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
inputText <- readFile "Advent20191206_1_input.txt"
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
Next follow the functions I use to read in the tree.
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.
