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Advent of Code 2020 Solution Megathread - Day 6: Custom Customs

Ryan Palo on December 06, 2020

Yesterday, a lot of you finished quickly and were looking for more to keep you occupied on a Saturday. All of the golfed one-liners were really fu...
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bgaster profile image
Benedict Gaster •

OK, today seemed pretty straightforward, although it took me to reach task 2 to have the insight that it was union and then intersect... oh well it is a Sunday :-)

Anyway Haskell was a nice fit and simply used set operations on lists to keep with my somewhat weak goal of using lists.

main  = do xs <- IOT.readFile "day6_input" <&> fmap (map unpack . T.lines) . T.splitOn "\n\n"
           print (sum $ fmap (length . foldr1 union) xs)
           print (sum $ fmap (length . foldr1 intersect) xs)
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kais_blog profile image
Kai • • Edited

I've created a step-by-step tutorial (TypeScript) again:

kais_blog image

[Advent of Code 2020] Day 6 Step-by-Step Tutorial (TypeScript)

Kai ・ Dec 6 ・ 6 min read

#adventofcode #tutorial #typescript #javascript

and I did something with bits, just for fun:

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mgasparel profile image
Mike Gasparelli •

Clever solution. I had an instinct that I could use binary to solve Day5, although couldn't quite flesh out a working solution (ended up solving it in a different way). It didn't even dawn on me that you could use binary for this challenge, although it seems obvious now. 👍

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meseta profile image
Yuan Gao • • Edited

Python one-liners, thanks to set theory and list comprehension, and map
Part 1:

sum([len(set(entry.replace("\n",""))) for entry in open("input.txt").read().split("\n\n")])
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Part 2:

sum(len(set.intersection(*map(set, entry.split()))) for entry in open("input.txt").read().split("\n\n"))
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I try to explain this more fully at dev.to/meseta/advent-of-code-2020-...

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willsmart profile image
willsmart •

Python impl today. Prob not great python but works.

print(reduce(
    lambda acc, v: acc + len(
        set(v.replace('\n', ''))
    ),
    open("6.txt").read().split('\n\n'),
    0
))

print(reduce(
    lambda acc, v: acc + len(reduce(
        lambda acc, v: acc & v,
        map(
            lambda v: set(v), 
            v.split('\n')
        )
    )),
    open("6.txt").read().split('\n\n'),
    0
))
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Tomorrow will try something like prolog?!
Failing that, maybe Haskell.

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readyready15728 profile image
readyready15728 •

Ruby, part 2:

require 'set'

groups = File.read('06.txt').split /\n\n/

puts (groups.map do |group|
  responses = group.split(/\n/).map { |response| Set.new(response.each_char) }
  responses.inject { |product, current| product.intersection(current) }.length
end).sum
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Benjamin Trent •

Rust!

use std::collections::HashSet;

#[aoc_generator(day6)]
fn to_vec(input: &str) -> Vec<Vec<Vec<char>>> {
    input
        .split("\n\n")
        .map(|i| {
            i.lines()
                .map(|s| s.chars().collect::<Vec<char>>())
                .collect()
        })
        .collect()
}

#[aoc(day6, part1)]
fn answer_count(input: &Vec<Vec<Vec<char>>>) -> usize {
    input
        .iter()
        .map(|group| {
            group
                .iter()
                .flat_map(|v| v.iter().map(|c| *c).collect::<HashSet<char>>())
                .collect::<HashSet<char>>()
                .len()
        })
        .sum()
}

#[aoc(day6, part2)]
fn abs_answer_count(input: &Vec<Vec<Vec<char>>>) -> usize {
    input
        .iter()
        .map(|group| {
            group
                .iter()
                .map(|v| v.iter().map(|c| *c).collect::<HashSet<char>>())
                .fold(Option::None, |l, g| {
                    if l.is_none() {
                        Some(g.clone())
                    } else {
                        Some(
                            l.unwrap_or(HashSet::new())
                                .intersection(&g)
                                .map(|c| *c)
                                .collect(),
                        )
                    }
                })
                .unwrap_or(HashSet::new())
                .len()
        })
        .sum()
}
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Christopher Kruse •

Ugh, made it too late for this day (but I had it done in time for the problem, honest!).

