--- title: Treetop Tree House - Advent of Code Day 8 with Ruby slug: treetop-tree-house-advent-of-code-day-8-with-ruby published_at: 2022-12-08 22:00:02 +0000 updated_at: 2026-03-04 20:15:11 +0000 summary: description: Treetop Tree House - Advent of Code Day 8 with Ruby Challenge: https://adventofcode.com/2022/day/8 Solution: https://gist.github.com/cjavdev/a152b93c3d2328a6b910eedce01c19ee #ruby #adventofcode tags: [cjav_dev, web development tutorials, web development for beginners, vim, ruby, advent of code, advent of code 2022, tree top treehouse, advent of code ruby, advent of code 2022 day 8, advent of code 2022 day 8 ruby, aoc 2022, aoc 2022 ruby, flat_map, advent of code solutions, ruby solutions, ruby tutorial] views: 367 author: CJ Avilla url: https://www.cjav.dev/videos/treetop-tree-house-advent-of-code-day-8-with-ruby youtube_url: https://www.youtube.com/watch?v=3SHQOyUVvgA youtube_id: 3SHQOyUVvgA embed_url: https://www.youtube.com/embed/3SHQOyUVvgA thumbnail_url: https://i.ytimg.com/vi/3SHQOyUVvgA/hqdefault.jpg type: video --- # Treetop Tree House - Advent of Code Day 8 with Ruby *Published: December 08, 2022* *Views: 367* ## Watch [Watch on YouTube](https://www.youtube.com/watch?v=3SHQOyUVvgA) [![Treetop Tree House - Advent of Code Day 8 with Ruby](https://i.ytimg.com/vi/3SHQOyUVvgA/hqdefault.jpg)](https://www.youtube.com/watch?v=3SHQOyUVvgA) ## Description Treetop Tree House - Advent of Code Day 8 with Ruby Challenge: https://adventofcode.com/2022/day/8 Solution: https://gist.github.com/cjavdev/a152b93c3d2328a6b910eedce01c19ee #ruby #adventofcode ## Transcript what's up welcome back in this episode you'll see how to solve day eight of the Advent of code for 2022 with Ruby this one is called Tree Top tree house and this is a pretty fun one I think there was a couple of challenges last year that were similar to this where you get a grid of numbers and then you have to do interesting things with the grid of numbers in this case the elves have a quadrocopter they're flying around some forest and measuring the heights of different trees and we have to figure some things out about the trees so this grid here is going to be our puzzle input and so this input represents the height of trees at different points so right here the tree is three high and then it's zero High then three High seven High three high so that's kind of like how tall the tree is each tree can only be one digit in height what we need to do is first figure out if a tree is visible meaning that like from the top left right or bottom we can see the tree because the other trees in front of it are smaller so right here for this top left five it is visible because if you were to look at the forest from the left you would see a tree of height two and then the next one you would see is a tree of height five we wouldn't be able to see the tree behind it from the left because it's five nor the other trees that are behind those because those are smaller we're talking about maybe looking from the left in this case we would see this middle five if we came from the right though we would not be able to see the four because the nine is just like this really tall tree so it gives us a bunch of test cases for this small grid the top left five is visible the top middle five is visible Etc so what I want to do is go through and determine whether or not a tree is visible and right now this like answer for the first part of the question is just how many of those are visible so what I think what we want to do is just transform this grid into a grid of ones and zeros one if it's visible zero if it's not and we'll just take the sum so the first thing we want to do here is just kind of like iterate over the input so we'll say data is data dot read lines data dot map Chomp and then dot map to get out the characters and that should get us kind of like some numbers right or some like I guess it's going to be string values but that'll give it that'll get us started so Ruby date so when we run this nothing is coming out because and then we'll finally print out those results so if we run these if we run this net right now we're getting the string values of the heights so let's also map each of those and convert them into integers all right so now we have an array of arrays each row in the array represents sort of a line in our grid and we can take this this is kind of like the trees okay so we want to take these trees and the first thing we want to do is like check to see if there are any invisible so let's make a new method called check visible and that's going to take in some trees and maybe it takes let's have this spit out some like result and the result is going to be a new array and the way that we can figure out that array is by mapping over I and J or kind of like the coordinates of the Matrix so what we can do here is say like trees dot length dot times do I and then trees DOT first dot length dot times do J and instead of just dot times we can do dot map I think and we should get back some new array so for now let's just put the number one in there and see what we get so well it's p check visible for trees and see what we get back all right so now we have just a bunch of ones so now what we want to do is maybe we'll change this to visibility and then we'll make another method check visible that takes in the trees but also I and J so we're going to check the visibility of a specific tree and we'll just have this map over and return the visibility of a specific tree so again the if the trees are at the edges then they're definitely visible because from the left you can see the first tree it's right at the edge so we want to return one if I is 0 or J is 0 or I or Z or I or J are kind of like at the boundaries okay