Create maths/pi_generator.py

[JulianStiebler]

Describe your change:

• Add an algorithm?
• Fix a bug or typo in an existing algorithm?
• Documentation change?

Checklist:

• I have read CONTRIBUTING.md.
• This pull request is all my own work -- I have not plagiarized.
• I know that pull requests will not be merged if they fail the automated tests.
• This PR only changes one algorithm file. To ease review, please open separate PRs for separate algorithms.
• All new Python files are placed inside an existing directory.
• All filenames are in all lowercase characters with no spaces or dashes.
• All functions and variable names follow Python naming conventions.
• All function parameters and return values are annotated with Python type hints.
• All functions have doctests that pass the automated testing.
• All new algorithms include at least one URL that points to Wikipedia or another similar explanation.
• If this pull request resolves one or more open issues then the commit message contains Fixes: #{$ISSUE_NO}.

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[JulianStiebler]

Kay all tests seem fine. Do i need to add a unittest myself?

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[JulianStiebler]

Hi! Thanks for your responses.

• added commentary as recommended by @rohan472000
• added math.pi comparison & math.isclose()-test as recommended by @cclauss
• added some more commentary since our output will be a str() and to use in math-functions, we need to convert to float. (Theres also a type-hint on calculate_pi() tho but just for clearness i also wrote down in the docstring
• I didn't remove the test against 50 and 100 preknown values of pi since the constant from math.pi is only 16 digits
• I also didn't touched the main-function call

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[cclauss]

Let's land this one! The only other thing that I could think of doing would be to see how low the tolerances on math.isclose() could be set and still pass the tests but that is clearly not vital.

Thanks for your contribution!

[JulianStiebler]

>>> math.isclose(float(calculate_pi(15)), math.pi, abs_tol = 0)
True

Trying:
    math.isclose(float(calculate_pi(15)), math.pi, abs_tol = 0)
Expecting:
    True
ok

>>> math.isclose(float(calculate_pi(50)), math.pi, abs_tol = 0.1)
True

Trying:
    math.isclose(float(calculate_pi(50)), math.pi, abs_tol = 0.1)
Expecting:
    True
ok

>>> math.isclose(float(calculate_pi(50)), math.pi, abs_tol = 0)
True

Trying:
    math.isclose(float(calculate_pi(50)), math.pi, abs_tol = 0)
Expecting:
    True
ok

Is that what you meant?
Thanks for your review!

Can i still commit it with those changed tests or how does it go from here when its merged?

[cclauss]

My sense is that when abs_tol = 0, the function goes with its default of 9 decimal places.

Perhaps if abs_tol was set to 0.00000000000000000000000001 then tests would start to fail.

[JulianStiebler]

>>>math.isclose(float(calculate_pi(100)), math.pi, abs_tol = 0.0000000000000001)
True

Trying:
    math.isclose(float(calculate_pi(50)), math.pi, abs_tol = 0.0000000000000001)
Expecting:
    True
ok

Since the constant only delivers 15 digits,
but for completeness:

    >>> math.isclose(float(calculate_pi(50)), float(3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679), abs_tol = 0.00000000000000000000000000000000000001)
    True


Trying:
    math.isclose(float(calculate_pi(50)), float(3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679), abs_tol = 0.00000000000000000000000000000000000001)
Expecting:
    True
ok

Since we didnt interrupted the series but rather finished it with a limit it should always match pi to the exact digit we chose
I also tested it with abs_tol = 0.000000000000000000000000000000000000000000000001 (50 digits excluding the .) on calculate_pi(50) and it passed too. The arbitrary taken Pi-number is not from the algorithm! I searched google for 100digits of pi and compared 2-3 different ones (were all the same but to make sure :D)

[cclauss]

Nice work! Thanks.