 x_2 (j_1 j_2) | i_1 i_2 y) &= \sum_{x_1} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \sum_{x_1} C(x_1 x_2 j_2 | i_1 i_2 y) \nonumber \\
&= \sum_{x_1} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \frac{1}{k_1'} \sum_{j_1' \in [k_1] - S} \sum_{x_1} P(x_1 x_2 (j_1' j_2) | i_1 i_2 y) \nonumber \\
&= \sum_{x_1} P(x_1 x_2 (j_1 j_2) | i_1' i_2 y) + \frac{1}{k_1'} \sum_{j_1' \in [k_1] - S} \sum_{x_1} P(x_1 x_2 (j_1' j_2) | i_1' i_2 y) 
\end{align} 
$$

since $P$ is non-signaling.

$$ = \sum_{x_1} P'(x_1 x_2 (j_1 j_2) | i_1' i_2 y). \tag{4.31} $$

b) Then with $x_2$:

$$ 
\begin{align} 
\sum_{x_2} P'(x_1 x_2 (j_1 j_2) | i_1 i_2 y) &= \sum_{x_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \sum_{x_2} C(x_1 x_2 j_2 | i_1 i_2 y) \nonumber \\
&= \sum_{x_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2 y) + \frac{1}{k_1'} \sum_{j_1' \in [k_1] - S} \sum_{x_2} P(x_1 x_2 (j_1' j_2) | i_1 i_2 y) \nonumber \\
&= \sum_{x_2} P(x_1 x_2 (j_1 j_2) | i_1 i_2' y) + \frac{1}{k_1'} \sum_{j_1' \in [k_1] - S} \sum_{x_2} P(x_1 x_2 (j_1' j_2) | i_1 i_2' y) 
\end{align} 
$$

since $P$ is non-signaling.

$$ = \sum_{x_2} P'(x_1 x_2 (j_1 j_2) | i_1 i_2' y). \tag{4.32} $$
c) Finally with $(j_1j_2)$, we have that $\sum_{j_1 \in S, j_2} P'(x_1x_2(j_1j_2)|i_1i_2y)$ is equal to:

$$
\begin{align*}
& \sum_{j_2} \sum_{j_1 \in S} P(x_1x_2(j_1j_2)|i_1i_2y) + \sum_{j_2} \sum_{j_1 \in S} C(x_1x_2j_2|i_1i_2y) \\
&= \sum_{j_2} \sum_{j_1 \in S} P(x_1x_2(j_1j_2)|i_1i_2y) + \sum_{j_2} \sum_{j_1 \in S} \frac{1}{k'_1} \sum_{j'_1 \in [k_1]-S} P(x_1x_2(j'_1j_2)|i_1i_2y) \\
&= \sum_{j_2} \sum_{j_1 \in S} P(x_1x_2(j_1j_2)|i_1i_2y) + \sum_{j_2} \sum_{j'_1 \in [k_1]-S} P(x_1x_2(j'_1j_2)|i_1i_2y) \\
&= \sum_{j_1, j_2} P(x_1x_2(j_1j_2)|i_1i_2y) \\
&= \sum_{j_1, j_2} P(x_1x_2(j_1j_2)|i_1i_2y') \quad \text{since P is non-signaling.} \\
&= \sum_{j_1 \in S, j_2} P'(x_1x_2(j_1j_2)|i_1i_2y'). \tag{4.33}
\end{align*}
$$

Thus $P'$ is a correct solution of the program computing $S^{\text{NS}}(W, k'_1, k_2)$, and it
leads to the value:

$$
\begin{align*}
S^{\text{NS}}(W, k'_1, k_2) &\ge \frac{1}{k'_1 k_2} \sum_{x_1, x_2, y, i_1 \in S, i_2} W(y|x_1 x_2) P'(x_1 x_2(i_1 i_2)|i_1 i_2 y) \\
&\ge \frac{1}{k'_1 k_2} \sum_{x_1, x_2, y, i_1 \in S, i_2} W(y|x_1 x_2) P(x_1 x_2(i_1 i_2)|i_1 i_2 y) \quad (4.34) \\
&= \frac{1}{k'_1} \sum_{i_1 \in S} f(i_1) \ge \frac{1}{k_1} \sum_{i_1 \in [k_1]} f(i_1) = S^{\text{NS}}(W, k_1, k_2).
\end{align*}
$$

4. Consider optimal non-signaling probability distributions $P$ and $P'$ reaching respec-
tively the values $S^{\text{NS}}(W, k_1, k_2)$ and $S^{\text{NS}}(W', k'_1, k'_2)$. Then by Proposition 2.11,
$P \otimes P'$ is a non-signaling probability distribution on $(\mathcal{X}_1 \times \mathcal{X}'_1) \times (\mathcal{X}_2 \times \mathcal{X}'_2) \times
(([(k_1] \times [k'_1]) \times ([k_2] \times [k'_2])) \times ([k_1] \times [k'_1]) \times ([k_2] \times [k'_2]) \times (\mathcal{Y} \times \mathcal{Y}'))$, which is
trivially in bijection with $(\mathcal{X}_1 \times \mathcal{X}'_1) \times (\mathcal{X}_2 \times \mathcal{X}'_2) \times ([k_1 k'_1] \times [k_2 k'_2]) \times [k_1 k'_1] \times
[k_2 k'_2] \times (\mathcal{Y} \times \mathcal{Y}').$ This will be a feasible solution of the program computing
$S^{\text{NS}}(W \otimes W', k_1 k'_1, k_2 k'_2)$. Thus, we get that $S^{\text{NS}}(W \otimes W', k_1 k'_1, k_2 k'_2)$ is larger
than or equal to (where the first sums are over all indices $x_1 x'_1, x_2 x'_2, yy', i_1 i'_1, i_2 i'_2$):

$$
\begin{align*}
& \sum (W \otimes W') (yy'|x_1x'_1x_2x'_2) (P \otimes P') (x_1x'_1x_2x'_2(i_1i'_1i_2i'_2)|i_1i'_1, i_2i'_2yy') \\
&= \sum (W(y|x_1x_2) \cdot W'(y'|x'_1x'_2)) (P(x_1x_2(i_1i_2)|i_1i_2y) \cdot P'(x'_1x'_2(i'_1i'_2)|i'_1i'_2y')) \\
&= \left( \sum_{i_1,i_2,x_1,x_2,y} W(y|x_1x_2)P(x_1x_2(i_1i_2)|i_1i_2y) \right) \\
& \quad \times \left( \sum_{x'_1,x'_2,y',i'_1,i'_2} W'(y'|x'_1x'_2)P'(x'_1x'_2(i'_1i'_2)|i'_1i'_2y') \right) \\
&= S^{\text{NS}}(W, k_1, k_2) \cdot S^{\text{NS}}(W', k'_1, k'_2). \tag{4.35}
\end{align*}
$$

In particular, applying this $n$ times on the same MAC $W$ gives the first corollary, and
the second one comes from the fact that $S^{\text{NS}}(W \otimes W', k_1, k_2) \ge S^{\text{NS}}(W, k_1, k_2)$.
$S^{\text{NS}}(W', 1, 1) = S^{\text{NS}}(W, k_1, k_2)$, since $S^{\text{NS}}(W', 1, 1) = 1$ by the first property of Proposition 4.9. $\square$

**Corollary 4.10.** 1. $C^{\text{NS}}(W)$ is convex.

2. If $(R_1, R_2)$ is achievable with non-signaling assistance, then we have $R_1 \le \log_2 |\mathcal{X}_1|$, $R_2 \le \log_2 |\mathcal{X}_2|$ and $R_1 + R_2 \le \log_2 |\mathcal{Y}|$.

3. If $(R_1, R_2)$ is achievable with non-signaling assistance, then for all $R'_i \le R_i$, $(R'_1, R'_2)$ is achievable with non-signaling assistance.

*Proof.* 1. It comes directly from Proposition 2.15 and the fourth property of Proposition 4.9.

2. By the second property of Proposition 4.9, we have $S^{\text{NS}}(W^{\otimes n}, k_1^n, k_2^n) \le \frac{|\mathcal{X}_1^n|}{k_1^n}$. In particular, if $R_1 > \log_2 |\mathcal{X}_1|$, then $k_1 > |\mathcal{X}_1|$ and we get that $S^{\text{NS}}(W^{\otimes n}, k_1^n, k_2^n) \le (\frac{|\mathcal{X}_1|}{k_1})^n \underset{n \to +\infty}{\to} 0$, so $R_1 > \log_2 |\mathcal{X}_1|$ is not achievable with non-signaling assistance. Symmetrically, $R_2 > \log_2 |\mathcal{X}_2|$ is not achievable with non-signaling assistance.

Furthermore, if one takes $R_1 + R_2 > \log_2 |\mathcal{Y}|$, then in particular $k_1 k_2 > |\mathcal{Y}|$, so by the second property of Proposition 4.9, $S^{\text{NS}}(W^{\otimes n}, k_1^n, k_2^n) \le \frac{|\mathcal{Y}^n|}{k_1^n k_2^n} = (\frac{|\mathcal{Y}|}{k_1 k_2})^n \underset{n \to +\infty}{\to} 0$. Thus, $R_1 + R_2 > \log_2 |\mathcal{Y}|$ is not achievable with non-signaling assistance.

3. Since $(R_1, R_2)$ is achievable with non-signaling assistance, we have by definition that $S^{\text{NS}}(W^{\otimes n}, \lceil 2^{nR_1} \rceil, \lceil 2^{nR_2} \rceil) \underset{n \to +\infty}{\to} 1$. But, for all positive integers $n$, we have that $\lceil 2^{nR'_1} \rceil \le \lceil 2^{nR_1} \rceil$ and $\lceil 2^{nR'_2} \rceil \le \lceil 2^{nR_2} \rceil$, so by the third property of Proposition 4.9, we have that $S^{\text{NS}}(W^{\otimes n}, \lceil 2^{nR'_1} \rceil, \lceil 2^{nR'_2} \rceil) \ge S^{\text{NS}}(W^{\otimes n}, \lceil 2^{nR_1} \rceil, \lceil 2^{nR_2} \rceil)$. Thus $S^{\text{NS}}(W^{\otimes n}, \lceil 2^{nR'_1} \rceil, \lceil 2^{nR'_2} \rceil) \underset{n \to +\infty}{\to} 1$ since it is upper bounded by 1, and so $(R'_1, R'_2)$ is achievable with non-signaling assistance. $\square$

**Proposition 4.11.** $C_0^{\text{NS}}(W)$ is the closure of the set of rate pairs $(R_1, R_2)$ such that:

$$ \exists n \in \mathbb{N}^*, S^{\text{NS}}(W^{\otimes n}, \lceil 2^{R_1 n} \rceil, \lceil 2^{R_2 n} \rceil) = 1. $$

*Proof.* If $(R_1, R_2)$ is such that $\exists n_0 \in \mathbb{N}^*, \forall n \ge n_0$, $S^{\text{NS}}(W^{\otimes n}, \lceil 2^{R_1 n} \rceil, \lceil 2^{R_2 n} \rceil) = 1$, then in particular $\exists n \in \mathbb{N}^*, S^{\text{NS}}(W^{\otimes n}, \lceil 2^{R_1 n} \rceil, \lceil 2^{R_2 n} \rceil) = 1$. So, $C_0^{\text{NS}}(W)$, which is the closure of the former rate pairs, is in particular included in the closure of the latter rate pairs.

For the other inclusion, consider a rate pair $(R_1, R_2)$ and let us assume that there exists some positive integer $n$ such that $S^{\text{NS}}(W^{\otimes n}, \lceil 2^{R_1 n} \rceil, \lceil 2^{R_2 n} \rceil) = 1$. Let us show that for any $(R'_1, R'_2)$ such that $R'_1 < R_1$ and $R'_2 < R_2$:

$$ \exists n_0 \in \mathbb{N}^*, \forall n \ge n_0, S^{\text{NS}}(W^{\otimes n}, \lceil 2^{R'_1 n} \rceil, \lceil 2^{R'_2 n} \rceil) = 1, $$

which is enough to conclude, since we consider only closure of such sets.

First, for all positive integers $m$, we have that $S^{\text{NS}}(W^{\otimes nm}, \lceil 2^{R_1 nm} \rceil, \lceil 2^{R_2 nm} \rceil) = 1$. By the fourth property of Proposition 4.9, we have that $S^{\text{NS}}((W^{\otimes n})^{\otimes m}, \lceil 2^{R_1 n} \rceil^m, \lceil 2^{R_2 n} \rceil^m) \ge$
$[S^{\text{NS}}(W^{\otimes n}, [2^{R_1 n}], [2^{R_2 n}])]^m = 1$, so $S^{\text{NS}}((W^{\otimes n})^{\otimes m}, [2^{R_1 n}]^m, [2^{R_2 n}]^m) = 1$ since $S^{\text{NS}}(W, k_1, k_2) \le 1$ by the first property of Proposition 4.9. But $(W^{\otimes n})^{\otimes m} = W^{\otimes nm}$, and $[2^{R_1 n}]^m \ge [2^{R_1 nm}]$, $[2^{R_2 n}]^m \ge [2^{R_2 nm}]$, so by the third property of Proposition 4.9, we have $S^{\text{NS}}(W^{\otimes nm}, [2^{R_1 nm}], [2^{R_2 nm}]) \ge 1$, so $S^{\text{NS}}(W^{\otimes nm}, [2^{R_1 nm}], [2^{R_2 nm}]) = 1$.

Then, consider some $r \in \{0, \dots, n-1\}$. By the fourth property of Proposition 4.9, we have that:

$$ S^{\text{NS}}(W^{\otimes (nm+r)}, [2^{R_1 nm}], [2^{R_2 nm}]) = S^{\text{NS}}(W^{\otimes nm} \otimes W^{\otimes r}, [2^{R_1 nm}], [2^{R_2 nm}]) \tag{4.36} \\ \ge S^{\text{NS}}(W^{\otimes nm}, [2^{R_1 nm}], [2^{R_2 nm}]) = 1, $$

$$ \text{so } S^{\text{NS}}(W^{\otimes (nm+r)}, [2^{R_1 nm}], [2^{R_2 nm}]) = 1. $$

But $[2^{R_1 nm}] = [2^{\frac{R_1 nm}{nm+r}}(nm+r)] = [2^{\frac{R_1}{1+\delta}(nm+r)}]$ with $\delta = \frac{r}{nm} \le \frac{1}{m}$, and symmetrically $[2^{R_1 nm}] = [2^{\frac{R_1}{1+\delta}(nm+r)}]$. Thus in particular, for all $R'_1 \le \frac{R_1}{1+\frac{1}{m}}$ and $R'_2 \le \frac{R_2}{1+\frac{1}{m}}$, we have that for all $n' \ge nm$, $S^{\text{NS}}(W^{\otimes n'}, [2^{R'_1 n'}], [2^{R'_2 n'}]) = 1$. So for any $(R'_1, R'_2)$ such that $R'_1 < R_1$ and $R'_2 < R_2$, there is large enough $m$ such that $R'_1 \le \frac{R_1}{1+\frac{1}{m}}$ and $R'_2 \le \frac{R_2}{1+\frac{1}{m}}$, and thus we get the expected property on $(R'_1, R'_2)$ for $n_0 := nm$. $\square$

### 4.3.3 Linear Program With Reduced Size for Structured Channels

Although $S^{\text{NS}}(W, k_1, k_2)$ can be computed in polynomial time in $W$, $k_1$ and $k_2$, a channel of the form $W^{\otimes n}$ has exponential size in $n$. Thus, the linear program for $S^{\text{NS}}(W^{\otimes n}, k_1, k_2)$ grows exponentially with $n$. However, using the invariance of $W^{\otimes n}$ under permutations, one can find a much smaller linear program computing $S^{\text{NS}}(W^{\otimes n}, k_1, k_2)$.

**Definition 4.2.** Let $G$ a group acting on $\mathcal{X}_1, \mathcal{X}_2, \mathcal{Y}$. We say that a MAC $W : \mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}$ is invariant under the action of $G$ if:

$$ \forall g \in G, W(g \cdot y | g \cdot x_1 g \cdot x_2) = W(y | x_1 x_2). $$

In particular, for $W^{\otimes n} : \mathcal{X}_1^n \times \mathcal{X}_2^n \to \mathcal{Y}^n$, the symmetric group $G := \mathfrak{S}_n$ acts in a natural way in any set $\mathcal{A}$ raised to power $n$. So for $\sigma \in \mathfrak{S}_n$, we have that:

$$ W^{\otimes n}(\sigma \cdot y^n | \sigma \cdot x_1^n \sigma \cdot x_2^n) = \prod_{i=1}^{n} W(y_{\sigma(i)} | x_{1,\sigma(i)} x_{2,\sigma(i)}) = \prod_{i=1}^{n} W(y_i | x_{1,i} x_{2,i}) \tag{4.37} \\ = W^{\otimes n}(y^n | x_1^n x_2^n), $$

and so $W^{\otimes n}$ is invariant under the action of $\mathfrak{S}_n$.

Let $\mathcal{Z} := \{\mathcal{X}_1, \mathcal{X}_2, \mathcal{Y}, \mathcal{X}_1 \times \mathcal{Y}, \mathcal{X}_2 \times \mathcal{Y}, \mathcal{X}_1 \times \mathcal{X}_2, \mathcal{X}_1 \times \mathcal{X}_2 \times \mathcal{Y}\}$. Let us call $\mathcal{O}_G(\mathcal{A})$ the set of orbits of $\mathcal{A}$ under the action of $G$. Then, one can find an equivalent smaller linear program for $S^{\text{NS}}(W, k_1, k_2)$:

**Theorem 4.12.** Let $W : \mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}$ a MAC invariant under the action of $G$. Let us name systematically $w \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{X}_2 \times \mathcal{Y})$, $u \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{X}_2)$, $u^1 \in \mathcal{O}_G(\mathcal{X}_1)$, $u^2 \in \mathcal{O}_G(\mathcal{X}_2)$, $v^1 \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{Y})$, $v^2 \in \mathcal{O}_G(\mathcal{X}_2 \times \mathcal{Y})$, $v \in \mathcal{O}_G(\mathcal{Y})$. We will also call $z_A$ the projection of $z \in \mathcal{O}_G(\mathcal{B})$ on $\mathcal{A}$, for $\mathcal{A}, \mathcal{B} \in \mathcal{Z}$ and $\mathcal{A}$ projection of $\mathcal{B}$; note that $z_A \in \mathcal{O}_G(\mathcal{A})$, since by definition of the action, the projection of an orbit is an orbit. Let us finally call
$W(w) := W(y|x_1x_2)$ for any $(x_1, x_2, y) \in w$, which is well-defined since $W$ is invariant under $G$. We have that $S^{\text{NS}}(W, k_1, k_2)$ is the solution of the following linear program:

$$
\begin{array}{ll}
\underset{r,r^1,r^2,p}{\text{maximize}} & \displaystyle \frac{1}{k_1k_2} \sum_{w \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{X}_2 \times \mathcal{Y})} W(w)r_w \\
\text{subject to} & \displaystyle \sum_{w:w_{\mathcal{X}_2}\mathcal{Y}=v} r_w = |v|, \forall v \in \mathcal{O}_G(\mathcal{Y}) \\
& \displaystyle \sum_{w:w_{\mathcal{X}_2}\mathcal{Y}=v^2} r_w^1 = k_1 \sum_{w:w_{\mathcal{X}_2}\mathcal{Y}=v^2} r_w, \forall v^2 \in \mathcal{O}_G(\mathcal{X}_2 \times \mathcal{Y}) \\
& \displaystyle \sum_{w:w_{\mathcal{X}_1}\mathcal{Y}=v^1} r_w^2 = k_2 \sum_{w:w_{\mathcal{X}_1}\mathcal{Y}=v^1} r_w, \forall v^1 \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{Y}) \\
& \displaystyle \sum_{u:u_{\mathcal{X}_2}=v_{\mathcal{X}_2}^2} p_u = \frac{|v_{\mathcal{X}_2}^2|}{|v^2|} k_1 \sum_{w:w_{\mathcal{X}_2}\mathcal{Y}=v^2} r_w^2, \forall v^2 \in \mathcal{O}_G(\mathcal{X}_2 \times \mathcal{Y}) \\
& \displaystyle \sum_{u:u_{\mathcal{X}_1}=v_{\mathcal{X}_1}^1} p_u = \frac{|v_{\mathcal{X}_1}^1|}{|v^1|} k_2 \sum_{w:w_{\mathcal{X}_1}\mathcal{Y}=v^1} r_w^1, \forall v^1 \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{Y}) \\
& 0 \le r_w \le r_w^1, r_w^2 \le \frac{|w|}{|w_{\mathcal{X}_1\mathcal{X}_2}|} p_{w_{\mathcal{X}_1}\mathcal{X}_2}, \forall w \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{X}_2 \times \mathcal{Y}) \\
& \frac{|w|}{|w_{\mathcal{X}_1\mathcal{X}_2}|} p_{w_{\mathcal{X}_1}\mathcal{X}_2} - r_w^1 - r_w^2 + r_w \ge 0, \forall w \in \mathcal{O}_G(\mathcal{X}_1 \times \mathcal{X}_2 \times \mathcal{Y}).
\end{array}
\tag{4.38}
$$

**Corollary 4.13.** For a channel $W: \mathcal{X}_1 \times \mathcal{X}_2 \rightarrow \mathcal{Y}$, $S^{\text{NS}}(W^{\otimes n}, k_1, k_2)$ is the solution of a linear program of size bounded by $O(n^{|\mathcal{X}_1|+|\mathcal{X}_2|+|\mathcal{Y}|-1})$, thus it can be computed in polynomial time in $n$.

*Proof.* We use the linear program obtained in Theorem 4.12 with $G := \mathfrak{S}_n$ acting on $W^{\otimes n}$ as described before. The number of variables and constraints is linear in the number of orbits of the action of $\mathfrak{S}_n$ on the different sets $A \in \mathcal{Z}$, where here $\mathcal{Z} := \{\mathcal{X}_1^n, \mathcal{X}_2^n, \mathcal{Y}^n, \mathcal{X}_1^n \times \mathcal{Y}^n, \mathcal{X}_2^n \times \mathcal{Y}^n, \mathcal{X}_1^n \times \mathcal{X}_2^n, \mathcal{X}_1^n \times \mathcal{X}_2^n \times \mathcal{Y}^n\}$. For example, for $A \in \mathcal{X}_1^n \times \mathcal{X}_2^n \times \mathcal{Y}^n$, we have that:

$$ |\mathcal{O}_{\mathfrak{S}_n}(\mathcal{X}_1^n \times \mathcal{X}_2^n \times \mathcal{Y}^n)| = \left( \frac{n + |\mathcal{X}_1||\mathcal{X}_2||\mathcal{Y}|-1}{|\mathcal{X}_1||\mathcal{X}_2||\mathcal{Y}|-1} \right) \le (n + |\mathcal{X}_1||\mathcal{X}_2||\mathcal{Y}|-1)^{|\mathcal{X}_1||\mathcal{X}_2||\mathcal{Y}|-1}. $$

So the number of variables and constraints is $O(n^{|\mathcal{X}_1|+|\mathcal{X}_2|+|\mathcal{Y}|-1})$. Note also that all the numbers occurring in this linear program are integers or fractions of integers, with those integers ranging in $[|\mathcal{X}_1||\mathcal{X}_2||\mathcal{Y}|-1]^n$, thus of size $O(n \log(|\mathcal{X}_1||\mathcal{X}_2||\mathcal{Y}|))$. So the size of this linear program is bounded by $O(n^{|\mathcal{X}_1|+|\mathcal{X}_2|+|\mathcal{Y}|-1})$, and thus $S^{\text{NS}}(W^{\otimes n}, k_1, k_2)$ can be computed in polynomial time in $n$; see for instance Section 7.1 of [GM07]. $\square$

In order to prove Theorem 4.12, we will need several lemmas. For all of them, $\mathcal{A}$ and $\mathcal{B}$ will denote finite sets on which a group $G$ is acting, and $x^G$ will denote the orbit of $x$ under $G$:

**Lemma 4.14.** Let $\tau \in \mathcal{O}_G(\mathcal{A} \times \mathcal{B})$, and call $\nu := \tau_{\mathcal{A}}$ and $\mu := \tau_{\mathcal{B}}$. For $x \in \nu$, let us call $B_{\tau}^x := \{y : (x, y) \in \tau\}$. Then, $|B_{\tau}^x| = |B_{\tau}^{x'}| =: c_{\tau}^{\nu}$ for any $x, x' \in \nu$, and furthermore, we
have that $c_{\tau}^{\nu} = \frac{|\tau|}{|\nu|}$. Symmetrically, the same occurs for $A_{\tau}^{y} := \{x : (x, y) \in \tau\}$ with $y \in \mu$, where one gets that $|A_{\tau}^{y}| = |A_{\tau}^{y'}| =: c_{\tau}^{\mu} = \frac{|\tau|}{|\mu|}$ for $y, y' \in \mu$.

*Proof.* Let $x, x' \in \nu$. Thus there exists $g \in G$ such that $x' = g \cdot x$. Let:

$$f : \begin{array}{ccc} B_{\tau}^{x} & \rightarrow & B_{\tau}^{x'} \\ y & \mapsto & g \cdot y. \end{array}$$

First, $f$ is well defined. Indeed, if $y \in B_{\tau}^{x} = \{y : (x, y) \in \tau\}$, then we have that $g \cdot y \in \{y : (g \cdot x, y) \in \tau\} = B_{\tau}^{x'}$, since $\tau \in O_G(A \times B)$. Let us show that $f$ is injective. If $g \cdot y = g \cdot y'$, then $g^{-1} \cdot (g \cdot y) = (g^{-1}g) \cdot y = y$, $g^{-1} \cdot (g \cdot y') = y'$, so $y = y'$. Thus we get that $|B_{\tau}^{x}| \le |B_{\tau}^{x'}|$. By a symmetric argument with $x'$ replacing $x$ and $g^{-1}$ replacing $g$, we get that $|B_{\tau}^{x'}| \le |B_{\tau}^{x}|$, and so $|B_{\tau}^{x}| = |B_{\tau}^{x'}| =: c_{\tau}^{\nu}$.

