我正在尝试实现一个实现深度优先搜索和广度优先搜索的序言程序,解决了以下问题
Rowena has three unmarked glasses of different sizes: 3 ounces, 5 ounces, and 8 ounces. The largest glass is full. What can Rowena do to get 4 ounces of liquid into each of the larger two glasses?
我会有(大、中、小)
所以初始状态是 (8,0,0),目标状态是 (4,4,0)。
现在我知道我在状态空间中有 6 个可用的移动。
(1,2) 将大号倒入中号或小号
(3,4) 倒入中号或大号或小号
(5,6) 将小倒入中或大中
现在我只需要第一条规则的帮助,其余的我会弄清楚。所以我只能在大> 0和中未满的情况下将大倒入中,新大变成旧大减去倒入中的量,新中变成旧中加上量那倒进去了,小当然永远不会改变。
这是我尝试的。
%move rule #1: Pour Large into Medium (L--> M)
%move(oldstate,newstate)
move([L, M, S], [NewLarge,NewMedium,S]) :-
L > 0, %We can't move large into medium if Large has nothing
M < 5, %We can't pour into the medium if medium is full
Diff = 5 - M,
NewLarge is L - Diff, %calculate the new Large
NewMedium is M + (L - NewLarge). %calculate the new Medium
这是第一个可用移动(大到中)的正确实现吗?我在那里得到了正确的逻辑吗?
最佳答案
我认为逻辑应该是
move([L, M, S], [NewLarge, NewMedium, S]) :-
Diff is min(5 - M, L),
Diff > 0,
NewLarge is L - Diff, %calculate the new Large
NewMedium is M + Diff. %calculate the new Medium
关于prolog - 在状态之间移动(Prolog 实现),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33598922/