I'm working on vision project with OPENCV : how to find point x position in an image B
knowing it's position in image A?
and that the two projections are of the same plane ( the floor).
thanks for your answers.
Assume that you have a Point A with homogeneous coordinate :
A = [x, y, 1]
Maybe you also have an homography H like this:
H = [h00 h01 h02;
h10 h11 h12;
h20 h21 h22]
Then do the product H * A' where A' is the transpose of A
You will get something like this
Bs = [X, Y, s]
You have to normalize the result by s in order to have
B = [Xb, Yb, 1]
Related
I am currently developing a grid for a simple simulation and I have been tasked with interpolating some values tied to vertices of a triangle.
So far I have this:
let val1 = 10f
let val2 = 15f
let val3 = 12f
let point1 = Vector2(100f, 300f), val1
let point2 = Vector2(300f, 102f), val2
let point3 = Vector2(100f, 100f), val3
let points = [point1; point2; point3]
let find (points : (Vector2*float32) list) (pos : Vector2) =
let (minX, minXv) = points |> List.minBy (fun (v, valu) -> v.X)
let (maxX, maxXv) = points |> List.maxBy (fun (v, valu)-> v.X)
let (minY, minYv) = points |> List.minBy (fun (v, valu) -> v.Y)
let (maxY, maxYv) = points |> List.maxBy (fun (v, valu) -> v.Y)
let xy = (pos - minX)/(maxX - minX)*(maxX - minX)
let dx = ((maxXv - minXv)/(maxX.X - minX.X))
let dy = ((maxYv - minYv)/(maxY.Y - minY.Y))
((dx*xy.X + dy*xy.Y)) + minXv
Where you get a list of points forming a triangle. I find the minimum X and Y and the max X and Y with the corresponding values tied to them.
The problem is this approach only works with a right sided triangle. With an equilateral triangle the mid point will end up having a higher value at its vertex than the value that is set.
So I guess the approach is here to essentially project a right sided triangle and create some sort of transformation matrix between any triangle and this projected triangle?
Is this correct? If not, then any pointers would be most appreciated!
You probably want a linear interpolation where the interpolated value is the result of a function of the form
f(x, y) = a*x + b*y + c
If you consider this in 3d, with (x,y) a position on the ground and f(x,y) the height above it, this formula will give you a plane.
To obtain the parameters you can use the points you have:
f(x1, y1) = x1*a + y1*b * 1*c = v1 ⎛x1 y1 1⎞ ⎛a⎞ ⎛v1⎞
f(x2, y2) = x2*a + y2*b * 1*c = v2 ⎜x2 y2 1⎟ * ⎜b⎟ = ⎜v2⎟
f(x3, y3) = x3*a + y3*b * 1*c = v3 ⎝x3 y3 1⎠ ⎝c⎠ ⎝v3⎠
This is a 3×3 system of linear equations: three equations in three unknowns.
You can solve this in a number of ways, e.g. using Gaussian elimination, the inverse matrix, Cramer's rule or some linear algebra library. A numerics expert may tell you that there are differences in the numeric stability between these approaches, particularly if the corners of the triangle are close to lying on a single line. But as long as you're sufficiently far away from that degenerate situation, it probably doesn't make a huge practical difference for simple use cases. Note that if you want to interpolate values for multiple positions relative to a single triangle, you'd only compute a,b,c once and then just use the simple linear formula for each input position, which might lead to a considerable speed-up.
Advanced info: For some applications, linear interpolation is not good enough, but to find something more appropriate you would need to provide more data than your question suggests is available. One example that comes to my mind is triangle meshes for 3d rendering. If you use linear interpolation to map the triangles to texture coordinates, then they will line up along the edges but the direction of the mapping can change abruptly, leading to noticeable seams. A kind of projective interpolation or weighted interpolation can avoid this, as I learned from a paper on conformal equivalence of triangle meshes (Springborn, Schröder, Pinkall, 2008), but for that you need to know how the triangle in world coordinates maps to the triangle in texture coordinates, and your also need the triangle mesh and the correspondence to the texture to be compatible with this mapping. Then you'd map in such a way that you not only transport corners to corners, but also circumcircle to circumcircle.
Say you have two column vectors vv and ww, each with 7 elements (i.e., they have dimensions 7x1). Consider the following code:
z = 0;
for i = 1:7
z = z + v(i) * w(i)
end
A) z = sum (v .* w);
B) z = w' * v;
C) z = v * w;
D) z = w * v;
According to the solutions, answers (A) AND (B) are the right answers, can someone please help me understand why?
Why is z = v * w' which is similar to answer (B) but only the order of the operation changes, is false? Since we want a vector that by definition only has one column, wouldn't we need a matrix of this size: 1x7 * 7x1 = 1x1 ? So why is z = v' * w false ? It gives the same dimension as answer (B)?
z = v'*w is true and is equal to w'*v.
