在学习了一些 Scala 和 FP 的好处之后,我正在重新实现我之前的一些 CS 作业,以更好地理解 FP。然而,我得到了一项似乎用 FP 实现的任务(或至少是微不足道的翻译)。
在解决简单的 2D 迷宫时,有必要记住访问过哪些节点。但是,如果没有共享状态,每个递归调用如何知道其他递归调用检查了哪些节点?我可以将迷宫作为参数传递给每个递归调用,并返回一个包含访问过的地方的新迷宫,但这似乎计算量太大,无法在每个递归调用中复制整个迷宫。是否需要更高级的方法来实现不可变迷宫求解器?
在学习了一些 Scala 和 FP 的好处之后,我正在重新实现我之前的一些 CS 作业,以更好地理解 FP。然而,我得到了一项似乎用 FP 实现的任务(或至少是微不足道的翻译)。
在解决简单的 2D 迷宫时,有必要记住访问过哪些节点。但是,如果没有共享状态,每个递归调用如何知道其他递归调用检查了哪些节点?我可以将迷宫作为参数传递给每个递归调用,并返回一个包含访问过的地方的新迷宫,但这似乎计算量太大,无法在每个递归调用中复制整个迷宫。是否需要更高级的方法来实现不可变迷宫求解器?
您可以传递包含访问节点的集合(如果节点本身在您的设置中具有可比性,则可以传递它们的 ID/名称)。将项目添加到不可变集合通常需要O(log n)
,因此检查元素是否包含在集合中。所以这比复制迷宫要便宜得多。
也许您注意到我之前的答案已被删除。虽然我是在开玩笑,但同时也只是暗示“计算机显示所有死角为红色,连接入口和出口的路径为绿色”,同时也隐喻了我所理解的功能范式 - 一种包含预先计算的确定性。鉴于我的理解和知识有限,我在 Haskell 中研究了一个避免递归深度搜索的示例,计算 4x5 迷宫的路径,给定一个数组,其中迷宫中的每个单元(即每个数组元素)仅包含它可以连接到的单元格;-1 表示入口,-2 表示出口。(您可以在代码部分的顶部看到迷宫的轮廓。)我知道,更有经验的程序员可以做得更多更好。
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import Data.List
import Data.Maybe
--Each element in the maze lists the indexes of connected cells, '-1' for entrance, '-2' for exit
maze = [[-1,1], [0,2,5], [1,3], [2],
[5], [4,6,1,9], [5,7], [6,11],
[12], [5,13,10], [9], [7,15],
[8,16], [14,9,17], [13,15], [14,11],
[12,17], [13,16,18], [17,19], [18,-2]]
maze' = [[-1,1], [0,2], [1,3], [2,7],
[8,5], [4,6], [5,7], [3,6],
[4,9], [8,10], [9,11], [10,15],
[16,13], [12,14], [13,15], [11,14],
[12,17], [16,18], [17,19], [18,-2]]
index a = fromJust $ elemIndex a maze
indexes a = map (index) a
areConnected index_a index_b = elem index_a (maze !! index_b)
isStart a --(a :: cell)
| elem (-1) a = True
| otherwise = False
isEnd a --(a :: cell)
| elem (-2) a = True
| otherwise = False
hasStart a --(a :: [cell])
| isStart (head a) = True
| otherwise = False
hasEnd a --(a :: [cell])
| isEnd (last a) = True
| otherwise = False
isSequenced (w:x:xs) (y:z:zs) --includes possibility of overlap since we do not know how many cells comprise the solution
| areConnected (index $ last xs) (index y)
|| last xs == y
|| let (b:c:cs) = reverse (w:x:xs) in [c,b] == [y,z] = True
| otherwise = False
removeBacktracks (x:xs)
| (x:xs) == [] = []
| xs == [] = [x]
| x == head xs = removeBacktracks xs
| length xs > 1 && x == let (y:ys) = xs in head ys = removeBacktracks (tail xs)
| otherwise = x : removeBacktracks xs
--list dead ends
dead_ends = filter (\x -> length x==1 && find (==(-1)) x == Nothing) maze
dead_ends_indexes = map (index) dead_ends
connectedToDeadEnd (x:xs)
| x `elem` dead_ends_indexes = True
| not (x `elem` dead_ends_indexes) && xs == [] = False
| otherwise = connectedToDeadEnd xs
--list first from dead ends
first_from_dead_ends = filter (\x -> length x==2 && find (==(-1)) x == Nothing && connectedToDeadEnd x) maze
--create sequences
filtered = [l | l <- maze, not (elem l dead_ends) && not (elem l first_from_dead_ends)]
sequences_3 = [[a,b,c] | a <- filtered, not (isEnd a),
b <- filtered, not (isEnd b || isStart b), areConnected (index a) (index b),
c <- filtered, not (isStart c), a /= c, areConnected (index b) (index c)]
sequences_4 = [a ++ [b] | a <- sequences_3, not (hasEnd a), b <- filtered,
last a /= b, areConnected (index $last a) (index b)]
paths = take 1 [indexes $ concat [a, b, c, d, e] | a <- sequences, hasStart a,
b <- sequences, not (hasStart b || hasEnd b),
isSequenced a b,
c <- sequences, b /= c, not (hasStart c || hasEnd c),
isSequenced b c,
d <- sequences, c /= d, not (hasStart d || hasEnd d),
isSequenced c d,
e <- sequences, hasEnd e,
isSequenced d e]
where sequences
| length filtered < 16 = sequences_3
| otherwise = sequences_4
path = removeBacktracks $ head paths
main = print path
--outputs: [0,1,5,9,13,17,18,19]