day 12 part 1 (using recursive BFS, with tons of debug info)
This commit is contained in:
@@ -24,7 +24,7 @@ LDLIB := -l$(LIB)
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export LD_LIBRARY_PATH = $(LIBDIR)
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CFLAGS += -std=gnu99
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CFLAGS += -O2
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#CFLAGS += -O2
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CFLAGS += -g
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CFLAGS += -Wall
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CFLAGS += -Wextra
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@@ -1,100 +1,46 @@
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--- Day 12: Passage Pathing ---
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--- Part Two ---
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With your submarine's subterranean subsystems subsisting suboptimally, the only way you're getting out of this cave anytime soon is by finding a path yourself. Not just a path - the only way to know if you've found the best path is to find all of them.
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After reviewing the available paths, you realize you might have time to visit a single small cave twice. Specifically, big caves can be visited any number of times, a single small cave can be visited at most twice, and the remaining small caves can be visited at most once. However, the caves named start and end can only be visited exactly once each: once you leave the start cave, you may not return to it, and once you reach the end cave, the path must end immediately.
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Fortunately, the sensors are still mostly working, and so you build a rough map of the remaining caves (your puzzle input). For example:
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start-A
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start-b
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A-c
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A-b
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b-d
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A-end
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b-end
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This is a list of how all of the caves are connected. You start in the cave named start, and your destination is the cave named end. An entry like b-d means that cave b is connected to cave d - that is, you can move between them.
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So, the above cave system looks roughly like this:
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start
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/ \
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c--A-----b--d
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\ /
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end
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Your goal is to find the number of distinct paths that start at start, end at end, and don't visit small caves more than once. There are two types of caves: big caves (written in uppercase, like A) and small caves (written in lowercase, like b). It would be a waste of time to visit any small cave more than once, but big caves are large enough that it might be worth visiting them multiple times. So, all paths you find should visit small caves at most once, and can visit big caves any number of times.
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Given these rules, there are 10 paths through this example cave system:
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Now, the 36 possible paths through the first example above are:
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start,A,b,A,b,A,c,A,end
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start,A,b,A,b,A,end
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start,A,b,A,b,end
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start,A,b,A,c,A,b,A,end
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start,A,b,A,c,A,b,end
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start,A,b,A,c,A,c,A,end
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start,A,b,A,c,A,end
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start,A,b,A,end
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start,A,b,d,b,A,c,A,end
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start,A,b,d,b,A,end
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start,A,b,d,b,end
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start,A,b,end
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start,A,c,A,b,A,b,A,end
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start,A,c,A,b,A,b,end
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start,A,c,A,b,A,c,A,end
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start,A,c,A,b,A,end
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start,A,c,A,b,d,b,A,end
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start,A,c,A,b,d,b,end
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start,A,c,A,b,end
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start,A,c,A,c,A,b,A,end
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start,A,c,A,c,A,b,end
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start,A,c,A,c,A,end
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start,A,c,A,end
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start,A,end
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start,b,A,b,A,c,A,end
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start,b,A,b,A,end
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start,b,A,b,end
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start,b,A,c,A,b,A,end
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start,b,A,c,A,b,end
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start,b,A,c,A,c,A,end
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start,b,A,c,A,end
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start,b,A,end
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start,b,d,b,A,c,A,end
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start,b,d,b,A,end
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start,b,d,b,end
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start,b,end
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(Each line in the above list corresponds to a single path; the caves visited by that path are listed in the order they are visited and separated by commas.)
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The slightly larger example above now has 103 paths through it, and the even larger example now has 3509 paths through it.
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Note that in this cave system, cave d is never visited by any path: to do so, cave b would need to be visited twice (once on the way to cave d and a second time when returning from cave d), and since cave b is small, this is not allowed.