Furthermore, $\{B_{\tau}^{x}\}_{x \in \nu}$ is a partition of $\tau$, so $\sum_{x \in \nu} |B_{\tau}^{x}| = |\nu|c_{\tau}^{\nu} = |\tau|$, and $c_{\tau}^{\nu} = \frac{|\tau|}{|\nu|}$. $\square$

**Lemma 4.15.** For any $(x, y) \in A \times B$ and $v_{(x,y)G}$ variable indexed by orbits of $A \times B$, let us define the variable $v_{x,y} := \frac{v_{(x,y)G}}{|(x,y)G|}$. We have:

$$\sum_{x \in A} v_{x,y} = \frac{1}{|y^G|} \sum_{\tau \in O_G(A \times B) : \tau_B = y^G} v_{\tau}, \forall y \in B.$$

*Proof.*

$$\begin{align} \sum_{x \in A} v_{x,y} &= \sum_{\tau \in O_G(A \times B): \tau_B = y^G} \sum_{x \in A:(x,y) \in \tau} v_{x,y} \nonumber \\ &= \sum_{\tau \in O_G(A \times B): \tau_B = y^G} \sum_{x \in A:(x,y) \in \tau} \frac{v_{\tau}}{|\tau|} \quad \text{since } (x,y)^G = \tau \nonumber \\ &= \sum_{\tau \in O_G(A \times B): \tau_B = y^G} c_{\tau}^{y^G} \frac{v_{\tau}}{|\tau|} \quad \text{by Lemma 4.14, since } y \in \tau_B \tag{4.39} \\ &= \sum_{\tau \in O_G(A \times B): \tau_B = y^G} \frac{|\tau|}{|y^G|} \frac{v_{\tau}}{|\tau|} = \frac{1}{|y^G|} \sum_{\tau \in O_G(A \times B): \tau_B = y^G} v_{\tau}. \nonumber \end{align}$$

**Lemma 4.16.** For any $\tau \in O_G(A \times B)$, $\mu \in O_G(B)$ and $v_{x,y}$ variable indexed by elements of $A \times B$, let us define $v_{\tau} := \sum_{(x,y) \in \tau} v_{x,y}$. We have:

$$\sum_{\tau \in O_G(A \times B): \tau_B = \mu} v_{\tau} = \sum_{y \in \mu} \sum_{x \in A} v_{x,y}.$$

*Proof.*

$$\sum_{\tau \in O_G(A \times B): \tau_B = \mu} v_{\tau} = \sum_{\tau \in O_G(A \times B): \tau_B = \mu} \sum_{(x,y) \in \tau} v_{x,y} = \sum_{y \in \mu} \sum_{x \in A} v_{x,y}.$$
**Proof of Theorem 4.12.** Let $r_{x_1,x_2,y}, r_{x_1,x_2,y}^1, r_{x_1,x_2,y}^2, p_{x_1,x_2}$ a feasible solution of the program defined in Proposition 4.7, and call $S := \frac{1}{k_1k_2} \sum_{x_1,x_2,y} W(y|x_1x_2)r_{x_1,x_2,y}$ its value. Define:

$$ 
\begin{align} 
r_w &:= \sum_{(x_1,x_2,y) \in w} r_{x_1,x_2,y}, & r_w^1 &:= \sum_{(x_1,x_2,y) \in w} r_{x_1,x_2,y}^1, \tag{4.40} \\ 
r_w^2 &:= \sum_{(x_1,x_2,y) \in w} r_{x_1,x_2,y}^2, & p_u &:= \sum_{(x_1,x_2) \in u} p_{x_1,x_2}. \nonumber 
\end{align} 
$$

Let us show that $r_w, r_w^1, r_w^2, p_u$ is a feasible solution of the program defined in Theorem 4.12, and that its value $S^* := \frac{1}{k_1k_2} \sum_w W(w)r_w = S$.

First, we have $S^* = S$. Indeed:

$$ 
\begin{align} 
S^* &= \frac{1}{k_1k_2} \sum_w W(w)r_w = \frac{1}{k_1k_2} \sum_w W(w) \sum_{(x_1,x_2,y) \in w} r_{x_1,x_2,y} \nonumber \\ 
&= \frac{1}{k_1k_2} \sum_w \sum_{(x_1,x_2,y) \in w} W(y|x_1x_2)r_{x_1,x_2,y} \tag{4.41} \\ 
&= \frac{1}{k_1k_2} \sum_{x_1,x_2,y} W(y|x_1x_2)r_{x_1,x_2,y} = S. \nonumber 
\end{align} 
$$

Then, all the constraints are satisfied. Indeed, thanks to Lemma 4.16, we have for the first constraint:

$$ \sum_{w: w_y=v} r_w = \sum_{y \in v} \sum_{x_1, x_2} r_{x_1, x_2, y} = \sum_{y \in v} 1 = |v|. $$

For the second constraint (and symmetrically for the third constraint), we have:

$$ \sum_{w: w_{\mathcal{X}_2} y = v^2} r_w^1 = \sum_{(x_2, y) \in v^2} \sum_{x_1} r_{x_1, x_2, y}^1 = \sum_{(x_2, y) \in v^2} k_1 \sum_{x_1} r_{x_1, x_2, y} = k_1 \sum_{w: w_{\mathcal{X}_2} y = v^2} r_w. $$

For the fourth (and symmetrically for the fifth), we have:

$$ 
\begin{align} 
\sum_{w: w_{\mathcal{X}_2} y = v^2} r_w^2 &= \sum_{(x_2, y) \in v^2} \sum_{x_1} r_{x_1, x_2, y}^2 = \sum_{(x_2, y) \in v^2} \frac{1}{k_1} \sum_{x_1} p_{x_1, x_2} \nonumber \\ 
&= \frac{1}{k_1} \sum_{x_2 \in v_{\mathcal{X}_2}^2} \sum_{y: (x_2, y) \in v^2} \sum_{x_1} p_{x_1, x_2} \nonumber \\ 
&= \frac{1}{k_1} \sum_{x_2 \in v_{\mathcal{X}_2}^2} \frac{|v^2|}{|v_{\mathcal{X}_2}^2|} \sum_{x_1} p_{x_1, x_2} \quad \text{thanks to Lemma 4.14} \tag{4.42} \\ 
&= \frac{1}{k_1} \frac{|v^2|}{|v_{\mathcal{X}_2}^2|} \sum_{u: u_{\mathcal{X}_2} = v_{\mathcal{X}_2}^2} p_u. \nonumber 
\end{align} 
$$

Finally for the last constraints, we only need to compute:

$$ 
\begin{align} 
\sum_{(x_1, x_2, y) \in w} p_{x_1, x_2} &= \sum_{(x_1, x_2) \in w_{\mathcal{X}_1 \mathcal{X}_2}} \sum_{y: (x_1, x_2, y) \in w} p_{x_1, x_2} = \sum_{(x_1, x_2) \in w_{\mathcal{X}_1 \mathcal{X}_2}} \frac{|w|}{|w_{\mathcal{X}_1 \mathcal{X}_2}|} p_{x_1, x_2} \nonumber \\ 
&= \frac{|w|}{|w_{\mathcal{X}_1 \mathcal{X}_2}|} p_{w_{\mathcal{X}_1 \mathcal{X}_2}}, \tag{4.43} 
\end{align} 
$$
which implies that the linear inequalities on $p_{x_1,x_2}, r_{x_1,x_2,y}, r_{x_1,x_2,y}^1, r_{x_1,x_2,y}^2$ get transposed respectively to the values $\frac{|w|}{|w_{X_1 X_2}|} p_{w_{X_1 X_2}}, r_w, r_w^1, r_w^2$. Indeed, for instance, one has for any $x_1, x_2, y$ that $p_{x_1,x_2} - r_{x_1,x_2,y}^1 - r_{x_1,x_2,y}^2 + r_{x_1,x_2,y} \ge 0$. Thus for some orbit $w$:

$$ \sum_{(x_1,x_2,y) \in w} \left( p_{x_1,x_2} - r_{x_1,x_2,y}^1 - r_{x_1,x_2,y}^2 + r_{x_1,x_2,y} \right) \ge 0, $$

and then $\frac{|w|}{|w_{X_1 X_2}|} p_{w_{X_1 X_2}} - r_w^1 - r_w^2 + r_w \ge 0$, which was what we wanted to show.

Now let us consider a feasible solution $r_w, r_w^1, r_w^2, p_u$ of the program defined in Theorem 4.12, with a value $S^* := \frac{1}{k_1 k_2} \sum_w W(w) r_w$. Define:

$$ \begin{align} r_{x_1,x_2,y} &:= \frac{r_{(x_1,x_2,y)^G}}{|(x_1, x_2, y)^G|}, & r_{x_1,x_2,y}^1 &:= \frac{r_{(x_1,x_2,y)^G}}{|(x_1, x_2, y)^G|}, \tag{4.44} \\ r_{x_1,x_2,y}^2 &:= \frac{r_{(x_1,x_2,y)^G}}{|(x_1, x_2, y)^G|}, & p_{x_1,x_2} &:= \frac{p_{(x_1,x_2)^G}}{|(x_1, x_2)^G|}. \end{align} $$

Let us show that $r_{x_1,x_2,y}, r_{x_1,x_2,y}^1, r_{x_1,x_2,y}^2, p_{x_1,x_2}$ is a feasible solution of the program defined in Proposition 4.7, and that its value $S := \frac{1}{k_1 k_2} \sum_{x_1,x_2,y} W(y|x_1 x_2) r_{x_1,x_2,y} = S^*$.

First we have $S = S^*$. Indeed:

$$ \begin{align} S &= \frac{1}{k_1 k_2} \sum_{x_1,x_2,y} W(y|x_1 x_2) r_{x_1,x_2,y} = \frac{1}{k_1 k_2} \sum_{x_1,x_2,y} W(y|x_1 x_2) \frac{r_{(x_1,x_2,y)^G}}{|r_{(x_1,x_2,y)^G}|} \nonumber \\ &= \frac{1}{k_1 k_2} \sum_w \sum_{(x_1,x_2,y) \in w} W(y|x_1 x_2) \frac{r_w}{|w|} = \frac{1}{k_1 k_2} \sum_w \sum_{(x_1,x_2,y) \in w} W(w) \frac{r_w}{|w|} \tag{4.45} \\ &= \frac{1}{k_1 k_2} \sum_w |w| W(w) \frac{r_w}{|w|} = \frac{1}{k_1 k_2} \sum_w W(w) r_w = S^*. \end{align} $$

Then, all the constraints are satisfied. Indeed, thanks to Lemma 4.15, we have for the first constraint:

$$ \sum_{x_1,x_2} r_{x_1,x_2,y} = \frac{1}{|y^G|} \sum_{w:w_{Y}=y^G} r_w = \frac{|y^G|}{|y^G|} = 1. $$

For the second constraint (and symmetrically for the third constraint), we have:

$$ \begin{align} \sum_{x_1} r_{x_1,x_2,y}^1 &= \frac{1}{|(x_2, y)^G|} \sum_{w:w_{X_2 Y}=(x_2, y)^G} r_w^1 = \frac{k_1}{|(x_2, y)^G|} \sum_{w:w_{X_2 Y}=(x_2, y)^G} r_w \tag{4.46} \\ &= k_1 \sum_{x_1} r_{x_1,x_2,y}. \end{align} $$
For the fourth (and symmetrically for the fifth), we have:

$$ 
\begin{aligned}
\sum_{x_1} r_{x_1, x_2, y}^2 &= \frac{1}{|(x_2, y)^G|} \sum_{w:w_{\mathcal{X}_2\mathcal{Y}}=(x_2, y)^G} r_w^2 = \frac{1}{|(x_2, y)^G|} \frac{1}{k_1} \frac{|(x_2, y)^G|}{|(x_2, y)^G|} \sum_{u:u_{\mathcal{X}_2}=(x_2, y)^G} p_u \\
&= \frac{1}{k_1} \frac{1}{|(x_2, y)^G|} \sum_{u:u_{\mathcal{X}_2}=(x_2, y)^G} p_u \\
&= \frac{1}{k_1} \frac{1}{|x_2^G|} \sum_{u:u_{\mathcal{X}_2}=x_2^G} p_u \quad \text{since } (x_2, y)^G_{\mathcal{X}_2} = x_2^G \\
&= \frac{1}{k_1} \sum_{x_1} p_{x_1, x_2}.
\end{aligned}
\tag{4.47} $$

Finally, to conclude with the last constraints, one has only to see that for any $x_1, x_2, y$:

$$ \frac{|(x_1, x_2, y)^G|}{|(x_1, x_2, y)^G|} p(x_1, x_2, y)^G_{\mathcal{X}_1 \mathcal{X}_2} = \frac{|(x_1, x_2, y)^G|}{|(x_1, x_2)^G|} p(x_1, x_2)^G = |(x_1, x_2, y)^G| p_{x_1, x_2}, $$

which implies that the linear inequalities on $\frac{|w|}{|w_{\mathcal{X}_1 \mathcal{X}_2}|} p_{w_{\mathcal{X}_1 \mathcal{X}_2}}, r_w, r_w^1, r_w^2$ get transposed respectively to the values $p_{x_1, x_2}, r_{x_1, x_2, y}, r_{x_1, x_2, y}^1, r_{x_1, x_2, y}^2$. Indeed, for instance, one has for any $w$ that $\frac{|w|}{|w_{\mathcal{X}_1 \mathcal{X}_2}|} p_{w_{\mathcal{X}_1 \mathcal{X}_2}} - r_w^1 - r_w^2 + r_w \ge 0$. But for any $(x_1, x_2, y) \in w$, one has that $r_{x_1, x_2, y} = \frac{r_w}{|w|}, r_{x_1, x_2, y}^1 = \frac{r_w^1}{|w|}, r_{x_1, x_2, y}^2 = \frac{r_w^2}{|w|}$. Thanks to the previous inequality, we have that $p_{x_1, x_2} = \frac{p_{w_{\mathcal{X}_1 \mathcal{X}_2}}}{|w_{\mathcal{X}_1 \mathcal{X}_2}|}$, and thus:

$$ p_{x_1, x_2} - r_{x_1, x_2, y}^1 - r_{x_1, x_2, y}^2 + r_{x_1, x_2, y} = \frac{p_{w_{\mathcal{X}_1 \mathcal{X}_2}}}{|w_{\mathcal{X}_1 \mathcal{X}_2}|} - \frac{r_w^1}{|w|} - \frac{r_w^2}{|w|} + \frac{r_w}{|w|} \ge 0, $$

which was what we wanted to show. □

# 4.4 Non-Signaling Achievability Bounds

## 4.4.1 Zero-Error Non-Signaling Assisted Achievable Rate Pairs

We will now present a numerical method to find efficiently inner bounds on $C_0^{\text{NS}}(W)$. Thanks to Corollary 4.13, we know how to decide in polynomial time in $n, k_1, k_2$ whether $S^{\text{NS}}(W^{\otimes n}, k_1, k_2) = 1$. However, by Proposition 4.11, if $S^{\text{NS}}(W^{\otimes n}, k_1, k_2) = 1$, then we have that $(\frac{\log(k_1)}{n}, \frac{\log(k_2)}{n}) \in C_0^{\text{NS}}(W)$, which describes a way of computing achievable points for that capacity region. More precisely, this leads to the following result:

**Proposition 4.17** (Inner Bounds on $C_0^{\text{NS}}(W)$). Let us define the zero-error non-signaling assisted $n$-shots capacity region $C_{0, \le n}^{\text{NS}}(W)$ in the following way:

$$ C_{0, \le n}^{\text{NS}}(W) := \left\{ \left( \frac{\log(k_1)}{n}, \frac{\log(k_2)}{n} \right) : S^{\text{NS}}(W^{\otimes n}, k_1, k_2) = 1 \right\}. $$

Then, we have that $\forall n \in \mathbb{N}, C_{0, \le n}^{\text{NS}}(W) \subseteq C_0^{\text{NS}}(W)$, and that one can decide in polynomial time in $n, k_1, k_2$ if $(\frac{\log(k_1)}{n}, \frac{\log(k_2)}{n}) \in C_{0, \le n}^{\text{NS}}(W)$.

This implies that we can find efficiently achievable rate pairs for MACs.
**Application to the binary adder channel** The binary adder channel $W_{BAC}$ is defined by:

$$ \forall x_1, x_2 \in \{0, 1\}, \forall y \in \{0, 1, 2\}, W_{BAC}(y|x_1x_2) := \mathbb{1}_{y=x_1+x_2}. $$

Its classical capacity region $C(W_{BAC})$ is well known and consists of all $(R_1, R_2)$ such that $R_1 \le 1, R_2 \le 1, R_1 + R_2 \le \frac{3}{2}$, as a consequence of Theorem 4.1. Its zero-error classical capacity $C_0(W_{BAC})$ is not yet characterized. A lot of work has been done in finding outer and inner bounds on this region [Lin69, vT78, KL78, Jr.78, KLWY83, vdBvT85, BB98, UL98, AB99, MÖ05, OS15]. To date, the best lower bound on the sum-rate capacity is $\log_2(240/6) \approx 1.3178$ [MÖ05].

Thanks to Proposition 4.17, we were able to compute the regions $C_{0, \le n}^{NS}(W)$ for $n$ going up to 7, which led to Figure 4.4. The code can be found on GitHub. It uses Mosek linear programming solver [AA00].

Note that the linear program from Theorem 4.12 has still a large number of variables and constraints although polynomial in $n$. Specifically, for $n = 2$, it has 244 variables and 480 constraints; for $n = 3$, it has 1112 variables and 2054 constraints; for $n = 7$, it has 95592 variables and 162324 constraints; finally, for $n = 8$, it has 226911 variables and 383103 constraints.

The first noticeable result coming from these curves is that the zero-error non-signaling assisted sum-rate capacity beats with only 7 copies the classical sum-rate capacity of $\frac{3}{2}$, even without a zero-error constraint, with a value of $\frac{2\log_2(42)}{7} \approx 1.5406$, coming from the fact that $S^{NS}(W_{BAC}^{\otimes 7}, 42, 42) = 1$ and Proposition 4.11. This implies that $C_0^{NS}(W_{BAC})$ has larger sum-rate pairs than $C(W_{BAC})$, and that $C^{NS}(W_{BAC})$ is strictly larger than $C(W_{BAC})$. This sum-rate can even be increased up to $\frac{\log_2(72)}{4} \approx 1.5425$, since we have computed $S^{NS}(W_{BAC}^{\otimes 8}, 72, 72) = 1$, which is the largest number of copies we have been able to manage with our efficient version of the linear program from Theorem 4.12. This should be compared with the upper bound on the non-signaling assisted sum-rate capacity coming from Proposition 4.26, which is $\log_2(3) \approx 1.5850$ for $R_1 = R_2$.

Another surprising property is the speed at which one obtains efficient zero-error non-signaling assisted codes compared to classical zero-error codes. Indeed, with only three copies of the binary adder channel, one gets that $S^{NS}(W_{BAC}^{\otimes 3}, 4, 5) = 1$, which corresponds to a sum-rate of $\frac{2+\log_2(5)}{3} \approx 1.4406$, which already largely beats the best known zero-error achieved sum-rate of $\log_2(240/6) \approx 1.3178$ [MÖ05]. These results are summarized in the following theorem:

**Theorem 4.18.** We have that $(\frac{\log_2(72)}{8}, \frac{\log_2(72)}{8}) \in C_0^{NS}(W_{BAC})$ but $(\frac{\log_2(72)}{8}, \frac{\log_2(72)}{8}) \notin C(W_{BAC})$, and as a consequence, we have that $C(W_{BAC}) \subsetneq C^{NS}(W_{BAC})$.

*Proof.* Since $2^{\frac{\log_2(72)}{8}} = 72$ and numerically $S^{NS}(W_{BAC}^{\otimes 8}, 72, 72) = 1$ thanks to Corollary 4.13, we get that $(\frac{\log_2(72)}{8}, \frac{\log_2(72)}{8}) \in C_0^{NS}(W_{BAC})$ by Proposition 4.11. However, $\frac{\log_2(72)}{8} + \frac{\log_2(72)}{8} > \frac{3}{2}$ so $(\frac{\log_2(72)}{8}, \frac{\log_2(72)}{8}) \notin C(W_{BAC})$ by Theorem 4.1 applied to $W_{BAC}$. Since $C(W_{BAC}) \subseteq C^{NS}(W_{BAC})$ and $C_0^{NS}(W_{BAC}) \subseteq C^{NS}(W_{BAC})$, we thus get that $C(W_{BAC}) \subsetneq C^{NS}(W_{BAC})$. $\square$
<figure><img src="image_18.png" /><figcaption>Figure 4.4 – Capacity regions of the binary adder channel $W_{BAC}$. The black dashed curve depicts the classical capacity region $C(W_{BAC})$, whereas the grey dashed curve shows the best known inner bound border on the zero-error classical capacity region $C_0(W_{BAC})$, made from results by [MÖ05, vdBvT85, KLWY83]; see [MÖ05] for a description of this border. On the other hand, the continuous curves depict the best zero-error non-signaling assisted achievable rate pairs for respectively 2, 3 and 7 copies of the binary adder channel.</figcaption></figure>

### 4.4.2 Non-Signaling Assisted Achievable Rate Pairs With Non-Zero Error

We have analyzed the non-signaling assisted capacity region through zero-error strategies and applied it to the BAC. However, if some noise is added to that channel, its zero-error non-signaling assisted capacity region becomes trivial (see Proposition 4.21). Thus, the previous method fails to find significant inner bounds on the non-signaling assisted capacity region of noisy MACs.

In this section, we use concatenated codes to obtain achievable rate pairs, and apply it to a noisy version of the BAC:

**Definition 4.3 (Concatenated Codes).** Given a MAC $W$ and a non-signaling assisted code $P$, let $W[P] : [k_1] \times [k_2] \rightarrow [\ell]$ with $W[P](j|i_1i_2) := \sum_{x_1,x_2,y} W(y|x_1x_2)P(x_1x_2j|i_1i_2y)$:
<figure><img src="image_19.png" /></figure>

Note that $W[P]$ is a MAC since $W[P](j|i_1i_2) \ge 0$ and:

$$ 
\begin{aligned}
\sum_j W[P](j|i_1i_2) &= \sum_{x_1,x_2,y} W(y|x_1x_2) \sum_j P(x_1x_2j|i_1i_2y) \\
&= \sum_{x_1,x_2} \left( \sum_y W(y|x_1x_2) \right) P(x_1x_2|i_1i_2) \quad \text{since P is non-signaling} \\
&= \sum_{x_1,x_2} P(x_1x_2|i_1i_2) = 1.
\end{aligned} 
\qquad (4.48) $$

The following proposition states that combining a classical code to a non-signaling strategy leads to inner bounds on the non-signaling assisted capacity region of a MAC:

**Proposition 4.19.** If $P$ is a non-signaling assisted code for the MAC $W$, we have that $C(W[P]) \subseteq C^{\text{NS}}(W)$.

*Proof.* Let $(R_1, R_2) \in C(W[P])$. Then, by definition, we have that:

$$ \lim_{n \to +\infty} S(W[P]^{\otimes n}, \lceil 2^{R_1 n} \rceil, \lceil 2^{R_2 n} \rceil) = 1. $$

Let us fix $\varepsilon > 0$. For a large enough $N$, we have $S(W[P]^{\otimes N}, \lceil 2^{R_1 N} \rceil, \lceil 2^{R_2 N} \rceil) \ge 1 - \varepsilon$. Let us call $\ell_1 := \lceil 2^{R_1 N} \rceil$ and $\ell_2 := \lceil 2^{R_2 N} \rceil$. Thus, there exists encoders $e_1 : [\ell_1] \to [k_1]$, $e_2 : [\ell_2] \to [k_2]$ and a decoder $d : [\ell] \to [\ell_1] \times [\ell_2]$ such that:

$$ \frac{1}{\ell_1 \ell_2} \sum_{i_1, i_2, j} W[P](j|i_1 i_2) \sum_{a_1 \in [\ell_1], a_2 \in [\ell_2]} e_1(i_1|a_1) e_2(i_2|a_2) d(a_1 a_2|j) \ge 1 - \varepsilon. $$

In particular, we have:

$$ \frac{1}{\ell_1 \ell_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) \left[ \sum_{i_1, i_2, j, a_1, a_2} P(x_1 x_2 j | i_1 i_2 y) e_1(i_1|a_1) e_2(i_2|a_2) d(a_1 a_2|j) \right] \ge 1 - \varepsilon. $$

Define $\hat{P}(x_1x_2(b_1b_2)|a_1a_2y) := \sum_{i_1, i_2, j} P(x_1x_2j|i_1i_2y)e_1(i_1|a_1)e_2(i_2|a_2)d(b_1b_2|j)$. One can easily check that $\hat{P}$ is non-signaling, and thus:

$$ S^{\text{NS}}(W^{\otimes N}, \ell_1, \ell_2) \ge \frac{1}{\ell_1 \ell_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) \sum_{a_1, a_2} \hat{P}(x_1 x_2(a_1, a_2)|a_1 a_2 y) \ge 1 - \varepsilon. $$
This implies that $\lim_{n \to +\infty} S^{\text{NS}}(W^{\otimes n}, [2^{R_1 n}], [2^{R_2 n}]) = 1$, i.e. $(R_1, R_2) \in \mathcal{C}^{\text{NS}}(W)$. $\square$

Thanks to Proposition 4.19, we have for any non-signaling assisted code $P$, $\mathcal{C}(W^{\otimes n}[P]) \subseteq \mathcal{C}^{\text{NS}}(W^{\otimes n})$. But if $(R_1, R_2) \in \mathcal{C}^{\text{NS}}(W^{\otimes n})$, we have that $(\frac{R_1}{n}, \frac{R_2}{n}) \in \mathcal{C}^{\text{NS}}(W)$. Thus, applying Theorem 4.1 on $W^{\otimes n}[P]$ leads to inner bounds on $\mathcal{C}^{\text{NS}}(W)$:

**Proposition 4.20 (Inner Bounds on $\mathcal{C}^{\text{NS}}(W)$).** For any number of copies $n$, number of inputs $k_1 \in [|X_1|^n]$ and $k_2 \in [|X_2|^n]$, non-signaling assisted codes $P$ on inputs in $[k_1]$, $[k_2]$ for $W^{\otimes n}$, and distributions $q_1, q_2$ on $[k_1]$, $[k_2]$, we have that the rate pairs $(R_1, R_2)$ satisfying the following constraints are in $\mathcal{C}^{\text{NS}}(W)$:

$$R_1 \le \frac{I(I_1:J|I_2)}{n}, \quad R_2 \le \frac{I(I_2:J|I_1)}{n}, \quad R_1 + R_2 \le \frac{I((I_1,I_2):J)}{n},$$

for $(I_1, I_2) \in [k_1] \times [k_2]$ following the product law $q_1 \times q_2$, and $J \in [\ell]$ the outcome of $W^{\otimes n}[P]$ on inputs $I_1, I_2$. In particular, the corner points of this capacity region are given by:

$$\left(\frac{I(I_1:J|I_2)}{n}, \frac{I(I_2:J)}{n}\right) \text{ and } \left(\frac{I(I_1:J)}{n}, \frac{I(I_2:J|I_1)}{n}\right).$$

*Proof.* The achievable region comes from the previous discussion. We just need to prove that the corner points are of the given form. If $R_1 = \frac{I(I_1:J|I_2)}{n}$, constraints on $R_2$ and $R_1 + R_2$ leads to a maximum $R_2 = \min\left(\frac{I(I_2:J|I_1)}{n}, \frac{I((I_1,I_2):J)}{n} - \frac{I(I_1:J|I_2)}{n}\right)$. However, $I((I_1, I_2):J) - I(I_1:J|I_2) = I(I_2:J)$ by the chain rule. We only need to show that $I(I_2:J) \le I(I_2:J|I_1)$ and the proof will be complete, since the other corner point is symmetric. We have:

$$\begin{aligned}
I(I_2:J) &= H(I_2) - H(I_2|J) = H(I_2|I_1) - H(I_2|J) \tag{4.49} \\
&\le H(I_2|I_1) - H(I_2|JI_1) = I(I_2:J|I_1),
\end{aligned}$$

the second equality coming from the fact that $I_1$ and $I_2$ are independent, and the inequality coming from the fact that $H(A|BC) \le H(A|B)$ for any $A, B, C$. $\square$

**Application to the Noisy Binary Adder Channel** We will now apply this strategy to a noisy version of the BAC. We will consider flip errors $\varepsilon_1, \varepsilon_2$ of inputs $x_1, x_2$ on $W_{\text{BAC}}$, which leads to the following definition of $W_{\text{BAC},\varepsilon_1,\varepsilon_2}$:

$$\begin{aligned}
\forall y, x_1, x_2, W_{\text{BAC},\varepsilon_1,\varepsilon_2}(y|x_1x_2) := & (1-\varepsilon_1)(1-\varepsilon_2)W_{\text{BAC}}(y|x_1x_2) \\
& + \varepsilon_1(1-\varepsilon_2)W_{\text{BAC}}(y|\overline{x_1x_2}) \\
& + (1-\varepsilon_1)\varepsilon_2W_{\text{BAC}}(y|x_1\overline{x_2}) \\
& + \varepsilon_1\varepsilon_2W_{\text{BAC}}(y|\overline{x_1x_2}).
\end{aligned} \quad (4.50)$$

First, let us note that the zero-error non-signaling assisted capacity region of $W_{\text{BAC},\varepsilon_1,\varepsilon_2}$ is trivial for $\varepsilon \in (0, 1)$:

**Proposition 4.21.** If $\varepsilon_1, \varepsilon_2 \in (0, 1)$, then $\mathcal{C}_0^{\text{NS}}(W_{\text{BAC},\varepsilon_1,\varepsilon_2}) = \{(0, 0)\}$.
Proof. If $S^{\text{NS}}(W^{\otimes n}, k_1, k_2) = 1$, then $\forall y^n, x_1^n, x_2^n: W^{\otimes n}(y^n|x_1^n x_2^n) > 0 \implies r_{x_1^n, x_2^n, y^n} = p_{x_1^n, x_2^n}$. Indeed, we have for an optimal $p, r$ that:

$$ \begin{aligned} S^{\text{NS}}(W^{\otimes n}, k_1, k_2) &= \frac{1}{k_1 k_2} \sum_{x_1^n, x_2^n, y^n} W^{\otimes n}(y^n|x_1^n x_2^n) r_{x_1^n, x_2^n, y^n} \\ &\le \frac{1}{k_1 k_2} \sum_{x_1^n, x_2^n, y^n} W^{\otimes n}(y^n|x_1^n x_2^n) p_{x_1^n, x_2^n} = 1, \end{aligned} \quad (4.51) $$

which implies the previous statement. But, for $W^{\otimes n}_{\text{BAC},\varepsilon_1,\varepsilon_2}$, one can easily check that for all $y^n, x_1^n, x_2^n, W^{\otimes n}(y^n|x_1^n x_2^n) > 0$ since $\varepsilon_1, \varepsilon_2 \in (0, 1)$. Indeed, one can flip the inputs to a valid preimage of the output. Thus if $S^{\text{NS}}(W^{\otimes n}_{\text{BAC},\varepsilon_1,\varepsilon_2}, k_1, k_2) = 1$, we have that $\forall y^n, x_1^n, x_2^n, r_{x_1^n, x_2^n, y^n} = p_{x_1^n, x_2^n}$. In particular, this implies that $\sum_{x_1^n, x_2^n} r_{x_1^n, x_2^n, y^n} = \sum_{x_1^n, x_2^n} p_{x_1^n, x_2^n}$, therefore $1 = k_1 k_2$, so $k_1 = 1$ and $k_2 = 1$. Thus $S^{\text{NS}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_2}) = 1$ implies that $(R_1, R_2) = (0, 0)$. $\square$

We have then applied the numerical method described in Proposition 4.20 to $W_{\text{BAC},\varepsilon_1,\varepsilon_2}$ for the symmetric case $\varepsilon_1 = \varepsilon_2 = \varepsilon := 10^{-3}$. Since it is hard to go through all non-signaling assisted codes $P$ and product distributions $q_1, q_2$, we have applied the heuristic of using non-signaling assisted codes obtained while optimizing $S^{\text{NS}}(W^{\otimes n}, k_1, k_2)$ in the symmetrized linear program. We have combined them with uniform $q_1, q_2$, as the form of those non-signaling assisted codes coming from our optimization program is symmetric. We have evaluated the achievable corner points for all $k_1, k_2 \le 2^n$ for $n \le 5$ copies which led to Figure 4.5: Compared to the noiseless binary adder channel, we can first notice that the classical capacity region is slightly smaller, with a classical sum-rate capacity of 1.478 at most. On the other hand, although the zero-error non-signaling assisted capacity of $W_{\text{BAC},\varepsilon,\varepsilon}$ is completely trivial, we have with our concatenated codes strategy found significant rate pairs achievable with non-signaling assistance. In particular, we have reached a non-signaling assisted sum-rate capacity of 1.493 which beats the best classical sum-rate capacity. Thus, it shows that non-signaling assistance can improve the capacity of the noisy binary adder channel as well.

