Set inversion

In mathematics, set inversion is the problem of characterizing the preimage X of a set Y by a function f, i.e., X = f&thinsp;&hairsp;−1(Y&thinsp;) = {x ∈ Rn | f(x) ∈ Y&thinsp;}. It can also be viewed as the problem of describing the solution set of the quantified constraint "Y(f&hairsp;(x))", where Y(&hairsp;y) is a constraint, e.g. an inequality, describing the set Y.

In most applications, f is a function from Rn to Rp and the set Y is a box of Rp (i.e. a Cartesian product of p intervals of R).

When f is nonlinear the set inversion problem can be solved using interval analysis combined with a branch-and-bound algorithm.

The main idea consists in building a paving of Rp made with non-overlapping boxes. For each box [x], we perform the following tests:
 * 1) if f&hairsp;([x]) ⊂ Y we conclude that [x] ⊂ X;
 * 2) if f&hairsp;([x]) ∩ Y = ∅ we conclude that [x] ∩ X = ∅;
 * 3) Otherwise, the box [x] the box is bisected except if its width is smaller than a given precision.

To check the two first tests, we need an interval extension (or an inclusion function) [f&thinsp;] for f. Classified boxes are stored into subpavings, i.e., union of non-overlapping boxes. The algorithm can be made more efficient by replacing the inclusion tests by contractors.

Example
The set X = f&thinsp;&hairsp;−1([4,9]) where f&hairsp;(x1, x2) = x$2 1$ + x$2 2$ is represented on the figure.

For instance, since [−2,1]2 + [4,5]2 = [0,4] + [16,25] = [16,29] does not intersect the interval [4,9], we conclude that the box [−2,1]&thinsp;×&thinsp;[4,5] is outside X. Since [−1,1]2 + [2,√5]2 = [0,1] + [4,5] = [4,6] is inside [4,9], we conclude that the whole box [−1,1]&thinsp;×&thinsp;[2,√5] is inside X.

Application
Set inversion is mainly used for path planning, for nonlinear parameter set estimation, for localization or for the characterization of stability domains of linear dynamical systems.