Let $S$ be a set of ordered vectors. Find $min_{x in S} | y-x|$. Find closest ordered vector.
$begingroup$
Let
$S={ (x_1,x_2,x_3,) : x_1 le x_2 le x_3 }$
For a given by ${bf y}$ how to solve the following problem
begin{align}
min_{ {bf x} in S} | {bf y}-{bf x}|
end{align}
The question asks what is the closest ordered vector any other vector.
If ${bf y} in S$, then the minimizer is given by ${bf x}={bf y}$. Therefore, the interesting case occurs when ${bf y} in S^c$.
geometry optimization euclidean-geometry
asked Dec 20 '18 at 1:31
LisaLisa
643313
$endgroup$
add a comment |
$begingroup$
Let
$S={ (x_1,x_2,x_3,) : x_1 le x_2 le x_3 }$
For a given by ${bf y}$ how to solve the following problem
begin{align}
min_{ {bf x} in S} | {bf y}-{bf x}|
end{align}
The question asks what is the closest ordered vector any other vector.
If ${bf y} in S$, then the minimizer is given by ${bf x}={bf y}$. Therefore, the interesting case occurs when ${bf y} in S^c$.
geometry optimization euclidean-geometry
asked Dec 20 '18 at 1:31
LisaLisa
643313
$endgroup$
add a comment |
$begingroup$
Let
$S={ (x_1,x_2,x_3,) : x_1 le x_2 le x_3 }$
For a given by ${bf y}$ how to solve the following problem
begin{align}
min_{ {bf x} in S} | {bf y}-{bf x}|
end{align}
The question asks what is the closest ordered vector any other vector.
If ${bf y} in S$, then the minimizer is given by ${bf x}={bf y}$. Therefore, the interesting case occurs when ${bf y} in S^c$.
geometry optimization euclidean-geometry
asked Dec 20 '18 at 1:31
LisaLisa
643313
$endgroup$
Let
$S={ (x_1,x_2,x_3,) : x_1 le x_2 le x_3 }$
For a given by ${bf y}$ how to solve the following problem
begin{align}
min_{ {bf x} in S} | {bf y}-{bf x}|
end{align}
The question asks what is the closest ordered vector any other vector.
If ${bf y} in S$, then the minimizer is given by ${bf x}={bf y}$. Therefore, the interesting case occurs when ${bf y} in S^c$.
geometry optimization euclidean-geometry
geometry optimization euclidean-geometry
asked Dec 20 '18 at 1:31
LisaLisa
643313
asked Dec 20 '18 at 1:31
LisaLisa
643313
asked Dec 20 '18 at 1:31
LisaLisa
643313
asked Dec 20 '18 at 1:31
LisaLisa
643313
asked Dec 20 '18 at 1:31
LisaLisa
643313
643313
add a comment |
add a comment |
1 Answer
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$begingroup$
I think the solution should start with an observation that solution to your problem, for any $mathbf{y}notin S$, should be $mathbf{x}$ with $x_{1}=x_{2}=x_{3}$ (suppose not, then one can improve on the objective of the $min$ problem).
With that, I think you are looking for a solution to $min_{z}sqrt{(y_{1}-z)^{2}+(y_{2}-z)^{2}+(y_{3}-z)^{2}}$, which is probably $(y_{1}+y_{2}+y_{3})/3$?
answered Dec 20 '18 at 1:44
JanJan
3,455518
$endgroup$
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
add a comment |
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1 Answer
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1 Answer
1
active
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$begingroup$
I think the solution should start with an observation that solution to your problem, for any $mathbf{y}notin S$, should be $mathbf{x}$ with $x_{1}=x_{2}=x_{3}$ (suppose not, then one can improve on the objective of the $min$ problem).
With that, I think you are looking for a solution to $min_{z}sqrt{(y_{1}-z)^{2}+(y_{2}-z)^{2}+(y_{3}-z)^{2}}$, which is probably $(y_{1}+y_{2}+y_{3})/3$?
answered Dec 20 '18 at 1:44
JanJan
3,455518
$endgroup$
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
add a comment |
$begingroup$
I think the solution should start with an observation that solution to your problem, for any $mathbf{y}notin S$, should be $mathbf{x}$ with $x_{1}=x_{2}=x_{3}$ (suppose not, then one can improve on the objective of the $min$ problem).
With that, I think you are looking for a solution to $min_{z}sqrt{(y_{1}-z)^{2}+(y_{2}-z)^{2}+(y_{3}-z)^{2}}$, which is probably $(y_{1}+y_{2}+y_{3})/3$?
answered Dec 20 '18 at 1:44
JanJan
3,455518
$endgroup$
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
add a comment |
$begingroup$
I think the solution should start with an observation that solution to your problem, for any $mathbf{y}notin S$, should be $mathbf{x}$ with $x_{1}=x_{2}=x_{3}$ (suppose not, then one can improve on the objective of the $min$ problem).
With that, I think you are looking for a solution to $min_{z}sqrt{(y_{1}-z)^{2}+(y_{2}-z)^{2}+(y_{3}-z)^{2}}$, which is probably $(y_{1}+y_{2}+y_{3})/3$?
answered Dec 20 '18 at 1:44
JanJan
3,455518
$endgroup$
I think the solution should start with an observation that solution to your problem, for any $mathbf{y}notin S$, should be $mathbf{x}$ with $x_{1}=x_{2}=x_{3}$ (suppose not, then one can improve on the objective of the $min$ problem).
With that, I think you are looking for a solution to $min_{z}sqrt{(y_{1}-z)^{2}+(y_{2}-z)^{2}+(y_{3}-z)^{2}}$, which is probably $(y_{1}+y_{2}+y_{3})/3$?
answered Dec 20 '18 at 1:44
JanJan
3,455518
answered Dec 20 '18 at 1:44
JanJan
3,455518
answered Dec 20 '18 at 1:44
JanJan
3,455518
answered Dec 20 '18 at 1:44
JanJan
3,455518
3,455518
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
add a comment |
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Can you elaborate more on why the solution requires $x_1=x_2=x_3$? I don't quite follow that step. Is it because it is a boundary of the set $S$?
$endgroup$
– Lisa
Dec 20 '18 at 1:49
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
$begingroup$
Let me try. Suppose $mathbf{x}=(x_{1},x_{2},x_{3})$ is a solution. Then $mathbf{x}in S$, that is $x_{1}leq x_{2}leq x_{3}$. Suppose, towards a contradiction, that $x_{1}<x_{2}$. Then $mathbf{x}'=(x_{1},x_{2}-epsilon,x_{3})$, for some small $epsilon>0$, improves over $mathbf{x}$ in your problem. Similarly for $x_{2}<x_{3}$.
$endgroup$
– Jan
Dec 20 '18 at 1:52
add a comment |
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