## 4.5 Relaxed Non-Signaling Assisted Capacity Region and Outer Bounds

A natural question that arises when studying the strength of non-signaling assistance is whether a result similar to Theorem 4.1 can be found to describe by a single-letter formula the non-signaling assisted capacity region of MACs. In particular, dropping the constraint that $(X_1, X_2)$ is in product form in Theorem 4.1 seems to be a particularly good candidate to characterize the non-signaling assisted capacity region of MACs, as this looks quite similar to allowing correlations between parties.

We have not been able to show the equivalence between this region and the non-signaling assisted capacity region; however, it turns out to be equivalent to the capacity region defined by a slight relaxation of non-signaling assistance, which we call $S^{\text{NS}}(W, k_1, k_2)$. In particular, this will give us the best known outer bound on the non-signaling capacity.
<figure><img src="image_20.png" /><figcaption>Figure 4.5 – Capacity regions of the noisy binary adder channel $W_{BAC,ε,ε}$ for $ε = 10^{-3}$. The black dashed curve depicts the classical capacity region $C(W_{BAC,ε,ε})$ which was found numerically using Theorem 4.1. The red point depicts the zero-error non-signaling assisted capacity region (Proposition 4.21). The blue curve depicts achievable non-signaling assisted rates pairs obtained from $C(W_{BAC,ε,ε}^{\otimes 5}[P])$ through the numerical method described in Proposition 4.20.</figcaption></figure>

**Definition 4.4.**

$$
\begin{align}
S^{\overline{\text{NS}}}(W, k_1, k_2) &:= \underset{r,p}{\text{maximize}} \quad \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y|x_1 x_2) r_{x_1, x_2, y} \nonumber \\
&\text{subject to} \quad \sum_{x_1, x_2} r_{x_1, x_2, y} \le 1 \nonumber \\
&\quad \sum_{x_1, x_2} p_{x_1, x_2} = k_1 k_2 \tag{4.52} \\
&\quad \sum_{x_1} p_{x_1, x_2} \ge k_1 \sum_{x_1} r_{x_1, x_2, y} \nonumber \\
&\quad \sum_{x_2} p_{x_1, x_2} \ge k_2 \sum_{x_2} r_{x_1, x_2, y} \nonumber \\
&\quad 0 \le r_{x_1, x_2, y} \le p_{x_1, x_2} \nonumber
\end{align}
$$

The following proposition shows that this is indeed a relaxation of the non-signaling constraint.

**Proposition 4.22.** $S^{\text{NS}}(W, k_1, k_2) \le S^{\overline{\text{NS}}}(W, k_1, k_2)$.
*Proof.* Let us take a solution $(p_{x_1,x_2}, r_{x_1,x_2,y}, r_{x_1,x_2,y}^1, r_{x_1,x_2,y}^2)_{x_1 \in \mathcal{X}_1, x_2 \in \mathcal{X}_2, y \in \mathcal{Y}}$ of the linear program computing $S^{\text{NS}}(W, k_1, k_2)$. Let us show that $(p_{x_1,x_2}, r_{x_1,x_2,y})_{x_1 \in \mathcal{X}_1, x_2 \in \mathcal{X}_2, y \in \mathcal{Y}}$ is a solution of the linear program computing $S^{\overline{\text{NS}}}(W, k_1, k_2)$ with a same objective value, from which the proposition follows.

They have indeed the same value, since the definition which is the same for both programs depends only on $r_{x_1,x_2,y}$. Let us show that all constraints are satisfied for the solution $(p_{x_1,x_2}, r_{x_1,x_2,y})_{x_1 \in \mathcal{X}_1, x_2 \in \mathcal{X}_2, y \in \mathcal{Y}}$.

We have $\sum_{x_1,x_2} r_{x_1,x_2,y} = 1 \le 1$ so the first constraint is satisfied. We have then that:

$$ \sum_{x_1,x_2} p_{x_1,x_2} = k_1 \sum_{x_1,x_2} r_{x_1,x_2,y}^2 = k_1 k_2 \sum_{x_1,x_2} r_{x_1,x_2,y} = k_1 k_2, $$

so the second constraint is satisfied.

For the third constraint (and symmetrically the fourth constraint), we have:

$$ \sum_{x_1} p_{x_1,x_2} = k_1 \sum_{x_1} r_{x_1,x_2,y}^2 \ge k_1 \sum_{x_1} r_{x_1,x_2,y}. $$

Finally, we have directly $0 \le r_{x_1,x_2,y} \le p_{x_1,x_2}$, so the last constraint is satisfied. $\square$

We can now define the relaxed non-signaling assisted capacity region of a MAC $W$ using Definition 2.12 by $C^{\overline{\text{NS}}}(W) := C[S^{\overline{\text{NS}}}](W)$.

**Proposition 4.23.** $C^{\overline{\text{NS}}}(W)$ is convex.

*Proof.* We only need to show that for all MACs $W, W'$ and integers $k_1, k_2, k'_1, k'_2$:

$$ S^{\overline{\text{NS}}}(W \otimes W', k_1 k'_1, k_2 k'_2) \ge S^{\overline{\text{NS}}}(W, k_1, k_2) \cdot S^{\overline{\text{NS}}}(W', k'_1, k'_2), $$

thanks to Proposition 2.15. Let us consider optimal solutions $(p_{x_1,x_2}, r_{x_1,x_2,y})_{x_1 \in \mathcal{X}_1, x_2 \in \mathcal{X}_2, y \in \mathcal{Y}}$ and $(p'_{x'_1,x'_2}, r'_{x'_1,x'_2,y'})_{x'_1 \in \mathcal{X}'_1, x'_2 \in \mathcal{X}'_2, y' \in \mathcal{Y}'}$ of the programs computing $S^{\text{NS}}(W, k_1, k_2)$ and $S^{\overline{\text{NS}}}(W', k'_1, k'_2)$ respectively. We construct a solution of the program computing the success probability $S^{\overline{\text{NS}}}(W \otimes W', k_1 k'_1, k_2 k'_2)$ by defining:

$$ p_{x_1 x'_1 x_2 x'_2}^* := p_{x_1, x_2} p_{x'_1, x'_2} \text{ and } r_{x_1 x'_1 x_2 x'_2, y y'}^* := r_{x_1, x_2, y} r_{x'_1, x'_2, y'}^*, $$

as $W \otimes W' : (\mathcal{X}_1 \times \mathcal{X}'_1) \times (\mathcal{X}_2 \times \mathcal{X}'_2) \to (\mathcal{Y} \times \mathcal{Y}')$. It is straightforward to check that it is a valid solution. Indeed, linear constraints are preserved by product, except for coefficients $k_1, k_2$ which are replaced by products $k_1 k'_1$ and $k_2 k'_2$. Therefore, we have that $S^{\text{NS}}(W \otimes W', k_1 k'_1, k_2 k'_2)$ is larger than or equal to:

$$ \begin{align} & \frac{1}{k_1 k'_1 k_2 k'_2} \sum_{x_1, x'_1, x_2, x'_2, y, y'} (W \otimes W') (y y' | x_1 x'_1 x_2 x'_2) r_{x_1 x'_1, x_2 x'_2, y y'}^* \nonumber \\ &= \frac{1}{k_1 k'_1 k_2 k'_2} \sum_{x_1, x'_1, x_2, x'_2, y, y'} W(y | x_1 x_2) W'(y' | x'_1 x'_2) r_{x_1, x_2, y} r_{x'_1, x'_2, y'}^* \tag{4.53} \\ &= \left( \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} W(y | x_1 x_2) r_{x_1, x_2, y} \right) \cdot \left( \frac{1}{k'_1 k'_2} \sum_{x'_1, x'_2, y'} W'(y' | x'_1 x'_2) r'_{x'_1, x'_2, y'} \right) \nonumber \\ &= S^{\overline{\text{NS}}}(W, k_1, k_2) \cdot S^{\overline{\text{NS}}}(W', k'_1, k'_2). \end{align} $$
<figure><img src="image_21.png" /></figure>

A direct property that follows from this definition and Proposition 4.22 is the fact that the non-signaling assisted capacity region is included in the relaxed non-signaling assisted capacity region.

**Corollary 4.24.** $C^{\overline{\text{NS}}}(W) \subseteq C^{\overline{\text{NS}}}(W)$.

We present now the main result of this section, the characterization of $C^{\overline{\text{NS}}}(W)$ by a single-letter formula.

**Theorem 4.25 (Characterization of $C^{\overline{\text{NS}}}(W)$).** $C^{\overline{\text{NS}}}(W)$ is the closure of the convex hull of all rate pairs $(R_1, R_2)$ satisfying:

$$R_1 < I(X_1 : Y | X_2), R_2 < I(X_2 : Y | X_1), R_1 + R_2 < I((X_1, X_2) : Y),$$

for $(X_1, X_2)$ following some law $P_{X_1 X_2}$ on $\mathcal{X}_1 \times \mathcal{X}_2$, and $Y \in \mathcal{Y}$ the outcome of $W$ on inputs $X_1, X_2$.

*Remark.* Note that the only difference with the classical capacity region of MACs in Theorem 4.1 is that the joint distribution of $X_1$ and $X_2$ does not have a product form constraint here.

The proof of Theorem 4.25 will be divided in Proposition 4.31 (outer bound part) and Proposition 4.33 (achievability part). But first, let us apply these results to the binary adder channel.

**Application to the Binary Adder Channel** Let us determine the relaxed non-signaling assisted capacity of the binary adder channel which will be plotted in Figure 4.6.

**Proposition 4.26.** $C^{\overline{\text{NS}}}(W_{BAC})$ has the following description:

$$C^{\overline{\text{NS}}}(W_{BAC}) = \bigcup_{q \in [\frac{1}{2}, \frac{2}{3}]} \{(R_1, R_2) : R_1 \le h(q), R_2 \le h(q), R_1 + R_2 \le q + h(q)\}.$$

*Remark.* Note that for $q = \frac{1}{2}$, the bound becomes $R_1 \le 1$, $R_2 \le 1$, $R_1 + R_2 \le \frac{3}{2}$ and when $q = \frac{2}{3}$ the bound becomes $R_1 \le \log_2(3) - \frac{2}{3}$, $R_2 \le \log_2(3) - \frac{2}{3}$, $R_1 + R_2 \le \log_2(3)$.

*Proof.* We use the characterization of $C^{\overline{\text{NS}}}$ provided by Theorem 4.25.

Let us consider an arbitrary $P_{X_1 X_2} = (p_{00}, p_{01}, p_{10}, p_{11})$. First, we have that $I((X_1, X_2) : Y) = H(Y) - H(Y | X_1 X_2) = H(Y)$ since $Y$ is a deterministic function of $(X_1, X_2)$. Then, we have that $I(X_1 : Y | X_2) = H(Y | X_2) - H(Y | X_1 X_2) = H(Y | X_2)$ for the same reason. Furthermore, given $X_2$, $Y$ is a deterministic function of $X_1$, so we have $I(X_1 : Y | X_2) = H(Y | X_2) - H(Y | X_1 X_2) = H(X_1 | X_2)$. Symmetrically we have as well $I(X_2 : Y | X_1) = H(X_2 | X_1)$. In all, $C^{\overline{\text{NS}}}(W_{BAC})$ is equal to:

$$\bigcup_{P_{X_1 X_2}} \{(R_1, R_2) : R_1 \le H(X_1 | X_2), R_2 \le H(X_2 | X_1), R_1 + R_2 \le H(X_1 + X_2)\}.$$
Let us call $B_1(P_{X_1X_2}) := H(X_1|X_2)$, $B_2(P_{X_1X_2}) := H(X_2|X_1)$ and $B_{12}(P_{X_1X_2}) := H(X_1 + X_2)$ the three bounds. Let us call $P_{\overline{X_1}\overline{X_2}} = (p_{11}, p_{10}, p_{01}, p_{00})$. One can notice that:

$$
\begin{align}
B_1(P_{\overline{X_1} \overline{X_2}}) &= H(\overline{X}_1 | \overline{X}_2) = H(1 - X_1 | 1 - X_2) = H(X_1 | X_2) = B_1(P_{X_1 X_2}), \nonumber \\
B_2(P_{\overline{X_1} \overline{X_2}}) &= H(1 - X_2 | 1 - X_1) = H(X_2 | X_1) = B_2(P_{X_1 X_2}), \nonumber \\
B_{12}(P_{\overline{X_1} \overline{X_2}}) &= H(\overline{X}_1 + \overline{X}_2) = H(1 - X_1 + 1 - X_2) \nonumber \\
&= H(2 - (X_1 + X_2)) = H(X_1 + X_2) = B_{12}(P_{X_1 X_2}). \tag{4.54}
\end{align}
$$

Since $B_{12}(P_{X_1X_2}) = H(X_1 + X_2) = H(p_{00}, p_{11}, p_{01} + p_{10})$, it is concave in $P_{X_1X_2}$ as $H$ is concave and $(p_{00}, p_{11}, p_{01} + p_{10})$ is linear in $P_{X_1X_2}$. Also, $B_1(P_{X_1X_2}) = H(X_1|X_2) = -D(P_{X_1X_2}||I \otimes P_{X_2})$ is concave in $P_{X_1X_2}$ as the divergence $D$ is jointly convex and $I \otimes P_{X_2}$ is linear in $P_{X_1X_2}$. By symmetry, $B_2(P_{X_1X_2})$ is as well concave in $P_{X_1X_2}$. Let us consider any of those three bounds, which we call $B$. We have by concavity of $B$ and the fact that $B(P_{X_1X_2}) = B(P_{\overline{X_1}\overline{X_2}})$:

$$
\begin{align}
B(P_{X_1 X_2}) &= \frac{B(P_{X_1 X_2}) + B(P_{\overline{X_1} \overline{X_2}})}{2} \le B\left(\frac{P_{X_1 X_2} + P_{\hat{X}_1 \hat{X}_2}}{2}\right) \tag{4.55} \\
&= B\left(\frac{q}{2}, \frac{1-q}{2}, \frac{1-q}{2}, \frac{q}{2}\right), \nonumber
\end{align}
$$

with $q = p_{00} + p_{11}$. This holds for the three bounds at the same time, so we can restrict ourselves to the distributions of the form $(\frac{q}{2}, \frac{1-q}{2}, \frac{1-q}{2}, \frac{q}{2})$ for some $q \in [0, 1]$, i.e., $P_{X_1X_2}(0,0) = P_{X_1X_2}(1,1) = \frac{q}{2}$ and $P_{X_1X_2}(0,1) = P_{X_1X_2}(1,0) = \frac{1-q}{2}$.

We have $P_Y(0) = P_Y(2) = \frac{q}{2}$ and $P_Y(1) = 1 - q$, so:

$$
\begin{align}
B_{12}(P_{X_1 X_2}) &= H(Y) = -q \log\left(\frac{q}{2}\right) - (1-q) \log(1-q) \nonumber \\
&= -q (\log(q) - 1) - (1-q) \log(1-q) \tag{4.56} \\
&= q + h(q). \nonumber
\end{align}
$$

We have $P_{X_2}(0) = P_{X_1X_2}(0,0) + P_{X_1X_2}(1,0) = \frac{q}{2} + \frac{1-q}{2} = \frac{1}{2}$ so $P_{X_2}(1) = \frac{1}{2}$. Thus:

$$
B_1(P_{X_1 X_2}) = H(X_1 | X_2) = \frac{1}{2} H(X_1 | X_2 = 0) + \frac{1}{2} H(X_1 | X_2 = 1) .
$$

We have $P_{X_1|X_2=0}(0) = \frac{P_{X_1X_2(0,0)}}{P_{X_2(0)}} = q$ so $H(X_1|X_2=0) = h(q)$. On the other hand, we have $P_{X_1|X_2=1}(1) = \frac{P_{X_1X_2(1,1)}}{P_{X_2(1)}} = q$ so we get as well $H(Y|X_2=1) = h(q)$, and $B_1(P_{X_1X_2}) = H(X_1|X_2) = h(q)$. Symmetrically, we also get $B_2(P_{X_1X_2}) = h(q)$. Therefore, we get that $C^{\overline{\text{NS}}}(W_{\text{BAC}})$ is the closure of the convex hull of:

$$
\bigcup_{q \in [0,1]} \{(R_1, R_2) : R_1 < h(q), R_2 < h(q), R_1 + R_2 < q + h(q)\}.
$$

However this set is already convex, so we have:

$$
\mathcal{C}^{\overline{\text{NS}}}(W_{\text{BAC}}) = \bigcup_{q \in [0,1]} \{(R_1, R_2) : R_1 \le h(q), R_2 \le h(q), R_1 + R_2 \le q + h(q)\}.
$$
Finally, we can restrict ourselves to $q \in [\frac{1}{2}, \frac{2}{3}]$, since $h$ is increasing from 0 to $\frac{1}{2}$ (thus $q \mapsto q+h(q)$ as well), and the fact that $q \mapsto q+h(q)$ achieves its maximum for $q = \frac{2}{3}$ with $\frac{2}{3}+h(\frac{2}{3}) = \log_2(3)$ and then decreases (whereas $h$ is decreasing from $\frac{1}{2}$ to 1), which completes the proof. $\square$

As before, one can also define a symmetrized version of the relaxed linear program computing the value $S^{\overline{\text{NS}}}(W^{\otimes n}, k_1, k_2)$ in polynomial time in $n$ and compute the zero-error $n$-shots capacity region by looking at the rates where $S^{\overline{\text{NS}}}(W^{\otimes n}, k_1, k_2) = 1$. We have computed this up to 7 copies of the binary adder channel, which led to Figure 4.6:

<figure><img src="image_22.png" /><figcaption>Figure 4.6 – Comparison of relaxed and regular non-signaling assisted capacity regions of the binary adder channel. The black dashed curve depicts the classical capacity region $C(W_{\text{BAC}})$, whereas the grey dotted curve depicts the relaxed non-signaling assisted capacity region $C^{\overline{\text{NS}}}(W_{\text{BAC}})$ as described in Proposition 4.26. In particular, the curved corners are obtained by taking $R_1 = h(R_2)$ for $R_2 \in [\frac{1}{2}, \frac{2}{3}]$ and symmetrically by switching the roles played by $R_1$ and $R_2$. The continuous blue (respectively red) curve depicts the zero-error (respectively relaxed) non-signaling assisted achievable rate pairs for 7 copies of the binary adder channel.</figcaption></figure>

The first noticeable result coming from these curves is that the values $S^{\overline{\text{NS}}}$ and $S^{\text{NS}}$ differ. While the highest sum-rate of $\frac{2\log_2(42)}{7} \approx 1.5406$ is achieved on 7 copies of the binary adder channel with zero-error and non-signaling assistance, coming from the fact that $S^{\text{NS}}(W_{\text{BAC}}^{\otimes 7}, 42, 42) = 1$, we have that $S^{\overline{\text{NS}}}(W_{\text{BAC}}^{\otimes 7}, 44, 44) = 1 > S^{\text{NS}}(W_{\text{BAC}}^{\otimes 7}, 44, 44) \approx$
0.9581 which implies that a sum-rate of $\frac{2\log_2(44)}{7} \approx 1.5598$ is achieved on 7 copies of the binary adder channel with zero-error and relaxed non-signaling assistance. It also largely beats the best found sum-rate of $\frac{\log_2(72)}{4} \approx 1.5425$ achieved on 8 copies with the regular version. However the fact that the non-signaling assisted capacity region is strictly contained in the relaxed one is still open, as the same rates could potentially be achieved by the cost of using more copies of the channel.

### 4.5.1 Outer Bound Part of Theorem 4.25

In order to prove Proposition 4.31, we use a connection between hypothesis testing and relaxed non-signaling assisted codes as established in [Mat12] for point-to-point channels.

**Definition 4.5 (Hypothesis Testing).** Given distributions $P^{(0)}$ and $P^{(1)}$ on the same space $C$, we define $\beta_{1-\varepsilon}(P^{(0)}, P^{(1)})$ to be the minimum type II error $\sum_{r \in C} T_r P^{(1)}(r)$ that can be achieved by statistical tests $T$ which give a type I error no greater than $\varepsilon$, i.e. $\sum_{r \in C} T_r P^{(0)}(r) \ge 1 - \varepsilon$.

In other words, we have that:

$$ \beta_{1-\varepsilon}(P^{(0)}, P^{(1)}) = \begin{array}{ll} \underset{T_r}{\operatorname{minimize}} & \displaystyle \sum_{r \in C} T_r P^{(1)}(r) \\ \text{subject to} & \displaystyle \sum_{r \in C} T_r P^{(0)}(r) \ge 1 - \varepsilon \\ & 0 \le T_r \le 1. \end{array} \quad (4.57) $$

**Lemma 4.27.** For any relaxed non-signaling assisted code $(p_{x_1,x_2}, r_{x_1,x_2,y})_{x_1 \in \mathcal{X}_1, x_2 \in \mathcal{X}_2, y \in \mathcal{Y}}$ with $(k_1, k_2)$ messages and a probability of success $1 - \varepsilon$, if $P_{X_1 X_2}(x_1, x_2) = \frac{p_{x_1,x_2}}{k_1 k_2}$ and $Y \in \mathcal{Y}$ is the outcome of $W$ on inputs $X_1, X_2$, we have:

$$ \begin{align} \beta_{1-\varepsilon} (P_{X_1 X_2 Y}, P_{X_1 X_2} \times P_{Y|X_2}) &\le \frac{1}{k_1} \notag \\ \beta_{1-\varepsilon} (P_{X_1 X_2 Y}, P_{X_1 X_2} \times P_{Y|X_1}) &\le \frac{1}{k_2} \tag{4.58} \\ \beta_{1-\varepsilon} (P_{X_1 X_2 Y}, P_{X_1 X_2} \times P_Y) &\le \frac{1}{k_1 k_2}. \notag \end{align} $$

*Remark.* These three bounds are actually achieved with the same statistical test.

*Proof.* This result is a direct generalization of Theorem 9 in [Mat12] for point-to-point channels, itself a generalization of Theorem 27 in [PPV10] without non-signaling assistance.

Let us name $W_0 := W$ and $W_1$ a MAC yet to be defined. The coding strategy described by $r_{x_1,x_2,y}$ and $p_{x_1,x_2}$ leads to a probability of success on channel $i \in \{0, 1\}$ given by:

$$ \begin{align} 1 - \varepsilon_i &= \frac{1}{k_1 k_2} \sum_{x_1, x_2, y} r_{x_1, x_2, y} W_i(y|x_1 x_2) \notag \\ &= \sum_{x_1, x_2, y: p_{x_1, x_2} > 0} \frac{r_{x_1, x_2, y}}{p_{x_1, x_2}} W_i(y|x_1 x_2) \frac{p_{x_1, x_2}}{k_1 k_2} \quad \text{since } 0 \le r_{x_1, x_2, y} \le p_{x_1, x_2} \tag{4.59} \\ &= \sum_{x_1, x_2, y} T_{x_1, x_2, y} W_i(y|x_1 x_2) \frac{p_{x_1, x_2}}{k_1 k_2}, \notag \end{align} $$
with $T_{x_1,x_2,y} := \frac{r_{x_1,x_2,y}}{p_{x_1,x_2}}$ if $p_{x_1,x_2} > 0$, and $T_{x_1,x_2,y} := 0$ otherwise.

