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AoC Day 1: Chronal Calibration

Ryan Palo on December 01, 2018

Overview My Solution DEV Leaderboards Overview After my post last night, @aspittel suggested that we start a daily discussion post ch...

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Ali Spittel • Dec 1 '18 • Edited

So excited about this 🙌🏻!

Python

Part 1

data = open('input.txt', 'r')
total = 0
for line in data:
    total += int(line)
print(total)

Part 2

data = [int(i) for i in open('input.txt', 'r')]

def get_first_duplicate_total(data):
    total = 0
    prev_totals = set([0])
    while True:
        for i in data:
            total += i
            if total in prev_totals:
                return total
            prev_totals.add(total)
    return total

print(get_first_duplicate_total(data))

I also learned about itertools.cycle through reading the Reddit solutions, that would make it so that I don't need the while True:

Ryan Palo • Dec 1 '18

Ah! So good. The clean-ness of Python never stops making me happy!

Using a set was a good idea. I didn't think of that, but it would be a lot faster for checking whether or not an item was present.

I'm going to offer unsolicited suggestions, but feel free to totally ignore them if you already knew about them (probable) or don't like them, since your solution already looks really nice.

For part one, generator expressions could be your friend:

data = open('input.txt', 'r')
total = sum(int(line) for line in data)
print(total)

For the second part, check out itertools.cycle.

Ali Spittel • Dec 1 '18

Awesome! Yes, thank you! Found out about itertools.cycle this morning -- feels super niche but still really cool.

Part one could also be a one-liner!

print(sum(int(line for line in open('input.txt', 'r'))))

Florian Rohrer • Dec 1 '18 • Edited

I'd also suggest to use a context manager (the with keyword) for clean opening and closing of files.

Part 1:

with open("input.txt") as f:
    freq = sum(int(i.strip()) for i in f)
freq

Part 2:

from itertools import cycle

with open("input.txt") as f:
    freqs = [int(i.strip()) for i in f]

seen = set()
current = 0
for f in cycle(freqs):
    if current in seen:
        print(current)
        break
    else:
        seen.add(current)
        current += f

Also: "Hey checkout my Github Repo!"

Ali Spittel • Dec 1 '18

True! Always forget to do that since I really only do file handling for code challenges at this point.

jess unrein • Dec 2 '18 • Edited

Golang solution

Part 1

package main

import (
    "bufio"
    "os"
    "strconv"
)

func freq(f *os.File)(sum int) {
    scanner := bufio.NewScanner(f)
    for scanner.Scan() {
        i, err := strconv.Atoi(scanner.Text())
        if err != nil {
            panic(err)
        }
        sum += i
    }
    return
}

func main() {
    f, err := os.Open("input1.txt")
    if err != nil {
        panic(err)
    }
    freq(f)
}

Part 2

package main

import (
    "io/ioutil"
    "strconv"
    "container/ring"
    "strings"
)

func dup(x string)(sum int) {
    list := strings.Split(x, "\n")
    r := ring.New(len(list))
    seen := map[int]bool{0: true}
    for i := 0; i < r.Len(); i++ {
        num, err := strconv.Atoi(list[i])
        if err != nil {
            panic(err)
        }
        r.Value = num
        r = r.Next()
    }
    for true {
        sum += r.Value.(int)
        if (seen[sum] == true) {
            return sum
        }
        seen[sum] = true
        r = r.Next()
    }
    return
}

func main() {
    f, err := ioutil.ReadFile("input1.txt")
    if err != nil {
        panic(err)
    }
    dup(string(f))
}

I also wrote solutions in Python that look very similar to Ali's, so I decided to run a little benchmark test using today's input from the Advent of Code website. Here are the results

`python3 1_1.py` 100 times
real    0m4.261s
user    0m2.845s
sys 0m0.936s

`go run 1_1.go` 100 times
real    0m34.169s
user    0m26.057s
sys 0m13.171s

go build 1_1.go

`./1_1` 100 times
real    0m0.641s
user    0m0.216s
sys 0m0.253s

`python3 1_2.py` 100 times
real    0m8.347s
user    0m6.126s
sys 0m1.477s

`go run 1_2.go` 100 times
real    0m38.925s
user    0m31.556s
sys 0m14.668s

go build 1_2.go

`./1_2` 100 times
real    0m3.891s
user    0m3.067s
sys 0m0.657s

Ryan Palo • Dec 2 '18

Woah, nice! Am I reading this right, that the Python is significantly faster than Go? Or is this including compilation each time?

jess unrein • Dec 2 '18 • Edited

Python is significantly faster than running go run myfile.go. However, that's not Go's intended use case, really. That's more for debugging along the way, since go run does include the compilation each time.

I think it's interesting to compare the difference between go run and executing the compiled go executable, so the third set of each (ex: ./1_1) is go running the resulting compiled executable 100 times.

Ryan Palo • Dec 2 '18

Ooooh ok that makes sense. Cool!

jess unrein • Dec 2 '18

Also, I am not a regular Go programmer. I've started doing challenges in Go just for my own amusement, but I would definitely not assume that whatever solution I come up with is optimal 😅

Tiago Pires • Dec 2 '18

Hi jess!

Just a small advert: you should defer f.Close() after check if there's an error.

Also: Check out my GitHub Repo.

jess unrein • Dec 2 '18

Oh, if this were production code I totally would, but I'm really only building these solutions to solve the problem, and not looking to make them infinitely extensible.

rhymes • Dec 1 '18 • Edited

Is it normal that part 2 is killing my CPU :D ? It's been running for minutes in Elixir but nada.

Anyhow, I've tried day 1 with all three languages which is not a great idea!

