--- title: Spaceship operator - Day 07 - Advent of Code 2023 slug: spaceship-operator-day-07-advent-of-code-2023 published_at: 2023-12-07 18:00:29 +0000 updated_at: 2026-03-04 20:15:17 +0000 summary: description: This video tackles the \"Camel Cards\" poker puzzle. We’ll parse the input data into Hand objects, modeling ranks, and bids. We can sort hands by type and strength by including the Comparable module and implementing spaceship logic. For part 1, we’ll build a card counter and hand typing system, assigning numeric ranks. This lets us multiply bids by position to get total winnings. For part 2's joker rules, we’ll tweak the ranks and add joker offset logic to promote hands. The core solution focuses on custom classes and comparable spaceships for sorting. I polish with enumerable pipelines for a concise answer. While not golf code, we end up with readable Ruby leveraging objects, sorting, and clever tricks. 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, comparable, spaceship, space ship, space ship operator, rocket ship operator, rocket ship] views: 291 author: CJ Avilla url: https://www.cjav.dev/videos/spaceship-operator-day-07-advent-of-code-2023 youtube_url: https://www.youtube.com/watch?v=bASPI9cV7R4 youtube_id: bASPI9cV7R4 embed_url: https://www.youtube.com/embed/bASPI9cV7R4 thumbnail_url: https://i.ytimg.com/vi/bASPI9cV7R4/hqdefault.jpg type: video --- # Spaceship operator - Day 07 - Advent of Code 2023 *Published: December 07, 2023* *Views: 291* ## Watch [Watch on YouTube](https://www.youtube.com/watch?v=bASPI9cV7R4) [![Spaceship operator - Day 07 - Advent of Code 2023](https://i.ytimg.com/vi/bASPI9cV7R4/hqdefault.jpg)](https://www.youtube.com/watch?v=bASPI9cV7R4) ## Description This video tackles the "Camel Cards" poker puzzle. We’ll parse the input data into Hand objects, modeling ranks, and bids. We can sort hands by type and strength by including the Comparable module and implementing spaceship logic. For part 1, we’ll build a card counter and hand typing system, assigning numeric ranks. This lets us multiply bids by position to get total winnings. For part 2's joker rules, we’ll tweak the ranks and add joker offset logic to promote hands. The core solution focuses on custom classes and comparable spaceships for sorting. I polish with enumerable pipelines for a concise answer. While not golf code, we end up with readable Ruby leveraging objects, sorting, and clever tricks. 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 seven for the Advent of code in 2023 this one's called camel cards it is a simple version of Poker so we've taken an Airship up we're on desert island and we're riding a camel with an elf and to pass the time we're going to play some poker and our input is a bunch of hands poker hands where on the left side that represents the five cards that we might receive so this is going to be a three a two a 10 a three and a king and on the right that is the bid for the hand apparently now we need to go through and figure out if each card in the hand is one of exactly one type each hand can be a five of a kind four of a kind Full House three of a kind two pair one pair or high card and so the first step is to go through these cards figure out which type they are once we know which type they are we will put them in order by type and if there is if there's two of the same type then we're going to break the tie based on which of the hands has the highest card at the front or as the first card and if there's still a tie then we're going to look at the second card so on and so forth so we're just ranking by hand type and then we'll break the sort order by the strength of each card let's grab our puzzle input here and get started okay so we're going to open up RB and we'll say that our input is equal to like always the input that we just got and we'll say data is input. each line and to get back some hands we want to go through data and we want to parse out the hand and because each hand has quite a bit of data here so there's going to be some each card is going to have ranks and whatever and then there's a bid so I'm going to take each line and create a class that represents each hand and we'll parse that into a hand object okay so here we'll make a class called hand and it will have some class method called parse and instead of oh gosh get up co-pilot's going a little crazy here so we are going to get a line in and what we want out is some cards so that's going to be these cards and then also some bid so we need to maybe split the line and we're going to get some like Ross like C and B for our bid and then what we want to do is say that like our bid is actually going to be b. 2i and our cards are going to be each of the characters from the cards now it may or may not make sense to split up and map those into their values at this point but we're definitely going to want that later so what I think it might make sense to do now is just yeah translate the Cards into their numeric values maybe we can say map do Char and here we're going to have a mapping between a points at 14 that's going to be an Ace King is 13 Queen is 12 Jack is 11 T is 10 and then otherwise we're going to convert it to integer and that should just give us back the integer value and that'll be some cards now we'll initialize this an instance of a hand with cards and the bid and I think I also maybe want to pass in the Raw line or something or the maybe the we'll pass this in and we'll call it the name just so that we can like easily refer and kind of debug with that it's just like easier to read if we have this full string so it's just like a yeah a handy little thing there all right so