If now $Y$ is the output of the channel $W_i$, the joint distribution of $X_1, X_2, Y$ is given by
$$P_{X_1X_2Y}^{(i)}(x_1, x_2, y) = W_i(y|x_1x_2)P_{X_1X_2}(x_1, x_2) = W_i(y|x_1x_2)\frac{p_{x_1,x_2}}{k_1k_2}.$$

On the other hand, we have that for all $x_1, x_2, y$, $0 \le T_{x_1,x_2,y} \le 1$ since $0 \le r_{x_1,x_2,y} \le p_{x_1,x_2}$. So we get that:

$$1 - \varepsilon_i = \sum_{x_1,x_2,y} T_{x_1,x_2,y} P_{X_1X_2Y}^{(i)}(x_1, x_2, y).$$

Since $\sum_{x_1,x_2,y} T_{x_1,x_2,y} P_{X_1X_2Y}^{(0)}(x_1, x_2, y) \ge 1 - \varepsilon_0$ and $0 \le T_{x_1,x_2,y} \le 1$, we have:

$$\beta_{1-\varepsilon_0}(P^{(0)}, P^{(1)}) \le \sum_{x_1,x_2,y} T_{x_1,x_2,y} P_{X_1X_2Y}^{(1)}(x_1, x_2, y) = 1 - \varepsilon_1.$$

Let us now consider three general cases, depending on the fact that $W_1$ does not depend on $x_1, x_2$ or both: $W_1(y|x_1x_2) := Q^{(1)}(y|x_2)$; $W_1(y|x_1x_2) := Q^{(2)}(y|x_1)$; $W_1(y|x_1x_2) := Q^{(0)}(y)$. These will give respectively the three bounds we want.

First, let us consider the case where $W_1(y|x_1x_2) := Q^{(1)}(y|x_2)$ (the second case where $W_1(y|x_1x_2) := Q^{(2)}(y|x_1)$ being symmetric), we have that:

$$\begin{align}
1 - \varepsilon_1 &= \sum_{x_1,x_2,y} T_{x_1,x_2,y} Q^{(1)}(y|x_2) \frac{p_{x_1,x_2}}{k_1k_2} = \frac{1}{k_1k_2} \sum_{x_2,y} Q^{(1)}(y|x_2) \sum_{x_1} T_{x_1,x_2,y} p_{x_1,x_2} \nonumber \\
&= \frac{1}{k_1k_2} \sum_{x_2,y} Q^{(1)}(y|x_2) \sum_{x_1} r_{x_1,x_2,y} \le \frac{1}{k_1k_2} \sum_{x_2,y} Q^{(1)}(y|x_2) \frac{1}{k_1} \sum_{x_1} p_{x_1,x_2} \nonumber \\
&= \frac{1}{k_1} \sum_{x_1,x_2} \frac{p_{x_1,x_2}}{k_1k_2} \sum_{y} Q^{(1)}(y|x_2) = \frac{1}{k_1} \sum_{x_1,x_2} \frac{p_{x_1,x_2}}{k_1k_2} = \frac{1}{k_1}. \tag{4.60}
\end{align}$$

For the third case, when $W_1(y|x_1x_2) := Q^{(0)}(y)$, we have:

$$\begin{align}
1 - \varepsilon_1 &= \sum_{x_1,x_2,y} T_{x_1,x_2,y} Q^{(0)}(y) \frac{p_{x_1,x_2}}{k_1k_2} = \frac{1}{k_1k_2} \sum_{y} Q^{(0)}(y) \sum_{x_1,x_2} T_{x_1,x_2,y} p_{x_1,x_2} \tag{4.61} \\
&= \frac{1}{k_1k_2} \sum_{y} Q^{(0)}(y) \sum_{x_1,x_2} r_{x_1,x_2,y} \le \frac{1}{k_1k_2} \sum_{y} Q^{(0)}(y) = \frac{1}{k_1k_2}. \nonumber
\end{align}$$

In those three cases, we have respectively $P_{X_1X_2Y}^{(1)} = P_{X_1X_2} \times Q_{Y|X_2}^{(1)}; P_{X_1X_2} \times Q_{Y|X_1}^{(2)}$; and $P_{X_1X_2} \times Q_{Y}^{(0)}$. Specializing those cases with $Q_{Y|X_2}^{(1)} := P_{Y|X_2}; Q_{Y|X_1}^{(2)} := P_{Y|X_1}; Q_{Y}^{(0)} := P_Y$ and using the fact that $\beta_{1-\varepsilon_0}(P^{(0)}, P^{(1)}) \le 1 - \varepsilon_1$ concludes the proof. $\square$

**Lemma 4.28.** For any relaxed non-signaling assisted code $(p_{x_1,x_2}, r_{x_1,x_2,y})_{x_1 \in \mathcal{X}_1, x_2 \in \mathcal{X}_2, y \in \mathcal{Y}}$ with $(k_1, k_2)$ messages and a probability of success $1 - \varepsilon$, if $P_{X_1X_2}(x_1, x_2) = \frac{p_{x_1,x_2}}{k_1k_2}$ and $Y \in \mathcal{Y}$ is the outcome of $W$ on inputs $X_1, X_2$, we have:

$$\begin{align}
\log(k_1) &\le \frac{I(X_1:Y|X_2) + h(\varepsilon)}{1 - \varepsilon}, \nonumber \\
\log(k_2) &\le \frac{I(X_2:Y|X_1) + h(\varepsilon)}{1 - \varepsilon}, \tag{4.62} \\
\log(k_1) + \log(k_2) &\le \frac{I((X_1, X_2):Y) + h(\varepsilon)}{1 - \varepsilon}. \nonumber
\end{align}$$
*Proof.* Thanks to Lemma 4.27, with the fact that $P_{X_1X_2} = P_{X_1|X_2} \times P_{X_2} = P_{X_2|X_1} \times P_{X_1}$, we have already:

$$ 
\begin{align} 
\beta_{1-\epsilon} (P_{X_1 X_2 Y}, (P_{X_1|X_2} \times P_{Y|X_2}) \times P_{X_2}) &\le \frac{1}{k_1}, \notag \\
\beta_{1-\epsilon} (P_{X_1 X_2 Y}, (P_{X_2|X_1} \times P_{Y|X_1}) \times P_{X_1}) &\le \frac{1}{k_2}, \tag{4.63} \\
\beta_{1-\epsilon} (P_{X_1 X_2 Y}, P_{X_1 X_2} \times P_Y) &\le \frac{1}{k_1 k_2}. \notag
\end{align} 
$$

Following the steps of section G in [PPV10], since any hypothesis test is a binary-output transformation, by data-processing inequality for divergence, we have that:

$$ 
\begin{align} 
d \left( 1 - \epsilon \middle| \beta_{1-\epsilon} \left( P_{X_1 X_2 Y}, (P_{X_1|X_2} \times P_{Y|X_2}) \times P_{X_2} \right) \right) & \notag \\
= d \left( \beta_{1-\epsilon} \left( P_{X_1 X_2 Y}, P_{X_1 X_2 Y} \right) \middle| \beta_{1-\epsilon} \left( P_{X_1 X_2 Y}, (P_{X_1|X_2} \times P_{Y|X_2}) \times P_{X_2} \right) \right) \tag{4.64} \\
\le D \left( P_{X_1 X_2 Y} \middle| \left( P_{X_1|X_2} \times P_{Y|X_2} \right) \times P_{X_2} \right) &= I(X_1 : Y | X_2), \notag 
\end{align} 
$$

where the binary divergence is defined by $d(a||b) := a \log(\frac{a}{b}) + (1-a) \log(\frac{1-a}{1-b})$ and satisfies $d(a||b) \ge -h(a) - a \log(b)$, and thus:

$$ \log\left(\frac{1}{b}\right) \le \frac{d(a||b) + h(a)}{a} = \frac{d(a||b) + h(1-a)}{a}. $$

This leads to:

$$ \log(k_1) \le \frac{1}{\log\left(\left(\beta_{1-\epsilon}(P_{X_1 X_2 Y}, (P_{X_1|X_2} \times P_{Y|X_2}) \times P_{X_2})\right)\right)} \le \frac{I(X_1:Y|X_2) + h(\epsilon)}{1-\epsilon}. $$

Similarly for the two other inequalities, since $D(P_{X_1 X_2 Y} || (P_{X_2|X_1} \times P_{Y|X_1}) \times P_{X_1}) = I(X_2:Y|X_1)$ and $D(P_{X_1 X_2 Y} || P_{X_1 X_2} \times P_Y) = I((X_1, X_2):Y)$, we get:

$$ 
\begin{align} 
\log(k_1) &\le \frac{I(X_1:Y|X_2) + h(\epsilon)}{1-\epsilon}, \notag \\
\log(k_2) &\le \frac{I(X_2:Y|X_1) + h(\epsilon)}{1-\epsilon}, \tag{4.65} \\
\log(k_1) + \log(k_2) &\le \frac{I((X_1,X_2):Y) + h(\epsilon)}{1-\epsilon}. \notag
\end{align} 
$$

□

In order to show additivity of the outer bound, we use the following lemma.

**Lemma 4.29.** For any distribution $P_{X_1^n X_2^n}$ of $(X_1^n, X_2^n)$, if $Y^n \in \mathcal{Y}^n$ is the outcome of $W^n$ on inputs $X_1^n, X_2^n$, we have:

$$ 
\begin{align} 
I(X_1^n:Y^n|X_2^n) &\le \sum_{i=1}^n I(X_{1,i}:Y_i|X_{2,i}) \notag \\
I(X_2^n:Y^n|X_1^n) &\le \sum_{i=1}^n I(X_{2,i}:Y_i|X_{1,i}) \tag{4.66} \\
I((X_1^n, X_2^n):Y^n) &\le \sum_{i=1}^n I((X_{1,i}, X_{2,i}):Y_i). \notag
\end{align} 
$$
*Proof.* Consider $n$ copies of $W$. Let us write $X_{1,-i} := X_{1,1} \dots X_{1,i-1} X_{1,i+1} \dots X_{1,n}$ and $Z^n := Z_1 \dots Z_n$. We have:

$$
\begin{align}
I(X_1^n : Y^n | X_2^n) &= I(X_1^n : Y^n | X_2^n) \nonumber \\
&= \sum_{i=1}^n I(X_1^n : Y_i | X_2^n Y^{i-1}) \quad \text{by the chain rule} \nonumber \\
&= \sum_{i=1}^n I(X_{1,i} : Y_i | X_2^n Y^{i-1}) + \sum_{i=1}^n I(X_{1,-i} : Y_i | X_2^n Y^{i-1} X_{1,i}) \tag{4.67} \\
&= \sum_{i=1}^n I(X_{1,i} : Y_i | X_2^n Y^{i-1}), \nonumber
\end{align}
$$

where the last equality comes from Lemma 4.30. As a result,

$$
\begin{align}
I(X_1^n : Y^n | X_2^n) &= \sum_{i=1}^{n} H(Y_i | X_2^n Y^{i-1}) - H(Y_i | X_2^n Y^{i-1} X_{1,i}) \nonumber \\
&= \sum_{i=1}^{n} H(Y_i | X_2^n Y^{i-1}) - H(Y_i | X_{2,i} X_{1,i}) \tag{4.68} \\
&\text{since } X_{2,-i} Y^{i-1} \to (X_{1,i}, X_{2,i}) \to Y_i \text{ Markov chain.} \nonumber \\
&\le \sum_{i=1}^{n} H(Y_i | X_{2,i}) - H(Y_i | X_{2,i} X_{1,i}) = \sum_{i=1}^{n} I(X_{1,i} : Y_i | X_{2,i}). \nonumber
\end{align}
$$

Symmetrically, by switching the roles of $X_1$ and $X_2$, we get the second part of Lemma 4.29.
For the sum-rate case:

$$
\begin{align*}
I((X_1^n, X_2^n) : Y^n) &= \sum_{i=1}^{n} I((X_1^n, X_2^n) : Y_i | Y^{i-1}) \text{ by the chain rule} \\
&= \sum_{i=1}^{n} I((X_{1,i}, X_{2,i}) : Y_i | Y^{i-1}) + \sum_{i=1}^{n} I((X_{1,-i}, X_{2,-i}) : Y_i | Y^{i-1} X_{1,i} X_{2,i}) \\
&= \sum_{i=1}^{n} I((X_{1,i}, X_{2,i}) : Y_i | Y^{i-1}) \\
&\quad \text{since } (X_{1,-i}, X_{2,-i}) \to Y^{i-1} X_{1,i} X_{2,i} \to Y_i \text{ Markov chain.} \\
&= \sum_{i=1}^{n} H(Y_i | Y^{i-1}) - H(Y_i | Y^{i-1} X_{1,i} X_{2,i}) \\
&= \sum_{i=1}^{n} H(Y_i | Y^{i-1}) - H(Y_i | X_{1,i} X_{2,i}) \text{ since } Y^{i-1} \to (X_{1,i}, X_{2,i}) \to Y_i \text{ Markov chain.} \\
&\le \sum_{i=1}^{n} H(Y_i) - H(Y_i | X_{2,i} X_{1,i}) = \sum_{i=1}^{n} I((X_{1,i}, X_{2,i}) : Y_i).
\end{align*}
$$

<figure><img src="image_23.png" /></figure>

We next prove a technical lemma that was used in the previous proof.

**Lemma 4.30.** For any distribution $P_{X_1^n X_2^n}$ of $(X_1^n, X_2^n)$, if $Y^n \in \mathcal{Y}^n$ is the outcome of $W^n$ on inputs $X_1^n, X_2^n$, we have $I(X_{1,-i} : Y_i | X_{1,i} X_2^n Y^{i-1}) = 0$.
Proof. Let us show that, conditioned on any particular instance of $X_{1,i} = x_{i,1}$, $X_2^n = x_2^n$, $Y_1^{i-1} = y^{i-1}$, $X_{1,-i}$ and $Y_i$ are independent.

We have:

$$ 
\begin{aligned}
\mathbb{P}(Y_i = y_i | X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1}) &= \mathbb{P}(Y_i = y_i | X_{1,i} = x_{i,1}, X_{2,i} = x_{2,i}) \\
&= W(y_i | x_{1,i}x_{2,i}),
\end{aligned} 
\quad (4.69) $$

by definition of the law of $Y_i$. On the other hand, we have that:

$$ \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n, Y^n = y_n) = \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{n} W(y_j | x_{1,j}x_{2,j}). $$

Thus, we have that $\mathbb{P}(X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1})$ is equal to:

$$ 
\begin{aligned}
& \sum_{x_{1,-i}, x_2^n, y_i, \dots, y_n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{n} W(y_j | x_{1,j}x_{2,j}) \\
&= \sum_{x_{1,-i}, x_2^n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{i-1} W(y_j | x_{1,j}x_{2,j}) \prod_{j=i}^{n} \left( \sum_{y_j} W(y_j | x_{1,j}x_{2,j}) \right) \quad (4.70) \\
&= \sum_{x_{1,-i}, x_2^n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{i-1} W(y_j | x_{1,j}x_{2,j}).
\end{aligned} $$

And then, $\mathbb{P}(X_{1,-i} = x_{1,-i} | X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1})$ is equal to:

$$ \frac{\sum_{x_2^n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{i-1} W(y_j | x_{1,j}x_{2,j})}{\sum_{x_{1,-i}, x_2^n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{i-1} W(y_j | x_{1,j}x_{2,j})}. $$

But:

$$ 
\begin{aligned}
& \mathbb{P}(X_{1,-i} = x_{1,-i}, Y_i = y_i | X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1}) \\
&= \frac{\sum_{x_2^n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{i} W(y_j | x_{1,j}x_{2,j})}{\sum_{x_{1,-i}, x_2^n} \mathbb{P}(X_1^n = x_1^n, X_2^n = x_2^n) \prod_{j=1}^{i-1} W(y_j | x_{1,j}x_{2,j})} \\
&= \mathbb{P}(X_{1,-i} = x_{1,-i} | X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1}) W(y_i | x_{1,i}x_{2,i}) \quad (4.71) \\
&= \mathbb{P}(X_{1,-i} = x_{1,-i} | X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1}) \\
&\quad \times \mathbb{P}(Y_i = y_i | X_{1,i} = x_{i,1}, X_2^n = x_2^n, Y_1^{i-1} = y^{i-1}).
\end{aligned} $$

Thus, conditioned on any particular instance of $X_{1,i} = x_{i,1}$, $X_2^n = x_2^n$, $Y_1^{i-1} = y^{i-1}$, $X_{1,-i}$ and $Y_i$ are independent, and so $I(X_{1,-i} : Y_i | X_{1,i}, X_2^n Y_1^{i-1}) = 0$. $\square$

Combining the previous results gives the desired outer bound.
**Proposition 4.31** (Outer bound part of Theorem 4.25). If a rate pair is achievable with relaxed non-signaling assistance then it is in the closure of the convex hull of all $(R_1, R_2)$ satisfying:

$$R_1 < I(X_1:Y|X_2), R_2 < I(X_2:Y|X_1), R_1+R_2 < I((X_1, X_2):Y),$$

for $(X_1, X_2)$ following some law $P_{X_1X_2}$ on $\mathcal{X}_1 \times \mathcal{X}_2$, and $Y \in \mathcal{Y}$ the outcome of $W$ on inputs $X_1, X_2$.

*Proof.* Consider $(R_1, R_2)$ achievable with relaxed non-signaling assistance: we have a sequence of relaxed non-signaling assisted codes for $n$ copies of the MAC $W$ with $k_1 = 2^{nR_1}$, $k_2 = 2^{nR_2}$ messages and an error probability $\epsilon_n \xrightarrow{n \to +\infty} 0$, along with associated distributions of $X_1^n X_2^n Y^n$.

Thus combining Lemma 4.28 and Lemma 4.29, we have that:

$$R_1 \le \frac{1}{n} \frac{I(X_1^n:Y^n|X_2^n) + h(\epsilon_n)}{1 - \epsilon_n} \le \frac{1}{n} \frac{\sum_{i=1}^n I(X_{1,i}:Y_i|X_{2,i}) + h(\epsilon_n)}{1 - \epsilon_n},$$

$$R_2 \le \frac{1}{n} \frac{I(X_2^n:Y^n|X_1^n) + h(\epsilon_n)}{1 - \epsilon_n} \le \frac{1}{n} \frac{\sum_{i=1}^n I(X_{2,i}:Y_i|X_{1,i}) + h(\epsilon_n)}{1 - \epsilon_n},$$

$$R_1 + R_2 \le \frac{1}{n} \frac{I((X_1^n, X_2^n):Y^n) + h(\epsilon_n)}{1 - \epsilon_n} \le \frac{1}{n} \frac{\sum_{i=1}^n I((X_{1,i}, X_{2,i}):Y_i) + h(\epsilon_n)}{1 - \epsilon_n}. \quad (4.72)$$

Then let us consider some random variable $Q$ uniform on $[n]$ and independent of $(X_1^n, X_2^n, Y^n)$. Then we can write:

$$\sum_{i=1}^{n} I(X_{1,i}: Y_i|X_{2,i}) = \sum_{i=1}^{n} I(X_{1,i}: Y_i|X_{2,i}, Q = i) = nI(X_{1,Q}: Y_Q|X_{2,Q}, Q).$$

Since $Y_Q$ conditioned on $X_{1,Q}$ and $X_{2,Q}$ still follows the law of the MAC $W(y|x_1x_2)$, we can take $X_1 = X_{1,Q}$, $X_2 = X_{2,Q}$, and then the output of the channel $Y$ satisfies $Y = Y_Q$, and thus we obtain:

$$R_1 \le \frac{I(X_1:Y|X_2,Q) + \frac{h(\epsilon_n)}{n}}{1 - \epsilon_n}.$$

Doing this similarly on the other conditional mutual informations, we get:

$$R_1 \le \frac{I(X_1:Y|X_2,Q) + \frac{h(\epsilon_n)}{n}}{1 - \epsilon_n}, \\ R_2 \le \frac{I(X_2:Y|X_1,Q) + \frac{h(\epsilon_n)}{n}}{1 - \epsilon_n}, \quad (4.73)$$

$$R_1 + R_2 \le \frac{I((X_1, X_2):Y|Q) + \frac{h(\epsilon_n)}{n}}{1 - \epsilon_n}.$$

By taking the limit as $n$ goes to infinity, since the limit of $\epsilon_n$ is 0, then the limit of $\frac{h(\epsilon_n)}{n}$ is 0 as well and we get that $(R_1, R_2)$ must be in the set of rate pairs such that:

$$R_1 \le I(X_1:Y|X_2,Q), \\ R_2 \le I(X_2:Y|X_1,Q), \quad (4.74)$$

$$R_1 + R_2 \le I((X_1, X_2):Y|Q),$$
for some uniform $Q$ in a finite set, $(X_1, X_2)$ any joint law depending on $Q$, and $Y$ the output of $W$ on inputs $(X_1, X_2)$.

Finally, in order to show that this is the right region, one has only to see that the corner points of this region, such as for instance $(I(X_1 : Y|Q), I(X_2 : Y|X_1, Q))$, are finite convex combination of the points $(I(X_1 : Y|Q = q), I(X_2 : Y|X_1, Q = q))$ which are all in the capacity region of the theorem by taking $(X_1 X_2) \sim P_{X_1 X_2|Q=q}$. This implies that $(R_1, R_2)$ is in the convex hull of that region, so we can drop the random variable $Q$ and the proof is completed. $\square$

The main consequence of that outer bound on the relaxed non-signaling assisted capacity region is that it holds also for the non-signaling assisted capacity region thanks to Corollary 4.24:

**Corollary 4.32 (Outer Bound on the Non-Signaling Assisted Capacity Region).** If a rate pair is achievable with non-signaling assistance, then it is in the closure of the convex hull of all $(R_1, R_2)$ satisfying:

$$R_1 < I(X_1 : Y | X_2), R_2 < I(X_2 : Y | X_1), R_1 + R_2 < I((X_1, X_2) : Y),$$

for $(X_1, X_2)$ following any law $P_{X_1 X_2}$ on $\mathcal{X}_1 \times \mathcal{X}_2$, and $Y \in \mathcal{Y}$ the outcome of $W$ on inputs $X_1, X_2$.

### 4.5.2 Achievability Part of Theorem 4.25

In order to construct the relaxed non-signaling assisted code for achievability, we will use the notions of jointly and conditional typical sets described in Section 2.2.2:

**Proposition 4.33 (Achievability part of Theorem 4.25).** If a rate pair is in the closure of the convex hull of $(R_1, R_2)$ satisfying:

$$R_1 < I(X_1 : Y | X_2), R_2 < I(X_2 : Y | X_1), R_1 + R_2 < I((X_1, X_2) : Y),$$

for $(X_1, X_2)$ following some law $P_{X_1 X_2}$ on $\mathcal{X}_1 \times \mathcal{X}_2$, and $Y \in \mathcal{Y}$ the outcome of $W$ on inputs $X_1, X_2$, then it is in $C^{\mathcal{N}\mathcal{S}}(W)$.

*Proof.* Let us fix $\varepsilon, \varepsilon' \in (0, 1)$ such that $\varepsilon' < \varepsilon \le \frac{1}{2}$. Let $n \in \mathbb{N}$ which will be chosen large enough during the proof.

We consider $n$ independent random variables $(X_{1,i}X_{2,i}Y_i) \sim P_{X_1 X_2 Y}$, following the distribution $P_{X_1 X_2 Y}(x_{1,i}, x_{2,i}, y_i) = W(y_i|x_{1,i}x_{2,i})P_{X_1 X_2}(x_{1,i}, x_{2,i})$. We call $P_{X_1^n X_2^n Y^n}$ the law of their product. We have then $P_{X_1^n X_2^n}(x_1^n, x_2^n) := \prod_{i=1}^n P_{X_1 X_2}(x_{1,i}, x_{2,i})$. If $\hat{Y}$ is the output of $W^{\otimes n}$ on $X_1^n X_2^n$, we have that:

$$
\begin{align}
P_{X_1^n X_2^n \hat{Y}}(x_1^n, x_2^n, y^n) &= W^{\otimes n}(y^n | x_1^n x_2^n) P_{X_1^n X_2^n}(x_1^n, x_2^n) \nonumber \\
&= W^{\otimes n}(y^n | x_1^n x_2^n) \prod_{i=1}^n P_{X_1 X_2}(x_{1,i}, x_{2,i}) \nonumber \\
&= \prod_{i=1}^n W(y_i | x_{1,i} x_{2,i}) P_{X_1 X_2}(x_{1,i}, x_{2,i}) \tag{4.75} \\
&= \prod_{i=1}^n P_{X_1 X_2 Y}(x_{1,i}, x_{2,i}, y_i). \nonumber
\end{align}
$$
Thus, $\hat{Y}$ follows the product law of $Y_i$, i.e. $\hat{Y} = Y^n$.

Let us consider $C_1, C_2, C_3$ some positive numbers independent of $n$ and $\epsilon$ which we will define later, $k_1 = 2^{nR_1}$, $k_2 = 2^{nR_2}$ integers with $(R_1, R_2)$ positive rates such that:

$$ 
\begin{align} 
R_1 &\le I(X_1:Y|X_2) - \frac{1}{n} - C_1\epsilon, \nonumber \\ 
R_2 &\le I(X_2:Y|X_1) - \frac{1}{n} - C_2\epsilon, \tag{4.76} 
\end{align} 
$$

$$R_1 + R_2 \le I((X_1, X_2):Y) - \frac{1}{n} - C_3\epsilon.$$

We define a solution of $\overline{S^{\text{NS}}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_2})$ in the following way:

$$p_{x_1^n, x_2^n} := \begin{cases} \frac{2^{n(R_1+R_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} & \text{if } (x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2), \\ 0 & \text{otherwise.} \end{cases}$$

and:

$$r_{x_1^n, x_2^n, y^n} := \begin{cases} p_{x_1^n, x_2^n} & \text{if } (x_1^n, x_2^n, y^n) \in \mathcal{T}_{\epsilon'}^n(X_1, X_2, Y), \\ 0 & \text{otherwise.} \end{cases}$$

By construction, the constraint $0 \le r_{x_1^n, x_2^n, y^n} \le p_{x_1^n, x_2^n}$ is satisfied. We have also that:

$$\sum_{x_1^n, x_2^n} p_{x_1^n, x_2^n} = \sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2)} \frac{2^{n(R_1+R_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} = 2^{n(R_1+R_2)},$$

which is equal to $k_1k_2$. If $(x_1^n, x_2^n, y^n) \in \mathcal{T}_{\epsilon'}^n(X_1, X_2, Y)$, we have that $(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon'}^n(X_1, X_2) \subseteq \mathcal{T}_{\epsilon}^n(X_1, X_2)$, so in that case:

$$r_{x_1^n, x_2^n, y^n} = \frac{2^{n(R_1+R_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}.$$

If $y^n \notin \mathcal{T}_{\epsilon'}^n(Y)$, then for all $(x_1^n, x_2^n), (x_1^n, x_2^n, y^n) \notin \mathcal{T}_{\epsilon'}^n(X_1, X_2, Y)$, so $\sum_{x_1^n, x_2^n} r_{x_1^n, x_2^n, y^n} = 0 \le 1$ in that case.

