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Diff for: ‎2023/day10/inp.txt

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Diff for: ‎2023/day10/main.go

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package main
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import (
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"fmt"
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"image"
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"math"
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"os"
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"regexp"
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"slices"
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"strings"
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)
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func main() {
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input, _ := os.ReadFile("inp.txt")
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grid, start := map[image.Point]rune{}, image.Point{}
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for y, s := range strings.Fields(string(input)) {
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for x, r := range s {
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grid[image.Point{x, y}] = r
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if r == 'S' {
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start = image.Point{x, y}
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}
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}
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}
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for s, r := range map[string]rune{
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`[7F|].[JL|].`: '|', `.[-7J].[-FL]`: '-', `[7F|][-7J]..`: 'L',
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`[7F|]..[-FL]`: 'J', `..[JL|][-FL]`: '7', `.[-7J][JL|].`: 'F',
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} {
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if ok, _ := regexp.MatchString(s, string([]rune{
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grid[start.Add(image.Point{0, -1})], grid[start.Add(image.Point{1, 0})],
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grid[start.Add(image.Point{0, 1})], grid[start.Add(image.Point{-1, 0})],
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})); ok {
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grid[start] = r
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}
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}
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path, area := []image.Point{start}, 0
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for p, n := start, start; ; path = append(path, n) {
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p, n = n, start
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for _, d := range map[rune][]image.Point{
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'|': {{0, -1}, {0, 1}}, '-': {{1, 0}, {-1, 0}}, 'L': {{0, -1}, {1, 0}},
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'J': {{0, -1}, {-1, 0}}, '7': {{0, 1}, {-1, 0}}, 'F': {{0, 1}, {1, 0}},
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}[grid[p]] {
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if !slices.Contains(path, p.Add(d)) {
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n = p.Add(d)
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}
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}
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area += p.X*n.Y - p.Y*n.X
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if n == start {
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break
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}
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}
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fmt.Println(len(path) / 2)
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fmt.Println(int(math.Abs(float64(area)))/2 - len(path)/2 + 1)
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}

Diff for: ‎2023/day11/inp.txt

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Diff for: ‎2023/day11/main.go

