Связать последний элемент списка с равным первым элементом другого списка

Как насчет:

def path(a, b):
    if not b:
        return a
    for n, (p, q) in enumerate(b):
        if p == a[-1][1]:
            return path(a + [(p, q)], b[:n] + b[n+1:])
        if q == a[-1][1]:
            return path(a + [(q, p)], b[:n] + b[n+1:])
    raise ValueError("no path", a, b)

L = [(8, 2), (8, 5), (2, 12), (12, 13), (5, 6), (6, 7), (13, 14), (14, 3), (7, 4), (3, 4)]
print path([L[0]], L[1:])
#[(8, 2), (2, 12), (12, 13), (13, 14), (14, 3), (3, 4), (4, 7), (7, 6), (6, 5), (5, 8)]

Вот решение, которое находит все возможные цепочки:

def find_chains(points):
    """For a list of points, take the first point and finds all chains that
    could be made from the following points.

    Returns a list of lists of tuples. Each sublist is a possible chain.
    """

    def find_solutions(start, choices):
        """For a starting point and a list of choices, return all choices that
        might succeed that point.

        Returns a list of tuples. These tuples contain the next choice, and all
        of the remaining points should that choice be used.
        """

        ret = []
        for i, possible_choice in enumerate(choices):
            # Determine whether or not the choice is appropriate.
            added_choice = None
            if start[1] == possible_choice[0]:
                added_choice = possible_choice
            elif start[1] == possible_choice[-1]:
                added_choice = possible_choice[::-1]

            # If it is, add a tuple of that choice and all other choices to the
            # return list.
            if added_choice:
                ret.append((
                    added_choice,
                    [
                        k
                        for j, k
                        in enumerate(choices)
                        if i != j
                    ]
                ))

        return ret

    solutions = []
    start = points[0]
    other_points = points[1:]
    if not other_points:
        # If there aren't any remaining points, then the only possible path is
        # that of just the first point.
        return [[start]]

    # Find all chains through `other_points` and add them to our solutions.
    for next_point, remaining_points in find_solutions(start, other_points):
        for subchain in find_chains([next_point] + remaining_points):
            solutions.append([start] + subchain)
    return solutions

Разница между этим решением и другими более короткими решениями заключается в том, что другие просто берут первую возможную точку и находят оттуда подцепь, тогда как это решение берет все возможные точки. Пример:

test = [(1, 2), (2, 3), (2, 4), (3, 4)]
print(find_chains(test))
# Finds two possible solutions:
# [[(1, 2), (2, 3), (3, 4), (4, 2)],
#  [(1, 2), (2, 4), (4, 3), (3, 2)]]

test = [(8, 2), (8, 5), (2, 12), (12, 13), (5, 6), (6, 7), (13, 14),
        (14, 3), (7, 4), (3, 4)]
print(find_chains(test))
# Finds the only solution:
# [[(8, 2), (2, 12), (12, 13), (13, 14), (14, 3), (3, 4), (4, 7), (7, 6), (6, 5), (5, 8)]]

test = [(1, 2), (3, 4)]
print(find_chains(test))
# There weren't any solutions:
# []