如何在Haskell中创建函数minimax？
如何在Haskell中创建函数minimax？
提问于 2018-09-29 01:24:12
我一直在努力在Haskell中创建一个minimax函数，但在SO上找不到任何好的替代方案，因此我问：
我想创建以下函数：
minimax :: Player -> Rose Board -> Rose Int我想得到一个整数的RoseTree，它们必须是1，0或-1 (对于有回合的玩家来说，走法可以是好的，中性的，也可以是坏的。
(root :> leaves) -- constructor of a Rose
(board :> boards) -- constructor of a Rose Board
(Int :> Ints) -- constructor of a Rose Int我已经编写了函数hasWinner，minimum‘and maximum’让事情变得更简单：
hasWinner :: Board -> Maybe Player
minimum' :: [Int] -> Int
minimum' (x:xs) | x == -1 = -1
| otherwise = minimum' xs
maximum' :: [Int] -> Int
maximum' (x:xs) | x == 1 = 1
| otherwise = maximum' xs此外，我认为我的基本情况如下：
minimax player (board :> []) = (0 :> [])目前，我所处的位置是：
minimax player (board :> boards)| maximum' [(isWinner player b (minimax' player)) | (b :> bs) <- boards] == 1 = _
| minimum' [(isWinner player b (minimax' player)) | (b :> bs) <- boards] == -1 = _
| otherwise = _
where minimax' player | player == P1 = P2
| otherwise = P1
isWinner p1 board p2 | hasWinner board == Just p1 = 1
| hasWinner board == Just p2 = -1
| otherwise = 0我只是不知道当我发现一块棋盘要么赢了，要么输了，要么没有后果时，会发生什么。那么，如何使用正确的Rose Board再次调用minimax函数呢？
任何帮助都是非常感谢的！
雷蒙
回答 1
Stack Overflow用户
发布于 2018-09-29 02:06:51
实际上，我找到了另一篇讨论我在这里遇到的相同问题的帖子：Haskell Recursive Minimax Tree
答案是这样的：
minimax :: Player -> Rose Board -> Rose Int
minimax p (r :> []) | hasWinner r == Just P1 = 1 :> []
| hasWinner r == Just P2 = (-1) :> []
| otherwise = 0 :> []
minimax P1 (r :> rs) = maximum (map root xs) :> xs
where xs = map (minimax (nextPlayer P1)) rs
minimax P2 (r :> rs) = minimum (map root xs) :> xs
where xs = map (minimax (nextPlayer P2)) rs我希望这至少能在未来帮助其他人。此答案中使用的函数包括：
root :: Rose a -> a
root (a :> bs) = a
nextPlayer :: Player -> Player
nextPlayer P1 = P2
nextPlayer P2 = P1
hasWinner :: Board -> Maybe Player
minimum :: Ord a => [a] -> a
maximum :: Ord a => [a] -> a干杯!
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52560126
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