They both makes 1*1 matrix, which is a number value in octave.
See this:
octave:5> v = rand(7, 1);
octave:6> w = rand(7, 1);
octave:7> v'*w
ans = 1.3110
octave:8> w'*v
ans = 1.3110
octave:9> sum(v.*w)
ans = 1.3110
Answers A and B both perform a dot product of the two vectors, which yields the same result as the code provided. Answer A first performs the element-wise product (.*) of the two column vectors, then sums those intermediate values. Answer B performs the same mathematical operation but does so via a dot product (i.e., matrix multiplication).
Answer C is incorrect because it would be performing a matrix multiplication on misaligned matrices (7x1 and 7x1). The same is true for D.
z = v * w', which was not one of the options, is incorrect because it would yield a 7x7 matrix (instead of the 1x1 scalar value desired). The point is that order matters when performing matrix multiplication. (1xN)X(Nx1) -> (1x1), whereas (Nx1)X(1xN) -> (NxN).
z = v' * w is actually a correct solution but was simply not provided as one of the options.
I saw u,v image coordinates at bottom of here. I downloaded the data and one sample is [214.65 222.52 145.72 165.42 96.492 114.22 64.985 71.877 43.323
33.477 128.98 173.29 120.12 160.49 121.11 134.89 128. 98.462
175.26 177.23 177.23 151.63 178.22 130.95 177.23 98.462 212.68
175.26 214.65 118.15 215.63 80.738 208.74 68.923 249.11 173.29
242.22 122.09 237.29 86.646 234.34 48.246].
I did search but did not find explanation of u,v image coordinates and how to convert to x-y coordinates. It is not UV mapping, because the data is not between [0, 1]. I may be wrong.
any comments welcomed. Thanks
To be more confident, we can plot these points using Matlab/Octave or OpenCV on the corresponding color image and see if their positions match to the labeled joints. For joint structure we can look at the same README file W, T0, T1, T2, T3, I0, I1, I2, I3, M0, M1, M2, M3, R0, R1, R2, R3, L0, L1, L2, L3. Every joint has 2 coordinates so the sequence of 42 numbers correspond to u, v (X, Y) coordinates of corresponding joints in the sequence.
I tried to directly plot an image and 2D points in Matlab/Octave using this code:
clc; clear;
im = imread('0001_color_composed.png');
data = csvread('0001_joint2D.txt');
x = zeros(length(data)/2,1);
y = x;
for i = 1: length(data)/2
x(i) = data(2*i-1);
y(i) = data(2*i);
end
imshow(im);
hold on;
plot(x, y, 'go');
and these image and annotation. As you can see in the resulting image below all u, v coordinates correspond to pixel coordinates in X and Y counted from the top left corner of image in pixels, i.e. u = X, v = Y (as if image shown using imshow(), the origin of the coordinate frame for consecutive plots is set to the image coordinate frame origin which is the top left corner).
Suppose I have an image A, I applied Gaussian Blur on it with Sigam=3 So I got another Image B. Is there a way to know the applied sigma if A,B is given?
Further clarification:
Image A:
Image B:
I want to write a function that take A,B and return Sigma:
double get_sigma(cv::Mat const& A,cv::Mat const& B);
Any suggestions?
EDIT1: The suggested approach doesn't work in practice in its original form(i.e. using only 9 equations for a 3 x 3 kernel), and I realized this later. See EDIT1 below for an explanation and EDIT2 for a method that works.
EDIT2: As suggested by Humam, I used the Least Squares Estimate (LSE) to find the coefficients.
I think you can estimate the filter kernel by solving a linear system of equations in this case. A linear filter weighs the pixels in a window by its coefficients, then take their sum and assign this value to the center pixel of the window in the result image. So, for a 3 x 3 filter like
the resulting pixel value in the filtered image
result_pix_value = h11 * a(y, x) + h12 * a(y, x+1) + h13 * a(y, x+2) +
h21 * a(y+1, x) + h22 * a(y+1, x+1) + h23 * a(y+1, x+2) +
h31 * a(y+2, x) + h32 * a(y+2, x+1) + h33 * a(y+2, x+2)
where a's are the pixel values within the window in the original image. Here, for the 3 x 3 filter you have 9 unknowns, so you need 9 equations. You can obtain those 9 equations using 9 pixels in the resulting image. Then you can form an Ax = b system and solve for x to obtain the filter coefficients. With the coefficients available, I think you can find the sigma.
In the following example I'm using non-overlapping windows as shown to obtain the equations.
You don't have to know the size of the filter. If you use a larger size, the coefficients that are not relevant will be close to zero.
Your result image size is different than the input image, so i didn't use that image for following calculation. I use your input image and apply my own filter.
I tested this in Octave. You can quickly run it if you have Octave/Matlab. For Octave, you need to load the image package.
