The Fast Fourier Transform (FFT): Most Ingenious Algorithm Ever?

the world of algorithms is vast but we can often split them into two distinct classes the first class is those that are inherently useful think of your standard graph algorithms like dfs or bfs these algorithms show up all over the place they are efficient and as a result we like to understand them the second class of algorithms we study are ones that are just purely beautiful think of the first time you saw the incredibly simple recursive implementation of towers of annoy if you have a soul you feel a sense of wonder a sense of awe at the elegance of such an algorithm it happens to not actually be that useful or efficient as a matter of fact but we still study it just as we like to read a work of fiction it inspires us and motivates out-of-the-box thinking today i want to talk about an algorithm that rightfully belongs in both classes and my personal favorite algorithm the fast fourier transform or fft is without exaggeration one of the most important algorithms created in the last century so much of the modern technology that we have today such as wireless communication gps and in fact anything related to the vast field of signal processing relies on the insights of the fft but it's also one of the most beautiful albums you'll ever see the depth and sheer number of brilliant ideas that went into it is just astounding it's easy to miss the beauty aspect of the fft since it's often introduced in fairly complex settings that require a lot of prerequisite knowledge such as the discrete fourier transform time domain to frequency domain conversions and much more and to be fair to get a full appreciation of the applications of the fft you can't really avoid any of these ideas but i want to do something a little different we'll take a discovery based approach to learning about the fft in a context that you are all familiar with polynomial multiplication the way i see the structure of this video it's all about starting with some common ground and then slowly asking questions that will hopefully prompt you to discover the truly ingenious ideas behind the fast fourier transform all alright let's get started the setup here is simple we're given two polynomials and want to find the product our task will be to design an efficient algorithm for this problem mathematically we know one algorithm to multiply polynomials by repeatedly applying the distributive property all of you have perhaps instinctively been applying this algorithm since any algebra class before we try this idea though the first question we have to address is representation of polynomials in a computer the most natural way to represent them is through coefficients where we map coefficients to lists it helps to arrange our coefficients in the following order mainly because now the coefficient of the kth index is mapped to degree term we will refer to this representation as the coefficient representation of in general given 2d degree polynomials the product will have a degree of 2 times d and the running time of this algorithm if we actually went about implementing it with the most natural distributive property approach will be o of d squared since each term in polynomial a will have to be multiplied by all terms in polynomial b the question now is can we do better the first really clever idea comes from thinking about polynomials a little bit differently to see the key insight here let's take a look at one of the simplest polynomials a degree one polynomial or a linear function we can represent any line with exactly two coefficients one for the degree zero term and one term the key part that makes this representation valid is that every representation has a one-to-one mapping to a unique line what other representations of a line have this property there are actually several reasonable representations but the one that we are going to use is the two-point representation we know from geometry that any line can be represented by two distinct points and turns out that there is a neat extension of this for general polynomials which i will state here any polynomial of degree d can be uniquely represented by d plus one points for example if i gave you three random points this means that there is exactly one quadratic function that goes through all three of these points if i give you four points there is exactly one cubic function that goes through all these points this statement is perhaps a little surprising so deserves a proof suppose i give you d plus one unique points of a d degree polynomial p of x we want to show that for these set of points there's only one set of coefficients so if we actually evaluate our polynomial at each of these points

we get the following set of system of whenever you have a system of equations writing it as a matrix vector product is almost always helpful for analysis one nice property of this matrix is that if each of our original points is unique as it is in this case the matrix will always be invertible the easiest way to show this is by calculating the determinant which you will find is nonzero but i'll link a nice linear algebraic proof of this fact in the description as well for those interested anyways what this means is that for every set of points there exists a unique set of coefficients and consequently a unique polynomial taking a step back what this means is that there are actually two ways to represent polynomials of degree d the first of which is the coefficient representation and the second with just d plus one points which we'll call the value representation a nice property of using the value representation is multiplication of polynomials is much easier let's say i ask you to multiply these two polynomials a of x and b of x we know the resulting polynomial c of x will be of degree four so we'll need five points to uniquely represent the product what we can now do is take five points from each of the two polynomials and then simply multiply the function values one by one to get the value representation of the product of the two polynomials following our earlier rule there's only one degree 4 polynomial that passes through these points that polynomial happens to be the following in the coefficient representation and this is indeed the product of our original a of x and b of x polynomials with the multiplication operation being performed using the value representation we've now reduced the time for multiplication