---
title: ASCII values with ord and delete_if - Day 15 - Advent of Code 2023
slug: ascii-values-with-ord-and-delete_if-day-15-advent-of-code-2023
published_at: 2023-12-15 17:15:02 +0000
updated_at: 2026-03-04 20:12:55 +0000
summary: 
description: In this video, you’ll learn how to solve day 15 of the Advent of Code 2023. The first part involves writing a hash function to convert initialization sequences like \"rn=1\" into decimal values between 0-255. The hash function works by:  1. Getting the ASCII code for each character using .ord 2. Multiplying the current value by 17 3. Taking modulo 256 to get the final hash value  For part 2, the initialization sequences now can contain \"=\" to add lenses or \"-\" to remove them from \"boxes\", which are arrays. The boxes are stored in a hash keyed by the hash value of the label on the lense. We’ll parse the complex input into easier-to-work with structures. Then we’ll iterate through the instructions, either adding/updating lenses or removing them if the instruction contains \"-\". At the end, we’ll sum each lens's \"focusing power\" based on the box number and slot number.  After getting the right answer, we’ll spend some time refactoring some of the code, showing useful Ruby idioms like inject, delete_if, and pattern matching.  Advent of Code: https://adventofcode.com/ My Solutions: https://gist.github.com/cjavdev/d15a2a4ffed6c840c2fb28a093e9f927/ Playlist https://www.youtube.com/playlist?list=PLS6F722u-R6KYlGyUv65EFpGKl2Esmurr  #adventofcode  #ruby
tags: [cjav_dev, Learn to code, Beginner ruby, Advent of code, Advent of code 2023, Advent of code ruby, Aoc ruby, Aoc 2023, Vim, Advent of code vim, Advent of code explainer, Advent of code challenge, Code challenge, Advent of code tutorial, Web development tutorial, ord, ascii, ruby delete_if, ruby ord, ruby ascii]
views: 134
author: CJ Avilla
url: https://www.cjav.dev/videos/ascii-values-with-ord-and-delete_if-day-15-advent-of-code-2023
youtube_url: https://www.youtube.com/watch?v=hPa5hfJeFlU
youtube_id: hPa5hfJeFlU
embed_url: https://www.youtube.com/embed/hPa5hfJeFlU
thumbnail_url: https://i.ytimg.com/vi/hPa5hfJeFlU/hqdefault.jpg
type: video
---

# ASCII values with ord and delete_if - Day 15 - Advent of Code 2023

*Published: December 15, 2023*
*Views: 134*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=hPa5hfJeFlU)

[![ASCII values with ord and delete_if - Day 15 - Advent of Code 2023](https://i.ytimg.com/vi/hPa5hfJeFlU/hqdefault.jpg)](https://www.youtube.com/watch?v=hPa5hfJeFlU)

## Description

In this video, you’ll learn how to solve day 15 of the Advent of Code 2023. The first part involves writing a hash function to convert initialization sequences like "rn=1" into decimal values between 0-255. The hash function works by:

1. Getting the ASCII code for each character using .ord
2. Multiplying the current value by 17
3. Taking modulo 256 to get the final hash value

For part 2, the initialization sequences now can contain "=" to add lenses or "-" to remove them from "boxes", which are arrays. The boxes are stored in a hash keyed by the hash value of the label on the lense. We’ll parse the complex input into easier-to-work with structures. Then we’ll iterate through the instructions, either adding/updating lenses or removing them if the instruction contains "-". At the end, we’ll sum each lens's "focusing power" based on the box number and slot number.

After getting the right answer, we’ll spend some time refactoring some of the code, showing useful Ruby idioms like inject, delete_if, and pattern matching.

