Generate random circles in the square box with different diameters
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I was able to generate random circles inside the square box of dimension L=1 with the same diameter without overlapping, where X and Y are between -0.5 to 0.5.
Now, I want to generate random circles with 3 different diameters and I need to save the location of X,Y and diameter for each circle. It would be nice if we have the same number of circles per each diameter.
If you have any idea how to do so, please help me. Thanks
1 个评论
TRAN Quang Vu
2020-7-22
Hello,
I want to generate random circles inside the square box of dimension L=1 with the same diameter without overlapping, where X and Y are between -0.5 to 0.5. Can you help me? Thansk you.
采纳的回答
Brendan Hamm
2015-3-13
If you have k circles, then you can create three different areas (and thus different diameters) using:
k = 30;
X = rand(k,1) - 0.5;
Y = rand(k,1) - 0.5;
a = randi([1 3],k,1); %Random integers from discrete U([1,3])
s = scatter(X,Y,a);
If you want the same number of each diameter:
n = k/3; % May need to take floors if 3 is not a divisor of k
a = [5*ones(n,1); 10*ones(n,1); 15*ones(n,1)]; % Equal number of fives, tens, and fifteens
a = a(randperm(k)); % Shuffle the vector
s = scatter(X,Y,a);
randperm(k) permutes the integers 1:k, so I use this to randomly shuffle the values, although arguably if you were to specify the first ten observations were to be labeled size 5, and the next ten size 10, ... the (x,y) values are still random and there is no real need to do the permutation.
2 个评论
DINESH CHANDRA
2016-8-16
编辑:Image Analyst
2016-8-16
Is it possible that circles overlapping prior circles can be discarded?
Image Analyst
2016-8-16
Of course. Just keep track of radii and centers. If the distance from the center of the next/proposed circle is within the sum of the two radii, then just don't add it. If x and y are arrays of the centers, then do something like:
% Get distances of (xNew, yNew) from all other circles
distances = sqrt((xNew-x).^2+(yNew-y).^2);
% Get sums of radii
radiiSums = radiusNew + r;
% See if we can add this new one
okToAdd = all(distances > radiiSums);
if okToAdd
x(end+1) = xNew;
y(end+1) = yNew;
r(end+1) = radiusNew;
end
更多回答(1 个)
Image Analyst
2015-3-13
See my attached demo. Adapt it as needed with the rand() function to make random radii.
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