Day 2: Red-Nosed Reports
Megathread guidelines
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FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/22323136
- Where do I participate?: https://adventofcode.com/
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#Rust
initially, for part two I was trying to ignore a bad pair not a bad value - read the question!
Only installed Rust on Sunday, day 1 was a mess, today was more controlled. Need to look at some of the rust solutions for std library methods I don’t know about.
very focussed on getting it to actually compile/work over making it short or nice!
long!
`
pub mod task_2 {
pub fn task_1(input: &str) -> i32{ let mut valid_count = 0; let reports = process_input(input); for report in reports{ let valid = is_report_valid(report); if valid{ valid_count += 1; } } println!("Valid count: {}", valid_count); valid_count } pub fn task_2(input: &str) -> i32{ let mut valid_count = 0; let reports = process_input(input); for report in reports{ let mut valid = is_report_valid(report.clone()); if !valid { for position_to_delete in 0..report.len() { let mut updated_report = report.clone(); updated_report.remove(position_to_delete); valid = is_report_valid(updated_report); if valid { break; } } } if valid{ valid_count += 1; } } println!("Valid count: {}", valid_count); valid_count } fn is_report_valid(report:Vec<i32>) -> bool{ let mut increasing = false; let mut decreasing = false; let mut valid = true; for position in 1..report.len(){ if report[position-1] > report[position] { decreasing = true; } else if report[position-1] < report[position] { increasing = true; } else { valid = false; break; } if (report[position-1] - report[position]).abs() > 3 { valid = false; break; } if increasing && decreasing { valid = false; break; } } return valid; } pub fn process_input(input: &str) -> Vec<Vec<i32>>{ let mut reports: Vec<Vec<i32>> = Vec::new(); for report_string in input.split("\n"){ let mut report: Vec<i32> = Vec::new(); for value in report_string.split_whitespace() { report.push(value.parse::<i32>().unwrap()); } reports.push(report); } return reports; }
}
`
Elixir
defmodule Day02 do defp part1(reports) do reports |> Enum.map(fn report -> levels = report |> String.split() |> Enum.map(&String.to_integer/1) cond do sequence_is_safe?(levels) -> :safe true -> :unsafe end end) |> Enum.count(fn x -> x == :safe end) end defp part2(reports) do reports |> Enum.map(fn report -> levels = report |> String.split() |> Enum.map(&String.to_integer/1) sequences = 0..(length(levels) - 1) |> Enum.map(fn i -> List.delete_at(levels, i) end) cond do sequence_is_safe?(levels) -> :safe Enum.any?(sequences, &sequence_is_safe?/1) -> :safe true -> :unsafe end end) |> Enum.count(fn x -> x == :safe end) end defp all_gaps_within_max_diff?(numbers) do numbers |> Enum.chunk_every(2, 1, :discard) |> Enum.all?(fn [a, b] -> abs(b - a) <= 3 end) end defp is_strictly_increasing?(numbers) do numbers |> Enum.chunk_every(2, 1, :discard) |> Enum.all?(fn [a, b] -> a < b end) end defp is_strictly_decreasing?(numbers) do numbers |> Enum.chunk_every(2, 1, :discard) |> Enum.all?(fn [a, b] -> a > b end) end defp sequence_is_safe?(numbers) do (is_strictly_increasing?(numbers) or is_strictly_decreasing?(numbers)) and all_gaps_within_max_diff?(numbers) end def run(data) do reports = data |> String.split("\n", trim: true) p1 = part1(reports) p2 = part2(reports) IO.puts(p1) IO.puts(p2) end end data = File.read!("input.in") Day02.run(data)
Factor
: get-input ( -- reports ) "vocab:aoc-2024/02/input.txt" utf8 file-lines [ split-words [ string>number ] map ] map ; : slanted? ( report -- ? ) { [ [ > ] monotonic? ] [ [ < ] monotonic? ] } || ; : gradual? ( report -- ? ) [ - abs 1 3 between? ] monotonic? ; : safe? ( report -- ? ) { [ slanted? ] [ gradual? ] } && ; : part1 ( -- n ) get-input [ safe? ] count ; : fuzzy-reports ( report -- reports ) dup length <iota> [ remove-nth-of ] with map ; : tolerable? ( report -- ? ) { [ safe? ] [ fuzzy-reports [ safe? ] any? ] } || ; : part2 ( -- n ) get-input [ tolerable? ] count ;
Quite the interesting language choice. It’s so clean. I love it!
