python广度搜索解决八数码难题
作者：胡乱huluan
这篇文章主要介绍了python广度搜索解决八数码难题。想了解算法和数据结构的同学，一定要看一下
—— 八数码难题 ——
1.题目描述
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。
代码
使用算法：广度搜索算法
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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