我正在编写一个python类来找到8个皇后问题的解决方案。如何在solve
方法中正确实现回溯?我认为递归应该起作用,但是,在第一次尝试找不到解决方案后,程序将停止,并且不会发生回溯。所有辅助方法都可以正常工作。
EMPTY = 0
QUEEN = 1
RESTRICTED = 2
class Board:
# initializes a 8x8 array
def __init__ (self):
self.board = [[EMPTY for x in range(8)] for y in range(8)]
# pretty prints board
def printBoard(self):
for row in self.board:
print(row)
# places a queen on a board
def placeQueen(self, x, y):
# restricts row
self.board[y] = [RESTRICTED for i in range(8)]
# restricts column
for row in self.board:
row[x] = RESTRICTED
# places queen
self.board[y][x] = QUEEN
self.fillDiagonal(x, y, 0, 0, -1, -1) # restricts top left diagonal
self.fillDiagonal(x, y, 7, 0, 1, -1) # restructs top right diagonal
self.fillDiagonal(x, y, 0, 7, -1, 1) # restricts bottom left diagonal
self.fillDiagonal(x, y, 7, 7, 1, 1) # restricts bottom right diagonal
# restricts a diagonal in a specified direction
def fillDiagonal(self, x, y, xlim, ylim, xadd, yadd):
if x != xlim and y != ylim:
self.board[y + yadd][x + xadd] = RESTRICTED
self.fillDiagonal(x + xadd, y + yadd, xlim, ylim, xadd, yadd)
# recursively places queens such that no queen shares a row or
# column with another queen, or in other words, no queen sits on a
# restricted square. Should solve by backtracking until solution is found.
def solve(self, queens):
if queens == 8:
return True
for i in range(8):
if self.board[i][queens] == EMPTY:
self.placeQueen(queens, i)
if self.solve(queens - 1):
return True
self.board[i][queens] = RESTRICTED
return False
b1 = Board()
b1.solve(7)
b1.printBoard()
我的问题是在添加女王之后缺少董事会的深层副本吗?还是缺少回溯功能?
答案 0 :(得分:1)
两者都是:您在整个程序中只有一块木板。您尽其所能地填充它,直到所有正方形都被占据或限制为止。搜索失败,您从solve
返回。没有复位主板的机制,程序结束。
回溯将使此过程变得简单,但需要多个中间板。而不是拥有一个单独的棋盘对象...制作一个深副本,放置女王/王后,标记适当的RESTRICTED正方形,然后将更改后的副本传递到下一个级别。如果失败返回,则该副本作为局部变量自然消失。