I just googled the parts of the various languages I needed

Clojure

part 1

(def numbers-as-strings (clojure.string/split (slurp "input.txt") #"\n"))
(def numbers (map read-string numbers-as-strings))
(defn sum [coll] (reduce + coll))
(println (sum numbers))

I kind of hated it, I still don't have syntax highlighting nor formatting for some reason in VSCode and the REPL is quite slow to start (I didn't try with ClojureScript)

part 2

(def repeated-numbers (cycle numbers))

(loop
    [known-totals (set nil), total 0]
    (if (contains? known-totals total)
        (println total)
        (recur (conj known-totals total) (+ total (first (take 1 repeated-numbers))))))

I'm not sure it's correct, I had to kill it because it was hogging the CPU

Elixir

part 1

numbers_as_strings = String.split(String.trim(File.read!("input.txt")), "\n")
numbers = Enum.map(numbers_as_strings, fn x -> String.to_integer(x) end)
IO.puts(Enum.sum(numbers))

Well, this was the easiest one

part 2

defmodule Part2 do
    def find_total(repeated_numbers, totals, sum) do
        unless MapSet.member?(totals, sum) do
            MapSet.put(totals, sum)
            sum = sum + hd(Enum.take(repeated_numbers, 1))
            find_total(repeated_numbers, totals, sum)
        end

        IO.puts sum
    end
end

Part2.find_total(Stream.cycle(numbers), MapSet.new, 0)

Again, I'm not sure it's correct, it just kills my computer

Rust

part 1

use std::fs;

fn main() {
    let data = fs::read_to_string("input.txt").expect("Unable to read file");
    let numbers_as_strings = data.split("\n").collect::<Vec<&str>>();
    let numbers = numbers_as_strings.iter().filter_map(|n| n.parse::<i32>().ok()).collect::<Vec<i32>>();
    let sum: i32 = numbers.iter().sum();
    println!("{}", sum);
}

There's a bit of fiddling with types and syntax but the compiler it's quite helpful when you type stuff that doesn't make sense. It can get in the way of me doing the excercises :D

part 2

Well, this ran for half a second a produced the correct answer

use std::fs;
use std::collections::HashSet;

fn main() {
    // part 1
    let data = fs::read_to_string("input.txt").expect("Unable to read file");
    let numbers_as_strings = data.split("\n").collect::<Vec<&str>>();
    let numbers = numbers_as_strings.iter().filter_map(|n| n.parse::<i32>().ok()).collect::<Vec<i32>>();
    let sum: i32 = numbers.iter().sum();
    println!("{}", sum);

    // part 2
    let mut totals: HashSet<i32> = HashSet::new();
    let repeated_numbers = numbers.iter().cycle();
    let mut repeated_sum: i32 = 0;
    for num in repeated_numbers {
        if totals.contains(&repeated_sum) {
            println!("{}", repeated_sum);
            break;
        }
        totals.insert(repeated_sum);
        repeated_sum += num
    }
}

What am I doing wrong with Clojure and Elixir :D ?

I've put Day 1 on Github: github.com/rhymes/aoc2018/tree/mas...

Ryan Will • Dec 2 '18

At a glance, in the Elixir part two solution it looks like the list of numbers is just being passed through as is. So, it is never moving on to the rest of the numbers and instead the sum is just adding the first number in the list over and over.

Because repeated_numbers is never reassigned the recursive call find_total(repeated_numbers, totals, sum) is passing the unchanged list of all the numbers through to the next iteration.

rhymes • Dec 2 '18

At a glance, in the Elixir part two solution it looks like the list of numbers is just being passed through as is. So, it is never moving on to the rest of the numbers and instead the sum is just adding the first number in the list over and over.

That's probably it, I thought that:

hd(Enum.take(repeated_numbers, 1))

would take 1 number (and hence move the cursor).

Ryan Will • Dec 2 '18

It does, but because Elixir is immutable it returns a new list with the first element from repeated_numbers instead of mutating (changing in place) repeating_numbers

Also, this line hd(Enum.take(repeated_numbers, 1)) could be rewritten as hd(repeated_numbers) to get the same functionality. The more "Elixir way to write that would be to use pattern matching. Something like [head | tail] = repeated_numbers Here is some more info on pattern matching, if you're interested. I think it is super cool!

rhymes • Dec 2 '18

Thank you! :-)

I have to get used again to immutability and transformation instead of mutability and assignment :D

Pavel Gurkov • Dec 2 '18

On the Clojure pt 2, you're effectively running an endless loop. You always take the first element from the cycled number collection, so you'll never hit the exit condition.
To avoid this while using the approach you chose, add a third numbers binding to the loop, and pass (rest numbers) when recuring, while computing total using (first numbers).

rhymes • Dec 2 '18

Thanks! I then solved it thanks to this post

Advent of Code 2018 - Day 1

Christopher Kruse

#clojure#challenges#adventofcode

exactly like you said. You can see my version here github.com/rhymes/aoc2018/blob/mas...

Max Cerrina • Dec 2 '18

I totally thought my elixir one was dead at first too, so I kept killing it. Only about a 15 second run time though I'm just impatient. lemme look at yours!

rhymes • Dec 2 '18

Please do, I found a solution at the end, using reduce_while, instead of recursion:

repeated_numbers = Stream.cycle(numbers)
repeated_sum = Enum.reduce_while(repeated_numbers, {0, MapSet.new([0])}, fn i, {current, totals} ->
  sum = current + i

  if MapSet.member?(totals, sum) do
    {:halt, sum}
  else
    {:cont, {sum, MapSet.put(totals, sum)}}
  end
end)
IO.puts(repeated_sum)

I feel like I should get through the tutorial at least :D

Harriet • Dec 1 '18

I wanted to solve these ones in bash, which I've been focussing on learning lately. First challenge was great, found a simple oneliner:

echo $(( 0$(tr -d '\n' < ./day1/input.txt) ))

But part 2 was a nightmare. Super slow. Ended up using JS in the end, but would still love to know anyone's thoughts on how this could be optimised. Wasn't watching the clock but think it took >20 mins to run!

It's collecting previously computed frequencies in an array, and checking for the frequency in the array each time a new frequency is computed. Is it the maths that's likely to be so slow, or the looping, or both?

declare -i total=0
seen=()
found=

array_contains () {
  for i in "${seen[@]}"; do
    if [[ "$i" == "$1" ]]; then
      return 0
    fi
  done
  return 1
}

while [[ ! $found ]]; do
  for line in $(cat $1); do
    total=$(( ${total}${line} ))

    if array_contains "$total"; then
      echo "FOUND " $total
      found=1
      break
    else
      echo 'not found'
    fi

    seen+=($total)
  done
done

# ./main.sh input.txt

Ryan Palo • Dec 1 '18

Woah, this is awesome! Yeah the second parts always seem to need some fancier algorithmic trick to speed them up.

You might look into using an associative array, as those provide a much faster lookup time and you don’t have to loop through every value each time? I don’t know if that will be enough though.

Paula Gearon • Dec 1 '18

Unfortunately, associative arrays only appeared in Bash4. That's fine for Linux, but doesn't appear on Macs unless you manually install it (boo!).

Ryan Palo • Dec 1 '18

Good caveat to note, thanks!

