Tag Archives: divide and conquer

A Quasipolynomial Time Algorithm for Graph Isomorphism: The Details

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Update 2017-01-09: Laci claims to have found a workaround to the previously posted error, and the claim is again quasipolynoimal time! Updated arXiv paper to follow.

Update 2017-01-04: Laci has posted an update on his paper. The short version is that one small step of his analysis was not quite correct, and the result is that his algorithm is sub-exponential, but not quasipolynomial time. The fact that this took over a year to sort out is a testament to the difficulty of the mathematics and the skill of the mathematicians involved. Even the revised result is still a huge step forward in the analysis of graph isomorphism. Finally, this should reinforce how we should think about progress in mathematics: it comes in fits and starts, with occasional steps backward.

Update 2015-12-13: Laci has posted a preprint on arXiv. It’s quite terse, but anyone who is comfortable with the details sketched in this article should have a fine time (albeit a long time)  reading it.

Update 2015-11-21: Ken Regan and Dick Lipton posted an article with some more details, and a high level overview of how the techniques fit into the larger picture of CS theory.

Update 2015-11-16: Laci has posted the talk on his website. It’s an hour and a half long, and I encourage you to watch it if you have the time 🙂

Laszlo Babai has claimed an astounding theorem, that the Graph Isomorphism problem can be solved in quasipolynomial time (now outdated; see Update 2017-01-04 above). On Tuesday I was at Babai’s talk on this topic (he has yet to release a preprint), and I’ve compiled my notes here. As in Babai’s talk, familiarity with basic group theory and graph theory is assumed, and if you’re a casual (i.e., math-phobic) reader looking to understand what the fuss is all about, this is probably not the right post for you. This post is research level theoretical computer science. We’re here for the juicy, glorious details.

Note: this blog post will receive periodic updates as my understanding of the details improve.

Laci during his lecture.

Laci during his lecture. Photo taken by me.

Standing room only at Laci's talk.

Standing room only at Laci’s talk. My advisor in the bottom right, my coauthor mid-left with the thumbs. Various famous researchers spottable elsewhere.

Background on Graph Isomorphism

I’ll start by giving a bit of background into why Graph Isomorphism (hereafter, GI) is such a famous problem, and why this result is important. If you’re familiar with graph isomorphism and the basics of complexity theory, skip to the next section where I get into the details.

GI is the following problem: given two graphs $ G = (V_G, E_G), H = (V_H, E_H)$, determine whether the graphs are isomorphic, that is, whether there is a bijection $ f : V_G \to V_H$ such that $ u,v$ are connected in $ G$ if and only if $ f(u), f(v)$ are connected in $ H$. Informally, GI asks whether it’s easy to tell from two drawings of a graph whether the drawings actually represent the same graph. If you’re wondering why this problem might be hard, check out the following picture of the same graph drawn in three different ways.

GI-example

Indeed, a priori the worst-case scenario is that one would have to try all $ n!$ ways to rearrange the nodes of the first graph and see if one rearrangement achieves the second graph. The best case scenario is that one can solve this problem efficiently, that is, with an algorithm whose worst-case runtime on graphs with $ n$ nodes and $ m$ edges is polynomial in $ n$ and $ m$ (this would show that GI is in the class P). However, nobody knows whether there is a polynomial time algorithm for GI, and it’s been a big open question in CS theory for over forty years. This is the direction that Babai is making progress toward, showing that there are efficient algorithms. He didn’t get a polynomial time algorithm, but he got something quite close, which is why everyone is atwitter.

It turns out that telling whether two graphs are isomorphic has practical value in some applications. I hear rumor that chemists use it to search through databases of chemicals for one with certain properties (one way to think of a chemical compound is as a graph). I also hear that some people use graph isomorphism to compare files, do optical character recognition, and analyze social networks, but it seems highly probable to me that GI is not the central workhorse (or even a main workhorse) in these fields. Nevertheless, the common understanding is that pretty much anybody who needs to solve GI on a practical level can do so efficiently. The heuristics work well. Even in Babai’s own words, getting better worst-case algorithms for GI is purely a theoretical enterprise.

So if GI isn’t vastly important for real life problems, why are TCS researchers so excited about it?

Well it’s known that GI is in the class NP, meaning if two graphs are isomorphic you can give me a short proof that I can verify in polynomial time (the proof is just a description of the function $ f : V_G \to V_H$). And if you’ll recall that inside NP there is this class called NP-complete, which are the “hardest” problems in NP. Now most problems in NP that we care about are also NP-complete, but it turns out GI is not known to be NP-complete either. Now, for all we know P = NP and then the question about GI is moot, but in the scenario that most people believe P and NP are different, so it leaves open the question of where GI lies: does it have efficient algorithms or not?

So we have a problem which is in NP, it’s not known to be in P, and it’s not known to be NP-complete. One obvious objection is that it might be neither. In fact, there’s a famous theorem of Ladner that says if P is different from NP, then there must be problems in NP, not in P, and not NP-complete. Such problems are called “NP-intermediate.” It’s perfectly reasonable that GI is one of these problems. But there’s a bit of a catch.

