--- title: Valve flows and tunnels through a graph - Advent of Code 2022 Day 16 with Ruby slug: valve-flows-and-tunnels-through-a-graph-advent-of-code-2022-day-16-with-ruby published_at: 2022-12-19 22:00:06 +0000 updated_at: 2026-03-04 20:15:15 +0000 summary: description: Flows and tunnels through a graph - Advent of Code 2022 Day 16 with Ruby Challenge: https://adventofcode.com/2022/day/16 Solution: https://gist.github.com/cjavdev/b6f1f78bad76f69130cef4532742c97a #ruby #adventofcode tags: [cjav_dev, web development tutorials, web development for beginners, vim, ruby, advent of code, advent of code 2022, advent of code 2022 day 16, ruby tutorial, ruby solution] views: 373 author: CJ Avilla url: https://www.cjav.dev/videos/valve-flows-and-tunnels-through-a-graph-advent-of-code-2022-day-16-with-ruby youtube_url: https://www.youtube.com/watch?v=LzDRS7igO1k youtube_id: LzDRS7igO1k embed_url: https://www.youtube.com/embed/LzDRS7igO1k thumbnail_url: https://i.ytimg.com/vi/LzDRS7igO1k/hqdefault.jpg type: video --- # Valve flows and tunnels through a graph - Advent of Code 2022 Day 16 with Ruby *Published: December 19, 2022* *Views: 373* ## Watch [Watch on YouTube](https://www.youtube.com/watch?v=LzDRS7igO1k) [![Valve flows and tunnels through a graph - Advent of Code 2022 Day 16 with Ruby](https://i.ytimg.com/vi/LzDRS7igO1k/hqdefault.jpg)](https://www.youtube.com/watch?v=LzDRS7igO1k) ## Description Flows and tunnels through a graph - Advent of Code 2022 Day 16 with Ruby Challenge: https://adventofcode.com/2022/day/16 Solution: https://gist.github.com/cjavdev/b6f1f78bad76f69130cef4532742c97a #ruby #adventofcode ## Transcript hey what's up welcome back in this episode you'll learn how to solve part one of day 16 for the Advent of code this one is about all of these valves and tunnels and flows and we're going to go through some tunnels to turn on some Valves and we want to see if we can optimize our like traversal through this graph so that we have the highest flow rate at the end and so the puzzle the way that the puzzle is presented is that you have 30 minutes before the volcano that you're inside of is going to explode and we don't have time to to get out and so what we need to do is go around and change or open all of these different valves so the valves are going to start off as closed and this is the input we receive is we're going to get the name of a valve its flow rate and then some tunnels from that valve that lead to other valves and we want to figure out the best way to go through each of these different tunnels in order to turn on all of the valves and once a valve is open it will release some pressure and the pressure is the flow rate times however many minutes that it's open and so again because we only have 30 minutes it's constrained to that 30 minute window we need to figure out some algorithm to Traverse this graph and in fact the first thing we want to do is actually just build a graph out of this so I'm going to start off with dropping our input here at the bottom and if arc V is empty we're just running this sort of as the the examples or this this test mode I want to say data is data.readlines otherwise we'll say data is file dot read lines and in both cases we will Chomp to make sure that we don't have that new line at the end and we should be good all right so the next part is that we want to parse out all of our valves into some graph or some map of the tunnels we'll just say that our like our valves is going to be some dictionary and we'll use a constant so we can write a couple different methods and still have access to it and we also want to have a graph so we're going to have we're going to have two dictionaries here one is going to hold the valve and its flow rates it'll just be a mapping of it to zero BB to 13 cc to two and then we're also going to keep a graph which is going to be a a to its adjacent valves all of the different valves that are near we'll go through data we'll say data dot each line and we want to map the line so that we can pull out the details here we're just going to use we'll use regular Expressions again to figure out what we want now there's one interesting thing about the regular expression for this use case and that is that in some of these for some of these we have plural and some we don't so we're going to use Ruby alert again today and what we want to do is say we grabbed one of these inputs and that's matching perfectly we want to grab off the name of the valves I'm going to use a capture group here we'll say that this is the name and it's actually like just a to z and they there has to be two of them so that gives us one capture group for a then we want to okay so now it says it has the flow rate we want to grab the the flow rate we're going to say rate and this is going to be an integer there is no negative rates and then here at the end we have we have tunnels lead to Valves and then some list of valves at the end so let's just replace all of those with something else called I don't know valves and this is going to be I don't know some some words and actually let's just grab everything to the end and we should be good to go okay and then later on we'll split the valves on comma and space to figure out what they actually are there's probably some way to do that with regular Expressions if please let me know down in the comments how do you split yeah how can you how can you