Writing a program for generating random numbers involving a circle and a square, help please??

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Reelz 2012-4-17

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A circle of radius 1 is centered in a square target with sides of length 2. The are of the circle to the are of the square is pi/4, so if you throw darts at random, they will have an equal chance of hitting any part of the target. I know that the probability is pi/4, and that pi is 4 times the ratio of: # of throws inside the cirlce (this is divide>) ---------------------------------- total # of throws

What I am trying to accomplish is generating random coordinate pairs (x,y) using x = -1 + 2*rand(n,1)), and plotting pi as a function of the number n of throws by first: throwing 1 dart for the 1st estimate, and throwing 2 darts to get the second, 3 for the third, etc.

I have some coding but a student told me it was all wrong and to try asking on this website. help please!!!

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Reelz 2012-4-17

Program is on a school computer, which I won't be able to access for a couple of days. I was trying to create a square matrix with some sort of circulant matrix inside, but got so confused, help please

Walter Roberson 2012-4-17

You can just throw one more virtual dart each time.

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回答（1 个）

Geoff 2012-4-18

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You are generating your x- and y-values between -1 and 1, which means you're automatically sampling randomly within the square. That part is fine.

But like Walter suggests, you can just do one dart at a time. That is, you decide the maximum number of throws N you want to do and generate those N co-ordinates only once.

You need an in/out test for your circle and a way to calculate how your pi ratio evolves as the number of throws increases. You will end up with a N-length vector showing (in theory) progressive convergence on pi.

Functions that will help you achieve this are hypot and cumsum.

By the way, get in the habit of posting your code - especially if you think it's wrong. You are more likely to receive help if there's hard evidence that you are also trying to solve it on your own. =)

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Reelz 2012-4-24

Here is some code that might help.

It is just I am having trouble adjusting it so fit my needs of 1) throwing 1 dart for the 1st estimate, and throwing 2 darts to get the second, 3 for the third, etc. and plotting pi as a function of n throws and (2) throwing 1 dart to get the first estimate, a second one and using the two darts thrown together to get the 2nd estimate, and throwing another one using those three to get a 3rd estimate

CODE:

function mycir = approxcir(n)

% Input: n = number of points to generate

% Default is n = 1e6

% Larger values of n should perform better

if nargin < 1

n = 1e6;

end

% Generate uniformly distributed points in

% [0, 1] x [0, 1]

xy = rand(n, 2);

% Compute distance from (0.5, 0.5)

r = sqrt((xy(:,1)-0.5).^2+(xy(:,2)-0.5).^2);

% Count fraction of points within 1/2 unit of (0.5, 0.5)

frac = sum(r <= 0.5) / n;

% Since square has side 1, circle has radius (1/2)

% and should have area of pi*(1/2)^2

% frac is approximately pi/4 so pi is approximately 4*frac

mypi = 4*frac;

Walter Roberson 2012-5-1

You write,

% Count fraction of points within 1/2 unit of (1, 1)

frac = sum(r <= 1) / n;

but that code is for counting the fraction within 1 unit of (1,1)

Geoff 2012-5-1

What do you mean by "not working at all"? What happens when you run the code? Be more helpful in your description. Why are you computing distance to (1,1)? You have obviously decided to perform the experiment on only one quadrant of the circle/square, so why not centre it at (0,0) to make your code clearer? It makes no difference. Please look up the 'hypot' function in MatLab help. I've suggested it to you at least twice. I notice that your function's return value is 'mycir', but at the bottom where you do the calculation you put it into 'mypi'. The calculation itself looks fine. Works for me. I got pi = 3.14 with just 1000 random numbers.

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