Some work on AOC script. Needs more
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[1]Advent of Code
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• [2][About]
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• [3][Events]
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• [4][Shop]
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• [5][Settings]
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• [6][Log Out]
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br0xen [7](AoC++) 36*
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$year=[8]2023;
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• [9][Calendar]
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• [10][AoC++]
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• [11][Sponsors]
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• [12][Leaderboard]
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• [13][Stats]
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Our [14]sponsors help make Advent of Code possible:
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[15]McGraw Hill - Transforming education, one line of code at a time.
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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 [16]observatory. The Elf within turns out to be a researcher studying
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cosmic expansion using the giant telescope here.
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He doesn't know anything about the missing machine parts; he's only
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visiting for this research project. However, he confirms that the hot
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springs are the next-closest area likely to have people; he'll even take
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you straight there once he's done 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
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single giant image (your puzzle input). The image includes empty space (.)
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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
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shortest path between every pair of galaxies. However, there's a catch:
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the universe expanded in the time it took the light from those galaxies to
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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
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should 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
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expansion 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
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of galaxies can be found. It can help to assign every galaxy a unique
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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; order
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within the pair doesn't matter. For each pair, find any shortest path
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between the two galaxies using only steps that move up, down, left, or
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right exactly one . or # at a time. (The shortest path between two
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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
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from galaxy 5 to galaxy 9 (the eight locations marked # plus the step onto
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galaxy 9 itself). 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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Your puzzle answer was 9957702.
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--- Part Two ---
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The galaxies are much older (and thus much farther apart) than the
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researcher initially estimated.
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Now, instead of the expansion you did before, make each empty row or
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column one million times larger. That is, each empty row should be
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replaced with 1000000 empty rows, and each empty column should be replaced
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with 1000000 empty columns.
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(In the example above, if each empty row or column were merely 10 times
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larger, the sum of the shortest paths between every pair of galaxies would
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be 1030. If each empty row or column were merely 100 times larger, the sum
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of the shortest paths between every pair of galaxies would be 8410.
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However, your universe will need to expand far beyond these values.)
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Starting with the same initial image, expand the universe according to
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these new rules, then find the length of the shortest path between every
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pair of galaxies. What is the sum of these lengths?
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Your puzzle answer was 512240933238.
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Both parts of this puzzle are complete! They provide two gold stars: **
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At this point, you should [17]return to your Advent calendar and try
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another puzzle.
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If you still want to see it, you can [18]get your puzzle input.
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You can also [Shareon [19]Twitter [20]Mastodon] this puzzle.
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References
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Visible links
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1. https://adventofcode.com/
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2. https://adventofcode.com/2023/about
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3. https://adventofcode.com/2023/events
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4. https://teespring.com/stores/advent-of-code
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5. https://adventofcode.com/2023/settings
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6. https://adventofcode.com/2023/auth/logout
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7. Advent of Code Supporter
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https://adventofcode.com/2023/support
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8. https://adventofcode.com/2023
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9. https://adventofcode.com/2023
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10. https://adventofcode.com/2023/support
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11. https://adventofcode.com/2023/sponsors
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12. https://adventofcode.com/2023/leaderboard
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13. https://adventofcode.com/2023/stats
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14. https://adventofcode.com/2023/sponsors
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15. https://mheducation.com/
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16. https://en.wikipedia.org/wiki/Observatory
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17. https://adventofcode.com/2023
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18. https://adventofcode.com/2023/day/11/input
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19. https://twitter.com/intent/tweet?text=I%27ve+completed+%22Cosmic+Expansion%22+%2D+Day+11+%2D+Advent+of+Code+2023&url=https%3A%2F%2Fadventofcode%2Ecom%2F2023%2Fday%2F11&related=ericwastl&hashtags=AdventOfCode
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20. javascript:void(0);
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@ -0,0 +1,200 @@
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[1]Advent of Code
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• [2][About]
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• [3][Events]
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• [4][Shop]
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• [5][Settings]
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• [6][Log Out]
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||||
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||||
br0xen [7](AoC++) 36*
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var y=[8]2023;
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• [9][Calendar]
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• [10][AoC++]
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• [11][Sponsors]
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• [12][Leaderboard]
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• [13][Stats]
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Our [14]sponsors help make Advent of Code possible:
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[15]Bank of America - We use technology, models and data to make financial
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lives better for our clients and communities.
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--- Day 17: Clumsy Crucible ---
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The lava starts flowing rapidly once the Lava Production Facility is
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operational. As you leave, the reindeer offers you a parachute, allowing
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you to quickly reach Gear Island.
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As you descend, your bird's-eye view of Gear Island reveals why you had
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trouble finding anyone on your way up: half of Gear Island is empty, but
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the half below you is a giant factory city!
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You land near the gradually-filling pool of lava at the base of your new
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lavafall. Lavaducts will eventually carry the lava throughout the city,
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but to make use of it immediately, Elves are loading it into large
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[16]crucibles on wheels.
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The crucibles are top-heavy and pushed by hand. Unfortunately, the
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crucibles become very difficult to steer at high speeds, and so it can be
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hard to go in a straight line for very long.
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To get Desert Island the machine parts it needs as soon as possible,
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you'll need to find the best way to get the crucible from the lava pool to
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the machine parts factory. To do this, you need to minimize heat loss
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while choosing a route that doesn't require the crucible to go in a
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straight line for too long.
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Fortunately, the Elves here have a map (your puzzle input) that uses
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traffic patterns, ambient temperature, and hundreds of other parameters to
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calculate exactly how much heat loss can be expected for a crucible
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entering any particular city block.
