2

我有这个水壶问题要解决，我必须通过深度优先算法来解决。我有两个水壶，一个 4 加仑和一个 3 加仑，它们都没有标记。你如何才能将 2 加仑的水准确地倒入 4 加仑的水壶中？最初两个水壶都是空的。

规则是：

capacity(C,JC), for C equal j1 or j2; 
jugs(C1,C2), where C1 and C2 give the current contents of the jugs

初始起始事实：jugs(0,0) 目标：jugs(2,0) 或 jugs(0,2)

initial_state(jugs,jugs(0,0)). 

final_state(jugs(2,C2)). 
final_state(jugs(C1,2)).  

transition(jugs(C1,C2),fill(j1)). 
transition(jugs(C1,C2),fill(j2)). 
transition(jugs(C1,C2),empty(j1)) :-  
    C1 > 0. 
transition(jugs(C1,C2),empty(j2)) :-  
    C2 > 0. 
transition(jugs(C1,C2),transfer(j2,j1)). 
transition(jugs(C1,C2),transfer(j1,j2)).  


update(jugs(C1,C2),empty(j1),jugs(0,C2)). 
update(jugs(C1,C2),empty(j2),jugs(C1,0)). 
update(jugs(C1,C2),fill(j1),jugs(Capacity,C2)) :-  
    capacity(j1,Capacity). 
update(jugs(C1,C2),fill(j2),jugs(C1,Capacity)) :-  
    capacity(j2,Capacity). 
update(jugs(C1,C2),transfer(j2,j1),jugs(W1,W2)) :-  
    capacity(j1,Capacity), 
    Water is C1 + C2, 
    Overhang is Water - Capacity , 
    adapt(Water,Overhang,W1,W2). 
update(jugs(C1,C2),transfer(j1,j2),jugs(W1,W2)) :-  
    capacity(j2,Capacity), 
    Water is C1 + C2, 
    Overhang is Water - Capacity , 
    adapt(Water,Overhang,W2,W1).  

adapt(Water,Overhang,Water,0) :- Overhang =< 0. 
adapt(Water,Overhang,C,Overhang) :- 
    Overhang > 0, 
    C is Water - Overhang. 




legal(jugs(C1,C2)).  

capacity(j1,4). 
capacity(j2,3).

先感谢您！

4

0 回答 0