# A Quasipolynomial Time Algorithm for Graph Isomorphism: The Details

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. 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. Photo taken by me.

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.

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

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 $X$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 Codes of Solomon, Reed, and Muller

Last time we defined the Hamming code. We also saw that it meets the Hamming bound, which is a measure of how densely a code can be packed inside an ambient space and still maintain a given distance. This time we’ll define the Reed-Solomon code which optimizes a different bound called the Singleton bound, and then generalize them to a larger class of codes called Reed-Muller codes. In future posts we’ll consider algorithmic issues behind decoding the codes, for now we just care about their existence and optimality properties.

## The Singleton bound

Recall that a code $C$ is a set of strings called codewords, and that the parameters of a code $C$ are written $(n,k,d)_q$. Remember $n$ is the length of a codeword, $k = \log_q |C|$ is the message length, $d$ is the minimum distance between any two codewords, and $q$ is the size of the alphabet used for the codewords. Finally, remember that for linear codes our alphabets were either just $\{ 0,1 \}$ where $q=2$, or more generally a finite field $\mathbb{F}_q$ for $q$ a prime power.

One way to motivate for the Singleton bound goes like this. We can easily come up with codes for the following parameters. For $(n,n,1)_2$ the identity function works. And to get a $(n,n-1,2)_2$-code we can encode a binary string $x$ by appending the parity bit $\sum_i x_i \mod 2$ to the end (as an easy exercise, verify this has distance 2). An obvious question is can we generalize this to a $(n, n-d+1, d)_2$-code for any $d$? Perhaps a more obvious question is: why can’t we hope for better? A larger $d$ or $k \geq n-d+1$? Because the Singleton bound says so.

Theorem [Singleton 64]: If $C$ is an $(n,k,d)_q$-code, then $k \leq n-d+1$.

Proof. The proof is pleasantly simple. Let $\Sigma$ be your alphabet and look at the projection map $\pi : \Sigma^n \to \Sigma^{k-1}$ which projects $x = (x_1, \dots, x_n) \mapsto (x_1, \dots, x_{k-1})$. Remember that the size of the code is $|C| = q^k$, and because the codomain of $\pi,$ i.e. $\Sigma^{k-1}$ has size $q^{k-1} < q^k$, it follows that $\pi$ is not an injective map. In particular, there are two codewords $x,y$ whose first $k-1$ coordinates are equal. Even if all of their remaining coordinates differ, this implies that $d(x,y) < n-k+1$.

$\square$

It’s embarrassing that such a simple argument can prove that one can do no better. There are codes that meet this bound and they are called maximum distance separable (MDS) codes. One might wonder how MDS codes relate to perfect codes, but they are incomparable; there are perfect codes that are not MDS codes, and conversely MDS codes need not be perfect. The Reed-Solomon code is an example of the latter.

## The Reed-Solomon Code

Irving Reed (left) and Gustave Solomon (right).

The Reed-Solomon code has a very simple definition, especially for those of you who have read about secret sharing.

Given a prime power $q$ and integers $k \leq n \leq q$, the Reed-Solomon code with these parameters is defined by its encoding function $E: \mathbb{F}_q^k \to \mathbb{F}_q^n$ as follows.

1. Generate $\mathbb{F}_q$ explicitly.
2. Pick $n$ distinct elements $\alpha_i \in \mathbb{F}_q$.
3. A message $x \in \mathbb{F}_q^k$ is a list of elements $c_0 \dots c_{k-1}$. Represent the message as a polynomial $m(x) = \sum_j c_jx^j$.
4. The encoding of a message is the tuple $E(m) = (m(\alpha_1), \dots, m(\alpha_n))$. That is, we just evaluate $m(x)$ at our chosen locations in $\alpha_i$.

Here’s an example when $q=5, n=3, k=3$. We’ll pick the points $1,3,4 \in \mathbb{F}_5$, and let our message be $x = (4,1,2)$, which is encoded as a polynomial $m(x) = 4 + x + 2x^2$. Then the encoding of the message is

$\displaystyle E(m) = (m(1), m(3), m(4)) = (2, 0, 0)$

Decoding the message is a bit more difficult (more on that next time), but for now let’s prove the basic facts about this code.

