Advent Of Code

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Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

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console.log('Hello World')

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❄️ - 2023 DAY 3 SOLUTIONS -❄️ (programming.dev)

submitted 1 year ago* (last edited 1 year ago) by Ategon@programming.dev to c/advent_of_code@programming.dev

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Day 3: Gear Ratios

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[–] sjmulder@lemmy.sdf.org 2 points 1 year ago* (last edited 1 year ago)

Language: C

Part 2 stumped me for a little bit, it wasn't an obvious extension of part 1. Part 1 was about numbers (with one or more ...) while part 2 worked from the symbols (with exactly two ...). Going the other way would require more bookkeeping to avoid double counting.

And for the implementation: if you loop over the grid and check surrounding cells for digits you'd have to account for a bunch of cases, e.g. NW/N or N/NE being part of the same number or NW and NE being part of separate numbers. And you'd have to parse the numbers again. But building a graph or reference list of some sort is both unergonomic with C and not necessarily any simpler.

I ended up just writing out the cases, and honestly it didn't turn out too bad.

GitHub link

Abridged code

int main(int argc, char **argv)
{
	static char G[GSZ][GSZ];
	static int N[GSZ][GSZ];
	int p1=0,p2=0, h=0, x,y, dx,dy, n=0,sym=0,r;
	
	for (h=0; fgets(&amp;G[h+1][1], GSZ-1, stdin); h++)
		assert(h &lt; GSZ);

	/*
	 * Pass 1: parse numbers and solve part 1. For every digit in
	 * G, the full number it is part of is stored in N.
	 */
	for (y=1; y&lt;=h; y++)
	for (x=1; G[y][x]; x++)
		if (isdigit(G[y][x])) {
			n = n*10 + G[y][x]-'0';

			for (dy=-1; dy&lt;2; dy++)
			for (dx=-1; dx&lt;2; dx++)
				sym = sym || (x &amp;&amp; y &amp;&amp;
				    G[y+dy][x+dx] != '.' &amp;&amp;
				    ispunct(G[y+dy][x+dx]));
		} else {
			for (dx=-1; isdigit(G[y][x+dx]); dx--)
				N[y][x+dx] = n;
			if (sym)
				p1 += n;
			n = sym = 0;
		}

	/*
	 * Pass 2: solve part 2 by finding all '*', then counting and
	 * multiplying adjecent numbers.
	 *
	 * Horizontal adjecency is trivial but vertical/diagonal has
	 * two situations: if there's a digit directly North of the '+',
	 * it must be a single number: NW and NE would connect to it.
	 * If N isn't a digit, digits in NW and NE belong to separate
	 * numbers.
	 */
	for (y=1; y&lt;=h; y++)
	for (x=1; G[y][x]; x++) {
		if (G[y][x] != '*')
			continue;

		n = 0; r = 1;

		if (N[y][x-1]) { n++; r *= N[y][x-1]; }
		if (N[y][x+1]) { n++; r *= N[y][x+1]; }

		if (N[y-1][x]) { n++; r *= N[y-1][x]; } else {
			if (N[y-1][x-1]) { n++; r *= N[y-1][x-1]; }
			if (N[y-1][x+1]) { n++; r *= N[y-1][x+1]; }
		}

		if (N[y+1][x]) { n++; r *= N[y+1][x]; } else {
			if (N[y+1][x-1]) { n++; r *= N[y+1][x-1]; }
			if (N[y+1][x+1]) { n++; r *= N[y+1][x+1]; }
		}

		if (n == 2)
			p2 += r;
	}

	printf("%d %d\n", p1, p2);
	return 0;
}