Mixing
Determine if a point is in a hyperellipsoid
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I'm implementing an algorithm for a scientific application that requires determining if a set of points is inside an n-dimensional ellipsoid (hyper-ellipsoid). I have this working for the 2-d and 3-d cases, but I am not sure of how to proceed for cases of 4 or more dimensions.
Can a hyper-ellipsoid of n-dimensions be treated as a set of all the unique 3-d ellipsoid combinations?
This seems like it would not be the case given that two ellipses in 3-dimensions (each sharing a dimension) do not include all the points of the 3-d ellipsoid that covers their boundaries.
For the 3-d case, I am using the criterion:
$$
left( {x over a} right) ^2 + left({y over b}right) ^2 + left({z over c}right) ^2 le 1
$$
for inclusion, where:
- a, b, and c are the half-lengths of the ellipsoid's axes
- x, y, and z are the coordinates of a point
Note: In this implementation, points on the surface are counted as inside the ellipsoid.
geometry
asked Jan 30 at 15:38
FiverFiver
1105
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add a comment |
$begingroup$
I'm implementing an algorithm for a scientific application that requires determining if a set of points is inside an n-dimensional ellipsoid (hyper-ellipsoid). I have this working for the 2-d and 3-d cases, but I am not sure of how to proceed for cases of 4 or more dimensions.
Can a hyper-ellipsoid of n-dimensions be treated as a set of all the unique 3-d ellipsoid combinations?
This seems like it would not be the case given that two ellipses in 3-dimensions (each sharing a dimension) do not include all the points of the 3-d ellipsoid that covers their boundaries.
For the 3-d case, I am using the criterion:
$$
left( {x over a} right) ^2 + left({y over b}right) ^2 + left({z over c}right) ^2 le 1
$$
for inclusion, where:
- a, b, and c are the half-lengths of the ellipsoid's axes
- x, y, and z are the coordinates of a point
Note: In this implementation, points on the surface are counted as inside the ellipsoid.
geometry
asked Jan 30 at 15:38
FiverFiver
1105
$endgroup$
$begingroup$
Is there anything stopping you from using an inequality like $left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1?$
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– Adrian Keister
Jan 30 at 16:06
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@AdrianKeister Thanks, does the pattern of summing these squares hold true for n-dimenstions?
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– Fiver
Jan 30 at 16:19
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Yep! It's what we call Euclidean distance, and it generalizes to $n$ dimensions.
$endgroup$
– Adrian Keister
Jan 30 at 16:21
1
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Excellent, that's easy then. If you add this as an answer, I'll accept. Thanks again!
$endgroup$
– Fiver
Jan 30 at 16:28
add a comment |
$begingroup$
I'm implementing an algorithm for a scientific application that requires determining if a set of points is inside an n-dimensional ellipsoid (hyper-ellipsoid). I have this working for the 2-d and 3-d cases, but I am not sure of how to proceed for cases of 4 or more dimensions.
Can a hyper-ellipsoid of n-dimensions be treated as a set of all the unique 3-d ellipsoid combinations?
This seems like it would not be the case given that two ellipses in 3-dimensions (each sharing a dimension) do not include all the points of the 3-d ellipsoid that covers their boundaries.
For the 3-d case, I am using the criterion:
$$
left( {x over a} right) ^2 + left({y over b}right) ^2 + left({z over c}right) ^2 le 1
$$
for inclusion, where:
- a, b, and c are the half-lengths of the ellipsoid's axes
- x, y, and z are the coordinates of a point
Note: In this implementation, points on the surface are counted as inside the ellipsoid.
geometry
asked Jan 30 at 15:38
FiverFiver
1105
$endgroup$
I'm implementing an algorithm for a scientific application that requires determining if a set of points is inside an n-dimensional ellipsoid (hyper-ellipsoid). I have this working for the 2-d and 3-d cases, but I am not sure of how to proceed for cases of 4 or more dimensions.
Can a hyper-ellipsoid of n-dimensions be treated as a set of all the unique 3-d ellipsoid combinations?
This seems like it would not be the case given that two ellipses in 3-dimensions (each sharing a dimension) do not include all the points of the 3-d ellipsoid that covers their boundaries.
For the 3-d case, I am using the criterion:
$$
left( {x over a} right) ^2 + left({y over b}right) ^2 + left({z over c}right) ^2 le 1
$$
for inclusion, where:
- a, b, and c are the half-lengths of the ellipsoid's axes
- x, y, and z are the coordinates of a point
Note: In this implementation, points on the surface are counted as inside the ellipsoid.
geometry
geometry
asked Jan 30 at 15:38
FiverFiver
1105
asked Jan 30 at 15:38
FiverFiver
1105
edited Jan 30 at 16:18
Fiver
asked Jan 30 at 15:38
FiverFiver
1105
asked Jan 30 at 15:38
FiverFiver
1105
asked Jan 30 at 15:38
FiverFiver
1105
1105
$begingroup$
Is there anything stopping you from using an inequality like $left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1?$
$endgroup$
– Adrian Keister
Jan 30 at 16:06
$begingroup$
@AdrianKeister Thanks, does the pattern of summing these squares hold true for n-dimenstions?
