Given a positive integer $n$, a constant $k=2/3$, and $7$ variables $x_1, x_2, x_3, x_{12}, x_{13}, x_{23}, x_{123}$ (non-negative reals or integers) I would like to find:

$$\min \binom{x_1}2$$

subject to:

$$\binom{x_1}2 \ge \binom{x_2}2 \ge \binom{x_3}2 \tag{1}$$

$$\binom{n}2 \ge \binom{x_1}2+\binom{x_2}2+\binom{x_3}2-\binom{x_{12}}2-\binom{x_{23}}2-\binom{x_{13}}2 \ge k\binom{n-1}2 \tag{2}$$

$$\binom{x_1}2 \ge \binom{x_{12}}2+\binom{x_{13}}2 \ge k\binom{x_1-1}2 \tag{3}$$

$$\binom{x_2}2 \ge \binom{x_{12}}2+\binom{x_{23}}2 \ge k\binom{x_2-1}2 \tag{4}$$

$$\binom{x_3}2 \ge \binom{x_{13}}2+\binom{x_{23}}2 \ge k\binom{x_3-1}2 \tag{5}$$

$$x_1+x_2+x_3-x_{12}-x_{13}-x_{23}+x_{123} = n \tag{6}$$

This is a starting example, because from this I would like to solve the natural extension of the problem with $2^q-1$ variables, $q \gt 3$.

If we remove $(6)$ from the problem and we could replace $\binom{x_1-1}2, \binom{x_2-1}2, \binom{x_3-1}2$ with $\binom{x_1}2, \binom{x_2}2, \binom{x_3}2$ on the RHS of $(3), (4), (5)$ and then set $y_1=\binom{x_1}2, y_2=\binom{x_2}2, y_3=\binom{x_3}2, y_{12}=\binom{x_{12}}2, y_{13}=\binom{x_{13}}2, y_{23}=\binom{x_{23}}2, y_{123}=\binom{x_{123}}2$ we would get a linear program. Unfortunately, we can't do that, that's why we have a "nearly" linear program.

One idea that came to mind is that e.g.:

$$k\binom{x_1-1}2 = k\frac{x_1-2}{x_1}\binom{x_1}2 \ge \frac{2}{3}k\binom{x_1}2$$

when $x_1 \ge 6$, therefore we could replace the RHS for $x_1 \ge 6$ and enumerate all the possible values of $x_1$ for $x_1 \lt 6$, but we would have another $6$ linear programs to solve. If we fix $\binom{x_2-1}2$ and $\binom{x_3-1}2$ in the same way, the additional programs to solve would be $6^3 = 216$. That will be much bigger for $q \gt 3$.

Another idea is keep $(6)$, then use the identity:


however, with this one we would need a way to enforce the relation between $y_1 = \binom{x_1}2$ and $x_1$, i.e. linearize:

$$y_1 = \frac{x_1(x_1-1)}2$$

Any idea or suggestion?

  • 2
    $\begingroup$ I guess nearness is in the eyes of the beholder, because I wouldn't call this almost linear. $\endgroup$ Jan 9, 2023 at 0:04
  • 1
    $\begingroup$ What is the motivating problem for this formulation? $\endgroup$
    – RobPratt
    Jan 9, 2023 at 15:01
  • $\begingroup$ @RobPratt, yet another formulation related to finite union-closed families of sets. When I have some time, I will add an explanation, however it's a little long. $n$ is the number of sets of the family, $x_1$ the number of sets containing element $1$, $x_{12}$ the number of sets containing element $1$ and $2$, and so on. We are considering here the case where the universe is with max $3$ elements. $\endgroup$ Jan 9, 2023 at 15:21
  • $\begingroup$ $(2)$ enforces that at least $2/3$ of couples of (non-empty) sets have non empty intersection (see here), $(3),(4),(5)$ is the same as $(2)$ but for the subfamilies of all sets containing $1,2,3$ respectively. And finally we are searching the minimum of the maximum frequency of elements. The hope is having $\min x_1 \ge n/2$. $\endgroup$ Jan 9, 2023 at 15:21

1 Answer 1


Here's a linearization that you might find useful. Suppose integer variable $x\in\{0,\dots,M\}$. Then you can enforce $y = \binom{x}{2}$ by introducing binary variables $z_j$ and linear constraints: \begin{align} \sum_{j=0}^M z_j &= 1 \\ \sum_{j=0}^M j z_j &= x \\ \sum_{j=0}^M \binom{j}{2} z_j &= y \\ \end{align} More generally, you can enforce any functional relationship $y=f(x)$ by replacing $\binom{j}{2}$ with $f(j)$.


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