I am trying to generate a cubic grid in Matlab so that I can produce a grid of M x N x Q cubes with M, N and Q being integer numbers. I don't need to plot it but rather to produce a B-Rep of the grid (vertex matrix and faces matrix - with no duplicate internal faces). I have tried two approaches:
Copy and translate points in the X, Y, Z direction, eliminate duplicate points and try to generate the new topology (I have no idea how).
Use the Matlab Neuronal Neural Network toolbox, specifically the gridplot function that produces a 3D grid of points but the faces matrix cannot be generated from this function.
Any suggestions?
Thank you.
Update
The vertex matrix contains all 8 points of each cube and the faces matrix all 6 faces of each cube. I can generate that with the following code:
clc
clear
fac = [1 2 3 4;
4 3 5 6;
6 7 8 5;
1 2 8 7;
6 7 1 4;
2 3 5 8];
vert_total = [];
face_total = fac;
for x = 0 : 1
for y = 0 : 1
for z = 0 : 1
vert = [1 1 0;
0 1 0;
0 1 1;
1 1 1;
0 0 1;
1 0 1;
1 0 0;
0 0 0];
vert(:,1) = vert(:,1) + x;
vert(:,2) = vert(:,2) + y;
vert(:,3) = vert(:,3) + z;
vert_total = [vert_total; vert];
face = face_total(end-5:end,:);
face_total = [face_total; face+8];
end
end
end
The problem with this code is that it contains duplicate vertex and duplicate faces. Eliminating the repeated vertex is pretty straightforward using the unique function, but I don't know how to handle the topology (faces matrix) when I eliminate the repeated points (obviously, some of the faces should be eliminated as well).
Any sugestions with that?
You can create the 3D grid, and then keep only those at 6 faces. Someone else may point a better way than this.
M = 5; N = 6; Q = 7;
[X, Y, Z] = ndgrid(1:M, 1:N, 1:Q); % 3D
faces = X==1 | X==M | Y==1 | Y==N | Z==1 | Z==Q;
X = X(faces);
Y = Y(faces);
Z = Z(faces);
Now [X Y Z] are coordinates for faces.
Related
I need to find a way to mesh triangles then to refine using refine.
My vertices of my original triangles are stored inside a matrix of size nb_points * 2.
My faces are stored in a nb_faces * 3 matrix.
The data for each face is stored in a nb_face * 1 matrix.
The meshing is done by diving the area using the middles of the segments of the triangles.
Example :
Origin :
vertices = [0 1 ;
2 3 ;
4 1 ;
4 5];
faces = [1 2 3;
2 3 4];
data = [1 2];
Result expected after meshing :
vertices = [0 1;
2 3;
4 1;
4 5;
1 2;
3 2;
2 1;
3 4;
4 3];
faces = [1 5 7;
2 5 6;
5 6 7;
7 6 3;
2 6 8;
6 8 9;
6 9 3;
8 4 9];
data = [1 1 1 1 2 2 2 2];
I am displaying using :
FV.Vertices = vertices;
FV.Faces = faces;
FV.FaceVertexCData = data;
figure; hold on; axis equal; grid on;
patch(FV,'FaceColor','flat');
Precision :
I do not want to use the following functions which gives way too many vertices and faces :
generateMesh()
refinemesh()
The data are temperatures since this is a simulation of heat transfer.
With a for loop it can be done quite easily, here is one solution:
% Dummy data
vertices = [0 1 ;
2 3 ;
4 1 ;
4 5];
faces = [1 2 3;
2 3 4];
data = [1 2];
% Number of vertices
vnum = size(vertices,1);
% new faces empty vector
nfaces = [];
% triangular shift
tshift = [2,-1,-1].';
% Run the algorithm
for ii = 1:size(faces,1)
% For each triangle get the 3 pairs of vertices
nsk = [faces(ii,1), faces(ii,2);faces(ii,2), faces(ii,3);faces(ii,3), faces(ii,1)];
% Compute the center of each pair of vertices
cmiddle = (vertices(nsk(:,1),:)+vertices(nsk(:,2),:))/2
% Compute the new faces
nfaces = [nfaces;[nsk(:,1),vnum+(ii*3-3+[1:3].'),vnum+(ii*3-3+[1:3].')+tshift];[(vnum+(ii*3-2)):(vnum+(ii*3))]]
% Add the new vertices
vertices = [vertices;cmiddle];
end
% Delete the duplicate vertices
[vertices,~,ind] = unique(vertices,'rows');
faces = ind(nfaces);
% Plot
figure; hold on; axis equal; grid on;
patch('Faces',faces,'Vertices',vertices,'FaceVertexCData',kron(data,ones(1,4)).','FaceColor','flat')
colorbar
If you find a way to generate the nsk vector without for loop you could even get rid of the loop. This code will only work on triangle, but it can be adapted if needed.
