On the 1st of December 2018, I decided to give it a try to Advent of Code. AoC is, basically, a programming challenge website where you get two puzzles unlocked every day of December from 1st to 25th – hence the name. It has a ranking system which scores are based on the absolute time you took to solve the puzzles – i.e. spent time as soon as the puzzles got unlocked. As a French living in Paris, I find this a bit unfair (we get puzzles unlocked at around 5:00 AM!) and then I just realized I could do the puzzles for fun only.

This blog post sums up what I did with AoC#18, my thoughts about the puzzles and even a meta-discussion about programming challenges. I used Haskell for most challenges and switched to Rust for no specific reason. Just fun.

Also, because the puzzles’ texts are somewhat long, I will just give a link to the text instead of copying it and will just make a very short description of the problem each time.

Finally, don’t expect to find all the puzzles: I stopped at day 15. I will explain the reasons why at the end of this article. My adventure can be found here. Most puzzles have input data. You will find them in day directories as input.txt files.

Enjoy the reading!

If you want to try and take the challenges, I advise you not to read any further as this article would spoil you solutions! You will need a Github account and… a lot of time.

Day 1: Chronal Calibration

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Part 1

This puzzle is about summing arrays and finding duplicates (without counting).

First puzzle I discovered and first AoC challenge I ever took, I was surprised at the simplicity. Basically, given lines containing a single number preceded by either + or - , we must compute the sum of all the numbers.

The first thing to do is to parse the file into a stream of numbers. In Haskell, this was almost a one liner:

main :: IO () main = do numbers <- fmap (map parse . lines) getContents where parse ('+':xs) = read xs -- (1) parse n = read n

If I removed the + from the file, I could even have gone with a one liner:

numbers <- fmap (map read . lines) getContents

The read function is not safe and I don’t recommend using it for real, production projects. In the context of this challenge, though, it was alright.

I don’t have much to say about this puzzle as I found it not very interesting.

Part 2

We just want to get the first cumulative sum that appears twice. That is, each time a new number is summed, we get a given “current sum”. We add positive and negative integers, so we might end up, at some time, with the same number. Because we don’t know when that will happen, we need to turn the input stream into an infinite stream of numbers that repeat over and over. Haskell has the cycle function that does exactly that: it takes a list and makes it repeat infinitely.

What is great about cycle in Haskell is that it could be implemented in a very naive way and yet yield good performance – thanks GHC!:

cycle :: [a] -> [a] cycle x = let x' = x ++ x' in x'

Or, prefered (personal opinon):

cycle :: [a] -> [a] cycle = fix . (++)

fix is a very powerful function any Haskell should know. Typically, when you want to corecursively build something, it’s very likely you’ll need fix . I advise you to have a look at my netwire tutorial in which I used fix pretty much everywhere.

There are several ways to fix this problem. I chose to use a set to hold all the known intermediate sums. Because we’re working with integers, the Haskell IntSet type is perfect for that. The algorithm is then just a special fold over an infinite list that stops when a sum is seen twice by looking it up in the integer set:

findFreq _ freq [] = freq -- this shouldn’t ever happen since the list is infinite findFreq knownFreq freq (change:xs) | newFreq `member` knownFreq = newFreq | otherwise = findFreq (insert newFreq knownFreq) newFreq xs where newFreq = freq + change

Haskell solution

Day 2: Inventory Management System

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Part 1

This puzzle requires you to recognize and count characters in strings. For each string, you must check whether they have any letter that appears exactly once and any letter that appears exactly three times. Adding those counts provides a checksum for a list of strings.

I did that in Haskell as well. My idea was to use a count-array of size 26 (since the strings are only in [a-z] . So the first part of my algorithm is to count, for a given string, the number of time each letter occurs. This is done really easily with a fold that increments the proper value value in the array at the index representing the character folded on in the string. That index is obtained by turning a character into a numerical representation and subtracting it the numerical value of 'a' :

treatHash :: (Natural, Natural) -> String -> (Natural, Natural) treatHash occurs hash = let result :: State = foldl' replace initialState hash in addPair occurs (foldl' incrOccur (0, 0) result) where replace st l = accum (+) st [(ord l - ord 'a', 1)] incrOccur i@(twos, threes) x | x == 2 = (1, threes) | x == 3 = (twos, 1) | otherwise = i addPair (a, b) (c, d) = (a + c, b + d)

The replace function performs the update in an input vector (point-free function). This will effectively increment the vector’s value at index 3 if the character is d (because ord 'd' - ord 'a' = 3 . incrOccur , here, is used to fold over the resulting array ( Vector ) and compute if a letter appears twice and if a letter appears three times. You will notice that this function doesn’t sum. For instance, if you have aabb , you get no letter that appears three times but you have two that appear exactly twice. However, incrOccur will give you (1, 0) , because the puzzle states it should be done this way. Finally, addPair add the number of twos and threes to a pre-defined input – that will be handy for an outer fold.

The algorithm is then run on the input and we get the final result:

hashes <- fmap lines getContents let checksum = uncurry (*) $ foldl' treatHash (0, 0) hashes

We transform the standard input into lines – i.e. [String] – and then fold them with our treatHash function. Finally, uncurry (*) is a little oneliner to simply multiply the left part of a pair with its right part.

Part 2

This part is about finding strings that are almost the same and differ and only by one letter. As doing a programming challenge, the first thing I want to find is a solution, not the best solution. I also think people should really learn from this:

Make it work. Make it clean. Make it fast.

Especially on programming challenges, you’ll be very, very unlikely to implement anything more than (1.), because the inputs are often two small to really benefit from a real optimization. Your time is a real important resource, don’t waste it. Measure whether you really need an optimization. I’m not saying that you should be doing something bad because it’s easier. I’m saying that if it takes you N minutes to write a solution that runs in M milliseconds, if you know that you could do it in, for instance, 10N minutes to write a solution that runs in only 0,7M milliseconds for the foreseen milliseconds, will, you’re going to waste your time.

So, in my case, I started with a naive approach that runs in O(N²): comparing all strings to all others:

searchCommon :: [String] -> String searchCommon [] = "" -- the empty list has nothing in common searchCommon (h:hs) = search h hs -- for each string, search in all remaining where search h [] = searchCommon hs -- exhausted, search with the next string search h (x:xs) | almostSame h x = commonLetters h x -- grab the common letters if almost the same | otherwise = search h xs -- if not, just try the next string in the sub-list

The algorithm is pretty straight-forward. The searchCommon and search functions are mutually recursive functions that go from, respectively, the whole list of strings to test and the local tail as we advance. almostSame is defined as follows:

almostSame :: String -> String -> Bool almostSame = go 0 where go diffNb (l:ls) (r:rs) | diffNb > 1 = False | l /= r = go (succ diffNb) ls rs | otherwise = go diffNb ls rs go diffNb _ _ = diffNb == 1

This function is a special zip that short-circuits if it knows there are two many differences. When both the input strings are exhausted, if diffNb == 1 , then only one character has changed, so the the overwhole function evaluates to True .

