Let's say i want to generate a set of coordinates(x,y), using rand. Take any one point of the generated random set, the distance between this point and those points( nearest/the first layer) surround it could be limited in certain range. The overview effect looks 'these random points are uniformly distributed.
Limiting random co-ordinates about a point (x,y) with in a distance 'r' is more like selecting random points in a circle with centre (x,y) and radius 'r'.
Below should help
http://www.mathworks.com/matlabcentral/answers/294-generate-random-points-inside-a-circle
All n points must be inside a circle. The diameter of this circle is the maximum distance (d_max) between any two points. Use polar coordinates: r = (d_max / 2) * sqrt(rand(n, 1)); phi = 2 * pi * rand(n, 1); The square root is needed to get uniform areal density. Convert polar coordinates to Descartes coordinates the usual way: x = x0 + r .* cos(phi); y = y0 + r .* sin(phi); where (x0, y0) gives the center of the circle.
The result for d_max = 10:
The histogram of the distances between points:
Use Modulus
mod([1:5],3)
ans =
1 2 0 1 2
http://www.mathworks.co.uk/help/techdoc/ref/mod.html
Related
I want N points on a circle, set apart by a Euclidean distance (straight line, not around the circumference) of 50. The radius is determined by the number of points and the distance between consecutive points.
However, when I choose one of the points as a reference and calculate the distances to other points, I do not get any distance values equal to 50.
Below is my code:
N = 100; % number of points
eclddst = 50; % euclidean distance between adjacent points
r = eclddst/(2*sin(pi/N)); % radius of the circle
cord = r*exp((0:1/(N-1):1)*pi*2*1i)'; % coordinates
XCor = real(cord);
YCor = imag(cord);
N_COORD = [XCor YCor];
% taking location 3 as the reference point to check the distance between the points
DSTNT = sqrt( (N_COORD(3,1)-N_COORD(:,1)).^2 + ( N_COORD(3,2)- N_COORD(:,2)).^2)';
The distance values around the third point which I obtain are:
100.959
50.505
0.000
50.505
100.959
151.311
The points adjacent to point 3 should have 50 as their distance value and not 50.505.
Why do I get this error?
Thanks in advance.
Your issue is in the number of points which your formula generates, because of a misleading statement at the link you were following. Note that it was stated:
r = A*exp((0:1/300:1)*pi*2j); % 300 point circle, radius A
However, this will give 301 points not 300. If you plotted this though, you would see only 300 points (much easier to see if you use, say, 4 points). The first and last points are identical since
exp(0*pi*2j) = exp(1*pi*2j) = 1
To get around this, the simplest options are to change 0:1/300:1 to not reach 1, or simply create and remove an extra point as shown here:
N = 100; % number of points
d = 50; % Euclidean distance between adjacent points
% Euclidean distance around circle is essentially side length of N-sided polygon.
% Triangular sector angle within polygon (rads): 2*pi/N
% By bisecting the triangle to get a right-triangle, we can deduce that
% sin((2*pi/N)/2) = (d/2)/r => r = (d/2)/sin(pi/N)
r = (d/2)/sin(pi/N);
% Use linspace as it's clearer than colon array, create N+1 points
% Note that point 1 and point N+1 will be identical!
complexcoords = r*exp(linspace(0,1,N+1)*pi*2*1i).';
% Remove the last point as you put it in an array
coords = [real(complexcoords(1:N)) imag(complexcoords(1:N))];
As a check:
% Euclidean distances from point 3
dists = sqrt((coords(3,1)-coords(:,1)).^2 + (coords(3,2)-coords(:,2)).^2);
dists(1:5)
>> 99.951
50 % Neighbouring points are distance 50!
0
50 % Ditto
99.951
Note that you should be careful using '. This is the complex conjugate transpose, meaning that x + yi becomes x - yi. You won't notice this on a circle centred around 0 and an even number of points, but when you want to transpose something in MATLAB always use .', otherwise you may get some hard to diagnose issues! I have corrected this in my above code.
