Unreadable regex chain creative solution for Digit Stack by ale1ster - python coding challenges

Enable Javascript in your browser and then refresh this page, for a much enhanced experience.

Unreadable regex chain solution in Creative category for Digit Stack by ale1ster

def digit_stack(cs, s=__import__('re').sub):
    fix = lambda f,c,n: c if n == c else fix(f, f(c), c)
    return eval('0' + s(r'(\d)', r'+\1', s(r'(POP|PEEK|PUSH\d)', r'',
        fix (lambda x: s(r'PUSH(\d)(\d*)PEEK', r'PUSH\1\2\1',
        s(r'PUSH(\d)(\d*)POP', r'\1\2', x)), s(' ', '', ''.join(cs)), ''))))

July 3, 2014

Comments:

nickie on Sept. 20, 2014, 8:37 a.m.
Holy cow...

ale1ster on Sept. 20, 2014, 12:04 p.m.
The title says it all :) It is, however, broken down in lines for readability ( :P )

Legat on Feb. 27, 2016, 12:03 a.m.
This is just unbelievable!! :)

gssanyam12 on March 2, 2018, 12:38 a.m.
can anyone please explain this code?

ale1ster on Nov. 9, 2019, 5:04 p.m.
Here you go :) <pre class='brush: python'>def digit_stack(cs, s=__import__('re').sub): # Define fixpoint fix = lambda f,c,n: c if n == c else fix(f, f(c), c) # Join all cmds and remove whitespace #(["PUSH d"] -> "PUSHd"), (["PUSH d", "POP"] -> "PUSHdPOP") clean_str = s(' ', '', ''.join(cs)) # Continue operating until nothing can be done in the following manner: # 1. If there is a Push and a Peek follows, it is equivalent to adding the digit # to the sum, while leaving it also in the stack, so the Peek is consumed and the Push remains. # 2. If there is a Push and then a Pop, it is equivalent to pushing the digit to the stack and # then popping it and adding it to the sum, so both are consumed. # X. These are the only operations (and order of operations) that affect the sum total. # X. So the rest of the operations (Push, Pops, Peeks) leave the sum unchanged, so we remove them later. # =. In essence, this is a form of term rewritting. fix_sub_fun = lambda x: s(r'PUSH(\d)(\d*)PEEK', r'PUSH\1\2\1', s(r'PUSH(\d)(\d*)POP', r'\1\2', x)) # Nothing more can be done now fix_result = fix (fix_sub_fun, clean_str, '') # Remove remaining commands, and we have a clean string of digits that # have been popped or peeked from the stack. # They are not overlapping, so there is no need for fixpoint in this case. sub_remaining_cmds = s(r'(POP|PEEK|PUSH\d)', r'', fix_result) # Add '+' in front of each digit and '0' in the beginning, and we have a valid arithmetic expression add_ops = '0' + s(r'(\d)', r'+\1', sub_remaining_cmds) # Evaluate the resulting valid arithmetic expression return eval(add_ops) </pre>

Ice Base

0 %

PyCharm The Python IDE for Professional Developers. Enjoy productive Python, web and scientific development with PyCharm. Download it now!