I am using the following code for Pi approximation, using the rejection sampling method.
% DISPLAY A CIRCLE INSCRIBED IN A SQUARE
figure;
a = 0:.01:2*pi;
x = cos(a); y = sin(a);
hold on
plot(x,y,'k','Linewidth',2)
t = text(0.5, 0.05,'r');
l = line([0 1],[0 0],'Linewidth',2);
axis equal
box on
xlim([-1 1])
ylim([-1 1])
title('Unit Circle Inscribed in a Square')
pause;
rand('seed',12345)
randn('seed',12345)
delete(l); delete(t);
% DRAW SAMPLES FROM PROPOSAL DISTRIBUTION
samples = 2*rand(2,100000) - 1;
% REJECTION
reject = sum(samples.^2) > 1;
% DISPLAY REJECTION CRITERION
scatter(samples(1,~reject),samples(2,~reject),'b.')
scatter(samples(1,reject),samples(2,reject),'rx')
hold off
xlim([-1 1])
ylim([-1 1])
The expected result should be a blue circle inside of a red square. When I run the code the red point for building the square are displayed, but the blue point not.
The expected result should be as the picture found at this Link
But I get the following result:
Does anybody know what I can possibly do in order to visualize the blue points as well? Thanks in advance.
The scatter function is probably not the best choice for such huge number of points. Try this instead:
t = 0:.01:2*pi;
x = cos(t); y = sin(t);
samples = 2*rand(2,100000) - 1;
reject = sum(samples.^2) > 1;
props = {'LineStyle','none', 'Marker','.', 'MarkerSize',1};
line(x, y, 'Color','k', 'LineWidth',2)
line(samples(1,~reject), samples(2,~reject), props{:}, 'Color','b')
line(samples(1,reject), samples(2,reject), props{:}, 'Color','r')
axis equal; axis([-1 1 -1 1])
box on
Related
I try to solve the following 2D elliptic PDE electrostatic problem by fixing the Parallel plate Capacitors code. But I have problem to plot the circle region. How can I plot a circle region rather than the square?
% I use following two lines to label the 50V and 100V squares
% (but it should be two circles)
V(pp1-r_circle_small:pp1+r_circle_small,pp1-r_circle_small:pp1+r_circle_small) = 50;
V(pp2-r_circle_big:pp2+r_circle_big,pp2-r_circle_big:pp2+r_circle_big) = 100;
% Contour Display for electric potential
figure(1)
contour_range_V = -101:0.5:101;
contour(x,y,V,contour_range_V,'linewidth',0.5);
axis([min(x) max(x) min(y) max(y)]);
colorbar('location','eastoutside','fontsize',10);
xlabel('x-axis in meters','fontsize',10);
ylabel('y-axis in meters','fontsize',10);
title('Electric Potential distribution, V(x,y) in volts','fontsize',14);
h1=gca;
set(h1,'fontsize',10);
fh1 = figure(1);
set(fh1, 'color', 'white')
% Contour Display for electric field
figure(2)
contour_range_E = -20:0.05:20;
contour(x,y,E,contour_range_E,'linewidth',0.5);
axis([min(x) max(x) min(y) max(y)]);
colorbar('location','eastoutside','fontsize',10);
xlabel('x-axis in meters','fontsize',10);
ylabel('y-axis in meters','fontsize',10);
title('Electric field distribution, E (x,y) in V/m','fontsize',14);
h2=gca;
set(h2,'fontsize',10);
fh2 = figure(2);
set(fh2, 'color', 'white')
You're creating a square due to the way you're indexing (see this post on indexing). You've specified the rows to run from pp1-r_circle_small to pp1+r_circle_small and similar for the columns. Given that Swiss cheese is not an option, you're creating a complete square.
From geometry we know that all points within distance sqrt((X-X0)^2 - (Y-Y0)^2) < R from the centre of the circle at (X0,Y0) with radius R are within the circle, and the rest outside. This means that you can simply build a mask:
% Set up your grid
Xsize = 30; % Your case: 1
Ysize = 30; % Your case: 1
step = 1; % Amount of gridpoints; use 0.001 or something
% Build indexing grid for circle search, adapt as necessary
X = 0:step:Xsize;
Y = 0:step:Ysize;
[XX,YY] = meshgrid(X, Y);
V = zeros(numel(X), numel(Y));
% Repeat the below for both circles
R = 10; % Radius of your circle; your case 0.1 and 0.15
X0 = 11; % X coordinate of the circle's origin; your case 0.3 and 0.7
Y0 = 15; % Y coordinate of the circle's origin; your case 0.3 and 0.7
% Logical check to see whether a point is inside or outside
mask = sqrt( (XX - X0).^2 + (YY - Y0).^2) < R;
V(mask) = 50; % Give your circle the desired value
imagesc(V) % Just to show you the result
axis equal % Use equal axis to have no image distortion
mask is a logical matrix containing 1 where points are within your circle and 0 where points are outside. You can then use this mask to logically index your potential grid V to set it to the desired value.
