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Let $|Ax|$ be the element-wise absolute value of $A x$, i.e., $|Ax|_i = |A(i,:)x|$. The inequalities are element-wise inequalities, i.e., $|A(:,i)x| \geq b(i)$. Also, let $\|x\|$ denote the $2$-norm of $x$.

The constraint $|Ax| \leq b$ is convex. However, $|Ax| \geq b$ is not convex. Is there a way to solve the following optimization problem?

$$\begin{array}{ll} \text{minimize} & \|x\|^2\\ \text{subject to} & |Ax| \geq b\end{array}$$

I have used $\|x\|$ so that the solution can be bounded. How can the above problem be solved?

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    $\begingroup$ $Ax$ is a vector, so what does $|Ax|$ denote? You're using a different symbol $\| \cdot \|$ to denote a norm. $\endgroup$ – littleO Apr 19 at 10:36
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    $\begingroup$ @TonyS.F. But which norm? Am I allowed to assume it's the Euclidean norm? $\endgroup$ – littleO Apr 19 at 10:54
  • $\begingroup$ That's a fair point, it's ambiguous whether it's the $\ell^1$ or the euclidean. $\endgroup$ – Tony S.F. Apr 19 at 10:55
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    $\begingroup$ $|Ax|$ is taking elementwise absolute values of the vector $Ax$. I will add this in the question. $\endgroup$ – vi11 Apr 19 at 11:26
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Suppose we are given $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm b \in \mathbb R^m$. Let $\mathrm a_k \in \mathbb R^n$ denote the $k$-th row of matrix $\rm A$, i.e.,

$$\mathrm A = \begin{bmatrix} — \mathrm a_1^\top —\\ — \mathrm a_2^\top —\\ \vdots\\ — \mathrm a_m^\top — \end{bmatrix}$$

Thus, $|\rm A x| \geq b$ is equivalent to

$$\bigwedge_{i=1}^m \left(|\mathrm a_i^\top \mathrm x| \geq b_i\right) \equiv \bigwedge_{i=1}^m \left( \left( \mathrm a_i^\top \mathrm x \geq b_i \right) \lor \left( \mathrm a_i^\top \mathrm x \leq -b_i \right) \right)$$

which is in CNF. Converting it to DNF, we then obtain a union of convex polytopes. The optimization problem we are given seeks the point in this union that is closest to the origin in the Euclidean sense.

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  • $\begingroup$ You will then have to find the closest point to the origin in $2^m$ different polytopes though -- one for each choice of each of the $m$ disjunctions. So this would be very expensive for $m>20$ or so. $\endgroup$ – Rahul Apr 20 at 13:34
  • $\begingroup$ @Rahul I am wondering what happens when vector $\rm b$ has non-positive entries. For example, let us suppose that $b_1 \leq 0$; then, $|\mathrm a_1^\top \mathrm x| \geq b_1$ is satisfied for all $\mathrm x \in \mathbb R^n$ and we are left with $m-1$ conjuncts. Hence, there should be at most $2^p$ polytopes, where $p$ is the number of positive entries of vector $\rm b$. However, there may be symmetry to be exploited and, thus, the exponential may not be all that bad. $\endgroup$ – Rodrigo de Azevedo Apr 21 at 17:11
  • $\begingroup$ @RodrigodeAzevedo Can I know how exactly to obtain the DNF forms in the convex polytopes and further how is it solved? $\endgroup$ – vi11 Apr 22 at 6:34
  • $\begingroup$ @vi11 I will update my answer when I find time. How big is your $m$? Are all $b_i$'s positive? $\endgroup$ – Rodrigo de Azevedo Apr 22 at 19:36
  • $\begingroup$ Yes all $b_i$'s are positive. $m$ could be large too. $\endgroup$ – vi11 Apr 23 at 5:00

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