6
$\begingroup$

Given two ellipsoids \begin{align}\mathcal{E}_1 &= \{ X \mid X^\top A_1 X + 2B_1^\top X + C_1 \leq 0\}\\\mathcal{E}_2 &= \{ X \mid X^\top A_2 X + 2 B_2^\top X + C_2 \leq 0\}\end{align} are both non-empty, it is possible to test if $\mathcal{E}_1 \subseteq \mathcal{E}_2$. Indeed, by the use of the so called $S$-procedure, $\mathcal{E}_1 \subseteq \mathcal{E}_2\iff\exists \lambda > 0$ such that $$ \begin{bmatrix} A_2 &B_2\\ B_2^\top &C_2\end{bmatrix} \preceq \lambda \begin{bmatrix} A_1 &B_1 \\ B_1^\top &C_1\end{bmatrix}$$ See https://web.stanford.edu/~boyd/cvxbook/.

Question

Assume that $\mathcal{E}_1 \not\subseteq \mathcal{E}_2$. I want to find a point in $\mathcal{E}_1 \setminus \mathcal{E}_2$. How can I do that? I think one should follow the proof of the $S$-procedure (the necessity part) and eventually construct one on those lines. Can someone help?

Edit

This paper might have an answer: Ye, Y. (n.d.). Quadratic Programming Over an Ellipsoid. Encyclopedia of Optimization, 2112–2116. doi:10.1007/0-306-48332-7_408

$\endgroup$

2 Answers 2

5
$\begingroup$

Define quadratic functions $g_1(X)$ and $g_2(X)$ such that \begin{align}\mathcal{E}_1 &= \{ X \mid g_1(X) \le 0\}\\\mathcal{E}_2 &= \{ X \mid g_2(X) \le 0\}\end{align}

It follows from the definition that you can find a point in $\mathcal{E}_1 \setminus \mathcal{E}_2$ by solving $$\begin{align} \min &\quad -g_2(X)\\\text{s.t.} &\quad g_1(X) \le 0.\end{align}$$ Informally, you find the point in $\mathcal{E}_1$ with the maximum (scaled) distance from the center of $\mathcal{E}_2$. If the optimal value of $-g_2(X) < 0$, then this point is not in $\mathcal{E}_2$, as required.

The problem above is not obviously convex: if $g_1(X)$ is convex, then the objective $-g_1(X)$ is typically not convex.

However, in this special case (and under some technical assumptions) the S-lemma can be used to reformulate this problem as a convex problem!

For details, see Theorem 2.2 in the survey by Pólik and Terlaky, for example. I personally find the lecture notes of lecture 12 provided here very helpful.


Edit: As pointed out by C Marius and Mark L. Stone, the approach above does not actually result in a point $X \in \mathcal{E}_1 \setminus \mathcal{E}_2$. Instead, it answers whether such a point exists. I looked further into this, and it turns out you can actually recover a point $X$ after solving the reformulated problem.

In this paper, Tuy and Tuan present the S-lemma as a strong duality theorem for specific non-convex problems, including the one above. In this specific case, we have (ignoring some technical details):

$$\inf_{X\in \mathbb{R}^n} \sup_{\lambda\ge 0} \{-g_2(X) + \lambda g_1(X)\} = \sup_{\lambda\ge 0} \inf_{X\in \mathbb{R}^n} \{-g_2(X) + \lambda g_1(X)\}.$$

The left-hand side is the primal problem, which is equivalent to the optimization problem stated at the top ($g_1(X) \le 0$ is optimal, because otherwise $\lambda \rightarrow \infty$ would send the objective to $+\infty$). The right-hand side is the dual problem, which is equivalent to the convex reformulation mentioned above.

After solving the dual problem (the convex reformulation), we obtain an optimal dual value $\bar{\lambda}$. You then recover a primal solution in a similar way as for linear programming: by using the optimality conditions (Tuy and Tuan, Theorem 3):

$$-\nabla g_2(X) + \bar{\lambda} \nabla g_1(X) = 0 \wedge \bar{\lambda} g_1(X) = 0 \wedge g_1(X) \le 0.$$


Example: Consider two ellipses, given by $A_1 = I$, $B_1 = (-1, 0)^\top$, $C_1 = 0$ (blue circle), and $A_2 = I$, $B_1 = 0$, $C_2 = -1$ (red circle)

Ellipse example

The dual is given by $$\begin{align} \max_{V, \lambda} &\quad V\\ \text{s.t.} &\quad \begin{bmatrix} \lambda A_1 - A_2 & \lambda B_1 - B_2 \\ (\lambda B_1 - B_2)^\top & \lambda C_1 - C_2 - V \end{bmatrix} \succeq 0 \\ ~&~\lambda \ge 0.\end{align}$$

Solving this SDP yields the solution $V = -3$ and $\bar{\lambda} = 2$. As $V < 0$, there exists an $X \in \mathcal{E}_1 \setminus \mathcal{E}_2$.

