Related
We are trying an alteration optimization strategy to solve Lyapunov problems.
We break down our decision variables into two sets, Set 1 and Set 2.
We were perplexed how it was possible that, after getting a solution to Set 1, and plugging in those solved variables into the optimization over Set 2, the transferred variables would not be feasible.
The constraints that fail are those due to the SOS coefficient matching equality constraints.
Here, we print in each row the constraint that failed, and the value of our Initial Guess. We can see that the Initial Guess is off only a very small amount compared to the constraints.
LinearEqualityConstraint
(2 * Symmetric(97,40) + 2 * Symmetric(96,41)) == 9.50028
[9.50027496]
LinearEqualityConstraint
(2 * Symmetric(97,47) + 2 * Symmetric(96,48)) == 234.465
[234.4647013]
LinearEqualityConstraint
(2 * Symmetric(97,54) + 2 * Symmetric(96,55)) == -234.463
[-234.46336504]
LinearEqualityConstraint
(2 * Symmetric(97,61) + 2 * Symmetric(96,62)) == 12.7962
[12.79618825]
LinearEqualityConstraint
(2 * Symmetric(97,68) + 2 * Symmetric(96,69)) == -12.7964
[-12.79637068]
LinearEqualityConstraint
(2 * Symmetric(97,75) + 2 * Symmetric(96,76)) == -51.4061
[-51.40605828]
LinearEqualityConstraint
(2 * Symmetric(97,81) + 2 * Symmetric(96,82)) == 51.406
[51.40604213]
LinearEqualityConstraint
(2 * Symmetric(97,86) + 2 * Symmetric(96,87)) == 192.794
[192.79430158]
LinearEqualityConstraint
(2 * Symmetric(97,90) + 2 * Symmetric(96,91)) == -141.924
[-141.92366183]
LinearEqualityConstraint
(2 * Symmetric(97,93) + 2 * Symmetric(96,94)) == -37.6674
[-37.66740401]
InitialGuess V_sos and
Our guess for what's happening is:
When you extract the solution from one optimization using result.GetSolution(var), you lose some precision.
Or, when you set the previous solution using prog.SetInitialGuess(np_array) you lose some precision.
What's the solution here? Should we just keep feeding the solution back in even though it says infeasible?
This is a partial cookbook when I debug SOS problem, especially when working with Lyapunov problems:
Choose the right monomial basis
The main idea is to remove the 0-th order monomial 1 from the monomial basis of the sos polynomial. Here is a quick explanation:
The mathematical problem is
Find λ(x)
−Vdot − λ(x) * (ρ − V) is sos
λ(x) is sos
Namely you want to prove that V≤ρ ⇒ Vdot ≤ 0
So first I would suggest to re-write your dynamics to make sure that 0 is the goal state (you can always shift your state).
Second you can see that since x=0 is the equilibrium point, then both V(0) = 0 and Vdot(0) = 0 (Because x=0 is the global minimal of V(x), hence ∂V/∂x=0 at x=0, indicating Vdot(0) = 0), now your sos polynomial p(x) = −Vdot − λ(x) * (ρ − V) must satisfy p(0) = -λ(0) * ρ. But λ(x) >= 0 and ρ > 0, so we know λ(0) = 0.
Lemma
If a sos polynomial s(x) satisfies s(0) = 0, then its monomial basis cannot contain the 0-th order monomial (namely 1).
Proof
Remember that s(x) is a sos polynomial, namely
s(x) = m(x)ᵀQm(x)
where m(x) contains the monomial basis, and Q is a psd matrix. Now let's decompose the monomial basis m(x) into two parts, the 0-th order monomial 1 and the remaining monomials mbar(x). For example, if m(x) = [x1, x2, 1], then mbar(x) = [x1, x2]. We also decompose the psd matrix Q accordingly
s(x) = [mbar(x)]ᵀ [Q11 Q10] [mbar(x)]
[ 1] [Q10 Q00] [ 1]
Since s(0) = Q00 = 0, we also know that Q10 = 0, so now we can use a smaller psd matrix Q11 rather than Q. Equivalently we write s(x) = mbar(x)ᵀ * Q11 * mbar(x), where mbar(x) is the monomial basis that doesn't contain the 0-th order monomial, QED.
