Spread a random user in hexagonal cell

Question

Ahmed Hassaan 2012-1-27

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/27191-spread-a-random-user-in-hexagonal-cell

评论： Walter Roberson 2016-7-10

Hello,i drawn one hexagonal cells and I want to spread a random users in it and specify the coordinates of each user in an array. thats my trial :

clear all;
close all;
clc;
t=linspace(0,2*pi,7);
x=0+1*cos(t);
y=0+1*sin(t);
plot(x,y);
hold on
grid on;
 for i=1:100
     xa=-0.8+1.6*rand(1,1);  
     ya=-0.8+1.6*rand(1,1);  
     xra(i)=xa;  
     yra(i)=ya;  
    figure(1)  
    plot(xra(i),yra(i),'k.');  
    hold on
 end

as u see the whole cell cannot be filled by the users ,

Thanks for ur help :)

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Answer 1

Walter Roberson 2012-1-27

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https://ww2.mathworks.cn/matlabcentral/answers/27191-spread-a-random-user-in-hexagonal-cell#answer_35391

When you are choosing xa and ya, you are choosing the coordinates from a square from [-0.8, 0.8] in both x and y. Your hexagon extends a bit past +/- 0.8 in y, and extends to +/- 1 in x, and because of the angled lines there are parts of the square (whose coordinates you are choosing from) that are not within the hexagon.

The square: because -0.8+1.6*rand(1,1) ranges from -0.8 + 1.6*0 to -0.8 + 1.6 * 1 which is -0.8 to +0.8, and you use that calculation for both x and y.

You need to find a way of choosing exactly withinin the hexagon, or you need to find a way of choosing over a larger area and discarding the points that are not in the hexagon.

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Walter Roberson 2012-1-28

http://matlab.wikia.com/wiki/FAQ#How_can_I_create_variables_A1.2C_A2.2C....2CA10_in_a_loop.3F

Walter Roberson 2012-1-28

Hint: http://www.mathworks.com/help/techdoc/ref/inpolygon.html

There is even a relevant example.

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Answer 2

Ahmed Ibrahim 2016-5-6

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/27191-spread-a-random-user-in-hexagonal-cell#answer_221110

编辑：Ahmed Ibrahim 2016-5-6

在 MATLAB Online 中打开

this is an update for your code:

-----------------------------------

function [xra,yra]=hex_rand(R,i)

% R is the hexagon side length

% i is the number of samples

t=linspace(0,2*pi,7);

x=0+R*cos(t);

y=0+R*sin(t);

plot(x,y);

hold on

grid on;

for m=1:i

xa=x(1)*2*(rand()-1/2);

ya=y(2)*2*(rand()-1/2);

     if ((xa>(R/2)) && (ya>(-sqrt(3)*xa+(R*sqrt(3))))) %%the areas where the problem occur
        ya=-sqrt(3)*xa+(R*sqrt(3));
     elseif ((xa>(R/2)) && (ya<(sqrt(3)*xa-(R*sqrt(3)))))
        ya=sqrt(3)*xa-(R*sqrt(3));
     elseif ((xa<(-R/2)) && (ya>(sqrt(3)*xa+(R*sqrt(3)))))
        ya=sqrt(3)*xa+(R*sqrt(3));
     elseif ((xa<(-R/2)) && (ya<(-sqrt(3)*xa-(R*sqrt(3)))))
        ya=-sqrt(3)*xa-(R*sqrt(3));
     end
     xra(m)=xa;  
     yra(m)=ya;  
    figure(1)   
 end  
 plot(xra,yra,'k.');  
   hold on
end

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AMAL ALGEDIR 2016-7-10

编辑：AMAL ALGEDIR 2016-7-10

how I can change this code if the center of the hexagon is not (0,0). how I can make work with more than one cell? I could draw more than one cell but I could not fix the position od users?