Haskell 中无限列表的笛卡尔积 [英] Cartesian product of infinite lists in Haskell

问题描述

我有一个用于有限列表的函数

I have a function for finite lists

> kart :: [a] -> [b] -> [(a,b)]
> kart xs ys = [(x,y) | x <- xs, y <- ys]

但是如何为无限列表实现它?我听说过康托尔和集合论.

but how to implement it for infinite lists? I have heard something about Cantor and set theory.

我还发现了一个类似的功能

I also found a function like

> genFromPair (e1, e2) = [x*e1 + y*e2 | x <- [0..], y <- [0..]]

但我不确定它是否有帮助，因为 Hugs 只提供配对而不会停止.

But I'm not sure if it helps, because Hugs only gives out pairs without ever stopping.

感谢您的帮助.

推荐答案

你的第一个定义，kart xs ys = [(x,y) |x <- xs, y <- ys]，等价于

Your first definition, kart xs ys = [(x,y) | x <- xs, y <- ys], is equivalent to

kart xs ys  =  xs >>= (x ->
               ys >>= (y -> [(x,y)]))

哪里

(x:xs) >>= g  =  g x ++ (xs >>= g)
(x:xs) ++ ys  =  x : (xs ++ ys)

是顺序操作.将它们重新定义为交替操作，

are sequential operations. Redefine them as alternating operations,

(x:xs) >>/ g  =  g x +/ (xs >>/ g)
(x:xs) +/ ys  =  x : (ys +/ xs)
[]     +/ ys  =  ys

并且您的定义也应该适用于无限列表:

and your definition should be good to go for infinite lists as well:

kart_i xs ys  =  xs >>/ (x ->
                 ys >>/ (y -> [(x,y)]))

测试，

Prelude> take 20 $ kart_i [1..] [101..]
[(1,101),(2,101),(1,102),(3,101),(1,103),(2,102),(1,104),(4,101),(1,105),(2,103)
,(1,106),(3,102),(1,107),(2,104),(1,108),(5,101),(1,109),(2,105),(1,110),(3,103)]

由 理性策划者"提供

/a>.(另见conda、condi、conde、condu).

另一种更明确的方法是创建单独的子流并将它们组合起来:

another way, more explicit, is to create separate sub-streams and combine them:

kart_i2 xs ys = foldr g [] [map (x,) ys | x <- xs]
  where
     g a b = head a : head b : g (tail a) (tail b)

这实际上产生了完全相同的结果.但是现在我们可以更好地控制如何组合子流.我们可以更加倾斜:

this actually produces exactly the same results. But now we have more control over how we combine the sub-streams. We can be more diagonal:

kart_i3 xs ys = g [] [map (x,) ys | x <- xs]
  where                                       -- works both for finite
  g [] [] = []                                --  and infinite lists
  g a  b  = concatMap (take 1) a
            ++ g (filter (not . null) (take 1 b ++ map (drop 1) a))
                 (drop 1 b)

所以现在我们得到

Prelude> take 20 $ kart_i3 [1..] [101..]
[(1,101),(2,101),(1,102),(3,101),(2,102),(1,103),(4,101),(3,102),(2,103),(1,104)
,(5,101),(4,102),(3,103),(2,104),(1,105),(6,101),(5,102),(4,103),(3,104),(2,105)]

通过一些搜索SO我还发现了一个Norman Ramsey 的回答 似乎是另一种生成序列的方法，将这些子流分为四个区域 - 左上角，顶行，左列，递归其余列.他的merge和我们这里的+/是一样的.

With some searching on SO I've also found an answer by Norman Ramsey with seemingly yet another way to generate the sequence, splitting these sub-streams into four areas - top-left tip, top row, left column, and recursively the rest. His merge there is the same as our +/ here.

你的第二个定义，

genFromPair (e1, e2) = [x*e1 + y*e2 | x <- [0..], y <- [0..]]

相当于只是

genFromPair (e1, e2) = [0*e1 + y*e2 | y <- [0..]]

因为列表 [0..] 是无限的，所以 x 的任何其他值都没有机会发挥作用.这是以上定义都想避免的问题.

Because the list [0..] is infinite there's no chance for any other value of x to come into play. This is the problem that the above definitions all try to avoid.

这篇关于Haskell 中无限列表的笛卡尔积的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！