Meaning “unique nearest neighbour” (k-NN)
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I'm doing an exercise about k-NN, k-Neighbor classifier. And I don't understand the following sentence:
Show that for all x ∈ $R^d$ which have a unique nearest neighbor
amongst the points in {x1, . . . , xn} there exists an $h_0 > 0$ such that for all $h < h_0$ the resulting SVM prediction is the same as the prediction made by a Nearest Neighbor (1-NN) classifier.
What is meant by unique nearest neighbor?
I know what k-Neighbour classifier is, but what is the nearest neighbor?
Happy Holidays
machine-learning
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
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add a comment |
$begingroup$
I'm doing an exercise about k-NN, k-Neighbor classifier. And I don't understand the following sentence:
Show that for all x ∈ $R^d$ which have a unique nearest neighbor
amongst the points in {x1, . . . , xn} there exists an $h_0 > 0$ such that for all $h < h_0$ the resulting SVM prediction is the same as the prediction made by a Nearest Neighbor (1-NN) classifier.
What is meant by unique nearest neighbor?
I know what k-Neighbour classifier is, but what is the nearest neighbor?
Happy Holidays
machine-learning
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
$endgroup$
$begingroup$
The nearest neighbor... is the neighbor that is nearest. Seriously, if you know what a $k-$(nearest)-neighbor classifier is, you should know what a nearest neighbor is. I'm nor sure what is your doubt.
$endgroup$
– leonbloy
Dec 25 '18 at 15:39
add a comment |
$begingroup$
I'm doing an exercise about k-NN, k-Neighbor classifier. And I don't understand the following sentence:
Show that for all x ∈ $R^d$ which have a unique nearest neighbor
amongst the points in {x1, . . . , xn} there exists an $h_0 > 0$ such that for all $h < h_0$ the resulting SVM prediction is the same as the prediction made by a Nearest Neighbor (1-NN) classifier.
What is meant by unique nearest neighbor?
I know what k-Neighbour classifier is, but what is the nearest neighbor?
Happy Holidays
machine-learning
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
$endgroup$
I'm doing an exercise about k-NN, k-Neighbor classifier. And I don't understand the following sentence:
Show that for all x ∈ $R^d$ which have a unique nearest neighbor
amongst the points in {x1, . . . , xn} there exists an $h_0 > 0$ such that for all $h < h_0$ the resulting SVM prediction is the same as the prediction made by a Nearest Neighbor (1-NN) classifier.
What is meant by unique nearest neighbor?
I know what k-Neighbour classifier is, but what is the nearest neighbor?
Happy Holidays
machine-learning
machine-learning
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
asked Dec 25 '18 at 15:33
Tommaso BendinelliTommaso Bendinelli
14110
14110
$begingroup$
The nearest neighbor... is the neighbor that is nearest. Seriously, if you know what a $k-$(nearest)-neighbor classifier is, you should know what a nearest neighbor is. I'm nor sure what is your doubt.
$endgroup$
– leonbloy
Dec 25 '18 at 15:39
add a comment |
$begingroup$
The nearest neighbor... is the neighbor that is nearest. Seriously, if you know what a $k-$(nearest)-neighbor classifier is, you should know what a nearest neighbor is. I'm nor sure what is your doubt.
$endgroup$
– leonbloy
Dec 25 '18 at 15:39
$begingroup$
The nearest neighbor... is the neighbor that is nearest. Seriously, if you know what a $k-$(nearest)-neighbor classifier is, you should know what a nearest neighbor is. I'm nor sure what is your doubt.
$endgroup$
– leonbloy
Dec 25 '18 at 15:39
$begingroup$
The nearest neighbor... is the neighbor that is nearest. Seriously, if you know what a $k-$(nearest)-neighbor classifier is, you should know what a nearest neighbor is. I'm nor sure what is your doubt.
$endgroup$
– leonbloy
Dec 25 '18 at 15:39
add a comment |
1 Answer
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Sometimes it is possible to have neighbors that are equidistance.
The question is describing points of which there is exactly one nearest neighbors, those points do not have two neighbors that share the minimum distance from it.
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
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add a comment |
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1 Answer
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$begingroup$
Sometimes it is possible to have neighbors that are equidistance.
The question is describing points of which there is exactly one nearest neighbors, those points do not have two neighbors that share the minimum distance from it.
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Sometimes it is possible to have neighbors that are equidistance.
The question is describing points of which there is exactly one nearest neighbors, those points do not have two neighbors that share the minimum distance from it.
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
add a comment |
$begingroup$
Sometimes it is possible to have neighbors that are equidistance.
The question is describing points of which there is exactly one nearest neighbors, those points do not have two neighbors that share the minimum distance from it.
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
Sometimes it is possible to have neighbors that are equidistance.
The question is describing points of which there is exactly one nearest neighbors, those points do not have two neighbors that share the minimum distance from it.
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
answered Dec 25 '18 at 15:49
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
add a comment |
add a comment |
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$begingroup$
The nearest neighbor... is the neighbor that is nearest. Seriously, if you know what a $k-$(nearest)-neighbor classifier is, you should know what a nearest neighbor is. I'm nor sure what is your doubt.
$endgroup$
– leonbloy
Dec 25 '18 at 15:39