我正在Python中编写一个数独求解器,它需要部分填充棋盘并使用回溯和前向检查来填充其余部分并解决难题。正向检查是每次向空白单元格赋值时检查其行,列和框未赋值“邻居”在赋值后是否仍有非空域的情况。回数在数独求解失败
为了表示我的电路板(尺寸:N x N电路板,带有P x Q框),我使用了一个2D列表,其中每个条目的形式为[value,[domain]],其中value是分配的单元格的数量(如果未分配,则为0),域是可以保持数独拼图一致的单元格的可能值。
我相信我做错了我的递归,但无法弄清楚什么。该函数总是失败并返回False。下面是我的递归求解器函数的一部分,并带有预处理代码。如有必要,我可以发布整个函数及其辅助函数。
def fc_recursive_solve(board, N, P, Q, row, col, outputFile, timeout):
###some preprocessing here to check if puzzle is solved and find next empty cell if not
vals = board[row][col][1] #domain/possible values for the empty cell
for num in vals:
#if num doesn't violate the row, col, and box sudoku constraints
if constraintCheck(row, col, num, P, N, Q, board):
#make copy of cell's domain for backtracking
tempDomain = copy.deepcopy(board[row][col][1])
board[row][col][0] = num #assign num to the cell
#remove num from domains of neighbors,
#if an empty domain results after removing num,
#return False and the original board,
#else return True and the updated board
noEmptyDomains, board = propagate_fc(board, N, P, Q, row, col, num)
if noEmptyDomains:
board[row][col][1] = [num] #update domain to be num and then recurse
if fc_recursive_solve(board, N, P, Q, row, col, outputFile, timeout):
return True
#backtrack -- reset value and domain of assigned cell
board[row][col][1] = tempDomain
board[row][col][0] = 0
else:
board[row][col][0] = 0
return False
编辑:更多的代码,并尝试Blckknght的解决方案
def fc_recursive_solve(board, N, P, Q, row, col, outputFile, timeout):
if time.clock() >= timeout:
return "timeout"
while row < N and board[row][col][0] != 0: #find next blank
if col == N-1:
row = row + 1
col = 0
else:
col = col + 1
if row == N: #solved
return True
for num in vals:
if constraintCheck(row, col, num, P, N, Q, board):
board[row][col][0] = num
noEmptyDomains, new_board = propagate_fc(board, N, P, Q, row, col, num) # new var
if noEmptyDomains:
new_board[row][col][1] = [num] # this doesn't modify board, only new_board
if fc_recursive_solve(new_board, N, P, Q, row, col, outputFile, timeout):
return True
board[row][col][0] = 0 # the only thing that's required to backtrack
return False
def propagate_fc(board, N, P, Q, row, col, num):
origBoard = copy.deepcopy(board)
#row propagate
for x in range(N):
if board[x][col][0] == 0:
if num in board[x][col][1]:
board[x][col][1].remove(num)
if len(board[x][col][1]) == 0:
return False, origBoard #domain is empty; return original board
#col propagate
for y in range(N):
if board[row][y][0] == 0:
if num in board[row][y][1]:
board[row][y][1].remove(num)
if len(board[row][y][1]) == 0:
return False, origBoard #domain is empty
#box propagate
rDiv = row/P
cDiv = col/P
for i in range((rDiv * P), ((rDiv + 1) * P)):
for j in range((cDiv * Q), ((cDiv + 1) * Q)):
if board[i][j][0] == 0:
if num in board[i][j][1]:
board[i][j][1].remove(num)
if len(board[i][j][1]) == 0:
return False, origBoard #domain is empty
return True, board #success; return board with updated domains
它看起来并不像你回溯撤销对董事会propagate_fc'的'行动的任何地方。 – Blckknght