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 <[EMAIL PROTECTED]>wrote: > On 11/23/08 13:52, Luke Palmer wrote: > >> 2008/11/23 Larry Evans <[EMAIL PROTECTED]>: >> >>> 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 > > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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