---
title: Point in polygon - Day 10 - Advent of Code
slug: point-in-polygon-day-10-advent-of-code
published_at: 2023-12-11 01:15:00 +0000
updated_at: 2026-03-04 20:15:21 +0000
summary: 
description: Join me to solve Day 10 of Advent of Code 2023 - The Pipe Maze - in Ruby!   In this coding challenge, we'll navigate a grid map of different pipe connections to: Parse the input grid data Find starting points and valid neighbors Traverse around the pipe loop to map its path Determine enclosed tiles inside the loop Use raycasting logic to count wall intersections Handle tricky cases like squeezed openings   The solution builds a search to map out the full pipe structure. We'll refine the directional rules and fix bugs in the graph traversal logic.   Mapping polygons and counting enclosed regions employs some novel techniques like raycasting and parity checks.  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: 273
author: CJ Avilla
url: https://www.cjav.dev/videos/point-in-polygon-day-10-advent-of-code
youtube_url: https://www.youtube.com/watch?v=017epvQWPtc
youtube_id: 017epvQWPtc
embed_url: https://www.youtube.com/embed/017epvQWPtc
thumbnail_url: https://i.ytimg.com/vi/017epvQWPtc/hqdefault.jpg
type: video
---

# Point in polygon - Day 10 - Advent of Code

*Published: December 11, 2023*
*Views: 273*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=017epvQWPtc)

[![Point in polygon - Day 10 - Advent of Code](https://i.ytimg.com/vi/017epvQWPtc/hqdefault.jpg)](https://www.youtube.com/watch?v=017epvQWPtc)

## Description

Join me to solve Day 10 of Advent of Code 2023 - The Pipe Maze - in Ruby!


In this coding challenge, we'll navigate a grid map of different pipe connections to:
Parse the input grid data
Find starting points and valid neighbors
Traverse around the pipe loop to map its path
Determine enclosed tiles inside the loop
Use raycasting logic to count wall intersections
Handle tricky cases like squeezed openings


The solution builds a search to map out the full pipe structure. We'll refine the directional rules and fix bugs in the graph traversal logic.


Mapping polygons and counting enclosed regions employs some novel techniques like raycasting and parity checks.