The final part we need is commonLetters , that is pretty straight-forward, too:

commonLetters :: String -> String -> String commonLetters (l:ls) (r:rs) | l == r = l : commonLetters ls rs | otherwise = ls -- all the rest is identic, smartass

We construct a list that has the same characters as the input lists as long as they’re equal. As soon as they differ, we discard the character and just return any of the input lists – they’re equal. This function only works, obviously, if both the input strings are almost identical (only one character differs and they have the same length). The otherwise branch is a short-circuit optimization that prevents us from traversing the whole inputs after the difference.

Haskell solution

Day 3: No Matter How You Slice It

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Part 1

This problem was the first “challenging” as it required more thinking. The idea is that a very large area – the text gives us the hint it is at least 1000 inches on each side but truly, we do not need this information – must be sliced into several rectangular smaller areas (with edges aligned with X and Y axis). Obviously, some rectangular area might overlap and the aim of the first part of this problem is to find out the total overlapping area.

My idea was simple: we can trick and change the problem by discretizing it. This is something I have said already but most of AoC problems have hidden properties and hidden hints. By thinking more about the form of the solution, the area to find is expressed in inches². The rectangular areas are all lying on a 2D grid (1000×1000 at least, remember?) and their coordinates are always natural numbers (they belong to [0;+∞] ). Then, my idea was to state that 1 inch² is actually a single “cell” at any coordinate in that grid – imagine that as a pixel. 4 inches² would be a 2×2 region in that grad (yielding 4 cells).

So instead of using the general idea of an area (M×N for a rectangle which sides are M and N long wide), we can simply break a rectangular area into its most atomic components (1×1 cells)… and sum them up! We will effectively end up with the area of this area.

A more interesting property of my way of solving it: we can now have a big array mapping to each 1×1 cell the number of times a rectangle lies on it. When iterating through all the rectangles, we just break them into a list of 1×1 cells, look ingg up and updating the big count array. Once we’re done, we just have to filter that big array to remove any element which count is less or equal to 1. The remaining elements are 1×1 cells that contain at least two overlapping rectangles – we don’t really care about the number. We don’t actually care about those elements: the length of that filtered array is the area we are looking for, because it is the sum of 1×1 elements!

Converting a rectangular area – called claim in the text – into 1×1 cells is very easy in Haskell:

type Plan = Map (Int, Int) Natural -- the big array I told you type Claim = (Id, Pos, Area) type Id = Text -- the “name” of the fabric the rectangular area is about type Pos = (Int, Int) type Area = (Int, Int) claim1x1 :: Claim -> [Pos] claim1x1 (_, (line, row), (width, height)) = [(x, y) | x <- [line .. line + width - 1], y <- [row .. row + height - 1]]

As you can see, I chose to use a Map (Int, Int) Natural instead of an array to store the number of overlaps per 1×1 cell. That enables us not to care at all about the size of the grid (remember when I told you we don’t need the 1000 hint?).

The function that updates the Plan to take a claim into account is:

checkClaims :: [Claim] -> Plan checkClaims = foldl' (\p (x, y) -> checkInch x y p) mempty . concatMap claim1x1 checkInch :: Int -> Int -> Plan -> Plan checkInch line row = insertWith (+) (line, row) 1

Given a list of claims ( [Claim] ), checkClaims maps the claim1x1 function and concats the result, yielding a [Pos] list. That list is then folded over with the checkInch x y p function, that takes an empty map as initial value. checkInch just increment the value found in the map if it already exists; otherwise, it sets that value to 1 .

Finally, we need to compute the area:

overlappingInches :: Plan -> Int overlappingInches = length . M.filter (> 1)

As I told you, that is crystal clear: it’s just the length of the filtered map.

Part 2

This part is interesting also: you need to find out the Id of the claim that doesn’t overlap with any other claim. I will not go into too much details about the algorithm as it’s very similar to the previous one: instead of storing the number of overlaps by 1×1 cell, we store a Set Id , giving all claims that are overlapping – we can see that as a more general form of the first part. We also need a Map Id Natural that maps a fabric and the number of times it overlaps another. The fabric that doesn’t overlap any other is then easily identifiable within that map: it has its associated value set to 0 :

searchNonOverlapped :: [Claim] -> Maybe Id searchNonOverlapped claims = case M.toList filtered of [(i, _)] -> Just i -- the text supposes there’s only one _ -> Nothing where (_, overlapped) = indexClaims claims filtered = M.filter (== 0) overlapped -- hello

Haskell solution

Day 4: Repose Record

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Part 1

Aaaah, day 4… I really don’t get why I got so annoyed by this one. It is quite simple in theory. But that made me realize something I don’t really like about AoC: the hidden rules. You will see, if you read this whole article, that some puzzles require you to make very important assumptions about the input – and there’re a lot of assumptions you could make, so you have to make the right ones!

This puzzle is about guards that must protect a prototype manufacturing lab. You are given an unordered list of events that occur about the guards:

When they begin their shifts (you are given the ID of the guard here).

When they fall asleep.

When they wake up.

Each action gives you a timestamp so that you can re-create the historical events.

The goal of the part 1 is to find the guard that has the most minutes asleep, and especially, find the minute it is asleep the most – we must multiply the ID of the guard by the minute.

Obviously, the first part of this puzzle is to re-order the input to have it chronologically. Here is the setup code:

data Entry = Entry { timestamp :: LocalTime, action :: Text } deriving (Show) type GuardianId = Natural data Action = BeginShift GuardianId | WakeUp | FallAsleep deriving (Eq, Ord, Show) entryFromString :: Text -> Maybe Entry entryFromString s = case split (== ']') s of [timestamp, action] -> Just $ Entry (parse . unpack $ T.drop 1 timestamp) action _ -> Nothing where parse = parseTimeOrError False defaultTimeLocale "%Y-%-m-%-d %H:%M"

The parsing part is not really interesting as it’s just challenge code: nasty but working parsing code. :D

I then re-ordered the input with:

entries <- fmap (fromJust . traverse entryFromString . T.lines) T.getContents let byTimeSorted = sortBy (comparing timestamp) entries

By the way, that code made me want to tweet about how Haskell is actually pretty easy to read and reason about. Anyway.

