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""" | |
Run the algorithm below using CPython, Cython, PyPy and Numba and compare | |
their performance. (This is implementing a spigot algorithm by A. Sale, | |
D. Saada, S. Rabinowitz, mentioned on | |
http://mail.python.org/pipermail/edu-sig/2012-December/010721.html). | |
""" | |
def pi_digits(n): | |
"Generate n digits of Pi." | |
k, a, b, a1, b1 = 2, 4, 1, 12, 4 | |
while n > 0: | |
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 and n > 0: | |
yield int(d) | |
n -= 1 | |
a, a1 = 10 * (a % b), 10 * (a1 % b1) | |
d, d1 = a / b, a1 / b1 | |
# >>> list(pi_digits(20)) | |
# [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4] |
I've been trying to wrap my head around changing this algorithm to return a specific digit but not have it use a list to read off of (if this makes sense of course). I would really appreciate any help that I can get.
You can replace the yield int(d)
above with if n == 1: return int(d)
to get the nth digit only (renaming it to pi_digit(i)
would then be more appropriate, too):
>>> pi_digit(5)
5
>>> pi_digit(1)
3
If you want to follow Python's 0-based indexing giving the first digit for i=0 you'd also need to take care of that.
I've been trying to wrap my head around changing this algorithm to return a specific digit but not have it use a list to read off of (if this makes sense of course). I would really appreciate any help that I can get.
You can replace the
yield int(d)
above withif n == 1: return int(d)
to get the nth digit only (renaming it topi_digit(i)
would then be more appropriate, too):>>> pi_digit(5) 5 >>> pi_digit(1) 3If you want to follow Python's 0-based indexing giving the first digit for i=0 you'd also need to take care of that.
Thanks a lot man.
:)
Hello and apologies if this is a rather stupid question.
I've been trying to wrap my head around changing this algorithm to return a specific digit but not have it use a list to read off of (if this makes sense of course). I would really appreciate any help that I can get.
Cheers.