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Commits
| Author | SHA1 | Date | |
|---|---|---|---|
 Maya Herrscher
|
6d87bf0038 | ||
|
Maya Herrscher
|
731ce4b177 |
No files matched your search
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--- Day 5: Print Queue ---
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Satisfied with their search on Ceres, the squadron of scholars suggests subsequently scanning the stationery stacks of sub-basement 17.
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The North Pole printing department is busier than ever this close to Christmas, and while The Historians continue their search of this historically significant facility, an Elf operating a very familiar printer beckons you over.
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The Elf must recognize you, because they waste no time explaining that the new sleigh launch safety manual updates won't print correctly. Failure to update the safety manuals would be dire indeed, so you offer your services.
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Safety protocols clearly indicate that new pages for the safety manuals must be printed in a very specific order. The notation X|Y means that if both page number X and page number Y are to be produced as part of an update, page number X must be printed at some point before page number Y.
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The Elf has for you both the page ordering rules and the pages to produce in each update (your puzzle input), but can't figure out whether each update has the pages in the right order.
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For example:
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47|53
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97|13
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97|61
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97|47
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75|29
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61|13
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75|53
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29|13
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97|29
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53|29
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61|53
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97|53
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61|29
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47|13
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75|47
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97|75
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47|61
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75|61
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47|29
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75|13
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53|13
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75,47,61,53,29
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97,61,53,29,13
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75,29,13
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75,97,47,61,53
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61,13,29
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97,13,75,29,47
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The first section specifies the page ordering rules, one per line. The first rule, 47|53, means that if an update includes both page number 47 and page number 53, then page number 47 must be printed at some point before page number 53. (47 doesn't necessarily need to be immediately before 53; other pages are allowed to be between them.)
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The second section specifies the page numbers of each update. Because most safety manuals are different, the pages needed in the updates are different too. The first update, 75,47,61,53,29, means that the update consists of page numbers 75, 47, 61, 53, and 29.
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To get the printers going as soon as possible, start by identifying which updates are already in the right order.
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In the above example, the first update (75,47,61,53,29) is in the right order:
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75 is correctly first because there are rules that put each other page after it: 75|47, 75|61, 75|53, and 75|29.
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47 is correctly second because 75 must be before it (75|47) and every other page must be after it according to 47|61, 47|53, and 47|29.
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61 is correctly in the middle because 75 and 47 are before it (75|61 and 47|61) and 53 and 29 are after it (61|53 and 61|29).
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53 is correctly fourth because it is before page number 29 (53|29).
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29 is the only page left and so is correctly last.
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Because the first update does not include some page numbers, the ordering rules involving those missing page numbers are ignored.
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The second and third updates are also in the correct order according to the rules. Like the first update, they also do not include every page number, and so only some of the ordering rules apply - within each update, the ordering rules that involve missing page numbers are not used.
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The fourth update, 75,97,47,61,53, is not in the correct order: it would print 75 before 97, which violates the rule 97|75.
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The fifth update, 61,13,29, is also not in the correct order, since it breaks the rule 29|13.
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The last update, 97,13,75,29,47, is not in the correct order due to breaking several rules.
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For some reason, the Elves also need to know the middle page number of each update being printed. Because you are currently only printing the correctly-ordered updates, you will need to find the middle page number of each correctly-ordered update. In the above example, the correctly-ordered updates are:
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75,47,61,53,29
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97,61,53,29,13
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75,29,13
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These have middle page numbers of 61, 53, and 29 respectively. Adding these page numbers together gives 143.
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Of course, you'll need to be careful: the actual list of page ordering rules is bigger and more complicated than the above example.
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Determine which updates are already in the correct order. What do you get if you add up the middle page number from those correctly-ordered updates?
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Your puzzle answer was 5948.
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--- Part Two ---
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While the Elves get to work printing the correctly-ordered updates, you have a little time to fix the rest of them.
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For each of the incorrectly-ordered updates, use the page ordering rules to put the page numbers in the right order. For the above example, here are the three incorrectly-ordered updates and their correct orderings:
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75,97,47,61,53 becomes 97,75,47,61,53.
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61,13,29 becomes 61,29,13.
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97,13,75,29,47 becomes 97,75,47,29,13.
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After taking only the incorrectly-ordered updates and ordering them correctly, their middle page numbers are 47, 29, and 47. Adding these together produces 123.
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Find the updates which are not in the correct order. What do you get if you add up the middle page numbers after correctly ordering just those updates?
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Your puzzle answer was 3062.