Advent of Code: https://adventofcode.com/
My Solutions: https://gist.github.com/cjavdev/d15a2a4ffed6c840c2fb28a093e9f927/
Playlist https://www.youtube.com/playlist?list=PLS6F722u-R6KYlGyUv65EFpGKl2Esmurr

#adventofcode  #ruby

## Transcript

what's up welcome back in this episode you'll see how to solve day 15 for the Advent of code this one is called a lens Library we've got the first half of this is trying to figure out how to convert from this initialization sequence syntax into a hashed value so each of these are uh little like instructions or some initialization sequence we want to convert this or each of these comma separated things into a decimal value between 0 and 255 this is some example input we're going to use the asy value and then we're going to multiply that by 17 and then find the remainder by dividing by 256 so this first part here we have to basically go through each asky character and we're going to have some current value that's keeping track as we go through each of the characters so we're going to look at R first find what its asky value is in the asky table and then we are going to increase the current value by R's asky value then we'll take that current value and multiply by 17 and then we will modulo that by 256 and we're going to get back some value so in uh the first example here it lets us work with this uh thing called hash so let's uh open this up and we'll just say that we want to make some new method here called uh I don't know hash or something that's going to take some input and it will return a value so if we print out the hash of hash then we expect to get back the value 52 so yeah if we run this right now we get nil and 52 so let's go through and figure out how to um follow this algorithm so determine the asky code for each character so I guess first we want to say input. chars and then in Ruby there is a method called or so if we look at this real quick so if you have any character if you call. on it that will give you the asky character value for that character so if we look at in the example here it says the first character of H its asky code is 72 so if we look at h. ORD we get back 72 so we can use the concept of this or thing and say map we can pass in Ord and that will give us all of the asky values so if we run this now we should get an array 72 65 83 72 and if you go through here we're going to see 72 65 83 72 so that is the correct asue values for these now we need some like current value tracker thing so we'll call this C and it's going to start at zero and that's what we'll ultimately return from this method and the first thing we want to do is increase it by this asky value so maybe we'll just say do each do character and we want to increase C by the characters value I guess we can't use C twice can we okay so this is actually like the asky so we'll increase by the asky value and then what we want to do is multiply that current value by 17 C * equals 17 and then we want to multiply that value or then we want to divide by 256 and take the remainder so we can use this modulo operator to get the remainder so 256 and that should give us back our our value at the end here so let's see all right we got 52 and 52 so that's looking pretty cool we have a bunch of examples here and you'll notice there's like an equal sign and a Dash those also have asy values may if we just say our examples are this list here all right and see okay let's switch out comes for a closing quote and a comma and we'll insert a quote here all right so now we've got a bunch of examples and we can iterate over those examples. each do input and expected and we're going to just print out the yeah just to see what we're what we're actually getting here so we spelled this wrong okay so for rn1 we got 3030 253 253 since all these values match our hashing algorithm is working and so it says go through and sum up these results and it should be 1320 so we'll just get result is zero and then result plus equals that and we should get 1320 at the end here and we got 1320 so that's looking good so what would the sum of the results be for our initialization sequence so we got to grab our puzzle input which is this giant line line and it's all one line which is just it's wild actually I think yeah we could probably just put this in data equals this thing okay all right so we want to pop this maybe here okay and then we're going to say examples is now data do split on comma and in this case we don't actually have expected so let's let's just do a new thing here so this is going to be our like instructions or something in the initialization sequence so now we'll say instructions. map do uh instruction and we want to get the hash value of those instructions and we want to find the sum of those and that's going to give us some result we want to print out that result and let's see what we get here five 515 974 515 974 boom all right awesome so that is the solution for part one oh hey real quick a Shameless plug buildand learn. this is a podcast where me and my buddy Colin hang out and we talk about software development and also just things that we're building and we're learning so if that kind of stuff interests you head over to building. Dev you can listen on any of the podcast players that you're used to so yeah let's get back into add the code cheers all right let's talk about part two so it took me quite a while to understand the instructions for part two so I think it makes sense to go through and really deeply explain what's going on here we're going to have a set of boxes and we'll have 256 of those boxes each box will have an ID of the box and then it will have some set of lenses our input rather than just hashing the entire thing we now need to like actually break out parts of the