How to genetate random number under constraint

Asked by Muhammad Nabeel Hussain on 26 Sep 2019

Latest activity Commented on by Adam Danz on 26 Sep 2019

Accepted Answer by Adam Danz

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I want to genetate two vector X(size=37x1) and Y(size=37x1) of random numbers between

100 to 1900 such that difference between any two numbers of one vector X or Y should be greater than 200.

I have tried generating random number again and again while rejecting if they dont met constraint.

but it make infinte loop.

Actually X and Y are cordinates of turbines. Constraint is no turbine can be in the 200 meter radius of any other turbine. Any idea for doing this very fast? Thanks

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Bruno Luong on 26 Sep 2019

Then your description "such that difference between any two numbers of one vector X or Y should be greater than 200." is incorrect. The two points

(100,1000)
(150, 100)

are separated by a distance larger than 200. But it is not satisfied what you have originally stated: since the difference of X is 50.

Please clarify your question.

Adam Danz on 26 Sep 2019

This is actually fairly easy to do if you just monitor the distance between the turbines rather than the distance between each x and y coordinate.

Your question states "any two numbers of one vector X or Y should be greater than 200." In that case, the turbines cannot be any closer than 200*sqrt(2).

Instead of choosing values of x and y individually that have a minimum distance of 200, you just need to monitor the distance between the turbine coordinates.

Adam Danz on 26 Sep 2019

@Muhammad Nabeel Hussain, I'm curous how you plotted the plane symbols. Mind explaining?

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Answer by Adam Danz on 26 Sep 2019

Edited by Adam Danz on 26 Sep 2019

Accepted Answer

This takes milliseconds.

The main idea is to start off with a set of random coordinates and continually replace coordinates that are too close to another coordinate until either 1) all distances are less than the minimum distance requirement or 2) 100 attempts were made per coordinate.

%Define parameters
nPoints = 37;       %number of coordinates
lim = [100,1900];   %bounds of random numbers
minDist = 200;      %minimum distance between turbines
% Create random coordinates and continually replace coordinates
% that are too close to another point.  Stop when minimum distance
% is satisfied or after making nPoints*100 attempts.
xy = nan(nPoints,2);
c = 0; %Counter
while any(isnan(xy(:))) && c<(nPoints*100)
    % Fill NaN values with new random coordinates
    xy(isnan(xy)) = rand(1,sum(isnan(xy(:)))) * (lim(2)-lim(1)) + lim(1);
    % Identify rows that are too close to another point
    [~,isTooClose] = find(triu(squareform(pdist(xy)) < minDist,1));
    % Replace too-close coordinates with NaN
    xy(isTooClose,:) = NaN; 
    c = c+1;
end
% Throw error if the loop had to quit and missing values remain
if any(isnan(xy(:)))
    error('The while-loop gave up. There are %d coordinates with missing values.',sum(isnan(xy(:,1))))
end
% Display number of attempts 
fprintf('%d number of attempts.\n', c)
% Show the minimum distance 
distances = squareform(pdist(xy)); 
fprintf('Min distance = %.2f\n', min(distances(distances~=0)))
% Plot results
 figure() 
 plot(xy(:,1),xy(:,2), 'ks', 'MarkerSize', 10)
 grid on

These data in the figure above were produced in <0.08 sec and the minimum distance between coordinates is 205.52.

2 Comments

Muhammad Nabeel Hussain on 26 Sep 2019

Thankyou Adam Danz. It solved my problem.

Adam Danz on 26 Sep 2019

Glad I could help.

@Muhammad Nabeel Hussain, I'm curous how you plotted the plane symbols. Mind explaining?

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Answer by Bruno Luong on 26 Sep 2019

Similar question has been answered in this thread

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Answer by Jos (10584) on 26 Sep 2019

Brute force attempt:

N = 20 ;
xyRange = [100 1900] ;
minimumDistance = 200 ;
attempt_counter = 1 ; 
Distances = 0 ;
while any(Distances < minimumDistance) && attempt_counter < 10000
    attempt_counter = attempt_counter + 1 ;
    Pxy = randi(xyRange, N, 2) ; 
    Distances = pdist(Pxy) ;
end
if attempt_counter < 1000
    plot(Pxy(:,1), Pxy(:,2),'bo') ;
else
    disp('No positions found.') ;
end

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