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，每個棋子上標有1至8的某一數(shù)字，不同棋子上標的數(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)和一個目標狀態(tài)，找出一種從初始狀態(tài)轉變成目標狀態(tài)的移動棋子步數(shù)最少的移動步驟。
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)
