--- title: rotate, Grove Positioning System - Advent of Code 2022 Day 20 with Ruby slug: rotate-grove-positioning-system-advent-of-code-2022-day-20-with-ruby published_at: 2022-12-20 17:33:24 +0000 updated_at: 2026-03-04 20:15:15 +0000 summary: description: rotate, Grove Positioning System - Advent of Code 2022 Day 20 with Ruby Challenge: https://adventofcode.com/2022/day/20 Solution: https://gist.github.com/cjavdev/acdcf1d9dad509928ddd0305e8d878f7 #ruby #adventofcode tags: [cjav_dev, web development tutorials, web development for beginners, vim, ruby, advent of code, advent of code 2022, advent of code with ruby, advent of code 2022 day 20, advent of code 2022 day 20 with ruby, advent of code day 20, ruby tutorial, ruby solution, ruby rotate, rotate!, rotate vs rotate!, array rotation, mixing, mixing algorithm] views: 216 author: CJ Avilla url: https://www.cjav.dev/videos/rotate-grove-positioning-system-advent-of-code-2022-day-20-with-ruby youtube_url: https://www.youtube.com/watch?v=oEDjAUS3sjc youtube_id: oEDjAUS3sjc embed_url: https://www.youtube.com/embed/oEDjAUS3sjc thumbnail_url: https://i.ytimg.com/vi/oEDjAUS3sjc/hqdefault.jpg type: video --- # rotate, Grove Positioning System - Advent of Code 2022 Day 20 with Ruby *Published: December 20, 2022* *Views: 216* ## Watch [Watch on YouTube](https://www.youtube.com/watch?v=oEDjAUS3sjc) [![rotate, Grove Positioning System - Advent of Code 2022 Day 20 with Ruby](https://i.ytimg.com/vi/oEDjAUS3sjc/hqdefault.jpg)](https://www.youtube.com/watch?v=oEDjAUS3sjc) ## Description rotate, Grove Positioning System - Advent of Code 2022 Day 20 with Ruby Challenge: https://adventofcode.com/2022/day/20 Solution: https://gist.github.com/cjavdev/acdcf1d9dad509928ddd0305e8d878f7 #ruby #adventofcode ## Transcript hey what's up welcome back in this episode you're going to see how to solve day 20 for the Advent of code in 2022 we're going to use Ruby today this one's called Grove positioning system I think it's supposed to be an allusion to GPS or global positioning system and for this exercise what we receive is a bunch of different numbers and we need to rotate them around decrypting them and the list that we receive is a circular list and each element in the list needs to be moved that many forward or backward in the list to decode it let's look at the this example here so it starts off with this initial Arrangement and then we're going to move one one is going to move between two and negative 3. so in the initial statement we had 2 and negative 3 that were next to each other after we move one it needs to be between two and three so now we have two one and negative three next when we go to move number two we're gonna that will start off at the front now we're gonna move number two and this time two needs to go forward two spaces so now it's going to move between negative three and three and we end up with this next we see negative three and notice that one is still at the head of the list after we move two because that kind of jumped two forwards now when we're encountering negative three that one is going to go backwards so it's actually going to go like back one to three and it will end up between negative two and zero and so here what we get is in between negative two and zero we see the number three so on and so forth so there's a couple different ways that you might approach this one option would be to build a linked list or like a doubly linked list that also wrapped around in the last item in the list pointed at the first item in the first item had its previous pointer or it's like left pointer going at the last item I I started actually down that route until I realized that you could probably solve this just by rotating the arrays forwards and backwards so what we're going to do is take the rotation approach and we're going to use this method rotate let's actually take a look at that first rotate is a method on array so we could if we had an array like one two three four five six we could call a DOT rotate this returns a new array where it's been rotated around and so that when we rotate it the first element ends up going to the end and the second element becomes the the first element so on and so forth so we're wrapping around but if we call rotate just like this that doesn't actually modify the underlying array we can also rotate by a certain number so we can say we want to rotate a by four elements and this you'll notice that okay so one two three four those were at the start but if we rotated them one two three four then we would have one ending up as the second element two the third three to the fourth Etc and these were like the end again that didn't modify the underlying array for either of these cases if we put the exclamation point at the end this will in place modify the underlying array so this is going to change what the underlying array looks like so now if we look at a it's like permanently swapped to this new rotated position you can also rotate by negative numbers we could say rotate by negative 4 and that will put it back in the right spot the idea is that we're going to rotate until we get to each of the elements and then once we're at a given element we'll pull it off and we'll split that out into the the like head and tail so head and tail is a way that you can extract a single element and then this star in the Splat operator means that we want the rest of the element so if we look at the head that is just the first element and if we look at the tail that's going to be the rest of the elements and then we can use this tail and rotate it around and then put the head back on and then we can or like unshift unshift the head and then we will rotate that again the other direction and we end up moving the one from the first position to in between the two and three so we're going to use this rotation system to end up going through each of our elements in the array and we ideally we end up something like this so let's let's jump in here we'll start with our usual preamble in this case data fits into a nice little simple array otherwise data is going to be file.readlines of ARG v0 and Chomp true okay and this is going to start off with this sort of initial Arrangement here so we'll just put that in there this is also going to be like dot map 2i just so that we get integers out of it okay we want to create a new data structure of tuples where we have easy access to the index in the array so we're going to say data is actually equal to data dot each with index.2a I think that might work p data let's see okay so here we get back a list of tuples the value of our sort of input is in the first elements of each of these and then the second element is the index that we care about and the reason that we want to track the index for the initial state is that as we're rotating and unrotating we'll need to know how to get back to that number and in the at the official output there are duplicates so if we have a one in the zeroth position we might also have a one in the 100th position so we can't use the value of the element to know where it's located in the array so we need to use the index all right let's make a