Modeling combinatorial disjunctive constraints via junction trees

Abstract

We introduce techniques to build small ideal mixed-integer programming (MIP) formulations of combinatorial disjunctive constraints (CDCs) via the independent branching scheme. We present a novel pairwise IB-representable class of CDCs, CDCs admitting junction trees, and provide a combinatorial procedure to build MIP formulations for those constraints. Generalized special ordered sets (${\text {SOS}}k$) can be modeled by CDCs admitting junction trees and we also obtain MIP formulations of ${\text {SOS}}k$. Furthermore, we provide a novel ideal extended formulation of any combinatorial disjunctive constraints with fewer auxiliary binary variables with an application in planar obstacle avoidance.

Appendices

A Proof of Lemma 5

Proof

Since k is a finite positive integer, then there exists an integer $b'$ such that $k \le 2^{b'} + 1$. Thus, the total number of the bicliques is no larger than

$$\begin{aligned} \sum _{i=0}^{b-1} \min \left\{ 2^i, \Bigg \lceil \frac{k-1+2^{b-i-1}}{2^{b-i}} \Bigg \rceil \right\}&\le \sum _{i=0}^{b-1} \Bigg \lceil \frac{k-1+2^{b-i-1}}{2^{b-i}} \Bigg \rceil \end{aligned}$$

(22a)

$$\begin{aligned}&= \sum _{i=1}^{b} \Bigg \lceil \frac{k-1+2^{i-1}}{2^{i}} \Bigg \rceil \end{aligned}$$

(22b)

$$\begin{aligned}&= b + \sum _{i=1}^{b} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil \end{aligned}$$

(22c)

$$\begin{aligned}&\le b + \sum _{i=1}^{b'} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil \end{aligned}$$

(22d)

We switch the value of i to $b-i$ in (22b). In order to obtain (22d), we prove it in two cases. If $b' \ge b$, since $\Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil \ge \Bigg \lceil -\frac{1}{2} \Bigg \rceil$ for all positive integer i, it is trivial to show the correctness. If $b'< b$,

$$\begin{aligned} \sum _{i=1}^{b} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil&= \sum _{i=1}^{b'} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil + \sum _{i=b'+1}^{b} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil \end{aligned}$$

(23a)

$$\begin{aligned}&\le \sum _{i=1}^{b'} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil + \sum _{i=b'+1}^{b} \Bigg \lceil \frac{k-1}{2^{b'+1}} - \frac{1}{2} \Bigg \rceil \end{aligned}$$

(23b)

$$\begin{aligned}&= \sum _{i=1}^{b'} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil , \end{aligned}$$

(23c)

where (23c) is obtained by the fact that $k\le 2^{b'} +1$.

Since $k \le 2^{b'} + 1$, then we can find $a_j \in \{0, 1\}$ for $j \in \{0, \ldots , b'-1\}$ such that $k-2 = \sum _{j=0}^{b'-1} a_j 2^j$. Then,

$$\begin{aligned} \sum _{i=1}^{b'} \Bigg \lceil \frac{k-1-2^{i-1}}{2^{i}} \Bigg \rceil&= \sum _{i=1}^{b'} \Bigg \lceil \frac{\left( \sum _{j=0}^{b'-1} a_j 2^j\right) + 1 -2^{i-1}}{2^{i}} \Bigg \rceil \end{aligned}$$

(24a)

$$\begin{aligned}&= \sum _{i=1}^{b'} \Bigg \lceil \left( \sum _{j=0}^{b'-1} a_j 2^{j-i}\right) + 2^{-i} -\frac{1}{2} \Bigg \rceil \end{aligned}$$

(24b)

$$\begin{aligned}&= \sum _{i=1}^{b'-1} \sum _{j=i}^{b'-1} a_j 2^{j-i}+ \sum _{i=1}^{b'} \Bigg \lceil \left( \sum _{j=0}^{i-1} a_j 2^{j-i}\right) + 2^{-i} -\frac{1}{2} \Bigg \rceil \end{aligned}$$

(24c)

$$\begin{aligned}&= \sum _{i=1}^{b'-1} \sum _{j=i}^{b'-1} a_j 2^{j-i} + \sum _{i=1}^{b'}a_{i-1}. \end{aligned}$$

