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)
Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ or pastebin (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)
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
🔒This post will be unlocked when there is a decent amount of submissions on the leaderboard to avoid cheating for top spots
I like it, it's simple and to the point. I've learned that one of the most helpful things to do when solving these puzzles is to not make it more complicated than it needs to be, and you certainly succeeded better at that today than I did.
My solution for day 1 part 1 was simple and to the point. The other ones are getting increasingly less so. You're right that sometimes it's best not to get too fancy, but I think soon I may have to break out such advanced programming techniques as "functions" and maybe "objects", instead of writing increasingly convoluted piles of nested loops. xD
Hi there! Looks like you linked to a Lemmy community using a URL instead of its name, which doesn't work well for people on different instances. Try fixing it like this: [email protected]
I wrote today's program in Python. (I am going to do a different language each day.) The only thing that gave me a little trouble was I was counting "\n" as a part label. Once I realized that I was able to get both problems done very quickly.
I've been using Regexes for every day so far, this time it helped in finding numbers along with their start and end position in a line. For the second part I mostly went with the approach of part 1 which was to look at all numbers and then figure out if it has a part symbol around it. Only in part 2 I saved all numbers next to a gear * in a hash table that maps each gear position to a list of adjacent numbers. Then in the end I can just look at all gears with exactly 2 numbers attached.
Also it has to be said, multiplying two numbers is the exact opposite of getting their ratio!
import collections
import re
from .solver import Solver
class Day03(Solver):
def __init__(self):
super().__init__(3)
self.lines = []
def presolve(self, input: str):
self.lines = input.rstrip().split('\n')
def solve_first_star(self):
adjacent_to_symbols = set()
for i, line in enumerate(self.lines):
for j, sym in enumerate(line):
if sym in ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.'):
continue
for di in (-1, 0, 1):
for dj in (-1, 0, 1):
adjacent_to_symbols.add((i + di, j + dj))
numbers = []
for i, line in enumerate(self. lines):
for number_match in re.finditer(r'\d+', line):
is_adjacent_to_symbol = False
for j in range(number_match.start(), number_match.end()):
if (i, j) in adjacent_to_symbols:
is_adjacent_to_symbol = True
if is_adjacent_to_symbol:
numbers.append(int(number_match.group()))
return sum(numbers)
def solve_second_star(self):
gear_numbers = collections.defaultdict(list)
adjacent_to_gears = {}
for i, line in enumerate(self.lines):
for j, sym in enumerate(line):
if sym == '*':
for di in (-1, 0, 1):
for dj in (-1, 0, 1):
adjacent_to_gears[(i + di, j + dj)] = (i, j)
for i, line in enumerate(self. lines):
for number_match in re.finditer(r'\d+', line):
adjacent_to_gear = None
for j in range(number_match.start(), number_match.end()):
if (i, j) in adjacent_to_gears:
adjacent_to_gear = adjacent_to_gears[(i, j)]
if adjacent_to_gear:
gear_numbers[adjacent_to_gear].append(int(number_match.group()))
ratios = []
for gear_numbers in gear_numbers.values():
match gear_numbers:
case [a, b]:
ratios.append(a * b)
return sum(ratios)
[Rust] Harder one today, for part 1 I ended up getting stuck for a bit since I wasnt taking numbers at the end of lines into account and in part 2 I defined my gears vector in the wrong spot and spent a bit debugging that
Code
(lemmy removes some chars, all chars are in code link)
use std::fs;
fn part1(input: String) -> u32 {
let lines = input.lines().collect::>();
let mut sum = 0;
for i in 0..lines.len() {
let mut num = 0;
let mut valid = false;
let chars = lines[i].chars().collect::>();