The next part of the algorithm is to transform the entries into a list of timed action. I actually decided to stream it so that I could benefit from Haskell’s stream fusion – and because it’s so simple and transparent:

readGuardianId :: Text -> Natural readGuardianId = readT . T.drop 1 treatEntries :: [Entry] -> [(LocalTime, Action)] treatEntries = map $ \entry -> let time = timestamp entry in case T.words (action entry) of ["Guard", ident, "begins", "shift"] -> (time, BeginShift $ readGuardianId ident) ("falls":_) -> (time, FallAsleep) ("wakes":_) -> (time, WakeUp) _ -> error "lol"

That is like mixing streaming and parsing at the same time. Then, the core of my algorithm: dispatch the actions by guard. That is mandatory if we want to actually accumulate the “state” of a guardian (when they’re sleeping, waking up, etc.). Otherwise, we get interleaved results.

dispatchActions :: [(LocalTime, Action)] -> Map GuardianId [(LocalTime, Action)] dispatchActions = go mempty Nothing where go guardians _ ((t, action@(BeginShift gid)):xs) = go (insertAction gid t action guardians) (Just gid) xs go guardians jgid@(Just gid) ((t, action):xs) = go (insertAction gid t action guardians) jgid xs go guardians _ [] = fmap V.toList guardians go _ _ _ = error "dispatchActions: the impossible fucking occurred!" insertAction gid t action guardians = M.insertWith (flip (<>)) gid (V.singleton (t, action)) guardians

This is a by-hand fold that just applies the rule of beginning a shift (storing the ID of the guardian that went napping so that we can correctly dispatch the remaining events).

Then the tricky part:

type Minute = Natural type Minutes = [Minute] minutesCounts :: [(LocalTime, Action)] -> Minutes minutesCounts = go zeroMinutes Nothing where zeroMinutes = replicate 60 0 -- (1) asMinutes = todMin . localTimeOfDay -- the guard was sleeping go minutes (Just sleepTime) ((t, action):xs) = case action of BeginShift _ -> go minutes Nothing xs FallAsleep -> go minutes (Just t) xs -- not sure if that would even occur in the input WakeUp -> go (addSleepCount minutes (asMinutes sleepTime) (asMinutes t)) Nothing xs -- the guard was awake, so we’re only interested in when they go to sleep go minutes Nothing ((t, action):xs) = case action of FallAsleep -> go minutes (Just t) xs _ -> go minutes Nothing xs go minutes _ [] = minutes addSleepCount minutes sleepTime t = zipWith (+) minutes range -- (2) where -- this function is a bit hacky but it generates, for a given range of time, a list of 60 -- elements where the time period has 1 and all the other has 0 (I leave you to the -- exercise of making that a better function) range :: Minutes range = replicate sleepTime 0 <> replicate (fromIntegral t - sleepTime) 1 <> replicate (60 - t) 0

This big function generates a list which length is 60 – mapping the number of times a guard has passed sleeping at a given minute from midnight to 1:00 AM (see (1) and (2) ).

Finally, what we need is a way to compute frequencies – or counts. That is, given a list of anyting, compute that number of time a given anything happens in the list. I wrote a small utility function for that – I got inspired by [@jle], thanks!:

freqTable :: (Ord a) => [a] -> Map a Count freqTable = M.fromListWith (+) . map (,1)

Then, finding the guard that has slept the more and the minute is easy:

findMostOccurring :: Map a Count -> (a, Count) findMostOccurring = maximumBy (comparing snd) . M.toList -- and Haskell is hard?! ;) findSleepiest :: Map GuardianId [(LocalTime, Action)] -> (GuardianId, (Minute, Count)) findSleepiest = fmap (findMostOccurring . freqTable . spanIndex) . maximumBy (comparing $ sum . snd) . M.toList . fmap minutesCounts where spanIndex = concatMap (\(i, x) -> replicate (fromIntegral x) i) . zip [0..]

We first find the guard that has the most time asleep ( maximumBy (comparing $ sum . snd) . Then, we find the minutes at which they were asleep the most ( findMostOccurring ). We are given the guard ID, the given minute and the number of times they were asleep at that minute. Yay!

Part 2

For this part, we would like to know which guard is most frequently asleep on the same minute? We already have written all the code needed for that:

findMostFrequentlySleepy :: Map GuardianId [(LocalTime, Action)] -> (GuardianId, Minute) findMostFrequentlySleepy = fmap findMin . maximumBy (comparing $ maximum . snd) . M.toList . fmap minutesCounts where findMin = fst . maximumBy (comparing snd) . zip [0..]

Instead of summing, we find the maximum time a guard was asleep. Pretty easy.

Haskell solution

Day 5: Alchemical Reduction

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Part 1

That puzzle is very natural to solve in Haskell. You are given an ASCII string that contains only letters (lower case and upper case) that represent polymers. You must compute their final reduction by following some basic rules:

Two adjacent letters will cancel them out if they don’t have the same case (lower vs. upper) but are the same letter. For instance, aA and Bb cancel each other but not aa nor BB .

and cancel each other but not nor . This rule happens until no more reduction is possible. For instance, AbBa will first get reduced to Aa because aB will cancel out, but then Aa will also cancel out and we will be left with nothing.

You must give the number of units left in the final reducted polymer after all reductions have occurred.

As I said, that is very simple and elegant in Haskell:

reduce :: String -> String reduce = go [] where go [] (x:xs) = go [x] xs go a [] = a go (a:xs) (b:bs) | not (isLetter b) = go (a:xs) bs | (toLower a /= toLower b) || (isLower a && isLower b) || (isUpper a && isUpper b) = go (b:a:xs) bs | otherwise = go xs bs

I decided to use a zipper-like traversal. My idea is the following:

Read a character. If we have nothing previously seen, just place it in the previous list.

If we have something previously seen, check whether it reacts with the last previously seen character. If so, just drop it and drop the previously seen character, and go on with the next character.

When we have exhausted the input list, the previously seen list of characters is the final form of the reduction.

This algorithm allows me to reduce by doing a forwards-and-backwards kind of sweeping, yielding nice performance. Also, notice that the resulting list is reversed because of how we accumulate the seen characters. Because we don’t care about the order, we will not reverse it back to its original order.

The result is just the length of the output list.

Part 2

This part asks us to find the polymer that is the smaller if we remove one kind of unit (a single letter type). So if we remove a for instance, we must remove all a and A .

As there’re only 26 possible solutions (from a to z ), and because my solution to part 1 was already fast, I decided to go brute-force with this one: reducing the input string without a’s, reducing the input string without b’s, reducing without c’s, etc. And then simply take the shorter one.

bruteForce :: String -> Int bruteForce polymers = minimum (map length allReduced) where types = ['a' .. 'z'] allReduced = map (\c -> reduce $ filter (\x -> toLower x /= c) polymers) types

Winner winner chicken dinner.

Haskell solution

Day 6: Chronal Coordinates

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Part 1

That puzzle was one of the funniest I did. The idea is that, given an infinite 2D map, you are given a list of several points of interest (POIs) in the form of (x, y) coordinates. The goal, for this first part, is to find the largest zone in which all points have the same POI. What it means is that, given several POIs, every positions on the map has a nearest POI (it can have several if it’s at equal distance to several POIs – those must be discarded by the algorithm so that they do not count into any zone). Several positions with the same nearest POI and adjacent to each others form a zone, so that anyone in that zone knows that the nearest POI is the same accross all spanning positions of the zone – you can picture the zone easily as discs centered on the POIs, but deformed by other POIs.

