Advent of Code 2019-08 with R & JavaScript

colinfay

Colin Fay

Posted on December 9, 2019

Advent of Code 2019-08 with R & JavaScript

Solving Advent of Code 2019-08 with R and JavaScript.

[Disclaimer] Obviously, this post contains a big spoiler about Advent of Code, as it gives solutions for solving day 8.

Instructions

Find the instructions at: https://adventofcode.com/2019/day/8

R solution

Part one

library(magrittr)
library(purrr)

ipt <- read.delim("input8.txt", header = FALSE, colClasses = "character")$V1
ipt <- strsplit(ipt, "")[[1]] %>% as.numeric()

layers_size <- 6 * 25
l <- list()

for (i in 1: (length(ipt)/layers_size)){
  l[[i]] <- ipt[1:150]
  ipt <- ipt[151:length(ipt)]
}

mn <- l %>% 
  lapply(table) %>% 
  map_dbl("0") %>% 
  which.min()

l[[mn]] %>% 
  table()

## .
## 0 1 2 
## 7 14 129

14 * 129

## [1] 1806
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Part two

v <- c()
for (i in seq_len(layers_size)){
  idx <- map_dbl(l, i) 
  v[i] <- idx[idx %in% c(0,1)][1]
}

library(dplyr)
library(tidyr)
library(ggplot2)
library(tibble)
matrix(v, ncol = 6) %>% 
  as.data.frame() %>%
  rowid_to_column() %>%
  gather(key = key, value = value, V1:V6) %>%
  mutate(key = gsub("V(.)", "\\1", key) %>% as.numeric()) %>%
  ggplot(aes(rowid, key, fill = as.factor(value))) + 
  geom_tile() + 
  coord_fixed() + 
  scale_fill_viridis_d() + 
  scale_y_reverse()
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JS solution

var ipt = fs.readFileSync("input8.txt", 'utf8').split("").filter(x => x.length != 0 & x != '\n').map(x => parseInt(x));

var layers_size = 6 * 25;
var layer_n = ipt.length / layers_size;
var res = [];

function table(vec){
  var tbl = {};
  vec.map(function(x){
    if (tbl[x]){
      tbl[x] = tbl[x] + 1;
    } else {
      tbl[x] = 1;
    }
  })
  return tbl;
}

for (var i = 0; i < layer_n; i ++){
  res[i] = ipt.splice(0, layers_size);
}

var res_b = res.map(x => table(x));
var minim = Math.min.apply(Math, res_b.map(x => x['0']));
var smallest = res_b.filter(x => x['0'] == minim);

smallest[0]["1"] * smallest[0]["2"];

## 1806

var v = [];

for (var i = 0; i < layers_size; i ++){
  var idx = res.map(x => x[i]);
  v[i] = idx.find(z => z== 0 | z == 1);
}

var nn = [];

for (var i = 0; i < 6; i ++){
  nn[i] = v.splice(0, 25).join(" ").replace(/0/g, " ");
}

nn

## [ '    1 1     1 1     1 1 1 1   1 1 1       1 1    ',
##   '      1   1     1   1         1     1   1     1  ',
##   '      1   1     1   1 1 1     1     1   1     1  ',
##   '      1   1 1 1 1   1         1 1 1     1 1 1 1  ',
##   '1     1   1     1   1         1   1     1     1  ',
##   '  1 1     1     1   1         1     1   1     1  ' ]
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colinfay
Colin Fay

Posted on December 9, 2019

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