# Interesting shapes using probability and discrete view of a problem

Suppose we have a circle of radius $r$, we show the distance between a point and the center of the circle by $d$. We then choose each point inside the circle with probability $\frac{d}{r}$ , and turn it black (note that $\frac{d}{r}<1$). With these rules we get shapes like this: (With help of Java)

The shapes made were pretty interesting to me. I decided to add a few lines of code to my program to keep count of the drawn points, and then divide them by the area of the circle, to see what percentage of the circle is full. All I got for multiple number of tests was a number getting close to $2/3$ as the circle got bigger.

Problem A: Almost what percentage of the circle is black?

I found an answer as following:
As all the points with distance $l$ from the center lie on a circle of radius $l$ and the same center, to find the "probable" number of black points, we shall consider all circles with radius $l$ ($0<l<r$). Consider one of these circles with radius $l$. the circumference of that circle is $2\pi l$ and by how we choose black points, almost $\frac{l}{r}*2\pi l=\frac{2\pi l^2}{r}$ of that circle is black. Taking the sum of all these circles means integrating the circumferences of these circles from $l=0$ to $r$. So: $$\int_0^r{\frac{2\pi l^2}{r}dl}=\frac{2\pi r^2}{3}$$ and so the percentage of all black points shall be $$\frac{\frac{2\pi r^2}{3}}{\pi r^2}=\frac{2}{3}$$.

The other question that came up in my mind is:

Problem B: For given number $x\geq 1$, what is the probability $P$ that $1/x$ of the circle is black?

I think the answer is $P=0$ for any given $x$. We can say if there exists an area $A>0$ full of black points, there exists an infinite number of points so that their distance to the center of the circle is equal, and the probability of all of them being black is $0$ (because there is an infinite number of them, $d$ and $r$ are constant, $\frac{d}{r}<1$ and $\lim_{n\to\infty}{(\frac{d}{r})}^n=0$).

It gets more interesting with a discrete view of the problem. Suppose we have a grid full of $1*1$ squares with the size $(2r+1)*(2r+1)$.We define the reference square with length $2l+1$ (we call $l$ it's radius) as the set of squares "around" the reference square with length $2l-1$, and the reference square with length $1$ is the set of $8$ squares around the central square. We define the distance $d$ of a square from the central square by the radius of it's reference square. Now we are ready to propose a problem similar to problem A:

Problem C: Suppose each square with distance $d$ to the central square turns black with a probability $\frac{d}{r}$. Prove that Almost $2/3$ of the squares turn black, as $r\to \infty$

We begin our proof by proving a reference square of radius $l$ contains exactly $8l$ squares. This is easy because there are $(2l+1)^2-(2l-1)^2=8l$ squares in a reference square with radius $l$. Now, each square in the reference square is black with probability $\frac{l}{r}$, so almost $\frac{l}{r}*8l=\frac{8l^2}{r}$ of the reference square is black (Similar to the proof of Problem A). By summing up all the reference squares with radius $l=1$ to $r$ we get: $$\sum_{l=1}^{r} \frac{8l^2}{r} = \frac{8}{r}*\frac{(r)(r+1)(2r+1)}{6}=\frac{4(r+1)(2r+1)}{3}$$ and so the percentage of the whole square that is black (as $r$ tends to infinity) is:$$\lim_{r\to\infty} \frac{\frac{4(r+1)(2r+1)}{3}}{(2r+1)^2}=\frac{4(r+1)}{3(2r+1)}=\frac{2}{3}$$ as expected.
Some problems, though, still remain.

First, I would appreciate the verification of my solutions to problems A, B and C. The program I wrote in java only fills out a finite number of "pixels", does this represent Problem A or Problem C? would there be a difference if we draw the circle for infinite number of "points"?

Second,The result of Problem B seems a little strange because as $X$ tends to $\frac{3}{2}$, $P$ should go to $1$ because $\frac{2}{3}$ of the circle is black (as proved in Problem A). Then Why do we get $P=0$ for all values of $X$? How can we explain this?

Third, What is the connection between the discrete view of the problem and the main problem? Can someone generalize the proof of Problem C to prove Problem A? I think one can easily generalize Problem C to "circles" full of tiny squares and prove the fact similarly, but going to an infinite number of points instead of "pixels" (which are equivalent to tiny squares in problem C) is still another matter.

I would appreciate any help.

• When you write "choose each point inside the circle", do you mean pixels instead of points? If you pick a finite number of "points", the disc remains almost-all-white...
– A.G.
Commented Jan 5, 2016 at 5:40
• I rewrote my questions to avoid ambiguity.
– CODE
Commented Jan 5, 2016 at 7:58

Your approaches to Problems A and C are correct. To eliminate the ambiguity between 0-dimensional points and 2-dimensional pixels and probability, we can determine the expected proportion $s$ of shaded pixels at radius $d$:

$$E(s)=\lim_{n\to\infty} (\sum_{i=1}^n {1 \over n} * {d \over r}) = {d \over r}$$

That allows us to integrate from $0$ to $r$ to find the expected shaded area of the entire circle, based on each individual thin ring, just as you demonstrated. $${\int_0^r E(s) * 2 \pi d \; dd \over \int_0^r 2 \pi d \; dd } \\ = {\int_0^r {d \over r} * 2 \pi d \; dd \over \int_0^r 2 \pi d \; dd } \\= {{2 \over 3} \pi r^2 \over \pi r^2 } ={2 \over 3}$$

For Problem B, I think the approach may be flawed. You're trying to determine the probability of the aggregate proportional area, which is a single value and can be determined empirically based on the probability of any given point being shaded. Essentially, $P(E(s)={1 \over x})$. Imagine that you have an urn with a single white ball:

$$P(white) = 1\\ P(not\,white) = 0$$

Just as this theoretical urn has only one potential outcome, a white ball, so does your circle have only one potential value for the overall ratio of shaded area.

$$P(Shaded \, area = {2 \over 3} \pi r ^2) = 1\\ P(Shaded \, area \ne {2 \over 3} \pi r^2) = 0$$