# The Many Faces of Set Cover

A while back Peter Norvig posted a wonderful pair of articles about regex golf. The idea behind regex golf is to come up with the shortest possible regular expression that matches one given list of strings, but not the other.

“Regex Golf,” by Randall Munroe.

In the first article, Norvig runs a basic algorithm to recreate and improve the results from the comic, and in the second he beefs it up with some improved search heuristics. My favorite part about this topic is that regex golf can be phrased in terms of a problem called set cover. I noticed this when reading the comic, and was delighted to see Norvig use that as the basis of his algorithm.

The set cover problem shows up in other places, too. If you have a database of items labeled by users, and you want to find the smallest set of labels to display that covers every item in the database, you’re doing set cover. I hear there are applications in biochemistry and biology but haven’t seen them myself.

If you know what a set is (just think of the “set” or “hash set” type from your favorite programming language), then set cover has a simple definition.

Definition (The Set Cover Problem): You are given a finite set $U$ called a “universe” and sets $S_1, \dots, S_n$ each of which is a subset of $U$. You choose some of the $S_i$ to ensure that every $x \in U$ is in one of your chosen sets, and you want to minimize the number of $S_i$ you picked.

It’s called a “cover” because the sets you pick “cover” every element of $U$. Let’s do a simple. Let $U = \{ 1,2,3,4,5 \}$ and

$\displaystyle S_1 = \{ 1,3,4 \}, S_2 = \{ 2,3,5 \}, S_3 = \{ 1,4,5 \}, S_4 = \{ 2,4 \}$

Then the smallest possible number of sets you can pick is 2, and you can achieve this by picking both $S_1, S_2$ or both $S_2, S_3$. The connection to regex golf is that you pick $U$ to be the set of strings you want to match, and you pick a set of regexes that match some of the strings in $U$ but none of the strings you want to avoid matching (I’ll call them $V$). If $w$ is such a regex, then you can form the set $S_w$ of strings that $w$ matches. Then if you find a small set cover with the strings $w_1, \dots, w_t$, then you can “or” them together to get a single regex $w_1 \mid w_2 \mid \dots \mid w_t$ that matches all of $U$ but none of $V$.

Set cover is what’s called NP-hard, and one implication is that we shouldn’t hope to find an efficient algorithm that will always give you the shortest regex for every regex golf problem. But despite this, there are approximation algorithms for set cover. What I mean by this is that there is a regex-golf algorithm $A$ that outputs a subset of the regexes matching all of $U$, and the number of regexes it outputs is such-and-such close to the minimum possible number. We’ll make “such-and-such” more formal later in the post.

What made me sad was that Norvig didn’t go any deeper than saying, “We can try to approximate set cover, and the greedy algorithm is pretty good.” It’s true, but the ideas are richer than that! Set cover is a simple example to showcase interesting techniques from theoretical computer science. And perhaps ironically, in Norvig’s second post a header promised the article would discuss the theory of set cover, but I didn’t see any of what I think of as theory. Instead he partially analyzes the structure of the regex golf instances he cares about. This is useful, but not really theoretical in any way unless he can say something universal about those instances.

I don’t mean to bash Norvig. His articles were great! And in-depth theory was way beyond scope. So this post is just my opportunity to fill in some theory gaps. We’ll do three things:

1. Show formally that set cover is NP-hard.
2. Prove the approximation guarantee of the greedy algorithm.
3. Show another (very different) approximation algorithm based on linear programming.

Along the way I’ll argue that by knowing (or at least seeing) the details of these proofs, one can get a better sense of what features to look for in the set cover instance you’re trying to solve. We’ll also see how set cover depicts the broader themes of theoretical computer science.

## NP-hardness

The first thing we should do is show that set cover is NP-hard. Intuitively what this means is that we can take some hard problem $P$ and encode instances of $P$ inside set cover problems. This idea is called a reduction, because solving problem $P$ will “reduce” to solving set cover, and the method we use to encode instance of $P$ as set cover problems will have a small amount of overhead. This is one way to say that set cover is “at least as hard as” $P$.

