A post-quantum associative memory

We will focus on the simplest type of incomplete perfect memory, whose general working principle is as follows. We begin by storing in it an m-bit string $x\in \{0,1\}^m$ by means of a suitable encoding. The value of x is then forgotten, with the only remaining record being stored in the memory. Later, we are given Nm-bit strings $Y = \{y_1,\dots,y_N\}$, with $y_i\in \{0,1\}^m,\,\, i = 1,\dots,N$, with the promise that one of these matches the original string, $y_i = x$. Our task is to determine x by making a suitable measurement on our device. The memory is called incomplete if the largest achievable N satisfies $N_{\max}\lt2^m$, and perfect if the recovery can be achieved with unit success probability for all choices of x and Y, with the constraint that Y has cardinality N.

If we model physical systems in terms of GPTs, the problem can be equivalently seen as asking for a GPT A and an encoding function $\rho:\{0,1\}^m \to \Omega_A$, with $\Omega_A$ being the state space on A, such that any N distinct states $\rho(x_1),\ldots, \rho(x_N) \in \Omega_A$ are perfectly distinguishable. Equivalently, we could demand the existence of 2^m states $\rho_1,\ldots,\rho_{2^m}\in \Omega_A$ that are N-wise mutually distinguishable, meaning that any N of them are (jointly) perfectly distinguishable. We will formalise our notions of perfect distinguishability and mutual N-wise distinguishability in definitions 8 and 9. Especially the former concept has attracted considerable interest recently [17, 18].

In order to assess the capacity of a memory in system A, we will want to quantify the effective compression operated by the encoding ρ. This presents a problem; we cannot count the number of bits or qubits in the system A, because this will be modelled by a GPT that is in general neither classical nor quantum. However, there is a universal way to quantify how 'large' a GPT is: its dimension. Since a classical m-bit system can be represented by a GPT of dimension 2^m , we could employ the logarithm $\log_2 d$ of the dimension $d = \dim V$ of a certain GPT as an effective measure of its memory capacity. Along the same lines, we could employ the compression factor

$$\begin{equation} \kappa = \frac{\#\ \textrm{encoded bits}}{\log_2 \dim V} = \frac{m}{\log_2 d} \end{equation} \tag{ 1 }$$

to assess the quality of the scheme. The N-wise compression factor, denoted with $\kappa(N, m)$, is the maximum such κ that is achievable with all possible GPTs. It is a universal function of the pair (N, m), as the optimisation does away with the degree of freedom represented by the choice of the underlying GPT. Clearly, it is given by $\kappa(N,m) = \frac{m}{\log_2 d(N,m)}$, where $d(N,m)$ is the minimum d such that a d-dimensional GPT hosting an N-wise mutually distinguishable set of states of cardinality 2^m can be found. This discussion allows us to precisely state our problem as follows.

Problem. For all pairs of positive integers $N,m$, compute $\kappa(N,m)$, i.e. determine the minimum dimension of a GPT that can host an N-wise mutually distinguishable set of states of cardinality 2^m .

Remark 1. Given n strings encoded into a system A, and granted that those states are all pairwise perfectly distinguishable, it will take (at most) n − 1 measurements to uniquely identify the desired state via a tournament-like method, granted that the measurements are non-disturbing (see [19]). Alternatively, if the measurements are disturbing, we would require an equivalent number of copies of the system A. If instead those states are mutually N-wise distinguishable, these numbers reduce to $\left\lceil\frac{n-1}{N-1}\right\rceil$.

Before we get to the formalism of GPTs, by means of which we will explore more exotic theories, we can first examine the performance of the most familiar ones: quantum and classical mechanics. For these examples we choose N = 2, so that we are finding the maximum number of pairwise distinguishable states which a system can store.

• Classical theory. If two classical probability distributions over an alphabet $\mathcal{X}$ are pairwise perfectly distinguishable it means that they have disjoint supports inside $\mathcal{X}$. If 2ⁿ probability distributions on $\mathcal{X}$ are pairwise perfectly distinguishable, we deduce that their supports Y_i are all disjoint, and therefore $d = |\mathcal{X}| \unicode{x2A7E} \sum_{i = 1}^{2^n} |Y_i| \unicode{x2A7E} 2^n$. Expressed in words, this entails that in order to accommodate 2ⁿ pairwise perfectly distinguishable states, a classical system must have dimension at least 2ⁿ . This lower bound is trivially tight, so that the N = 2 compression factor of classical theories is precisely 1.
• Quantum theory. If 2ⁿ quantum states are pairwise perfectly distinguishable, their supports must be pairwise orthogonal. This means that the total dimension of the Hilbert space is at least 2ⁿ . Since the dimension of quantum mechanics as a GPT is the square of the Hilbert space dimension (cf (2)), we see that a quantum system capable of accommodating 2ⁿ pairwise perfectly distinguishable states must have dimension at least $2^{2n}$. Again, this lower bound is easily seen to be tight, entailing that the N = 2 compression factor for quantum theory is precisely $1/2$.

Since their compression factors are at most 1, classical as well as quantum theory perform rather poorly at the task we are interested in here.

In [11] Popescu and Rohrlich famously showed that a hypothetical 'super-quantum' theory could outperform quantum mechanics at non-local tasks. However, results in [20, 21] indicate that such exotic theories may not beat quantum theory in terms of computational capacity. Here we will see how other theories fare at the task of implementing an associative memory and, in particular, seek out the optimal theory—that with the highest compression ratio defined above.

Example 4 States in a classical probability theory are simply probability distributions over some finite alphabet $\mathcal{X}$. The corresponding GPT will have dimension $d = |\mathcal{X}|$, where $|\mathcal{X}|$ is the size of $\mathcal{X}$. Formally, it can be defined as a triple $\big(\unicode{x0211D}^{d},\, \unicode{x0211D}^{d}_{+},\, u\big)$, where $\unicode{x0211D}^{d}_{+}\,: = \,\{x\in \unicode{x0211D}^{d}:\, x_{i}\unicode{x2A7E} 0\ \forall\, i = 1,\ldots, d\}$ is just the positive orthant, and the unit effect is a functional acting as $u(y) = \sum_{i = 1}^{d} y_i$ for all $y\in \unicode{x0211D}^{d}$. The state space is therefore formed by all non-negative vectors $x\in \unicode{x0211D}_+^d$ such that $u(x) = \sum_i x_i = 1$; geometrically, this set is shaped as a simplex with d vertices, which we denote by $\mathcal{S}_d$.

Example 5 The quantum mechanical theory of a k-level system can also be phrased in the GPT language. Formally, we can define it as the triple $\left( \mathrm{H}_k,\, \mathrm{PSD}_k,\, Tr \right)$, where $\mathrm{H}_k$ is the real vector space of k × k Hermitian matrices, $\mathrm{PSD}_k$ is the cone of k × k positive semidefinite matrices, and Tr is the trace functional. Observe that the real dimension of k-level quantum mechanics is

$$\begin{equation}\begin{aligned}\hspace{0pt} \dim \mathrm{H}_k = k^2. \end{aligned} \end{equation} \tag{ 2 }$$

Example 6  n-gon theories (sometimes referred to as polygon theories), are those in which the state space is described by a regular n-sided polygon. These theories are well studied [32, 34–37], and contain the local structure of Popescu–Rohrlich boxes as a particular case (n = 4). Interestingly, there is a general difference between those in which n is odd and those in which n is even: for odd n, the theories are strongly self-dual, meaning that the dual cone $C^*$ is isomorphic to C via an isomorphism mediated by a positive definite scalar product. For even n, the theories are only weakly self-dual, meaning that C and $C^*$ are merely linearly isomorphic.

Remark 7. One particularly nice property of n-gon theories is that they give a (restricted) version of both quantum and classical theories in limiting cases. In the limiting case of n = 2, the polygon collapses to the line segment; this can be taken to represent a stochastic classical bit (such as a coin). In the other extreme, at $n = \infty$, the 'polygon' describes a circle—which can be thought of representing a slice through the Bloch sphere, such as the slice of states with real-valued coefficients $|\psi\rangle = \alpha|0\rangle+\beta|1\rangle$, with $\alpha, \beta \in \unicode{x0211D}$ and $\alpha^2+\beta^2 = 1$.

