Choose n points randomly from a circle, how to calculate the probability that all the points are in one semicircle?

I want to create a graphic monte carlo simulation that solves this problem, The mathematical answer is : n/2^(n-1)

maybe someone have the algorithm code for that question ?

[Guillaume]

It's not entirely clear what you need help with. The problem itself is simple, so we don't know what's blocking you. You'd just run a monte-carlo simulation which is easy enough to implement.

The one difficultly I can see is the generation of points in a circular uniform distribution. You have to be careful not to use the naive approach of just picking uniform random radii and angles. However, less than 5 minutes of searching on the internet would give you simple and reliable algorithms for that, if it's not already built in matlab.

[elroi berkovits]

Can u try to help with the code?

i have this so far:

clear all

clc

clear

% Plot a circle.

angles = linspace(0, 2*pi, 720); % 720 is the total number of points

radius = 20;

xCenter = 50;

yCenter = 40;

x = radius * cos(angles) + xCenter;

y = radius * sin(angles) + yCenter;

% Plot circle.

plot(x, y, 'b-', 'LineWidth', 2);

% Plot center.

hold on;

plot(xCenter, yCenter, 'k+', 'LineWidth', 2, 'MarkerSize', 16);

grid on;

axis equal;

xlabel('X', 'FontSize', 16);

ylabel('Y', 'FontSize', 16);

% Now get random locations along the circle.

s1 = 2; % Number of random points to get.

n=1000; % Number of sim

xRandom = zeros(n,s1);

yRandom = zeros(n,s1);

for i=0:n-1

randomIndexes = randperm(length(angles), s1);

xRandom(i+1,:) = x(randomIndexes);

yRandom(i+1,:) = y(randomIndexes);

plot(xRandom, yRandom, 'ro', 'LineWidth', 2, 'MarkerSize', 16);

end

P = nnz(all(xRandom >=50,2))

R = nnz(all(xRandom <50,2))

PROB = (P+R)/n

Answer 1

[KSSV]

% Generate circle 
r = 1. ;   % radius 
th = linspace(0,2*pi) ; 
x = r*cos(th) ; 
y = r*sin(th) ; 
% Geenrate random points inside the circle 
N = 5000 ;
a = -1 ; b = 1 ; 
xr = (b-a).*rand(N,1) + a;
yr = (b-a).*rand(N,1) + a;
% pick points lying inside the circle 
idx1 = inpolygon(xr,yr,x,y) ; 
xr = xr(idx1) ; 
yr = yr(idx1) ; 
% random points inside circle 
NP1 = nnz(idx1) ; 
% points lying in semicircle 
idx2 = inpolygon(xr,yr,x(1:50),y(1:50)) ; 
NP2 = nnz(idx2) ; 
iwant = NP2/NP1 ;   % probability that points lie in semicircle 

plot(x,y) 
hold on
plot(xr,yr,'.r')
plot(xr(idx2),yr(idx2),'ob')

[Guillaume]

Considering that it's trivial to check if a point is inside a circle centered on the origin (sqrt(x.^2 + y.^2) <= radius) why approximate the circle to a polygon and rely on inpolygon for that?

[elroi berkovits]

It is very nice and efficient, but it does not answer the statistical question, for example at 3 points the probability is 0.75, at 4 points the probability is 0.5, at 5 points the probability is 0.3125, etc.

[KSSV]

@Guillaume

Yes you are right..I made the problem complex instead of using the inequality.

@elroi berkovits please noe that, you need not to use inpolygon. We have given you a idea, you should try the one you want.