[Haskell-cafe] Hand calculation of Bird's definition of zip using foldr

[Miguel Mitrofanov]

zip [1,2,3] [4,5,6] = zip (1:2:3:[]) (4:5:6:[]) = foldr f e (1:2:3:[])  
(4:5:6:[]) = f 1 (foldr f e (2:3:[])) (4:5:6:[]) = (1, 4) : foldr f e  
(2:3:[]) (5:6:[]) = (1, 4) : f 2 (foldr f e (3:[])) (5:6:[]) = (1,  
4) : (2, 5) : foldr f e (3:[]) (6:[]) = (1, 4) : (2, 5) : f 3 (foldr f  
e []) (6:[]) = (1, 4) : (2, 5) : (3, 6) : foldr f e [] [] = (1, 4) :  
(2, 5) : (3, 6) : e [] = (1, 4) : (2, 5) : (3, 6) : [] = [(1, 4), (2,  
5), (3, 6)]

On 12 Mar 2009, at 20:01, R J wrote:

> Can someone provide a complete hand calculation of zip [1,2,3]  
> [4,5,6] using the following definition of zip, based on foldr:
>
> zip    ::    [a] -> [b] -> [(a, b)]
> zip    =    foldr f e
>     where
>     e ys    =    []
>     f x g [ ]    =    []
>     f x g (y : ys)    =    (x , y) : g ys
>
>
> foldr    ::    (a -> b -> b) -> b -> ([a] -> b)
> foldr _ e []    =    e
> foldr f e (x : xs)    =    f x (foldr f e xs)
>
>
> This implementation of zip produces the expected result [(1, 4), (2,  
> 5), (3, 6)], but I'm unable to do the hand calculation and don't  
> understand why it works.  Part of my problem is that "e" is defined  
> as a function that takes one argument, I don't see how that fits in  
> with the usual scheme for foldr, which, as I understand it, is:
>
> foldr f e [x1, x2, ...] = f x1 (f x2 (f x3 ...(f xn e)))...
>
> Thanks, as always, to all in this great community.
>
>
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