Error in intersection_ia

[kaushik]

Hi:

I am trying to find the internal approximation ellipsoid for the
intersection of an Ellipse and Polytpe.

First, I verified that the intersection is there using intersect(E,P)
and distance(E,P). The first function gave 1 and the second 0.

Now, when I use intersection_ia(E,P); I get the following error:

??? Error using ==> ellipsoid.intersection_ia at 88
INTERSECTION_IA: ellipsoids and hyperplanes must be of the same
dimension.

Error in ==> ellipsoid.intersection_ia>l_polyintersect at 234
EA = intersection_ia(EA, HA(i));

Error in ==> ellipsoid.intersection_ia at 126
e = [e l_polyintersect(E1(i, j), X)];

The size of the ellipsoid and polytope both match (in my case in
R^15).

~Kaushik

[Alex]

Kaushik,

it would be helpful for debugging if you provided your specific
example - ellipsoid and hyperplane parameters.

Thanks,
Alex.

[Kaushik]

I am attaching the m file which generates the ellipse and polytope.

~Kaushik

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[Alex]

Kaushik,

Thanks for the file and sorry it takes some time for me to respond.
Now I know what the problem is.

The way the intersection of an ellipsoid with a polytope is computed
right now is that compute ellipsoidal approximations of intersections
of this ellipsoid and the halfspaces, intersection of which forms the
polytope.

Suppose, you have N halfspaces, H_1, ..., H_N.
So, here is what happens, given original ellipsoid E and polytope
P(H_1,...,H_N):
E_1 = intersection of E and H_1;
E_2 = intersection of E_1 and H_2;
...
E_N = intersection of E_(N-1) and H_N.

E_N is the result - it approximates the intersection of E and P.

Now, the problem is that for some k, 1<k<N,
E_k = intersection of E_(k-1) and H_k returned empty ellipsoid
(ellipsoid of dimension 0).
In theory, it should not happen (if the intersection of E and P is
nonempty).
In practice, it can be caused by the rounding error of some sort of
the optimization solver or something like that.

Anyway, I would like to rewrite the ellipsoid and polytope
intersection using different algorithm, which should produce less
conservative result.
If you bear with me, I'll let you know once I'm done.

best,
Alex.