Day 9: Mirage Maintenance

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  • itslilith@lemmy.blahaj.zone
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    11 months ago

    (Cursed) Python

    I solved the actual thing recursively in Rust, but I decided that wasn’t cursed enough, so I present: Polynomial fitting!

    import numpy.polynomial.polynomial as pol
    
    with open("input.txt") as f:
      lines = list(map(lambda l: list(map(int, l.split(" "))), f.read().split("\n")))
    
    lo, hi = 0, 0
    
    for line in lines:
      for i in range(len(line)):
        poly, (r, *_) = pol.Polynomial.fit(range(len(line)), line, full=True, deg=i)
        if r < 0.0000000001:
          break
    
      lo += int(round(poly(-1)))
      hi += int(round(poly(len(line))))
    
    print(f"Part 1: {hi}")
    print(f"Part 2: {lo}")
    
  • corristo@programming.dev
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    11 months ago

    APL

    I finally managed to make use of ⍣ :D

    input←⊃⎕NGET'inputs/day9.txt'1
    p←{⍎('¯'@((⍸'-'∘=)⍵))⍵}¨input
    f←({⍵⍪⊂2-⍨/⊃¯1↑⍵}⍣{∧/0=⊃¯1↑⍺})
    ⎕←+/{+/⊢/¨f⊂⍵}¨p ⍝ part 1
    ⎕←+/{-/⊣/¨f⊂⍵}¨p ⍝ part 2
    
  • janAkali@lemmy.one
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    11 months ago

    Nim

    Part 1:
    The extrapolated value to the right is just the sum of all last values in the diff pyramid. 45 + 15 + 6 + 2 + 0 = 68
    Part 2:
    The extrapolated value to the left is just a right-folded difference (right-associated subtraction) between all first values in the pyramid. e.g. 10 - (3 - (0 - (2 - 0))) = 5

    So, extending the pyramid is totally unneccessary.

    Total runtime: 0.9 ms
    Puzzle rating: Easy, but interesting 6.5/10
    Full Code: day_09/solution.nim
    Snippet:

    proc solve(lines: seq[string]): AOCSolution[int] =
      for line in lines:
        var current = line.splitWhitespace().mapIt(it.parseInt())
        var firstValues: seq[int]
    
        while not current.allIt(it == 0):
          firstValues.add current[0]
          block p1:
            result.part1 += current[^1]
    
          var nextIter = newSeq[int](current.high)
          for i, v in current[1..^1]:
            nextIter[i] = v - current[i]
          current = nextIter
    
        block p2:
          result.part2 += firstValues.foldr(a-b)
    
  • cacheson@kbin.social
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    11 months ago

    Nim

    Pretty easy one today. Made a Pyramid type to hold the values and their layers of diffs, and an extend function to predict the next value. For part 2 I just had to make an extendLeft version of it that inserts and subtracts instead of appending and adding.

  • mykl@lemmy.world
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    11 months ago

    I even have time to knock out a quick Uiua solution before going out today, using experimental recursion support. Bleeding edge code:

    # Experimental!
    {"0 3 6 9 12 15"
     "1 3 6 10 15 21"
     "10 13 16 21 30 45"}
    StoInt ← /(+×10)▽×⊃(≥0)(≤9).-@0
    NextTerm ← ↬(
      ↘1-↻¯1..      # rot by one and take diffs
      (|1 ↫|⊢)=1⧻⊝. # if they're all equal grab else recurse
      +⊙(⊢↙¯1)      # add to last value of input
    )
    ≡(⊜StoInt≠@\s.⊔) # parse
    ⊃(/+≡NextTerm)(/+≡(NextTerm ⇌))
    
  • cvttsd2si@programming.dev
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    11 months ago

    Scala3

    def diffs(a: Seq[Long]): List[Long] =
        a.drop(1).zip(a).map(_ - _).toList
    
    def predictNext(a: Seq[Long], combine: (Seq[Long], Long) => Long): Long =
        if a.forall(_ == 0) then 0 else combine(a, predictNext(diffs(a), combine))
    
    def predictAllNexts(a: List[String], combine: (Seq[Long], Long) => Long): Long = 
        a.map(l => predictNext(l.split(raw"\s+").map(_.toLong), combine)).sum
    
    def task1(a: List[String]): Long = predictAllNexts(a, _.last + _)
    def task2(a: List[String]): Long = predictAllNexts(a, _.head - _)
    
  • mykl@lemmy.world
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    11 months ago

    Dart

    I was getting a bad feeling when it explained in such detail how to solve part 1 that part 2 was going to be some sort of nightmare of traversing all those generated numbers in some complex fashion, but this has got to be one of the shortest solutions I’ve ever written for an AoC challenge.