Now that we have a rigorous definition of GPT in place, we can also give a precise meaning to the various notions of perfect distinguishability employed in this paper. We start with the basic definition of perfect distinguishability for a set of states in a GPT. For additional details and further motivation we refer the reader to [17, 18]. (Perfect distinguishability).

Definition 8 Let $(V,C,u)$ be a GPT with state space Ω. We say that some finitely many states $\{\omega_i\}_{i\in I}\subseteq \Omega$ are perfectly distinguishable if there exists a measurement $(e_i)_{i\in I}$ such that $e_i(\omega_j) = \delta_{i,j}$ for $i,j\in I$.

We can now give a notion of mutual distinguishability for sets of states. (Mutual N -wise distinguishability).

Definition 9 Let $(V,C,u)$ be a GPT with state space Ω. A set of states $\mathcal{S}\subseteq \Omega$ is said to be mutually N-wise distinguishable if every subset $S\in\mathcal{S}$ of cardinality $|S| = N$ is perfectly distinguishable as per definition 8. If N = 2 we also say that the states in $\mathcal{S}$ are pairwise perfectly distinguishable.

Remark 10. The fact that $\{\omega_1,\omega_2\}$ and $\{\omega_2,\omega_3\}$ are separately perfectly distinguishable does not imply, in general, that $\{\omega_1,\omega_2,\omega_3\}$ are perfectly distinguishable. More generally; the union of some sets which are N-wise distinguishable is not necessarily mutually N-wise distinguishable itself.

In what follows we will be interested in the minimal GPT dimension that is needed in order to achieve mutually N-wise distinguishable sets with a prescribed number of elements, or, vice versa, in the maximal number of elements that a mutually N-wise distinguishable set of states can have in GPTs of a fixed dimension. We thus formalise the following definition.

Definition 11. For two positive integers $N,m$, we denote with $d(N,m)$ the minimum dimension $\dim V_A$ among all GPTs $A = (V_A,C_A,u_A)$ having the property that the corresponding state space $\Omega_A = C_A\cap u_A^{-1}(1)$ contains a set of mutually N-wise distinguishable states of cardinality 2^m . The corresponding compression factor is defined by

$$\begin{equation} \kappa\left(N,m\right) \,: = \,\frac{m}{\log_2 d\left(N,m\right)}\, . \end{equation} \tag{ 3 }$$

If we accept the assumptions leading to the GPT framework as we have defined it above, calculating or estimating $\kappa(N,m)$ from above (equivalently, calculating or estimating $d(N,m)$ from below) amounts to establishing the ultimate physical bounds to the compression of information realised by an incomplete but perfect memory. The rest of the paper is devoted to the understanding of these quantities and to their exact computation in a few interesting cases.

We start by looking at the most extreme case, that where the memory is in fact complete, i.e. the information can be retrieved. This corresponds to setting $N = 2^m$. In this case, even GPTs do not grant any advantage over classical probability theory.

Lemma 12. For all positive integers m, it holds that $d(2^m,m) = 2^m$ and hence $\kappa(2^m,m) = 1$. In other words, there exists a GPT (namely, classical probability theory) of dimension 2^m hosting 2^m perfectly distinguishable states, but no GPT of smaller dimension enjoying that same property.

Proof. Since perfectly distinguishable states must be linearly independent, the dimension of the host vector space of any GPT accommodating 2^m perfectly distinguishable states must be at least 2^m .

The above lemma 12 is slightly disappointing, as it tells us that even GPTs cannot perform better than classical probability theory at the implementation of a perfect and complete memory. However, this state of affairs changes dramatically when we consider smaller values of N, i.e. when we look instead at perfect but incomplete memories. We will see how this is possible in the next section.

Before commencing, a note on geometric terminology. We say that a hyperplane $V\subset \unicode{x0211D}^n$ supports a set X in a point $x\in V\cap X$ if V touches X in x without 'cutting through' it, in other words, if the whole X lies in one of the two closed half-spaces determined by V, with $x\in V\cap X$. Formally:

Definition 14. Let $X\subseteq \unicode{x0211D}^n$ be a subset of a Euclidean space. We say that a hyperplane $V\subset \unicode{x0211D}^n$ supports X in a point $x\in X$ if: (i) $x\in V$; and (ii) X is entirely contained inside one of the closed half-spaces determined by V.

Let us consider a d-dimensional GPT with state space $\Omega\subset \unicode{x0211D}^{d-1}$ and the set of (distinct) states $\{\rho_i\}_{i = 1,\ldots, 2^m}\subset \Omega$. Assume that any pair $\{\rho_i, \rho_j\}$ with i ≠ j is perfectly distinguishable by means of a measurement $(e_{ij},\, u-e_{ij})$, as per definition 8. Explicitly, this means that $e_{ij}(\rho_i) = 1$ and $e_{ij}(\rho_j) = 0$. Note that the set of vectors v such that $e_{ij}(v) = 0$ and the set of vectors w such that $e_{ij}(w) = 1$ form two parallel hyperplanes V and W. Note that $\rho_j\in V$ and $\rho_i\in W$. Clearly, since $0\unicode{x2A7D} e_{ij}(\omega)\unicode{x2A7D} 1$ for all states ω, the whole Ω lies between V and W. We can say that W and V support the state space Ω in ρ_i and ρ_j , respectively. Vice versa, this condition is entirely equivalent to ρ_i and ρ_j being perfectly distinguishable. To get a clear geometric intuition it is instructive to explore the special case where the state space is 2-dimensional; with our convention, this corresponds to the case where d = 3, because the global GPT will feature a 3-dimensional cone whose section is our 2-dimensional state space.

We thus consider a 3-dimensional GPT with states confined to a set $\Omega\subset \unicode{x0211D}^{2}$. The situation is as depicted in figure 1. The two states $\rho_i,\rho_j\in \Omega$ in figure 1 are indeed perfectly distinguishable, because the entire set Ω is enclosed between two parallel lines supporting it in ρ_i and ρ_j , respectively. However, one can see that not all pairs among the 6 states marked with black dots can be perfectly distinguishable ⁶ .

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Let us make this discussion a bit more rigorous. Assume that we are given k states $\rho_1,\ldots, \rho_k\in \Omega$, with the promise that they are pairwise perfectly distinguishable. We can ask ourselves: how large can k be? The convex hull of $\rho_1,\ldots, \rho_k$ will naturally form a polygon $P\subseteq \Omega$. In fact, we have that every ρ_i must correspond to a vertex of P in order for the perfect distinguishability condition to be obeyed. Consider now two neighbouring vertices $\rho_i,\rho_{i+1}$ of P, as well as the edge connecting them. Call $\alpha_i,\alpha_{i+1}$ the internal angles of P at vertices $\rho_i,\rho_{i+1}$. It can be shown that, in order for $\rho_i, \rho_{i+1}$ to be perfectly distinguishable, it has to hold that $\alpha_i+\alpha_{i+1}\unicode{x2A7D} \pi$ (cf figure 2). Summing over $i = 1,\ldots, k$, with the convention that $k+1\equiv 1$, we obtain that

$$\begin{equation}\begin{aligned}\hspace{0pt} k\pi \unicode{x2A7E} \sum_{i = 1,\ldots, k} \left(\alpha_i+\alpha_{i+1}\right) = 2 \sum_i \alpha_i\, . \end{aligned} \end{equation} \tag{ 5 }$$

The sum on the right-hand side is just the sum of all internal angles of a convex polygon with k vertices. From elementary geometry, this is well known to be $(k-2)\pi$. Therefore, we obtain the inequality

$$\begin{equation}\begin{aligned}\hspace{0pt} k\pi \unicode{x2A7E} 2\left(k-2\right)\pi\, , \end{aligned} \end{equation} \tag{ 6 }$$

which yields immediately $k\unicode{x2A7D} 4 = 2^2$, in line with theorem 13. This bound is tight, because the four vertices of a square state space correspond to pairwise perfectly distinguishable states—a more general version of this latter statement will be proved in the next section.

