Why does fundamental matrix have 7 degrees of freedom? - image-processing

There are 9 parameters in the fundamental matrix to relate the pixel co-ordinates of left and right images but only 7 degrees of freedom (DOF).
The reasoning for this on several pages that I've searched says :
Homogenous equations means we lose a degree of freedom
The determinant of F = 0, therefore we lose another degree of freedom.
I don't understand why those 2 reasons mean we lose 2 DOF - can someone explain it?

We initially have 9 DOF because the fundamental matrix is composed of 9 parameters, which implies that we need 9 corresponding points to compute the fundamental matrix (F). But because of the following two reasons, we only need 7 corresponding points.
Reason 1
We lose 1 DOF because we are using homogeneous coordinates. This basically is a way to represent nD points as a vector form by adding an extra dimension. ie) A 2D point (0,2) can be represented as [0,2,1], in general [x,y,1]. There are useful properties when using homogeneous coordinates with 2D/3D transformation, but I'm going to assume you know that.
Now given the expression p and p' representing pixel coordinates:
p'=[u',v',1] and p=[u,v,1]
the fundamental matrix:
F = [f1,f2,f3]
[f4,f5,f6]
[f7,f8,f9]
and fundamental matrix equation:
(transposed p')Fp = 0
when we multiple this expression in algebra form, we get the following:
uu'f1 + vu'f2 + u'f3 + uv'f4 + vv'f5 + v'f6 + uf7 + vf8 + f9 = 0.
In a homogeneous system of linear equation form Af=0 (basically the factorization of the above formula), we get two components A and f.
A:
[uu',vu',u', uv',vv',v',u,v,1]
f (f is essentially the fundamental matrix in vector form):
[f1,f2'f3,f4,f5,f6,f7,f8,f9]
Now if we look at the components of vector A, we have 8 unknowns, but one known value 1 because of homogeneous coordinates, and therefore we only need 8 equations now.
Reason 2
det F = 0.
A determinant is a value that can be obtained from a square matrix.
I'm not entirely sure about the mathematical details of this property but I can still infer the basic idea, and, hopefully, you can as well.
Basically given some matrix A
A = [a,b,c]
[d,e,f]
[g,h,i]
The determinant can be computed using this formula:
det A = aei+bfg+cdh-ceg-bdi-afh
If we look at the determinant using the fundamental matrix, the algebra would look something like this:
F = [f1,f2,f3]
[f4,f5,f6]
[f7,f8,f9]
det F = (f1*f5*f8)+(f2*f6*f7)+(f3*f4*f8)-(f3*f5*f7)-(f2*f4*f9)-(f1*f6*f8)
Now we know the determinant of the fundamental matrix is zero:
det F = (f1*f5*f8)+(f2*f6*f7)+(f3*f4*f8)-(f3*f5*f7)-(f2*f4*f9)-(f1*f6*f8) = 0
So, if we work out only 7 of the 9 parameters of the fundamental matrix, we can work out the last parameter using the above determinant equation.
Therefore the fundamental matrix has 7DOF.

The reasons why F has only 7 degrees of freedom are
F is a 3x3 homogeneous matrix. Homogeneous means there is a scale ambiguity in the matrix, so the scale doesn't matter (as shown in #Curator Corpus 's example). This drops one degree of freedom.
F is a matrix with rank 2. It is not a full rank matrix, so it is singular and its determinant is zero (Proof here). The reason why F is a matrix with rank 2 is that it is mapping a 2D plane (image1) to all the lines (in image 2) that pass through the epipole (of image 2).
Hope it helps.

As for the highest votes answer by nbro, I think it can be interpreted as this way where we have reason two, matrix F has a rank2, so its determinant is zero as a constraint to the f variable function. So, we only need 7 points to determine the rest of variables (f1-f8), with the previous constriant. And 8 equations, 8 variables, leaving only one solution. So there is 7 DOF.

Related

Why do we normalize homography or fundamental matrix?