Otherwise, if $y^n \in \mathcal{T}_{\epsilon'}^n(Y)$, then:

$$ 
\begin{align} 
\sum_{x_1^n, x_2^n} r_{x_1^n, x_2^n, y^n} &= 2^{n(R_1+R_2)} \frac{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon'}^n(X_1, X_2|y^n)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} \nonumber \\ 
&\le 2^{n(R_1+R_2)} \frac{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2|y^n)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} \tag{4.77} \\ 
&\le 2^{n(R_1+R_2)} \frac{2^{-n(1-\epsilon)H(X_1, X_2)}}{2^{-n(1+\epsilon)H(X_1, X_2)}} \frac{|\mathcal{T}_{\epsilon}^n(X_1, X_2|y^n)|}{|\mathcal{T}_{\epsilon}^n(X_1, X_2)|} \quad \text{since } (x_1^n, x_2^n) \in \mathcal{T}_{\epsilon}^n(X_1, X_2) \nonumber \\ 
&= 2^{n(R_1+R_2+2\epsilon H(X_1, X_2))} \frac{|\mathcal{T}_{\epsilon}^n(X_1, X_2|y^n)|}{|\mathcal{T}_{\epsilon}^n(X_1, X_2)|}. \nonumber 
\end{align} 
$$
But $|\mathcal{T}_{\varepsilon}^n(X_1, X_2|y^n)| \le 2^{n(1+\varepsilon)H(X_1, X_2|Y)}$ and for a large enough $n$, $|\mathcal{T}_{\varepsilon}^n(X_1, X_2)| \ge (1-\varepsilon)2^{n(1-\varepsilon)H(X_1, X_2)} \ge 2^{n((1-\varepsilon)H(X_1, X_2)-\frac{1}{n})}$, so in that case:

$$ \sum_{x_1^n, x_2^n} r_{x_1^n, x_2^n, y^n} \le 2^{n(R_1+R_2+2\varepsilon H(X_1, X_2))} \frac{2^{n(1+\varepsilon)H(X_1, X_2|Y)}}{2^{n((1-\varepsilon)H(X_1, X_2)-\frac{1}{n})}} \quad (4.78) \\ = 2^{n(R_1+R_2-I(X_1, X_2:Y)+\frac{1}{n}+C_3\varepsilon)} \le 1, $$

since $I(X_1, X_2:Y) = H(X_1, X_2) - H(X_1, X_2|Y)$ and $R_1+R_2 \le I(X_1, X_2:Y) - \frac{1}{n} - C_3\varepsilon$, with $C_3 := H(X_1, X_2|Y) + 3H(X_1, X_2)$.

Let us focus on the constraint $\sum_{x_1^n} p_{x_1^n, x_2^n} \ge k_1 \sum_{x_1^n} r_{x_1^n, x_2^n, y^n}$ (the symmetric constraint $\sum_{x_2^n} p_{x_1^n, x_2^n} \ge k_2 \sum_{x_2^n} r_{x_1^n, x_2^n, y^n}$ will be achieved for symmetric reasons).

Let us fix $(x_2^n, y^n)$. If $(x_2^n, y^n) \notin \mathcal{T}_{\varepsilon'}^n(X_2, Y)$, then for all $x_1^n$, $(x_1^n, x_2^n, y^n) \notin \mathcal{T}_{\varepsilon'}^n(X_1, X_2, Y)$, thus $r_{x_1^n, x_2^n, y^n} = 0$ and the constraint is fulfilled. Let us assume that $(x_2^n, y^n) \in \mathcal{T}_{\varepsilon'}^n(X_2, Y)$. Since $r_{x_1^n, x_2^n, y^n} > 0$ implies that $(x_1^n, x_2^n, y^n) \in \mathcal{T}_{\varepsilon'}^n(X_1, X_2, Y)$, we have that:

$$ \sum_{x_1^n} r_{x_1^n, x_2^n, y^n} = \sum_{x_1^n \in \mathcal{T}_{\varepsilon'}^n(X_1|x_2^n, y^n)} r_{x_1^n, x_2^n, y^n} = \sum_{x_1^n \in \mathcal{T}_{\varepsilon'}^n(X_1|x_2^n, y^n)} p_{x_1^n, x_2^n} . $$

Thus:

$$ \begin{aligned} \frac{\sum_{x_1^n} p_{x_1^n, x_2^n}}{k_1 \sum_{x_1^n} r_{x_1^n, x_2^n, y^n}} &\ge \frac{1}{k_1} \frac{\sum_{x_1 \in \mathcal{T}_{\varepsilon}^n(X_1|x_2^n)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{x_1 \in \mathcal{T}_{\varepsilon'}^n(X_1|x_2^n, y^n)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} \\ &\ge \frac{1}{k_1} \frac{\sum_{x_1 \in \mathcal{T}_{\varepsilon}^n(X_1|x_2^n)} P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{x_1 \in \mathcal{T}_{\varepsilon'}^n(X_1|x_2^n, y^n)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} \\ &\ge \frac{1}{k_1} \frac{2^{-n(1+\varepsilon)H(X_1, X_2)}}{2^{-n(1-\varepsilon)H(X_1, X_2)}} \frac{|\mathcal{T}_{\varepsilon}^n(X_1|x_2^n)|}{|\mathcal{T}_{\varepsilon}^n(X_1|x_2^n, y^n)|} \\ &\ge 2^{n(-R_1-2\varepsilon H(X_1, X_2))} \frac{|\mathcal{T}_{\varepsilon}^n(X_1|x_2^n)|}{|\mathcal{T}_{\varepsilon}^n(X_1|x_2^n, y^n)|} . \end{aligned} \quad (4.79) $$

But $|\mathcal{T}_{\varepsilon}^n(X_1|x_2^n, y^n)| \le 2^{n(1+\varepsilon)H(X_1|X_2Y)}$ and for a large enough $n$ we have $\forall x_2^n \in \mathcal{T}_{\varepsilon'}^n(X_2), |\mathcal{T}_{\varepsilon}^n(X_1|x_2^n)| \ge (1-\varepsilon)2^{n(1-\varepsilon)H(X_1|X_2)} \ge 2^{n((1-\varepsilon)H(X_1|X_2)-\frac{1}{n})}$, so we get with $C_1 := 2H(X_1, X_2) + H(X_1|X_2Y) + H(X_1|X_2)$ (symmetrically $C_2 := 2H(X_1, X_2) + H(X_2|X_1Y) + H(X_2|X_1)$):

$$ \begin{aligned} \frac{\sum_{x_1^n} p_{x_1^n, x_2^n}}{k_1 \sum_{x_1^n} r_{x_1^n, x_2^n, y^n}} &\ge 2^{n(H(X_1|X_2)-\frac{1}{n}-H(X_1|X_2Y)-R_1-C_1\varepsilon)} \quad (4.80) \\ &= 2^{n(I(X_1:Y|X_2)-\frac{1}{n}-C_1\varepsilon-R_1)} \ge 1. \end{aligned} $$

For a large enough $n$, all constraints are satisfied, thus $(p_{x_1^n, x_2^n}, r_{x_1^n, x_2^n, y^n})$ is a valid solution.
Then:

$$
\begin{align}
\overline{S^{\text{NS}}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_2}) &\ge \frac{1}{2^{n(R_1+R_2)}} \sum_{(x_1^n, x_2^n, y^n) \in \mathcal{T}_{\varepsilon'}^n(X_1, X_2, Y)} W^{\otimes n}(y^n | x_1^n, x_2^n) r_{x_1^n, x_2^n, y^n} \nonumber \\
&= \sum_{(x_1^n, x_2^n, y^n) \in \mathcal{T}_{\varepsilon'}^n(X_1, X_2, Y)} \frac{P_{X_1^n X_2^n Y^n}(x_1^n, x_2^n, y^n)}{P_{X_1^n X_2^n}(x_1^n, x_2^n)} \frac{P_{X_1^n X_2^n}(x_1^n, x_2^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\varepsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} \nonumber \\
&= \frac{\sum_{(x_1^n, x_2^n, y^n) \in \mathcal{T}_{\varepsilon'}^n(X_1, X_2, Y)} P_{X_1^n X_2^n Y^n}(x_1^n, x_2^n, y^n)}{\sum_{(x_1^n, x_2^n) \in \mathcal{T}_{\varepsilon}^n(X_1, X_2)} P_{X_1^n X_2^n}(x_1^n, x_2^n)} \xrightarrow{n \to +\infty} 1, \tag{4.81}
\end{align}
$$

since typical sets cover asymptotically the whole probability mass. Thus, we get that $S^{\overline{\text{NS}}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_2}) \xrightarrow{n \to +\infty} 1$ since $S^{\overline{\text{NS}}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_2}) \le 1$. Therefore, for sufficiently large $n$ we can achieve a rate pair arbitrarily close to the outer bound. Finally, since $C^{\overline{\text{NS}}}(W)$ is closed and convex, a rate pair that is in the closure of the convex hull of the initial region is also in $C^{\overline{\text{NS}}}(W)$, and thus the proof is completed. $\square$

## 4.6 Independent Non-Signaling Assisted Capacity Region

The goal of this section is to show that independent non-signaling assistance does not change the capacity region of a MAC $W$, i.e. that $C^{\text{NSSR}}(W) = C(W)$. In order to prove this result, we will need some properties in the one-sender one-receiver case from [BF18] recalled in Section 2.3.

Let us state our result on independent non-signaling assistance, which says that even in one-shot scenarios, the success probability with and without that assistance are close:

**Theorem 4.34.** For any $\ell_1, k_1, \ell_2, k_2$, $S_{\text{sum}}(W, \ell_1, \ell_2)$ is larger than or equal to:

$$ \min \left( \frac{k_1}{\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right), \frac{k_2}{\ell_2} \left( 1 - \left(1 - \frac{1}{k_2}\right)^{\ell_2} \right) \right) S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2). $$

In particular, this will imply that the capacity regions are the same:

**Corollary 4.35.** $C^{\text{NSSR}}(W) = C(W)$.

*Proof.* We will show that $C_{\text{sum}}^{\text{NSSR}}(W) = C_{\text{sum}}(W)$, which is enough to conclude thanks to Proposition 4.2 and Proposition 4.3. We apply Theorem 4.34 on the MAC $W^{\otimes n}$.

Let us fix $k_1 = 2^{nR_1}$, $k_2 = 2^{nR_2}$ and $\ell_1 = \frac{2^{nR_1}}{n}$, $\ell_2 = \frac{2^{nR_2}}{n}$. Since:

$$ \frac{k}{\ell} \left( 1 - \left(1 - \frac{1}{k}\right)^{\ell} \right) \ge \frac{k}{\ell} \left( 1 - e^{-\frac{\ell}{k}} \right) \ge 1 - \frac{\ell}{2k}, $$

and $1 - \frac{\ell_1}{2k_1} = 1 - \frac{\ell_2}{2k_2} = 1 - \frac{1}{2n}$, we get:

$$ \left(1 - \frac{1}{2n}\right) S_{\text{sum}}^{\text{NSSR}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_2}) \le S_{\text{sum}}\left(W^{\otimes n}, \frac{2^{nR_1}}{n}, \frac{2^{nR_2}}{n}\right). $$
As $1 - \frac{1}{2n}$ tends to 1 when $n$ tends to infinity, we get that $\forall \varepsilon > 0, \exists N \in \mathbb{N}, \forall n \ge N$:

$$ (1 - \varepsilon)S_{\text{sum}}^{\text{NSSR}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_1}) \le S_{\text{sum}}(W^{\otimes n}, 2^{n(R_1 - \frac{\log(n)}{n})}, 2^{n(R_2 - \frac{\log(n)}{n})}). $$

Thus, if $\lim_{n \to +\infty} S_{\text{sum}}^{\text{NSSR}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_1}) = 1$, we have that for all $R'_1 < R_1$ and $R'_2 < R_2$:

$$ \lim_{n \to +\infty} S_{\text{sum}}(W^{\otimes n}, 2^{nR'_1}, 2^{nR'_1}) \ge 1 - \varepsilon. $$

Since this is true for all $\varepsilon > 0$, we get in fact that $\lim_{n \to +\infty} S_{\text{sum}}(W^{\otimes n}, 2^{nR'_1}, 2^{nR'_1}) = 1$. This implies that $C_{\text{sum}}^{\text{NSSR}}(W) \subseteq C_{\text{sum}}(W)$, and thus that the capacity regions are equal as the other inclusion is always satisfied. $\square$

In order to prove Theorem 4.34, we will need the following lemma:

**Lemma 4.36.** If $S_1, S_2$ are classical codes (i.e. multisets with elements in $\mathcal{X}_1, \mathcal{X}_2$) of size $\ell_1, \ell_2$:

$$ S_{\text{sum}}(W, \ell_1, \ell_2) \ge \frac{1}{2} \left( \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} + \frac{f_{W_{S_1, \ell_1}^2}(S_2)}{\ell_2} \right), $$

where $W_{S_2, \ell_2}^1$ is the channel defined by $W_{S_2, \ell_2}^1(y|x_1) = \frac{1}{\ell_2} \sum_{i_2=1}^{\ell_2} W(y|x_1 S_2^{i_2})$ and similarly for $W_{S_1, \ell_1}^2$.

*Proof.* Let us define $e_1(x_1|\ell_1) := \mathbb{1}_{S_1^{i_1}=x_1}$ and $e_2(x_2|\ell_2) := \mathbb{1}_{S_2^{i_2}=x_2}$. Then for fixed $y$, let us take $j_1^y \in \text{argmax}_{i_1}\{\sum_{i_2=1}^{\ell_2} W(y|S_1^{i_1} S_2^{i_2})\}$, $j_2^y \in \text{argmax}_{i_2}\{\sum_{i_1=1}^{\ell_1} W(y|S_1^{i_1} S_2^{i_2})\}$ and then define $d(j_1|y) := \mathbb{1}_{j_1=j_1^y}$, $d(j_2|y) := \mathbb{1}_{j_2=j_2^y}$. We have then:

$$
\begin{align}
S_{\text{sum}}(W, \ell_1, \ell_2) &\ge \frac{1}{\ell_1 \ell_2} \sum_{i_1, i_2, x_1, x_2, y} W(y|x_1 x_2) \mathbb{1}_{S_1^{i_1}=x_1} \mathbb{1}_{S_2^{i_2}=x_2} \frac{\mathbb{1}_{i_1=i_1^y} + \mathbb{1}_{i_2=i_2^y}}{2} \nonumber \\
&= \frac{1}{\ell_1 \ell_2} \sum_{i_1, i_2, y} W(y|S_1^{i_1} S_2^{i_2}) \frac{\mathbb{1}_{i_1=i_1^y} + \mathbb{1}_{i_2=i_2^y}}{2} \nonumber \\
&= \frac{1}{\ell_1 \ell_2} \sum_y \frac{1}{2} \left( \sum_{i_2} W(y|S_1^{i_1} S_2^{i_2}) + \sum_{i_1} W(y|S_1^{i_1} S_2^{i_2}) \right) \nonumber \\
&= \frac{1}{\ell_1 \ell_2} \sum_y \frac{1}{2} \left( \max_{i_1} \sum_{i_2} W(y|S_1^{i_1} S_2^{i_2}) + \max_{i_2} \sum_{i_1} W(y|S_1^{i_1} S_2^{i_2}) \right) \tag{4.82} \\
&= \frac{1}{2} \left( \frac{\sum_y \max_{i_1} \left[ \frac{1}{\ell_2} \sum_{i_2} W(y|S_1^{i_1} S_2^{i_2}) \right]}{\ell_1} + \frac{\sum_y \max_{i_2} \left[ \frac{1}{\ell_1} \sum_{i_1} W(y|S_1^{i_1} S_2^{i_2}) \right]}{\ell_2} \right) \nonumber \\
&= \frac{1}{2} \left( \frac{\sum_y \max_{i_1} W_{S_2, \ell_2}^1(y|S_1^{i_1})}{\ell_1} + \frac{\sum_y \max_{i_2} W_{S_1, \ell_1}^2(y|S_2^{i_2})}{\ell_2} \right) \nonumber \\
&= \frac{1}{2} \left( \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} + \frac{f_{W_{S_1, \ell_1}^2}(S_2)}{\ell_2} \right). \nonumber
\end{align}
$$

<figure><img src="image_24.png" /></figure>
We have now all the tools to prove Theorem 4.34:

*Proof of Theorem 4.34.* Let us consider an optimal solution $r^1, r^2, p^1, p^2$ of the program of Proposition 4.8 computing $S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2)$.

Let us fix some multiset $S_2$ with elements in $\mathcal{X}_2$ of size $\ell_2$. Note that $r^1$ and $p^1$ are a feasible solution of the program of Proposition 2.17 computing $S^{\text{NS}}(W_{S_2, \ell_2}^1, k_1)$. As a result, we can apply Theorem 2.18 and get the following statement. For the multiset $S_1$ obtained by choosing $\ell_1$ elements of $\mathcal{X}_1$ independently according to the distribution $\left(\frac{p_{x_1}^1}{k_1}\right)_{x_1 \in \mathcal{X}_1}$, we have:

$$ \mathbb{E}_{S_1} \left[ \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} \right] \geq \frac{k_1}{\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right) \cdot \frac{1}{k_1} \sum_{x_1, y} W_{S_2, \ell_2}^1(y|x_1) r_{x_1, y}^1. $$

Now, let $S_2$ be the multiset obtained by choosing $\ell_2$ elements of $\mathcal{X}_2$ independently according to the distribution $\left(\frac{p_{x_2}^2}{k_2}\right)_{x_2 \in \mathcal{X}_2}$. We have:

$$ \begin{aligned} \mathbb{E}_{S_2} \left[ \frac{1}{k_1} \sum_{x_1, y} W_{S_2, \ell_2}^1(y|x_1) r_{x_1, y}^1 \right] &= \mathbb{E}_{S_2} \left[ \frac{1}{k_1} \sum_{x_1, y} \frac{1}{\ell_2} \sum_{i_2=1}^{\ell_2} W(y|x_1 S_2^{i_2}) r_{x_1, y}^1 \right] \\ &= \frac{1}{\ell_2} \sum_{i_2=1}^{\ell_2} \mathbb{E}_{X_2 \sim \frac{p_{x_2}^2}{k_2}} \left[ \frac{1}{k_1} \sum_{x_1, y} W(y|x_1 X_2^{i_2}) r_{x_1, y}^1 \right] = \mathbb{E}_{X_2 \sim \frac{p_{x_2}^2}{k_2}} \left[ \frac{1}{k_1} \sum_{x_1, y} W(y|x_1 X_2) r_{x_1, y}^1 \right] \\ &= \frac{1}{k_1} \sum_{x_1, x_2, y} \frac{p_{x_2}^2}{k_2} W(y|x_1 x_2) r_{x_1, y}^1 = \frac{1}{k_1} \sum_{x_1, y} W_{p^2, k_2}^1(y|x_1) r_{x_1, y}^1. \end{aligned} \quad (4.83) $$

Thus in all, we have:

$$ \begin{aligned} \mathbb{E}_{S_2} \left[ \mathbb{E}_{S_1} \left[ \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} \right] \right] &\geq \mathbb{E}_{S_2} \left[ \frac{k_1}{\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right) \cdot \frac{1}{k_1} \sum_{x_1, y} W_{S_2, \ell_2}^1(y|x_1) r_{x_1, y}^1 \right] \\ &= \frac{k_1}{\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right) \cdot \mathbb{E}_{S_2} \left[ \frac{1}{k_1} \sum_{x_1, y} W_{S_2, \ell_2}^1(y|x_1) r_{x_1, y}^1 \right] \\ &\geq \frac{k_1}{\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right) \cdot \frac{1}{k_1} \sum_{x_1, y} W_{p^2, k_2}^1(y|x_1) r_{x_1, y}^1, \end{aligned} \quad (4.84) $$

and symmetrically for $\mathbb{E}_{S_1} \left[ \mathbb{E}_{S_2} \left[ \frac{f_{W_{S_1, \ell_1}^2}(S_2)}{\ell_2} \right] \right]$. Since there exists codes $S_1^*, S_2^*$ such that:

$$ \frac{1}{2} \left( \frac{f_{W_{S_2, \ell_2}^1}(S_1^*)}{\ell_1} + \frac{f_{W_{S_1, \ell_1}^2}(S_2^*)}{\ell_2} \right) \geq \mathbb{E}_{S_1, S_2} \left[ \frac{1}{2} \left( \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} + \frac{f_{W_{S_1, \ell_1}^2}(S_2)}{\ell_2} \right) \right], $$
and by applying Lemma 4.36, we get:

$$
\begin{align}
S_{\text{sum}}(W, \ell_1, \ell_2) &\ge \frac{1}{2} \left( \frac{f_{W_{S_2^*, \ell_2}^1}(S_1^*)}{\ell_1} + \frac{f_{W_{S_1^*, \ell_1}^2}(S_2^*)}{\ell_2} \right) \nonumber \\
&\ge \mathbb{E}_{S_1, S_2} \left[ \frac{1}{2} \left( \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} + \frac{f_{W_{S_1, \ell_1}^2}(S_2)}{\ell_2} \right) \right] \nonumber \\
&= \frac{1}{2} \left( \mathbb{E}_{S_2} \left[ \mathbb{E}_{S_1} \left[ \frac{f_{W_{S_2, \ell_2}^1}(S_1)}{\ell_1} \right] \right] + \mathbb{E}_{S_1} \left[ \mathbb{E}_{S_2} \left[ \frac{f_{W_{S_1, \ell_1}^2}(S_2)}{\ell_2} \right] \right] \right) \tag{4.85} \\
&\ge \frac{k_1}{2\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right) \cdot \frac{1}{k_1} \sum_{x_1, y} W_{p^2}^1(y|x_1) r_{x_1, y}^1 \nonumber \\
&\quad + \frac{k_2}{2\ell_2} \left( 1 - \left(1 - \frac{1}{k_2}\right)^{\ell_2} \right) \cdot \frac{1}{k_2} \sum_{x_2, y} W_{p^1}^2(y|x_2) r_{x_2, y}^2 \nonumber \\
&\ge \min \left( \frac{k_1}{\ell_1} \left( 1 - \left(1 - \frac{1}{k_1}\right)^{\ell_1} \right), \frac{k_2}{\ell_2} \left( 1 - \left(1 - \frac{1}{k_2}\right)^{\ell_2} \right) \right) S_{\text{sum}}^{\text{NSSR}}(W, k_1, k_2), \nonumber
\end{align}
$$

which concludes the proof. □

<figure><img src="image_25.png" /></figure>

*Remark.* In the whole proof of Theorem 4.34, as well as the properties it depends on, we have never used the fact that the output of the channel *y* was the same for both decoders *d*₁ and *d*₂. This implies that the result also holds for interference channels, i.e. two-sender two-receiver channels *W*(*y*₁*y*₂|*x*₂*x*₂). Specifically, non-signaling assistance shared between the first sender and the first receiver and independently shared between the second sender and the second receiver does not change the capacity region of interference channels.

## 4.7 Conclusion

In this chapter, we have addressed the computational complexity of the multiple-access channel coding problem, by showing that no polynomial-time constant approximation exists if a complexity hypothesis on random *k*-SAT formulas is true. Then, we studied the impact of non-signaling assistance on the capacity of multiple-access channels. We have developed an efficient linear program computing the success probability of the best non-signaling assisted code for a finite number of copies of a multiple-access channel. In particular, this gives lower bounds on the zero-error non-signaling assisted capacity of multiple-access channels. Applied to the binary adder channel, these results were used to prove that a sum-rate of $\frac{\log_2(72)}{4} \approx 1.5425$ can be reached with zero error, which beats the maximum classical sum-rate capacity of $\frac{3}{2}$. For noisy channels, we have developed a technique giving lower bounds through the use of concatenated codes. Applied to the noisy binary adder channel, this technique was used to show that non-signaling assistance still improves the sum-rate capacity. We have also found an outer bound on the non-signaling assisted capacity region through a relaxed notion of non-signaling assistance, whose capacity region was characterized by a single-letter formula. Finally, we have shown that independent non-signaling assistance does not change the capacity region.

Our results suggest that quantum entanglement may also increase the capacity of such channels. However, even for the binary adder channel, this question remains open. One
could also ask if such efficient methods to compute the best non-signaling assisted codes
can be extended to Gaussian multiple-access channels. Finally, establishing a single-letter
formula for the non-signaling assisted capacity of multiple-access channels is the main
open question left here. It remains open even for the binary adder channel. Proving that
non-signaling assistance and relaxed non-signaling assistance coincide asymptotically
would directly answer this question and show that the capacity region is described in
Theorem 4.25.
<figure><img src="image_26.png" /></figure>
# Broadcast Channel Coding With Non-Signaling Correlations

Broadcast channels, introduced by Cover in [Cov72], describe the simple network communication setting where one sender aims to transmit individual messages to two receivers. Contrary to point-to-point channels [Sha48] or multiple-access channels [Lia73, Ahl73], the capacity region of broadcast channels is known only for particular classes such as the degraded [Ber73, Gal74, AK75], deterministic [Mar77, Pin78] and semi-deterministic [GP80]. Only inner bounds [Cov75, vdM75, Mar79] and outer bounds [Sat78, Mar79, NG07, GN20] on the capacity region are known in the general setting.

On the one hand, it is natural to ask whether quantum-entanglement or even non-signaling correlations change the capacity region of broadcast channels. It is known that gaps exist between the classical, quantum-entanglement assisted and non-signaling assisted capacity regions of two-sender two-receiver interference channels [QS17] and MACs [LALS20, SLSS22]; see Chapter 4 for a detailed analysis of non-signaling assistance on MACs. However, the influence of nonlocal resources on broadcast channels has been comparably less studied. We only know that quantum entanglement shared between decoders does not change the capacity region [PDB21].

On the other hand, from an algorithmic point of view, a crucial question in information theory is the complexity of the channel coding problem, which entails maximizing the success probability that can be achieved by sending a fixed number of messages over a channel. However, as solving exactly this problem is NP-hard, a natural question that arises is its approximability. For point-to-point channels, Barman and Fawzi found in [BF18] a $(1 - e^{-1})$-approximation algorithm running in polynomial time. They showed that it is NP-hard to approximate the channel coding problem in polynomial time for any strictly better ratio. For $\ell$-list-decoding, where the decoder is allowed to output a list of $\ell$ guesses, a polynomial-time approximation algorithm achieving a $1 - \frac{\ell^\ell e^{-\ell}}{\ell!}$ ratio was found in [BFGG20], and it was shown to be NP-hard to do better in Chapter 3; see Theorem 3.1. For multiple-access channel coding, the complexity of the problem can be linked to the bipartite densest subgraph problem [FKP01], which cannot be approximated within any constant ratio under a complexity hypothesis on random $k$-SAT formulas [AAM$^{+}$11]; see Theorem 4.5 from Chapter 4. However, the approximability of broadcast channel coding has not been addressed in the literature.
In the point-to-point scenario studied in [BF18], the existence of a constant-ratio ap-
proximation algorithm is linked to the equality of the capacity regions with and without
non-signaling assistance. Indeed, giving non-signaling assistance to the channel coding
problem turns it into a linear program, thus computable in polynomial time. In fact, it is
equal to its natural linear relaxation, which is a common strategy towards approximating
an integer linear program. Showing that this approximation strategy guarantees a constant
ratio is the key ingredient in proving the equality of the capacity regions with and without
non-signaling assistance. This raises the following questions on broadcast channels: Does
the capacity region of the broadcast channel change when non-signaling resources between
parties are allowed? What is the best approximability ratio of the broadcast channel coding
problem? How those two questions are related?

**Contributions** In this chapter, as a first result, we prove that the sum success probabilities of the broadcast channel coding problem are the same with and without non-signaling assistance between decoders; see Theorem 5.4. This strengthens a result by [PDB21] establishing that entanglement between the decoders does not change the capacity region.

The main focus of this chapter is to study the influence of sharing a non-signaling resource between the three parties. Our main result shows that for the class of deterministic broadcast channels, non-signaling resources shared between the three parties does not change the capacity region; see Theorem 5.10 and Corollary 5.11. In order to prove this result, we consider the algorithmic problem of optimal channel coding for a deterministic broadcast channel. For this problem, we describe a $(1 - e^{-1})^2$-approximation algorithm running in polynomial time. This is achieved through a graph interpretation of the problem, where one aims at partitioning a bipartite graph into $k_1$ and $k_2$ parts, such that the resulting quotient graph is the densest possible; see Proposition 5.7 and Theorem 5.8. To prove our result on the limitations of non-signaling assistance for deterministic broadcast channels, we use the same ideas as the ones involved in the analysis of the approximation algorithm.