I'm using the following kernel to blur the image:
h =
0.10963 0.11184 0.10963
0.11184 0.11410 0.11184
0.10963 0.11184 0.10963
When I estimate it using a window size 5, I get the following. As I said, the coefficients that are not relevant are close to zero.
g =
9.5787e-015 -3.1508e-014 1.2974e-015 -3.4897e-015 1.2739e-014
-3.7248e-014 1.0963e-001 1.1184e-001 1.0963e-001 1.8418e-015
4.1825e-014 1.1184e-001 1.1410e-001 1.1184e-001 -7.3554e-014
-2.4861e-014 1.0963e-001 1.1184e-001 1.0963e-001 9.7664e-014
1.3692e-014 4.6182e-016 -2.9215e-014 3.1305e-014 -4.4875e-014
EDIT1:
First of all, my apologies.
This approach doesn't really work in the practice. I've used the filt = conv2(a, h, 'same'); in the code. The resulting image data type in this case is double, whereas in the actual image the data type is usually uint8, so there's loss of information, which we can think of as noise. I simulated this with the minor modification filt = floor(conv2(a, h, 'same'));, and then I don't get the expected results.
The sampling approach is not ideal, because it's possible that it results in a degenerated system. Better approach is to use random sampling, avoiding the borders and making sure the entries in the b vector are unique. In the ideal case, as in my code, we are making sure the system Ax = b has a unique solution this way.
One approach would be to reformulate this as Mv = 0 system and try to minimize the squared norm of Mv under the constraint squared-norm v = 1, which we can solve using SVD. I could be wrong here, and I haven't tried this.
Another approach is to use the symmetry of the Gaussian kernel. Then a 3x3 kernel will have only 3 unknowns instead of 9. I think, this way we impose additional constraints on v of the above paragraph.
I'll try these out and post the results, even if I don't get the expected results.
EDIT2:
Using the LSE, we can find the filter coefficients as pinv(A'A)A'b. For completion, I'm adding a simple (and slow) LSE code.
Initial Octave Code:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = conv2(a, h, 'same');
% use non-overlapping windows to for the Ax = b syatem
% NOTE: boundry error checking isn't performed in the code below
s = floor(size(a)/2);
y = s(1);
x = s(2);
w = k*k;
y1 = s(1)-floor(w/2) + r;
y2 = s(1)+floor(w/2);
x1 = s(2)-floor(w/2) + r;
x2 = s(2)+floor(w/2);
b = [];
A = [];
for y = y1:k:y2
for x = x1:k:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
A = [A; f(:)'];
end
end
% estimated filter kernel
g = reshape(A\b, k, k)
LSE method:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = floor(conv2(a, h, 'same'));
s = size(a);
y1 = r+2; y2 = s(1)-r-2;
x1 = r+2; x2 = s(2)-r-2;
b = [];
A = [];
for y = y1:2:y2
for x = x1:2:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
f = f(:)';
A = [A; f];
end
end
g = reshape(A\b, k, k) % A\b returns the least squares solution
%g = reshape(pinv(A'*A)*A'*b, k, k)
I searched around and it turns out the answer to this is surprising hard to find. Theres algorithm out there that can generate a random orientation in quaternion form but they involve sqrt and trig functions. I dont really need a uniformly distributed orientation. I just need to generate (many) quaternions such that their randomness in orientation is "good enough." I cant specify what is "good enough" except that I need to be able to do the generation quickly.
Quoted from http://planning.cs.uiuc.edu/node198.html:
Choose three points u, v, w ∈ [0,1] uniformly at random. A uniform, random quaternion is given by the simple expression:
h = ( sqrt(1-u) sin(2πv), sqrt(1-u) cos(2πv), sqrt(u) sin(2πw), sqrt(u) cos(2πw))
From Choosing a Point from the Surface of a Sphere by George Marsaglia:
Generate independent x, y uniformly in (-1..1) until z = x²+y² < 1.
Generate independent u, v uniformly in (-1..1) until w = u²+v² < 1.
Compute s = √((1-z) / w).
Return the quaternion (x, y, su, sv). It's already normalized.
This will generate a uniform random rotation because 4D spheres, unit quaternions and 3D rotations have equivalent measures.
The algorithm uses one square root, one division, and 16/π ≈ 5.09 random numbers on average. C++ code:
Quaternion random_quaternion() {
double x,y,z, u,v,w, s;
do { x = random(-1,1); y = random(-1,1); z = x*x + y*y; } while (z > 1);
do { u = random(-1,1); v = random(-1,1); w = u*u + v*v; } while (w > 1);
s = sqrt((1-z) / w);
return Quaternion(x, y, s*u, s*v);
}
Simplest way to generate it, just generate 4 random float and normalize it if required. If you want to produce rotation matrices later , than normalization can be skipped and convertion procedure should note nonunit quaternions.