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For example:
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2413432311323
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3215453535623
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3255245654254
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3446585845452
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4546657867536
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1438598798454
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4457876987766
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3637877979653
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4654967986887
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4564679986453
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1224686865563
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2546548887735
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4322674655533
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Each city block is marked by a single digit that represents the amount of
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heat loss if the crucible enters that block. The starting point, the lava
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pool, is the top-left city block; the destination, the machine parts
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factory, is the bottom-right city block. (Because you already start in the
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top-left block, you don't incur that block's heat loss unless you leave
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that block and then return to it.)
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Because it is difficult to keep the top-heavy crucible going in a straight
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line for very long, it can move at most three blocks in a single direction
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before it must turn 90 degrees left or right. The crucible also can't
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reverse direction; after entering each city block, it may only turn left,
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continue straight, or turn right.
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One way to minimize heat loss is this path:
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2>>34^>>>1323
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32v>>>35v5623
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32552456v>>54
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3446585845v52
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4546657867v>6
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14385987984v4
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44578769877v6
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36378779796v>
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465496798688v
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456467998645v
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12246868655<v
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25465488877v5
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43226746555v>
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This path never moves more than three consecutive blocks in the same
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direction and incurs a heat loss of only 102.
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Directing the crucible from the lava pool to the machine parts factory,
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but not moving more than three consecutive blocks in the same direction,
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what is the least heat loss it can incur?
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Your puzzle answer was 1001.
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--- Part Two ---
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The crucibles of lava simply aren't large enough to provide an adequate
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supply of lava to the machine parts factory. Instead, the Elves are going
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to upgrade to ultra crucibles.
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Ultra crucibles are even more difficult to steer than normal crucibles.
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Not only do they have trouble going in a straight line, but they also have
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trouble turning!
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Once an ultra crucible starts moving in a direction, it needs to move a
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minimum of four blocks in that direction before it can turn (or even
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before it can stop at the end). However, it will eventually start to get
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wobbly: an ultra crucible can move a maximum of ten consecutive blocks
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without turning.
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In the above example, an ultra crucible could follow this path to minimize
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heat loss:
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2>>>>>>>>1323
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32154535v5623
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32552456v4254
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34465858v5452
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45466578v>>>>
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143859879845v
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445787698776v
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363787797965v
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465496798688v
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456467998645v
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122468686556v
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254654888773v
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432267465553v
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In the above example, an ultra crucible would incur the minimum possible
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heat loss of 94.
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Here's another example:
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111111111111
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999999999991
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999999999991
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999999999991
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999999999991
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Sadly, an ultra crucible would need to take an unfortunate path like this
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one:
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1>>>>>>>1111
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9999999v9991
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9999999v9991
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9999999v9991
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9999999v>>>>
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This route causes the ultra crucible to incur the minimum possible heat
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loss of 71.
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Directing the ultra crucible from the lava pool to the machine parts
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factory, what is the least heat loss it can incur?
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Your puzzle answer was 1197.
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Both parts of this puzzle are complete! They provide two gold stars: **
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At this point, you should [17]return to your Advent calendar and try
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another puzzle.
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If you still want to see it, you can [18]get your puzzle input.
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You can also [Shareon [19]Twitter [20]Mastodon] this puzzle.
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References
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Visible links
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1. https://adventofcode.com/
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2. https://adventofcode.com/2023/about
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3. https://adventofcode.com/2023/events
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4. https://teespring.com/stores/advent-of-code
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5. https://adventofcode.com/2023/settings
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6. https://adventofcode.com/2023/auth/logout
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7. Advent of Code Supporter
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https://adventofcode.com/2023/support
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8. https://adventofcode.com/2023
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9. https://adventofcode.com/2023
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10. https://adventofcode.com/2023/support
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11. https://adventofcode.com/2023/sponsors
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12. https://adventofcode.com/2023/leaderboard
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13. https://adventofcode.com/2023/stats
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14. https://adventofcode.com/2023/sponsors
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15. https://careers.bankofamerica.com/
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16. https://en.wikipedia.org/wiki/Crucible
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17. https://adventofcode.com/2023
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18. https://adventofcode.com/2023/day/17/input
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19. https://twitter.com/intent/tweet?text=I%27ve+completed+%22Clumsy+Crucible%22+%2D+Day+17+%2D+Advent+of+Code+2023&url=https%3A%2F%2Fadventofcode%2Ecom%2F2023%2Fday%2F17&related=ericwastl&hashtags=AdventOfCode
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20. javascript:void(0);
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11
sbin/aoc
11
sbin/aoc
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@ -35,11 +35,12 @@ getproblem() {
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DY=${PWD:5:5}
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DY=${DY/day0/}
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DY=${DY/day/}
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DIRDY=$DY
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fi
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if [[ ! -d "$AOCROOT/$YR/day$DIRDY" ]]; then
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mkdir "$AOCROOT/$YR/day$DIRDY"
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if [[ ! -d "$AOCROOT/$YR/day$DY" ]]; then
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mkdir "$AOCROOT/$YR/day$DY"
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fi
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cd "$AOCROOT/$YR/day$DIRDY"
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cd "$AOCROOT/$YR/day$DY"
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if [[ -f "problem" ]]; then
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echo "Found problem file."
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if [[ $1 != "-f" ]]; then
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@ -50,8 +51,8 @@ getproblem() {
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fi
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# Remove any zero padding from day
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DY=$(echo "$CURRDAY"|awk '$0*=1')
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echo "Getting problem at $CURRYEAR/day/$DY"
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DY=$(echo "$DY"|awk '$0*=1')
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echo "Getting problem at $YR/day/$DY"
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elinks -dump https://adventofcode.com/$CURRYEAR/day/$DY > problem
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}
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