pin.py

pin.py

import re

def does_pin_match(pin):
    
    # FROM https://docs.python.org/3/library/re.html
    # Matches the end of the string or just before the newline at the end of the
    # string, and in MULTILINE mode also matches before a newline. foo matches
    # both ‘foo’ and ‘foobar’, while the regular expression foo$ matches only
    # ‘foo’. More interestingly, searching for foo.$ in 'foo1\nfoo2\n' matches
    # ‘foo2’ normally, but ‘foo1’ in MULTILINE mode; searching for a single $ in
    # 'foo\n' will find two (empty) matches: one just before the newline, and one
    # at the end of the string.
    #
    # So, we can't use `^` and `$`, because `$` does not catch trailing newline.
    # "123456\n" will match this:
    # pattern = re.compile(r'^[0-9]{4}$|^[0-9]{6}$')
    #
    # Instead, we use `\A` and `\Z`.
    pattern = re.compile(r'\A[0-9]{4}\Z|\A[0-9]{6}\Z')
    return bool(pattern.match(pin))

print("Three digits:")
print(does_pin_match('123'))

print("Four digits:")
print(does_pin_match('1234'))

print("Five digits:")
print(does_pin_match('12345'))

print("Six digits:")
print(does_pin_match('123456'))

print("Six digits with newline:")
print(does_pin_match("123456\n"))

print("Six digits with newline, then digit:")
print(does_pin_match("123456\n1"))

Running:

$ python3 foo.py
Three digits:
False
Four digits:
True
Five digits:
False
Six digits:
True
Six digits with newline:
False
Six digits with newline, then digit:
False