---
title: Recursion with Ruby and inline rspec
slug: recursion-with-ruby-and-inline-rspec
published_at: 2021-07-28 14:00:21 +0000
updated_at: 2026-03-04 20:14:03 +0000
summary: 
description: In this episode you'll learn all about recursion and see the implementation of fibonacci, factorial, and towers of hanoi. You'll also see how to write inline rspec tests in the same file you're experimenting with! You'll even get. alook at memoization / caching to speed up recursive calls.  Related video: https://www.youtube.com/watch?v=WPSeyjX1-4s
tags: [Recursion, Recursion with Ruby, Writing rspecs inline, Rspec in the same file, rspec test in same file, How to write simple rspec tests to autorun, fib with ruby, fibonacci with ruby, factorial with ruby, towers of hanoi with ruby, how to implement recursion with ruby, ruby tutorial, ruby for beginners]
views: 1492
author: CJ Avilla
url: https://www.cjav.dev/videos/recursion-with-ruby-and-inline-rspec
youtube_url: https://www.youtube.com/watch?v=SAfpPJkHRec
youtube_id: SAfpPJkHRec
embed_url: https://www.youtube.com/embed/SAfpPJkHRec
thumbnail_url: https://i.ytimg.com/vi/SAfpPJkHRec/hqdefault.jpg
type: video
---

# Recursion with Ruby and inline rspec

*Published: July 28, 2021*
*Views: 1492*

## Watch

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

[![Recursion with Ruby and inline rspec](https://i.ytimg.com/vi/SAfpPJkHRec/hqdefault.jpg)](https://www.youtube.com/watch?v=SAfpPJkHRec)

## Description

In this episode you'll learn all about recursion and see the implementation of fibonacci, factorial, and towers of hanoi. You'll also see how to write inline rspec tests in the same file you're experimenting with! You'll even get. alook at memoization / caching to speed up recursive calls.

Related video: https://www.youtube.com/watch?v=WPSeyjX1-4s

## Transcript

hey welcome back in this episode we're going to talk about recursion so first up what is recursion and then we're going to dive into some examples of some really common computer science problems or you know functions that we would write with recursion and we're going to do that in ruby so again some more beginner ruby stuff um and talking about this concept of recursion so what is recursion google if you google recursion and you'll get this definition here which is from the oxford languages it tells you that it is the repeated application of a recursive procedure or definition and you're seeing like recursive here inside of the definition of recursion and i don't know like defining uh defining a word with the same word is like the definition of recur it's like a meta thing like recursion and recursion if you go down a little further you'll see that wikipedia has it defined as this like in computer science recursion is a method of solving a problem where the solution depends on solutions to smaller instances of the same problem so such problems can generally be solved by iteration but it needs to identify and index the smaller instances at programming time all right so what does that actually mean who knows hard to say uh but yeah this is often referred to as like the divide and conquer approach to solving a problem where like you are trying to break a problem down into smaller problems that look exactly the same they're sort of the same shape it's when a method usually calls itself and as long as that method has one base case it can break out so there's a bunch of fun images on google images if you just search for recursion so you get a bunch of these like funny here we go again here we go again here we go again all the way all the way through so that's like an example of recursion here's another example of recursion where you've got like triangles that are made up of smaller triangles that are made up of even smaller triangles that are made up of even smaller triangles all the way down until you get like the smallest size triangle you can think of there's also a bunch of other examples like when you're kind of looking at yourself in a video editor and you see like yourself inside of the video inside of the video inside the video a bunch of different times that's another example of recursion but at the end of the day recursion is all about solving a smaller version of the same problem and kind of like adding on one layer it's like a russian doll that russian doll thing where you have like kind of a big doll and then you open it up there's a smaller doll inside and you open that up and there's a smaller doll inside and it's all like the process of opening the doll is like the going to be the sort of the same action that you're going to mirror across all of the dolls as you work your way down the the like the i don't know shell of all of the russian dolls um this video on youtube six numbers uh episode six or lecture six recursion and dictionaries from the mit open courseware is excellent there's a bunch of really awesome um visualizations and stuff that this professor goes through they use python in this uh in this course and he kind of goes through a little bit of the code but doesn't go too too deep in depth so um the content that we cover today will be sort of the ruby version that is the actual live demo of some of the solutions to the problems that were presented in this episode there's also a really cool um visualization of towers of hanoi which is shown both here on the side and also that's this like fisher price game where you have the towers and you try to move the rings from one tower to the other tower without ever putting a big ring on top of a smaller ring and we'll talk about that process in just a bit so that's what recursion is let's jump into some code so i'm gonna open up a new file here called recursion the other thing i wanted to do was like show how to write inline r spec files so i'm going to require r spec here at the top and in practice if you wrote like a method called foo and you wanted to test it out you could say like oh like puts foo and then you'll see the output of foo and then later on if you want to write another method um then you