Day 4: Ceres Search
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FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
Haskell
Popular language this year :)
I got embarrassingly stuck on this one trying to be clever with list operations. Then I realized I should just use an array…
import Data.Array.Unboxed (UArray)
import Data.Array.Unboxed qualified as A
import Data.Bifunctor
readInput :: String -> UArray (Int, Int) Char
readInput s =
let rows = lines s
n = length rows
in A.listArray ((1, 1), (n, n)) $ concat rows
s1 `eq` s2 = s1 == s2 || s1 == reverse s2
part1 arr = length $ filter isXmas $ concatMap lines $ A.indices arr
where
isXmas ps = all (A.inRange $ A.bounds arr) ps && map (arr A.!) ps `eq` "XMAS"
lines p = [take 4 $ iterate (bimap (+ di) (+ dj)) p | (di, dj) <- [(1, 0), (0, 1), (1, 1), (1, -1)]]
part2 arr = length $ filter isXmas innerPoints
where
innerPoints =
let ((i1, j1), (i2, j2)) = A.bounds arr
in [(i, j) | i <- [i1 + 1 .. i2 - 1], j <- [j1 + 1 .. j2 - 1]]
isXmas p = up p `eq` "MAS" && down p `eq` "MAS"
up (i, j) = map (arr A.!) [(i + 1, j - 1), (i, j), (i - 1, j + 1)]
down (i, j) = map (arr A.!) [(i - 1, j - 1), (i, j), (i + 1, j + 1)]
main = do
input <- readInput <$> readFile "input04"
print $ part1 input
print $ part2 input
I struggled a lot more when doing list slices that I would’ve liked to
Haskell
import Data.List qualified as List
collectDiagonal :: [String] -> Int -> Int -> String
collectDiagonal c y x
| length c > y && length (c !! y) > x = c !! y !! x : collectDiagonal c (y+1) (x+1)
| otherwise = []
part1 c = do
let forwardXMAS = map (length . filter (List.isPrefixOf "XMAS") . List.tails) $ c
let backwardXMAS = map (length . filter (List.isPrefixOf "XMAS") . List.tails . reverse) $ c
let downwardXMAS = map (length . filter (List.isPrefixOf "XMAS") . List.tails ) . List.transpose $ c
let upwardXMAS = map (length . filter (List.isPrefixOf "XMAS") . List.tails . reverse ) . List.transpose $ c
let leftSideDiagonals = map (\ y -> collectDiagonal c y 0) [0..length c]
let leftTopDiagonals = map (\ x -> collectDiagonal c 0 x) [1..(length . List.head $ c)]
let leftDiagonals = leftSideDiagonals ++ leftTopDiagonals
let rightSideDiagonals = map (\ y -> collectDiagonal (map List.reverse c) y 0) [0..length c]
let rightTopDiagonals = map (\ x -> collectDiagonal (map List.reverse c) 0 x) [1..(length . List.head $ c)]
let rightDiagonals = rightSideDiagonals ++ rightTopDiagonals
let diagonals = leftDiagonals ++ rightDiagonals
let diagonalXMAS = map (length . filter (List.isPrefixOf "XMAS") . List.tails) $ diagonals
let reverseDiagonalXMAS = map (length . filter (List.isPrefixOf "XMAS") . List.tails . reverse) $ diagonals
print . sum $ [sum forwardXMAS, sum backwardXMAS, sum downwardXMAS, sum upwardXMAS, sum diagonalXMAS, sum reverseDiagonalXMAS]
return ()
getBlock h w c y x = map (take w . drop x) . take h . drop y $ c
isXBlock b = do
let diagonal1 = collectDiagonal b 0 0
let diagonal2 = collectDiagonal (map List.reverse b) 0 0
diagonal1 `elem` ["SAM", "MAS"] && diagonal2 `elem` ["SAM", "MAS"]
part2 c = do
let lineBlocks = List.map (getBlock 3 3 c) [0..length c - 1]
let groupedBlocks = List.map (flip List.map [0..(length . head $ c) - 1]) lineBlocks
print . sum . map (length . filter isXBlock) $ groupedBlocks
return ()
main = do
c <- lines <$> getContents
part1 c
part2 c
return ()
J
Unsurprisingly this is the kind of problem that J is really good at. The dyadic case (table) of the adverb /
is doing all the heavy lifting here: it makes a higher rank tensor by traversing items of the specified rank on each side and combining them according to the remaining frame of each side’s shape. The hard part is arranging the arguments so that your resulting matrix has its axes in the correct order.
