---
title: Enumerator, CRT display - Advent of Code - Day 10 with Ruby
slug: enumerator-crt-display-advent-of-code-day-10-with-ruby
published_at: 2022-12-10 22:30:00 +0000
updated_at: 2026-03-04 20:15:12 +0000
summary: 
description: Enumerator, CRT display - Advent of Code - Day 10 with Ruby  Challenge: https://adventofcode.com/2022/day/10 Solution: https://gist.github.com/cjavdev/9f0aab41191b1050886257c0ce052be3  #ruby #adventofcode
tags: [cjav_dev, web development tutorials, web development for beginners, vim, ruby, enumerator, advent of code, advent of code 2022, advent of code ruby, advent of code 2022 day 10, advent of code day 10 ruby, ruby solutions, ruby solution, ruby tutorial, ruby enumerator]
views: 268
author: CJ Avilla
url: https://www.cjav.dev/videos/enumerator-crt-display-advent-of-code-day-10-with-ruby
youtube_url: https://www.youtube.com/watch?v=5Qjktd13QUQ
youtube_id: 5Qjktd13QUQ
embed_url: https://www.youtube.com/embed/5Qjktd13QUQ
thumbnail_url: https://i.ytimg.com/vi/5Qjktd13QUQ/hqdefault.jpg
type: video
---

# Enumerator, CRT display - Advent of Code - Day 10 with Ruby

*Published: December 10, 2022*
*Views: 268*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=5Qjktd13QUQ)

[![Enumerator, CRT display - Advent of Code - Day 10 with Ruby](https://i.ytimg.com/vi/5Qjktd13QUQ/hqdefault.jpg)](https://www.youtube.com/watch?v=5Qjktd13QUQ)

## Description

Enumerator, CRT display - Advent of Code - Day 10 with Ruby

Challenge: https://adventofcode.com/2022/day/10
Solution: https://gist.github.com/cjavdev/9f0aab41191b1050886257c0ce052be3

