Here's an easier, pythonic way to limit the number of digits, given the
original, non-terminating pi generator:
>>> import itertools
>>> print(list( itertools.islice(pi_digits(), 10) ))
[3, 1, 4, 1, 5, 9, 2, 6, 5, 3]
David H
-----Original Message-----
From: "michel paul" <pythonic.m...@gmail.com>
Sent: Sunday, December 23, 2012 6:49pm
To: "Kirby Urner" <kur...@oreillyschool.com>
Cc: da...@handysoftware.com, "edu-sig@python.org" <edu-sig@python.org>, "michel
paul" <pythonic.m...@gmail.com>
Subject: Re: [Edu-sig] generate digits of pi
I realized something. This was the original version:
def pi_digits():
k, a, b, a1, b1 = 2, 4, 1, 12, 4
while True:
p, q, k = k*k, 2*k+1, k+1
a, b, a1, b1 = a1, b1, p*a+q*a1, p*b+q*b1
d, d1 = a/b, a1/b1
while d == d1:
yield int(d)
a, a1 = 10*(a%b), 10*(a1%b1)
d, d1 = a/b, a1/b1
I forget where it came from. Like I had mentioned, I had a really bright
student awhile back who was really intrigued by this, and he at one point
edited it to produce the digits in binary. In the original form the generator
never terminates. Somewhere along the line an edit was made to try to get it to
terminate at n digits. Probably to make calling it easy to call as in
list(pi_digits(n)).
- Michel
On Sat, Dec 22, 2012 at 5:34 PM, Kirby Urner <[mailto:kur...@oreillyschool.com]
kur...@oreillyschool.com> wrote:
Got it, no wrong digits just not always exactly the number you asked for.
This happens often in 3.2 as well:
>>> exp = ((n,len(list(pi_digits(n)))) for n in range(10000)) # (number asked,
>>> number got)
>>> exp2 = ((a,b) for a,b in exp if a != b) # filter on "not same"
>>> for i in range(10): print(next(exp2), end=", ")
(2, 3), (4, 5), (10, 11), (16, 17), (18, 19), (22, 23), (28, 31), (29, 31),
(30, 31), (34, 36),
Kirby
On Sat, Dec 22, 2012 at 4:40 PM, <[mailto:da...@handysoftware.com]
da...@handysoftware.com> wrote:
In each case I asked for only 79 digits, but got 79, 82, and 83 digits
depending on whether I was using python 3.2, python 2.6, or python 2.6 with
-Qnew, respectively. The digits all seem to be correct, but the algorithm for
stopping at digit n seems to be very sensitive.
David H
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