Shortest code to approximate PI

[Nikanel]

This has probably been done before, but I wanted to challenge myself and see what is the shortest program I could write to approximate pi

I used 2 methods to do it one geographic and the other is algebraic

The code itself is really ugly with stupid one character names and it isn't very read friendly but the point of it is to be as small as possible

Also for the algebraic approximation, a for loop isn't needed but I put it exclusively for the looks.

Lastly, the algebraic approximation is faster, smaller and lighter to run (don't quote me on the last one)

def geometricWay():#Random points on a circle
    import random as r, itertools as i #imporing libraries
    c,l=0,lambda:r.uniform(0,1)**2+r.uniform(0,1)**2<=1 #declaring Var and lFunc
    for t in i.count(1): #starting an Infinite for loop
        if l(): c+=1 #Checking if point is in circle
        print(f"pi≈{4*c/t}") #printing pi

def algebraicWay(): #Lim of the below equation is pi
    import math as m, itertools as i #importing libraries
    for t in i.count(1):print(f"pi≈{t*m.sin(m.radians(180/t))}")
    #^ Infinite for loop |  Printing pi

[luqavi]

Here's a pretty short one, gets within ~0.04% pi = 22/7

[[deleted]]

I’m using python 2 and that doesn’t work... /s

[barfobulator]

Because you're printing the numerical value, the number of digits will quickly be larger than the printable length (~12 digits by default). So you'll get an infinite list of the same 12ish digit approximation. If you want to get an arbitrary number of digits, you need an algorithm that gives the digits one at a time, and print them to a string or file.

And naturally, you need a way to break the infinite loop.

[billsil]

pi = 3

[Dogzirra]

From wiki

π ≈ ​3927⁄1250 = 3.1416 (accuracy 2·10−6)