Set inversion
Mathematical problem of finding the set mapped by a specified function to a certain range From Wikipedia, the free encyclopedia
In mathematics, set inversion is the problem of characterizing the preimage X of a set Y by a function f, i.e., X = f −1(Y ) = {x ∈ Rn | f(x) ∈ Y }. 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 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[1] using interval analysis combined with a branch-and-bound algorithm.[2]
The main idea consists in building a paving of Rp made with non-overlapping boxes. For each box [x], we perform the following tests:
- if f ([x]) ⊂ Y we conclude that [x] ⊂ X;
- if f ([x]) ∩ Y = ∅ we conclude that [x] ∩ X = ∅;
- 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 ] 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 −1([4,9]) where f (x1, x2) = x2
1 + x2
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] × [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] × [2,√5] is inside X.

Application
Set inversion is mainly used for path planning, for nonlinear parameter set estimation,[3][4] for localization[5][6] or for the characterization of stability domains of linear dynamical systems.[7]
References
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