would have to just like keep writing put statements or keep writing inspect statements to to test your code as you go but for these recursive methods i wanted to write some tests for them and so we can actually just require our spec directly at the top of any old ruby file and then at the bottom we can say rspec dot describe recursion problems and then we can actually just like start writing um tests in here so i'm gonna say it uh works for a simple base case and the very first the very first exercise that i wanted to go through was implementing the code for the fibonacci sequence so the fibonacci sequence is this sequence that looks like this 0 1 1 2 three five and can you see the pattern at this point if you're not familiar with the fibonacci sequence try to identify the pattern here um so we've got zero one one two three five eight thirteen twenty four this is like a fibonacci this is a sequence that's used in a lot of um different things i think it was originally developed by trying to figure out how many bunnies would exist in a box or something after they reproduce and um the the example goes in more depth in that video for mit but this is a really famous sequence where each number is the sum of it the previous two numbers so 13 here is the sum of 5 plus eight and if we look at three that is the sum of one plus two and twenty-four is the sum of eight plus thirteen that's actually not true it's twenty-one okay uh so this is it's always like whatever the number is it's like the sum of the previous two numbers so what would be after 13 or i'm sorry what would be after 21 well it's going to be the sum of 21 and 13. what is that 34 i think right so we could we could go on forever and try to figure out what the the fibonacci sequences but we could also write a method to figure out these fibonacci numbers and this like determining the nth fibonacci number is a problem that lends itself really well to recursion so let's look at how we might implement this so we have let's make our method called fib of n where we're going to pass in some number and it's going to give us back the nth fibonacci number so there's two steps that you need when you're writing a recursive function first you're going to need a base case and the base case is going to be checking something about the input and it's usually the base case is like the smallest possible case that you can look at or maybe the first two or three or four possible cases and then after you have your base case set up then you're gonna like actually um recurse or like call the same i like call myself basically right you're gonna call the same method over with a one smaller version of that same problem so if we're looking at this problem right and we wanted to pass in some number maybe we wanted to pass in one and we wanted to give us back the number one or we wanted to pass in five or six or something and it gives us back the number five so what we might do here is say that first we need to implement a base case and the base cases in the case of fibonacci sequence is if we get in a zero so if we receive n is one maybe we want to return zero so return zero if n is equal to one so that's going to be like one base case then the next base case we could say return one if n is equal to 2 because that's going to be this number here and then we could also say return 1 if n is equal to 3 and return 2 if n is equal to 4 right so at this point we have like several base cases that are implemented um and we can actually at this point also maybe we'll just start with this works for a simple base case so fib expect fib of uh of 1 to equal zero and then here i'm going to say r spec recursion and that should run our r spec test over here and we see one example and zero failure so let's comment out our case there just so that we have a failure because we want to start off with a red test we want to start off with a failing test we'll comment our base case back in and we'll run this again okay so it works for a simple base case um it works for a second base case and we'll say expect fib of let's say two dot two equal one and then we will run this again we'll run our our spec test and we see that we expected one but we got nil that's because we haven't actually commented in our base case here and we'll run this again so now we have two examples that are working right and if we wanted to go even further and say that it works for a third base case then we can say fib of 3 is also going to be 1. okay we'll run our test and we get our failure and then we comment in our base case then we run this test again and so on and so forth right okay so we've got our base cases those all work now let's try a more complex base case or a more complex case so it works for a let's say a complex case and we'll pass in fibonacci of i don't know seven okay and so we have one two three four five six seven so we expect it to give us back the number eight so this is going to be the number eight now if we run our test and it's going to fail okay so what i'm going to do is i'm actually going to move um these base cases up so that we can keep all of the test code on the page while we have or i'm sorry the implementation code on the page while we're looking at our tests so that it makes it a little bit easier to follow but in practice i would split kind of split these up a bit more okay now we need to to start on the process where we're calling ourselves so in order to get the fibonacci of n recall that in order to find the the like the next fibonacci number we can add the previous two and so fibonacci of uh of seven or i'm sorry yeah seven here right one two three four five six seven fibonacci of 7 is the same as the sum of fibonacci of 6 plus the fibonacci of 5. so if we say fib of 6 plus the fib of five that should give us back our our case down here for fibonacci of seven because it's the fibonacci of the previous one and the one that's like before the previous one so if we run this again um oh we actually have stack level two deep so shoot my uh my example did not work