Used HashSet to my benefit here, as I could apply set theory to the problem and get it done easily.

As always, in Github.

use aoc_runner_derive::{aoc, aoc_generator};
use std::collections::HashSet;

#[aoc_generator(day6)]
fn parse_input_day6(input: &str) -> Vec<String> {
    input
        .split("\n\n")
        .map(|group| String::from(group))
        .collect()
}

#[aoc(day6, part1)]
fn sum_group_questions(input: &Vec<String>) -> usize {
    let answers: Vec<HashSet<char>> = input
        .iter()
        .map(|group| group.chars().filter(|x| *x != '\n').collect())
        .collect();
    answers.iter().map(|group| group.len()).sum()
}

#[aoc(day6, part2)]
fn sum_group_common_questions(input: &Vec<String>) -> usize {
    input
        .iter()
        .map(|group| {
            let mut first_run = true;
            group
                .lines()
                .map(|answers| answers.chars().collect::<HashSet<char>>())
                .fold(HashSet::new(), |memo, ans| {
                    if memo.is_empty() && first_run {
                        first_run = false;
                        ans.clone()
                    } else {
                        let intersect = memo.intersection(&ans).cloned().collect();
                        intersect
                    }
                })
                .len()
        })
        .sum()
}
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galoisgirl profile image
Anna •

COBOL

   IDENTIFICATION DIVISION.
   PROGRAM-ID. AOC-2020-06-2.
   AUTHOR. ANNA KOSIERADZKA.

   ENVIRONMENT DIVISION.
   INPUT-OUTPUT SECTION.
   FILE-CONTROL.
       SELECT INPUTFILE ASSIGN TO "d6.input"
       ORGANIZATION IS LINE SEQUENTIAL.

   DATA DIVISION.
   FILE SECTION.
     FD INPUTFILE
     RECORD IS VARYING IN SIZE FROM 1 to 99
     DEPENDING ON REC-LEN.
     01 INPUTRECORD PIC X(99).
   WORKING-STORAGE SECTION.
     01 FILE-STATUS PIC 9 VALUE 0.
     01 REC-LEN PIC 9(2) COMP.
     01 WS-GROUP-ANSWERS PIC 9 OCCURS 26 TIMES.
     01 WS-CHAR PIC X.

   LOCAL-STORAGE SECTION.
     01 I UNSIGNED-INT VALUE 1.
     01 C UNSIGNED-INT VALUE 1.
     01 X UNSIGNED-INT VALUE 1.
     01 GROUP-SIZE UNSIGNED-INT VALUE 0.
     01 GROUP-TOTAL UNSIGNED-INT VALUE 0.
     01 TOTAL UNSIGNED-INT VALUE 0.

   PROCEDURE DIVISION.
   001-MAIN.
       PERFORM 004-INIT-VARIABLES.
       OPEN INPUT INPUTFILE.
       PERFORM 002-READ UNTIL FILE-STATUS = 1.
       CLOSE INPUTFILE.
       PERFORM 004-NEXT-GROUP.
       DISPLAY TOTAL.
       STOP RUN.

   002-READ.
        READ INPUTFILE
            AT END MOVE 1 TO FILE-STATUS
            NOT AT END PERFORM 003-PROCESS-RECORD
        END-READ.

   003-PROCESS-RECORD.
       IF REC-LEN = 0 THEN
          PERFORM 004-NEXT-GROUP
       ELSE 
          PERFORM 005-PROCESS-ROW
       END-IF.

   004-INIT-VARIABLES.
       PERFORM VARYING I FROM 1 BY 1 UNTIL I > 26
          MOVE 0 TO WS-GROUP-ANSWERS(I)
       END-PERFORM.
       MOVE 0 TO GROUP-SIZE.
       MOVE 0 TO GROUP-TOTAL.

   004-NEXT-GROUP.
       IF GROUP-SIZE > 0 THEN
          PERFORM 006-TALLY-GROUP-TOTAL
       END-IF.
       ADD GROUP-TOTAL TO TOTAL.
       PERFORM 004-INIT-VARIABLES.