(24d)

We take out all the integer parts to get (24c). For (24d), we can show the correctness by proving that $\Bigg \lceil \left( \sum _{j=0}^{i-1} a_j 2^{j-i}\right) + 2^{-i} -\frac{1}{2} \Bigg \rceil = a_{i-1}$. We prove it by cases. If $a_{i-1} = 0$,

$$\begin{aligned} -\frac{1}{2} \le \left( \sum _{j=0}^{i-1} a_j 2^{j-i}\right) + 2^{-i} -\frac{1}{2} \le \sum _{j=0}^{i-2} 2^{j-i} + 2^{-i} - \frac{1}{2} = 0. \end{aligned}$$

If $a_{i-1} = 1$, then

$$\begin{aligned} 1 \ge \left( \sum _{j=0}^{i-1} a_j 2^{j-i}\right) + 2^{-i} -\frac{1}{2} \ge \frac{1}{2} + 2^{-i} - \frac{1}{2} = 2^{-i} > 0. \end{aligned}$$

Then, by reordering the summation,

$$\begin{aligned} \sum _{i=1}^{b'-1} \sum _{j=i}^{b'-1} a_j 2^{j-i} + \sum _{i=1}^{b'}a_{i-1}&= \sum _{j=1}^{b'-1} a_j \sum _{i=1}^{j} 2^{j-i} + \sum _{j=0}^{b'-1}a_{j} \end{aligned}$$

(25a)

$$\begin{aligned}&= a_0 + \sum _{j=1}^{b'-1}\left( a_j + a_j \sum _{i=0}^{j-1} 2^i\right) \end{aligned}$$

(25b)

$$\begin{aligned}&= a_0 + \sum _{j=1}^{b'-1}a_j 2^j \end{aligned}$$

(25c)

$$\begin{aligned}&= k - 2. \end{aligned}$$

(25d)

$\square$

B Proofs of Propositions 15 and 16

Proof of Proposition 15

Given $N > k \ge 2$, we have

$$\begin{aligned} \lceil \log _2(\lceil N / k \rceil - 1) \rceil + 3k&= \lceil \log _2(\lceil (N - k) / k \rceil ) \rceil + 3k \end{aligned}$$

(26a)

$$\begin{aligned}&\ge \lceil \log _2((N - k) / k) \rceil + 3k \end{aligned}$$

(26b)

$$\begin{aligned}&= \lceil \log _2(N - k) - \log _2 (k) \rceil + 3k \end{aligned}$$

(26c)

$$\begin{aligned}&\ge \lceil \log _2(N - k)\rceil - \lceil \log _2 (k) \rceil + 3k \end{aligned}$$

(26d)

$$\begin{aligned}&\ge \lceil \log _2(N - k + 1)\rceil - 1 - \lceil \log _2 (k) \rceil + 3k \end{aligned}$$

(26e)

$$\begin{aligned}&\ge \lceil \log _2(N - k + 1)\rceil - 1 + 2k \end{aligned}$$

(26f)

$$\begin{aligned}&> \lceil \log _2(N - k + 1) \rceil + k - 2 , \end{aligned}$$

(26g)

where (26d) is obtained by the fact that $\lceil x - y \rceil \ge \lceil x - \lceil y \rceil \rceil = \lceil x \rceil - \lceil y \rceil$ for any x, y and (26e) is because $\lceil \log _2(x)\rceil \ge \lceil \log _2(x + 1)\rceil - 1$ for any $x \ge 1$.                  $\square$

Proof of Proposition 16 Furthermore, we suppose that $k > C \lceil \log _2(N) \rceil$ for some $C > \frac{1}{2}$. Then,

$$\begin{aligned} \frac{\lceil \log _2(N - k + 1) \rceil + k - 2}{\lceil \log _2(\lceil N / k \rceil - 1) \rceil + 3k}&\le \frac{\lceil \log _2(N) \rceil + k}{3 k} \end{aligned}$$

(27a)

$$\begin{aligned}&< \frac{\frac{k}{C} + k}{3 k} \end{aligned}$$

(27b)

$$\begin{aligned}&= \frac{1 + C}{3 C} \end{aligned}$$

(27c)

$$\begin{aligned}&< 1. \end{aligned}$$

(27d)

$\square$