Let $S$ be a set of ordered vectors. Find $min_{x in S} | y-x|$. Find closest ordered vector.












1












$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 20 '18 at 1:31









      LisaLisa

      643313




      643313






















          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 $(y_{1}+y_{2}+y_{3})/3$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          1 Answer
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          0












          $begingroup$

          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 $(y_{1}+y_{2}+y_{3})/3$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          0












          $begingroup$

          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 $(y_{1}+y_{2}+y_{3})/3$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          0












          0








          0





          $begingroup$

          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 $(y_{1}+y_{2}+y_{3})/3$?






          share|cite|improve this answer









          $endgroup$



          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 $(y_{1}+y_{2}+y_{3})/3$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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