---
title: combination, filter_map - Day 11 - Advent of Code
slug: combination-filter_map-day-11-advent-of-code
published_at: 2023-12-11 21:00:09 +0000
updated_at: 2026-03-04 20:15:21 +0000
summary: 
description: Join me as we traverse the cosmos while solving Day 11 of Advent of Code 2023 - Cosmic Expansion - in Ruby!   In this coding galaxy quest, we'll: Parse an image grid of galaxies and empty space Expand the grid by inserting extra rows and columns Find all galaxy locations and calculate distances Generate combinations of galaxies to compare Refactor distance to handle cosmic doubling Scale empty spaces to multiply distances   The solution handles parsing the image input and initially expanding it. We then pivot to a calculation-based approach without materializing massive arrays. Some elegant Ruby built in methods like combination and inject help compare galaxy pairs.  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: 254
author: CJ Avilla
url: https://www.cjav.dev/videos/combination-filter_map-day-11-advent-of-code
youtube_url: https://www.youtube.com/watch?v=Oh5tAPD4adQ
youtube_id: Oh5tAPD4adQ
embed_url: https://www.youtube.com/embed/Oh5tAPD4adQ
thumbnail_url: https://i.ytimg.com/vi/Oh5tAPD4adQ/hqdefault.jpg
type: video
---

# combination, filter_map - Day 11 - Advent of Code

*Published: December 11, 2023*
*Views: 254*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=Oh5tAPD4adQ)

[![combination, filter_map - Day 11 - Advent of Code](https://i.ytimg.com/vi/Oh5tAPD4adQ/hqdefault.jpg)](https://www.youtube.com/watch?v=Oh5tAPD4adQ)

## Description

Join me as we traverse the cosmos while solving Day 11 of Advent of Code 2023 - Cosmic Expansion - in Ruby!


In this coding galaxy quest, we'll:
Parse an image grid of galaxies and empty space
Expand the grid by inserting extra rows and columns
Find all galaxy locations and calculate distances
Generate combinations of galaxies to compare
Refactor distance to handle cosmic doubling
Scale empty spaces to multiply distances


The solution handles parsing the image input and initially expanding it. We then pivot to a calculation-based approach without materializing massive arrays. Some elegant Ruby built in methods like combination and inject help compare galaxy pairs.