Fact: The Reed-Solomon code is linear. This is just because polynomials of a limited degree form a vector space. Adding polynomials is adding their coefficients, and scaling them is scaling their coefficients. Moreover, the evaluation of a polynomial at a point is a linear map, i.e. it’s always true that $m_1(\alpha) + m_2(\alpha) = (m_1 + m_2)(\alpha)$, and scaling the coefficients is no different. So the codewords also form a vector space.

Fact: $d = n - k + 1$, or equivalently the Reed-Solomon code meets the Singleton bound. This follows from a simple fact: any two different single-variable polynomials of degree at most $k-1$ agree on at most $k-1$ points. Indeed, otherwise two such polynomials $f,g$ would give a new polynomial $f-g$ which has more than $k-1$ roots, but the fundamental theorem of algebra (the adaptation for finite fields) says the only polynomial with this many roots is the zero polynomial.

So the Reed-Solomon code is maximum distance separable. Neat!

One might wonder why one would want good codes with large alphabets. One reason is that with a large alphabet we can interpret a byte as an element of $\mathbb{F}_{256}$ to get error correction on bytes. So if you want to encode some really large stream of bytes (like a DVD) using such a scheme and you get bursts of contiguous errors in small regions (like a scratch), then you can do pretty powerful error correction. In fact, this is more or less the idea behind error correction for DVDs. So I hear. You can read more about the famous applications at Wikipedia.

## The Reed-Muller code

The Reed-Muller code is a neat generalization of the Reed-Solomon code to multivariable polynomials. The reason they’re so useful is not necessarily because they optimize some bound (if they do, I haven’t heard of it), but because they specialize to all sorts of useful codes with useful properties. One of these is properties is called local decodability, which has big applications in theoretical computer science.

Anyway, before I state the definition let me remind the reader about compact notation for multivariable polynomials. I can represent the variables $x_1, \dots, x_n$ used in the polynomial as a vector $\mathbf{x}$ and likewise a monomial $x_1^{\alpha_1} x_2^{\alpha_2} \dots x_n^{\alpha_n}$ by a “vector power” $\mathbf{x}^\alpha$, where $\sum_i \alpha_i = d$ is the degree of that monomial, and you’d write an entire polynomial as $\sum_\alpha c_\alpha x^{\alpha}$ where $\alpha$ ranges over all exponents you want.

Definition: Let $m, l$ be positive integers and $q > l$ be a prime power. The Reed-Muller code with parameters $m,l,q$ is defined as follows:

1. The message is the list of multinomial coefficients of a homogeneous degree $l$ polynomial in $m$ variables, $f(\mathbf{x}) = \sum_{\alpha} c_\alpha x^\alpha$.
2. You encode a message $f(\mathbf{x})$ as the tuple of all polynomial evaluations $(f(x))_{x \in \mathbb{F}_q^m}$.

Here the actual parameters of the code are $n=q^m$, and $k = \binom{m+l}{m}$ being the number of possible coefficients. Finally $d = (1 - l/q)n$, and we can prove this in the same way as we did for the Reed-Solomon code, using a beefed up fact about the number of roots of a multivariate polynomial:

Fact: Two multivariate degree $\leq l$ polynomials over a finite field $\mathbb{F}_q$ agree on at most an $l/q$ fraction of $\mathbb{F}_q^m$.

For messages of desired length $k$, a clever choice of parameters gives a good code. Let $m = \log k / \log \log k$, $q = \log^2 k$, and pick $l$ such that $\binom{m+l}{m} = k$. Then the Reed-Muller code has polynomial length $n = k^2$, and because $l = o(q)$ we get that the distance of the code is asymptotically $d = (1-o(1))n$, i.e. it tends to $n$.

A fun fact about Reed-Muller codes: they were apparently used on the Voyager space missions to relay image data back to Earth.

## The Way Forward

So we defined Reed-Solomon and Reed-Muller codes, but we didn’t really do any programming yet. The reason is because the encoding algorithms are very straightforward. If you’ve been following this blog you’ll know we have already written code to explicitly represent polynomials over finite fields, and extending that code to multivariable polynomials, at least for the sake of encoding the Reed-Muller code, is straightforward.

