Solution to problem 12 part 1 in Python
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src/Year_2021/P12.py
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src/Year_2021/P12.py
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# --- Day 12: Passage Pathing ---
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# With your submarine's subterranean subsystems subsisting suboptimally, the
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# only way you're getting out of this cave anytime soon is by finding a path
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# yourself. Not just a path - the only way to know if you've found the best
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# path is to find all of them.
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# Fortunately, the sensors are still mostly working, and so you build a rough
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# 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
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# named start, and your destination is the cave named end. An entry like b-d
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# means that cave b is connected to cave d - that is, you can move between
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# 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
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# end, and don't visit small caves more than once. There are two types of
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# caves: big caves (written in uppercase, like A) and small caves (written in
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# lowercase, like b). It would be a waste of time to visit any small cave more
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# than once, but big caves are large enough that it might be worth visiting
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# them multiple times. So, all paths you find should visit small caves at most
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# 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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# start,A,b,A,c,A,end
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# start,A,b,A,end
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# start,A,b,end
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# start,A,c,A,b,A,end
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# start,A,c,A,b,end
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# start,A,c,A,end
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# start,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,end
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# (Each line in the above list corresponds to a single path; the caves visited
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# by that path are listed in the order they are visited and separated by
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# commas.)
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# Note that in this cave system, cave d is never visited by any path: to do so,
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# cave b would need to be visited twice (once on the way to cave d and a second
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# time when returning from cave d), and since cave b is small, this is not
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# 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
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# most once?
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from collections import defaultdict
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from typing import DefaultDict
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caves_map = defaultdict(list)
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with open("files/P12.txt") as f:
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for row in f:
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start, end = row.strip().split("-")
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caves_map[start].append(end)
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caves_map[end].append(start)
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def dfs(
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node: str,
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graph: DefaultDict[str, list[str]],
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visited: set[str],
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already_visited: bool,
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counter: int = 0,
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):
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if node == "end":
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return 1
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for neighbour in graph[node]:
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if neighbour.isupper():
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counter += dfs(neighbour, graph, visited, already_visited)
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else:
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if neighbour not in visited:
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counter += dfs(
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neighbour, graph, visited | {neighbour}, already_visited
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)
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elif already_visited and neighbour not in {"start", "end"}:
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counter += dfs(neighbour, graph, visited, False)
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return counter
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def part_1() -> None:
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ans = dfs("start", caves_map, {"start"}, already_visited=False)
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print(f"There are {ans} paths visiting small caves")
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if __name__ == "__main__":
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part_1()
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