The hard problem we’ll reduce to set cover is called 3-satisfiability (3-SAT). In 3-SAT, the input is a formula whose variables are either true or false, and the formula is expressed as an OR of a bunch of clauses, each of which is an AND of three variables (or their negations). This is called 3-CNF form. A simple example:

$\displaystyle (x \vee y \vee \neg z) \wedge (\neg x \vee w \vee y) \wedge (z \vee x \vee \neg w)$

The goal of the algorithm is to decide whether there is an assignment to the variables which makes the formula true. 3-SAT is one of the most fundamental problems we believe to be hard and, roughly speaking, by reducing it to set cover we include set cover in a class called NP-complete, and if any one of these problems can be solved efficiently, then they all can (this is the famous P versus NP problem, and an efficient algorithm would imply P equals NP).

So a reduction would consist of the following: you give me a formula $\varphi$ in 3-CNF form, and I have to produce (in a way that depends on $\varphi$!) a universe $U$ and a choice of subsets $S_i \subset U$ in such a way that

$\varphi$ has a true assignment of variables if and only if the corresponding set cover problem has a cover using $k$ sets.

In other words, I’m going to design a function $f$ from 3-SAT instances to set cover instances, such that $x$ is satisfiable if and only if $f(x)$ has a set cover with $k$ sets.

Why do I say it only for $k$ sets? Well, if you can always answer this question then I claim you can find the minimum size of a set cover needed by doing a binary search for the smallest value of $k$. So finding the minimum size of a set cover reduces to the problem of telling if theres a set cover of size $k$.

Now let’s do the reduction from 3-SAT to set cover.

If you give me $\varphi = C_1 \wedge C_2 \wedge \dots \wedge C_m$ where each $C_i$ is a clause and the variables are denoted $x_1, \dots, x_n$, then I will choose as my universe $U$ to be the set of all the clauses and indices of the variables (these are all just formal symbols). i.e.

$\displaystyle U = \{ C_1, C_2, \dots, C_m, 1, 2, \dots, n \}$

The first part of $U$ will ensure I make all the clauses true, and the last part will ensure I don’t pick a variable to be both true and false at the same time.

To show how this works I have to pick my subsets. For each variable $x_i$, I’ll make two sets, one called $S_{x_i}$ and one called $S_{\neg x_i}$. They will both contain $i$ in addition to the clauses which they make true when the corresponding literal is true (by literal I just mean the variable or its negation). For example, if $C_j$ uses the literal $\neg x_7$, then $S_{\neg x_7}$ will contain $C_j$ but $S_{x_7}$ will not. Finally, I’ll set $k = n$, the number of variables.

Now to prove this reduction works I have to prove two things: if my starting formula has a satisfying assignment I have to show the set cover problem has a cover of size $k$. Indeed, take the sets $S_{y}$ for all literals $y$ that are set to true in a satisfying assignment. There can be at most $n$ true literals since half are true and half are false, so there will be at most $n$ sets, and these sets clearly cover all of $U$ because every literal has to be satisfied by some literal or else the formula isn’t true.

The reverse direction is similar: if I have a set cover of size $n$, I need to use it to come up with a satisfying truth assignment for the original formula. But indeed, the sets that get chosen can’t include both a $S_{x_i}$ and its negation set $S_{\neg x_i}$, because there are $n$ of the elements $\{1, 2, \dots, n \} \subset U$, and each $i$ is only in the two $S_{x_i}, S_{\neg x_i}$. Just by counting if I cover all the indices $i$, I already account for $n$ sets! And finally, since I have covered all the clauses, the literals corresponding to the sets I chose give exactly a satisfying assignment.

Whew! So set cover is NP-hard because I encoded this logic problem 3-SAT within its rules. If we think 3-SAT is hard (and we do) then set cover must also be hard. So if we can’t hope to solve it exactly we should try to approximate the best solution.

