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I'm trying to understand how to compute the value of Pi by means of the Monte Carlo simulation. I have a circle inside a square where the sides of the square are tangent to the circle. As data I have the number of random points and the ratio between the lenght of one side of the square and the diameter.

$$\displaystyle \frac{l}{d} = r_{a}$$

According to what I've read I have to count the number of points lying inside the circle and the total of points. The area of a circle is $A_{circle} = \pi r^2$ and the area of a square is $A_{square} = l^2$. To check if a point is inside the circle I have to check $\sqrt{x^2+y^2} <= r$

I don't know how to relate the ratio $r_a$ with the radius so to check if the point is inside or outside the circle. Any suggestion?

UPDATE

This has been a really frustrating stuff and is a really simple example of Monte Carlo approximation.

Assuming $d=1$ then I have that $\displaystyle \frac{l}{d} = l = r_a$ Therefore, to check if a random point is inside the circle I have to check only that

$$\sqrt{x^2 + y^2} ≤ \frac{1}{2}$$

or it has to be related to the ratio $r_a$ ??

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That's very simple: suppose that your circle have an unitary radius, then its area equal to $\pi$. Area of square equal to 4, so to calculate $\pi$ you simply need to count points inside a circle (say, $N_c$), divide it to total number of points (say, $N_t$) and multipy by 4: $\pi = 4\cdot\frac{N_c}{N_t}$. I did such test in my simple raytracing. Really, it gives $\pi$ with accuracy of $n$ digits, where $n = \sqrt{N_t}$. –  Eddy_Em Feb 19 '13 at 11:52
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Thanks @Eddy_Em that part I know it already. The problem is that I don't know what's the radius of the circle. The only data available is the ratio between one side of the square and the diameter of the circle, $\frac{l}{d} = r_a$. From there I don't know how to calculate the radius. –  BRabbit27 Feb 19 '13 at 12:14
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There's something unclear to me then. Can you explain/show a situation with r_a different from 1? –  Dr_Sam Feb 19 '13 at 12:28
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Let's say we have a square and a circle where $d = l$ that means that the ratio $r_a = 1$. Now suppose I make my circle smaller but the square is the same size as before then the ratio $r_a > 1$. From here, I want to compute the value of $\pi$ but I only know the value of $r_a$, the values of $l$ and $d$ are unknown –  BRabbit27 Feb 19 '13 at 12:35
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@BRabbit27 If you make the circle smaller, it will not be tangent anymore to the sides of the square. –  Dr_Sam Feb 19 '13 at 12:53
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2 Answers 2

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Suppose, that your circle have unitary radius: $r=1$, then length of square's side equal to $l = r_a \cdot d = 2\cdot r_a$.

So, area of circle equals to $S_c = \pi$ and area of square equals to $S_s = 4 r_a^2$.

If you "throw" $N_t$ random points inside of square, a part of them (say, $N_c$) will fall into the circle. Relation $\frac{N_c}{N_t}$ equals to proportions of their areas, $\frac{S_c}{S_s}$:

$$\frac{N_c}{N_t} = \frac{S_c}{S_s}$$

Now, to calculate $\pi$ you should divide $N_c$ to $N_t$ and multipy by $S_s$:

$$\pi = 4r_a^2\cdot\frac{N_c}{N_t}$$

In case of unitary $l$ you should set $r = \frac{1}{2 r_a}$.

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I don't know why but the math formulas are not readable. I don't know if it's something wrong with my browser or you missed something. –  BRabbit27 Feb 19 '13 at 13:06
    
@BRabbit27, set r = 1 / (2 r_a) –  Eddy_Em Feb 19 '13 at 13:07
    
I understand all that. What I don't is how to check if a given point $(x,y)$ is inside the circle. It should be something like $\sqrt{x^2+y^2} <= radiusOfCircle$ but what would be that radius for any arbitrary circle inside a square? –  BRabbit27 Feb 19 '13 at 13:20
    
@BRabbit27, I've told you upper: if you "throw" your random numbers into square of 1x1, then check hits to circle simply: $x^2+y^2 < \frac{1}{4 r_a^2}$. –  Eddy_Em Feb 19 '13 at 13:40
    
(Sorry for formating: I didn't know that scicomp uses mathjax) –  Eddy_Em Feb 19 '13 at 13:43
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I don't think that you can find the radius $r$ with the informations that you provide. For example, you can imagine having

  • a circle with diameter $d=1$ centered in a square with side $l=1$
  • a circle with diameter $d=2$ centered in a square with side $l=2$

Both circles will be tangent to their respective squares and both yield $r_a=1$. So, if your data consists only in $r_a$, you cannot distinguish between the two situations.

In this context, the data $r_a$ seems even useless, since to obtain a circle tangent to the sides of the square, you need $r_a = 1$.

If you want to forget about the hypothesis that the circle is tangent to the square, you can still approximate $\pi$ for different values of $r_a$ but you need to modify your formula and this does not fix your problem with finding $r$.

One last point is that actually, $r$ does not really matter: if you have a set of random points in the square with size 1, you can scale them to make them appear in a square of size $l$ (whatever is $l$). Applying the formula to the square of size 1 or size $l$ will give you exactly the same approximation of $\pi$.

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I don't understand when you say $r$ does not really matter, if it does not, how can I compute if a point $P(x,y)$ is inside the circle? What I'm trying to understand is how to compute if the point is inside the circle, I have the formula $\sqrt{x^2 + y^2} ≤ r$ so I need something to count the number of points inside the circle... –  BRabbit27 Feb 20 '13 at 7:42
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It does not matter in the sense that you can choose $r$, then take $l=d=2r$ and make your computations. Whatever value for $r$ you have taken, the result will be exactly the same. –  Dr_Sam Feb 20 '13 at 7:55
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