---
title: partition and inject intersection &:& 👯‍♂️ - Day 13 - Advent of Code 2023
slug: partition-and-inject-intersection-day-13-advent-of-code-2023
published_at: 2023-12-14 00:01:08 +0000
updated_at: 2026-03-04 20:15:20 +0000
summary: 
description: In this episode, we solve day 13 of Advent of Code 2023, called \"Point of Incidence\", where we're trying to find lines of symmetry in patterns of dots and pound signs representing mirrors on Lava Island.  First, we parse the input into arrays of character arrays representing each pattern. Then we write some methods to check if a line is symmetrical around a given point, find all points of symmetry on a line, and ultimately find lines of symmetry in a full pattern.  The key is finding the intersection of symmetry points across all lines to pinpoint the overall symmetry. We handle both horizontal and vertical lines of symmetry.  For part 1, we report the number of rows above or columns left of the symmetry lines.  For part 2, we consider smudges - swapping one character in the pattern and rechecking for other valid lines of symmetry. We brute force check all possibilities, collect the new lines, and subtract out ones we already had.  In the end, we arrive at working solutions for both parts, matching the provided examples.    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]
views: 138
author: CJ Avilla
url: https://www.cjav.dev/videos/partition-and-inject-intersection-day-13-advent-of-code-2023
youtube_url: https://www.youtube.com/watch?v=bKO93YzC0Co
youtube_id: bKO93YzC0Co
embed_url: https://www.youtube.com/embed/bKO93YzC0Co
thumbnail_url: https://i.ytimg.com/vi/bKO93YzC0Co/hqdefault.jpg
type: video
---

# partition and inject intersection &:& 👯‍♂️ - Day 13 - Advent of Code 2023

*Published: December 14, 2023*
*Views: 138*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=bKO93YzC0Co)

[![partition and inject intersection &:& 👯‍♂️ - Day 13 - Advent of Code 2023](https://i.ytimg.com/vi/bKO93YzC0Co/hqdefault.jpg)](https://www.youtube.com/watch?v=bKO93YzC0Co)

## Description

In this episode, we solve day 13 of Advent of Code 2023, called "Point of Incidence", where we're trying to find lines of symmetry in patterns of dots and pound signs representing mirrors on Lava Island.

First, we parse the input into arrays of character arrays representing each pattern. Then we write some methods to check if a line is symmetrical around a given point, find all points of symmetry on a line, and ultimately find lines of symmetry in a full pattern.

The key is finding the intersection of symmetry points across all lines to pinpoint the overall symmetry. We handle both horizontal and vertical lines of symmetry.

For part 1, we report the number of rows above or columns left of the symmetry lines.

For part 2, we consider smudges - swapping one character in the pattern and rechecking for other valid lines of symmetry. We brute force check all possibilities, collect the new lines, and subtract out ones we already had.

In the end, we arrive at working solutions for both parts, matching the provided examples. 


