Rust
Part 1 was super simple with wrapping_add
and wrapping_mul
on a u8
. Building an actual hash map in Part 2 was nice.
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console.log('Hello World')
Part 1 was super simple with wrapping_add
and wrapping_mul
on a u8
. Building an actual hash map in Part 2 was nice.
That array initialisation is pure poetry! π
I'm not fluent in Rust, but is this something like the C++ placement new? Presumably just declaring a table of Vecs won't automatically call the default constructor? (Sorry for my total ignorance -- pointers to appropriate reading material appreciated)
You can create an array filled with all the same values in Rust, but only if all values have the same memory representation because they will be copied. That just doesn't work with Vec's, because they must all have their own unique pointer. And to have uninitialized values at first (think NULL-pointers for every Vec) while creating each Vec, something like this is apparently needed.
The appropriate way would certainly have been to store the map as a Vec>
instead of an array, but I just wanted to see if could.
Ah, I see! Thank you.
Took a while to figure out what part 2 was all about. Didn't have the energy to golf this one further today, so looking forward to seeing the other solutions!
Solution
0.3 line-seconds
import Data.Char
import Data.List
import Data.List.Split
import qualified Data.Vector as V
hash :: String -> Int
hash = foldl' (\a c -> ((a + ord c) * 17) `rem` 256) 0
hashmap :: [String] -> Int
hashmap = focus . V.toList . foldl' step (V.replicate 256 [])
where
focus = sum . zipWith focusBox [1 ..]
focusBox i = sum . zipWith (\j (_, z) -> i * j * z) [1 ..] . reverse
step boxes s =
let (label, op) = span isLetter s
i = hash label
in case op of
['-'] -> V.accum (flip filter) boxes [(i, (/= label) . fst)]
('=' : z) -> V.accum replace boxes [(i, (label, read z))]
replace ls (n, z) =
case findIndex ((== n) . fst) ls of
Just j ->
let (a, _ : b) = splitAt j ls
in a ++ (n, z) : b
Nothing -> (n, z) : ls
main = do
input <- splitOn "," . head . lines <$> readFile "input15"
print $ sum . map hash $ input
print $ hashmap input
def hash(s: String): Long = s.foldLeft(0)((h, c) => (h + c)*17 % 256)
extension [A] (a: List[A])
def mapAtIndex(idx: Long, f: A => A): List[A] =
a.zipWithIndex.map((e, i) => if i == idx then f(e) else e)
def runProcedure(steps: List[String]): Long =
@tailrec def go(boxes: List[List[(String, Int)]], steps: List[String]): List[List[(String, Int)]] =
steps match
case s"$label-" :: t =>
go(boxes.mapAtIndex(hash(label), _.filter(_._1 != label)), t)
case s"$label=$f" :: t =>
go(boxes.mapAtIndex(hash(label), b =>
val slot = b.map(_._1).indexOf(label)
if slot != -1 then b.mapAtIndex(slot, (l, _) => (l, f.toInt)) else (label, f.toInt) :: b
), t)
case _ => boxes
go(List.fill(256)(List()), steps).zipWithIndex.map((b, i) =>
b.zipWithIndex.map((lens, ilens) => (1 + i) * (b.size - ilens) * lens._2).sum
).sum
def task1(a: List[String]): Long = a.head.split(",").map(hash).sum
def task2(a: List[String]): Long = runProcedure(a.head.split(",").toList)
Yes, it's a hash table. Did I pick a language with built in hash tables? Of course I didn't. Could I have used one of the many libraries implementing one? Sure. But the real question is, can we make do with stuffing things into a few static arrays at nearly zero memory and runtime cost? Yes!
In the spirit of Fred Brooks, itβll suffice here to show my data structures:
struct slot { char label[8]; int lens; };
struct box { struct slot slots[8]; int nslots; };
static struct box boxes[256];
0.248 line-seconds (sixth simplest so far after days 6, 2, 1, 4 and 9).