See, Ladner’s theorem doesn’t provide a natural problem which is NP intermediate; what Ladner did in his theorem was assume P is not NP, and then use that assumption to invent a new problem that he could prove is NP intermediate. If you come up with a problem whose only purpose is to prove a theorem, then the problem is deemed unnatural. In fact, there is no known “natural” NP-intermediate problem (assuming P is not NP). The pattern in CS theory is actually that if we find a problem that might be NP-intermediate, someone later finds an efficient algorithm for it or proves it’s NP-complete. There is a small and dwindling list of such problems. I say dwindling because not so long ago the problem of telling whether an integer is prime was in this list. The symptoms are that one first solves the problem on many large classes of special cases (this is true of GI) or one gets a nice quasipolynomial-time algorithm (Babai’s claimed new result), and then finally it falls into P. In fact, there is even stronger evidence against it being NP-complete: if GI were NP-complete, the polynomial hierarchy would collapse. To the layperson, the polynomial hierarchy is abstruse complexity theoretic technical hoo-hah, but suffice it to say that most experts believe the hierarchy does not collapse, so this counts as evidence.

So indeed, it could be that GI will become the first ever problem which is NP-intermediate (assuming P is not NP), but from historical patterns it seems more likely that it will fall into P. So people are excited because it’s tantalizing: everyone believes it should be in P, but nobody can prove it. It’s right at the edge of the current state of knowledge about the theoretical capabilities and limits of computation.

This is the point at which I will start assuming some level of mathematical maturity.

The Main Result

The specific claim about graph isomorphism being made is the following:

Theorem: There is a deterministic algorithm for GI which runs in time $ 2^{O(\log^c(n))}$ for some constant $ c$.

This is an improvement over the best previously known algorithm which had runtime $ 2^{\sqrt{n \log n}}$. Note the $ \sqrt{n}$ in the exponent has been eliminated, which is a huge difference. Quantities which are exponential in some power of a logarithm are called “quasipolynomial.”

But the main result is actually a quasipolynomial time algorithm for a different, more general problem called string automorphism. In this context, given a set $ X$ a string is a function from $ X$ to some finite alphabet (really it is a coloring of $ X$, but we are going to use colorings in the usual sense later so we have to use a new name here). If the set $ X$ is given a linear ordering then strings on $ X$ really correspond to strings of length $ |X|$ over the alphabet. We will call strings $ x,y \in X$.

Now given a set $ X$ and a group $ G$ acting on $ X$, there is a natural action of $ G$ on strings over $ X$, denoted $ x^\sigma$, by permuting the indices $ x^{\sigma}(i) = x(\sigma(i))$. So you can ask the natural question: given two strings $ x,y$ and a representation of a group $ G$ acting on $ X$ by a set of generating permutations of $ G$, is there a $ \sigma \in G$ with $ x^\sigma = y$? This problem is called the string isomorphism problem, and it’s clearly in NP.

Now if you call $ \textup{ISO}_G(x,y)$ the set of all permutations in $ G$ that map $ x$ to $ y$, and you call $ \textup{Aut}_G(x) = \textup{ISO}_G(x,x)$, then the actual theorem Babai claims to have proved is the following.

Theorem: Given a generating set for a group $ G$ of permutations of a set $ X$ and a string $ x$, there is a quasipolynomial time algorithm which produces a generating set of the subgroup $ \textup{Aut}_G(x)$ of $ G$, i.e. the string automorphisms of $ x$ that lie in in $ G$.

It is not completely obvious that GI reduces to the automorphism problem, but I will prove it does in the next section. Furthermore, the overview of Babai’s proof of the theorem follows an outline laid out by Eugene Luks in 1982, which involves a divide-and-conquer method for splitting the analysis of $ \textup{Aut}_G(x)$ into simpler and simpler subgroups as they are found.

Luks’s program

Eugene Luks was the first person to incorporate “serious group theory” (Babai’s words) into the study of graph isomorphism. Why would group theory help in a question about graphs? Let me explain with a lemma.

Lemma: GI is polynomial-time reducible to the problem of computing, given a graph $ X$, a list of generators for the automorphism group of $ G$, denoted $ \textup{Aut}(X)$.

Proof. Without loss of generality suppose $ X_1, X_2$ are connected graphs. If we want to decide whether $ X_1, X_2$ are isomorphic, we may form the disjoint union $ X = X_1 \cup X_2$. It is easy to see that $ X_1$ and $ X_2$ are isomorphic if and only if some $ \sigma \in \textup{Aut}(X)$ swaps $ X_1$ and $ X_2$. Indeed, if any automorphism with this property exists, every generating set of $ \textup{Aut}(G)$ must contain one.

$ \square$

Similarly, the string isomorphism problem reduces to the problem of computing a generating set for $ \textup{Aut}_G(x)$ using a similar reduction to the one above. As a side note, while $ \textup{ISO}_G(x,y)$ can be exponentially large as a set, it is either the empty set, or a coset of $ \textup{Aut}_G(x)$ by any element of $ \textup{ISO}_G(x,y)$. So there are group-theoretic connections between the automorphism group of a string and the isomorphisms between two strings.