from our original d squared operations to the order of only degree d operations okay so let's propose a plan for an improvement to polynomial multiplication we are given two polynomials in the coefficient representation of degree d each we know multiplication is faster using the value representation so what we'll do is evaluate each of these polynomials at 2d plus 1 points multiply each of these points pairwise to get the value representation of the product polynomial and then finally somehow convert the value representation back to the final coefficient representation this is the grand plan but there are several pieces of the puzzle we haven't figured out what we're missing is really some sort of magic box that could take polynomials in the coefficient representation and convert them to the value representation and then vice versa that magic box my friends and trust me it is truly magical is the fast fourier transform let's first focus on taking polynomials from the coefficient representation to the value representation which we will call evaluation we have a degree d polynomial and we want to evaluate the polynomial at n points where n is some number greater than d plus one let's think about the most straightforward way to do this we can pick n random x coordinates and simply calculate the respective y coordinate this works but when we deconstruct what's actually going on here we run into our old foe each evaluation will take of d operations making this method run in o of n times d time which ends up being o of d squared to evaluate all endpoints so back to where we started can we find a way to optimize this let's try to simplify the problem let's say instead of considering a general polynomial we wanted to instead just evaluate a simple polynomial like p of x is equal to x squared at eight points the question now is which points should we pick is there any set of points when knowing the value of one point immediately implies the value of another in fact there is if we pick the point x equals one we immediately know the value of the point x equals negative one similarly if we pick x equals two we automatically know x equals negative two will have the same value extending this idea the key property we want here is that our eight points should be positive negative pairs the reason this works is due to the property of even functions where a function evaluated at negative x is going to equal the function evaluated at positive x okay but then the next immediate question is what about functions like x cubed does the same trick work it actually kind of does but with one caveat each positive x value will have the same value as the negative x value but with the sign flipped so in these two cases of odd and even degree single term polynomials instead of evaluating eight individual points we can actually get

away with evaluating exactly four positive points from which we immediately know the value of the respective negative points let's see if we can extend this idea to a more general polynomial taking inspiration from our early exploration let's split our polynomial into even and odd degree terms if we factor an x from the odd degree terms we end up having two new polynomials where these new polynomials have only even degree terms let's actually give these polynomials formal names the first representing the even terms and the second representing odd terms after factoring out x this fact allows us to recycle a lot of computation between positive and negative pairs of points a bonus fact is that since these even and odd polynomials are functions of x squared they are polynomials of degree two which is a much simpler problem than our original degree five polynomial generalizing these observations if we have an n minus one degree polynomial that we want to evaluate at n points we can split the polynomial into even and odd terms with these two smaller polynomials now having degree n over two minus one so how do we evaluate these polynomials with half the degree of our original polynomial well what's beautiful here is that this is just another evaluation problem but this time we need to evaluate the polynomials at each of our original inputs squared and this works out nicely since our original points were positive negative pairs so if we originally had n points we now only end up having n over two points this is starting to smell like the start of a recursive algorithm let's take a look at the bigger picture we want to evaluate a polynomial p of x at endpoints where the endpoints are positive negative paired we split the polynomial into odd and even degree components where we now have two simpler polynomials of degree n over two minus one that only need n over two points to once we recursively evaluate these smaller polynomials we can then go through every point in our original set of endpoints and calculate the respective values by utilizing the relationship between the positive and negative paired points this gives us the value representation of our original polynomial if we can get this to work this means we have an o of n log n recursive algorithm since the two recursive sub problems have half the size of the original problem and take linear time to evaluate the endpoints this would be a huge improvement from our earlier quadratic running time but there is one major problem can you spot the issue the problem occurs at the recursive step the entire scheme relies on the fact that the polynomial will have positive and negative paired points for evaluation this works at the top level but the next level we are evaluating n over two points where each is a squared value these all end up being positive so the recursion breaks so then the natural question is can we make these new set of points positive negative paired some of you may already see it but this actually leads the third absolutely ingenious idea behind the fft the only way this is possible is if we expand the domain of possible initial points to include for special choice of complex numbers the recursive relation works perfectly where every subsequent set of points will contain positive negative pairs what possible set of initial endpoints has this property this is a hard question and to answer it we're going to do a little bit of reverse engineering with an example let's say we have a degree 3 polynomial which requires at least n equals 4 points for its value representation these points need to be positive negative pairs so we can write them as x1 negative x1 x2 and negative x2 we know that the recursive step will require that we evaluate the odd and even splits of the polynomial at two points x1 squared and x2 squared now the key constraint here is that for the recursion to work these two points also have to be positive negative pairs so we have an equivalence between x2 squared and negative x1 squared at the bottom level of the recursion will have a single point x1 to the power 4. now what's nice is that we get to choose these points let's see what happens if we pick our initial x1 to be 1. this means two of our initial points are 1 and negative 1 which at the next level of recursion means that x1 squared and negative x1 also have to be 1 and negative 1 respectively and at the bottom layer we have only one point which ends up being 1. now the question becomes what x2 should we choose so that when we

