Day 7: Laboratories
Megathread guidelines
- Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
- You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL
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
Kotlin
Part 1 is easily solved by simulating the beams and modifying the grid in-place.
This does not work for part 2, however. Here I opted for a BFS algorithm, moving down row-by-row.
Similar to other solutions, I store the amount of incoming paths to a splitter, so that I can reference it later. Practically, the two beams from each splitter are representatives of all incoming beams. This reduces the search complexity by a lot!
The final count of timelines is the sum of the array that collects the incoming beam counts for each bottom-row of the diagram.Code
class Day07 : AOCSolution { override val year = 2025 override val day = 7 override fun part1(inputFile: String): String { val diagram = readResourceLines(inputFile) .mapArray { line -> line.mapArray { char -> Cell.byChar(char) } } .toGrid() var count = 0 for (y in 1 until diagram.height) { for (x in 0 until diagram.width) { // Search for beam sources in the preceding row if (diagram[x, y - 1] in Cell.beamSourceTypes) { // Simulate the beam moving down when (diagram[x, y]) { Cell.Empty -> diagram[x, y] = Cell.Beam Cell.Splitter -> { // Split the beam and count this splitter diagram[x - 1, y] = Cell.Beam diagram[x + 1, y] = Cell.Beam count++ } else -> continue } } } } return count.toString() } override fun part2(inputFile: String): String { val diagram = readResourceLines(inputFile) .mapArray { line -> line.mapArray { char -> Cell.byChar(char) } } .toGrid() val height = diagram.height.toLong() val startPosition = diagram.positionOfFirst { it == Cell.Start } // Working stack of beam origin and split origin positions val stack = ArrayDeque<Pair<Position, Position>>() stack.add(startPosition to startPosition) // Splitter positions mapped to the count of timelines to them // Start with the start position and 1 timeline. val splitters = mutableMapOf<Position, Long>(startPosition to 1) // Keep track of all splitters for which new beams have been spawned already // Could be used to solve part 1, as well val spawnedSplitters = mutableSetOf<Position>() // Count the timelines per diagram exit, which is the bottom-most row val diagramExits = LongArray(diagram.width) while (stack.isNotEmpty()) { // Breadth first search for memorizing the amount of paths to a splitter val (beamOrigin, splitOrigin) = stack.poll() val originPathCount = splitters.getValue(splitOrigin) val nextPosition = beamOrigin + Direction.DOWN if (nextPosition.y < height) { if (diagram[nextPosition] == Cell.Splitter) { if (nextPosition !in spawnedSplitters) { // Only spawn new beams, if they weren't spawned already stack.add((nextPosition + Direction.LEFT) to nextPosition) stack.add((nextPosition + Direction.RIGHT) to nextPosition) spawnedSplitters.add(nextPosition) // Initialize the count splitters[nextPosition] = originPathCount } else { splitters.computeIfPresent(nextPosition) { _, v -> v + originPathCount } } } else { // Just move down stack.add(nextPosition to splitOrigin) } } else { diagramExits[nextPosition.x.toInt()] += originPathCount } } // Sum the count of timelines leading to the bottom row, i.e. leaving the diagram for each position return diagramExits.sum().toString() } private companion object { enum class Cell(val char: Char) { Start('S'), Empty('.'), Splitter('^'), Beam('|'); override fun toString(): String { return char.toString() } companion object { fun byChar(char: Char) = entries.first { it.char == char } val beamSourceTypes = arrayOf(Start, Beam) } } } }Rust
Dynamic programming? Nah, just keep track of the number of overlapping beams and part 2 becomes no different to part 1.
