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 .
Related
Background:
Thanks for your attention! I am learning the basic knowledge of 2D convolution, linear algebra and PyTorch. I encounter the implementation problem about the psedo-inverse of the convolution operator. Specifically, I have no idea about how to implement it in an efficient way. Please see the following problem statements for details. Any help/tip/suggestion is welcomed.
(Thanks a lot for your attention!)
The Original Problem:
I have an image feature x with shape [b,c,h,w] and a 3x3 convolutional kernel K with shape [c,c,3,3]. There is y = K * x. How to implement the corresponding pseudo-inverse on y in an efficient way?
There is [y = K * x = Ax], how to implement [x_hat = (A^+)y]?
I guess that there should be some operations using torch.fft. However, I still have no idea about how to implement it. I do not know if there exists an implementation previously.
import torch
import torch.nn.functional as F
c = 32
K = torch.randn(c, c, 3, 3)
x = torch.randn(1, c, 128, 128)
y = F.conv2d(x, K, padding=1)
print(y.shape)
# How to implement pseudo-inverse for y = K * x in an efficient way?
Some of My Efforts:
I may know that the 2D convolution is a linear operator. It is equivalent to a "matrix product" operator. We can actually write out the matrix form of the convolution and calculate its psedo-inverse. However, I think this type of operation will be inefficient. And I have no idea about how to implement it in an efficient way.
According to Wikipedia, the psedo-inverse may satisfy the property of A(A_pinv(x))=x, where A is the convolutional operator, A_pinv is its psedo-inverse, and x may be any image feature.
(Thanks again for reading such a long post!)
This takes the problem to another level.
The convolution itself is a linear operation, you can determine the matrix of the operation and solve a least square problem directly [1], or compute the pseudo-inverse as you mentioned, and then apply to different outputs and predicting a projection of the input.
I am changing your code to using padding=0
import torch
import torch.nn.functional as F
# your code
c = 32
K = torch.randn(c, c, 1, 1)
x = torch.randn(4, c, 128, 128)
y = F.conv2d(x, K, bias=torch.zeros((c,)))
Also, as you probably already suggested the convolution can be computed as ifft(fft(h)*fft(x)). However, the conv2d function is a cross-correlation, so you have to conjugate the filter leading to ifft(fft(h)*fft(x)), also you have to apply this to two axes, and you have to make sure the FFT is calcuated using the same representation (size), since the data is real, we can apply multi-dimensional real FFT. To be complete, conv2d works on multiple channels, so we have to calculate summations of convolutions. Since the FFT is linear, we can simply compute the summations on the frequency domain
using einsum.
s = y.shape[-2:]
K_f = torch.fft.rfftn(K, s)
x_f = torch.fft.rfftn(x, s)
y_f = torch.einsum('jkxy,ikxy->ijxy', K_f.conj(), x_f)
y_hat = torch.fft.irfftn(y_f, s)
Except for the borders it should be accurate (remember FFT computes a cyclic convolution).
torch.max(abs(y_hat[:,:,:-2,:-2] - y[:,:,:,:]))
Now, notice the pattern jk,ik->ij on the einsum, that means y_f[i,j] = sum(K_f[j,k] * x_f[i,k]) = x_f # K_f.T, if # is the matrix product on the first two dimensions. So to invert this operation we have to can interpret the first two dimensions as matrices. The function pinv will compute pseudo-inverses on the last two axes, so in order to use that we have to permute the axes. If we right multiply the output by the pseudo-inverse of transposed K_f we should invert this operation.
s = 128,128
K_f = torch.fft.rfftn(K, s)
K_f_inv = torch.linalg.pinv(K_f.T).T
y_f = torch.fft.rfftn(y_hat, s)
x_f = torch.einsum('jkxy,ikxy->ijxy', K_f_inv.conj(), y_f)
x_hat = torch.fft.irfftn(x_f, s)
print(torch.mean((x - x_hat)**2) / torch.mean((x)**2))
Nottice that I am using the full convolution, but the conv2d actually cropped the images. Let's apply that
y_hat[:,:,128-(k-1):,:] = 0
y_hat[:,:,:,128-(k-1):] = 0
Repeating the calculation you will see that the input is not accurate anymore, so you have to be careful about what you do with your convolution, but in some situations where you can get this to work it will be in fact efficient.
s = 128,128
K_f = torch.fft.rfftn(K, s)
K_f_inv = torch.linalg.pinv(K_f.T).T
y_f = torch.fft.rfftn(y_hat, s)
x_f = torch.einsum('jkxy,ikxy->ijxy', K_f_inv.conj(), y_f)
x_hat = torch.fft.irfftn(x_f, s)
print(torch.mean((x - x_hat)**2) / torch.mean((x)**2))
I'm writing a k-means algorithm. At each step, I want to compute the distance of my n points to k centroids, without a for loop, and for d dimensions.