But more importantly, computing the automorphism group of a graph reduces to computing the automorphism subgroup of a particular group for a given string in the following way. Given a graph $ X$ on a vertex set $ V = \{ 1, 2, \dots, v \}$ write $ X$ as a binary string on the set of unordered pairs $ Z = \binom{V}{2}$ by mapping $ (i,j) \to 1$ if and only if $ i$ and $ j$ are connected by an edge. The alphabet size is 2. Then $ \textup{Aut}(X)$ (automorphisms of the graph) induces an action on strings as a subgroup $ G_X$ of $ \textup{Aut}(Z)$ (automorphisms of strings). These induced automorphisms are exactly those which preserve proper encodings of a graph. Moreover, any string automorphism in $ G_X$ is an automorphism of $ X$ and vice versa. Note that since $ Z$ is larger than $ V$ by a factor of $ v^2$, the subgroup $ G_X$ is much smaller than all of $ \textup{Aut}(Z)$.

Moreover, $ \textup{Aut}(X)$ sits inside the full symmetry group $ \textup{Sym}(V)$ of $ V$, the vertex set of the starting graph, and $ \textup{Sym}(V)$ also induces an action $ G_V$ on $ Z$. The inclusion is

$ \displaystyle \textup{Aut}(X) \subset \textup{Sym}(V)$

induces

$ \displaystyle G_X = \textup{Aut}(\textup{Enc}(X)) \subset G_V \subset \textup{Aut}(Z)$

I.e.,

Call $ G = G_V$ the induced subgroup of permutations of strings-as-graphs. Now we just have some subgroup of permutations $ G$ of $ \textup{Aut}(Z)$, and we want to find a generating set for $ \textup{Aut}_G(x)$ (where $ x$ happens to be the encoding of a graph). That is exactly the string automorphism problem. Reduction complete.

Now the basic idea to compute $ \textup{Aut}_G(x)$ is to start from the assumption that $ \textup{Aut}_G(x) = G$. We know it’s a subgroup, so it could be all of $ G$; in terms of GI if this assumption were true it would mean the starting graph was the complete graph, but for string automorphism in general $ G$ can be whatever. Then we try to refute this belief by finding additional structure in $ \textup{Aut}_G(x)$, either by breaking it up into smaller pieces (say, orbits) or by constructing automorphisms in it. That additional structure allows us to break up $ G$ in a way that $ \textup{Aut}_G(x)$ is a subgroup of the product of the corresponding factors of $ G$.

The analogy that Babai used, which goes back to graphs, is the following. If you have a graph $ X$ and you want to know its automorphisms, one thing you can do is to partition the vertices by degree. You know that an automorphism has to preserve the degree of an individual vertex, so in particular you can break up the assumption that $ \textup{Aut}(X) = \textup{Sym}(V)$ into the fact that $ \textup{Aut}(X)$ must be a subgroup of the product of the symmetry groups of the pieces of the partition; then you recurse. In this way you’ve hugely reduced the total number of automorphisms you need to consider. When the degrees get small enough you can brute-force search for automorphisms (and there is some brute-force searching in putting the pieces back together). But of course this approach fails miserably in the first step you start with a regular graph, so one would need to look for other kinds of structure.

One example is an equitable partition, which is a partition of vertices by degree relative to vertices in other blocks of the partition. So a vertex which has degree 3 but two degree 2 neighbors would be in a different block than a vertex with degree 3 and only 1 neighbor of degree 2. Finding these equitable partitions (which can be done in polynomial time) is one of the central tools used to attack GI. As an example of why it can be very helpful: in many regimes a Erdos-Renyi random graph has asymptotically almost surely a coarsest equitable partition which consists entirely of singletons. This is despite the fact that the degree sequences themselves are tightly constrained with high probability. This means that, if you’re given two Erdos-Renyi random graphs and you want to know whether they’re isomorphic, you can just separately compute the coarsest equitable partition for each one and see if the singleton blocks match up. That is your isomorphism.

Even still, there are many worst case graphs that resist being broken up by an equitable partition. A hard example is known as the Johnson graph, which we’ll return to later.

For strings the sorts of structures to look for are even more refined than equitable partitions of graphs, because the automorphism group of a graph can be partitioned into orbits which preserve the block structure of an equitable partition. But it still turns out that Johnson graphs admit parts of the automorphism group that can’t be broken up much by orbits.

The point is that when some useful substructure is found, it will “make progress” toward the result by breaking the problem into many pieces (say, $ n^{\log n}$ pieces) where each piece has size $ 9/10$ the size of the original. So you get a recursion in the amount of time needed which looks like $ f(n) \leq n^{\log n} f(9n/10)$. If you call $ q(n) = n^{\log n}$ the quasipolynomial factor, then solving the recurrence gives $ f(n) \leq q(n)^{O(\log n)}$ which only adds an extra log factor in the exponent. So you keep making progress until the problem size is polylogarithmic, and then you brute force it and put the pieces back together in quasipolynomial time.