use std::collections::VecDeque; fn parse_input(input: &str) -> (Vec<Vec<bool>>, (usize, usize)) { let splits = input .lines() .map(|l| l.chars().map(|c| c == '^').collect()) .collect(); // Assume start is on first row let start = (input.chars().position(|c| c == 'S').unwrap(), 0); (splits, start) } fn solve(input: String) { let (splits, start) = parse_input(&input); let mut nsplits = 0u32; let mut timelines = 1u64; let mut frontier = VecDeque::from([(start, 1)]); while let Some((pos, multiplicity)) = frontier.pop_front() { let (x, y) = (pos.0, pos.1 + 1); if y == splits.len() { // Falls out of bottom continue; } if splits[y][x] { nsplits += 1; timelines += multiplicity; if let Some((b, m2)) = frontier.back_mut() && *b == (x - 1, y) { *m2 += multiplicity; } else { frontier.push_back(((x - 1, y), multiplicity)); } frontier.push_back(((x + 1, y), multiplicity)); } else if let Some((b, m2)) = frontier.back_mut() && *b == (x, y) { *m2 += multiplicity; } else { frontier.push_back(((x, y), multiplicity)); } } println!("Part 1: {nsplits}"); println!("Part 2: {timelines}"); } fn main() -> std::io::Result<()> { let (input, _) = util::get_input("day7.txt")?; solve(input); Ok(()) }Ahh I knew there would be some simple combined representation like this but couldn’t be bothered. Nice!
QIR (Quantum Intermediate Representation)
Nah, just kidding - I used Rust. The only tricky part seemed to be finding time on a Sunday to get it coded!
Part 1 I swept down with a bool array for beams and part 2 I swept up with a score array and summed when it split (joined).
struct Teleporter(String); impl Teleporter { fn new(s: String) -> Self { Self(s) } fn start_pos(line: &str) -> Result<usize> { line.find('S').ok_or_eyre("Start not found") } fn splits(&self) -> Result<usize> { let mut input = self.0.lines(); let start_line = input.next().ok_or_eyre("No start line")?; let start = Self::start_pos(start_line)?; let mut beams = vec![false; start_line.len()]; beams[start] = true; let mut splits = 0; for line in input { for (i, ch) in line.bytes().enumerate() { if beams[i] && ch == b'^' { splits += 1; beams[i] = false; beams[i - 1] = true; beams[i + 1] = true; } } } Ok(splits) } fn timelines(&self) -> Result<usize> { let mut input = self.0.lines(); let start_line = input.next().ok_or_eyre("No start line")?; let start = Self::start_pos(start_line)?; let mut num_paths = vec![1; start_line.len()]; for line in input.rev() { for (i, c) in line.bytes().enumerate() { if c == b'^' || c == b'S' { num_paths[i] = num_paths[i - 1] + num_paths[i + 1]; } } } Ok(num_paths[start]) } }Uiua
A bit late getting to this, but happy with this solution, despite it being nothing more than just building/summing all paths. As usual, you can drop your own file onto that solution.
D ← ".......S.......\n...............\n.......^.......\n...............\n......^.^......\n...............\n.....^.^.^.....\n...............\n....^.^...^....\n...............\n...^.^...^.^...\n...............\n..^...^.....^..\n...............\n.^.^.^.^.^...^.\n..............." # D ← &fras"AOC2025day07.txt" # <- Uncomment and drop file here. Parse ← ⊃↘↙1≠@.⊜∘⊸≠@\n Flow ← ⊃(+⊃↻₁↻₋₁×|׬)⊙⊸⊣ P₁ ← /+>0♭⬚0×⟜∧(˜⊂↥Flow) P₂ ← /+⊣∧(˜⊂+Flow) ⊃P₁ P₂ Parse DMan, your solution is so much cleaner than the hell I was trying (and failing) to coerce into doing what I wanted. Curse my inability to think bottom-up and array-oriented at the same time!
When I compare it to some other Uiua solutions on the Discord I feel like I’m still just banging rocks together.