The tricky part is that the map is infinite and POIs are scattered around it.

My idea was based on something I do a lot on my spare time with my 3D projects: compute AABBs. An AABB is an enclosing box in which all points lie. The idea is that its size must be as minimal as possible. An AABB, which stands for Axis-Aligned Bounding Box, is the minimal bounding volume that enclose a set of points and which has its edged aligned with the axis (in our case, the X and Y axis). By the way, an AABB in 2D is also called MBR, which stands for Minimum Bounding Rectangle. I chose AABB instead of MBR because I’m more used to work in 3D, but they’re essentially the same thing.

So, the first thing I wanted to do is to compute the AABB of all the POIs for a two reasons:

The most important one: it helped me reduce the problem space from an infinite space to a finite one: instead of having an infinite 2D map, I now have a finite rectangular 2D region to explore. Which is, you’ll have to admit, much more comfortable. ;)

If you read the puzzle’s text, you’ll have notice this: “Your goal is to find the size of the largest area that isn’t infinite.” What it means is that all the points outside of the AABB are in an area that is always infinite, so we will never be interested in those zones.



Ok, let’s see my code:

type Point = (Int, Int) -- Note to readers: this the way I like to encode AABB because it eases some operations on points. -- The lower point is a point that belongs to the AABB that satisfies the rule that no other point -- with different coordinate are lower than it. Same thing for upper. I also like that encoding -- because generating an AABB from a set of points is trivial. data AABB = AABB { aabbLower :: Point, aabbUpper :: Point } deriving (Eq, Show) findAABB :: [Point] -> AABB findAABB [] = error "nein" -- this will never be called, so we don’t care about type safety here findAABB (a:ls) = foldl' updateAABB (AABB a a) ls where updateAABB (AABB (lx, ly) (ux, uy)) (x, y) = AABB { aabbLower = (min (min lx x) lx, min (min ly y) ly), aabbUpper = (max (max ux x) ux, max (max uy y) uy) } -- This function gives me a list of points that are in the AABB. It actually gives me all the points -- the AABB wraps. aabbToStream :: AABB -> [Point] aabbToStream (AABB (lx, ly) (ux, uy)) = [(x, y) | x <- [lx .. ux], y <- [ly .. uy]] -- Test whether a point lies on any edges of the AABB. You’ll get why this function is important -- later. liesOnAABB :: Point -> AABB -> Bool liesOnAABB (x, y) (AABB (lx, ly) (ux, uy)) = x == lx || x == ux || y == ly || y == uy

I annotated the code with comments so that you can get what it’s for.

So, I create the AABB and then I call aabbToStream in order to get all points. You might already have guessed the next step: we are going to find the nearest POI to all the points. For this, I just went naive and just computed the Manhattan distance to all POI and kept the smallest. If we map that function to all coordinates generated by the AABB, we get the first part of the solution.

manhDist :: Point -> Point -> Int manhDist (a, b) (c, d) = abs (a - c) + abs (b - d) nearest :: Point -> [(Int, Point)] -> Maybe Int nearest p points = case sortBy (comparing snd) $ map (\(i, x) -> (i, manhDist p x)) points of [a] -> Just (fst a) a:b:_ -> if snd a == snd b then Nothing else Just (fst a) -- (1) _ -> error "nearest"

Here, (1) applies the rule I described earlier about at least two POIs at the same distance: we just discard the point and it doesn’t participate in creating any zone.

Then, how do we find the biggest area? Easy: we re-use our freqTable function from Day 4 to compute a frequency table! In my case, I just renamed that function freqs :

freqs :: (Ord a) => [a] -> Map a Natural freqs = fromListWith (+) . map (,1)

If we call that function on a list of [Int] , we end up with Map (Maybe Int) Natural that gives us the number of positions a given POI is the nearest. It’s perfect, because it’s exactly what we are looking for!

biggestArea :: [Maybe Int] -> Natural biggestArea = snd . maximumBy (comparing snd) . M.toList . freqs . catMaybes

Here, catMaybes just remove the Nothing case so that we go from [Maybe Int] to [Int] . We then find out which POI has the biggest number of nearest positions and we simply return it. Because those are positions, their sum is then the area of the zone: we’re done. Or almost. Remember that some zones have an infinite areas. Thanks to the AABB, it’s actually easy to find which ones: those have at least one point that lies on the AABB’s edges. We just have to iterate through all the points and black list some points:

blackListPoints :: [(Point, Maybe Int)] -> AABB -> Set Int blackListPoints points aabb = foldl' blacklist mempty points where blacklist blist (p, Just i) = if liesOnAABB p aabb then S.insert i blist else blist blacklist blist _ = blist

Part 2

The part 2 asks something different: now we want to find the area of the region containing all locations for which the total distance to all POI is less than a given constant ( 10000 ). My solution was actually way easier than expected, surprisingly:

safeArea = filter (\p -> sum (map (manhDist p) coords) <= 10000) points

Done. :)

Haskell solution

Day 7: The Sum of Its Parts

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Part 1

Here we go again: graph theory. Fortunately for us, it’s not a hard graph puzzle. That first part is to simply display a string that shows the order in which a graph must be traversed. If two nodes can be traversed at the same time, the node which letter comes first alphabetically is traversed first.

I’ll just show the traversal because the rest is not really interesting for that puzzle:

-- The graph encodes the relationship in reverse: it maps each node its list of dependencies. -- So if we have something like A -> [], it means that the A node doesn’t have any dependency. type Graph = Map Step (Set Step) type Step = Char -- Get the list of all available steps; i.e. they don’t have any dependency. getAvailable :: Graph -> [Step] getAvailable gr = [step | (step, set) <- M.toList gr, S.null set] -- Traverse the graph and get the ordered steps to go through. stepAvailable :: Graph -> [Step] stepAvailable gr = case sort (getAvailable gr) of [] -> [] (s:sx) -> s : stepAvailable (removeStep s gr) removeStep :: Step -> Graph -> Graph removeStep s = purgeDep s . M.delete s where purgeDep = fmap . S.delete

It’s a typical functional problem that gets solved very easily in Haskell.

Part 2

The second part is pretty interesting. Instead of stepping through all the steps sequentially, you ar given a pool of workers. It will take a given amount of time for a given worker to complete a task and able us to visit a given node in the graph. We have to guess how many time it will take to complete all of the steps.

I won’t post the code (it’s on GitHub if you want to have a look at it) as it’s a bit boring and the idea of my solution is enough. The concept is to have a stepped simulation (i.e. you perform a set of action in a given “round”, then repeat). In my case, each round is composed of several steps:

First, partition the current work load into a set of done tasks and running tasks. This is quite easy to do by just checking at the remaining time of each task. If the remaining time is 0 , then it’s done, otherwise it’s still running. Generate the time increment. This is the minimal duration until a next interesting action occurs (i.e. a task gets done). Nothing can happen below that duration. That value can easily be found by looking up the remaining duration of the running tasks and taking the minimum. If we still have running tasks, step forward (i.e. recursively call the same function) by advancing the current time by the time increment and removing the done tasks. The backlog, that is implicit, can be created by monoidal operations and is detailed in the code on GitHub. When the backlog gets empty, we have the final time and the answer to the initial question.