So why removing the 0-th order monomial from the monomial basis and use the smaller psd matrix Q11 is a good idea when your sos polynomial s(x) satisfies s(0) = 0? The reason is that if the monomial basis contains 1, then your psd matrix Q has to be on the boundary of the psd cone, namely your SDP problem doesn't have a strict interior. This could leads to violation of Slater's condition, which also breaks the strong duality. One example is that if your s(x) = x², by including the 0'th order monomial, it is written as
x² = [x] [1 0] [x]
[1] [0 0] [1]
And you see that the Gram matrix [[1 0], [0, 0]] is on the boundary of the psd cone (with one eigen value equal to 0). But if you remove 1 from the monomial basis, then its Gram matrix is just Q11=1, strictly in the interior of the psd cone.
In Drake, after removing 1 from the monomial basis, you can create your sos polynomial λ(x) as
lambda_poly, lambda_gram = prog.NewSosPolynomial(monomial_basis)
whee monomial_basis doesn't contain the 0-th order monomial.
Backoff during bilinear alternation
This is a typical problem in bilinear alternation. The issue is that when you solve a conic optimization problem with an objective function, the optimal solution always occurs at the boundary of the cone, namely it is very close to being infeasible. Then when you fix some variables to this solution at the cone boundary, in the next iteration the problem is very likely infeasible due to numerical roundoff error.
A typical solution is that after solving the optimization problem on variable Set 1 with an objective, now "backoff" a little bit by solving a feasibility problem on variable Set 1. This new solution is often strictly feasible (namely it is inside the strict interior of the cone), now pass this strictly feasible solution Set 1 to the next iteration and search for Set 2.
More concretely, suppose at one iteration you solve the following optimization problem
min c'*x
s.t constraint_on_x
and denote the optimal cost as p. Now solve a new feasibility problem
find x
s.t c'*x <= p + epsilon
constraint_on_x
where epsilon can be a small positive number. This new solution will be used in the next iteration to search for a different set of variables.
You can check if your solution is on the boundary of the positive semidefinite cone by checking the Eigen value of your psd matrix. Here is the pseudo-code
for binding : prog.positive_semidefinite_constraints():
psd_sol = result.GetSolution(binding.variables())
psd_sol.reshape((binding.evaluator().matrix_rows(), binding.evaluator().matrix_rows()))
print(f"minimal eigenvalue {np.linalg.eig(psd_sol)[0].min()}")
You should see that before doing this "backoff" some of the minimal eigen value is almost 0. After "backoff" the minimal eigen value gets larger.
I understand well perceptron so put accent only on kernel but I am not familiar with matemathic expressions so please give me an numerical example and a guide on kernel.
For example:
My hyperplane of perceptron is x1*w1+x2*w2+x3*w3+b=0; The RBF kernel formula: k(x,z) = exp((-|x-z|^2)/2*variance^2) where takes action the radial basis function kernel here. Is x an input and what is z variable here?
Or of what I have to calculate variance if it is variance in the formula?
Somewhere I have understood so that I have to plug this formula in perceptron decision function x1*w1+x2*w2+x3*w3+b=0; but how does it look look like If I plug in?
I would like to ask a numerical example to avoid confusion.
Linear Perceptron
As you know linear perceptrons can be trained for binary classification. More precisely, if there is n features, x1, x2, ..., xn in n-dimensional space, Rn, and you want to label them in 2 categories, y1 & y2 (usually -1 and +1), you can use linear perceptron which defines a hyperplane w1*x1 + ... + wn*xn + b = 0 to do so.
w1*x1 + ... + wn*xn + b > 0 or W.X + b > 0 ==> class = y1
w1*x1 + ... + wn*xn + b < 0 or W.X + b < 0 ==> class = y2
Linear perceptron will work well, only if the problem is linearly separable in Rn. For example, in 2D space, this means that one line can separate the 2 sets of points.
