i want x,y coordinates which are randomly generated between 1 to 300 ,condition is distance between every coordinate is >20. how to proceed further

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M.Prasanna kumar
M.Prasanna kumar 2018-12-14
评论： Image Analyst 2019-1-7
i=5 ; %% no. of random coordinates
xc = round((1 + (60-1)*rand(i,1)),0); %% random X- coordinates
yc = round((1 + (40-1)*rand(i,1)),0); %% random Y- coordinates
c = [xc,yc] ; %% storing random coordinates in a matrix
d = pdist(c); %% pdist finds distance between each and every coordinate (all possible distances)
z= squareform(d);%% it gives a 5*5 matrix containing distances
v = 500; %%
n = size(z,1);
z(1:(n+1):end) = v; %% replacing diagonal elements by some scalar because all diag elements wiil be zero
index = find(z<20) %% find which index number in matrix which are less than 20

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Akira Agata
Akira Agata 2018-12-14
How about the following?
BWremain = true(300);
% Number of random coordinates
N = 5;
% Selected coordinates are stored these variables
row = nan(N,1);
col = nan(N,1);
for kk = 1:N
BW = false(300);
p = find(BWremain);
p = p(randperm(numel(p),1));
BW(p) = true;
BW = bwdist(BW) > 20;
BWremain = BWremain & BW;
[row(kk),col(kk)] = ind2sub(size(BWremain),p);
end
Selected coordinates and circles (R = 20) around them looks like this.
selectedPoints.png
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Akira Agata
Akira Agata 2019-1-7
Sorry for my late response.
I believe "so that distance ... is < 20" in your question must be a typo of "... is > 20" (because answer becomes Inf). If so, the problem can be treated as an well-known "Packing problem".
Packing problem (Wikipedia)
https://en.wikipedia.org/wiki/Packing_problems
Image Analyst
Image Analyst 2019-1-7
So it could be very easy - if your radii are a constant. Just do a normal circle packing in a honeycomb pattern.
If your radii vary, and they can go all the way down to zero, then you can fit an infinite number in the box because the circles can get infinitesimally small so an unlimited number will fit into the remaining spaces.
If you have some other case, like radii are variable and can range from 3 to 50, then it's much harder. To solve that, I'd probably use a Monte Carlo simulation though it will give a practical solution, not the solution with the most discs in it. You could do a simulation with a high resolution digital image where you call bwdist() after each disc is placed to find out what the max radius possible at that time is, and then place the next circle at the smallest location where that disc's radius could fit (which is at the place where the Euclidian distance transform equals that disc's radius). (If you're a beginner, I don't blame you if you don't understand a word I said.)

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Image Analyst
Image Analyst 2018-12-14
You can use a strategy where you get a point, then use sqrt() to get the distances from all prior points. If it's farther away than 20 from all of them, then keep it, otherwise throw it away and try a new point.
pointsToPlace = 5 ; % # of random coordinates we need to place
% Preallocate points
x = zeros(1, pointsToPlace);
y = zeros(1, pointsToPlace);
loopCounter = 1;
maxIterations = 10000; % Number of tries before giving up.
numberPlaced = 0; % No points placed yet.
while numberPlaced < pointsToPlace && loopCounter < maxIterations
% Get new coordinate
xProposed = round((1 + (60-1)*rand()),0); %% random X- coordinates
yProposed = round((1 + (40-1)*rand()),0); %% random Y- coordinates
if loopCounter == 1
% First one automatically gets added of course.
numberPlaced = 1;
x(numberPlaced) = xProposed;
y(numberPlaced) = yProposed;
else
% Compute distance to all prior coordinates.
distances = sqrt((xProposed - x(1:numberPlaced)) .^ 2 + (yProposed - y(1:numberPlaced)) .^2);
% If less than 20, add it
if min(distances > 20)
numberPlaced = numberPlaced + 1;
x(numberPlaced) = xProposed;
y(numberPlaced) = yProposed;
end
end
loopCounter = loopCounter + 1;
end
fprintf('Placed %d points after %d iterations\n', numberPlaced, loopCounter-1);
plot(x, y, 'b*', 'LineWidth', 2, 'MarkerSize', 14);
grid on;
xlim([0, 60]);
ylim([0, 40]);
xlabel('X', 'FontSize', 20);
ylabel('Y', 'FontSize', 20);
0001 Screenshot.png

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