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

hey what's up welcome back in this episode you'll learn how to solve day 13 of the Advent of code for 2023 this one's called point of incidence and we've now landed on lava island we don't see any lava instead what we see is a bunch of rocks and Ash and there are mirrors in lava island somewhere we're making notes of patterns that are happening on the ground and this is our puzzle input is our notes and it's going to be in these chunks of patterns so this first one is a chunk and then the second one is a chunk and they're separated by two new lines and in the pattern somewhere there is a line of symmetry so we need to find the line of symmetry that is where kind of the mirror is if you close your eyes a little bit you'll start to notice the forms that take shape from the different pound sign and Dot patterns so in this first pattern you'll notice that the the the first five columns here are a mirror or are a mirror of the the last four and the first column doesn't have a matching column on the other side and that's okay so we just care about patterns being symmetrical about some point or some like row or column and then in the second pattern we have one where it is horizontal like this the Symmetry is horizontal around this point four and five the solution for part one is to to summarize our pattern notes so if we have a vertical mirror then we want to count the number of columns that is on the left side of the mirror if we have a horizontal mirror then we're going to count the number of rows that is above the line of symmetry so let's grab our example input here and get started open up main. RB and we'll start with input here being equal to something like this all right the first part is we want to parse out our input so we can say I don't know the patterns are input dot yeah we'll split on two new lines and then we'll map each of those chunks of patterns and we'll split that on new line and then we want to map that to characters and I think this should give us back something reasonable let's check take a look okay so this looks like what we want so we have an array of patterns where each pattern is an array of arrays each row in the pattern is all of the different characters that are in that thing so the way that I thought about this was to first kind of like break down the problem into individual lines so like what is the point of symmetry or what are the points of symmetry for this first top line so if you look at it and you go okay the First Column can't really be symmetrical with itself so we really need to start with the pair of the first two columns and those are not there's no symmetry there and then if we keep going technically we don't have a line of symmetry around one point so it's always going to be there's always going to be a pair of elements that are the same next to each other so that one doesn't work here these four are not symmetrical these five are not symmetrical but if you look at all the way over to the right here um we have dot pound pound Dot and that is symmetrical with DOT pound pound dot so we are looking for ideally the number here which is the number of 1 2 3 4 five we want to get back five and that's the number of columns that are to the left of the point of symmetry for this specific row so we can grab onto that that let's actually also require rpec auto run we'll describe some method um like around or something yeah we'll just call the method around it returns true if a pattern is symmetrical around a given point or something I don't know so it's want to expect that this around method of some pattern and we'll give it characters and if we say is it symmetrical around the number one then we don't we expect that to be false right we don't think it's symmetrical around 1 2 3 4 6 seven or eight we do think it's symmetrical around five so we want to basically just say we'll be pretty explicit here okay and then around five we think it is true okay so then now we'll go run our thing here around is not available that's because we haven't defined it yet so we'll make a new method here around and this is going to take in some line and an index all right expected false got nil we need to partition the line around that index so we're going to pass in 1 2 3 4 5 6 7 and that's going to let us like split up the line into two different sections and the way that we can do that is say we're going to have a left and a right and there is a method on um there's a method on array called Partition so if we look at that for a second so if we have a is 1 2 3 4 5 6 A.P partition takes in a block and if it evaluates the true those end up in one bucket if it evaluates to false the other ones end up in the other bucket in our case we want to use this method with the index of the element so again we can use with index on the end here and pass that in so now we have some index and we want to say if the index is greater than three or something then partition um or I guess technically we want it to be less than so that we have our left on the left and we have the right on the right so this by passing in three here that splits it at the third point where we'll have three columns on the left so again if we partition on four then we end up with four columns on the left um so yeah let's do partition with index and we'll have our character and we'll have our local IND index where the index is less than index and now we end up with left and right you'll notice again though if we are partitioning at a point that's not smack in the middle of the index then we end up with one part of the array that's longer than the other part we want to take the minimum of the two and just check we want to check only that length that kind of gives us our midpoint we want to get some length which is going to be um the minimum of left and right now what we want to do is we want to go um left is equal to left. reverse okay so as we're comparing these we want to flip it around and then we'll check from that point up to the other so in this specific case if this was our left we would have dot pound pound Dot and then this pound sign would be at the end and what we're going to do is trim off any leftover chunks that don't match up to and including that point by using this array sub indexing for instance here if we had left and right and we said left from zero to two um zero not including two then we get the sub array that is just one two technically we want 34 so we'll say reverse and that would be what we're looking for 43 okay if the left is equal to the right I'm not sure if that's what we want but let's start there and just see if our tests pass okay expected false but got true for number seven seven one two three four five six seven yeah I guess technically seven is also correct so this should be true because you can you could this is like a point of symmetry if we chop off the rest of this stuff which we do want to do and then eight should be false okay oops all right so we've got a passing test and we have a method that will tell us whether or not something is symmetrical around a given point now what we want to do is go through and find all the points where uh the a line is symmetrical I called this one maybe points of symmetry and this will take in a line and this is going to return an array of points so now we want to do something like one up to line. length select uh that might actually work so now let's do another little test here so it finds points of symmetry expect that points of symmetry for that is going to be equal to 5 and seven let's see see okay that's passing so yeah so when we pass this in we should get back five and seven the idea is that we'll go through each row and find the points of symmetry and then we'll have to find all of the points of symmetry for all the rows and then we will find the uh intersection of all of those and if it's symmetrical that should give us the point that is symmetrical so we've got the points of symmetry now what we want to do is find the lines of symmetry so line lines of symmetry and this is going to take in a full pattern and lines of symmetry is going to go over each of the rows or each of the lines of the pattern so pattern. each do line and or yeah I guess we can map this onto um points of symmetry so let's see if we can get something out of that finds lines of symmetry okay so then we'll just use this as our part of our test here okay and pattern is input. spit. map let's see if that gives us what we expect okay that looks good so now what we want to do is pass this pattern in so we expect that line of symmetry for the pattern is equal to the element five I think that should maybe work okay pattern uh pattern We spelled this wrong