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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'') = . It can also be viewed as the problem of describing the solution set of the quantified constraint "∃ y . (∧ Y(y) )", where Y(y) is a constraint, for example, an inequality, describing the set Y. In most applications, ''f'' is a function from R''n'' to R''p'' and the set ''Y'' is a box of R''p'' (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 (), we perform the following tests: # if ''f''(()) ⊂ ''Y'' we conclude that () ⊂ ''X''; # if ''f''(()) ∩ ''Y'' = ∅ we conclude that () ∩ ''X'' = ∅; # Otherwise, the box () 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) () 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(()) where ''f''(''x''1, ''x''2) = ''x''12 + ''x''22 is represented on the figure. For instance, since ()2+()2=()+()=() does not intersect the interval (), we conclude that the box ()×() is outside ''X''. Since ()2+()2=()+()=() is inside (), we conclude that the whole box ()×() is inside ''X''. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「set inversion」の詳細全文を読む スポンサード リンク
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