考虑到用户输入n，它代表玩家的数量，我计算所有可能的玩家联盟（元组），然后计算所有可能的联盟组合，以便在每组中每个玩家属于一个并且只有一个联盟。
示例：n = 3
#coalitions
[(1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
#combinations of coalitions that respect the condition
[((1, 2, 3),), ((1,), (2, 3)), ((2,), (1, 3)), ((3,), (1, 2)), ((1,), (2,), (3,))]
为了做到这一点，我写道：
n = int(raw_input())
players = xrange(1,n+1)
players_comb = [coal for length in players for coal in list(combinations(players,length))]
coal_comb = [coalset for l in players for coalset in list(combinations(players_comb,l)) if sum(map(len,coalset)) == n and set(player for coal in coalset for player in coal) == set(players)]
问题是coal_comb非常低效。对于istance，它将计算长度n的所有组合，即使只有一个（即((1 ... n),)）尊重条件。
我认为函数是O（2 ^（2 ^ n））。 我的电脑甚至无法计算n = 6。
关于如何提高效率的任何想法？
更新 我提出了这个解决方案：
combs = [comb for length in xrange(2,n) for comb in combinations_with_replacement(players,length) if sum(comb) == n]
H = [tuple(players_comb[:n]),(players_comb[-1],)]
for comb in combs:
layouts = set(frozenset(layout) for layout in product(*[[coal for coal in players_comb if len(coal) == length] for length in comb]) if set(player for coal in layout for player in coal) == set(players))
H.extend(tuple(tuple(layout) for layout in layouts))
但这只是一个微小的改进：现在我可以计算n = 8（运行时间超过50秒）。