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Both parts of this puzzle are complete! They provide two gold stars: **
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Executable
+37
@@ -0,0 +1,37 @@
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#!/usr/bin/env python3
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import sys
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from functools import cmp_to_key
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if __name__ == '__main__':
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lines = [line.strip('\n') for line in open(sys.argv[1])]
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rules = [tuple([int(y) for y in x.split('|')]) for x in lines[:lines.index('')]]
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updates = [[int(x) for x in line.split(',')] for line in lines[lines.index('')+1:]]
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def compare(a,b):
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for rule in rules:
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if a in rule and b in rule:
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return -1 if a == rule[0] else 1
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return 0
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res1, res2 = 0, 0
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for update in updates:
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correct = True
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for i, u in enumerate(update):
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for u2 in update[i+1:]:
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if compare(u,u2) == 1:
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correct = False
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break
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if not correct:
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update.sort(key=cmp_to_key(compare))
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res2 += update[int(len(update)/2)]
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break
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if correct: res1 += update[int(len(update)/2)]
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# challenge 1
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print(f"challenge 1:\n{res1}\n")
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# challenge 2
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print(f"challenge 2:\n{res2}")
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@@ -0,0 +1,28 @@
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47|53
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97|13
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97|61
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97|47
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75|29
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61|13
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75|53
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29|13
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97|29
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53|29
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61|53
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97|53
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61|29
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47|13
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75|47
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97|75
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47|61
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75|61
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47|29
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75|13
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53|13
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75,47,61,53,29
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97,61,53,29,13
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75,29,13
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75,97,47,61,53
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61,13,29
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97,13,75,29,47
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+1371
File diff suppressed because it is too large.
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--- Day 6: Guard Gallivant ---
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The Historians use their fancy device again, this time to whisk you all away to the North Pole prototype suit manufacturing lab... in the year 1518! It turns out that having direct access to history is very convenient for a group of historians.
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You still have to be careful of time paradoxes, and so it will be important to avoid anyone from 1518 while The Historians search for the Chief. Unfortunately, a single guard is patrolling this part of the lab.
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Maybe you can work out where the guard will go ahead of time so that The Historians can search safely?
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You start by making a map (your puzzle input) of the situation. 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 map shows the current position of the guard with ^ (to indicate the guard is currently facing up from the perspective of the map). Any obstructions - crates, desks, alchemical reactors, etc. - are shown as #.
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Lab guards in 1518 follow a very strict patrol protocol which involves repeatedly following these steps:
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If there is something directly in front of you, turn right 90 degrees.
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Otherwise, take a step forward.
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Following the above protocol, the guard moves up several times until she reaches an obstacle (in this case, a pile of failed suit prototypes):
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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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Because there is now an obstacle in front of the guard, she turns right before continuing straight in her new facing direction:
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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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Reaching another obstacle (a spool of several very long polymers), she turns right again and continues downward:
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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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.#......v.
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........#.
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#.........
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......#...
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This process continues for a while, but the guard eventually leaves the mapped area (after walking past a tank of universal solvent):
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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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......#v..
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By predicting the guard's route, you can determine which specific positions in the lab will be in the patrol path. Including the guard's starting position, the positions visited by the guard before leaving the area are marked with an X:
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....#.....
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....XXXXX#
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....X...X.
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..#.X...X.
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..XXXXX#X.
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..X.X.X.X.
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.#XXXXXXX.
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.XXXXXXX#.
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#XXXXXXX..
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......#X..
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In this example, the guard will visit 41 distinct positions on your map.
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Predict the path of the guard. How many distinct positions will the guard visit before leaving the mapped area?
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Your puzzle answer was 5212.
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--- Part Two ---
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While The Historians begin working around the guard's patrol route, you borrow their fancy device and step outside the lab. From the safety of a supply closet, you time travel through the last few months and record the nightly status of the lab's guard post on the walls of the closet.
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Returning after what seems like only a few seconds to The Historians, they explain that the guard's patrol area is simply too large for them to safely search the lab without getting caught.
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Fortunately, they are pretty sure that adding a single new obstruction won't cause a time paradox. They'd like to place the new obstruction in such a way that the guard will get stuck in a loop, making the rest of the lab safe to search.
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To have the lowest chance of creating a time paradox, The Historians would like to know all of the possible positions for such an obstruction. The new obstruction can't be placed at the guard's starting position - the guard is there right now and would notice.
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In the above example, there are only 6 different positions where a new obstruction would cause the guard to get stuck in a loop. The diagrams of these six situations use O to mark the new obstruction, | to show a position where the guard moves up/down, - to show a position where the guard moves left/right, and + to show a position where the guard moves both up/down and left/right.
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Option one, put a printing press next to the guard's starting position:
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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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.#.O^---+.
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........#.
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#.........
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......#...
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Option two, put a stack of failed suit prototypes in the bottom right quadrant of the mapped area:
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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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......O.#.
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#.........
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......#...
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Option three, put a crate of chimney-squeeze prototype fabric next to the standing desk in the bottom right quadrant:
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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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.+----+O#.
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#+----+...
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......#...
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Option four, put an alchemical retroencabulator near the bottom left corner:
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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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#O+---+...
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......#...
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Option five, put the alchemical retroencabulator a bit to the right instead:
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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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#..O+-+...
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......#...
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Option six, put a tank of sovereign glue right next to the tank of universal solvent:
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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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#+----++..
|
||||
......#O..
