## 5 Simple Steps for Solving Any Recursive Problem

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today we are going to talk about

recursion I think most computer science

students would agree that recursion felt

computer science topic in this video I'm

gonna show you five simple steps that

recursive problem and hopefully these

reality recursion isn't actually that

I'm gonna show you how to apply these

steps in the context of solving three

specific recursive problems each of them

progressively more difficult than the

last so make sure you stick to the end

of the video and as always I think

you'll get the most out of this video if

you take the time to pause and ponder

some of the problems that we work

through and the steps we went through to

solve them so without further ado let's

get started

suppose we want to solve the following

problem recursively write a function

that given an input end sums all the

non-negative integers up to n here's a

couple quick examples of inputs to this

function to give you a sense for what we

are trying to implement a lot of you may

have seen an iterative version of this

function or even a mathematical version

of it that we went during a previous

video but if you had to solve it

recursively how would you do it

recursion is all about taking a problem

and solving it using simpler versions of

the problem so the first step in solving

a recursive problem requires you to ask

yourself what's the simplest possible

input for the function here the simplest

possible case is n equal 0 where we know

that the result should also be 0 the

simplest case often translates itself

into a base case for our recursive

problem the base case of a recursive

function is the only input where we

provide an explicit answer all other

solutions to the problem we'll build on

the base case the second key step to

solving recursive problems is trying

examples and visualizing how the inputs

and outputs of these functions interact

in this problem one fun way to visualize

things is to treat our sum as building a

triangle with the base as the input

value that way

some is a total number of individual

blocks in this triangle you can try as

many examples as you want once we have a

set of examples we now want to move to

the third step which involves relating

larger examples to smaller examples take

a second to see if you see any

relationships between the cases here

that pop out to you put in more specific

terms with these examples ask yourself

hey if I was given the answer for the N

equals 4 case could I solve the N equals

5 case if I was given the ancestor N

equals 3 case could I saw the N equals 4

case one interesting relationship that

you may have noticed is that if we know

the answer to the N equals 4 case all we

get the solution to our N equals 5 case

and similarly if we know the answer to

the N equals 3 case we can get the

4 it's not too hard to convince yourself

that this should be true in general but

if necessary feel free to try some

larger cases to verify this now brings

us to the fourth key step which is to

generalize this pattern let's say we

want to figure out the sum for the input

N equals K we can find the sum by first

taking the sum up to N equals K minus 1

and then all we have to do is add K to

this value once we've generalize a

pattern the last step now involves

writing code by combining this recursive

pattern with the base case this actually

turns out to be not too hard since the

code almost writes itself from our base

cases and recursive pattern let's dig a

little bit deeper to see how this really

works for a particular input let's say

we run this function for input N equals

5 what actually ends up happening well

we go through the initial function and

then hit a recursive case which adds 5

to the result of calling out some

function on the input 4 so we know that

this should theoretically give us the

10 but technically this function hasn't

been evaluated yet what actually ends up

happening is we have to go back into our

function and repeat the process we hit

the recursive case again which then

causes us to now add 4 to the value of

sum called on 3 which once again kicks

off the same process as before let's

take a moment to see how this unravels

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this continues until we hit the base

case which then is used to softs the sum

function called on n equals 1 which then

solves a sum function called on n equals

2 and so on and so forth until we

eventually build up the solution for our

original N equals 5 case it's always ok

to take some time to unravel the

recursion to ensure that things work but

as you get more experience solving these

type of problems there's a concept

called the recursive leap of faith that

I would like you to think about the

recursive leap of faith makes solving

recursive problems much faster it's all

about assuming that easier versions of

the problem will be correct in this

particular problem the recursive leap of

faith would say that if I wanted to

solve some called on n equals 5 let me

just assume that some on N equals 4 is

going to work and then if I add 5 to

that some call on N equals 5 will also

be correct don't worry if you're a

little hesitant to believe in the

recursive leap of faith but it is a

useful problem-solving trick that helps

in more complex recursive problems let's

now explore a more involved recursive

problem here's the problem write a

function that takes two inputs N and M

and outputs the number of unique paths

from the top left corner to the bottom

right corner of an N by M grid one key

constraint is that you can only move

down or write one unit at a time before

we apply our five steps

let's first familiarize ourselves with a

few inputs and outputs for this function

here's what happens when N equals two

and M is equal to four

[Music]

we end up with exactly four paths and

here's another example with N equals

three and M equals three

[Music]

