[Haskell-beginners] list, map, sequence - stack overflow and performance issues

[Julian Ong]

 Sorry, corrected some typos below in the number of apostrophes.
    On Saturday, January 16, 2021, 04:14:47 PM PST, Julian Ong <julian_ong at yahoo.com> wrote:  
 
 Hi Haskellers - I'm learning Haskell and attempting to solve the Advent of Code 2020 puzzles using Haskell. I'm stuck on part 2 of Day 15 and have been for a while now, so I'm reaching out.
The puzzle asks you to find the nth element in a list of integers. Here's how the list is constructed:
Start with a seed list of integers, like [0,3,6]. Then, referring to the last element (6), the next element is given by these rules:
   
   - If the last element was the first time the element has appeared in the list, then the next element is 0.
   - Otherwise, the next element is the age, or distance in the number of index positions, between the last element and when it last appeared before that.

For example, starting with [0,3,6], the next elements are 0, 3, 3, 1, 0, 4, 0, etc.
Part 1 of the puzzle asks you to find the 2020th element in the list.
You can do this by constructing increasingly longer lists like this (using Data.List):
nextNum :: [Int] -> [Int]nextNum l@(x:xs) = if not (x `elem` xs) then 0 : l else age l : l    where        age (x:xs) = let Just i = elemIndex x xs                           in  i+1
Then:
 head $ (iterate nextNum [6,3,0]) !! 2017
will give you the 2020th element of 436.
Note that you provide the starting list in reverse order and iterate so that it will keep adding new elements to the head of the list, which is more efficient than adding to the end.
You can also use unfoldr to generate the list element by element like this:
 nextNum' :: [Int] -> IntnextNum' (x:xs) = if not (x `elem` xs) then 0 else age x xs    where        age x xs = let Just i = elemIndex x xs                               in  i+1
Then:
(unfoldr (\l -> Just (nextNum' l, nextNum' l : l)) slist) !! 2016

will give you the 2020th element of 436.
---
Part 2 of the puzzle asks you to find the 30000000th element given starting list [9,3,1,0,8,4].
I cannot find a way to do this without stack overflow and performance issues (I've run my attempts overnight with no answer generated). I've tried using Data.Map and Data.Sequence because my Stack Overflow searching suggested these might be more efficient data structures for this sort of task. Here are my attempts:
-- Uses Data.Map to avoid duplicate numbers thereby shortening the list. The dictionary entry (k, v) gives the element and the last position of that element.
 nextNum'' :: (IntMap Int, (Int, Int)) -> (IntMap Int, (Int, Int))nextNum'' (mp, (k, v)) = case IntMap.lookup k mp of    Nothing  -> (IntMap.insert k v mp, (0, v+1))    Just pos -> (IntMap.insert k v mp, (v-pos, v+1))

Then:
snd $ (iterate nextNum'' (IntMap.fromList [(0,1),(3,2)],(6,3))) !! 2017

provides the answer for the 2020th element but either stack overflows or runs for hours (if I use a strict version of iterate) trying to figure out the 30000000th element.
Similarly, using Data.Sequence, I tried:
 nextNum''' :: Seq Int -> IntnextNum''' (xs :|> x) = if not (x `elem` xs) then 0 else age x xs    where        age x xs = let Just i = Seq.elemIndexR x xs                          in  Seq.length xs - i
 aoc15b' :: Seq Int -> Int -> Intaoc15b' slist tnum = (\(xs :> x) -> x) $ Seq.viewr (Seq.unfoldr (\l -> if Seq.length l == tnum then Nothing else let nnum = force (nextNum''' l) in Just (nnum, force (l |> nnum))) slist)
I found that I needed to fix stack overflow problems by using "force" from Control.DeepSeq. Despite seemingly fixing stack overflow issues though, the calculation just takes too long, and in fact, I have never been able to actually output a solution.
I thought that using Data.Map or Data.Sequence would speed things up based on my Stack Overflow searching, but I'm unable to come up with a Haskell solution that runs in reasonable time.
I'm at a loss for different strategies at this point and would appreciate any ideas from the community.
Thanks, Julian








  
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