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We now set out to generalise the analysis to any dimension ⁷ . In light of the geometric construction discussed in section 3.1 (the situation is entirely analogous to that depicted in figure 1 for d = 3), we can reformulate our problem as follows:

Problem (reformulation). Determine the minimum $d_m = d(2,m)$ such that there exists a set $X = \{\rho_1,\ldots,\rho_{2^m}\}\subset \unicode{x0211D}^{d_m-1}$ with the following property: for any two distinct $\rho_i,\rho_j\in X$, there are two parallel hyperplanes $W,V\subset \unicode{x0211D}^{d_m-1}$ of which one supports X in ρ_i and the other supports X in ρ_j .

We now explain how to achieve a construction with the above properties in dimension $d = m+1$. The argument is quite simple, and it is worthwhile explaining it in words before delving into the mathematical formalism. The state space of the GPT we pick to achieve the bound is shaped as a hypercube of dimension m. Since the whole theory includes also multiples of normalised spaces, its dimension is in fact m + 1. The 2^m states we choose correspond to the vertices of the hypercube. The crucial point now is that any two distinct vertices will be sitting each on one of two parallel hyperplanes that enclose the whole state space. Those hyperplanes, that are spanned by two opposite faces of the hypercube, will define the binary measurement needed to discriminate the states in question. This bit of reasoning already shows that any two vertices of the hypercube indeed represent perfectly distinguishable states.

We now make this argument rigorous. Construct the GPT $\left(\unicode{x0211D}^{m+1}, C_{g,m}, u\right)$, where

$$\begin{equation}\begin{aligned}\hspace{0pt} C_{g,n} \,: = \,\left\{\left(x_0,x_1,\ldots, x_n\right)^\intercal\in \unicode{x0211D}^{m+1}:\ x_0 \unicode{x2A7E} \max_{1\unicode{x2A7D} i\unicode{x2A7D} n} |x_i| \right\} \end{aligned} \end{equation} \tag{ 7 }$$

and moreover $u\left( (x_0,x_1,\ldots, x_m)^\intercal \right) \,: = \,x_0$. The state space of this GPT is clearly a hypercube of dimension m. Now, for $\epsilon\in \{\pm 1\}^n$, define $\rho_\epsilon\in \unicode{x0211D}^{m+1}$ by

$$\begin{equation}\begin{aligned}\hspace{0pt} \left(\rho_\epsilon\right)_i \,: = \,\left\{\begin{array}{ll} 1 & \text{if $i = 0$,} \\ \epsilon_i & \text{if $i\unicode{x2A7E} 1$.} \end{array}\right. \end{aligned} \end{equation} \tag{ 8 }$$

Note that there are exactly 2^m distinct choices for ε. We deduce that the 2^m states ρ_ε are pairwise perfectly distinguishable. To see why, consider $\epsilon,\epsilon^{^{\prime}}\in \{\pm 1\}^n$ that are distinct. Then, they will differ at some position $i\in \{1,\ldots, m\}$. Without loss of generality, we can assume that $\epsilon_i = +1$ and $\epsilon^{^{\prime}}_i = -1$. Now, consider the two-element collection $(e_i,u-e_i)$, where the functional e_i is defined by

$$\begin{equation}\begin{aligned}\hspace{0pt} e_i \left( \left(x_0,x_1,\ldots, x_n\right)^\intercal \right) \,: = \,\frac{x_0+x_i}{2}\, . \end{aligned} \end{equation} \tag{ 9 }$$

Note that for all $x\in C_{g,m}$ we have that $0\unicode{x2A7D} e_i(x)\unicode{x2A7D} u(x)$; hence, the collection $(e_i,u-e_i)$ defines a binary measurement. It is now elementary to verify that

$$\begin{equation}\begin{aligned}\hspace{0pt} e_i\left(\rho_\epsilon\right) = 1\, , \quad \left(u-e_i\right)\left(\rho_\epsilon\right) = 0\, , \quad e_i\left(\rho_{\epsilon^{^{\prime}}}\right) = 0\, , \quad \left(u-e_i\right)\left(\rho_{\epsilon^{^{\prime}}}\right) = 1\, . \end{aligned} \end{equation} \tag{ 10 }$$

These are precisely the conditions needed to ensure that ρ_ε and $\rho_{\epsilon^{^{\prime}}}$ are perfectly distinguishable ⁸ .

We have therefore constructed a GPT of dimension m + 1 which is capable of accommodating 2^m pairwise perfectly distinguishable states; hence,

$$\begin{equation} d\left(2,m\right) \unicode{x2A7D} m+1\, ,\qquad \text{or}\qquad \kappa\left(2,m\right)\unicode{x2A7E} \frac{m}{\log_2\left(m+1\right)}\, . \end{equation} \tag{ 11 }$$

What is remarkable here is that those limits far exceed the capabilities of both classical and quantum theory—each of these have an exponential scaling in the number of dimensions required to store m bits, whereas hypercubic theories scale only linearly. Equivalently, the compression factor for the case N = 2 of pairwise perfect distinguishability is at most 1 for classical and quantum theory, but scales almost linearly in m (up to logarithmic factors) for the best conceivable GPT. This demonstrates a sort of exponential advantage of general GPTs over classical and quantum theories.

It remains to show that the above construction is optimal. From the mathematical standpoint, this is highly non-trivial. To overcome this hurdle, we exploit the reformulation of the problem presented in section 3.2: in that form, the problem was posed for the first time by Klee [40] and was solved not long after by Danzer and Grünbaum [16]. Their solution shows that $d_m = m+1$ is a minimum for any m. We restate their result for our convenience below ⁹ .(Danzer–Grünbaum [16]).

Theorem 15 For a positive integer n, the maximum cardinality of a set $X\subset \unicode{x0211D}^n$ such that for any two distinct $x_1,x_2\in X$ there are two parallel hyperplanes $V_1,V_2\subset \unicode{x0211D}^n$ with the property that V_i supports X in x_i ($i = 1,2$) is precisely 2ⁿ . This cardinality is achieved by the set of vertices of a hypercube. Moreover, up to affine operations the set of vertices of a hypercube is the only set of points with this property having maximal cardinality.

For the interested reader, in appendix A we present a brief but self-contained account of, and homage to, the beautiful proof by Danzer and Grünbaum [16]; see also [41, chapter 17]. We can now formally deduce the proof of theorem 13 as a simple corollary of the above result.

Proof. By theorem 15, the dimension dʹ of any space capable of hosting 2^m points with the property discussed in the problem reformulation on p. 10 satisfies that $d^{^{\prime}}\unicode{x2A7E} m$. The dimension of the corresponding GPT is obtained by adding one, so that $d(2,m)\unicode{x2A7E} m+1$. The above example, also on p. 10, achieves this bound. Hence $d(2,m) = m+1$, completing the proof.

At first, we could hope that a simple generalisation of the hypercube construction may work. To explain how to obtain such a generalisation, we start by observing that from the geometric standpoint a hypercube can equivalently be seen as a Cartesian product of segments. Indeed, the extreme points of a simple line segment can be thought of as having coordinates $(\pm 1)$. A square, of having all four combinations of $(\pm 1, \pm 1)$, and so on for cubes, and hypercubes in any dimension. This operation of combining vertices by concatenating their coordinates corresponds precisely to the geometric construction of the Cartesian product. Such a construction can be translated into the world of GPTs in a fully general fashion, giving rise to the notion of prism theories, which we explore in more detail in appendix B.