all right so let's actually say this is fibonacci of five sorry fibonacci of five so one two three four five so we're trying to find this number as three so we're gonna say three and then here the fibonacci of five is going to be fibonacci of four plus the fibonacci of three right so now if we run this again we have a passing test so that works for a more complex case now um we also want this to work for fibonacci of seven which should give us eight right yes okay so now at this point right like the way that we have written our tests they sort of work for those base cases and they work for one complex case but now we need to figure out like how do we derive four and three um instead of passing four and three we wanna pass something else down to fibonacci so instead of four and three we can actually pass n minus one so if we're trying to figure out you know the seventh fibonacci number then we wanna pass six and we can figure out six based on you know 7 minus 1 and then we also want to pass in 7 minus 2 and that's going to give us back the the n minus 2 version so now we are going to call fibonacci with n minus 1 and fibonacci with n minus 2 until we get to our base cases so if we run this recursion again now we have all of our examples passing so this is going to work for both a our both of our base cases all of our base cases and a more complex case let's even like make this this more like go off further so one two three four five six seven eight nine ten the tenth one should give us 34. okay so if we run this all right now we are we're still passing all of our tests that is fantastic and looks great so we have uh we have a working solution for finding the fibonacci number and it is recur this is using recursion because we are calling fibonacci from inside of itself now one thing that's important is that every single time that the fibonacci function is called the n is going to be associated with that function call so it's going to be tied to that function call and we're going to have frames so every single time we have a method call we're going to be pushing a frame onto the stack which has reference to n so when we call fibonacci of n with some number that's going to push a stack onto the frame with that n so in the case of 4 we pushed on a uh or i guess case of five so uh the other thing too is that like it turns out that we can actually figure out two based on uh the previous two numbers here so two and three so we can actually just start removing some of these base cases and our code will continue to work right and furthermore we can actually figure out one based on zero and one so we actually don't even need that base case so we can kind of like just keep going down further and further and in fact uh let's call that good but like yeah there's there's ways that you can make it even fancier okay so every time that you call the function you're getting a new frame and that frame is going to have reference to some number and that number is going to be associated with the frame that's getting pushed onto the stack so if we call this with fibonacci of 4 then a new frame is going to be pushed onto the stack and n is going to be equal to 4 and then as soon as we call fibonacci here with n minus 1 we're going to be calling fibonacci with 3 that's going to push another frame onto the stack so we have fibonacci of 3 and at that on that frame n is going to be equal to three that's a separate frame from the one below it which has n equals four so n equals three is going to be pushed onto the stack then as soon as we get we enter the uh the function with n equals three we're going to get down to this call again and when n is equal to 2 we're going to push another frame onto the stack and then it's going to see that n equals 2 is here and it's going to return 1. so that very first or like that last frame that we pushed onto the stack where n was equal to 2 that's going to pop off the stack and return and then we're going to have the return value here is going to be 1 we're going to have 1 plus and then we're going to have to call n minus 2 again and that's going to give us back 1 and that will return zero and so one plus zero is going to be returned from fib of three so it's kind it's a little bit tricky to follow but keep in mind that every single time that we call the function a brand new frame is being added to the stack to the function call stack and every single frame has reference to its own variables um or like it's its own like arguments here so this is not we're not using a closure here like this this function is not closing over any other any other variables every single variable that we're accessing here was either passed in or defined inside of the function we didn't actually define any in here but um the only reference that we're using that is like getting closed over is the reference to the fib function name itself so that's the fibonacci sequence this is how you solve it and this is a really common problem that you might solve with recursion so also down in my r-spec method here i want to say describe describe and then i'm going to say fib or just fib and i'm going to put these into a describe block because i want to next go on to another recursive function called factorial so we're going to say describe fact okay factorial finding the factorial of a number is another really common use case for for using recursion and so let's make a new method here called fact and then talk about what what the factorial is so factorial is when you have some number you might have like five bang this is like a way to write the factorial function is like the exclamation point at the end um and it means like if you have if you have five bang or whatever it means five times 4 times 3 times 2 times 1 all the way down right and so um if you if you look at this right like 5 times 4 times 3 times 2 times 1 is really like whatever the the number is times the factorial of the the next lower number so this is actually like we can rewrite 5 factorial as 5 times 4 factorial right because 4 