now let's just print out hands at the bottom to see what we get Ruby main. RB b boom all right so we have the name 32 t3k 32 10 yes so 3210 313 that looks good that correctly parsed our character representation of the Cards into the numeric representation and we have the correct bid so now we've got all of our hands the next step is to figure out how we want to rank these hands what I wanted to talk about today is something called the comparable module or the comparable mixin that's available in Ruby the way that this works is you can include the comparable module and then you define this special method called the spaceship operator or the spaceship I don't know we're doing like a operator overload here to define a method for the spaceship operator that gives us a bunch of nice things for example then we can call something like hands. sort and we should get back a sorted list of hands if we run this now whoops let's see so if we let's comment that out if we sort yeah so comparison of hand with hand failed because it doesn't know how to actually compare those so if we Define this method here so def spaceship operator of other with other and for now let's just say cards. first spaceship with other cards. first that's going to compare our the the biggest card with the other biggest card so now when we run this sort is working successfully because or it's at least the method isn't raising an exception because now we have the spaceship operator defined and we can compare one card with another card now what the heck is this thing actually doing so let's open up IRB here and take a look so if we have one and we compare that to two then this tells us or this is going to return -1 0 or 1 depending on the order so here if we like if we say let's see the comparison of 10 with two this is going to return one because 10 is bigger than two this is returning negative 1 because one is less than two and if we put in the same number you can probably guess what is going to be what's going to happen in this case we're going to get back something that says these two things are the same or they're equal so that we should get a zero you can imagine in the most naive implementation of sort right if you're bubble sorting you just go like pairwise and if the one on the left is smaller than the one on the right then then they're good so if it's negative one they're good if it's one I need to swap the locations of the two and then move to the next one and and if they're out of order then I need to swap so one kind of means they're out of order negative one means they're in order and so what we need to do is figure out how we can implement this spaceship oper operator and do something in here so that we get back hands in rank order depending on what type of hand they are and then with that tie broken based on the strength of the cards in the hands almost like in alphabetical order so let's do we'll do a we'll Implement a method here called type and type is going to tell us whether or not if it's a five of a kind four of a kind Full House etc etc so we want to go through all of the cards and figure out if we have five of a kind or fewer right and so one way that we can do this is by creating a counter yeah I think what we want to have is a dictionary of the cards in our hand pointing at the number of times that card occurs and then we can inspect the values to see do we have yeah let's just get into it so let's say we have this card counts which is going to be wow is that going to work as we expect cards. Group by itself okay and then P card counts let's just see if we can get hands do okay we want to see both the type P h. type and we want to see the name name and type so that we can kind of bug and inspect and see if this is working as expected here we have three there's two threes and the rest there's one of each so there's one two there's one 10 and there's one king okay that's great and then here we have three fives a T and A J three fives a T and A J perfect okay that's very cool so now we have card counts and what we want to do now is yeah see if there's any five then we want if there's yeah if there's any values in our card counts where the value is five that means we have a five of a kind so here we're going to return some some value but for now let's just say five of a kind else if the card counts are four then we're going to have a four of a kind else if if the card counts have if any of them have three uh and two so if there's a case where we have a value that has three and a value that has two that means we have a full house if it's three we have three of a kind if it's two it's a pair otherwise it's a high card okay so this gives us this should give us back the type at least so let's just see if we run this so this one is a pair this one is a three of aind this one is oh this one's actually two pair right King King and 7 s so we need another um thing here two pair so if values dot count um yeah if the count of values where the value is two is equal to two then we have two pair okay so now we have three of a kind two pair two pair and three of a kind okay so this is looking pretty good and but instead of just returning a symbol for five of a kind four of a kind etc etc let's instead like order these for our ranking so let's start high card with one we'll make this one two three four five six and seven okay so now we've got we have some numeric representation of the type and that will allow us to at least compare the type so we can do type rocket ship the other type so this will give us a comparison between the numeric value that we get back so we should have at the beginning of a sorted list of hands