Noticing this, we could be tempted to conjecture that theorem 13 could be extended to any N in the naïve way, i.e. that the extreme states of a GPT with state space $\mathcal{S}_N^{\times l}$, the l-fold Cartesian product of the N-vertex simplex, form a mutually N-wise distinguishable set. However, we can quickly see that this is not the case, and that the relationship between simplex structure and the size of mutual distinguishability does not extend beyond N = 2. We show this with an example:

Example 16. Call $\rho_1,\ldots, \rho_q$ the vectors of the canonical basis of $\unicode{x0211D}^q$, thought of as states in the classical GPT $\big(\unicode{x0211D}^q,\unicode{x0211D}_+^q, u\big)$ described in example 4. Explicitly, we will have $\rho_i = \left(0,\ldots, 1, \ldots 0\right)^\intercal$, where the single non-zero entry is in the $i\text{th}$ position. Then the extremal (pure) states of the l-fold product $\mathcal{S}_q^{\times l}$ are of the form $(\rho_{i_1},\ldots, \rho_{i_l})$, where $i_1,\ldots, i_l\in \{1,\ldots, q\}$. Consider the 3 states

$$\begin{align} \omega_1\,: = \,\left(\rho_1,\ldots,\rho_1\right)\, ,\qquad \omega_2\,: = \,\left(\rho_1,\rho_2,\rho_1,\ldots,\rho_1\right)\, ,\qquad \omega_3\,: = \,\left(\rho_2,\rho_1,\ldots,\rho_1\right)\, . \end{align} \tag{ 12 }$$

Then $\omega_1,\omega_2,\omega_3$ are not jointly distinguishable. In fact, note that

$$\begin{equation} \omega_2+\omega_3 = \left(\rho_1+\rho_2,\rho_2+\rho_1,\rho_1,\ldots,\rho_1\right) \unicode{x2A7E} \omega_1\, . \end{equation} \tag{ 13 }$$

Thus, if $e_1\cdot \omega_2 = e_1\cdot \omega_3 = 0$, then also $e_1\cdot \omega_1 = 0$. In other words, there cannot be a measurement singling out ω₁ from this triple of states.

Although no simple generalisations of the exact computation in theorem 13 are available, we can obtain a general result that guarantees that for fixed N and very large m, the scaling of the compression factor $\kappa(N,m)$ in m is exactly the same as that given by theorem 13. In other words, the scaling of $\kappa(N,m)$ in m for a fixed N does not depend on N. To prove this somewhat surprising result, we will make use of a probabilistic argument, while we leave open the task of finding a constructive proof of the result below.

Theorem 17. For all fixed integers $N\unicode{x2A7E} 2$, it holds that

$$\begin{equation}\begin{aligned}\hspace{0pt} \lim_{m\to\infty} \frac{\kappa\left(N,m\right)}{\kappa\left(2,m\right)} = \lim_{m\to\infty} \frac{\kappa\left(N,m\right)}{m/\log_2\left(m\right)} = 1\, . \end{aligned} \end{equation} \tag{ 14 }$$

Equivalently, for every fixed $N\unicode{x2A7E} 2$ we have that

$$\begin{equation}\begin{aligned}\hspace{0pt} d\left(N,m\right)\unicode{x2A7D} m^{1 + o_N\left(1\right)} \qquad \left(m\to\infty\right)\, . \end{aligned} \end{equation} \tag{ 15 }$$

Proof. Clearly, if a set of GPT states is mutually N-wise distinguishable for some $N\unicode{x2A7E} 2$, it is also 2-wise distinguishable (i.e. we have pairwise perfect distinguishability). Hence, $\kappa(N,m)\unicode{x2A7D} \kappa(2,m) = \frac{m}{\log_2(m+1)}$. Hence, inequalities (14) and (15) are clearly equivalent, because

$$\begin{equation}\begin{aligned}\hspace{0pt} 1\unicode{x2A7E} \frac{\kappa\left(N,m\right)}{\kappa\left(2,m\right)} = \frac{\kappa\left(N,m\right)}{m/\log_2\left(m\right)} = \frac{\log_2\left(m\right)}{\log_2 d\left(N,m\right)}\, , \end{aligned} \end{equation} \tag{ 16 }$$

and the right-hand side tends to 1 as $m\to\infty$ if and only if $d(N,m) \unicode{x2A7D} m^{1+o_N(1)}$. Now, to establish (15) we need to find, for fixed N and very large m, an example of a GPT of dimension $m^{1+o_N(1)}$ that can accommodate a mutually N-wise distinguishable set of states of cardinality approximately 2^m . To this end, consider the GPT with state space $\mathcal{S}_{q}^{\times l} = \mathcal{S}_{q(m)}^{\times l(m)}$ of example 16, where $q = q(m)$ and $l = l(m)$ are defined by

$$\begin{equation}\begin{aligned}\hspace{0pt} q\left(m\right) \,: = \,\left\lfloor\left(\log_2\left(m\right)\right)^2\right\rfloor\,,\qquad l\left(m\right) \,: = \,\left\lfloor\frac{2Nm}{\log_2\max\left\{\frac{2 q\left(m\right)}{N\left(N-1\right)}, 2\right\}}\right\rfloor\, . \end{aligned} \end{equation} \tag{ 17 }$$

Note that with these choices we have that

$$\begin{equation}\begin{aligned}\hspace{0pt} \log_2 \frac{N\left(N-1\right)}{2q\left(m\right)} = - \frac{2Nm}{l\left(m\right)}\, \left(1+o_N\left(1\right)\right)\, . \end{aligned} \end{equation} \tag{ 18 }$$

Let us now draw states at random in an i.i.d. fashion from $\mathcal{S}_{q(m)}^{\times l(m)}$. Every state, of the form $\omega = (\rho_{i_1},\ldots, \rho_{i_l})$, is in turn constructed by drawing $i_1,\ldots, i_l\in \{1,\ldots, q(m)\}$ uniformly at random, again in an i.i.d. manner. We now ask ourselves: given N random states $\omega^{(1)}, \ldots, \omega^{(N)}\in \mathcal{S}_{q(m)}^{\times l(m)}$, with $\omega^{(j)} = \big(\rho_{i_{1,j}}, \ldots, \rho_{i_{l,j}}\big)$, when are they perfectly distinguishable by looking only at the first components of each $\omega^{(j)}$, i.e. the states $\rho_{i_{1,j}}$, for $j = 1,\ldots, N$? The answer to the above question is clear: whenever the first components of $\omega^{(1)}, \ldots, \omega^{(N)}$, i.e. the states $\rho_{i_{1,1}},\ldots, \rho_{i_{1,N}}$, are all different. This happens with probability

$$\begin{equation}\begin{aligned}\hspace{0pt} \mathrm{Pr}\left\{\text{$1^\text{in on line}$ component discriminates $\omega^{(1)}, \ldots, \omega^{(N)}$} \right\} = \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right) , \end{aligned} \end{equation} \tag{ 19 }$$

because $\prod_{k = 1}^N (1-\frac{k}{q(m)})$ is the probability that N random numbers between 1 and q(m), in our construction $i_{1,1},\ldots, i_{1,N}$, are all different. Hence,

$$\begin{equation}\begin{aligned}\hspace{0pt} \mathrm{Pr}\left\{\text{$1^\text{in on line}$ component does not discriminate $\omega^{(1)}, \ldots, \omega^{(N)}$} \right\} = 1 - \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right) . \end{aligned} \end{equation} \tag{ 20 }$$

Since we can look at any component of choice, there are l(m) of them, and these are all independent,

$$\begin{equation}\begin{aligned}\hspace{0pt} &\mathrm{Pr}\left\{\text{no component discriminates $\omega^{(1)}, \ldots, \omega^{(N)}$} \right\} \\ &\qquad = \left(\mathrm{Pr}\left\{\text{$1^\text{in on line}$ component does not discriminate $\omega^{(1)}, \ldots, \omega^{(N)}$} \right\}\right)^{l\left(m\right)} \\ &\qquad = \left(1 - \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right)\right)^{l\left(m\right)}\, . \end{aligned} \end{equation} \tag{ 21 }$$