As far as hardness is concerned, we consider the subproblem of broadcast channel coding where the number of messages one decoder is responsible of is maximum. This subproblem can be interpreted as a social welfare maximization problem. In the theory of fair division [BT96, Mou03], social welfare maximization entails partitioning a set of goods among agents in order to maximize the sum of their utilities. The social welfare problem has been extensively studied through black box approach [BN05], which led to a precise analysis of achievable approximation ratio as well as hardness results [DS06, MSV08], depending on the class of utility functions considered and the type of black box access to them. We refine the hardness result for the class of fractionally sub-additive utility functions to the subclass coming from the broadcast channel coding subproblem interpretation. Specifically, we show that in the value query model, we cannot achieve a better approximation ratio than $\Omega(\frac{1}{\sqrt{m}})$ in polynomial time, with $m$ the size of one of the outputs of the channel: see Theorem 5.14. Following the previous discussion on the links between constant-ratio approximation algorithms and non-signaling capacity regions, this hardness result is a first step towards showing that sharing a non-signaling resource between the three parties of a broadcast channel can enlarge its capacity region.

**Organization** In Section 5.1, we define precisely the different versions of the broadcast channel coding problem depending on the choice of objective value, and show that they all lead to the same capacity region. In Section 5.2, we define the different non-signaling
assisted versions of the broadcast channel coding problem. In particular, we show that the sum success probabilities with and without non-signaling assistance shared between decoders are the same, and that it implies that the related capacity regions are the same. In Section 5.3, we address both algorithmic aspects and capacity considerations of deterministic broadcast channels. Specifically, we describe a $(1-e^{-1})^2$-approximation algorithm running in polynomial time for that class, and we show that the capacity region for that class is the same with or without non-signaling assistance. Finally, in Section 5.4, we show that in the value query model, we cannot achieve a better approximation ratio than $\Omega(\frac{1}{\sqrt{m}})$ in polynomial time for the general broadcast channel coding problem, with $m$ the size of one of the outputs of the channel.

# 5.1 Broadcast Channel Coding

## 5.1.1 Broadcast Channels

The coding problem for a broadcast channel $W: \mathcal{X} \to \mathcal{Y}_1 \times \mathcal{Y}_2$ can be stated in the following way. We want to encode a pair of messages belonging to $[k_1] \times [k_2]$ into $\mathcal{X}$. The pair is given as input to $W$, which results in two random outputs in $\mathcal{Y}_1$ and $\mathcal{Y}_2$. From the output in $\mathcal{Y}_1$ (resp. $\mathcal{Y}_2$), we want to decode back the original message in $[k_1]$ (resp. $[k_2]$). We will call $e: [k_1] \times [k_2] \to \mathcal{X}$ the encoder, $d_1: \mathcal{Y}_1 \to [k_1]$ the first decoder and $d_2: \mathcal{Y}_2 \to [k_2]$ the second decoder. The scenario is depicted in Figure 5.1.

<figure><img src="image_27.png" /><figcaption>Figure 5.1 – Coding for a broadcast channel $W$.</figcaption></figure>

We will call $p_1(W, e, d_1)$ (resp. $p_2(W, e, d_2)$) the probability of successfully decoding the first (resp. second) message, i.e. that $j_1 = i_1$ (resp. $j_2 = i_2$), given that the encoder is $e$ and the decoder is $d_1$ (resp. $d_2$). We will also consider $p(W, e, d_1, d_2)$, the probability of successfully decoding both messages, i.e. that $j_1 = i_1$ and $j_2 = i_2$, given that the encoder is $e$ and the decoders are $d_1, d_2$.

We aim to find the best encoder and decoders according to some figure of merit. However, to do so, we need a one-dimensional real-valued objective to optimize. This leads to two different quantities.

## 5.1.2 The Sum Success Probability $S_{\text{sum}}(W, k_1, k_2)$

We will focus first on maximizing $\frac{p_1(W,e,d_1)+p_2(W,e,d_2)}{2}$ over all encoders $e$ and decoders $d_1, d_2$. We will call $S_{\text{sum}}(W, k_1, k_2)$ the resulting maximum sum probability of successfully encoding and decoding the messages through $W$, given that the input pair of messages is
taken uniformly in $[k_1] \times [k_2]$. $S_{sum}(W, k_1, k_2)$ is the solution of the following optimization program:

$$
\begin{array}{ll}
\underset{e,d_1,d_2}{\operatorname{maximize}} & \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) \frac{d_1(i_1 | y_1) + d_2(i_2 | y_2)}{2} \\
\text { subject to } & \sum_{x \in \mathcal{X}} e(x | i_1 i_2) = 1, \forall i_1 \in [k_1], i_2 \in [k_2] \\
& \sum_{j_1 \in [k_1]} d_1(j_1 | y_1) = 1, \forall y_1 \in \mathcal{Y}_1 \\
& \sum_{j_2 \in [k_2]} d_2(j_2 | y_2) = 1, \forall y_2 \in \mathcal{Y}_2 \\
& e(x | i_1 i_2), d_1(j_1 | y_1), d_2(j_2 | y_2) \geq 0
\end{array}
\qquad (5.1)
$$

*Proof.* One should note that we allow in fact non-deterministic encoders and decoders for generality reasons, although the value of the program is not changed as it is convex. Besides that remark, let us name $I_1, I_2, J_1, J_2, X, Y_1, Y_2$ the random variables corresponding to $i_1, i_2, j_1, j_2, x, y_1, y_2$ in the coding and decoding process. Then, given $e, d_1, d_2$ and $W$, the objective value of the previous program comes from:

$$
\begin{align}
p_1(W, e, d_1) &= \mathbb{P}(J_1 = I_1) = \frac{1}{k_1 k_2} \sum_{i_1, i_2} \mathbb{P}(J_1 = i_1 | I_1 = i_1, I_2 = i_2) \nonumber \\
&= \frac{1}{k_1 k_2} \sum_{i_1, i_2, x} e(x | i_1 i_2) \mathbb{P}(J_1 = i_1 | i_1, i_2, X = x) \nonumber \\
&= \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) \mathbb{P}(J_1 = i_1 | i_1, i_2, x, Y_1 = y_1, Y_2 = y_2) \tag{5.2} \\
&= \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) d_1(i_1 | y_1), \nonumber
\end{align}
$$

and symmetrically for $p_2(W, e, d_2)$, which leads to the announced objective value. $\square$

<figure><img src="image_28.png" /></figure>

One can rewrite this optimization program in a more convenient way, proving that $S_{sum}(W, k_1, k_2)$ depends only on the marginals of $W$:

**Proposition 5.1.**

$$
\begin{align}
S_{sum}(W, k_1, k_2) = & \underset{e,d_1,d_2}{\operatorname{maximize}} & \frac{1}{2k_1k_2} \sum_{i_1,x,y_1} W_1(y_1|x) d_1(i_1|y_1) \sum_{i_2} e(x|i_1i_2) \nonumber \\
& & + \frac{1}{2k_1k_2} \sum_{i_2,x,y_2} W_2(y_2|x) d_2(i_2|y_2) \sum_{i_1} e(x|i_1i_2) \nonumber \\
\text{subject to} \quad & \sum_{x \in \mathcal{X}} e(x|i_1i_2) = 1, \forall i_1 \in [k_1], i_2 \in [k_2] \tag{5.3} \\
& \sum_{j_1 \in [k_1]} d_1(j_1|y_1) = 1, \forall y_1 \in \mathcal{Y}_1 \nonumber \\
& \sum_{j_2 \in [k_2]} d_2(j_2|y_2) = 1, \forall y_2 \in \mathcal{Y}_2 \nonumber \\
& e(x|i_1i_2), d_1(j_1|y_1), d_2(j_2|y_2) \ge 0 \nonumber
\end{align}
$$
*Proof.* It follows from the definitions $W_1(y_1|x) := \sum_{y_2} W(y_1y_2|x)$ and $W_2(y_2|x) := \sum_{y_1} W(y_1y_2|x)$. $\square$

### 5.1.3 The Joint Success Probability $S(W, k_1, k_2)$

We will now focus on maximizing $p(W, e, d_1, d_2)$ over all encoders $e$ and decoders $d_1, d_2$. We will call $S(W, k_1, k_2)$ the resulting maximum probability of successfully encoding and decoding the messages through $W$, given that the input pair of messages is taken uniformly in $[k_1] \times [k_2]$. $S(W, k_1, k_2)$ is the solution of the following optimization program:

$$ 
\begin{aligned}
S(W, k_1, k_2) &:= \max_{e,d_1,d_2} \frac{1}{k_1 k_2} \sum_{i_1,i_2,x,y_1,y_2} W(y_1 y_2 | x) e(x | i_1 i_2) d_1(i_1 | y_1) d_2(i_2 | y_2) \\
&\quad \text{subject to} \\
&\quad \sum_{x \in \mathcal{X}} e(x | i_1 i_2) = 1, \forall i_1 \in [k_1], i_2 \in [k_2] \\
&\quad \sum_{j_1 \in [k_1]} d_1(j_1 | y_1) = 1, \forall y_1 \in \mathcal{Y}_1 \\
&\quad \sum_{j_2 \in [k_2]} d_2(j_2 | y_2) = 1, \forall y_2 \in \mathcal{Y}_2 \\
&\quad e(x | i_1 i_2), d_1(j_1 | y_1), d_2(j_2 | y_2) \ge 0
\end{aligned} 
\quad (5.4) $$

The proof is the same as in the sum probability scenario. We define the (resp. sum) capacity region using Definition 2.12 by $C(W) := C[S](W)$ (resp. $C_{sum}(W) := C[S_{sum}](W)$).

The objective values of those two optimization programs are not the same, but $S(W, k_1, k_2)$ and $S_{sum}(W, k_1, k_2)$ still characterize the same capacity region [Wil90]:

**Proposition 5.2.** For any broadcast channel $W$, $C(W) = C_{sum}(W)$.

*Proof.* Let us focus on error probabilities rather than success ones. Call them respectively $E(W, k_1, k_2) := 1 - S(W, k_1, k_2)$ and $E_{sum}(W, k_1, k_2) := 1 - S_{sum}(W, k_1, k_2)$. Let us fix a solution $e, d_1, d_2$ of the optimization program computing $S(W, k_1, k_2)$. Let us remark first that:

$$ \sum_{i_1,i_2,x,y_1,y_2} W(y_1 y_2 | x) e(x | i_1 i_2) = k_1 k_2, $$

thus, the value of the maximum error for those encoder and decoders is:

$$ 
\begin{aligned}
E(W, k_1, k_2, e, d_1, d_2) &:= 1 - \frac{1}{k_1 k_2} \left( \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) d_1(i_1 | y_1) d_2(i_2 | y_2) \right) \\
&= \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) \\
&\quad - \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) d_1(i_1 | y_1) d_2(i_2 | y_2) \\
&= \frac{1}{k_1 k_2} \left( \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) [1 - d_1(i_1 | y_1) d_2(i_2 | y_2)] \right).
\end{aligned} 
\quad (5.5) $$
Similarly, the value of the sum error $E_{\text{sum}}(W, k_1, k_2, e, d_1, d_2)$ is equal to:

$$ 
\begin{aligned}
& 1 - \frac{1}{k_1 k_2} \left( \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) \frac{d_1(i_1 | y_1) + d_2(i_2 | y_2)}{2} \right) \\
& = \frac{1}{k_1 k_2} \left( \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2) e(x | i_1 i_2) \left[ 1 - \frac{d_1(i_1 | y_1) + d_2(i_2 | y_2)}{2} \right] \right) .
\end{aligned} 
\quad (5.6) $$

However, for $x, y \in [0, 1]$, we have that:

$$ 1 - xy \ge \max(1-x, 1-y) \ge 1 - \frac{x+y}{2}, $$

and:

$$ 1 - xy \le (1-x) + (1-y) = 2 \left( 1 - \frac{x+y}{2} \right). $$

This means that, for all $e, d_1, d_2$, we have:

$$ E_{\text{sum}}(W, k_1, k_2, e, d_1, d_2) \le E(W, k_1, k_2, e, d_1, d_2) \le 2E_{\text{sum}}(W, k_1, k_2, e, d_1, d_2), $$

so, maximizing over all $e, d_1, d_2$, we get:

$$ E_{\text{sum}}(W, k_1, k_2) \le E(W, k_1, k_2) \le 2E_{\text{sum}}(W, k_1, k_2). $$

Thus, up to a multiplicative factor 2, the error is the same. In particular, when one of those errors tends to zero, the other one tends to zero as well. This implies that the capacity regions are the same. $\square$

## 5.2 Non-Signaling Assistance

In this section, we will consider the broadcast channel coding problem with additional resources, in order to determine how these resources affect its success probabilities as well as the capacity regions that can be defined from them.

### 5.2.1 Non-Signaling Assistance Between Decoders

Here, we consider the case where the receivers are given non-signaling assistance. This resource, which is a theoretical but easier to manipulate generalization of quantum entanglement, can be characterized as follows. A non-signaling box $d(j_1j_2|y_1y_2)$ is any joint conditional probability distribution such that the marginal from one party is independent of the other party's input, i.e. we have:

$$ 
\begin{aligned}
& \forall j_1, y_1, y_2, y'_2, \quad \sum_{j_2} d(j_1j_2|y_1y_2) = \sum_{j_1} d(j_1j_2|y_1y'_2), \\
& \forall j_2, y_1, y_2, y'_1, \quad \sum_{j_1} d(j_1j_2|y_1y_2) = \sum_{j_1} d(j_1j_2|y'_1y_2).
\end{aligned} 
\quad (5.7) $$
Thus, when receivers are given non-signaling assistance, the product $d_1(j_1|y_1)d_2(j_2|y_2)$ is replaced by the non-signaling box $d(j_1j_2|y_1y_2)$. Thus, we define the joint and sum success probabilities $S^{\text{NSdec}}(W, k_1, k_2)$ (resp. $S^{\text{NSdec}}_{\text{sum}}(W, k_1, k_2)$) by:

$$ \begin{array}{ll} \underset{e,d_1,d_2}{\text{maximize}} & \displaystyle \frac{1}{k_1 k_2} \sum_{i_1,i_2,x,y_1,y_2} W(y_1 y_2|x) e(x|i_1 i_2) d(i_1 i_2|y_1 y_2) \\[1.5em] \text{(resp.} \quad & \displaystyle \underset{e,d_1,d_2}{\text{maximize}} \quad \frac{1}{2k_1 k_2} \sum_{i_1,i_2,x,y_1,y_2} W(y_1 y_2|x) e(x|i_1 i_2) \sum_{j_2} d(i_1 j_2|y_1 y_2) \\ & \quad + \frac{1}{2k_1 k_2} \sum_{i_1,i_2,x,y_1,y_2} W(y_1 y_2|x) e(x|i_1 i_2) \sum_{j_1} d(j_1 i_2|y_1 y_2) \\[1em] \text{subject to} & \sum_x e(x|i_1 i_2) = 1 \qquad (5.8) \\ & \sum_{j_2} d(j_1 j_2|y_1 y_2) = \sum_{j_1} d(j_1 j_2|y_1 y_2') \\ & \sum_{j_1} d(j_1 j_2|y_1 y_2) = \sum_{j_1} d(j_1 j_2|y_1 y_2') \\ & \sum_{j_1,j_2} d(j_1 j_2|y_1 y_2) = 1 \\ & e(x|i_1 i_2), d(j_1 j_2|y_1 y_2) \ge 0 \end{array} $$

The (resp. sum) capacity region with non-signaling assistance between decoders is defined using Definition 2.12 by $C^{\text{NSdec}}(W) := C[S^{\text{NSdec}}](W)$ (resp. $C^{\text{NSdec}}_{\text{sum}}(W) := C[S^{\text{NSdec}}_{\text{sum}}](W)$).

We will now show that sum and joint capacity regions with non-signaling assistance between decoders are the same.

**Proposition 5.3.** For any broadcast channel $W$, $C^{\text{NSdec}}_{\text{sum}}(W) = C^{\text{NSdec}}(W)$.

*Proof.* Given an encoder $e$ and a non-signaling decoding box $d$, the maximum success probability of encoding and decoding correctly with those is given by:

$$ S^{\text{NSdec}}(W, k_1, k_2, e, d) := \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2|x) e(x|i_1 i_2) d(i_1 i_2|y_1 y_2). $$

This should be compared to the sum success probability $S^{\text{NSdec}}_{\text{sum}}(W, k_1, k_2, e, d)$ of encoding and decoding correctly with those:

$$ \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2|x) e(x|i_1 i_2) \left[ \frac{\sum_{j_2} d(i_1 j_2|y_1 y_2) + \sum_{j_1} d(j_1 i_2|y_1 y_2)}{2} \right]. $$

Similarly to what was done in Proposition 5.2, we focus on error probabilities rather than success probabilities. This leads again to:

$$ E^{\text{NSdec}}(W, k_1, k_2, e, d) = \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2|x) e(x|i_1 i_2) [1 - d(i_1 i_2|y_1 y_2)], $$
and $E_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2, e, d)$ equal to:

$$ \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) e(x | i_1 i_2) \left[ \frac{1 - \sum_{j_2} d(i_1 j_2 | y_1 y_2)}{2} + \frac{1 - \sum_{j_1} d(j_1 i_2 | y_1 y_2)}{2} \right]. $$

But we have that:

$$ \begin{aligned} 1 - d(i_1 i_2 | y_1 y_2) &\ge \max \left( 1 - \sum_{j_2} d(i_1 j_2 | y_1 y_2), 1 - \sum_{j_1} d(j_1 i_2 | y_1 y_2) \right) \quad (5.9) \\ &\ge \frac{1 - \sum_{j_2} d(i_1 j_2 | y_1 y_2)}{2} + \frac{1 - \sum_{j_1} d(j_1 i_2 | y_1 y_2)}{2}, \end{aligned} $$

since $d(j_1 j_2 | y_1 y_2) \in [0, 1]$, and we have that:

$$ \begin{aligned} & 1 - \sum_{j_2} d(i_1 j_2 | y_1 y_2) + 1 - \sum_{j_1} d(j_1 i_2 | y_1 y_2) \\ &= 1 - d(i_1 i_2 | y_1 y_2) + 1 - \sum_{(j_1, j_2) \in S} d(j_1 j_2 | y_1 y_2) \quad (5.10) \\ &\ge 1 - d(i_1 i_2 | y_1 y_2), \end{aligned} $$

with $S := \{(i_1, j_2) : j_2 \in [k_2] - \{i_2\}\} \cup \{(j_1, i_2) : j_1 \in [k_1] - \{i_1\}\}$.

Thus, this implies that:

$$ E_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2, e, d) \le E^{\text{NSdec}}(W, k_1, k_2, e, d) \le 2E_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2, e, d), $$

and by maximizing over all $e$ and $d$:

$$ E_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2) \le E^{\text{NSdec}}(W, k_1, k_2) \le 2E_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2). $$

As before, this implies that the capacity regions are the same. $\square$

<figure><img src="image_29.png" /></figure>

We will now prove that sum success probabilities of the broadcast channel coding problem are the same with and without non-signaling assistance between decoders. In particular, this implies that non-signaling resources shared between decoders does not change the capacity region. Note that, after the publication of [PDB21], Pereg et al. added a remark to the arXiv version of their paper that their result stating that entanglement shared between decoders does not change the capacity of a broadcast channel could be generalized to non-signaling assistance. The theorem below strengthens this result showing that non-signaling assistance between the decoders cannot increase the sum success probability even in the one-shot setting and for arbitrary broadcast channels.

**Theorem 5.4.** For any $W, k_1, k_2$, we have $S_{\text{sum}}(W, k_1, k_2) = S_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2)$. As a consequence, $C(W) = C^{\text{NSdec}}(W)$.

*Proof.* In the sum scenario, since the objective function does not depend on the product $d_1(j_1|y_1)d_2(j_2|y_2)$ but only on the marginals $d_1(j_1|y_1)$ and $d_2(j_2|y_2)$, the non-signaling
box won't give additional decoding power. Indeed, for any encoder e and non-signaling decoding box d, we have that:

$$ S_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2, e, d) := \frac{1}{2k_1k_2} \sum_{i_1,x,y_1} W_1(y_1|x) \left( \sum_{j_2} d(i_1j_2|y_1y_2) \right) \sum_{i_2} e(x|i_1i_2) \\ + \frac{1}{2k_1k_2} \sum_{i_2,x,y_2} W_2(y_2|x) \left( \sum_{j_1} d(j_1i_2|y_1y_2) \right) \sum_{i_1} e(x|i_1i_2). \tag{5.11} $$

Thus, by choosing $d_1(j_1|y_1) := \sum_{j_2} d(j_1j_2|y_1y_2)$ and $d_2(j_2|y_2) := \sum_{j_1} d(j_1j_2|y_1y_2)$, which are well-defined since d is a non-signaling box, we have $S_{\text{sum}}(W, k_1, k_2, e, d_1, d_2) = S_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2, e, d)$. By optimizing over all e and d, $S_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2) \le S_{\text{sum}}(W, k_1, k_2)$. Since the inequality is obvious in the other direction, as $d(j_1j_2|y_1y_2) := d_1(j_1|y_1)d_2(j_2|y_2)$ is always a non-signaling box, we have that $S_{\text{sum}}(W, k_1, k_2) = S_{\text{sum}}^{\text{NSdec}}(W, k_1, k_2)$. This implies in particular that the capacity regions are the same, i.e. $C_{\text{sum}}(W) = C_{\text{sum}}^{\text{NSdec}}(W)$

Finally, since $C(W) = C_{\text{sum}}(W)$ by Proposition 5.2 and $C_{\text{sum}}^{\text{NSdec}}(W) = C^{\text{NSdec}}(W)$ by Proposition 5.3, we get that $C(W) = C^{\text{NSdec}}(W)$. $\square$

### 5.2.2 Full Non-Signaling Assistance

In this section, we will consider the case where the sender and the receivers are given non-signaling assistance. This means that a three-party non-signaling box $P(x_{j_1j_2}|(i_1i_2)y_1y_2)$ will replace the product $e(x|i_1i_2)d_1(j_1|y_1)d_2(j_2|y_2)$ in the previous objective values. A joint conditional probability $P(x_{j_1j_2}|(i_1i_2)y_1y_2)$ is a non-signaling box if the marginal from any two parties is independent of the removed party's input:

$$ \begin{align} \forall j_1, j_2, i_1, i_2, y_1, y_2, i'_1, i'_2, \quad & \sum_x P(x_{j_1j_2}|(i_1i_2)y_1y_2) = \sum_x P(x_{j_1j_2}|(i'_1i'_2)y_1y_2), \nonumber \\ \forall x, j_2, i_1, i_2, y_1, y_2, y'_1, \quad & \sum_{j_1} P(x_{j_1j_2}|(i_1i_2)y_1y_2) = \sum_{j_1} P(x_{j_1j_2}|(i_1i_2)y'_1y_2), \nonumber \\ \forall x, j_1, i_1, i_2, y_1, y_2, y'_2, \quad & \sum_{j_2} P(x_{j_1j_2}|(i_1i_2)y_1y_2) = \sum_{j_2} P(x_{j_1j_2}|(i_1i_2)y'_1y'_2). \tag{5.12} \end{align} $$

The scenario is depicted in Figure 5.2.

The cyclicity of Figure 5.2 is at first sight counter-intuitive. Note first that P being a non-signaling box is completely independent of W: in particular, the variables $y_1, y_2$ do not need to follow any laws in the definition of P being a non-signaling box. Therefore, the remaining ambiguity is the apparent need to encode and decode at the same time. However, since P is a non-signaling box, we won't need to do both at the same time, although the global correlation between the sender and the receivers will be characterized only by $P(x_{j_1j_2}|(i_1i_2)y_1y_2)$. Indeed, $\forall y_1, y_2, P(x|(i_1i_2)) = P(x|(i_1i_2)y_1y_2)$ by the non-signaling property of P. Thus, one can get the output x on input $(i_1i_2)$ without access to $y_1, y_2$, as that knowledge won't affect the law of x. Then $(y_1, y_2)$ follows the law given by W given that x. Finally, given access to $y_1, y_2$, the decoding process is described by:

$$ P(j_1j_2|(i_1i_2)y_1y_2x) = \frac{P(x_{j_1j_2}|(i_1i_2)y_1y_2)}{P(x|(i_1i_2)y_1y_2)} = \frac{P(x_{j_1j_2}|(i_1i_2)y_1y_2)}{P(x|(i_1i_2))}, $$
<figure><img src="image_30.png" /><figcaption>Figure 5.2 – A non-signaling box $P$ replacing $e, d_1, d_2$ in the coding problem for the broadcast channel $W$.</figcaption></figure>

so we recover $P(j_1j_2|(i_1i_2)y_1y_2x)) \times P(x|(i_1i_2)) = P(xj_1j_2|(i_1i_2)y_1y_2)$, and therefore, the process is not cyclic. Non-signaling boxes define exactly the conditional probability distributions where it is possible to consider the conditional probabilities of each party independently. This clarifies how one can effectively encode and then decode messages through a non-signaling box.

We will call the maximum sum success probability $S_{\text{sum}}^{\text{NS}}(W, k_1, k_2)$, which is given by the following linear program, where the constraints translate precisely the fact that $P$ is a non-signaling box:

$$
\begin{align}
S_{\text{sum}}^{\text{NS}}(W, k_1, k_2) &:= \max_{P} \frac{1}{2k_1k_2} \sum_{i_1,x,y_1} W_1(y_1|x) \sum_{i_2,j_2} P(xi_1j_2|(i_1i_2)y_1y_2) \nonumber \\
&\quad + \frac{1}{2k_1k_2} \sum_{i_2,x,y_2} W_2(y_2|x) \sum_{i_1,j_1} P(xj_1i_2|(i_1i_2)y_1y_2) \nonumber \\
&\text{subject to} \quad \sum_x P(xj_1j_2|(i_1i_2)y_1y_2) = \sum_x P(xj_1j_2|(i'_1i'_2)y_1y_2) \nonumber \\
&\quad \sum_{j_1} P(xj_1j_2|(i_1i_2)y_1y_2) = \sum_{j_1} P(xj_1j_2|(i_1i_2)y'_1y_2) \nonumber \\
&\quad \sum_{j_2} P(xj_1j_2|(i_1i_2)y_1y_2) = \sum_{j_2} P(xj_1j_2|(i_1i_2)y_1y'_2) \nonumber \\
&\quad \sum_{x,j_1,j_2} P(xj_1j_2|(i_1i_2)y_1y_2) = 1 \nonumber \\
&\quad P(xj_1j_2|(i_1i_2)y_1y_2) \ge 0 \tag{5.13}
\end{align}
$$

Since it is given as a linear program, the complexity of computing $S_{\text{sum}}^{\text{NS}}(W, k_1, k_2)$ is polynomial in the number of variables and constraints (see for instance Section 7.1 of [GM07]), which is a polynomial in $|\mathcal{X}|, |\mathcal{Y}_1|, |\mathcal{Y}_2|, k_1$ and $k_2$.