## The greedy approach

The method that Norvig uses in attacking the meta-regex golf problem is the greedy algorithm. The greedy algorithm is exactly what you’d expect: you maintain a list $L$ of the subsets you’ve picked so far, and at each step you pick the set $S_i$ that maximizes the number of new elements of $U$ that aren’t already covered by the sets in $L$. In python pseudocode:

def greedySetCover(universe, sets):
chosenSets = set()
leftToCover = universe.copy()
unchosenSets = sets

covered = lambda s: leftToCover & s

while universe != 0:
if len(chosenSets) == len(sets):
raise Exception("No set cover possible")

nextSet = max(unchosenSets, key=lambda s: len(covered(s)))
unchosenSets.remove(nextSet)
leftToCover -= nextSet

return chosenSets


This is what theory has to say about the greedy algorithm:

Theorem: If it is possible to cover $U$ by the sets in $F = \{ S_1, \dots, S_n \}$, then the greedy algorithm always produces a cover that at worst has size $O(\log(n)) \textup{OPT}$, where $\textup{OPT}$ is the size of the smallest cover. Moreover, this is asymptotically the best any algorithm can do.

One simple fact we need from calculus is that the following sum is asymptotically the same as $\log(n)$:

$\displaystyle H(n) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = \log(n) + O(1)$

Proof. [adapted from Wan] Let’s say the greedy algorithm picks sets $T_1, T_2, \dots, T_k$ in that order. We’ll set up a little value system for the elements of $U$. Specifically, the value of each $T_i$ is 1, and in step $i$ we evenly distribute this unit value across all newly covered elements of $T_i$. So for $T_1$ each covered element gets value $1/|T_1|$, and if $T_2$ covers four new elements, each gets a value of 1/4. One can think of this “value” as a price, or energy, or unit mass, or whatever. It’s just an accounting system (albeit a clever one) we use to make some inequalities clear later.

In general call the value $v_x$ of element $x \in U$ the value assigned to $x$ at the step where it’s first covered. In particular, the number of sets chosen by the greedy algorithm $k$ is just $\sum_{x \in U} v_x$. We’re just bunching back together the unit value we distributed for each step of the algorithm.

Now we want to compare the sets chosen by greedy to the optimal choice. Call a smallest set cover $C_{\textup{OPT}}$. Let’s stare at the following inequality.

$\displaystyle \sum_{x \in U} v_x \leq \sum_{S \in C_{\textup{OPT}}} \sum_{x \in S} v_x$

It’s true because each $x$ counts for a $v_x$ at most once in the left hand side, and in the right hand side the sets in $C_{\textup{OPT}}$ must hit each $x$ at least once but may hit some $x$ more than once. Also remember the left hand side is equal to $k$.

Now we want to show that the inner sum on the right hand side, $\sum_{x \in S} v_x$, is at most $H(|S|)$. This will in fact prove the entire theorem: because each set $S_i$ has size at most $n$, the inequality above will turn into

$\displaystyle k \leq |C_{\textup{OPT}}| H(|S|) \leq |C_{\textup{OPT}}| H(n)$

And so $k \leq \textup{OPT} \cdot O(\log(n))$, which is the statement of the theorem.

So we want to show that $\sum_{x \in S} v_x \leq H(|S|)$. For each $j$ define $\delta_j(S)$ to be the number of elements in $S$ not covered in $T_1, \cup \dots \cup T_j$. Notice that $\delta_{j-1}(S) - \delta_{j}(S)$ is the number of elements of $S$ that are covered for the first time in step $j$. If we call $t_S$ the smallest integer $j$ for which $\delta_j(S) = 0$, we can count up the differences up to step $t_S$, we get

$\sum_{x \in S} v_x = \sum_{i=1}^{t_S} (\delta_{i-1}(S) - \delta_i(S)) \cdot \frac{1}{T_i - (T_1 \cup \dots \cup T_{i-1})}$

The rightmost term is just the cost assigned to the relevant elements at step $i$. Moreover, because $T_i$ covers more new elements than $S$ (by definition of the greedy algorithm), the fraction above is at most $1/\delta_{i-1}(S)$. The end is near. For brevity I’ll drop the $(S)$ from $\delta_j(S)$.