So far we have only considered one N-tuple of states. If we draw a subset $\mathcal{M}\subset \mathcal{S}_{q(m)}^{\times l(m)}$ of $M = |\mathcal{M}| = 2^m$ states in total, there are $\binom{M}{N}$ distinct such N-tuples (up to re-ordering). Therefore, the probability that at least one of them is such that no component discriminates it is at most

$$\begin{equation}\begin{aligned}\hspace{0pt} &\mathrm{Pr} \left(\bigcup_{\omega^{\left(1\right)},\ldots, \omega^{\left(N\right)}\in \mathcal{M}\ \text{distinct}} \left\{\text{no component discriminates $\omega^{(1)}, \ldots, \omega^{(N)}$} \right\} \right) \\ &\qquad \unicode{x2A7D} \binom{M}{N}\, \mathrm{Pr} \left\{\text{no component discriminates $\omega^{(1)}, \ldots, \omega^{(N)}$} \right\} \\ &\qquad = \binom{M}{N} \left(1 - \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right)\right)^{l\left(m\right)} . \end{aligned} \end{equation} \tag{ 22 }$$

As long as we can guarantee that the rightmost side of (22) stays below 1, we will know that there exists a choice of $\mathcal{M}$ such that for every distinct $\omega^{(1)},\ldots, \omega^{(N)}\in \mathcal{M}$, some component will discriminate them. Hence, we will have implicitly constructed a mutually N-wise distinguishable set $\mathcal{M}$ — this is, of course, an instance of the celebrated probabilistic method [42, 43]. And indeed, it is not difficult to show that the rightmost side of (22) goes to 0 as $m\to\infty$. Indeed, since $q(m)\xrightarrow[\! m\to\infty \!]{\mathrm{}} \infty$ and N is fixed one sees that

$$\begin{equation}\begin{aligned}\hspace{0pt} 1 - \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right) = \left(1+o_N\left(1\right)\right) \sum_{k = 1}^{N-1} \frac{k}{q\left(m\right)} = \left(1+o_N\left(1\right)\right)\, \frac{N\left(N-1\right)}{2q\left(m\right)}\, . \end{aligned} \end{equation} \tag{ 23 }$$

Thus,

$$\begin{equation}\begin{aligned}\hspace{0pt} &\log_2 \left(\binom{M}{N} \left(1 - \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right)\right)^{l\left(m\right)}\right) \\ &\qquad \unicode{x2A7D} Nm \left\{1 + \frac{l\left(m\right)}{Nm} \log_2 \left(1 - \prod_{k = 1}^{N-1} \left(1-\frac{k}{q\left(m\right)}\right)\right) \right\} \\ &\qquad = Nm \left\{1 + \frac{l\left(m\right)}{Nm} \log_2 \frac{N\left(N-1\right)}{2q\left(m\right)} + \frac{l\left(m\right)}{Nm} \log_2\left(1+o_N\left(1\right)\right) \right\} \\ &\qquad = Nm \left\{1 - 2 + o_N\left(1\right) + o_N\left(\frac{l\left(m\right)}{m}\right) \right\} \\ &\qquad = Nm \left\{-1 + o_N\left(1\right) \right\} . \end{aligned} \end{equation} \tag{ 24 }$$

where in the second line we used the crude approximation $\binom{M}{N}\unicode{x2A7D} M^N$, in the fourth we employed (18), and in the last we noted that $l(m)/m \xrightarrow[\! m\to\infty \!]{\mathrm{}} 0$.

This proves that for every fixed N and all sufficiently large m, the GPT $\mathcal{S}_{q(m)}^{\times l(m)}$ can accommodate a mutually N-wise distinguishable set of states of cardinality 2^m . Since the dimension of that GPT is $l(m)\big(q(m)-1\big) +1$, we deduce that

$$\begin{align} \frac{\kappa(N,m)}{m/\log_2(m)} &\unicode{x2A7E} \frac{\frac{m}{\log_2(l(m)(q(m)-1) +1)}}{m/\log_2(m)} = \frac{\log_2(m)}{\log_2(l(m)(q(m)-1) +1)} \nonumber\\ & = \frac{\log_2(m)}{\log_2(m) + O_N\,(\!\log_2 \!(\log_2 m))} \xrightarrow[\! m\to\infty \!]{\mathrm{}} 1\, . \end{align} \tag{ 25 }$$

This concludes the proof.

At this point, it is wise to pause for a moment our search for mutual N-distinguishable sets and ask ourselves a basic question: how do we decide whether a given a set of states is jointly perfectly distinguishable?

We record here the simple observation that not only the question of perfect distinguishability, but actually the calculation of the minimal error probability in joint discrimination of a set of states in a given GPT is in fact a convex program [44]. This is particularly interesting and useful, as in many situations arising naturally in applications the underlying cone admits an efficient description in terms of linear inequalities, or else in terms of inequalities in the Löwner partial order, i.e. the one determined by positive semi-definiteness. The former is the case, for instance, for classical theories (example 4). A description in terms of positive semi-definite constraints, instead, can be formulated not only for quantum theory itself (example 5), but also for several GPTs that are of great interest in entanglement theory [45]. Notable examples in this context include the theory of NPT entanglement [46–50] and that of extendibility [51–55].

Lemma 18. Let $(V,C,u)$ be a d-dimensional GPT with state space Ω. Given states $\{\omega_i\}_{i = 1}^N \subset \Omega$ and a priori probabilities $\{p_i\}_{i = 1}^N$, the maximal success probability in the associated task of state discrimination is given by the convex program

$$\begin{equation}\begin{aligned}\hspace{0pt} \begin{array}{llll} P_s^{\max}\left(\left\{p_i,\omega_i\right\}_{i = 1}^N\right) & \!\!\!\!=\!\!\!\! & \max & \sum_{i = 1}^N p_i\, e_i \cdot \omega_i \\ && \mathrm{s.t.} & e_1,\ldots, e_N \in C^*,\ \sum_{i = 1}^N e_i = u\, . \end{array} \end{aligned} \end{equation} \tag{ 26 }$$

If C is polyhedral with M extremal rays, i.e. if there exist finitely many $v_1,\ldots, v_M\in V$ such that $C = \left\{\sum_{j = 1}^M a_j v_j:\, a_j\unicode{x2A7E} 0\ \, \forall\, j\right\}$, then (26) can be rephrased as a linear program, namely,

$$\begin{equation}\begin{aligned}\hspace{0pt} \begin{array}{llll} P_s^{\max}\left(\left\{p_i,\omega_i\right\}_{i = 1}^N\right) & \!\!\!\!=\!\!\!\! & \max & \sum_{i = 1}^N p_i\, e_i \cdot \omega_i\nonumber \\ && \mathrm{s.t.} & e_i\cdot v_j\unicode{x2A7E} 0\quad \forall\ i = 1,\ldots, N,\quad \forall\ j = 1,\ldots, M,\ \sum_{i = 1}^N e_i = u\, .\nonumber\\ \end{array} \end{aligned} \end{equation} \tag{ 27 }$$

The above program can be solved efficiently, in time $O\left( d(d+M)^{3/2} N^{5/2}\right)$.

Proof. The most general state discrimination procedure consists of making a measurement $(e_i)_{i = 1,\ldots, N}$, and guessing the unknown state to be ω_i upon having obtained outcome i. The average probability of success of this strategy is precisely $\sum_i p_i\, e_i\cdot \omega_i$. The constraints in (26) are those required to make sure that $(e_i)_{i = 1,\ldots, N}$ is in fact a valid measurement in the GPT $(V,C,u)$.

If C is polyhedral with M extremal rays spanned by vectors $v_1,\ldots, v_M$, then naturally $e\in V^*$ satisfies that $e\in C^*$ if and only if $e\cdot v_j\unicode{x2A7E} 0$ for all $j = 1,\ldots,M$. In this way one derives (27) from (26). Finally, the estimates on the efficiency of the linear program solution are taken from the work by Vaidya [56]. To make the comparison precise, note that in our case $n = d(N-1) \sim dN$ is the number of real variables ¹⁰ and m = NM is the number of constraints.

