Tim Roberts wrote:
On Sun, 23 Jan 2005 10:28:47 -0800 (PST), Ali Polatel <[EMAIL PROTECTED]> wrote:
I found a code which calculates pi with an interesting algorithm the programme code is below:
from sys import stdout
def f((q,r,t,k)):
n = (3*q+r) / t
if (4*q+r) / t == n:
return (10*q,10*(r-n*t),t,k,n)
else:
return (q*k, q*(4*k+2)+r*(2*k+1),t*(2*k+1),k+1)
# Call pi(20) for first 20 digits, or pi() for all digits
def pi(n=-1):
printed_decimal = False
r = f((1,0,1,1))
while n != 0:
if len(r) == 5:
stdout.write(str(r[4]))
if not printed_decimal:
stdout.write('.')
printed_decimal = True
n -= 1
r = f(r[:4])
#stdout.write('\n')
if __name__ == '__main__':
from sys import argv
try:
digit_count = long(argv[1])
except:
digit_count=int(raw_input('How many digits? :'))
pi(digit_count)
This code gives the number in an unusual format like "3.1415'None'" it has a number part and a string part .
Your problem is that the function "pi" doesn't actually return anything. If you do this, for example:
print pi(5)
you will see 3.1415None. The first 5 digits of pi come from the sys.stdout.write calls in the function, and the "None" comes from the print statement when it tries to display the result of the function.
I can certainly tell you how to change this code so that it returns a value instead of always printing to stdout, but first let me ask what you plan to do with it. You said:
Any idea how to seperate this 'None' from the number and make it a real normal number on which I can do operations like +1 -1 or like that
This algorithm is useless as a way to generate a "real normal number", because there are no "real normal numbers" that keep an infinite number of floating point digits. This function is fun for generating 120 decimal digits for amusement purposes, but it has no practical value. If all you want is a "pi" that you can use in computation, you cannot do better than this:
from math import pi
one word: decimal
Double-precision floating point arithmetic on x86 machines has about 15 digits of precision. Even if you generate pi(80) using your function, as soon as you convert it to a float to use it in a function, it will lose all but the first 15 digits.
Here is your code, changed so that pi() returns a string value:
import sys
def f((q,r,t,k)): n = (3*q+r) / t if (4*q+r) / t == n: return (10*q,10*(r-n*t),t,k,n) else: return (q*k, q*(4*k+2)+r*(2*k+1),t*(2*k+1),k+1)
# Call pi(20) for first 20 digits, or pi() for all digits def pi(n=-1): s = [] r = f((1,0,1,1)) while n != 0: if len(r) == 5: s.append( str(r[4]) ) n -= 1 r = f(r[:4]) s.insert( 1, '.' ) return ''.join( s )
if __name__ == '__main__': try: digit_count = long(sys.argv[1]) except: digit_count = int(raw_input('How many digits? :')) print pi(digit_count) print repr(float(pi(digit_count)))
and here is what happens when you try to use it in arithmetic:
C:\Tmp>x.py 5 3.1415 3.1415000000000002
C:\Tmp>x.py 10 3.141592653 3.141592653
C:\Tmp>x.py 15 3.14159265358979 3.14159265358979
C:\Tmp>x.py 20 3.1415926535897932384 3.1415926535897931
C:\Tmp>
:-)
regards-ly y'rs -- Steve Holden http://www.holdenweb.com/ Python Web Programming http://pydish.holdenweb.com/ Holden Web LLC +1 703 861 4237 +1 800 494 3119
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