Haskell
That was a fun little problem.
import Data.Map qualified as Map import Data.Set qualified as Set import Data.Tuple (swap) readInput s = Map.fromDistinctAscList [((i, j), c) | (i, l) <- zip [0 ..] $ lines s, (j, c) <- zip [0 ..] l] beamPaths input = scanl step (Map.singleton startX 1) [startY .. endY] where Just (startY, startX) = lookup 'S' $ map swap $ Map.assocs input Just ((endY, _), _) = Map.lookupMax input step beams y = Map.unionsWith (+) $ [ if input Map.!? (y + 1, j) == Just '^' then Map.fromList [(j - 1, n), (j + 1, n)] else Map.singleton j n | (j, n) <- Map.assocs beams ] part1 = sum . map Set.size . (zipWith (Set.\\) <*> tail) . map Map.keysSet . beamPaths part2 = sum . last . beamPaths main = do input <- readInput <$> readFile "input07" print $ part1 input print $ part2 inputAnd here’s a super-simple version, because why not.
import Data.List (elemIndex, elemIndices) import Data.Map qualified as Map import Data.Maybe (fromJust) import Data.Set qualified as Set main = do (start : rows) <- lines <$> readFile "input07" let splitsByRow = zipWith ( \row beams -> Set.intersection (Map.keysSet beams) . Set.fromDistinctAscList $ elemIndices '^' row ) rows beamsByRow beamsByRow = scanl ( \beams splits -> let unsplit = beams `Map.withoutKeys` splits split = beams `Map.restrictKeys` splits splitLeft = Map.mapKeysMonotonic pred split splitRight = Map.mapKeysMonotonic succ split in Map.unionsWith (+) [unsplit, splitLeft, splitRight] ) (Map.singleton (fromJust $ elemIndex 'S' start) 1) splitsByRow print . sum $ map Set.size splitsByRow print . sum $ last beamsByRow
Nim
Another simple one.
Part 1: count each time a beam crosses a splitter.
Part 2: keep count of how many particles are in each column in all universes
(e.g. with a simple 1d array), then sum.Runtime:
116 μs95 µs86 µsold version
type AOCSolution[T,U] = tuple[part1: T, part2: U] proc solve(input: string): AOCSolution[int, int] = var beams = newSeq[int](input.find '\n') beams[input.find 'S'] = 1 for line in input.splitLines(): var newBeams = newSeq[int](beams.len) for pos, cnt in beams: if cnt == 0: continue if line[pos] == '^': newBeams[pos-1] += cnt newBeams[pos+1] += cnt inc result.part1 else: newbeams[pos] += cnt beams = newBeams result.part2 = beams.sum()Update: found even smaller and faster version that only needs a single array.
Update #2: small optimizationtype AOCSolution[T,U] = tuple[part1: T, part2: U] proc solve(input: string): AOCSolution[int, int] = var beams = newSeq[int](input.find '\n') beams[input.find 'S'] = 1 for line in input.splitLines(): for pos, c in line: if c == '^' and beams[pos] > 0: inc result.part1 beams[pos-1] += beams[pos] beams[pos+1] += beams[pos] beams[pos] = 0 result.part2 = beams.sum()Full solution at Codeberg: solution.nim
Ah, it took me looking at your updated Codeberg version to understand this - you looked at part two in the opposite way than I did but it comes out the same in the end (and yours is much more efficient).
