Day 19 - Linen Layout
Megathread guidelines
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FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
Rust
Definitely not Aho–Corasick cool or genius, but does get ~11ms on both parts. Gains over the other Rust solves are partly a less naïve approach to towel checking and partly getting it to work with 0
String
allocations, which involves both an instructive lifetime annotation and the clever use of a niche standard library function.Code
pub fn day19(input: &str) -> (u64, u64) { fn recur_helper<'a>(pat: &'a str, towels: &[&str], map: &mut HashMap<&'a str, u64>) -> u64 { let mut hits = 0; let mut towel_subset = towels; for l in 1..=pat.len() { let test_pat = &pat[0..l]; match towel_subset.binary_search(&test_pat) { Ok(idx) => { if pat[l..].is_empty() { return hits + 1; } else if let Some(&v) = map.get(&pat[l..]) { hits += v; } else { let v = recur_helper(&pat[l..], towels, map); map.insert(&pat[l..], v); hits += v; } towel_subset = &towel_subset[idx..]; let end = towel_subset.partition_point(|&x| x.starts_with(test_pat)); towel_subset = &towel_subset[..end]; if towel_subset.is_empty() { return hits; } } Err(idx) => { towel_subset = &towel_subset[idx..]; let end = towel_subset.partition_point(|&x| x.starts_with(test_pat)); towel_subset = &towel_subset[..end]; if towel_subset.is_empty() { return hits; } } } } hits } let mut part1 = 0; let mut part2 = 0; let mut lines = input.lines(); let mut towels = lines.next().unwrap().split(", ").collect::<Vec<_>>(); towels.sort(); let mut map = HashMap::new(); for pat in lines.skip(1) { let tmp = recur_helper(pat, &towels, &mut map); part2 += tmp; if tmp > 0 { part1 += 1; } } (part1, part2) }
nice job! now do it without recursion. something I always attempt to shy away from as I think it is dirty to do, makes me feel like it is the “lazy man’s” loop.
Aho-Corasick is definitely meant for this kind of challenge that requires finding all occurrences of multiple patterns, something worth reading up on! If you are someone who builds up a util library for future AoC or other projects then that would likely come in to use sometimes.
Another algorithm that is specific to finding one pattern is the Boyer-Moore algorithm. something to mull over: https://softwareengineering.stackexchange.com/a/183757
have to remember we are all discovering new interesting ways to solve similar or same challenges. I am still learning lots, hope to see you around here and next year!
Python3
Solver uses trie just like the other python solve here, I try to not look at solve threads until after it is solved. yet, I somehow got a solve that only takes ~7 ms. previously I would rebuild the trie every time and made it take ~55 ms.
Code
from collections import deque def profiler(method): from time import perf_counter_ns def wrapper_method(*args: any, **kwargs: any) -> any: start_time = perf_counter_ns() ret = method(*args, **kwargs) stop_time = perf_counter_ns() - start_time time_len = min(9, ((len(str(stop_time))-1)//3)*3) time_conversion = {9: 'seconds', 6: 'milliseconds', 3: 'microseconds', 0: 'nanoseconds'} print(f"Method {method.__name__} took : {stop_time / (10**time_len)} {time_conversion[time_len]}") return ret return wrapper_method def build_aho_corasick(towels): # Each node: edges dict, failure link, output (which towels end here) trie = [{'fail': 0, 'out': []}] # Build trie of towels for index, word in enumerate(towels): node = 0 for char in word: if char not in trie[node]: trie[node][char] = len(trie) trie.append({'fail': 0, 'out': []}) node = trie[node][char] trie[node]['out'].append(len(word)) # store length of matched towel # Build fail links (BFS) queue = deque() # Initialize depth-1 fail links for c in trie[0]: if c not in ('fail', 'out'): nxt = trie[0][c] trie[nxt]['fail'] = 0 queue.append(nxt) # BFS build deeper fail links while queue: r = queue.popleft() for c in trie[r]: if c in ('fail', 'out'): continue nxt = trie[r][c] queue.append(nxt) f = trie[r]['fail'] while f and c not in trie[f]: f = trie[f]['fail'] trie[nxt]['fail'] = trie[f][c] if c in trie[f] else 0 trie[nxt]['out'] += trie[trie[nxt]['fail']]['out'] return