mrz.py

import re
def permutations(s, before="", replace="O", replace_by="0"):
    """
    This function will replace every occurance of a character or string with another character or string.
    In every possible setup, e.g. 'xxx' as input, where 'y' replaces 'x' will be return ['yxx', 'xyx', 'xxy', 'yyx', 'yxy', ... ]
    Written by Jurien Vegter in order to fix an issue where the passporteye ocr mistakes '0's for 'O's
    """
    if len(s) == len(before):
        result = []
    elif replace in s[len(before):]:
        pattern = r'(^{before}[^{replace}]*)({replace})(.*$)'.format(before=before, replace=replace)
        replaced = re.sub(pattern, r'\1[]\3', s).replace("[]", replace_by)
        result = [s, replaced]
        before = re.sub(pattern, r'\1', s)
        result.extend(permutations(s, before + replace, replace, replace_by))
        result.extend(permutations(replaced, before + replace_by, replace, replace_by))
    else:
        result = [s]

    return result

Replace doubtful character with its alternative possible value and then re-compute check digit?