Based on the above result, we can state its implications for the problem of perfect discrimination, which is of interest here:

Corollary 19. Let $(V,C,u)$ be a d-dimensional GPT with state space Ω. Deciding whether the states $\{\omega_i\}_{i = 1}^N \subset \Omega$ are perfectly distinguishable is a convex feasibility problem [44]:

$$\begin{equation}\begin{aligned}\hspace{0pt} \begin{array}{ll} \mathrm{find} & e_1,\ldots, e_N \in V^* \\ \mathrm{s.t.} & e_1,\ldots, e_N \in C^*,\quad \sum_{i = 1}^N e_i = u,\quad e_i\cdot \omega_i = 1\ \ \forall\ i = 1,\ldots, N\, . \end{array} \end{aligned} \end{equation} \tag{ 28 }$$

If C is polyhedral with M extremal rays then the above program becomes linear, and can be solved in time at most $O\left( d(d+M)^{3/2} N^{5/2}\right)$.

Consider for simplicity a GPT $(V,C,u)$ whose cone is polyhedral. Thanks to lemma 18, we know that whether a given set of states $\{\omega_i\}_i$ can be discriminated perfectly can be decided efficiently. But how do we start searching for a maximal mutually N-wise distinguishable set of states? Before we proceed to answer this, we first show that the search can be restricted to pure states, i.e. to extremal points of the state space.

Lemma 20. Let $(V,C,u)$ be a d-dimensional GPT with state space Ω. Given some $N\unicode{x2A7E} 2$, a mutually N-wise distinguishable set can be searched among pure states, i.e. extremal points of Ω.

Proof. Let us assume that a mutually N-wise distinguishable set $\mathcal{S}\subseteq \Omega$ has been found. Every $\omega\in \mathcal{S}$ will admit a (not necessarily unique) decomposition of the form $\omega = \sum_{i = 1}^d p_i^\omega \varphi_i^\omega$, where $\varphi_i^\omega \in \Omega$ are pure states. Let us pick i such that $p_i^\omega\gt0$, and consider the associated pure state $\varphi_i^\omega$. Repeating this procedure for every $\omega\in \mathcal{S}$, we can form a set of pure states $\mathcal{S}^{^{\prime}} = \{\varphi_i^\omega:\, \omega\in \mathcal{S},\, p_i^\omega\gt0 \}$.

We claim that also $\mathcal{S}^{^{\prime}}$ is mutually N-wise distinguishable. To see why, pick some pure states $\varphi_{i_1}^{\omega_1},\ldots, \varphi_{i_N}^{\omega_N}\in \mathcal{S}^{^{\prime}}$, and consider the corresponding states $\omega_1,\ldots, \omega_N\in \mathcal{S}$. Let $(e_j)_{j = 1,\ldots, N}$ be the measurement that achieves perfect discrimination of the set $\{\omega_j\}_j$, i.e. such that $1 = e_j\cdot \omega_j = \sum_i p_i^{\omega_i} e_j \cdot \varphi_i^{\omega_j}$ for all j. We immediately deduce that $e_j\cdot \varphi_i^{\omega_j} = 1$ for all i and j such that $p_i^{\omega_j}\gt0$, and in particular $e_j\cdot \varphi_{i_j}^{\omega_j} = 1$ for all j. This implies that the states $\varphi_{i_1}^{\omega_1},\ldots, \varphi_{i_N}^{\omega_N}$ are perfectly distinguishable by means of the measurement $(e_j)_{j = 1,\ldots, N}$.

We could wonder whether a similar restriction applies to the measurements as well, i.e. whether it suffices to restrict the search to extremal effects. After all, if $e\cdot \omega = 1$ and $e = \sum_i p_i^e f_i$ with f_i extremal effects, it follows that $f_i \cdot \omega = 1$ whenever $p_i\gt0$; we could therefore imagine to replace e with any f_i such that $p_i\gt0$. The reason why this does not work, however, is that doing so in general alters the sum of all the effects, which needs to be equal to the order unit. This means that in general restricting to extremal effects is not guaranteed to yield all possible feasible measurements. We construct an example to demonstrate this in appendix C.

Nonetheless, the fact that we can restrict ourselves to the finite set of pure states makes our search for the largest N-wise mutually distinguishable set of states much easier to approach. Restricting ourselves to extremal effects would have been useful in that it would have enabled us to simplify the search for distinguishing measurements, but the convex approach described in section 5.1 serves perfectly well to that purpose. The restriction to pure states, on the other hand, means that the next component of our search can take place on the terrain of a finite, rather than infinite, set.

Our search can be split into two distinct steps:

• Joint distinguishability: For a given GPT with a state space Ω, discover all subsets of $\mathrm{ext}(\Omega)$ which are N-wise distinguishable. For example, if we had N = 3, we would be finding all triples of pure states which were distinguishable by a tripartite measurement. We call such sets θⁱ , and the set of such sets $\Theta = \{\theta^i\}$.
• Mutual distinguishability: Find the largest set $\Phi\subseteq \textrm{ext}(\Omega)$ such that every subset $\phi \subset \Phi$ of cardinality $|\phi| = N$ is also an element of Θ. This means that any N size subset of Φ is N-wise distinguishable—or, equivalently, that Φ is N-wise mutually distinguishable.

The first of these steps can be straightforwardly achieved using the methods described in section 5.1. Once the set of jointly distinguishable sets Θ is in hand, we can proceed to finding the largest N-wise mutually distinguishable set Φ. Given that we know the elements of Θ, we know all groups of states which are N-wise distinguishable. We can think of this relationship between states—that of being N-wise jointly distinguishable—as a connection between them. In fact, we can take this logic literally; we can construct a hypergraph overlaying our state space. Formally, recall that an (undirected) hypergraph is a pair (V, E), where V is a (finite) set of so-called nodes, and E is a subset of the power set of V, i.e. a collection of subsets of V. We refer to the elements of E as hyperedges, and to E itself as the hyperedge set. A hypergraph is called N-regular if each hyperedge has cardinality precisely N.

In the hypergraph we construct, the vertices correspond to the states we are considering (typically the pure states of the theory), and the hyperedges are all the subsets of N states that are perfectly distinguishable. Note that if N = 2 then every edge connects two vertices, yielding an ordinary graph. (Distinguishablity hypergraphs).

Definition 21 Given some integer $N\unicode{x2A7E} 2$ and a GPT with state space Ω and finitely many pure states, i.e. such that $|\mathrm{ext} (\Omega)|\lt\infty$, the N-distinguishability hypergraph of Ω, denoted $\mathcal{G}(\Omega; N) = \big(\mathrm{ext}(\Omega),\Theta\big)$, is the N-regular hypergraph with node set $\mathrm{ext}(\Omega)$ and set of hyperedges Θ given by all subsets of $\mathrm{ext}(\Omega)$ of cardinality N which are perfectly distinguishable in Ω according to definition 8.

Above, we described the task of finding Φ as follows:

Find the largest set $\Phi\subseteq \textrm{ext}(\Omega)$ such that every subset $\phi \subset \Phi$ of cardinality $|\phi| = N$ is also an element of Θ.

With the graph-theoretic view of our problem in mind, we can re-formulate this problem as follows:

What is the largest sub-graph $\mathcal{G}^{^{\prime}}$ of $\mathcal{G}(\Omega; N)$ which is N-complete, in the sense that every subset of nodes of $\mathcal{G}^{^{\prime}}$ of cardinality N is a hyperedge?

This being a particular phrasing of the well-known maximum clique problem. Or, to be more precise; in our case we seek the maximum N-clique on an N-regular hypergraph. On ordinary graphs (ordinary in the sense that they are not hypergraphs), the problem is well studied [57–61], and algorithms are known both for exact solutions, and for faster, inexact solutions—a review appears in [62]. This problem is known to be NP -complete [63].

In the case of hypergraphs, however, less is known. The problem can be tackled by adapting an existing algorithm called hClique [64]. In our notation, the procedure works by examining each edge in turn, and finding the largest clique Q branching out from the nodes on that edge. In order to do this, we begin with an edge θ, and set the initial clique Q to the nodes connected by that edge θ. We then examine the set of nodes not included in θ, $\Omega^{^{\prime}} = \mathrm{ext}(\Omega) \setminus \theta$. Then, for each $\omega\in \Omega^{^{\prime}}$, we check if ω is fully connected to Q. If it is, then it can be added to Q, and the clique can grow. (Fully connected cliques).