trie def count_possible_designs_aho(trie, design): n = len(design) dp = [0] * (n + 1) dp[0] = 1 # State in the trie automaton state = 0 for i, char in enumerate(design, 1): # Advance in trie while state and char not in trie[state]: state = trie[state]['fail'] if char in trie[state]: state = trie[state][char] else: state = 0 # For every towel match that ends here: for length_matched in trie[state]['out']: dp[i] += dp[i - length_matched] return dp[n] @profiler def main(input_data): towels,desired_patterns = ( sorted(x.split(), key=len, reverse=True) for x in input_data.replace(',', ' ').split('\n\n') ) part1 = 0 part2 = 0 # Build Aho-Corasick structure trie = build_aho_corasick(towels) for pattern in desired_patterns: count = count_possible_designs_aho(trie, pattern) if count: part1 += 1 part2 += count return part1,part2 if __name__ == "__main__": with open('input', 'r') as f: input_data = f.read().replace('\r', '').strip() result = main(input_data) print("Part 1:", result[0], "\nPart 2:", result[1])
Python
Approach: Recursive memoized backtracking with a Trie
I get to use one of my favorite data structures here, a Trie! It helps us figure out whether a prefix of the design is a valid pattern in linear time.
I use backtracking to choose potential component patterns (using the Trie), kicking off matching the rest of the design down the stack. We can continue matching longer patterns immediately after the recursion stack unwinds.
In addition, I use global memoization to keep track of the feasibility (part 1) or the number of combinations (part 2) for designs and sub-designs. This way, work done for earlier designs can help speed up later ones too.I ended up combining part 1 and 2 solutions into a single function because part 1 is a simpler variant of part 2 where we count all designs with the number of possible pattern combinations > 0.
Reading Input
import os here = os.path.dirname(os.path.abspath(__file__)) # read input def read_data(filename: str): global here filepath = os.path.join(here, filename) with open(filepath, mode="r", encoding="utf8") as f: return f.read()
Trie Implementation
class Trie: class TrieNode: def __init__(self) -> None: self.children = {} # connections to other TrieNode self.end = False # whether this node indicates an end of a pattern def __init__(self) -> None: self.root = Trie.TrieNode() def add(self, pattern: str): node = self.root # add the pattern to the trie, one character at a time for color in pattern: if color not in node.children: node.children[color] = Trie.TrieNode() node = node.children[color] # mark the node as the end of a pattern node.end = True
Solution
def soln(filename: str): data = read_data(filename) patterns, design_data = data.split("\n\n") # build the Trie trie = Trie() for pattern in patterns.split(", "): trie.add(pattern) designs = design_data.splitlines() # saves the design / sub-design -> number of component pattern combinations memo = {} def backtrack(design: str): nonlocal trie # if design is empty, we have successfully # matched the caller design / sub-design if design == "": return 1 # use memo if available if design in memo: return memo[design] # start matching a new pattern from here node = trie.root # number of pattern combinations for this design pattern_comb_count = 0 for i in range(len(design)): # if design[0 : i+1] is not a valid pattern, # we are done matching characters if design[i] not in node.children: break # move along the pattern node = node.children[design[i]] # we reached the end of a pattern if node.end: # get the pattern combinations count for the rest of the design / sub-design # all of them count for this design / sub-design pattern_comb_count += backtrack(design[i + 1 :]) # save the pattern combinations count for this design / sub-design memo[design] = pattern_comb_count return pattern_comb_count pattern_comb_counts = [] for design in designs: pattern_comb_counts.append(backtrack(design)) return pattern_comb_counts assert sum(1 for dc in soln("sample.txt") if dc > 0) == 6 print("Part 1:", sum(1 for dc in soln("input.txt") if dc > 0)) assert sum(soln("sample.txt")) == 16 print("Part 2:", sum(soln("input.txt")))