G’MIC solution
spoiler
it day2 crop. 0,0,0,{h#-1-2} split. -,{_'\n'} foreach { replace_str. " ",";" ({t}) rm.. } safe_0,safe_1=0 foreach { ({h}) a[-2,-1] y num_of_attempts:=da_size(#-1)+1 store temp repeat $num_of_attempts { $temp if $> eval da_remove(#-1,$>-1) fi eval " safe=1; i[#-1,1]>i[#-1,0]?( for(p=1,p<da_size(#-1),++p, if(!inrange(i[#-1,p]-i[#-1,p-1],1,3,1,1),safe=0;break();); ); ):( for(p=1,p<da_size(#-1),++p, if(!inrange(i[#-1,p-1]-i[#-1,p],1,3,1,1),safe=0;break();); ); ); safe;" rm if $> if ${} safe_1+=1 break fi else if ${} safe_0,safe_1+=1 break fi fi } } echo Day" "2:" "${safe_0}" :: "${safe_1}
Kotlin:
import kotlin.math.abs import kotlin.math.sign data class Report(val levels: List<Int>) { fun isSafe(withProblemDampener: Boolean): Boolean { var orderSign = 0.0f // - 1 is descending; +1 is ascending levels.zipWithNext().forEachIndexed { index, level -> val difference = (level.second - level.first).toFloat() if (orderSign == 0.0f) orderSign = sign(difference) if (sign(difference) != orderSign || abs(difference) !in 1.0..3.0) { // With problem dampener: Drop either element in the pair or the first element from the original list and check if the result is now safe. return if (withProblemDampener) { Report(levels.drop(1)).isSafe(false) || Report(levels.withoutElementAt(index)).isSafe(false) || Report(levels.withoutElementAt(index + 1)).isSafe(false) } else false } } return true } } fun main() { fun part1(input: List<String>): Int = input.map { Report(it.split(" ").map { it.toInt() }).isSafe(false) }.count { it } fun part2(input: List<String>): Int = input.map { Report(it.split(" ").map { it.toInt() }).isSafe(true) }.count { it } // Or read a large test input from the `src/Day01_test.txt` file: val testInput = readInput("Day02_test") check(part1(testInput) == 2) check(part2(testInput) == 4) // Read the input from the `src/Day01.txt` file. val input = readInput("Day02") part1(input).println() part2(input).println() }
The Report#isSafe method essentially solves both parts.
I’ve had a bit of a trip up in part 2:
I initially only checked, if the report was safe, if either elements in the pair were to be removed. But in the edge case, that the first pair has different monotonic behaviour than the rest, the issue would only be detected by the second pair with indices (2, 3), whilst removing the first element in the list would yield a safe report.
Uiua
Took me a bit longer to get this one but still quite simple overall.
Spent quite some time on getting to know thetry
andassert
operators better.Run with example input here
# Get the indices matching the ascending/ # descending criteria CheckAsc ← ≡°□⍚(⍣(⊸⍤.≍⍆.)⍣(⊸⍤.≍⇌⍆.)0) # Get the indices matching the distance criteria CheckDist ← ≡°□⍚(⍣(⊸⍤.≠1∈:0)0×⊓≥≤1,3⌵⧈-) Split ← ⊙(▽≠1)▽,, PartOne ← ( &rs ∞ &fo "input-2.txt" ⊜(□⊜⋕≠@ .)≠@\n. CheckAsc. ▽ CheckDist ⧻⊚ ) PartTwo ← ( &rs ∞ &fo "input-2.txt" ⊜(□⊜⋕≠@ .)≠@\n. CheckAsc. Split CheckDist. Split ⊙(⊂) ⧻ : ⍚(≡(▽:°⊟)⍜¤⊞⊟:≠1⊞=.⇡⧻.) ≡(⧻⊚CheckDist▽CheckAsc.°□) +⧻◴⊚ ) &p "Day 2:" &pf "Part 1: " &p PartOne &pf "Part 2: " &p PartTwo
C#
using MathNet.Numerics.LinearAlgebra; public class Day02 : Solver { private ImmutableList<Vector<Double>> data; public void Presolve(string input) { data = input.Trim().Split("\n") .Select( line => Vector<Double>.Build.DenseOfEnumerable(line.Split(' ').Select(double.Parse)) ).ToImmutableList(); } private bool IsReportSafe(Vector<Double> report) { Vector<Double> delta = report.SubVector(1, report.Count - 1) .Subtract(report.SubVector(0, report.Count - 1)); return (delta.ForAll(x => x > 0) || delta.ForAll(x => x < 0)) && Vector<Double>.Abs(delta).Max() <= 3; } private bool IsDampenedReportSafe(Vector<Double> report) { for (Double i = 0; i < report.Count; ++i) { var dampened = Vector<Double>.Build.DenseOfEnumerable( report.EnumerateIndexed() .Where(item => item.Item1 != i) .Select(item => item.Item2)); if (IsReportSafe(dampened)) return true; } return false; } public string SolveFirst() => data.Where(IsReportSafe).Count().ToString(); public string SolveSecond() => data.Where(IsDampenedReportSafe).Count().ToString(); }