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 11 in the Advent of code for 2023 this one is called Cosmic expansion have you ever been to a planetarium or an observatory and gotten to lay down and pretend that you're flying out into the universe the Boston Museum of Science in Massachusetts has this really amazing experience where can go to the Planetarium and one of the shows is they'll kind of like Get take you on a tour through the universe and you can see all these different galaxies and how they're spread out and what's really cool is that our exercise today is about space and galaxies and Cosmic expansion and so we are given some input and our input is an image that contains some galaxies out in the distance so each of these pound signs represents a Galaxy and each dot rep represents just empty space and what we're going to do is we're going to start with this input here and we need to find the shortest path between every pair of galaxies so we're going to go from this galaxy to this one and this one to this one and this one to this one and so on and so forth each unique pair but part of the challenge is that any row or column in our image actually needs to be expanded and every single time that we find a row or a column that is all dots that represents a space between galaxies that's twice as big as the result of cosmic expansion so when we took the image Cosmic expansion happened and it's actually twice as big as that so we need to take that input and explode it out where there's spaces we need to make it bigger so this is what we would get from this input so taking this first input that has maybe 10 across and we're going to end up with this output that has 13 across or something so each of these arrows or these carrots represents a row or a column that needs to be expanded then once we have the expanded universe we can label each of the galaxies and we want to find for every single pair what is the shortest distance between galaxies and the distance between galaxies is just going to use up down left and right so if you have a Keen Eye this is just like the rise and run or whatever the number of times that we go to the right or the number of times that we go down is going to be the same if we if we go all the way across here and then down or all the way down and then all the way across we're going to end up with the same number of steps in our path what's really nice about that is we can use this thing called The Manhattan distance which is just the absolute value of this location's x minus this location's X plus the absolute value of this location y minus this location Y and then you end up with the number of steps that are here let's jump in and grab this input we'll crack open main. RB and get to work first we want to break out input into a grid so we're going to say data is equal to input. each line. map Chomp and then we'll say grid is data. map uh line two chars sure P grid and that's going to give us an output that looks like this okay so the first thing we want to do is find which rows and which columns are empty so we're going to say empty rows is grid. map with index row and x and we're going to say um if the row is all periods then we want to return X now if we do this we end up with some Nils right P empty rows okay empty [Music] rows did I spell it wrong what's going on here so here our empty Row's output is nil nil3 nil nil nil 7 I know what you're thinking and I know what GitHub co-pilot is thinking do compact that'll remove the Nils but what I wanted to show you was this method in Ruby called filter map which is just exactly for this purpose filter map will only like will just exclude the Nils for you automatically which is so handy also did you know that you could pop with index onto the end of these innumerables and then you just get an index for free amazing okay so that's our empty rows now let's find our empty columns is going to be basically the same thing but we're going to transpose the grid and filter map again and with index and if all the call colums are dots then we should end up with some empty columns so we'll we'll print out empty rows and empty columns just to make sure that we have something that works here so 3 S should be our empty rows so three and seven are our empty rows and then it should be like two six and 9ish or something like that two five and eight that's roughly correct okay now what we want to do is we want to go through our grid and expand the grid if we want to start expanding the number of rows or the number of columns we don't necessarily want to just iterate in order from top to bottom because then our index is going to shift so for instance if we start with this number three and we insert a row below it then our index of our last one gets moved forward right so if we're iterating over three and seven and we want to insert a new row here um if we insert a row after three then seven is no longer the correct pointer at the next row that needs to be expanded so the way I solve this is just to go backwards so if we start with row seven and we pop in a row after seven and then we go to three and we pop a row in after three then we should be good so we're just going to go through the rows and columns the empty rows and empty columns backwards and that will tell us where we need to insert a row so now we want to say grid. insert at row some new empty row so array. new that is the same length as the first grid and then we want to do the same thing for columns so empty calls. reverse. each do column and then we're going to say grid. insert uh column and that I think actually I think this is opposite this one we want the length of we want the length of the column to be equal to oh no that that was right okay yeah the length of the column should be equal to the length of the thing and this one should be the length of any yeah any given row okay so if we if we go through the grid and we print out row. jooin what does it look like now oh this looks funky okay this is wrong okay so maybe this was right am I backwards here there we go now we have a square but it has this extra dangly guy right here what is the deal with that that seems wrong okay what is going on row. reverse so P row and let's see row is this one and oh grid dot okay so then we need to actually go over right we need to go over each of the rows and for each row we need to insert into the column that is missing okay all right there we go all right let's just do some back of the napkin double checking here to see if this looks like the same as what we have in the example so I'm going to just search for that okay so this is when the rows are twice as big that looks correct and that matches that okay and then the bottom row should be correctly this okay great so now we have an expanded grid so we've expanded the rows and columns to be twice as big cool now what we want to do is find all the galaxies so where's all the galaxies on the list so comment out our Printing and we want to find some galaxies so galaxies are equal to we want to go through the grid and we want to find the cells where the cell is equal to a pound sign and if that is if it is a pound sign we're going to keep track of the X and Y and now if we print out galaxies we should get some list of galaxies 04 1 920 etc etc etc and that looks pretty good we want to grab each pair of galaxies and figure out the shortest distance between those two galaxies there is a method in Ruby that you can call on array so consider an array like 1 2 3 4 we can say a do combination of two and this will give us in enumerator if we do 2 a this will give us all of the combinations of two elements from this array without repeating a combination so this gives us it's missing any duplicates so 1: 2 1: 3 1: 4 2: 3 2: 4 and 3: 4 and and then this is so this is all the different combinations of all the different elements that are here we can do the same thing with our pairs of Galaxy or our our galaxies so we want to do galaxies do combination of do each do X and Y and what I want to do is build a dictionary that looks like this where we have some Galaxy position so 0 0 is the from Galaxy and then 1 one is the two Galaxy and then the distance would be one and then what we can do is we can go