   005-PROCESS-ROW.
       ADD 1 TO GROUP-SIZE.
       PERFORM VARYING I FROM 1 BY 1 UNTIL I > REC-LEN
          MOVE INPUTRECORD(I:1) TO WS-CHAR
          COMPUTE C = FUNCTION ORD(WS-CHAR)
          COMPUTE X = WS-GROUP-ANSWERS(C - 97) + 1
          MOVE X TO WS-GROUP-ANSWERS(C - 97)
       END-PERFORM.

   006-TALLY-GROUP-TOTAL.
       PERFORM VARYING I FROM 1 BY 1 UNTIL I > 26
          IF WS-GROUP-ANSWERS(I) = GROUP-SIZE THEN 
             ADD 1 TO GROUP-TOTAL
          END-IF 
       END-PERFORM.
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particleflux profile image
Stefan Linke •

2 Solutions again.

Go:

package main

import (
    "bytes"
    "fmt"
    "io/ioutil"
    "os"
)

func countQuestions(group []byte) (int, int) {
    counts := map[byte]int{}
    numPeople := 1
    for _, question := range group {
        if question == '\n' {
            numPeople++
        } else {
            counts[question]++
        }
    }

    numAllYes := 0
    for _, c := range counts {
        if c == numPeople {
            numAllYes++
        }
    }

    return len(counts), numAllYes
}

func main() {
    input, _ := ioutil.ReadAll(os.Stdin)
    groups := bytes.Split(input, []byte("\n\n"))

    sum1, sum2 := 0, 0
    for _, group := range groups {
        a, b := countQuestions(group)
        sum1 += a
        sum2 += b
    }

    fmt.Println(sum1, sum2)
}
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And tweet-sized PHP:

<?for($i=0,$z=explode("

",file_get_contents('input'));$z[$i];){$a+=count($f=count_chars($z[$i++],1))-(($n=$f[10])>0);foreach($f as$k=>$v)$b+=$k!=10&$v==$n+1;}echo"$a $b";
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flwidmer •

I decided to do this year in Haskell. Perhaps if it gets too crazy, I'll revert to what I know better, but until now it's been fun.

I might start a library with the groups and addAll functions, seems I'm going to use them for every puzzle...

import Data.List.Split ( splitOn )
import Data.List ( intersect, nub )

main :: IO ()
main = do
    input <- readFile "input"
    putStrLn "day6"
    print $ solve1 input
    print $ solve2 input

solve1 :: String -> Int
solve1 a = 
  let grouped = groups a
      answers = map (filter (\x -> x `elem` ['a'..'z']) . nub) grouped
  in  addAll answers

solve2 :: String -> Int
solve2 a =
  let grouped = map lines $ groups a
      intersected = map (foldr1 intersect) grouped 
  in  addAll intersected

addAll :: [[a]] -> Int 
addAll = sum . map length 

groups :: String -> [String]
groups = splitOn "\n\n"
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mgasparel profile image
Mike Gasparelli •

I thought the most obvious way to do this would be to intersect a bunch of HashSets, but seemed to be more short & sweet to just check that every character in the first line was contained in every other line 🤷

Part 1 

    public class Part1 : Puzzle<IEnumerable<PlaneGroup>, int>
    {
        public override int SampleAnswer => 11;

        public override IEnumerable<PlaneGroup> ParseInput(string rawInput)
            => rawInput
                .Split(Environment.NewLine + Environment.NewLine)
                .Where(line => line.Length > 0)
                .Select(group =>
                    new PlaneGroup(
                        group
                            .Split(Environment.NewLine)
                            .Where(x => x.Length > 0)));

        public override int Solve(IEnumerable<PlaneGroup> input)
            => input.Sum(x => x.CountDistinctAnswers());
    }
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Part 2 

    public class Part2 : Part1
    {
        public override int SampleAnswer => 6;

        public override int Solve(IEnumerable<PlaneGroup> input)
            => input.Sum(x => x.CountIntersectingAnswers());
    }
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PlaneGroup 

    public class PlaneGroup
    {
        List<string> answers;

        public PlaneGroup(IEnumerable<string> answers)
        {
            this.answers = answers.ToList();
        }

        public int CountDistinctAnswers()
            => new HashSet<char>(answers.SelectMany(a => a))
                .Count();

        public int CountIntersectingAnswers()
            => answers.First()
                .Count(c => answers.All(a => a.Contains(c)));
    }
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Derk-Jan Karrenbeld •