This part was interesting because it made me write a parallel graph traversal (a graph task scheduler) that could be used as base for a real and parallelized (I/O) task scheduler. Interesting stuff.

Haskell solution

Day 8: Memory Maneuver

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Part 1

In its simple form, this puzzle is not really interesting and I could definitely present you the solution in a single paragraph. However, I found it pretty fun to do so I’ll go a bit further.

The problem is the following: we are given a list of numbers that represent a data structure. That data structure is basically a tree with tagged metadata. The goal is to parse the list of numbers to generate a memory representation of that tree and compute checksums on it. The structure is:

A header, that has the number of children of a given node and the number of metadata entries.

Zero or many children.

At least one or many metadata entries.

The file format is made so that the data are nested. I’ll copy and explain an example:

2 3 0 3 10 11 12 1 1 0 1 99 2 1 1 2 A---------------------------------- B----------- C----------- D-----

Here, only the first line is present in the input file. The first 2 means that the first ( A ) node has two children (we don’t know anything about them yet) and the 3 means it has three metadata. Those are the header. Then, since it has two children, the next 0 is the start of the header of its first children ( B ), which has no child and three metadata ( 3 ). The next 10 , 11 and 12 are then those metadata (since it doesn’t have any child). This node is then complete. If you go back up in the tree, you know that A still has another child. So the next number, 1 , is the number of child of C and 1 its number of metadata. The next number 0 is the number of child of D and it has 1 metadata, which is 99 . C , as seen above, has one metadata, so 2 is C ’s metadata. Then, since A has three metadata, 1 , 1 and 2 are its.

Pfiou. Seems hard to read for a human, right? However, if you’re used a bit to recursive data structure and more specifically recursive parsers, this kind of encoding is actually pretty neat!

Let’s go and implement the parser of that tree. First, the structure. We will not need the header in the output (it’s only used for parsing), so we will not encode that directly (it’s still available as the length of the list of children and length of the list of metadata entries):

data Node = Node { nodeChildren :: [Node], nodeMetadata :: NonEmpty Natural } deriving (Eq, Show)

Pretty simple, right? This is a self-recursing data structure that is pretty simple and basic to any functional programmer.

The NonEmpty a data type, in Haskell, is a list that cannot have zero element. That is enforced at compilation as it’s impossible to create such a list without explicitly giving at least one element. All the operations defined on NonEmpty a respect that rule (for instance, removing an element from it might not remove anything if it has only one element – otherwise you’d break its invariant and Haskell is not Javascript. Lol.

So, how do we parse that? I’ve already spoiled you the solution: we need to implement a recursive parser. I know that because I’ve been using parsec for like 7 years now, so I’m pretty used to that kind of parsing and as you use it, you will quickly recognize when you can use such an idiom.

However, instead of using parsec directly, I will implement it myself with some very basic types every Haskellers know – if you don’t: go learn them! I’ll be using the State type only, which is basically just a recursive function used in a monadic fancy way:

-- A possible representation of the State monad is just a function that takes a value 's' and -- returns a new, altered 's' (we call that a state, but as you can see, it’s completely pure code, -- no mutation happens at all) and an output value 'a'. data State s a = State { runState :: s -> (s, a) }

The real State type is defined in the mtl Haskell library.

So here, you have the impression or illusion of s being a state, but what it truly is is just a value that is passed around to a recursive function – and plot twist: recursion is the way to implement locally-defined states, but I will not explain that (read my netwire tutorial and the Auto type, if you are interested).

The functor / monadic part:

instance Functor (State s) where fmap f = State . fmap (fmap f) . runState instance Applicative (State s) where pure x = State (, x) p <*> q = State $ \s -> let (s', f) = runState p s (s'', a) = runState q s' in (s'', f a) instance Monad (State s) where return = pure q >>= f = State $ \s -> let (s', a) = runState q s in runState (f a) s'

All of this can be generated automatically by GHC with deriving data annotation.

And some combinators we’ll need:

-- Get the current value of the “state”. get :: State s s get = State $ \s -> (s, s) -- Get the current value of the “state” with a function pre-applied to it. gets :: (s -> a) -> State s a gets = flip fmap get -- Change the value of the “state” by applying a function to the state. modify :: (s -> s) -> State s () modify f = State $ \s -> (f s, ()) -- Just a convenient method to just get the output value and discard the final state. You need the -- initial value to use as state. evalState :: State s a -> s -> a evalState st = snd . runState st

So basically, since this is a very basic and simple code (I think all Haskellers should write that in their first month using Haskell, it’s a good exercise), I just included the mtl library and used its State type to write my recursive parser.

This is my parser:

newtype Parser a = Parser { runParser :: State [Natural] a } deriving (Applicative, Functor, Monad)

So basically, a Parser a generates value of type a and maintains a list of Natural around. Those Natural are the numbers from the input we are going to parse. Let’s write the actual parser now.

-- Turns the (string-encoded) list of numbers and generates the root node, that contains all of -- the children. parse :: String -> Node parse = evalState (runParser parseNode) . map read . words -- Read a single number from the input and consume it from the state. readInput :: Parser Natural readInput = Parser $ gets head <* modify tail parseNode :: Parser Node parseNode = do -- We read the two first numbers (header) childrenNb <- readInput metadataNb <- readInput -- Recursive parser! The NE.fromList is an unsafe function that is used for convenience for this -- puzzle part. children <- replicateM (fromIntegral childrenNb) parseNode metadata <- fmap NE.fromList (replicateM (fromIntegral metadataNb) readInput) pure $ Node children metadata

As you can see, the parser code is extremely simple with a recursive combinator parser! And we’re actually done for the first part. The checksum is simple and is:

checksum :: Node -> Natural checksum node = metadataChecksum (nodeMetadata node) + sum (map checksum $ nodeChildren node) metadataChecksum :: NonEmpty Natural -> Natural metadataChecksum = sum . NE.toList

Part 2

The second part is not interesting as it just requires a new method to compute the “value” of a given node:

nodeValue :: Node -> Natural nodeValue (Node [] metadata) = metadataChecksum metadata nodeValue (Node children metadata) = sum [nodeValue n | Just n <- map index (NE.toList metadata)] where index i = let i' = fromIntegral i - 1 in if i' < 0 || i' >= length children then Nothing else Just (children !! i')

Haskell solution

Day 9: Marble Mania

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Part 1 & 2

This puzzle was the first one when I decided to go full Rust! All the remaining puzzles were solved in Rust – if you were reading only for Haskell, sorry for your loss. :(

This puzzle is not really interesting as it’s just a fancy algorithm that adds element to a collection and sometimes removes from it. There was a trick, though: the second part requires to run our algorithm on an input that was a hundred times larger.