for j in 0..chars.len() {
let character = chars[j];
let parts = ['*', '#', '+', '$', '/', '%', '=', '-', '&', '@'];
if character.is_digit(10) {
num = num * 10 + character.to_digit(10).unwrap();
if i > 0 {
if parts.contains(&lines[i - 1].chars().collect::>()[j]) {
valid = true;
}
if j > 0 {
if parts.contains(&lines[i - 1].chars().collect::>()[j - 1]) {
valid = true;
}
}
if j < chars.len() - 1 {
if parts.contains(&lines[i - 1].chars().collect::>()[j + 1]) {
valid = true;
}
}
}
if i < lines.len() - 1 {
if parts.contains(&lines[i + 1].chars().collect::>()[j]) {
valid = true;
}
if j > 0 {
if parts.contains(&lines[i + 1].chars().collect::>()[j - 1]) {
valid = true;
}
}
if j < chars.len() - 1 {
if parts.contains(&lines[i + 1].chars().collect::>()[j + 1]) {
valid = true;
}
}
}
if j > 0 {
if parts.contains(&lines[i].chars().collect::>()[j - 1]) {
valid = true;
}
}
if j < chars.len() - 1 {
if parts.contains(&lines[i].chars().collect::>()[j + 1]) {
valid = true;
}
}
}
else {
if valid == true {
sum += num;
}
num = 0;
valid = false;
}
if j == chars.len() - 1 {
if valid == true {
sum += num;
}
num = 0;
valid = false;
}
}
}
return sum;
}
fn part2(input: String) -> u32 {
let lines = input.lines().collect::>();
let mut gears: Vec<(usize, usize, u32)> = Vec::new();
let mut sum = 0;
for i in 0..lines.len() {
let mut num = 0;
let chars = lines[i].chars().collect::>();
let mut pos: (usize, usize) = (0, 0);
let mut valid = false;
for j in 0..chars.len() {
let character = chars[j];
let parts = ['*'];
if character.is_digit(10) {
num = num * 10 + character.to_digit(10).unwrap();
if i > 0 {
if parts.contains(&lines[i - 1].chars().collect::>()[j]) {
valid = true;
pos = (i - 1, j);
}
if j > 0 {
if parts.contains(&lines[i - 1].chars().collect::>()[j - 1]) {
valid = true;
pos = (i - 1, j - 1);
}
}
if j < chars.len() - 1 {
if parts.contains(&lines[i - 1].chars().collect::>()[j + 1]) {
valid = true;
pos = (i - 1, j + 1);
}
}
}
if i < lines.len() - 1 {
if parts.contains(&lines[i + 1].chars().collect::>()[j]) {
valid = true;
pos = (i + 1, j);
}
if j > 0 {
if parts.contains(&lines[i + 1].chars().collect::>()[j - 1]) {
valid = true;
pos = (i + 1, j - 1);
}
}
if j < chars.len() - 1 {
if parts.contains(&lines[i + 1].chars().collect::>()[j + 1]) {
valid = true;
pos = (i + 1, j + 1);
}
}
}
if j > 0 {
if parts.contains(&lines[i].chars().collect::>()[j - 1]) {
valid = true;
pos = (i, j - 1);
}
}
if j < chars.len() - 1 {
if parts.contains(&lines[i].chars().collect::>()[j + 1]) {
valid = true;
pos = (i, j + 1);
}
}
}
else {
if valid == true {
let mut current_gear = false;
for gear in &gears {
if gear.0 == pos.0 && gear.1 == pos.1 {
sum += num * gear.2;
current_gear = true;
break;
}
}
if !current_gear {
let tuple_to_push = (pos.0.clone(), pos.1.clone(), num.clone());
gears.push((pos.0.clone(), pos.1.clone(), num.clone()));
}
}
num = 0;
valid = false;
}
if j == chars.len() - 1 {
if valid == true {
let mut current_gear = false;
for gear in &gears {
if gear.0 == pos.0 && gear.1 == pos.1 {
sum += num * gear.2;
current_gear = true;
break;
}
}
if !current_gear {
let tuple_to_push = (pos.0.clone(), pos.1.clone(), num.clone());
gears.push((pos.0.clone(), pos.1.clone(), num.clone()));
}
}
num = 0;
valid = false;
}
}
}
return sum;
}
fn main() {
let input = fs::read_to_string("data/input.txt").unwrap();
println!("{}", part1(input.clone()));
println!("{}", part2(input.clone()));
}
Efficiency? Elegant Code? Nope but It works.
Luckily I did part 1 by looking at the symbols first anyway, so extending to part two was trivial. Also originally had a bug where I treated all symbols as cogs, not only '*'. Interestingly it worked anyway as only '*'s had two adjacent numbers in my data. It is fixed in this version.
Hacked together combined code (originally I did each part as separate programs but they shared so much that I ended up combining then so the post is shorter): https://pastebin.com/Dij2XSYe
Edit: anything in angle brackets is not displaying even with backslashes, idk why but i have moved the code to a pastebin.
