There is a noisy image as fig a. by reducing image noise finally a smoothspline could fit to the dots as shown in fig b. now it's desire to find two axillary lines both side of the original line like offset (in this case it's better to say outline). how can these lines(yellow and green) position be found?
if there is a simple straight line it would be easy but here is spline.
any idea would be appreciated.
I think this is what you wanted:
% generate random curve
xy = randi(5,[2,3]);
t = 1:3;
tq = linspace(1,3,100);
xyq = interp1(t',xy',tq','spline');
xx = xyq(:,1);
yy = xyq(:,2);
% get curve's approx. angle in each point
theta = atan2(xyq(2:end,2) - xyq(1:end-1,2),xyq(2:end,1) - xyq(1:end-1,1));
theta(end+1) = theta(end);
% add or subtract 90 degrees to get downward or upward normal angle
tp1 = theta + pi/2;
tp2 = theta - pi/2;
% distance from original curve
d = 0.1;
% compute x-y additions
[xa1,ya1] = pol2cart(tp1,d);
[xa2,ya2] = pol2cart(tp2,d);
% plot curve and its axillary lines
plot(xx,yy,'g')
hold on
plot(xyq(:,1) + xa1,xyq(:,2) + ya1,'b')
plot(xyq(:,1) + xa2,xyq(:,2) + ya2,'r')
legend('orig.','axil._1','axil._2');
and you get this:
Related
How to distribute the points to be like Fig.A
This matlab code for Fig. B :
N = 30; % number of points
r = 0.5; % r = radius
d = 50; % dimension
C_point = 0; % center point
figure, clf
C = ones(1, d) * C_point;
C_rep = repmat( C,N,1);
X = randn(N,d);
s2 = sum(X.^2,2) ;
radius = r * (rand(N,1).^(1/d));
X = X.*repmat(radius./sqrt(s2),1,d) + C_rep;
%% Plot 2D
t = linspace(0, 2*pi, 100);
x = r*cos(t) + C(1);
y = r*sin(t) + C(2);
plot(x,y,'b')
hold on
plot(C(1),C(2),'b.', 'MarkerSize', 10) % center point
hold on
plot(X(:,1), X(:,2),'r.','markersize',10);
axis equal;rotate3d off; rotate3d on;drawnow;shg;
hold on
ax = axis;
Source of the code
What I should change to be like fig. A
The OP's code computes points uniformly distributed within a d-dimensional box, projects those onto a d-dimensional sphere, then samples the radius to move them inside the d-dimensional ball. This is perfect except that the points inside the box, when projected onto the sphere, do not form a uniform distribution on that sphere. If instead you find random points distributed in a Gaussian distribution, you are guaranteed uniform angle distribution.
First compute points with a Gaussian distribution in d dimensions (I do all here with minimal changes to the OP's code):
N = 1000; % number of points
r = 0.5; % r = radius
d = 3; % dimension
C_point = 0; % center point
C = ones(1,d) * C_point;
C_rep = repmat(C,N,1);
X = randn(N,d);
Note that I use randn, not rand. randn creates a Gaussian distribution.
Next we normalize the vectors so the points move to the sphere:
nX = sqrt(sum(X.^2,2));
X = X./repmat(nX,1,d);
These points are uniformly distributed, which you can verify by scatter3(X(:,1),X(:,2),X(:,3)); axis equal and turning the display around (a 2D rendering doesn't do it justice). This is the reason I set d=3 above, and N=1000. I wanted to be able to plot the points and see lots of them.
Next we compute, as you already did, a random distance to the origin, and correct it for the dimensionality:
radius = r * (rand(N,1).^(1/d));
X = X.*repmat(radius,1,d) + C_rep;
X now is distributed uniformly in the ball. Again, scatter3(X(:,1),X(:,2),X(:,3)); axis equal shows this.
However, if you set d=50 and then plot only two dimensions of your data, you will not see the data filling the circle. And you will not see a uniform distribution either. This is because you are projecting a 50-D ball onto 2 dimensions, this simply does not work. You either have to trust the math, or you have to slice the data:
figure, hold on
t = linspace(0, 2*pi, 100);
x = r*cos(t) + C(1);
y = r*sin(t) + C(2);
plot(x,y,'b')
plot(C(1),C(2),'b.', 'MarkerSize', 10) % center point
axis equal
I = all(abs(X(:,3:d))<0.1,2);
plot(X(I,1), X(I,2),'r.','markersize',10);
The I there indexes points that are close to the origin in dimensions perpendicular to the first two shown. Again, with d=50 you will have very few points there, so you will need to set N very large! To see the same density of points as in the case above, for every dimension you add, you need to multiply N by 10. So for d=5 you'd have N=1000*10*10=1e5, and for d=50 you'd need N=1e50. That is totally impossible to compute, of course.