\displaystyle \begin{aligned} \sum_{x \in S} v_x & \leq \sum_{i=1}^{t_S} (\delta_{i-1} - \delta_i) \frac{1}{\delta_{i-1}} \\ & \leq \sum_{i=1}^{t_S} (\frac{1}{1 + \delta_i} + \frac{1}{2+\delta_i} \dots + \frac{1}{\delta_{i-1}}) \\ & = \sum_{i=1}^{t_S} H(\delta_{i-1}) - H(\delta_i) \\ &= H(\delta_0) - H(\delta_{t_S}) = H(|S|) \end{aligned}

And that proves the claim.

$\square$

I have three postscripts to this proof:

1. This is basically the exact worst-case approximation that the greedy algorithm achieves. In fact, Petr Slavik proved in 1996 that the greedy gives you a set of size exactly $(\log n - \log \log n + O(1)) \textup{OPT}$ in the worst case.
2. This is also the best approximation that any set cover algorithm can achieve, provided that P is not NP. This result was basically known in 1994, but it wasn’t until 2013 and the use of some very sophisticated tools that the best possible bound was found with the smallest assumptions.
3. In the proof we used that $|S| \leq n$ to bound things, but if we knew that our sets $S_i$ (i.e. subsets matched by a regex) had sizes bounded by, say, $B$, the same proof would show that the approximation factor is $\log(B)$ instead of $\log n$. However, in order for that to be useful you need $B$ to be a constant, or at least to grow more slowly than any polynomial in $n$, since e.g. $\log(n^{0.1}) = 0.1 \log n$. In fact, taking a second look at Norvig’s meta regex golf problem, some of his instances had this property! Which means the greedy algorithm gives a much better approximation ratio for certain meta regex golf problems than it does for the worst case general problem. This is one instance where knowing the proof of a theorem helps us understand how to specialize it to our interests.

Norvig’s frequency table for president meta-regex golf. The left side counts the size of each set (defined by a regex)

## The linear programming approach

So we just said that you can’t possibly do better than the greedy algorithm for approximating set cover. There must be nothing left to say, job well done, right? Wrong! Our second analysis, based on linear programming, shows that instances with special features can have better approximation results.

In particular, if we’re guaranteed that each element $x \in U$ occurs in at most $B$ of the sets $S_i$, then the linear programming approach will give a $B$-approximation, i.e. a cover whose size is at worst larger than OPT by a multiplicative factor of $B$. In the case that $B$ is constant, we can beat our earlier greedy algorithm.

The technique is now a classic one in optimization, called LP-relaxation (LP stands for linear programming). The idea is simple. Most optimization problems can be written as integer linear programs, that is there you have $n$ variables $x_1, \dots, x_n \in \{ 0, 1 \}$ and you want to maximize (or minimize) a linear function of the $x_i$ subject to some linear constraints. The thing you’re trying to optimize is called the objective. While in general solving integer linear programs is NP-hard, we can relax the “integer” requirement to $0 \leq x_i \leq 1$, or something similar. The resulting linear program, called the relaxed program, can be solved efficiently using the simplex algorithm or another more complicated method.

The output of solving the relaxed program is an assignment of real numbers for the $x_i$ that optimizes the objective function. A key fact is that the solution to the relaxed linear program will be at least as good as the solution to the original integer program, because the optimal solution to the integer program is a valid candidate for the optimal solution to the linear program. Then the idea is that if we use some clever scheme to round the $x_i$ to integers, we can measure how much this degrades the objective and prove that it doesn’t degrade too much when compared to the optimum of the relaxed program, which means it doesn’t degrade too much when compared to the optimum of the integer program as well.

If this sounds wishy washy and vague don’t worry, we’re about to make it super concrete for set cover.

We’ll make a binary variable $x_i$ for each set $S_i$ in the input, and $x_i = 1$ if and only if we include it in our proposed cover. Then the objective function we want to minimize is $\sum_{i=1}^n x_i$. If we call our elements $X = \{ e_1, \dots, e_m \}$, then we need to write down a linear constraint that says each element $e_j$ is hit by at least one set in the proposed cover. These constraints have to depend on the sets $S_i$, but that’s not a problem. One good constraint for element $e_j$ is

$\displaystyle \sum_{i : e_j \in S_i} x_i \geq 1$

In words, the only way that an $e_j$ will not be covered is if all the sets containing it have their $x_i = 0$. And we need one of these constraints for each $j$. Putting it together, the integer linear program is

The integer program for set cover.

