4
$\begingroup$

Let $S\subseteq \mathbb{R}^n$ be a convex region defined by $$g_i(x)\leq 0, ~~i\in 1,\ldots,m,$$ where $g_i$ are convex functions. The goal is to decide whether $S$ is empty, and if not - find a point $x\in S$.

In the book of Boyd and Vandenberghe (chapter 11), I found the following algorithm. Solve the following optimization problem:

$$\text{ mininize } ~~p~~ \text{ subject to } ~~g_i(x)\leq p, ~~i\in 1,\ldots,m.$$

Denote the optimal solution by $(p^*,x^*)$. If $p^*>0$, then $S$ is empty. If $p^*\leq 0$, then $x^*$ is a point in $S$ (in particular, if $p^* < 0$ then $x^*$ is in the interior of $S$; if $p^*=0$ then interior of $S$ is empty and $x^*$ is a boundary point).

The problem is that, when $p^*$ approaches 0, the runtime complexity of solving this auxiliary problem approaches infinity; here is the exact runtime from the book of Boyd and Vandenberghe:

enter image description here

As you can see, $p^*$ is in the denominator. They also say that:

enter image description here

I would like to know what is known about the complexity of this decision problem. In particular:

  • Suppose the functions $g_i$ are given explicitly (e.g. they are all polynomials with rational coefficients; this also means that they are continuously differentiable infinitely many times), and every basic arithmetic operation on real numbers takes unit time. Is there a proof that the decision problem cannot be solved in a finite number of arithmetic operations?
  • Are there sub-classes of convex functions (except linear functions) for which the decision problem can be solved in finite time? In polynomial time?

EDIT: Based on the comments, it seems the question is non-trivial even in the case of quadratic programming, where all functions $g_i$ are convex quadratic functions. The Wikipedia page mentions that the problem can be solved in weakly-polynomial time using the ellipsoid method, but to use the ellipsoid method we need a lower bound on the volume of the feasible region, which is not always guaranteed.

$\endgroup$
11

1 Answer 1

4
$\begingroup$

Warning: As one of the comments points out, the sum of squares is not necessarily convex, so the hardness reduction suggested below does not work. The problem still lies in $\exists\mathbb{R} \subseteq \mathrm{PSPACE}$ though.

Deciding whether there is an $x \in \mathbb{R}^n$ such that $f_i(x) = 0$ for a family of quadratic polynomials is complete for the existential theory of the reals, $\exists\mathbb{R}$. So testing whether $f(x) = \sum_i (f_i(x))^2 \leq 0$ has a solution is also complete for $\exists\mathbb{R}$, and, as a sum of squares, $f$ is a convex function. So solving the non-emptiness problem of a convex set is as hard as deciding truth in the existential theory of the reals.

The hardness result goes back to Blum, Shub, and Smale. On a Theory of Complexity over the Real Numbers, it follows from the main theorem, but it's stated in the BSS-model. A statement in the $\exists\mathbb{R}$-framework can be found as Lemma 3.2 in Schaefer, Stefankovic. Fixed Points, Nash Equilibria, and the Existential Theory of the Reals.

$\endgroup$
13
  • 1
    $\begingroup$ A simple Google search gives this useful description on Wikipedia. en.wikipedia.org/wiki/Existential_theory_of_the_reals $\endgroup$ Commented Dec 15, 2023 at 6:02
  • 1
    $\begingroup$ @ErelSegal-Halevi The original algorithm is due to Canny, Some algebraic and geometric computations in PSPACE, doi.org/10.1145/62212.62257 . A follow-up improvement: doi.org/10.1093/comjnl/36.5.409 $\endgroup$ Commented Dec 15, 2023 at 7:46
  • 3
    $\begingroup$ For this direction, the trivial reduction works: $\exists\vec x\,\bigwedge_ig_i(\vec x)\le0$ is, literally, an existential statement in the theory of the reals. $\endgroup$ Commented Dec 17, 2023 at 15:51
  • 3
    $\begingroup$ @ErelSegal-Halevi The quoted passage doesn't focus on polynomials, does it? The existential theory of the reals does, so this might be were the disconnect lies. Also it says "in practice". I don't think that a PSPACE algorithm is very practical, in general. $\endgroup$
    – Tassle
    Commented Dec 18, 2023 at 13:47
  • 2
    $\begingroup$ @NealYoung I don’t even need that they are convex. I only need that they are polynomials with rational (or: real algebraic, if represented in a suitable way) coefficients, as I wrote above. $\endgroup$ Commented Dec 18, 2023 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.