人工智能—状态空间法（state space）[四]

这个题目是课上老师留下来的问题，然后借用之前传教士和狼的套路，将这个问题用编程实现。

题目描述：农夫需要把狼、羊、菜和自己运到河对岸去，只有农夫能够划船，而且船比较小，除农夫之外每次只能运一种东西，还有一个棘手问题，就是如果没有农夫看着，羊会偷吃菜，狼会吃羊。请考虑一种方法，让农夫能够安全地安排这些东西和他自己过河。

最终得出两套方案：

#以左岸为例
a=1#农夫在左岸
b=1#狼在左岸
c=1#羊在左岸
d=1#白菜在左岸
class nongfu():
    def __init__(self):
        pass
    def operations(self,s):
        p1 =[s[0]-1,s[1]-1,s[2],s[3],0]
        p2 =[s[0]-1,s[1],s[2]-1,s[3],0]
        p3 =[s[0]-1,s[1],s[2],s[3]-1,0]
        p4 =[s[0]-1,s[1],s[2],s[3],0]
        p = [p1,p2,p3,p4]
        q1 =[s[0]+1,s[1]+1,s[2],s[3],1]
        q2 =[s[0]+1,s[1],s[2]+1,s[3],1]
        q3 =[s[0]+1,s[1],s[2],s[3]+1,1] 
        q4 =[s[0]+1,s[1],s[2],s[3],1]   
        q = [q1,q2,q3,q4]
        operators =[q,p]
        return operators 
    def enable(self,operator):

        if operator[0]<0 or operator[1]<0 or operator[2]<0 or operator[3]<0:
            return False
        if a-operator[0]<0 or b-operator[1]<0 or c-operator[2]<0 or d-operator[3]<0:
            return False
        if ((b-operator[1])==1 and (c-operator[2])==1) and operator[0]==1:
            return False
        if ((c-operator[2])==1 and (d -operator[3])==1) and operator[0]==1:
            return False
        if operator[1] ==1 and operator[2] ==1 and operator[0]==0:
            return False
        if operator[2] ==1 and operator[3]==1 and operator[0]==0:
            return False        
        return True  

    def trun(self,states,operatorindex,roads,e):
            indexs =[]
            sign =0
            a =0
            tem = []
            for s in states:             
                for operator in self.operations(s)[e]: 
                    if self.enable(operator) :
                        if operator not in roads:
                           indexs.append(self.operations(s)[e].index(operator))
                           tem.append(operator)
                           roads.append(operator)
                           a =1
                if a ==1:
                   pass
                else:
                   roads.append(['no way'])      
            operatorindex =indexs
            states =tem
            e = abs(e-1)
            return states,operatorindex,roads,e

e =1#初始状态左岸
operatorindex =[' ']
states =[[1,1,1,1,1]]
roads = [[1,1,1,1,1]]
nongfupro=nongfu()
while roads[-1] !=[0,0,0,0,0]:
      [states,operatorindex,roads,e]=nongfupro.trun(states,operatorindex,roads,e)
print roads