    int nextTerm(Iterable ns) {
      var diffs = ns.window(2).map((e) => e.last - e.first);
      return ns.last +
          ((diffs.toSet().length == 1) ? diffs.first : nextTerm(diffs.toList()));
    }
    
    List> parse(List lines) => [
          for (var l in lines) [for (var n in l.split(' ')) int.parse(n)]
        ];
    
    part1(List lines) => parse(lines).map(nextTerm).sum;
    part2(List lines) => parse(lines).map((e) => nextTerm(e.reversed)).sum;
    
  • pnutzh4x0r@lemmy.ndlug.org
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    11 months ago

    Language: Python

    Part 1

    Pretty straightforward. Took advantage of itertools.pairwise.

    def predict(history: list[int]) -> int:
        sequences = [history]
        while len(set(sequences[-1])) > 1:
            sequences.append([b - a for a, b in itertools.pairwise(sequences[-1])])
        return sum(sequence[-1] for sequence in sequences)
    
    def main(stream=sys.stdin) -> None:
        histories   = [list(map(int, line.split())) for line in stream]
        predictions = [predict(history) for history in histories]
        print(sum(predictions))
    
    Part 2

    Only thing that changed from the first part was that I used functools.reduce to take the differences of the first elements of the generated sequences (rather than the sum of the last elements for Part 1).

    def predict(history: list[int]) -> int:
        sequences = [history]
        while len(set(sequences[-1])) > 1:
            sequences.append([b - a for a, b in itertools.pairwise(sequences[-1])])
        return functools.reduce(
            lambda a, b: b - a, [sequence[0] for sequence in reversed(sequences)]
        )
    
    def main(stream=sys.stdin) -> None:
        histories   = [list(map(int, line.split())) for line in stream]
        predictions = [predict(history) for history in histories]
        print(sum(predictions))
    

    GitHub Repo

  • hades@lemm.ee
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    3 months ago

    Python

    from .solver import Solver
    
    class Day09(Solver):
    
      def __init__(self):
        super().__init__(9)
        self.numbers: list[list[int]] = []
    
      def presolve(self, input: str):
        lines = input.rstrip().split('\n')
        self.numbers = [[int(n) for n in line.split(' ')] for line in lines]
        for line in self.numbers:
          stack = [line]
          while not all(x == 0 for x in stack[-1]):
            diff = [stack[-1][i+1] - stack[-1][i] for i in range(len(stack[-1]) - 1)]
            stack.append(diff)
          stack.reverse()
          stack[0].append(0)
          stack[0].insert(0, 0)
          for i in range(1, len(stack)):
            stack[i].append(stack[i-1][-1] + stack[i][-1])
            stack[i].insert(0, stack[i][0] - stack[i-1][0])
    
      def solve_first_star(self) -> int:
        return sum(line[-1] for line in self.numbers)
    
      def solve_second_star(self) -> int:
        return sum(line[0] for line in self.numbers)
    
      • Sekoia@lemmy.blahaj.zone
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        11 months ago
        1. Setting up boilerplate beforehand, I only need to fill in the functions (and the return types)
        2. Really good parsing library (aoc_parse). Today my entire parsing code was parser!(lines(repeat_sep(i64, " ")))
        3. Iterators! Actually really ideal for AoC, where pipelines of data are really common. Today both the main part (sum of lines) and inner part (getting a vec of differences) can be done pretty easily through iterators

        Today was pretty ideal for my setup. In general I think Rust is really good for later days, because the safety and explicitness make small mistakes rarer (like if you get an element from a HashMap that doesn’t exist, you don’t get a None later down the road (unless you want it, in which case it’s explicit), you get an exception where it happened.

        I just really like Rust :3

      • morrowind@lemmy.ml
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        11 months ago

        I guess I’ll have to take rustaceans who claim they’re more productive in rust than python seriously now

  • morrowind@lemmy.ml
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    11 months ago

    Crystal

    recursion is awesome! (sometimes)

    input = File.read("input.txt")
    
    seqs = input.lines.map &.split.map &.to_i
    
    sums = seqs.reduce({0, 0}) do |prev, sequence|
    	di = diff(sequence)
    	{prev[0] + sequence[0] - di[0], prev[1] + di[1] + sequence[-1]}
    end
    puts sums
    
    
    def diff(sequence)
    	new = Array.new(sequence.size-1) {|i| sequence[i+1] - sequence[i]}
    
    	return {0, 0} unless new.any?(&.!= 0)
    
    	di = diff(new)
    	{new[0] - di[0], di[1] + new[-1]}
    end
    
  • JRaccoon@discuss.tchncs.de
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    11 months ago

