hough transformation using the normal equation of a line.
For calculating the value of r while keeping the values of x and y same for different θ the following formula is used.
r = sin(θ)y + cos(θ)x
But the results are not the same as shown in the slides. Am I missing something? I am a newbie please be gentle.
As #Ash explained in the comment section above the angles are expected to be expressed in radians not degree.
So r = sin(θ*pi/180)y + cos(θ*pi/180)x will give the correct answer.
Related
I'm studying a course on vision systems and one of the questions posed was;
For the accumulator shown;
Determine the most likely r,θ combination representing the straight line of the greatest strength in the original image.
From my understanding of the accumulator this would be r = 60, θ = 150 as the 41 votes is the highest number of votes in this cluster of large votes. Am I correct with this combination?
And hence calculate the equation of this line in the form y = mx + c
I'm not sure of the conversion steps required to convert the r = 60, θ = 150 to y = mx + c with the information given since r = 60, θ = 150 denotes 1 point on the line.
State the resolution of your answer and give your reasoning
I assume the resolution is got to do with some of the steps in the auscultation and not the actual resolution of the original image since that's irrelevant to the edges detected in the image.
Any guidance on the above 3 points would be greatly appreciated!
Yes, this is correct.
This is asking you what the slope and intercept are of the line given r and theta. r and theta are not one point on the line, they are one point of the accumulator. r and theta describe a line using the line equation in polar coordinates: . This is the cool thing about the hough transform, every line in one space, (i.e. image space) can be described by a point in another space (r, theta). This could be done with m and b from the line equation , but as we all know, m is undefined for vertical lines. This is the reason the polar line equation is used. It is important to note that the line described by the HT r and theta refers to a line from the origin extending to the actual line in the image. This means your image line y = mx + b equation will need to be orthogonal to the polar equation. The wiki article on the HT describes this well and shows examples. I would recommend drawing a diagram of your r and theta extending to a line like this:
Then use trig to get two points on the red line. Two points are enough to give you m and b from the line equation.
I'm not entirely sure what "resolution" refers to in this context. But it does seem like your line estimator will have some precision loss since r is every 20 mm and theta is every 15 degrees. Perhaps it is asking what degree of error you could get given an accumulator of this resolution.
This question already has an answer here:
Understanding Hough Transform [closed]
(1 answer)
Closed 3 years ago.
I want to understand the Hough Transformation for school.
I know that we can not represent vertical lines which are parallel to the Y-axis (with the y = m*x+b). But we can do this with the polar coordinates r and theta with (y= - cos(theta)/sin(theta) * x + r/sin(theta)).
But lets say I have a line which goes trough this two points. P1(0,0) and P2(0, 100). So this is a line which is exactly like the Y-Axis.
How can this be represented by the polar coordinates r and theta?
since r is 0 so theta is also 0. I don't understand how this line can be represented in the hough space... :/
Can someone explain this to me?
Your equation for the Hough transform can also be written as (is more commonly written as):
r = x*cos(theta) + y*sin(theta)
This can still be solved if you set r=0. In fact, this represents all the lines that go through the pixel at (0,0).
For the case of the vertical line through (0,0), we have r=0 and theta=pi/2. This leads to:
0 = x*1 + y*0
This is satisfied for x=0 and any y. So all pixels (0,y) form this line.
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.
I have three points: (1,1), (2,3), (3, 3.123). I assume the hypothesis is , and I want to do linear regression on the three points. I have two methods to calculate θ:
Method-1: Least Square
import numpy as np
# get an approximate solution using least square
X = np.array([[1,1],[2,1],[3,1]])
y = np.array([1,3,3.123])
theta = np.linalg.lstsq(X,y)[0]
print theta
Method-2: Matrix multiplication
We have the following derivation process:
# rank(X)=2, rank(X|y)=3, so there is no exact solution.
print np.linalg.matrix_rank(X)
print np.linalg.matrix_rank(np.c_[X,y])
theta = np.linalg.inv(X.T.dot(X)).dot(X.T.dot(y))
print theta
Both method-1 and method-2 can get result [ 1.0615 0.25133333], it seems that method-2 is equivalent to least square. But, I don't know why, can anyone reveal the underlying principle of their equivalence?
Both approaches are equivalent, because least squares method is
θ = argmin (Xθ-Y)'(Xθ-Y) = argmin ||(Xθ-Y)||^2 = argmin ||(Xθ-Y)||, that means that you try to minimize length of vector (Xθ-Y), so you try to minimize distance between Xθ and Y. X is a constant matrix, so Xθ is vector from column space of X. That means the shortest distance between these two vectors is when Xθ is equal to projection of vector Y to column space of X (can be easy observed from picture). That results to Y^(hat) = Xθ = X(X'X)^(-1)X'Y, where X(X'X)^(-1)X' is the projection matrix to column space of X. After some changes you can observe that this is equivalent with (X'X)θ = X'Y. You can find exact proof in any linear algebra book.
I recently started a CV course and am going through old homeworks (the current ones aren't released). I've implemented a Hough Lines function, I loop through each point, if it's an edge, then I loop through 0-180 (or -90 to 90) theta values, and calculate rho, and finally store in an array.
When I tried to convert back from Polar Coordinates, I can find an X,Y pair (using rho * sin(theta), and rho * cos(theta)), however I don't understand how to convert that to a line in Cartesian space. To have a line you need either 2 points or a point and a direction (assuming ray then of course)
I just understand where the point is.
I've done some searching but can't seem to quite find the answer, folks tend to say, polar tells you x, then bam you have a line in cartesian, but I seem to be missing that connection where the "bam" was.
What I mean is described here;
Explain Hough Transformation
Also Vector/line from polar coordinates
Where it's asked how do I draw a line from polar coords, which the response was well here's x and y. but to me never mentions rest of that solution.
Is the line somehow related to y = mx+b where m is theta and b is rho?
If not how do I convert back to a line in cartesian space.
EDIT:
After reviewing Sunreef's answer, and trying to convert so y was on it's own side, I discovered this answer as well:
How to convert coordinates back to image (x,y) from hough transformation (rho, theta)?
It appears what I think I'm looking for is this
m = -cotθ
c = p*cosecθ
EDIT#2
I found some other examples on the net. Basically yes I'll need rho * sin(theta) and rho*cos(theta)
The other part that was messing me up was that I needed to convert to radians, once i did that, I started getting good results.
You are right that you can get some base point at the line as
(X0, Y0) = (rho * cos(theta), rho * sin(theta))
and you can find (unit) direction vector of this line as perpendicular to normal:
(dx, dy) = ( -sin(theta), cos(theta))
Taken from Wikipedia:
The non-radial line that crosses the radial line ϕ = ɣ perpendicularly at the point (r0, ɣ) has the equation: r(ϕ) = r0 * sec(ϕ - ɣ).
If I suppose that the coordinates you have for your line are ɣ and r0, then you can rewrite this equation like this:
r(ϕ) * cos(ϕ) * cos(ɣ) + r(ϕ) * sin(ϕ) * sin(ɣ) - r0 = 0
And we know that when translating polar to cartesian coordinates, if we have a point P(r, ϕ) in the polar plane, then its coordinates in the cartesian plane will be:
x = r * cos(ϕ)
y = r * sin(ϕ)
So the equation above becomes a line equation as follows:
x * cos(ɣ) + y * sin(ɣ) - r0 = 0
This is the equation of your line in cartesian coordinates.
(Tell me if you see some mistakes, I did that quickly)