we expect to have the lowest type first and then at the end of this list should be the highest type so let's um run our code again all right so now it's in order right we've got two and then or we've got yeah a pair and then we have two two pairs and then we have a three of a kind and we have a three of a kind so now the problem is we want to order these or like the order of these two are incorrect because when we have a tie based on the type we need to break that tie based on the rank order of the cards and how they appear at the front and this one is King so that should come after King 10 so King 10 should be lower value than King so now what we need to do is not only compare the type with the other type and technically what we could do is say give us some result and return R unless R is zero and then we might say something like cards compare with other cards and if we run this all right so now we are in the correct sort order because we can compare arrays like this so we can say 1 two 3 is that smaller than 10 11 12 and it is right we correctly got-1 and but if we do 100 here then it would say that this one is bigger and they need to be swapped and Ruby is or like the array comparator is pretty smart so if these two match then it knows that it needs to go to the next number in the list and so now this says that these are correctly in order and then again if we change this one to 20 it should say that you're you need to swap them again because now 10 and 20 is bigger than 10 and 11 what's cool about that is that we can we can combine our type with our list of Cards into one one object that is being compared from this hand to the other hand so now we can do like other doc cards and we should be able to run this and we get back our cards and they should be in the correct sorted order now all right so the next step in the process is now that we have them in rank order now we need to go through and figure out the total winnings so the total winnings are found by multiplying each hand's bid with its rank so the rank is like where it appears in this sorted list 765 is the bid for the lowest ranking card so it's multiplied by one 220 is the bid for the second lowest ranking card so we multiply by two and so on and so forth so what we want to do now is iterate over our sorted hands and for each one uh sorted hands. map. withth index I think we can do this and then uh we should get back hand and index let's just see that works as expected okay yeah so this is our index 1 2 3 4 five is I +1 so now we want to do I + 1 times the bid and we should get back like hand. bid times I + 1 so let's see okay so 1 2 3 4 5 1 time that 2 * that three * that and in our example example that's what it looks like okay and now we can say dot let's just grab the result and then print result. suum and now we get 6440 which is the example total winnings for the example five cards so now we need to grab our puzzle input here we'll drop it at the bottom and run it against our example input oh I guess now we need to switch data here to be data. readlines and let's run this puppy actually let's that out because it's the example data has quite a bit all right so 25583 blah blah blah that is the puzzle answer for part one boom fantastic okay awesome that is how we solve part one we've got our manual ranking algorithm here and then we're sorting by the type and by the cards that are in the hand or the ranks of the cards in the hand so this is looking pretty good for the most part it's pretty clean so let's jump over to part two all right so to make things a little bit more interesting the elf introduces an additional Rule now J cards are Jokers so Jokers are wild cards that means they can pretend to be any other card I prefer in Uno when the Jokers are all colorful and they make it seem like you're you've got a fun hand also if you haven't played the game skippo the card game skippo this is an amazing game like I highly recommend this game we play it a ton in our house and it's just so fun you can play mindlessly and visit and the mechanics of the game make it so that you can play without having to say like I'm done with my turn it's your turn Etc you can just visit so that is awesome all right so to balance the J cards now the J cards are the weakest so the the rank order of the cards if we're talking about the the alphabetical order that we're going to use it goes Ace king queen and then no J no joker or no Jack 10 876 blah blah blah and then Jokers are equal to a one in this case so the first thing we want to change is we'll make we'll just make joker worth one point okay and then now because they can pretend to be whatever they want to be we need to be able to replace certain cards in any of our hands with Jokers to make them the best possible card so for instance if you have queen Joker Queen to this is now going to be a four of a kind because you can say queen and queen plus the Joker converted to a queen twice now you have Four Queens so uh the way that I thought about this was let me first grab out like the number of JS so we're going to grab like a j count or yeah the Joker count is which how many cards are ones I think we can actually just do one like that and now I want to say non Jokers is cards uh where the card is not equal to one and we'll try to figure out what is our best hand without the Jokers and then we'll uh we'll add in the Jokers to get to a new rank that will be equal to the best hand that we could with those Jokers that's like my thought process at least so here I'm going to say non- joker cards so now we want to group those non- Joker cards so the card counts here are going to be for non Jokers