Python广度优先搜索实现八数码问题求解

温以宁
温以宁
发布时间:2026-10-14 | 2人浏览过
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本文介绍如何使用Python语言结合广度优先搜索(BFS)算法解决经典的八数码(九宫格)难题。通过定义状态类、生成子状态及路径回溯,实现从初始布局到目标布局的最少步数寻优,提供完整的代码逻辑与运行示例。

1.题目描述

八数码问题,亦称九宫格问题,是在一个3×3的棋盘上进行的。棋盘内放置了八个标有1至8数字的棋子,且每个棋子上的数字互不相同。棋盘上还留有一个空格,允许将与空格相邻的棋子移动至该空格位置。问题的核心目标是:给定一个初始状态和一个目标状态,寻找一种移动方案,使得从初始状态转换到目标状态所需的移动步数最少。

代码

使用算法:广度优先搜索算法

python


import numpy as np

class 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, None

if __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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