My computer crashed right most of the way through and I lost everything, so this was even more frustrating than it should have been
Also damn, lemmy's tabs are massive
will post part 2 when I get to it
input = File.read("input.txt")
lines = input.lines.map(&.chars)
sum = 0
num_marker = nil
lines.each_with_index do |line, y|
line.each_with_index do |char, x|
num_marker ||= x if char.number?
if (!char.number? || x == line.size-1) && num_marker
if check_symbol(y, (num_marker-1)..x, lines)
sum += lines[y][(char.number? ? num_marker..x : num_marker...x)].join.to_i
end
num_marker = nil
end end end
puts sum
def check_symbol(y, rx, lines)
carr = [ lines[y][rx.begin]?, lines[y][rx.end]? ]
carr += rx.map {|x| lines[y-1][x]? } if y > 0
carr += rx.map {|x| lines[y+1][x]? } if y < lines.size-1
carr.each {|c| return true if c && c != '.' && !c.number?}
false
end
Hi there! Looks like you linked to a Lemmy community using a URL instead of its name, which doesn't work well for people on different instances. Try fixing it like this: [email protected]
I get the feeling that I should include some default types for handling 2D maps in my boilerplate, it's a very recurring problem in AoC after all.
My solution is reasonably simplistic - and therefore also a bit slow, but the design meant I could do part 2 with just a few extra lines of code on the already processed data, here's the functional part of it; (I push the previous days solution as part of my workflow for starting with the current day so the full code won't be up until tomorrow)
Ruby
The code has been compressed for brevity.
Point = Struct.new('Point', :x, :y)
PartNumber = Struct.new('PartNumber', :number, :adjacent) do
def adjacent?(to); adjacent.include?(to); end
def irrelevant?; adjacent.empty?; end
def to_i; number; end
end
class Implementation
def initialize
@map = []; @dim = { width: 0, height: 0 }; @symbols = []; @numbers = []
end
def input(line)
@dim[:width] = line.size; @dim[:height] += 1
@map += line.chars
end
def calc
for y in (0..@dim[:height]-1) do
for x in (0..@dim[:width]-1) do
chr = get(x, y); next if chr =~ /\d/ || chr == '.'
@symbols << Point.new(x, y)
end
end
for y in (0..@dim[:height]-1) do
buf = ""; adj = []
for x in (0..@dim[:width]) do # Going one over, to fake a non-number as an end char on all lines
chr = get(x, y)
if chr =~ /\d/
buf += chr
(-1..1).each do |adj_x|
(-1..1).each do |adj_y|
next if adj_x == 0 && adj_y == 0 ||
(x + adj_x < 0) || (x + adj_x >= @dim[:width]) ||
(y + adj_y < 0) || (y + adj_y >= @dim[:height])
sym = Point.new(x + adj_x, y + adj_y)
adj << sym if @symbols.any? sym
end
end
elsif !buf.empty?
@numbers << PartNumber.new(buf.to_i, adj)
buf = ""; adj = []
end
end
end
end
def output
part1 = @numbers.reject(&:irrelevant?).map(&:to_i).sum
puts "Part 1:", part1
gears = @symbols.select do |sym|
next unless get(sym) == '*'
next unless @numbers.select { |num| num.adjacent? sym }.size == 2
true
end
part2 = gears.sum { |gear| @numbers.select { |num| num.adjacent? gear }.map(&:to_i).inject(:*) }
puts "Part 2:", part2
end
private
def get(x, y = -1)
y = x.y if x.is_a?(Point)
x = x.x if x.is_a?(Point)
return unless (0..@dim[:width]-1).include?(x) && (0..@dim[:height]-1).include?(y)
@map[y * @dim[:width] + x % @dim[:width]]
end
end
-- SPDX-FileCopyrightText: 2023 Jummit
--
-- SPDX-License-Identifier: GPL-3.0-or-later
local lines = {}
for line in io.open("3.input"):lines() do
table.insert(lines, "."..line..".")
