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.
Related
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.
for a machine vision project I am trying to search image data for quadratic surfaces (f(x,y) = Ax^2+Bx+Cy^2+Dy+Exy+F). My plan is to iterate through regions of data and perform a surface-fit, look at the error, see if it's a continuous surface (which would probably indicate a feature in the image).
I was previously able to find quadratic curves (f(x) = Ax^2+Bx+C) in the image data by sampling lines, by using the equations on this site
Link
this worked well, was promising, but it would be much more useful for my task to find 2-D regions that form continuous surfaces.
I see lots of articles indicating that least squares regressions scales up to multiple dimensions, but I'm not able to find code for this Hopefully there is a "closed form" (non-iterative, just compute from your data points) solution, like described above for 1D data. Does anybody know of some source or pseudocode that accomplishes this? Thanks.
(Sorry if my terminology is a bit off.)
I'm not sure what your background is, but if you know some linear algebra you will find linear least squares on wikipedia useful.
Lets take the following example. Say we have the following image
and we want to know how well this fits to a 2D quadratic function in a least squares sense.
Probably the most straightforward way to solve the problem is to compute the optimal coefficients in a least squares sense, then check the error.
First we need to describe the matrices.
Let X be a matrix containing every x,y coordinate in the image, taking the form
X = [x1 x1^2 y1 y1^2 x1*y1 1;
x2 x2^2 y2 y2^2 x2*y2 1;
...
xN xN^2 yN yN^2 xN*yN 1];
For the example image above, X would be a 100x6 matrix.
Let y be the image intensity values in a vector of the form
y = [img(x1,y1);
img(x2,y2);
...
img(xN,yN)]
In this case y is a 100 element column vector.
We want to minimize the least squares objective function S with respect to the vector of coefficients b
S(b) = |y - X*b|^2
where |.| is the L2 norm and b is the desired coefficients
b = [A;
B;
C;
D;
E;
F]
Taking the vector derivative of S(b) with respect to b, setting to zero, and solving for b leads to the standard least squares solution.
b = inv(X'X)*X'*y
where inv is the matrix inverse, ' is transpose, and * is matrix multiplication.
MATLAB example.
% Generate an image
% define x,y coordinates for each location in the image
[x,y] = meshgrid(1:10,1:10);
% true coefficients
b_true = [0.1 0.5 0.3 -0.4 0.4 124];
% magnitude of noise
P = 2;
% create image
img = b_true(1).*x + b_true(2).*x.^2 + b_true(3).*y + b_true(4).*y.^2 + b_true(5).*x.*y + b_true(6);
noise = P*randn(10,10);
img = img + noise;
% Begin least squares optimization
% create matrices
X = [x(:) x(:).^2 y(:) y(:).^2 x(:).*y(:) ones(size(x(:)))];
y = img(:);
% estimated coefficients
b = (X.'*X)\(X.')*y
% mean square error (expected to be near P^2)
E = 1/numel(y) * sum((y - X*b).^2)
Output
b =
0.0906
0.5093
0.1245
-0.3733
0.3776
124.5412
E =
3.4699
In your application you would probably want to define some threshold such that when E < threshold you accept the image (or image region) as a quadratic polynomial.
Hi when i am reading about Stanford's Machine Learning materials about autoencoder, found a formula hard to prove by myself. Link to Material
Question is:
" What input image x would cause ai to be maximally activated? "
Screen shot of the Question and Context:
Many thanks to your answers in advance!
While this can be rigorously solved using KLT conditions and Lagrange multipliers, there is a more intuitive way to figure the result out. I assume that f(.) is a monotone increasing, sigmoid type of nonlinearity (ReLU is also valid). So, finding the maximum of w1x1+...+w100x100 + b under the constraint (x1)^2+...+(x100)^2 <= 1 is equivalent to finding the maximum of f(w1x1+...+w100x100 + b) with the same constraint.
Note that g = w1x1+...+w100x100 + b is a linear function of x terms (Name it as g, so later we can refer it by that). So, the direction of largest increase at any point (x1,...,x100) in the domain of that function is the same, which is the gradient. The gradient is simply (w1,w2,...,w100) at any point in the domain, which means if we go in the direction of (w1,w2,...,w100), independent from where we start, we obtain the largest increase in the function. To make things simplier and to allow us to visualize, assume that we are in the R^2 space and the function is w1x1 + w2x2 + b:
The optimum x1 and x2 are constrained to lie in or on the circle C:(x1)^2 + (x2)^2 =1. Assume that we are on the origin (0.0). If we go in the direction of the gradient (blue arrow) (w1,w2), we are going to attain the largest value of the function where the blue arrow intersects with the circle. That intersection has the coordinates c*(w1,w2) and it is c^2(w1^2 + w2^2) = 1, where c is a scalar coefficient. c is easily solved as c= 1 / sqrt(w1^2 + w2^2). Then at the intersection we have x1=w1/sqrt(w1^2 + w2^2) and x2=w2/sqrt(w1^2 + w2^2), which the solution we seek. This can be extended in the same way to 100 dimensional case.
You may ask why we started at the origin and not any other point in the circle. Note that the red line is perpendicular to the gradient vector and the function is constant along that line. Draw that (u1,u2) line, preserving its orientation, arbitrarily with the constraint that it intersects the circle C. Then choose any point on the line, such that it lies within the circle. On the (u1,u2) line, you start at the same value of the function g, wherever you are. Then as you go in the (w1,w2) direction, the longest path taken within the circle always goes through the origin, which means the path you increase the function g the most.
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)
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.