About Matlab Code

I want to ask about matlab's code of this clip >>http://www.youtube.com/watch?v=EbC0eP0Dq9A<<

Step for Algorithm .. 1. fix number of point for random . 2. when it ploted, separate which point in a circle and another out of circle. 3. after point are random finish, calculate ratio in a picture. by Ratio = (Rectangle's areas * all of point in a circle) / number of point for random

Sorry, If I spell grammar wrong..

[Techit]

yeah, thx for all comments :)

but i need some guild line for a code

because i can't understand that youtube comments

thanks agian

and i'm sorry for my newbie in this matlab

Answer 1

[Paulo Silva]

clf
r=1;
rectangle('Position',[-1,-1,2*r,2*r],'Curvature',[1,1])
axis([-1 1 -1 1])
hold on
n=1000;
x=-1+2*r*rand(n,1);
y=-1+2*r*rand(n,1);
plot(x,y,'r.');
in=sum((x.^2+y.^2)<r);
(2*r)^2*in/n  %Thanks Sean, (2*r)^2 is the area of the circle

Increasing n gives better approximations for the value of pi.

[Techit]

Paulo, Thx for your answer !

I want to know about code of this >> Monte Carlo-method is used here to approximate pi <<

[Paulo Silva]

It's fixed now, thanks Sean for pointing out the obvious error on the code

[Techit]

@Paulo Silva , Can u show me about code, If i want differnce color between point in/out a circle,

[Paulo Silva]

I don't know what you want, just guessing, add this code to the end of the code I provided in my answer

clf

idx=(x.^2+y.^2)<r;

hold on

plot(x(~idx),y(~idx),'r')

plot(x(idx),y(idx))

[Techit]

Yess, u understand that i want to ask, I have last Question .. how i display output(Pi), like cout or printf?

[Paulo Silva]

fprintf('The aproximated value of pi is %g\n',(2*r)^2*in/n)

[Techit]

I want to show Pi on Figure, Can I do?

if it can, what's a code ??

[Paulo Silva]

text(-0.5,1.1,['aproximmated \pi value=' num2str((2*r)^2*in/n)])

[Techit]

Thank you very much for your help.

> ขอบคุณครับ < I'm from Thailand :)

Answer 2

[Sean de Wolski]

I think the answer to your question is already answered in the video comments.

[Oleg Komarov]

Agree

[Techit]

I want to know about Code of this clip :)

[Techit]

i can't imagine it cus i'm just a beginner

[Paulo Silva]

Sean and Oleg are right, the answer is there and it's easy enough for anyone to replicate the code in just a few simple lines.

[Sean de Wolski]

okay,

help rand

help le

help power

is LITERALLY all you need.

[Oleg Komarov]

Division? (troll)

[Sean de Wolski]

still need the divide though... darn.

[Walter Roberson]

Shouldn't that be

help rdivide

rather than ldivide ?

[Sean de Wolski]

hmmm lt or le? The video says lt, but points on the circumference are part of the circle...

Answer 3

[Sean de Wolski]

Alright - darkness golf

f=@(n)etc

I can do it with 38 characters (lt); 39 (le).