[Tutor] How to calculate pi with another formula?

[Dick Moores]

Gregor Lingl wrote at 11:27 10/29/2004:
>Hi Dick!
>
>Accidentally I just was tinkering around with the new
>decimal module of Python2.4. (By the way: it also works
>with Python 2.3 - just copy it into /Python23/Lib)
>
>The attached program uses a very elementary (and inefficient)
>formula to calculate pi, namely as the area of a 6*2**n-sided
>polygon (starting with n=0), inscribed into a circle of radius 1.
>(Going back to Archimedes, if I'm right ...)
>
>Nevertheless it calculates pi with a precision of (nearly)
>100 digits, and the precision can be  arbitrarily enlarged.
>In the output of this program only the last digit is not correct.
>
>import decimal
>
>decimal.getcontext().prec = 100
>
>def calcpi():
>    s = decimal.Decimal(1)
>    h = decimal.Decimal(3).sqrt()/2
>    n = 6
>    for i in range(170):
>        A = n*h*s/2  # A ... area of polygon
>        print i,":",A
>        s2 = ((1-h)**2+s**2/4)
>        s = s2.sqrt()
>        h = (1-s2/4).sqrt()
>        n = 2*n
>
>calcpi()
>
>Just for fun ...
>
>Gregor

This works great, and if I change the precision to, say, 2000, and the 
range to 2000, I get pi accurate to the 1,205th digit (this took 66 
minutes, with psyco employed), when I compare with the pi pages on the web.

Now to my new question. I have an artist friend who knows an artist who 
needs pi expressed in base 12. I don't know how many digits he needs, but 
I think he'll take what he can get. Is there a way to use math.log(x, 
base) with the decimal module to accomplish this? Or is there another 
way? Or is there no way?

Thanks,

Dick Moores
rdm at rcblue.com