Subsequence From Madhava–Leibniz Series As Fast As Possible With Python

Madhava–Leibniz series: All I need is to create a subsequence from element k to element n: from this series to be created and returned as fast as possible. I started from 'l

Solution 1:

I slightly changed the madhava() function by using itertools.cycle:

def madhava(k, n):  # Optimized for speed any way you like.
    # you don't need to allocate the array in advance, so comment it out
    # series = [0.0] * (n - k + 1)
    first_divisor = 2 * k + 1
    last_divisor_plus_1 = 2 * n + 2

    if k & 1:
        series = [what / divisor for what, divisor in zip(cycle([-1, 1]), range(first_divisor, last_divisor_plus_1, 2))]
    else:
        series = [what / divisor for what, divisor in zip(cycle([1, -1]), range(first_divisor, last_divisor_plus_1, 2))]
    return series

Original version on my machine (AMD 2400G, Ubuntu 18.04):

('3', '6', '7')
Linux-5.0.20-050020-generic-x86_64-with-Ubuntu-18.04-bionic
Python ('default', 'Oct 22 2018 11:32:17') GCC 8.2.0
Executing in 64bit

Time Pi 3.141592653589793 Error of calculation   function
m * n = 4 * 25 = 100
 0.000  3.131592903558554 0.00999975003123943 madhava
 0.000  3.131592903558554 0.00999975003123943 leibniz
m * n = 4 * 250 = 1,000
 0.000  3.140592653839794 0.000999999749998981 madhava
 0.000  3.140592653839794 0.000999999749998981 leibniz
m * n = 4 * 2,500 = 10,000
 0.001  3.141492653590034 9.99999997586265e-05 madhava
 0.002  3.141492653590034 9.99999997586265e-05 leibniz
m * n = 4 * 25,000 = 100,000
 0.009  3.141582653589720 1.0000000073340232e-05 madhava
 0.016  3.141582653589720 1.0000000073340232e-05 leibniz
m * n = 4 * 250,000 = 1,000,000
 0.091  3.141591653589774 1.0000000187915248e-06 madhava
 0.150  3.141591653589774 1.0000000187915248e-06 leibniz
m * n = 4 * 2,500,000 = 10,000,000
 0.890  3.141592553589792 1.0000000161269895e-07 madhava
 1.546  3.141592553589792 1.0000000161269895e-07 leibniz
m * n = 4 * 25,000,000 = 100,000,000
 9.002  3.141592643589326 1.0000467121074053e-08 madhava
15.699  3.141592643589326 1.0000467121074053e-08 leibniz

The version using itertools.cycle:

('3', '6', '7')
Linux-5.0.20-050020-generic-x86_64-with-Ubuntu-18.04-bionic
Python ('default', 'Oct 22 2018 11:32:17') GCC 8.2.0
Executing in 64bit

Time Pi 3.141592653589793 Error of calculation   function
m * n = 4 * 25 = 100
 0.000  3.131592903558554 0.00999975003123943 madhava
 0.000  3.131592903558554 0.00999975003123943 leibniz
m * n = 4 * 250 = 1,000
 0.000  3.140592653839794 0.000999999749998981 madhava
 0.000  3.140592653839794 0.000999999749998981 leibniz
m * n = 4 * 2,500 = 10,000
 0.001  3.141492653590034 9.99999997586265e-05 madhava
 0.002  3.141492653590034 9.99999997586265e-05 leibniz
m * n = 4 * 25,000 = 100,000
 0.007  3.141582653589720 1.0000000073340232e-05 madhava
 0.016  3.141582653589720 1.0000000073340232e-05 leibniz
m * n = 4 * 250,000 = 1,000,000
 0.061  3.141591653589774 1.0000000187915248e-06 madhava
 0.152  3.141591653589774 1.0000000187915248e-06 leibniz
m * n = 4 * 2,500,000 = 10,000,000
 0.608  3.141592553589792 1.0000000161269895e-07 madhava
 1.530  3.141592553589792 1.0000000161269895e-07 leibniz
m * n = 4 * 25,000,000 = 100,000,000
 6.167  3.141592643589326 1.0000467121074053e-08 madhava
15.617  3.141592643589326 1.0000467121074053e-08 leibniz

Another version, using itertools.starmap and operator.truediv:

def madhava_leibniz_starmap(k, n):
    return starmap(truediv, zip(cycle([1, -1] if k & 1 else [1, -1]), range(2*k+1, 2*n+2, 2)))

On my computer this yields:

('3', '6', '7')
Linux-5.0.20-050020-generic-x86_64-with-Ubuntu-18.04-bionic
Python ('default', 'Oct 22 2018 11:32:17') GCC 8.2.0
Executing in 64bit

chunks * elements = m * n = 4 * 25,000,000 = 100,000,000

Time Pi 3.141592653589793 Error of calculation   function
15.569  3.141592643589326 1.0000467121074053e-08 leibniz
 8.968  3.141592643589326 1.0000467121074053e-08 madhava
 6.182  3.141592643589326 1.0000467121074053e-08 madhava_leibniz
 5.238  3.141592643589326 1.0000467121074053e-08 madhava_leibniz_starmap

Which is roughly one second faster than list comprehension.