The typical trap is that when you add value in the middle of a collection, the complexity in terms of memory and CPU can largely vary. Everything depends on what you do. For very rare additions / deletions, it’s possible that you can accept O(n) complexities. However, if you often insert stuff, you might want something else. In the same spirit, some data structure can efficiently add in O(1) at the beginning or end of the collection or might require a complete copy.

Even though the puzzle is not interesting in itself, it reminds us how crucial and critical it is that a programmer must know what structure to use depending on the inputs and operations that will be performed on the data. In our case, we are going to add and remove a lot at arbitrary places in the collection. Vectors are really bad candidates at that kind of operations, because they will require a complete scan of the right part of the collection, which is O(n), every time you add or delete something (to shift right / left, respectively). This is bad. Vectors are also bad when you want to add at its beginning (it requires the same right shift as the random case).

Double-ended queue ( VecDeque in Rust) are a solution to the problem to insert at the beginning. That insertion is O(1) amortized.

My solution to this was to use a zipper to stay focus on a given number in the collection but also “move around” in O(1). The idea is the following:

struct Zipper { left: VecDeque<isize>, middle: isize, right: VecDeque<isize>, }

When you want to move to the left, you just take the middle number and push_front it to the right double-ended queue and you pop_back the left one and that value becomes the new middle . You do the opposite thing to go to the right.

To insert an element at the current location, you just push_front middle to the right and then middle is assigned the new value. To remove, you just pop_front right into middle .

Then, the all puzzle is just adding and removing according to a predicate. Since all the operations I mentioned above run in O(1) amortized (they might allocate if the buffer is too small), we will not suffer from the typical O(n²) complexity a Vec implementation has.

Rust solution

Day 10: The Stars Align

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Part 1

This is the kind of problem I suck the most at. Not because they’re hard. Because they’re easier than expected. As an engineer, I tend to overthink about the context, the input’s hidden properties, the possible errors, the heuristics, what could go wrong, etc. On a regular job basis, that is actually a good thing – it’s better to foresee things than to have to repair them. However, at a given extreme, that way of thinking will make you go nuts and will make you think of an easy and straightforward problem as a complex and convoluted one. I know that and the main thing my mind does when solving a problem is to think about how really hard a problem is. I just completely failed on this one, haha. :D

So, you are given a short list (~360) of 2D points. You know nothing about how they’re spread nor the limits they lie in. You only have ~360 points on a 2D plane. Those points, however, come with two attributes:

Their positions, as a pair of relative numbers ([ -∞; +∞] ).

). Their (constant) velocities, as a pair of relative numbers as well ([ -∞; +∞] ).

So basically, each point starts at a given position and goes into a straight line forever. The unit of the velocity is not known and is not really needed – even though it might be unit/s . Nevertheless, the text gives the hinting that, at a given (unknown) time, the points form a message that can be visually witness. The goal is to give the message.

In the first 15 minutes, I went through several thoughts. “Whoa, they want us to write an OCR?! Really?!”. Nah, you dumbass. Since we all have a unique input (and hence, a unique expected output), we don’t have to write an algorithm that can recognize any text. We just have to get to visualize our input.

However, when does the text appear? At t = 0 , we have a large cloud of spread points. We don’t know when the text will form. Also, the challenge explicitly states that the text forms only once: the points will never gather into text afterwards. We must not miss it then.

My idea was that to find hidden properties of the overall text first. By being able to extract a useful information telling me whether or not I’m far or close from having a visualizable text, I was able to run a loop-like simulation, moving each points by its respectiv velocities, until that hidden information reaches a local minimum. As an engineer, I was annoyed by that, because I had no idea whether the first local minimum was the right one – the puzzle’s text doesn’t state anything about that and I had not found any information to help with that in the input. I could also use the wrong criteria (maybe we’re looking for a local maximum?). I got stuck with those ideas for long minutes.

Finally, I decided to implement a specific criteria:

At each simulation step, we compute the AABB that encloses all the input points that have moved.

If that AABB’s area is less then the previous one, we store the area, the width and height of the AABB and the current time of the simulation. We also print those values on the standard output.

If not, we continue.

When I ran that loop, I got the first local minimum in 10011 seconds. Clearly, if you tried to actually run that simulation with the real time, you’d be waiting for a long time – 10011 seconds is 2 hours, 46 minutes and 51 seconds.

The size of the AABB at t = 10011 was also pretty small (around 60×60). I then decided to display the message directly in the console. In order to do that, I had to transform my 2D points (expressed in the natural ℝ² basis we use in world space coordinates) into a space that I could easily use to display (basically, [0; w] and [0; h] ). That transformation is done with the following code:

// The rendered “map” let mut lol = vec!['.'; w as usize * h as usize]; for p in points { let x = (p.position.0 - aabb.lower.0) * (w - 1) / w; let y = (p.position.1 - aabb.lower.1) * (h - 1) / h; let o = x + y * w; lol[o as usize] = '#'; }

Then, we just need to iterate on all the points and render them to the terminal to finish the challenge:

for row in 0 .. h { for col in 0 .. w { print!("{}", lol[(col + row * w) as usize]); } println!(""); }

Part 2

Part 2 was almost a joke: we were asked to give the time at which the text appeared. As this was a hidden property to find in the first place, completing part 2 took a few seconds: 10011 .

Rust solution

Day 11: Chronal Charge

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Part 1

A pretty common algorithm to implement: sliding window. Basically, you are given a matrix of numbers and you have to compute several sums using a sliding kernel which size is 3×3. The size of the matrix is 300×300 and you just want to compute the biggest 3×3 square (and give its index in the matrix as row / column).

This was my solution:

let mut largest = (0, i8::min_value()); // (index, power) // 298 is 300 - 2: we want to stop there so that the 3×3 square won’t overflow for row in 0 .. 298 { for col in 0 .. 298 { let mut power = 0; // sum the square for i in 0 .. 3 { for k in 0 .. 3 { power += grid[index(col + i, row + k)]; } } let i = index(col, row); // if its power is any larger, store it along with its index if (power == largest.1 && i < largest.0) || power > largest.1 { largest = (i, power); } } } println!("Largest fuel cell: ({}, {})", 1 + largest.0 % 300, 1 + largest.0 / 300);

That’s pretty much it. Second part is more interesting.

Part 2

For this part, the problem changes a bit: we still want to sum squares, but we want to get the find the square that has the largest total power of any size comprised between 1×1 and 300×300 – we want its index and its size.

That problem can be solved in several ways, with different complexities. It’s easy to see that you can quickly go with a bad complexity if you decide to refactor the previous algorithm to take a dimension (that will be squared) and call it 300 times. Maybe that would be enough.