Definition 22 Let Ω be the state space for a GPT $(V,C,u)$, and assume that Ω has finitely many pure states, i.e. that $|\mathrm{ext}(\Omega)|\lt\infty$. Let $\mathcal{G}(\Omega; N) = \big(\mathrm{ext}(\Omega),\Theta\big)$ be the N-distinguishability hypergraph of Ω, as per definition 21. Let $\theta\in \Theta$ be a hyperedge of $\mathcal{G}(\Omega; N)$. A node $\omega\in \mathrm{ext}(\Omega)\setminus \theta$ is fully N-connected to θ if for all $a\in \theta$ it holds that $(\theta\setminus a)\cup \{\omega\}\in \Theta$.

This process will discover the largest clique which can be built out from each edge; this is the set of maximal cliques. The largest of these will be the maximum clique, our object of interest.

If we take N = 2, we have the simpler problem of finding the maximum clique on a (non-hyper) graph. Note that this is the case even for high-dimensional state spaces, because the dimensionality of the hypergraph described in definition 21 depends only upon the number of states connected by each edge (perfectly distinguishable through a single measurement), not upon the dimension of V itself. The case depicted in figure 3 is two dimensional in two senses: the original state space occupies a two-dimensional surface embedded in $\unicode{x0211D}^3$, and the graph formed from it can be represented on the plane, since each edge is a line.

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As stated above, the problem of discovering the largest set of N-wise mutually distinguishable states for a given GPT—equivalent to finding the maximum N-clique for an undirected hypergraph—may not be amenable to a closed analytical solution in general. Even though we do not yet know which theories would be optimal for associative memories in the case that N > 2, our methods in this section reveal an exact approach for probing candidate theories, in any dimension and for any N.

Before delving into the proof, we need to fix some terminology. A convex body in the Euclidean space $\unicode{x0211D}^n$ is a compact convex subset $A \subset \unicode{x0211D}^n$ with non-empty interior, in formula $int(A)\neq \emptyset$. We say that two convex bodies $A,B\subset \unicode{x0211D}^n$ touch each other if $A\cap B \neq \emptyset$ but $int(A)\cap int(B) = \emptyset$, which corresponds to the intuitive notion of two solids touching only at their surfaces.

Two sets $A,B\subseteq \unicode{x0211D}^n$ can be added together via the Minkowski addition, defined by

$$\begin{equation} A+B\,: = \,\left\{a+b:\, a\in A,\, b\in B\right\} . \end{equation} \tag{ A.1 }$$

In what follows, for some $x\in \unicode{x0211D}^n$ we will often write x + A instead of $\{x\}+A$. We can also multiply a given set by any real number $\lambda\in \unicode{x0211D}$, by setting

$$\begin{equation} \lambda A \,: = \,\left\{\lambda a:\, a\in A\right\} . \end{equation} \tag{ A.2 }$$

Naturally, the Minkowski difference between two sets $A,B\subseteq \unicode{x0211D}^n$ can now be constructed as $A-B\,: = \,A + (-B)$. If A and B are convex then also A + B and $\lambda A$ are such. If they are convex bodies and λ ≠ 0, then also A + B and $\lambda A$ are convex bodies. A special type of Minkowski addition is the Minkowski symmetrisation. For $A\subseteq \unicode{x0211D}^n$, this is defined by

$$\begin{equation} \widetilde{A} \,: = \,\frac{A-A}{2}\, . \end{equation} \tag{ A.3 }$$

Clearly, if A is a convex body then so is $\widetilde{A}$. In what follows we will need the following standard lemma, whose proof is included only for the sake of completeness (it follows e.g. from [69, corollary 6.6.2]).

Lemma 23. Let $A\subset \unicode{x0211D}^n$ be a convex body. Then

$$\begin{equation}\begin{aligned}\hspace{0pt} \widetilde{\mathrm{int} A} = \mathrm{int\,} \widetilde{A}\, . \end{aligned} \end{equation} \tag{ A.4 }$$

Proof. We start by showing that $\widetilde{\mathrm{int\,} A} \subseteq \mathrm{int\,} \widetilde{A}$. First, note that $\widetilde{\mathrm{int\,} A} \subseteq \widetilde{A}$, simply because $\mathrm{int}(A)\subseteq A$. Second, observe that $\widetilde{\mathrm{int\,} A}$ is open. To show this, pick some $a\in \widetilde{\mathrm{int\,} A}$, so that $a = \frac{b-c}{2}$ with $b,c\in \mathrm{int\,} A$. Let ε > 0 be such that $\|\delta\|\lt\epsilon$ implies that $b+\delta,\,c+\delta\in \mathrm{int\,} A$, where $\|\cdot\|$ denotes the Euclidean norm. Then as long as $\|\delta\|\unicode{x2A7D} \epsilon$ we also have that $a+\delta = \frac{(b+\delta) - (c-\delta)}{2}\in \widetilde{\mathrm{int\,} A}$. This confirms that $\widetilde{\mathrm{int\,} A}$ is indeed open. Since the interior of a set is nothing but its largest open subset, from this and the inclusion $\widetilde{\mathrm{int\,} A} \subseteq \widetilde{A}$ we deduce that $\widetilde{\mathrm{int\,} A} \subseteq \mathrm{int\,} \widetilde{A}$.

For the other inclusion, take $a\in \mathrm{int\,} \widetilde{A}$, and some sufficiently small ε > 0 such that $\frac{a}{1-\epsilon} = \frac{b-c}{2} \in \widetilde{A}$, where $b,c\in A$ (note that the left-hand side converges to a as $\epsilon\to 0^+$ and is thus eventually in $\widetilde{A}$). Consider a point $p\in \mathrm{int\,} A$; we now claim that $(1-\epsilon)b+\epsilon p,\, (1-\epsilon) c +\epsilon p\in \mathrm{int} A$ for all $0\lt\epsilon\lt1$. To see this geometrically intuitive fact, fix ε > 0 and pick η > 0 such that $\|\delta\|\unicode{x2A7D} \eta$ implies that $p+\delta\in A$. Then as soon as $\left\|\delta^{^{\prime}}\right\|\unicode{x2A7D} \epsilon\eta$ we have that for example $(1-\epsilon)b+\epsilon p + \delta^{^{\prime}} = (1-\epsilon)b+\epsilon (p + \delta)\in A$, where $\delta\,: = \,\delta^{^{\prime}}/\epsilon$. This proves that $(1-\epsilon)b+\epsilon p,\, (1-\epsilon) c +\epsilon p\in \mathrm{int} A$, as claimed. Now,

$$\begin{equation}\begin{aligned}\hspace{0pt} a = \left(1-\epsilon\right)\frac{b-c}{2} = \frac{\left( \left(1-\epsilon\right) b+\epsilon p\right) - \left( \left(1-\epsilon\right) c + \epsilon p\right)}{2} \in \frac12 \left( \mathrm{int} A - \mathrm{int} A\right) = \widetilde{\mathrm{int} A}\, , \end{aligned} \end{equation} \tag{ A.5 }$$

concluding the proof.

We are now ready to present Danzer and Grünbaum's argument [16], in a slightly simplified form.

Proof of theorem 15 and therefore of theorem 13. For a positive integer n, some finite subset $X\subset \unicode{x0211D}^n$, and a convex body $A\subset \unicode{x0211D}^n$, we define the following properties:

• $P(n,X)$: X is not contained in any hyperplane of $\unicode{x0211D}^n$ (in other words, its affine hull has dimension n) and for all distinct $x_1,x_2\in X$ there are parallel hyperplanes $V_1,V_2\subset \unicode{x0211D}^n$ such that V_i supports X in x_i , for $i = 1,2$.
• $Q(n,A,X)$: For all $x_1,x_2\in X$, the convex bodies $x_1+A$ and $x_2+A$ touch each other.
• $Q^*(n,A,X)$: Same as $Q(n,A,X)$, but we additionally require that $A = -A$ (i.e. that A be centrally symmetric).

