Haskell

This was fun and (fairly) easy! Off-by-one errors are a likely source of bugs here.

import Control.Monad
import Data.List
import Data.List.Split
import Data.Maybe

score d pat = ((100 *) <$> search pat) `mplus` search (transpose pat)
  where
    search pat' = find ((d ==) . rdiff pat') [1 .. length pat' - 1]
    rdiff pat' i =
      let (a, b) = splitAt i pat'
       in length $ filter (uncurry (/=)) $ zip (concat $ reverse a) (concat b)

main = do
  input <- splitOn [""] . lines <$> readFile "input13"
  let go d = print . sum . map (fromJust . score d) $ input
  go 0
  go 1

Line-seconds score: 0.102 😉

Python

from .solver import Solver


def is_mirrored_x(pattern: set[tuple[int, int]], max_x: int, max_y: int,
                  x_mirror: int, desired_errors: int = 0) -> bool:
  min_x = max(0, 2 * x_mirror - max_x)
  max_x = min(max_x, 2 * x_mirror)
  errors = 0
  for y in range(max_y):
    for x in range(min_x, x_mirror):
      mirrored = 2 * x_mirror - x - 1
      if (x, y) in pattern and (mirrored, y) not in pattern:
        errors += 1
      if (x, y) not in pattern and (mirrored, y) in pattern:
        errors += 1
      if errors > desired_errors:
        return False
  return errors == desired_errors

def is_mirrored_y(pattern: set[tuple[int, int]], max_x: int, max_y: int,
                  y_mirror: int, desired_errors: int = 0) -> bool:
  min_y = max(0, 2 * y_mirror - max_y)
  max_y = min(max_y, 2 * y_mirror)
  errors = 0
  for x in range(max_x):
    for y in range(min_y, y_mirror):
      mirrored = 2 * y_mirror - y - 1
      if (x, y) in pattern and (x, mirrored) not in pattern:
        errors += 1
      if (x, y) not in pattern and (x, mirrored) in pattern:
        errors += 1
      if errors > desired_errors:
        return False
  return errors == desired_errors

def find_mirror_axis(pattern: set[tuple[int, int]], max_x: int, max_y: int,
                     desired_errors: int = 0) -> tuple[None, int]|tuple[int, None]:
  for possible_x_mirror in range(1, max_x):
    if is_mirrored_x(pattern, max_x, max_y, possible_x_mirror, desired_errors):
      return possible_x_mirror, None
  for possible_y_mirror in range(1, max_y):
    if is_mirrored_y(pattern, max_x, max_y, possible_y_mirror, desired_errors):
      return None, possible_y_mirror
  raise RuntimeError('No mirror axis found')

class Day13(Solver):

  def __init__(self):
    super().__init__(13)
    self.patterns: list[set[tuple[int, int]]] = []
    self.dimensions: list[tuple[int, int]] = []

  def presolve(self, input: str):
    patterns = input.rstrip().split('\n\n')
    for pattern in patterns:
      lines = pattern.splitlines()
      points: set[tuple[int, int]] = set()
      max_x = 0
      max_y = 0
      for y, line in enumerate(lines):
        max_y = max(max_y, y)
        for x, char in enumerate(line):
          max_x = max(max_x, x)
          if char == '#':
            points.add((x, y))
      self.patterns.append(points)
      self.dimensions.append((max_x + 1, max_y + 1))

  def solve_first_star(self) -> int:
    sum = 0
    for pattern, (max_x, max_y) in zip(self.patterns, self.dimensions, strict=True):
      mirror_x, mirror_y = find_mirror_axis(pattern, max_x, max_y)
      sum += (mirror_x or 0) + (mirror_y or 0) * 100
    return sum

  def solve_second_star(self) -> int:
    sum = 0
    for pattern, (max_x, max_y) in zip(self.patterns, self.dimensions, strict=True):
      mirror_x, mirror_y = find_mirror_axis(pattern, max_x, max_y, 1)
      sum += (mirror_x or 0) + (mirror_y or 0) * 100
    return sum

Scala3

// i is like
//  # # # # #
//   1 2 3 4
def smudgesAround(i: Int, s: List[List[Char]]): Long =
    val toEdge = math.min(i, s.size - i)
    (0 until toEdge).map(e => s(i - e - 1).lazyZip(s(i + e)).count(_ != _)).sum

def symmetries(g: List[List[Char]], smudges: Int) =
    val rows = (1 until g.size).filter(smudgesAround(_, g) == smudges)
    val g2 = g.transpose
    val cols = (1 until g2.size).filter(smudgesAround(_, g2) == smudges)
    100*rows.sum + cols.sum

def task1(a: List[String]): Long = a.chunk(_ == "").map(g => symmetries(g.map(_.toList), 0)).sum
def task2(a: List[String]): Long = a.chunk(_ == "").map(g => symmetries(g.map(_.toList), 1)).sum

Rust

Part 2 turned out easier than I thought initially. My code assumes that there is only 1 mirror in each field which means I don’t have to remember where the smudge is, I just have to find a mirror line where the two halves differ at exactly one spot.

