how to generate random number in a rectangle like this?

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Daniel Niu 2022-11-2

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评论： William Rose 2022-11-3

the inscribed rectangle has length 4 and width 2. The shadow part can appear randomly. How to achieve that using MATLAB?

Your help would be highly appreciated!

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Daniel Niu 2022-11-2

Dear Jan,

Sorry for confusing you. Actually, I just generate random for example 1000 points within the circle(r=sqrt(5)), and I want to know how much points are locate in the non-grey rectangle region. the original point is the center of the rectangle.

thank you!

William Rose 2022-11-3

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@Daniel Niu,

Let us assume the center of the goal is at 0,0. The radius is R=sqrt(5). Generate 1000 points, uniformly distributed within this circle:

N=1000; R=sqrt(5);

r=R*sqrt(rand(1,N));

theta=2*pi*rand(1,N);

x=r.*cos(theta);

y=r.*sin(theta);

plot(x,y,'rx');

axis equal; grid on

xticks([-3:3]); yticks([-3:3])

Now lets find how many shots are in gray1: x=(-1,1), y=(-1,1)

gray1=sum((x>=-1 & x<1) & (y>=-1 & y<1));
gray2=sum((x>=-2 & x<-1) & (y>=0 & y<1)) + sum((x>=-1 & x<0) & (y>=-1 & y<0));
gray3=sum((x>=1 & x<2) & (y>=0 & y<1)) + sum((x>=0 & x<1) & (y>=-1 & y<0));
gray4=sum((x>=-2 & x<0) & (y>=-1 & y<0));
gray5=sum((x>=0 & x<2) & (y>=-1 & y<0));
fprintf('Gray1=%d, gray2=%d, gray3=%d, gray4=%d, gray5=%d\n',...
    gray1,gray2,gray3,gray4,gray5)
Gray1=265, gray2=137, gray3=112, gray4=139, gray5=117

Since the shots are uniformly distributed, we expect twice as many shots in gray1 than in each of the other regions, since gray1 has twice the area of each other region. The results appear to be what we expect.

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Answer 1

William Rose 2022-11-2

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@Daniel Niu, I will assume that there are 5 equally likely options, and that for each option, you want to pick a random point in the gray region.

Then you can do it with two random steps:

1. Pick a random integrer in [1,5]. The tells you which of the 5 cases you are currently using.

2. Pick a random 2D point in the gray region selected by step 1.

When doing step 2, you may want to do a random coin flip if you are in case 2 or 3, because the gray region in case 2 and case 3 has two non-contiguous squares. Alternatively, if you are in case 2 or case 3, you can pick a pont in the "superset" rectangle of case 2 or 3, then reject the point and try again if the point you got is not in one of the two gray squares that are active for that case.

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William Rose 2022-11-2

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@Daniel Niu,

I assume the outer rectangle goes from (0,0) to (4,2). Here is pseudo code. You can fil in the details.

N=1000; %number of points to find
x=zeros(N,2); %allocate array for the 2D points we are going to find
for i=1:N  
    j=randi(5);  %pick a case randomly from 5 possible cases
    switch j
        case 1
            %find random point in case 1 region
        case 2
            %find random point in case 2 region
        case 3
            %find random point in case 3 region
        case 4
            %find random point in case 4 region
        case 5
            %find random point in case 5 region
    end
end

Example: To find a random point, x(1,:), in the range x=3-4, y=1-2 (which is half of case 3), do:

x(1,:)=[3+rand(),1+rand()]; disp(x(1,:))
    3.8307    1.9839

You can show, with a bit of thinking, that the expected probabilities in each of the 8 locations are given below, if we assume the 5 cases are equally likely:

10% | 5% | 5% | 10%

----|-----|-----|-----

10% | 25% | 25% | 10%

When I did it, with N=1000, I got the plot below:

Looks reaosnable.

Daniel Niu 2022-11-2

在 MATLAB Online 中打开

Thank you,William.

Actually, I just generate random for example 1000 points within the circle(r=sqrt(5)), and I want to know how much points are locate in the non-grey rectangle region. the original point is the center of the rectangle.

I write some code following your instruction, but I am a little confusing. How to achieve my goal. I want to know how much point locate at the non grey rectangle region.

clear
N=1000;
r = sqrt(5);
x0 = 0; % Center of the circle in the x direction.
y0 = 0; % Center of the circle in the y direction.
% Now create the set of points.
t = 2*pi*rand(N,1);
r = r*sqrt(rand(N,1));
x = x0 + r.*cos(t);
y = y0 + r.*sin(t);
% Now display our random set of points in a figure.
plot(x,y, 'bo', 'MarkerSize', 5)
hold on
xx=zeros(N,2); %allocate array for the 2D points we are going to find
for i=1:N  
    j=randi(5);  %pick a case randomly from 5 possible cases
    switch j
        case 1
            xx(:)=[-2+4.*rand(N,1),-1+2.*rand(N,1)]; %find random point in case 1 region
        case 2
            %find random point in case 2 region
        case 3
            %find random point in case 3 region
        case 4
            %find random point in case 4 region
        case 5
            %find random point in case 5 region
    end
end
axis square;
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
fontSize = 15;
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
title('Random Locations Within a Circle', 'FontSize', fontSize);
legend('boxoff')

Daniel Niu 2022-11-2

编辑：Daniel Niu 2022-11-2

@William Rose

the grey shadow represent a goal keeper and only the point in the non-grey region, it means the score. it just simulate the penalty kick.

the 5 cases are equally likely options as you assumed.

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how to generate random number in a rectangle like this?

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how to generate random number in a rectangle like this?

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