Question

I’m very new with Haskell, only starting to learn it.
I’m using “Learn You a Haskell for Great Good!” tutorial for start, and saw example of solving “3n+1” problem:

chain :: (Integral a) => a -> [a]  
chain 1 = [1]  
chain n  
    | even n =  n:chain (n `div` 2)  
    | odd n  =  n:chain (n*3 + 1)  

numLongChains :: Int  
numLongChains = length (filter isLong (map chain [1..100]))  
    where isLong xs = length xs > 15  

so, numLongChains counts all chains that longer 15 steps, for all numbers from 1 to 100.

Now, I wanna my own:

numLongChains' :: [Int]  
numLongChains' = filter isLong (map chain [1..100])
    where isLong xs = length xs > 15    

so now, I wanna not to count these chains, but return filtered list with these chains.
But now I get error when compiling:

Couldn't match expected type `Int' with actual type `[a0]'
Expected type: Int -> Bool
  Actual type: [a0] -> Bool
In the first argument of `filter', namely `isLong'
In the expression: filter isLong (map chain [1 .. 100])

What can be the problem?

Answer 1

The type signature of numLongChains is probably not correct. Depending on what you want to do, one of the following is needed:

• You simply want to count those chains, your function numLongChains obviously shall return a number, change the first line to length $ filter isLong (map chain [1..100]) and the type to Int
• You want to return a list of lengths of the long chains. In this case, the type signature is fine, but you need to return a length. I’d suggest you, the calculate the length before filtering and filter on it. The function’s body becomes filter (>15) (map (length . chain) [1..100]).
• You want to return all chains that are longer than 15 chars. Just change the signature to [[Int]] (A list of chains (lists) of Ints) and you’re fine.