Dart

Just banging strings together again. Simple enough once I understood that the original reflection may also be valid after desmudging. I don’t know if we were supposed to do something clever for part two, but my part 1 was fast enough that I could just try every possible smudge location and part 2 still ran in 80ms

bool reflectsH(int m, List p) => p.every((l) {
      var l1 = l.take(m).toList().reversed.join('');
      var l2 = l.skip(m);
      var len = min(l1.length, l2.length);
      return l1.take(len) == l2.take(len);
    });

bool reflectsV(int m, List p) {
  var l1 = p.take(m).toList().reversed.toList();
  var l2 = p.skip(m).toList();
  var len = min(l1.length, l2.length);
  return 0.to(len).every((ix) => l1[ix] == l2[ix]);
}

int findReflection(List p, {int butNot = -1}) {
  var mirrors = 1
      .to(p.first.length)
      .where((m) => reflectsH(m, p))
      .toSet()
      .difference({butNot});
  if (mirrors.length == 1) return mirrors.first;

  mirrors = 1
      .to(p.length)
      .where((m) => reflectsV(m, p))
      .toSet()
      .difference({butNot ~/ 100});
  if (mirrors.length == 1) return 100 * mirrors.first;

  return -1; //never
}

int findSecondReflection(List p) {
  var origMatch = findReflection(p);
  for (var r in 0.to(p.length)) {
    for (var c in 0.to(p.first.length)) {
      var pp = p.toList();
      var cells = pp[r].split('');
      cells[c] = (cells[c] == '#') ? '.' : '#';
      pp[r] = cells.join();
      var newMatch = findReflection(pp, butNot: origMatch);
      if (newMatch > -1) return newMatch;
    }
  }
  return -1; // never
}

Iterable> parse(List lines) => lines
    .splitBefore((e) => e.isEmpty)
    .map((e) => e.first.isEmpty ? e.skip(1).toList() : e);

part1(lines) => parse(lines).map(findReflection).sum;

part2(lines) => parse(lines).map(findSecondReflection).sum;

C

Implementing part 1 with a bunch of for loops made me wonder about elegant NumPy solutions but then part 2 came along and it was a perfect fit! Just change a flag to a counter and remove the if-match-early-exit.

https://github.com/sjmulder/aoc/blob/master/2023/c/day13.c

int main()
{
	static char g[32][32];
	int p1=0,p2=0, w,h, x,y,i, nmis;
	
	while (!feof(stdin)) {
		for (h=0; ; h++) {
			assert(h < (int)LEN(*g));
			if (!fgets(g[h], LEN(*g), stdin)) break;
			if (!g[h][0] || g[h][0]=='\n') break;
		}

		assert(h>0); w = strlen(g[0])-1;
		assert(w>0);

		for (x=1; x

I decided to use string comparison for some reason, which meant part two wasn’t as quick as it would have been.

use std::{cmp, fs, iter};

fn transpose(rows: &[&str]) -> Vec {
    (0..rows[0].len())
        .map(|i| {
            let bytes: Vec<_> = (0..rows.len()).map(|j| rows[j].as_bytes()[i]).collect();
            String::from_utf8(bytes).unwrap()
        })
        .collect()
}

fn reflection_value(rows: &[&str]) -> Option {
    'row_loop: for i in 0..(rows.len() - 1) {
        if rows[i] != rows[i + 1] {
            continue;
        }
        // we have an initial match
        let other_matches = cmp::min(i, rows.len() - 2 - i);
        for j in 1..=other_matches {
            if rows[i - j] != rows[i + 1 + j] {
                continue 'row_loop;
            }
        }
        return Some(i as u32 + 1);
    }
    None
}

fn summary(file_path: &str) -> u32 {
    fs::read_to_string(file_path)
        .expect("Can't read input file")
        .split("\n\n")
        .map(|s| {
            let rows: Vec<&str> = s.split('\n').collect();

            if let Some(v) = reflection_value(&rows) {
                v * 100
            } else {
                let cols_owned: Vec = transpose(&rows);
                let cols: Vec<&str> = cols_owned.iter().map(|s| s.as_str()).collect();
                reflection_value(&cols).expect("No reflections found")
            }
        })
        .sum()
}

fn str_diff(str1: &str, str2: &str) -> u32 {
    iter::zip(str1.chars(), str2.chars())
        .map(|(s1, s2)| if s1 == s2 { 0 } else { 1 })
        .sum()
}

fn smudged_reflection_value(rows: &[&str]) -> Option {
    for i in 0..(rows.len() - 1) {
        let num_cmps = cmp::min(i, rows.len() - 2 - i);
        let errs: u32 = (0..=num_cmps)
            .map(|j| str_diff(rows[i - j], rows[i + 1 + j]))
            .sum();

        if errs != 1 {
            continue;
        }
        return Some(i as u32 + 1);
    }
    None
}

fn smudged_summary(file_path: &str) -> u32 {
    fs::read_to_string(file_path)
        .expect("Can't read input file")
        .split("\n\n")
        .map(|s| {
            let rows: Vec<&str> = s.split('\n').collect();

            if let Some(v) = smudged_reflection_value(&rows) {
                v * 100
            } else {
                let cols_owned: Vec = transpose(&rows);
                let cols: Vec<&str> = cols_owned.iter().map(|s| s.as_str()).collect();
                smudged_reflection_value(&cols).expect("No reflections found")
            }
        })
        .sum()
}

fn main() {
    println!(" normal: {}", summary("d13/input.txt"));
    println!("smudged: {}", smudged_summary("d13/input.txt"));
}

Nim

Part 1: For part 1 I just compare rows, then transpose matrix and compare rows again.
Part 2: For part 2 I compare rows and get symmetry type for each comparison: symmetric(==), almost symmetric (1 smudge), not symmetric (2+ difference). Then I just use two booleans to make sure we found exactly one smudge in each mirror.

Total runtime: 0.3 ms
Puzzle rating: 6/10
Code: day_13/solution.nim

Nim

This one was a nice change of pace after the disaster of day 12. For part 1 I kept a list of valid column indexes, and checked those columns in each row for reflections, eliminating columns from the list as I went. To find vertical reflections, I just transposed the grid first.

Part 2 looked daunting at first, but I just needed to add a smudges counter to each column candidate, eliminating them when their counter reaches 2. For scoring, just count the 1-smudge columns.

• Day 13, part 1
• Day 13, part 2