The real interesting algorithms come when you’re trying to decode. For example, in the Reed-Solomon code we’d take as input a bunch of points in a plane (over a finite field), only some of which are consistent with the underlying polynomial that generated them, and we have to reconstruct the unknown polynomial exactly. Even worse, for Reed-Muller we have to do it with many variables!

We’ll see exactly how to do that and produce working code next time.

Until then!

# Finding the majority element of a stream

Problem: Given a massive data stream of $n$ values in $\{ 1, 2, \dots, m \}$ and the guarantee that one value occurs more than $n/2$ times in the stream, determine exactly which value does so.

Solution: (in Python)

def majority(stream):
held = next(stream)
counter = 1

for item in stream:
if item == held:
counter += 1
elif counter == 0:
held = item
counter = 1
else:
counter -= 1

return held

Discussion: Let’s prove correctness. Say that $s$ is the unknown value that occurs more than $n/2$ times. The idea of the algorithm is that if you could pair up elements of your stream so that distinct values are paired up, and then you “kill” these pairs, then $s$ will always survive. The way this algorithm pairs up the values is by holding onto the most recent value that has no pair (implicitly, by keeping a count how many copies of that value you saw). Then when you come across a new element, you decrement the counter and implicitly account for one new pair.

Let’s analyze the complexity of the algorithm. Clearly the algorithm only uses a single pass through the data. Next, if the stream has size $n$, then this algorithm uses $O(\log(n) + \log(m))$ space. Indeed, if the stream entirely consists of a single value (say, a stream of all 1’s) then the counter will be $n$ at the end, which takes $\log(n)$ bits to store. On the other hand, if there are $m$ possible values then storing the largest requires $\log(m)$ bits.

Finally, the guarantee that one value occurs more than $n/2$ times is necessary. If it is not the case the algorithm could output anything (including the most infrequent element!). And moreover, if we don’t have this guarantee then every algorithm that solves the problem must use at least $\Omega(n)$ space in the worst case. In particular, say that $m=n$, and the first $n/2$ items are all distinct and the last $n/2$ items are all the same one, the majority value $s$. If you do not know $s$ in advance, then you must keep at least one bit of information to know which symbols occurred in the first half of the stream because any of them could be $s$. So the guarantee allows us to bypass that barrier.

This algorithm can be generalized to detect $k$ items with frequency above some threshold $n/(k+1)$ using space $O(k \log n)$. The idea is to keep $k$ counters instead of one, adding new elements when any counter is zero. When you see an element not being tracked by your $k$ counters (which are all positive), you decrement all the counters by 1. This is like a $k$-to-one matching rather than a pairing.

# Hamming’s Code

## Or how to detect and correct errors

Last time we made a quick tour through the main theorems of Claude Shannon, which essentially solved the following two problems about communicating over a digital channel.

1. What is the best encoding for information when you are guaranteed that your communication channel is error free?
2. Are there any encoding schemes that can recover from random noise introduced during transmission?

The answers to these questions were purely mathematical theorems, of course. But the interesting shortcoming of Shannon’s accomplishment was that his solution for the noisy coding problem (2) was nonconstructive. The question remains: can we actually come up with efficiently computable encoding schemes? The answer is yes! Marcel Golay was the first to discover such a code in 1949 (just a year after Shannon’s landmark paper), and Golay’s construction was published on a single page! We’re not going to define Golay’s code in this post, but we will mention its interesting status in coding theory later. The next year Richard Hamming discovered another simpler and larger family of codes, and went on to do some of the major founding work in coding theory. For his efforts he won a Turing Award and played a major part in bringing about the modern digital age. So we’ll start with Hamming’s codes.

We will assume some basic linear algebra knowledge, as detailed our first linear algebra primer. We will also use some basic facts about polynomials and finite fields, though the lazy reader can just imagine everything as binary $\{ 0,1 \}$ and still grok the important stuff.

Richard Hamming, inventor of Hamming codes. [image source]

## What is a code?

The formal definition of a code is simple: a code $C$ is just a subset of $\{ 0,1 \}^n$ for some $n$. Elements of $C$ are called codewords.