Two main lemmas, which are theorems in their own right

This is where the details start to get difficult, in part because Babai jumped back and forth between thinking of the object as a graph and as a string. The point of this in the lecture was to illustrate both where the existing techniques for solving GI (which were in terms of finding canonical graph substructures in graphs) break down.

The central graph-theoretic picture is that of “individualizing” a vertex by breaking it off from an existing equitable partition, which then breaks the equitable partition structure so you need to do some more (polytime) work to further refine it into an equitable partition again. But the point is that you can take all the vertices in a block, pick all possible ways to individualize them by breaking them into smaller blocks. If you traverse these possibilities in a canonical order, you will eventually get down to a partition of singletons, which is your “canonical labeling” of the graph. And if you do this process with two different graphs and you get to different canonical labelings, you had to have started with non-isomorphic graphs.

The problem is that when you get to a coarsest equitable partition, you may end up with blocks of size $ \sqrt{n}$, meaning you have an exponential number of individualizations to check. This is the case with Johnson graphs, and in fact if you have a Johnson graph $ J(m,t)$ which has $ \binom{m}{t}$ vertices and you individualize fewer than $ m/10t$ if them, then you will only get down to blocks of size polynomially smaller than $ \binom{m}{t}$, which is too big if you want to brute force check all individualizations of a block.

The main combinatorial lemma that Babai proves to avoid this problem is that the Johnson graphs are the only obstacle to finding efficient partitions.

Theorem (Babai 15): If $ X$ is a regular graph on $ m$ vertices, then by individualizing a polylog number of vertices we can find one of the three following things:

  1. A canonical coloring with each color class having at most 90% of all the nodes.
  2. A canonical equipartition of some subset of the vertices that has at least 90% of the nodes (i.e. a big color class from (1)).
  3. A canonically embedded Johnson graph on at least 90% of the nodes.

[Edit: I think that what Babai means by a “canonical coloring” is an equitable partition of the nodes (not to be confused with an equipartition), but I am not entirely sure. I have changed the language to reflect more clearly what was in the talk as opposed to what I think I understood from doing additional reading.]

The first two are apparently the “easy” cases in the sense that they allow for simple recursion that has already been known before Babai’s work. The hard part is establishing the last case (and this is the theorem whose proof sketch he has deferred for two more weeks). But once you have such a Johnson graph your life is much better, because (for a reason I did not understand) you can recurse on a problem of size roughly the square root of the starting size.

In discussing Johnson graphs, Babai said they were a source of “unspeakable misery” for people who want to solve GI quickly. At the same time, it is a “curse and a blessing,” as once you’ve found a Johnson graph embedded in your problem you can recurse to much smaller instances. This routine to find one of these three things is called the “split-or-Johnson” routine.

The analogue for strings (I believe this is true, but I’m a bit fuzzy on this part) is to find a “canonical” $ k$-ary relational structure (where $ k$ is polylog in size) with some additional condition on the size of alternating subgroups of the automorphism group of the $ k$-ary relational structure. Then you can “individualize” the points in the base of this relational structure and find analogous partitions and embedded Johnson schemes (a special kind of combinatorial design).

One important fact to note is that the split-or-Johnson routine breaks down at $ \log^3(n)$ size, and Babai has counterexamples that say his result is tight, so getting GI in P would have to bypass this barrier with a different idea.

The second crucial lemma has to do with giant homomorphisms, and this is the method by which Babai constructs automorphisms that bound $ \textup{Aut}_G(x)$ from below. As opposed to the split-or-Johnson lemma, which finds structure to bound the group from above by breaking it into simpler pieces. Fair warning: one thing I don’t yet understand is how these two routines interact in the final algorithm. My guess is they are essentially run in parallel (or alternate), but that guess is as good as wild.

Definition: A homomorphism $ \varphi: G \to S_m$ is called giant if the image of $ G$ is either the alternating group $ A_n$ or else all of $ S_m$. I.e. $ \varphi$ is surjective, or almost so. Let $ \textup{Stab}_G(x)$ denote the stabilizer subgroup of $ x \in G$. Then $ x$ is called “affected” by $ \varphi$ if $ \varphi|_{\textup{Stab}_g(x)}$ is not giant.

The central tool in Babai’s algorithm is the dichotomy between points that are affected and those that are not. The ability to decide this property in quasipolynomial time is what makes the divide and conquer work.

There is one deep theorem he uses that relates affected points to giant homomorphisms:

Theorem (Unaffected Stabilizer Theorem): Let $ \varphi: G \to S_m$ be a giant homomorphism and $ U \subset G$ the set of unaffected elements. Let $ G_{(U)}$ be the pointwise stabilizer of $ U$, and suppose that $ m > \textup{max}(8, 2 + \log_2 n)$. Then the restriction $ \varphi : G_{(U)} \to S_m$ is still giant.

Babai claimed this was a nontrivial theorem, not because the proof is particularly difficult, but because it depends on the classification of finite simple groups. He claimed it was a relatively straightforward corollary, but it appears that this does not factor into the actual GI algorithm constructively, but only as an assumption that a certain loop invariant will hold.

