Defining a cost function to find a rotation
$begingroup$
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
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
$endgroup$
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
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
asked Dec 4 '18 at 9:11
user8469759user8469759
1,4311617
1,4311617
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$C_1$ 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
answered Dec 4 '18 at 9:34
user622715user622715
261
$endgroup$
$begingroup$
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
|
show 2 more comments
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
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oldest
votes
$begingroup$
$C_1$ 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
answered Dec 4 '18 at 9:34
user622715user622715
261
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
$C_1$ 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
answered Dec 4 '18 at 9:34
user622715user622715
261
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
$C_1$ 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
answered Dec 4 '18 at 9:34
user622715user622715
261
$endgroup$
$C_1$ 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
answered Dec 4 '18 at 9:34
user622715user622715
261
answered Dec 4 '18 at 9:34
user622715user622715
261
answered Dec 4 '18 at 9:34
user622715user622715
261
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
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