Kotlin
Tried recursive for part1, didn’t work. LUCKILY for once I was smart and committed anyway, as it was the right solution for part2! I was losing my mind for a bit though as I had originally written my method with Integers…
Solution
class Day07 : Puzzle { val grid = mutableListOf<MutableList<Char>>() val partTwoCache = mutableMapOf<Pair<Int, Int>, Long>() override fun readFile() { val input = readInputFromFile(2025, 7, false) for (line in input.lines().filter { it.isNotBlank() }) { grid.add(line.toCharArray().toMutableList()) } } override fun solvePartOne(): String { grid[1][grid[0].indexOf('S')] = '|' var splits = 0 for (r in 1..<grid.size - 1) { for (c in 0..<grid[r].size) { if (grid[r][c] == '|') { if (grid[r+1][c] == '.') { grid[r+1][c] = '|' } else if (grid[r+1][c] == '^') { grid[r+1][c-1] = '|' grid[r+1][c+1] = '|' splits++ } } } } return splits.toString() } override fun solvePartTwo(): String { val start = grid[0].indexOf('S') return (1 + processBeamPartTwo(1, start)).toString() // don't forget to count the original timeline } private fun processBeamPartTwo(row: Int, column: Int): Long { if (partTwoCache.contains(Pair(row, column))) { return partTwoCache[Pair(row, column)]!! } if (row == grid.size) return 0L if (column == grid[row].size || column < 0) return 0L val out = if (grid[row][column] == '^') { // splitter 1L + processBeamPartTwo(row, column - 1) + processBeamPartTwo(row, column + 1) } else { processBeamPartTwo(row + 1, column) } partTwoCache[Pair(row, column)] = out return out } }full code on Codeberg
I didn’t do recursion on part 1, so my part 1 and 2 were fairly different.
const val start = 'S' const val empty = '.' const val splitter = '^' const val beam = '|' var width: IntRange = IntRange(0, 0) var height: IntRange = IntRange(0, 0) val cache: MutableMap<Pair<Int, Int>, Long> = mutableMapOf() var map: List<List<Char>> = listOf() fun main() { val input = getInput(7) map = parseInput1(input) height = map.indices width = map[0].indices val startLocation = map[0].indexOf(start) to 0 val splits = moveBeam(startLocation) + 1 println(splits) } fun parseInput1(input: String): List<List<Char>> = input.lines() .filter { it.isNotBlank() } .map { it.toCharArray().toList() } fun moveBeam(beamLocation: Pair<Int, Int>): Long { if (cache.containsKey(beamLocation)) { return cache[beamLocation]!! } val belowLocation = beamLocation.first to beamLocation.second + 1 if (belowLocation.second !in height) { return 0L } if (cache.containsKey(belowLocation)) { return cache[belowLocation]!! } val below = map[belowLocation.second][belowLocation.first] var splits = 0L if (below == empty) { splits = moveBeam(belowLocation) } else if (below == splitter) { splits++ val leftLocation = belowLocation.first - 1 to belowLocation.second val left = if (leftLocation.first in width) map[leftLocation.second][leftLocation.first] else '!' if (left == empty || left == splitter) { splits += moveBeam(leftLocation) } val rightLocation = belowLocation.first + 1 to belowLocation.second val right = if (rightLocation.first in width) map[rightLocation.second][rightLocation.first] else '!' if (right == empty || right == splitter) { splits += moveBeam(rightLocation) } } cache[beamLocation] = splits return splits }
Uiua
Heavily
copiedahem, inspired by @mykl@lemmy.world’s solution :)Parse ← °⊂ ▽⊸≡/↥≠@.⊜∘⊸≠@\n Flow ← +⊃(/+≡↻1_¯1¤×|×⊙¬) Propagate ← ˜∧(⊸˜Flow) Part₁ ← /+♭↧ ⊸Propagate Part₂ ← /+⊣Propagate ⊙(˜⊂0) $ .......S....... $ ............... $ .......^....... $ ............... $ ......^.^...... $ ............... $ .....^.^.^..... $ ............... $ ....^.^...^.... $ ............... $ ...^.^...^.^... $ ............... $ ..^...^.....^.. $ ............... $ .^.^.^.^.^...^. $ ............... &fras "7.txt" ⊃(Part₁|Part₂) ParsePython
Not too difficult, especially since there are no edge cases (particles leaving the grid, adjacent splitters).