I want to know about why do we normalize the homography or fundamental matrix? Here is the code in particular.
H = H * (1.0 / H[2, 2]) # Normalization step. H is [3, 3] matrix.
I can understand that we have to normalize the data before computing SVD because of instability caused by linear least squares but why do we normalize it in end?
A homography in 3D space has 8 degrees of freedom by definition, mapping from one plane to another using perspective. Such a homography can be defined by giving four points, which makes eight coordinates (scalars).
A 3x3 matrix has 9 elements, so it has 9 degrees of freedom. That is one degree more than needed for a homography.
The homography doesn't change when the matrix is scaled (multiplied by a scalar). All the math works the same. You don't need to normalize your homography matrix.
It is a good idea to normalize.
For one, it makes the arithmetic somewhat tamer. Have some wikipedia links to fields of study because weaving all these into a coherent sentence... doesn't add anything:
Numerical analysis, Condition number, Floating-point arithmetic, Numerical error, Numerical stability, ...
Also, normalization makes the matrix easier for humans to interpret. The most common normalization is to scale the matrix such that the last element becomes 1. That is convenient because this whole math happens in a projective space, where the projection causes points to be mapped to the w=1 plane, making vectors have a 1 for the last element.
How is the homography matrix provided to you?
For example, in the scene that some library function calculates and provides the homography matrix to you,
if the function specification doesn't mention about the scale...
In an extreme case, the function can be implemented as:
Matrix3x3 CalculateHomographyMatrix( some arguments )
{
Matrix3x3 H = ...; //Homogoraphy Calculation
return Non_Zero_Random_Value * H; //Wow!
}
Element values may become very large or very small and using such values to your process may cause problems (floating point computation errors).

Why convolution in spatial domain equal to multiplication in frequency domain?

Why it is said that "convolution of an image in spatial domain is equal to multiplication in frequency domain" ?
Could anyone please explain it briefly?
StackOverflow, unfortunately, doesn't support MathJaX hence it is hard to show the math here.
One way to explain is that Convolution is Linear Invariant Operator.
As you know, Linear Time / Spatially Invariant Systems basically do one thing - Delay and Scaling.
The Eigen Functions of Delay and Scaling are the Harmonic Functions.
Which means that give a signal described by harmonic signals (Practically its Fourier Transform) Linear Time / Spatially Invariant Operator only scales it by complex number (Scaling and shifting by phase) which is what you do in the Fourier Domain.
It is similar to Diagonalization in Linear Algebra.
For instance let's thing of the Filter we apply on the image as an operator - A.
So the output of the system is y = A x.
If A is diagonalizable as A = P^T D P where D is diagonal matrix and P P^T = I, namely Unitary Matrix.
So y = A x = P^T D P x hence by defining z = P x and t = P y we get t = D z namely we only need to multiply each element in t and not the whole matrix multiplication.
If you think about P as the Fourier Transom operator then instead of doing Matrix Multiplication you can have element wise multiplication in other domain - Fourier Domain.

3D reconstruction from two calibrated cameras - where is the error in this pipeline?