end
local width = #lines[1]
local height = #lines
local function at(x, y, w)
if y < 1 or y > height then return nil end
return lines[y]:sub(x, x + w - 1)
end
local sum = 0
local gears = {}
for y, line in ipairs(lines) do
local start = 1
local outLine = line
while true do
local newStart, numEnd = line:find("%d+", start)
if not newStart then break end
local symbol = false
local num = tonumber(line:sub(newStart, numEnd))
for y = y - 1, y + 1 do
local surrounding = at(newStart - 1, y, numEnd - newStart + 3)
if surrounding then
if surrounding and surrounding:match("[^.%d]") then
symbol = true
end
for i = 1, #surrounding do
local gear = surrounding:sub(i, i) == "*"
if gear then
if not gears[y] then
gears[y] = {}
end
local x = i + newStart - 2
if not gears[y][x] then
gears[y][i + newStart - 2] = {}
end
table.insert(gears[y][x], num)
end
end
end
end
if symbol then
sum = sum + num
end
start = numEnd + 1
end
end
print(sum)
local ratio = 0
for _, line in pairs(gears) do
for _, gears in pairs(line) do
if #gears == 2 then
ratio = ratio + gears[1] * gears[2]
end
end
end
print(ratio)
Hare (Part one only)
// SPDX-FileCopyrightText: 2023 Jummit
//
// SPDX-License-Identifier: GPL-3.0-or-later
use strings;
use regex;
use fmt;
use os;
use bufio;
use io;
use strconv;
use types;
fn star_in(lines: []str, x: uint, y: uint, w: uint) bool = {
let start = y;
if (start > 0) start -= 1;
let end = y + 1;
if (end >= len(lines)) end -= 1;
const re = regex::compile(`[^.0-9]`)!;
for (let h = start; h <= end; h += 1) {
fmt::println(strings::sub(lines[h], x, x + w))!;
if (regex::test(&re, strings::sub(lines[h], x, x + w))) {
fmt::println("")!;
return true;
};
};
fmt::println("")!;
return false;
};
export fn main() void = {
const file = os::open("3.input")!;
defer io::close(file)!;
const buf = bufio::newscanner(file, types::SIZE_MAX);
let lines: []str = [];
defer strings::freeall(lines);
for (true) {
match (bufio::scan_line(&buf)!) {
case io::EOF =>
break;
case let line: const str =>
append(lines, strings::dup(line));
};
};
const height = len(lines);
const width = len(lines[0]);
let sum: uint = 0;
let gears: [](uint, uint) = [];
const num_re = regex::compile(`[0-9]+`)!;
for (let y = 0u; y < len(lines); y += 1) {
let nums = regex::findall(&num_re, lines[y]);
defer regex::result_freeall(nums);
for (let i = 0z; i < len(nums); i += 1) {
for (let j = 0z; j < len(nums[i]); j += 1) {
const find = nums[i][j];
const num = strconv::stou(find.content)!;
let start = find.start: uint;
let w = len(find.content): uint + 2;
if (start > 0) {
start -= 1;
} else {
w -= 1;
};
if (star_in(lines, start, y, w)) {
sum += num;
};
};
};
};
fmt::printfln("{}", sum)!;
};
Part 2 stumped me for a little bit, it wasn't an obvious extension of part 1. Part 1 was about numbers (with one or more ...) while part 2 worked from the symbols (with exactly two ...). Going the other way would require more bookkeeping to avoid double counting.
And for the implementation: if you loop over the grid and check surrounding cells for digits you'd have to account for a bunch of cases, e.g. NW/N or N/NE being part of the same number or NW and NE being part of separate numbers. And you'd have to parse the numbers again. But building a graph or reference list of some sort is both unergonomic with C and not necessarily any simpler.
I ended up just writing out the cases, and honestly it didn't turn out too bad.
int main(int argc, char **argv)
{
static char G[GSZ][GSZ];
static int N[GSZ][GSZ];
int p1=0,p2=0, h=0, x,y, dx,dy, n=0,sym=0,r;
for (h=0; fgets(&G[h+1][1], GSZ-1, stdin); h++)
assert(h < GSZ);
/*
* Pass 1: parse numbers and solve part 1. For every digit in
* G, the full number it is part of is stored in N.
