Day 21: Keypad Conundrum
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
Finally got this one done very late last night!
I kept getting stuck reasoning about the recursion here. For some reason I got myself convinced that after a move, the state of the ‘upper’ dpads could make it more advantageous to pick one followup move over another - i.e. steps aren’t independent.
It took a bunch of manually working through sequences to convince myself that, after every move, every dpad above it would be on A. With that, it’s ‘just’ recursive pathfinding for independent moves.
Since there are relatively few types of moves needed on the dpad, I just sketched them out and wrote the routes in code directly (up to two options per move, e.g. left,up or up,left).
Code
#include "common.h" static int64_t dpmem[26][8][2]; enum {NW,NN,NE,EE,SE,SS,SW,WW}; /* * We can sufficiently describe npad/dpad movements as: "move N x, N y, * then press A N times" (a 'move'). It's never faster to take a detour. * After pressing A on one dpad, dpads on all levels above will also be * on A, so these moves can be considered in isolation. * * This function gives the cost of executing a move with the dpad in * direction 'd' at level 'l', excluding the A presses for actually * tapping the selected directions, but including returning the cursor * back to the A button. * * E.g., moving 2 up and 3 left (NW), the steps could be <AAv<AAA>>^. * Excluding the A presses, that's 6 steps. Pulling out the A presses * lets us drop the dx and dy arguments and simplify the memoization. * They're easily calculated: it's |dx|+|dy|. * * The gaps present a problem: picking the lowest-cost option (e.g. * left-then-up vs. up-then-left) may cause us to cross a gap. The * "mind the gap" argument 'mtg' must be set to 1 if-and-only-if there * is such potential, e.g. a move from < to ^ on the dpad (up,right * crosses the gap) or a move from A to 1 on the npad (left,up crosses * the gap). With the flag set, these orderings will be avoided. */ static int64_t dp(int d, int mtg, int l) { int64_t ret, alt=INT64_MAX; if (l<=0) return 0; assert(l >= 0); assert(l < (int)LEN(dpmem)); assert(mtg==0 || mtg==1); if ((ret = dpmem[l][d][mtg])) return ret; /* routes avoiding gaps where necessary */ ret = d==SE ? 1+dp(SS,0,l-1) + 1+dp(WW,0,l-1) + 2+dp(SE,0,l-1) : d==SW ? 2+dp(SW,0,l-1) + 1+dp(WW,0,l-1) + 3+dp(NE,1,l-1) : d==SS ? 2+dp(SW,0,l-1) + 2+dp(NE,0,l-1) : d==NE ? 1+dp(SS,0,l-1) + 2+dp(NW,0,l-1) + 1+dp(EE,0,l-1) : d==NW ? 1+dp(WW,0,l-1) + 2+dp(SW,1,l-1) + 3+dp(NE,1,l-1) : d==NN ? 1+dp(WW,0,l-1) + 1+dp(EE,0,l-1) : d==EE ? 1+dp(SS,0,l-1) + 1+dp(NN,0,l-1) : d==WW ? 3+dp(SW,1,l-1) + 3+dp(NE,1,l-1) : (assert(!"bad dir"), -1); /* alternatives crossing the gaps */ alt = mtg ? INT64_MAX : d==SE ? 2+dp(SW,0,l-1) + 1+dp(EE,0,l-1) + 1+dp(NN,0,l-1) : d==SW ? 3+dp(SW,1,l-1) + 1+dp(EE,0,l-1) + 2+dp(NE,0,l-1) : d==NE ? 1+dp(WW,0,l-1) + 2+dp(SE,0,l-1) + 1+dp(NN,0,l-1) : d==NW ? 3+dp(SW,1,l-1) + 2+dp(NE,1,l-1) + 1+dp(EE,0,l-1) : INT64_MAX; return dpmem[l][d][mtg] = MIN(ret, alt); } static int64_t npcost(const char *s, int lv) { int64_t cost=0; int x=2,y=3, x1,y1, mtg, dir; for (; *s == 'A' || (*s >= '0' && *s <= '9'); s++) { x1 = *s=='A' ? 2 : *s=='0' ? 1 : 2-(9-*s+'0') % 3; y1 = *s=='A' ? 3 : *s=='0' ? 3 : (9-*s+'0') / 3; /* potentially crossing a gap? */ mtg = (x==0 && y1==3) || (x1==0 && y==3); dir = y1>y ? (x1>x ? SE : x1<x ? SW : SS) : y1<y ? (x1>x ? NE : x1<x ? NW : NN) : (x1>x ? EE : x1<x ? WW : 0); cost += dp(dir, mtg, lv) + abs(x1-x) + abs(y1-y) +1; x = x1; y = y1; } return cost; } int main(int argc, char **argv) { char buf[8]; int digits; int64_t p1=0,p2=0; if (argc > 1) DISCARD(freopen(argv[1], "r", stdin)); while (fgets(buf, sizeof(buf), stdin)) { digits = atoi(buf); p1 += digits * npcost(buf, 2); p2 += digits * npcost(buf, 25); } printf("21: %"PRId64" %"PRId64"\n", p1, p2); return 0; }
https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day21.c
Rust
Like many it seems, this one also broke my brain. Its basically the same as Day 19, but something about it mentally broke me.
