Defining a cost function to find a rotation












0












$begingroup$


Suppose I have pairs of vector $left{(v_i,w_i)right}_{1 leq i leq n}$, and I want to find an angle $theta$ that describes an optimal rotation that aligns all the pairs.



Now two possible cost functions came to my mind the first one is



$$
C_1(theta) = frac{1}{2n} sum_i (theta - theta_i)^2
$$



where $theta_i$ is the angle between $v_i$ and $w_i$.
On the other end however since $leftlangle v_i,w_i rightrangle = leftlVert v_i rightrVert leftlVert w_i rightrVert cos theta_i$



Clearly the two vectors are aligned if $cos theta_i = 1$ therefore I was thinking of minimizing



$$
C_2(theta) = frac{1}{2n} sum_i left(1 - cos left( theta -theta_iright) right)^2
$$



For both functions I'd assume all the angles, including the unknown are in the interval $[-pi,pi]$.



It is clear that if there's a perfect rigid transformations the solution would be the same for both cases, however in general which one is better?



Is there a known similar problem that uses $C_2$ instead of something like $C_1$?



Computationally speaking $C_1$ has a closed form solution



(Assume you have many many pairs $(v_i,w_i)$...)






calculus linear-algebra reference-request optimization







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Suppose I have pairs of vector $left{(v_i,w_i)right}_{1 leq i leq n}$, and I want to find an angle $theta$ that describes an optimal rotation that aligns all the pairs.



    Now two possible cost functions came to my mind the first one is



    $$
    C_1(theta) = frac{1}{2n} sum_i (theta - theta_i)^2
    $$



    where $theta_i$ is the angle between $v_i$ and $w_i$.
    On the other end however since $leftlangle v_i,w_i rightrangle = leftlVert v_i rightrVert leftlVert w_i rightrVert cos theta_i$



    Clearly the two vectors are aligned if $cos theta_i = 1$ therefore I was thinking of minimizing



    $$
    C_2(theta) = frac{1}{2n} sum_i left(1 - cos left( theta -theta_iright) right)^2
    $$



    For both functions I'd assume all the angles, including the unknown are in the interval $[-pi,pi]$.



    It is clear that if there's a perfect rigid transformations the solution would be the same for both cases, however in general which one is better?



    Is there a known similar problem that uses $C_2$ instead of something like $C_1$?



    Computationally speaking $C_1$ has a closed form solution



    (Assume you have many many pairs $(v_i,w_i)$...)






    calculus linear-algebra reference-request optimization







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose I have pairs of vector $left{(v_i,w_i)right}_{1 leq i leq n}$, and I want to find an angle $theta$ that describes an optimal rotation that aligns all the pairs.



      Now two possible cost functions came to my mind the first one is



      $$
      C_1(theta) = frac{1}{2n} sum_i (theta - theta_i)^2
      $$



      where $theta_i$ is the angle between $v_i$ and $w_i$.
      On the other end however since $leftlangle v_i,w_i rightrangle = leftlVert v_i rightrVert leftlVert w_i rightrVert cos theta_i$



      Clearly the two vectors are aligned if $cos theta_i = 1$ therefore I was thinking of minimizing



      $$
      C_2(theta) = frac{1}{2n} sum_i left(1 - cos left( theta -theta_iright) right)^2
      $$



      For both functions I'd assume all the angles, including the unknown are in the interval $[-pi,pi]$.



      It is clear that if there's a perfect rigid transformations the solution would be the same for both cases, however in general which one is better?



      Is there a known similar problem that uses $C_2$ instead of something like $C_1$?



      Computationally speaking $C_1$ has a closed form solution



      (Assume you have many many pairs $(v_i,w_i)$...)






      calculus linear-algebra reference-request optimization







      share|cite|improve this question









      $endgroup$




      Suppose I have pairs of vector $left{(v_i,w_i)right}_{1 leq i leq n}$, and I want to find an angle $theta$ that describes an optimal rotation that aligns all the pairs.



      Now two possible cost functions came to my mind the first one is



      $$
      C_1(theta) = frac{1}{2n} sum_i (theta - theta_i)^2
      $$



      where $theta_i$ is the angle between $v_i$ and $w_i$.
      On the other end however since $leftlangle v_i,w_i rightrangle = leftlVert v_i rightrVert leftlVert w_i rightrVert cos theta_i$



      Clearly the two vectors are aligned if $cos theta_i = 1$ therefore I was thinking of minimizing



      $$
      C_2(theta) = frac{1}{2n} sum_i left(1 - cos left( theta -theta_iright) right)^2
      $$



      For both functions I'd assume all the angles, including the unknown are in the interval $[-pi,pi]$.



      It is clear that if there's a perfect rigid transformations the solution would be the same for both cases, however in general which one is better?



      Is there a known similar problem that uses $C_2$ instead of something like $C_1$?



