文件夹不返回无限列表

Question

我已阅读https://www.haskell.org/haskellwiki/Foldl_as_foldr和其他一些有关foldl和foldr之间差异的博客文章。 现在，我尝试使用文件夹将斐波那契数列编写为无限列表，并提出了以下解决方案：

fibs2 :: [Integer]
fibs2 = foldr buildFibs [] [1..]
  where
    buildFibs :: Integer -> [Integer] -> [Integer]
    buildFibs _ [] = [0]
    buildFibs _ [0] = [1,0]
    buildFibs _ l@(x:s:z) = (x + s):l

但是当我take 3 fibs2 ，该函数不会返回。 我认为文件夹递归可以在这些情况下与无限列表一起使用。 为什么这不适用于我的解决方案？

Answer 1

问自己：哪个斐波那契数字将是列表中的第一个？ 我对您的代码的理解是，该问题的答案是“最大的问题”（概念上，每次buildFibs迭代buildFibs在结果列表的buildFibs添加一个稍大的数字）。 由于有无数斐波纳契数，因此计算需要一段时间！

Answer 2

这是进行方程式推理的好方法：

fibs2 = foldr buildFibs [] [1..]

foldr f z [] = z
foldr f z (x:xs) = f x (foldr f z xs)

foldr buildFibs [] [1..] =
    buildFibs 1 (foldr buildFibs [] [2..]) =
    buildFibs 1 (buildFibs 2 (foldr buildFibs [] [3..])) =
    buildFibs 1 (buildFibs 2 (buildFibs 3 (foldr buildFibs [] [4..]))) =
    ...

我希望现在您可以看到问题所在： foldr在返回之前试图遍历整个列表。 如果我们改用foldl会怎样？

foldl f z [] = z
foldl f z (x:xs) = foldl f (f z x) xs

buildFibs' = flip buildFibs

foldl buildFibs' [] [1..] =
    foldl buildFibs' (buildFibs 1 []) [2..] = 
    foldl buildFibs' [0] [2..] =
    foldl buildFibs' (buildFibs 2 [0]) [3..] =
    foldl buildFibs' [0,1] [3..] =
    foldl buildFibs' (buildFibs 3 [0,1]) [4..] =
    foldl buildFibs' (0+1 : [0,1]) [4..] =
    foldl buildFibs' [1,0,1] [4..] =
    foldl buildFibs' (buildFibs 4 [1,0,1]) [5..] =
    foldl buildFibs' (1+0 : [1,0,1]) [5..] =
    foldl buildFibs' [1,1,0,1] [5..] =
    foldl buildFibs' (buildFibs 5 [1,1,0,1]) [6..] =
    foldl buildFibs' [2,1,1,0,1] [6..] =
    -- For brevity I'll speed up the substitution
    foldl buildFibs' [3,2,1,1,0,1] [7..] =
    foldl buildFibs' [5,3,2,1,1,0,1] [8..] =
    foldl buildFibs' [8,5,3,2,1,1,0,1] [9..] =
    ...

因此，如您所见，您实际上可以使用buildFibs和foldl计算斐波那契数，但是不幸的是，您向后建立了一个无限的列表，您将永远无法计算列表中的特定元素，因为foldl永远不会终止。 但是，您可以计算有限数量的它们：

> take 10 $ foldl buildFibs' [] [1..10]
[34,21,13,8,5,3,2,1,1,0]