My initial solution mutated the input for the simulation which I cleaned up after by creating an array that would record the number of particle paths at every column location + some other optimizations. I chose to implement both parts in the same function because they share the majority of the logic.
click to view code
def solve(data: str): grid = data.splitlines() m, n = len(grid), len(grid[0]) # find the first particle particle_paths = [0] * n # count of particle paths that will reach this column for j in range(n): if grid[0][j] == 'S': particle_paths[j] = 1 break # count the number of splits for part 1 splits = 0 # simulate the particle moving down the grid # optimization 1: we can start from the 3rd row (index 2) because that's where the first splitter is # optimization 2: we can skip alternating rows because every other row is empty for i in range(2, m, 2): # particle paths per column after this row is processed next_particle_paths = [0] * n for j in range(n): if particle_paths[j] == 0: # skip if there are no particle paths coming from above in this column continue if grid[i][j] == '.': # no splitter here, the number of paths in this column remains the same # make sure to use += to account for neighboring splitters dumping additional paths into this column next_particle_paths[j] += particle_paths[j] else: # splitter activated here, any particle arriving here can end up in the left or right column # this can be simulated by adding the number of paths to the columns on either side splits += 1 next_particle_paths[j-1] += particle_paths[j] next_particle_paths[j+1] += particle_paths[j] # update vars for next iteration particle_paths = next_particle_paths # return both # the number of splits AND # the count of timelines a particle would create return splits, sum(particle_paths) sample = """.......S....... ............... .......^....... ............... ......^.^...... ............... .....^.^.^..... ............... ....^.^...^.... ............... ...^.^...^.^... ............... ..^...^.....^.. ............... .^.^.^.^.^...^. ...............""" assert solve(sample) == (21, 40)My choice of representation for part 1 ended up being fortunate, as all I needed to change for part 2 was three
setfinpropagatethat becameincf. This was also another good use case for multiple return values, as it’s the beam vector you want toreduceon, but for part 1 the split count needs to come along for the ride.(defun parse-line (line) (let ((result (make-array (length line) :initial-element 0))) (loop for i from 0 to (1- (length line)) if (member (char line i) '(#\^ #\S)) do (setf (aref result i) 1)) result)) (defun read-inputs (filename) (let ((input-lines (uiop:read-file-lines filename))) (mapcar #'parse-line input-lines))) (defun propagate (in-beams splitters) (let ((out-beams (make-array (length in-beams) :initial-element 0)) (splits 0)) (loop for i from 0 to (1- (length in-beams)) if (not (zerop (aref in-beams i))) do (if (not (zerop (aref splitters i))) (progn (incf (aref out-beams (1- i)) (aref in-beams i)) (incf (aref out-beams (1+ i)) (aref in-beams i)) (incf splits)) (incf (aref out-beams i) (aref in-beams i)))) (values out-beams splits))) (defun propagate-all (initial-beam-vector splitter-map) (let* ((total-splits 0) (final-beam-vector (reduce #'(lambda (in-beams splitters) (multiple-value-bind (out-beams this-splits) (propagate in-beams splitters) (incf total-splits this-splits) out-beams)) splitter-map :initial-value initial-beam-vector))) (values final-beam-vector total-splits))) (defun main-1 (filename) (let ((grid (read-inputs filename))) (multiple-value-bind (final-beam-vector total-splits) (propagate-all (car grid) (cdr grid)) total-splits))) (defun main-2 (filename) (let ((grid (read-inputs filename))) (multiple-value-bind (final-beam-vector total-splits) (propagate-all (car grid) (cdr grid)) (reduce #'+ (coerce final-beam-vector 'list)))))(Browser-based) Javascript
I got lazy for part 1 and stuffed the beam head’s coordinates into a string instead of reusing my implementation that puts them into an int, gambling on part 1 not needing that extra bit of speed.
I did a similar mistake for part 2 as I did on day 6; I wrote a(n overkill) solution for keeping track of the entire path each beam/split follows - which involved cloning each array-storing-a-beam’s-path at each split. Whereas on day 6 I got an “out of memory” error, this time I locked up my computer 💪 🔥 (Control+C doesn’t exist in the browser console, and Control+W to close the tab wasn’t getting through). Was very disappointed to realize we only need to keep track of the number of beams present at each position/index/abscissa without caring about which path they took to reach said index.