Once we understand this formulation of set cover, the relaxation is trivial. We just replace the last constraint with inequalities.

For a given candidate assignment $x$ to the $x_i$, call $Z(x)$ the objective value (in this case $\sum_i x_i$). Now we can be more concrete about the guarantees of this relaxation method. Let $\textup{OPT}_{\textup{IP}}$ be the optimal value of the integer program and $x_{\textup{IP}}$ a corresponding assignment to $x_i$ achieving the optimum. Likewise let $\textup{OPT}_{\textup{LP}}, x_{\textup{LP}}$ be the optimal things for the linear relaxation. We will prove:

Theorem: There is a deterministic algorithm that rounds $x_{\textup{LP}}$ to integer values $x$ so that the objective value $Z(x) \leq B \textup{OPT}_{\textup{IP}}$, where $B$ is the maximum number of sets that any element $e_j$ occurs in. So this gives a $B$-approximation of set cover.

Proof. Let $B$ be as described in the theorem, and call $y = x_{\textup{LP}}$ to make the indexing notation easier. The rounding algorithm is to set $x_i = 1$ if $y_i \geq 1/B$ and zero otherwise.

To prove the theorem we need to show two things hold about this new candidate solution $x$:

1. The choice of all $S_i$ for which $x_i = 1$ covers every element.
2. The number of sets chosen (i.e. $Z(x)$) is at most $B$ times more than $\textup{OPT}_{\textup{LP}}$.

Since $\textup{OPT}_{\textup{LP}} \leq \textup{OPT}_{\textup{IP}}$, so if we can prove number 2 we get $Z(x) \leq B \textup{OPT}_{\textup{LP}} \leq B \textup{OPT}_{\textup{IP}}$, which is the theorem.

So let’s prove 1. Fix any $j$ and we’ll show that element $e_j$ is covered by some set in the rounded solution. Call $B_j$ the number of times element $e_j$ occurs in the input sets. By definition $B_j \leq B$, so $1/B_j \geq 1/B$. Recall $y$ was the optimal solution to the relaxed linear program, and so it must be the case that the linear constraint for each $e_j$ is satisfied: $\sum_{i : e_j \in S_i} x_i \geq 1$. We know that there are $B_j$ terms and they sums to at least 1, so not all terms can be smaller than $1/B_j$ (otherwise they’d sum to something less than 1). In other words, some variable $x_i$ in the sum is at least $1/B_j \geq 1/B$, and so $x_i$ is set to 1 in the rounded solution, corresponding to a set $S_i$ that contains $e_j$. This finishes the proof of 1.

Now let’s prove 2. For each $j$, we know that for each $x_i = 1$, the corresponding variable $y_i \geq 1/B$. In particular $1 \leq y_i B$. Now we can simply bound the sum.

\displaystyle \begin{aligned} Z(x) = \sum_i x_i &\leq \sum_i x_i (B y_i) \\ &\leq B \sum_{i} y_i \\ &= B \cdot \textup{OPT}_{\textup{LP}} \end{aligned}

The second inequality is true because some of the $x_i$ are zero, but we can ignore them when we upper bound and just include all the $y_i$. This proves part 2 and the theorem.