factorial is going to be 4 times 3 times 2 times 1 and then like 4 factorial can be written as like um this one can be like 4 times 3 factorial right and so on and so forth and so the smallest the smallest base case here is going to be one as soon as n gets down to one we just return one and then we're going to multiply that by the previous number so let's try to let's try to implement this so it works for a simple base case and we're going to say like expect that factorial of 1 to equal 1 and we'll run our code and it's failing because we are not returning anything yet so we're gonna say return one if n equals one the other thing that's worth the other thing that's worth mentioning here is that like i'm not handling um invalid input so if someone passes in negative one or if they pass in banana or if they pass in some object it's not going to work it's going to crash i'm assuming that like some validation some input validation is happening before this and we're just talking about recursion all right so factorial of n uh return one if n is one and then um otherwise we wanna return n or so this is this is where we need to figure out like our recurse right so this is going to be our base case and then we need to call our call myself call my self and so earlier when i made a bunch of base cases right when i was writing out i wrote out all four base cases for the fibonacci sequence this can be a really helpful exercise when you're trying to just think through the model of what um what the pattern is for some sequence of numbers and so in in the case of factorial right like we could write okay return two if n is equal to two and then return like six if n is equal to three um but it's a little bit harder in the case of factorial to sort of visualize it without going really far or the other way you could do this is like return one times two and then return one times two times three that might be like another way to sort of think about how we wanna return from this function let's also add a couple more base cases here so we're going to add like two and then three we expect this to be two and six and if we just run this it should pass and it should totally work right because we've written them all out as base cases now we can say it works for a more complex case and here i'm just going to write like expect um 4 2 equal what is that 24 i think 6 times 4 24. all right so now we've got to figure out how to like actually call ourselves so we know we're going to call a fact of um probably one smaller recall that like up here we were we were using like oh 5 factorial is the same as 5 times 4 factorial so here we were passing in if we were passing in 5 then it would be like call 5 times and then 4 factorial so 4 factorial here we have it written with an exclamation point but really this exclamation point is the same as like the function call passing in n so if we say factorial of n minus 1 this gives us 4 factorial that gives us like if we were passing in 5 as our number this gives us 4 factorial right and so then we want to say like okay n times right because that's going to give us the n times that number and so if we now run in fact like this might even be an even better way to do this is like what if we do like factorial of 1 times 2 and then we want to do like factorial of 1 times factorial of 2 times 3. that's like a better even better way to visualize it right like we have some number and then we have like the factorial of this one smaller times the times the number or the factorial of one smaller times the factorial of two smaller times the fact times the number um so that sort of gives us the pattern so now if we run our test we have five passing examples so we know this is working and we can actually start removing some of these base cases so if we run it now it should continue to work and i think we actually only need this one base case return one if n is equal to one and then everything else should just sort of work and it's so elegant it's so beautiful right just like two very simple statements right return one if n is equal to one great and then otherwise return n times n times the smaller the next smaller case and then we kind of like work our way all the way down so that is factorial i'm going to remove these from the base cases because um yeah so let's let's uh let's also like come up with a more fun case i guess factorial of six so what is that going to be that's like 24 times 5 uh oh gosh 24 times 5 times 6 is going to be 720 i think all right let's run this okay we have five passing examples and that that looks great um okay so we have now two examples of recursion that are really useful and tell you or show you how to like call a function from itself let's keep going let's do one more we'll do towers of hanoi so in the towers of hanoi you have you have three towers and on one tower it starts off with the discs in order of smallest to the biggest on the bottom and your task is to move the disks from one tower to another tower and the limitation is that you're not allowed to put a disc that is bigger on top of a disc that is smaller so you can you can put any size smaller disc on top of a huge disc but you can't put a big disk on top of a small disk and you want to like move all of the disks over and so there is a there is like a small algorithm which is like if you're trying to move all of the disks from tower 1 to tower three you're going to use tower two the tower that's in the middle as like sort of your alternative tower and you'll move maybe if you start off with just one disk right that's going to be your base case so you want to move one disk from some tower to some other tower right and so you can move directly if you have one disc you can move directly if you have two discs you need to move the smallest one first to the alternative tower then the one that was underneath it to the big tower and then the small one to the big tower on top of the bigger one and then if you have uh if you have three discs then you need to move the small one to the big tower