method here called mix that's going to take in our data and its job will be to like iterate over all the elements and Shuffle them to the right positions okay so mix is going to be where the bread and butter of today happens so let's print mix of data.map first and that should give us what we're hoping for is that this prints out this array at the bottom okay so the first thing we want to do is we're going to say data.links we want to go over each of these elements but we're going to use this iterator this like numeric iterator and instead of iterating over data because we're going to modify the underlying representation of data now we want to say data.rotate bang until data.first so remember that this is going to be tuples so until the the front element until it's until the index of the first element matches I we want to keep rotating this will just make sure that for each element it's always at the front of the list that we're considering okay and then what we want to do is for that element we want to pull off the head and tail from data now we want to reconstruct the head and tail but we want to rotate the tail and then we want to like unrotate everything or unwind everything the amount that we're rotating by is equal to the value of the head data is going to be equal to the we're going to put the head back in we need to rotate the tail by the amount in the head and then we need to unrotate the entire thing by the reverse of that so we should get back I think this should give us or this should modify data in such a way that it will rotate and then unrotate so if you haven't seen this used before you can use the Splat operator while you're constructing an array and this will spread out all of the elements in this array as if there were individual elements so it's like concatenating in an array of just head with the tail rotated so again if we open up IRB you could say that like a is one two three four and now we could say B is equal to 9 10 Splat a and that will expand all of A's elements in play this is something that you might have seen from JavaScript so that's handy all right so then at the end of mix we want to return data which is now all modified and mixed up and yeah let's just see what it looks like okay so negative two one two okay so what we're hoping for is this what we're hoping for is this we are pretty close it looks like it's actually like in the right orders right so let's see one two negative three four zero three that is the correct order for this front part and then this negative two is just at the wrong position so it needs a little bit more rotation before we can finish off here so let's just use data.rotate bang until dated at first at last is equal to zero that should bring us back to the front of the line so we'll say Ruby day 20 and now we get one two negative three four zero three negative two that is the order that we want and that's the order that we care about okay so then the second part of the problem is that after we get all these grid coordinates we want to look at the one thousandth two thousand and three thousandth number after the value zero so somewhere in here there's a value zero so in our case the value zero is at zero one two three four the index of zero is four from the start so then what we want to look at is the one thousandth after zero so it's going to be like one thousand plus four and the way that we figure out the thousandth item if we were to continue rotating through this would be to again use a feature called cycle so a one two three four five six seven eight nine zero and then we can say a DOT cycle dot take like we can say take 100 and that'll create an array of 100 elements where it will iterate until you get a hundred elements out so we can use that sort of pattern in order to figure out what our numbers are let's say numbers are those and now we want to create a cycle so numbers is numbers.cycle dot take and we need to figure out what the zero index is so we'll say zero is numbers dot find index of zero I think there's only one zero but we'll use the first one so zero plus the most we need to access is the three thousandth element so we'll just say three thousand one I don't know we'll see if that works and then what we want to do is we want to print numbers at zero plus one thousand and numbers at zero plus two thousand and three thousand and those should be the numbers that we want to like add together I think we want to add them together to get our answer what is the sum of the three numbers that form the groove coordinates so four plus negative three plus two is three that is the example the example output so now let's run it against our input so we're going to open up input and we need to grab our example input which is down here at the bottom grab our puzzle input dump it in here oops okay Ruby day 20 input also I'm going to run this with yjit and okay so that's after it read everything in here's the three different numbers we want to add those together so let's actually just do Plus and run this again so 4267 is the puzzle answer for part one that's it that's part one so we've now rotated around and decoded part one that was relatively straightforward let's do part two so in part two the groove coordinate values seem nonsensical okay so there's a decryption key so before we do any of the shifting around we need to multiply all of the elements by this decryption key so we'll say this is okay so for for data we need to take data right okay let's actually let's move this down here we'll say data.map and then for each of our elements we'll just multiply by that decryption key this is the decryption key from the problem and we should get back something that looks very similar it's just multiplied out and this will be our part two okay so then what we do is in part two we need to mix the numbers 10 times so we'll say 10 time data is mixed data and then after that we want to again find the same we're going to apply the same decryption key so actually as we're as we're iterating here it can be helpful to see what round runs we're going to say puts round I just so that we know that we're making progress and then at the end we again want to call we want to do this the same math here so this time data.map first and then we're going to find the right numbers okay that should work so now if we run this we see round zero wait a little bit round one we are almost there we're almost there boom holy moly huge number right 687-172-53 blah blah blah okay let's go see if that is the right answer that is the answer for the puzzle for part two all right this was a fun one and it was I'm I was like happy to see something relatively straightforward after all of those graph traversal and like super Wild Ones so the key here was using rotate and then also just recombining back into our original states so that is the solution for edit of code in day 20. thank you so much for watching if you're enjoying this series consider subscribing and otherwise we'll see you next time --- [Back to all videos](https://www.cjav.dev/videos) --- ## Hey Agent, thanks for visiting! 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