If we use the first of the optimality conditions, we have $$-\nabla g_2(X) + \bar{\lambda} \nabla g_1(X) = 2 (\bar{\lambda} A_1 - A_2)X + 2(\bar{\lambda} B_1 - B_2) = 2IX - (-4, 0)^\top = 0,$$ which solves for $\bar{X} = (2,0)^\top$. It is easy to check that the other conditions are also satisfied, which means that $\bar{X}$ is a primal optimal solution. Indeed, $\bar{X} \in \mathcal{E}_1 \setminus \mathcal{E}_2$.

In the picture, the yellow circle corresponds to the ellipse $g_2(X) = -V$. This ellipse intersects $\mathcal{E}_1$ at the maximum (weighted) distance from the center of $\mathcal{E}_2$. Their intersection is exactly the point $\bar{X}$.

I have omitted various technical details, for which I refer to the references.

$\endgroup$
7
  • 1
    $\begingroup$ I may be wrong, but at a first glance I see that the method described in lecture 12 of the provided link, solves the optimization problem, but the result is the minimum value of $-g_2$ on $g_1 \leq 0$, not the point for which this happens. Is this correct? $\endgroup$
    – C Marius
    Commented Apr 22, 2020 at 9:18
  • $\begingroup$ If $V$ is this optimal value, and $V < 0$, then I think you can show there exists a point you are interested in with $g_1(x)=0$ and $g_2(x)=-V$. That is, the point is in the intersection of two ellipsoids. I imagine there exist alghorithms for this, but I have no experience here. $\endgroup$ Commented Apr 22, 2020 at 12:57
  • 1
    $\begingroup$ Isn't finding such a point the "point" of the question? The approaches in my answer, although unfortunately non-convex, do that. $\endgroup$ Commented Apr 22, 2020 at 15:50
  • 1
    $\begingroup$ The approach above allows you to answer the question of existence in polynomial time. Finding a feasible point is reduced to solving a system of two quadratic equality constraints. I am guessing that this would be easier than dealing with two non-convex inequalities, but I have to admit I don't know. $\endgroup$ Commented Apr 22, 2020 at 17:04
  • 1
    $\begingroup$ Thank you for your time! What I am not very sure thou, is how easy is this (fiding a point) in $\mathbb{R}^n$. Indeed those optimality conditions force 2 quadratic equalities and an inequality, which may be dificult to solve in general ... I think $\endgroup$
    – C Marius
    Commented Apr 27, 2020 at 18:01
2
$\begingroup$

This is a feasibility problem, find $X$ such that $$X^TA_1 X + 2B_1^TX + C_1 \leq 0, X^T A_2 X + 2 B_2^T X + C_2 \gt 0$$.In general it will be non-convex (although it is convex if $A_1$ is positive semidefinite and $A_2$ is negative semidefinite; however, given mention of "ellipsoid", it seems reasonable to presume that $A_1$ and $A_2$ are both positive semidefinite). It's also a bit tricky due to the strict inequality in the 2nd constraint. In practice, you will need to change that to $$X^T A_2 X + 2 B_2^T X + C_2 +\text{small_positive_number} \ge 0,$$and deal with the possibility of the only (what should be) feasible point requiring a smaller positive number. Alternatively, you could solve a QCQP which is to maximize the LHS (without small_positive_number) of the 2nd constraint, subject to the first constraint; and determine the original problem to be solved if this problem has positive optimal objective function - but this generally involves more computing than solving the original feasibility problem.

In the convex case ($A_1$ positive semidefinite and $A_2$ negative semidefinite, which given my previous remark, seems unlikely to occur, unless $A_2$ is degenerate), you can use a convex QCQP or SOCP solver, and can formulate the problem using a convex optimization tool.

In the non-convex case (which from a practical perspective, you should always be in), you can always try using a local optimizer, and if it succeeds, you're done. However, it could fail even when there truly is a feasible point. In order to "guarantee" finding a feasible value if one exists, you should use a global solver, such as BARON or Gurobi 9.x (which can find global optimum of non-convex quadratics) or CPLEX with optimalitytarget = 3 (if the alternative optimization formulation is used, because that moves the non-convexity from the constraints to the objective function, where it is allowed by CPLEX).

$\endgroup$
2
  • 1
    $\begingroup$ Testing ellipsoid containment is a convex problem according to Boyds book. Therefore, it can be decided solving a convex optimization problem whether one ellipsoid is contained in the other. I was wondering if in case, say, $\mathcal{E}_1$ is not included in $\mathcal{E}_2$ is easy, in the same spirit, to obtain a point which proves this. $\endgroup$
    – C Marius
    Commented Apr 21, 2020 at 16:39
  • $\begingroup$ Convexity doesn't know "spirit" and can be a cruel mistress. The (nontrivial) complement of a nonlinear convex region is gernerally non-convex. You are trying to find a feasible point in a non-convex region, that is a non-convex problem. $\endgroup$ Commented Apr 21, 2020 at 16:48

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.