otherwise we want to return false at the bottom so let's see what this gives us uh okay check oh right this should now be visibility okay run that all right so now we can see like the top row and the bottom row are all ones and then the rest are just showing us like okay yeah we're properly getting the outside now we need to do a part of this where we are going to figure out whether or not um a tree is actually visible that's in the inside and so for this we want to say like something like return one if the um if the height at a certain point so let's get the height the height is going to be tree trees at i j that's the value of the height and let's also grab the row so row is going to be trees of I the column is going to be trees dot transpose and then give us J so let's look at transpose for just a second you can take any uh 2D array so let's say that we start with one two three four right and you can call a DOT transpose we can call transpose on that and that will like swap the rows and columns so you'll see that the rows of a are one and two across the top and then three and four across the bottom and then when we call transpose that transposes the Matrix so that now we have one and three which was the column and then two and four which was the second column so transpose can come in handy anytime we're working with these grids and we want to say something like if the height is greater than the max of rho from zero up to but not including J and then we want to return one if the height of this specific tree is greater than like J plus one to the end and then we also want to return it if the column is greater than zero to I and then the height is greater than I plus 1 to negative 1. I think that's right I don't know let's see let's run this okay I don't think this is working as expected so what I want to do now is pull in our spec and just write some quick tests so we're going to say if our V Dot empty require our spec auto run and then we're going to say our spec.describe it works for the example case so expect visibility of trees to equal and then here we're going to like we'll figure out kind of what we need to actually write out okay so from the example instructions here so the top left five in the middle that's visible so this should be one and then the top middle five is visible that should be one and then the top right one is not visible the top left or the left middle five is visible and the center three is not visible from any direction but the right middle three is visible from the right and then the bottom row the middle five is visible so then this one should be one okay so let's see if that works as expected and it does not we're expecting that this one is visible but it's not so let's go look at our let's go look at our math here so the max oh you know what maybe that needs to be three three dots right up to but not including J okay getting closer getting closer is our test correct here oh no it's not okay so this one should be visible I got that one wrong okay so here right 5 is visible because you can see it from the left but not the right and three is not visible because this 3 is bigger than it so I had the test wrong okay let's test this out boom okay we have a passing test and now what I want to do is let's actually convert this yeah so like we have this working test we're just going to assume that this continues working and then the answer is just going to be something like P visibility of trees dot flatten dot sum and that should give us some number 21 and that matches the example use case so 21 trees are visible and then if we grab our test input which is this giant thing and we're going to open up day 8 input drop that oh drop that in and then here instead of data.redlines we'll just we'll say data is file.readlines of arcv.0 so that we read it in from the uh from our input and we run it and we get 1719 which is our puzzle answer for part one all right hooray nice okay part two content with the amount of tree cover so now in part two what we want to do is try to figure out which tree has the best scenic view and the way that you define a scenic view is how far you can see or how many trees you can see basically the way that you define a scenic view is you find a given tree and then from that tree you look outwards and you try to figure out um how many trees you can see so from this tree here this number five if we look to the left we only see one tree we see a five and if we look up we see one tree if we look to the right we see one two trees and if we look down we're gonna see one two and so the way that you figure out the score is you multiply the scores for each Direction so one times one times two times two and for this specific five we get back the value four and then if we wanted to look at the bottom five here this one is actually just a little bit better because we can see two trees to the right two trees to the left and then two trees up and one tree down so we end up with two times two times one times two which is eight and so that is the scenic value for this sort of this five here so let's go back into our thing here let's write like another little test so it like it works on the scenic stuff I don't know uh okay and so here what we want to do is something like expect scenic score for a given tree so we'll say trees of um Row one so zero one and then column two we expect the scenic score for that to be four so let's change this to four and then we expect the scenic score for zero one two three Row three and column zero one two again to be eight and let's just see if we can get this some like Scenic score method working so Scenic score is going to take in some trees and I and J and this is going to be interesting okay so what we want