P.S. Getting Bash 4 on MacOS is a very good idea 😬

Max Cerrina • Dec 2 '18 • Edited

First version, with Elixir, was ~15 seconds, to do what ended up being 130+ iterations on part two.

After some careful matching down to a more or less more-than-binary tree, I got it to about 10 seconds.

... then I went nuts and decided to see if I could do it with JUST sending messages aaaaaaand boy does Elixir do well with that, because I got it to 1.5 seconds total.

I just doublechecked and it is still hitting that loop the same number of times. 137991.

JP • Dec 2 '18

Did it with Elixir too. I found the Stream module to be quite useful, especially for the first part. For the second part, I'm not sure my approach is best, but at least it is working.
If you want to check my repo: github.com/JPYamamoto/advent_of_co...

Max Cerrina • Dec 2 '18

Yeah, that's what I started with and definitely what I would like actually use for anything. It's soooooooo much more readable like how you have it! The trick with named Processes I did totally works here but I suspect would become insanely complex or fail as soon as you needed data that relied on the other data too, instead of just "has this been seen before" :D

JP • Dec 2 '18

Also, I suspect there's a way to do the second part with Streams and avoid having so much information in memory as I did with my approach. Maybe something that takes advantage of Stream.cycle and Stream.scan. But not sure how to implement it. I'll let you know if I figure it out. 😀

Max Cerrina • Dec 2 '18

Yeah, I kept my memory footprint relatively low I think even in version one because I read everything into the IO stream and then just kept throwing that list around, which wasn't greaaaaaaaaaaaat but.

Ryan Palo • Dec 2 '18

Wow, nice!

Max Cerrina • Dec 2 '18

Thanks! I rechecked it like five times because I was so surprised.

Max Cerrina • Dec 2 '18

github.com/aleph-naught2tog/aoc_18...

Ryan Will • Dec 1 '18

Thanks for making this post and sharing your solutions!

Here are my solutions in Elixir and JavaScript.

To start a couple functions to parse the input.

def parse(input) do
  input
  |> String.split("\n")
  |> Enum.map(&(String.trim(&1)
    |> String.to_integer())
  )
end

function parse() {
  return input()
    .split('\n')
    .map(Number);
}

Part one's solutions.

def part_1 do
  input() 
  |> parse() 
  |> Enum.sum()
end

function partOne() {
  return parse()
    .reduce((total, current) => total + current, 0);
}

Part two's solutions.

This solution takes advantage of Elixir's recursion, pattern matching, and multiple function clauses.

def part_2 do
 input() 
 |> parse() 
 |> part_2(0, %{})
end

def part_2([current | rest] = _frequencies, total, record) do
  total = current + total
  record = Map.update(record, total, 1, &(&1 + 1))

  case record[total] do
    n when n > 1 ->
      total

    _ ->
      part_2(rest, total, record)
  end
end

def part_2([], total, record) do
  input()
  |> parse()
  |> part_2(total, record)
end

I made two solutions to part two in JavaScript the first uses a more imperative approach, and the second was meant to mimic the Elixir solution using recursion. The issue with that is JavaScript not being tail call optimized. In order to get around that I used a trampoline function and a while loop. Though, that approach is insanely slow.

function partTwoImperative() {
  let record = {};
  let frequency = 0;
  let index = 0;
  let frequencyList = parse();
  let totalLoops = 0;

  const found = (record, frequency) => 
    (record[frequency] || 0) > 1;
  const getFrequencyRecordValue = (record, frequency) =>
    (record[frequency] || 0) + 1;

  while (!found(record, frequency)) {
    if (index === frequencyList.length) index = 0;

    frequency += frequencyList[index];
    record[frequency] = getFrequencyRecordValue(record, frequency);
    index += 1;
    totalLoops += 1;
  }

  return [frequency, totalLoops];
}

function partTwoTrampoline() {
  const found = (record, frequency) => (record[frequency] || 0) > 1;

  const getFrequencyRecordValue = (record, frequency) =>
    (record[frequency] || 0) + 1;

  const partTwoRecursive = ([current, ...rest], total, record) => {
    const newTotal = total + current;

    const newRecord = {
      ...record,
      [newTotal]: getFrequencyRecordValue(record, newTotal)
    };

    return !found(newRecord, newTotal)
      ? () =>
          partTwoRecursive(
            rest.length ? rest : [3, 3, 4, -2, -4],
            newTotal,
            newRecord
          )
      : newTotal;
  };

  let ret = partTwoRecursive([3, 3, 4, -2, -4], 0, {});

  while (typeof ret === 'function') {
    ret = ret();
  }

  return ret;
}

Andre Myrick • Dec 2 '18

Wow, this is really cool! I completed day one too, but my code is nowhere near that clean. Could you recommend any resources that helped you get into Elixir?

Ryan Will • Dec 2 '18

Thanks! This sounds like a cop-out answer but the Elixir docs are a great resource. elixir-lang.org/

I'd also be happy to answer any questions you have.

Yordi Verkroost • Dec 1 '18

Awesome, this is a nice place to discuss the solutions! This is mine, in Elixir:

Common code (used by both part 1 and part 2):

defmodule AoC.DayOne.Common do
  def read_input(path) do
    File.stream!(path)
    |> Stream.map(&String.trim/1)
    |> Stream.filter(fn x -> x != "" end)
    |> Stream.map(&String.trim_trailing/1)
    |> Stream.map(&String.to_integer/1)
    |> Enum.to_list()
  end
end

Part 1


defmodule AoC.DayOne.PartOne do
  alias AoC.DayOne.Common

  def main() do
    Common.read_input("lib/day1/input.txt")
    |> calculate_sum()
    |> IO.puts()
  end

  def calculate_sum(numbers) do
    Enum.sum(numbers)
  end
end

Part 2

defmodule AoC.DayOne.PartTwo do
  alias AoC.DayOne.Common

  def main() do
    Common.read_input("lib/day1/input.txt")
    |> calculate_result()
    |> IO.puts()
  end

  def calculate_result(numbers, frequencies \\ [], base \\ 0, index \\ 0)

  def calculate_result(numbers, frequencies, base, index) when index == length(numbers) do
    calculate_result(numbers, frequencies, base, 0)
  end

  def calculate_result(numbers, frequencies, base, index) do
    base = base + Enum.at(numbers, index)
    if (Enum.member?(frequencies, base)) do
      base
    else
      frequencies = [base | frequencies]
      calculate_result(numbers, frequencies, base, index + 1)
    end
  end
end

If you want, you could check out my full repo on GitHub.