Algorithm
One common algorithm to train the perceptron, i.e., find weights and bias, w's & b, based on N data points, X1, ..., XN, and their labels, Y1, ..., YN is the following:
Initialize: W = zeros(n,1); b = 0
For i=1 to N:
Calculate F(Xi) = W.Xi + b
If F(Xi)*Yi <= 0:
W <--- W + Xi*Yi
b <--- b + Yi
This will give the final value for W & b. Besides, based on the training, W will be a linear combination of training points, Xi's, more precisely, the ones that were misclassified. So W = a1*X1 + ... + ...aN*XN where a's are in {0,y1,y2}.
Now, if there is a new point, let's say Z, to label, we check the sign of F(Z) = W.Z + b = a1*(X1.Z) + ... + aN*(XN.Z) + b. It is interesting that only the inner product of new point and training points take part in it.
Kernel Perceptron
Now, if the problem is not linearly separable, one may try to go to a higher dimensional space in which a hyperplane can do the classification. As an example, consider a circle in 2D space. The points inside and outside of the circle can't be separated by a line. However, if you find a transformation that can take the points to 3D space such that the first 2 coordinates remain the same for all points, and the 3rd coordinate become +1 and -1 for the points inside and outside of the circle respectively, then a plane defined as 3rd coordinate = 0 can separate the points.
Finding such transformations can be difficult and computationally heavy, so the kernel trick is introduced. Notice that we only used the inner product of new points with the training points. Kernel trick employs this fact and defines the inner product of the transformed points without actually finding the transformation.
If the unknown transformation is P(X) then Kernel function will be:
K(Xi,Xj) = <P(Xi),P(Xj)>. So instead of finding P, kernel functions are defined which represent the scalar result of the inner product in high-dimensional space. There are also theorems about what functions can be kernel functions, i.e., correspond to inner product in another space.
After choosing a kernel function, the algorithm will be modified as follows:
Initialize: F(X) = 0
For i=1 to N:
Calculate F(Xi)
If F(Xi)*Yi <= 0:
F(.) <--- F(.) + K(.,Xi)*Yi + Yi
At the end, F(.) = a1*K(.,X1) + ... + ...aN*K(.,XN) + b where a's are in {0,y1,y2}.
RBF Kernel
Radial basis function is one type of kernel function that is actually computing the inner product in an infinite-dimensional space. It can be written as
K(Xi,Xj) = exp(- norm2(Xi-Xj)^2 / (2*sigma^2))
Sigma is some parameter that you can work with to find an optimum value for. For example, you can train the model with different values of sigma and then find the best value based on the performance. You can start with sigma = 1
After training the model to find F(.), for a new data Z, the sign of F(Z) = a1*K(Z,X1) + ... + ...aN*K(Z,XN) + b will determine the class.
Remarks:
Regarding to your question about variance, you don't need to find any variance.
About x and z in your question, in each iteration, you should find the kernel output for the current data point and all the previously added points (the points that were misclassified and hence were added to F).
I couldn't come up with a simple instructive numerical example.
References:
I borrowed some notation from
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjVu-fXo8DOAhVDxCYKHQkcDDAQFggoMAE&url=http%3A%2F%2Falex.smola.org%2Fteaching%2Fpune2007%2Fpune_3.pdf&usg=AFQjCNHlxy9TnY8xNe2-QDERipN_GycSqQ&bvm=bv.129422649,d.eWE
I am trying to extract Rotation matrix and Translation vector from the essential matrix.
<pre><code>
SVD svd(E,SVD::MODIFY_A);
Mat svd_u = svd.u;
Mat svd_vt = svd.vt;
Mat svd_w = svd.w;
Matx33d W(0,-1,0,
1,0,0,
0,0,1);
Mat_<double> R = svd_u * Mat(W).t() * svd_vt; //or svd_u * Mat(W) * svd_vt;
Mat_<double> t = svd_u.col(2); //or -svd_u.col(2)
</code></pre>
However, when I am using R and T (e.g. to obtain rectified images), the result does not seem to be right(black images or some obviously wrong outputs), even so I used different combination of possible R and T.
I suspected to E. According to the text books, my calculation is right if we have:
E = U*diag(1, 1, 0)*Vt
In my case svd.w which is supposed to be diag(1, 1, 0) [at least in term of a scale], is not so. Here is an example of my output:
svd.w = [21.47903827647813; 20.28555196246256; 5.167099204708699e-010]
Also, two of the eigenvalues of E should be equal and the third one should be zero. In the same case the result is:
eigenvalues of E = 0.0000 + 0.0000i, 0.3143 +20.8610i, 0.3143 -20.8610i
As you see, two of them are complex conjugates.
