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Working part two!
Classic off by one error
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‎2019/19/README.md‎

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# Answers
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| Part 1 | Part 2 |
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|--------|---------|
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| `234` | ` ` |
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| Part 1 | Part 2 |
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|--------|-----------|
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| `234` | `9290812` |
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## --- Day 19: Tractor Beam ---
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Unsure of the state of Santa's ship, you borrowed the tractor beam technology from Triton. Time to test it out.
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When you're safely away from anything else, you activate the tractor beam, but nothing happens. It's hard to tell whether it's working if there's nothing to use it on. Fortunately, your ship's drone system can be configured to deploy a drone to specific coordinates and then check whether it's being pulled. There's even an [Intcode](9) program (your puzzle input) that gives you access to the drone system.
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When you're safely away from anything else, you activate the tractor beam, but nothing happens. It's hard to tell whether it's working if there's nothing to use it on. Fortunately, your ship's drone system can be configured to deploy a drone to specific coordinates and then check whether it's being pulled. There's even an [Intcode](https://adventofcode.com/2019/day/9) program (your puzzle input) that gives you access to the drone system.
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The program uses two input instructions to request the _X and Y position_ to which the drone should be deployed. Negative numbers are invalid and will confuse the drone; all numbers should be _zero or positive_.
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@@ -30,7 +30,7 @@ To better understand the tractor beam, it is important to _get a good picture_ o
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9........##
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In this example, the _number of points affected by the tractor beam_ in the 10x10 area closest to the emitter is `_27_`.
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In this example, the _number of points affected by the tractor beam_ in the 10x10 area closest to the emitter is _`27`_.
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However, you'll need to scan a larger area to _understand the shape_ of the beam. _How many points are affected by the tractor beam in the 50x50 area closest to the emitter?_ (For each of X and Y, this will be `0` through `49`.)
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@@ -44,43 +44,43 @@ The beam gets wider as it travels away from the emitter; you'll need to be a min
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For example, suppose you have the following tractor beam readings:
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#.......................................
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.#......................................
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..##....................................
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...###..................................
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....###.................................
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.....####...............................
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......#####.............................
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......######............................
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.......#######..........................
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........########........................
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.........#########......................
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..........#########.....................
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...........##########...................
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...........############.................
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............############................
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.............#############..............
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..............##############............
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...............###############..........
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................###############.........
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................#################.......
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.................########OOOOOOOOOO.....
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..................#######OOOOOOOOOO#....
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...................######OOOOOOOOOO###..
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....................#####OOOOOOOOOO#####
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.....................####OOOOOOOOOO#####
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.....................####OOOOOOOOOO#####
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......................###OOOOOOOOOO#####
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.......................##OOOOOOOOOO#####
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........................#OOOOOOOOOO#####
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.........................OOOOOOOOOO#####
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..........................##############
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..........................##############
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...........................#############
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............................############
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.............................###########
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In this example, the _10x10_ square closest to the emitter that fits entirely within the tractor beam has been marked `O`. Within it, the point closest to the emitter (the only highlighted `_O_`) is at X=`25`, Y=`20`.
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<pre><code>#.......................................
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.#......................................
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..##....................................
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...###..................................
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....###.................................
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.....####...............................
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......#####.............................
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......######............................
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.......#######..........................
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........########........................
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.........#########......................
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..........#########.....................
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...........##########...................
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...........############.................
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............############................
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.............#############..............
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..............##############............
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...............###############..........
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................###############.........
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................#################.......
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.................########<b>O</b>OOOOOOOOO.....
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..................#######OOOOOOOOOO#....
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...................######OOOOOOOOOO###..
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....................#####OOOOOOOOOO#####
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.....................####OOOOOOOOOO#####
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.....................####OOOOOOOOOO#####
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......................###OOOOOOOOOO#####
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.......................##OOOOOOOOOO#####
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........................#OOOOOOOOOO#####
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.........................OOOOOOOOOO#####
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..........................##############
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..........................##############
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...........................#############
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............................############
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.............................###########
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</code></pre>
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In this example, the _10x10_ square closest to the emitter that fits entirely within the tractor beam has been marked `O`. Within it, the point closest to the emitter (the only highlighted **`O`**) is at X=`25`, Y=`20`.
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Find the _100x100_ square closest to the emitter that fits entirely within the tractor beam; within that square, find the point closest to the emitter. _What value do you get if you take that point's X coordinate, multiply it by `10000`, then add the point's Y coordinate?_ (In the example above, this would be `250020`.)