Dāvis Namsons • Dec 1 '18

That's a good idea !

Ruby

Part 1

input = File.open('./freqs.in').read

frequency = 0

input.each_line do |line|
  frequency += line.to_i
end

puts frequency

Part 2

input = File.open('./freqs.in').read

frequency_changes = input.scan(/[-+]\d+/).collect! &:to_i

frequency = 0
frequencies = []
i = 0

until frequencies.include?(frequency)
  frequencies << frequency
  frequency += frequency_changes[i]
  i == frequency_changes.length - 1 ? i = 0 : i += 1
end

puts frequency

Arjun Rajkumar • Dec 9 '18 • Edited

Hi Davis.. Just starting on the #adventofcode series.

Just curious how you are reading the inputs. E.g. first problem says the input is on adventofcode.com/2018/day/1/input

Will you be using Nokigiri or something similar to read the URL and parse the page to get the inputs?
Sorry if this sounds dumb.. But am wondering how you going to download the file?

Thanks

Matt Morgis • Dec 2 '18 • Edited

Node.js

First, I created an async generator that read the input file stream chunk by chunk and yield each number line by line.

async function* streamToFrequencies(stream) {
  let previous = "";
  for await (const chunk of stream) {
    previous += chunk;
    let eolIndex;
    while ((eolIndex = previous.indexOf("\n")) >= 0) {
      // exclude the EOL
      const number = previous.slice(0, eolIndex);
      yield parseInt(number);
      previous = previous.slice(eolIndex + 1);
    }
  }
  if (previous.length > 0) {
    yield parseInt(previous);
  }
}

Then part 1 was pretty straight forward:

const addFrequencies = async frequencies => {
  let sum = 0;
  for await (const frequency of frequencies) {
    sum += frequency;
  }
  return sum;
};

const sum = stream => {
  return addFrequencies(streamToFrequencies(stream));
};

This approach made part 2 ugly because I had to find a way to re-open the file stream and loop back through the inputs. Using a set over an array really helped with performance.

const calibrate = async stream => {
  let currentFrequency = 0;
  const frequenciesFound = new Set([0]);

  while (true) {
    // clone stream and put in cold storage
    // in case we need to re-read inputs.
    let frozenStream = clone(stream);

    for await (const frequency of streamToFrequencies(stream)) {
      currentFrequency += frequency;
      if (frequenciesFound.has(currentFrequency)) {
        return currentFrequency;
      }

      frequenciesFound.add(currentFrequency);
    }
    stream = frozenStream;
  }
};

Putting it all together:

const frequencyStream = () => {
  return fs.createReadStream(__dirname + "/input.txt", {
    encoding: "utf-8",
    highWaterMark: 256
  });
};

const main = async () => {
  const frequencySum = await sum(frequencyStream());
  console.log({frequencySum})
  const frequencyCalibration = await calibrate(frequencyStream());
  console.log({frequencyCalibration});
};

main();

Full code: github.com/MattMorgis/Advent-Of-Co...

Tiago Romero • Dec 2 '18 • Edited

I also did mine in JS but I decided to use the readline interface to read each line individually and spend less memory by not loading the entire file in the memory at once.

I haven't trying using for await (... of ...) with the readline interface. Maybe I'll try that next. If anyone would like to try it please post it here.

Here are my solutions:

My solution in JavaScript / Node 11, using the readline interface:

readLines.js

const fs = require('fs');
const readline = require('readline');

const readLines = (file, onLine) => {
    const reader = readline.createInterface({
        input: fs.createReadStream(file),
        crlfDelay: Infinity
    });

    reader.on('line', onLine);

    return new Promise(resolve => reader.on('close', resolve));
};

const readFile = async file => {
    const lines = [];
    await readLines(file, line => lines.push(line));  
    return lines;
}

module.exports = {
    readLines,
    readFile
};

01a.js

const { readFile } = require('./readLines');

(async () => {
    const lines = await readFile('01-input.txt');

    const frequency = lines.reduce((frequency, line) => frequency + Number(line), 0);

    console.log(`The final frequency is ${frequency}`);
})();

01b.js

const { readFile } = require('./readLines');

(async () => {
    const lines = await readFile('01-input.txt');

    const frequencySet = new Set();

    let frequency = 0;
    let didAFrequencyReachTwice = false;

    while (!didAFrequencyReachTwice) {
        for (let line of lines) {
            frequency += Number(line);
            if (frequencySet.has(frequency)) {
                didAFrequencyReachTwice = true;
                break;
            }
            else {
                frequencySet.add(frequency);
            }
        }
    }

    console.log(`The first frequency reached twice is ${frequency}`);
})();

Matt Morgis • Dec 2 '18

Readline doesn't work with async iterators and for await yet, but it just landed in 11.x staging.

Once it is released, my streamToFrequencies generator won't be needed.

Also, createReadStream only reads the file in 256 byte chunks at a time (or whatever you set the highwatermark to be, it does not read the entire file into memory. readFile would, however.

To read more about async iterators and generators and the for await syntax, check out 2ality.com/2018/04/async-iter-node...

Tiago Romero • Dec 3 '18

Thanks a bunch @mattmorgis !

Milton Mazzarri • Dec 2 '18

Thanks for sharing, here are my solutions in Elixir and Racket.

I'm not sure if the Racket code is idiomatic, so, any advice is more than welcome.