Note: This will, obviously, not create a perfect circle, given you cannot plot a perfect circle on a square grid. The finer the grid, the more circle-like your "circle" will be. This shows the result with step = 0.01
Note 2: You'll need to tweek your definition of X, Y, X0, Y0 and R to match your values.
Applying 3D rotation matrix to the x,y,z values obtained from surface function object. The error I get is due to the matrix not being nonconforment but how can I adjust the matrix correctly?
I know hgtransform / makehgtform can do rotations but I need to use rotation matrices since I plan on testing it using matrices created from quaternions.
I've created a little plane out of cylinders and the surface functions.
See code below:
clear all,clf
ax=axes('XLim',[-2 2],'YLim', [-2 10],'ZLim',[-1.5 1.5]);
grid on;
%axis equal;
xlabel('x');
ylabel('y');
zlabel('z');
ax
% rotate around
rot_mat = [.707 -.707 0;.707 .707 0; 0 0 1] %rotation matrix
[xc yc zc] = cylinder([0.1 0.0]); %cone
[x y z]= cylinder([0.2 0.2]);
h(1) = surface(xc,zc,-yc,'FaceColor', 'red'); %noise cone
h(2) = surface(z,y,0.5*x,'FaceColor', 'blue'); %right wing
h(3) = surface(-z,y,0.5*x,'FaceColor', 'yellow');%left wing
h(4) = surface(x,-1.5*z,0.5*y,'FaceColor', 'green'); %main body
h(5) = surface(xc,(1.5*yc)-1.3,z*.5,'FaceColor', 'red'); %tail
view(3);
x_temp = get(h(1),'xdata'); % get x values
y_temp = get(h(1),'ydata');
z_temp =get(h(1),'zdata');
xc_new=x_temp.*rot_mat;
%zc_new=
%yc_new=
I can get the x,y, and z value by using the commands
x_temp = get(h(1),'xdata');
y_temp = get(h(1),'ydata');
z_temp = get(h(1),'zdata');
The error I get is due to the matrix being nonconforment but how can I adjust the matrix correctly?
error: test_object_matrix_rot: product: nonconformant arguments (op1 is 2x21, op2 is 3x3).
The error is with the line xc_new=x_temp.*rot_mat;
PS: I'm using Octave 5.0.91 which is like Matlab
YOu are messing up a lot of things......in fact I would say, you have made your work complex. YOu should straight away work on matrices to rotate to new positons instead of arrays and picking them from the figure.
This line:
x_temp = get(h(1),'xdata'); % get x values
giving you a 2*21 array and your rot_mat is 3X3.....you cannot multiply them. YOu need to pick (x,y,z) and multiply this point with rotation matrix to get the point shifted. Check the below pseudo code.....yo can develop your logic with the below example code.
t = 0:0.1:1;
[X,Y,Z] = cylinder((t));
%% Rotation
th = pi/2 ;
Rx = [1 0 0 ; 0 cos(th) -sin(th) ; 0 sin(th) cos(th)] ;
P0 = [X(:) Y(:) Z(:)] ;
P1 = P0*Rx ;
X1 = reshape(P1(:,1),size(X)) ;
Y1 = reshape(P1(:,2),size(X)) ;
Z1 = reshape(P1(:,3),size(X)) ;
figure
hold on
surf(X,Y,Z)
surf(X1,Y1,Z1)
view(3)
I was trying to produce the following figure:
as similar as possible. I was having difficulties because I wasn't sure how to plot the unit circle. How does one do that? I tried using the polar function but I couldn't then make it also plot the blue crosses and red crosses. Anyone has an idea?
Edited answer
Here is a part of the code:
% --- Plot everything on the same axes
hold on
% --- Plot the unit circle:
theta = linspace(0, 2*pi, 360);
plot(cos(theta), sin(theta), 'k--');
% --- Plot the blue points
x = [0.2 0.4 0.4 0.8];
y = [0.4 0.2 0.8 0.4];
scatter(x, y, 'bs');
% --- Plot the red points:
x = [0.4 0.8];
y = [0.4 0.8];
scatter(x, y, 'ro');
There are still many modifications to do to get the final plots, but at least this is a starting point.
Second edit
To answer your question about the intersection of a line and a circle, you have to start with the maths behind. Line and circle are defined by the following equations:
y = a.x + b % Cartesian definition of a line
(x-x0)² + (y-y0)² = r² % Cartesian definition of a circle
If you combine the two, you realize that finding the intersection is similar to finding the solution of:
(a²+1).x² + 2(a(b-y0)-x0).x + x0²+(b-y0)²-r² = 0
i.e. the roots of a polynom. As this is a trinom, there are 3 possibilities: 0 solution (no intersection), 1 solution (line is tangent to the circle) and 2 solutions (line crossing the circle).