There are many posts about 3D reconstruction from stereo views of known internal calibration, some of which are excellent. I have read a lot of them, and based on what I have read I am trying to compute my own 3D scene reconstruction with the below pipeline / algorithm. I'll set out the method then ask specific questions at the bottom.
0. Calibrate your cameras:
This means retrieve the camera calibration matrices K1 and K2 for Camera 1 and Camera 2. These are 3x3 matrices encapsulating each camera's internal parameters: focal length, principal point offset / image centre. These don't change, you should only need to do this once, well, for each camera as long as you don't zoom or change the resolution you record in.
Do this offline. Do not argue.
I'm using OpenCV's CalibrateCamera() and checkerboard routines, but this functionality is also included in the Matlab Camera Calibration toolbox. The OpenCV routines seem to work nicely.
1. Fundamental Matrix F:
With your cameras now set up as a stereo rig. Determine the fundamental matrix (3x3) of that configuration using point correspondences between the two images/views.
How you obtain the correspondences is up to you and will depend a lot on the scene itself.
I am using OpenCV's findFundamentalMat() to get F, which provides a number of options method wise (8-point algorithm, RANSAC, LMEDS).
You can test the resulting matrix by plugging it into the defining equation of the Fundamental matrix: x'Fx = 0 where x' and x are the raw image point correspondences (x, y) in homogeneous coordinates (x, y, 1) and one of the three-vectors is transposed so that the multiplication makes sense. The nearer to zero for each correspondence, the better F is obeying it's relation. This is equivalent to checking how well the derived F actually maps from one image plane to another. I get an average deflection of ~2px using the 8-point algorithm.
2. Essential Matrix E:
Compute the Essential matrix directly from F and the calibration matrices.
E = K2TFK1
3. Internal Constraint upon E:
E should obey certain constraints. In particular, if decomposed by SVD into USV.t then it's singular values should be = a, a, 0. The first two diagonal elements of S should be equal, and the third zero.
I was surprised to read here that if this is not true when you test for it, you might choose to fabricate a new Essential matrix from the prior decomposition like so: E_new = U * diag(1,1,0) * V.t which is of course guaranteed to obey the constraint. You have essentially set S = (100,010,000) artificially.
4. Full Camera Projection Matrices:
There are two camera projection matrices P1 and P2. These are 3x4 and obey the x = PX relation. Also, P = K[R|t] and therefore K_inv.P = [R|t] (where the camera calibration has been removed).
The first matrix P1 (excluding the calibration matrix K) can be set to [I|0] then P2 (excluding K) is R|t
Compute the Rotation and translation between the two cameras R, t from the decomposition of E. There are two possible ways to calculate R (U*W*V.t and U*W.t*V.t) and two ways to calculate t (±third column of U), which means that there are four combinations of Rt, only one of which is valid.
Compute all four combinations, and choose the one that geometrically corresponds to the situation where a reconstructed point is in front of both cameras. I actually do this by carrying through and calculating the resulting P2 = [R|t] and triangulating the 3d position of a few correspondences in normalised coordinates to ensure that they have a positive depth (z-coord)
5. Triangulate in 3D
Finally, combine the recovered 3x4 projection matrices with their respective calibration matrices: P'1 = K1P1 and P'2 = K2P2
And triangulate the 3-space coordinates of each 2d point correspondence accordingly, for which I am using the LinearLS method from here.
QUESTIONS:
Are there any howling omissions and/or errors in this method?
My F matrix is apparently accurate (0.22% deflection in the mapping compared to typical coordinate values), but when testing E against x'Ex = 0 using normalised image correspondences the typical error in that mapping is >100% of the normalised coordinates themselves. Is testing E against xEx = 0 valid, and if so where is that jump in error coming from?
The error in my fundamental matrix estimation is significantly worse when using RANSAC than the 8pt algorithm, ±50px in the mapping between x and x'. This deeply concerns me.
'Enforcing the internal constraint' still sits very weirdly with me - how can it be valid to just manufacture a new Essential matrix from part of the decomposition of the original?
Is there a more efficient way of determining which combo of R and t to use than calculating P and triangulating some of the normalised coordinates?
My final re-projection error is hundreds of pixels in 720p images. Am I likely looking at problems in the calibration, determination of P-matrices or the triangulation?
The error in my fundamental matr1ix estimation is significantly worse
when using RANSAC than the 8pt algorithm, ±50px in the mapping between
x and x'. This deeply concerns me.
Using the 8pt algorithm does not exclude using the RANSAC principle.
When using the 8pt algorithm directly which points do you use? You have to choose 8 (good) points by yourself.
In theory you can compute a fundamental matrix from any point correspondences and you often get a degenerated fundamental matrix because the linear equations are not independend. Another point is that the 8pt algorithm uses a overdetermined system of linear equations so that one single outlier will destroy the fundamental matrix.
Have you tried to use the RANSAC result? I bet it represents one of the correct solutions for F.
My F matrix is apparently accurate (0.22% deflection in the mapping
compared to typical coordinate values), but when testing E against
x'Ex = 0 using normalised image correspondences the typical error in
that mapping is >100% of the normalised coordinates themselves. Is
testing E against xEx = 0 valid, and if so where is that jump in error
coming from?
Again, if F is degenerated, x'Fx = 0 can be for every point correspondence.
Another reason for you incorrect E may be the switch of the cameras (K1T * E * K2 instead of K2T * E * K1). Remember to check: x'Ex = 0
'Enforcing the internal constraint' still sits very weirdly with me -
how can it be valid to just manufacture a new Essential matrix from
part of the decomposition of the original?
It is explained in 'Multiple View Geometry in Computer Vision' from Hartley and Zisserman. As far as I know it has to do with the minimization of the Frobenius norm of F.
You can Google it and there are pdf resources.
Is there a more efficient way of determining which combo of R and t to
use than calculating P and triangulating some of the normalised
coordinates?
No as far as I know.
My final re-projection error is hundreds of pixels in 720p images. Am
I likely looking at problems in the calibration, determination of
P-matrices or the triangulation?
Your rigid body transformation P2 is incorrect because E is incorrect.