Result:
You can repeat the operation:
I have 4 known (corner) values of an matrix:
grid = [2 5
5 8];
This values are my corners and I want to interpolate between them, that I get a non-squared matrix, e.g. a 350x250 matrix.
I searched some familiar questions but I couldnt find a solution or wasnt able to transfer it to my problem. I found the interp2 function but not how to tell that I only have the 4 corner values?
I made an easy example to show what I need and what doesnt work (or what I am doing wrong):
test = [2 0 0 0 8
0 0 0 0 0
6 0 0 0 12 ];
[X, Y] = meshgrid(1:5,1:3);
M = interp2(X,Y,test,X,Y);
The resulting matrix is exactly the same as the input matrix test. How can I interpolate the zeros? In general my Input is only this:
grid = [2 8; 6 12];
After long thinking I found a solution:
x = 350;
y = 250;
V = [2 8
6 12];
[X, Y] = meshgrid(1:2,1:2);
a = linspace(1,2,x);
Xq = a(ones(1,y),:); % oder repmat([1 n], y);
b = linspace(1,2,y)';
Yq = b(:,ones(1,x));
M = interp2(X,Y,V,Xq,Yq);
I hope it helps anybody who has the same problem, I didnt understood that the final size is only in the last two parameters present.
How to vectorize this code in MATLAB?
n = 3;
x = zeros(n);
y = x;
for i = 1:n
x(:,i) = i;
y(i,:) = i;
end
I am not able to vectorize it. Please help.
You can use meshgrid :
n = 3;
[x,y] = meshgrid(1:n,1:n)
x =
1 2 3
1 2 3
1 2 3
y =
1 1 1
2 2 2
3 3 3
n=3;
[x,y]=meshgrid(1:n);
This uses meshgrid which does this automatically.
Or you can use bsxfun as Divakar suggests:
bsxfun(#plus,1:n,zeros(n,1))
Just as a note on your initial looped code: it's bad practise to use i as a variable
If I can add something to the mix, create a row vector from 1 to n, then use repmat on this vector to create x. After, transpose x to get y:
n = 3;
x = repmat(1:n, n, 1);
y = x.';
Running this code, we get:
>> x
x =
1 2 3
1 2 3
1 2 3
>> y
y =
1 1 1
2 2 2
3 3 3
How can I calculate in MatLab similarity transformation between 4 points in 3D?
