[問題] node, actions, visited=fringe.pop()
不好意思怕有人誤會
得先說明這是作業，但沒要求解釋以下的問題
最近在寫berkelyey pacman
有3個步驟有點不太明白
想請問
(1)
node, actions, visited = fringe.pop()
一次把右邊的pop assign給左邊的多個variables是什麼意思呢？
(2)
fringe.push((coord, actions+[direction], visited+[node]))
push定義如下，我的理解是依序把coord, actions+[direction], visited+[node]
加入到list中，不知是否正確呢？
(3)
for coord, direction, steps in problem.getSuccessors(node):
我的程度只到 for i in list:
有點不太明白for 後面接多個variables跑的意思是什麼
#Code:
def depthFirstSearch(problem)
fringe = util.Stack()
fringe.push( (problem.getStartState(), [], []) )
while not fringe.isEmpty():
node, actions, visited = fringe.pop()
for coord, direction, steps in problem.getSuccessors(node):
if not coord in visited:
if problem.isGoalState(coord):
return actions + [direction]
fringe.push((coord, actions + [direction], visited + [node] ))
return []
＃util
class Stack:
"A container with a last-in-first-out (LIFO) queuing policy."
def __init__(self):
self.list = []
def push(self,item):
"Push 'item' onto the stack"
self.list.append(item)
def pop(self):
"Pop the most recently pushed item from the stack"
return self.list.pop()
def isEmpty(self):
"Returns true if the stack is empty"
return len(self.list) == 0
--
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※ 編輯: veriaw (140.112.49.174), 03/23/2016 12:22:29
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