I am new to Haskell and currently trying to port my solutions for the 2019 installment of the coding challenge AdventOfCode to Haskell. So, I would very much appreciate any suggestions how to make the code more readable and, in particular, more idiomatic.

This post shows my solution of day 6 part 2, but also includes the function totalDecendantCount used to solve part 1. If you have not solved these problems and still intend to do so, stop reading immediately.

For both problems, you get a file with an orbit specification on each line of the form A)B, which tells you that B orbits A. This describes a tree of bodies orbiting each other with root COM.

In part 1, you have to compute a check sum. More precisely, you have to compute the sum of the number of direct and indirect orbits of each body, which is the same as the sum of the number of descendants of each body in the tree.

In part 2, which you cannot see if you have not finished part 1, you have to compute the minimal number of transfers between orbits from you (YOU) to Santa (SAN).

I have kept the entire solution for each part of each day in a single module with a single exported function that prints the solution. For day 6 part 2 it starts as follows.

module AdventOfCode20191206_2
( distanceToSanta
) where

import System.IO
import Data.List.Split
import Data.List
import Data.Maybe
import Data.Hashable
import qualified Data.HashMap.Strict as Map

distanceToSanta :: IO ()
distanceToSanta = do
let orbitList = (map orbit . lines) inputText
let orbits = orbitMap $catMaybes orbitList let pathToSanta = fromJust$ path orbits "COM" "YOU" "SAN"
let requiredTransfers = length pathToSanta - 3
print requiredTransfers


We subtract 3 from the length of the path because it consists of the bodies on the path and you only have to transfer from the body you already orbit to the body Santa orbits.

To store the tree, I use a HashMap.Strict and introduce the following type aliases and helper function to make things a bit more descriptive.

type OrbitSpecification = (String,String)
type ChildrenMap a = Map.HashMap a [a]

children :: (Eq a, Hashable a) => ChildrenMap a -> a -> [a]
children childrenMap = fromMaybe [] . flip Map.lookup childrenMap


orbit :: String -> Maybe OrbitSpecification
orbit str =
case orbit_specification of
[x,y] -> Just (x,y)
_ -> Nothing
where orbit_specification = splitOn ")" str

orbitMap :: [OrbitSpecification] -> ChildrenMap String
orbitMap = Map.fromListWith (++) . map (applyToSecondElement toSingleElementList)

applyToSecondElement :: (b -> c) -> (a,b) -> (a,c)
applyToSecondElement f (x,y) = (x, f y)

toSingleElementList :: a -> [a]
toSingleElementList x = [x]


To solve part 1, I introduce two general helper function to generate aggregates over children or over all descendents.

childrenAggregate :: (Eq a, Hashable a) => ([a] -> b) -> ChildrenMap a -> a -> b
childrenAggregate aggregatorFnc childrenMap = aggregatorFnc . children childrenMap

decendantAggregate :: (Eq a, Hashable a) => (b -> b -> b) -> (ChildrenMap a -> a -> b) -> ChildrenMap a -> a -> b
decendantAggregate resultFoldFnc nodeFnc childrenMap node =
foldl' resultFoldFnc nodeValue childResults
where
nodeValue = nodeFnc childrenMap node
childFnc = decendantAggregate resultFoldFnc nodeFnc childrenMap
childResults = map childFnc $children childrenMap node  The descendantAggragate recursively applies a function nodeFnc to a node node and all its descendants and folds the results using some function resultFoldFnc. This allows to define the necessary functions to count the total number of descendants of a node as follows. childrenCount :: (Eq a, Hashable a) => ChildrenMap a -> a -> Int childrenCount = childrenAggregate length decendantCount :: (Eq a, Hashable a) => ChildrenMap a -> a -> Int decendantCount = decendantAggregate (+) childrenCount totalDecendantCount :: (Eq a, Hashable a) => ChildrenMap a -> a -> Int totalDecendantCount = decendantAggregate (+) decendantCount  For part 2, we use that between two points in a tree, there is exactly one path (without repetition). First, we define a function to get a path from the root of a (sub)tree to the destination, provided it exists. pathFromRoot :: (Eq a, Hashable a) => ChildrenMap a -> a -> a -> Maybe [a] pathFromRoot childrenMap root destination | destination == root = Just [root] | null childPaths = Nothing | otherwise = Just$ root:(head childPaths)
where
rootChildren = children childrenMap root
pathFromNewRoot newRoot = pathFromRoot childrenMap newRoot destination
childPaths = mapMaybe pathFromNewRoot rootChildren


