126 lines
4.4 KiB
Python
126 lines
4.4 KiB
Python
# --- Day 8: Memory Maneuver ---
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# The sleigh is much easier to pull than you'd expect for something its weight.
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# Unfortunately, neither you nor the Elves know which way the North Pole is
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# from here.
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# You check your wrist device for anything that might help. It seems to have
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# some kind of navigation system! Activating the navigation system produces
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# more bad news: "Failed to start navigation system. Could not read software
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# license file."
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# The navigation system's license file consists of a list of numbers (your
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# puzzle input). The numbers define a data structure which, when processed,
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# produces some kind of tree that can be used to calculate the license number.
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# The tree is made up of nodes; a single, outermost node forms the tree's root,
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# and it contains all other nodes in the tree (or contains nodes that contain
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# nodes, and so on).
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# Specifically, a node consists of:
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# A header, which is always exactly two numbers:
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# The quantity of child nodes.
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# The quantity of metadata entries.
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# Zero or more child nodes (as specified in the header).
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# One or more metadata entries (as specified in the header).
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# Each child node is itself a node that has its own header, child nodes, and
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# metadata. For example:
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# 2 3 0 3 10 11 12 1 1 0 1 99 2 1 1 2
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# A----------------------------------
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# B----------- C-----------
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# D-----
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# In this example, each node of the tree is also marked with an underline
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# starting with a letter for easier identification. In it, there are four
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# nodes:
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# A, which has 2 child nodes (B, C) and 3 metadata entries (1, 1, 2).
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# B, which has 0 child nodes and 3 metadata entries (10, 11, 12).
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# C, which has 1 child node (D) and 1 metadata entry (2).
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# D, which has 0 child nodes and 1 metadata entry (99).
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# The first check done on the license file is to simply add up all of the
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# metadata entries. In this example, that sum is 1+1+2+10+11+12+2+99=138.
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# What is the sum of all metadata entries?
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from copy import copy
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from typing import List
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with open("files/P8.txt") as f:
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tree = [int(num) for num in f.read().strip().split()]
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def sum_metadata(lst: List[int]) -> int:
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childs = lst.pop(0)
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metadata = lst.pop(0)
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ans = sum(sum_metadata(lst) for _ in range(childs)) + sum(
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lst.pop(0) for _ in range(metadata)
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)
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return ans
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def part_1() -> None:
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tree_p1 = copy(tree)
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print(f"The sum of all metadata entries is {sum_metadata(tree_p1)}")
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# --- Part Two ---
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# The second check is slightly more complicated: you need to find the value of
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# the root node (A in the example above).
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# The value of a node depends on whether it has child nodes.
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# If a node has no child nodes, its value is the sum of its metadata entries.
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# So, the value of node B is 10+11+12=33, and the value of node D is 99.
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# However, if a node does have child nodes, the metadata entries become indexes
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# which refer to those child nodes. A metadata entry of 1 refers to the first
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# child node, 2 to the second, 3 to the third, and so on. The value of this
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# node is the sum of the values of the child nodes referenced by the metadata
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# entries. If a referenced child node does not exist, that reference is
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# skipped. A child node can be referenced multiple time and counts each time it
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# is referenced. A metadata entry of 0 does not refer to any child node.
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# For example, again using the above nodes:
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# Node C has one metadata entry, 2. Because node C has only one child node,
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# 2 references a child node which does not exist, and so the value of node C is
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# 0.
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# Node A has three metadata entries: 1, 1, and 2. The 1 references node A's
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# first child node, B, and the 2 references node A's second child node, C.
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# Because node B has a value of 33 and node C has a value of 0, the value of
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# node A is 33+33+0=66.
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# So, in this example, the value of the root node is 66.
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# What is the value of the root node?
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def find_root_value(lst: List[int]) -> int:
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childs = lst.pop(0)
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metadata = lst.pop(0)
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values = [find_root_value(lst) for _ in range(childs)]
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metadatas = [lst.pop(0) for _ in range(metadata)]
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if childs == 0:
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return sum(metadatas)
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return sum(
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values[child - 1] for child in metadatas if child - 1 in range(childs)
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)
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def part_2() -> None:
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tree_p2 = copy(tree)
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print(f"The value of the root node is {find_root_value(tree_p2)}")
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if __name__ == "__main__":
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part_1()
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part_2()
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