this post was submitted on 11 Dec 2024
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Advent Of Code

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Day 11: Plutonian Pebbles

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[โ€“] [email protected] 3 points 3 weeks ago (1 children)

I had a very similar take on this problem, but I was not caching the results of a blink for a single stone, like youre doing with subtree_pointers. I tried adding that to my solution, but it didn't make an appreciable difference. I think that caching the lengths is really the only thing that matters.

C#

    static object Solve(Input i, int numBlinks)
    {
        // This is a cache of the tuples of (stoneValue, blinks) to
        // the calculated count of their child stones.
        var lengthCache = new Dictionary<(long, int), long>();
        return i.InitialStones
            .Sum(stone => CalculateUltimateLength(stone, numBlinks, lengthCache));
    }

    static long CalculateUltimateLength(
        long stone,
        int numBlinks,
        IDictionary<(long, int), long> lengthCache)
    {
        if (numBlinks == 0) return 1;
        
        if (lengthCache.TryGetValue((stone, numBlinks), out var length)) return length;

        length = Blink(stone)
            .Sum(next => CalculateUltimateLength(next, numBlinks - 1, lengthCache));
        lengthCache[(stone, numBlinks)] = length;
        return length;
    }

    static long[] Blink(long stone)
    {
        if (stone == 0) return [1];

        var stoneText = stone.ToString();
        if (stoneText.Length % 2 == 0)
        {
            var halfLength = stoneText.Length / 2;
            return
            [
                long.Parse(stoneText.Substring(0, halfLength)),
                long.Parse(stoneText.Substring(halfLength)),
            ];
        }

        return [stone * 2024];
    }
[โ€“] [email protected] 1 points 3 weeks ago

I think that caching the lengths is really the only thing that matters.

Yep, it is just a dynamic programming problem really.