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我正在尝试使用 BFS 搜索、DFS、贪婪和 A* 等技术解决 8tile 难题,并使用曼哈顿距离作为我的启发式解决方案。

问题是,虽然我可以解决很多问题,但问题在于一些谜题,我扩展父节点时得到的子节点可能已经在较旧的节点中。

我不知道我是否能够很好地解释自己,但我的主要问题是我试图查看我创建的新节点是否已经在旧节点上。

有了这个问题,我通常会进入深度 9,然后我的程序不会推进或给出解决方案。

我的一个想法是使用代码:

if node in prev:
    continue
prev.append(node)

但我想我走错路了。

我在 python 上做这个,这是我的代码,以防有人可以帮助我。

#!/usr/bin/python

import sys
import copy


class Board:
    def __init__(self, matrix, whitepos=None):
        self.matrix = matrix
        self.whitepos = whitepos
        if not whitepos:
            for y in xrange(3):
                for x in xrange(3):
                    if board[y][x] == 0:
                        self.whitepos = (x, y)


def is_final_state(board):
    final = [[1, 2, 3], [8, 0, 4], [7, 6, 5]]
    for y in xrange(3):
        for x in xrange(3):
            if board.matrix[y][x] != final[y][x]:
                return False
    return True


def get_whitepos(board):
    return board.whitepos


def move(board, x, y, dx, dy):
    b = copy.deepcopy(board.matrix)
    b[y][x] = b[y + dy][x + dx]
    b[y + dy][x + dx] = 0
    return Board(b, (x + dx, y + dy))


def manhattan_heur(board):
    finalpos = [(1, 1), (0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (1, 2), (0, 2),
                (0, 1)]
    cost = 0
    for y in xrange(3):
        for x in xrange(3):
            t = board.matrix[y][x]
            xf, yf = finalpos[t]
            cost += abs(xf - x) + abs(yf - y)
    return cost


def wrongplace_heur(board):
    finalpos = [(1, 1), (0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (1, 2), (0, 2),
                (0, 1)]
    cost = 0
    for y in xrange(3):
        for x in xrange(3):
            t = board.matrix[y][x]
            if finalpos[t] != (x, y):
                cost += 1
    return cost


def heuristic(board):
    return manhattan_heur(board)


class Node:
    def __init__(self, board, parent):
        self.state = board
        self.parent = parent
        if not parent:
            self.g = 0
        else:
            self.g = parent.g + 1
        self.h = heuristic(board)

    def test_goal(self):
        return is_final_state(self.state)

    def expand(self):
        children = []
        b = self.state
        x, y = get_whitepos(b)
        if x > 0:
            children.append(Node(move(b, x, y, -1, 0), self))
        if x < 2:
            children.append(Node(move(b, x, y, +1, 0), self))
        if y > 0:
            children.append(Node(move(b, x, y, 0, -1), self))
        if y < 2:
            children.append(Node(move(b, x, y, 0, +1), self))
        return children


class Solution:
    def __init__(self, node, mem_needed, steps):
        self.node = node
        self.mem_needed = mem_needed
        self.iterations = steps

    def inc(self, other):
        self.node = other.node
        self.mem_needed = max(self.mem_needed, other.mem_needed)
        self.iterations += other.iterations


def search(board, queue_fn, queue_arg=None):
    max_nodes = 1
    steps = 0
    nodes = [Node(Board(board), None)]
    prev = []
    depth = 0
    while nodes:
        node = nodes.pop(0)

        if node.g > depth:
            depth = node.g
            print depth

        if node in prev:
            continue
        prev.append(node)

        if node.test_goal():
            return Solution(node, max_nodes, steps)
        new_nodes = node.expand()
        nodes = queue_fn(nodes, new_nodes, queue_arg)

        max_nodes = max(max_nodes, len(nodes))
        steps += 1
    return Solution(None, max_nodes, steps)


def fifo_queue(nodes, new_nodes, _):
    nodes.extend(new_nodes)
    return nodes


def bl_search(board):
    return search(board, fifo_queue)


def lifo_queue(nodes, new_nodes, _):
    new_nodes.extend(nodes)
    return new_nodes


def dfs_search(board):
    return search(board, lifo_queue)


def bpl_queue(nodes, new_nodes, max_depth):
    def f(n):
        return n.g <= max_depth

    new_nodes = filter(f, new_nodes)
    new_nodes.extend(nodes)
    return new_nodes


def bpi_search(board):
    solution = Solution(None, 0, 0)
    for max_depth in xrange(0, sys.maxint):
        sol = search(board, bpl_queue, max_depth)
        solution.inc(sol)
        if solution.node:
            return solution


def sort_queue(nodes, new_nodes, cmp):
    nodes.extend(new_nodes)
    nodes.sort(cmp)
    return nodes


def guloso2_search(board):
    def cmp(n1, n2):
        return n1.h - n2.h

    return search(board, sort_queue, cmp)