input into um component pieces each of these instructions that are in the initialization sequence have either an equal sign or a dash if they have an equal sign then we're going to add something or set something in the box if it has a minus sign we're going to remove a lens from that box with that label the way that I thought about this entire structure was we're going to have a hash where each of the keys in the hash is just a numeric D technically we could use an array we'll have 256 of them from 0 to 255 and each key will point at an array of lenses so the array of lenses might look something like this where we'll have the label of the lens and then the focal length of the lens so when we get an instruction like RN equals 1 this means that the label of the lens is RN and the focal length of the lens lens is one and because it's an equals that means that we want to find which box the lens goes in or is related to based on the label so we're going to use our hashing algorithm to find out the box ID and then because it's an equal sign that means that we want to either add or update the lens with this label in this box so RN equals 1 means go through the box and find if there is a lens with a label RN and if there is update it to one if there isn't then add a new one that has one now if we encounter an instruction that has a minus this means we want to remove any lens with CM from any of the boxes we're going to Hash cm and find the box that goes with CM and then we'll look through all of the lenses that are in that box and remove the one with the CM label so given this example input here of all of these different instructions so RN is hashed to zero and in this case we're setting one so after we encounter the RN equals 1 instruction our box zero will look like this after we encounter this instruction it's a no op since this doesn't we don't have any CM lenses anywhere so nothing happens and then when we encounter QP QP hashes to one and because QP of 3 isn't in here yet we add it we keep going we keep going when we encounter OT equal 9 we add otal 9 to box 3 and then what's interesting is that later on after we encounter OT equals 7 so before OT was nine and now rather than adding ot7 we're going to find OT in box 3 and update it from 9 to 7 and at the end then we're going to go through figure out the focusing power of each lens based on how close it is to the light source and we get some number back we want to parse out each of these instructions and I want to reformat it so that it's a little easier to work with so if we come down here we're going to comment out all of this and let's just start from scratch here so we're going to have our um instructions are now going to be examples. map first and we'll go through instructions. each do ins and let's actually map this so we want to like restructure these so that we have it split basically by that symbol and we want to still know what what symbol it was there's a couple different ways to do this but the way that I thought was we'll grab off the label and then in A and B and we'll do instruction. chars and then yeah then we can just say if B is equal to a minus sign then our list is going to be uh it's just going to be the label on the left so we'll have instruction up to from 0 to1 and then we will have the actual symbol that we want to use so the minus sign and then we'll have nil as our final argument otherwise we're going to have the instruction is going to be yeah up to -2 and then um in this case a will be our equal sign and B will be our number our focal length so then this will be like two integer because we want those to be integer values so let's see what we get back here print R so if we run this okay so now we have it broken down but this is incorrect so we don't want the equal sign in there so instruction um up two but not including that okay so now we have the label this one still also has that so let's go up two but not including negative 1 all right so now we have our label our instruction and our focal length label instruction focal length label instruction focal length so we have nil in the case of that we're removing it but that's okay all right so now that we've got it in this new format what we can do is start to build up our dictionary let let's go through let's call this yeah we'll just change that to instructions this is an interesting pattern right like instead of setting equal to we're like flowing through and then ending up down here and then we can keep flowing instructions and then we'll maybe we'll have some like dictionary or yeah we'll have our boxes and that's going to be hash do new it'll be one of these hashes that automatically initializes to an empty array because we want to have a container for each of our lenses so now we want to go through each of our instructions and it will have a l label it will have an instruction and it'll have the focal length of each of these things so we want to index into boxes with the hashed label so we're going to use our hashing algorithm for our label and this should give us back the list of lenses that are at that location so what's cool about that is when we're using this default hash syntax if it's never encountered this label then it's going to create a new array otherwise it'll give us the array that's at that key now what we want to do is say if the instruction is minus then we want to remove the lens that has this label from the list of lenses so we can do lenses. delete if and we can pass a block and each lens will have a label and a focal length and if the lens is label equals label then we'll remove it otherwise the instruction will be an equal sign so for now we can just pop in the label and the focal length and see what we end up with for our uh boxes at