#[cfg(test)] mod tests { use std::collections::HashMap; static NUMPAD: [[char; 3]; 4] = [ ['7', '8', '9'], ['4', '5', '6'], ['1', '2', '3'], ['X', '0', 'A'], ]; static DPAD: [[char; 3]; 2] = [['X', '^', 'A'], ['<', 'v', '>']]; fn valid_path(pad: &[[char; 3]], start: (isize, isize), path: &str) -> bool { let mut pos = (start.0, start.1); for c in path.chars() { match c { '^' => pos.0 -= 1, 'v' => pos.0 += 1, '<' => pos.1 -= 1, '>' => pos.1 += 1, 'A' => {} _ => unreachable!(), }; if pad[pos.0 as usize][pos.1 as usize] == 'X' { return false; } } true } fn move_pad(pad: &[[char; 3]], start: char, end: char) -> Vec<String> { let mut start_coord = (0, 0); let mut end_coord = (0, 0); for i in 0..pad.len() { for j in 0..3 { if pad[i][j] == end { end_coord = (i as isize, j as isize); } if pad[i][j] == start { start_coord = (i as isize, j as isize); } } } let delta_i = end_coord.0 - start_coord.0; let vert = match delta_i { -3 => "^^^", -2 => "^^", -1 => "^", 0 => "", 1 => "v", 2 => "vv", 3 => "vvv", _ => unreachable!(), }; let delta_j = end_coord.1 - start_coord.1; let horz = match delta_j { -2 => "<<", -1 => "<", 0 => "", 1 => ">", 2 => ">>", _ => unreachable!(), }; let vert_path = horz.to_string() + vert + "A"; let horz_path = vert.to_string() + horz + "A"; if !valid_path(pad, start_coord, &vert_path) { return vec![horz_path]; } if !valid_path(pad, start_coord, &horz_path) { return vec![vert_path]; } vec![vert_path, horz_path] } fn dpad_seq_len(p0: &str, depth: usize, cache: &mut HashMap<(String, usize), usize>) -> usize { if depth == 0 { return p0.len(); } if let Some(cost) = cache.get(&(p0.to_string(), depth)) { return *cost; } let mut first = 'A'; let mut length = 0; for second in p0.chars() { let moves = move_pad(&DPAD, first, second); let mut min = usize::MAX; for m in moves { let l = dpad_seq_len(&m, depth - 1, cache); if l < min { min = l; } } length += min; first = second; } cache.insert((p0.to_string(), depth), length); length } fn numpad_seq_len( p0: &str, depth: usize, cache: &mut HashMap<(String, usize), usize>, ) -> usize { let mut first = 'A'; let mut length = 0; for second in p0.chars() { let moves = move_pad(&NUMPAD, first, second); let mut min = usize::MAX; for m in moves { let l = dpad_seq_len(&m, depth, cache); if l < min { min = l; } } length += min; first = second; } length } #[test] fn test_numpad2dpad() { let mut cache = HashMap::new(); assert_eq!(68, numpad_seq_len("029A", 2, &mut cache)); assert_eq!(60, numpad_seq_len("980A", 2, &mut cache)); assert_eq!(68, numpad_seq_len("179A", 2, &mut cache)); assert_eq!(64, numpad_seq_len("456A", 2, &mut cache)); assert_eq!(64, numpad_seq_len("379A", 2, &mut cache)); } #[test] fn day21_part1_test() { let mut cache = HashMap::new(); let input = std::fs::read_to_string("src/input/day_21.txt").unwrap(); let codes = input.split('\n').collect::<Vec<&str>>(); let mut total = 0; for code in codes { let min_length = numpad_seq_len(code, 2, &mut cache); println!("{code}: {min_length}"); total += min_length * code[0..3].parse::<usize>().unwrap(); } println!("{}", total); } #[test] fn day21_part2_test() { let mut cache = HashMap::new(); let input = std::fs::read_to_string("src/input/day_21.txt").unwrap(); let codes = input.split('\n').collect::<Vec<&str>>(); let mut total = 0; for code in codes { let min_length = numpad_seq_len(code, 25, &mut cache); println!("{code}: {min_length}"); total += min_length * code[0..3].parse::<usize>().unwrap(); } println!("{}", total); } }
Rust
For me this was the hardest puzzle so far this year. First I did a bunch of things that turned out not to work properly. For example I tried to solve it with a greedy algorithm that always moved horizontally first then vertically, but that ignores the gaps that need to be avoided (what a sneaky requirement) and also somehow doesn’t guarantee the shortest sequence.
