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find invariant coordination | find invariant coordination | ||
<math>\varphi : \Re ^k \rightarrow \Re ^n </math> --[[User:Han47|Han47]] 10:41, 10 March 2008 (EDT) | <math>\varphi : \Re ^k \rightarrow \Re ^n </math> --[[User:Han47|Han47]] 10:41, 10 March 2008 (EDT) | ||
− | such that <math>\varphi (x) = \varphi (\bar x) </math> which are related by a rotation & translation | + | such that <math>\varphi (x) = \varphi (\bar x) </math> for all <math>x, \bar x</math> which are related by a rotation & translation |
+ | |||
+ | Do not trivialize! | ||
+ | |||
+ | e.g.) <math>\varphi (x) =0 </math> gives us invariant coordinate but lose separation | ||
+ | |||
+ | Want <math>\varphi (x) = \varphi (\bar x) </math> | ||
+ | <math>\Leftrightarrow x, \bar x</math> are related by a rotation and translation | ||
+ | |||
+ | Example |
Revision as of 10:44, 10 March 2008
Nearest Neighbors Clarification Rule(Alternative Approaches) --Han47 10:34, 10 March 2008 (EDT)
Alternative Approach
find invariant coordination $ \varphi : \Re ^k \rightarrow \Re ^n $ --Han47 10:41, 10 March 2008 (EDT) such that $ \varphi (x) = \varphi (\bar x) $ for all $ x, \bar x $ which are related by a rotation & translation
Do not trivialize!
e.g.) $ \varphi (x) =0 $ gives us invariant coordinate but lose separation
Want $ \varphi (x) = \varphi (\bar x) $ $ \Leftrightarrow x, \bar x $ are related by a rotation and translation
Example