Doc links: ctranspose/', transpose/.'.
How can I efficiently find all pixels (u',v') within a distance x from a given pixel (u,v). Specifically all (u',v') such that sqrt((u-u')^2 +(v-v')^2) <= x
To find all pixels at distance x you can use this indexing trick (for Manhattan distance)
u=10;v=10;
x=3.4;
pixels=img([floor(u-x) ceil(u+x)],[floor(v-x) ceil(v+x)]);
However, note that this will add any pixel that is within the distance, even if its just a piece of the pixel! E.g. (6,6) is inside!
For Euclidean distance, you need to define a circle around the point
[xi,yi]=meshgrid(1:size(img,1),1:size(img,2));
mask=sqrt((xi-u).^2+(yi-v).^2)<x; % or ceil(x)
pixel_values=img(mask);
[pixel_indices_u pixel_indices_v]=find(mask);
u = rand(1e2,1); % Create a 100 random points in the unit square
v = rand(1e2,1); % Create a 100 random points in the unit square
a = 0.3; % Choosen pixel
b = 0.4;
x = 0.1; % radius from the choosen pixel
idx = (sqrt((u-a).^2 +(v-b).^2) <= x); % Create a logical index array
locs = [u(idx) v(idx)]; % Index the locations
This is basically exactly what you describe in the question, with the points being labelled as (u,v) and the "destination point" so to speak labelled with (a,b) as a prime (') denotes the complex conjugate transpose in MATLAB.
Read more on logical indexing in this very insightful answer by Luis Mendo
Assume that I have a vector with 6 distance elements as
D = [10.5 44.8 30.01 37.2 23.4 49.1].
I'm trying to create random pair positions of a given distances, inside a 200 meters circle. Note that the distance D created by using (b - a).*rand(6,1) + a, with a = 10 and b = 50 in Matlab. I do not know how to generate the random pairs with given the distances.
Could anybody help me in generating this kind of scenario?
This is an improvement to Alessiox's answer. It follows same logic, first generate a set of points ([X1 Y1]) that have at least distance D from the main circle border, then generate the second set of points ([X2 Y2]) that have exact distance D from the first set.
cx = 50; cy = -50; cr = 200;
D = [10.5 44.8 30.01 37.2 23.4 49.1]';
n = numel(D);
R1 = rand(n, 1) .* (cr - D);
T1 = rand(n, 1) * 2 * pi;
X1 = cx+R1.*cos(T1);
Y1 = cy+R1.*sin(T1);
T2 = rand(n, 1) * 2 * pi;
X2 = X1+D.*cos(T2);
Y2 = Y1+D.*sin(T2);
You can tackle the problem using a two-steps approach. You can
randomly generate the first point and then you can consider the circle whose center is that point and whose radius is the distance in D
once you did draw the circle, every point lying on that circle will have distance D from the first point previously created and by random selection of one of these candidates you'll have the second point
Let's see with an example: let's say you main circle has radius 200 and its center is (0,0) so we start by declaring some main variables.
MAINCENTER_x=0;
MAINCENTER_y=0;
MAINRADIUS=200;
Let's now consider the first distance, D(1)=10.5 and we now generate the first random point which (along with its paired point - I reckon you don't want one point inside and the other outside of the main circle) must lie inside the main circle
r=D(1); % let's concentrate on the first distance
while true
x=((MAINRADIUS-2*r) - (-MAINRADIUS+2*r))*rand(1,1) + (-MAINRADIUS+2*r);
y=((MAINRADIUS-2*r) - (-MAINRADIUS+2*r))*rand(1,1) + (-MAINRADIUS+2*r);
if x^2+y^2<=(MAINRADIUS-2*r)^2
break;
end
end
and at the end of this loop x and y will be our first point coordinates.