data_file_name =: '4.data'
NB. cutopen yields boxed lines, so unbox them and ravel items to make a letter matrix
grid =: ,. > cutopen fread data_file_name
NB. pad the grid on every side with #'XMAS' - 1 spaces
hpadded_grid =: ((' ' & ,) @: (, & ' '))"1 grid
padded_grid =: (3 1 $ ' ') , hpadded_grid , (3 1 $ ' ')
NB. traversal vectors
directions =: 8 2 $ 1 0 1 1 0 1 _1 1 _1 0 _1 _1 0 _1 1 _1
NB. rpos cpos matches rdir cdir if the string starting at rpos cpos in
NB. direction rdir cdir is the string we want
matches =: 4 : 0
*/ ,'XMAS' -: padded_grid {~ <"1 x +"1 y *"1 0 i. 4
)"1
positions =: (3 + i. 0 { $ grid) ,"0/ (3 + i. 1 { $ grid)
result1 =: +/, positions matches/ directions
NB. pairs of traversal vectors
x_directions =: 4 2 2 $ 1 1 _1 1 1 1 1 _1 _1 _1 _1 1 _1 _1 1 _1
NB. rpos cpos x_matches 2 2 $ rdir1 cdir1 rdir2 cdir2 if there is an 'A' at
NB. rpos cpos and the string in each of dir1 and dir2 centered at rpos cpos
NB. is the string we want
x_matches =: 4 : 0
NB. (2 2 $ rdir1 cdir1 rdir2 cdir2) *"1 0/ (_1 + i.3) yields a matrix
NB. 2 3 $ (_1 * dir1) , (0 * dir1) , (1 * dir1) followed by the same for dir2
*/ ,'MAS' -:"1 padded_grid {~ <"1 x +"1 y *"1 0/ _1 + i. 3
)"1 2
result2 =: +/, positions x_matches/ x_directions
Uiua
Just part1 for now as I need to walk the dog :-)
[edit] Part 2 now added, and a nicer approach than Part 1 in my opinion, if you’re able to keep that many dimensions straight in your head :-)
[edit 2] Tightened it up a bit more.
Grid ← ⊜∘⊸≠@\n "MMMSXXMASM\nMSAMXMSMSA\nAMXSXMAAMM\nMSAMASMSMX\nXMASAMXAMM\nXXAMMXXAMA\nSMSMSASXSS\nSAXAMASAAA\nMAMMMXMMMM\nMXMXAXMASX"
≡⍉⍉×⇡4¤[1_0 0_1 1_1 1_¯1] # Use core dirs to build sets of 4-offsets.
↯∞_2⇡△ Grid # Get all possible starting points.
&p/+♭⊞(+∩(≍"XMAS")⇌.⬚@.⊡:Grid≡+¤) # Part 1. Join the two into a table, use to pick 4-elements, check, count.
Diags ← [[¯. 1_1] [¯. 1_¯1]]
BothMas ← /×≡(+∩(≍"MS")⇌.)⬚@.⊡≡+Diags¤¤ # True if both diags here are MAS.
&p/+≡BothMas⊚="A"⟜¤Grid # Part 2. For all "A"s in grid, check diags, count where good.
The operators have all got ascii names you can type, and the formatter converts them to the symbols. It’s a bit odd but really worthwhile, as you get access to the powerful array handling functionality that made solving today’s challenges so much more straightforward than in other languages.
Nim
Could be done more elegantly, but I haven’t bothered yet.
proc solve(input: string): AOCSolution[int, int] =
var lines = input.splitLines()
block p1:
# horiz
for line in lines:
for i in 0..line.high-3:
if line[i..i+3] in ["XMAS", "SAMX"]:
inc result.part1
for y in 0..lines.high-3:
#vert
for x in 0..lines[0].high:
let word = collect(for y in y..y+3: lines[y][x])
if word in [@"XMAS", @"SAMX"]:
inc result.part1
#diag \
for x in 0..lines[0].high-3:
let word = collect(for d in 0..3: lines[y+d][x+d])
if word in [@"XMAS", @"SAMX"]:
inc result.part1
#diag /
for x in 3..lines[0].high:
let word = collect(for d in 0..3: lines[y+d][x-d])
if word in [@"XMAS", @"SAMX"]:
inc result.part1
block p2:
for y in 0..lines.high-2:
for x in 0..lines[0].high-2:
let diagNW = collect(for d in 0..2: lines[y+d][x+d])
let diagNE = collect(for d in 0..2: lines[y+d][x+2-d])
if diagNW in [@"MAS", @"SAM"] and diagNE in [@"MAS", @"SAM"]:
inc result.part2