I have a piecewise function, where domain changes for each case. The function is as follows:
For
(x,y)greater than Divider v= f(x,y) (A1)
(x,y)less than Divider v = g(x,y) (A2)
The location of the divider changes with tilt angle of the rectangle given in figures 1 and 2.Figure 1 & 2 The divider will always be a bisector of the rectangle. For example, the divider makes an angle (alpha + 90) with the horizontal.
If the rectangle makes an angle 0, it's easy to implement above functions as I can create meshgrid from
x =B to C & y = A to D for A1
x =A to B & y = A to D for A2
However, when the angles for the rectangle are different, I can't figure out how to create the mesh to calculate the function v using the algorithm A1 and A2 above.
I was thinking of using some inequality and using the equation of the line (as I have the co-ordinates for the center of the rectangle and the angle of tilt). But, I can't seem to think of a way to do it for all angles (for example , slope of pi/2 as in the first figure, yields infinity). Even if I do create some kind of inequality, I can't create a mesh.
1Please help me with this problem. I have wasted a lot of time on this. It seems to be out of my reach
%% Constants
Angle1=0;
Angle1=Angle1.*pi./180;
rect_center=0; % in m
rect_length=5; % in m
rect_width=1; % in m
rect_strength=1.8401e-06;
Angle2=0;
Angle2 =Angle2.*pi./180;
%% This code calculates the outer coordinates of the rectangle by using the central point
% the following code calculates the vertices
vertexA=rect_center+(-rect_width./2.*exp(1i.*1.5708)-rect_length./2).*exp(1i.*Angle2);
vertexA=[vertexA,vertexA+2.*(rect_width./2.*exp(1i.*1.5708)).*exp(1i.*Angle2)];
vertexB=rect_center+(-rect_width./2.*exp(1i.*1.5708)+rect_length./2).*exp(1i.*Angle2);
vertexB=[vertexB,vertexB+2.*(rect_width./2.*exp(1i.*1.5708)).*exp(1i.*Angle2)];
za1=vertexA(1:numel(vertexA)/2);
za2=vertexA(1+numel(vertexA)/2:numel(vertexA));
zb1=vertexB(1:numel(vertexB)/2);
zb2=vertexB(1+numel(vertexB)/2:numel(vertexB));
arg1=exp(-1i.*Angle2);
%% This Section makes the two equations necessary for making the graphs
syms var_z
% Equation 1
Eqn1(var_z)=1.5844e-07.*exp(-1i.*Angle1).*var_z./9.8692e-13;
% subparts of the Equation 2
A = 1.0133e+12.*(-1i.*rect_strength.*exp(-1i*Angle2)./(2*pi.*rect_length.*rect_width*0.2));
ZA1 = var_z+za1-2*rect_center;
ZA2 = var_z+za2-2*rect_center;
ZB1 = var_z+zb1-2*rect_center;
ZB2 = var_z+zb2-2*rect_center;
ZAA2 = log(abs(ZA2)) + 1i*mod(angle(ZA2),2*pi);
ZAA1 = log(abs(ZA1)) + 1i*mod(angle(ZA1),2*pi);
ZBB1 = log(abs(ZB1)) + 1i*mod(angle(ZB1),2*pi);
ZBB2 = log(abs(ZB2)) + 1i*mod(angle(ZB2),2*pi);
%Equation 2 ; this is used for the left side of the center
Eqn2= A*(ZA2*(log(ZA2)-1)-(ZA1*(log(ZA1)-1))+(ZB1*(log(ZB1)-1))-(ZB2*(log(ZB2)-1)));
%Equation 3 ; this is used for the right side of the center
Eqn3 = A.*(ZA2*(ZAA2-1)-(ZA1*(ZAA1-1))+(ZB1*(ZBB1-1))-(ZB2*(ZBB2-1)));
%Equation 4 :Add Equation 2 and Equation 1; this is used for the left side of the center
Eqn4 = matlabFunction(Eqn1+Eqn2,'vars',var_z);
%Equation 5: Add Equation 3 and Equation 1; this is used for the right side of the center
Eqn5 = matlabFunction(Eqn1+Eqn3,'vars',var_z);
%% Prepare for making the plots
minx=-10; %min x coordinate
maxx=10; %max x coordinate
nr_x=1000; %nr x points
miny=-10; %min y coordinate
maxy=10; %max y coordinate
nr_y=1000; %nr y points