Similarly, we define the maximum joint success probability $S^{\text{NS}}(W, k_1, k_2)$ in the following
way:

$$
\begin{equation} \tag{5.14}
\begin{aligned}
S^{\text{NS}}(W, k_1, k_2) &:= \underset{P}{\text{maximize}} \quad \frac{1}{k_1 k_2} \sum_{i_1, i_2, x, y_1, y_2} W(y_1 y_2 | x) P(x i_1 i_2 | (i_1 i_2) y_1 y_2) \\
&\text{subject to} \quad \sum_x P(x j_1 j_2 | (i_1 i_2) y_1 y_2) = \sum_x P(x j_1 j_2 | (i'_1 i'_2) y_1 y_2) \\
& \qquad \sum_{j_1} P(x j_1 j_2 | (i_1 i_2) y_1 y_2) = \sum_{j_1} P(x j_1 j_2 | (i_1 i_2) y'_1 y_2) \\
& \qquad \sum_{j_2} P(x j_1 j_2 | (i_1 i_2) y_1 y_2) = \sum_{j_2} P(x j_1 j_2 | (i_1 i_2) y_1 y'_2) \\
& \qquad \sum_{x, j_1, j_2} P(x j_1 j_2 | (i_1 i_2) y_1 y_2) = 1 \\
& \qquad P(x j_1 j_2 | (i_1 i_2) y_1 y_2) \ge 0
\end{aligned}
\end{equation}
$$

We can rewrite both these programs in more convenient and smaller linear programs:

**Proposition 5.5.**

$$
\begin{equation}
\begin{aligned}
S_{\text{sum}}^{\text{NS}}(W, k_1, k_2) &= \underset{p,r,r^1,r^2}{\text{maximize}} \quad \frac{1}{2k_1k_2} \left( \sum_{x,y_1} W_1(y_1|x) r_{x,y_1}^1 + \sum_{x,y_2} W_2(y_2|x) r_{x,y_2}^2 \right) \\
&\quad \text{subject to} \\
& \quad \sum_x r_{x,y_1,y_2} = 1 \\
& \quad \sum_x r_{x,y_1}^1 = k_2 \\
& \quad \sum_x r_{x,y_2}^2 = k_1 \\
& \quad \sum_x p_x = k_1 k_2 \\
& \quad 0 \le r_{x,y_1,y_2} \le r_{x,y_1}^1, r_{x,y_2}^2 \le p_x \\
& \quad p_x - r_{x,y_1}^1 - r_{x,y_2}^2 + r_{x,y_1,y_2} \ge 0
\end{aligned}
\tag{5.15}
\end{equation}
$$

$$
\begin{equation}
\begin{aligned}
S^{\text{NS}}(W, k_1, k_2) &= \underset{p,r,r^1,r^2}{\text{maximize}} \quad \frac{1}{k_1 k_2} \sum_{x,y_1,y_2} W(y_1 y_2|x) r_{x,y_1,y_2} \\
&\quad \text{subject to} \\
& \quad \sum_x r_{x,y_1,y_2} = 1 \\
& \quad \sum_x r_{x,y_1}^1 = k_2 \\
& \quad \sum_x r_{x,y_2}^2 = k_1 \\
& \quad \sum_x p_x = k_1 k_2 \\
& \quad 0 \le r_{x,y_1,y_2} \le r_{x,y_1}^1, r_{x,y_2}^2 \le p_x \\
& \quad p_x - r_{x,y_1}^1 - r_{x,y_2}^2 + r_{x,y_1,y_2} \ge 0
\end{aligned}
\tag{5.16}
\end{equation}
$$
*Proof.* One can check that given a solution of the original program, the following choice of variables is a valid solution of the second program achieving the same objective value:

$$
\begin{align}
r_{x,y_1,y_2} &:= \sum_{i_1,i_2} P(xi_1i_2|(i_1i_2)y_1y_2), \nonumber \\
r_{x,y_1}^1 &:= \sum_{j_2,i_1,i_2} P(xi_1j_2|(i_1i_2)y_1y_2), \nonumber \\
r_{x,y_2}^2 &:= \sum_{j_1,i_1,i_2} P(xj_1i_2|(i_1i_2)y_1y_2), \nonumber \\
p_x &:= \sum_{j_1,j_2,i_1,i_2} P(xj_1j_2|(i_1i_2)y_1y_2). \tag{5.17}
\end{align}
$$

For the other direction, given those variables, a non-signaling probability distribution $P(xj_1j_2|(i_1i_2)y_1y_2)$ is given by, for $j_1 \neq i_1$ and $j_2 \neq i_2$:

$$
\begin{align}
P(xi_1i_2|(i_1i_2)y_1y_2) &= \frac{r_{x,y_1,y_2}}{k_1k_2}, \nonumber \\
P(xj_1i_2|(i_1i_2)y_1y_2) &= \frac{r_{x,y_2}^2 - r_{x,y_1,y_2}}{k_1k_2(k_1-1)}, \nonumber \\
P(xi_1j_2|(i_1i_2)y_1y_2) &= \frac{r_{x,y_1}^1 - r_{x,y_1,y_2}}{k_1k_2(k_2-1)}, \nonumber \\
P(xj_1j_2|(i_1i_2)y_1y_2) &= \frac{p_x - r_{x,y_1}^1 - r_{x,y_2}^2 + r_{x,y_1,y_2}}{k_1k_2(k_1-1)(k_2-1)}. \tag{5.18}
\end{align}
$$

□

As before, we define the (resp. sum) capacity region with non-signaling assistance using
Definition 2.12 by $C^{\text{NS}}(W) := C[\text{S}^{\text{NS}}](W)$ (resp. $C_{\text{sum}}^{\text{NS}}(W) := C[\text{S}_{\text{sum}}^{\text{NS}}](W)$).

**Proposition 5.6.** For any broadcast channel $W$, $C^{\text{NS}}(W) = C_{\text{sum}}^{\text{NS}}(W)$.

*Proof.* Let us show that:

$$
2S_{\text{sum}}^{\text{NS}}(W, k_1, k_2) - 1 \le S^{\text{NS}}(W, k_1, k_2) \le S_{\text{sum}}^{\text{NS}}(W, k_1, k_2).
$$

This will imply in particular that:

$$
\lim_{n \to +\infty} S_{\text{sum}}^{\text{NS}}(W^{\otimes n}, [2^{R_1 n}], [2^{R_2 n}]) = 1 \iff \lim_{n \to +\infty} S_{\text{sum}}^{\text{NS}}(W^{\otimes n}, [2^{R_1 n}], [2^{R_2 n}]) = 1,
$$

thus define the same capacity region.

Let us consider an optimal solution $p_x, r_{x,y_1,y_2}, r_{x,y_1}^1, r_{x,y_2}^2$ of the program computing $S_{\text{sum}}^{\text{NS}}(W, k_1, k_2)$. We have:

$$
S_{\text{sum}}^{\text{NS}}(W, k_1, k_2) = \frac{1}{k_1 k_2} \left( \sum_{x, y_1, y_2} W(y_1 y_2 | x) \frac{r_{x, y_1}^1 + r_{x, y_2}^2}{2} \right).
$$
However $r_{x,y_1}^1 + r_{x,y_2}^2 \le p_x + r_{x,y_1,y_2}$ so we get that:

$$ 
\begin{aligned}
S_{\text{sum}}^{\text{NS}}(W, k_1, k_2) &\le \frac{1}{2k_1k_2} \left( \sum_{x,y_1,y_2} W(y_1y_2|x) (p_x + r_{x,y_1,y_2}) \right) \\
&= \frac{1}{2} + \frac{1}{2} \left[ \frac{1}{k_1k_2} \left( \sum_{x,y_1,y_2} W(y_1y_2|x) r_{x,y_1,y_2} \right) \right] \\
&\le \frac{1}{2} + \frac{1}{2} S^{\text{NS}}(W, k_1, k_2),
\end{aligned}
\quad (5.19) $$

since $p_x, r_{x,y_1,y_2}, r_{x,y_1}^1, r_{x,y_2}^2$ is a valid solution of the program computing $S^{\text{NS}}(W, k_1, k_2)$.

On the other hand, consider now $p_x, r_{x,y_1,y_2}, r_{x,y_1}^1, r_{x,y_2}^2$ an optimal solution of the program computing $S^{\text{NS}}(W, k_1, k_2)$. We have that $r_{x,y_1,y_2} \le r_{x,y_1}^1, r_{x,y_2}^2$ so we have that $r_{x,y_1,y_2} \le \frac{r_{x,y_1}^1+r_{x,y_2}^2}{2}$ and thus:

$$ 
\begin{aligned}
S^{\text{NS}}(W, k_1, k_2) &= \frac{1}{k_1k_2} \left( \sum_{x,y_1,y_2} W(y_1y_2|x) r_{x,y_1,y_2} \right) \\
&\le \frac{1}{k_1k_2} \left( \sum_{x,y_1,y_2} W(y_1y_2|x) \frac{r_{x,y_1}^1 + r_{x,y_2}^2}{2} \right) \\
&\le S_{\text{sum}}^{\text{NS}}(W, k_1, k_2),
\end{aligned}
\quad (5.20) $$

since $p_x, r_{x,y_1,y_2}, r_{x,y_1}^1, r_{x,y_2}^2$ is a valid solution of the program computing $S_{\text{sum}}^{\text{NS}}(W, k_1, k_2)$. This prove the inequalities $2S_{\text{sum}}^{\text{NS}}(W, k_1, k_2) - 1 \le S^{\text{NS}}(W, k_1, k_2) \le S_{\text{sum}}^{\text{NS}}(W, k_1, k_2)$, and thus concludes the proof. $\square$

## 5.3 Approximation Algorithm for Deterministic Broadcast Channel Coding

In this section, we will address the question of the approximability of $S(W, k_1, k_2)$, in the restricted scenario of a deterministic broadcast channel $W$. Specifically, we study the problem of finding a code $e : [k_1] \times [k_2] \to \mathcal{X}$, $d_1 : \mathcal{Y}_1 \to [k_1]$, $d_2 : \mathcal{Y}_2 \to [k_2]$ that maximizes the program computing $S(W, k_1, k_2)$. Note that the restriction to deterministic codes does not affect the value of the objective of the program which is convex, and that the problem is as hard as finding any code maximizing the program computing $S(W, k_1, k_2)$, as a deterministic code with a better or equal value can be retrieved easily from any code.

We say that $W$ is deterministic if $\forall x, y_1, y_2, W(y_1y_2|x) \in \{0, 1\}$. We can then define $(W_1(x), W_2(x))$ as the only pair $(y_1, y_2)$ such that $W(y_1y_2|x) = 1$, which exists uniquely as $W$ is a conditional probability distribution. Thus, the deterministic broadcast channel coding problem can be defined in the following way:

**Definition 5.1 (DETBCCH).** Given a deterministic channel $W$ and integers $k_1$ and $k_2$, the deterministic broadcast channel coding problem, which we call DETBCCH, entails maximizing

$$ S(W, k_1, k_2, e, d_1, d_2) := \frac{1}{k_1k_2} \sum_{i_1, i_2} \mathbb{1}_{d_1(W_1(e(i_1i_2)))=i_1} \mathbb{1}_{d_2(W_2(e(i_1i_2)))=i_2} $$

over all functions $e : [k_1] \times [k_2] \to \mathcal{X}$, $d_1 : \mathcal{Y}_1 \to [k_1]$, $d_2 : \mathcal{Y}_2 \to [k_2]$.
### 5.3.1 Reformulation as a Bipartite Graph Problem

We will reformulate DETBCC as a bipartite graph problem. But first, let us introduce some notations:

**Definition 5.2 (Graph notations).** Consider a bipartite graph $G = (V_1 \sqcup V_2, E \subseteq V_1 \times V_2)$:

1. $G^{P_1, P_2}$, the quotient of $G$ by partitions $P_1, P_2$ of respectively $V_1, V_2$, is defined by:

$$G^{P_1, P_2} := (P_1 \sqcup P_2, \{(p_1, p_2) \in P_1 \times P_2 : E \cap (p_1 \times p_2) \neq \emptyset\}) .$$

2. $e_G(P_1, P_2) := |E^{G^{P_1, P_2}}|$ is the number of edges of $G^{P_1, P_2}$.

3. $N_G^{P_1, P_2}(p) := N_{G^{P_1, P_2}}(p)$ is the set of neighbors of $p$ in the graph $G^{P_1, P_2}$.

4. Similarly, $\deg_G^{P_1, P_2}(p) := \deg_{G^{P_1, P_2}}(p)$ is the degree, i.e. the number of neighbors, of $p$ in the graph $G^{P_1, P_2}$.

5. We will use $V_1, V_2$ in previous notations when we do not partition on the left and right part respectively (or identify those to trivial partitions in singletons). For instance, $G^{V_1, V_2} = G$.

6. We will use the notations $e(P_1, P_2)$, $N_{P_1, P_2}(p)$ and $\deg_{P_1, P_2}(p)$ when the graph $G$ considered is clear from context.

Now, let us remark that a deterministic channel $W$, up to a permutation of elements of $\mathcal{X}$, is characterized by the following bipartite graph:

**Definition 5.3 (Bipartite Graph $G_W$ associated with the deterministic channel $W$).**

$$G_W := (\mathcal{Y}_1 \sqcup \mathcal{Y}_2, E = \{(y_1, y_2) \in \mathcal{Y}_1 \times \mathcal{Y}_2 : \exists x \in \mathcal{X}, y_1 = W_1(x) \text{ and } y_2 = W_2(x)\}) .$$

Indeed, permuting the elements of $\mathcal{X}$ does not change $G_W$ nor $S(W, k_1, k_2)$. As a consequence, up to a multiplicative factor $k_1k_2$, we will show that DETBCC is equivalent to the following bipartite graph problem:

**Definition 5.4 (DENSESTQUOTIENTGRAPH).** Given a bipartite graph $G = (V_1 \sqcup V_2, E)$ and integers $k_1, k_2$, the problem DENSESTQUOTIENTGRAPH entails maximizing $e_G(P_1, P_2)$, the number of edges of the quotient graph of $G$ by $P_1, P_2$, over all partitions $P_1$ of $V_1$ in $k_1$ parts and $P_2$ of $V_2$ in $k_2$ parts.

**Proposition 5.7.** Given a deterministic channel $W$ and integers $k_1, k_2$, it is equivalent to solve DETBCC on $W, k_1, k_2$ or DENSESTQUOTIENTGRAPH on $G_W, k_1, k_2$. That is to say, given an optimal solution of one of those problems, one can efficiently construct an optimal solution of the other. Furthermore, their optimal values satisfy $k_1k_2DETBCC(W, k_1, k_2) = DENSESTQUOTIENTGRAPH(G_W, k_1, k_2)$.

*Proof.* Consider an optimal solution $e, d_1, d_2$ of DETBCC. Note that $d_1$ defines a partition $P_1$ of $\mathcal{Y}_1$ in $k_1$ parts and $d_2$ defines a partition $P_2$ of $\mathcal{Y}_2$ in $k_2$ parts, with $P_b^{ib} := \{y_b \in \mathcal{Y}_b :$
$d_b(y) = i_b$} for $b \in \{1, 2\}$. Then we have:

$$
\begin{align}
k_1 k_2 S(W, k_1, k_2, e, d_1, d_2) &= \sum_{i_1, i_2} \mathbb{1}_{i_1=d_1(W_1(e(i_1, i_2)))) \mathbb{1}_{i_2=d_2(W_2(e(i_1, i_2)))} \nonumber \\
&= \sum_{i_1, i_2} \mathbb{1}_{W_1(e(i_1, i_2)) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(e(i_1, i_2)) \in \mathcal{P}_2^{i_2}}. \tag{5.21}
\end{align}
$$

However, since we consider an optimal solution, we have that:

$$
\mathbb{1}_{W_1(e(i_1, i_2)) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(e(i_1, i_2)) \in \mathcal{P}_2^{i_2}} = \max_{x \in \mathcal{X}} \mathbb{1}_{W_1(x) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(x) \in \mathcal{P}_2^{i_2}},
$$

as $e(i_1, i_2)$ appears only here in the objective value. Thus:

$$
\begin{align}
k_1 k_2 S(W, k_1, k_2, e, d_1, d_2) &= \sum_{i_1, i_2} \max_{x \in \mathcal{X}} \mathbb{1}_{W_1(x) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(x) \in \mathcal{P}_2^{i_2}} \nonumber \\
&= \sum_{i_1, i_2} \mathbb{1}_{\exists (y_1, y_2) \in E^{GW} : y_1 \in \mathcal{P}_1^{i_1} \text{ and } y_2 \in \mathcal{P}_2^{i_2}} \nonumber \\
&= \sum_{i_1, i_2} \mathbb{1}_{E^{GW} \cap (\mathcal{P}_1^{i_1} \times \mathcal{P}_2^{i_2}) \neq \emptyset} \nonumber \\
&= e_{GW}(\mathcal{P}_1, \mathcal{P}_2), \tag{5.22}
\end{align}
$$

which proves that given an optimal solution of DETBCC, one can efficiently construct a
solution $\mathcal{P}_1, \mathcal{P}_2$ of DENSESTQUOTIENTGRAPH such that:

$$
e_{GW}(\mathcal{P}_1, \mathcal{P}_2) = k_1 k_2 \mathrm{DETBCC}(W, k_1, k_2).
$$

For the other direction, consider an optimal solution $\mathcal{P}_1, \mathcal{P}_2$ of DENSESTQUOTIENTGRAPH.
We have as before that:

$$
e_{G_W}(\mathcal{P}_1, \mathcal{P}_2) = \sum_{i_1, i_2} \max_{x \in \mathcal{X}} \mathbb{1}_{W_1(x) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(x) \in \mathcal{P}_2^{i_2}} .
$$

Now, let us define $e(i_1, i_2) \in \underset{x \in \mathcal{X}}{\operatorname{argmax}} \mathbb{1}_{W_1(x) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(x) \in \mathcal{P}_2^{i_2}}$ and $d_b(y_b)$ the index $i_b$ such that $y_b \in \mathcal{P}_b^{ib}$, for $b \in \{1, 2\}$. With those definitions, we get again that:

$$
\begin{align}
\max_{x \in \mathcal{X}} \mathbb{1}_{W_1(x) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(x) \in \mathcal{P}_2^{i_2}} &= \mathbb{1}_{W_1(e(i_1, i_2)) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(e(i_1, i_2)) \in \mathcal{P}_2^{i_2}} \tag{5.23} \\
&= \mathbb{1}_{i_1=d_1(W_1(e(i_1, i_2)))) \mathbb{1}_{i_2=d_2(W_2(e(i_1, i_2)))}, \nonumber
\end{align}
$$

and thus we have:

$$
\begin{align}
e_{G_W}(\mathcal{P}_1, \mathcal{P}_2) &= \sum_{i_1, i_2} \max_{x \in \mathcal{X}} \mathbb{1}_{W_1(x) \in \mathcal{P}_1^{i_1}} \mathbb{1}_{W_2(x) \in \mathcal{P}_2^{i_2}} \nonumber \\
&= \sum_{i_1, i_2} \mathbb{1}_{i_1=d_1(W_1(e(i_1, i_2)))} \mathbb{1}_{i_2=d_2(W_2(e(i_1, i_2)))} \tag{5.24} \\
&= k_1 k_2 S(W, k_1, k_2, e, d_1, d_2), \nonumber
\end{align}
$$

which proves that given an optimal solution of DENSESTQUOTIENTGRAPH, one can effi-
ciently construct a solution e, d₁, d₂ of DETBCC such that:

$$
k_1 k_2 S(W, k_1, k_2, e, d_1, d_2) = \text{DENSESTQUOTIENTGRAPH}(G_W, k_1, k_2).
$$
In particular, this implies that the optimal objective values satisfy:

$$k_1 k_2 \text{DETBCC}(W, k_1, k_2) = \text{DENSESTQUOTIENTGRAPH}(G_W, k_1, k_2).$$

Therefore, the solutions of both problems constructed throughout the proof are in fact optimal. □

*Remark.* Note that all bipartite graphs can be written as $G_W$ for some deterministic broadcast channel $W$, with $W$ unique up to a permutation of $\mathcal{X}$.

### 5.3.2 Approximation Algorithm for DENSESTQUOTIENTGRAPH

In this section, we will sort out how hard is DENSESTQUOTIENTGRAPH, and thanks to Proposition 5.7, how hard is it to solve DETBCC.

**Theorem 5.8.** There exists a polynomial-time $(1-e^{-1})^2$-approximation algorithm for DENSESTQUOTIENTGRAPH. Furthermore, it is NP-hard to solve exactly DENSESTQUOTIENTGRAPH.

**Corollary 5.9.** There exists a polynomial-time $(1-e^{-1})^2$-approximation algorithm for DETBCC. Furthermore, it is NP-hard to solve exactly DETBCC.

The approximation algorithm is a two-step process. First, we consider the problem of maximizing $\sum_{i_2=1}^{k_2} \min(k_1, \deg_{V_1, P_2}(P_2^{i_2}))$ over all partitions $P_2$ of $V_2$ in $k_2$ parts. We will show that this is a special case of the submodular welfare problem, which can be approximated within a factor $1-e^{-1}$ in polynomial time [Von08]. We then choose the partition $P_1$ on $V_1$ in $k_1$ parts uniformly at random. This partition pair will give an objective value $e(P_1, P_2)$ within a $(1-e^{-1})^2$ factor from the optimal solution in expectation.

*Proof of Theorem 5.8.* Consider first the hardness result. Let us show that the decision version of DENSESTQUOTIENTGRAPH is NP-complete. It is in NP, the certificate being the two partitions and the selection of edges between those partitions. It is NP-hard as one of its particular cases is the SETSPLITTING problem (see for instance [GJ79]), in the case where $k_1 = 2$ and $k_2 = |V_2|$, by interpreting the neighbors of $v_2 \in V_2$ as a set covering elements of $V_1$.

We will show nonetheless that this problem can be approximated within a factor $(1-e^{-1})^2$ in polynomial time. First we consider the case where $k_2 = |V_2|$. We can then always assume that the right partition is $P_2 := \{\{v_2\} : v_2 \in V_2\}$, which leads necessarily to a greater or equal number of edges in the quotient graph that with any other right partition. So, in that setting, we only need to find a partition of $V_1$ in $k_1$ parts maximizing the number of edges between vertices in the right part and the quotient of the left vertices.

First, the maximum value we can get is upper bounded by $\sum_{v_2 \in V_2} \min(k_1, \deg(v_2))$. Indeed, each vertex of $v_2$ van be connected at most to the $k_1$ parts of $V_1$, so its contribution is bounded by $k_1$, and there needs to be an edge to each part it is connected, so its contribution is also bounded by $\deg(v_2)$. Let us show that if we take a partition $P_1$ of $V_1$ uniformly at random, we get:

$$ \begin{aligned} \mathbb{E}_{P_1}[e(P_1, V_2)] & \ge \left(1 - \left(1 - \frac{1}{k_1}\right)^{k_1}\right) \sum_{v_2 \in V_2} \min(k_1, \deg(v_2)) \\ & \ge (1 - e^{-1}) \max_{P_1} e(P_1, V_2). \end{aligned} \quad (5.25) $$
We have $e(\mathcal{P}_1, V_2) = \sum_{v_2 \in V_2} \deg_{\mathcal{P}_1, V_2}(v_2)$, so by linearity of expectation $\mathbb{E}_{\mathcal{P}_1}[e(\mathcal{P}_1, V_2)] = \sum_{v_2 \in V_2} \mathbb{E}_{\mathcal{P}_1}[\deg_{\mathcal{P}_1, V_2}(v_2)]$. However $\deg_{\mathcal{P}_1, V_2}(v_2) = |\{i_1 \in [k_1] : N(v_2) \cap \mathcal{P}_1^{i_1} \neq \emptyset\}|$. Recall also that for any $v_1$, $\mathbb{P}(v_1 \in \mathcal{P}_1^{i_1}) = \frac{1}{k_1}$ since the partition is taken uniformly at random. Thus, we get:

$$
\begin{align}
\mathbb{E}_{\mathcal{P}_1}[\deg_{\mathcal{P}_1, V_2}(v_2)] &= \mathbb{E}_{\mathcal{P}_1}[|\{i_1 \in [k_1] : N(v_2) \cap \mathcal{P}_1^{i_1} \neq \emptyset\}|] = \mathbb{E}_{\mathcal{P}_1}\left[\sum_{i_1=1}^{k_1} \mathbb{1}_{N(v_2) \cap \mathcal{P}_1^{i_1} \neq \emptyset}\right] \nonumber \\
&= \sum_{i_1=1}^{k_1} \mathbb{E}_{\mathcal{P}_1}\left[\mathbb{1}_{N(v_2) \cap \mathcal{P}_1^{i_1} \neq \emptyset}\right] = \sum_{i_1=1}^{k_1} \mathbb{P}(N(v_2) \cap \mathcal{P}_1^{i_1} \neq \emptyset) \nonumber \\
&= \sum_{i_1=1}^{k_1} \left(1 - \mathbb{P}(N(v_2) \cap \mathcal{P}_1^{i_1} = \emptyset)\right) \nonumber \\
&= \sum_{i_1=1}^{k_1} \left(1 - \prod_{v_1 \in N(v_2)} \mathbb{P}(v_1 \notin \mathcal{P}_1^{i_1})\right) \nonumber \\
&= \sum_{i_1=1}^{k_1} \left(1 - \prod_{v_1 \in N(v_2)} \mathbb{P}(v_1 \notin \mathcal{P}_1^{i_1})\right) = k_1 \left(1 - \left(1 - \frac{1}{k_1}\right)^{\deg(v_2)}\right), \tag{5.26}
\end{align}
$$

since $\mathbb{P}(v_1 \notin \mathcal{P}_1^{i_1}) = 1 - \frac{1}{k_1}$ and $|N(v_2)| = \deg(v_2)$. So, in all:

$$
\mathbb{E}_{\mathcal{P}_1}[e(\mathcal{P}_1, V_2)] = \sum_{v_2 \in V_2} \mathbb{E}_{\mathcal{P}_1}[\deg_{\mathcal{P}_1, V_2}(v_2)] = k_1 \sum_{v_2 \in V_2} \left(1 - \left(1 - \frac{1}{k_1}\right)^{\deg(v_2)}\right).
$$

However, the function $f: x \mapsto 1 - \left(1 - \frac{1}{k_1}\right)^x$ is nondecreasing concave with $f(0) = 0$, so $\frac{f(x)}{x} \ge \frac{f(y)}{y}$ for $x \le y$. In particular, we have that:

$$
f(\min(k_1, \deg(v_2))) \geq \frac{\min(k_1, \deg(v_2))}{k_1} f(k_1),
$$

and thus:

$$
\begin{align}
\mathbb{E}_{\mathcal{P}_1}[e(\mathcal{P}_1, V_2)] &\ge k_1 \sum_{v_2 \in V_2} \left(1 - \left(1 - \frac{1}{k_1}\right)^{\min(k_1, \deg(v_2))}\right) \nonumber \\
&\ge k_1 \frac{\sum_{v_2 \in V_2} \min(k_1, \deg(v_2))}{k_1} \left(1 - \left(1 - \frac{1}{k_1}\right)^{k_1}\right) \tag{5.27} \\
&\ge \left(1 - \left(1 - \frac{1}{k_1}\right)^{k_1}\right) \sum_{v_2 \in V_2} \min(k_1, \deg(v_2)) \nonumber \\
&\ge (1 - e^{-1}) \max_{\mathcal{P}_1} e(\mathcal{P}_1, V_2). \nonumber
\end{align}
$$

Let us now consider the general case with $k_2$ unconstrained. We apply the previous discussion on the graph $G^{V_1, \mathcal{P}_2}$ for some fixed partition $\mathcal{P}_2$ of $V_2$. Since $e_{G^{V_1, \mathcal{P}_2}}(\mathcal{P}_1, \mathcal{P}_2) = e(\mathcal{P}_1, \mathcal{P}_2)$, we have the upper bound:

$$
\max_{\mathcal{P}_1} e(\mathcal{P}_1, \mathcal{P}_2) \le \sum_{i_2=1}^{k_2} \min \left( k_1, \deg_{V_1, \mathcal{P}_2} (\mathcal{P}_2^{i_2}) \right),
$$
and the previous algorithm gives us a partition $\mathcal{P}_1$ of $V_1$ such that:

$$e(\mathcal{P}_1, \mathcal{P}_2) \geq (1 - e^{-1}) \sum_{i_2=1}^{k_2} \min (k_1, \deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) .$$

Therefore, let us focus on the following optimization problem:

$$\max_{\mathcal{P}_2} \sum_{i_2=1}^{k_2} \min (k_1, \deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) ,$$

We will give a $(1 - e^{-1})$-approximation algorithm running in polynomial time for this problem. In all, this will allow us to get in polynomial time a partition pair $(\mathcal{P}_1, \mathcal{P}_2)$ such that:

$$\begin{align}
e(\mathcal{P}_1, \mathcal{P}_2) &\geq (1 - e^{-1}) \sum_{i_2=1}^{k_2} \min (k_1, \deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) \nonumber \\
&\geq (1 - e^{-1})^2 \max_{\mathcal{P}_2} \sum_{i_2=1}^{k_2} \min (k_1, \deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) \tag{5.28} \\
&\geq (1 - e^{-1})^2 \max_{\mathcal{P}_1, \mathcal{P}_2} e(\mathcal{P}_1, \mathcal{P}_2) . \nonumber
\end{align}$$

The problem $\max_{\mathcal{P}_2} \sum_{i_2=1}^{k_2} \min (k_1, \deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))$ is a particular instance of the submodular welfare problem from [Von08]. Note that $\deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) = \deg_{V_1, \{\mathcal{P}_2^{i_2}, V_2 - \mathcal{P}_2^{i_2}\}}(\mathcal{P}_2^{i_2})$, as the degree of $\mathcal{P}_2^{i_2}$ does not depend on the rest of the partition $\mathcal{P}_2$. Then, $h(S_2) := \min (k_1, \deg_{V_1, \{S_2, V_2 - S_2\}}(S_2))$, for $S_2 \subseteq V_2$, is a nondecreasing submodular function, as $S_2 \mapsto \deg_{V_1, \{S_2, V_2 - S_2\}}(S_2)$ is a nondecreasing submodular function on $V_2$ and $\min(k_1, \cdot)$ is nondecreasing concave. Thus, we want to maximize $\sum_{i_2=1}^{k_2} h(S_{i_2})$ where $(S_{i_2})_{i_2 \in [k_2]}$ is a partition of items in $V_2$ among $k_2$ bidders. It is a particular case of the submodular welfare problem where each nondecreasing submodular utility weight is the same for all bidders and equal to $h$. Thus, thanks to [Von08], there exists a polynomial-time $(1 - e^{-1})$-approximation of $\max_{\mathcal{P}_2} \sum_{i_2=1}^{k_2} \min (k_1, \deg_{V_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))$. $\square$

### 5.3.3 Non-Signaling Assisted Capacity Region for Deterministic Channels

Thanks to Theorem 5.8 and Proposition 5.7, there exists a constant-factor approximation algorithm for the broadcast channel coding problem running in polynomial time. We aim to show here that the non-signaling assisted value is linked by a constant factor to the unassisted one. Indeed, the hope is that the non-signaling assisted program is linked to the linear relaxation of the unassisted problem, thus is likely a good approximation since the broadcast channel coding problem can be approximated in polynomial time.