This is deceptively simple, but here’s the intuition. Say we know we want to send messages of length $k$, so that our messages are in $\{ 0,1 \}^k$. Then we’re really viewing a code $C$ as the image of some encoding function $\textup{Enc}: \{ 0,1 \}^k \to \{ 0,1 \}^n$. We can define $C$ by just describing what the set is, or we can define it by describing the encoding function. Either way, we will make sure that $\textup{Enc}$ is an injective function, so that no two messages get sent to the same codeword. Then $|C| = 2^k$, and we can call $k = \log |C|$ the message length of $C$ even if we don’t have an explicit encoding function.

Moreover, while in this post we’ll always work with $\{ 0,1 \}$, the alphabet of your encoded messages could be an arbitrary set $\Sigma$. So then a code $C$ would be a subset of tuples in $\Sigma^n$, and we would call $q = |\Sigma|$.

So we have these parameters $n, k, q$, and we need one more. This is the minimum distance of a code, which we’ll denote by $d$. This is defined to be the minimum Hamming distance between all distinct pairs of codewords, where by Hamming distance I just mean the number of coordinates that two tuples differ in. Recalling the remarks we made last time about Shannon’s nonconstructive proof, when we decode an encoded message $y$ (possibly with noisy bits) we look for the (unencoded) message $x$ whose encoding $\textup{Enc}(x)$ is as close to $y$ as possible. This will only work in the worst case if all pairs of codewords are sufficiently far apart. Hence we track the minimum distance of a code.

So coding theorists turn this mess of parameters into notation.

Definition: A code $C$ is called an $(n, k, d)_q$-code if

• $C \subset \Sigma^n$ for some alphabet $\Sigma$,
• $k = \log |C|$,
• $C$ has minimum distance $d$, and
• the alphabet $\Sigma$ has size $q$.

The basic goals of coding theory are:

1. For which values of these four parameters do codes exist?
2. Fixing any three parameters, how can we optimize the other one?

In this post we’ll see how simple linear-algebraic constructions can give optima for one of these problems, optimizing $k$ for $d=3$, and we’ll state a characterization theorem for optimizing $k$ for a general $d$. Next time we’ll continue with a second construction that optimizes a different bound called the Singleton bound.

## Linear codes and the Hamming code

A code is called linear if it can be identified with a linear subspace of some finite-dimensional vector space. In this post all of our vector spaces will be $\{ 0,1 \}^n$, that is tuples of bits under addition mod 2. But you can do the same constructions with any finite scalar field $\mathbb{F}_q$ for a prime power $q$, i.e. have your vector space be $\mathbb{F}_q^n$. We’ll go back and forth between describing a binary code $q=2$ over $\{ 0,1 \}$ and a code in $\mathbb{F}_q^n$. So to say a code is linear means:

• The zero vector is a codeword.
• The sum of any two codewords is a codeword.
• Any scalar multiple of a codeword is a codeword.

Linear codes are the simplest kinds of codes, but already they give a rich variety of things to study. The benefit of linear codes is that you can describe them in a lot of different and useful ways besides just describing the encoding function. We’ll use two that we define here. The idea is simple: you can describe everything about a linear subspace by giving a basis for the space.

Definition: generator matrix of a $(n,k,d)_q$-code $C$ is a $k \times n$ matrix $G$ whose rows form a basis for $C$.

There are a lot of equivalent generator matrices for a linear code (we’ll come back to this later), but the main benefit is that having a generator matrix allows one to encode messages $x \in \{0,1 \}^k$ by left multiplication $xG$. Intuitively, we can think of the bits of $x$ as describing the coefficients of the chosen linear combination of the rows of $G$, which uniquely describes an element of the subspace. Note that because a $k$-dimensional subspace of $\{ 0,1 \}^n$ has $2^k$ elements, we’re not abusing notation by calling $k = \log |C|$ both the message length and the dimension.

For the second description of $C$, we’ll remind the reader that every linear subspace $C$ has a unique orthogonal complement $C^\perp$, which is the subspace of vectors that are orthogonal to vectors in $C$.

Definition: Let $H^T$ be a generator matrix for $C^\perp$. Then $H$ is called a parity check matrix.