%This vector starts from left corner (minx) to the middle of the plot surface,
%The middle of the plot surface lies at the center of the rectange
%created earlier
xvec1=minx:(rect_center-minx)/(0.5*nr_x-1):rect_center;
%This vector starts from middle to the right corner (maxx) of the plot surface,
%The middle of the plot surface lies at the center of the rectange
%created earlier
xvec2=rect_center:(maxx-rect_center)/(0.5*nr_x-1):maxx;
%the y vectors start from miny to maxy
yvec1=miny:(maxy-miny)/(nr_y-1):maxy;
yvec2=miny:(maxy-miny)/(nr_y-1):maxy;
% create mesh from above vectors
[x1,y1]=meshgrid(xvec1,yvec1);
[x2,y2]=meshgrid(xvec2,yvec2);
z1=x1+1i*y1;
z2=x2+1i*y2;
% Calculate the above function using equation 4 and equation 5 using the mesh created above
r1 = -real(Eqn5(z1));
r2 = -real(Eqn4(z2));
%Combine the calculated functions
Result = [r1 r2];
%Combine the grids
x = [x1 x2];
y = [y1 y2];
% plot contours
[c,h]=contourf(x,y,Result(:,:,1),50,'LineWidth',1);
% plot the outerboundary of the rectangle
line_x=real([vertexA;vertexB]);
line_y=imag([vertexA;vertexB]);
line(line_x,line_y,'color','r','linestyle',':','linewidth',5)
The final Figure is supposed to look like this.Final Expected Figure.
I'm not sure which angle defines the dividing line so I assume it's Angle1. It looks like logical indexing is the way to go here. Instead of creating two separate mesh grids we simply create the entire mesh grid then partition it into two sets and operate on each independently.
%% Prepare for making the plots
minx=-10; %min x coordinate
maxx=10; %max x coordinate
nr_x=1000; %nr x points
miny=-10; %min y coordinate
maxy=10; %max y coordinate
nr_y=1000; %nr y points
% create full mesh grid
xvec=linspace(minx,maxx,nr_x);
yvec=linspace(miny,maxy,nr_y);
[x,y]=meshgrid(xvec,yvec);
% Partition mesh based on divider line
% Assumes the line passes through (ox,oy) with normal vector defined by Angle1
ox = rect_center;
oy = rect_center;
a = cos(Angle1);
b = sin(Angle1);
c = -(a*ox + b*oy);
% use logical indexing to opperate on the appropriate parts of the mesh
idx1 = a*x + b*y + c < 0;
idx2 = ~idx1;
z = zeros(size(x));
z(idx1) = x(idx1) + 1i*y(idx1);
z(idx2) = x(idx2) + 1i*y(idx2);
% Calculate the above function using equation 4 and equation 5
% using the mesh created above
Result = zeros(size(z));
Result(idx1) = -real(Eqn5(z(idx1)));
Result(idx2) = -real(Eqn4(z(idx2)));
For example with Angle1 = 45 and Angle2 = 45 we get the following indexing
>> contourf(x,y,idx1);
>> line(line_x,line_y,'color','r','linestyle',':','linewidth',5);
where the yellow region uses Eqn5 and the blue region uses Eqn4. This agrees with the example you posted but I don't know what the resulting contour map for other cases is supposed to look like.
Hope this helps.
I'm trying to make a surf plot that looks like:
So far I have:
x = [-1:1/100:1];
y = [-1:1/100:1];
[X,Y] = meshgrid(x,y);
Triangle1 = -abs(X) + 1.5;
Triangle2 = -abs(Y) + 1.5;
Z = min(Triangle1, Triangle2);
surf(X,Y,Z);
shading flat
colormap winter;
hold on;
[X,Y,Z] = sphere();
Sphere = surf(X, Y, Z + 1.5 );% sphere with radius 1 centred at (0,0,1.5)
hold off;
This code produces a graph that looks like :
A pyramid with square base ([-1,1]x[-1,1]) and vertex at height c = 1.5 above the origin (0,0) is erected.
The top of the pyramid is hollowed out by removing the portion of it that falls within a sphere of radius r=1 centered at the vertex.