‎2019/19/input.js‎

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const input = [
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109,424,203,1,21102,1,11,0,1106,0,282,21101,0,18,0,1105,1,259,2102,1,1,221,203,1,21102,31,1,0,1106,0,282,21101,38,0,0,1106,0,259,21002,23,1,2,21202,1,1,3,21102,1,1,1,21102,57,1,0,1106,0,303,2102,1,1,222,21001,221,0,3,21002,221,1,2,21101,0,259,1,21102,1,80,0,1106,0,225,21102,1,93,2,21102,1,91,0,1106,0,303,2101,0,1,223,21001,222,0,4,21102,1,259,3,21101,225,0,2,21101,225,0,1,21101,118,0,0,1106,0,225,20101,0,222,3,21102,1,120,2,21102,1,133,0,1106,0,303,21202,1,-1,1,22001,223,1,1,21101,0,148,0,1106,0,259,2102,1,1,223,21001,221,0,4,20102,1,222,3,21102,1,23,2,1001,132,-2,224,1002,224,2,224,1001,224,3,224,1002,132,-1,132,1,224,132,224,21001,224,1,1,21102,195,1,0,106,0,108,20207,1,223,2,20101,0,23,1,21101,-1,0,3,21102,1,214,0,1106,0,303,22101,1,1,1,204,1,99,0,0,0,0,109,5,2101,0,-4,249,21201,-3,0,1,21201,-2,0,2,21202,-1,1,3,21101,0,250,0,1105,1,225,21202,1,1,-4,109,-5,2106,0,0,109,3,22107,0,-2,-1,21202,-1,2,-1,21201,-1,-1,-1,22202,-1,-2,-2,109,-3,2106,0,0,109,3,21207,-2,0,-1,1206,-1,294,104,0,99,22102,1,-2,-2,109,-3,2106,0,0,109,5,22207,-3,-4,-1,1206,-1,346,22201,-4,-3,-4,21202,-3,-1,-1,22201,-4,-1,2,21202,2,-1,-1,22201,-4,-1,1,21201,-2,0,3,21102,343,1,0,1106,0,303,1106,0,415,22207,-2,-3,-1,1206,-1,387,22201,-3,-2,-3,21202,-2,-1,-1,22201,-3,-1,3,21202,3,-1,-1,22201,-3,-1,2,21201,-4,0,1,21101,0,384,0,1106,0,303,1105,1,415,21202,-4,-1,-4,22201,-4,-3,-4,22202,-3,-2,-2,22202,-2,-4,-4,22202,-3,-2,-3,21202,-4,-1,-2,22201,-3,-2,1,21202,1,1,-4,109,-5,2106,0,0
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]
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const path = require('path');
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const fs = require('fs');
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const input = fs
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.readFileSync(path.join(__dirname, 'input.txt'), 'utf8')
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.toString()
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.trim()
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.split(',')
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.map(v => parseInt(v, 10));
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module.exports = {
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input,

‎2019/19/part-one.js‎

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const { TractorBeam } = require('./tractor-beam');
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let tractor_beam = new TractorBeam(input);
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console.log(tractor_beam.partOne());
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let answer = tractor_beam.partOne();
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console.log(tractor_beam.grid.toString());
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console.log(answer);

‎2019/19/part-two.js‎

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let tractor_beam = new TractorBeam(input);
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console.log(tractor_beam.partTwo());
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// 9370819 - too high

‎2019/19/tractor-beam.js‎

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/**
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* Once we have a row that is wide enough, we can move onto the next phase.
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* Take the left most point in the row (bottom_edge.x). Calculate the value
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* at the point + `square_size` down from there. If we hit a `0`, go to B.
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* at the `point + square_size - 1` down from there. If we hit a `0`, go to B.
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*
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* If we hit a `1`, then move right `square_size` amount from that point.
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* If we hit a `1`, then move right `square_size - 1` amount from that point.
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* If _that_ is a `1`, we found our top-left point. Exit.
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*
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* B. If it's a `0`, then move inward `square_size - row_width` number of times,
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* continually checking the point + `square_size` down. If we iterate through
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* continually checking the `point + square_size - 1` down. If we iterate through
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* the available squares in the row, then the top-left point doesn't exist
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* in that row. Calculate the next set of edges and start again.
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*/
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const row_width = this.getWidth(bottom_edge, top_edge);
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for (
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let x = bottom_edge.x;
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x < bottom_edge.x + row_width - square_size;
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x <= bottom_edge.x + row_width - square_size;
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x++
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) {
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let y = bottom_edge.y + square_size;
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/**
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* If I want a 2x2 square, and I'm at 0,0, I don't jump to 0,2 / 2,2 / 0,2
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*
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* 01234
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* 0 x.x--
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* 1 ...
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* 2 x.x
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* 3 |
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* 4 |
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*
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* That'd make a 3x3 square! So I jump forward the square size minus 1.
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*/
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let y = bottom_edge.y + square_size - 1;
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let output = this.computeAt(x, y);
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this.grid.set(x, y, output);
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if (output === 1) {
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let bottom_right = this.computeAt(x + square_size, y);
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let bottom_right = this.computeAt(x + square_size - 1, y);
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if (bottom_right === 1) {
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found_square = { x, y: bottom_edge.y };
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break;

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