How to create a circle filled with equidistant points inside it?

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Rushil Asthana 2021-10-6

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评论： Chunru 2021-10-7

I want to create circle filled with equidistant points inside it. I have tried making the following program but it's taking too long to give output

radius = 100;
xc = 1;
yc = xc+1;
jc= 1;
jy=jc+1;
area = pi*radius*radius;
for i=xc:yc:area
    theta = i*(2*pi);
    r = sqrt(area)*radius;
    for j=jc:jy:i
        x = xc + r.*cos(theta);
        y = yc + r.*sin(theta);
    end
    plot(x,y,'.')
end

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Stephen23 2021-10-6

编辑：Stephen23 2021-10-6

@Rushil Asthana: because you did not mention anything about an optimal solution or any specific boundary requirements then this task is trivially easy: first generate a square grid of points large enough to cover the required circle, then simply discard those outside the circle. Reasonably efficient and should take only a few lines of code.

That will provide you with exactly what you asked for: points with exactly equal spacing all within a circle.

Rushil Asthana 2021-10-6

@John D'Errico What will the code look like essentially? Cause I need help with exactly what you mentioned.

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Chunru 2021-10-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1468201-how-to-create-a-circle-filled-with-equidistant-points-inside-it#answer_803321

编辑：Chunru 2021-10-7

在 MATLAB Online 中打开

Here is a sub-optimal (or almost optimal) solution. Filled dots will form the hexagonal formation.

radius = 10;

d = 1; % distance

xall=[]; yall=[];

dy = sqrt(3)/2 * d;

ny = floor(radius/dy);

for i=-ny:ny

y = dy*i;

if rem(i, 2)==0

nx = floor((sqrt(radius^2 - y^2))/d);

x = (-nx:nx)'*d;

else

nx = floor((sqrt(radius^2 - y^2)-d/2)/d);

x = (-nx-0.5:nx+0.5)'*d;

end

xall = [xall; x];

yall = [yall; y*ones(size(x))];

end

plot(xall(:), yall(:), '.');

hold on

theta = 0:360;

plot(radius*cosd(theta), radius*sind(theta), 'r')

axis equal

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Image Analyst 2021-10-7

@Rushil Asthana, then could you please click the "Accept this answer" link to grant @Chunru the reputation points he deserves for helping you. Thanks in advance. 🙂