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In his paper Divisors in Residue Classes, H.W. Lenstra describes the following (Proposition 2.1 in the paper).

Proposition 2.1 (H.W Lenstra) If $V$ is a finite set with $w$ a finite measure on the power-set of $V$, and $\alpha > 1/4$ is any real number, then for any collection $\mathscr{D}$ of subsets of $V$ satisfying $$\max\left( w\left(D_1\setminus D_2\right), w\left(D_2\setminus D_1\right)\right)\geq \alpha w\left(V\right)$$ for all $D_1,D_2 \in \mathscr{D}$, then $\#\mathscr{D}\leq c(\alpha)$ where $c(\alpha)$ is a finite constant only depending on $\alpha$.

The proof of this proposition is straightforward and makes sense to me, but I have a question about a statement made later on in the paper. Lenstra later says,

For an integer $k\geq 2$, let $\alpha(k)$ be the largest possible value of $\alpha$ for which the hypotheses of Proposition (2.1) can be satisfied with $\#\mathscr{D}=k$. It is not difficult to see that $\alpha(k)$ exists and that, for given $k$, it can be computed by solving a linear programming problem with $2^k + 1$ variables.

Does anyone have some insight into how to set up the linear program that Lenstra is describing here? I'm familiar with the trick of adding an extra variable to a linear programming question in order to describe the maximum of two variables, (like this question here) but can't see how to set up this linear program to maximize $\alpha$.

Edit: Here's an idea I've been playing around with, set $\mathscr{D} = \{D_1,\dots,D_k\}$ and let $\mathscr{P}$ be the power set of $\mathscr{D}$ and create the linear program with the $2^{k}+1$ variables $x_{A}$ for $A \in \mathscr{P}$ and $\alpha$. The variables $0\leq x_{A}\leq 1$ will represent the value of $w\left(\bigcup_{D_i \in A}D_i\right)$ since wlog we can take $w(V) = 1$. Then the condition that $\max\left( w\left(D_1\setminus D_2\right), w\left(D_2\setminus D_1\right)\right)\geq \alpha w\left(V\right)$ can be given by $x_{\{D_i,D_j\}} \geq x_{\{D_i\}} + \alpha$ for all $i < j$ (since we can also assume wlog that we order the elements of $A$ according to weight so that $x_{\{D_1\}} \leq \dots \leq x_{\{D_k\}}$. Such a linear system can be maximized for $\alpha$, but I still am not sure if this particular linear program will give the optimal $\alpha$.

MathManiac5772
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  • I think you mean $\ge x_{D_i} + \alpha$. – RobPratt Feb 11 '25 at 00:40
  • Also, $x_1 \le \dots \le x_{2^k}$ should instead be $x_{D_1} \le \dots \le x_{D_k}$. – RobPratt Feb 11 '25 at 02:00
  • @RobPratt, yes you're absolutely right. I've made the appropriate edits. – MathManiac5772 Feb 11 '25 at 15:01
  • I suspect you have the right variables, but those constraints are not sufficient. You can always get $\alpha=1$ by taking $x_{{D_i}}=0$ and $x_A=1$ otherwise. – RobPratt Feb 11 '25 at 15:21
  • @RobPratt, correct. Two extra criterion that will fix that is $x_{A} \leq \sum_{D_i \in A}x_{{D_i}}$, which is just the union bound, and $x_{A} \geq x_{{D_i}}$ for all $D_i \in A$. – MathManiac5772 Feb 11 '25 at 17:33
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    Including both the union bound constraint and $x_A \ge x_B$ for $A \supseteq B$ yields $\alpha=1$ for $n=2$ and $\alpha=1/2$ for $n>2$. – RobPratt Feb 11 '25 at 19:31

1 Answers1

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Here's what I suspect is the approach that was used to compute the values in Table 1. As you proposed, there is a decision variable for each subset of $\{D_1,\dots,D_k\}$, and $\alpha$ is another decision variable, yielding $2^k+1$ variables. The phrase "with linear programming techniques" does not necessarily mean that a compact LP formulation was used. Instead, the following sentence hints at an indirect approach:

In all cases except $k = 9$ the inequalities from (2.7) were sufficient to obtain these upper bounds.

Note that (2.7) consists of an infinite set of inequalities (one for every $t\in \mathbb{Z}$). A common technique in that situation is to dynamically generate violated inequalities only as needed ("row generation" or "cutting plane method"). That is, solve the LP with only a finite subset of the inequalities imposed, "separate" the resulting optimal solution (that is, implicitly find a violated inequality), include that inequality, and repeat until there are no more violations. Each solve yields an upper bound on the maximum $\alpha$ because we are relaxing the problem by ignoring some constraints. It sounds like, except for $k=9$, a matching lower bound for $\alpha$ was found, as in (2.5). Presumably for $k=9$ some additional ad hoc constraints were imposed to reduce the upper bound to a known lower bound.

RobPratt
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