python - 高效地生成长度为 N 的列表中长度为 K 的循环移位的所有排列

Question

How can I generate all permutations of cyclic shifts of length k in a list of length n. Here shift is cyclic and right. Notice that:

如果 K==1 ，则没有移位。因此，这些 0 类次没有排列。
如果 K==2 ，这相当于交换元素。因此所有n!可以生成排列。

例如。如果列表是 [1 4 2],K=2(因此从 0 到 N-K，循环)

P1: [1,4,2] #Original list. No shift.
P2: [4,1,2] #Shift from 0 of [1,4,2]
P3: [4,2,1] #Shift from 1 of [4,1,2] as 0 gives P1
P4: [2,4,1] #Shift from 0 of [4,2,1]  
P5: [2,1,4] #Shift from 1 of [1,4,2] as 0 of P4=P3
P6: [1,2,4] #Shift from 0 of [2,1,4]

如果 K==3，事情会变得有趣，因为一些排列被排除在外。

例如。 if list=[1,3,4,2],K=3(因此从索引 0 到 4-3，循环)

P1 : [1,3,4,2] #Original list. No shift. 
P2 : [4,1,3,2] #Shift from 0th of [1,3,4,2]  
P3 : [3,4,1,2] #Shift from 0th of [4,1,3,2]  
P4 : [3,2,4,1] #Shift from 1th of [3,4,1,2] as 0th gives P1
P5 : [4,3,2,1] #Shift from 0th of [3,2,4,1] 
P6 : [2,4,3,1] #Shift from 0th of [4,3,2,1] 
P7 : [2,1,4,3] #Shift from 1th of [2,4,3,1] as 0th gives P3
P8 : [4,2,1,3] #Shift from 0th of [2,1,4,3] 
P9 : [1,4,2,3] #Shift from 0th of [4,2,1,3] 
P10: [2,3,1,4] #Shift from 1th of [2,1,4,3] as 0 from P9=P7,1 from P9=P1,1 from P8=P5  
P11: [1,2,3,4] #Shift from 0th of [2,3,1,4] 
P12: [3,1,2,4] #Shift from 0th of [1,2,3,4] 

#Now,all have been generated, as moving further will lead to previously found values.

Notice,that these permutations are half (12) of what should've been (24). To, implement this, algorithm, I am currently using backtracking. Here's what I have tried so far (in Python)

def get_possible_cyclic(P,N,K,stored_perms): #P is the original list
    from collections import deque  

    if P in stored_perms:
        return    #Backtracking to the previous

    stored_perms.append(P)

    for start in xrange(N-K+1):
        """
        Shifts cannot wrap around. Eg. 1,2,3,4 ,K=3
        Recur for  (1,2,3),4 or 1,(2,3,4) where () denotes the cycle
        """
        l0=P[:start]                    #Get all elements that are before cycle ranges
        l1=deque(P[start:K+start])      #Get the elements we want in cycle
        l1.rotate()                     #Form their cycle
        l2=P[K+start:]                  #Get all elements after cycle ranges

        l=l0+list(l1)+l2                #Form the required list
        get_possible_cyclic(l,N,K,stored_perms)

    for index,i in enumerate(stored_perms):    
        print i,index+1

get_possible_cyclic([1,3,4,2],4,3,[])
get_possible_cyclic([1,4,2],3,2,[])

This produces the output

[1, 3, 4, 2] 1
[4, 1, 3, 2] 2
[3, 4, 1, 2] 3
[3, 2, 4, 1] 4
[4, 3, 2, 1] 5
[2, 4, 3, 1] 6
[2, 1, 4, 3] 7
[4, 2, 1, 3] 8
[1, 4, 2, 3] 9
[2, 3, 1, 4] 10
[1, 2, 3, 4] 11
[3, 1 ,2, 4] 12

[1, 4, 2] 1
[4, 1, 2] 2
[4, 2, 1] 3
[2, 4, 1] 4
[2, 1, 4] 5
[1, 2, 4] 6

This is exactly what I want, but a lot lot slower,since here the recursion depth exceeds for N>7. I hope, I have explained myself clearly. Anyone, with any optimizations?

Answer 1

支票

if P in stored_perms:

随着 stored_perms 的增长变得越来越慢，因为它需要一次将 P 与 stored_perms 的元素进行比较，直到找到副本或遇到列表末尾。由于每个排列都会被添加到 stored_perms 一次，与 P 的比较次数至少是找到的排列数的二次方，通常是所有可能的排列或一半，取决于 k 是偶数还是奇数(假设 1 < k < N)。

使用 set 效率更高. Python 的集合基于哈希表，因此成员检查通常是 O(1) 而不是 O(N)。但是，有一些限制:

1. 加入集合的元素需要是hashable , 并且 Python 列表不可散列。幸运的是，元组是可散列的，所以一个小的改变就解决了这个问题。

2. 迭代一个集合是不可预测的。特别是，您无法在迭代集合时可靠地修改它。

除了将 P 更改为元组并将 stored_perms 更改为集合之外，值得考虑基于工作队列而不是递归搜索的搜索。我不知道它是否会更快，但它避免了递归深度的任何问题。

将所有这些放在一起，我将以下内容放在一起:

def get_cyclics(p, k):
  found = set()      # set of tuples we have seen so far
  todo = [tuple(p)]  # list of tuples we still need to explore
  n = len(p)
  while todo:
    x = todo.pop()
    for i in range(n - k + 1):
      perm = ( x[:i]                    # Prefix
             + x[i+1:i+k] + x[i:i+1]    # Rotated middle
             + x[i+k:]                  # Suffix
             )
      if perm not in found:
        found.add(perm)
        todo.append(perm)
  for x in found:
    print(x)