$\square$

1. The proof works equally well when the sets are weighted, i.e. your cost for picking a set is not 1 for every set but depends on some arbitrarily given constants $w_i \geq 0$.
2. We gave a deterministic algorithm rounding $y$ to $x$, but one can get the same result (with high probability) using a randomized algorithm. The idea is to flip a coin with bias $y_i$ roughly $\log(n)$ times and set $x_i = 1$ if and only if the coin lands heads at least once. The guarantee is no better than what we proved, but for some other problems randomness can help you get approximations where we don’t know of any deterministic algorithms to get the same guarantees. I can’t think of any off the top of my head, but I’m pretty sure they’re out there.
3. For step 1 we showed that at least one term in the inequality for $e_j$ would be rounded up to 1, and this guaranteed we covered all the elements. A natural question is: why not also round up at most one term of each of these inequalities? It might be that in the worst case you don’t get a better guarantee, but it would be a quick extra heuristic you could use to post-process a rounded solution.
4. Solving linear programs is slow. There are faster methods based on so-called “primal-dual” methods that use information about the dual of the linear program to construct a solution to the problem. Goemans and Williamson have a nice self-contained chapter on their website about this with a ton of applications.

Williamson and Shmoys have a large textbook called The Design of Approximation Algorithms. One problem is that this field is like a big heap of unrelated techniques, so it’s not like the book will build up some neat theoretical foundation that works for every problem. Rather, it’s messy and there are lots of details, but there are definitely diamonds in the rough, such as the problem of (and algorithms for) coloring 3-colorable graphs with “approximately 3″ colors, and the infamous unique games conjecture.

I wrote a post a while back giving conditions which, if a problem satisfies those conditions, the greedy algorithm will give a constant-factor approximation. This is much better than the worst case $\log(n)$-approximation we saw in this post. Moreover, I also wrote a post about matroids, which is a characterization of problems where the greedy algorithm is actually optimal.

Set cover is one of the main tools that IBM’s AntiVirus software uses to detect viruses. Similarly to the regex golf problem, they find a set of strings that occurs source code in some viruses but not (usually) in good programs. Then they look for a small set of strings that covers all the viruses, and their virus scan just has to search binaries for those strings. Hopefully the size of your set cover is really small compared to the number of viruses you want to protect against. I can’t find a reference that details this, but that is understandable because it is proprietary software.

Until next time!

# Markov Chain Monte Carlo Without all the Bullshit

I have a little secret: I don’t like the terminology, notation, and style of writing in statistics. I find it unnecessarily complicated. This shows up when trying to read about Markov Chain Monte Carlo methods. Take, for example, the abstract to the Markov Chain Monte Carlo article in the Encyclopedia of Biostatistics.

Markov chain Monte Carlo (MCMC) is a technique for estimating by simulation the expectation of a statistic in a complex model. Successive random selections form a Markov chain, the stationary distribution of which is the target distribution. It is particularly useful for the evaluation of posterior distributions in complex Bayesian models. In the Metropolis–Hastings algorithm, items are selected from an arbitrary “proposal” distribution and are retained or not according to an acceptance rule. The Gibbs sampler is a special case in which the proposal distributions are conditional distributions of single components of a vector parameter. Various special cases and applications are considered.

I can only vaguely understand what the author is saying here (and really only because I know ahead of time what MCMC is). There are certainly references to more advanced things than what I’m going to cover in this post. But it seems very difficult to find an explanation of Markov Chain Monte Carlo without all any superfluous jargon. The “bullshit” here is the implicit claim of an author that such jargon is needed. Maybe it is to explain advanced applications (like attempts to do “inference in Bayesian networks”), but it is certainly not needed to define or analyze the basic ideas.

So to counter, here’s my own explanation of Markov Chain Monte Carlo, inspired by the treatment of John Hopcroft and Ravi Kannan.

## The Problem is Drawing from a Distribution

Markov Chain Monte Carlo is a technique to solve the problem of sampling from a complicated distribution. Let me explain by the following imaginary scenario. Say I have a magic box which can estimate probabilities of baby names very well. I can give it a string like “Malcolm” and it will tell me the exact probability $p_{\textup{Malcolm}}$ that you will choose this name for your next child. So there’s a distribution $D$ over all names, it’s very specific to your preferences, and for the sake of argument say this distribution is fixed and you don’t get to tamper with it.