the end so let's just P boxes and we'll run this okay so we end up with zero points at rn1 cm2 one is empty and that's fine and then at the end we have box three has ot9 ab5 pc6 and ot7 so we shouldn't have two lenses with the same label in this box instead of adding ot7 to the end here what we want to do is update ot9 so what we need to do is say if if we already have a lens with that same label so maybe we say lens is equal to lenses. find so if we have a lens with that label we want to update it otherwise we want to add it so if we're updating the lens then yeah we're going to say update the lens to have that same focal length and if we run this again we get back this value and now we have ot7 is correctly updated here rn1 cm2 at index zero should match what we see in the example here so rn1 and cm2 at index zero that looks good and then we should see o ot7 ab5 and pc6 ot7 ab5 pc6 those are at uh number three so we're looking we're looking good so we've got boxes that have the right data in the right places now we want to go through the list of boxes and figure out uh this value here this is going to be like the focusing power of each lens let's first print out like one one one and maybe the maybe even the product of that just so that we can like double check the way that we're breaking this down we want to go through boxes. each do the key and then the value is going to be the label and the focal length of each lens and let's print out the k + one because the box number is actually the one indexed version of that so we'll do k+ one and then we want to say times the first slot so we want to know what index in the list of oh actually this isn't label and focal length right this is lenses so now we need to do lenses. each with index actually do uh label and focal length and then slot number and now let's just print out the slot + one and we'll print out the focal length Okay so now we should have the box number the slot number and the focal length and then we also want like the product of those so we need like maybe like components I don't know is this and then we'll P comp and and comp. inject uh multiplication and let's see what we get here all right so 111 122 and then we have 14 28 40 okay I think this looks right this matches what we're seeing in the example so now what we want to do is collect these up so we'll have some sum and we want to do sum plus equals comp. inject and then we'll PE out the sum at the end we should get 145 we see 145 there so that is looking good now we want to do this against our instructions from data let's run this and we get 2658 94 265 894 boom that's the answer so that is the answer for day 15 of the Advent of code part two all right this was pretty fun breaking down the instructions here like this this is like an intermediate step because we wanted to get it into a data structure that worked better for us and I like doing that in a in sort of like a separate place than when we actually like iterate and do the actual instruction set so this is like a preemptive step to format and parse out the things that we want we could probably actually make a method here called parse that would take in some instruction and then return that value and we can break this out into this parse method and we can say map do instruction parsey instruction and then we can just run each against the result of that and we should get back we should still get back the same answer okay we're looking good another thing that we might want to do is yeah like we could clean up some of this here we can just do sum plus equals this whole thing do inject multiplication okay same answer that's looking good technically I'm sure we could do some fancy inject thing that added all these up but that's probably clean enough for now also our our hashing algorithm here we use the the variable name c not super yeah it's not uh it's not very descriptive so we could the other thing is that like when you have a method that has a variable that's being set at the top and then the same thing is being returned at the bottom often times times there's a way to rewrite it so maybe we can do inject start with zero and now we have C and perhaps we can actually get away with this maybe and uh let's just see here okay same answer so that's cool we could also rewrite this to use pattern matching if we wanted but that's all right I liked this was like one of the things that I hadn't used a ton before is this delete if and this is a way to delete elements from an array with a block which is nice and here we're destructuring into label and focal length I don't actually know if we need to yeah in fact we don't actually need to destructure because it's implied because each lens is a tuple of the lens in focal length what I'm really I don't know I'm curious to see the new in the Ruby 3.4 you can say it and it will be the reference to the first block argument which will be cool so here we could do like it. first or something and that would be our label but for now we can just do underscore one there was an argument that underscore seemed like it was an optional thing yeah I don't know that's cool what else can we do here this syntax you might not have ever seen before we are assigning something while in a conditional so if we are not able to find a lens and this ends up being nil then we end up going to the next block so that is the same thing as saying if lens then keep going right but instead of having to have it on two different lines you can keep that on the same line yeah all right fun Ruby stuff for the day that's it for this one thanks so much for watching if you're enjoying the series please hit that subscribe button and we'll see you in the next one cheers

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