day 6 part 1 (simple algorithm) + part 2 README
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Initial state: 3,4,3,1,2
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3,4,3,1,2
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@@ -51,3 +51,14 @@ Each day, a 0 becomes a 6 and adds a new 8 to the end of the list, while each ot
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In this example, after 18 days, there are a total of 26 fish. After 80 days, there would be a total of 5934.
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In this example, after 18 days, there are a total of 26 fish. After 80 days, there would be a total of 5934.
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Find a way to simulate lanternfish. How many lanternfish would there be after 80 days?
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Find a way to simulate lanternfish. How many lanternfish would there be after 80 days?
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Your puzzle answer was 362666.
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The first half of this puzzle is complete! It provides one gold star: *
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--- Part Two ---
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Suppose the lanternfish live forever and have unlimited food and space. Would they take over the entire ocean?
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After 256 days in the example above, there would be a total of 26984457539 lanternfish!
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How many lanternfish would there be after 256 days?
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142
2021/day06/aoc-c.c
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142
2021/day06/aoc-c.c
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/* aoc-c: Advent2021 game, day 6 parts 1 & 2
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*
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* Copyright (C) 2021 Bruno Raoult ("br")
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* Licensed under the GNU General Public License v3.0 or later.
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* Some rights reserved. See COPYING.
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*
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* You should have received a copy of the GNU General Public License along with this
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* program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
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*
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* SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <unistd.h>
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#include <string.h>
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#include <malloc.h>
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#include "debug.h"
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#include "bits.h"
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#include "list.h"
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#include "pool.h"
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typedef struct fish {
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s16 value;
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struct list_head list;
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} fish_t;
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static pool_t *pool;
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LIST_HEAD(fish_head);
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static int nfish=0;
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//#ifdef DEBUG
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static void print_fish()
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{
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fish_t *fish;
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int i = 0;
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printf("fish # = %d\n", nfish);
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list_for_each_entry(fish, &fish_head, list) {
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printf("%s%d", i? ",":"", fish->value);
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i++;
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}
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printf("\n");
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}
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//#endif
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static struct list_head *read_fish()
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{
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char *buf, *token;
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size_t alloc = 0;
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fish_t *fish;
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if (getline(&buf, &alloc, stdin) < 0)
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return NULL;
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if (!(pool = pool_init("fish", 1024, sizeof (struct fish))))
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return NULL;
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token = strtok(buf, ",\n");
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while (token) {
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//printf("token=[%s]\n", token);
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if (!(fish = pool_get(pool)))
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return NULL;
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fish->value = atoi(token);
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list_add_tail(&fish->list, &fish_head);
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nfish ++;
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token = strtok(NULL, ",\n");
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//print_fish();
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}
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free(buf);
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return &fish_head;
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}
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static int doit(int part)
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{
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fish_t *fish, *new;
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int toadd;
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int iter = part == 1? 80: 256;
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/* initial algorithm surely does not work for part 2:
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* Too many memory allocation, nneed to rethink the whole...
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*
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*/
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for (; iter; --iter) {
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//printf("iter = %2d: ", iter);
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toadd = 0;
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list_for_each_entry(fish, &fish_head, list) {
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fish->value--;
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if (fish->value < 0) { /* was zero: create new fish */
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fish->value = 6;
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toadd++;
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nfish++;
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}
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}
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while (toadd--) {
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if (!(new = pool_get(pool)))
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return -1;
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new->value = 8;
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list_add_tail(&new->list, &fish_head);
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}
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}
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return nfish;
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}
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static int usage(char *prg)
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{
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fprintf(stderr, "Usage: %s [-d debug_level] [-p part]\n", prg);
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return 1;
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}
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int main(int ac, char **av)
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{
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int opt;
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u32 exercise = 1, res;
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while ((opt = getopt(ac, av, "d:p:")) != -1) {
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switch (opt) {
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case 'd':
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debug_level_set(atoi(optarg));
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break;
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case 'p': /* 1 or 2 */
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exercise = atoi(optarg);
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if (exercise < 1 || exercise > 2)
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return usage(*av);
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break;
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default:
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return usage(*av);
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}
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}
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if (optind < ac)
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return usage(*av);
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read_fish();
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print_fish();
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res = doit(exercise);
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//print_fish();
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printf ("%s : res=%d\n", *av, res);
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/* TODO: free board/mem pool */
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exit (0);
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}
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