This function only finds paths down from the root of a (sub)tree. General paths come in three variations: path from the root of a (sub)tree, the inverse of such a path or the concatenation of a path to the root of a subtree and one from that root to the end point. Thus, we get the path as follows.

path :: (Eq a, Hashable a) => ChildrenMap a -> a -> a -> a -> Maybe [a]
path childrenMap root start end =
let maybeStartEndPath = pathFromRoot childrenMap start end
in if isJust maybeStartEndPath
then maybeStartEndPath
else let maybeEndStartPath = pathFromRoot childrenMap end start
in case maybeEndStartPath of
Just endStartPath -> Just $reverse endStartPath Nothing -> let rootPathToStart = pathFromRoot childrenMap root start rootPathToEnd = pathFromRoot childrenMap root end in if isNothing rootPathToStart || isNothing rootPathToEnd then Nothing else connectedPath (fromJust rootPathToStart) (fromJust rootPathToEnd)  To connect the paths in the last alternative, we follow both paths from the root to the last common point and then build it by concatenation the reverse of the path to the start with the path to the destination. connectedPath :: Eq a => [a] -> [a] -> Maybe [a] connectedPath rootToStart rootToEnd = case pathPieces of Nothing -> Nothing Just (middle, middleToStart, middleToEnd) -> Just$ (reverse middleToStart) ++ [middle] ++ middleToEnd
where pathPieces = distinctPathPieces rootToStart rootToEnd

distinctPathPieces :: Eq a => [a] -> [a] -> Maybe (a, [a], [a])
distinctPathPieces [x] [y] = if x == y then Just (x, [], []) else Nothing
distinctPathPieces (x1:y1:z1) (x2:y2:z2)
| x1 /= x2 = Nothing
| y1 /= y2 = Just (x1, y1:z1, y2:z2)
| otherwise = distinctPathPieces (y1:z1) (y2:z2)
distinctPathPieces _ _ = Nothing


This solution heavily depends on the input describing a tree. In case a DAG is provided, a result will be produced that is not necessary correct. For totalDescendantCount, nodes after joining branches will be counted multiple times and path will find a path, but not nescesarily the shortest one. If there are cycles in the graph provided, the recursions in the functions will not terminate.

## Simplification

In path, notice how the code gets more nested as you try each possible path (either from start to end, or end to start, or from end to root and root to start). You can use the Alternative instance for Maybe to simplify this code:

let maybeStartEndPath = pathFromRoot childrenMap start end
maybeEndStartPath = pathFromRoot childrenMap end start
maybeRootPath = [...] -- see below
in
maybeStartEndPath
<|> fmap reverse maybeEndStartPath
<|> maybeRootPath



This code will try maybeStartEndPath first. If it returns Nothing, it will move on to the next option and so on.

For your final case (which I've named maybeRootPath), you do the following check:

if isNothing rootPathToStart || isNothing rootPathToEnd
then Nothing
else connectedPath (fromJust rootPathToStart) (fromJust rootPathToEnd)



This is more consicely done with liftA2 from Control.Applicative. liftA2 lifts a binary function into an applicative context:

λ :set -XTypeApplications
λ :t liftA2 @Maybe
liftA2 @Maybe :: (a -> b -> c) -> (Maybe a -> Maybe b -> Maybe c)


Then, if either argument is Nothing, the function will return Nothing without having to pattern match. So we can fill in maybeRootPath above with