use a capture group to grab an array of things I don't know maybe it's like a fancy real expression thing all right now you'll notice that we haven't actually matched all of the results because down here there is a singular tunnel leads to valve GG in that case tunnels is plural but we this S at the end is optional and if we have an S for leads we want to make that optional and finally we want to make the S on valves optional which we can do by dropping on the the question mark there so now we're able to match them all and we have the right sort of rates and stuff so we'll use this as our regular expression here and we're going to match on line I'm going to say.match line and this is going to give us back our match and now what we want to do is say yeah valves at the match name is equal to the rate and we want to say our graph at the match name is the valves.split wow that's really nice okay so now we'll just say p Valves and P graph and we should be good to go all right so we'll say Ruby day 16 and okay so this is our flow rates this is all of our flow rates for our valves and this is all of our adjacencies for our valves now what we want to do is Traverse the tree and at each point we need to figure out whether or not we want to open the valve and then we also need to go through all of the neighboring Valves and see if there's one of them that's going to return the best flow rate at the end here's here's how we're going to do this let's make a method called Max flow here we'll pass in the value of the current valve that we're looking at we'll pass in all of the opened valves and then we'll also pass in like how much time we have left and the idea is that we're going to recursively call Max flow with something something something time minus one and then ultimately we will execute this with like Max flow of a a with an empty list and then we have 30 minutes so we'll pass in 30 and this should give us back hopefully this should give us back some amount that gives us like the maximum possible flow through the this through all of the opened valves if we went through in the optimal with the optimal approach so as we're going through each of the valves we can we estimated it's going to take one minute to open a single valve and one minute to follow any tunnel from one valve to another here what we need to figure out is what is the max of opening the current valve and then what is the max if we were to open or if we were to travel to another tunnel here we want to say something like if we're going to open if we open the current valve we want to say something like if opened does not include current like if if we haven't already opened this one then we can consider opening it and if we open it then what we want to do is add the current valve into the list of opened opened valves then what we want to do is now we need to calculate how much of how much pressure will be released as a result of opening this current valves current valves times the amount of time that there is left this is the time left otherwise we're going to recursively call Max flow passing the time down to the next one okay otherwise if or even if we didn't even if we didn't open the current valve we want to go through all of the neighboring tunnels and figure out if any of those will ultimately lead to a better Max flow we want to go through graph of the current at each and for each one of these we want to see the max is going to be Max flow of the next valve passing down opened and okay in in both because we want to do we want to open this we want to consider opening the valve as us spending a step and going to the next way and then we also want to consider going through all of our adjacencies we need both of these to happen for the same time and then we want to take the better of the two right here we want to start with Max is equal to zero and then ultimately we're going to return Max at the end of this instead of just grabbing Max is equal to the max flow and Max is equal to the max flow here what we need to do is grab the maximum of Max or Max flow another way to say that this would be something like current Max is Max flow of current and we want to say Max is equal to current max if current Max is greater than Max we want to do this in in both cases in this case we're going to say we're passing down the next value and instead of instead of modifying opened here by shoveling in current we don't actually want to modify that underlying array instead what we want to do is pass down some new current open is going to be opened plus this new current value and the reason is that the reason is that when we recurse we want to only consider the this new current opened if we've opened a valve here if we were to modify that then we would be passing down the modified opened for each of our adjacent for each of our adjacent neighbors okay this is going to be really slow because Max flow in this case we're not caching anything we're just constantly recursively calling but we will as we're going through all the tunnels we're going to visit the same valve multiple times and what our base case here is going to be return 0 if time is less than or equal to zero okay and now if we run Max flow for a a through this tunnel let's just see what we get okay no next value but it's right there what are you talking about next oh valve next valve I said next value okay now we're getting stack