Elixir

# https://adventofcode.com/2018/day/1
#
# To run each exercise, you can do the following:
#
# elixir -r exercise.exs -e "IO.inspect(Frequency.first_exercise())"
# elixir -r exercise.exs -e "IO.inspect(Frequency.second_exercise())"
#
defmodule Frequency do
  def first_exercise, do: frequency(process_file())

  def second_exercise, do: first_frequency_reached_twice(process_file())

  @spec frequency([integer]) :: integer
  def frequency(freq_changes), do: Enum.sum(freq_changes)

  @spec first_frequency_reached_twice([integer], {integer, MapSet.t()} | integer) :: integer
  def first_frequency_reached_twice(frequency_changes, acc \\ {0, MapSet.new([0])})

  def first_frequency_reached_twice(_, acc) when is_integer(acc), do: acc

  def first_frequency_reached_twice(frequency_changes, acc) do
    result =
      Enum.reduce_while(frequency_changes, acc, fn digit, {current, past} ->
        next = current + digit

        if MapSet.member?(past, next) do
          {:halt, next}
        else
          {:cont, {next, MapSet.put(past, next)}}
        end
      end)

    first_frequency_reached_twice(frequency_changes, result)
  end

  defp process_file do
    "input"
    |> File.stream!()
    |> Stream.map(fn x -> x |> String.trim() |> String.to_integer() end)
    |> Enum.to_list()
  end
end

ExUnit.start()

defmodule FrequencyTest do
  use ExUnit.Case

  import Frequency

  test "should calculate frequency" do
    test_cases = [
      {[1, -2, 3, 1], 3},
      {[1, 1, 1], 3},
      {[1, 1, -2], 0},
      {[-1, -2, -3], -6}
    ]

    Enum.each(test_cases, fn {changes, expected} ->
      assert frequency(changes) == expected
    end)
  end

  test "should stop when a frequency is reached twice" do
    test_cases = [
      {[1, -2, 3, 1, 1, -2], 2},
      {[1, -1], 0},
      {[3, 3, 4, -2, -4], 10},
      {[-6, 3, 8, 5, -6], 5},
      {[7, 7, -2, -7, -4], 14}
    ]

    Enum.each(test_cases, fn {changes, expected} ->
      assert first_frequency_reached_twice(changes) == expected
    end)
  end
end

Racket

#lang racket/base

#|
Reference: https://adventofcode.com/2018/day/1

If you want to test the results:

λ racket
> (require (file "exercise.rkt"))
> (frequency (file->list "input"))
520
> (first-frequency-reached-twice (file->list "input"))
394
|#

(require racket/set)
(require racket/match)

(provide frequency first-frequency-reached-twice)

(define (frequency changes)
  (foldl + 0 changes))

(define (first-frequency-reached-twice changes)
  (find-frequency (cons 0 (set 0)) changes))

(define (find-frequency acc changes)
  (match acc
    [result
     #:when (integer? result)
     result]
    [(cons current previous-frequencies)
     (find-frequency (reduce-while changes previous-frequencies current) changes)]))

(define (reduce-while changes previous-frequencies current)
  (match changes
    [changes
     #:when (eq? '() changes)
     (cons current previous-frequencies)]
    [(cons head tail)
     (define next (+ current head))
     (if (set-member? previous-frequencies next)
         next
         (reduce-while tail (set-add previous-frequencies next) next))]))

Unit tests for Racket:

#lang racket/base

(require rackunit
         "exercise.rkt")

(define exercise-tests
  (test-suite
   "Tests for exercise day 1"

   (test-case
    "Should calculate frequency"

    (check-equal? (frequency '(1 -2 3 1)) 3)
    (check-equal? (frequency '(1 1 1)) 3)
    (check-equal? (frequency '(1 1 -2)) 0)
    (check-equal? (frequency '(-1 -2 -3)) -6))

   (test-case
    "Should find first frequency reached twice"

    (check-equal? (first-frequency-reached-twice '(1 -1)) 0)
    (check-equal? (first-frequency-reached-twice '(1 -2 3 1 1 -2)) 2)
    (check-equal? (first-frequency-reached-twice '(3 3 4 -2 -4)) 10)
    (check-equal? (first-frequency-reached-twice '(-6 3 8 5 -6)) 5)
    (check-equal? (first-frequency-reached-twice '(7 7 -2 -7 -4)) 14))))

(require rackunit/text-ui)
(run-tests exercise-tests)

Shritesh Bhattarai • Dec 1 '18 • Edited

I'm doing AoC with Rust too and the solution I came up with looks almost exactly like yours.

Part 1

pub fn callibrate(input: &[&str]) -> i32 {
    input.iter().map(|s| s.parse::<i32>().unwrap()).sum()
}

Part 2: I used a HashSet

pub fn twice(input: &[&str]) -> i32 {
    let mut numbers = input.iter().map(|s| s.parse::<i32>().unwrap()).cycle();

    let mut seen_results = HashSet::new();
    seen_results.insert(0);

    let mut sum = 0;

    loop {
        let num = numbers.next().unwrap();
        sum += num;

        if seen_results.contains(&sum) {
            return sum;
        } else {
            seen_results.insert(sum);
        }
    }
}

Instead of changing the main.rs file every day, I'll suggest creating a separate file for each day inside src/bin directory. That way, you can execute them using cargo run --bin day_1. Small but well thought out features like this is why I love Rust so much. For example, you can look at my repo at github.com/shritesh/advent-of-code

I hope to get through all of the challenges this year and learn from everyone here.

Ryan Palo • Dec 1 '18

Oooh! Good idea! I’m definitely doing that. Thanks!

Ben Lovy • Dec 1 '18 • Edited

F# - this is my first brush with F# and .NET in general, so pointers are more'n welcome!

Common code:

let getFrequencies fileName =
  let lines = IO.File.ReadLines(fileName)
  lines
    |> Seq.map Convert.ToInt32
    |> Seq.toList

Part 1:

let rec addFreq acc s =
  match s with
    | [] -> acc
    | freq::freqs -> addFreq (acc + freq) freqs

let day1Part1 fileName =
  getFrequencies fileName |> addFreq 0

part 2:

let rec addFreqWithState acc visited whole remaining =
  match remaining with
    | [] -> addFreqWithState acc visited whole whole
    | head::tail ->
      let newval = acc + head
      if Array.contains newval visited then
        newval
      else
        addFreqWithState newval (Array.append visited [| newval |]) whole tail

let day1Part2 fileName =
  let freqs = getFrequencies fileName
  addFreqWithState 0 [| |] freqs freqs

Entry point:

open System
open Library

[<EntryPoint>]
let main argv =
    day1Part1 argv.[0] |> printfn "Part 1 result: %i"
    day1Part2 argv.[0] |> printfn "Part 2 result: %i"

    0 // return an integer exit code

Part 2 is sloooow. How would you optimize this?

Ali Spittel • Dec 1 '18 • Edited

Nice!! For optimizing part 2, I would use a set! O(1) to check if an item is in it!

Ben Lovy • Dec 1 '18

Aha! Thanks so much, it's blazingly fast now:

let rec addFreqWithState acc visited whole remaining =
  match remaining with
    | [] -> addFreqWithState acc visited whole whole
    | head::tail ->
      let newval = acc + head
      if Set.contains newval visited then
        newval
      else
        addFreqWithState newval (Set.add newval visited) whole tail

let day1Part2 fileName =
  let freqs = getFrequencies fileName
  addFreqWithState 0 (new Set<int> (Seq.empty)) freqs freqs

I guess the Tour of F# page shouldn't be my only learning resource :)

Jon Bristow • Dec 1 '18 • Edited

Kotlin Solution!

