#Any Queens puzzle
def share_diagonal(x0, y0, x1, y1):
""" Is (x0, y0) on a shared diagonal with (x1, y1)? """
dy = abs(y1 - y0)
dx = abs(x1 - x0)
return dx == dy
def col_clashes(bs, c):
"""
Return True if the queen at column c clashes
with any queen to its left.
"""
for i in range(c):
if share_diagonal(i, bs[i], c, bs[c]):
return True
return False
def has_clashes(the_board):
"""
Determine whether we have any queens clashing on the diagonals.
We're assuming here that the_board is a permutation of column
numbers, so we're not explicitly checking row or column clashes.
If it has clashes, return True.
"""
for col in range(1, len(the_board)):
if col_clashes(the_board, col):
return True
return False
def interchange_list(j, k, list):
temp = list[j]
list[j] = list[k]
list[k] = temp
def generating_next_permutation_in_lexicographic_order(per_list):
n = len(per_list) - 1
j = n - 1
while per_list[j] > per_list[j + 1]:
j = j - 1
if j < 0:
return 0
k = n
while per_list[j] > per_list[k]:
k = k - 1
interchange_list(j, k, per_list)
r = n
s = j + 1
while r > s:
interchange_list(r, s, per_list)
r = r - 1
s = s + 1
return per_list
def main(num):
per_list = list(range(0, num))
tries = 0
num_found = 0
result = []
while per_list != 0:
tries += 1
if not has_clashes(per_list):
#print("Found solution {0} in {1} tries.".format(per_list, tries))
list1 = per_list
result.append(list1)
#print(result)
num_found += 1
per_list = generating_next_permutation_in_lexicographic_order(per_list)
print(num_found)
print(result)
main(8)
印出結果為
92
[[7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1 , 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3 , 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5 , 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7 , 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0 ], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2 , 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4 , 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6 , 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1 , 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3 , 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5 , 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7 , 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0 ], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2 , 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4 , 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6 , 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1 , 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3 , 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5 , 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7 , 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0 ], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2 , 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4 , 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6 , 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1 , 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3 , 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5 , 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7 , 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0], [7, 6, 5, 4, 3, 2, 1, 0 ], [7, 6, 5, 4, 3, 2, 1, 0]]
[Finished in 0.2s]
為啥result都變成一樣的了?
滿天的星座2017-05-18 11:02:11
問題出在generating_next_permutation_in_lexicographic_order這個函數。
Python裡List是可變型,所以你全域事實上只操作了一個List,然後不斷把同一個List的引用放入result裡面當然會是這樣。
一種簡單的修改:
generating_next_permutation_in_lexicographic_order(per_list):
import copy
per_list = copy.deepcopy(per_list)
#剩下是你原来的代码