#ruby #adventofcode

## Transcript

hey what's up welcome back in this episode you'll learn how to solve day 10 of the Advent of code for 2022 with Ruby and this one is called cathode ray tube and we are going to like start rendering stuff out to a CRT monitor but in the beginning we are just working on figuring out some Cycles as we are reading through some instructions so we're going to receive a small program that supports two different commands one is no op and one is ADD X we're also going to need to keep track of a clock and there are cycles that are related to each of these commands so for a no op that is going to take one cycle add X is going to take two cycles and add X will update a register by manipulating X based on the number that is passed as the argument to X and that only happens after the two cycles complete when we encounter this argument so we have this big input that I'm going to grab and we'll drop into the bottom of some file to start we'll read all of those lines and you can pass Chomp true here and that will achieve the same thing as passing like dot map Chomp and then what we want to do is split each of these lines on a space so that we can get the command and the argument so we're going to map and split on space and then we want to map that command in the ARG and we want to get back the command and the ARG as an integer if it exists and for now let's just print out what we would get so let's say Ruby day 10 day 10 and we get back at X and 15 and so on and so forth for the no Ops we're going to get 0 and that's okay because that's that doesn't actually impact what we're going to do with this now we can say dot each do for our Command and our argument and if the command is a no op then we're going to do something otherwise if the command add X then we're going to do something else so what we need to do is keep track of some Cycles so we have our cycles and that's going to be just a counter that we are going to increment as we encounter each of our commands here and we're also going to have something we're going to keep track of but for now let's just say like Cycles Plus equals one and then when we encounter add X we're actually going to increment Cycles twice so we need to increment it once that will complete a whole cycle and we're going to increment it twice that'll complete a whole cycle and only after we have incremented the Cycles twice do we want to increment our register X so our register X starts at one and then we're going to modify the register based on the argument to add X in this case we're just like modifying that so this is how Cycles relate to the commands so for a no op cycle we're going to increment by one for in add X we're going to increment by two now we need to keep track of the value of the register X for every single cycle and so if we look at this it says during the 20th cycle register X has the value 21 so this new thing called a signal strength is the cycle number times the value of register X and so 420 is being stored inside of there so every single time that we cycle we need to update some like cycle tracker or something cycle tracker and we'll keep that in a dictionary so here we'll just say cycle tracker at Cycles is equal to x times Cycles the value in the register x times the Cycles gives us the signal strength so I guess maybe we could call that variable signal strength but whatever so what's important is that when we're calling add X we also want to update our cycle tracker after every single cycle ends so even though we're not incrementing X until down here we still want to update the cycle tracker twice before before incrementing the register so if we run it against our example input here and just say Ruby day one oh I guess we're gonna get back we're getting back in the numerator so let's change this to data equals and then we'll P data just just to see what we get back and or we want to print out the cycle tracker so that we can see what's inside the cycle tracker so we have the mapping of the cycle to the signal strength for that point now the next instruction says that we want to find the signal strength during the 20th 60th hundredth etc etc so one thing we could do is just say give me the cycle tracker value at 20 uh you know like plus the one at 60 plus the one at 100 whatever but instead what I wanted to do was just play around a little bit with a ruby class called an enumerator so there is let's open up IRB you can say enumerator dot produce and you can give it some initial value and then a block and it will execute the block and the first argument to the block is whatever value was the previous value so we can just actually just call this like previous and then you can do whatever you want on that previous maybe say previous Plus I don't know three you can use this enumerator in order to generate lots of random numbers so for instance what we could do is we could say give us like create a new enumerator e and if we say e dot next it should give us 0 because that's the initial value and now we should see three because every time we call next on it it's going to execute this block so e dot next e dot next e dot next e dot next and in fact if you have an array like one two three and you just call Dot each but you don't pass it a block the thing you get back is also an enumerator now in the case of each you're going to get back this enumerator and you could call 2A on it and that would give you back an array with all of the elements that were yielded to the block right we can take an enumerator and use that to generate a series of numbers and so for this specific series of numbers it's saying like start with the 20th cycle and then give us the 60th the 100th the 140th so every 40th cycle after the 20th those are the ones that we want we could start our enumerator at number 20 and then we could say okay we want to increment whatever the previous value was by 40. and give us back some new enumerator e and we could say e dot take and we only want one two three four five six of those so take six and that gives us the 20th 60th hundredth one forty one eighty one or two twenty so we can use this enumerator so we can say enumerator dot produce 20 up to something dot take six dot each do n and this is where we're going to sort of like reach into cycle tracker at n and in fact what we could do is use inject here with an initial value of zero and we're going to use that to increment our sum so we're going to take cycle tracker at n plus the sum and this will give us back our result and now we can print out our result and we get back some value so this should be 13 920 and if we run this we get back 13 140 that's wrong okay so are we doing the right thing oh no 13 140 was okay so 13 140 was for the example input 13920 is for my test input so let's grab all of our test input and we will add that to day 10 input we have to update our code here so that it works for if ARG V otempty otherwise data is file.readlines and you can also pass Chomp true into here I didn't know that before so that's kind of Handy and so now if we pass day 10 with input we are going to get back 13 920 that is going to be my part one answer all right let's take a look at part two so in part two it says that it seems like the register X controls the horizontal position of a Sprite and the Sprite is three pixels wide so if we look down here at the example this is an example of a Sprite so when X is equal to one so the row here is zero indexed when X is equal to one the Sprite is going to cover zero one and two so this is the Sprite when X is equal to one now when X is equal to 16 it's going to cover 15 16 and 17. so that's like what x is doing is it's moving this Sprite back and forth within a range of like 40 pixels now the next step of this is that every single cycle we are going to print a pixel on the screen and whether we print the pound sign or the period is going to be determined on whether or not the Sprite is in the range when we're printing that point out so X is going to kind of like bounce around a ton and then the Cycles are just going to go from left to right to print out this grid let's start by iterating over our cycle tracker so we're going to get the cycle and we're also going to get some value X and for now what I want to do is just print out a grid so something like we'll just print out the pound sign for now now if we run Ruby day 10 we get this giant list so we don't actually want puts we want to use print so that it doesn't add a new line after it and we get a bunch of octothorps so what I because the width of the screen is 40 pixels and our cycle is one based we need to figure out what like the column is based on cycle mod 40. and if the column is equal to zero we want to print a new line so puts we can just say puts and that'll that'll create a new line all right let's run this okay this is pretty close we have an off by one error the top right corner is down here in the bottom so I think this is actually cycle -1 because it's one based let's run this again all right now we have our screen rendering so the next step is going to be to correctly print out what you know whether we want an octothorpe or a period so the x value is going to be determining whether or not we're printing out a DOT right so in the beginning X is equal to one so here's the Sprite value and because the cycle is 1 we are like in range of the Sprite so we print this out then when we cycle to 2 we're in range still of where that Sprite position is so we print out again now when we finish cycle two we added we just added 15 to register X so the Sprite slid all the way over to the middle here so when we get to cycle number three cycle three does not have any part of the Sprite showing at that point so we're just going to render a period so let's do that so what we want to do is say something like if the column is between x minus 1 and X plus one then we want to print that out else we want to print the period out okay and if we scroll to the very bottom here this is kind of what we're going for this is what we're expecting that's what we want to see so let's run this again boom okay so this is exactly what we're seeing from the example right now it says use your use your input and run it again so we're going to use it with our input day 10 input and now that's actually printing out some letters they're kind of hard to see so I wanted to like just if we print out a space let's see if that makes it easier here boom okay egl HBL FJ and that was my answer here if you get it right it's really legible if you get it wrong it's impossible to read this is kind of a fun little project there was one similar to this in last year's challenge but this one was kind of interesting because yeah the x is sliding around and there's all these different pieces but I thought it would be fun to play with this enumerator situation and you know enumerator.produced to generate a series of numbers so if you haven't seen that that can be kind I also initially thought like oh maybe it might be fun to have this yield the cycle and also yield the signal and whatever the whatever the value of x is at that register at that point but we'll do that in another episode this was a another fun one thanks again so much for watching and we'll see in the next one foreign

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