*/
for (y=1; y<=h; y++)
for (x=1; G[y][x]; x++)
if (isdigit(G[y][x])) {
n = n*10 + G[y][x]-'0';
for (dy=-1; dy<2; dy++)
for (dx=-1; dx<2; dx++)
sym = sym || (x && y &&
G[y+dy][x+dx] != '.' &&
ispunct(G[y+dy][x+dx]));
} else {
for (dx=-1; isdigit(G[y][x+dx]); dx--)
N[y][x+dx] = n;
if (sym)
p1 += n;
n = sym = 0;
}
/*
* Pass 2: solve part 2 by finding all '*', then counting and
* multiplying adjecent numbers.
*
* Horizontal adjecency is trivial but vertical/diagonal has
* two situations: if there's a digit directly North of the '+',
* it must be a single number: NW and NE would connect to it.
* If N isn't a digit, digits in NW and NE belong to separate
* numbers.
*/
for (y=1; y<=h; y++)
for (x=1; G[y][x]; x++) {
if (G[y][x] != '*')
continue;
n = 0; r = 1;
if (N[y][x-1]) { n++; r *= N[y][x-1]; }
if (N[y][x+1]) { n++; r *= N[y][x+1]; }
if (N[y-1][x]) { n++; r *= N[y-1][x]; } else {
if (N[y-1][x-1]) { n++; r *= N[y-1][x-1]; }
if (N[y-1][x+1]) { n++; r *= N[y-1][x+1]; }
}
if (N[y+1][x]) { n++; r *= N[y+1][x]; } else {
if (N[y+1][x-1]) { n++; r *= N[y+1][x-1]; }
if (N[y+1][x+1]) { n++; r *= N[y+1][x+1]; }
}
if (n == 2)
p2 += r;
}
printf("%d %d\n", p1, p2);
return 0;
}
Holy moley, if this year is intended to be impervious to AI solution, it's also working pretty well as a filter to this paltry Human Intelligence.
Find interesting symbols, look around for digits, and expand these into numbers. A dirty hacky solution leaning on a re-used Grid class. Not recommended.
late ListGrid grid;
Index look(Index ix, Index dir) {
var here = ix;
var next = here + dir;
while (grid.isInBounds(next) && '1234567890'.contains(grid.at(next))) {
here = next;
next = here + dir;
}
return here;
}
/// Return start and end indices of a number at a point.
(Index, Index) expandNumber(Index ix) =>
(look(ix, Grid.dirs['L']!), look(ix, Grid.dirs['R']!));
int parseNumber((Index, Index) e) => int.parse([
for (var i = e.$1; i != e.$2 + Grid.dirs['R']!; i += Grid.dirs['R']!)
grid.at(i)
].join());
/// Return de-duplicated positions of all numbers near the given symbols.
nearSymbols(Set syms) => [
for (var ix in grid.indexes.where((i) => syms.contains(grid.at(i))))
{
for (var n in grid
.near8(ix)
.where((n) => ('1234567890'.contains(grid.at(n)))))
expandNumber(n)
}.toList()
];
part1(List lines) {
grid = ListGrid([for (var e in lines) e.split('')]);
var syms = lines
.join('')
.split('')
.toSet()
.difference('0123456789.'.split('').toSet());
// Find distinct number locations near symbols and sum them.
return {
for (var ns in nearSymbols(syms))
for (var n in ns) n
}.map(parseNumber).sum;
}
part2(List lines) {
grid = ListGrid([for (var e in lines) e.split('')]);
// Look for _pairs_ of numbers near '*' and multiply them.
var products = [
for (var ns in nearSymbols({'*'}).where((e) => e.length == 2))
ns.map(parseNumber).reduce((s, t) => s * t)
];
return products.sum;
}
My Python solution for part 1 and part 2. I really practice my regex skills.
spoiler
#!/usr/bin/python3
import re
value_re = '(\d+)'
symbol_re = '[^\d.]'
