Data.Graph 中的 path 函数如何在 Haskell 中工作？

Question

我已阅读有关路径函数的文档（对于有向图）。搜索引擎中没有太多内容。

我找到的所有内容基本上都归结为this link。

它真正说的是路径函数的作用（我已经知道），以及输出示例。我想弄清楚的是路径功能如何工作或如何手动实现。到目前为止，我已经走到了这一步，但现在卡住了（或者可能完全走错了路，我不能确定）：

member _ [] = False
member n (x:xs)
    | x == n = True
    | otherwise = membre n xs
    
getX :: (a, b) -> a
getX (x,_) = x

getY :: (a, b) -> b
getY (_,y) = y

getAllY :: [(a , b)] -> [b]
getAllY [] = [] 
getAllY (x:xs) = (getY x):(getAllY xs)

fltrX [] _ = []
fltrX (x:xs) n 
    | (getX x) == n = x:(fltrX xs n)
    | otherwise = fltrX xs n
    
fltrY [] _ = []
fltrY (x:xs) n 
    | (getY x) == n = x:(fltrY xs n)
    | otherwise = fltrY xs n

path' :: [(a , a)] -> a -> a -> Bool
path' [] _ _ = False
path' (xs) n m 
    | (member m (getAllY (fltrX xs n))) = True
    | otherwise = *recursive statement giving me a headache*

但我无法理解如何正确地进行递归。我想我只是盯着它太久了，不能再清楚地看到整个问题。逻辑应该是这样的：

• 它从 x = n 的元组中获取 y 值
• 它会在上面的 Y 值列表中找到所有 x 的元组
• 如果没有 Y 值 = m，则对具有 n =（每个 X 值）且 m = m 的新元组列表执行路径函数
• 如果在 Y 值中找到 m，则返回 true

谁能指出我正确的方向（双关语），最好不要为我输入代码？

谢谢！

Answer 1

感谢评论者提供帮助我编写以下解决方案的资源：

member _ [] = False
member n (x:xs)
    | x == n = True
    | otherwise = member n xs
    
unique [] = []
unique (x:xs)
    | member x xs == True = unique xs
    | otherwise = x:(unique xs)


findAllEndsForX :: [(Int , Int)] -> Int -> [Int]
findAllEndsForX [] _ = []
findAllEndsForX ((a,b):xs) y
  | xs == [] = a:[b]
  | a == y = a:(findAllEndsForX xs b)
  | otherwise = (findAllEndsForX xs b)

accessible :: [(Int , Int)] -> Int -> [Int]
accessible [] _ = []
accessible xs n = unique (findAllEndsForX xs n)

path' :: [(Int , Int)] -> Int -> Int -> Bool
path' [] _ _ = False
path' xs v w = member w (accessible xs v)

我的最终目标是模仿路径功能的功能，而不一定是实现。我想在不使用任何预定义函数的情况下完成这项工作。