square x2 we end up with negative 1. the answer to that is the complex number i which means that the four points that we need to evaluate this polynomial at r 1 negative 1 i and negative i an alternate perspective to what we just did here is that we essentially just solved the equation x to the power of 4 equals 1. since at the bottom layer of the recursion the value of any of our original points to the power of four was one we know this equation has four solutions all of which are encompassed by a special set of points called the fourth roots of unity let's see if this generalizes if given a degree 5 polynomial we'll need n is greater than or equal to 6 points since our cursive method is splitting each problem in half it's convenient to just pick a power of two so let's pick n equals eight now what we need to do is to find eight points that are positive negative paired such that each of these points when raised to the eighth power is equal to one we see that the right points are the eighth roots of unity generalizing this to any d degree polynomial what we will do is pick n is greater than or equal to d plus one points such that n is the power of two and the points that we should choose are the nth roots of unity this fact deserves a little bit more explanation why does this work before we answer that question let's formalize a few things the nth roots of unity are the solution to the following equation and they are best visualized as equally spaced points on the unit circle the angle between these points is 2 pi over n with this fact a nice way to compactly write these points is with complex exponential notation through euler's formula one standard way to define the roots of unity is by defining this omega term as e to the power of two pi i over n and then what this allows us to do is define individual roots of unity quite compactly here are some examples so now when we say we want to evaluate a polynomial at the nth roots of unity what that really means is we want to evaluate it at omega to the power of 0 1 so on and so forth until omega to the power of n minus 1. so going back to our original question of why evaluating the polynomial p of x at the nth root of unity works for our recursive algorithm there are two key properties at play ear for one our original set of points are positive negative paired where for the jth root omega to the power of j plus n over 2 is going to be the corresponding pair now in our recursive step we will be squaring each of these points and passing them on to the even and odd degree polynomials this is what happens when we square our original n roots of unity this reveals the second key property of the nth roots of unity when we square the nth roots of unity we end up with the n over two roots of unity which are also positive negative paired and are just the right number of points for the two new polynomials of half the degree this same pattern holds at every level of the recursion until we end up with just one point how beautiful is that all right we are now ready to outline the core fast fourier transform algorithm the fft will take in a coefficient representation of a degree n minus one polynomial where n is the power of two we will define omega as e to the power of 2 pi i over n to allow us to define roots of unity easily the first case we need to handle is the base case which is going to be when n is equal to 1. all this means is that we are evaluating the polynomial at one point a recursive case is two calls to the f of t one on even degree terms and one on odd the intention is that these polynomials are now half the degree of our original polynomial so they only need to be evaluated at n over 2 roots of unity assuming the recursion works the output of these calls will be the corresponding value representation of these even and odd degree term polynomials which we will label as y e and y o now on to the tricky part which is to take the output from these recursive calls and combine them to get the value representation of our original degree n minus 1 polynomial we saw earlier that the key idea was to use the relationship between positive and negative pairs but now we have to slightly modify this logic for our roots of unity inputs as a quick note yes i did modify the indexing to zero indexing because we're getting ready to write some code we know the j-th input point will correspond to j root of unity which results in the following relationship

we also saw earlier that the paired point negative omega to the power of j is equal to plus n over 2 due to the properties of the roots of unity using this fact we can modify the second equation as follows and lastly one more fact that's nice is that the j index in our y e and y o list correspond to the even and odd polynomials evaluated at omega to the power of 2 times j what this allows us to do is rewrite our equations as follows which makes it much easier to implement code as mentioned this part is tricky so i encourage you to take your time and verify that each of these steps is indeed true the final step in the fft algorithm is to then return the values of a polynomial p evaluated at the nth roots of unity let's now translate this outline logic into code our function f of t will take an input p which is the coefficient representation of a polynomial p we first define n as the length of p and we will assume that n is a power of two just to be clear there are implementations of the f t that can handle n not being a power of two but those are way more complicated the power of two cases encompass the core ideas