Now, the questions are:
Is the decomposition of E and calculation of R and T done in a right way?
If the calculation is right, why the internal rules of essential matrix are not satisfied by the results?
If everything about E, R, and T is fine, why the rectified images obtained by them are not correct?
I get E from fundamental matrix, which I suppose to be right. I draw epipolar lines on both the left and right images and they all pass through the related points (for all the 16 points used to calculate the fundamental matrix).
Any help would be appreciated.
Thanks!
I see two issues.
First, discounting the negligible value of the third diagonal term, your E is about 6% off the ideal one: err_percent = (21.48 - 20.29) / 20.29 * 100 . Sounds small, but translated in terms of pixel error it may be an altogether larger amount.
So I'd start by replacing E with the ideal one after SVD decomposition: Er = U * diag(1,1,0) * Vt.
Second, the textbook decomposition admits 4 solutions, only one of which is physically plausible (i.e. with 3D points in front of the camera). You may be hitting one of non-physical ones. See http://en.wikipedia.org/wiki/Essential_matrix#Determining_R_and_t_from_E .
If I have three points that create an angle, what would be the best way to determine if a fourth point resides within the angle created by the previous three?
Currently, I determine the angle of the line to all three points from the origin point, and then check to see if the test angle is in between the two other angles but I'm trying to figure out if there's a better way to do it. The function is run tens of thousands of times an update and I'm hoping that there's a better way to achieve what I'm trying to do.
Let's say you have angle DEF (E is the "pointy" part), ED is the left ray and EF is the right ray.
* D (Dx, Dy)
/
/ * P (Px, Py)
/
/
*---------------*
E (Ex, Ey) F (Fx, Fy)
Step 1. Build line equation for line ED in the classic Al * x + Bl * y + Cl = 0 form, i.e. simply calculate
Al = Dy - Ey // l - for "left"
Bl = -(Dx - Ex)
Cl = -(Al * Ex + Bl * Ey)
(Pay attention to the subtraction order.)
Step 2. Build line equation for line FE (reversed direction) in the classic Ar * x + Br * y + Cr = 0 form, i.e. simply calculate
Ar = Ey - Fy // r - for "right"
Br = -(Ex - Fx)
Cr = -(Ar * Ex + Br * Ey)
(Pay attention to the subtraction order.)
Step 3. For your test point P calculate the expressions
Sl = Al * Px + Bl * Py + Cl
Sr = Ar * Px + Br * Py + Cr
Your point lies inside the angle if and only if both Sl and Sr are positive. If one of them is positive and other is zero, your point lies on the corresponding side ray.
That's it.
Note 1: For this method to work correctly, it is important to make sure that the left and right rays of the angle are indeed left and right rays. I.e. if you think about ED and EF as clock hands, the direction from D to F should be clockwise. If it is not guaranteed to be the case for your input, then some adjustments are necessary. For example, it can be done as an additional step of the algorithm, inserted between steps 2 and 3
Step 2.5. Calculate the value of Al * Fx + Bl * Fy + Cl. If this value is negative, invert signs of all ABC coefficients:
Al = -Al, Bl = -Bl, Cl = -Cl
Ar = -Ar, Br = -Br, Cr = -Cr
Note 2: The above calculations are made under assumption that we are working in a coordinate system with X axis pointing to the right and Y axis pointing to the top. If one of your coordinate axes is flipped, you have to invert the signs of all six ABC coefficients. Note, BTW, that if you perform the test described in step 2.5 above, it will take care of everything automatically. If you are not performing step 2.5 then you have to take the axis direction into account from the very beginning.
As you can see, this a precise integer method (no floating point calculations, no divisions). The price of that is danger of overflows. Use appropriately sized types for multiplications.
This method has no special cases with regard to line orientations or the value of the actual non-reflex angle: it work immediately for acute, obtuse, zero and straight angle. It can be easily used with reflex angles (just perform a complementary test).