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/*
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--- Day 11: Cosmic Expansion ---
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You continue following signs for "Hot Springs" and eventually come across
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an observatory. The Elf within turns out to be a researcher studying cosmic
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expansion using the giant telescope here.
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He doesn't know anything about the missing machine parts; he's only visiting for
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this research project. However, he confirms that the hot springs are the next-closest
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area likely to have people; he'll even take you straight there once he's done
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with today's observation analysis.
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Maybe you can help him with the analysis to speed things up?
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The researcher has collected a bunch of data and compiled the data into a single
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giant image (your puzzle input).
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The image includes empty space (.) and galaxies (#). For example:
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...#......
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.......#..
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#.........
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..........
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......#...
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.#........
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.........#
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..........
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.......#..
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#...#.....
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The researcher is trying to figure out the sum of the lengths of the shortest path
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between every pair of galaxies. However, there's a catch: the universe expanded
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in the time it took the light from those galaxies to reach the observatory.
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Due to something involving gravitational effects, only some space expands.
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In fact, the result is that any rows or columns that contain no galaxies should
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all actually be twice as big.
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In the above example, three columns and two rows contain no galaxies:
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v v v
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...#......
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.......#..
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#.........
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>..........<
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......#...
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.#........
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.........#
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>..........<
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.......#..
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#...#.....
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^ ^ ^
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These rows and columns need to be twice as big; the result of cosmic expansion
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therefore looks like this:
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....#........
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.........#...
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#............
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.............
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.............
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........#....
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.#...........
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............#
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.............
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.............
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.........#...
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#....#.......
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Equipped with this expanded universe, the shortest path between every pair of
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galaxies can be found. It can help to assign every galaxy a unique number:
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....1........
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.........2...
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3............
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.............
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.............
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........4....
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.5...........
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............6
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.............
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.............
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.........7...
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8....9.......
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In these 9 galaxies, there are 36 pairs. Only count each pair once;
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order within the pair doesn't matter. For each pair, find any shortest
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path between the two galaxies using only steps that move up, down, left,
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or right exactly one . or # at a time. (The shortest path between
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two galaxies is allowed to pass through another galaxy.)
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For example, here is one of the shortest paths between galaxies 5 and 9:
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....1........
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.........2...
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3............
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.............
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.............
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........4....
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.5...........
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.##.........6
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..##.........
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...##........
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....##...7...
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8....9.......
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This path has length 9 because it takes a minimum of nine steps to get from galaxy
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5 to galaxy 9 (the eight locations marked # plus the step onto galaxy 9 itself).
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Here are some other example shortest path lengths:
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Between galaxy 1 and galaxy 7: 15
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Between galaxy 3 and galaxy 6: 17
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Between galaxy 8 and galaxy 9: 5
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In this example, after expanding the universe, the sum of the shortest
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path between all 36 pairs of galaxies is 374.
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Expand the universe, then find the length of the shortest path between
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every pair of galaxies. What is the sum of these lengths?
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--- Part Two ---
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The galaxies are much older (and thus much farther apart)
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than the researcher initially estimated.
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Now, instead of the expansion you did before, make each empty row or column one million times larger.
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That is, each empty row should be replaced with 1000000 empty rows,
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and each empty column should be replaced with 1000000 empty columns.
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(In the example above, if each empty row or column were merely 10 times larger,
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the sum of the shortest paths between every pair of galaxies would be 1030.
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If each empty row or column were merely 100 times larger, the sum of the shortest
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paths between every pair of galaxies would be 8410. However, your universe will
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need to expand far beyond these values.)
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Starting with the same initial image, expand the universe according to these new rules,
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then find the length of the shortest path between every pair of galaxies.
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What is the sum of these lengths?
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*/
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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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"math"
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"os"
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)
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func Day11() [2]int {
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return [2]int{
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d11p1(),
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d11p2(),
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}
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}
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func main() {
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fmt.Println(Day11())
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}
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func GetShortestDistordedDistance(distortion int) int {
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file, err := os.Open("inp.txt")
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if err != nil {
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log.Fatal(err)
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}
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scanner := bufio.NewScanner(file)
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//get all data from input file
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starChart := [][]rune{}
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for scanner.Scan() {
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starChart = append(starChart, []rune(scanner.Text()))
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}
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if scanner.Err() != nil {
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log.Fatal(scanner.Err())
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}
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//find all rows that doesn't have a '#'
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rowsWithoutHash := []int{}
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for y := 0; y < len(starChart); y++ {
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found := false
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for x := 0; x < len(starChart[y]); x++ {
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if starChart[y][x] == '#' {
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found = true
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break
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}
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}
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if !found {
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rowsWithoutHash = append(rowsWithoutHash, y)
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}
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}
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//find all columns that doesn't have a '#'
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colsWithoutHash := []int{}
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for x := 0; x < len(starChart[0]); x++ {
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found := false
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for y := 0; y < len(starChart); y++ {
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if starChart[y][x] == '#' {
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found = true
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break
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}
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}
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if !found {
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colsWithoutHash = append(colsWithoutHash, x)
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}
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}
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//get the distoreded coordinates of all # symbols
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starPos := [][2]int{}
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for y := 0; y < len(starChart); y++ {
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for x := 0; x < len(starChart[y]); x++ {
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if starChart[y][x] == '#' {
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cumulDistortionX := 0
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cumulDistortionY := 0
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for i := 0; i < len(rowsWithoutHash); i++ {
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if y > rowsWithoutHash[i] {
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cumulDistortionY += distortion - 1
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}
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}
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for i := 0; i < len(colsWithoutHash); i++ {
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if x > colsWithoutHash[i] {
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cumulDistortionX += distortion - 1
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}
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}
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starPos = append(starPos, [2]int{x + cumulDistortionX, y + cumulDistortionY})
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}
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}
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}
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//get the sum of shortest path
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sum := 0
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for i := 0; i < len(starPos)-1; i++ {
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for j := i + 1; j < len(starPos); j++ {
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sum += ManhattanDistance(starPos[i], starPos[j])
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}
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}
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return sum
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}
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func d11p1() int {
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return GetShortestDistordedDistance(2)
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}
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func d11p2() int {
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return GetShortestDistordedDistance(1000000)
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}
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// calculate the Manhattan Distance between A and B and return the distance
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func ManhattanDistance(posA [2]int, posB [2]int) int {
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return int(math.Abs(float64(posB[0])-float64(posA[0])) + math.Abs(float64(posB[1])-float64(posA[1])))
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}

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