After reaching part 2 it became clear that a recursive function (with memoization) is needed again, and of course in the end it turned out a lot simpler than what I had cooked up before (you don’t want to see that). Now even part 2 takes just 1.6ms.
Solution
use euclid::{default::*, point2, vec2}; use rustc_hash::FxHashMap; use std::iter; type Move = Option<Vector2D<i32>>; const KEYPAD_GAP: Point2D<i32> = point2(0, 3); const DPAD_GAP: Point2D<i32> = point2(0, 0); fn keypad_pos(n: u8) -> Point2D<i32> { match n { b'7' => point2(0, 0), b'8' => point2(1, 0), b'9' => point2(2, 0), b'4' => point2(0, 1), b'5' => point2(1, 1), b'6' => point2(2, 1), b'1' => point2(0, 2), b'2' => point2(1, 2), b'3' => point2(2, 2), b'0' => point2(1, 3), b'A' => point2(2, 3), other => panic!("Invalid keypad symbol {other}"), } } // `None` is used for A fn dpad_pos(d: Move) -> Point2D<i32> { match d { Some(Vector2D { x: 0, y: -1, .. }) => point2(1, 0), None => point2(2, 0), Some(Vector2D { x: -1, y: 0, .. }) => point2(0, 1), Some(Vector2D { x: 0, y: 1, .. }) => point2(1, 1), Some(Vector2D { x: 1, y: 0, .. }) => point2(2, 1), other => panic!("Invalid dpad symbol {other:?}"), } } fn moves_for_diff(diff: Vector2D<i32>, pos: Point2D<i32>, gap: Point2D<i32>) -> Vec<Vec<Move>> { let horizontal = iter::repeat_n( Some(vec2(diff.x.signum(), 0)), diff.x.unsigned_abs() as usize, ); let vertical = iter::repeat_n( Some(vec2(0, diff.y.signum())), diff.y.unsigned_abs() as usize, ); if pos + vec2(diff.x, 0) == gap { // Must not move horizontal first, or we hit the gap vec![vertical.chain(horizontal).chain(iter::once(None)).collect()] } else if pos + vec2(0, diff.y) == gap { vec![horizontal.chain(vertical).chain(iter::once(None)).collect()] } else { // Try both horizontal first and vertical first vec![ horizontal .clone() .chain(vertical.clone()) .chain(iter::once(None)) .collect(), vertical.chain(horizontal).chain(iter::once(None)).collect(), ] } } fn dpad_sequence_len( start: Move, end: Move, rounds: u32, cache: &mut FxHashMap<(Move, Move, u32), u64>, ) -> u64 { if rounds == 0 { return 1; } if let Some(len) = cache.get(&(start, end, rounds)) { return *len; } let start_pos = dpad_pos(start); let end_pos = dpad_pos(end); let diff = end_pos - start_pos; let possible_paths = moves_for_diff(diff, start_pos, DPAD_GAP); let shortest_sequence = possible_paths .iter() .map(|moves| { moves .iter() .fold((0, None), |(cost, prev), &m| { (cost + dpad_sequence_len(prev, m, rounds - 1, cache), m) }) .0 }) .min() .unwrap(); cache.insert((start, end, rounds), shortest_sequence); shortest_sequence } fn keypad_sequence_len(start: u8, end: u8, rounds: u32) -> u64 { let start_pos = keypad_pos(start); let end_pos = keypad_pos(end); let diff = end_pos - start_pos; let possible_paths = moves_for_diff(diff, start_pos, KEYPAD_GAP); let mut cache = FxHashMap::default(); possible_paths .iter() .map(|moves| { moves .iter() .fold((0, None), |(cost, prev), &m| { (cost + dpad_sequence_len(prev, m, rounds, &mut cache), m) }) .0 }) .min() .unwrap() } fn solve(input: &str, rounds: u32) -> u64 { let mut sum: u64 = 0; for l in input.lines() { let mut prev = b'A'; let mut len = 0; for b in l.bytes() { len += keypad_sequence_len(prev, b, rounds); prev = b; } let code_n: u64 = l.strip_suffix('A').unwrap().parse().unwrap(); sum += code_n * len; } sum } fn part1(input: String) { println!("{}", solve(&input, 2)); } fn part2(input: String) { println!("{}", solve(&input, 25)); } util::aoc_main!();
Also on github
For a challenge that was conceptually very similar to Day 19, this one completely broke me all over again. I think it was just near impossible to mentally visualize. I only got it with lots of reading your code, so thanks!