To recall, we started with this assumption that $ \textup{Aut}_G(x)$ was the entire symmetry group we started with, which is in particular an assumption that the inclusion $ \varphi \to S_m$ is giant. Now you want to refute this hypothesis, but you can’t look at all of $ S_m$ because even the underlying set $ m$ has too many subsets to check. But what you can do is pick a test set $ A \subset [m]$ where $ |A|$ is polylogarithmic in size, and test whether the restriction of $ \varphi$ to the test set is giant in $ \textup{Sym}(A)$. If it is giant, we call $ A$ full.

Theorem (Babai 15): One can test the property of a polylogarithmic size test set being full in quasipolynomial time in m.

Babai claimed it should be surprising that fullness is a quasipolynomial time testable property, but more surprising is that regardless of whether it’s full or not, we can construct an explicit certificate of fullness or non-fullness. In the latter case, one can come up with an explicit subgroup which contains the image of the projection of the automorphism group onto the symmetry group of the test set. In addition, given two test sets $ A,B$, one can efficiently compare the action between the two different test sets. And finding these non-full test sets is what allows one to construct the $ k$-ary relations. So the output of this lower bound technique informs the upper bound technique of how to proceed.

The other outcome is that $ A$ could be full, and coming up with a certificate of fullness is harder. The algorithm sketched below claims to do it, and it involves finding enough “independent” automorphisms to certify that the projection is giant.

Now once you try all possible test sets, which gives $ \binom{m}{k}^2$ many certificates (a quasipolynomial number), one has to aggregate them into a full automorphism of $ x$, which Babai assured us was a group theoretic exercise.

The algorithm to test fullness (and construct a certificate) he called the Local Certificates Algorithm. It was sketched as follows: you are given as input a set $ A$ and a group $ G_A \subset G$ being the setwise stabilizer of $ A$ under $ \psi_A : G_A \to \textup{Sym}(A)$. Now let $ W$ be the group elements affected by $ \psi_A$. You can be sure that at least one point is affected. Now you stabilize on $ W$ and get a refined subgroup of $ G_A$, which you can use to compute newly affected elements, growing $ W$ in each step. By the unaffected stabilizer theorem, this preserves gianthood. Furthermore, in each step you get layers of $ W$, and all of the stabilizers respect the structure of the previous layers. Babai described this as adding on “layers of a beard.”

The termination of this is either when $ W$ stops growing, in which case the projection is giant and $ W$ is our certificate of fullness (i.e. we get a rich family of automorphisms that are actually in our target automorphism group), or else we discover the projected ceases to be giant and $ W$ is our certificate of non-fullness. Indeed, the subgroup generated by these layers is a subgroup of $ \textup{Aut}_G(x)$, and the subgroup generated by the elements of a non-fullness certificate contain the automorphism group.

Not enough details?

This was supposed to be just a high-level sketch of the algorithm, and Babai is giving two more talks elaborating on the details. Unfortunately, I won’t be able to make it to his second talk in which he’ll discuss some of the core group theoretic ideas that go into the algorithm. I will, however, make it to his third talk in which he will sketch the proof of the split-or-Johnson routine. That is in two weeks from the time of this writing, and I will update this post with any additional insights then.

Babai has not yet released a preprint, and when I asked him he said “soon, soon.” Until then 🙂

This blog post is based on my personal notes from Laszlo Babai’s lecture at the University of Chicago on November 10, 2015. At the time of this writing, Babai’s work has not been peer reviewed, and my understanding of his lectures has large gaps and may be faulty. Do not put your life in danger based on information in this post.

The Fast Fourier Transform

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John Tukey, one of the developers of the Cooley-Tukey FFT algorithm.

It’s often said that the Age of Information began on August 17, 1964 with the publication of Cooley and Tukey’s paper, “An Algorithm for the Machine Calculation of Complex Fourier Series.” They published a landmark algorithm which has since been called the Fast Fourier Transform algorithm, and has spawned countless variations. Specifically, it improved the best known computational bound on the discrete Fourier transform from $ O(n^2)$ to $ O(n \log n)$, which is the difference between uselessness and panacea.

Indeed, their work was revolutionary because so much of our current daily lives depends on efficient signal processing. Digital audio and video, graphics, mobile phones, radar and sonar, satellite transmissions, weather forecasting, economics and medicine all use the Fast Fourier Transform algorithm in a crucial way. (Not to mention that electronic circuits wouldn’t exist without Fourier analysis in general.) Before the Fast Fourier Transform algorithm was public knowledge, it simply wasn’t feasible to process digital signals.

Amusingly, Cooley and Tukey’s particular algorithm was known to Gauss around 1800 in a slightly different context; he simply didn’t find it interesting enough to publish, even though it predated the earliest work on Fourier analysis by Joseph Fourier himself.

In this post we will derive and implement a Fast Fourier Transform algorithm, and explore a (perhaps naive) application to audio processing. In particular, we will remove white noise from a sound clip by filtering the frequency spectrum of a noisy signal.

As usual, all of the resources used for this post are available on this blog’s Github page.

