Rewrite recursion function as a pipeline functions composition

Question

I'm writing my homework (CIS194 Haskell course).

I must rewrite the following recursive function to pipeline functions (without obvious recursion).

fun2 :: Integer -> Integer
fun2 1 = 0
fun2 n 
    | even n = n + fun2 ( n ‘div‘ 2 )
    | otherwise = fun2 (3 * n + 1) 

My first try is here:

fun2''' = sum 
          . (filter (even)) 
          . unfoldr (\x -> if (even x) 
                          then Just (x, x `div` 2)
                          else if (x==1) then Nothing 
                               else Just (x, x * 3 + 1))

Is this a normal solution or is it weird?

And how can I rewrite fun2 better?

Now i try write version with takeWhile and iterate

my 2nd try:

fun2'' :: Integer -> Integer
fun2'' = sum 
         . (filter even) 
         . takeWhile (/=1) 
         . iterate (\x -> if even x 
                          then x `div` 2
                          else x * 3 + 1 )

i have little problems with until version now.

Answer 1

Looks not bad, the only thing here that's a bit of a red flag in Haskell is else if. In this case, it can be rewritten nicely in applicative style:

{-# LANGUAGE TupleSections     #-}

import Control.Applicative
import Control.Monad (guard)

fun2''' = sum 
          . filter even
          . unfoldr ( \x -> fmap (x,) $
                   x`div`2 <$ guard(even x)
               <|> x*3 + 1 <$ guard( x/=1 )
             )

Answer 2

Nested ifs can now be written with multi-way IF:

g :: Integer -> Integer
g = sum . 
     unfoldr (\x-> 
          if | even x    -> Just (x, x `div` 2) 
             | x==1      -> Nothing 
             | otherwise -> Just (0, x * 3 + 1))

Or you can define your own if operator,

(??) t (c,a) | t = c | otherwise = a

g = sum . unfoldr (\x-> even x ?? (Just (x, x `div` 2) ,
                        (x==1) ??  (Nothing, Just (0, x * 3 + 1))))

Same function with until, with sum and filter fused into it:

g = fst . until ((==1).snd) 
            (\(s,n) -> if even n then (s+n,n`div`2) else (s,3*n+1)) 
    . ((,)0)

or

g = sum . filter even . f

f :: Integer -> [Integer]
f = (1:) . fst . until ((==1).snd) 
            (\(s,n) -> if even n then (n:s,n`div`2) else (n:s,3*n+1)) 
    . ((,)[])

The last function, f, shows the whole Collatz sequence for a given input number, reversed.