Python数独回溯
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【中文标题】Python数独回溯【英文标题】:Python sudoku backtracking 【发布时间】:2021-01-02 11:29:14 【问题描述】:我正在通过一个简单的示例(数独)尝试回溯算法。我首先尝试了另一种方法,取消了更多可能性,但在遇到同样的错误后,我切换到了更简单的解决方案。
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寻找第一个未解决的点
填写 1 到 9 之间的每个数字,如果有效则回溯新字段
当我运行它并输出无效字段时,我可以看到当算法退出递归调用时,该递归调用中的位置仍然是 9(因此算法找不到任何东西点)
例如前两行看起来像这样(它试图解决一个空字段):
[1, 2, 3, 4, 6, 9, 9, 9, 9]
[9, 9, 9, 9, 9, 9, 9, 0, 0]
我认为这是一个参考错误并插入
[e for e in field]
在回溯调用中,以便旧字段不会被更改,尽管这似乎没有帮助。
这是我的代码:
for i in range(9):
a = [field[i][j] for j in range(9) if field[i][j] != 0]
if len(a) != len(set(a)):
return False
for i in range(9):
a = [field[j][i] for j in range(9) if field[j][i] != 0]
if len(a) != len(set(a)):
return False
for x in range(3):
for y in range(3):
a = []
for addX in range(3):
for addY in range(3):
spot = field[x * 3 + addX][y * 3 + addY]
if spot != 0:
a.append(spot)
if len(a) != len(set(a)):
return False
return True
def findEmpty(field):
for i in range(9):
for j in range(9):
if field[i][j] == 0:
return i, j
def backtracking(field):
find = findEmpty(field)
if not find:
return True, field
else:
x, y = find
for i in range(1, 10):
print(f"Trying i at x y")
field[x][y] = i
if isValid(field):
s = backtracking([e for e in field])
if s[0]:
return s
else:
print("Not valid")
for row in field:
print(row)
return False, None
field = [[0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]]
solution = backtracking(field)
if solution[0]:
print("There was a solution. The field is:")
for row in solution[1]:
print(row)
else:
print("No solution was found")
【问题讨论】:
【参考方案1】:好的,所以根据我在日志中看到的情况,当代码到达 9 并且仍然没有得到答案时,它会回溯,但将值保持在 9。
所以发生的情况是,每次程序回溯时,它都会将值保留为 9,然后转到上一个值,这也可能会转到 9,这是无效的,因为我们回溯的值已经是 9。正如您在示例中看到的那样,这会导致程序直接回溯到开始并使大多数插槽为 9 的循环。
所以解决方案是在 backtrack() 中添加几行,如下所示。简而言之,额外的 2 行检查无效答案是否为 9,如果是,则将其重置为 0 并回溯到之前的值,直到得到有效答案。
def backtracking(field):
find = findEmpty(field)
if not find:
return True, field
else:
x, y = find
for i in range(1, 10):
print(f"Trying i at x y")
field[x][y] = i
if isValid(field):
s = backtracking(field)
if s[0]:
return s
else:
print("Not valid")
if field[x][y] == 9:
field[x][y] = 0
for row in field:
print(row)
return False, None
它给出的解决方案:
[2, 3, 4, 5, 1, 6, 7, 8, 9]
[1, 5, 6, 7, 8, 9, 2, 3, 4]
[7, 8, 9, 2, 3, 4, 1, 5, 6]
[3, 1, 2, 4, 5, 7, 6, 9, 8]
[4, 6, 5, 1, 9, 8, 3, 2, 7]
[8, 9, 7, 3, 6, 2, 4, 1, 5]
[5, 2, 8, 6, 4, 1, 9, 7, 3]
[6, 7, 3, 9, 2, 5, 8, 4, 1]
[9, 4, 1, 8, 7, 3, 5, 6, 2]
【讨论】:
没错。尽管一种更简单的方法(我也发现)是在原始代码的 for 循环之后添加field[x][y] = 0
。我的问题更多是为什么会发生这种情况,因为我的代码应该可以工作。
哦,也可以。当你进行回溯时,你几乎是在告诉程序只是让它飞起来,因为它看不到前方,所以这样的事情经常发生。【参考方案2】:
我做了一些研究,显然这确实是一个参考错误。对我来说,导入 python 复制库并分配每个新字段说
f = copy.deepcopy(field)
解决了这个问题(这也适用于复杂的示例)。
【讨论】:
【参考方案3】:class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
n=len(board)
# create state variables,keep track of rows, cols and boxes
rows=[ for _ in range(n)]
cols=[ for _ in range(n)]
boxes=[ for _ in range(n)]
# get the initial state of the grid
for r in range(n):
for c in range(n):
if board[r][c]!='.':
val=board[r][c]
box_id=self.get_box_id(r,c)
boxes[box_id][val]=True
rows[r][val]=True
cols[c][val]=True
# this backtracking just tells if shoul move to the next cell or not
self.backtrack(board,boxes,rows,cols,0,0)
def backtrack(self,board,boxes,rows,cols,r,c):
# base case. If I hit the last row or col, means all digits were correct so far
if r>=len(board) or c>=len(board[0]):
return True
# situation when cell is empty, fill it with correct value
if board[r][c]==".":
for num in range(1,10):
box_id=self.get_box_id(r,c)
box=boxes[box_id]
row=rows[r]
col=cols[c]
str_num=str(num)
# check rows, cols and boxes make sure str_num is not used before
if self.is_valid(box,col,row,str_num):
board[r][c]=str_num
boxes[box_id][str_num]=True
cols[c][str_num]=True
rows[r][str_num]=True
# if I am the last col and I placed the correct val, move to next row. So first col of the next row
if c==len(board)-1:
if self.backtrack(board,boxes,rows,cols,r+1,0):
return True
# if I am in col between 0-8, move to the col+1, in the same row
else:
if self.backtrack(board,boxes,rows,cols,r,c+1):
return True
# If I got a wrong value, then backtrack. So clear the state that you mutated
del box[str_num]
del row[str_num]
del col[str_num]
board[r][c]="."
# if cell is not empty just call the next backtracking
else:
if c==len(board)-1:
if self.backtrack(board,boxes,rows,cols,r+1,0):
return True
else:
if self.backtrack(board,boxes,rows,cols,r,c+1):
return True
return False
def is_valid(self,box,row,col,num):
if num in box or num in row or num in col:
return False
else:
return True
# a helper to get the id of the 3x3 sub grid, given row and column
def get_box_id(self,r,c):
row=(r//3)*3
col=c//3
return row+col
【讨论】:
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