human
Solve the following python programming problem given the following code:
@not_implemented_for('undirected')
@not_implemented_for('multigraph')
def is_reachable(G, s, t):
assistant
#Here is the Python program you requested.

    """Decides whether there is a path from `s` to `t` in the
    tournament.
    This function is more theoretically efficient than the reachability
    checks than the shortest path algorithms in
    :mod:`networkx.algorithms.shortest_paths`.
    The given graph **must** be a tournament, otherwise this function\s
    behavior is undefined.
    Parameters
    G : NetworkX graph
    A directed graph representing a tournament.
    s : node
    A node in the graph.
    t : node
    A node in the graph.
    Returns
    bool
    Whether there is a path from `s` to `t` in `G`.
    Notes
    Although this function is more theoretically efficient than the
    generic shortest path functions, a speedup requires the use of
    parallelism. Though it may in the future, the current implementation
    does not use parallelism, thus you may not see much of a speedup.
    This algorithm comes from [1].
    References
    .. [1] Tantau, Till.
    "A note on the complexity of the reachability problem for
    tournaments."
    *Electronic Colloquium on Computational Complexity*. 2001.
    <http://eccc.hpi-web.de/report/2001/092/>
    """
    def two_neighborhood(G, v):
        'Returns    the    set    of    nodes    at    distance    at    most    two    from    `v`.\n\n                                `G`    must    be    a    graph    and    `v`    a    node    in    that    graph.\n\n                                The    returned    set    includes    the    nodes    at    distance    zero    (that    is,\n                                the    node    `v`    itself),    the    nodes    at    distance    one    (that    is,    the\n                                out-neighbors    of    `v`),    and    the    nodes    at    distance    two.\n\n                                '
        return {x for x in G if ((x == v) or (x in G[v]) or any((is_path(G, [v, z, x]) for z in G)))}
    def is_closed(G, nodes):
        'Decides    whether    the    given    set    of    nodes    is    closed.\n\n                                A    set    *S*    of    nodes    is    *closed*    if    for    each    node    *u*    in    the    graph\n                                not    in    *S*    and    for    each    node    *v*    in    *S*,    there    is    an    edge    from\n                                *u*    to    *v*.\n\n                                '
        return all(((v in G[u]) for u in (set(G) - nodes) for v in nodes))
    neighborhoods = [two_neighborhood(G, v) for v in G]
    return all(((not (is_closed(G, S) and (s in S) and (t not in S))) for S in neighborhoods))