Uiua
Uiua is still developing very quickly, and this code uses the experimental
tuples
function, hence the initial directive.# Experimental! "7 6 4 2 1\n1 2 7 8 9\n9 7 6 2 1\n1 3 2 4 5\n8 6 4 4 1\n1 3 6 7 9" ⊜(⊜⋕⊸≠@\s)⊸≠@\n # Partition at \n, then at space, parse ints. IsSorted ← +⊃(≍⇌⍆.|≍⍆.) # Compare with sorted array. IsSmall ← /××⊃(>0|<4)⌵↘¯1-↻1. # Copy offset by 1, check diffs. IsSafe ← ×⊃IsSmall IsSorted # Safe if Small steps and Ordered. IsSafer ← ±/+≡IsSafe ⧅<-1⧻. # Choose 4 from 5, check again. &p/+≡IsSafe . # Part1 : Is each row safe? &p/+≡(±+⊃IsSafe IsSafer) # Part2 : Is it safe or safer?
How do you write this, not conceptually but physically. Do you have a char picker open at all times?
Haha, you can do it that way, in fact the online Uiua Pad editor has all the operators listed along the top.
But all the operators have ascii names, so you can type e.g.
IsSmall = reduce mul mul fork(>0|<4) abs drop neg 1 - rot 1 dup
and the formatter will reduce that toIsSmall ← /××⊃(>0|<4)⌵↘¯1-↻1.
whenever you save or execute code.That works in the Pad, and you can enable similar functionality in other editors.
i can only imagine doing it with a drawing tablet
I like to assume people using array programming languages just have a crystal ball that they use to call upon magic runes on the screen
This looks so alien! Does it work with the full set? The comment says 5, choose 4, but I guess it’s written as n, choose n-1?
Yes, it should do. I do run the solutions against the live data, but sometimes tweak the solutions afterwards, so can’t always guarantee them :-). I left the comment as 5 choose 4 as it felt clearer in the context of the test data.
It does still feel very alien at times, but I do love being able to think about how to adopt a more arrays-based approach to solving these problems.
TypeScript
Solution
import { AdventOfCodeSolutionFunction } from "./solutions"; /** * this function evaluates the * @param levels a list to check * @returns -1 if there is no errors, or the index of where there's an unsafe event */ export function EvaluateLineSafe(levels: Array<number>) { // this loop is the checking every number in the line let isIncreasing: boolean | null = null; for (let levelIndex = 1; levelIndex < levels.length; levelIndex++) { const prevLevel = levels[levelIndex - 1]; // previous const level = levels[levelIndex]; // current const diff = level - prevLevel; // difference const absDiff = Math.abs(diff); // absolute difference // check if increasing too much or not at all if (absDiff == 0 || absDiff > 3) return levelIndex; // go to the next report // set increasing if needed if (isIncreasing === null) { isIncreasing = diff > 0; continue; // compare the next numbers } // check if increasing then decreasing if (!(isIncreasing && diff > 0 || !isIncreasing && diff < 0)) return levelIndex; // go to the next report } return -1; } export const solution_2: AdventOfCodeSolutionFunction = (input) => { const reports = input.split("\n"); let safe = 0; let safe_damp = 0; // this loop is for every line main: for (let i = 0; i < reports.length; i++) { const report = reports[i].trim(); if (!report) continue; // report is empty const levels = report.split(" ").map((v) => Number(v)); const evaluation = EvaluateLineSafe(levels); if(evaluation == -1) { safe++; continue; } // search around where it failed for (let offset = evaluation - 2; offset <= evaluation + 2; offset++) { // delete an evaluation in accordance to the offset let newLevels = [...levels]; newLevels.splice(offset, 1); const newEval = EvaluateLineSafe(newLevels); if(newEval == -1) { safe_damp++; continue main; } } } return `Part 1: ${safe} Part 2: ${safe + safe_damp}`; }
God, I really wish my solutions weren’t so convoluted. Also, this is an O(N^3) solution…
I don’t think your solution is O(N^3). Can you explain your reasoning?