gear_re = '(\*)'
def main():
input = list()
with open("input.txt", 'r') as in_file:
for line in in_file:
input.append(line.strip('\n'))
length = len(input)
width = len(input[0])
value_sum = 0
for idx, line in enumerate(input):
for match in re.finditer(value_re, line):
for line_mask in input[max(idx - 1, 0):min(idx + 2, length)]:
valid_chars = line_mask[max(match.span()[0] - 1, 0):min(match.span()[1] + 1, width)]
if re.search(symbol_re, valid_chars):
value_sum += int(match[0])
break
print(f"Value sum = {value_sum}")
gear_ratio = 0
for idx, line in enumerate(input):
for match in re.finditer(gear_re, line):
valid_lines = input[max(idx - 1, 0):min(idx + 2, length)]
min_range = max(match.span()[0] - 1, 0)
max_range = min(match.span()[1], width)
num_of_adjacent = 0
temp_gear_ratio = 1
for valid_line in valid_lines:
for match in re.finditer(value_re, valid_line):
if match.span()[0] in range(min_range,max_range + 1) or match.span()[1] - 1 in range(min_range,max_range + 1):
num_of_adjacent += 1
temp_gear_ratio *= int(match[0])
if num_of_adjacent == 2:
gear_ratio += temp_gear_ratio
print(f"Gear ratio = {gear_ratio}")
if __name__ == '__main__':
main()
Classic AoC grid problem... Tedious as usual, but very doable. Took my time and I'm pretty happy with the result. :]
Part 1
For the first part, I decided to break the problem into: 1. Reading the schematic, 2. Finding the numbers, 3. Finding the parts. This was useful for Part 2 as I could re-use my read_schematic and find_numbers functions.
Two things I typically do for grid problems:
Pad the grid so you can avoid annoying boundary checks.
I have a DIRECTIONS list I loop through so I can check easily check the neighbors.
Schematic = list[str]
Number = tuple[int, int, int]
DIRECTIONS = (
(-1, -1),
(-1, 0),
(-1, 1),
( 0, -1),
( 0, 1),
( 1, -1),
( 1, 0),
( 1, 1),
)
def read_schematic(stream=sys.stdin) -> Schematic:
schematic = [line.strip() for line in stream]
columns = len(schematic[0]) + 2
return [
'.'*columns,
*['.' + line + '.' for line in schematic],
'.'*columns,
]
def is_symbol(s: str) -> bool:
return not (s.isdigit() or s == '.')
def find_numbers(schematic: Schematic) -> Iterator[Number]:
rows = len(schematic)
columns = len(schematic[0])
for r in range(1, rows):
for number in re.finditer(r'[0-9]+', schematic[r]):
yield (r, *number.span())
def find_parts(schematic: Schematic, numbers: Iterator[Number]) -> Iterator[int]:
for r, c_head, c_tail in numbers:
part = int(schematic[r][c_head:c_tail])
for c in range(c_head, c_tail):
neighbors = (schematic[r + dr][c + dc] for dr, dc in DIRECTIONS)
if any(is_symbol(neighbor) for neighbor in neighbors):
yield part
break
def main(stream=sys.stdin) -> None:
schematic = read_schematic(stream)
numbers = find_numbers(schematic)
parts = find_parts(schematic, numbers)
print(sum(parts))
Part 2
For the second part, I just found the stars, and then I found the gears by checking if the stars are next to two numbers (which I had found previously).
def find_stars(schematic: Schematic) -> Iterator[Star]:
rows = len(schematic)
columns = len(schematic[0])
for r in range(1, rows):
for c in range(1, columns):
token = schematic[r][c]
if token == '*':
yield (r, c)
def find_gears(schematic: Schematic, stars: Iterator[Star], numbers: list[Number]) -> Iterator[int]:
for star_r, star_c in stars:
gears = [
int(schematic[number_r][number_c_head:number_c_tail])
for number_r, number_c_head, number_c_tail in numbers
if any(star_r + dr == number_r and number_c_head <= (star_c + dc) < number_c_tail for dr, dc in DIRECTIONS)
]
if len(gears) == 2:
yield gears[0] * gears[1]
def main(stream=sys.stdin) -> None:
schematic = read_schematic(stream)
numbers = find_numbers(schematic)
stars = find_stars(schematic)
gears = find_gears(schematic, stars, list(numbers))
print(sum(gears))
Another day of the 2023 Advent of Code, and another day where I hate looking
at my code. This year just seems like it is starting off a lot more complex
than I remember in previous years. This one was a little tricky, but I got
there without any major setbacks. Another one I am excited to come back to and
clean up, but this first pass is all about getting a solution, and this one
works.
I kept trying to create clever solutions, but ended up falling back on regex when it was taking to long. THE TLDR is we scan the list of strings for a symbol, then parse the three lines above, below and inline with the symbol for digits. Then we try and match the indexes of the match and the area around the symbol. Part 2 was a small modification, and was mostly about getting the existing code to conform the data into a pattern for each of the three lines.