P.S. The four possible combinations of +/- signs for Sl and Sr correspond to four sectors, into which the plane is divided by lines ED and EF.
* D
/
(-,+) / (+,+)
/
-------*------------* F
/ E
(-,-) / (+,-)
/
By using this method you can perform the full "which sector the point falls into" test. For an angle smaller than 180 you just happen to be interested in only one of those sectors: (+, +). If at some point you'll need to adapt this method for reflex angles as well (angles greater than 180), you will have to test for three sectors instead of one: (+,+), (-,+), (+,-).
Describe your origin point O, and the other 2 points A and B then your angle is given as AOB. Now consider your test point and call that C as in the diagram.
Now consider that we can get a vector equation of C by taking some multiple of vector OA and some multiple of OB. Explicitly
C = K1 x OA + K2 OB
for some K1,K2 that we need to calculate. Set O to the origin by subtracting it (vectorially) from all other points. If coordinates of A are (a1,a2), B = (b1,b2) and C = (c1,c2) we have in matrix terms
[ a1 b1 ] [ K1 ] = [ c1 ]
[ a2 b2 ] [ K2 ] = [ c2 ]
So we can solve for K1 and K2 using the inverse of the matrix to give
1 / (a1b2 - b1a2) [ b2 -b1 ] [ c1 ] = [ K1 ]
[ -a2 a1 ] [ c2 ] = [ K2 ]
which reduces to
K1 = (b2c1 - b1c2)/(a1b2 - b1a2)
K2 = (-a2c1 + a1c2)/(a1b2 - b1a2)
Now IF the point C lies within your angle, the multiples of the vectors OA and OB will BOTH be positive. If C lies 'under' OB, then we need a negative amount of OA to get to it similarly for the other direction. So your condition is satisfied when both K1 and K2 are greater than (or equal to) zero. You must take care in the case where a1b2 = b1a2 as this corresponds to a singular matrix and division by zero. Geometrically it means that OA and OB are parallel and hence there is no solution. The algebra above probably needs verifying for any slight typo mistake but the methodology is correct. Maybe long winded but you can get it all simply from point coordinates and saves you calculating inverse trig functions to get angles.
The above applies to angles < 180 degrees, so if the your angle is greater than 180 degrees, you should check instead for
!(K1 >= 0 && K2 >= 0)
as this is exterior to the segment less than 180 degree. Remember that for 0 and 180 degrees you will have a divide by zero error which must be checked for (ensure a1b2 - b1a2 != 0 )
Yes, I meant the smallest angle in my comment above. Look at this thread for an extensive discussion on cheap ways to find the measure of the angle between two vectors. I have used the lookup-table approach on many occasions with great success.
Triangle O B C has to be positive oriented and also triangle O C A. To calaculate orientation, just use Shoelace formula. Both values has to be positive.
I'm trying to develop an application using SOM in analyzing data. However, after finishing training, I cannot find a way to visualize the result. I know that U-Matrix is one of the method but I cannot understand it properly. Hence, I'm asking for a specific and detail example how to construct U-Matrix.
I also read an answer at U-matrix and self organizing maps but it only refers to 1 row map, how about 3x3 map? I know that for 3x3 map:
m(1) m(2) m(3)
m(4) m(5) m(6)
m(7) m(8) m(9)
a 5x5 matrix must me created:
u(1) u(1,2) u(2) u(2,3) u(3)
u(1,4) u(1,2,4,5) u(2,5) u(2,3,5,6) u(3,6)
u(4) u(4,5) u(5) u(5,6) u(6)
u(4,7) u(4,5,7,8) u(5,8) u(5,6,8,9) u(6,9)
u(7) u(7,8) u(8) u(8,9) u(9)
but I don't know how to calculate u-weight u(1,2,4,5), u(2,3,5,6), u(4,5,7,8) and u(5,6,8,9).
Finally, after constructing U-Matrix, is there any way to visualize it using color, e.g. heat map?
Thank you very much for your time.
Cheers
I don't know if you are still interested in this but I found this link
http://www.uni-marburg.de/fb12/datenbionik/pdf/pubs/1990/UltschSiemon90
which explains very speciffically how to calculate the U-matrix.
Hope it helps.