3 nested for loops
It’s not as simple as that. You can have 20 nested for loops with complexity of O(1) if all of them only ever finish one iteration.
Or you can have one for loop that iterates 2^N times.
What do you think my complexity is?
I think it could be maybe O(n^2) because the other for loop which tries elements around the first error will only execute a constant of 5 times in the worst case? I’m unsure.
It really depends on what your parameter n is. If the only relevant size is the number of records (let’s say that is n), then this solution takes time in O(n), because it loops over records only once at a time. This ignores the length of records by considering it constant.
If we also consider the maximum length of records (let’s call it m), then your solution, and most others I’ve seen in this thread, has a time complexity in O(n * m^2) for part 2.
It’s O(n).
If you look at each of the levels of all reports, you will access it a constant number of times: at most twice in each call to
EvaluateLineSafe
, and you will callEvaluateLineSafe
at most six times for each report.
Elixir
defmodule AdventOfCode.Solution.Year2024.Day02 do use AdventOfCode.Solution.SharedParse @impl true def parse(input) do for line <- String.split(input, "\n", trim: true), do: String.split(line) |> Enum.map(&String.to_integer/1) end def part1(input) do Enum.count(input, &is_safe(&1, false)) end def part2(input) do Enum.count(input, &(is_safe(&1, true) or is_safe(tl(&1), false))) end def is_safe([a, b, c | rest], can_fix) do cond do (b - a) * (c - b) > 0 and abs(b - a) in 1..3 and abs(c - b) in 1..3 -> is_safe([b, c | rest], can_fix) can_fix -> is_safe([a, c | rest], false) or is_safe([a, b | rest], false) true -> false end end def is_safe(_, _), do: true end
Nim
import strutils, times, sequtils, sugar # check if level transition in record is safe proc isSafe*(sign:bool, d:int): bool = sign == (d>0) and d.abs in 1..3; #check if record is valid proc validate*(record:seq[int]): bool = let sign = record[0] > record[1]; return (0..record.len-2).allIt(isSafe(sign, record[it] - record[it+1])) # check if record is valid as-is # or if removing any item makes the record valid proc validate2*(record:seq[int]): bool = return record.validate or (0..<record.len).anyIt(record.dup(delete(it)).validate) proc solve*(input:string): array[2,int] = let lines = input.readFile.strip.splitLines; let records = lines.mapIt(it.splitWhitespace.map(parseInt)); result[0] = records.countIt(it.validate); result[1] = records.countIt(it.validate2);
I got stuck on part 2 trying to check everything inside a single loop, which kept getting more ugly. So then I switched to just deleting one item at a time and re-checking the record.
Reworked it after first finding the solution to compress the code a bit, though the range iterators don’t really help with readability.
I did learn about the
sugar
import, which I used to make the sequence duplication more compact:record.dup(delete(it)
.Cool to see another solution in Nim here =)
(0..<record.len).anyIt(record.dup(delete(it)).validate)
That’s smart. I haven’t thought of using iterators to loop over indexes (except in a
for loop
).I got stuck on part 2 trying to check everything inside a single loop, which kept getting more ugly.
Yeah I’ve thought of simple ways to do this and found none. And looking at the input - it’s too easy to bruteforce, especially in compiled lang like Nim.
