2018 Day08 Complete
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package main
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import (
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"bufio"
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"fmt"
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"log"
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"os"
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"strconv"
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"strings"
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)
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var top *Node
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func main() {
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inp := StdinToString()
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hackTheGibson(inp)
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part1()
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part2()
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}
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var nodeIndex byte
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func hackTheGibson(inp string) {
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parts := strings.Split(inp, " ")
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var rem []int
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for i := range parts {
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rem = append(rem, Atoi(parts[i]))
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}
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top, _ = parseNode(rem)
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}
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func parseNode(vals []int) (*Node, []int) {
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id := nodeIndex
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nodeIndex = nodeIndex + 1
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retN := &Node{
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id: id,
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ChildCount: vals[0],
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MetaCount: vals[1],
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}
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rem := vals[2:]
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children := retN.ChildCount
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if children == 0 {
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retN.Metadata = rem[:retN.MetaCount]
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return retN, rem[retN.MetaCount:]
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} else {
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var cN *Node
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for children > 0 {
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cN, rem = parseNode(rem)
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retN.Children = append(retN.Children, cN)
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children--
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}
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retN.Metadata = rem[:retN.MetaCount]
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rem = rem[retN.MetaCount:]
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}
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return retN, rem
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}
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func part1() {
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fmt.Println("= Part 1 =")
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fmt.Println("Sum:", top.recurMetaSum())
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}
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func part2() {
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fmt.Println("= Part 2 =")
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fmt.Println(top.value())
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}
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type Node struct {
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id byte
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ChildCount int
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MetaCount int
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Metadata []int
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Children []*Node
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}
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var indent int
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func (n *Node) value() int {
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var ret int
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if n.ChildCount == 0 {
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for _, v := range n.Metadata {
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ret += v
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}
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} else {
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indent++
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for _, v := range n.Metadata {
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if v > 0 && v <= n.ChildCount {
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ret += n.Children[v-1].value()
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}
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}
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indent--
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}
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return ret
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}
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func (n *Node) recurMetaSum() int {
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var ret int
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for _, v := range n.Children {
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ret += v.recurMetaSum()
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}
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for _, v := range n.Metadata {
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ret += v
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}
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return ret
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}
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func (n *Node) string() string {
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ret := fmt.Sprintf("[ %d: %d %d ]", n.id, n.ChildCount, n.MetaCount)
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return ret
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}
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func StdinToString() string {
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var input string
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scanner := bufio.NewScanner(os.Stdin)
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for scanner.Scan() {
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input = input + scanner.Text()
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}
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return input
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}
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func Atoi(i string) int {
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var ret int
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var err error
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if ret, err = strconv.Atoi(i); err != nil {
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log.Fatal("Invalid Atoi")
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}
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return ret
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}
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File diff suppressed because one or more lines are too long
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@ -0,0 +1,90 @@
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Advent of Code
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--- Day 8: Memory Maneuver ---
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The sleigh is much easier to pull than you'd expect for something its weight. Unfortunately, neither you nor the Elves know which way the North Pole is from here.
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You check your wrist device for anything that might help. It seems to have some kind of navigation system! Activating the navigation system produces more bad news: "Failed to start
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navigation system. Could not read software license file."
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The navigation system's license file consists of a list of numbers (your puzzle input). The numbers define a data structure which, when processed, produces some kind of tree that can
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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, and it contains all other nodes in the tree (or contains nodes that contain 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 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 starting with a letter for easier identification. In it, there are four 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 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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Your puzzle answer was 36891.
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--- Part Two ---
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The second check is slightly more complicated: you need to find the value of 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. 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 which refer to those child nodes. A metadata entry of 1 refers to the first child node, 2 to the second,
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3 to the third, and so on. The value of this node is the sum of the values of the child nodes referenced by the metadata entries. If a referenced child node does not exist, that
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reference is skipped. A child node can be referenced multiple time and counts each time it 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, 2 references a child node which does not exist, and so the value of node C is 0.
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* Node A has three metadata entries: 1, 1, and 2. The 1 references node A's first child node, B, and the 2 references node A's second child node, C. Because node B has a value of 33
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and node C has a value of 0, the value of 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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Your puzzle answer was 20083.
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Both parts of this puzzle are complete! They provide two gold stars: **
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References
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Visible links
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. https://adventofcode.com/
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. https://adventofcode.com/2018/about
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. https://adventofcode.com/2018/events
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. https://adventofcode.com/2018/settings
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. https://adventofcode.com/2018/auth/logout
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. Advent of Code Supporter
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https://adventofcode.com/2018/support
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. https://adventofcode.com/2018
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. https://adventofcode.com/2018
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. https://adventofcode.com/2018/support
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. https://adventofcode.com/2018/sponsors
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. https://adventofcode.com/2018/leaderboard
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. https://adventofcode.com/2018/stats
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. https://adventofcode.com/2018/sponsors
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. https://en.wikipedia.org/wiki/Tree_(data_structure)
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. https://adventofcode.com/2018
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. https://adventofcode.com/2018/day/8/input
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@ -0,0 +1 @@
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2 3 0 3 10 11 12 1 1 0 1 99 2 1 1 2
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