I can calculate transform matrix from
T*X = Xp,
but it will give me affine matrix due to small errors in points coordinates. How can I fit that matrix to similarity one? I need something like fitgeotrans, but in 3D
Thanks
If I am interpreting your question correctly, you seek to find all coefficients in a 3D transformation matrix that will best warp one point to another. All you really have to do is put this problem into a linear system and solve. Recall that warping one point to another in 3D is simply:
A*s = t
s = (x,y,z) is the source point, t = (x',y',z') is the target point and A would be the 3 x 3 transformation matrix that is formatted such that:
A = [a00 a01 a02]
[a10 a11 a12]
[a20 a21 a22]
Writing out the actual system of equations of A*s = t, we get:
a00*x + a01*y + a02*z = x'
a10*x + a11*y + a12*z = y'
a20*x + a21*y + a22*z = z'
The coefficients in A are what we need to solve for. Re-writing this in matrix form, we get:
[x y z 0 0 0 0 0 0] [a00] [x']
[0 0 0 x y z 0 0 0] * [a01] = [y']
[0 0 0 0 0 0 x y z] [a02] [z']
[a10]
[a11]
[a12]
[a20]
[a21]
[a22]
Given that you have four points, you would simply concatenate rows of the matrix on the left side and the vector on the right
[x1 y1 z1 0 0 0 0 0 0] [a00] [x1']
[0 0 0 x1 y1 z1 0 0 0] [a01] [y1']
[0 0 0 0 0 0 x1 y1 z1] [a02] [z1']
[x2 y2 z2 0 0 0 0 0 0] [a10] [x2']
[0 0 0 x2 y2 z2 0 0 0] [a11] [y2']
[0 0 0 0 0 0 x2 y2 z2] [a12] [z2']
[x3 y3 z3 0 0 0 0 0 0] * [a20] = [x3']
[0 0 0 x3 y3 z3 0 0 0] [a21] [y3']
[0 0 0 0 0 0 x3 y3 z3] [a22] [z3']
[x4 y4 z4 0 0 0 0 0 0] [x4']
[0 0 0 x4 y4 z4 0 0 0] [y4']
[0 0 0 0 0 0 x4 y4 z4] [z4']
S * a = T
S would now be a matrix that contains your four source points in the format shown above, a is now a vector of the transformation coefficients in the matrix you want to solve (ordered in row-major format), and T would be a vector of target points in the format shown above.
To solve for the parameters, you simply have to use the mldivide operator or \ in MATLAB, which will compute the least squares estimate for you. Therefore:
a = S^{-1} * T
As such, simply build your matrix like above, then use the \ operator to solve for your transformation parameters in your matrix. When you're done, reshape T into a 3 x 3 matrix. Therefore:
S = ... ; %// Enter in your source points here like above
T = ... ; %// Enter in your target points in a right hand side vector like above
a = S \ T;
similarity_matrix = reshape(a, 3, 3).';
With regards to your error in small perturbations of each of the co-ordinates, the more points you have the better. Using 4 will certainly give you a solution, but it isn't enough to mitigate any errors in my opinion.
Minor Note: This (more or less) is what fitgeotrans does under the hood. It computes the best homography given a bunch of source and target points, and determines this using least squares.
Hope this answered your question!
The answer by #rayryeng is correct, given that you have a set of up to 3 points in a 3-dimensional space. If you need to transform m points in n-dimensional space (m>n), then you first need to add m-n coordinates to these m points such that they exist in m-dimensional space (i.e. the a matrix in #rayryeng becomes a square matrix)... Then the procedure described by #rayryeng will give you the exact transformation of points, you then just need to select only the coordinates of the transformed points in the original n-dimensional space.
As an example, say you want to transform the points:
(2 -2 2) -> (-3 5 -4)
(2 3 0) -> (3 4 4)
(-4 -2 5) -> (-4 -1 -2)
(-3 4 1) -> (4 0 5)
(5 -4 0) -> (-3 -2 -3)
Notice that you have m=5 points which are n=3-dimensional. So you need to add coordinates to these points such that they are n=m=5-dimensional, and then apply the procedure described by #rayryeng.
I have implemented a function that does that (find it below). You just need to organize the points such that each of the source-points is a column in a matrix u, and each of the target points is a column in a matrix v. The matrices u and v are going to be, thus, 3 by 5 each.
WARNING:
the matrix A in the function may require A LOT of memory for moderately many points nP, because it has nP^4 elements.
To overcome this, for square matrices u and v, you can simply use T=v*inv(u) or T=v/u in MATLAB notation.
The code may run very slowly...
