生成所有可能的排列函数

时间:2016-11-30 03:24:06

标签: python list set combinatorics itertools

我感兴趣的是在集合S = {1,2,...,n}中找到排列p:S-> S.特别是,所有能够置换i和j的函数:p(i)= j和p(j)= i;或保持不变p(i)= i或p(j)= j。

例如,如果S = {1,2,3},我应该得到类似的东西:

p0 = [(1), (2), (3)] # p(1)=1, p(2)=2, p(3)=3
p1 = [(1,2), (3)] # p(1)=2, p(2)=1, p(3)=3
p2 = [(1,3), (2)]
p3 = [(2,3), (1)]

如果S = {1,2,3,4}:

p0 = [(1), (2), (3), (4)]
p1 = [(1,2), (3,4)]
p2 = [(1,2), (3), (4)]  # p(1)=2, p(2)=1, p(3)=3, p(4)=4
p3 = [(1,3), (2,4)]
p4 = [(1,3), (2), (4)]
p5 = [(1,4), (2,3)]
p6 = [(1,4), (2), (3)]
p7 = [(1), (3), (2,4)] 
p8 = [(1), (4), (2,3)]
p9 = [(1), (2), (3,4)]

感谢。

2 个答案:

答案 0 :(得分:0)

不确定如何以建设性的方式执行此操作,但构建所有排列并过滤掉那些不符合标准的排列非常简单。没有评论这个效率:

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答案 1 :(得分:0)

假设目标是找到仅涉及二进制交换的排列。

from itertools import combinations

def opairs(li):
    if not li:
        yield []
        return
    li_cpy = li.copy()
    for h in range(1,len(li)):
        li_cpy = li[1:]
        del(li_cpy[h-1])
        for subli in opairs(li_cpy):
            yield [(li[0], li[h])] + subli

def swaps(n):
    assert n%2==0
    yield list(map(lambda _: (_,), range(n)))
    for subsize in range(1, n//2+1):
        for head in combinations(range(n), subsize*2):
            tail = []
            ihead = iter(head)
            ihead_next = next(ihead)
            for i in range(n):
                if i==ihead_next:
                    try:
                        ihead_next = next(ihead)
                    except: continue
                else:
                    tail.append((i,))
            for phead in opairs(list(head)):
                yield phead+tail

for p in swaps(4): print(p)

输出:

[(0,), (1,), (2,), (3,)]
[(0, 1), (2,), (3,)]
[(0, 2), (1,), (3,)]
[(0, 3), (1,), (2,)]
[(1, 2), (0,), (3,)]
[(1, 3), (0,), (2,)]
[(2, 3), (0,), (1,)]
[(0, 1), (2, 3)]
[(0, 2), (1, 3)]
[(0, 3), (1, 2)]