# Estimate $\pi$ using Picks Theorem.

This $$\pi$$ estimate is derived from Pick's theorem and the approximation of a circle with a lattice polygon

$$\begin{equation*} \pi \approx \lim_{r\to\infty} \frac{i + \frac{b}{2} - 1}{r^2} \end{equation*}$$

To derive it, Pick's theorem states

$$A = i + \frac{b}{2} - 1$$

where $$A$$ is the area of the lattice polygon.

For a circle with radius $$r$$, the area is given by:

$$A = \pi r^2$$

When the lattice polygon closely approximates the circle, their areas $$A$$ are nearly equal:

$$\pi r^2 \approx i + \frac{b}{2} - 1$$

$$\pi \approx \frac{i + \frac{b}{2} - 1}{r^2}$$

As $$r \to \infty$$, the lattice polygon becomes an increasingly accurate representation of the circle, and the area ratio converges to the value of pi:

$$\begin{equation*} \pi \approx \lim_{r\to\infty} \frac{i + \frac{b}{2} - 1}{r^2} \end{equation*}$$

Examples

For $$r=10$$, the estimated value of $$\pi$$ is $$4.645000$$ For $$r=100$$, the estimated value of $$\pi$$ is $$3.156050$$ For $$r=10000$$, the estimated value of $$\pi$$ is $$3.141592$$ (C Code)

Is this a valid method to estimate $$\pi$$?

• – lhf
Jun 3 at 12:00

The main thing you need to prove to establish that your technique works is that the relative error in area converges to $$0$$ as $$r\to \infty$$. You didn't say exactly how you compute the lattice polygon, and depending on the details this error may be somewhat tricky to bound.
Another approach is to compute two estimates: one that's an upper bound (by ensuring the lattice polygon strictly contains the circle) and one a lower bound (by building a lattice polygon strictly inside the circle). For each $$r$$ this will give you both an estimate for $$\pi$$ and a bound on the error of the estimate.