In MATLAB:
u = [2 2 -4 -3 5;-2 3 -2 4 -4;2 0 5 1 0]; % setting the set of source points
v = [-3 3 -4 4 -3;5 4 -1 0 -2;-4 4 -2 5 -3]; % setting the set of target points
T = findLinearTransformation(u,v); % calculating the transformation
You can verify that T is correct by:
I = eye(5);
uu = [u;I((3+1):5,1:5)]; % filling-up the matrix of source points so that you have 5-d points
w = T*uu; % calculating target points
w = w(1:3,1:5); % recovering the 3-d points
w - v % w should match v ... notice that the error between w and v is really small
The function that calculates the transformation matrix:
function [T,A] = findLinearTransformation(u,v)
% finds a matrix T (nP X nP) such that T * u(:,i) = v(:,i)
% u(:,i) and v(:,i) are n-dim col vectors; the amount of col vectors in u and v must match (and are equal to nP)
%
if any(size(u) ~= size(v))
error('findLinearTransform:u','u and v must be the same shape and size n-dim vectors');
end
[n,nP] = size(u); % n -> dimensionality; nP -> number of points to be transformed
if nP > n % if the number of points to be transform exceeds the dimensionality of points
I = eye(nP);
u = [u;I((n+1):nP,1:nP)]; % then fill up the points to be transformed with the identity matrix
v = [v;I((n+1):nP,1:nP)]; % as well as the transformed points
[n,nP] = size(u);
end
A = zeros(nP*n,n*n);
for k = 1:nP
for i = ((k-1)*n+1):(k*n)
A(i,mod((((i-1)*n+1):(i*n))-1,n*n) + 1) = u(:,k)';
end
end
v = v(:);
T = reshape(A\v, n, n).';
end
I have an intensity/greyscale image, and I have chosen a pixel inside this image. I want to send vectors starting from this pixel in all directions/angles, and I want to sum all the intensities of the pixels touching one vector, for all vectors.
After this step I would like to plot a histogram with the intensities on one axis and the angle on the other axis. I think I can do this last step on my own, but I don't know how to create these vectors inside my greyscale image and how to get the coordinates of the pixels a vector touches.
I previously did this in C++, which required a lot of code. I am sure this can be done with less effort in MATLAB, but I am quite new to MATLAB, so any help would be appreciated, since I haven't found anything helpful in the documentation.
It might not be the best way to solve it, but you can do it using a bit of algebra, heres how...
We know the Point-Slope formula of a line passing through point (a,b) with angle theta is:
y = tan(theta) * (x-a) + b
Therefore a simple idea is to compute the intersection of this line with y=const for all const, and read the intensity values at the intersection. You would repeat this for all angles...
A sample code to illustrate the concept:
%% input
point = [128 128]; % pixel location
I = imread('cameraman.tif'); % sample grayscale image
%% calculations
[r c] = size(I);
angles = linspace(0, 2*pi, 4) + rand;
angles(end) = [];
clr = lines( length(angles) ); % get some colors
figure(1), imshow(I), hold on
figure(2), hold on
for i=1:length(angles)
% line equation
f = #(x) tan(angles(i))*(x-point(1)) + point(2);
% get intensities along line
x = 1:c;
y = round(f(x));
idx = ( y<1 | y>r ); % indices of outside intersections
vals = diag(I(x(~idx), y(~idx)));
figure(1), plot(x, y, 'Color', clr(i,:)) % plot line
figure(2), plot(vals, 'Color', clr(i,:)) % plot profile
end
hold off
This example will be similar to Amro's, but it is a slightly different implementation that should work for an arbitrary coordinate system assigned to the image...
Let's assume that you have matrices of regularly-spaced x and y coordinates that are the same size as your image, such that the coordinates of pixel (i,j) are given by (x(i,j),y(i,j)). As an example, I'll create a sample 5-by-5 set of integer coordinates using MESHGRID:
>> [xGrid,yGrid] = meshgrid(1:5)
xGrid =
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
yGrid =
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
Next we can define a line y = m*(x - a) + b passing through the coordinate system by selecting some values for the constants and computing y using the x coordinates of the grid:
>> a = 0;
>> b = 1;
>> m = rand
m =
0.5469
>> y = m.*(xGrid(1,:)-a)+b
y =
1.5469 2.0938 2.6406 3.1875 3.7344
Finally, we find the y points in the grid that differ from the points computed above by less than the grid size:
>> index = abs(yGrid-repmat(y,size(yGrid,1),1)) <= yGrid(2,1)-yGrid(1,1)
index =
1 0 0 0 0
1 1 1 0 0
0 1 1 1 1
0 0 0 1 1
0 0 0 0 0
and use this index matrix to get the x and y coordinates for the pixels crossed by the line:
>> xCrossed = xGrid(index);
>> yCrossed = yGrid(index);