Geometric representation of Perceptrons (Artificial neural networks)

I am taking this course on Neural networks in Coursera by Geoffrey Hinton (not current).
I have a very basic doubt on weight spaces.
https://d396qusza40orc.cloudfront.net/neuralnets/lecture_slides%2Flec2.pdf
Page 18.
If I have a weight vector (bias is 0) as [w1=1,w2=2] and training case as {1,2,-1} and {2,1,1}
where I guess {1,2} and {2,1} are the input vectors. How can it be represented geometrically?
I am unable to visualize it? Why is training case giving a plane which divides the weight space into 2? Could somebody explain this in a coordinate axes of 3 dimensions?
The following is the text from the ppt:
1.Weight-space has one dimension per weight.
2.A point in the space has particular setting for all the weights.
3.Assuming that we have eliminated the threshold each hyperplane could be represented as a hyperplane through the origin.
My doubt is in the third point above. Kindly help me understand.
It's probably easier to explain if you look deeper into the math. Basically what a single layer of a neural net is performing some function on your input vector transforming it into a different vector space.
You don't want to jump right into thinking of this in 3-dimensions. Start smaller, it's easy to make diagrams in 1-2 dimensions, and nearly impossible to draw anything worthwhile in 3 dimensions (unless you're a brilliant artist), and being able to sketch this stuff out is invaluable.
Let's take the simplest case, where you're taking in an input vector of length 2, you have a weight vector of dimension 2x1, which implies an output vector of length one (effectively a scalar)
In this case it's pretty easy to imagine that you've got something of the form:
input = [x, y]
weight = [a, b]
output = ax + by
If we assume that weight = [1, 3], we can see, and hopefully intuit that the response of our perceptron will be something like this:
With the behavior being largely unchanged for different values of the weight vector.
It's easy to imagine then, that if you're constraining your output to a binary space, there is a plane, maybe 0.5 units above the one shown above that constitutes your "decision boundary".
As you move into higher dimensions this becomes harder and harder to visualize, but if you imagine that that plane shown isn't merely a 2-d plane, but an n-d plane or a hyperplane, you can imagine that this same process happens.
Since actually creating the hyperplane requires either the input or output to be fixed, you can think of giving your perceptron a single training value as creating a "fixed" [x,y] value. This can be used to create a hyperplane. Sadly, this cannot be effectively be visualized as 4-d drawings are not really feasible in browser.
Hope that clears things up, let me know if you have more questions.
I have encountered this question on SO while preparing a large article on linear combinations (it's in Russian, https://habrahabr.ru/post/324736/). It has a section on the weight space and I would like to share some thoughts from it.
Let's take a simple case of linearly separable dataset with two classes, red and green:
The illustration above is in the dataspace X, where samples are represented by points and weight coefficients constitutes a line. It could be conveyed by the following formula:
w^T * x + b = 0
But we can rewrite it vice-versa making x component a vector-coefficient and w a vector-variable:
x^T * w + b = 0
because dot product is symmetrical. Now it could be visualized in the weight space the following way:
where red and green lines are the samples and blue point is the weight.
More possible weights are limited to the area below (shown in magenta):
which could be visualized in dataspace X as:
Hope it clarifies dataspace/weightspace correlation a bit. Feel free to ask questions, will be glad to explain in more detail.
The "decision boundary" for a single layer perceptron is a plane (hyper plane)
where n in the image is the weight vector w, in your case w={w1=1,w2=2}=(1,2) and the direction specifies which side is the right side. n is orthogonal (90 degrees) to the plane)
A plane always splits a space into 2 naturally (extend the plane to infinity in each direction)
you can also try to input different value into the perceptron and try to find where the response is zero (only on the decision boundary).
Recommend you read up on linear algebra to understand it better:
https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces
For a perceptron with 1 input & 1 output layer, there can only be 1 LINEAR hyperplane. And since there is no bias, the hyperplane won't be able to shift in an axis and so it will always share the same origin point. However, if there is a bias, they may not share a same point anymore.
I think the reason why a training case can be represented as a hyperplane because...
Let's say
[j,k] is the weight vector and
[m,n] is the training-input
training-output = jm + kn
Given that a training case in this perspective is fixed and the weights varies, the training-input (m, n) becomes the coefficient and the weights (j, k) become the variables.
Just as in any text book where z = ax + by is a plane,
training-output = jm + kn is also a plane defined by training-output, m, and n.
Equation of a plane passing through origin is written in the form:
ax+by+cz=0
If a=1,b=2,c=3;Equation of the plane can be written as:
x+2y+3z=0
So,in the XYZ plane,Equation: x+2y+3z=0
Now,in the weight space;every dimension will represent a weight.So,if the perceptron has 10 weights,Weight space will be 10 dimensional.
Equation of the perceptron: ax+by+cz<=0 ==> Class 0
ax+by+cz>0 ==> Class 1
In this case;a,b & c are the weights.x,y & z are the input features.
In the weight space;a,b & c are the variables(axis).
So,for every training example;for eg: (x,y,z)=(2,3,4);a hyperplane would be formed in the weight space whose equation would be:
2a+3b+4c=0
passing through the origin.
I hope,now,you understand it.
Consider we have 2 weights. So w = [w1, w2]. Suppose we have input x = [x1, x2] = [1, 2]. If you use the weight to do a prediction, you have z = w1*x1 + w2*x2 and prediction y = z > 0 ? 1 : 0.
Suppose the label for the input x is 1. Thus, we hope y = 1, and thus we want z = w1*x1 + w2*x2 > 0. Consider vector multiplication, z = (w ^ T)x. So we want (w ^ T)x > 0. The geometric interpretation of this expression is that the angle between w and x is less than 90 degree. For example, the green vector is a candidate for w that would give the correct prediction of 1 in this case. Actually, any vector that lies on the same side, with respect to the line of w1 + 2 * w2 = 0, as the green vector would give the correct solution. However, if it lies on the other side as the red vector does, then it would give the wrong answer.
However, suppose the label is 0. Then the case would just be the reverse.
The above case gives the intuition understand and just illustrates the 3 points in the lecture slide. The testing case x determines the plane, and depending on the label, the weight vector must lie on one particular side of the plane to give the correct answer.

