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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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🦌 - 2024 DAY 16 SOLUTIONS -🦌 (programming.dev)

submitted 2 days ago by [email protected] to c/[email protected]

24 comments fedilink hide all child comments

Day 16: Reindeer Maze

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[–] [email protected] 4 points 1 day ago (1 children)

C

Yay more grids! Seemed like prime Dijkstra or A* material but I went with an iterative approach instead!

I keep an array cost[y][x][dir], which is seeded at 1 for the starting location and direction. Then I keep going over the array, seeing if any valid move (step or turn) would yield to a lower best-known-cost for this state. It ends when a pass does not yield changes.

This leaves us with the best-known-costs for every reachable state in the array, including the end cell (bit we have to take the min() of the four directions).

Part 2 was interesting: I just happend to have written a dead end pruning function for part 1 and part 2 is, really, dead-end pruning for the cost map: remove any suboptimal step, keep doing so, and you end up with only the optimal steps. 'Suboptimal' here is a move that yields a higher total cost than the best-known-cost for that state.

It's fast enough too on my 2015 i5:

day16  0:00.05  1656 Kb  0+242 faults

Code

#include "common.h"

#define GZ 145

enum {NN, EE, SS, WW};

static const int dx[]={0,1,0,-1}, dy[]={-1,0,1,0};

static char g[GZ][GZ];		/* with 1 tile border */
static int cost[GZ][GZ][4];	/* per direction, starts at 1, 0=no info */

static int traversible(char c) { return c=='.' || c=='S' || c=='E'; }

static int
minat(int x, int y)
{
	int acc=0, d;

	for (d=0; d<4; d++)
		if (cost[y][x][d] && (!acc || cost[y][x][d] < acc))
			acc = cost[y][x][d];

	return acc;
}


static int
count_exits(int x, int y)
{
	int acc=0, i;

	assert(x>0); assert(x<GZ-1);
	assert(y>0); assert(y<GZ-1);

	for (i=0; i<4; i++)
		acc += traversible(g[y+dy[i]][x+dx[i]]);
	
	return acc;
}

/* remove all dead ends */
static void
prune_dead(void)
{
	int dirty=1, x,y;

	while (dirty) {
		dirty = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
			if (g[y][x]=='.' && count_exits(x,y) < 2)
				{ dirty = 1; g[y][x] = '#'; }
	}
}

/* remove all dead ends from cost[], leaves only optimal paths */
static void
prune_subopt(void)
{
	int dirty=1, x,y,d;

	while (dirty) {
		dirty = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
		for (d=0; d<4; d++) {
			if (!cost[y][x][d])
				continue;

			if (g[y][x]=='E') {
				if (cost[y][x][d] != minat(x,y))
					{ dirty = 1; cost[y][x][d] = 0; }
				continue;
			}

			if (cost[y][x][d]+1 > cost[y+dy[d]][x+dx[d]][d] &&
			    cost[y][x][d]+1000 > cost[y][x][(d+1)%4] &&
			    cost[y][x][d]+1000 > cost[y][x][(d+3)%4])
				{ dirty = 1; cost[y][x][d] = 0; }
		}
	}
}

static void
propagate_costs(void)
{
	int dirty=1, cost1, x,y,d;

	while (dirty) {
		dirty = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
		for (d=0; d<4; d++) {
			if (!traversible(g[y][x]))
				continue;

			/* from back */
			if ((cost1 = cost[y-dy[d]][x-dx[d]][d]) &&
			    (cost1+1 < cost[y][x][d] || !cost[y][x][d]))
				{ dirty = 1; cost[y][x][d] = cost1+1; }

			/* from right */
			if ((cost1 = cost[y][x][(d+1)%4]) &&
			    (cost1+1000 < cost[y][x][d] || !cost[y][x][d]))
				{ dirty = 1; cost[y][x][d] = cost1+1000; }

			/* from left */
			if ((cost1 = cost[y][x][(d+3)%4]) &&
			    (cost1+1000 < cost[y][x][d] || !cost[y][x][d]))
				{ dirty = 1; cost[y][x][d] = cost1+1000; }
		}
	}
}

int
main(int argc, char **argv)
{
	int p1=0,p2=0, sx=0,sy=0, ex=0,ey=0, x,y;
	char *p;

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

	for (y=1; fgets(g[y]+1, GZ-1, stdin); y++) {
		if ((p = strchr(g[y]+1, 'S'))) { sy=y; sx=p-g[y]; }
		if ((p = strchr(g[y]+1, 'E'))) { ey=y; ex=p-g[y]; }
		assert(y+1 < GZ-1);
	}

	cost[sy][sx][EE] = 1;

	prune_dead();
	propagate_costs();
	prune_subopt();

	p1 = minat(ex, ey) -1;	/* costs[] values start at 1! */

	for (y=1; y<GZ-1; y++)
	for (x=1; x<GZ-1; x++)
		p2 += minat(x,y) > 0;

	printf("16: %d %d\n", p1, p2);
	return 0;
}

[–] mykl 3 points 1 day ago

Very interesting approach. Pruning deadends by spawning additional walls is a very clever idea.