this post was submitted on 06 Dec 2024
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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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Day 6: Guard Gallivant

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[โ€“] [email protected] 0 points 2 weeks ago

C

Got super stumped on part 2. I'd add an obstacle for every tile on the path of part 1 but I kept getting wrong results, even after fixing some edge cases. Spent too much time looking at terminal dumps and mp4 visualisations.

Eventually I gave up and wrote a for(y) for(x) loop, trying an obstacle in every possible tile, and that gave the correct answer. Even that brute force took only 2.5 ish seconds on my 2015 PC! But having that solution allowed me to narrow it down again to a reasonably efficient version similar to what I had before. Still I don't know where I went wrong the first time.

Code

#include "common.h"

#define GZ 134

struct world { char g[GZ][GZ]; int x,y, dir; };

static const char carets[] = "^>v<";
static const int dx[4] = {0,1,0,-1}, dy[4] = {-1,0,1,0};

static inline char *ahead(struct world *w) {
    return &w->g[w->y+dy[w->dir]][w->x+dx[w->dir]]; }
static inline int visited(char t) { return t && strchr(carets, t); }
static inline int traversible(char t) { return t=='.' || visited(t); }

/* new tile, previously visited tile, in a loop, out of bounds */
enum { ST_NEW, ST_SEEN, ST_LOOP, ST_END };

static int
step(struct world *w)
{
	char *cell;
	int is_new;

	assert(w->x >= 0); assert(w->x < GZ);
	assert(w->y >= 0); assert(w->y < GZ);

	cell = &w->g[w->y][w->x];

	if (!traversible(*cell))	/* out of bounds? */
		return ST_END;
	while (*ahead(w) == '#')	/* turn if needed */
		w->dir = (w->dir +1) %4;
	if (*cell == carets[w->dir])	/* walked here same dir? */
		return ST_LOOP;

	is_new = *cell == '.';
	*cell = carets[w->dir];
	w->x += dx[w->dir];
	w->y += dy[w->dir];

	return is_new ? ST_NEW : ST_SEEN;
}

int
main(int argc, char **argv)
{
	static struct world w0,w1,w2;
	int p1=0,p2=0, x,y, r,i;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));

	for (y=1; y<GZ && fgets(w0.g[y]+1, GZ-1, stdin); y++)
		;

	for (y=0; y<GZ; y++)
	for (x=0; x<GZ; x++)
	for (i=0; i<4; i++)
		if (w0.g[y][x] == carets[i])
			{ w0.x=x; w0.y=y; w0.dir=i; goto found_start; }
found_start:
	w0.g[y][x] = '.';

	/* keep the clean copy of the grid (needed for part 2) */
	memcpy(&w1, &w0, sizeof(w1));

	/* part 1: trace the path and count unseen tiles */
	while ((r = step(&w1)) <= ST_SEEN)
		p1 += r == ST_NEW;

	/* part 2: try putting obstacles on each tile seen in p1 */
	for (y=0; y<GZ; y++)
	for (x=0; x<GZ; x++)
		if (visited(w1.g[y][x])) {
			memcpy(&w2, &w0, sizeof(w2));
			w2.g[y][x] = '#';
			while ((r = step(&w2)) <= ST_SEEN) ;
			p2 += r == ST_LOOP;
		}

	printf("06: %d %d\n", p1, p2);
	return 0;
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day06.c