The problem is I have a hard time splitting on my number of dimensions with the Matlab functions I know. Here is my current code, with x being my n 2D-points and y my k centroids (also 2D-points of course), and with the points distributed along dimension 1, and the spatial coordinates along the dimension 2:
dist = #(a,b) (a - b).^2;
dx = bsxfun(dist, x(:,1), y(:,1)'); % x is (n,1) and y is (1,k)
dy = bsxfun(dist, x(:,2), y(:,2)'); % so the result is (n,k)
dists = dx + dy; % contains the square distance of each points to the k centroids
[_,l] = min(dists, [], 2); % we then argmin on the 2nd dimension
How to vectorize furthermore ?
First edit 3 days later, searching on my own
Since asking this question I made progress on my own towards vectorizing this piece of code.
The code above runs in approximately 0.7 ms on my example.
I first used repmat to make it easy to do broadcasting:
dists = permute(permute(repmat(x,1,1,k), [3,2,1]) - y, [3,2,1]).^2;
dists = sum(dists, 2);
[~,l] = min(dists, [], 3);
As expected it is slightly slower since we replicate the matrix, it runs at 0.85 ms.
From this example it was pretty easy to use bsxfun for the whole thing, but it turned out to be extremely slow, running in 150 ms so more than 150 times slower than the repmat version:
dist = #(a, b) (a - b).^2;
dists = permute(bsxfun(dist, permute(x, [3, 2, 1]), y), [3, 2, 1]);
dists = sum(dists, 2);
[~,l] = min(dists, [], 3);
Why is it so slow ? Isn't vectorizing always an improvement on speed, since it uses vector instructions on the CPU ? I mean of course simple for loops could be optimized to use it aswell, but how can vectorizing make the code slower ? Did I do it wrong ?
Using a for loop
For the sake of completeness, here's the for loop version of my code, surprisingly the fastest running in 0.4 ms, not sure why..
for i=1:k
dists(:,i) = sum((x - y(i,:)).^2, 2);
endfor
[~,l] = min(dists, [], 2);
Note: This answer was written when the question was also tagged MATLAB. Links to Octave documentation added after the MATLAB tag was removed.
You can use the pdist2MATLAB/Octave function to calculate pairwise distances between two sets of observations.
This way, you offload the bother of vectorization to the people who wrote MATLAB/Octave (and they have done a pretty good job of it)
X = rand(10,3);
Y = rand(5,3);
D = pdist2(X, Y);
D is now a 10x5 matrix where the i, jth element is the distance between the ith X and jth Y point.
You can pass it the kind of distance you want as the third argument -- e.g. 'euclidean', 'minkowski', etc, or you could pass a function handle to your custom function like so:
dist = #(a,b) (a - b).^2;
D = pdist2(X, Y, dist);
As saastn mentions, pdist2(..., 'smallest', k) makes things easier in k-means. This returns just the smallest k values from each column of pdist2's result. Octave doesn't have this functionality, but it's easily replicated using sort()MATLAB/Octave.
D_smallest = sort(D);
D_smallest = D_smallest(1:k, :);
I am trying to find a simple algorithm to find the correspondence between two sets of 2D points (registration). One set contains the template of an object I'd like to find and the second set mostly contains points that belong to the object of interest, but it can be noisy (missing points as well as additional points that do not belong to the object). Both sets contain roughly 40 points in 2D. The second set is a homography of the first set (translation, rotation and perspective transform).
I am interested in finding an algorithm for registration in order to get the point-correspondence. I will be using this information to find the transform between the two sets (all of this in OpenCV).
Can anyone suggest an algorithm, library or small bit of code that could do the job? As I'm dealing with small sets, it does not have to be super optimized. Currently, my approach is a RANSAC-like algorithm:
Choose 4 random points from set 1 and from set 2.
Compute transform matrix H (using openCV getPerspective())
Warp 1st set of points using H and test how they aligned to the 2nd set of points
Repeat 1-3 N times and choose best transform according to some metric (e.g. sum of squares).
Any ideas? Thanks for your input.
With python you can use Open3D librarry, wich is very easy to install in Anaconda. To your purpose ICP should work fine, so we'll use the classical ICP, wich minimizes point-to-point distances between closest points in every iteration. Here is the code to register 2 clouds:
import numpy as np
import open3d as o3d
# Parameters:
initial_T = np.identity(4) # Initial transformation for ICP
distance = 0.1 # The threshold distance used for searching correspondences
(closest points between clouds). I'm setting it to 10 cm.
