Re: [Haskell-cafe] Re: howto tuple fold to do n-ary cross product?

It's more natural to consider the cross product of no sets to be [[]] so your crossr becomes: crossr [] = [[]] crossr (x:xs) = concat (map (\h ->map (\t -> h:t) (crossr tail)) hd) which we can rewrite with list comprehensions for conciseness: crossr [] = [[]] crossr (x:xs) = [ a:as | a <- x, as <- crossr xs ] then look at the definition of foldr: foldr f z [] = z foldr f z (x:xs) = f x (foldr f z xs) and, considering (foldr f z) == crossr, you should derive the definition of f and z. On Mon, Nov 24, 2008 at 5:43 AM, Larry Evans wrote:

On 11/23/08 13:52, Luke Palmer wrote:

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2008/11/23 Larry Evans :

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http://www.muitovar.com/monad/moncow.xhtml#list

contains a cross function which calculates the cross product of two lists. That attached does the same but then used cross on 3 lists. Naturally, I thought use of fold could generalize that to n lists; however, I'm getting error:

You should try writing this yourself, it would be a good exercise. To begin with, you can mimic the structure of cross in that tutorial, but make it recursive. After you have a recursive version, you might try switching to fold or foldM.

Thanks. The recursive method worked with: -{--cross.hs-- crossr::[[a]] -> [[a]]

crossr lls = case lls of { [] -> [] ; [hd] -> map return hd ; hd:tail -> concat (map (\h ->map (\t -> h:t) (crossr tail)) hd) } -}--cross.hs--

However, I'm not sure fold will work because fold (or rather foldr1) from: http://haskell.org/ghc/docs/latest/html/libraries/base/Prelude.html#12

has signature:

(a->a->a)->[a]->a

and in the cross product case, a is [a1]; so, the signature would be

([a1]->[a1]->[a1]->[[a1]]->[a1]

but what's needed as the final result is [[a1]].

Am I missing something?

-Larry

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