I like your short cut of converting the keys from a grid to a set of coordinates, which really simplifies how you get the path.
Glad to be able to help. This one really was a doozy.
Haskell
I get the feeling this solution is more complicated than necessary, which means I probably haven’t understood the problem properly. Anyway, dynamic programming saves the day again!
import Control.Monad import Data.List import Data.Map (Map) import Data.Map qualified as Map type Pos = (Int, Int) makeKeypad :: [[Char]] -> Map Char Pos makeKeypad rows = Map.fromList [(c, (i, j)) | (i, r) <- zip [0 ..] rows, (j, c) <- zip [0 ..] r, c /= '_'] numericKeypad = makeKeypad ["789", "456", "123", "_0A"] directionalKeypad = makeKeypad ["_^A", "<v>"] movesToButton :: Map Char Pos -> Pos -> Pos -> [[Char]] movesToButton keypad (i1, j1) (i2, j2) = let di = i2 - i1 dj = j2 - j1 v = replicate (abs di) $ if di > 0 then 'v' else '^' h = replicate (abs dj) $ if dj > 0 then '>' else '<' hv = guard ((i1, j2) `elem` keypad) >> return (h ++ v) vh = guard ((i2, j1) `elem` keypad) >> return (v ++ h) in (++ ['A']) <$> nub (hv ++ vh) indirectLength :: Int -> [Char] -> Int indirectLength levels = (minimum . map (go levels)) . inputMoves numericKeypad where mapInput keypad f = (zipWith f <*> drop 1) . map (keypad Map.!) . ('A' :) inputMoves keypad = fmap concat . sequence . mapInput keypad (movesToButton keypad) go 0 = length go l = sum . mapInput directionalKeypad (\p1 p2 -> lengths Map.! (l, p1, p2)) lengths = let ps = Map.elems directionalKeypad in Map.fromList [((l, p1, p2), bestLength l p1 p2) | l <- [1 .. levels], p1 <- ps, p2 <- ps] bestLength l p1 p2 = minimum . map (go (l - 1)) $ movesToButton directionalKeypad p1 p2 complexity :: Int -> String -> Int complexity bots code = indirectLength bots code * read (init code) main = do input <- lines <$> readFile "input21" mapM_ (print . sum . flip map input . complexity) [2, 25]
Dart
The secret ingredient is a big cache.
import 'dart:math'; import 'package:collection/collection.dart'; import 'package:more/more.dart'; Map<String, Point<num>> mapper(List<String> list) => { for (var r in list.indexed()) for (var c in r.value.split('').indexed()) c.value: Point<num>(c.index, r.index) }; var dirMap = mapper(['_^A', '<v>']); var numMap = mapper(['789', '456', '123', '_0A']); var lenCache = <(String, int), int>{}; int len(String code, int level, isNum) => lenCache.putIfAbsent((code, level), () => _len(code, level, isNum)); int n(p, isNum, level) => len(getMoves(p[0], p[1], isNum), level - 1, false); int _len(String code, int level, isNum) => (level == 0) ? code.length : 'A$code'.split('').window(2).map((p) => n(p, isNum, level)).sum; String getMoves(String f, String t, bool isNum) { var map = isNum ? numMap : dirMap; var (from, to) = (map[f]!, map[t]!); var mv = to - from; var s = <String>{}; var x = ''.padRight(mv.x.abs() as int, mv.x.sign == 1 ? '>' : '<'); var y = ''.padRight(mv.y.abs() as int, mv.y.sign == 1 ? 'v' : '^'); // avoid '_', dislike '<' if (Point(from.x, to.y) != map['_']!) s.add('$y${x}A'); if (Point(to.x, from.y) != map['_']!) s.add('$x${y}A'); return (s.length < 2 || mv.x > 0) ? s.first : s.last; } int solve(String code, int level) => len(code, level + 1, true) * int.parse(code.skipLast(1)); part1(List<String> lines) => lines.map((e) => solve(e, 2)).sum; part2(List<String> lines) => lines.map((e) => solve(e, 25)).sum;