Equality constrained least squares problem: How to minimize the distance from a set of points to a point on a...
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Equality constrained least squares problem: Given is a set of N points Pi and a line in R^3. Find a point P on the line that minimizes Minimizing this sum.
linear-algebra numerical-methods computer-science
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
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Equality constrained least squares problem: Given is a set of N points Pi and a line in R^3. Find a point P on the line that minimizes Minimizing this sum.
linear-algebra numerical-methods computer-science
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
$endgroup$
add a comment |
$begingroup$
Equality constrained least squares problem: Given is a set of N points Pi and a line in R^3. Find a point P on the line that minimizes Minimizing this sum.
linear-algebra numerical-methods computer-science
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
$endgroup$
Equality constrained least squares problem: Given is a set of N points Pi and a line in R^3. Find a point P on the line that minimizes Minimizing this sum.
linear-algebra numerical-methods computer-science
linear-algebra numerical-methods computer-science
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
asked Jan 11 at 19:06
Sebastian RedlSebastian Redl
1
1
add a comment |
add a comment |
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Note that your expression means you're minimizing the sum of the squares of the distances, not the sum of the distances
Without the constraint, the point in R^3 that would minimize the sum is simply the mean of the points, the centroid. When you move away from the centroid, you start to induce a penalty. Expanding squares quickly shows that it doesn't matter which way you deviate from the mean, just by how much. So, "move" your solution onto the line.
That is really say, the solution: compute the mean of your points. Project that to the line (i.e. the point on the line closest to the mean). This is your optimum.
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
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$begingroup$
Note that your expression means you're minimizing the sum of the squares of the distances, not the sum of the distances
Without the constraint, the point in R^3 that would minimize the sum is simply the mean of the points, the centroid. When you move away from the centroid, you start to induce a penalty. Expanding squares quickly shows that it doesn't matter which way you deviate from the mean, just by how much. So, "move" your solution onto the line.
That is really say, the solution: compute the mean of your points. Project that to the line (i.e. the point on the line closest to the mean). This is your optimum.
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
$endgroup$
add a comment |
$begingroup$
Note that your expression means you're minimizing the sum of the squares of the distances, not the sum of the distances
Without the constraint, the point in R^3 that would minimize the sum is simply the mean of the points, the centroid. When you move away from the centroid, you start to induce a penalty. Expanding squares quickly shows that it doesn't matter which way you deviate from the mean, just by how much. So, "move" your solution onto the line.
That is really say, the solution: compute the mean of your points. Project that to the line (i.e. the point on the line closest to the mean). This is your optimum.
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
$endgroup$
add a comment |
$begingroup$
Note that your expression means you're minimizing the sum of the squares of the distances, not the sum of the distances
Without the constraint, the point in R^3 that would minimize the sum is simply the mean of the points, the centroid. When you move away from the centroid, you start to induce a penalty. Expanding squares quickly shows that it doesn't matter which way you deviate from the mean, just by how much. So, "move" your solution onto the line.
That is really say, the solution: compute the mean of your points. Project that to the line (i.e. the point on the line closest to the mean). This is your optimum.
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
$endgroup$
Note that your expression means you're minimizing the sum of the squares of the distances, not the sum of the distances
Without the constraint, the point in R^3 that would minimize the sum is simply the mean of the points, the centroid. When you move away from the centroid, you start to induce a penalty. Expanding squares quickly shows that it doesn't matter which way you deviate from the mean, just by how much. So, "move" your solution onto the line.
That is really say, the solution: compute the mean of your points. Project that to the line (i.e. the point on the line closest to the mean). This is your optimum.
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
answered Jan 11 at 19:13
Alex MeiburgAlex Meiburg
1,820617
1,820617
add a comment |
add a comment |
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