R (R-Wasm)
input = file('input2024day2.txt',open='r') lines = readLines(input) library(stringr) safe = 0 safe2 = 0 for (ln in lines){ vals = as.numeric(unlist(str_split(ln,' '))) diffs = diff(vals) cond1 = min(diffs) > 0 || max(diffs) < 0 cond2 = max(abs(diffs)) < 4 if (cond1 && cond2){ safe = safe + 1 } else { #Problem Dampener dampen = FALSE for (omit in -1:-length(vals)){ diffs = diff(vals[omit]) cond1 = min(diffs) > 0 || max(diffs) < 0 cond2 = max(abs(diffs)) < 4 if (cond1 && cond2){ dampen = TRUE } } if (dampen){ safe2 = safe2 + 1} } } print (safe) #Part 1 print (safe + safe2) #Part 2
J
There is probably a way to write this more point-free. You can definitely see here the friction involved in the way J wants to regard lists as arrays: short rows of the input matrix are zero padded, so you have to snip off the padding before you process each row, and that means you can’t lift some of the operations back up to the parent matrix because it will re-introduce the padding as it reshapes the result; this accounts for a lot of the
"1
everywhere (you can interpretv"1
as “force the verbv
to operate on rank 1 subarrays of the argument”).data_file_name =: '2.data' data =: > 0 ". each cutopen toJ fread data_file_name NB. {. take, i. index of; this removes trailing zeros remove_padding =: {.~ i.&0 NB. }. behead, }: curtail; this computes successive differences diff =: }. - }: NB. a b in_range y == a <: y <: b in_range =: 4 : '(((0 { x) & <:) * (<: & (1 { x))) y' NB. a row is safe if either all successive differences are in [1..3] or all in [_3.._1] NB. +. or ranges =: 2 2 $ 1 3 _3 _1 row_safe =: (+./"1) @: (*/"1) @: (ranges & (in_range"1 _)) @: diff @: remove_padding result1 =: +/ safe"1 data NB. x delete y is y without the xth element delete =: 4 : '(x {. y) , ((>: x) }. y)'"0 _ modified_row =: 3 : 'y , (i.#y) delete y' modified_row_safe =: 3 : '+./"1 row_safe"1 modified_row"1 y' result2 =: +/ modified_row_safe data
def is_safe(report: list[int]) -> bool: global removed acceptable_range = [_ for _ in range(-3,4) if _ != 0] diffs = [] if any([report.count(x) > 2 for x in report]): return False for i, num in enumerate(report[:-1]): cur = num next = report[i+1] difference = cur - next diffs.append(difference) if difference not in acceptable_range: return False if len(diffs) > 1: if diffs[-1] * diffs[-2] <= 0: return False return True with open('input') as reports: list_of_reports = reports.readlines()[:-1] count = 0 failed_first_pass = [] failed_twice = [] for reportsub in list_of_reports: levels = [int(l) for l in reportsub.split()] original = levels.copy() if is_safe(levels): safe = True count += 1 else: failed_first_pass.append(levels) for report in failed_first_pass: print(report) working_copy = report.copy() for i in range(len(report)): safe = False working_copy.pop(i) print("checking", working_copy) if is_safe(working_copy): count += 1 safe = True break else: working_copy = report.copy() print(count)
this took me so fucking long and in the end i just went for brute force anyway. there are still remnants of some of previous, overly complicated, failed attempts, like the hideous
global removed
. In the end, I realized I was fucking up by using remove() instead of pop(), it was causing cases with duplicates where the removal of one would yield a safe result to count as unsafe.
I forgot that this started yesterday, so I’m already behind. I quite like my solution for part one,
but part two will have to waitedit: part 2 was a lot simpler than I thought after a night’s sleep.Rust
use color_eyre::eyre; use std::{fs, num, str::FromStr}; #[derive(Debug, PartialEq, Eq)] struct Report(Vec<isize>); impl FromStr for Report { type Err = num::ParseIntError; fn from_str(s: &str) -> Result<Self, Self::Err> { let v: Result<Vec<isize>, _> = s .split_whitespace() .map(|num| num.parse::<isize>()) .collect(); Ok(Report(v?)) } } impl Report { fn is_safe(&self) -> bool { let ascending = self.0[1] > self.0[0]; let (low, high) = if ascending { (1, 3) } else { (-3, -1) }; self.0.windows(2).all(|w| { let a = w[0]; let b = w[1]; b >= a + low && b <= a + high }) } fn is_dampsafe(&self) -> bool { if self.is_safe() { return true; } for i in 0..self.0.len() { let damped = { let mut v = self.0.clone(); v.remove(i); Self(v) }; if damped.is_safe() { return true; } } false } } fn main() -> eyre::Result<()> { color_eyre::install()?; let part1 = part1("d02/input.txt")?; let part2 = part2("d02/input.txt")?; println!("Part 1: {part1}\nPart 2: {part2}"); Ok(()) } fn part1(filepath: &str) -> eyre::Result<isize> { let mut num_safe = 0; for l in fs::read_to_string(filepath)?.lines() { if Report::from_str(l)?.is_safe() { num_safe += 1; } } Ok(num_safe) } fn part2(filepath: &str) -> eyre::Result<isize> { let mut num_safe = 0; for l in fs::read_to_string(filepath)?.lines() { if Report::from_str(l)?.is_dampsafe() { num_safe += 1; } } Ok(num_safe) }
Tests
#[cfg(test)] mod tests { use super::*; #[test] fn sample_part1() { assert_eq!(part1("test.txt").unwrap(), 2); } #[test] fn sample_part2() { assert_eq!(part2("test.txt").unwrap(), 4); } }