我必须定义一个无限的自行车手
enumInts::Cyclist Integer
包含自然顺序中的所有整数，其中零是当前元素。
我所做的是：
data Cyclist a=Elem (Cyclist a) a (Cyclist a)
enumInts:: Cyclist Integer
enumInts=Elem prev 0 next
where
prev=help2 enumInts 0
next=help1 enumInts 0
-- Create positive part
help1::Cyclist Integer -> Integer -> Cyclist Integer
help1 prev n=present
where present=Elem prev (n+1) next
where next=help1 present (n+1)
-- Create negative part
help2::Cyclist Integer -> Integer -> Cyclist Integer
help2 next n=present
where present=Elem prev (n-1) next
where prev=help2 present (n-1)
它正在编译自己。但我不确定它是否正常工作...所以我想看看它的结果，例如。11个单位。它应该是 ：-5 -4 -3 -2 -1 0 1 2 3 4 5 个值。有可能看到吗？(我知道它是无限的(但例如。在斐波那契数列中，我们可以使用"取 11 个 fibs"，它给了它们。这里选项"采取n..."不起作用(嗯，或者它有效，但我不知道如何使用它(。我会感谢您的帮助..
我敢肯定，现在已经超过了你的截止日期，所以我在处理双无限整数时玩得很开心：
允许有限部分
要制作一个 take 函数，我必须编辑你的类型，以便它可以是有限的：
data Cyclist a=Elem (Cyclist a) a (Cyclist a) | Empty
deriving Show
takeToDepth :: Int -> Cyclist a -> Cyclist a
takeToDepth 0 _ = Empty
takeToDepth n (Elem c1 a c2)
| n >0 = Elem (takeToDepth (n-1) c1) a (takeToDepth (n-1) c2)
| otherwise = Empty
takeToDepth n Empty = Empty
但是现在我们可以看到数据类型中的一个错误：
*Main> takeToDepth 1 enumInts
Elem Empty 0 Empty
0 -- I've drawn the tree
和
*Main> takeToDepth 2 enumInts
Elem (Elem Empty (-1) Empty) 0 (Elem Empty 1 Empty)
0
| -- looks OK
--- -- see the end of the answer for how I pretty printed
/
-1 1
到目前为止看起来还不错，但是：
*Main> takeToDepth 3 enumInts
Elem (Elem (Elem Empty (-2) Empty) (-1) (Elem Empty 0 Empty))
0 (Elem (Elem Empty 0 Empty) 1 (Elem Empty 2 Empty))
这不是我们想要的结构 - 它有三个零！
0
|
-----
/
-1 1
| |
--- --
/ /
-2 0 0 2 -- oops! We've re-created zero for 1 and -1
最后有两个0 s和每个数字的两个。如果我们更深入，情况会更糟
*Main> takeToDepth 4 enumInts
Elem (Elem (Elem (Elem Empty (-3) Empty) (-2) (Elem Empty (-1) Empty)) (-1)
(Elem (Elem Empty (-1) Empty) 0 (Elem Empty 1 Empty))) 0
(Elem (Elem (Elem Empty (-1) Empty) 0 (Elem Empty 1 Empty)) 1
(Elem (Elem Empty 1 Empty) 2 (Elem Empty 3 Empty)))
0
|
--------------------------
/
-1 1
| |
------------- -----------
/ /
-2 0 0 2
| | | |
------- ----- ----- -----
/ / / /
-3 -1 -1 1 -1 1 1 3
| | | | | | | |
--- --- --- -- --- -- -- --
/ / / / / / / /
-4 -2 -2 0 -2 0 0 2 -2 0 0 2 0 2 2 4
我们不需要中间的所有这些东西。我们想要的更像
this = Elem (Elem (Elem (Elem Empty (-3) Empty) (-2) Empty) (-1) Empty)
0 (Elem Empty 1 (Elem Empty 2 (Elem Empty 3 Empty)))
0
|
---
/
-1 1
| |
-2 2
| |
-3 3
这很好，但是Empty太多了，令人困惑。
创建执行预期操作的数据类型。
我们真正需要的是一个当前元素，类似于向右延伸的列表，以及向左向后延伸的列表。编译器没有方向感，因此我们将对两者使用相同的结构，但请记住将左侧向后打印到右侧。
首先，我们需要一个绝对无限的列表：
data InfiniteList a = IL a (InfiniteList a) deriving Show
tailIL (IL _ therest) = therest
headIL (IL a _ ) = a
fromList [] = error "fromList: finite list supplied"
fromList (x:xs) = IL x (fromList xs)
toList (IL a therest) = a:toList therest
现在我们可以在两个方向上使其无限：
data DoublyInfiniteList a = DIL {left :: InfiniteList a,
here :: a,
right :: InfiniteList a}
deriving Show
enumIntsDIL = DIL {left = fromList [-1,-2..], here = 0, right = fromList [1..]}
看起来像这样：
0
|
---
/
-1 1
| |
-2 2
| |
-3 3
| |
-4 4
只有无限多个元素，而不仅仅是 9 个。
让我们做一种移动的方式。这可以通过使用reverse，toList和fromList来提高效率，但是通过这种方式，您可以看到如何弄乱其中的各个部分：
go :: Int -> DoublyInfiniteList a -> DoublyInfiniteList a
go 0 dil = dil
go n dil | n < 0 = go (n+1) DIL {left = tailIL . left $ dil,
here = headIL . left $ dil,
right = IL (here dil) (right dil)}
go n dil | n > 0 = go (n-1) DIL {left = IL (here dil) (left dil),
here = headIL . right $ dil,
right = tailIL . right $ dil}
现在，每当我们想要有限时，我们都可以转换为另一种数据类型。
data LeftRightList a = LRL {left'::[a],here'::a,right'::[a]} -- deriving Show
toLRL :: Int -> DoublyInfiniteList a -> LeftRightList a
toLRL n dil = LRL {left' = take n . toList . left $ dil,
here' = here dil,
right' = take n . toList . right $ dil}
这给了
*Main> toLRL 10 enumIntsDIL
LRL {left' = [-1,-2,-3,-4,-5,-6,-7,-8,-9,-10], here' = 0, right' = [1,2,3,4,5,6,7,8,9,10]}
但你可能想打印它，所以它看起来像你的意思：
import Data.List -- (Put this import at the top of the file, not here.)