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Here is a slightly larger example:
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dc-end
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HN-start
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start-kj
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dc-start
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dc-HN
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LN-dc
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HN-end
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kj-sa
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kj-HN
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kj-dc
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The 19 paths through it are as follows:
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start,HN,dc,HN,end
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start,HN,dc,HN,kj,HN,end
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start,HN,dc,end
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start,HN,dc,kj,HN,end
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start,HN,end
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start,HN,kj,HN,dc,HN,end
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start,HN,kj,HN,dc,end
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start,HN,kj,HN,end
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start,HN,kj,dc,HN,end
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start,HN,kj,dc,end
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start,dc,HN,end
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start,dc,HN,kj,HN,end
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start,dc,end
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start,dc,kj,HN,end
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start,kj,HN,dc,HN,end
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start,kj,HN,dc,end
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start,kj,HN,end
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start,kj,dc,HN,end
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start,kj,dc,end
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Finally, this even larger example has 226 paths through it:
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fs-end
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he-DX
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fs-he
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start-DX
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pj-DX
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end-zg
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zg-sl
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zg-pj
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pj-he
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RW-he
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fs-DX
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pj-RW
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zg-RW
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start-pj
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he-WI
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zg-he
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pj-fs
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start-RW
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How many paths through this cave system are there that visit small caves at most once?
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Given these new rules, how many paths through this cave system are there?
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295
2021/day12/aoc-c.c
Normal file
295
2021/day12/aoc-c.c
Normal file
@@ -0,0 +1,295 @@
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/* aoc-c.c: Advent of Code 2021, day 11 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 <ctype.h>
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#include "debug.h"
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#include "pool.h"
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#include "bits.h"
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#include "list.h"
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/* graph representation used :
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*
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* NODE ARRAY LINK (doubly linked)
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* ------------- --------------
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* | node[x] | | link x_1 |
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* prev link <----> list_head <--------> list head <----> next link <----> ...
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* | ... | | |
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* ------------- ----< node y ptr |
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* | | ... |
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* ... | --------------
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* |
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* ------------- <--- --------------
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* | node[y] | | link y_1 |
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* prev link <----> list_head <--------> list head <----> next link
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* | ... | | |
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* ------------- | node ptr |
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* | ... |
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* --------------
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*
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*/
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#define MAX_LINKS 10
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#define MAX_NODES 128
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#define START 0 /* start is always nodes[0] */
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#define END 1 /* end is always nodes[1] */
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typedef struct node {
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char token[16]; /* node name */
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int val; /* unused ? */
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int passed; /* number of passages */
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int multiple; /* can pass multiple times ? */
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int nlinks; /* number of links */
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struct list_head list; /* links list */
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} node_t;
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typedef struct link { /* links between nodes */
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node_t *node; /* linked node */
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int val; /* unused ? */
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struct list_head list; /* next/prev link */
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} link_t;
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static node_t nodes[MAX_NODES]; /* nodes list */
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static int nnodes; /* ... and size */
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static pool_t *links_pool; /* links memory pool */
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static node_t* stack[1024]; /* moves stack */
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static int nstack; /* ... and size */
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static int npaths; /* number of solutions */
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static void print_tokens()
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{
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log_f(3, "nodes = %d\n", nnodes);
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for (int i = 0; i < nnodes; ++i) {
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log(3, "%2d: %6s mult=%d passed=%d\n", i, nodes[i].token,
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nodes[i].multiple, nodes[i].passed);
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}
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}
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static void print_links()
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{
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struct link *link;
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node_t *node;
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log_f(3, "links = %d\n", nnodes);
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for (int i = 0; i < nnodes; ++i) {
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node = nodes+i;
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log(3, "[%6s]", node->token);
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list_for_each_entry(link, &node->list, list) {
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printf(" --> %s", link->node->token);
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}
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log(3, "\n");
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}
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}
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/* find node from token value
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* return @node or NULL
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*/
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static node_t *find_token(char *token)
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{
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log_f(2, "token=[%s]\n", token);
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for (int i = 0; i < nnodes; ++i) {
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log_i(4, "compare=[%s]\n", nodes[i].token);
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if (!strcmp(nodes[i].token, token)) {
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return nodes+i;
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}
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}
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return NULL;
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}
