human
Complete this python program:
def GetServiceVersions(namespace):
assistant
#Here is the Python program you requested.

    """Get all the versions for the service with specified namespace (partially) ordered
    by compatibility (i.e. any version in the list that is compatible with some version
    v in the list will preceed v)
    """
    def compare(a, b):
        if (a == b):
            return 0
        if (b in parentMap[a]):
            return (-1)
        if (a in parentMap[b]):
            return 1
        return ((a > b) - (a < b))
    if PY3:
        return sorted([v for (v, n) in iteritems(serviceNsMap) if (n == namespace)], key=cmp_to_key(compare))
    else:
        return sorted([v for (v, n) in iteritems(serviceNsMap) if (n == namespace)], compare)