Encoding the position information in real distance metric
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The Euclidean distance doesn't preserve the exact position information. For example, the distance of the points (3,1) and (1,3) would be the same from the origin. Is there any distance metric which can encode the position information in a single real value, meaning the distance would be different for each point with respect to origin?
geometric-invariant-theory mahalanobis-distance hausdorff-distance
asked Dec 11 '18 at 10:52
anonanon
95110
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$begingroup$
The Euclidean distance doesn't preserve the exact position information. For example, the distance of the points (3,1) and (1,3) would be the same from the origin. Is there any distance metric which can encode the position information in a single real value, meaning the distance would be different for each point with respect to origin?
geometric-invariant-theory mahalanobis-distance hausdorff-distance
asked Dec 11 '18 at 10:52
anonanon
95110
$endgroup$
add a comment |
$begingroup$
The Euclidean distance doesn't preserve the exact position information. For example, the distance of the points (3,1) and (1,3) would be the same from the origin. Is there any distance metric which can encode the position information in a single real value, meaning the distance would be different for each point with respect to origin?
geometric-invariant-theory mahalanobis-distance hausdorff-distance
asked Dec 11 '18 at 10:52
anonanon
95110
$endgroup$
The Euclidean distance doesn't preserve the exact position information. For example, the distance of the points (3,1) and (1,3) would be the same from the origin. Is there any distance metric which can encode the position information in a single real value, meaning the distance would be different for each point with respect to origin?
geometric-invariant-theory mahalanobis-distance hausdorff-distance
geometric-invariant-theory mahalanobis-distance hausdorff-distance
asked Dec 11 '18 at 10:52
anonanon
95110
asked Dec 11 '18 at 10:52
anonanon
95110
asked Dec 11 '18 at 10:52
anonanon
95110
asked Dec 11 '18 at 10:52
anonanon
95110
asked Dec 11 '18 at 10:52
anonanon
95110
95110
add a comment |
add a comment |
1 Answer
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Yes.
Pick your favourite bijection (or at least injection) $fcolon Bbb R^2to [0,infty)$ such that $fbigl((0,0)bigr)=0$ and declare $d(a,b)=|f(a)-f(b)|$.
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes.
Pick your favourite bijection (or at least injection) $fcolon Bbb R^2to [0,infty)$ such that $fbigl((0,0)bigr)=0$ and declare $d(a,b)=|f(a)-f(b)|$.
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
$endgroup$
add a comment |
$begingroup$
Yes.
Pick your favourite bijection (or at least injection) $fcolon Bbb R^2to [0,infty)$ such that $fbigl((0,0)bigr)=0$ and declare $d(a,b)=|f(a)-f(b)|$.
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
$endgroup$
add a comment |
$begingroup$
Yes.
Pick your favourite bijection (or at least injection) $fcolon Bbb R^2to [0,infty)$ such that $fbigl((0,0)bigr)=0$ and declare $d(a,b)=|f(a)-f(b)|$.
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
$endgroup$
Yes.
Pick your favourite bijection (or at least injection) $fcolon Bbb R^2to [0,infty)$ such that $fbigl((0,0)bigr)=0$ and declare $d(a,b)=|f(a)-f(b)|$.
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
answered Dec 11 '18 at 12:09
Hagen von EitzenHagen von Eitzen
280k23272505
280k23272505
add a comment |
add a comment |
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