level two deep okay we're for some reason we are recursively calling too much right we're going way too far oh so instead of time plus one we actually want to subtract time here and we're going to start at 30 and we're counting down down to zero so that was going the wrong direction there okay it's hanging it's hanging in this case if you go to run something and it hangs for more than a couple seconds you want to kill it and then give yourself some some output here let's just we'll put like current open in the current time and just see what this looks like all right so this if we scroll back and look at this you can tell all right it's looking at you know what if we had opened AAA then DD then CC then BB at time one and it would tell us like okay this would this is the the max possible value at this point let's actually we'll print Max down here are we making it here yet so we're not actually let's see no we're not making it there yet okay one thing that we need to do is if we were to look at all of this output we would find that there are several look at this there's several calls to with the opened being aaddcc if we were to search for this then we would find that there are a ton of repeating versions of this aaddcc with repeating so the current was D the current was D and we had to open up that that sort of several times or we went through that same flow several times what I want to do is create a cache which will store as a dictionary and we need to create a key for this cache I think we can pass current opened and time as our key and now we want to return the cache key if or we want to return the cache at that key if there is a key in it with that value so here we're going to say cash if we found a Max let's store that off in the cache at this key and we'll run it again looking at our key this is still too slow looking at our key current is a string opened is an array and time is a number now when we think about this opened array so if we were to just like print out let's P key for a second and if we think about this array you'll notice that a lot of times we're going to be like visiting the same nodes several times and then if we were to go through so here you'll notice that we've already gone through Ada aaddcc JJ and we've already talked about or in let's see if we look at this opened value here you'll notice that we have a d c so D and C are not ordered and so it's possible that we we somehow went to C before D in this case notice that we are at ee now but we we could have already seen ee so one way that we can improve our cache key is by sorting opened before we encounter it let's run this again all right we got zero back which seems wrong so our current Max is going to be pressure plus the max flow when we recurse and then in this case we're just spending we're not we're not increasing the current Max by anything because we're spending the time to go to that new valve in this case we are increasing the current Max because we are opening a valve so let's try this again all right we got 1732 and well let's look at our answer here the the puzzle answer was 1986. oh actually let's see yeah okay so for we're still doing a test approach so we're looking for 16.51 so that was too high right here I forgot that when you open a valve it does not increase the pressure for for the current minute so we need to do time minus one here I think this might be the problem let's see we want to do the current we want to release the current valve at time minus one okay there we go 1651 that was the answer for part one and then we or that's like the answer for the test input we can grab our puzzle input here and open up input drop it in and then run Ruby day 16. against the input this is another one of the puzzles where the it takes a really long time for the input to run in fact for part two it took my solution a really really long time to run we're not actually going to solve part two today but the solution is very similar there's a couple different small optimizations that you need to make but hopefully this solving of part one gets you far enough and we've like learned about one more approach to traversing through a graph and also calculating in this case so using dynamic programming to build up an optimal path through the graph all right this is taking too long I forgot one other optimization that we can make so let's kill that come back over here now if the pressure does not equal zero if the if the pressure is zero we don't actually care about going down this path and so we don't need to make this recursive call and we can kill this entire Branch if the pressure is zero let's add that in and we'll rerun our code all right there we go 1986 we come back over here 1986 was the puzzle answer for part two or for part that's the puzzle answer for part one now in part two it gets a little bit different the time goes down to 26 minutes and now you're going to have an elephant work with you to open different Valves and so the the logic becomes a little bit different but we're not going to get into that today because part two takes so long to run I'm actually just going to skip over part two and head right into day 17. thanks so much for watching hope you enjoyed it and we'll see in the next one --- [Back to all videos](https://www.cjav.dev/videos) --- ## Hey Agent, thanks for visiting! 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