python广度搜索解决八数码难题

八数码问题也称为九宫问题。在3×3的棋盘，摆有八个棋子，每个棋子上标有1至8的某一数字，不同棋子上标的数字不相同。棋盘上还有一个空格，与空格相邻的棋子可以移到空格中。要求解决的问题是：给出一个初始状态和一个目标状态，找出一种从初始状态转变成目标状态的移动棋子步数最少的移动步骤。

使用算法：广度搜索算法

python

import numpy as npclass State: def __init__(self, state, directionFlag=None, parent=None): self.state = state self.direction = [’up’, ’down’, ’right’, ’left’] if directionFlag: self.direction.remove(directionFlag) self.parent = parent self.symbol = ’ ’ def getDirection(self): return self.direction def showInfo(self): for i in range(3): for j in range(3): print(self.state[i, j], end=’ ’) print('n') print(’->n’) return def getEmptyPos(self): postion = np.where(self.state == self.symbol) return postion def generateSubStates(self): if not self.direction: return [] subStates = [] boarder = len(self.state) - 1 row, col = self.getEmptyPos() if ’left’ in self.direction and col > 0: s = self.state.copy() temp = s.copy() s[row, col] = s[row, col-1] s[row, col-1] = temp[row, col] news = State(s, directionFlag=’right’, parent=self) subStates.append(news) if ’up’ in self.direction and row > 0: s = self.state.copy() temp = s.copy() s[row, col] = s[row-1, col] s[row-1, col] = temp[row, col] news = State(s, directionFlag=’down’, parent=self) subStates.append(news) if ’down’ in self.direction and row < boarder: s = self.state.copy() temp = s.copy() s[row, col] = s[row+1, col] s[row+1, col] = temp[row, col] news = State(s, directionFlag=’up’, parent=self) subStates.append(news) if self.direction.count(’right’) and col < boarder: s = self.state.copy() temp = s.copy() s[row, col] = s[row, col+1] s[row, col+1] = temp[row, col] news = State(s, directionFlag=’left’, parent=self) subStates.append(news) return subStates def solve(self): openTable = [] closeTable = [] openTable.append(self) steps = 1 while len(openTable) > 0: n = openTable.pop(0) closeTable.append(n) subStates = n.generateSubStates() path = [] for s in subStates: if (s.state == s.answer).all(): while s.parent and s.parent != originState: path.append(s.parent) s = s.parent path.reverse() return path, steps+1 openTable.extend(subStates) steps += 1 else: return None, Noneif __name__ == ’__main__’: symbolOfEmpty = ’ ’ State.symbol = symbolOfEmpty originState = State(np.array([[2, 8, 3], [1, 6 , 4], [7, symbolOfEmpty, 5]])) State.answer = np.array([[1, 2, 3], [8, State.symbol, 4], [7, 6, 5]]) s1 = State(state=originState.state) path, steps = s1.solve() if path: for node in path: node.showInfo() print(State.answer) print('Total steps is %d' % steps)

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