Part 1

Please exuse my tricked out kotlin, I've added some personal extension functions to let me think about list comprehensions more simply.

private fun String.parseN() = let { (head, tail) ->
    when (head) {
        '+' -> 1
        '-' -> -1
        else -> throw Exception("Illegal input!")
    } * tail.toInt()
}

fun answer1(input: List<String>) = input.sumBy(String::parseN)

Part 2

I made a silly mistake, and lost about an hour trying to figure out where I went wrong. I should set up my test runners ahead of day 2 so I don't end up with this same kind of typo.

fun answer2(input: List<String>): Int {
    val ns = input.map(String::parseN).scan(Int::plus)
    val ending = ns.last
    return (sequenceOf(listOf(0)) + generateSequence(ns) {
        it.map(ending::plus)
    }).findFirstDupe(emptySet())
}


tailrec fun Sequence<List<Int>>.findFirstDupe(seen: Set<Int>): Int {
    val (head, tail) = this
    val intersections = head intersect seen
    return when {
        intersections.isNotEmpty() -> head.find { it in intersections }!!
        else -> tail.findFirstDupe(seen + head)
    }
}

/**
 * My prewritten `scan` function... it's just a `fold` that keeps every step.
 */
fun <A, B> Sequence<A>.scan(initial: B, operation: (B, A) -> B) =
        fold(sequenceOf(initial)) { scanList, curr ->
            scanList + operation(scanList.last(), curr)
        }

Part 2 is "slow", but it's still under ~~a minute~~ 5 seconds to brute force through 150 iterations (or so) of the cycle.

Christopher Kruse • Dec 1 '18

Ali's tweet and the Advent of Code project inspired me to update a long-latent blog, and make my first actual post on DEV! Thanks to both of you for the inspiration, and here's a not-quite-fully successful implementation in Clojure w/ write-up: dev.to/ballpointcarrot/advent-of-c...

Paula Gearon • Dec 1 '18

I made my way here by following @ASpittel. Thanks Ali.

I just finished at Clojure/conj, so it made sense to do today in Clojure. I should consider another language though, since AoC is always a good way to stretch yourself in something new.

Clojure

As per Ryan's suggestions, I pulled the read-file/split operation into it's own lines operation. I also pulled each puzzle into it's own function.

(ns day1
  (:require [clojure.string :refer [split]]))

(defn lines [input-file]
  (split (slurp input-file) #"\n"))

(defn star
  [input-file]
  (->> (lines input-file)
       (map #(Integer/parseInt %))
       (apply +)))

(println (star "input1.txt"))

This highlights something that bugs me about Clojure... there is no built-in function to read a number from a string. It's simple to call out to Java like I just did, but I think there should be something built into the language.

(defn star2
  [input-file]
  (let [numbers (->> (lines input-file) (map #(Integer/parseInt %)))]
    (loop [[n & rn] (cycle numbers) f 0 acc #{}]
      (let [f (+ f n)]
        (if (acc f)
          f
          (recur rn f (conj acc f)))))))

(println (star2 "input1.txt"))

This second puzzle needed something more than a single operation. My first thought was using a reduce function, but then I realized that I needed to keep going through the input over and over until completion, and reduce works on the entire input... unless I created a transducer, which is too much work for something this simple. I probably should have thought about using loop in the first place.

Carly Ho 🌈 • Dec 2 '18

PHP

Part 1:

<?php
$freq = 0;
$list = file_get_contents($argv[1]);
$changes = explode("\n", trim($list));
foreach ($changes as $change) {
    $freq += intval($change);
}
echo $freq;
die(1);

Part 2:

<?php
$freq = 0;
$seen = array(0);
$current = 0;
$newfreq = null;
$list = file_get_contents($argv[1]);
$changes = explode("\n", trim($list));
do {
    if ($newfreq) {
        array_push($seen, $newfreq);
    }
    $freq += intval($changes[$current]);
    $newfreq = $freq;
    $current++;
    if ($current == count($changes)) {
        $current = 0;
    }
} while (!in_array($newfreq, $seen));
echo $freq;
die(1);

Pavel Gurkov • Dec 2 '18 • Edited

Clojure:

(utils/read-file just slurps and splits by newlines)

Part 1:

(->>
  (utils/read-file (str utils/resources-path "day1.txt"))
  (map #(Integer. ^String %))
  (apply +))

Part 2:

(->>
   (utils/read-file (str utils/resources-path "day1.txt"))
   (map #(Integer. ^String %))
   (cycle)
   (reduce
     (fn [{:keys [freq seen]} freq-change]
       (if (seen freq)
         (reduced freq)
         {:freq (+ freq freq-change) :seen (conj seen freq)}))
     {:freq 0 :seen #{}}))

Ashley Maria • Dec 10 '18

Pt 2 JS/Node Solution:

const fs = require('fs');

const datas = fs.readFileSync('input.txt', 'utf-8');
let currFrequency = new Set();

const changes = datas.split(/\n|\r/m)
    .map(Number);

let frequency = 0;
let i = 0;

while (true) {
    frequency += changes[i];
    if (currFrequency.has(frequency)) {
        console.log(frequency);
        break;
    }
    currFrequency.add(frequency);
    i = i === changes.length - 1 ? 0 : i + 1;
}

Neil Gall • Dec 2 '18

I'm enjoying reading everyone's solutions. After doing last year's in Haskell I'm using Kotlin to try to improve my knowledge of the language and libraries. I'm also avoiding using an IDE, as auto-suggest and magic building don't teach you what's really going on under the hood.

package adventofcode2018.day1

import java.io.File

/** Infinitely repeat a sequence */
fun <T> Sequence<T>.repeat() = generateSequence { asIterable() }.flatten()

/** Like fold() but returns a sequence of all the accumulator values,
 *  not just the last one. `seq.scanl(i, f).last() == seq.fold(i, f)` */
fun <T, U> Sequence<T>.scanl(initial: U, f: (U, T) -> U): Sequence<U> {
    var acc: U = initial
    return map { x -> acc = f(acc, x); acc }
}

fun main(args: Array<String>) {
    val input: List<Int> = File("input.txt").readLines().map(String::toInt)

    // part 1
    val result = input.sum()
    println("Part 1 result: ${result}")

    // part 2
    val repeatedInput = input.asSequence().repeat()
    val accumulatedInput = repeatedInput.scanl(0, Int::plus)
    val unconsumed = accumulatedInput.dropWhile(mutableSetOf<Int>()::add)
    println("Part 2 result: ${unconsumed.first()}")
}

I know the idea is to avoid everyone's github but I'm writing notes on my solutions at github.com/neilgall/adventofcode2018

Bjarne Magnussen • Dec 2 '18

I am using Advent of Code to learn Golang, and here is the solution I came up with. Suggestions for improvements are always welcome!