Derivation

It turns out that there are a number of different ways to speed up the naive Fourier Transform computation. As we saw in our primer on the discrete Fourier transform, the transform itself can be represented as a matrix. With a bit of nontrivial analysis, one can factor the Fourier transform matrix into a product of two sparse matrices (i.e., which contain mostly zeros), and from there the operations one can skip are evident. The intuitive reason why this should work is that the Fourier transform matrix is very structured, and the complex exponential has many useful properties. For more information on this method, see these lecture notes (p. 286).

We will take a different route which historically precedes the matrix factorization method. In fact, the derivation we’ll trace through below was Cooley and Tukey’s original algorithm. The method itself is a classical instance of a divide and conquer algorithm. Familiar programmers would recognize the ideas in common sorting algorithms: both mergesort and quicksort are divide and conquer algorithms.

The difficulty here is to split a list of numbers into two lists which are half in size in such a way that the Fourier transforms of the smaller pieces can be quickly combined to form the Fourier transform of the original list. Once we accomplish that, the rest of the algorithm falls out from recursion; further splitting each piece into two smaller pieces, we will eventually reach lists of length two or one (in which case the Fourier transform is completely trivial).

For now, we’ll assume that the length of the list is a power of 2. That is,

$ f = (f[1], f[2], \dots, f[n]), n = 2^k$ for some $ k$.

We will also introduce the somewhat helpful notation for complex exponentials. Specifically, $ \omega[p,q] = e^{2\pi i q/p}$. Here the $ p$ will represent the length of the list, and $ q$ will be related to the index. In particular, the complex exponential that usually shows up in the discrete Fourier transform (again, refer to our primer) is $ \omega[n, -km] = e^{-2 \pi i mk/n}$. We write the discrete Fourier transform of $ f$ by specifying its $ m$-th entry:

$ \displaystyle \mathscr{F}f[m] = \sum_{k=0}^{n-1} f[k]\omega[n, -km]$

Now the trick here is to split up the sum into two lists each of half size. Of course, one could split it up in many different ways, but after we split it we need to be able to rewrite the pieces as discrete Fourier transforms themselves. It turns out that if we split it into the terms with even indices and the terms with odd indices, things work out nicely.

Specifically, we can write the sum above as the two sums below. The reader should study this for a moment (or better yet, try to figure out what it should be without looking) to ensure that the indices all line up. The notation grows thicker ahead, so it’s good practice.

$ \displaystyle \mathscr{F}f[m] = \sum_{k=0}^{\frac{n}{2} – 1} f[2k] \omega[n,-2km] + \sum_{k=0}^{\frac{n}{2} – 1} f[2k+1]\omega[n,-(2k+1)m]$.

Now we need to rewrite these pieces as Fourier transforms. Indeed, we must replace the occurrences of $ n$ in the complex exponentials with $ n/2$. This is straightforward in the first summation, since

$ \omega[n, -2km] = e^{-2 \pi i km/n} = \omega[\frac{n}{2}, -km]$.

For the second summation, we observe that

$ \displaystyle \omega[n, -(2k+1)m] = \omega[n, -2km] \cdot \omega[n,-m]$.

Now if we factor out the $ \omega[n,-m]$, we can transform the second sum in the same way as the first, but with that additional factor out front. In other words, we now have

$ \displaystyle \mathscr{F}f[m] = \sum_{k=0}^{\frac{n}{2}-1} f[2k] \omega[n/2, -km] + \omega[n, -m] \sum_{k=0}^{\frac{n}{2}-1}f[2k+1] \omega[n/2, -km]$.

If we denote the list of even-indexed entries of $ f$ by $ f_{\textup{even}}$, and vice versa for $ f_{\textup{odd}}$, we see that this is just a combination of two Fourier transforms of length $ n/2$. That is,

$ \displaystyle \mathscr{F}f = \mathscr{F}f_{\textup{even}} + \omega[n,-m] \mathscr{F}f_{\textup{odd}}$

But we have a big problem here. Computer scientists will recognize this as a type error. The thing on the left hand side of the equation is a list of length $ n$, while the thing on the right hand side is a sum of two lists of length $ n/2$, and so it has length $ n/2$. Certainly it is still true for values of $ m$ which are less than $ n/2$; we broke no algebraic laws in the way we rewrote the sum (just in the use of the $ \mathscr{F}$ notation).

Recalling our primer on the discrete Fourier transform, we naturally extended the signals involved to be periodic. So indeed, the length-$ n/2$ Fourier transforms satisfy the following identity for each $ m$.