$endgroup$
– Fiver
Jan 30 at 16:19
$begingroup$
Yep! It's what we call Euclidean distance, and it generalizes to $n$ dimensions.
$endgroup$
– Adrian Keister
Jan 30 at 16:21
1
$begingroup$
Excellent, that's easy then. If you add this as an answer, I'll accept. Thanks again!
$endgroup$
– Fiver
Jan 30 at 16:28
add a comment |
$begingroup$
Is there anything stopping you from using an inequality like $left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1?$
$endgroup$
– Adrian Keister
Jan 30 at 16:06
$begingroup$
@AdrianKeister Thanks, does the pattern of summing these squares hold true for n-dimenstions?
$endgroup$
– Fiver
Jan 30 at 16:19
$begingroup$
Yep! It's what we call Euclidean distance, and it generalizes to $n$ dimensions.
$endgroup$
– Adrian Keister
Jan 30 at 16:21
1
$begingroup$
Excellent, that's easy then. If you add this as an answer, I'll accept. Thanks again!
$endgroup$
– Fiver
Jan 30 at 16:28
$begingroup$
Is there anything stopping you from using an inequality like $left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1?$
$endgroup$
– Adrian Keister
Jan 30 at 16:06
$begingroup$
Is there anything stopping you from using an inequality like $left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1?$
$endgroup$
– Adrian Keister
Jan 30 at 16:06
$begingroup$
@AdrianKeister Thanks, does the pattern of summing these squares hold true for n-dimenstions?
$endgroup$
– Fiver
Jan 30 at 16:19
$begingroup$
@AdrianKeister Thanks, does the pattern of summing these squares hold true for n-dimenstions?
$endgroup$
– Fiver
Jan 30 at 16:19
$begingroup$
Yep! It's what we call Euclidean distance, and it generalizes to $n$ dimensions.
$endgroup$
– Adrian Keister
Jan 30 at 16:21
$begingroup$
Yep! It's what we call Euclidean distance, and it generalizes to $n$ dimensions.
$endgroup$
– Adrian Keister
Jan 30 at 16:21
1
1
$begingroup$
Excellent, that's easy then. If you add this as an answer, I'll accept. Thanks again!
$endgroup$
– Fiver
Jan 30 at 16:28
$begingroup$
Excellent, that's easy then. If you add this as an answer, I'll accept. Thanks again!
$endgroup$
– Fiver
Jan 30 at 16:28
add a comment |
1 Answer
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You can use Euclidean distance and generalize the formula you already have:
$$left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1.$$
More dimensions will simply add more terms: variable squared over half-axis-lengths squared.
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
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$begingroup$
You can use Euclidean distance and generalize the formula you already have:
$$left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1.$$
More dimensions will simply add more terms: variable squared over half-axis-lengths squared.
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
add a comment |
$begingroup$
You can use Euclidean distance and generalize the formula you already have:
$$left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1.$$
More dimensions will simply add more terms: variable squared over half-axis-lengths squared.
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
add a comment |
$begingroup$
You can use Euclidean distance and generalize the formula you already have:
$$left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1.$$
More dimensions will simply add more terms: variable squared over half-axis-lengths squared.
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
You can use Euclidean distance and generalize the formula you already have:
$$left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1.$$
More dimensions will simply add more terms: variable squared over half-axis-lengths squared.
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
answered Jan 30 at 16:37
Adrian KeisterAdrian Keister
5,26971933
5,26971933
add a comment |
add a comment |
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$begingroup$
Is there anything stopping you from using an inequality like $left(dfrac{x}{a}right)^{!2}+left(dfrac{y}{b}right)^{!2}+left(dfrac{z}{c}right)^{!2}+left(dfrac{w}{d}right)^{!2}le 1?$
$endgroup$
– Adrian Keister
Jan 30 at 16:06
$begingroup$
@AdrianKeister Thanks, does the pattern of summing these squares hold true for n-dimenstions?
$endgroup$
– Fiver
Jan 30 at 16:19
$begingroup$
Yep! It's what we call Euclidean distance, and it generalizes to $n$ dimensions.
$endgroup$
– Adrian Keister
Jan 30 at 16:21
1
$begingroup$
Excellent, that's easy then. If you add this as an answer, I'll accept. Thanks again!
$endgroup$
– Fiver
Jan 30 at 16:28