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

this one is called the pipe Maze and you land on the this floating metal Island and it's covered in all types of different pipes and your job is to we're going to figure out some stuff about these different pipes and we're supposed to be tracking down a critter some metal Critter that like scurried off and the pipes are in a map that is the result of our note taking and we've drawn vertical pipes like this vertical bar symbol then we have a Dash as a horizontal pipe and L j7f these are all different types of pipes and then there is s which is our starting position and if we encounter a period that is just like the ground and there's no pipe on that tile and we need to figure out how to navigate through this grid of pipes and it turns out that there is one giant large continuous loop and our first step is to figure out what is the furthest point from the starting point that is inside the loop so what I did was just draw my way around the loop count up all the different positions in the loop and then divide by two and that kind of gives us the farthest distance in the loop from our starting position so let's grab this example here and we'll open up main. RB and we'll start that as our input here we go okay the first thing we want to do is split this up so we'll say data is input. each line. map Chomp and then for each of those we're going to get some grid and our grid is going to be data. map to characters I think that should give us something to work with here that we want to see all right so now we've got a bunch of characters our grid is an array of arrays where each row has all of the different pipe connectors for that row so we're encountering ground ground then an F pipe which means like the pipe connects from the bottom and from the right and then we have a seven pipe which connects from the bottom into the left and then we have ground so on and so forth as you're reading through this instructions the set of instructions you'll notice that like an L is a 90° bend J 7 and F these all represent 90° bends pointing to different directions and then vertical is just up and down so it's going to connect North and South horizontal is going to connect east and west and first of all we need to figure out our starting point so start is going to be grid. each with index find row row include or row. index or something of s does that give us our start nope so let's just do it the let's do it the manual way so grid. each withth index do row and then row. each withth index I don't know why but I like using X and Y even though they're not really X and Y so everything is transposed we'll just say if the cell is equal to X then the XY or then start is equal to XY all right so two Z so that would give us row number two and column zero so that looks good so this is going to give us our starting point then from our starting point we need to figure out how to Traverse so from the start technically we don't know whether or not we're going to be able to go north south east or west so need to figure out what is connected to the start and is a valid outpath what we can do is start with all of the different cardinal directions here and create sort of like a start exits so we can create a list of start exits from the start if there is a j to the right then that is a valid exit because J can connect like to the left and for this start because we have a pipe right below us that can connect to the north so we could connect from this from start to the South so these two are technically like known as the directions we're going to go from the start but for our actual input we might have an S that's plopped in the middle of nowhere so we need all the valid directions from any given point and then we can go around all the neighbors of s and find the ones where s is a valid destination from the neighbors so let's let's just say like de neighbors and we're going to give it the the grid and we'll give it some point and here we can go through all of the directions and uh grab the ones that are valid Neighbors in fact like depending on what kind of point it is it's going to have some set of directions again if it's a J then it can go we can go up so that means that we want to go negative-1 rows and zero columns or it can go to the left which is 0 row changes and a negative one column change so these are the directions that are valid for J we'll go through and fill them out for l so L can go up and in the same column or it can go stay in the same row and move to the right dot can go nowhere and then the s for now we're we're going to say that this the S can go nowhere and we'll have to like manually manage how s works so what we want to do is go through the valid directions for a given point in this case we want to grab like the pipe at grid of x y and then we want to index into dur at that pipe and map over xdx and ydy to get back the neighbors so we need to go through all possible directions around s and then find if their neighbors include this starts but we'll just go through each of these and say if so this is actually going to give us like a an S an S neighbor which is the start plus DX and the start plus Dy if um if the neighbors of the S neighbor include the start then add those to the start exits and now we have some start exits okay so can we leave at three and 21 that's the question can we leave at three Z or 2 one yes those are the correct ones because we can either leave through this J or through this pipe as we're leaving the start point there are some other examples in here too which we could grab on to so like this could be another input I believe yeah so let's pop this in here okay one one that is the start point and then our exits could [Music] be uh zero one actually that's not valid so L does l have the right stuff here so L can go to the north this is actually just negative negative 1 because we can go north or 0 one which means we can go to the right okay let's try this again all right 2112 so from here we can go to 21 or one two all right so that's correct all right so we might have to go back through here and double check all of our different directions but for now that's going to give us our start exits now what we want to do is just grab one of those start exit points and start like traversing around and it should all be connected into a giant Loop so until we come back around to the beginning so now that we have these start exits let's say that like our start point is start exits. first and we want to go yeah we're going to have some list of like our actual pipe and our pipe is going to start with the start start point and the start and we're going to use a set here just so that we have a little bit faster operation when we're checking to see if something is still in the pipe we're going to start with the start point so while the uh start Point does not equal um start exits. last so that should be like the other connector into s right so s is going to have two things that are valid and that are connected to it one is going to be the start point that we're going to start traversing the pipe around the other will be the the end connector that goes back to start so we're saying while the start point doesn't equal the end point keep going so we're going to grab the neighbors of um passing in the grid and the current start point do