Only took a few minutes today; here is what I got in Ruby:

require 'benchmark'

class GroupAnswers
  def self.from_lines(lines)
    GroupAnswers.new(lines.map(&:chomp))
  end

  def initialize(people)
    questions = people.join('').chars.uniq
    self.count = questions.count { |q| people.all? { |l| l.include?(q) } }
  end

  def to_i
    self.count
  end

  private

  attr_accessor :count
end

groups = File
  .read('input.txt')
  .split(/\n\n/)

Benchmark.bmbm do |b|
  b.report do
    puts groups.sum { |group| GroupAnswers.from_lines(group.split(/\n/)).to_i }
  end
end
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Matt Ellen-Tsivintzeli •

I spent a long time trying to get a regex to work for part 2, but I have given up for the time being. Got a short couple of answers for this:

function testScoresp1()
{
  const input = document.getElementsByTagName('pre')[0].innerHTML;
  const groups = input.split('\n\n');
  return groups.map(group => new Set(group.replaceAll('\n', ''))).reduce((sum, group) => sum + group.size, 0);
}

function testScoresp2()
{
  const input = document.getElementsByTagName('pre')[0].innerHTML;
  const groups = input.split('\n\n');
  const alphabet = 'abcdefghijklmnopqrstuvwxyz'.split('');
  return alphabet.reduce((sum, letter) =>
  {
    return sum + groups.filter(group => group.trim().split('\n').every(line => line.indexOf(letter) > -1)).length;
  }, 0);
}
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Erik Bäckman • • Edited

Haskell:

module Main where

import Control.Arrow (Arrow((&&&)))
import Data.List.Split (splitOn)
import Control.Monad (join)
import Data.List (nub, intersect)

parseInput :: String -> [[String]]
parseInput = fmap lines . splitOn "\n\n"

solveP1 :: [[String]] -> Int
solveP1 = sum . fmap (length . nub . join)

solveP2 :: [[String]] -> Int
solveP2 = sum . fmap answers
  where
    answers [x]    = length x
    answers (x:xs) = length . foldr intersect x $ xs

main :: IO ()
main = print . (solveP1 &&& solveP2) . parseInput =<< readFile "./day6inp.txt"
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vncz profile image
Vincenzo Chianese • 
(ns day6 (:require [clojure.string :refer [split]]
                   [clojure.set :refer [intersection]]))

(def input (-> "./day6input.txt"
               (clojure.core/slurp)
               (split #"\n\n")))

(def any-answers (->> input
                      (map #(re-seq #"\w" %))
                      (map set)
                      (reduce #(+ %1 (count %2)) 0))) ; Part 1

(def all-answers (->> input
                      (map #(split % #"\n"))
                      (map #(map (fn [x] (set (clojure.core/char-array x))) %))
                      (map #(apply intersection %))
                      (reduce #(+ %1 (count %2)) 0))) ; Part 2
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Harry Gibson •

Python. A bit less elegant than some of the other python versions already posted but the same idea using set theory

n = 0
checkset = set()
with open('input.txt', 'r') as in_file:
    for l in in_file:
        if l.strip() == '':
            n += len(checkset)
            checkset.clear()
        else:
            checkset.update((c for c in l.strip()))
    if l.strip() != '': n += len(checkset)
print(f"Part 1: total is {n}")

checkset = set()
new_group = True
n = 0
with open('input.txt', 'r') as in_file:
    for l in in_file:
        if l.strip() == '':
            n += len(checkset)
            checkset.clear()
            new_group = True
        else:
            if new_group:
                checkset.update((c for c in l.strip()))
                new_group = False
            else:
                # must be in every person: equivalent to set intersection 
                checkset.intersection_update((c for c in l.strip()))
    if l.strip() != '': n += len(checkset)
print(f"Part 2: total is {n}")
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Kudos Beluga • • Edited