okay I'm just going to close that because I keep navigating over there all right so we expected five but we got 5 S 1 15 1 15 1 15 1 15 1 1357 and 5 so you'll notice there is something in in here and that is that the pattern is the same one really cool thing you can do is if you have an array of arrays where we have maybe one two and two and 2 3 and we want to find the intersection of all of these arrays we could say B is the first A and C is the second a and then we can do B Amper C that will give us the intersection so that tells us like what's the same in both of them so this operator here is something that we want to apply to all of the elements so we can say a. inject um and pass in like a or I don't know d and e as our arguments and then we can say d % e and this needs to be an array and then we're going to get some like accumulator no we don't want to pass an array okay so if we pass an initial argument it uses that as the first D or the first like element but if you don't pass any arguments then it uses the first element of the collection as the first argument in this case we want to apply this intersection operator against all of these elements so we can do this but inject recall that inject has this shortcut where we can say amp colon and then the name of a method and in this case the method name is also called ampers this is the first time I've seen this and I think it looks pretty cool it's two erss holding hands or something holding a colon all right so we're say ERS ENT and that should do something for us oh look at that boom we got a passing test okay all right so now we have lines of symmetry now we can use lines of symmetry for the pattern for both rows rows and columns because now what we can do is just pass in the transposed grid to get out any columns so what we want to do now is go over each of the patterns so we'll comment this out and say patterns. each do pattern and for each one we want to get the lines of symmetry rows maybe we'll just call it like Sim rows and that'll be that and Sim calls will be that and we can print out sim rows actually I think this is backwards I think this one is going to do the columns and this one is going to do rows okay so now we'll just say rows is rows and columns is columns and we'll run it and okay one of the patterns is symmetrical around the fifth column and the other pattern is symmetrical over the fourth row which does match up with what we're expecting so we can say that our answer here starts at zero and what we want to do is say answer plus equals I think it's there's a special pattern here so 100 multiplied by the number of rows we'll do Sim rows. 2i * 100 plus Sim columns 2 I and that should give us a number ah because we only have one uh we'll just grab off the first result there of our lines and it says no examples found but really what we want to see is puts part one answer is answer and what do we got what do we got 405 okay that is what the example is expecting let's grab our input our puzzle input looks like this we'll drop it here at at the end and then instead of input being input we'll override input to be data. read we'll rerun this and we get 3488 9 3 4889 is boom that is the answer for part one okay we're able to check where the mirrors are and find the answer based on the Symmetry that we're finding on these rows in columns okay part two you we keep walking through we're walking through Lava Island and boom we run into a mirror that's because some of the mirrors have smudges on them I hate smudges smudges smudgy smudgy smudge and so it's possible that a single dot in the pattern is incorrect and so we need to figure out now if the mirror was in a different place because one of these notes was incorrect we need to swap exactly one period to a pound sign or one pound side to a period and see if it's valid now there's a bunch of different ways that you can do this and a lot of optimizations I did not take any optimized approach I just said okay let's take the entire grid and one by one we will replace each of these things with the opposite we'll check to see what lines of symmetry we end up with okay so what we want to do is while we're iterating here we're going to go through all possible rows and Columns of the pattern so we want to say pattern. each with index do row and sure the row index and then row. each2 index column index okay we have to go through every single cell right so we need this this doubly nested Loop here and what we want to do is like at row index column index swap from uh dot to period or period to pound sign right so let's do this let's because we're going to be modifying or like working with another array inside of here rather than updating pattern itself we want a sub pattern which is pattern. map dup we want like a duplicate of that and we want the subpattern at row index column index to be swapped so this should give us the swapped element at that single index then what we want to do is get the sub rows and that's going to be lines of symmetry and then sub columns which is going to be those lines of symmetry this time instead of grabbing off the first one what I found was as we're going through and changing these different smudges we might end up getting back the same line of symmetry that we already had when we had like a pristine smudge-free mirror and so what we want to do is grab all the possible lines of symmetry for both of these and we want to collect them up and then remove the valid lines that are that are here we actually want to keep track per pattern sub rows is going to be some actually let's use a set because this is going to be unique for the pattern so we'll have sub rows and sub columns and then here we can do plus equals and then we should get some sub rows and sub columns per pattern and eventually we're going to want to remove the sim row and Sim column so let's take a look at what this looks like against the example input so let's just run it okay so sub rows is three sub columns is five it'll also help here to print out the rows and columns for the pattern and at the beginning of a pattern let's also just like print out something so we can know the separation between patterns okay so for this first pattern there are no rows that are were symmetrical around and we're symmetrical around column 5 this is when we're smudge-free now when we start to introduce smudges we can see symmetry around the third row and you'll notice that we still see symmetry around the fifth column and that's because it's it's picking this up when we smudge something that's not even in scene before we check that similar down here we have rows when it's UNS smudged and we have this other one so we basically want to remove our our existing valid ones from the sub rows so here what we can do is say sub rows uh sub rows minus equals Sim rows and sub columns minus equals Sim columns and now we should see something different so now we just see the set with three and the set with one and that is correct so our answer for part two I'm going to move this up here that's all our part one stuff and then our part two stuff is down here so our let's say answer two is equal to zero and we want answer two to be equal to the same same sort of deal as here so let's actually just copy this and paste it here and then in instead of sim rows this is going to be sub rows. first. 2i and sub columns. first. 2i and we're expecting 400 and we got 805 ah because we're incrementing answer that should be answer two and then we want to print out the part two answer answer two boom okay so for part one our answer is 405 for part two our answer is 400 and and that's looking pretty good okay if we run it against our actual input uh that's super messy and loud but ah whatever we'll just let it run okay I think there are 100 patterns in the example input we get this huge output all right three4 224 in the yeah so that is our answer for part two awesome okay definitely another really tricky one and yeah I think there was another there's technically another approach to where instead of checking to see if the left and right are equal here you can check to see if the left is different from the right by one if it's off by just like one one of the bits is flipped basically then you can say yeah there's a smudge in this line but then you have to pass that knowledge back up through the chain so it's a little easier to Brute Force and just say I'm going to try every single possible sub pattern and throw it at this lines of symmetry business and get back some answers compare those to what we already had etc etc we're making progress got 12 days left thank you so much for watching and we'll see you in the next one

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