Note $H$ has the basis for $C^\perp$ as columns. This means it has dimensions $n \times (n-k)$. Moreover, it has the property that $x \in C$ if and only if the left multiplication $xH = 0$. Having zero dot product with all columns of $H$ characterizes membership in $C$.

The benefit of having a parity check matrix is that you can do efficient error detection: just compute $yH$ on your received message $y$, and if it’s nonzero there was an error! What if there were so many errors, and just the right errors that $y$ coincided with a different codeword than it started? Then you’re screwed. In other words, the parity check matrix is only guarantee to detect errors if you have fewer errors than the minimum distance of your code.

So that raises an obvious question: if you give me the generator matrix of a linear code can I compute its minimum distance? It turns out that this problem is NP-hard in general. In fact, you can show that this is equivalent to finding the smallest linearly dependent set of rows of the parity check matrix, and it is easier to see why such a problem might be hard. But if you construct your codes cleverly enough you can compute their distance properties with ease.

Before we do that, one more definition and a simple proposition about linear codes. The Hamming weight of a vector $x$, denoted $wt(x)$, is the number of nonzero entries in $x$.

Proposition: The minimum distance of a linear code $C$ is the minimum Hamming weight over all nonzero vectors $x \in C$.

Proof. Consider a nonzero $x \in C$. On one hand, the zero vector is a codeword and $wt(x)$ is by definition the Hamming distance between $x$ and zero, so it is an upper bound on the minimum distance. In fact, it’s also a lower bound: if $x,y$ are two nonzero codewords, then $x-y$ is also a codeword and $wt(x-y)$ is the Hamming distance between $x$ and $y$.

$\square$

So now we can define our first code, the Hamming code. It will be a $(n, k, 3)_2$-code. The construction is quite simple. We have fixed $d=3, q=2$, and we will also fix $l = n-k$. One can think of this as fixing $n$ and maximizing $k$, but it will only work for $n$ of a special form.

We’ll construct the Hamming code by describing a parity-check matrix $H$. In fact, we’re going to see what conditions the minimum distance $d=3$ imposes on $H$, and find out those conditions are actually sufficient to get $d=3$. We’ll start with 2. If we want to ensure $d \geq 2$, then you need it to be the case that no nonzero vector of Hamming weight 1 is a code word. Indeed, if $e_i$ is a vector with all zeros except a one in position $i$, then $e_i H = h_i$ is the $i$-th row of $H$. We need $e_i H \neq 0$, so this imposes the condition that no row of $H$ can be zero. It’s easy to see that this is sufficient for $d \geq 2$.

Likewise for $d \geq 3$, given a vector $y = e_i + e_j$ for some positions $i \neq j$, then $yH = h_i + h_j$ may not be zero. But because our sums are mod 2, saying that $h_i + h_j \neq 0$ is the same as saying $h_i \neq h_j$. Again it’s an if and only if. So we have the two conditions.

• No row of $H$ may be zero.
• All rows of $H$ must be distinct.

That is, any parity check matrix with those two properties defines a distance 3 linear code. The only question that remains is how large can $n$  be if the vectors have length $n-k = l$? That’s just the number of distinct nonzero binary strings of length $l$, which is $2^l - 1$. Picking any way to arrange these strings as the rows of a matrix (say, in lexicographic order) gives you a good parity check matrix.

Theorem: For every $l > 0$, there is a $(2^l - 1, 2^l - l - 1, 3)_2$-code called the Hamming code.

Since the Hamming code has distance 3, we can always detect if at most a single error occurs. Moreover, we can correct a single error using the Hamming code. If $x \in C$ and $wt(e) = 1$ is an error bit in position $i$, then the incoming message would be $y = x + e$. Now compute $yH = xH + eH = 0 + eH = h_i$ and flip bit $i$ of $y$. That is, whichever row of $H$ you get tells you the index of the error, so you can flip the corresponding bit and correct it. If you order the rows lexicographically like we said, then $h_i = i$ as a binary number. Very slick.

Before we move on, we should note one interesting feature of linear codes.

Definition: A code is called systematic if it can be realized by an encoding function that appends some number $n-k$ “check bits” to the end of each message.

