Post-Christmas Advent of Code In Haskell - Day 2

Today’s post is about Day 2: “Inventory Management System” . We are given a file containing random looking strings and are asked to calculate some checksums and also find a certain pair among them..

Day 2 / Part 1

We have to calculate a checksum for the strings (IDs) in the input file. The checksum algorithm works as follows:

Checksum = Twos * Threes Twos = number of IDs that contain a letter exactly 2 times Threes = number of IDs that contain a letter exactly 3 times

It also provides an example:

"abcdef" -> no letters that appear exactly two or three times. "bababc" -> two a and three b, so it counts for both. "abbcde" -> two b, but no letter appears exactly three times. "abcccd" -> three c, but no letter appears exactly two times. "aabcdd" -> two a and two d, but it only counts once. "abcdee" -> two e. "ababab" -> three a and three b, but it only counts once. Checksum = Twos * Threes = 4 * 3 = 12

getOccurrences

In order to determine the checksum we have to look at each string separately and test both properties separately: (1) Does it contain any letter exactly twice? (2) Does it contain any letter exactly three times?

So getOccurrences should take a String and return something that conveys whether or not the String fulfills either or both of those properties..

getOccurrences :: String -> ( Bool , Bool )

Admittedly (Bool, Bool) is a type you should usually be avoiding since it is entirely meaningless when it appears without context. Luckily we aren’t writing production code today so I am just going to pretend this never happened and carry on.

We will assume the first value refers to letters appearing twice ( True –> appearing twice), and the second one to letters appearing three times ( True –> appearing three times).

How do we go about finding out if a string has any re-occurring characters? We can use some handy functions from Data.List , namely sort and group:

> import Data.List > (group . sort) "bababc" ["aa", "bbb", "c"]

We turned our String into a sorted list grouped by characters. In order to get to our desired result type we just have to check if this list contains strings of length 2 or 3 respectively:

ofLength n = filter (( == n ) . length )

We can use ofLength on our intermediate value from above:

> ofLength 2 ["aa", "bbb", "c"] ["aa"] > ofLength 3 ["aa", "bbb", "c"] ["bbb"]

(&&&)

What we want to do now is apply two functions ( ofLength 2 and ofLength 3 ) to one value (our intermediate result) and collect both results in a tuple. It just so happens that the Arrow function (&&&) does exactly what we want:

> import Control.Arrow > (+1) &&& (*1) 0 (1,0)

Note: I don’t know the first thing about the whole Arrow abstraction but I’ve seen (&&&) used here and there and it fits our purposes right now. Furthermore Advent of Code is the perfect opportunity to just play round with things like that. Anyhow, moving on:

ofLength 2 &&& ofLength 3 $ ["aa", "bbb", "c"] (["aa"], ["bbb"])

We are getting closer. We now have a tuple, but the type is still wrong. We are at ([String], [String]) instead of (Bool, Bool) . The values of our tuple should be True or False depending on whether or not the list is empty. Well that is easy enough:

> (not . null) ["foo"] True

Now we just have to apply this to both values of the tuple:

> fmap (not . null) (["aa"], ["bbb"]) (["aa"], True) -- Oops, right, this doesn't work..

Bifunctor

Functor won’t do since we want to map over both values of the tuple. bimap to the rescue:

> import Data.Bifunctor > :t bimap bimap :: (a -> b) -> (c -> d) -> p a c -> p b d > bimap (not . null) (not . null) (["aa"], ["bbb"]) (True, True)

Putting The Pieces Together

getOccurrences :: String -> ( Bool , Bool ) getOccurrences = bimap ( not . null ) ( not . null ) . ( ofLength 2 &&& ofLength 3 ) . group . sort

Now we just have to write a function that applies getOccurrences to all IDs, sums up, and finally multiplies the values.

calcChecksum

Let’s assume that calcChecksum will get the contents of the input file as input and we want to return the final checksum:

calcChecksum :: String -> Int

Our input is a file containing newline separated strings so we can start with something like this:

f :: String -> [( Bool , Bool )] f = fmap getOccurrences . lines

We split the String and apply our getOccurrences function to all IDs.