Part 1:

package main

import (
    "bufio"
    "fmt"
    "os"
    "strconv"
)

// readLines reads a whole file into memory
// and returns a slice of its lines.
func readLines(path string) ([]int, error) {
    file, err := os.Open(path)
    if err != nil {
        return nil, err
    }
    defer file.Close()

    var lines []int
    scanner := bufio.NewScanner(file)
    for scanner.Scan() {
        // string to int
        i, err := strconv.Atoi(scanner.Text())
        if err != nil {
            return nil, err
        }
        lines = append(lines, i)
    }
    return lines, scanner.Err()
}

func main() {
    lines, err := readLines("input")
    if err != nil {
        panic(err)
    }

    var sum int
    for _, line := range lines {
        sum += line
    }
    fmt.Printf("Sum is: %d\n", sum)
}

Part 2:

package main

import (
    "bufio"
    "fmt"
    "os"
    "strconv"
)

// readLines reads a whole file into memory
// and returns a slice of its lines.
func readLines(path string) ([]int, error) {
    file, err := os.Open(path)
    if err != nil {
        return nil, err
    }
    defer file.Close()

    var lines []int
    scanner := bufio.NewScanner(file)
    for scanner.Scan() {
        // string to int
        i, err := strconv.Atoi(scanner.Text())
        if err != nil {
            return nil, err
        }
        lines = append(lines, i)
    }
    return lines, scanner.Err()
}

func main() {
    lines, err := readLines("input")
    if err != nil {
        panic(err)
    }

    occurences := map[int]int{0: 1}
    var freq, j int
    for occurence := 1; occurence < 2; j++ {
        freq += lines[j%len(lines)]
        _, ok := occurences[freq]
        if ok {
            occurence = 2
        }
        occurences[freq] = 1
    }
    fmt.Printf("%d\n", freq)
}

I am also using Python that I have more experience with to cross check solutions.

Part 1:

def chronal_calibration(fname):
    with open(fname) as f:
        content = f.readlines()
    return sum([int(x.strip()) for x in content])


chronal_calibration('input')

Part 2

import itertools

def chronal_calibration_cycle(fname):
    with open(fname) as f:
        content = f.readlines()
    content = [int(x.strip()) for x in content]
    freq = 0
    occurences = {freq: 1}
    for change in itertools.cycle(content):
        freq += change
        occurences[freq] = occurences.get(freq, 0) + 1
        if occurences[freq] > 1:
            break
    return freq


print(chronal_calibration_cycle('input'))

Josh Wisenbaker • Dec 2 '18

Hmm... I just took the input and stuck it in an array rather than reading it from disk. That's not hard to do in Swift though, maybe I'll go back and add in the file loading.

In any case, here is what I did in my Playground!

Swift

Part 1

This one was super easy in Swift via it's reduce(_:_:) function. Given that I had the Ints in an Array, it was a one liner. I added some context to the output because why not?

print("The final calculated frequency is: \(frequencyInput.reduce(0,+))")

For those of you new to Swift, this version of reduce has two inputs. The first one is the starting value for the accumulator and the second is the closure of what to do with the values from the collection. This is simply a shorthand syntax for adding all the values in a collection together and returning it. The long version looks like this:

let frequency = frequencyInput.reduce(0, { x, y in
    x + y
})

Part 2

This part was a bit tougher and my first attempt had all kinds of bad smells like breaking to function labels and such. Eventually I wrapped it all up in a closure that takes an Array of Int values and then returns a single Int with the first repeated number.

/// A closure that takes an `Array` of `Int` values and adds them together in a loop. Upon finding the first repeated
/// value it breaks and returns.
let findRepeatedFrequency = {(_ inputArray: [Int]) -> Int in
    var accumilated = 0
    var frequencies = Set<Int>()
    while true {
        for skew in inputArray {
            accumilated += skew
            if frequencies.contains(accumilated) {
                return accumilated
            }
            frequencies.insert(accumilated)
        }
    }
}

print("The first repeated frequency is: \(findRepeatedFrequency(frequencyInput))")

Essentially this is a for-loop version of the reduce operation, but then I also maintain a history of the values in a Set and check to see if this is a value we've run into before.

I solved the second one with a closure instead of a regular function because I've been teaching myself more about them lately.

Ryan Palo • Dec 2 '18

Thanks! The explanations are really interesting too!

Antonello Galipò • Dec 2 '18 • Edited

Hi there!
I've used Javascript.
Here's my part one solution:

const fs = require('fs');

// Read the input
let contents = fs.readFileSync('frequenciesDiff.txt','utf8');

// Parse to a int array
let diffs = contents.split("\n").map(item => parseInt(item));

// Reduce
let result = diffs.reduce((a,b) => a+b);


// Profit!
console.log(result);

I just read the input from file, parsed to a int array and used the reduce function.