Furthermore, let us set

$$\begin{align} p_n &\,: = \,\sup\left\{|X|:\, \exists\, X\subset \unicode{x0211D}^n~\text{finite:}\ P\left(n,X\right) \right\} , \end{align} \tag{ A.6 }$$

$$\begin{align} q_n &\,: = \,\sup\left\{|X|:\, \exists\, X\subset \unicode{x0211D}^n~\text{finite}, A\subset \unicode{x0211D}^n~\text{convex body:}\ Q\left(n,X,A\right) \right\} , \end{align} \tag{ A.7 }$$

$$\begin{align} q_n^* &\,: = \,\sup\left\{|X|:\, \exists\, X\subset \unicode{x0211D}^n~\text{finite}, A = -A\subset \unicode{x0211D}^n \text{convex body:}\ Q^*\left(n,X,A\right) \right\} , \end{align} \tag{ A.8 }$$

where $|\cdot|$ denotes the cardinality of a finite set, i.e. the number of elements it contains. The geometrically intuitive fact that the 2ⁿ vertices of the hypercube satisfy $P(n,X)$ — and hence $p_n\unicode{x2A7E} 2^n$ — has been discussed in section 3.2, so we will not dwell on it further. The problem is to show that $p_n\unicode{x2A7D} 2^n$. The proof can be broken down into the following chain of inequalities:

$$\begin{equation} 2^n \unicode{x2A7D} p_n \stackrel{{\scriptsize \mbox{(i)}}}{\unicode{x2A7D}} q_n \stackrel{{\scriptsize \mbox{(ii)}}}{ = } q_n^* \stackrel{{\scriptsize \mbox{(iii)}}}{\unicode{x2A7D}} 2^n\, . \end{equation} \tag{ A.9 }$$

We now justify one by one the three crucial steps (i)–(iii):

• i  
  In fact, for all n and for all sets $X\subseteq \unicode{x0211D}^n$ we have that $P(n,X)\Longrightarrow Q\left(n,-\mathrm{conv}(X),X\right)$, where $\mathrm{conv}$ denotes the convex hull. To see this, assume that $P(n,X)$ holds. Then, for $x_1,x_2\in X$ with $x_1\neq x_2$ there exists a hyperplane $V\subset \unicode{x0211D}^n$ such that the set X, and hence also the convex body $\mathrm{conv}(X)$, is entirely contained between $x_1+V$ and $x_2+V$. Multiplying by −1 and translating, we see that $x_1-\mathrm{conv}(X)$ is entirely contained between V and $x_1-x_2+V$, and analogously $x_2-\mathrm{conv}(X)$ is entirely contained between V and $x_2-x_1+V$. Since $x_1\neq x_2$, we see that the convex bodies $x_1-\mathrm{conv}(X)$ and $x_2-\mathrm{conv}(X)$ are entirely contained into each of the two closed half-spaces determined by V. This implies that their interiors, which are instead contained into the corresponding open half-spaces, are disjoint. Remembering that $0\in (x_1-\mathrm{conv}(X))\cap (x_2-\mathrm{conv}(X))$, we see that in fact $x_1-\mathrm{conv}(X)$ and $x_2-\mathrm{conv}(X)$ touch each other.
• ii  
  We show that for all n, for all finite $X\subseteq \unicode{x0211D}^n$, and for all $A\subseteq \unicode{x0211D}^n$,

  $$\begin{equation} Q\left(n,A,X\right)\quad \Longleftrightarrow\quad Q\left(n,\widetilde{A},X\right) , \end{equation} \tag{ A.10 }$$

  so that naturally $q_n = q_n^*$. Start by noting the following: for a set $A\subseteq \unicode{x0211D}^n$ and two points $x,y\in \unicode{x0211D}^n$,

  $$\begin{equation} \left(x+A\right)\cap\left(y+A\right)\neq \emptyset\quad\Longleftrightarrow\quad \frac{x-y}{2}\in \widetilde{A}\, , \end{equation} \tag{ A.11 }$$

  where $\widetilde{A}$ is the Minkowski symmetrisation of A. Therefore, for fixed $x_1,x_2\in X$, we have that $(x_1+A)\cap (x_2+A)\neq \emptyset$ if and only if $\frac{x_1-x_2}{2}\in \widetilde{A}$. Since A and $\widetilde{A}$ have the same Minkowski symmetrisation, this is also equivalent to $(x_1+\widetilde{A})\cap (x_2+\widetilde{A})\neq \emptyset$. In other words,

  $$\begin{equation} \left(x_1+A\right)\cap \left(x_2+A\right)\neq \emptyset \quad \Longleftrightarrow\quad \left(x_1+\widetilde{A}\right)\cap \left(x_2+\widetilde{A}\right) \neq \emptyset\, . \end{equation} \tag{ A.12 }$$

  Applying this to $\mathrm{int} A$ instead of A, we get that

  $$\begin{equation} \begin{aligned} &\mathrm{int}\left(x_1+A\right)\cap \mathrm{int}\left(x_2+A\right) = \left(x_1+\mathrm{int}\left(A\right)\right) \cap \left(x_2+\mathrm{int}\left(A\right)\right) = \emptyset \\ &\qquad \Longleftrightarrow\quad \left(x_1+\widetilde{\mathrm{int} A}\right)\cap \left(x_2+\widetilde{\mathrm{int} A}\right) = \mathrm{int}\left(x_1+\widetilde{A}\right)\cap \mathrm{int}\left(x_2+\widetilde{A}\right) = \emptyset\, , \end{aligned} \end{equation} \tag{ A.13 }$$

  where the identity $x_i+\widetilde{\mathrm{int} A} = \mathrm{int}(x_i+\widetilde{A})$ follows from lemma 23. We have therefore proved that the convex bodies $x_1+A$ and $x_2+A$: (a) intersect if and only if so do $x_1+\widetilde{A}$ and $x_2+\widetilde{A}$; and (b) have disjoint interiors if and only if so do $x_1+\widetilde{A}$ and $x_2+\widetilde{A}$. In other words, $x_1+A$ and $x_2+A$ touch each other if and only if also $x_1+\widetilde{A}$ and $x_2+\widetilde{A}$ touch each other.
• iii  
  We now show that $q_n^*\unicode{x2A7D} 2^n$. To this end, pick a convex body $A = -A\subset \unicode{x0211D}^n$ and some set $X\subseteq \unicode{x0211D}^n$ such that $Q^*(n,A,X)$ holds. Set $B\,: = \,\mathrm{conv}(X)$. By (A.11), for all $x_1,x_2\in X$ it must hold that $\frac{x_1-x_2}{2}\in \widetilde{A} = A$. Then we claim that for all $x\in X$,

  $$\begin{equation} \frac{x+B}{2}\subseteq x+A\, . \end{equation} \tag{ A.14 }$$

  To see this, up to taking the convex hull it suffices to show that $\frac{x+y}{2}\subseteq x+A$ for all $y\in X$. And indeed, thanks to the above observation $\frac{x+y}{2} = x + \frac{y-x}{2}\in x+A$. This proves (A.14). As an immediate consequence of this together with $Q^*(n,A,X)$, observe that the interiors of the convex bodies $\frac{x+B}{2}$, indexed by $x\in X$, are all disjoint and moreover contained in B, because this is convex. Since convex bodies are well known to be Lebesgue measurable [70], we can now deduce that the volume of B is at least equal to the sum of the volumes of the bodies $\frac{x+B}{2}$, in formula

  $$\begin{equation} \mathrm{vol}\left(B\right) \unicode{x2A7E} \sum_{x\in X} \mathrm{vol}\left(\frac{x+B}{2}\right) = \sum_{x\in X} \frac{1}{2^n} \mathrm{vol}\left(x+B\right) = \sum_{x\in X} \frac{1}{2^n} \mathrm{vol}\left(B\right) = \frac{|X|}{2^n}\, \mathrm{vol}\left(B\right)\, . \end{equation} \tag{ A.15 }$$

  Since $\mathrm{vol}(B)\gt0$ because B is a convex body, we obtain that $|X|\unicode{x2A7D} 2^n$, as claimed.

This concludes the proof.