through and figure out what are all the distances between all these galaxies we can add up the values of this dictionary and that'll at least kind of like let us keep track of the distances between each one so if we have these combinations of two um then uh and we map over them we can get back yeah let's actually keep these combined as like a Galaxy a and Galaxy B and then we can make our first element of this array a comma B and the second element of the array could be distance of a to B okay and for now we'll just make a method called distance that takes in A and B and returns one just so that we can start to get an idea of what this actually looks like so we'll say our result is is some P result and we'll run this okay so 0419 goes to one 0420 goes to one okay great so now we actually have to implement this this shortest distance method okay so we're going to take in A and B for now we can just implement the shortest distance as being like ax a y is equal to a and then BX b y is equal to B so we're destructuring the X and Y locations for for the Galaxy and then what we want to do is take the absolute value of um ax minus BX that's going to give us yeah the absolute value of the x coordinates subtracted the difference of the y-coordinates subtracted we're going to add the absolute values and that should give us the distance so if we run this so from 04 to 1 9 we get six so let's actually print our grid out again and just see from 04 to 1 n it gets six so 1 2 3 4 five six so that looks correct and then let's look back at the example here and it says the this path has length nine because it takes a minimum of nine steps to go from this location to this location so this five is on 0 1 2 3 4 five 6 1 so we're going 61 to 11 something so if we run this again our 61 to 115 is nine so this is the answer in the example now what we can do is add up these distances so we might actually be able to just say inject some and then we'll take A and B and then maybe we don't actually even need a dictionary we can just say sum plus distance of a to B and and uh that didn't seem [Music] to do what we thought it was going to do oh my word 374 so 374 is look at that that is the answer from the example so in this case we're not actually keeping track of that dictionary like I talked about we are just figuring out the distance and printing it out all right like always let's grab our puzzle input pop it here at the bottom and use this okay and then we're going to make our data B dat. read lines. map Chom and run it against our actual input and what do we get we get a whole lot of nothing okay so we don't want to print out the grid anymore and we don't actually need this one what do we get for the result 955 6896 okay let's see if that matches hey that's our answer for part one of the puzzle amazing okay what what have we got going on here we are we're just finding the distance between A and B we're adding that up and getting some result back that we're then printing out so we've got our distances between galaxies we've expanded the galaxies let's head on to part two okay part part two the galaxies are actually much older and thus much further apart than the researcher initially estimated instead of expanding the way we did before now each empty column needs to be 1 million times larger so every empty row needs to be replaced with 1 million empty rows okay that's a lot of empty rows I don't think our expansion algorithm is going to work anymore right we don't want to go through and make a million bym million array and it's going to actually it would actually end up being like the number of empty spaces so 6 million by 6 million or whatever array so we've got to take a different approach here but we want to somehow expand it by a million so in the example above if each row or column was 10 times larger then the answer would be 130 so what we want to do is refactor this so that instead of expanding our grid like instead of doing this empty rows empty columns inserting business we actually don't want to insert right we want to use some other approach where we're going to use empty rows and empty columns in order to figure out our distance so before let's actually try to make it so that our result is equal to what it was before if this is the distance of it of like the galaxies between each other with no expansion then the expansion would be something like d plus e equals the number of rows that are between a uh A and B and plus equals the number of columns that are between empty columns so we need to go through the empty columns and the empty rows that are between A and B how do we do that I think we want to go okay so in this case when we're doing the absolute value of a minus B it doesn't matter whether a or b is larger right but we want to iterate in the correct direction basically so we need to take the minimum of ax and BX and the maximum X Min x max is equal to this I think so let's let's see so if we have 150 min max then we get back 1 and 50 and if we have 50 and one we get 150 okay so minmax is going to give us back this Two element array we want to do the same thing with Y and then we want it iterate from y do y or x x Min up to X Max and we want to say if empty rows include x then we want to D plus equals 1 right so as we're iterating over our X's we want to see if there's any empty rows and if there are we want to increase the Distance by one because remember we're trying to get back to our original answer here and that might need to be um that okay let's just replace all these with X with Y and then this is still X okay and then at the end we want to return D so let's see if we get the same number ah okay so we need to pass in we need to pass in empty rows and empty columns as arguments now to distance so we have those as variables up here so we can just pass them in down when we're calling this distance function and now if we run this we get back 96 27156 which was not the same as our answer it's a little bit higher than our answer so perhaps we need to go up to but in not including 96 okay H oh you know what we didn't change this to columns this shouldn't be empty rows this should be empty calls okay let's see here 955 6896 955 6896 okay so this is the this is like an alternative solution to part one so now what we want to do is instead of increasing it by one let's use this sample input and try to increase it by uh 10 so if it's 10 times bigger then let's just see if we increase by 10 and we run it okay so we get 1,112 but the answer should have been 1,30 and when we're thinking about doubling it went from one we like replaced one with two so now we're going to replace one with 10 and since uh in this scenario we're already counting up the empty row once we need this to be 10 minus 1 right so technically Ally like the original answer was 2 - 1 so now it's 10 - one and we get back 1030 okay and if it was a 100 times larger then it would be 8410 so if we make this 100 we should get 8410 and we do okay so now what we want to do is figure out what our puzzle answer is so we need to make this a million times bigger one really cool thing about Ruby is we can just put these underscores so that our numbers are legible and we run this and that is for the example input so what is it against our input huge number huge number and that is our puzzle answer awesome this was okay so this is this pretty fun little solution here turns out we actually didn't need to do this insert business so we'll remove that we're finding our galaxies we've got a distance method here that seems to work well and it supports Cosmic expansion for both part one and part two technically I guess we could put in a comment here that was like part one would be D+ = 2us 1 just to make it clear that we were doubling in size in that case this combination method is pretty cool you can you can even do like combinations of three and that will give you all the different combos of three that can be really help helpful anytime you want to do combination methods like combination permutation and pro like dot product things like that are really really useful on the array methods in Ruby yeah thanks so much for watching and we'll see you in the next one cheers

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