我必须定义一个无限骑车者
enumInts::Cyclist Integer
包含自然顺序中的所有整数，其中0为当前元素。
我做的是：
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)
正在编译自己。但我不确定它是否能够正常工作......所以我希望看到它的结果，例如。 11个单位。它应该是：-5 -4 -3 -2 -1 0 1 2 3 4 5值。 有可能看到它吗？ （我知道它是无限的）但是对于例如。在fibonacci序列中，我们可以使用'take 11 fibs'来给它们。这里选项'take n ..'不起作用（嗯或它有效，但我不知道如何使用它）。 我很感激你的帮助..
答案 0 :(得分：3)
现在已经过了你的截止日期，我确定，所以我很乐意处理双倍无限的整数：
要制作拍摄功能，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
但现在我们可以在您的数据类型中看到错误：
*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 - 它里面有三个零！
0
|
-----
/ \
-1 1
| |
--- --
/ \ / \
-2 0 0 2 -- oops! We've re-created zero for 1 and -1
最后有两个0和两个数字。如果我们更深入，那就更糟了
*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东西。我们想要的更像是
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，但有很多Empty令人困惑。
我们真正需要的是一个当前元素，类似于向右延伸的列表，以及向后延伸到左侧的列表。编译器没有方向感，因此我们将对两者使用相同的结构，但记得将左手侧向后打印到右侧。
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实现无限：
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
只有无限多的元素，而不仅仅是9。
让我们走一条路。使用reverse，toList和fromList可以提高效率，但这样您就可以了解如何处理其中的部分内容：
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}
我们现在可以在每次想要限定时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}
哪个给出了
*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]}
但你可能想打印出来，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
哪个给出了
*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，我们刚刚转换成一个列表并显示出来，但我们已经失去了表明我们所处位置的能力。
import Data.Tree
import Data.Tree.Pretty
有几种不同类型的树木等等，所以我给自己一个班级将它们各自转换成树：
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
我们这样使用：
*Main> see 5 $ go (-2) enumIntsDIL
-2
|
---
/ \
-3 -1
| |
-4 0
| |
-5 1
| |
-6 2
首先是你的自行车手：
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]
答案 1 :(得分：0)
如果您想要这些数字，为什么不使用
enumInts :: Integer -> [Integer]
enumInts n = [-(n`div`2)..(n`div`2)]
答案 2 :(得分：0)
虽然它是内置的，但您可以将列表类型想象为
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根据自己的类型调整take的定义。
答案 3 :(得分：0)
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
答案 4 :(得分：0)
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
其中：
label (Elem _ x _) = x
forward (Elem _ _ x) = x
backward (Elem x _ _) = x