Expansion of random $0/1$ polytopes

Let $S\subset V(P)$ with $|S|\leq|V(P)|/2$ and let $T:=V(P)\setminus S$. Set $s=|S|$. We need to show that there are at least $s/2c$ edges in $P$ which connect a vertex in $S$ to a vertex in $T$.

Observe that at least one of $\{x\in V(C^{k}):\pi_{k}^{-1}(x)\subset S\}$ or $\{x\in V(C^{k}):\pi_{k}^{-1}(x)\subset T\}$ has cardinality at most $2^{k-1}$. Assume that $\{x\in V(C^{k}):\pi_{k}^{-1}(x)\subset S\}$ has cardinality at most $2^{k-1}$. The proof for the other case is nearly the same. The projection of $S$, i.e. $\pi_{k}(S)$, is a subset of vertices of $C^{k}$ and by assumption 2, $|\pi_{k}(S)|\geq s/c$. There are two cases to consider.

Case 1: The cardinality of $M:=\pi_{k}(S)\cap\pi_{k}(T)$ is at least $s/2c$.

Case 2: The cardinality of $M$ is less than $s/2c$.

For Case 1, for each $x\in M$, $\pi_{k}^{-1}(x)$ is a face of $P$ which contains points from $S$ and points from $T$. Since graphs of polytopes are connected, there exists an edge in this face going from a point in $S$ to a point in $T$. Since $|M|\geq s/2c$, we have found $s/2c$ edges in $P$ from $S$ to $T$. Each of these edges is unique because the image of each edge by $\pi_{k}$ is a unique vertex in $M$.

For Case 2, let $U=\pi_{k}(S)\setminus M$. Since $|\pi_{k}(S)|\geq s/c$ and $|M|\leq s/2c$, we have that $|U|\geq s/2c$. By assumption 1, we know that $\pi_{k}(S\cup T)$ contains every vertex of $C^{k}$. Finally, recall that we are assuming that $\{x\in V(C^{k}):\pi_{k}^{-1}(x)\subset S\}$ has cardinality at most $2^{k-1}$. This means that $|U|\leq 2^{k-1}$. Now using the fact that the edge expansion of $C^{k}$ is 1, we know that there are at least $|U|\geq s/2c$ edges of $C^{k}$ going from a vertex in $U$ to a vertex in $V(C^{k})\setminus U$. Let $E$ be the set of those edges. We will show that each such edge lifts (through $\pi_{k}^{-1}$) to an edge of $P$ that has one point in $S$ and one point in $T$ as its endpoints. That is, for each edge $e\in E$, we consider the preimage $\pi_{k}^{-1}(e)$ and we will show that there exists some edge of $P$ that is contained in $\pi_{k}^{-1}(e)$ and which has one point in $S$ and one point in $T$ as its endpoints.

For each edge $e\in E$ we have $e=\operatorname{conv}(u,m)$ with $u\in U$ and $m\in V(C^{k})\setminus U$. The pre-image $\pi_{k}^{-1}(e)$ is a face (call it $F$) of $P$ which has two proper faces $\pi_{k}^{-1}(u),\pi_{k}^{-1}(m)$.²²2For those unfamiliar with polytope theory, here is an explanation of why these preimages are faces and/or proper faces: We know that $F$ is a face of $P$ because if $H$ is a hyperplane supporting $e$ as a face of $C^{k}$, then $\pi_{k}^{-1}(H)$ is a hyperplane that supports $F$ as a face of $P$. A similar argument shows that $\pi_{k}^{-1}(u),\pi_{k}^{-1}(m)$ are both faces of $F$ and they are proper because they do not contain all the vertices of $F$. Since $u\in U$, the face $\pi_{k}^{-1}(u)$ contains only points from $S$. Furthermore, by the way $U$ was constructed, for every $m\in V(C^{k})\setminus U$, we know that $\pi_{k}^{-1}(m)$ contains at least one point from $T$. Let $t$ be a point in $T\cap\pi_{k}^{-1}(m)$. We claim that there is an edge in the face $F$ which goes from $t$ to a point $s\in\pi_{k}^{-1}(u)$. Indeed, if this were not the case, all of the edges in $F$ incident to $t$ would be contained in the face $\pi_{k}^{-1}(m)$. Now if we consider $F$ as a full dimensional polytope in $\operatorname{aff}F$, because $\pi_{k}^{-1}(m)$ is a proper face of $F$, it is contained in a hyperplane in $\operatorname{aff}F$. This implies that the vertex $t$ has the property that all edges incident to $t$ are contained in a hyperplane in $\operatorname{aff}F$ which is not possible by Proposition 6. We have shown that for every edge $e\in E$, there is an edge $e^{\prime}$ in $P$ which goes from a point in $S$ to a point in $T$ and also that $\pi_{k}(e^{\prime})=e$. Since all of the edges $e\in E$ are unique, the fact that $\pi_{k}(e^{\prime})=e$ for all $e\in E$ implies that all of the edges $e^{\prime}$ that we construct are unique. Since $|E|\geq s/2c$ we have shown that there are at least this many edges in $P$ going from $S$ to $T$ and we are done. ∎

