How can I calculate Pi?

[goldenpeach]

Hi!

I do some programs in my free time(if you want to know,I use python) and my next goal is to calculate Pi.

The problem,is that I don't know what you have to do to calculate the digits of Pi.

What are the calculations I need to do?

Thank you for reading!

[Kerbart]

https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node12.html

[goldenpeach]

Thank you or the answer!

[K^2]

Bah. It's always best to start with something simple that you can really understand. The most basic way to compute Pi is a geometric one. Start with a regular hexagon inscribed in a circle of radius 1. Its perimeter is 6, which gives you first approximation Pi=3. Then drop a perpendicular bisector on each side to construct a regular dodecagon inscribed in the same circle. You should be able to use Pythagoras' Theorem to compute the new perimeter. Keep going. As the polygon gets rounder and rounder, your approximation will get closer to the value of Pi.

The bad news is that it converges ridiculously slowly, so it's not something you'd use for thousands of digits, but it's a good start to see how the algorithm works.

Then you can sit down and try to understand more advanced formulae and where they come from. Most of them are highly non-trivial, and programming them is much easier than understanding.

[falofonos]

103993/33102 is apparently a good approximation.

[Idobox]

There an even simpler way.

Build a large square matrix, of size r*r.

Now count how many points of this matrix are less than r away from the origin, you can do that by checking if x²+y²<=r².

The points that fit that rule are within a quarter disk of area pi*r²/4, and the whole matrix has an area r².

So take the number of points in the quarter circle, multiply by 4, divide by the number of points in the whole matrix, and you get an approximation of pi.

the more points in the matrix, the more accurate the result.

You can optimize by:

doing that only for one half of the matrix (it's symmetric along the diagonal)

measuring the surface of a ring with r1 > distance >r2. If you do it smartly, you will only have to check a few points (all points in the ring have neighbours that are also in the ring)

massively parallelisation of the task.

[goldenpeach]

Thank you,all of you, for the answers!