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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _compare_versions(v1, v2):
 """ Compare two version strings and return -1, 0 or 1 depending on the equality of the subset of matching version numbers. The implementation is inspired by the top answer at http://stackoverflow.com/a/1714190/997768. """

    def normalize(v):
        # strip trailing .0 or .00 or .0.0 or ...
        v = re.sub(r'(\.0+)*$', '', v)
        result = []
        for part in v.split('.'):
            # just digits
            m = re.match(r'^(\d+)$', part)
            if m:
                result.append(int(m.group(1)))
                continue
            # digits letters
            m = re.match(r'^(\d+)([a-zA-Z]+)$', part)
            if m:
                result.append(int(m.group(1)))
                result.append(m.group(2))
                continue
            # digits letters digits
            m = re.match(r'^(\d+)([a-zA-Z]+)(\d+)$', part)
            if m:
                result.append(int(m.group(1)))
                result.append(m.group(2))
                result.append(int(m.group(3)))
                continue
        return tuple(result)

    n1 = normalize(v1)
    n2 = normalize(v2)

    return (n1 > n2) - (n1 < n2)