---
title: (2/2) Beacon Exclusion - Advent of Code 2022 Day 15 with Ruby
slug: 2-2-beacon-exclusion-advent-of-code-2022-day-15-with-ruby
published_at: 2022-12-15 22:00:15 +0000
updated_at: 2026-03-04 20:15:15 +0000
summary: 
description: (2/2) Beacon Exclusion - Advent of Code 2022 Day 15 with Ruby  Challenge: https://adventofcode.com/2022/day/15 Solution: https://gist.github.com/cjavdev/9a736b1ed879882e17d50dc9684d790d  #ruby #adventofcode
tags: [cjav_dev, web development tutorials, web development for beginners, vim, ruby, advent of code, advent of code 2022, advent of code day 15, advent of code 2022 day 15, advent of code ruby, advent of code 2022 ruby, advent of code 2022 day 15 ruby, ruby solution, ruby tutorial, beacon exclusion]
views: 138
author: CJ Avilla
url: https://www.cjav.dev/videos/2-2-beacon-exclusion-advent-of-code-2022-day-15-with-ruby
youtube_url: https://www.youtube.com/watch?v=ij9WmVQfLgc
youtube_id: ij9WmVQfLgc
embed_url: https://www.youtube.com/embed/ij9WmVQfLgc
thumbnail_url: https://i.ytimg.com/vi/ij9WmVQfLgc/hqdefault.jpg
type: video
---

# (2/2) Beacon Exclusion - Advent of Code 2022 Day 15 with Ruby

*Published: December 15, 2022*
*Views: 138*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=ij9WmVQfLgc)

[![(2/2) Beacon Exclusion - Advent of Code 2022 Day 15 with Ruby](https://i.ytimg.com/vi/ij9WmVQfLgc/hqdefault.jpg)](https://www.youtube.com/watch?v=ij9WmVQfLgc)

## Description

(2/2) Beacon Exclusion - Advent of Code 2022 Day 15 with Ruby

Challenge: https://adventofcode.com/2022/day/15
Solution: https://gist.github.com/cjavdev/9a736b1ed879882e17d50dc9684d790d

