haskell - 解决狮子、狼和山羊的问题

Question

我正在尝试解决发现的狮狼和山羊问题 here

-- Lions Wolves Goats
-- Lion -> Wolf -> Goat :: -1, -1, +1
-- Lion -> Goat -> Wolf :: -1, +1, -1
-- Wolf -> Goat -> Lion :: +1, -1, -1
--                      -------------
--                      :: -1, -1, -1

这就是我的解决方案:

eat :: (Int, Int, Int) -> (Int, Int, Int)
eat (lions, wolves, goats) = eat (lions - 1, wolves - 1, goats - 1)
eat (0, wolves, goats) = eat (1, wolves - 1, goats - 1)
eat (lions, 0, goats) = eat (lions - 1, 1, goats - 1)
eat (lions, wolves, 0) = eat (lions - 1, wolves - 1, 1)
eat (0, 0, goats) = (0, 0, goats)

此代码可以编译，但代码会永远挂起。我很确定我错过了一场比赛，但我不知道是什么。请让我知道如何正确解决这个问题。

Answer 1

模式是从上到下、从左到右读取的。因此，您的第一个模式与所有其他模式相匹配。重新排序:

eat :: (Int, Int, Int) -> (Int, Int, Int)
eat (0     , 0      , goats) =     (0         , 0          , goats    )
eat (0     , wolves , goats) = eat (1         , wolves - 1 , goats - 1)
eat (lions , 0      , goats) = eat (lions - 1 , 1          , goats - 1)
eat (lions , wolves , 0)     = eat (lions - 1 , wolves - 1 , 1        )
eat (lions , wolves , goats) = eat (lions - 1 , wolves - 1 , goats - 1)

请注意，这假设所有值最初都是正数。如果其中之一为负，您仍然会出现循环。另请注意，您错过了其他“稳定”状态，并且您对狮子、狼和山羊的非零群的逻辑已关闭。如果有山羊和狼，狮子可能会吃掉它们中的任何一个。此外，狼可能会先于狮子吃掉山羊。

这就是非决定论发挥作用的部分。