def astar_search(board):
    def cmp(n1, n2):
        return (n1.g + n1.h) - (n2.g + n2.h)

    return search(board, sort_queue, cmp)


def print_solution(search, sol):
    print
    print "*", search
    node = sol.node
    if node:
        print "moves:", node.g
        while node:
            print "\t", node.state.matrix
            node = node.parent
    else:
        print "no solution found"
    print "nodes needed:", sol.mem_needed
    print "iterations:  ", sol.iterations


board = [[6, 5, 7], [2, 0, 1], [8, 4, 3]]

print_solution("bl", bl_search(board))
print_solution("dfs", dfs_search(board))
print_solution("bpi", bpi_search(board))
print_solution("guloso2", guloso2_search(board))
print_solution("astar", astar_search(board))
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1 回答 1

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看起来您正在以正确的方式进行操作,但是您需要在 Node 类中定义__eq__and__ne__方法;否则node in prev将始终返回False,因为 Python 不知道如何与node列表中的项目进行比较。查看Python 数据模型文档,了解更多关于如何在用户定义类型上进行比较的信息。

我抓住了你的代码并添加了几个(非常天真的)方法来进行相等性检查,它似乎不再挂起。还值得注意的是,您的类应该从基类继承object(见下文)。这些是我所做的更改(在上下文中):

class Board(object):
    def __init__(self, matrix, whitepos=None):
        self.matrix = matrix
        self.whitepos = whitepos
        if not whitepos:
            for y in xrange(3):
                for x in xrange(3):
                    if board[y][x] == 0:
                        self.whitepos = (x, y)
    def __eq__(self, o):
        # Note that comparing whitepos is not strictly necessary; but I left 
        # it in as a safety measure in case the board state gets corrupted.
        # If speed becomes an issue, take it out.
        return (self.matrix, self.whitepos) == (o.matrix, o.whitepos)

class Node(object):
    def __init__(self, board, parent):
        self.state = board
        self.parent = parent
        if not parent:
            self.g = 0
        else:
            self.g = parent.g + 1
        self.h = heuristic(board)

    def test_goal(self):
        return is_final_state(self.state)

    def expand(self):
        children = []
        b = self.state
        x, y = get_whitepos(b)
        if x > 0:
            children.append(Node(move(b, x, y, -1, 0), self))
        if x < 2:
            children.append(Node(move(b, x, y, +1, 0), self))
        if y > 0:
            children.append(Node(move(b, x, y, 0, -1), self))
        if y < 2:
            children.append(Node(move(b, x, y, 0, +1), self))
        return children

    def __eq__(self, o):
        # Note that you don't have to compare parents, since your goal
        # is to eliminate ANY nodes with the same position.
        return self.state == o.state

class Solution(object):
    def __init__(self, node, mem_needed, steps):
        self.node = node
        self.mem_needed = mem_needed
        self.iterations = steps

    def inc(self, other):
        self.node = other.node
        self.mem_needed = max(self.mem_needed, other.mem_needed)
        self.iterations += other.iterations

#...

print_solution("bl", bl_search(board))
# I commented out all but the first search to avoid cluttering up the output.

通过这些更改,代码确实产生了一个解决方案(我将由您来验证它是否正确,但这是我的输出)。

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* bl
moves: 20
    [[1, 2, 3], [8, 0, 4], [7, 6, 5]]
    [[1, 2, 3], [8, 6, 4], [7, 0, 5]]
    [[1, 2, 3], [8, 6, 4], [0, 7, 5]]
    [[1, 2, 3], [0, 6, 4], [8, 7, 5]]
    [[1, 2, 3], [6, 0, 4], [8, 7, 5]]
    [[1, 0, 3], [6, 2, 4], [8, 7, 5]]
    [[0, 1, 3], [6, 2, 4], [8, 7, 5]]
    [[6, 1, 3], [0, 2, 4], [8, 7, 5]]
    [[6, 1, 3], [2, 0, 4], [8, 7, 5]]
    [[6, 1, 3], [2, 7, 4], [8, 0, 5]]
    [[6, 1, 3], [2, 7, 4], [8, 5, 0]]
    [[6, 1, 3], [2, 7, 0], [8, 5, 4]]
    [[6, 1, 0], [2, 7, 3], [8, 5, 4]]
    [[6, 0, 1], [2, 7, 3], [8, 5, 4]]
    [[6, 7, 1], [2, 0, 3], [8, 5, 4]]
    [[6, 7, 1], [2, 5, 3], [8, 0, 4]]
    [[6, 7, 1], [2, 5, 3], [8, 4, 0]]
    [[6, 7, 1], [2, 5, 0], [8, 4, 3]]
    [[6, 7, 0], [2, 5, 1], [8, 4, 3]]
    [[6, 0, 7], [2, 5, 1], [8, 4, 3]]
    [[6, 5, 7], [2, 0, 1], [8, 4, 3]]
nodes needed: 44282
iterations:   59930

希望这可以帮助!

于 2013-03-06T17:08:13.117 回答