to do is figure out how many trees from that specific point going outwards so if we're in the middle and we need to go to the left we sort of need to start at this i j position and then work our way to the left until either we hit the edge or we hit a tree that is the same size or taller than us and then we need to stop and we'll include that tree so we need to start from J minus one down to zero dot each do K and then for for that we need to also get like I guess the row again and the column again and in this case we're going across the row backwards and I guess should this be I should be J it should be J okay so then we're gonna we need to count up how many scores there are so the score in this case is going to start at zero and then we're going to say score plus equals one for each tree that we encounter and then we want to break if row at K is greater than the height of the current tree um so height again is going to be trees at IJ and is it I think it's greater than or equal to we're going to break and then at the end of that we want to like collect up the scores into some list so say scores or shovel in the score and then at the very end we want to do like we want to get the multiplication uh applied across all the scores so we'll say scores dot inject times and what's I don't know if there's a better way to do this but I think we might need to do kind of like the same thing for all the different directions right so this is to the left let's do it like uh to the right which is going to be something like we got to reset the score down to zero and then we need to go from J plus one up to length from J plus 1 to trees Dot first dot length dot each and then if the row at K is greater than or equal to the height we add that up okay so for now let's just like P scores and see what we get so we'll just run this again and undefined for greater than or okay so this is probably up two but not including okay expected four got two so here we go we have our scores one and two so is that right for this one so to the left there's one and to the right there's two okay I think that's I don't know it's directionally correct and then yeah above is gonna be something like score is equal to zero and then I minus one yep that's the row above all the way up to yep okay and then we're gonna do score plus one break if the column is greater than or equal to height yep and then uh below and this is where GitHub copilot becomes like pretty good right because it knows now the pattern that we're trying to follow and I mean it's not always it's not always great right but in this case okay so I plus one is the row below us and we're going down to the final row and then we we want to increment our score and then after each of these we want to do scores shovel in the score and let's see what we get now all right so we got two one two one and we okay so we actually have a passing test so we can remove this print scores situation what do we want from the end what is the highest Scenic score possible for any tree so we want to map over all of these again so I guess like visibility and then like I don't know scenery deaf I we could probably just comment this out and say Scenic score and now our visibility becomes visibility dot flatten dot Max and [Music] okay uh our test failed but we got eight okay so then if we run this with our input day eight input okay five hundred and ninety thousand eight twenty four so that is the answer for part two okay so what we're doing now is we have this Scenic score thing which honestly if you have a better way to do this like this is not this is not clever at all so please drop a link to a gist down in the description below with your preferred approach to growing out in different directions in the grid so you don't have to write this out manually for all different uh directions because that's kind of annoying so um yeah show me how to do that better uh and then this visibility thing this is like really handy because we're just mapping over um one thing that is interesting is that down here both for the first case and the second case we called dot flatten so instead of calling flatten there we can use a method inside of our visibility called flat map because after we've like figured out the value that belongs there we actually don't care whether the array is flat or not and so by doing this now we should get just we just get like a single array and then we can just call Max on it right and so let's do this without the input so it's a little bit easier to read okay so the output here is just like one like single array right it's flattened out um so flat map can come in handy I don't think we need to actually flat map this one because I think the inner elements are all already yeah the inner elements are already flat arrays so we just call flat map on the top and that is going to look at all of the elements that are yielded to the block and it will call flatten on it after each one so that we end up with just this single thing here so if we don't need to preserve if we don't need to preserve the sort of the Grid or locations after we call something in our case we're either calling sum or we're calling or we're calling Max we only need the values that are in the array we don't actually need the entire array so flat map that can come in handy when you're talking about this stuff also I don't know if you've seen this like dot times.map thing but that's kind of like an interesting way to create another array of arrays yeah I don't know hopefully this was useful thanks so much for watching and we'll see in the next one cheers --- [Back to all videos](https://www.cjav.dev/videos) --- ## Hey Agent, thanks for visiting! 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