So I need to keep the part of the surface of the sphere that is inside the pyramid and delete the rest. Note that the y axis in each plot is different, that's why the second plot looks condensed a bit. Yes there is a pyramid going into the sphere which is hard to see from that angle.
I will use viewing angles of 70 (azimuth) and 35 (elevation). And make sure the axes are properly scaled (as shown). I will use the AXIS TIGHT option to get the proper dimensions after the removal of the appropriate surface of the sphere.
Here is my humble suggestion:
N = 400; % resolution
x = linspace(-1,1,N);
y = linspace(-1,1,N);
[X,Y] = meshgrid(x,y);
Triangle1 = -abs(X)+1.5 ;
Triangle2 = -abs(Y)+1.5 ;
Z = min(Triangle1, Triangle2);
Trig = alphaShape(X(:),Y(:),Z(:),2);
[Xs,Ys,Zs] = sphere(N-1);
Sphere = alphaShape(Xs(:),Ys(:),Zs(:)+2,2);
% get all the points from the pyramid that are within the sphere:
inSphere = inShape(Sphere,X(:),Y(:),Z(:));
Zt = Z;
Zt(inSphere) = nan; % remove the points in the sphere
surf(X,Y,Zt)
shading interp
view(70,35)
axis tight
I use alphaShape object to remove all unwanted points from the pyramid and then plot it without them:
I know, it's not perfect, as you don't see the bottom of the circle within the pyramid, but all my tries to achieve this have failed. My basic idea was plotting them together like this:
hold on;
Zc = Zs;
inTrig = inShape(Trig,Xs(:),Ys(:),Zs(:)+1.5);
Zc(~inTrig) = nan;
surf(Xs,Ys,Zc+1.5)
hold off
But the result is not so good, as you can't really see the circle within the pyramid.
Anyway, I post this here as it might give you a direction to work on.
An alternative to EBH's method.
A general algorithm from subtracting two shapes in 3d is difficult in MATLAB. If instead you remember that the equation for a sphere with radius r centered at (x0,y0,z0) is
r^2 = (x-x0)^2 + (y-y0)^2 + (z-z0)^2
Then solving for z gives z = z0 +/- sqrt(r^2-(x-x0)^2-(y-y0)^2) where using + in front of the square root gives the top of the sphere and - gives the bottom. In this case we are only interested in the bottom of the sphere. To get the final surface we simply take the minimum z between the pyramid and the half-sphere.
Note that the domain of the half-sphere is defined by the filled circle r^2-(x-x0)^2-(y-y0)^2 >= 0. We define any terms outside the domain as infinity so that they are ignored when the minimum is taken.
N = 400; % resolution
z0 = 1.5; % sphere z offset
r = 1; % sphere radius
x = linspace(-1,1,N);
y = linspace(-1,1,N);
[X,Y] = meshgrid(x,y);
% pyramid
Triangle1 = -abs(X)+1.5 ;
Triangle2 = -abs(Y)+1.5 ;
Pyramid = min(Triangle1, Triangle2);
% half-sphere (hemisphere)
sqrt_term = r^2 - X.^2 - Y.^2;
HalfSphere = -sqrt(sqrt_term) + z0;
HalfSphere(sqrt_term < 0) = inf;
Z = min(HalfSphere, Pyramid);
surf(X,Y,Z)
shading interp
view(70,35)
axis tight
I'm trying to find vanishing points of vanishing lines to estimate a depth map for a 2D image.
First, I detected the vanishing lines of the 2D image using hough transform. Here is my code in Matlab:
Img =imread('landscape.bmp'); %read the 2D image
%Convert the image to Grayscale
I=rgb2gray(Img);
%Edge Detection
Ie=edge(I,'sobel');
%Hough Transform
[H,theta,rho] = hough(Ie);
% Finding the Hough peaks (number of peaks is set to 5)
P = houghpeaks(H,5,'threshold',ceil(0.2*max(H(:))));
x = theta(P(:,2));
y = rho(P(:,1));
%Vanishing lines
lines = houghlines(I,theta,rho,P,'FillGap',170,'MinLength',350);
[rows, columns] = size(Ie);
figure, imshow(~Ie)
hold on
xy_1 = zeros([2,2]);
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
% Get the equation of the line
x1 = xy(1,1);
y1 = xy(1,2);
x2 = xy(2,1);
y2 = xy(2,2);
slope = (y2-y1)/(x2-x1);
xLeft = 1; % x is on the left edge
yLeft = slope * (xLeft - x1) + y1;
xRight = columns; % x is on the reight edge.