each do neighbor and for each neighbor if the pipe so if the set of connectors does not already include the neighbor we're going to add it to that to the pipe and then we're going to set the start point to that neighbor and then we're going to break and that should um actually I don't know if we want to break here or not so at the end we should just have P pipe and that should include all of the different things that are in the pipe okay so 21 one one all right let's let's also put input just so we can see so we're starting here at s and then we're going to 21 so we're going to this bottom one and then one one which means that we're going back to the start so maybe this is our start node this is our start yeah let's actually to make this a little bit easier to follow we'll put start and then start point in a different order there in the beginning okay so we start at 1 one that's our starting point then 21 2 one is this B this Bottom bar then 31 is the L then 32 is this pipe then 33 is J then 23 is this pipe then okay good so we're we went around the circle that looks great and now what we want to do is put like P dot or pipe do size so how long was this thing so eight so the answer for part one would be like size / two because that's like the farthest from that point so if we come down here and we look at the input here for this example it was four and for this other example let's grab this other one and see we should get eight three h we have something incorrect here all right so for this complex Loop have we got okay our answer is coming out as seven so let's print the input and then we'll P the pipe again and we end up with 20 two 0 and then 3 0 3 0 4 0 41 3 1 32 okay so then from here something happened when we went from 32 we ended up at 21 somehow but 21 is incorrect 32 to 21 so neighbor of 31 which is f gives us 32 and 21 okay so f is going the wrong way f is pointing the wrong way F if we're in the same row we go up one column or we go down a row yep okay oh there we go okay we got our we had our we had a bug in our data F again should have gone the yeah it should have gone down I think it was going up so now we can do p. size / two and we get eight fantastic okay we don't need to print the pipe anymore all right and we can take out this debugging okay great all right so now what we want to do is run this against our real input so now we can grab our puzzle input paste it at the bottom run Ruby main. RB actually now we have to say data is data. readines map Chomp and 6599 cross our fingers 6599 is part one okay so that is the solution for part one you you basically go around the entire Loop find out how long it is divide by two that gives you the furthest Steps From the starting position to the farthest point from the starting position along the pipe all right part two this is this got really tricky in my opinion it was hard to figure out our goal now is to figure out what is the area that is enclosed inside of the loop so in this case this example Loop has a start point and then it goes around and around and around and around and back to S and when you look at this you might wonder are all of these 10 dots that are inside of this upper section are those enclosed in the loop and the the answer is no but these two and these two are because they are in they're on the inside of this polygon that's defined by all these different points so if you look at this example here you'll see that the eyes are inside the loop and the O's are outside the loop and so in this larger example it shows okay so if this was the input um only there's a couple of tiles in here that are inside the loop and then there's one over here and one over here so there's one here and one here that are inside the loop but otherwise they're outside so in this case you've got to count up all of the tiles that are inside of the loop and that will give you your answer there is a an algorithm called the Ray casting algorithm or point in polygon and I think this is probably like the best example to for how to think about this is that you're going to we're going to look at all the different points on the entire grid and we're going to count from that point all the way to the right and we're going to look to see how many times we intersect with a wall and depending on the number of times we intersect with the wall that will let us know whether or not we're inside or outside because if we go from the left and we intersect you can kind of like think about a a circle right if you start on one side of the circle and you cross you cross the boundary once and then you cross the boundary again then you're on the outside of the circle if you start on the inside of the circle and you're going to the left then you'll only cross once right and so what you can basically do is count up how many times am I encountering a wall and for the number of times that I encounter a wall if it's odd then I was inside if it's even I was outside however this becomes a little trick trickier because you can squeeze through a pipe it doesn't have to have a full in this case it doesn't have to have a full tile path outside of the tiles to count as being inside or or outside of the loop so in this case if you look at this the seven and F are right next to each other these two pipes are right next to each other and this J and L are right next to each other but these are technically not called enclosed in the loop and so you could squeeze between here to find something outside the loop so my thought process was iterate over each like every single point and what I'm going to do is count the number of times that I encounter a wall in thisa if we if we look at this example here so this wall is north south this wall is really just south technically because if something could squeeze in between the top of this and something else it just really blocks the South and then we have a bunch of lines or like a bunch of connectors that are East West those AR those don't really count as walls and then we have this other wall that is technically blocking to the South but not to the North and then we have a north south block and then we have an end so if we started here we would count one North and one South then we would count one South and then another South so we have three South and then here we have another South and another North so we end up with two North and four south facing blockers and then because they're even um and really in that case we can just take the minimum and because they're even we know that we're outside if instead we take one of these inside points and we're going to go just to the the right so here we have north south north south north south so there's three North and three South blocks here because that's odd we know we're inside so that's the rationale is we're just going to count up how many walls we encounter take the minimum of the north facing walls versus the south facing walls and then if it's odd we know we're inside if it's not odd we are outside all right so how does this work so for this part um what I want to do is I want to iterate through um we'll just say like this is uh part one so we're going to continue reusing our same pipe structure but now what we want to do is we want to like just create a new grid that only has our pipe on the grid so I'm going to say like clean grid is going to be array. new of grid size