So, in practice you have to:
Get the parameters a, b, x0, y0 and r of your problem
Find the roots of the polynom (for instance, with the function roots)
Decide what to do based on the number of roots.
Hope this helps,
%% Construct polar grid (put inside another function)
range = 0.2:0.2:1;
n = 50;
for i=1:length(range)
ro = ones(1,n);
ro = ro.*range(i);
th = linspace(0,pi*2,n);
[xx,yy] = pol2cart(th,ro);
plot(xx,yy, 'color',[.9 .9 .9]), grid on, hold on
n = round(n * 1.5);
end
%% Plot like normal
x1 = [.2 .4 .4 .8];
y1 = [.4 .2 .8 .4];
x2 = [.4 .8];
y2 = [.4 .8];
plot(x1,y1, 'bx',...
x2,y2, 'ro');
xlim([-.05 1]);
ylim([-.05 1]);
I'd like to draw a curve on an empty (semilog-y) graph by clicking the points I want it to run through, on the X-Y plane.
Is there a function for this?
edit: I'm trying to do this by obtaining the position of last pointer click -
axis([0 3000 0 1000]);
co=get(gcf, 'CurrentPoint');
It seems to return the cursor position at the time of execution, but it does not change later.
edit2: Here's what works for me. The actual drawing I can do by using the arrays of points collected.
clear
clc
h=plot(0);
grid on;
xlim([0 3000]);
ylim([0 1000]);
datacursormode on;
% Enlarge figure to full screen.
screenSize = get(0,'ScreenSize');
set(gcf, 'units','pixels','outerposition', screenSize);
hold on;
% Print the x,y coordinates - will be in plot coordinates
x=zeros(1,10); y=zeros(1,10);
for p=1:10;
[x(p),y(p)] = ginput(1) ;
% Mark where they clicked with a cross.
plot(x(p),y(p), 'r+', 'MarkerSize', 20, 'LineWidth', 3);
% Print coordinates on the plot.
label = sprintf('(%.1f, %.1f)', x(p), y(p));
text(x(p)+20, y(p), label);
end
Not really, but now there is:
function topLevel
%// parameters
xrange = [0 100];
yrange = [1e-4 1e4];
%// initialize figure, plot
figure, clf, hold on
plot(NaN, NaN);
axis([xrange yrange]);
set(gca, 'YScale', 'log')
t = text(sum(xrange)/2, sum(yrange)/2, ...
'<< Need at least 3 points >>',...
'HorizontalAlignment', 'center');
%// Main loop
xs = []; p = [];
ys = []; P = [];
while true
%// Get new user-input, and collect all of them in a list
[x,y] = ginput(1);
xs = [xs; x]; %#ok<AGROW>
ys = [ys; y]; %#ok<AGROW>
%// Plot the selected points
if ishandle(p)
delete(p); end
p = plot(xs, ys, 'rx');
axis([xrange yrange]);
%// Fit curve through user-injected points
if numel(xs) >= 3
if ishandle(t)
delete(t); end
%// Get parameters of best-fit in a least-squares sense
[A,B,C] = fitExponential(xs,ys);
%// Plot the new curve
xp = linspace(xrange(1), xrange(end), 100);
yp = A + B*exp(C*xp);
if ishandle(P)
delete(P); end
P = plot(xp,yp, 'b');
end
end
%// Fit a model of the form y = A + B·exp(C·x) to data [x,y]
function [A, B, C] = fitExponential(x,y)
options = optimset(...
'maxfunevals', inf);
A = fminsearch(#lsq, 0, options);
[~,B,C] = lsq(A);
function [val, B,C] = lsq(A)
params = [ones(size(x(:))) x(:)] \ log(abs(y-A));
B = exp(params(1));
C = params(2);
val = sum((y - A - B*exp(C*x)).^2);
end
end
end
Note that as always, fitting an exponential curve can be tricky; the square of the difference between model and data is exponentially much greater for higher data values than for lower data values, so there will be a strong bias to fit the higher values better than the lower ones.