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|
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It doesn't really matter what you choose to use as an obstacle so long as you and The Historians can put it into position without the guard noticing. The important thing is having enough options that you can find one that minimizes time paradoxes, and in this example, there are 6 different positions you could choose.
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|
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You need to get the guard stuck in a loop by adding a single new obstruction. How many different positions could you choose for this obstruction?
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|
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Your puzzle answer was 1767.
|
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|
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Both parts of this puzzle are complete! They provide two gold stars: **
|
||||
Executable
+81
@@ -0,0 +1,81 @@
|
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#!/usr/bin/env python3
|
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|
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import sys
|
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from copy import deepcopy
|
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import numpy as np
|
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|
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|
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def show(maps):
|
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st = ''
|
||||
for line in maps:
|
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print(''.join(line))
|
||||
|
||||
|
||||
def find_guard(current):
|
||||
move = (0,0)
|
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for y in range(0, len(current)):
|
||||
for x, field in enumerate(current[y]):
|
||||
if field == '.' or field == '#' or field == 'X': continue
|
||||
elif field == '^': move = np.array((0, -1))
|
||||
elif field == '>': move = np.array((1, 0))
|
||||
elif field == 'v': move = np.array((0, 1))
|
||||
elif field == '<': move = np.array((-1, 0))
|
||||
guard = np.array((x, y))
|
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return(guard, move, field)
|
||||
|
||||
|
||||
def guard(m, guard, move):
|
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next = {(0,-1): '>', (1,0): 'v', (0,1): '<', (-1,0): '^'}
|
||||
nextm = {(0,-1): (1,0), (1,0): (0,1), (0,1): (-1, 0), (-1,0): (0,-1)}
|
||||
current = m
|
||||
|
||||
while (0 <= guard[0] + move[0] < len(m[0]) and 0 <= guard[1] + move[1] < len(m)):
|
||||
gx, gy = guard
|
||||
guard = guard + move
|
||||
x, y = guard
|
||||
if m[y][x] == '#':
|
||||
current[gy][gx] = next[tuple(move)]
|
||||
return current, True, np.array((gx, gy)), np.array(nextm[tuple(move)])
|
||||
else:
|
||||
current[gy][gx] = 'X'
|
||||
current[guard[1]][guard[0]] = 'X'
|
||||
return current, False, np.array((guard[0], guard[1])), np.array(nextm[tuple(move)])
|
||||
|
||||
|
||||
def test_loop(m, x, y, g, move, o):
|
||||
current = deepcopy(m)
|
||||
current[y][x] = '#'
|
||||
positions = set([])
|
||||
positions.add((g[0], g[1], o))
|
||||
running = True
|
||||
while running:
|
||||
current, running, g, move = guard(current, g, move)
|
||||
o = current[g[1]][g[0]]
|
||||
if (g[0], g[1], o) in positions:
|
||||
return True
|
||||
positions.add((g[0], g[1], o))
|
||||
return False
|
||||
|
||||
|
||||
if __name__ == '__main__':
|
||||
maps = [list(line.strip('\n')) for line in open(sys.argv[1])]
|
||||
sguard, smove, start = find_guard(maps)
|
||||
|
||||
# challenge 1
|
||||
running = True
|
||||
current = deepcopy(maps)
|
||||
g, m = sguard, smove
|
||||
while running:
|
||||
current, running, g, m = guard(current, g, m)
|
||||
|
||||
res1 = sum([m.count('X') for m in current])
|
||||
print(f"challenge 1:\n{res1}\n")
|
||||
|
||||
# challenge 2
|
||||
res2 = 0
|
||||
for y in range(0, len(current)):
|
||||
for x, field in enumerate(current[y]):
|
||||
if field == '#' or field == '.': continue
|
||||
res2 += test_loop(maps, x, y, sguard, smove, start)
|
||||
print(f"challenge 2:\n{res2}")
|
||||
|
||||
@@ -0,0 +1,10 @@
|
||||
....#.....
|
||||
.........#
|
||||
..........
|
||||
..#.......
|
||||
.......#..
|
||||
..........
|
||||
.#..^.....
|
||||
........#.
|
||||
#.........
|
||||
......#...
|
||||
@@ -0,0 +1,130 @@
|
||||
................#.................#.#...........................#.#..................#............................................
|
||||
..........#................#...........#...................#.#....................................................................
|
||||
....#......#......#............#.#........#..........#.......#.......#...#........#..........#.........................#..........
|
||||
..................#............#.....##...............#..#.....................#...........#......................................
|
||||
#.........#.............................................................#........#............#........#.....#...#...........#....
|
||||
.....#...#......#..................................................#........#..................#......#.......................#...
|
||||
#.................................................#......................#......#...................................#.............
|
||||
...#.#...#.....................................#.......................#.........#..................................#....#........
|
||||
..............................................#..............#...#...........##..........................#.......#............#...
|
||||
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Reference in new issue
Block a user