Now comes the problem: I want to efficiently draw a name from this distribution $D$. This is the problem that Markov Chain Monte Carlo aims to solve. Why is it a problem? Because I have no idea what process you use to pick a name, so I can’t simulate that process myself. Here’s another method you could try: generate a name $x$ uniformly at random, ask the machine for $p_x$, and then flip a biased coin with probability $p_x$ and use $x$ if the coin lands heads. The problem with this is that there are exponentially many names! The variable here is the number of bits needed to write down a name $n = |x|$. So either the probabilities $p_x$ will be exponentially small and I’ll be flipping for a very long time to get a single name, or else there will only be a few names with nonzero probability and it will take me exponentially many draws to find them. Inefficiency is the death of me.

So this is a serious problem! Let’s restate it formally just to be clear.

Definition (The sampling problem):  Let $D$ be a distribution over a finite set $X$. You are given black-box access to the probability distribution function $p(x)$ which outputs the probability of drawing $x \in X$ according to $D$. Design an efficient randomized algorithm $A$ which outputs an element of $X$ so that the probability of outputting $x$ is approximately $p(x)$. More generally, output a sample of elements from $X$ drawn according to $p(x)$.

Assume that $A$ has access to only fair random coins, though this allows one to efficiently simulate flipping a biased coin of any desired probability.

Notice that with such an algorithm we’d be able to do things like estimate the expected value of some random variable $f : X \to \mathbb{R}$. We could take a large sample $S \subset X$ via the solution to the sampling problem, and then compute the average value of $f$ on that sample. This is what a Monte Carlo method does when sampling is easy. In fact, the Markov Chain solution to the sampling problem will allow us to do the sampling and the estimation of $\mathbb{E}(f)$ in one fell swoop if you want.

But the core problem is really a sampling problem, and “Markov Chain Monte Carlo” would be more accurately called the “Markov Chain Sampling Method.” So let’s see why a Markov Chain could possibly help us.

## Random Walks, the “Markov Chain” part of MCMC

Markov Chain is essentially a fancy term for a random walk on a graph.

You give me a directed graph $G = (V,E)$, and for each edge $e = (u,v) \in E$ you give me a number $p_{u,v} \in [0,1]$. In order to make a random walk make sense, the $p_{u,v}$ need to satisfy the following constraint:

For any vertex $x \in V$, the set all values $p_{x,y}$ on outgoing edges $(x,y)$ must sum to 1, i.e. form a probability distribution.

If this is satisfied then we can take a random walk on $G$ according to the probabilities as follows: start at some vertex $x_0$. Then pick an outgoing edge at random according to the probabilities on the outgoing edges, and follow it to $x_1$. Repeat if possible.

I say “if possible” because an arbitrary graph will not necessarily have any outgoing edges from a given vertex. We’ll need to impose some additional conditions on the graph in order to apply random walks to Markov Chain Monte Carlo, but in any case the idea of randomly walking is well-defined, and we call the whole object $(V,E, \{ p_e \}_{e \in E})$Markov chain.

Here is an example where the vertices in the graph correspond to emotional states.

An example Markov chain; image source http://www.mathcs.emory.edu/~cheung/

In statistics land, they take the “state” interpretation of a random walk very seriously. They call the edge probabilities “state-to-state transitions.”

The main theorem we need to do anything useful with Markov chains is the stationary distribution theorem (sometimes called the “Fundamental Theorem of Markov Chains,” and for good reason). What it says intuitively is that for a very long random walk, the probability that you end at some vertex $v$ is independent of where you started! All of these probabilities taken together is called the stationary distribution of the random walk, and it is uniquely determined by the Markov chain.

However, for the reasons we stated above (“if possible”), the stationary distribution theorem is not true of every Markov chain. The main property we need is that the graph $G$ is strongly connected. Recall that a directed graph is called connected if, when you ignore direction, there is a path from every vertex to every other vertex. It is called strongly connected if you still get paths everywhere when considering direction. If we additionally require the stupid edge-case-catcher that no edge can have zero probability, then strong connectivity (of one component of a graph) is equivalent to the following property:

For every vertex $v \in V(G)$, an infinite random walk started at $v$ will return to $v$ with probability 1.