By the way, the site were I found the link has several resources referring to SOMs I leave it here in case anyone is interested:
http://www.ifs.tuwien.ac.at/dm/somtoolbox/visualisations.html
The essential idea of a Kohonen map is that the data points are mapped to a
lattice, which is often a 2D rectangular grid.
In the simplest implementations, the lattice is initialized by creating a 3D
array with these dimensions:
width * height * number_features
This is the U-matrix.
Width and height are chosen by the user; number_features is just the number
of features (columns or fields) in your data.
Intuitively this is just creating a 2D grid of dimensions w * h
(e.g., if w = 10 and h = 10 then your lattice has 100 cells), then
into each cell, placing a random 1D array (sometimes called "reference tuples")
whose size and values are constrained by your data.
The reference tuples are also referred to as weights.
How is the U-matrix rendered?
In my example below, the data is comprised of rgb tuples, so the reference tuples
have length of three and each of the three values must lie between 0 and 255).
It's with this 3D array ("lattice") that you begin the main iterative loop
The algorithm iteratively positions each data point so that it is closest to others similar to it.
If you plot it over time (iteration number) then you can visualize cluster
formation.
The plotting tool i use for this is the brilliant Python library, Matplotlib,
which plots the lattice directly, just by passing it into the imshow function.
Below are eight snapshots of the progress of a SOM algorithm, from initialization to 700 iterations. The newly initialized (iteration_count = 0) lattice is rendered in the top left panel; the result from the final iteration, in the bottom right panel.
Alternatively, you can use a lower-level imaging library (in Python, e.g., PIL) and transfer the reference tuples onto the 2D grid, one at a time:
for y in range(h):
for x in range(w):
img.putpixel( (x, y), (
SOM.Umatrix[y, x, 0],
SOM.Umatrix[y, x, 1],
SOM.Umatrix[y, x, 2])
)
Here img is an instance of PIL's Image class. Here the image is created by iterating over the grid one pixel at a time; for each pixel, putpixel is called on img three times, the three calls of course corresponding to the three values in an rgb tuple.
From the matrix that you create:
u(1) u(1,2) u(2) u(2,3) u(3)
u(1,4) u(1,2,4,5) u(2,5) u(2,3,5,6) u(3,6)
u(4) u(4,5) u(5) u(5,6) u(6)
u(4,7) u(4,5,7,8) u(5,8) u(5,6,8,9) u(6,9)
u(7) u(7,8) u(8) u(8,9) u(9)
The elements with single numbers like u(1), u(2), ..., u(9) as just the elements with more than two numbers like u(1,2,4,5), u(2,3,5,6), ... , u(5,6,8,9) are calculated using something like the mean, median, min or max of the values in the neighborhood.
It's a nice idea calculate the elements with two numbers first, one possible code for that is:
for i in range(self.h_u_matrix):
for j in range(self.w_u_matrix):
nb = (0,0)
if not (i % 2) and (j % 2):
nb = (0,1)
elif (i % 2) and not (j % 2):
nb = (1,0)
self.u_matrix[(i,j)] = np.linalg.norm(
self.weights[i //2, j //2] - self.weights[i //2 +nb[0], j // 2 + nb[1]],
axis = 0
)
In the code above the self.h_u_matrix = self.weights.shape[0]*2 - 1 and self.w_u_matrix = self.weights.shape[1]*2 - 1 are the dimensions of the U-Matrix. With that said, for calculate the others elements it's necessary obtain a list with they neighboors and apply a mean for example. The following code implements that's idea:
for i in range(self.h_u_matrix):
for j in range(self.w_u_matrix):
if not (i % 2) and not (j % 2):
nodelist = []
if i > 0:
nodelist.append((i-1,j))
if i < 4:
nodelist.append((i+1, j))
if j > 0:
nodelist.append((i,j -1))
if j < 4:
nodelist.append((i,j+1))
meanlist = [self.u_matrix[u_node] for u_node in nodelist]
self.u_matrix[(i,j)] = np.mean(meanlist)
elif (i % 2) and (j % 2):
meanlist = [
(i - 1, j),
(i + 1, j),
(i, j - 1),
(i, j + 1)]
self.u_matrix[(i,j)] = np.mean(meanlist)