Obtaining the convex hull of large number of convex sets

[gayanlanke]

I have 5 constrained sets, actually convex sets, and I need to obtain the convex hull of them. So I follow an approach as below.

clc;clear

sdpvar x y

F1 = [-1 <= x <= 1, -1 <= y <= 1];  
F2 = [-3 <= x <= -2, -2 <= y <= 2];
F3 = [-5 <= x <= -4, -3 <= y <= 3];
F4 = [-9 <= x <= -6, -5 <= y <= 5];

H=hull(F1,F2,F3,F4);  %% convex hull

This is perfectly fine, and I can get the convex hull without any problem. Imagine I have 100 constrained sets for example, do I need to still use the command hull(F1,F2,F3,....F100) (tedious of course!) to obtain the convex hull?

Instead of typing F1,F2,...F100, is it possible to define a vector or a cell array F_vec and get the convex hull as follows:

H=hull(F_vec)

Thanks in advance.

[johanlofberg]

You are generating them somewhere, so simply put all in a cell and then do hull(these{:})

Although it sounds like a very expensive representation if the sets actually look like this. Just compute all the vertices explicitly and then corresponding convex hull vertices, and then generate the constraint

[gayanlanke]

Thanks a lot, it worked.

For "Although it sounds like a very expensive representation if the sets actually look like this. Just compute all the vertices explicitly and then corresponding convex hull vertices, and then generate the constraint", although I represent them in this form, I already have the convex hull and then obtain A and b to represent each as hyperspaces given by A*x<=b. I guess that should be okay.

[johanlofberg]

Not sure what you are saying, but the constraints you show here have 4 vertices, and computing the convex hull of them is a simple call to convhull and then use a convex hull representation of those vertices, or if you want the facet-representation of the convex hull, it is also easily computed by said command as it is in 2D and thus tractable