The interesting feature is that all linear codes are systematic. The reason is as follows. The generator matrix $G$ of a linear code has as rows a basis for the code as a linear subspace. We can perform Gaussian elimination on $G$ and get a new generator matrix that looks like $[I \mid A]$ where $I$ is the identity matrix of the appropriate size and $A$ is some junk. The point is that encoding using this generator matrix leaves the message unchanged, and adds a bunch of bits to the end that are determined by $A$. It’s a different encoding function on $\{ 0,1\}^k$, but it has the same image in $\{ 0,1 \}^n$, i.e. the code is unchanged. Gaussian elimination just performed a change of basis.

If you work out the parameters of the Hamming code, you’ll see that it is a systematic code which adds $\Theta(\log n)$ check bits to a message, and we’re able to correct a single error in this code. An obvious question is whether this is necessary? Could we get away with adding fewer check bits? The answer is no, and a simple “information theoretic” argument shows this. A single index out of $n$ requires $\log n$ bits to describe, and being able to correct a single error is like identifying a unique index. Without logarithmically many bits, you just don’t have enough information.

## The Hamming bound and perfect codes

One nice fact about Hamming codes is that they optimize a natural problem: the problem of maximizing $d$ given a fixed choice of $n$, $k$, and $q$. To get this let’s define $V_n(r)$ denote the volume of a ball of radius $r$ in the space $\mathbb{F}_2^n$. I.e., if you fix any string (doesn’t matter which) $x$, $V_n(r)$ is the size of the set $\{ y : d(x,y) \leq r \}$, where $d(x,y)$ is the hamming distance.

There is a theorem called the Hamming bound, which describes a limit to how much you can pack disjoint balls of radius $r$ inside $\mathbb{F}_2^n$.

Theorem: If an $(n,k,d)_2$-code exists, then

$\displaystyle 2^k V_n \left ( \left \lfloor \frac{d-1}{2} \right \rfloor \right ) \leq 2^n$

Proof. The proof is quite simple. To say a code $C$ has distance $d$ means that for every string $x \in C$ there is no other string $y$ within Hamming distance $d$ of $x$. In other words, the balls centered around both $x,y$ of radius $r = \lfloor (d-1)/2 \rfloor$ are disjoint. The extra difference of one is for odd $d$, e.g. when $d=3$ you need balls of radius 1 to guarantee no overlap. Now $|C| = 2^k$, so the total number of strings covered by all these balls is the left-hand side of the expression. But there are at most $2^n$ strings in $\mathbb{F}_2^n$, establishing the desired inequality.

$\square$

Now a code is called perfect if it actually meets the Hamming bound exactly. As you probably guessed, the Hamming codes are perfect codes. It’s not hard to prove this, and I’m leaving it as an exercise to the reader.

The obvious follow-up question is whether there are any other perfect codes. The answer is yes, some of which are nonlinear. But some of them are “trivial.” For example, when $d=1$ you can just use the identity encoding to get the code $C = \mathbb{F}_2^n$. You can also just have a code which consists of a single codeword. There are also some codes that encode by repeating the message multiple times. These are called “repetition codes,” and all three of these examples are called trivial (as a definition). Now there are some nontrivial and nonlinear perfect codes I won’t describe here, but here is the nice characterization theorem.

Theorem [van Lint ’71, Tietavainen ‘73]: Let $C$ be a nontrivial perfect $(n,d,k)_q$ code. Then the parameters must either be that of a Hamming code, or one of the two:

• A $(23, 12, 7)_2$-code
• A $(11, 6, 5)_3$-code

The last two examples are known as the binary and ternary Golay codes, respectively, which are also linear. In other words, every possible set of parameters for a perfect code can be realized as one of these three linear codes.

So this theorem was a big deal in coding theory. The Hamming and Golay codes were both discovered within a year of each other, in 1949 and 1950, but the nonexistence of other perfect linear codes was open for twenty more years. This wrapped up a very neat package.

Next time we’ll discuss the Singleton bound, which optimizes for a different quantity and is incomparable with perfect codes. We’ll define the Reed-Solomon and show they optimize this bound as well. These codes are particularly famous for being the error correcting codes used in DVDs. We’ll then discuss the algorithmic issues surrounding decoding, and more recent connections to complexity theory.

Until then!