import collections
import re
from .solver import Solver
def _hash(string: str) -> int:
result = 0
for c in string:
result = (result + ord(c)) * 17 % 256
return result
def _assert_full_match(pattern: str, string: str):
m = re.fullmatch(pattern, string)
if not m:
raise RuntimeError(f'pattern {pattern} does not match {string}')
return m
class Day15(Solver):
input: list[str]
def __init__(self):
super().__init__(15)
def presolve(self, input: str):
self.input = input.rstrip().split(',')
def solve_first_star(self) -> int:
return sum(_hash(string) for string in self.input)
def solve_second_star(self) -> int:
boxes = [collections.OrderedDict() for _ in range(256)]
for instruction in self.input:
label, op, value = _assert_full_match(r'([a-z]+)([=-])(\d*)', instruction).groups()
box = boxes[_hash(label)]
match op:
case '-':
if label in box:
del box[label]
case '=':
box[label] = value
return sum((1 + box_idx) * (1 + lens_idx) * int(value)
for box_idx, box in enumerate(boxes)
for lens_idx, (_, value) in enumerate(box.items()))
Had to take a couple days off, but this was a nice one to come back to. Will have to find some time today to go back and do one or two of the 3 that I missed. I don't have much to say about this one - I had an idea almost immediately and it worked out without much struggle. There's probably some cleaner ways to write parts of this, but I'm not too disappointed with how it turned out.
https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day15.rs
use crate::Solver;
use std::collections::HashMap;
#[derive(Debug)]
struct Lens {
label: String,
focal_length: u32,
}
fn hash_algorithm(input: &str) -> u32 {
input
.chars()
.fold(0, |acc, ch| (acc + ch as u32) * 17 % 256)
}
pub struct Day15;
impl Solver for Day15 {
fn star_one(&self, input: &str) -> String {
input
.trim_end()
.split(',')
.map(hash_algorithm)
.sum::()
.to_string()
}
fn star_two(&self, input: &str) -> String {
let mut boxes: HashMap> = HashMap::new();
for instruction in input.trim_end().split(',') {
let (label, focal_length) = instruction
.split_once(|ch| char::is_ascii_punctuation(&ch))
.unwrap();
let box_number = hash_algorithm(label);
let lenses = boxes.entry(box_number).or_insert(vec![]);
if focal_length == "" {
lenses.retain(|lens| lens.label != label);
continue;
}
let new_lens = Lens {
label: label.to_string(),
focal_length: focal_length.parse().unwrap(),
};
if let Some(lens_index) = lenses.iter().position(|lens| lens.label == new_lens.label) {
lenses[lens_index].focal_length = new_lens.focal_length;
} else {
lenses.push(new_lens);
}
}
boxes
.iter()
.map(|(box_number, lenses)| {
lenses
.iter()
.enumerate()
.map(|(lens_index, lens)| {
(box_number + 1) * (lens_index as u32 + 1) * lens.focal_length
})
.sum::()
})
.sum::()
.to_string()
}
}
Just written as specced. If there's any underlying trick, I missed it totally.
9ms * 35 LOC ~= 0.35, so it'll do.
int decode(String s) => s.codeUnits.fold(0, (s, t) => ((s + t) * 17) % 256);
part1(List lines) => lines.first.split(',').map(decode).sum;
part2(List lines) {
var rules = lines.first.split(',').map((e) {
if (e.contains('-')) return ('-', e.skipLast(1), 0);
var parts = e.split('=');
return ('=', parts.first, int.parse(parts.last));
});
var boxes = Map.fromEntries(List.generate(256, (ix) => MapEntry(ix, [])));
for (var r in rules) {
if (r.$1 == '-') {
boxes[decode(r.$2)]!.removeWhere((l) => l.$1 == r.$2);
} else {
var box = boxes[decode(r.$2)]!;
var lens = box.indexed().firstWhereOrNull((e) => e.value.$1 == r.$2);
var newlens = (r.$2, r.$3);
(lens == null) ? box.add(newlens) : box[lens.index] = newlens;
}
}
return boxes.entries
.map((b) =>
(b.key + 1) *
b.value.indexed().map((e) => (e.index + 1) * e.value.$2).sum)
.sum;
}
9ms * 35 LOC is ~0.350 tho
Thatβs why I normally let computers do my sums for me. Corrected now.
This felt ... too simple. I think the hardest part of part two for me was reading comprehension. My errors were typically me not reading exactly was there.
Python
import re
import math
import argparse
import itertools
def int_hash(string:str) -> int:
hash = 0
for c in [*string]:
hash += ord(c)
hash *= 17
hash = hash % 256
return hash
class Instruction:
def __init__(self,string:str) -> None:
label,action,strength = re.split('([-=])',string)
self.label = label
self.action = action
if not strength:
strength = 0
self.strength = int(strength)
def __repr__(self) -> str:
return f"Instruction(l={self.label}, a={self.action}, s={self.strength})"
def __str__(self) -> str:
stren = str(self.strength if self.strength > 0 else '')
return f"{self.label}{self.action}{stren}"
class Lens:
def __init__(self,label:str,focal_length:int) -> None:
self.label:str = label
self.focal_length:int = focal_length
def __repr__(self) -> str:
return f"Lens(label:{self.label},focal_length:{self.focal_length})"
def __str__(self) -> str:
return f"[{self.label} {self.focal_length}]"
def main(line_list:str,part:int):
init_sequence = line_list.splitlines(keepends=False)[0].split(',')
sum = 0
focal_array = dict[int,list[Lens]]()
for i in range(0,256):
focal_array[i] = list[Lens]()
for s in init_sequence:
hash_value = int_hash(s)
sum += hash_value
# part 2 stuff
action = Instruction(s)
position = int_hash(action.label)
current_list = focal_array[position]
existing_lens = list(filter(lambda x:x.label == action.label,current_list))
if len(existing_lens) > 1:
raise Exception("multiple of same lens in box, what do?")