# Read your point clouds:
source = o3d.io.read_point_cloud("point_cloud_1.xyz")
target = o3d.io.read_point_cloud("point_cloud_0.xyz")
# Define the type of registration:
type = o3d.pipelines.registration.TransformationEstimationPointToPoint(False)
# "False" means rigid transformation, scale = 1
# Define the number of iterations (I'll use 100):
iterations = o3d.pipelines.registration.ICPConvergenceCriteria(max_iteration = 100)
# Do the registration:
result = o3d.pipelines.registration.registration_icp(source, target, distance, initial_T, type, iterations)
result is a class with 4 things: the transformation T(4x4), 2 metrict (rmse and fitness) and the set of correspondences.
To acess the transformation:
I used it a lot with 3D clouds obteined from Terrestrial Laser Scanners (TLS) and from robots (Velodiny LIDAR).
With MATLAB:
We'll use the point-to-point ICP again, because your data is 2D. Here is a minimum example with two point clouds random generated inside a triangle shape:
% Triangle vértices:
V1 = [-20, 0; -10, 10; 0, 0];
V2 = [-10, 0; 0, 10; 10, 0];
% Create clouds and show pair:
points = 5000
N1 = criar_nuvem_triangulo(V1,points);
N2 = criar_nuvem_triangulo(V2,points);
pcshowpair(N1,N2)
% Registrate pair N1->N2 and show:
[T,N1_tranformed,RMSE]=pcregistericp(N1,N2,'Metric','pointToPoint','MaxIterations',100);
pcshowpair(N1_tranformed,N2)
"criar_nuvem_triangulo" is a function to generate random point clouds inside a triangle:
function [cloud] = criar_nuvem_triangulo(V,N)
% Function wich creates 2D point clouds in triangle format using random
% points
% Parameters: V = Triangle vertices (3x2 Matrix)| N = Number of points
t = sqrt(rand(N, 1));
s = rand(N, 1);
P = (1 - t) * V(1, :) + bsxfun(#times, ((1 - s) * V(2, :) + s * V(3, :)), t);
points = [P,zeros(N,1)];
cloud = pointCloud(points)
end
results:
You may just use cv::findHomography. It is a RANSAC-based approach around cv::getPerspectiveTransform.
auto H = cv::findHomography(srcPoints, dstPoints, CV_RANSAC,3);
Where 3 is the reprojection threshold.
One traditional approach to solve your problem is by using point-set registration method when you don't have matching pair information. Point set registration is similar to method you are talking about.You can find matlab implementation here.
Thanks
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)
Is it possible to perform a Cascaded Hough Transform in OpenCV? I understand its just a HT followed by another one. The problem I'm facing is that the values returned are always rho and theta and never in y-intercept form.
Is it possible to convert these values back to y-intercept and split them into sub-spaces so I can detect vanishing points?
Or is it just better to program an implementation of HT myself in, say, Python?
you could try to populate the Hough domain with m and c parameters instead, so that y = mx + c can be re-written as c = y - mx so instead of the usual rho = x cos(theta) + y sin(theta), you have c = y - mx
normally, you'd go through the thetas and calculate the rho, then you increment the accumulator value for that pair of rho and theta. Here, you'd go through the value of m and calculate the values of c, then accumulate that m,c element in the accumulator. The bin with the most votes would be the right m,c
// going through the image looking for edge pixels
for (i = 0;i<numrows;i++)
{
for (j = 0;j<numcols;j++)
{
if (img[i*numcols + j] > 1)
{
for (n = first_m;n<last_m;n++)
{
index = i - n * j;
accum[n][index]++;
}
}
}
}
I guess where this becomes ineffective is that its hard to define the step size for going through m as they should technically go from -infinity to infinity so you'd kind of have trouble. yeah, so much for Hough transform in terms of m,c. Lol
I guess you could go the other way and isolate m so it would be m = (y-c)/x so that now, you cycle through a bunch of y values that make sense and its much more manageable though it's still hard to define your accumulator matrix because m still has no limit. I guess you could limit the values of m that you would be interested in looking for.
Yeah, much more sense to go with rho and theta and convert them into y = mx + c and then even making a brand new image and re-running the hough transform on it.
I don't think OpenCV can perform cascaded hough transforms. You should convert them to xy space yourself. This article might help you:
http://aishack.in/tutorials/converting-lines-from-normal-to-slopeintercept-form/