However, I wanted to implement something smarter on this one. It’s easy to see that a lot of spanning squares will overlap. For instance:

+-+-+-+-+-+ |0|1|2|3|4| +-+-+-+-+-+ |6|7|8|9|A| +-+-+-+-+-+ |B|C|D|E|F| +-+-+-+-+-+

If you consider the first, top-leftmost 2×2 square:

+-+-+ |0|1| +-+-+ |6|7| +-+-+

And the top-left-mostmost 3×3 square:

+-+-+-+ |0|1|2| +-+-+-+ |6|7|8| +-+-+-+ |B|C|D| +-+-+-+

You can see that a the smaller one is included in the bigger one. What it means is that each spanning square is a partial sum to spanning square of a higher dimension. My algorithm benefits from that in order to reduce the number of elements to sum at each given dimension.

Also, another thing I did that suprised people on IRC: I reversed the way the algorithm works in terms of traversal. Instead of traversing dimensions and then traversing the grid (for all dimensions then for all squares), I traverse the grid and then I traverse the dimensions (for all square in the grid, for all the dimensions). This gives me a more natural way to write the partial sums in my code.

Finally, I also work out some formalæ to know “what’s the biggest dimension we can go up to given the current grid cell.” Yeah, think twice: when you want to go through all dimensions from the top-leftmost cell, you will be able to sum squares from 1×1 up to 300×300. But which dimension can you go to when the starting (1×1) cell is in the middle of the grid? This is actually pretty easy. The formalæ can be found very quickly by thinking in terms of size of the grid (300×300) and the index of a grid cell. The biggest dimension is just the minimum of the maximal allowed row iterations and the maximal allowed column iterations. You can picture that mentally by “which edge I am the closest to?”. For rows, it’s simply 300 - current_row and for columns, 300 - current_column . The minimum value gives you the maximal spanning dimension you can go up to.

Finally, a word on how the partial sums are created: when you start computing the sum of the N dimension, you already have the partial sum of dimension N-1 (the algorithm starts with the first dimension set to a given value). Then, instead of summing N² element, since you already have the sum of (N-1)² , you just need to sum 2 × (N - 1) + 1 values. If you’re not convinced, at dimension 278 , 278² = 77284 sums while my algorithm is 2 × (278 - 1) + 1 = 555 sums. It’s around 139 times less.

In my code, I do that by adding – relative to the previous spanning square – the right column (which size is N - 1), the bottom line (N - 1 as well) and the single element in the diagonal. Hence 2 × (N - 1) + 1 . And that completes a new partial sum, that will be used for higher dimensions!

Here’s just a very quick schema to show you how to compute the sum at dimension 5 by using the sum of the spanning square of dimension 4 – · is already computed and R are the right column, B the bottom line and D the element in the diagonal:

+-+-+-+-+-+ |·|·|·|·|R| +-+-+-+-+-+ |·|·|·|·|R| +-+-+-+-+-+ |·|·|·|·|R| +-+-+-+-+-+ |·|·|·|·|R| +-+-+-+-+-+ |B|B|B|B|D| +-+-+-+-+-+

So, here’s the code:

let mut largest2 = (0, i64::min_value(), 0); // (index, power, dimension) // for all rows… for row in 0 .. 300 { let max_iter_row = 300 - row; // 300 -> 1 // for all columns… for col in 0 .. 300 { let max_iter_col = 300 - col; // 300 -> 1 let max_dim_squared = max_iter_row.min(max_iter_col); // 300x300 -> 1x1 // power used for nested dimensions let mut nested_power = grid[index(col, row)] as i64; // note: we don’t have to compute the first dimension because it’s set right away to the given // nested power // for all dimensions up to the max for d in 1 .. max_dim_squared { let mut power = nested_power; // compute the 2 × (N - 1) elements for k in 0 .. d { power += grid[index(col + d, row + k)] as i64; power += grid[index(col + k, row + d)] as i64; } // add the diagonal power += grid[index(col + d, row + d)] as i64; let i = index(col, row); if (power == largest2.1 && i < largest2.0) || power > largest2.1 { largest2 = (index(col, row), power, d + 1); } nested_power = power; } } } println!("Largest fuel cell of all: ({}, {}, {}, of power {})", 1 + largest2.0 % 300, 1 + largest2.0 / 300, largest2.2, largest2.1);

Rust solution

Day 12: Subterranean Sustainability

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Part 1

This puzzle looked a bit like the double-ended queue one from day 9. The extra bit of information is that you now have to apply a pattern on several values to know how they should be mutated. Given a list of flower pots and some rules that give you how a pot should grow flowers (or not) according to the state of itself and its neighbors, the goal is to predict the sum of the pots (their index in the list) for all pots that contain flowers after 20 generations.

In the first place, I had to recognize that I needed a double-ended queue. As always, the puzzle’s text doesn’t explicitly tell you that the pots at leftmost and rightmost positions can “spawn” new pots by applying the rules on empty pots (infinite). I was confused at that for a while.

My encoding of rules is pretty simple and wasteful: since a rule gives you a pattern (which pots) and an output (should have flowers / shouldn’t), a single byte should be enough for that (the length of a rule is five: it gives you the state of the two left neighbors, the state of the current pot and the state of the two right neighbors). However, I encoded those with a simple array of five binary states ( Rule::Empty and Rule::Pot ).

In order to apply a pattern, we must retreive the current pot and two at its left position and two at its right position (if pots are missing because we’re at edges, we must spawn empty pots with negative / positive indices). Then we can just apply the pattern by looking it up in a hashmap: we get the next generation value.

Nothing really interesting code-wise to show here.

Part 2

Part 2 is funny. We’re told that instead of finding the sums after 20 generations, we need to find it after fifty billion (50000000000) generations. Obviously, trying to run the above algorithm for 50000000000 generations will take ages, so we need to find a better way.

My first inital idea was that if I took a look at the actual sum value at each generation, I could – perhaps – see some kind of patterns. At first I was looking for cycles and hence cycling sums. I then run my algorithm and had a look at the output data. I was suprised to find that, very quickly, the flowers grow linearily. What it means is that, after a given number of generations, you can guess how many flowers there will be at a given future generation by applying a linear formula (typically, a simple multiplication and addition).

In my case, I noticed that at generation 100 , the sum was 6346 . At 101 , it was 6397 . At 102 , it was 6448 . At 200 , it was 16546 . You can see the pattern – if you don’t, compute the difference between the sum at 101 and the sum at 100 … and the difference of sum at 102 and 101 .

Hence, I came up with the following linear formula:

// O(1) get the score at a given generation – works only for gen ≥ 100. fn score_at(gen: u64) -> u64 { 6346 + (gen - 100) * 51 }

The actual implementation uses 101 instead of 100 because we want to get the sum after a given number of generations, not at.

That kind of linear optimization was really fun to write yet a bit tricky to find. :)

Rust solution

Day 13: Mine Cart Madness

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Part 1

I think this was the puzzle I enjoyed the most – among the ones I did. The goal is to parse is rails map on which wagons go and make wagons move around by respecting some specific rules: we must find wagon collision is report their positions 2D position.

