Day 19 - Linen Layout

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FAQ

  • Pyro@programming.dev
    4·
    8 days ago

    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")))