jim burton writes:

Paul Brown-4 wrote:

...

...

Cool idea! Can you post a link for the puzzles?

Thankyou! It's http://www.rubyquiz.com - They are mostly well suited to haskell, lot of mazes etc. I've done 5 or 6 with varying degrees of success but have learned a lot. This thing about strings in fifths is from #1, the solitaire cipher.

At a quick glance I can't see which bit needs it. The only mention of five is where it asks to split the string into groups of five characters (not into five equal parts), padded with Xs. You can do that like this: splitAtMb n l = let p = splitAt n l in if null $ fst p then Nothing else Just p in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (length l `mod` 5) 'X') To break a string into five equal parts with the last padded with Xs, try this: fifths l = let len = length l part_length = (len+4)`div`5 pad_length = 5*part_length - len in unfoldr (splitAtMb part_length) (l ++ replicate pad_length 'X') I haven't checked these at all carefully, but at least they illustrate the use of unfoldr. [aside: One might argue that the prelude ought to provide splitAtMb rather than splitAt.] -- Jón Fairbairn Jon.Fairbairn@cl.cam.ac.uk

tweak to in_fives

in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (5 - length l `mod` 5) 'X')

-- View this message in context: http://www.nabble.com/split-string-into-n-parts-tf2496941.html#a6961912 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

I wrote:

jim burton wrote:

...

Thankyou! It's http://www.rubyquiz.com - They are mostly well suited to haskell, lot of mazes etc. I've done 5 or 6 with varying degrees of success but have learned a lot. This thing about strings in fifths is from #1, the solitaire cipher.

At a quick glance I can't see which bit needs it. The only mention of five is where it asks to split the string into groups of five characters (not into five equal parts), padded with Xs.

You can do that like this:

splitAtMb n l = let p = splitAt n l in if null $ fst p then Nothing else Just p

Gah! Brain AWOL. I'm surprised no-one picked me up on that. Why didn't I use: splitAtMb n [] = Nothing splitAtMb n l = Just $ splitAt n l ?

in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (length l `mod` 5) 'X')

And using length makes this over-strict. maybe something like groups_of n = unfoldr (splitPad 5) where splitPad [] = Nothing splitPad l = Just $ mapFst (padwith 'X') (splitAt n l) padwith c l = take n $ l ++ replicate n c mapFst f (a,b) = (f a, b) -- in Data.Graph.Inductive.Query.Monad which is a little bit inefficient, but less clunky than checking for the end of list in order to apply padwith just once. -- Jón Fairbairn Jon.Fairbairn@cl.cam.ac.uk