Question

Can't seem to figure this out as a matlab file and was looking for help, you can't use any for or while loops and all functions used must be built into matlab.

Consider a square dartboard with each side having dimension equal to 2.

Centered inside this square place a circle having radius 1.

Now throw N = 1e4 darts at the dartboard in a random fashion so that they

are uniformly distributed over the surface of the dartboard.

The area of the circle is pi*radius^2. Given a radius of 1, the area of

the circle is equal to pi.

The area of the square is 2 * 2 = 4

The ratio of the circle area to the square area is pi/4. If this value is

multiplied by 4, the result is pi.

Since the darts are uniformly distributed over the dartboard, the ratio

of the darts that are in the circle to the total number must approximate

pi/4. So, multiplying this ratio by 4 will give an approximation to pi.

No For or While loops permitted.

No extraneous output to the Command Window.

Answer 1

Let us assume that the center of the board is at the origin. Every time a dart is thrown it will be at a location given by its x-coordinate and y-coordinate. So, we generate 1e4 such coordinates and then count the number that falls into the circle of radius 1.

The MATLAB code for generating the random coordinates (uniformly distributed) and counting the number of darts in the circle of radius 1, is given below:

x=-1+2*rand(1,1e4); % generate 1e4 random x-coordinates between [-1,1]
y=-1+2*rand(1,1e4); % generate 1e4 random y-coordinates between [-1,1]

z=x.*x+y.*y; % to check whether the coordinates are inside the unit circle
count=length(find(z<=1));

ratio=count/length(x);

pi_value=ratio*4;

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Running the above code in MATLAB yields count as 7862 and the value of $\pi$ as 3.1448

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