we end up with six unique paths all

right let's begin let's play around with

some simple inputs and see what happens

the simplest possible reasonable input

dysfunction is a one by one grid where

we end up with exactly one path if we

actually take a second to try some other

simple inputs you might notice something

interesting we can actually make a

fairly comprehensive base case by

noticing that if either one of these

dimensions is going to be one we have

only one unique path this can be written

quite concisely as a following base case

often with base cases you want to find a

way to be as comprehensive as possible

don't be afraid to adapt though it's

completely normal to come back later and

modify the base case let's now take a

look at a decent sample size of examples

to get a feel for the problem when

coming up with examples for recursive

problems it makes sense to do them in a

sweeping manner where a lot of the

examples are really close to each other

remember our ultimate goal is to find a

way to solve any specific case using

simpler versions of that case so it

makes sense to have examples that are

close to each other all right so now

that we have a sizable set of examples

take a look at these and see if you

notice any interesting relationships

between cases I know that's a lot to

look at so as a hint let's narrow down

our focus to just these three examples

do you notice anything how about now

there's actually a really nice

one-to-one correspondence between each

path and the two by three and three by

two cases two each path in the three by

three case for each path in the two by

three case we can take that path and go

one unit down to create a path in the

three by three case and for each path in

the three by two case we can move one

unit right to create a path in the three

by three case so at least in this

example we can confidently say that the

number of paths in the three by three

example is the sum of the number of

paths in the two by three example and

the number of paths in the three by two

example but is that just a coincidence

let's see if we can test this pattern by

working backwards let's try to find the

paths in the three by four case by

analyzing a similar set of easier

examples following the earlier pattern

let's see if the number of paths in the

three by four case is really the sum of

the Pat's in the three by three case and

two by four case we'll take each path

for the three by three case and move one

to the right to make a corresponding

path for the three by four case and then

we'll take each path in a two by four

case and go one step down to make a path

in the three by four case and it turns

out that the set of paths that are

created from this construction end up

being all the unique paths to a three by

four grid take a second to verify this

for yourself if you're skeptical

[Music]

so let's now generalize this pattern we

can find the total number of unique

paths in an N by n grid by first finding

all the unique paths from the N by M

minus 1 grid each of these paths

correspond to a path in the N by n grid

since all we do is we move one unit to

the right to create an Associated path

and the remaining paths can be found by

finding all the unique paths in an N

minus 1 by M grid where we now take each

path add one unit down and get the

corresponding path in the N by M case

once we found this general relationship

the hard part is now over the last thing

we do is take this general pattern bring

it together with a base case and now

write code the corresponding code for

such a seemingly overwhelming problem

ends up being remarkably simple and

elegant and you'll see this a lot in

recursive problems speaking of hard

problems let's take a look at a really

challenging problem here's the problem

write a function that counts the number

of ways you can partition and objects

using parts up to M as we've done so far

let's take a look at some examples to

get a sense of what this problem is even

asking for N equals 6 and M equals 4

here's how the partitions work out

[Music]

we end up with a total of nine unique

partitions let's take a look at one more

example to make sure we really

understand what's going on here's what

the partitions look like for inputs N

equals 5 and M equals 5 here we have

seven unique partitions and make sure

you take a second to really understand

why these are the only valid partitions

all right so this problem definitely

feels a little intimidating since it's

not really obvious how we would even go

about starting to solve it but this is

where we're gonna have to fall back on

our five steps

thought for what the simplest possible

input should be is when n is equal to

zero and M is equal to zero but what

does this mean what it's asking is how

many ways can you partition 0 objects

using parts up to 0 this may be a little

bit mind-blowing but there's actually

only one partition and that partition is

to include no parts and in fact trying

inputs N equals 0 with M equals 1 and

then N equals 0 with M equals 2 is

really no different this allows you to

make another fairly comprehensive base

case where the number of partitions is 1

if n is equal to 0 but there's some

other simple cases that we should also

check out if n is equal to 1 and M is

equal to 0 how many partitions do we

have how many ways can we partition one

object if we are only allowed to use up

to 0 parts well this is actually

impossible we need at least one part so

there are 0 ways to partition these

inputs and the story stays the same no

matter what the N is so this is the

first example of a problem where we will

need to separate base cases the second

one being that if M is equal to 0 we

have zero partitions all right let's now

visualize a sizeable set of examples as

before let's look at examples that are

relatively close to each other in

numbers so that it will

make it easier for us to find patterns

in the following steps once again if

don't want to be shy about writing out a

ton of examples not only will it make

you more comfortable with the problem

insights to the solution

[Music]