I just assumed a simple model and used a simple solution, but this gives a biased curve which might not be "optimal" in the sense that you need it to be. Any decent solution really depends on what you want specifically, and I'll leave that up to you ^_^
I am stuck with this problem for a long time now. I have a polygonal region (lets say, a hexagon). I can calculate the values of certain function at any point inside the polygon. Now I need to create a filled contour (using contourf in MATLAB) of this data. How do I go about it. I found some discussion on this topic at the link below (page 121)
http://www-personal.umich.edu/~jpboyd/eng403_chap4_contourplts.pdf
This works somewhat ok but it still produces jagged edges which I don't want. Anyone has any suggestion on this problem? Thanks. Here is my code
close all
Node = [ 1.0 0
0.5 0.8660
-0.5 0.8660
-1.0 0
-0.5 -0.8660
0.5 -0.8660];
[x,y] = meshgrid(-1:0.1:1,-1:0.1:1);
N = zeros(size(x));
for i=1:size(x,2)
for j=1:size(y,2)
p = [x(i,j) y(i,j)];
IN = inpolygon(p(1),p(2),Node(:,1),Node(:,2));
if IN
N(i,j)= rand;
else
N(i,j)= NaN;
end
end
end
figure
contourf(x,y,N,'LineStyle','none'), hold on;
xlabel('X'), ylabel('Y'), axis equal; axis off; colorbar;
line([Node(:,1);Node(1,1)],[Node(:,2);Node(1,2)],'Color',[1 1 1],'LineWidth',2.0)
clear IN i j p x y
You're using a square grid to sample a hexagonal area. This will indeed lead to boundary problems.
A better solution (but still not quite optimal) is to use a hexagonal grid:
%# define hexagon
Node = [ 1.0 0
0.5 0.8660
-0.5 0.8660
-1.0 0
-0.5 -0.8660
0.5 -0.8660];
%# Generate hexagonal grid
Rad3Over2 = sqrt(3) / 2;
[x, y] = meshgrid(0:51);
n = size(y,1);
x = Rad3Over2 * x;
y = y + repmat([0 0.5],[n,n/2]);
%# re-scale to -1..1
x = 2*x/max(x(:))-1;
y = 2*y/max(y(:))-1;
%# insode-polygon check
N = zeros(size(x));
for i=1:size(x,2)
for j=1:size(y,2)
p = [x(i,j) y(i,j)];
IN = inpolygon(p(1),p(2),Node(:,1),Node(:,2));
if IN
N(i,j)= rand;
else
N(i,j)= NaN;
end
end
end
%# make contour plot
figure(1)
contourf(x,y,N,'LineStyle','none'), hold on;
xlabel('X'), ylabel('Y'), axis equal; axis off; colorbar;
line([Node(:,1);Node(1,1)],[Node(:,2);Node(1,2)],'Color',[1 1 1],'LineWidth',2.0)
If you want even better coverage, you'll have to devise a grid that better covers your area and/or up the number of sample points.
For arbitrary irregular areas, you might want to experiment with irregular/random grids, distributed such that there are more points close to the boundaries, than there are in the middle of the area.
Suppose you have a function defined on a hexagon. Then take some square that contains the hexagon.
A simple first test could be to extend your function on the square to take the value = 0 for points outside of the hexagon.
Depending on the range of your function this may, or may not, give you a good answer.
If this doesn't work, then find the min of the function on the hexagon using min(min(A))) for an nxn matrix A. Then define the function to be min(min(A)) - 1.0 outside of the hexagon but in the square.
Then use contourf(x, y, z, v) where v is a vector of values for the output, so take v(1) = min(min(A)) - 1.0, and v(2:n) = linspace(min(min(A)), max(max(A)), n) for some integer n. You can probably specify the color of the level set min(min(A)) - 1.0 to be white, but I've never done this. I've had to do something like this before, and setting the function = 0 outside of he hexagon was enough.
If your function can only be evaluated inside the polygonal region, then my hexagonal grid answer still stands. However, if your function can be (or modified to be) evaluated outside the polygon, you might want to use a cheat:
figure(1), clf, hold on
%# External contour, square.
x1 = [-3 -3 +3 +3 -3]/2;
y1 = [-3 +3 +3 -3 -3]/2;
%# internal contour, some polygon
x2 = [1.0 0.5 -0.5 -1.0 -0.5 0.5];
y2 = [0 0.8660 0.8660 0 -0.8660 -0.8660];
%# convert to patches
[f, v] = poly2fv({x1, x2}, {y1, y2});
patch(...
'Faces' , f,...
'Vertices' , v,...
'FaceColor', get(0, 'defaultuicontrolbackgroundcolor'), ...
'EdgeColor', 'none'...
);
%# Generate function for contourplot
[x, y] = meshgrid(-2.8/2:0.1:2.8/2);
N = rand(size(x));
%# make contour plot
contourf(x,y,N,'LineStyle','none'), hold on;
xlabel('X'), ylabel('Y'), axis equal;
axis off; colorbar;
It basically creates your contour plot on a square region, and overlays a mask with a hexagonal hole in it, so it looks like it's only a contour of the hexagonal. I suspect it is a lot easier to tweak your function to allow function evaluations outside the region, then it is to (re-)invent some sort of grid that contains the boundaries.