Day 2: Red-Nosed Reports
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- 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)
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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/
- 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
C
First went through the input in one pass, number by number, but unfortunately that wouldn’t fly for part 2.
Code
#include "common.h" static int issafe(int *lvs, int n, int skip) { int safe=1, asc=0,prev=0, ns=0,i; for (i=0; safe && i<n; i++) { if (i == skip) { ns = 1; continue; } if (i-ns > 0) safe = safe && lvs[i] != prev && lvs[i] > prev-4 && lvs[i] < prev+4; if (i-ns == 1) asc = lvs[i] > prev; if (i-ns > 1) safe = safe && (lvs[i] > prev) == asc; prev = lvs[i]; } return safe; } int main(int argc, const char **argv) { char buf[64], *rest, *tok; int p1=0,p2=0, lvs[16],n=0, i; if (argc > 1) DISCARD(freopen(argv[1], "r", stdin)); while ((rest = fgets(buf, sizeof(buf), stdin))) { for (n=0; (tok = strsep(&rest, " ")); n++) { assert(n < (int)LEN(lvs)); lvs[n] = (int)strtol(tok, NULL, 10); } for (i=-1; i<n; i++) if (issafe(lvs, n, i)) { p1 += i == -1; p2++; break; } } printf("02: %d %d\n", p1, p2); }
What is this coding style? The function type, name and open brace placement made me think GNU at first, but the code in the body doesn’t look like GCS at all.
BSD more or less. Mostly K&R except for function declarations.
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
Haskell
This was quite fun! I got a bit distracted trying to rewrite
safe
in point-free style, but I think this version is the most readable. There’s probably a more monadic way of writinglessOne
as well, but I can’t immediately see it.safe xs = any gradual [diffs, negate <$> diffs] where diffs = zipWith (-) (drop 1 xs) xs gradual = all (`elem` [1 .. 3]) lessOne [] = [] lessOne (x : xs) = xs : map (x :) (lessOne xs) main = do input :: [[Int]] <- map (map read . words) . lines <$> readFile "input02" print . length $ filter safe input print . length $ filter (any safe . lessOne) input
Love to see your haskell solutions!
I am so far very amazed with the compactness of your solutions, your
lessOne
is very much mind-Bending. I have never used or seen<$>
before, is it a monadic?
Also I can’t seem to find your logic for this safety condition:
The levels are either all increasing or all decreasing
, did you figure that it wasn’t necessary?
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’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.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.
Rust
The function is_sorted_by on Iterators turned out helpful for compactly finding if a report is safe. In part 2 I simply tried the same with each element removed, since all reports are very short.
fn parse(input: String) -> Vec<Vec<i32>> { input.lines() .map(|l| l.split_whitespace().map(|w| w.parse().unwrap()).collect()) .collect() } fn is_safe(report: impl DoubleEndedIterator<Item=i32> + Clone) -> bool { let safety = |a: &i32, b: &i32| (1..=3).contains(&(b - a)); report.clone().is_sorted_by(safety) || report.rev().is_sorted_by(safety) } fn part1(input: String) { let reports = parse(input); let safe = reports.iter().filter(|r| is_safe(r.iter().copied())).count(); println!("{safe}"); } fn is_safe2(report: &[i32]) -> bool { (0..report.len()).any(|i| { // Try with each element removed is_safe(report.iter().enumerate().filter(|(j, _)| *j != i).map(|(_, n)| *n)) }) } fn part2(input: String) { let reports = parse(input); let safe = reports.iter().filter(|r| is_safe2(r)).count(); println!("{safe}"); } util::aoc_main!();
The
is_sorted_by
is a really nice approach. I originally tried using that function thinking that|a, b| a > b
or|a, b| a < b
would cut it but it didn’t end up working. I never thought to handle the check for the step being between 1 and 3 in the callback closure for that though.
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?
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?
Rust
use crate::utils::read_lines; pub fn solution1() { let reports = get_reports(); let safe_reports = reports .filter(|report| report.windows(3).all(window_is_valid)) .count(); println!("Number of safe reports = {safe_reports}"); } pub fn solution2() { let reports = get_reports(); let safe_reports = reports .filter(|report| { (0..report.len()).any(|i| { [&report[0..i], &report[i + 1..]] .concat() .windows(3) .all(window_is_valid) }) }) .count(); println!("Number of safe reports = {safe_reports}"); } fn window_is_valid(window: &[usize]) -> bool { matches!(window[0].abs_diff(window[1]), 1..=3) && matches!(window[1].abs_diff(window[2]), 1..=3) && ((window[0] > window[1] && window[1] > window[2]) || (window[0] < window[1] && window[1] < window[2])) } fn get_reports() -> impl Iterator<Item = Vec<usize>> { read_lines("src/day2/input.txt").map(|line| { line.split_ascii_whitespace() .map(|level| { level .parse() .expect("Reactor level is always valid integer") }) .collect() }) }
Definitely trickier than yesterday’s. I feel like the
windows
solution isn’t the best, but it was what came to mind and ended up working for me.