    TypeScript

    GitHub link

    It’s nice to have a quick easy one for a change

    Code
    import fs from "fs";
    
    const rows = fs.readFileSync("./09/input.txt", "utf-8")
        .split(/[\r\n]+/)
        .map(row => row.trim())
        .filter(Boolean)
        .map(row => row.split(/\s+/).map(number => parseInt(number)));
    
    console.info("Part 1: " + solve(structuredClone(rows)));
    console.info("Part 2: " + solve(structuredClone(rows), true));
    
    function solve(rows: number[][], part2 = false): number {
        let total = 0;
        for (const row of rows) {
            const sequences: number[][] = [row];
            while (sequences[sequences.length - 1].some(number => number !== 0)) { // Loop until all are zero
                const lastSequence = sequences[sequences.length - 1];
                const newSequence: number[] = [];
                for (let i = 0; i < lastSequence.length; i++) {
                    if (lastSequence[i + 1] !== undefined) {
                        newSequence.push(lastSequence[i + 1] - lastSequence[i]);
                    }
                }
                sequences.push(newSequence);
            }
    
            // For part two just reverse the sequences
            if (part2) {
                sequences.forEach(sequence => sequence.reverse());
            }
    
            // Add the first zero manually and loop the rest
            sequences[sequences.length - 1].push(0);
            for (let i = sequences.length - 2; i >= 0; i--) {
                sequences[i].push(part2
                    ? sequences[i][sequences[i].length - 1] - sequences[i + 1][sequences[i + 1].length - 1]
                    : sequences[i][sequences[i].length - 1] + sequences[i + 1][sequences[i + 1].length - 1]
                );
            }
        
            total += sequences[0].reverse()[0];
        }
    
        return total;
    }
    
  • purplemonkeymad@programming.dev
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    11 months ago

    Using a class here actually made part 2 super simple, just copy and paste a function. Initially I was a bit concerned about what part 2 would be, but looking at the lengths of the input data, there looked to be a resonable limit to how many additional rows there could be.

    python
    import re
    import math
    import argparse
    import itertools
    
    #https://stackoverflow.com/a/1012089
    def iter_item_and_next(iterable):
        items, nexts = itertools.tee(iterable, 2)
        nexts = itertools.chain(itertools.islice(nexts, 1, None), [None])
        return zip(items, nexts)
    
    class Sequence:
        def __init__(self,sequence:list) -> None:
            self.list = sequence
            if all([x == sequence[0] for x in sequence]):
                self.child:Sequence = ZeroSequence(len(sequence)-1)
                return
            
            child_sequence = list()
            for cur,next in iter_item_and_next(sequence):
                if next == None:
                    continue
                child_sequence.append(next - cur)
    
            if len(child_sequence) > 1:
                self.child:Sequence = Sequence(child_sequence)
                return
            
            # can't do diff on single item, use zero list
            self.child:Sequence = ZeroSequence(1)
    
        def __repr__(self) -> str:
            return f"Sequence([{self.list}], Child:{self.child})"
    
        def getNext(self) -> int:
            if self.child == None:
                new = self.list[-1]
            else: 
                new = self.list[-1] + self.child.getNext()
    
            self.list.append(new)
            return new
        
        def getPrevious(self) -> int:
            if self.child == None:
                new = self.list[0]
            else: 
                new = self.list[0] - self.child.getPrevious()
    
            self.list.insert(0,new)
            return new
    
    class ZeroSequence(Sequence):
        def __init__(self,count) -> None:
            self.list = [0]*count
            self.child = None
    
        def __repr__(self) -> str:
            return f"ZeroSequence(length={len(self.list)})"
    
        def getNext(self) -> int:
            self.list.append(0)
            return 0
        
        def getPrevious(self) -> int:
            self.list.append(0)
            return 0
    
    def parse_line(string:str) -> list:
        return [int(x) for x in string.split(' ')]
    
    def main(line_list):
        data = [Sequence(parse_line(x)) for x in line_list]
        print(data)
    
        # part 1
        total = 0
        for d in data:
            total += d.getNext()
        print("Part 1 After:")
        print(data)
        print(f"part 1 total: {total}")
    
        # part 2
        total = 0
        for d in data:
            total += d.getPrevious()
        print("Part 2 After:")
        print(data)
        print(f"part 2 total: {total}")
    
    
    if __name__ == "__main__":
        parser = argparse.ArgumentParser(description="day 1 solver")
        parser.add_argument("-input",type=str)
        parser.add_argument("-part",type=int)
        args = parser.parse_args()
        filename = args.input
        if filename == None:
            parser.print_help()
            exit(1)
        file = open(filename,'r')
        main([line.rstrip('\n') for line in file.readlines()])
        file.close()