instance Show a => Show (LeftRightList a) where
show lrl = (show'.reverse.left' $ lrl) -- doesn't work for infinite ones!
++ ", " ++ show (here' lrl) ++ " ,"
++ (show' $ right' lrl) where
show' = concat.intersperse "," . map show
这给了
*Main> toLRL 10 enumIntsDIL
-10,-9,-8,-7,-6,-5,-4,-3,-2,-1, 0 ,1,2,3,4,5,6,7,8,9,10
*Main> toLRL 10 $ go 7 enumIntsDIL
-3,-2,-1,0,1,2,3,4,5,6, 7 ,8,9,10,11,12,13,14,15,16,17
当然，我们可以转换为列表并显示它，但是我们将失去指示我们在哪里的能力。
附录：我如何漂亮地打印树木
import Data.Tree
import Data.Tree.Pretty
几种不同类型的树等，所以我给自己一个类将它们分别转换为树：
class TreeLike t where
toTree :: t a -> Tree a
treeTake :: Int -> Tree a -> Tree a
treeTake 1 (Node a _) = Node a []
treeTake n (Node a ts) | n > 1 = Node a (map (treeTake (n-1)) ts)
| otherwise = error "treeTake: attemt to take non-positive number of elements"
see :: (TreeLike t,Show a) => Int -> t a -> IO ()
see n = putStrLn.drawVerticalTree.fmap show.treeTake n.toTree
我们像这样使用：
*Main> see 5 $ go (-2) enumIntsDIL
-2
|
---
/
-3 -1
| |
-4 0
| |
-5 1
| |
-6 2
首先是您的骑行者：
instance TreeLike Cyclist where
toTree Empty = error "toTree: error - Empty"
toTree (Elem Empty a Empty) = Node a []
toTree (Elem Empty a c2) = Node a [toTree c2]
toTree (Elem c1 a Empty) = Node a [toTree c1]
toTree (Elem c1 a c2) = Node a [toTree c1,toTree c2]
接下来是双无限列表：
instance TreeLike InfiniteList where
toTree (IL a therest) = Node a [toTree therest]
instance TreeLike DoublyInfiniteList where
toTree dil = Node (here dil) [toTree $ left dil,toTree $ right dil]
然后是左右列表：
instance TreeLike [] where
toTree [] = error "toTree: can't make a tree out of an empty list"
toTree [x] = Node x []
toTree (x:ys) = Node x [toTree ys]
instance TreeLike LeftRightList where
toTree lrl = Node (here' lrl) [toTree $ left' lrl,toTree $ right' lrl]
当你想要这些数字时，为什么不直接使用
enumInts :: Integer -> [Integer]
enumInts n = [-(n`div`2)..(n`div`2)]
虽然它是内置的，但您可以将列表类型定义为
data [a] = [] | a : [a]
take可以这样定义：
take 0 xs = []
take n (x:xs) = x:take (n-1) xs
您应该尝试了解如何针对自己的类型调整take的定义。
我认为这可以解决为：
我们的无限数据
myList = ([0] ++ ) $ concat $ [[x] ++ [-x] | x <- [1..]]
从此列表中获取特定数量的元素：
takeOnly n = sort $ take n myList
这是我
的解决方案：
enumInts :: Cyclist Integer
enumInts = Elem (goLeft enumInts) 0 (goRight enumInts)
where
goLeft this@(Elem left n _) = let left = Elem (goLeft left) (n-1) this in left
goRight this@(Elem _ n right) = let right = Elem this (n+1) (goRight right) in right
你可以这样使用它：
label . forward . backward . forward . forward $ enumInts
哪里：
label (Elem _ x _) = x
forward (Elem _ _ x) = x
backward (Elem x _ _) = x