so now that you have a good set of

examples take a moment to see if you

notice any interesting relationships

among the partitions feel free to pause

the video if you're struggling to find

anything let's narrow our focus to a

subset of these examples all right so

the first interesting thing you might

have noticed by looking at these

examples is that a lot of the partitions

on the larger problems are repeated in

the easier version of a problem isn't

that a little bit wild these are the

types of patterns that show up a lot in

these types of recursive problems and we

definitely want to see if we can verify

if this holds for larger examples let's

take a look at the N equals 7 M equals 4

case we have a total of 11 partitions so

going off our previous example

we're expecting all the partitions of N

equals 7 M equals 3 to show up in these

partitions and in fact when we go

through the partitions we indeed find

that all 8 partitions from the N equals

7 M equals 3 case show up in the N

equals 7 and M equals 4 case how in the

world does that work what crazy sorcery

is going on here the reason this happens

is actually surprisingly simple if you

take a moment to think about it from a

perspective of how you are actually

creating these partitions you'll get an

intuitive sense for why this is the case

when you're trying to partition 7 using

parts up to 4 at some point this is

going to include all the partitions of 7

using parts up to 3 there's just no

other way around it more formally we can

assert that the partitions of n using

parts up to n minus 1 is a subset of the

partitions and end using parts up to M

okay so this seems like a pretty good

start we now know that a subset of the

partitions of n objects using parts up

to M can be found from the simpler

problem of partitioning and objects

using parts up to M minus 1 but what

about the remaining partitions where do

these partitions come from

let's analyze these cases what do they

all have in common do you see it they

all happen to use the actual M value as

part of each partition and this should

make sense because the subsets that we

were moved or partitions of n using

parts up to M minus 1 so the partitions

that are left here that use M represent

the only other case at this point you

might be thinking ok great

so the remaining partitions all use M

how does this actually help us well

let's think about what it actually means

to use em in a partition once we declare

hey I'm going to go ahead and use em as

a partition this is what we end up

having left in each of those cases all

that's going on here is we subtracted M

from our original n objects so we have a

new end value with the same and value as

before and check this out we actually

have another one-to-one correspondence

if we add em to each partition in the

smaller problem we end up with the same

partition in our original problem at

this point you might still be a little

skeptical because we tried this only on

relatively small examples so to really

verify this let's try one more large

example and this time we're gonna try

work backwards like we did in the grid

problem let's look at the case where n

is equal to 9 and M is equal to 5 based

on our current observations we expect

all the partitions in this case to

correspond to the partitions in the n is

equal to 4 and M is equal to 5 case and

then the N is equal to 9 and M is equal

to 4 case remember the first sub-problem

represents all the partitions that use

our original m as part of the partition

when the second sub-problem represents

partitions that use parts up to n minus

1 here are the partitions for the N is

equal to 4 and M equals 5 case by adding

M to each of these partitions we

generate the corresponding subset of

partitions for the N equals 9 and M

equals 5 case and combining this with

the 18 partitions from the N equals 9

and M equals 4 case we end up with 23

total partitions for the N equals 9 and

M equals 5 case and you can take the

time to verify this if you wish but

these do actually end up being all the

partitions for N equals 9 and M equals 5

case ok let's now generalize this

pattern if we want to solve the count

partitions problem called on a and B it

will end up being the sum of the count

partitions called on a minus B and B and

then count partitions called on a and B

minus 1 this is the key recursive

relation that will solve this problem

let's now execute the final step and put

together our recursive relation with the

base case one interesting wrinkle that

we will have to take account of is that

one of our recursive calls provides an

argument that is n minus M some of you

may have noticed that it's perfectly

valid to have cases where n is actually

less than M which then means that the

new kalt will have a negative argument

to count for this we have to include

this corner case in one of our base

cases if the N value provided is

negative we will have zero partitions

this problem is a good example of how

you might have to modify some of your

base cases

after finding a general recursive

pattern the cool thing about recursion

is that for such a tricky problem our

final solution actually ends up being

only six lines of code

[Music]

so to recap what we did in this video we

went over five key steps to solve any

recursive problem and then went about

applying these steps to three problems

of various difficulties you learned the

importance of starting simple and asking

yourself what the simplest possible case

is you found it helpful to work through

some examples to get a sense for what

the recursive problem was asking you

also went about relating hard examples

to easy examples to find a generalizable

recursive pattern and then with this

pattern in hand you went about writing

the code as with anything you learned

you're going to have to spend some time

to really contemplate and work through

recursive problems to really begin to

master solving them this video is all

about giving you a mental framework to

do that and I hope it was helpful in

that sense thanks for watching and as

always I'd appreciate it if you liked

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in the next video

[Music]