first then the medium one to the middle tower then the big one oh no then the small one back over to the the the medium or the middle tower and then the big one to the final tower then the small one back to the first tower then the middle one to the final tower then the small one to the final it's there's there's visualizations online like uh go go check it out but the way that we're going to do it is that we're going to have our towers are going to be arrays so we're going to have three different towers and to keep it simple we'll keep our arrays we'll keep the representation of the disks as integers in the array so let's describe describe let's see towers it like moves and or like it moves one disc right so this is going to be this is going to be our smallest case and so we need a from tower that's going to have one disc we need a 2 tower that's going to start off as empty and we need like an alt tower or like some temporary intermediate tower and then we need a method to call so we're going to expect that if we call towers and then we're going to pass it the number of disks we could just pass the from to an alternative but in this case let's pass the number of disks so we'll pass one as the number of disks then we're going to pass the from the 2 and the alt tower and uh actually we don't wanna like write expectation about the return value of that we're gonna ex we're gonna have an expectation that like this one disk was moved to the two tower so we're gonna say expect that from to equal the empty array and expect 2.2 equal one and expect alt to equal empty array so we want the the alternate after the end of this the alt should be empty the from should be empty and two should have one inside of it so let's run our tests okay they are failing undefined method towers so let's go make it a method called towers def towers and it's going to take the number of disks num disks uh i'd use okay whatever from to and alt okay so if the number of disks is 1 then we want to move from the from tower to the 2 tower so if num disks is equal to 1 then we want to like remove it from the end of the from tower also like for for reference um in the future we're gonna have our disks ordered like this like three two one so like the biggest disk is gonna be at the front and the smallest disk is gonna be at the back and so when there's one disk left we want to remove it from the right side we want to remove it from this side so in ruby in in a ruby's array we can call from.pop and that will return the um the last element and it will remove it from the array and so then we can say two dot push and that will push it onto the end of or like sort of the top of the stack for the two otherwise we need to do our recursion so like uh recurse so this is going to be this this part is our base case and i i hope that this works this should work i think um okay we've got six passing tests so it's it's able to move one disk so that's great um now let's try to get it to move two disks um so it moves two disks so we're gonna start off our from with 2 1 and we want our 2 to end up as 2 comma 1 and we need to pass in 2 disks and we want our alternate to end up as an empty array if we run this right now it's going to fail because um it's yeah it's empty because the number of disks was not one so in order to um to call myself call myself here we want to figure out what the smaller version of this is so if we call towers again we almost always are going to like call ourselves with one smaller like a one smaller version so let's just say like num disks of minus one so that gives us like n minus one basically now the trick is when you are calling yourself in the case of towers of hanoi with the one smaller version the the location of the from 2 and alternate are going to change so instead of passing from directly in here i think we want to pass in well we still want to take off from the from right and we still want to put it somewhere but in this case i think the 2 is going to be the alt and then the alternate is going to be the two i know it's confusing but let's let's just see if that works uh [Music] okay so it did not work we have a failure we expected empty array to be in from but we got the array of 2. so um right the other thing that can be helpful here is like every single time we get down to one disk we can print out like um like moving disk uh from and then we'll print out from to and then we can print out two and sometimes that can be helpful to see so let's print it out as we go and i don't see it num disks equals 1. i was expecting to see a little bit of output there all right so after we've called it with numdisks minus one um we also need to ultimately get it down so that we we are we're moving just one final disk from the from to the two and alt and then uh again we have to move so like the base case is that you have you have some tower with n disks and you move all of the disks that are on top of the smallest disk you want to move all of those to your alternate position then you want to take i'm sorry then your biggest disc at the bottom then you want to move all the discs that are smaller than the biggest disc to the alternate location then you want to take your biggest disc move it to the final location and then it's another problem of like okay now we have to take all the ones that were on the alternate location and move those to the final location so we we move all of the disks that are smaller than the biggest disk at the bottom from the from location to the alternate location then we're going to move our one bottom disk from the from to the two then we need to move all the disks again so again it's going to be num disks minus one from the from the from the alternate location to the two location with from being our new alternate location so i think this might do the dance okay so now we see our print statements coming out so we're moving disk from one okay so this is this is the more complex case where we have two right so we're moving from the from tower that has two and one to an empty and