$ \mathscr{F}f_{\textup{even}}[m] = \mathscr{F}f_{\textup{even}}[m+n/2] \\ \mathscr{F}f_{\textup{odd}}[m] = \mathscr{F}f_{\textup{odd}}[m+n/2]$

However, this does not mean we can use the same formula above! Indeed, for a length $ n$ Fourier transform, it is not true in general that $ \mathscr{F}f[m] = \mathscr{F}f[m+n/2]$. But the correct formula is close to it. Plugging in $ m + n/2$ to the summations above, we have

$ \displaystyle \mathscr{F}f[m+n/2] = \sum_{k=0}^{\frac{n}{2} – 1}f[2k] \omega[n/2, -(m+n/2)k] + \\ \omega[n, -(m+n/2)] \sum_{k=0}^{\frac{n}{2} – 1}f[2k+1] \omega[n/2, -(m+n/2)k]$

Now we can use the easy-to-prove identity

$ \displaystyle \omega[n/2, -(m+n/2)k] = \omega[n/2, -mk] \omega[n/2, -kn/2]$

And see that the right-hand-term is simply $ e^{2 \pi i k} = 1$. Similarly, one can trivially prove the identity

$ \displaystyle \omega[n, -(m+n/2)] = \omega[n, -m] \omega[n, -n/2] = -\omega[n, -m]$,

this simplifying the massive formula above to the more familiar

$ \displaystyle \mathscr{F}f[m + n/2] = \sum_k f[2k]\omega[n/2, -mk] – \omega[n, -m] \sum_k f[2k+1] \omega[n/2, -mk]$

Now finally, we have the Fast Fourier Transform algorithm expressed recursively as:

$ \displaystyle \mathscr{F}f[m] = \mathscr{F}f_{\textup{even}}[m] + \omega[n,-m] \mathscr{F}f_{\textup{odd}}[m]$
$ \displaystyle \mathscr{F}f[m+n/2] = \mathscr{F}f_{\textup{even}}[m] – \omega[n,-m] \mathscr{F}f_{\textup{odd}}[m]$

With the base case being $ \mathscr{F}([a]) = [a]$.

In Python

With all of that notation out of the way, the implementation is quite short. First, we should mention a few details about complex numbers in Python. Much to this author’s chagrin, Python represents $ i$ using the symbol 1j. That is, in python, $ a + bi$ is

a + b * 1j

Further, Python reserves a special library for complex numbers, the cmath library. So we implement the omega function above as follows.

import cmath

def omega(p, q):
   return cmath.exp((2.0 * cmath.pi * 1j * q) / p)

And then the Fast Fourier Transform algorithm is more or less a straightforward translation of the mathematics above:

def fft(signal):
   n = len(signal)
   if n == 1:
      return signal
   else:
      Feven = fft([signal[i] for i in xrange(0, n, 2)])
      Fodd = fft([signal[i] for i in xrange(1, n, 2)])

      combined = [0] * n
      for m in xrange(n/2):
         combined[m] = Feven[m] + omega(n, -m) * Fodd[m]
         combined[m + n/2] = Feven[m] - omega(n, -m) * Fodd[m]

      return combined

Here I use the awkward variable names “Feven” and “Fodd” to be consistent with the notation above. Note that this implementation is highly non-optimized. There are many ways to improve the code, most notably using a different language and cacheing the complex exponential computations. In any event, the above code is quite readable, and a capable programmer could easily speed this up by orders of magnitude.

Of course, we should test this function on at least some of the discrete signals we investigated in our primer. And indeed, it functions correctly on the unshifted delta signal

>>> fft([1,0,0,0])
[(1+0j), (1+0j), (1+0j), (1+0j)]
>>> fft([1,0,0,0,0,0,0,0])
[(1+0j), (1+0j), (1+0j), (1+0j), (1+0j), (1+0j), (1+0j), (1+0j)]

As well as the shifted verion (up to floating point roundoff error)

>>> fft([0,1,0,0])
[(1+0j), 
 (6.123233995736766e-17-1j), 
 (-1+0j), 
 (-6.123233995736766e-17+1j)]
>>> fft([0,1,0,0,0,0,0,0])
[(1+0j), 
 (0.7071067811865476-0.7071067811865475j), 
 (6.123233995736766e-17-1j), 
 (-0.7071067811865475-0.7071067811865476j), 
 (-1+0j), 
 (-0.7071067811865476+0.7071067811865475j), 
 (-6.123233995736766e-17+1j), 
 (0.7071067811865475+0.7071067811865476j)]

And testing it on various other shifts gives further correct outputs. Finally, we test the Fourier transform of the discrete complex exponential, which the reader will recall is a scaled delta.

>>> w = cmath.exp(2 * cmath.pi * 1j / 8)
>>> w
(0.7071067811865476+0.7071067811865475j)
>>> d = 4
>>> fft([w**(k*d) for k in range(8)])
[                           1.7763568394002505e-15j, 
 (-7.357910944937894e-16  + 1.7763568394002503e-15j), 
 (-1.7763568394002505e-15 + 1.7763568394002503e-15j), 
 (-4.28850477329429e-15   + 1.7763568394002499e-15j), 
 (8                       - 1.2434497875801753e-14j), 
 (4.28850477329429e-15    + 1.7763568394002505e-15j), 
 (1.7763568394002505e-15  + 1.7763568394002505e-15j), 
 (7.357910944937894e-16   + 1.7763568394002505e-15j)]

Note that $ n = 8$ occurs in the position with index $ d=4$, and all of the other values are negligibly close to zero, as expected.

