---
title: transpose, reverse, roll! - Day 14 - Advent of Code 2023
slug: transpose-reverse-roll-day-14-advent-of-code-2023
published_at: 2023-12-14 15:45:01 +0000
updated_at: 2026-03-04 20:15:20 +0000
summary: 
description: In this video, you’ll see how to solve Day 14 for the Advent of Code 2023 Part 1 and Part 2. The puzzle involves manipulating a grid representing a panel with rocks on it. The goal is to tilt the panel north, south, east and west to get the round rocks to roll around.   In Part 1, we’ll write a \"roll\" method to move the round rocks on a single row all the way to the left by swapping spaces. Then we’ll use \"transpose\" to tilt the whole grid north and calculate the \"weight\" or load on the north beam.   In Part 2, the puzzle introduces a \"spin cycle\" which tilts the panel in four directions repeatedly. We’ll write methods for tilting each direction. Then detect cycles in the grid states using a dictionary and set to count unique patterns. By moduloing the number of cycles by the cycle length, we’re able to index into the grid states and get the weight after 1 billion cycles.   The video demonstrates Ruby array manipulation, recursion, cycles and modulo to solve the Advent of Code puzzle.   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: 170
author: CJ Avilla
url: https://www.cjav.dev/videos/transpose-reverse-roll-day-14-advent-of-code-2023
youtube_url: https://www.youtube.com/watch?v=NtAEhk-QuLA
youtube_id: NtAEhk-QuLA
embed_url: https://www.youtube.com/embed/NtAEhk-QuLA
thumbnail_url: https://i.ytimg.com/vi/NtAEhk-QuLA/hqdefault.jpg
type: video
---

# transpose, reverse, roll! - Day 14 - Advent of Code 2023

*Published: December 14, 2023*
*Views: 170*

## Watch

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

[![transpose, reverse, roll! - Day 14 - Advent of Code 2023](https://i.ytimg.com/vi/NtAEhk-QuLA/hqdefault.jpg)](https://www.youtube.com/watch?v=NtAEhk-QuLA)

## Description

In this video, you’ll see how to solve Day 14 for the Advent of Code 2023 Part 1 and Part 2. The puzzle involves manipulating a grid representing a panel with rocks on it. The goal is to tilt the panel north, south, east and west to get the round rocks to roll around.


In Part 1, we’ll write a "roll" method to move the round rocks on a single row all the way to the left by swapping spaces. Then we’ll use "transpose" to tilt the whole grid north and calculate the "weight" or load on the north beam.


In Part 2, the puzzle introduces a "spin cycle" which tilts the panel in four directions repeatedly. We’ll write methods for tilting each direction. Then detect cycles in the grid states using a dictionary and set to count unique patterns. By moduloing the number of cycles by the cycle length, we’re able to index into the grid states and get the weight after 1 billion cycles.


The video demonstrates Ruby array manipulation, recursion, cycles and modulo to solve the Advent of Code puzzle.