Talking about the second part, I've user a dictionary to store frequencies and retrieve them quickly (access is O(1)), cycling in a circular manner through the provided diffs:

const fs = require('fs');

function duplicateFinder(diffs) {
    // Init an empty memo. We'll store here frequency values for easy retrieval
    let memo = {};

    let currentFrequency = 0;
    let i = 0;

    // Loop until currentFrequency is already in the memo.
    // When the loop breaks, it will be because currentFrequency got a value that had before. 
    // This means it's the second time it appears
    while (!(currentFrequency in memo)) {
        memo[currentFrequency] = currentFrequency.toString();

        currentFrequency += diffs[i];
        i = ++i % diffs.length
    }
    return currentFrequency;
}

// Read the input
let contents = fs.readFileSync('frequenciesDiff.txt', 'utf8');

// Parse to a int array
let diffs = contents.split("\n").map(item => parseInt(item));

// Do the work
let result = duplicateFinder(diffs);

// Profit!
console.log(result);

I'm planning to do this in Go too, since I'd love to practice this language, but I haven't much time for now :(

cloudyhug • Dec 2 '18

I'm doing this in C++ this year, but as we are proposing original solutions with very expressive languages, I'll show a solution in OCaml :)

OCaml

Part 1

let rec f freq ic =
  try f (freq + int_of_string (input_line ic)) ic
  with End_of_file -> freq

let () = Printf.printf "%d\n" (f 0 stdin)

Part 2

let rec read_changes changes ic =
  let next_line = try Some (input_line ic) with End_of_file -> None in
  match next_line with
  | None -> List.rev changes
  | Some c -> read_changes (int_of_string c :: changes) ic

let rec f freq freqs changes_done = function
  | [] -> f freq freqs [] (List.rev changes_done)
  | c :: r ->
    let freq' = freq + c in
    if List.mem freq' freqs then freq'
    else f freq' (freq' :: freqs) (c :: changes_done) r

let () = Printf.printf "%d\n" (f 0 [0] [] (read_changes [] stdin))

Using lists is not very efficient but the code is quite short, that is nice.

Pascal • Dec 2 '18

1.1

A bit overcomplicated. Didn't know parseInt parses signs too 😅

const data = document.querySelector('pre').innerText

const frequencies = data.split('\n')

function getFrequency (freqs) {
  return freqs.reduce((acc, val) => {
    const sign = val[0] === '+' ? 1 : -1
    const number = val.substring(1)

    return acc + sign * number
  }, 0)
}

const result = getFrequency(frequencies)
console.log(result)

1.2

The first result is the solution, not quite sure why it prints so many other results after that even if I set matchFound = true though 🤔

const data = document.querySelector('pre').innerText

const frequencies = data.split('\n')
let sums = []

function getFrequency (freqs, acc=0, callback) {
  return freqs.reduce((acc, val) => {
    const sign = val[0] === '+' ? 1 : -1
    const number = val.substring(1)
    const current = acc + sign * number
    callback(current)
    return current
  }, acc)
}

let startFreq = getFrequency(frequencies, 0, current => sums.push(current))

function isMatch (arr, el) {
  return arr.find(element => element === el)
}

function findMatch () {
    let matchFound = false

    while(!matchFound) {
    startFreq = getFrequency(frequencies, startFreq, current => {
      if (isMatch(sums, current)) {
        matchFound = true
        console.log('The match is', current)
      }
    })
    }
}

findMatch()

roeekl • Dec 3 '18 • Edited

C#

Part 1

Console.WriteLine(File.ReadAllLines(@".\day1_input.txt").Sum(s => int.Parse(s)));

Part 2

int[] input = File.ReadAllLines(@".\day1_input.txt").Select(s => int.Parse(s)).ToArray();
HashSet<int> set = new HashSet<int>();
int freq = 0;
for (int i = 0; set.Add(freq = freq + input[i % input.Length]); i++) ;
Console.WriteLine(freq);

Ian Kirker • Dec 4 '18

This seemed easiest to do in awk:

Part 1

BEGIN {
    f = 0
}

{
    f += $1
}

END {
    print f
}

awk -f 1.1.awk 1.1.input

Part 2

awk's arrays are basically hashes/associative arrays, so you can just use the frequencies as keys much like the Python dictionary solutions. Because awk is really focused on processing a list of files, this uses a hack on the built-in variable that tracks where on that list awk has gotten to, so it only stops once it's found the duplicate.

BEGIN {
    f = 0
    a[0] = 1
}

{
    f += $1
    if (f in a) {
        a[f] += 1
    } else {
        a[f] = 1
    }
    if (a[f] == 2) {
        print(f)
        exit 0
    }
    # Resets what file is the next to process
    ARGIND=0
}

awk -f 1.2.awk 1.2.input

Phil Nash • Dec 2 '18

Thanks for this Ryan! I've joined the leaderboard. Good luck to everyone!

Here's my solution in Crystal, both parts are included as one class:

class Device
  getter frequency : Int32
  getter duplicate : Int32 | Nil

  def initialize
    @frequency = 0
    @duplicate = nil
    @frequencies = Set{@frequency}
  end

  def update(input : Array(String))
    @frequency = input.map { |string| string.to_i }.reduce(@frequency) do |acc, i|
      frequency = acc + i
      @duplicate = frequency if @frequencies.includes?(frequency) && @duplicate.nil?
      @frequencies.add(frequency)
      frequency
    end
  end

  def find_duplicate(input : Array(String))
    while @duplicate.nil?
      update(input)
    end
  end
end


puts "--- Day 1: Chronal Calibration ---"
input = File.read_lines("./01/input.txt")
device = Device.new
device.update(input)
puts "Frequency result: #{device.frequency}"
device.find_duplicate(input)
puts "Frequency first duplicates at: #{device.duplicate}"

There are also tests here.

Arjun Rajkumar • Dec 9 '18 • Edited

Hi Ryan.. Just starting on the #adventofcode series.

Just curious how you are reading the inputs. E.g. first problem says the input is on adventofcode.com/2018/day/1/input

Do you parse the page to get the inputs?
Or copy/paste and store it as a file? Sorry if this sounds dumb.. But am wondering how you are downloading the file?

Thanks

Ryan Palo • Dec 9 '18

No problem! That would be slick to have it download the page for me, but I just go to the link and copy the text. I've got a directory in my project called data, and each day's input as day1.txt, day2.txt, etc. Then I just change the filename that I'm reading.

Let me know if this helps or if you have other questions :)

Arjun Rajkumar • Dec 9 '18

Thanks. The data folder makes sense.

Conlin Durbin • Dec 4 '18

A little late to starting on this, but I am excited to get going!

Part 1 (A little bit trolly)

var fs = require('fs');

fs.readFile('./numbers.txt', 'utf8', function(err, contents) {
  console.log(eval(contents));
});

Part 2 (could probably use some optimizations)

var fs = require('fs');

const hash = {};

fs.readFile('./numbers.txt', 'utf8', function(err, contents) {
  const lines = contents.split('\n');
  let current = 0;
  let i = 0;
  while(true) {
    if(i === lines.length-1) {
      i = 0;
    }
    const line = lines[i];
    const old = current;
    const num = Number(line)
    current += num;
    if(hash[current]) {
      console.log(current);
      break;
    }
    hash[current] = true;
    console.log(old, '+', num, '->', current);
    i++;
  }
});