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Monte-Carlo Method
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| − | for i = 1: | + | % Monte-carlo simulator for solving area of a quarter-circle of radius 1, |
| − | x = rand; | + | % dictated by bounds of x^2+y^2 = 1 |
| + | |||
| + | % Set up arrays and other constants | ||
| + | x = 0; % x coordinate | ||
| + | y = 0; % y coordinate | ||
| + | total = 10000; % amount of points to generate | ||
| + | count = 0; % number of points inside area | ||
| + | |||
| + | % monte-carlo solver | ||
| + | for i = 1:total | ||
| + | % generate a random point in 0<x<1 and 0<y<1 | ||
| + | x = rand; % random number between 0 and 1 | ||
y = rand; | y = rand; | ||
| − | if x^2+y^2 <=1 | + | % monte-carlo check if inside bounds |
| − | + | if (x^2+y^2 <= 1) | |
| + | count = count + 1; | ||
end | end | ||
| − | |||
end | end | ||
| − | area = | + | |
| + | % calculate 'true' area and monte-carlo solved area | ||
| + | circlearea = pi*1^2; % 'true' area | ||
| + | montearea = count/total*4; % monte-carlo area | ||
This, when run on my computer, produced the result of 0.783. The area of a quarter circle works out to be π/4 if the radius is 1. π/4 = 0.7854, so you can see that this is quite close. More iterations than 1000 would produce a closer answer. | This, when run on my computer, produced the result of 0.783. The area of a quarter circle works out to be π/4 if the radius is 1. π/4 = 0.7854, so you can see that this is quite close. More iterations than 1000 would produce a closer answer. | ||
Revision as of 22:00, 21 August 2008
This is a small, simple example of utilising the Monte-Carlo method to calculate the value of pi. It's easy because it is only a 2-dimensional problem and thus is easy to visualise, and is trivial in its code implementation.
Picture the graph shown in the parent article - that of a quarter circle plotted on a unit x-y graph:
Below is a small example piece of Matlab code that does the above simulation:
Code:
% Monte-carlo simulator for solving area of a quarter-circle of radius 1, % dictated by bounds of x^2+y^2 = 1
% Set up arrays and other constants x = 0; % x coordinate y = 0; % y coordinate total = 10000; % amount of points to generate count = 0; % number of points inside area
% monte-carlo solver
for i = 1:total
% generate a random point in 0<x<1 and 0<y<1
x = rand; % random number between 0 and 1
y = rand;
% monte-carlo check if inside bounds
if (x^2+y^2 <= 1)
count = count + 1;
end
end
% calculate 'true' area and monte-carlo solved area circlearea = pi*1^2; % 'true' area montearea = count/total*4; % monte-carlo area
This, when run on my computer, produced the result of 0.783. The area of a quarter circle works out to be π/4 if the radius is 1. π/4 = 0.7854, so you can see that this is quite close. More iterations than 1000 would produce a closer answer.