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

  • Quant@programming.dev
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    5 days ago

    Uiua

    Took me a bit longer to get this one but still quite simple overall.
    Spent quite some time on getting to know the try and assert 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
    
  • Rin@lemm.ee
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    10 days ago

    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…

    • hades@lemm.ee
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      9 days ago

      I don’t think your solution is O(N^3). Can you explain your reasoning?

        • hades@lemm.ee
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          9 days ago

          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.

          • Rin@lemm.ee
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            9 days ago

            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.

            • Gobbel2000@programming.dev
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              9 days ago

              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.

            • hades@lemm.ee
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              9 days ago

              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 call EvaluateLineSafe at most six times for each report.

  • sjmulder@lemmy.sdf.org
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    10 days ago

    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);
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day02.c

    • Faresh@lemmy.ml
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      10 days ago

      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.

  • mykl@lemmy.world
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    10 days ago

    Uiua

    Uiua is still developing very quickly, and this code uses the experimental tuples function, hence the initial directive.

    Try it Live!

    # 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?