that's going to be the alt then we're going to move two to so this is going to be our next case here now we're moving two from the from to the empty again so now the empty is the ultimate final destination this first empty is the alternate the second empty is the final destination and then um then we're getting down to our final call here where we're moving one from the alternate location to the final destination all right so i think we're at a place where we can say it moves like i don't know five disks or something and we'll just say like five four three two one and we're going to be moving five disks and we expect the two to have five four three two one and let's run this okay so let's take a look at this output here so it's a little bit hard to read but we do start with five four three two one and we're moving it to the two what we wanna do as we're looking at this output is ensure that they're always in descending order all of the arrays are in descending order because our constraint is that none of we should never be able to put a bigger disk on top of a smaller disk and so it looks like it's actually working totally as expected in that passes so this is another example of recursion we're solving towers of hanoi problem by sort of we've got our base case and we've got these bigger cases now i wanted to talk about one final piece and that is how we can optimize for recursion so if we go back to our first or how we can optimize like not re doing the same math over and over when we've already solved that problem so if we look at the case of fibonacci right if you put in some fibonacci number that is really large like let's say let's say even that we're going to do i know one two three four five six we're trying to figure out five in the process of figuring out five we're calling fib of n minus one so we're calling fib of four and when we figure out fib of four we're going to figure out we're also calling fib of three so when we call fib of four that will also then ultimately call fib of three so we're going to end up calling fib of three like six or seven times but we only really need to call fib of three one time because as soon as we know what the resulting value is for the fibonacci of three we can just use that in all future calls right and so there's a couple different approaches to uh optimizing this but what one thing i wanted to do was add a brand new style of test called a timeout test so we're gonna say require timeout this is another gem that you can use as part of your test suite and i'm going to come down to our fibonacci sequence thing and say like it is efficient do and and if we were to do like expect fibonacci of like 100 or something um to equal i don't know what it's going to equal so i'm just going to put in like garbage for now i i assume that the calculation is actually correct but uh now if we run our tests you'll see that it's pausing right here like the execution is stuck it's hung because calculating the fibonacci of 100 is doing those like fib of three figure three fib of three number three there's like a thousand times just trying to or like way more than a thousand times just trying to figure it out so i'm gonna cancel that um so that uh i think that ran for 19 seconds let's let's see what some like decent upper bound is like maybe 20 does fib of 20 hang um fib of 20 okay so fib of 20 wasn't big enough fib of 40 maybe um there fib of 40 is hanging for sure okay so what we want to do is ensure that our tests are passing in okay there we go we got some number great so let's let's use that as our result let's assume that's like correct um but the problem is that took nine seconds nine seconds to figure out what the fibonacci of 40 was and that nine seconds like if you think about the fibonacci this this method or like you think about the calls to this method as like a tree right like every time you call fib of a number we have two new branches of the tree so if we call with fib of five it's going to call a fib of four and fib of three and then if in fib of four it's going to call fib of three and fib of two and then fib of three is going to call fib of two and fib of one and fib of two is going to call fib of one and fib of zero like it's it's basically we're building up this giant tree and we really only need to solve the very left most branches of the tree and the rest of the tree we've already figured out the answers to those questions and so we need to like reuse those same answers and so there is a term called caching or memoization memoization it's not memorization you're not memorizing it it's kind of like that it's like basically the same thing but memoization is when you're storing that result so that you can use it again later so let's use this timeout library to write a little timeout thing down here so we're going to say like i think it's like timeout.timeout for one second um and we want to expect that a block runs and it doesn't um raise an exception for or raise error raise error for timeout error so we want this to not raise a timeout error when we call fib of 40. well actually i think we can i can't remember if we can keep both expectations in there but let's just see so timeout should actually time out after one second yeah okay so it timed out i think let's see um using expect not to raise uh risks false positives i know i know i know all right so i don't care so i'm going to like suppress that warning by adding this one liner that's in the output and then we see it hang and it fails because after one second it timeout it times out so execution expired so this timeout thing says like if your test takes longer than once or if like the block takes longer than one second then raise an exception and fail and so what we want to do now the problem now is we need to increase the efficiency of fibonacci or improve the efficiency of fibonacci so that it can calculate the fib of 40 in less than a second and the way that we're going to do that is with memoization so we sort of need to keep track of previous answers of previous answers