OpenCV CV findHomography assertion error - counter => 4

I'm currently finishing my evaluation-tool for interest point detectors. In the last steps I found a confusing error.
Mat findHomography(InputArray srcPoints, InputArray dstPoints, int method=0, double ransacReprojThreshold=3, OutputArray mask=noArray() )
The srcPoints and dstPoints are vector<Points2f> which stores the corresponding points of the matched keypoints. So far nothing special - It's like in the tutorials.
But when I use RANSAC and have a vector<Points2f> in range [0, ... , 4], I get an assertion error than the counter should be greater or equals four.
Question 1: Does the algorithm needs at least four points to describe what belongs to the current model or not and to create the consensus?
Question 2: Is there any documentation about this? (I took a look at the doc and the tutorials.)
Please note that I already have seen this question. But there is no satisfying answer for the behaviour of RANSAC. Or should I accept that this methods needs at least four points to find the homography?
Thanks for your help.
A homography cannot be computed with less than 4 pairs of points. That is because with only 3 points there is a perspective ambiguity. Picture a triangle
a
b c
in image 1. In image 2 the points have been transformed to look like this
a
b c
The distance between b and c has been cut in half. Unfortunately you don't know if that is because point c got closer to your or farther from you. With a 4th point the difference becomes clear.
a d
b c
Here is a square in image 1
d
a
b c
here d and c rotated towards you
a
d
b c
and here they rotated away from you.
I don't see this requirement in the openCV documentation but if you find any resources on homography calculation you won't have to read very far before you find this requirement and a more rigorous proof of 4 points being sufficient.
RANSAC is used to select 4 pairs of matching points in a greater set or correspondences (i.e. when srcPoints.size() >= 4). That's why you get an error if srcPoints.size() <=4.
You need at least 4 correspondences simply because a Homography matrix H has 8 degrees of freedom, hence 8 linear equations are required to find a solution. Since each pairs of points generates two linear equations (using x and y coordinates) you'll need a total of at least 4 correspondences.

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