# Advent of Code 2019-02 with R & JavaScript

**Colin Fay**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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Solving Advent of Code 2019-02 with R and JavaScript.

[Disclaimer] Obviously, this post contains a big spoiler about Advent

of Code, as it gives solutions for solving day 2.

[Disclaimer bis] I’m no JavaScript expert so this might not be the

perfect solution. And TBH, that’s also the case for the R solution.

## About the JavaScript code

The JavaScript code has been written in the same RMarkdown as the R

code. It runs thanks to the `{bubble}`

package:

https://github.com/ColinFay/bubble

## Instructions

Find the instructions at: https://adventofcode.com/2019/day/2

## R solution

When in doubt, use brute force.

Ken Thompson

### Part one

```
extract <- function(vec, idx) vec[as.character(idx)]
day_2 <- function(vec, one = 12, two = 2){
vec[2] <- one
vec[3] <- two
names(vec) <- 0:(length(vec) - 1)
start <- 0
repeat {
req <- extract(vec, start)
if (req == 99) break
if (req == 1) fun <- `+`
if (req == 2) fun <- `*`
vec[as.character(
extract(vec, start + 3)
)] <- fun(
extract(vec, extract(vec, start + 1)),
extract(vec, extract(vec, start + 2))
)
start <- start + 4
}
vec[1]
}
ipt <- scan( "input2.txt", what = numeric(), sep = ",")
day_2(ipt)
```

```
## 0
## 3409710
```

### Part two

```
x <- purrr::cross2(0:99, 0:99)
i <- 1
repeat{
res <- day_2(ipt, x[[i]][[1]], x[[i]][[2]])
if (res == 19690720) break
i <- i + 1
}
# Answer
100 * x[[i]][[1]] + x[[i]][[2]]
```

```
## [1] 7912
```

## JS solution

### Part one & Two

// Reading the

file

```
var res = fs.readFileSync("input2.txt", 'utf8').split(",").filter(x => x.length != 0);
var res = res.map(x => parseInt(x));
```

```
function day_2(vec, one = 12, two = 2){
var loc = vec.slice();
loc[1] = one;
loc[2] = two;
start = 0;
do {
var req = loc[start];
if (req === 99){
break;
}
pos1 = loc[start + 1];
pos2 = loc[start + 2];
pos3 = loc[start + 3];
if (req === 1){
loc[pos3] = loc[pos1] + loc[pos2];
} else if (req === 2){
loc[pos3] = loc[pos1] * loc[pos2];
}
start = start + 4;
} while (start < vec.length)
return loc[0]
}
```

```
day_2(res)
```

```
## 3409710
```

```
function make_array(l){
return Array.from({length: l}, (el, index) => index);
}
var x = make_array(100);
var y = make_array(100);
var cross = [];
for (var i = 0; i < x.length; i++){
for (var j = 0; j < y.length; j++){
cross.push(
[x[i], y[j]]
)
}
}
i = 0
do {
ans = day_2(res, cross[i][0], cross[i][1]);
if (ans == 19690720) break
i++
} while (i < cross.length)
```

```
100 * cross[i][0] + cross[i][1]
```

```
## 7912
```

## Day 2 takeaway

`Array.from({length: n}, (el, index) => index);`

is more or less the

equivalent of R`1:n`

When doing

`[] =`

in JS, we’re modifying the original objet. Compare

```
# R
x <- 1:3
x
```

```
## [1] 1 2 3
```

```
plop <- function(y){
y[1] <- 2
}
plop(x)
x
```

```
## [1] 1 2 3
```

to

```
// JS
var x = [1, 2, 3];
function yeay(ipt){
ipt[1] = 12
}
yeay(x)
```

```
x
```

```
## [ 1, 12, 3 ]
```

- JavaScript copies by reference. Compare:

```
# R
x <- 1:3
y <- x
x[1] <- 999
x
```

```
## [1] 999 2 3
```

```
y
```

```
## [1] 1 2 3
```

And

```
var x = make_array(3);
var y = x
x[1] = 999
```

```
x
y
```

```
## [ 0, 999, 2 ]
## [ 0, 999, 2 ]
```

- This can be prevented with
`obj.slice()`

```
var x = make_array(10);
var y = x.slice();
x[1] = 999
```

```
x
y
```

```
## [ 0, 999, 2, 3, 4, 5, 6, 7, 8, 9 ]
## [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
```

**leave a comment**for the author, please follow the link and comment on their blog:

**Colin Fay**.

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