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 14 for the Advent of code this one is called parabolic reflector dish we're still making our way trying to figure out how we can fix the environment with these elves and now we find ourselves in a pile of rocks or we have these panels that have a bunch of rocks on them and we can control if we're going to tilt the panel forward or back or left or right and our puzzle input is the status of the current rocks that are on the table or on the panel here the pound signs represent Cube shaped rocks and the O's represent round rocks and we have a controller that can tilt the table north south east or west when we tilt it North all of the round rocks will roll up to an edge or up to another rock or up to a cube-shaped rock but it will not roll past a cube Cube shaped Rock So by tilting it North forever then we end up with a result that looks like this so what we need to do is figure out the total load on the North support beams which is calculated based on how close the round rocks are to the north support beam so if it's right on the beam that's a 10 if it's one away that's 9 8 76 so it's like the distance from the beam tells us how much load is on it what we want to do is is grab our puzzle input and we're going to make like a one of those peg board type situations where when you roll a bunch of stuff along the board they may or may not get stuck on the on the pegs so let's grab this all right let's drop this in here we're going to say input is equal to this uh set of input here and our grid is going to be just an array of arrays with each of those elements in it so here we can say input. each line map Chomp and we'll grab out the characters and we'll pee the grid just so that we know what we're working with and if we look at main. RB this is what we get back okay now what we want to do is roll all of the rocks in part one we want to roll them all North and then calculate the sum caused by all of the rounded rocks so what I thought about was like we want to treat each column um individually and then ideally we can roll all the rocks in a certain direction and if I if we look at the Grid at zero. join then this gives us like this first the first row at the top and uh technically we want to work on the First Column instead of the first row so if we do grid. transpose that will give us the First Column we have two round rocks and then a space and then a round rock and then a space and then a round rock and then two spaces and then a pound sign and my first thought was let's treat each of these columns individually and try to get all the round rocks rolled all the way to the left so let's do that first is we'll just create a method here called roll or something and then this will take in a line or a row of rocks and what we can do is say something like row. length minus one do times do I so we want to look at each pair of BU rocks if we have a DOT on the left we want to replace that with an O because we want the rock to roll into place so uh if we see a DOT on the left and a round rock on the right then we want to swap them so we'll say row at I + 1 is equal to um row at I + one like this and I think now if we run roll of this we should see something this wants to return row okay let's just start with our row here and we'll P row do join and then we'll try one roll on the row and see what happens okay so we had one move closer and another one move closer but we still have the space in between so what we need to do is say some check if swap or swap here is going to be false and then swap is equal to true and at the end if we swapped we want to recursively call on it again otherwise we'll return the row okay so now we have completely rolled all of our round rocks to the left that is great let's try it with another row here okay that looks like it's working well and we'll try it with two oh fantastic okay so it even stopped right the rolling rocks stopped at the pound sign so I think this is working as we expect to roll a certain row now let's write a method called tilt that's going to tilt the whole grid and yeah we'll just for now we'll say map over this and roll roll each of the rows let's actually go grid. each do row and we'll put row. jooin and then we'll print out a a line and then we'll tilt the whole grid and maybe we'll just say grid is equal to the the Tilted the whole tilted grid and then we'll do the same print statement and let's take a look okay so we have our whole grid and then we tilted and they all slid to the left right we had all of the the Rocks they all rolled to the West they all rolled to the West um so technically what we want to do is we want to tilt uh North and so the way that we can do that is by transposing the grid and then transposing back let's see if we get the right answer here nice okay so now we are rolling up to the top um so that's like roll North basically so technically if we pass in a grid that is transposed tilt it and then transpose it back we get back uh the concept of rolling north so now we're able to tilt to the north and the next part of the problem for part one is we want to calculate the total uh load on the North support platform so we need to go through each of our rows and sum up the impact of a round rock being at the top so let's make a new method here called weight that's going to take in the grid and we'll have to go grid uh each with index do row and then row. each with index do cell and if the cell is a zero then we want some answer to be incremented so here we can say answer plus equals and the amount that we're increasing by is the length of the grid minus I because we're if we're at the Top If we're at the zeroth point this we want this value to be 10 which which is going to be the length of the grid minus 0 since there's 10 elements here so this should be grid. length minus I and I think that might give us a weight let's get our tilted grid and let's put print out the weight for the new grid okay we run this and we get back 136 which is the same answer as the part one example so now we'll grab our own puzzle input here here wow that is a big grid okay so we'll put that at the end as always and then here instead of input we will we'll make input be data. read and run this again we got this 11 3424 what does that look like that is the puzzle answer for part one okay awesome so we are now able to calculate the weight on the Northern support beams when we've tilted Ed North one time part two the parabolic reflector dish deforms but name may not be uh not able to focus the beam all right so for part two the idea is that uh we can click another button in this control panel called spin cycle and instead of just tilting North it will tilt North then West then South then East and that is one cycle and after the end of one cycle of the example input it goes from it will end up like looking like this so what we need to do the first thing we want to do is figure out how to make this cycle method and then at the end the question here is after a billion cycles or something what is the total load on the North support so