This turns out to be true, and will be proved through the following theorem:

**Theorem 5.10.** If $W$ is a deterministic broadcast channel, then for all $\ell_1 \leq k_1$ and $\ell_2 \leq k_2$:

$$\left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{k_1}\right) \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) S^{\text{NS}}(W, k_1, k_2) \leq S(W, \ell_1, \ell_2).$$
**Corollary 5.11.** For any deterministic broadcast channel $W$, $C^{\text{NS}}(W) = C(W)$.

*Proof.* We apply Theorem 5.10 on the deterministic broadcast channel $W^{\otimes n}$.

We fix $k_1 = 2^{nR_1}$, $k_2 = 2^{nR_2}$ and $\ell_1 = \frac{2^{nR_1}}{n}$, $\ell_2 = \frac{2^{nR_2}}{n}$. Since $1 - (1 - \frac{1}{\ell})^k \ge 1 - e^{-\frac{k}{\ell}}$, we get:

$$ \left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) (1 - e^{-n})^2 S^{\text{NS}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_1}) \le S\left(W^{\otimes n}, \frac{2^{nR_1}}{n}, \frac{2^{nR_2}}{n}\right). $$

As $\left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) (1 - e^{-n})^2$ tends to 1 when $n$ tends to infinity, we get $\forall \varepsilon > 0, \exists N \in \mathbb{N}$, $\forall n \ge N$:

$$ (1 - \varepsilon)S^{\text{NS}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_1}) \le S(W^{\otimes n}, 2^{n(R_1 - \frac{\log(n)}{n})}, 2^{n(R_2 - \frac{\log(n)}{n})}). $$

Thus, if $\lim_{n \to +\infty} S^{\text{NS}}(W^{\otimes n}, 2^{nR_1}, 2^{nR_1}) = 1$, we have that for all $R'_1 < R_1$ and $R'_2 < R_2$:

$$ \lim_{n \to +\infty} S(W^{\otimes n}, 2^{nR'_1}, 2^{nR'_1}) \ge 1 - \varepsilon. $$

Since this is true for all $\varepsilon > 0$, we get in fact that $\lim_{n \to +\infty} S(W^{\otimes n}, 2^{nR'_1}, 2^{nR'_1}) = 1$. This implies that $C^{\text{NS}}(W) \subseteq C(W)$, and thus that the capacity regions are equal as the other inclusion is always satisfied.

<figure><img src="image_31.png" /></figure>

Let us now prove the main result:

*Proof of Theorem 5.10.* The proof will be done in three parts. We will work on the graph $G_W$ (see Definition 5.3).

1. First, we prove that for any partition $\mathcal{P}_2$ of $\mathcal{Y}_2$ in $\ell_2$ parts:

$$ S(W, \ell_1, \ell_2) \ge \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{k_1}\right) \frac{\sum_{i_2=1}^{\ell_2} \min \left(k_1, \deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})\right)}{k_1 \ell_2}. $$

2. Then, we show that there exists a partition $\mathcal{P}_2$ such that:

$$ \frac{\sum_{i_2=1}^{\ell_2} \min \left(k_1, \deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})\right)}{k_1 \ell_2} \\ \ge \left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \frac{\min \left(k_1 k_2, \sum_{y_1} \min(k_2, \deg(y_1))\right)}{k_1 k_2}. \quad (5.29) $$

3. Finally, we prove that:

$$ \frac{\min \left(k_1 k_2, \sum_{y_1} \min(k_2, \deg(y_1))\right)}{k_1 k_2} \ge S^{\text{NS}}(W, k_1, k_2). $$
By combining these three inequalities, we get precisely the claimed result.

1. This part shares a lot of similarities with the proof of Theorem 5.8, which we will adapt to this particular situation. Let us show that if we take a partition $\mathcal{P}_1$ of $\mathcal{Y}_1$ of size $\ell_1$ uniformly at random, we get, for some fixed $\mathcal{P}_2$ of size $\ell_2$:

$$ \mathbb{E}_{\mathcal{P}_1}[e_{G_W}(\mathcal{P}_1, \mathcal{P}_2)] \geq \frac{\ell_1}{k_1} \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{k_1}\right) \sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) . $$

Since $\ell_1\ell_2 S(W, \ell_1, \ell_2) = \max_{\mathcal{P}_1 \text{ in } \ell_1 \text{ parts}, \mathcal{P}_2 \text{ in } \ell_2 \text{ parts}} e_{G_W}(\mathcal{P}_1, \mathcal{P}_2)$ by Proposition 5.7, this will imply that:

$$ \begin{aligned} S(W, \ell_1, \ell_2) &\geq \frac{1}{\ell_1 \ell_2} \mathbb{E}_{\mathcal{P}_1}[e_{G_W}(\mathcal{P}_1, \mathcal{P}_2)] \\ &\geq \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{k_1}\right) \frac{\sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))}{k_1 \ell_2}. \end{aligned} \quad (5.30) $$

We have that $e_{G_W}(\mathcal{P}_1, \mathcal{P}_2) = \sum_{i_2=1}^{\ell_2} \deg_{\mathcal{P}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})$, so by linearity of expectation, we have that $\mathbb{E}_{\mathcal{P}_1}[e_{G_W}(\mathcal{P}_1, \mathcal{P}_2)] = \sum_{i_2=1}^{\ell_2} \mathbb{E}_{\mathcal{P}_1}[\deg_{\mathcal{P}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})]$, so we will focus on the contribution of one particular $\mathcal{P}_2^{i_2}$.

Then, we have that $\deg_{\mathcal{P}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) = |\{i_1 \in [\ell_1] : N_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) \cap \mathcal{P}_1^{i_1} \neq \emptyset\}|$. Recall that $\mathbb{P}(v_1 \in \mathcal{P}_1^{i_1}) = \frac{1}{\ell_1}$ for any $v_1$ since the partition is taken uniformly at random. Thus:

$$ \begin{aligned} \mathbb{E}_{\mathcal{P}_1}[\deg_{\mathcal{P}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})] &= \mathbb{E}_{\mathcal{P}_1}[|\{i_1 \in [\ell_1] : N_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) \cap \mathcal{P}_1^{i_1} \neq \emptyset\}|] \\ &= \mathbb{E}_{\mathcal{P}_1}\left[\sum_{i_1=1}^{\ell_1} \mathbb{1}_{N_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) \cap \mathcal{P}_1^{i_1} \neq \emptyset}\right] = \sum_{i_1=1}^{\ell_1} \mathbb{E}_{\mathcal{P}_1}\left[\mathbb{1}_{N_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) \cap \mathcal{P}_1^{i_1} \neq \emptyset}\right] \\ &= \sum_{i_1=1}^{\ell_1} \mathbb{P}(N_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) \cap \mathcal{P}_1^{i_1} \neq \emptyset) = \sum_{i_1=1}^{\ell_1} \left(1 - \mathbb{P}(N_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) \cap \mathcal{P}_1^{i_1} = \emptyset)\right) \\ &= \sum_{i_1=1}^{\ell_1} \left(1 - \prod_{v_1 \in N(\mathcal{P}_2^{i_2})} \mathbb{P}(v_1 \notin \mathcal{P}_1^{i_1})\right) = \ell_1 \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{\deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})}\right). \end{aligned} \quad (5.31) $$

So, in all we have that:

$$ \begin{aligned} \mathbb{E}_{\mathcal{P}_1}[e_{G_W}(\mathcal{P}_1, \mathcal{P}_2)] &= \sum_{i_2=1}^{\ell_2} \mathbb{E}_{\mathcal{P}_1}[\deg_{\mathcal{P}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})] \\ &= \ell_1 \sum_{i_2=1}^{\ell_2} \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{\deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})}\right). \end{aligned} \quad (5.32) $$

However the function $f: x \mapsto 1 - \left(1 - \frac{1}{\ell_1}\right)^x$ is nondecreasing concave with $f(0) = 0$, so $\frac{f(x)}{x} \geq \frac{f(y)}{y}$ for $x \leq y$. In particular, we have that:

$$ f(\min(k_1, \deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))) \geq \frac{\min(k_1, \deg_{\mathcal{Y}_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})))}{k_1} f(k_1), $$
and thus:

$$
\begin{align}
\mathbb{E}_{\mathcal{P}_1}[e_{G_W}(\mathcal{P}_1, \mathcal{P}_2)] &\ge \ell_1 \sum_{i_2=1}^{\ell_2} \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{\min(k_1, \deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))}\right) \nonumber \\
&\ge \ell_1 \frac{\sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))}{k_1} \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{k_1}\right) \nonumber \\
&= \frac{\ell_1}{k_1} \left(1 - \left(1 - \frac{1}{\ell_1}\right)^{k_1}\right) \sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) , \tag{5.33}
\end{align}
$$

which concludes the first part of the proof.

2. Let us take $\mathcal{P}_2$ a partition of $\mathcal{Y}_2$ of size $\ell_2$ uniformly at random, and let us prove that:

$$
\mathbb{E} \left[ \sum_{i_2=1}^{\ell_2} \min \left( k_1, \deg_{y_1, \mathcal{P}_2} (\mathcal{P}_2^{i_2}) \right) \right]
$$

is greater than or equal to:

$$
\frac{\ell_2}{k_2} \left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \min \left(k_1 k_2, \sum_{y_1} \min(k_2, \deg(y_1))\right).
$$

First, $\sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) = \sum_{i_2=1}^{\ell_2} \varphi(\deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}))$ with $\varphi(j) := \min(k_1, j)$ which is a concave function. Recall the definition of the Poisson concavity ratio 3.1 $\alpha_{\varphi} = \inf_{x \in \mathbb{R}^+} \frac{\mathbb{E}[\varphi(\mathrm{Poi}(x))]}{\varphi(x)}$ which is equal to $1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}$ for that particular function. Let us find the law of $\deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})$:

$$
\begin{align}
\deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2}) &= \sum_{y_1} \mathbb{1}_{N(y_1) \cap \mathcal{P}_2^{i_2} \neq \emptyset} = \sum_{y_1} \left(1 - \mathbb{1}_{N(y_1) \cap \mathcal{P}_2^{i_2} = \emptyset}\right) \nonumber \\
&= \sum_{y_1} \left(1 - \mathbb{1}_{\forall y_2 \in N(y_1), y_2 \notin \mathcal{P}_2^{i_2}}\right) = \sum_{y_1} \mathrm{Ber}\left(1 - \left(1 - \frac{1}{\ell_2}\right)^{\deg(y_1)}\right) \tag{5.34}
\end{align}
$$

Thus:

$$
\begin{align}
\mathbb{E} \left[ \varphi(\deg_{y_1, \mathcal{P}_2}(\mathcal{P}_2^{i_2})) \right] &= \mathbb{E} \left[ \varphi \left( \sum_{y_1} \operatorname{Ber} \left( 1 - \left(1 - \frac{1}{\ell_2}\right)^{\deg(y_1)} \right) \right) \right] \nonumber \\
&\ge \mathbb{E} \left[ \varphi \left( \operatorname{Poi} \left( \sum_{y_1} \left( 1 - \left(1 - \frac{1}{\ell_2}\right)^{\deg(y_1)} \right) \right) \right) \right] \quad \text{by Proposition 2.2} \tag{5.35} \\
&\ge \alpha_{\varphi} \varphi \left( \sum_{y_1} \left( 1 - \left(1 - \frac{1}{\ell_2}\right)^{\deg(y_1)} \right) \right) \quad \text{by definition of } \alpha_{\varphi}. \nonumber
\end{align}
$$
But:

$$ \sum_{y_1} \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{\deg(y_1)}\right) \ge \sum_{y_1} \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{\min(k_2, \deg(y_1))}\right) \\ \ge \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \frac{1}{k_2} \sum_{y_1} \min(k_2, \deg(y_1)), \qquad (5.36) $$

as before. Since $\varphi$ is concave and $\varphi(0) = 0$, we have in particular that for all $0 \le c \le 1$ and $x \in \mathbb{R}$, $\varphi(cx) \ge c\varphi(x)$. We know also that $\varphi$ is nondecreasing. This implies that:

$$ \varphi\left(\sum_{y_1} \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{\deg(y_1)}\right)\right) \\ \ge \varphi\left(\left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \frac{1}{k_2} \sum_{y_1} \min(k_2, \deg(y_1))\right) \qquad (5.37) \\ \ge \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \varphi\left(\frac{1}{k_2} \sum_{y_1} \min(k_2, \deg(y_1))\right), $$

as $0 \le 1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2} \le 1$. Thus $\mathbb{E}[\varphi(\deg_{y_1}, \mathcal{P}_2(\mathcal{P}_2^{i_2}))]$ is larger than or equal to:

$$ \alpha_{\varphi} \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \min \left(k_1, \frac{1}{k_2} \sum_{y_1} \min(k_2, \deg(y_1))\right) \\ = \frac{1}{k_2} \left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \min \left(k_1 k_2, \sum_{y_1} \min(k_2, \deg(y_1))\right) \qquad (5.38) $$

since $\alpha_{\varphi} = 1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}$.

Finally, $\mathbb{E}[\sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{y_1}, \mathcal{P}_2(\mathcal{P}_2^{i_2}))] = \sum_{i_2=1}^{\ell_2} \mathbb{E}[\varphi(\deg_{y_1}, \mathcal{P}_2(\mathcal{P}_2^{i_2}))]$, so we get that:

$$ \mathbb{E}\left[\sum_{i_2=1}^{\ell_2} \min(k_1, \deg_{y_1}, \mathcal{P}_2(\mathcal{P}_2^{i_2}))\right] $$

is larger than or equal to:

$$ \frac{\ell_2}{k_2} \left(1 - \frac{k_1^{k_1} e^{-k_1}}{k_1!}\right) \left(1 - \left(1 - \frac{1}{\ell_2}\right)^{k_2}\right) \min \left(k_1 k_2, \sum_{y_1} \min(k_2, \deg(y_1))\right). $$

Thus, in particular, there exists some partition $\mathcal{P}_2$ that satisfies the same inequality, which concludes the second part of the proof.

3. Let us consider an optimal solution $r_{x,y_1,y_2}, p_x, r_{x,y_1}^1, r_{x,y_2}^2$ of the program computing $\mathbb{S}^{\mathbb{NS}}(W, k_1, k_2)$, so that $\mathbb{S}^{\mathbb{NS}}(W, k_1, k_2) = \frac{1}{k_1 k_2} \sum_x r_{x,W_1(x),W_2(x)}$.

a) It comes directly from $r_{x,y_1,y_2} \le p_x$ that:

$$ \sum_x r_{x,W_1(x),W_2(x)} \le \sum_x p_x = k_1 k_2. $$
b) $\sum_x r_{x,W_1(x),W_2(x)} = \sum_{y_1} \sum_{x:W_1(x)=y_1} r_{x,y_1,W_2(x)}$ and we have that:

i. $\sum_{x:W_1(x)=y_1} r_{x,y_1,W_2(x)} \le \sum_{x:W_1(x)=y_1} 1 = \deg(y_1)$,

ii. $\sum_{x:W_1(x)=y_1} r_{x,y_1,W_2(x)} \le \sum_{x:W_1(x)=y_1} r_{x,y_1}^1 \le \sum_x r_{x,y_1}^1 = k_2$,

so $\sum_{x:W_1(x)=y_1} r_{x,y_1,W_2(x)} \le \min(k_2, \deg(y_1))$, and thus:

$$ \sum_x r_{x,W_1(x),W_2(x)} \le \sum_{y_1} \min(k_2, \deg(y_1)). $$

In all, we get that:

$$ S^{\text{NS}}(W, k_1, k_2) = \frac{1}{k_1 k_2} \sum_x r_{x,W_1(x),W_2(x)} \le \frac{\min(k_1 k_2, \sum_{y_1} \min(k_2, \deg(y_1)))}{k_1 k_2}, $$

which concludes the third and last part of the proof.

<figure><img src="image_32.png" /></figure>

## 5.4 Hardness of Approximation for Broadcast Channel Coding

Since broadcast channels are more general than point-to-point channels (by defining $W_1(y_1|x) := \hat{W}(y_1|x)$ for $\hat{W}$ a point-to-point channel and taking $W_2(y_2|x) = \frac{1}{|\mathcal{Y}_2|}$ a completely trivial channel), computing a single value $S(W, k_1, k_2)$ is NP-hard, and it is even NP-hard to approximate within a better factor than $1 - e^{-1}$, as a consequence of the hardness result for point-to-point channels from [BF18].

The goal of this section is to give some evidence for the hardness of approximation of the general broadcast channel coding problem, specifically that it cannot be approximated in polynomial time within a $\Omega(1)$ factor. It will be a good insight that non-signaling assistance will enlarge the capacity region of the channel as discussed in the introduction.

Formally, one would want to show that it is NP-hard to approximate this problem within a $\Omega(1)$ factor in polynomial time. As a first step towards this goal, we will prove a $\Omega(\frac{1}{\sqrt{m}})$-approximation hardness in the value query model.

First, let us introduce formally the problem:

**Definition 5.5 (BCC).** Given a channel $W$ and integers $k_1, k_2$, the broadcast channel coding problem, which we call BCC, entails maximizing:

$$ S(W, k_1, k_2, e, d_1, d_2) := \frac{1}{k_1 k_2} \sum_{i_1, i_2, y_1, y_2} W(y_1 y_2 | e(i_1, i_2)) 1_{d_1(y_1)=i_1, d_2(y_2)=i_2}, $$

over all functions $e: [k_1] \times [k_2] \to \mathcal{X}, d_1: \mathcal{Y}_1 \to [k_1]$ and $d_2: \mathcal{Y}_2 \to [k_2]$.

As in the deterministic case, we restrict ourselves to deterministic encoders and decoders, which does not change the value nor the hardness of the problem. Also, it can be equivalently stated in terms of partitions corresponding to $d_1, d_2$ as:
**Proposition 5.12** (Equivalent formulation of BCC). Given a channel $W$ and integers $k_1$ and $k_2$, the broadcast channel coding problem, which we call BCC, entails maximizing:

$$ \frac{1}{k_1 k_2} \sum_{i_1, i_2} \max_x \sum_{y_1 \in \mathcal{P}_1^{i_1}, y_2 \in \mathcal{P}_2^{i_2}} W(y_1 y_2 | x), $$

over all partitions $\mathcal{P}_1$ of $\mathcal{Y}_1$ in $k_1$ parts and $\mathcal{P}_2$ of $\mathcal{Y}_2$ in $k_2$ parts.

### 5.4.1 Social Welfare Reformulation

The social welfare maximization problem is defined as follows: given a set $M$ of $m$ items as well as $k$ bidders with their associated utilities $(v_i : 2^M \to \mathbb{R}_+)$, the goal is to partition $M$ between the bidders to maximize the sum of their utilities. Formally, we want to compute:

$$ \underset{\mathcal{P} \text{ partition in } k \text{ parts of } M}{\text{maximize}} \sum_{i=1}^{k} v_i(\mathcal{P}^i). $$

Let us show that the subproblem of BCC restricted to $k_2 = |\mathcal{Y}_2|$ can be reformulated as a particular instance of the social welfare maximization problem. In that case, it is easy to see that $\mathcal{P}_2 = (\{y_2\})_{y_2 \in \mathcal{Y}_2}$ is always an optimal solution. Indeed, for any partition $\mathcal{P}_2$, we have:

$$ \begin{aligned} \frac{1}{k_1 |\mathcal{Y}_2|} \sum_{i_1, i_2} \max_x \sum_{y_1 \in \mathcal{P}_1^{i_1}, y_2 \in \mathcal{P}_2^{i_2}} W(y_1 y_2 | x) &\le \frac{1}{k_1 |\mathcal{Y}_2|} \sum_{i_1, i_2} \sum_{y_2 \in \mathcal{P}_2^{i_2}} \max_x \sum_{y_1 \in \mathcal{P}_1^{i_1}} W(y_1 y_2 | x) \\ &= \frac{1}{k_1 |\mathcal{Y}_2|} \sum_{i_1} \sum_{y_2 \in \mathcal{Y}_2} \max_x \sum_{y_1 \in \mathcal{P}_1^{i_1}} W(y_1 y_2 | x). \end{aligned} \tag{5.39} $$

Therefore, the objective function becomes:

$$ S^1(W, k_1, \mathcal{P}_1) := \frac{1}{k_1} \sum_{i_1=1}^{k_1} f_W^1(\mathcal{P}_1^{i_1}) \text{ with } f_W^1(S_1) := \frac{1}{|\mathcal{Y}_2|} \sum_{y_2} \max_x \sum_{y_1 \in S_1} W(y_1 y_2 | x). $$

Hence, up to a multiplicative factor $k_1$, maximizing $S^1(W, k_1, \mathcal{P}_1)$ over all partitions $\mathcal{P}_1$ of size $k_1$ is a particular case of the social welfare maximization problem with a common utility $f_W^1$ for all $k_1$ bidders.

### 5.4.2 Value Query Hardness

Let us first introduce the value query model. As described in [DS06, MSV08], a value query to a utility $v$ asks for the value of some input set $S \subseteq M$, and gets as response $v(S) \in \mathbb{R}_+$. In the value query model, we aim at solving the social welfare maximization problem accessing the data only through value queries to $(v_i)_{i \in [k]}$.

This is more restricted than using any algorithm, but in such a model, it is possible to show unconditional lower bounds on the number of queries needed to solve a given problem within an approximation rate. In the case of the social welfare maximization problem with XOS utility functions, the approximation rate achievable in polynomial time has
been proved in [DS06, MSV08] to be of the order of $\Theta(\frac{1}{\sqrt{m}})$. Specifically, in [DS06], a $\Omega(\frac{1}{m^{\frac{1}{2}}})$-approximation in polynomial time was given, and in [MSV08], it has been shown that any $\Omega(\frac{1}{m^{\frac{1}{2}-\epsilon}})$-approximation for $\epsilon > 0$ requires an exponential number of value queries. We will adapt their proof in the particular case of one common XOS utility function of the form $f_W^1$ for some broadcast channel $W$. But first, let us introduce the definition of XOS functions and prove that $f_W^1$ is one of those.

**Definition 5.6.** A linear valuation function (also known as additive) is a set function $a: 2^M \to \mathbb{R}_+$ that assigns a nonnegative value to every singleton $\{j\}$ for $j \in M$, and for all $S \subseteq M$ it holds that $a(S) = \sum_{j \in S} a(\{j\})$.

A fractionally sub-additive function (XOS) is a set function $f: 2^M \to \mathbb{R}_+$, for which there is a finite set of linear valuation functions $A = \{a_1, \dots, a_\ell\}$ such that $f(S) = \max_{i \in [\ell]} a_i(S)$ for every $S \subseteq M$.

*Remark.* Note that the size of $A$ is not bounded in the definition.

**Proposition 5.13.** $f_W^1$ is XOS.

*Proof.*

$$f_W^1(S) = \frac{1}{|\mathcal{Y}_2|} \sum_{y_2} \max_x \sum_{y_1 \in S_1} W(y_1 y_2 | x) = \max_{\lambda: \mathcal{Y}_2 \to \mathcal{X}} a_\lambda(S), \text{ where} \\ a_\lambda(S) = \frac{1}{|\mathcal{Y}_2|} \sum_{y_2} \sum_{y_1 \in S} W(y_1 y_2 | \lambda(y_2)) = \sum_{y_1 \in S} \left[ \frac{1}{|\mathcal{Y}_2|} \sum_{y_2} W(y_1 y_2 | m(y_2)) \right] \quad (5.40) \\ = \sum_{y_1 \in S} a_\lambda(\{y_1\}) \text{ with } a_\lambda(\{y_1\}) = \frac{1}{|\mathcal{Y}_2|} \sum_{y_2} W(y_1 y_2 | \lambda(y_2)) \in \mathbb{R}_+$$

So $f_W^1$ is the maximum of the set of $a_\lambda$ for $\lambda \in \mathcal{X}^{\mathcal{Y}_1}$, which are linear valuation functions, thus $f_W^1$ is XOS. $\square$

Let us now state the value query hardness of approximation of the broadcast channel problem:

**Theorem 5.14.** In the value query model, for any fixed $\epsilon > 0$, $a \Omega(\frac{1}{m^{\frac{1}{2}-\epsilon}})$-approximation algorithm for the broadcast channel coding problem on $W$, $k_1, k_2$, restricted to the case of $|\mathcal{Y}_2| = k_2$ and $m = |\mathcal{Y}_1| = k_1^2$, requires exponentially many value queries to $f_W^1$.

*Remark.* As our problem is a particular instance of the social welfare maximization problem with XOS functions, the polynomial-time $\Omega(\frac{1}{m^{\frac{1}{2}}})$-approximation from [DS06] works also here.

*Proof.* The proof is inspired by Theorem 3.1 of [MSV08]. We will show using probabilistic arguments that any $\Omega(\frac{1}{m^{\frac{1}{2}-\epsilon}})$-approximation algorithm requires an exponential number of value queries. Let us fix a small constant $\delta > 0$. We choose $k_1 \in \mathbb{N}$ as the number of