First, if $n\leq d$ then it is clear that $P_{n}^{d}$ has edge expansion at least $1/12d$ because $P_{n}^{d}$ has at most $d$ vertices. Indeed, given any subset $S\subset P_{n}^{d}$ with $|S|\leq|P_{n}^{d}|/2$, the fact that graphs of polytopes are connected implies that there is at least one edge connecting a vertex in $S$ to a vertex in $P_{n}^{d}\setminus S$. Since $|S|\leq d/2$, this is enough to show that the edge expansion of $P_{p}^{d}$ is at least $2/d\geq 1/12d$.

If $n\geq d2^{d}$, then we claim that $P_{n}^{d}=C^{d}$ with high probability. Indeed, the probability that there exists some vertex of $C^{d}$ that is not chosen once in $S_{n}^{d}$ is less than or equal to

and so the probability that $P_{n}^{d}\neq C^{d}$ goes to zero as $d\to\infty$. The fact that $P_{n}^{d}$ has edge expansion at least $1/12d$ with high probability now follows from the fact that the edge expansion of $C^{d}$ is 1.

Now assume that $d<n<d2^{d}$. Let $k$ be the largest integer such that $n\geq k2^{k}$. We will show that by considering the projection of $P_{n}^{d}$ to the first $k$ coordinates, we can apply Lemma 7 to show that the edge expansion of $P_{n}^{d}$ is at least $1/12d$ with high probability. Note that since $n<d2^{d}$, this means that $k<d$. Let $\pi_{k}:V(P_{n}^{d})\to\operatorname{\mathbb{R}}^{k}$ denote the orthogonal projection of $V(P_{n}^{d})$ to the first $k$ coordinates. We claim that two things hold with high probability:

Claim 1: Letting $C^{k}$ denote the $k$-cube in $\pi_{k}\operatorname{\mathbb{R}}^{d}=\operatorname{\mathbb{R}}^{k}$, every vertex of $C^{k}$ appears at least once in $\pi_{k}V(P_{n}^{d})$.

Claim 2: For every vertex $v\in V(C^{k})$, the cardinality of $\pi_{k}^{-1}(v)\subset V(P_{n}^{d})$ is at most $6d$.

To prove that the first claim holds with high probability, note that it suffices to prove that every vertex of $C^{k}$ appears at least once in $\pi_{k}S_{n}^{d}$ with high probability. Observe that $\pi_{k}S_{n}^{d}$ is the same as $S_{n}^{k}$. Therefore, the first claim is equivalent to the statement that every vertex of $C^{k}$ appears at least once in $S_{n}^{k}$. Since $n\geq k2^{k}$, as argued above, we have that the probability that there exists some vertex of $C^{k}$ that is not chosen once in $S_{n}^{k}$ is less than or equal to $(\frac{2}{e})^{k}$. Since we are assuming in this case that $(k+1)2^{k+1}>n>d$, we have that $k\to\infty$ as $d\to\infty$ and therefore the probability that there exists some vertex of $C^{k}$ that is not chosen once in $S_{n}^{k}$ goes to zero as $d\to\infty$. And so we have that that Claim 1 holds with high probability.

For the second claim, we use the well-known analysis of the classic \sayballs-into-bins problem from probability theory, see for example [15]. In our application, that balls are the points in $S_{n}^{d}$ and the bins are the vertices of $C^{k}$. That is, we have $n$ balls each of which is placed into one of $2^{k}$ bins uniformly at random. Using the fact that $k2^{k}\leq n\leq(k+1)2^{k+1}$, by [15, Theorem 1], each bin contains at most $6k$ balls with high probability. Since $k<d$, each bin contains at most $6d$ balls with high probability. In other words, Claim 2 holds with high probability.

Because Claims 1 and 2 hold with high probability, by Lemma 7 we have that the edge expansion of the graph of $P_{n}^{d}$ is at least $1/12d$ with high probability. ∎

First, if $p\leq d/2^{d}$, then it is clear that $P_{p}^{d}$ has edge expansion at least $1/12d$ with high probability because it has few vertices with high probability. Indeed, the cardinality of $S_{p}^{d}$ is a binomial random variable with number of trials $2^{d}$ and probability of success $p$ and so it has expected value $\mu:=p2^{d}$. Using the Chernoff bound, we have that $|S_{p}^{d}|$ is less than $3\mu=3p2^{d}\leq 3d$ with high probability. This means that $P_{p}^{d}$ has at most $3d$ vertices with high probability. Given any subset $S\subset P_{p}^{d}$ with $|S|\leq|P_{p}^{d}|/2$, the fact that graphs of polytopes are connected implies that there is at least one edge connecting a vertex in $S$ to a vertex in $P_{p}^{d}\setminus S$. Since $|S|\leq 3d/2$, this is enough to show that the edge expansion of $P_{p}^{d}$ is at least $2/3d\geq 1/12d$.