Where do we go from here? We have to count the True values from the first and the second value of all tuples in the list. Sure sounds like a fold to me, don’t you think? I tend to mess things up with folds so let’s sketch this out and have ghc help us out:

g = foldr _f ( 0 , 0 ) ( undefined :: [( Bool , Bool )])

ghc is going to report back:

• Found hole: _f :: (Bool, Bool) -> (a, b) -> (a, b) Where: ‘a’, ‘b’ are rigid type variables bound by the inferred type of it :: (Num a, Num b) => (a, b)

Alright, we can write something fulfilling that signature:

toNum x = if x then 1 else 0 sumUp ( f , s ) ( x , y ) = ( x + toNum f , y + toNum s )

Let’s try it:

> foldr sumUp (0,0) [(True, False), (True, False), (True, False)] (3,0)

Let’s add it to what we already have:

f :: String -> ( Int , Int ) f = foldr sumUp ( 0 , 0 ) . fmap getOccurrences . lines

Almost there! The only part missing is that we still need to multiply the first and second value of the tuple. We cannot just put multiplication in front of our composition because (*) expects 2 arguments where we just have a tuple. The answer to that is uncurry :

> :t (*) (*) :: Num a => a -> a -> a > :t uncurry uncurry :: (a -> b -> c) -> (a, b) -> c > :t uncurry (*) uncurry (*) :: Num c => (c, c) -> c

With that last bit we can complete our calcChecksum function:

calcChecksum :: String -> Int calcChecksum = uncurry ( * ) . foldr sumUp ( 0 , 0 ) . fmap getOccurrences . lines

Putting The Pieces Together

Here is the complete code for solving part 1 of this challenge:

ofLength n = filter (( == n ) . length ) toNum x = if x then 1 else 0 sumUp ( f , s ) ( x , y ) = ( x + toNum f , y + toNum s ) getOccurrences :: String -> ( Bool , Bool ) getOccurrences = bimap ( not . null ) ( not . null ) . ( ofLength 2 &&& ofLength 3 ) . group . sort calcChecksum :: String -> Int calcChecksum = uncurry ( * ) . foldr sumUp ( 0 , 0 ) . fmap getOccurrences . lines solvePart1 :: FilePath -> IO Int solvePart1 file = calcChecksum <$> readFile file

Day 2 / Part 2

The second part continues with the same input data. We are tasked with finding a pair of IDs differing in only 1 character. The result is said string with the differing character removed.

Let’s again start in the small and work our way up to the bigger picture. We are going to need a function to determine the distance between two strings - the number of differing characters between two strings:

strDist :: String -> String -> Int strDist [] _ = 0 strDist ( x : xs ) ( y : ys ) = if x == y then strDist xs ys else 1 + strDist xs ys

Let’s also quickly implement the function that we’ll need once we have our pair which drops the differing character:

dropEq :: String -> String -> String dropEq [] _ = [] dropEq ( x : xs ) ( y : ys ) = if x == y then x : dropEq xs ys else dropEq xs ys

Another simple, manual recursion. Now we need to start thinking about how we actually want to find our pair in the first place. The pair we are looking for could be between any two strings of our input. Thus let’s build a list of tuples representing all combinations. Haskell list comprehension comes in handy here:

getCombinations :: [ a ] -> [( a , a )] getCombinations xs = [( x , y ) | x <- xs , y <- xs ]

When fed with all IDs from the input file, the pair we are looking for is going to be one of the tuples in that list. We can find it by looking for the tuple where strDist yields 1 . Let’s put together what we have so far:

findPair :: String -> Maybe ( String , String ) findPair = find (( == 1 ) . uncurry strDist ) . getCombinations . lines

With that we can already find the tuple we are looking for! We only need to add one last transformation - we want a single string with the differing character omitted. We have already written dropEq :: String -> String -> String for that purpose. Note that we want to apply dropEq to two strings in a tuple so yet again we reach for uncurry.

findPair :: String -> Maybe ( String , String ) findPair = fmap ( uncurry dropEq ) . find (( == 1 ) . uncurry strDist ) . getCombinations . lines

Note how we need to use fmap as the tuple is wrapped in a Maybe

Putting The Pieces Together

Below is the full code for solving the second part of this challenge:

getCombinations :: [ b ] -> [( b , b )] getCombinations xs = [( x , y ) | x <- xs , y <- xs ] dropEq :: String -> String -> String dropEq [] _ = [] dropEq ( x : xs ) ( y : ys ) = if x == y then x : dropEq xs ys else dropEq xs ys strDist :: String -> String -> Int strDist [] _ = 0 strDist ( x : xs ) ( y : ys ) = if x == y then strDist xs ys else 1 + strDist xs ys findPair :: String -> Maybe String findPair = fmap ( uncurry dropEq ) . find (( == 1 ) . uncurry strDist ) . getCombinations . lines solvePart2 :: FilePath -> IO ( Maybe String ) solvePart2 file = findPair <$> readFile file

That’s All Folks

That’s it for Day 2. You can find the full code on github. If you have any feedback please don’t hesitate to reach out: @tpflug.