human
Complete the below
def grow_diag_final_and(srclen, trglen, e2f, f2e):

    ''''This module symmetrisatizes the source-to-target and target-to-source
    word alignment output and produces, aka. GDFA algorithm (Koehn, 2005).
    Step 1: Find the intersection of the bidirectional alignment.
    Step 2: Search for additional neighbor alignment points to be added, given
    these criteria: (i) neighbor alignments points are not in the
    intersection and (ii) neighbor alignments are in the union.
    Step 3: Add all other alignment points thats not in the intersection, not in
    the neighboring alignments that met the criteria but in the original
    foward/backward alignment outputs.
    >>> forw = (\'0-0 2-1 9-2 21-3 10-4 7-5 11-6 9-7 12-8 1-9 3-10 \'
    ...         \'4-11 17-12 17-13 25-14 13-15 24-16 11-17 28-18\')
    >>> back = (\'0-0 1-9 2-9 3-10 4-11 5-12 6-6 7-5 8-6 9-7 10-4 \'
    ...         \'11-6 12-8 13-12 15-12 17-13 18-13 19-12 20-13 \'
    ...         \'21-3 22-12 23-14 24-17 25-15 26-17 27-18 28-18\')
    >>> srctext = ("ãã® ãã ãª ãã­ãŒ çœè² ãã æ ã® ïŒ¬ é¢æ° "
    >>> trgtext = ("Therefore , we expect that the luminosity function "
    ...            "of such halo white dwarfs increases discontinuously "
    ...            "with the luminosity .")
    >>> srclen = len(srctext.split())
    >>> trglen = len(trgtext.split())
    >>> gdfa = grow_diag_final_and(srclen, trglen, forw, back)
    >>> gdfa == set([(28, 18), (6, 6), (24, 17), (2, 1), (15, 12), (13, 12),
    ...         (2, 9), (3, 10), (26, 17), (25, 15), (8, 6), (9, 7), (20,
    ...         13), (18, 13), (0, 0), (10, 4), (13, 15), (23, 14), (7, 5),
    ...         (25, 14), (1, 9), (17, 13), (4, 11), (11, 17), (9, 2), (22,
    ...         12), (27, 18), (24, 16), (21, 3), (19, 12), (17, 12), (5,
    ...         12), (11, 6), (12, 8)])
    True
    References:
    Koehn, P., A. Axelrod, A. Birch, C. Callison, M. Osborne, and D. Talbot.
    2005. Edinburgh System Description for the 2005 IWSLT Speech
    Translation Evaluation. In MT Eval Workshop.
    :type srclen: int
    :param srclen:...'''

=====
Give me a Python code to find the intersection of the bidirectional alignment?
assistant
#Here is Python code to find the intersection of the bidirectional alignment.
    e2f = [tuple(map(int, a.split('-'))) for a in e2f.split()]
    f2e = [tuple(map(int, a.split('-'))) for a in f2e.split()]
    neighbors = [((-1), 0), (0, (-1)), (1, 0), (0, 1), ((-1), (-1)), ((-1), 1), (1, (-1)), (1, 1)]
    alignment = set(e2f).intersection(set(f2e))
    union = set(e2f).union(set(f2e))
    aligned = defaultdict(set)
    for (i, j) in alignment:
        aligned['e'].add(i)
        aligned['j'].add(j)
    def grow_diag():
        '\n                                Search    for    the    neighbor    points    and    them    to    the    intersected    alignment\n                                points    if    criteria    are    met.\n                                '
        prev_len = (len(alignment) - 1)
        while (prev_len < len(alignment)):
            for e in range(srclen):
                for f in range(trglen):
                    if ((e, f) in alignment):
                        for neighbor in neighbors:
                            neighbor = tuple(((i + j) for (i, j) in zip((e, f), neighbor)))
                            (e_new, f_new) = neighbor
                            if (((e_new not in aligned) and (f_new not in aligned)) and (neighbor in union)):
                                alignment.add(neighbor)
                                aligned['e'].add(e_new)
                                aligned['f'].add(f_new)
                                prev_len += 1
    def final_and(a):
        '\n                                Adds    remaining    points    that    are    not    in    the    intersection,    not    in    the    \n                                neighboring    alignments    but    in    the    original    *e2f*    and    *f2e*    alignments\n                                '
        for e_new in range(srclen):
            for f_new in range(trglen):
                if ((e_new not in aligned) and (f_new not in aligned) and ((e_new, f_new) in a)):
                    alignment.add((e_new, f_new))
                    aligned['e'].add(e_new)
                   ...