Javascript answer, Part 1 and 2 in one script, change the part2 boolean to false if you want a part 1 answer.
Looking at other people's responses, mine feels very inefficient :(

let fs = require("fs"), part2 = false, n = 0;
fs.readFile("input.txt","utf8",(err,data) => {
if (err) throw err;
let veryuppercount = 0;
data.split(/\n\s/gi).map(e => {
  if (part2) {
  let uppercount = 0;
 e = e.split(/\n/g);
 e[0].split("").map(f => {
   let counter = 0;
   e.slice(1,e.length).map(g => {
   g.split("").map(h => {
     if (h == f) counter++
   })})
   if (counter == e.length-1) uppercount++
 })
  veryuppercount += uppercount;
  } else {
  e = e.replace(/\n/g,"")
  if (e) {
 let noduplicate = [...new Set(e.split(""))];
 n += noduplicate.length
  }}})
if (part2) console.log(veryuppercount)
else console.log(n)
})
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galoisgirl profile image
Anna •

JavaScript - I did it in COBOL first, but still wanted to practice reduce a bit.

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

const A_INDEX = "a".charCodeAt(0);

let lines = [];

function getClearAnswers() {
  return Array(26).fill(0)
}

const readInterface = readline.createInterface({
    input: fs.createReadStream('./example.txt'),
});

readInterface.on('line', function(line) {
  lines.push(line)
});

readInterface.on('close', function() {
  const result = lines.reduce((res, line) => {
    if (line.length == 0) {
      if (res.groupSize == 0) {
        return res
      }
      return {
        total: res.total + Object.values(res.groupAnswers).reduce((acc, x) => acc + (x == res.groupSize ? 1 : 0), 0),
        groupSize: 0,
        groupAnswers: getClearAnswers()
      }
    }
    const indices = line.split('').map(l => l.charCodeAt(0) - A_INDEX)
    return {
      total: res.total,
      groupSize: res.groupSize + 1,
      groupAnswers: indices.reduce((acc, x) => acc.fill(acc[x] + 1, x, x + 1), res.groupAnswers)
    }
  }, {
    total: 0,
    groupSize: 0,
    groupAnswers: getClearAnswers()
  });

  console.log(result.total)
})
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Patryk Woziński •

Yesterday I had no time to post my solution in Elixir, but there is!

defmodule AdventOfCode.Day6 do
  def part1(file_path) do
    file_path
    |> read_all_answers()
    |> Enum.map(&how_many_answers(&1, :step1))
    |> Enum.sum()
  end

  def part2(file_path) do
    file_path
    |> read_all_answers()
    |> Enum.map(&how_many_answers(&1, :step2))
    |> Enum.sum()
  end

  defp read_all_answers(file_path) do
    file_path
    |> File.read!()
    |> String.split("\n\n", trim: true)
    |> Enum.map(&String.split/1)
  end

  defp how_many_answers(list, :step1), do: how_many_answers(list, &MapSet.union/2)
  defp how_many_answers(list, :step2), do: how_many_answers(list, &MapSet.intersection/2)

  defp how_many_answers(list, f) do
    list
    |> Enum.map(&String.to_charlist/1)
    |> Enum.map(&MapSet.new/1)
    |> Enum.reduce(f)
    |> Enum.count()
  end
end
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vncz profile image
Vincenzo Chianese • • Edited

I've solved this using Clojure for the sake of. It seems to be the shortest one so far.

(ns day6 (:require [clojure.string :refer [split]]
                   [clojure.set :refer [intersection]]))

(def input (-> "./day6input.txt"
               (clojure.core/slurp)
               (split #"\n\n")))

(def any-answers (->> input
                      (map #(re-seq #"\w" %))
                      (map set)
                      (reduce #(+ %1 (count %2)) 0))) ; Part 1

(def all-answers (->> input
                      (map #(split % #"\n"))
                      (map #(map (fn [x] (set (clojure.core/char-array x))) %))
                      (map #(apply intersection %))
                      (reduce #(+ %1 (count %2)) 0))) ; Part 2
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rpalo profile image
Ryan Palo •

A little late to the party, but I got it done before bed time 😁

Day6.h:

#ifndef AOC2020_DAY6_H
#define AOC2020_DAY6_H

/// Day 6: Custom Customs
/// 
/// Search through lists of questionnaire answers.
/// Each questionnaire is 26 questions, so a yes to question 'b' would
/// be represented by the presence of 'b' in the output line.
/// People are one to a line, and parties are separated by a blank line.