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/* get node from token value (create if does not exist)
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* return @node
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*/
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static node_t *add_token_maybe(char *token)
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{
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node_t *node;
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if (!(node = find_token(token))) {
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node = nodes + nnodes;
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strcpy(node->token, token);
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node->multiple = isupper(*token)? 1: 0;
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node->passed = 0;
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node->nlinks = 0;
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INIT_LIST_HEAD(&node->list);
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nnodes++;
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}
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return node;
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}
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/* set all nodes with passed > pass (-1: all) "passed" value to zero
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*/
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static void clean_node_passed(int pass)
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{
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for (int i = 0; i < nnodes; ++i) {
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if (pass == -1 || nodes[i].passed >= pass)
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nodes[i].passed = 0;
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}
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}
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/* create a link between two nodes
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* return 1 if OK, 0 if error
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*/
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static int link_nodes(node_t *n1, node_t *n2)
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{
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struct link *link1, *link2;
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if (!((link1 = pool_get(links_pool)) && (link2 = pool_get(links_pool))))
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return 0;
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link1->node = n2;
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list_add_tail(&link1->list, &n1->list);
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link2->node = n1;
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list_add_tail(&link2->list, &n2->list);
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return 1;
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}
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/* read data and create tree.
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*/
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static int read_graph()
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{
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ssize_t len;
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size_t alloc = 0;
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char *buf = NULL, *token1, *token2;
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node_t *node1, *node2;
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if (!(links_pool = pool_init("links", 128, sizeof (struct link))))
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return -1;
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add_token_maybe("start");
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add_token_maybe("end");
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print_tokens();
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while ((len = getline(&buf, &alloc, stdin)) > 0) {
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token1 = strtok(buf, "-\n");
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token2 = strtok(NULL, "-\n");
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node1 = add_token_maybe(token1);
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node2 = add_token_maybe(token2);
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link_nodes(node1, node2);
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//log_f(2, "token1=[%s] token2=[%s]\n", token1, token2);
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}
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free(buf);
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print_tokens();
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clean_node_passed(-1);
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print_tokens();
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print_links();
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return nnodes;
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}
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/* push link ptr to stack
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*/
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inline static node_t *push(node_t *node)
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{
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return ((stack[nstack++] = node));
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}
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/* pop link ptr from stack
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*/
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inline static node_t *pop()
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{
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return nstack? stack[--nstack]: NULL;
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}
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/* traverse graph from a node, return
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*/
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static int bfs_run(node_t *node)
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{
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int i, res = 0;
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struct link *link;
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log_f(3, "node=%s npaths=%d\n", node->token, npaths);
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//log_f(3, "node=%p nodes=%p start=%p end=%p end1=%p\n",
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// node, nodes, nodes+START, nodes+END, &nodes[END]);
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/* we found a path */
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if (node == nodes+END) {
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npaths++;
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log(3, "solution :");
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for (i = 0; i < nstack; ++i)
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log(3, " -> %s", stack[i]->token);
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log(3, "\n");
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return 1;
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}
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if (list_empty(&node->list))
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return 0;
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i = 0;
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list_for_each_entry(link, &node->list, list) {
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if (link->node->multiple || !link->node->passed) {
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link->node->passed++;
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push(link->node);
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res += bfs_run(link->node);
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pop();
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link->node->passed--;
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}
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}
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return res;
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}
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/* run 1 full step.
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* return number of flashed octopuses
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*/
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static int part1()
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{
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nodes[START].passed = 1;
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push(nodes + START);
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return bfs_run(nodes);
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pop();
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}
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static int part2()
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{
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return 2;
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}
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static int doit(int part)
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{
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read_graph();
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return part == 1? part1(): part2();
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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, part = 1;
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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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part = atoi(optarg);
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if (part < 1 || part > 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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printf("%s : res=%d npaths=%d\n", *av, doit(part), npaths);
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exit (0);
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}
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Reference in New Issue
Block a user