match action.action:
case '-':
if len(existing_lens) == 1:
current_list.remove(existing_lens[0])
case '=':
if len(existing_lens) == 0:
current_list.append(Lens(action.label,action.strength))
if len(existing_lens) == 1:
existing_lens[0].focal_length = action.strength
case _:
raise Exception("unknown action")
print(f"Part1: {sum}")
#print(focal_array)
sum2 = 0
for i,focal_box in focal_array.items():
for l,lens in enumerate(focal_box):
sum2 += ( (i+1) * (l+1) * lens.focal_length )
print(f"Part2: {sum2}")
if __name__ == "__main__":
parser = argparse.ArgumentParser(description="template for aoc solver")
parser.add_argument("-input",type=str)
parser.add_argument("-part",type=int)
args = parser.parse_args()
filename = args.input
if filename == None:
parser.print_help()
exit(1)
part = args.part
file = open(filename,'r')
main(file.read(),part)
file.close()
Almost caught up. Not much to say about this one. Part 1 was a freebie. Part 2 had a convoluted description, but was still pretty easy.
import Data.Array
import qualified Data.ByteString.Char8 as BS
import Data.Char (isAlpha, isDigit)
import Relude
import qualified Relude.Unsafe as Unsafe
import Text.ParserCombinators.ReadP hiding (get)
hash :: String -> Int
hash = foldl' (\a x -> (a + x) * 17 `mod` 256) 0 . fmap ord
part1 :: ByteString -> Int
part1 = sum . fmap (hash . BS.unpack) . BS.split ',' . BS.dropEnd 1
-- Part 2
type Problem = [Operation]
type S = Array Int [(String, Int)]
data Operation = Set String Int | Remove String deriving (Show)
parse :: BS.ByteString -> Maybe Problem
parse = fmap fst . viaNonEmpty last . readP_to_S parse' . BS.unpack
where
parse' = sepBy parseOperation (char ',') <* char '\n' <* eof
parseOperation =
munch1 isAlpha
>>= \label -> (Remove label <$ char '-') +++ (Set label . Unsafe.read <$> (char '=' *> munch1 isDigit))
liftOp :: Operation -> Endo S
liftOp (Set label v) = Endo $ \s ->
let (b, a) = second (drop 1) $ span ((/= label) . fst) (s ! hash label)
in s // [(hash label, b <> [(label, v)] <> a)]
liftOp (Remove l) = Endo $ \s -> s // [(hash l, filter ((/= l) . fst) (s ! hash l))]
score :: S -> Int
score m = sum $ join [(* (i + 1)) <$> zipWith (*) [1 ..] (snd <$> (m ! i)) | i <- [0 .. 255]]
part2 :: ByteString -> Maybe Int
part2 input = do
ops <- appEndo . foldMap liftOp . reverse <$> parse input
pure . score . ops . listArray (0, 255) $ repeat []
Nice use of foldMap
!
My whole solution can be expressed in just two words: Ordered HashTable
Total runtime: 0.068 line-seconds (40 LOC * 1.7 ms)
Puzzle rating: exceptionally confusing description 4/10
Code: cleaned up solution with types
Snippet:
proc getHash(s: string): int =
for c in s:
result = ((result + c.ord) * 17) mod 256
proc solve(lines: seq[string]): AOCSolution[int] =
var boxes: array[256, OrderedTable[string, int]]
for line in lines:
block p1:
result.part1 += line.getHash()
block p2:
if line.endsWith('-'):
var name = line.strip(leading=false, chars={'-'})
boxes[getHash(name)].del(name)
else:
let (name, _, value) = line.partition("=")
boxes[getHash(name)][name] = value[0].ord - '0'.ord
for bi, box in boxes:
if box.len < 1: continue
for vi, val in enumerate(box.values):
result.part2 += (bi+1) * (vi+1) * val