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python廣度搜索解決八數(shù)碼難題

更新時間：2021年04月07日 14:18:20 作者：胡亂huluan

這篇文章主要介紹了python廣度搜索解決八數(shù)碼難題。想了解算法和數(shù)據(jù)結(jié)構(gòu)的同學(xué)，一定要看一下

—— 八數(shù)碼難題 ——

1.題目描述

八數(shù)碼問題也稱為九宮問題。在3×3的棋盤，擺有八個棋子，每個棋子上標(biāo)有1至8的某一數(shù)字，不同棋子上標(biāo)的數(shù)字不相同。棋盤上還有一個空格，與空格相鄰的棋子可以移到空格中。要求解決的問題是：給出一個初始狀態(tài)和一個目標(biāo)狀態(tài)，找出一種從初始狀態(tài)轉(zhuǎn)變成目標(biāo)狀態(tài)的移動棋子步數(shù)最少的移動步驟。

代碼

使用算法：廣度搜索算法

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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