#include <stdlib.h>
#include <stdio.h>

/// Return the number of letters that are present for any line in a group
int tally_anyone_yeses(FILE* fp);

/// Return the number of letters that are present in every line in a group
int tally_everyone_yeses(FILE* fp);

/// Run both 
int day6(void);
#endif
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Day6.c:

#include "Day6.h"

#include <stdio.h>
#include <string.h>

/// Number of letters in the alphabet
#define LETTERS 26

/// Max number of chars per line (yes to all plus newline plus NULL)
#define MAX_LINE_SIZE (LETTERS + 2)

int tally_anyone_yeses(FILE* fp) {
  int counts[LETTERS] = {0};
  char line[MAX_LINE_SIZE] = {0};

  // Log each occurrence of a character.
  while (!feof(fp)) {
    fgets(line, MAX_LINE_SIZE, fp);
    if (line[0] == '\n') break;

    for (int i = 0; line[i] != '\n' && line[i]; i++) {
      int index = line[i] - 'a';
      counts[index]++;
    }
    memset(line, 0, MAX_LINE_SIZE);
  }

  // Loop through and find the total count of those that appeared.
  int total = 0;
  for (int i = 0; i < LETTERS; i++) {
    if (counts[i] > 0) total++;
  }
  return total;
}

/// Run the part 1 code on the input file.
static int part1() {
  FILE* fp;
  fp = fopen("data/day6.txt", "r");
  if (fp == NULL) {
      printf("Couldn't open input file.\n");
      exit(EXIT_FAILURE);
  }

  // Loop through and add up the total for each group.
  int total = 0;
  while (!feof(fp)) {
      total += tally_anyone_yeses(fp);
  }

  fclose(fp);
  return total;
}

int tally_everyone_yeses(FILE* fp) {
  int counts[LETTERS] = {0};
  char line[MAX_LINE_SIZE] = {0};
  int people_count = 0;

  // Loop through the group and add up everybody's answers.
  while (!feof(fp)) {
    fgets(line, MAX_LINE_SIZE, fp);
    if (line[0] == '\n') break;

    people_count++;
    for (int i = 0; line[i] != '\n' && line[i]; i++) {
      int index = line[i] - 'a';
      counts[index]++;
    }
    memset(line, 0, MAX_LINE_SIZE);
  }

  // Count the number of letters where everyone answered it yes
  // i.e. number of yeses == number of people
  int total = 0;
  for (int i = 0; i < LETTERS; i++) {
    if (counts[i] == people_count) total++;
  }
  return total;
}

/// Run the part 2 code on the input file.
int part2() {
  FILE* fp;
  fp = fopen("data/day6.txt", "r");
  if (fp == NULL) {
      printf("Couldn't open input file.\n");
      exit(EXIT_FAILURE);
  }

  // Loop through and add up the total for each group.
  int total = 0;
  while (!feof(fp)) {
      total += tally_everyone_yeses(fp);
  }

  fclose(fp);
  return total;
}

int day6() {
  printf("====== Day 6 ======\n");
  printf("Part 1: %d\n", part1());
  printf("Part 2: %d\n", part2());
  return EXIT_SUCCESS;
}
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Paweł Świątkowski •

Re yesterday: actually my solution was in D, but I don't see it listed in the table.

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Ryan Palo •

OK cool, I wasn't 100% sure. I'll update that.

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Paweł Świątkowski •

Yeah, I should've mentioned that in the original. Those C-like languages look pretty much all the same if you don't know some particular details.

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rpalo profile image
Ryan Palo •

Haha! Yeah, I was thinking, "I'm pretty sure that's not Go, but maybe it's C#? Or C++?" I just thought of inspecting the HTML classes yesterday after running into a similar issue. 😂

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readyready15728 • • Edited

Ruby, part 1:

require 'set'

groups = File.read('06.txt').split /\n\n/

puts (groups.map do |group|
  Set.new(group.split(/\n/).join.each_char).length
end).sum
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Thibaut Patel •

Here is my walkthrough for day 6 in JavaScript:

Source code: github.com/tpatel/advent-of-code-2...