Now we shall generate all its neighbours, thus several candidates to be the second point in the pair. As said before, these candidates will be the points that lie on the circle whose center is (x,y) and whose radius is D(1)=10.5. We can create these candidates as follows:
% declare angular spacing
ang=0:0.01:2*pi;
% build neighbour points
xp=r*cos(ang)+x;
yp=r*sin(ang)+y;
Now xp and yp are two vectors that contain, respectively, the x-coordinates and y-coordinates of our candidates, thus we shall now randomly select one of these
% second point
idx=randsample(1:length(xp),1);
secondPoint_x=xp(idx);
secondPoint_y=yp(idx);
Finally, the pair (x,y) is the first point and the pair (secondPoint_x, secondPoint_y) is our second point. The following plot helps summarizing these steps:
The red circle is the main area (center in (0,0) and radius 200), the red asterisk is the first point (x,y) the blue little circle has center (x,y) and radius 10.5 and finally the black asterisk is the second point of the pair (secondPoint_x, secondPoint_y), randomly extracted amongst the candidates lying on the blue little circle.
You must certainly can repeat the same process for all elements in D or rely on the following code, which does the very same thing without iterating through all the elements in D.
MAINCENTER_x=0;
MAINCENTER_y=0;
MAINRADIUS=200;
D = [10.5 44.8 30.01 37.2 23.4 49.1];
% generate random point coordinates
while true
x=((MAINRADIUS-2*D) - (-MAINRADIUS+2*D)).*rand(1,6) + (-MAINRADIUS+2*D);
y=((MAINRADIUS-2*D) - (-MAINRADIUS+2*D)).*rand(1,6) + (-MAINRADIUS+2*D);
if all(x.^2+y.^2<=(MAINRADIUS-2*D).^2)
break;
end
end
% declare angular spacing
ang=0:0.01:2*pi;
% build neighbour points
xp=bsxfun(#plus, (D'*cos(ang)),x');
yp=bsxfun(#plus, (D'*sin(ang)),y');
% second points
idx=randsample(1:size(xp,2),length(D));
secondPoint_x=diag(xp(1:length(D),idx));
secondPoint_y=diag(yp(1:length(D),idx));
%plot
figure(1);
plot(MAINRADIUS*cos(ang)+MAINCENTER_x,MAINRADIUS*sin(ang)+MAINCENTER_y,'r'); %main circle
hold on; plot(xp',yp'); % neighbours circles
hold on; plot(x,y,'r*'); % first points (red asterisks)
hold on; plot(secondPoint_x,secondPoint_y,'k*'); %second points (black asterisks)
axis equal;
Now x and y (and secondPoint_x and secondPoint_y by extension) will be vector of length 6 (because 6 are the distances) in which the i-th element is the i-th x (or y) component for the first (or second) point.
I would like to find the two (2) intersection points when a linear line goes through a circle's centrum (x,y).
r = 13 radius
x = 0 x-coordinate
y = 7 y-coordinate
k = 9 slope value(?) y=kx+m y=9x+m
So first I'm drawing a circle with r=13 and a centrum of (0,7).
r=13
x=0
y=7
k=9
hold on
z = 0:pi/50:2*pi;
xunit = r * cos(z) + x;
yunit = r * sin(z) + y;
plot(xunit, yunit);
I'm wondering if it would be possible to plot up a circle in an easier way? Something like
(x−cx)^2 + (y −cy)^2 = r^2
(x-0)^2 + (y-7) = 13^2
I've tried this
plot((x−cx)^2 + (y −cy)^2 = r^2)
It doesn't do anything at all so the code must be incorrect.
Well, then I'm drawing the linear equation by calculating
y=kx+m
k=9
the line goes thorugh (0,7)
7=9*0+m
m=7
y=9x+7
so since I'm new to MatLab it took me a while to actually draw the line. I didn't find any easy function to plot it so I plotted a line like this:
I took some random values for x or y and calculated some coordinates.
(0,7)
(2,25)
(-2,-11)
plot([-2,2],[-11,25])
Result image: http://i.imgur.com/ag6HJlm.jpg
So now I just need to solve the intersection points with "solve" function. So well I would really appreciate some help!
best regards
Here is one approach to it:
%Place your lines and figures on the grid
linexypos = eye(100);
shapexypos = flipud(eye(100)) ;
% Guess where they come together
intersection = filter2(ones(3),linexypos + shapexypos);
[quality, loc] = max(intersection(:))
Note that you have to guess, as two lines with widths of 1 pixel may not have exactly the same location. (consider [1 0; 0 1] and [0 1;1 0], they cross but never exactly overlap).