grid size and then I'm just going to have spaces inside the grid to start out okay so this gives us a n byn array that matches the existing grid then we want to go through pipe. each do yeah and instead of X I mean we could do X now and just say pipe. each do row P or like puts row do jooin and this would give us an idea of what it looks like puts that's not what it looks like oh clean clean grid yeah what is here let's P what's our pipe again our pipe is this set pipe. each do XY clean Grid at XY p clean grid okay clean grid. each with index row. each with index print cell puts okay so this is what that example looks like where the X's are our pipe so we're really just in that case finding that one empty spot in the middle let's use this larger example that they've given us just to see if we can figure this out a little better all right so this is interesting looking right so we've got X's where our pipe went and then inside of here you see this like section which is going to be filled in with our insides right that's going to be like the stuff that's on the inside of the loop yeah okay so that's just to kind of like get an idea of what it looks like instead of putting in X though we actually need to know what pipe is there because we need to know whether or not it's blocking north and south or just south or just north North so what we can do in this case is set clean Grid at XY to Grid at XY and this shows us the pipe connectors where they are and we are doing well okay so now what we want to do is go through clean grid each with index actually we'll just use the same thing okay so for each element like each for each cell that we look at uh we want to skip it if it's in the pipe right we don't actually want to look at cells that are that are on the pipeline itself we only want to look at cells that are either outside or inside so we'll say next if pipe includes XY okay that will skip skip cells on the pipeline okay so now we're going to start from the point of the cell and we're going to work our way all the way to the end of that row so what we want to do here is do like y up to row. size. each and that's going to give us a new column right and as we're going through we want to count North facing blockers and count south facing blockers okay if if Grid or clean Grid at xy2 is in the list of North facing blockers so we'll have to make a list then we want to like increment North otherwise we'll increment South and then we're going to have some counter for each for every single cell we need a counter so North is equal to zero and South is equal to zero okay so North facing block blockers is going to be like J right because that's going to be blocked to the north L is blocked to the north pipe is blocked to the north and S is unknown we actually need to replace S with whatever connector type it would probably be but for now we'll just keep it as is or we can we can come back and and take a look later and then we'll count the South so the south facing ones are going to be f um 7 and the pipe okay so if the again remember that we can take the minimum of the two that's going to really be the number of things that we're encountering if the minimum is minimum so if north south do Min is odd that that means we're like inside inside plus equals 1 so it's like an inside count is going to be incremented by one let's also keep track of the actual points so inside cell is going to be XY okay so we'll just keep track of a list of inside cells here and we'll also keep track of the inside count and at the end we should print out inside and P inside cells and 151 that seems wrong ah okay what we kind of need to do is look at our start what I did was I just printed out start and start here is 04 so then I went to 0 1 2 3 4 where is start S oh zero okay 0 1 2 3 4 and then I like just looked at it and I said okay from here from this specific start we can go to the left or down so technically the start in this input example is a seven which means that it blocks to the South but not to the north so then I would add s into my South blockers here this is definitely not the way that yeah that's this is not a great solution okay okay so 151 is definitely wrong right so that's what we're getting though so what is the deal 151 it thinks there's 151 that match that that match inside cells that doesn't make sense okay maybe I'm off by one [Music] here one 61 I believe for this example it should just be 10 also why oh inside cells is being set okay we want to shovel in inside cells okay one Zer one 0er one 0er one 0 one 0 one zero why is it adding those a bunch of times H so we're going through our clean grid and let's print x with a space and then we'll print the cell and then at the end we'll print new line and a dollar sign just to know that we made it to the end of something okay [Music] so H print the cell does this also need to be self-facing no h let's assume it blocks both directions okay okay North is 1 2 3 4 5 oh you know what I think oh shoot this check here needs to be outside of this Loop okay let's see 11 all right getting warmer okay so it should be 10 why does it think it's 11 it for some reason it thinks 0 should be inside there but it definitely should not be inside there what okay so let's take out some of this print business okay three why does it think z0 should be in there so one two three oh because it thinks yeah so right all right so in this case in this specific puzzle input because s is a is equal to a s we only want it to be on south facing blockers so let's run it on our actual input here okay so our actual input let's look and try to find the start so 2631 so I actually did this for for my input I just found the S so I went here and in Vim I said 26 J 31 oops 31 l and I found my S right here and then I said okay how is this thing connected all right it looks like it can connect to the seven above it right it can go to the North or it can go to the South so s in the case of my actual input is a pipe which means it should be considered in both of these this is not sustainable or valid right we should probably build a way to figure out what are the valid paths out and if if it's if you can go up or down then replace it with a pipe if you can go left or right replace it with yeah just got to replace it with the correct connector but if we run it like this we get back 477 which was the puzzle answer for part two so this is super messy but I thought this is cool to look at right like you got your big old blob of a map and somewhere in here there's a bunch of inside spots that are filled in so that my friends is how you solve part 10 quite messy but I did learn a ton about this point in polygon raycasting algorithm so that was neat yeah this was tricky with the whole North and South blocker bits and I think I was I had an off by two error for a very long time and it was related to where one of these s's went so it is a very I don't know it's like a little bit tricky to get it just right and to understand like how the pipes are all connecting I've got a couple different drawings here of all tons of different loops and if you start inside and you count how many north south exits you end up with how to yeah figure out this bit so hopefully that was useful thank you so much for watching and we'll see you in the next one cheers

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