        maybeRootPath = join $liftA2 connectedPath rootPathToStart rootPathToEnd where rootPathToStart = pathFromRoot childrenMap root start rootPathToEnd = pathFromRoot childrenMap root end  The join is needed because connectedPath returns a Maybe already, and we've lifted it into Maybe, which leaves us with a return value of Maybe (Maybe [a]). join flattens nested monads, bringing us back to Maybe [a] ## Minor points Your function applyToSecondElement is second from Control.Arrow λ :t second @(->) second @(->) :: (b -> c) -> (d, b) -> (d, c)  toSingleElementList can also be written as (:[]) or return So orbitMap can be written orbitMap = Map.fromListWith (++) . map (second (:[]))  To your credit, your naming made both of these functions clear anyway, but it's more recognizable if you use functions that already exist. ## Algorithm I was going to suggest keeping each edge bidirectional instead of one-directional, so that you can directly check for a path from start to end instead of checking 3 cases. After reviewing the code, I think your approach is better from a functional perspective because it eliminates the need for you to check for cycles and keep a set as you search the graph. Good work. ## Revised Code import Control.Applicative import Control.Monad import Control.Arrow import System.IO import Data.List.Split import Data.List import Data.Maybe import Data.Hashable import qualified Data.HashMap.Strict as Map main :: IO () main = do inputText <- readFile "Advent20191206_1_input.txt" let orbitList = catMaybes$ (map orbit . lines) inputText
let orbits = orbitMap orbitList
let pathToSanta = fromJust $path orbits "COM" "YOU" "SAN" let requiredTransfers = length pathToSanta - 3 print requiredTransfers type OrbitSpecification = (String,String) type ChildrenMap a = Map.HashMap a [a] children :: (Eq a, Hashable a) => ChildrenMap a -> a -> [a] children childrenMap = fromMaybe [] . flip Map.lookup childrenMap orbit :: String -> Maybe OrbitSpecification orbit str = case orbit_specification of [x,y] -> Just (x, y) _ -> Nothing where orbit_specification = splitOn ")" str orbitMap :: [OrbitSpecification] -> ChildrenMap String orbitMap = Map.fromListWith (++) . map (second (:[])) childrenAggregate :: (Eq a, Hashable a) => ([a] -> b) -> ChildrenMap a -> a -> b childrenAggregate aggregatorFnc childrenMap = aggregatorFnc . children childrenMap decendantAggregate :: (Eq a, Hashable a) => (b -> b -> b) -> (ChildrenMap a -> a -> b) -> ChildrenMap a -> a -> b decendantAggregate resultFoldFnc nodeFnc childrenMap node = foldl' resultFoldFnc nodeValue childResults where nodeValue = nodeFnc childrenMap node childFnc = decendantAggregate resultFoldFnc nodeFnc childrenMap childResults = map childFnc$ children childrenMap node

childrenCount :: (Eq a, Hashable a) => ChildrenMap a -> a -> Int
childrenCount = childrenAggregate length

decendantCount ::  (Eq a, Hashable a) => ChildrenMap a -> a -> Int
decendantCount = decendantAggregate (+) childrenCount

totalDecendantCount :: (Eq a, Hashable a) => ChildrenMap a -> a -> Int
totalDecendantCount = decendantAggregate (+) decendantCount

pathFromRoot :: (Eq a, Hashable a) => ChildrenMap a -> a -> a -> Maybe [a]
pathFromRoot childrenMap root destination
| destination == root = Just [root]
| null childPaths = Nothing
| otherwise = Just $root:(head childPaths) where rootChildren = children childrenMap root pathFromNewRoot newRoot = pathFromRoot childrenMap newRoot destination childPaths = mapMaybe pathFromNewRoot rootChildren path :: (Eq a, Hashable a) => ChildrenMap a -> a -> a -> a -> Maybe [a] path childrenMap root start end = let maybeStartEndPath = pathFromRoot childrenMap start end maybeEndStartPath = pathFromRoot childrenMap end start maybeRootPath = join$ liftA2 connectedPath rootPathToStart rootPathToEnd
where
rootPathToStart = pathFromRoot childrenMap root start
rootPathToEnd = pathFromRoot childrenMap root end
in
maybeStartEndPath
<|> fmap reverse maybeEndStartPath
<|> maybeRootPath

connectedPath :: Eq a => [a] -> [a] -> Maybe [a]
connectedPath rootToStart rootToEnd =
case pathPieces of
Nothing -> Nothing
Just (middle, middleToStart, middleToEnd) ->
Just \$ (reverse middleToStart) ++ [middle] ++ middleToEnd
where pathPieces = distinctPathPieces rootToStart rootToEnd

distinctPathPieces :: Eq a => [a] -> [a] -> Maybe (a, [a], [a])
distinctPathPieces [x] [y] = if x == y then Just (x, [], []) else Nothing
distinctPathPieces (x1:y1:z1) (x2:y2:z2)
| x1 /= x2 = Nothing
| y1 /= y2 = Just (x1, y1:z1, y2:z2)
| otherwise = distinctPathPieces (y1:z1) (y2:z2)
distinctPathPieces _ _ = Nothing

• Thank you very much! I already had the feeling that there had to be some function to apply a function to the second element, but did not know Arrow. I will need some time to look things up in order to fully understand your simplifications. But now I know some directions to look into to improve my Haskell. Jan 13 '20 at 20:38