we're going to have to figure out how to make that work with less than a billion right but for now we can just create a new method called cycle let's actually move weight up here that will not change our concept of what a weight is and yeah let's make a new method here called like North and that's going to take in the grid and right now we already have this North method figured out right it's tilt tilt the grid. transpose. transpose that that will that will shoot us North and we can keep that as part of our part one input and maybe we can say part one is that and let's actually comment this out for now comment out part one down here okay so we want to go something like this North and then look what it yeah see what it looks like and let's also now we can because we we have a method that will tilt North we can remove the transpose but I do want to um make sure that uh we're not modifying the underlying grid so let's yeah let's create a new North grid and then we'll go through each of those and print them out so if we look at our puzzle input for the example uh we can see North and then we want to do that for each of these cardinal directions so we'll have north south east and west and then we'll have each of these north south east and west okay but if we run it they're not actually going in different directions right our methods for south east and west are going to be they're all going to be a little bit different in order to go East I think that's actually the default so if we just run tilt on the entire grid I believe that should get us east east oh no that's okay the default is West let's run it against West okay West is working east the in order to get a single row to tilt East we want to roll we want to roll each of the rows in Reverse so we actually want to like map um we want to map over each of the rows and call reverse on it pass that into tilt and then whatever we get back we'll uh re reverse that we're going to rewind it so that's going to get us East so now East we're rolling to the right and then South in order to go south what we want to do is we want to reverse the entire grid I think let's see South nope that didn't work we might need to also transpose and then transpose back and then reverse back and cool that gets us South so now we have north south east and west let's let's make a cycle method and that's going to take in a grid and the cycle is in a particular order here so we need to go north then West then South then East North West Southeast North and then West then South then East okay that's it looks funny all right so that should be a single cycle so if we run put cycle and we grab out our new um grid and then we print it out what does that look like okay so after one cycle it looks like this let's compare that against the input here so cycle should have yep a pound sign at the top we have this pattern of three at the bottom and oh we're missing we're missing one of these right so after one cycle we expect this to have four elements in it so we must are we like losing are we trimming something off somewhere all right let's break these up so that we can take a look grid is equal to north of grid grid is equal to South grid East North West South then East all right and we're going to dup it just so that we have a duplicate copy I am not convinced that this after cycle one is correct because it seems like our input is actually smaller or something I wonder if we have a different yeah maybe we have a different input what's going on how could there be hm so after one cycle it should look like this let's look again here oh you know what the grid that we're passing in was already modified maybe let's see yeah let's make this so that it always dupes the [Music] r okay there we go yeah it it keeps over it was overriding something that was being passed being passed down okay so now we have a cycle one that matches what what it looks like in the in the instructions now let's run it a couple of different Cycles so maybe we do I don't know 3 * do um n and then we'll print out oh this is cycle n and we'll keep going okay so after cycle two or after yeah so after two cycles here it should look like this which lines up with what we're seeing in the example and then after three Cycles it should look like this which it does so the process should work if you leave it running long enough but you're worried about the north support beams so if we if we run this for 300 times let's just take a look and see you'll notice that cycle 298 and 299 are actually different so 298 has this group of three round rocks here in the corner and on this one it has them all at the bottom they're not necessarily going to just settle right so if you look really closely at this they're not settling at all when I initially was looking at this I thought oh we can just instead of running it a million times we'll just run it until it settles and then we'll count the weight but instead we this whole cycle situation is our hint that we have to look for Cycles in our Loop every time we cycle we are going to get back a grid that is somewhat unique right and the thought is let's go through a cycle go through a cycle go through a cycle and just look at the data here so let's print out the weight of the grid and uh the current cycle that we're on basically so that yeah this n is going to be representing the number like the cycle that we're on we'll comment all these out and this one and okay so for cycle 299 our weight is 64 for 298 it's 65 297 it's 69 69 68 63 65 64 again so then we have 65 let's actually just take out the printing of the grid so that we can see just the weights and the cycle numbers if we look at the beginning when we first started running through these weights like the first couple Cycles it's not actually impact or we we yeah for the first at the end of the first cycle nothing has really settled there's still a bunch of weight on the North beam but then after one more cycle we're getting a lower number here 69 then after two cycles we're we're getting lower lower after four Cycles we're at 65 at five Cycles we're at 64 7 we get to 63 and then ultimately it like settles down right between 6 69 and 64 and somewhere in this process there is a loop it's going to go at the beginning it starts off traveling along some line of states that we are only going to actually encounter one time in the very beginning and then we enter a loop where we're going to see the same States over and over and over and what we want to do is kind of like detect the cycle and the length of the cycle and we also need to detect how long was that first section before we got to the loop the way that I thought about doing this was I'm going to serialize the grid into a key that we can use into a counter dictionary and we'll just count how many times we've seen that same grid pattern and that'll let us see we'll count up the ones that we only saw one time and then we'll for the ones that were more than that we can figure out what the cycle length is let's see here so we want to have some pattern counter thing that's going to be one of these counter cache