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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<sup>−1</sup>(Y&thinsp;) = {x ∈ R<sup>n</sup> | f(x) ∈ Y&thinsp;}. It can also be viewed as the problem of describing the solution set of the quantified constraint "Y(f(x))", where Y(y) is a constraint, e.g. an inequality, describing the set Y.

In most applications, f is a function from R<sup>n</sup> to R<sup>p</sup> and the set Y is a box of R<sup>p</sup> (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 R<sup>p</sup> made with non-overlapping boxes. For each box [x], we perform the following tests:

  1. if f([x]) ⊂ Y we conclude that [x] ⊂ X;
  2. if f([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<sup>−1</sup>([4,9]) where f(x<sub>1</sub>, x<sub>2</sub>) = x + x is represented on the figure.

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

References