Code
function part1(inputText) { let beams = new Set([inputText.indexOf('S')]); const beamSplits = new Set(); const lines = inputText.trimEnd().split('\n'); for (let y = 1; y < lines.length; y++) { const nextBeams = new Set(); for (const beamX of beams) { if (lines[y][beamX] === '.') { nextBeams.add(beamX); } else if (lines[y][beamX]) { beamSplits.add([beamX, y].toString()); nextBeams.add(beamX + 1); nextBeams.add(beamX - 1); } } beams = nextBeams; } return beamSplits.size; } { const start = performance.now(); const result = part1(document.body.textContent); const end = performance.now(); console.info({ day: 7, part: 1, time: end - start, result }); } function part2(inputText) { let beams = new Map([ [inputText.indexOf('S'), 1] ]); const lines = inputText.trimEnd().split('\n'); for (let y = 1; y < lines.length; y++) { // console.debug({ y, beams }) const nextBeams = new Map(); for (const [beamHeadX, beamCount] of beams) { if (lines[y][beamHeadX] === '.') { nextBeams.set(beamHeadX, (nextBeams.get(beamHeadX) ?? 0) + beamCount); } else if (lines[y][beamHeadX] === '^') { nextBeams.set(beamHeadX + 1, (nextBeams.get(beamHeadX + 1) ?? 0) + beamCount); nextBeams.set(beamHeadX - 1, (nextBeams.get(beamHeadX - 1) ?? 0) + beamCount); } } beams = nextBeams; } return beams.entries().reduce((accu, [_, count]) => accu + count, 0); } { const start = performance.now(); const result = part2(document.body.textContent); const end = performance.now(); console.info({ day: 7, part: 2, time: end - start, result }) }Surprised you didn’t trigger the OOM killer. Windows?
Naw, linux. I had a Youtube video running in the background that might have made the problem worse. I could switch desktops and move the mouse (albeit with a lot of lag and jitter) but not close the tab or even hard-close the browser.
Running the same code on my macbook triggered the “a script on this page is slowing down firefox, do you want to wait some more or pause and debug it?” prompt.
Either the OOM killer is not properly set up on my desktop, or it was actually a hot CPU loop that caused the lock-up and not memory pressure. Or some sort of hybrid cpu+mem problem? The code was cloning ever-longer arrays in a for-loop, and from my understanding the event loop for a browser page/tab is not only single-threaded but contains both UI events and script execution (unless you explicitly/specifically shell out to a web worker or the like).
Possibly using up swap space first?
This might be the actual culprit; I don’t think I have any swap space allocated (
htopalways displays0/0for it). I have 48 gigs of RAM, and the “out of memory” error I got on a previous day’s problem didn’t provoke a single slowdown, freeze, or program crash…
Go
Oh my scheduling God! Part 1 was easy. Then for part 2 I started tracing each particle one by one using goroutines, but spawning billions of goroutines seemed to make my poor laptop sad. So I implemented a whole thread pool with process management and stuff, but nothing worked. So at the end I started doing the unthinkable: using my brain.
It seems we can just reuse the same idea as part 1 one but with a clever counting scheme. Thing works and is as fast as part 1. I’m both happy and deeply sad not to have been able to leverage Go’s supposed killer features - which are actually rarely useful when programming things other than servers tbh.