Part 1
static char[] Symbols = { '@', '#', '$', '%', '&', '*', '/', '+', '-', '=' };
string pattern = @"\d+";
static List? list;
list = new List((await File.ReadAllLinesAsync(@".\Day 3\PuzzleInput.txt")));
int count = 0;
for (int row = 0; row < list.Count; row++)
{
for (int col = 0; col < list[row].Length; col++)
{
var c = list[row][col];
if (c == '.')
{
continue;
}
if (Symbols.Contains(c))
{
var res = Calculate(list[row - 1], col);
res += Calculate(list[row], col);
res += Calculate(list[row + 1], col);
count += res;
}
}
}
Console.WriteLine(count);
private static int Calculate(string line, int col)
{
List indexesToCheck = new List { col - 1, col, col + 1 };
int count = 0;
MatchCollection matches = Regex.Matches(line, pattern);
foreach (Match match in matches)
{
string number = match.Value;
if (AnyIndexInList(indexesToCheck, match.Index, match.Length))
{
count += Int32.Parse(number);
}
}
return count;
}
static bool AnyIndexInList(List list, int startIndex, int length)
{
for (int i = startIndex; i < startIndex + length; i++)
{
if (list.Contains(i))
{
return true;
}
}
return false;
}
Part 2:
list = new List((await File.ReadAllLinesAsync(@".\Day 3\PuzzleInput.txt")));
int count = 0;
for (int row = 0; row < list.Count; row++)
{
for (int col = 0; col < list[row].Length; col++)
{
var c = list[row][col];
if (c == '.')
continue;
if (c == '*')
{
var res1 = Calculate2(list[row - 1], col);
var res2 = Calculate2(list[row], col);
var res3 = Calculate2(list[row + 1], col);
count += (res1, res2, res3) switch
{
{res1: not null, res2: null, res3: null } when res1[1] != null => res1[0].Value * res1[1].Value,
{res1: null, res2: not null, res3: null } when res2[1] != null => res2[0].Value * res2[1].Value,
{res1: null, res2: null, res3: not null } when res3[1] != null => res3[0].Value * res3[1].Value,
{res1: not null, res2: not null, res3: null } => res1[0].Value * res2[0].Value,
{res1: not null, res2: null, res3: not null } => res1[0].Value * res3[0].Value,
{res1: null, res2: not null, res3: not null } => res2[0].Value * res3[0].Value,
{res1: not null, res2: not null, res3: not null } => res1[0].Value * res2[0].Value * res3[0].Value,
_ => 0
} ;
}
}
}
Console.WriteLine(count);
private static int?[]? Calculate2(string line, int col)
{
List indexesToCheck = new List { col - 1, col, col + 1 };
int?[]? count = null;
MatchCollection matches = Regex.Matches(line, pattern);
foreach (Match match in matches)
{
string number = match.Value;
if (AnyIndexInList(indexesToCheck, match.Index, match.Length))
{
if (count == null)
count = new int?[2] { Int32.Parse(number), null };
else {
count[1] = Int32.Parse(number);
};
}
}
return count;
}
I'll only post the actual parsing and solution. I have written some helpers which are in other files, as is the main function. For the full code, please see my github repo.
Here I used HashMap and HashSet, but that's just an optimization. I'm not even sure if they are faster than just using lists here...
Solution
import Lean.Data.HashSet
import Lean.Data.HashMap
namespace Day3
structure Coordinate : Type 0 where
x : Nat
y : Nat
deriving Repr, BEq, Ord, Hashable
def Coordinate.default : Coordinate := {x := 0, y := 0}
/--Returns the adjacent fields, row-major order (this is important)-/
def Coordinate.adjacents : Coordinate → List Coordinate
| { x := 0, y := 0} => [
⟨1,0⟩,
⟨0,1⟩, ⟨1,1⟩
]
| { x := 0, y := y} => [
⟨0,y.pred⟩, ⟨1,y.pred⟩,
⟨1,y⟩,
⟨0,y.succ⟩, ⟨1,y.succ⟩
]
| { x := x, y := 0} => [
⟨x.pred,0⟩, ⟨x.succ,0⟩,
⟨x.pred,1⟩, ⟨x,1⟩, ⟨x.succ,1⟩
]
| { x := x, y := y} => [
⟨x.pred,y.pred⟩, ⟨x,y.pred⟩, ⟨x.succ,y.pred⟩,
⟨x.pred,y⟩, ⟨x.succ,y⟩,
⟨x.pred,y.succ⟩, ⟨x,y.succ⟩, ⟨x.succ,y.succ⟩
]
structure Part : Type 0 where
symbol : Char
position : Coordinate
deriving Repr
structure PartNumber : Type 0 where
value : Nat
positions : List Coordinate
deriving Repr, BEq
-- Schematic is just using lists, because at this point it's not clear what kind of lookup
-- is needed in part 2... Probably some kind of HashMap Coordinate Whatever, but that's
-- guesswork for now...