of the algorithm we now handle the base case which is just a matter of returning our original p this makes sense since we only have one element making p a degree zero polynomial or constant otherwise we define omega as we have outlined and then proceed with the first part of the recursive step requires splitting the polynomial into even and odd degree terms which is quite easy to do then we recursively call our f of function on these polynomials that now have half the degree of our original polynomial we denote the outputs as y e and y o as we have done in the outline now it's time to put this all together we initialize our output list which will contain the final value representation then for all j up to n over 2 we calculate the value representations as we have outlined after populating all values in our list we then return that list and that's the fft overall pretty crazy how all the ideas we talked about end up coming together in 11 lines of truly elegant code let's now take a larger look at our original problem of polynomial multiplication and see where we are we now have a way to convert coefficient representations to value representations efficiently using the fft so now the only missing piece is the reverse process of converting from value representations to coefficient which is formally called interpolation this is where things get really wild on the surface the idea of reversing evaluation feels like a significantly harder task let's take a step back and look at this problem from another perspective evaluation and interpolation are closely connected and as we saw earlier we can express evaluation as a matrix vector product we have a vector of coefficients multiplied by a matrix of our evaluation points to give us the value representation now in the fft algorithm the kth evaluation point was a corresponding route of unity which allows us to rewrite the matrix vector product as follows this particular matrix has a special name the discrete fourier transform or dft matrix in most textbooks and references the fft at its core is an algorithm for calculating these types of matrix vector products efficiently polynomial evaluation at the roots of unity happens to be one case where this type of matrix vector product shows up so that's why we can use the fft anyways the nice fact about the fft and evaluation in this context is that interpolation is much easier to understand interpolation requires inversing this dft matrix for interpolation we are given a value representation of our polynomial and we want to find the coefficient representation which means we have to multiply the value representation by the inverse of the dft matrix so let me show you what the inverse of this matrix looks like i'm purposely skipping a lot of important linear algebra facts here since that would be an entirely different video but given that this is the inverse matrix what stands out to you it's really quite amazing but this inverse matrix looks almost the same as our original dft matrix in fact the only difference is that every single omega in our original dft matrix is now just replaced with omega to the power of negative one with a normalization factor of one over this indicates a potential to reuse the fft logic for interpolation since the matrix structure is basically the same

let's formalize the suspicion by doing a direct comparison in evaluation which involved the fft we are given a set of coefficients and evaluate the polynomial at the roots of unity to get a value representation this involved the following matrix vector product where we define omega as e to the power of 2 pi i divided by n looking at interpolation we now want to define what is formally called the inverse fft algorithm the inverse fft will take a value representation where each value was evaluated at the roots of unity and gives you a set of coefficients for the original polynomial basically reversing what the original fft did as we just saw this requires multiplying by the inverse of the dft matrix we noted that each omega in our original dft matrix now corresponds to 1 over n times omega to the power of negative 1. now the punch line here is that what this means is we can define the inverse f of t as the same fft function but now called on the value representation with omega defined as 1 over n times e to the negative two pi i divided by n that's it with those small changes we have an inverse fft that performs interpolation just so we are super clear on what sorcery just happened let me remind you of the original fft implementation and now let me show you the inverse fft implementation which takes the value representation as an input what we literally do is copy our fft implementation change the name of the recursive calls to match and then literally change one line of code one and that's all there is to it so if your mind isn't blown you haven't been paying attention let's take a look at what we just did we motivated the fft through the problem of polynomial multiplication where the first brilliant idea came from representing and multiplying polynomials using the value representation converting polynomials to a value representation required us to come up with an appropriate set of evaluation points our first attempts at solving this problem inspired the clever idea of using positive negative pairs but the recursion didn't quite work unless we expanded the domain to complex numbers the next brilliant idea came from using the nth roots of unity where the points at every level of recursion are positive negative paired this evaluation scheme using the roots of unity encompassed the essence of the fft algorithm when confronted with the problem of reversing the process using interpolation we discovered something truly astounding the inverse fft is the same algorithm but with one minor adjustment so if we take a look at what we just did here there's not one not two not three but four absolutely mind-blowing ideas that come together to make this work do i really need to say more on why this is my favorite algorithm that's all for this video and thanks for watching if you enjoyed the content please hit the like button so that this content will be recommended to more people if you want to see more content like this please don't forget to hit the subscribe button and if you want to more directly support the work of this channel please check out the patreon page linked in the description below i'll see you on the next video