Parsing is actually pretty simple: the input data is the 2D map that contains the rail system along with the initial position of wagons. About the map, I decided to store only crossings, not the actual rails, because I didn’t need them! So in order to do so, I changed a bit the regular way I encode 2D maps in memory and decided to use a hashmap!

#[derive(Clone, Copy, Debug, Eq, Hash, PartialEq)] enum Rail { Cross, // + RampRight, // / RampLeft // \ } struct Map(HashMap<(u32, u32), Rail>);

The rest of the code is actually pretty simple: there are several functions, one for moving carts, one for changing directions at cross, one for detecting collision. A main loop is responsible in moving carts and checking if there’s any collision. If no collision is detected, we just loop back up. If a detection is detected, we break the loop and display the position of the crash. The collision algorith returns the IDs of the carts that collided into each other.

let collision = loop { // sort the cart by y component carts.sort_by(|a, b| a.pos.cmp(&b.pos)); let collisions = move_carts(&map, &mut carts); if !collisions.is_empty() { break collisions[0]; } }; println!("First collision: {:?}", carts[collision.0].pos);

The sort_by is needed because of priority rules in the puzzle’s text.

Moving carts and detecting collision is pretty straightforward:

fn move_carts( map: &Map, carts: &mut Vec<Cart>, ) -> Vec<(usize, usize)> { let mut collisions: Vec<(usize, usize)> = Vec::new(); // no collision to begin with // move and check collision for all carts 'outer: for i in 0 .. carts.len() { // check that this cart hasn’t been collided into yet for &collision in &collisions { if i == collision.0 || i == collision.1 { // already collided, don’t move that continue 'outer; } } // move the cart and check if it’s collided into another move_cart(map, &mut carts[i]); let collision = find_collision(&carts, i); if let Some(collider) = collision { collisions.push((collider, i)); } } collisions }

This code is not really optimized – we redo the same thing very often – but it’s way than enough to solve that puzzle’s part. Finding collision is very simple: we just try to find a cart with the same position.

The tricky part is for moving at cross. The rules state that if you arrive at a cross, you have to turn in a given direction and change the future direction you will take at the future cross, if any. This was encoded inside each cart, so that they have a “memory” of turns to take.

#[derive(Debug)] struct Cart { pos: (u32, u32), dir: Direction, next_turn: Turn }

A cart starts by going on its (relative!) Turn::Left , then at the next turn it will go Turn::Straight , then Turn::Right and finaly will loop back to Turn::Left . Note how different it is to Direction : a Turn is relative to the current movement of a cart while a Direction is absolute (at first, I wanted to have North , East etc. for Direction so that confusion is not possible).

Part 2

In that part, we want to find the last standing cart, assuming that crashing carts are immediately removed from the map. The code is actually very similar: instead of breaking the loop at the first collision, we break it when there’s only one cart left on the map – we don’t forget te remove the crashed carts!

loop { // sort the cart by y component carts.sort_by(|a, b| a.pos.cmp(&b.pos)); for (ci, ck) in move_carts(&map, &mut carts) { carts = carts.into_iter().enumerate().filter(|&(i, _)| i != ci && i != ck).map(|(_, c)| c).collect(); } if carts.len() == 1 { break; } }; println!("Last standing cart: {:?} ", carts); // this contains only one cart

Rust solution

Day 14: Chocolate Charts

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Part 1

Very similar to the double-ended queue puzzle as well, this one doesn’t actually require any deletion, just indexing correctly into a growing buffer. Nothing really interesting to show about this problem, except maybe the way recipes are created.

To create new recipes, the two Elves combine their current recipes. This creates new recipes from the digits of the sum of the current recipes’ scores. With the current recipes’ scores of 3 and 7, their sum is 10, and so two new recipes would be created: the first with score 1 and the second with score 0. If the current recipes’ scores were 2 and 3, the sum, 5, would only create one recipe (with a score of 5) with its single digit.

In order to implement that, I recognized that I will always create at least one recipe: 0 + 0 = 0 , and as soon as I create two recipes, the other one will always be 1 , because the maximum value is 9 + 9 = 18 (1 and 8) . Here’s my code that gets those two recipe numbers:

fn create_new_recipe(a: usize, b: usize) -> (Option<usize>, usize) { let s = a + b; let x = s / 10; let y = s % 10; (if x == 1 { Some(x) } else { None }, y) }

Part 2

Part 2 is really not interesting as it’s just using ends_with to find a suffix. I’ll let you read the code if you’re interested.

Rust solution

Day 15: Beverage Bandits

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Part 1

Aaaaaaaaaand this is the last puzzle I attempted. I actually decided not to finish it because it was taking me time. I will tell you more about that in the conclusion.

The goal is to write a simulation of elves fighting goblins (or goblins fighting elves) and finding paths in a map that has holes / mountains in it. So most of the code to write is about Dijkstra or A*. The puzzle seemed interesting but it was way too much for my spare time to spend on. I advise you to have a look at the insanely long puzzle’s text – that will give you an idea of everything you need to implement in order to get the your solution working.

Conclusion

Ah, my first Advent of Code. It was both interesting, exciting, frustrating and time-consuming. I found several pros and drawbacks:

Pros, first:

It’s all fun. Since I live in France, I couldn’t compete with people who get the puzzles unlocked in the morning around 10:00 AM while they were unlocked around 5:00 AM in France – I have a job, I have to, you know… SLEEP! So I just did it for fun.

Writing algorithms to solve math and operational problems is always interesting and makes it a great training.

Comparing solutions on IRC was also pretty fun!

And drawbacks:

The fact west-europeans cannot compete (the timezone issue).

The difficulty of the puzzles is not correctly balanced. Some puzzle are insanely simple to solve and some others take you hours (if not days if you don’t have a lot of spare time). Sometimes I felt frustrated at this, because I’ve been lacking sleep time for weeks and had to go to bed in order not to destroy my physical health. I’m curious to hear about other west-europeans: how did you manage your time, social life and work life with AoC?

Some problems had a very long first part 2 and the second part was really stupid (read The Stars Align – Part 2). I would have preferred to have more consistency or more parts, for instance.

My general feeling is that it was fun, but I think that I won’t do it next year, because I had to put all my spare projects on hold for that. I didn’t learn anything new – all the algorithms had me write algorithms I already knew, except maybe the partial dimension squared algorithm I “invented”: someone told me that it’s very similar to a real and well-known algorithm! How funny is that! The algorithm is Summed-area table and my solution is, indeed, very similar to it. But the thing is: I came up with the idea, and this is priceless for training brains!

Now I’ll return to my graphics, Rust and over experiment projects of mine! I hope you liked that article, it was a bit long (it took me almost two weeks to write!) but I felt I needed to make it. To… well… scrap and forget about Advent of Code and all my spare time I didn’t use for my own projects. :)

Keep the vibes – especially you, @lqd.