Estimating the value of pi using random numbers

Pi ($\pi$) is a mathematical constant used in different applications in mathematics and physics. One of its common applications is that it represents the ratio of the circumference of a circle to its diameter as shown in the following equation:

$$\begin{equation} C = 2 \pi r \end{equation}$$

Here $C$ is the circumference and $r$ is the radius of the circle. It is approximated as $3.14$. $\pi$ is an irrational number. It means that it can’t be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. The commonly used interpretation of $\pi$ ($\frac{22}{7}$) is also an approximation.

In this blog, we’ll learn how to estimate the value of $\pi$ using random numbers.

Let’s start with drawing a circle of radius $r=1$ around the origin $(0,0)$ as shown below.

Let’s place the circle in a square whose one side has the same length as that of the diameter of the circle.

The following illustration shows the circle inscribed in a square.

Guess what would be the length of each side of the square? As the length of each side is $2r$ and $r=1$, therefore the length of each side is $2$.

Let’s compute the area of both shapes: the circle as well as the square. The area of the circle is $\pi r^2$. As $r=1$ in this setting, so, the area of this circle is $\pi$. The area of a square is computed as the square of the length of its side. The length of one side of the square is $2r$ and $r=1$. Therefore, the length of one side is $2$, and the area of the square is $2^2 = 4$.

Assume that the square is the total surface area available to us and we hang a circular dartboard on this area. The geometric properties of the available area and the dartboard are the same as those of our drawn shapes. We want to throw a dart and hit the circle. There are two possibilities:

• We hit the circle successfully.
• We miss the circle and the dart hits the outer square. Remember that is the total area available to us.

The probability of hitting the circle is the ratio of the area of the circle and that of the square. Let’s call it $P$.

$$\begin{equation}P = \frac{\pi}{4}\end{equation}$$

Now, let’s model and conduct an empirical experiment to throw the darts and examine whether they hit the circle or miss it. Each point hit by a dart is defined as $(x, y)$. The exact values of $x$ and $y$ are random as they are being determined by the physical experiment. We need to generate two random numbers notating $x$ and $y$. For this purpose, we use the rand() function of C++. It generates a positive random integer. If we reanalyze our surface area, both of its coordinates are bound between $-1$ and $1$. Our randomly generated number should be a real number between $-1$ and $1$.

• First of all, we limit the upper bound of the random number by dividing it over RAND_MAX which represents a significantly large integer (greater than or equal to the random integer).
• We multiply the random integer by $1.0$ before dividing it over RAND_MAX so that we get a real number as the output. This confines the random number between $0$ and $1$ (both inclusive).
• Next, we multiply it by $2$ to double the possible range between $0$ and $2$.
• At the end, we add $-1$ into it to shift its range between $-1$ and $1$ (both inclusive).

We perform the experiment twice for $x$ and $y$ coordinates.

Let’s understand how we can characterize a randomly generated point $(x, y)$ geometrically. Pythagoras’ theorem states that the hypotenuse can be computed using the other two sides in a right-angled triangle. A right-angled triangle that has a hypotenuse equal to the radius of the circle $(r)$ is shown in the following figure. We drop a perpendicular to get the projection on the x-axis. The perpendicular height gives the height $(y)$ and the projection length gives the base $(x)$ of the right-angled triangle.

According to Pythagoras’ theorem:

$$\begin{equation}r^2=x^2+y^2\end{equation}$$

The circle has $r=1$. So, $r^2=1$. The above equation becomes:

$$\begin{equation}1=x^2+y^2\end{equation}$$

If our randomly generated point $(x,y)$ falls exactly at the boundary of the circle, then $x^2+y^2 = 1$. If it falls within the circle then $x^2+y^2 < 1$. And if it falls outside the circle then $x^2+y^2 > 1$.

Let’s treat the outer square as a wall where we have hung a circular dartboard at its center. The radius of the circle is $1$. If we throw a dart then it would either hit the circle or miss it. If the dart hits the circle then $x^2+y^2\leq1$. If it misses the circle then it hits the part of the square that’s not covered by the circle. If we perform a physical experiment repeatedly—say 100,000 times and call it $\text{trials}$— we can count how many times the dart hits—let’s call it $\text{hits}$—the circle. Based on this empirical exercise, we can estimate the probability of hitting the circle as follows:

$$\begin{equation}P=\frac{\text{hits}}{\text{trials}}\end{equation}$$

Recall we computed $P$ using the geometric properties too. We can equate equations $(2)$ and $(5)$ as follows:

$$\begin{equation}\frac{\pi}{4}=\frac{\text{hits}}{\text{trials}}\end{equation}$$

$$\begin{equation}\pi=4 \times \frac{\text{hits}}{\text{trials}}\end{equation}$$

In the following program, we simulate the same experiment. We generate $(x,y)$ 100,000 times, count $\text{Hits}$, and estimate $\pi$.

Let’s go through the code:

• Lines 6–7: We initialize the number of trials as 100000 in trials and the number of hits as zero in hits. We set their data type to double so that we may preserve the calculations in the real domain later.
• Line 8: We configure the loop counter to run the trials iterations.
• Lines 10–11: We generate two random numbers between $-1$ and $1$.
• Lines 12–13: We characterize the generated random point $(x, y)$ to check if it falls within the circle or not. If it falls within the circle then we increment hits. Otherwise we treat it as a missed shot and the next iteration follows.
• Line 15: Finally, we estimate the value of pi based on the randomized experiment and geometric characteristics of the circle and square.

Run this program multiple times and you will see a fluctuation in the estimated value of pi each time you run it. This fluctuation will remain there because our estimation is based on the randomized experiment.

In this blog, we used a dartboard inscribed in a square to estimate the value of $\pi$. We used geometric characteristics and empirical experiments to estimate the probability of hitting the dartboard when a dart is thrown. Equating the probability computed using these two methods led us to estimate the value of $\pi$.

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