Here we goooooo:
day07.go
package main import ( "aoc/utils" ) func parseStartLine(line string) []bool { runes := []rune(line) beams := make([]bool, len(runes)) for idx, char := range runes { if char == 'S' { beams[idx] = true } } return beams } func stepOne(input chan string) (int, error) { beams := parseStartLine(<-input) splitCount := 0 for line := range input { runes := []rune(line) for idx, char := range runes { if char == '^' && beams[idx] { splitCount++ if idx > 0 { beams[idx-1] = true } if idx < len(runes)-1 { beams[idx+1] = true } beams[idx] = false } } } return splitCount, nil } func valueBeams(beams []bool) []int { valBeams := make([]int, len(beams)) for idx, b := range beams { val := 0 if b { val = 1 } valBeams[idx] = val } return valBeams } func stepTwo(input chan string) (int, error) { beams := valueBeams(parseStartLine(<-input)) for line := range input { runes := []rune(line) for idx, char := range runes { bc := beams[idx] if char == '^' && bc > 0 { beams[idx] = 0 if idx > 0 { beams[idx-1] += bc } if idx < len(runes)-1 { beams[idx+1] += bc } } } } sum := 0 for _, bc := range beams { sum += bc } return sum, nil } func main() { inputFile := utils.FilePath("day07.txt") utils.RunStep(utils.ONE, inputFile, stepOne) utils.RunStep(utils.TWO, inputFile, stepTwo) }~32B coroutines does sound like it might make any CPU a bit sad :D
Yeah, the process was terminated by Linux in about 2-3 mins x)
C
Accidentally solved part 2 first but had the foresight to save the code. Otherwise my solution looks similar to what other people are doing, just with more code 😅
Code
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <inttypes.h> #include <assert.h> #define GW 148 #define GH 74 static char g[GH][GW]; static uint64_t dp[2][GW]; static int w,h; /* recursively traces beam for part 1, marking visited splitters with 'X' */ static uint64_t solve_p1(int x, int y) { if (x<0 || x>=w || y<0 || y>=h || g[y][x] == 'X') return 0; else if (g[y][x] != '^') return solve_p1(x, y+1); else { g[y][x] = 'X'; return 1 + solve_p1(x-1, y) + solve_p1(x+1, y); } } /* DP for part 2 */ static uint64_t solve_p2(void) { int x,y, odd; for (y=h-1; y>=0; y--) for (x=1; x<w-1; x++) { /* only two rows relevant at a time, so don't store any more */ odd = y&1; if (g[y][x] == 'S') { printf("\n"); return dp[!odd][x] + 1; } dp[odd][x] = g[y][x] == '^' || g[y][x] == 'X' ? dp[!odd][x-1] + dp[!odd][x+1] + 1 : dp[!odd][x]; } return 0; } int main() { int x,y, sx,sy; for (h=0; ; ) { /* one bottom row of padding */ assert(h < GH-1); /* input already side padded, plus we have \n\0 */ if (!fgets(g[h], GW, stdin)) break; /* skip empty rows */ for (x=0; g[h][x]; x++) if (g[h][x] == 'S' || g[h][x] == '^') { h++; break; } } w = strlen(g[0])-1; /* strip \n */ for (y=0; y<h; y++) for (x=0; x<w; x++) if (g[y][x] == 'S') { sx=x; sy=y; break; } printf("07: %"PRIu64" %"PRIu64"\n", solve_p1(sx,sy), solve_p2()); }Go
Now I’m behind by 1 day, will try to catch up.
For part 2 I spent a good while thinking about it, then when I convinced myself my plan could work, struggled a bit with the implementation. But it worked in the end. Basically grid[i][j] is how many different ways you can reach a cell. Start at 1 on the S cell, then propagate the values down and keep adding up the nums when you reach cells through different paths. The answer is the sum of the nums in the last row.
spoiler
func part2() { // file, _ := os.Open("sample.txt") file, _ := os.Open("input.txt") defer file.Close() scanner := bufio.NewScanner(file) input := [][]rune{} for scanner.Scan() { line := []rune(scanner.Text()) input = append(input, line) } m := len(input) n := len(input[0]) grid := make([][]int, m) for i := range m { grid[i] = make([]int, n) } for i := range m { for j := range n { c := input[i][j] if i == 0 { if c == 'S' { grid[i][j] = 1 } continue } if c == '^' { grid[i][j-1] += grid[i-1][j] grid[i][j+1] += grid[i-1][j] } else { grid[i][j] = grid[i][j] + grid[i-1][j] } } } paths := 0 for j := range n { paths += grid[m-1][j] } fmt.Println(paths) }