If you want to visualize the situation, try contour(intersection)
I'm trying to generate a cloud of 2D points (uniformly) distributed within a triangle. So far, I've achieved the following:
The code I've used is this:
N = 1000;
X = -10:0.1:10;
for i=1:N
j = ceil(rand() * length(X));
x_i = X(j);
y_i = (10 - abs(x_i)) * rand;
E(:, i) = [x_i y_i];
end
However, the points are not uniformly distributed, as clearly seen in the left and right corners. How can I improve that result? I've been trying to search for the different shapes too, with no luck.
You should first ask yourself what would make the points within a triangle distributed uniformly.
To make a long story short, given all three vertices of the triangle, you need to transform two uniformly distributed random values like so:
N = 1000; % # Number of points
V = [-10, 0; 0, 10; 10, 0]; % # Triangle vertices, pairs of (x, y)
t = sqrt(rand(N, 1));
s = rand(N, 1);
P = (1 - t) * V(1, :) + bsxfun(#times, ((1 - s) * V(2, :) + s * V(3, :)), t);
This will produce a set of points which are uniformly distributed inside the specified triangle:
scatter(P(:, 1), P(:, 2), '.')
Note that this solution does not involve repeated conditional manipulation of random numbers, so it cannot potentially fall into an endless loop.
For further reading, have a look at this article.
That concentration of points would be expected from the way you are building the points. Your points are equally distributed along the X axis. At the extremes of the triangle there is approximately the same amount of points present at the center of the triangle, but they are distributed along a much smaller region.
The first and best approach I can think of: brute force. Distribute the points equally around a bigger region, and then delete the ones that are outside the region you are interested in.
N = 1000;
points = zeros(N,2);
n = 0;
while (n < N)
n = n + 1;
x_i = 20*rand-10; % generate a number between -10 and 10
y_i = 10*rand; % generate a number between 0 and 10
if (y_i > 10 - abs(x_i)) % if the points are outside the triangle
n = n - 1; % decrease the counter to try to generate one more point
else % if the point is inside the triangle
points(n,:) = [x_i y_i]; % add it to a list of points
end
end
% plot the points generated
plot(points(:,1), points(:,2), '.');
title ('1000 points randomly distributed inside a triangle');
The result of the code I've posted:
one important disclaimer: Randomly distributed does not mean "uniformly" distributed! If you generate data randomly from an Uniform Distribution, that does not mean that it will be "evenly distributed" along the triangle. You will see, in fact, some clusters of points.
You can imagine that the triangle is split vertically into two halves, and move one half so that together with the other it makes a rectangle. Now you sample uniformly in the rectangle, which is easy, and then move the half triangle back.
Also, it's easier to work with unit lengths (the rectangle becomes a square) and then stretch the triangle to the desired dimensions.
x = [-10 10]; % //triangle base
y = [0 10]; % //triangle height
N = 1000; %// number of points
points = rand(N,2); %// sample uniformly in unit square
ind = points(:,2)>points(:,1); %// points to be unfolded
points(ind,:) = [2-points(ind,2) points(ind,1)]; %// unfold them
points(:,1) = x(1) + (x(2)-x(1))/2*points(:,1); %// stretch x as needed
points(:,2) = y(1) + (y(2)-y(1))*points(:,2); %// stretch y as needed
plot(points(:,1),points(:,2),'.')
We can generalize this case. If you want to sample points from some (n - 1)-dimensional simplex in Euclidean space UNIFORMLY (not necessarily a triangle - it can be any convex polytope), just sample a vector from a symmetric n-dimensional Dirichlet distribution with parameter 1 - these are the convex (or barycentric) coordinates relative to the vertices of the polytope.