this post was submitted on 11 Dec 2024
13 points (93.3% liked)

Advent Of Code

987 readers
15 users here now

An unofficial home for the advent of code community on programming.dev!

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

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 1 year ago
MODERATORS
 

Day 11: Plutonian Pebbles

Megathread guidelines

  • 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)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

you are viewing a single comment's thread
view the rest of the comments
[โ€“] sjmulder@lemmy.sdf.org 1 points 2 weeks ago

C

Started out a bit sad that this problem really seemed to call for hash tables - either for storing counts for an iterative approach, or to memoize a recursive one.

Worried that the iterative approach would have me doing problematic O(n^2) array scans I went with recursion and a plan to memoize only the first N integers in a flat array, expecting low integers to be much more frequent (and dense) than higher ones.

After making an embarrassing amount of mistakes it worked out beautifully with N=1m (testing revealed that to be about optimal). Also applied some tail recursion shortcuts where possible.

day11  0:00.01  6660 Kb  0+1925 faults

Code

#include "common.h"

/* returns 1 and splits x if even-digited, 0 otherwise */
static int
split(uint64_t x, uint64_t *a, uint64_t *b)
{
	uint64_t p;
	int n, i;

	if (!x) return 0;
	for (n=0, p=1; p<=x;  n++, p*=10) ; if (n%2) return 0;
	for (i=0, p=1; i<n/2; i++, p*=10) ;
	*a = x/p;
	*b = x%p; return 1;
}

/*
 * recur() is memoized in mem[]. Testing found the size MEMZ to be optimal:
 * lowering siginificantly reduced hits, but raising tenfold didn't add a
 * single hit.
 */

#define MEMZ (1024*1024)
static uint64_t mem[MEMZ][76];

static uint64_t
recur(uint64_t v, int n)
{
	uint64_t a,b;

	if (n<1 ) return 1;
	if (v==0) return recur(1, n-1);
	if (v<10) return recur(v*2024, n-1);
	if (v<MEMZ && mem[v][n]) return mem[v][n];
	if (!split(v, &a, &b))   return recur(v*2024, n-1);

	return v<MEMZ ? mem[v][n] =
	    recur(a, n-1) + recur(b, n-1) :
	    recur(a, n-1) + recur(b, n-1);
}

int
main(int argc, char **argv)
{
	uint64_t p1=0,p2=0, val;

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

	while (scanf(" %"SCNu64, &val) == 1) {
		p1 += recur(val, 25);
		p2 += recur(val, 75);
	}

	printf("10: %"PRId64" %"PRId64"\n", p1, p2);
	return 0;
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day11.c