dictionary things okay and what we want to say is pattern counter at grid plus equals 1 but instead of pointing directly at the grid I want to serialize that into a string so that when later on I'm duping this grid because it's going to be a pointer to an array that has like different subarrays and things I want I want this to just always work and I think I can say so if we have an array of one two three and we do a.join yeah okay so that should be fine so grid. jooin can be our key into the pointer so we'll say grid. join here is our key and then we also maybe want another visited and this can be a set and this will just be like if visited it or not and we want to know how many unique patterns there are so this we can do visited and we'll shovel in the key and we should just get like a unique count so now at the end here if we print out pattern counter it's going to be gnarly because our keys are going to be huge but maybe uh dot values we'll just look at the values for a second and then we'll also print out the size of visited and see what we get here okay all right so put pattern counter and we'll put visited. length so the total visited like the total unique patterns here was nine and then our pattern counter values we got one one and then 43 43 43 43 42 42 yeah I think what we can do is remove the values that are one that we only saw one time and that will be kind of like the start of the or like the length of the offset so we can actually probably just do count ones and this is going to give us our offset and then our visited length is that and then our cycle length is or like the length of the loop is going to be visited minus offset and then yeah so let's print all this stuff out and just see what we get okay so this is telling us that our offset is two and our cycle length is seven for the example input and then what we want to do is we want to figure out what the weight is after the spin cycle runs for 1 billion cycles Run cycles is going to be a billion so in order to figure out our offset we want to take the Run cycles minus the offset so Run cycles minus the offset and that's going to give us like the starting point of the first or that's going to give us the number of Cycles where we're inside the loop and this is going to be some big number it's not going to be yeah okay so it's a billion minus the offset which is two so it's still a massive number now what we want to do is take that number in modulo with the cycle length and that is going to give us four now we should be able to use four to index into our to index in and grab out the weight but we don't have anything that's really actually tracking the weight we're just printing it out right now so let's do another thing weights which is going to be a dictionary here and we'll have our weights at n be equal to the weight of the grid and really we only care about the first few and after that it doesn't really matter so now this is going to be like our index and now we should be able to say puts the answer is um weights at index and let's see what we get back answer is 65 okay so we have we have an off by one somewhere I think because this is supposed to be 64 okay so the Run cycles minus the offset technically if we want to run a billion cycles and the the loop like the start of the loop is two we also I think need a set for the initial like the first run through this cycle because we're we're we're executing the cycle on the grid before we start counting so let's add another one here and I think oh 69 okay so plus one now we're just taking stabs here all right we get 64 which is the answer for the example input let's try it against our own input I know we don't want to just mess around there but should work okay all right let's grab our grid and run it it's taking a second uh oh okay Ruby main. RB all right it's cooking it's cooking so we are running through a bunch of Cycles do we actually need to run through all those Cycles probably not so let's open this up and let's decrease our cycle length to 200 and just see if we can get away with something that let's also run it this okay we finished 99 sort of test Cycles it says our offset is 102 and our cycle length is nine again so we get 102 569 and our oh our puzzle answer was 9603 so something is wrong something is wrong here let's take a look what is going on so let's take a closer look at the value values of this counter the pattern counter so let's let's like take this from the very beginning so we'll P pattern counter. values and let's use the example input to start just so we can get a better idea of what's going on here so we have we've encountered this first and second pattern twice and then we have 1 two 3 four five six seven is our cycle length and two is our offset I thought that maybe we can actually do partition or yeah let's just do count yeah if we count up one or maybe it needs to be like yeah count ones that should give us the number and then our cycle length should be pattern counter. values. count where it's greater than one is that the same as what we had before so cycle is two offset is or cycle is seven offset is okay and then we have our Run cycles and the index into weights is the number of Run cycles minus the offset so that's minus two plus one so that' be 99 actually this this must be minus this has got to be minus okay and if we run that we get 69 which is not the right answer for this one I am starting to think that our key into our weights is wrong or something do we need to dupe this again shouldn't have to 69 okay if our offset is two and our cycle length is seven minus 2 modulo 7 is four and our pattern values P pattern or weights at n or actually just P the weights so 0 is 87 1 is 69 2 is 69 3 4 5 is 6 for oh okay so we need to index at the offset plus the index right because yeah we don't want to we don't want to use three as our index we want to use five as our index and in order to get from three to five we need to skip over the offset so the offset was the first two and then yeah that makes sense okay all right so let's let's try this against our actual input and all right 96003 and that is our puzzle input for part two awesome okay that was another tough one I don't think we actually need this visited thing anymore I think we can get away without it and then I'm also not sure yeah I feel like we do still need this different um key into weights we want to collect up the weights just as we are we could make this more efficient instead of just blindly going 200 times we could say oh as soon as pattern counter at key is greater than yeah so I guess we could say break if pattern counter at key is greater than two or something just to be be safe and that should go much faster I think maybe okay it's a little faster we only ended up going through 120 cycles that time and we still got the correct answer at the end so that is cool for more videos about Ruby uh Stick Around by hitting that subscribe button thank you so much for watching really appreciate your time and attention and we'll see you in the next one cheers

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