我尝试在Python中使用预定义为的DepthFirstSearch类实现DepthFirstSearch算法：class Stack:
def __init__(self):
self.list = []
def push(self,item):
self.list.append(item)
def pop(self):
return self.list.pop()
def isEmpty(self):
return len(self.list) == 0
我还有一个功能：
^{pr2}$
它返回我们是否处于预定义的目标状态，
功能：def getStartState(self):
return self.startState
返回元组(int,int)，代理的位置，
和功能：def getSuccessors(self, state):
self._expanded += 1
return successors
它以((int,int),string,int)的形式返回代理的所有可用的下一个“移动”的元组，其中(int,int)是继任者的状态，string是方向(NSEW)，int是继任者的成本。在
到目前为止，我对实现GraphSearch的GraphSearch的实现是这样的：def depthFirstSearch(problem):
closed = []
fringe = util.Stack()
fringe.push(problem)
i = 0
while not fringe.isEmpty():
currentState = fringe.pop()
print "current's successors:", currentState.getSuccessors(currentState.getStartState())
if currentState.isGoalState(currentState.getStartState()):
return currentState
if not (currentState in closed):
closed.append(currentState)
print "closed now includes:", closed[i].getStartState()
children = currentState.getSuccessors(currentState.getStartState())
print "children:", children
while not children.isEmpty():
fringe.push(children.pop())
print "fringe:" fringe
i += 1
当我意识到这个问题并没有完成时。我完成了第一次迭代，然后第二次就停止了。以下是终端输出：current's successors: [((5, 4), 'South', 1), ((4, 5), 'West', 1)]
closed now includes: (5, 5)
children: [((5, 4), 'South', 1), ((4, 5), 'West', 1)]
Traceback (most recent call last):
...
...
...
AttributeError: 'list' object has no attribute isEmpty
所以很明显，我没有正确地遍历这个元组的列表，把它们作为两个附加的元组转移到边缘。在
有什么想法我应该如何穿越这个具体的实现？在