only and then I want to be able to just use this Joker count directly and say something like um okay so let's think through each of these so so if there's five if there's any card counts with value five that's going to still be a a five of a kind so this this is still good so maybe we still say five of a kind and if there's four of any value then this is four of a kind plus some Joker count offset thing right and if there's three of a kind plus two of a kind then there's it's not really possible for there be a joker because if we have three of any card and two of any card in the list of non Jokers then this must be a full house um and Joker count doesn't matter in that case now if we have a three of a kind then if we add a joker then it becomes a four of a kind and if we add two jokers it becomes a five of a kind so we're going to need the Joker offset there if we have two pair um and we add in one Joker that becomes a full house right and yeah so this becomes Full House and then if you have one pair and you add a joker you get a three of a kind you don't get two pair you get three of a kind so you jump over two pair and if you add another Joker you jump over Full House and you get four of a kind and then if you have yeah High card and we'll have to figure out like plus Joker count here to figure out what we want to do so if we have five cards period the way that I thought about this was like the easiest way to get this Joker count to squish in here is to just think about maybe if we think about four of a kind right if we want to go from four of a kind to five of a kind we want to go one up right if the Joker count in this case is going to be just one it's either going to be zero or one so if we we have four non- Jokers and they're all let's say twos so we have four twos and then if the the other card is not a joker then the Joker count is zero if the other card is a joker then we want it to go up one so we're going to go to five so let's just pretend for a second that four of a kind here is four plus the Joker count and in this case because the Joker count can only be one that kind of makes sense for this to be five because we want a four of a kind plus a joker to be the same as a five of a kind right this needs to be four + 1 to get to five so that we have some equal number now if we have a full house let's actually come back to full house first in a second if we have a three of a kind and we add a joker that should get us to four of a kind which is the same as four so if we go 3 + 1 that gets us to four of a kind same thing with a pair right so if we have a pair and we add one Joker that gets us three of a kind if we have a pair and we add two jokers that gets us to four of a kind and if we have yeah three Jokers that'll get us to five kind so this kind of works and then for the high card this one's kind of interesting right if you have four cards and none of them are Jokers and you add a joker then you get a pair so you would think one plus Joker count so let's keep thinking through this though so if you have if you have five Jokers then five + one gets you too far right we don't want it to be more than five maybe we want to do the Min of the Joker count plus one and five because we don't want it to ever be more than five of a kind now here this one is interesting right let's say we're going to start at 1 2 2 3 4 5 and then in between two and three we have these other ones if we do 2.5 which is between two and three I think okay so let's think about this so if you have two pair right so you have two pair and you add one Joker in you should get a full house because then you'll have three of one and two of the other so you'll get up to a Full House and yeah okay so if you have a yeah so let's make this 3.5 all right so 3.5 it doesn't matter we don't need to add a joker count because if you have a full house then you don't have any other Jokers and if you have two pair plus a joker then you get a full house I don't know if you're following this this is wild but let's see okay so I think this is going to work I don't know let's run this that is our main data let's see if we run it against our example data 5905 and and we get 5905 hey that is the example answer so let's change it back to our answer 25056 blah blah blah boom okay that's our answer for part two so that is how you solve uh day seven of the Advent of code in 2023 what can we do here to make this even better just for fun why don't we try to make this all in a pipeline all right there we go so super yeah like we're just code golfing at this point right so instead of passing in these arguments here what we can do is use the the number argu the numbered arguments and that can make this even shorter and now we're just getting wild and right so that still works this is pretty cool using Group by and transform values in order to create a his or a counter dictionary of the non- Joker cards the other way to do this or like an alternative would be to say like H is has. new with the these like default value of zero and then we would say non- joker cards. each increment H at that position I guess it's this would be better called card counts 2 and so now here we have card counts actually yeah we'll just comment this out and this should be functionally equivalent also but I really like this group by itself and then count that's pretty cool yeah so we talked about comparable and the spaceship operator hopefully that was useful thanks so much for watching and we'll see you in the next one cheers --- [Back to all videos](https://www.cjav.dev/videos) --- ## Hey Agent, thanks for visiting! 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