so if we have there's a couple different ways we can do memoization or caching the way that we're going to talk about today is by using a dictionary so we're going to use a hash where the keys in the hash are n so we're going to have some result maybe we're going to have 3 and that's going to point at the the answer for fibonacci of 3. so 0 1 2 3 so that should give us back 2 and then we're going to save the fibonacci of 4 and that's going to give us back 0 1 2 3 4. so we only ever have to calculate it once and then the next time if we have already figured out the answer to that like smaller version then we'll just return that so the first thing we need to do is figure out how we're going to like pass in or reference or use this this memoization container or this hash or whatever and so one way that i like to do this is i'm just going to pass in um a a dictionary here this is going to be a like an argument that has a default value of an empty hash and if we call fibonacci with n and don't pass anything it's just going to use this if we pass fibonacci of n with something then it will use that something that we pass in so now what we want to do is we're going to continue passing memo forward for both of our other calls and this is still going to fail like if we run the test now it's still going to like it should it shouldn't fail because we added an argument it's going to fail because it's still slow and so the next thing we need to do is actually speed it up and so what we want to do is we want to say return memo at n if memo dot has key n so if it has the answer already if it's already got the answer then we want to return the answer if it doesn't already have the answer we do need to calculate the answer and then go down that branch of the tree and we're going to store the result into memo memo of n is equal to the return value of fib of n minus 1 fib of n minus 2. all right so now if we run this we have a passing test and that was like boom super fast super fast so it's running in like way less than a second to calculate fib of 40. um and like we could now we can even do like fib of 100 probably and it will like just like nope nope like uh we don't know what the fib of 100 is well i don't have it memorized but like maybe some of you do but now we should be able to run it and it should like not just hang like it should just now it's just eating it up like no problem because we're only calculating a super small sliver of the tree and then we're reusing our answers right um also just for fun why don't we like print out p memo as we're like calculating it i'm gonna remove the put statements from towers just so that we can see it better so huh weird uh i expected oh you know what it needs to be after this p memo all right so now we've got like all of our answers right so like this is the fibonacci of 100. this is the fibonacci of 98 97 96 90 whatever uh this expectation didn't actually run but like the thing that we cared about was was this outer one um not raising an error as we're running it and so what's really cool is like you can see in our in our memo right three uh points at one four points at two five points at three six point five you'll notice also that it does not include zero one or two and that's because those are parts of the base cases right um cool so this is how you might set up memoization and improve the speed of a recursive function one thing to think about as you're trying to test these out and experiment with with recursion if you write a function that calls itself and you call it and there's no base case here let me show you what error you're going to get and you've you may have already seen this in uh in your experience with writing ruby okay here we go system stack error stack level two deep if you see stack level two deep it's almost it's like 99.99 going to be the case that you have a um a recursive call that has no base case okay so this is basically the same thing as a while loop like a wild true loop with no way to break out of it it is an infinite loop where the instead of like just looping forever you're going to uh recall that every single time we call the function we're pushing something onto a stack we're pushing a frame onto the stack so every time we call f we're pushing a frame onto the stack frame onto the stack frame onto the stack and there's a check at some point where you're like 11 000 stacks deep in ruby where the stack level is going to be just too deep it's not going to make any sense and so instead of just infinitely running on forever and taking up all your memory it pops and then just raises this exception that says hey the stack level is too deep so you might see this error as you're experimenting with recursion in this case um you'll you'll have to just know that like there's a missing base case and it's not breaking out of the recursive calls at some point so i think that's all i want to show this is my thoughts and experience with recursion a couple different solutions we talked about fibonacci sequence solving factorial towers of hanoi there's a bunch of other really common use cases for recursion like traversing a tree especially like depth first search of trees and graphs and things so i would really encourage you to check it out and experiment with recursion that said there are some really these are like really common use cases for recursion make it really elegant solutions but you can solve almost all or probably all the same problems that you can with recursion with iteration so just looping while loop for loop whatever so this is just like another handy tool to have in your toolkit thank you again so much for watching for your time and attention if you're liking the content on the channel i would love a thumbs up because that lets other people know that this is like good content and uh yeah if you'd like to see more ruby or web development content i would love to uh to have you around so hit that subscribe button thanks again and we'll see you next time

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