      Computationally speaking $C_1$ has a closed form solution



      (Assume you have many many pairs $(v_i,w_i)$...)






      calculus linear-algebra reference-request optimization




      calculus linear-algebra reference-request optimization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 9:11









      user8469759user8469759

      1,5381618




      1,5381618






















          1 Answer
          1






          active

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          1












          $begingroup$

          $C_1$ has a closed form if you know the angles $theta_i$ ,your vectors are in $2d$and don't have issues with $2pi$ type ambiguities. As for C_$2$ which is more standard, try this






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:39










          • $begingroup$
            By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:44










          • $begingroup$
            treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
            $endgroup$
            – user622715
            Dec 4 '18 at 9:55












          • $begingroup$
            Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
            $endgroup$
            – user8469759
            Dec 4 '18 at 10:24












          • $begingroup$
            $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
            $endgroup$
            – user622715
            Dec 4 '18 at 11:11













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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          $C_1$ has a closed form if you know the angles $theta_i$ ,your vectors are in $2d$and don't have issues with $2pi$ type ambiguities. As for C_$2$ which is more standard, try this






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:39










          • $begingroup$
            By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:44










          • $begingroup$
            treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
            $endgroup$
            – user622715
            Dec 4 '18 at 9:55












          • $begingroup$
            Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
            $endgroup$
            – user8469759
            Dec 4 '18 at 10:24












          • $begingroup$
            $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
            $endgroup$
            – user622715
            Dec 4 '18 at 11:11


















          1












          $begingroup$

          $C_1$ has a closed form if you know the angles $theta_i$ ,your vectors are in $2d$and don't have issues with $2pi$ type ambiguities. As for C_$2$ which is more standard, try this






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:39










          • $begingroup$
            By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:44










          • $begingroup$
            treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
            $endgroup$
            – user622715
            Dec 4 '18 at 9:55












          • $begingroup$
            Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
            $endgroup$
            – user8469759
            Dec 4 '18 at 10:24












          • $begingroup$
            $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
            $endgroup$
            – user622715
            Dec 4 '18 at 11:11
















          1












          1








          1





          $begingroup$

          $C_1$ has a closed form if you know the angles $theta_i$ ,your vectors are in $2d$and don't have issues with $2pi$ type ambiguities. As for C_$2$ which is more standard, try this






          share|cite|improve this answer









          $endgroup$



          $C_1$ has a closed form if you know the angles $theta_i$ ,your vectors are in $2d$and don't have issues with $2pi$ type ambiguities. As for C_$2$ which is more standard, try this







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 9:34









          user622715user622715

          261




          261












          • $begingroup$
            Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:39










          • $begingroup$
            By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:44










          • $begingroup$
            treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
            $endgroup$
            – user622715
            Dec 4 '18 at 9:55












          • $begingroup$
            Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
            $endgroup$
            – user8469759
            Dec 4 '18 at 10:24












          • $begingroup$
            $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
            $endgroup$
            – user622715
            Dec 4 '18 at 11:11




















          • $begingroup$
            Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:39










          • $begingroup$
            By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
            $endgroup$
            – user8469759
            Dec 4 '18 at 9:44










          • $begingroup$
            treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
            $endgroup$
            – user622715
            Dec 4 '18 at 9:55












          • $begingroup$
            Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
            $endgroup$
            – user8469759
            Dec 4 '18 at 10:24












          • $begingroup$
            $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
            $endgroup$
            – user622715
            Dec 4 '18 at 11:11


















          $begingroup$
          Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
          $endgroup$
          – user8469759
          Dec 4 '18 at 9:39




          $begingroup$
          Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison?
          $endgroup$
          – user8469759
          Dec 4 '18 at 9:39












          $begingroup$
          By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
          $endgroup$
          – user8469759
          Dec 4 '18 at 9:44




          $begingroup$
          By the way, in theory in my case $theta_i$'s are known in my problem, I can easily retrieve them.
          $endgroup$
          – user8469759
          Dec 4 '18 at 9:44












          $begingroup$
          treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
          $endgroup$
          – user622715
          Dec 4 '18 at 9:55






          $begingroup$
          treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $Omega$ on the $v$'s now becomes $Omega { v_1,cdots v_n}=Omega A$ and the square of all the distances $sum |Omega v_i-u_i|^2$ is $|Omega A-B|^2$
          $endgroup$
          – user622715
          Dec 4 '18 at 9:55














          $begingroup$
          Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
          $endgroup$
          – user8469759
          Dec 4 '18 at 10:24






          $begingroup$
          Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $sum |Omega v_i - u_i|^2$?
          $endgroup$
          – user8469759
          Dec 4 '18 at 10:24














          $begingroup$
          $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
          $endgroup$
          – user622715
          Dec 4 '18 at 11:11






          $begingroup$
          $v_i cdot u_i= |v||u|costheta_i$ so $Omega v_i cdot u_i=|v||u|cos(theta_i -theta_Omega)$
          $endgroup$
          – user622715
          Dec 4 '18 at 11:11




















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