Why is this Haskell expression so slow?

Question

I\'m working on Project Euler Problem 14. Here\'s my solution.

import Data.List

collatzLength :: Int->Int
collatzLength 1 = 1
collatzLength n | odd n = 1         


        

Answer 1

You're assuming that the collatzLength function will be memoized. Haskell does not do automatic memoization. You'll need to do that yourself. Here's an example using the data-memocombinators package.

import Data.List
import Data.Ord
import qualified Data.MemoCombinators as Memo

collatzLength :: Integer -> Integer
collatzLength = Memo.arrayRange (1,1000000) collatzLength'
  where
    collatzLength' 1 = 1
    collatzLength' n | odd n  = 1 + collatzLength (3 * n + 1)
                     | even n = 1 + collatzLength (n `quot` 2)

main = print $ foldl1' max $ [(collatzLength n, n) | n <- [1..1000000]]

This runs in about 1 second when compiled with -O2.

Answer 2

cL is short for collatzLength

cL!!n stands for collatzLength n

cL :: [Int]
cL = 1 : 1 : [ 1 + (if odd n then cL!!(3*n+1) else cL!!(n `div` 2)) | n <- [2..]]

Simple test:

ghci> cL !! 13
10

Answer 3

For being able to find a maximum of a list, the whole list needs to be evaluated.

So it will calculate collatzLength from 1 to 1000000 and collatzLength is recursive. The worst thing is, that your definition of collatzLength is even not tail-recursive.