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Consider the following set.

$S_n =\{ (x,y) ~|~x \in \mathbb{Z}_+~\wedge~ y \in \{0,1\}^n ~\wedge~x=\sum_{i=0}^{n-1} 2^i y_i \}$

$S_n$ is a collection of pairs $(x,y)$, where $x$ is an integer between 0 and $2^n-1$ and $y$ is its binary representation. I'm interested in the convex hull of $S_n$; you can call it $P_n$.

Has $P_n$ been studied before? Does it admit a compact extended formulation?

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1 Answer 1

up vote 6 down vote accepted

I think $S_n$ can be written in terms of inequalities in the obvious way. Let $$ Q_n = \{(x, y): x = \sum_{i = 0}^{n-1}{2^i y_i}, \forall i: 0 \leq y_i \leq 1\}. $$

I claim that $Q_n = S_n$. First, obviously all $(x, y) \in S_n$ are also in $Q_n$, so $S_n \subseteq Q_n$. Second, fix a point $(x^*, y^*) \in Q_n$. Consider the probability distribution over $\{0, 1\}^n$ induced by picking $y_i = 1$ with probability $y^*_i$, independently for each $i$. If $y$ is sampled from that distribution, then, by linearity of expectation, $$ \mathbb{E}\ x = \mathbb{E} \sum_{i = 0}^{n-1}{2^i y_i} = \sum_{i = 0}^{n-1}{2^i y^*_i} = x^*. $$ Therefore $(x^*, y^*)$ is in the convex hull of $S_n$, which proves $Q_n \subseteq S_n$.

BTW, this didn't really use anything special about $S_n$. Whenever you have the convex hull of the set $\{(x, y): y \in S, x = Ay\}$, and the convex hull of $S$ has a concise extended formulation, the same thing will work. The main point is that $x$ is a linear function of $y$. Here we used the fact that the cube $[0, 1]^n$ has a very easy formulation in terms of inequalities.

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