在状态之间移动（Prolog 实现）

Question

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 prolog 程序，它解决了以下问题

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?

我要（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 space 状态下有 6 个可用的移动。

(1,2) 将大号倒入中号或小号 (3,4) 将 Medium 倒入 large or small (5,6) 将小号倒入中号或大号

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 large > 0 且 medium 未满时将 large 倒入 medium，新的 large 变成旧的 large 减去倒入 medium 的量，新的 medium 变成旧的 medium 加上量那是倒进去的，小当然是不会变的。

这是我尝试过的。

%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（从大到中）的正确实施吗？我那里的逻辑正确吗？

Answer 1

我觉得逻辑应该是这样

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

在状态之间移动（Prolog 实现）

Moving between states (A Prolog implementation)

prolog

state-space