翻訳と辞書
Words near each other
・ "O" Is for Outlaw
・ "O"-Jung.Ban.Hap.
・ "Ode-to-Napoleon" hexachord
・ "Oh Yeah!" Live
・ "Our Contemporary" regional art exhibition (Leningrad, 1975)
・ "P" Is for Peril
・ "Pimpernel" Smith
・ "Polish death camp" controversy
・ "Pro knigi" ("About books")
・ "Prosopa" Greek Television Awards
・ "Pussy Cats" Starring the Walkmen
・ "Q" Is for Quarry
・ "R" Is for Ricochet
・ "R" The King (2016 film)
・ "Rags" Ragland
・ ! (album)
・ ! (disambiguation)
・ !!
・ !!!
・ !!! (album)
・ !!Destroy-Oh-Boy!!
・ !Action Pact!
・ !Arriba! La Pachanga
・ !Hero
・ !Hero (album)
・ !Kung language
・ !Oka Tokat
・ !PAUS3
・ !T.O.O.H.!
・ !Women Art Revolution


Dictionary Lists
翻訳と辞書 辞書検索 [ 開発暫定版 ]
スポンサード リンク

Separating axis theorem : ウィキペディア英語版
Hyperplane separation theorem

In geometry, the hyperplane separation theorem is either of two theorems about disjoint convex sets in ''n''-dimensional Euclidean space. In the first version of the theorem, if both these sets are closed and at least one of them is compact, then there is a hyperplane in between them and even two parallel hyperplanes in between them separated by a gap. In the second version, if both disjoint convex sets are open, then there is a hyperplane in between them, but not necessarily any gap. An axis which is orthogonal to a separating hyperplane is a separating axis, because the orthogonal projections of the convex bodies onto the axis are disjoint.
The hyperplane separation theorem is due to Hermann Minkowski. The Hahn–Banach separation theorem generalizes the result to topological vector spaces.
A related result is the supporting hyperplane theorem. In geometry, a maximum-margin hyperplane is a hyperplane which separates two 'clouds' of points and is at equal distance from the two. The margin between the hyperplane and the clouds is maximal. See the article on Support Vector Machines for more details.
== Statements and proof ==

The proof is based on the following lemma:
Proof of lemma: Let \delta = \inf \. Let x_j be a sequence in ''K'' such that |x_j| \to \delta. Note that (x_i + x_j)/2 is in ''K'' since ''K'' is convex and so |x_i + x_j|^2 \ge 4 \delta^2.
Since
:|x_i - x_j|^2 = 2|x_i|^2 + 2|x_j|^2 - |x_i + x_j|^2 \le 2|x_i|^2 + 2|x_j|^2 - 4\delta^2 \to 0
as i, j \to \infty, x_i is a Cauchy sequence and so has limit ''x'' in ''K''. It is unique since if ''y'' is in ''K'' and has norm δ, then|x - y|^2 \le 2|x|^2 + 2|y|^2 - 4\delta^2 = 0and ''x'' = ''y''. \square
Proof of theorem: Given disjoint nonempty convex sets ''A'', ''B'', let
:K = A + (-B) = \.
Since -B is convex and the sum of convex sets is convex, ''K'' is convex. By the lemma, the closure \overline of ''K'', which is convex, contains a vector ''v'' of minimum norm. Since \overline is convex, for any ''x'' in ''K'', the line segment
:v + t(x - v), \, 0 \le t \le 1
lies in \overline and so
:|v|^2 \le |v + t(x - v)|^2 = |v|^2 + 2 t \langle v, x - v \rangle + t^2|x-v|^2.
For 0 < t \le 1, we thus have:
:0 \le 2 \langle v, x \rangle - 2 |v|^2 + t|x-v|^2
and letting t \to 0 gives: \langle x, v \rangle \ge |v|^2. Hence, for any ''x'' in ''A'' and ''y'' in ''B'', we have: \langle x - y, v \rangle \ge |v|^2. Thus, if ''v'' is nonzero, the proof is complete since
:\inf_ \langle x, v \rangle \ge |v|^2 + \sup_ \langle y, v \rangle.
In general, if the interior of ''K'' is nonempty, then it can be exhausted by compact convex subsets K_n. Since 0 is not in ''K'', each K_n contains a nonzero vector v_n of minimum length and by the argument in the early part, we have: \langle x, v_n \rangle \ge 0 for any x \in K_n. We also normalize them to have length one; then the sequence v_n contains a convergent subsequence with limit ''v'', which has length one and is nonzero. We have \langle x, v \rangle \ge 0 for any ''x'' in the interior of ''K'' and by continuity the same holds for all ''x'' in ''K''. We now finish the proof as before. Finally, if ''K'' has empty interior, the affine set that it spans has dimension less than that of the whole space. Consequently ''K'' is contained in some hyperplane \langle \cdot, v \rangle = c; thus, \langle x, v \rangle \ge c for all ''x'' in ''K'' and we finish the proof as before. \square
The above proof also proves the first version of the theorem mentioned in the lede (to see it, note that ''K'' is closed under the hypothesis below.)
Here, the compactness in the hypothesis cannot be relaxed; see an example in the next section. Also,
This follows from the standard version since the separating hyperplane cannot intersect the interiors of the convex sets.

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「Hyperplane separation theorem」の詳細全文を読む



スポンサード リンク
翻訳と辞書 : 翻訳のためのインターネットリソース

Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.