-- Parts can refine the data further, anyhow.
structure Schematic : Type 0 where
parts : List Part
numbers : List PartNumber
deriving Repr
/-- nextByChar gives the coordinate of the **next** character. -/
private def Coordinate.nextByChar : Coordinate → Char → Coordinate
| {x := _, y := oldY}, '\n' => { x := 0, y := oldY + 1 }
| {x := oldX, y := oldY}, _ => { x := oldX + 1, y := oldY }
private def extractParts : List (Coordinate × Char) → List Part :=
(List.map (uncurry $ flip Part.mk)) ∘ (List.filter $ not ∘ λ (sc : Coordinate × Char) ↦ sc.snd.isWhitespace || sc.snd.isDigit || sc.snd == '.')
private def extractPartNumbers (input : List (Coordinate × Char)) : List PartNumber :=
let rec helper := λ
| [], none => []
| [], some acc => [acc] -- if we are still accumulating a number at the end, commit
| a :: as, none =>
if p: a.snd.isDigit then
--start accumulating a PartNumber
helper as $ some {value := a.snd.asDigit p, positions := [a.fst]}
else
--not accumulating, not a digit, not of interest.
helper as none
| a :: as, some acc =>
if p: a.snd.isDigit then
--keep accumulating
helper as $ some {value := acc.value * 10 + a.snd.asDigit p, positions := a.fst :: acc.positions }
else
-- we were accumulating, aren't on a number any more -> commit
acc :: helper as none
helper input none
def parse (schematic : String) : Option Schematic := do
-- I think this one is easier if we don't split the input in lines. Because:
let charsWithCoordinates ← match schematic.toList with
| [] => none
| c :: cs => pure $ cs.scan (λ s c ↦ (uncurry Coordinate.nextByChar s, c)) (Coordinate.default, c)
-- Whitespaces are **intentionally** left in. This makes extracting the numbers easier.
pure $ {
parts := extractParts charsWithCoordinates
numbers := extractPartNumbers charsWithCoordinates
}
def part1 (schematic : Schematic) : Nat :=
-- fast lookup: We need to know if a part is at a given coordinate
open Lean(HashSet) in
let partCoordinates := HashSet.insertMany HashSet.empty $ schematic.parts.map Part.position
let partNumbers := schematic.numbers.filter λnumber ↦
number.positions.any λposition ↦
position.adjacents.any partCoordinates.contains
partNumbers.foldl (· + PartNumber.value ·) 0
def part2 (schematic : Schematic) : Nat :=
-- and now it is obvious that keeping the parsed input free of opinions was a good idea.
-- because here we need quick lookup for the numbers, not the parts.
open Lean(HashMap) in
let numberCoordinates : HashMap Coordinate PartNumber :=
Lean.HashMap.ofList $ schematic.numbers.bind $ λ pn ↦ pn.positions.map (·, pn)
let gearSymbols := schematic.parts.filter (Part.symbol · == '*')
-- but the symbols aren't enough, they need to be adjacent to **exactly** 2 numbers
let numbersNextGearSymbols := List.dedup <$> gearSymbols.map λgs ↦
gs.position.adjacents.filterMap numberCoordinates.find?
let gearSymbols := numbersNextGearSymbols.filter (List.length · == 2)
let gearRatios := gearSymbols.map $ List.foldl (· * PartNumber.value ·) 1
gearRatios.foldl (· + ·) 0
Here's a tip: if you are using a language / standard library that doesn't have a set, you can mimic it with a map from your key to a nullary (in this case an empty struct)
Lemmy doesn't handle certain characters well currently such as left angle brackets and ampersands due to some sanitization in the back end to stop scripting attacks