yRight = slope * (xRight - x1) + y1;
plot([xLeft, xRight], [yLeft, yRight], 'LineWidth',1,'Color','blue');
%intersection of two lines (the current line and the previous one)
slopee = #(line) (line(2,2) - line(1,2))/(line(2,1) - line(1,1));
m1 = slopee(xy_1);
m2 = slopee(xy);
intercept = #(line,m) line(1,2) - m*line(1,1);
b1 = intercept(xy_1,m1);
b2 = intercept(xy,m2);
xintersect = (b2-b1)/(m1-m2);
yintersect = m1*xintersect + b1;
plot(xintersect,yintersect,'m*','markersize',8, 'Color', 'red')
xy_1 = xy;
% Plot original points on the lines .
plot(xy(1,1),xy(1,2),'x','markersize',8,'Color','yellow');
plot(xy(2,1),xy(2,2),'x','markersize',8,'Color','green');
end
Now I need to find the vanishing point to be able to estimate the depth map.
The vanishing point is chosen as the intersection point with the greatest number of intersections around it.
My question, in other words, is how can I find the intersection of a number of lines (vanishing lines) in Matlab? I guess one way to do it is to find the point whose sum of squared distances from all lines is minimal, but not sure how to do that in Matlab?
Any help would be appreciated.
Edit: I tried to find the intersection of the lines, but I could only find the intersection of each a line and the line after it. I don't know how to find the intersection of all the lines?
Here is an example of a picture I am using:
https://www.dropbox.com/s/mbdt6v60ug1nymb/landscape.bmp?dl=0
I am posting a link because I don't have enough reputations to post an image.
A simplistic approach:
You should be able to create an array with all the intersection points between the lines.
Pseudo code:
for i = 1:length(lines)-1
for j = i+1:length(lines)
//add intersection of lines i and j
If you have all the intersections, you could simply take the average.
OR, take the approach written up here:
https://math.stackexchange.com/questions/61719/finding-the-intersection-point-of-many-lines-in-3d-point-closest-to-all-lines
3d can be simplified to 2d :)
Given unit circle, and a set of M smaller circles of radius r. Find the maximum radius of the smaller circles that allows them all to fit inside the unit circle without overlap.
I have the following circles packing in polygon example link
I want to change equations that say that all circles are inside the polygon
theta = 2*pi/N; % Angle covered by each side of the polygon
phi = theta*(0:N-1)'; % Direction of the normal to the side of the polygon
polyEq = ( [cos(phi) sin(phi)]*x <= cdist-r );
to equations that say that all circles are inside the circle, but i don't know how. Can somebody help me?
Kind regards.
In your case, there are no "sides of a polygon" so there is no analogue to theta and you'll need to change every place theta is referenced, and all variables that reference theta (like phi).
Something like the following should work. I've just copy-pasted the code from your link, gotten rid of theta and phi, redefined cdist and polyEq, and made it plot a unit circle in the answer instead of a polygon. Please ask if any of these choices are unclear.
M = 19; % number of circles to fit
toms r % radius of circles
x = tom('x', 2, M); % coordinates of circle centers
clear pi % 3.1415...
cdist = 1; % radius of unit circle
%%%%%% equations saying all circles are inside of unit circle
polyEq = (sqrt(x(1,:).^2 + x(2,:).^2) + r <= cdist);
% create a set of equations that say that no circles overlap
circEq = cell(M-1,1);
for i=1:M-1
circEq{i} = ( sqrt(sum((x(:,i+1:end)-repmat(x(:,i),1,M-i)).^2)) >= 2*r );
end
% starting guess
x0 = { r == 0.5*sqrt(1/M), x == 0.3*randn(size(x)) };
% solve the problem, maximizing r
options = struct;
% try multiple starting guesses and choose best result
options.solver = 'multimin';
options.xInit = 30; % number of different starting guesses
solution = ezsolve(-r,{polyEq,circEq},x0,options);
% plot result
x_vals = (0:100)./100;
plot(sqrt(1 - x_vals.^2),'-') % top of unit circle
plot(-1.*sqrt(1 - x_vals.^2),'-') % bottom of unit circle
axis image
hold on
alpha = linspace(0,2*pi,100);
cx = solution.r*cos(alpha);
cy = solution.r*sin(alpha);
for i=1:M
plot(solution.x(1,i)+cx,solution.x(2,i)+cy) % circle number i
end
hold off
title(['Maximum radius = ' num2str(solution.r)]);