#ruby #adventofcode

## Transcript

what's up welcome back this is part two for day 15 of the Advent of code with Ruby and if you haven't seen part one you can head over and see how we implemented part one in the previous episode now in part two rather than just looking at a single Row in part two we need to look at this massive grid that is between 0 and 4 million so it's like a four million by four million grid we have our sensors and our beacons and again a sensor is tied to exactly one Beacon that is a certain distance away from the sensor and there's no beacons that are closer to that sensor and that Beacon is tied to exactly that one sensor what we get is this sort of like overlapping regions where the coverage or a sensor would have otherwise detected another Beacon that was closer they're all overlapping except on this four million by 4 million grid except one point so there's like this giant overlap of all of these different squares except there's one point where it's not overlapping and we need to find that exact point I actually a really hard time with this part too and so I looked for some hints got some ideas and so one of the ideas is to look at all of the edges for all of your squares and if you step one outside of that then you create a boundary around your sensor that is the distance plus one from the beacon and what that does is it creates this like one almost like a one pixel square around all of the squares and then what you can do is collect all of those points that are exactly distance plus one away from the sensor and you can look at every single one of those points to make sure that it is not already covered by another sensor and and at the end you'll end up with just one point it's worth mentioning that it would take like hours and hours or longer especially to look at a four million by 4 million grid and even with the approach I'm about to take it's going to take several gigabytes worth of memory let's jump into it okay so we we are still collecting up our sensors and beacons and we're going to still going to use this distance but I'm going to comment out I'm going to comment out part one here and so now what we want to do is find all of the points that are around a sensor that are D plus one from the sensor I'm actually going to use GitHub copilot here so I'm going to say it gets all the points around a point that are exactly D plus 1 away and we'll make a method here called get points that takes in the point and some D and then we'll let GitHub copilot write this okay I want to do this a little bit differently we're going to do like from minus D to d d is the distance from this point to this point right so this is 2 and we need to iterate over 2 and then one and then 0 and then negative 1 and then two again right if we're just looking at the x axis here we're just trying to figure out what is the X for all of these we need to go two then one right for this one then zero for this one then negative one for this one and then negative 2 for this one so we want to go from negative 2 to 2 and for each of those points in between we also want to go out in the other direction and so here we're going to say something like this is going to be this is going to be our DX and we need to see okay so d y is D minus the absolute value yeah so the distance minus the absolute value of the change that's going to give our d y because they're always going to be related right as you're like drawing out the line okay then what we can do is add into our list of points the going up in that direction and we're also going to add in the list of points going down in the other direction this will draw it like this and like this at the same time and then we'll come back in together at the same time so hopefully that makes sense we end up with this method called get points and that should give us that should give us all the points from a given point so now what we want to do is we want to iterate over all of the sensors and get all of the points that are D plus 1 away from the sensor so sensors dot each do sensor we need to collect up all the edge points so we'll say Edge points plus equals get points of the sensor D plus one and then at the end of this so Edge points is a set we should have like P Edge points dot size so this is going to be puts like finding Edge points and that many Edge points let's just see if that works against the example okay so finding the edge points it found 271 Edge points rather than using plus equals here because this is going to do like a concatenation which will require that we keep track of a giant list of points and also when we do this concatenation it's gonna have to move them over allocate memory and do all this other crazy stuff so I think it might be faster maybe I'm wrong if we pass in the edge points here as an argument so we say EP and then we say EP shovel and then we don't return anything it's just like we're passing in the set that's going to be modified we should get the same answer and I don't know if it's actually faster but okay now we need to iterate over each of those points and make sure that they're not near a beacon so now we have to say Edge points DOT each Edge point and now we want to go from each sensor and we're going to again have an exclude an exclusionary set so we'll say if the distance between The Edge point and the sensor is less than or equal to D that means it's like inside the range of another sensor right so we're going to have these overlapping regions where one square is going to be overlapped by another square and where that overlap happens the points that are just one outside of the edge of the other Square are going to be inside of the region for the other Square so at the end here we should have Edge points should be one more than excluded so let's just P Edge points that size minus excluded dot size and see what we get sensors.h okay 105 so that's too much we need to exclude any points that are outside of our range we haven't done that exclusion yet so let's actually do that we'll do that here we'll say like next if X Plus DX is greater than zero or X Plus DX is less or if it's less than zero or if it's greater than 4 million then we're going to skip it and then we also want to say next if y plus d y is out of range and then down here we want to skip if y minus d y I think let's see if that gives us better numbers 40 46 okay so now where are we having overlap let's just print all the edge points and see what's going on here okay these all seem reasonable all right so then okay so I jumped the gun for the actual input they have to be between zero and four million but for our for the example use case they have to be at most 20 so let's do like this like Max X Y thing in the example use case is 20 and then in the actual use case it's 4 million oops is that the right number 4 million okay so then instead of using this here we want to do Max x y x y Max x y and then we'll run this again nope so it's not in scope because we're using this method I'm just going to make it a constant called Max all right and then we'll change these all to oh look at that okay we got one one is what we want so now when we look at Edge points minus the excluded we should get exactly one point and we do and the set is 14 11 and that is the answer here 14 11. so we're going to now print out so this is going to give us an X and Y and then we want the answer is going to be let's see multiplying the x coordinate by this number so it puts the x coordinate times this number Plus the y-coordinate okay so we'll run that again part one in the test use case is that that's good so let's run it against our against our input and again we have to wait for a minute all right so I found 47 million Edge points that's a lot of edge points so it's now it's iterating over every single one of those Edge points around all of those different squares and checking to see if it's in the bounds of the other of the sensors and if it is then it's being excluded if it's not being excluded blah blah blah and then we're going to get some hopefully we get some point and an answer here at the end all right it looks like we got some points so this should be the easy part which is taking X and multiplying it with 4 million and then adding this we get back as Hugh Mungus number 12 trillion something so the national debt I don't know that's the puzzle answer for part two let's take a look at what's going on this was like a very mathy problem and yeah so there's a couple of things that I've tried to play around with and try to make it faster so one is removing all these function calls so instead of having a method get points I just move this stuff inside of where it's calling get points and then iterate over each of those points and then the same thing with distance like trying to pull that out of a method in here I don't know if that actually sped anything up too much there's probably a bunch of other little tweaks that you can do to make it faster but again this one was super tough for me but hopefully I explained the process and you can understand the code so that is day 15 of the Advent of code thank you so much for watching see you next time [Music]

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