So now assume that $p>d/2^{d}$.

Let $k$ be the largest integer such that $p2^{d}\geq k2^{k}$. We will show that by considering the projection of $P_{p}^{d}$ to the first $k$ coordinates, we can apply Lemma 7 to show that the edge expansion of $P_{p}^{d}$ is at least $1/12d$ with high probability. Note that since $p<1$, this means that $k<d$. Let $\pi_{k}:V(P_{p}^{d})\to\operatorname{\mathbb{R}}^{k}$ denote the orthogonal projection of $V(P_{p}^{d})$ to the first $k$ coordinates. We claim that two things hold with high probability:

Claim 1: Every vertex of $C^{k}$ appears at least once in $\pi_{k}V(P_{p}^{d})$.

Claim 2: For every vertex $v\in V(C^{k})$, the cardinality of $\pi_{k}^{-1}(v)\subset P_{p}^{d}$ is at most $6d$.

To prove the first claim holds with high probability, observe that for each $v\in V(C^{k})$, the set $\pi_{k}^{-1}(v)$ consists of $2^{d-k}$ vertices of $C^{d}$. Therefore, the probability that there is some vertex in $C^{k}$ that doesn’t appear in $\pi_{k}V(P_{p}^{d})$ is equal to $2^{k}(1-p)^{2^{d-k}}$. Now using the fact that $p2^{d}\geq k2^{k}$, we have that $1-p\leq 1-\frac{k2^{k}}{2^{d}}$. Therefore, the previously mentioned probability is at most $2^{k}(1-\frac{k}{2^{d-k}})^{2^{d-k}}$. This quantity is less than $(2/e)^{k}$. Since $k$ is the largest integer such that $p2^{d}\geq k2^{k}$, we know that $p2^{d}\leq(k+1)2^{k+1}$. Substituting $p>d/2^{d}$ into the previous inequality yields $d<(k+1)2^{k+1}$ and so $k\to\infty$ as $d\to\infty$. Now since the probability that there is some vertex that doesn’t appear in $\pi_{p}V(P_{n}^{d})$ is less than $(2/e)^{k}$, we have that this probability goes to zero as $d\to\infty$ and so Claim 1 holds with high probability.

For the second claim, observe that for each $v\in V(C^{k})$, $|\pi_{k}^{-1}(v)|$ is a binomial random variable with number of trials $2^{d-k}$ and probability of success $p$. This means that the expected value $\mu$ of each of these random variables is $p2^{d-k}$. Now by the fact that $k2^{k}\leq p2^{d}\leq(k+1)2^{k+1}$, we have that $k\leq\mu\leq 2(k+1)$. This means that $3\mu\leq 6(k+1)$. Therefore, using the Chernoff bound, we have

This means that the probability that there is some $v\in V(C^{k})$ such that $|\pi_{k}^{-1}(v)|\geq 6(k+1)$ is at most $\bigl{(}\frac{2e^{2}}{3^{3}}\bigr{)}^{k}$ which goes to zero as $k\to\infty$. Therefore, with high probability, for every vertex $v\in V(C^{k})$, the cardinality of $\pi_{k}^{-1}(v)$ is at most $6k+5<6d$.

Because Claims 1 and 2 hold with high probability, by Lemma 7 we have that the edge expansion of the graph of $P_{p}^{d}$ is at least $1/12d$ with high probability. ∎

For this proof, we will make use of the fact that the uniform model is in some sense very similar to the binomial model. Recall that in the proof of Theorem 9 we showed that with $\pi_{k}$ denoting the orthogonal projection to the first $k$ coordinates, two claims hold with high probability:

Claim 1: Every vertex of $C^{k}$ appears at least once in $\pi_{k}V(P_{p}^{d})$.

Claim 2: For every vertex $v\in V(C^{k})$, the cardinality of $\pi_{k}^{-1}(v)\subset P_{p}^{d}$ is at most $6d$.

Now it is easy to see that, considering $V(P_{p}^{d})$ as a subset of $\{0,1\}^{d}$, satisfying Claim 1 and Claim 2 is a convex property of subsets of $\{0,1\}^{d}$ as defined in [9, Section 1.3]. Therefore, by [9, Proposition 1.15], Claim 1 and 2 hold with high probability if we replace $P_{p}^{d}$ by $Q_{n}^{d}$ in the statements of these claims. Therefore, again using Lemma 7, we have that the edge expansion of $Q_{n}^{d}$ is at least $1/12d$ with high probability. ∎