So that’s great! It works! Unfortunately it’s not quite there in terms of usability. In particular, we require our signal to have length a power of two, and most signals don’t happen to be that long. It turns out that this is a highly nontrivial issue, and all implementations of a discrete Fourier transform algorithm have to compensate for it in some way.

The easiest solution is to simply add zeros to the end of the signal until it is long enough, and just work with everything in powers of two. This technique (called zero-padding) is only reasonable if the signal in question is actually zero outside of the range it’s sampled from. Otherwise, subtle and important things can happen. We’ll leave further investigations to the reader, but the gist of the idea is that the Fourier transform assumes periodicity of the data one gives it, and so padding with zeros imposes a kind of periodicity that is simply nonexistent in the actual signal.

Now, of course not every Fast Fourier transform uses zero-padding. Unfortunately the techniques to evaluate a non-power-of-two Fourier transform are relatively complex, and beyond the scope of this post (though not beyond the scope of this blog). The interested reader should investigate the so-called Chirp Z-transform, as discovered by Rader, et al. in 1968. We may cover it here in the future.

In any case, the code above is a good starting point for any technique, and as usual it is available on this blog’s Github page. Finally, the inverse transform has a simple implementation, since it can be represented in terms of the forward transform (hint: remember duality!). We leave the code as an exercise to the reader.

Experiments with Sound — I am no Tree!

We’ll manipulate a clip of audio from Treebeard, a character from Lord of the Rings.

For the remainder of this post we’ll use a more established Fast Fourier Transform algorithm from the Python numpy library. The reasons for this are essentially convenience. Being implemented in C and brandishing the full might of the literature on Fourier transform algorithms, the numpy implementation is lightning fast. Now, note that the algorithm we implemented above is still correct (if one uses zero padding)! The skeptical reader may run the code to verify this. So we are justified in abandoning our implementation until we decide to seriously focus on improving its speed.

So let’s play with sound.

The sound clip we’ll be using is the following piece of dialog from the movie The Lord of the Rings: The Two Towers. We include the original wav file with the code on this blog’s Github page.

If we take the Fourier transform of the sound sample, we get the following plot of the frequency spectrum. Recall, the frequency spectrum is the graph of the norm of the frequency values.

The frequency spectrum of an Ent speaking. The x-axis represents the index in the list (larger values correspond to larger frequencies), and the y-axis corresponds to intensity (larger values correspond to that frequency having a greater presence in the original signal). There is symmetry about the ~80,000 index, and we may consider the right half ‘negative’ frequencies because the Fourier Transform is periodic modulo its length.

Here we note that there is a symmetry to the graph. This is not a coincidence: if the input signal is real-valued, it will always be the case that the Fourier transform is symmetric about its center value. The reason for this goes back to our first primer on the Fourier series, in that the negative coefficients were complex conjugates of the positive ones. In any event, we only need concern ourselves with the first half of the values.

We will omit the details for doing file I/O with audio. The interested reader can inspect our source code, and they will find a very basic use of the scikits.audiolab library, and matplotlib for plotting the frequency spectrum.

Now, just for fun, let’s tinker with this audio bite. First we’ll generate some random noise in the signal. That is, let’s mutate the input signal as follows:

import random
inputSignal = [x/2.0 + random.random()*0.1 for x in inputSignal]

This gives us a sound bite with an annoying fuzziness in the background:

Next, we will use the Fourier transform to remove some of this noise. Of course, since the noise is random it inhabits all the frequencies to some degree, so we can’t eliminate it entirely. Furthermore, as the original audio clip uses some of the higher frequencies, we will necessarily lose some information in the process. But in the real world we don’t have access to the original signal, so we should clean it up as best we can without losing too much information.

To do so, we can plot the frequencies for the noisy signal:

Comparing this with our original graph (cheating, yes, but the alternative is to guess and check until we arrive at the same answer), we see that the noise begins to dominate the spectrum at around the 20,000th index. So, we’ll just zero out all of those frequencies.

def frequencyFilter(signal):
   for i in range(20000, len(signal)-20000):
      signal[i] = 0

And voila! The resulting audio clip, while admittedly damaged, has only a small trace of noise:

Inspecting the source code, one should note that at one point we halve the amplitude of the input signal (i.e., in the time domain). The reason for this is if one arbitrarily tinkers with the frequencies of the Fourier transform, it can alter the amplitude of the original signal. As one quickly discovers, playing the resulting signal as a wav file can create an unpleasant crackling noise. The amplitudes are clipped (or wrapped around) by the audio software or hardware for a reason which is entirely mysterious to this author.

In any case, this is clearly not the optimal way (or even a good way) to reduce white noise in a signal. There are certainly better methods, but for the sake of time we will save them for future posts. The point is that we were able to implement and use the Fast Fourier Transform algorithm to analyze the discrete Fourier transform of a real-world signal, and manipulate it in logical ways. That’s quite the milestone considering where we began!

Next up in this series we’ll investigate more techniques on sound processing and other one-dimensional signals, and we’ll also derive a multi-dimensional Fourier transform so that we can start working with images and other two-dimensional signals.

Until then!