In fact it will return infinitely often. This property is called the persistence of the state $v$ by statisticians. I dislike this term because it appears to describe a property of a vertex, when to me it describes a property of the connected component containing that vertex. In any case, since in Markov Chain Monte Carlo we’ll be picking the graph to walk on (spoiler!) we will ensure the graph is strongly connected by design.

Finally, in order to describe the stationary distribution in a more familiar manner (using linear algebra), we will write the transition probabilities as a matrix $A$ where entry $a_{j,i} = p_{(i,j)}$ if there is an edge $(i,j) \in E$ and zero otherwise. Here the rows and columns correspond to vertices of $G$, and each column $i$ forms the probability distribution of going from state $i$ to some other state in one step of the random walk. Note $A$ is the transpose of the weighted adjacency matrix of the directed weighted graph $G$ where the weights are the transition probabilities (the reason I do it this way is because matrix-vector multiplication will have the matrix on the left instead of the right; see below).

This matrix allows me to describe things nicely using the language of linear algebra. In particular if you give me a basis vector $e_i$ interpreted as “the random walk currently at vertex $i$,” then $Ae_i$ gives a vector whose $j$-th coordinate is the probability that the random walk would be at vertex $j$ after one more step in the random walk. Likewise, if you give me a probability distribution $q$ over the vertices, then $Aq$ gives a probability vector interpreted as follows:

If a random walk is in state $i$ with probability $q_i$, then the $j$-th entry of $Aq$ is the probability that after one more step in the random walk you get to vertex $j$.

Interpreted this way, the stationary distribution is a probability distribution $\pi$ such that $A \pi = \pi$, in other words $\pi$ is an eigenvector of $A$ with eigenvalue 1.

A quick side note for avid readers of this blog: this analysis of a random walk is exactly what we did back in the early days of this blog when we studied the PageRank algorithm for ranking webpages. There we called the matrix $A$ “a web matrix,” noted it was column stochastic (as it is here), and appealed to a special case of the Perron-Frobenius theorem to show that there is a unique maximal eigenvalue equal to one (with a dimension one eigenspace) whose eigenvector we used as a sort of “stationary distribution” and the final ranking of web pages. There we described an algorithm to actually find that eigenvector by iterated multiplication by $A$. The following theorem is essentially a variant of this algorithm but works under weaker conditions; for the web matrix we added additional “fake” edges that give the needed stronger conditions.

Theorem: Let $G$ be a strongly connected graph with associated edge probabilities $\{ p_e \}_e \in E$ forming a Markov chain. For a probability vector $x_0$, define $x_{t+1} = Ax_t$ for all $t \geq 1$, and let $v_t$ be the long-term average $v_t = \frac1t \sum_{s=1}^t x_s$. Then:

1. There is a unique probability vector $\pi$ with $A \pi = \pi$.
2. For all $x_0$, the limit $\lim_{t \to \infty} v_t = \pi$.

Proof. Since $v_t$ is a probability vector we just want to show that $|Av_t - v_t| \to 0$ as $t \to \infty$. Indeed, we can expand this quantity as

\displaystyle \begin{aligned} Av_t - v_t &=\frac1t (Ax_0 + Ax_1 + \dots + Ax_{t-1}) - \frac1t (x_0 + \dots + x_{t-1}) \\ &= \frac1t (x_t - x_0) \end{aligned}

But $x_t, x_0$ are unit vectors, so their difference is at most 2, meaning $|Av_t - v_t| \leq \frac2t \to 0$. Now it’s clear that this does not depend on $v_0$. For uniqueness we will cop out and appeal to the Perron-Frobenius theorem that says any matrix of this form has a unique such (normalized) eigenvector.

$\square$

One additional remark is that, in addition to computing the stationary distribution by actually computing this average or using an eigensolver, one can analytically solve for it as the inverse of a particular matrix. Define $B = A-I_n$, where $I_n$ is the $n \times n$ identity matrix. Let $C$ be $B$ with a row of ones appended to the bottom and the topmost row removed. Then one can show (quite opaquely) that the last column of $C^{-1}$ is $\pi$. We leave this as an exercise to the reader, because I’m pretty sure nobody uses this method in practice.

One final remark is about why we need to take an average over all our $x_t$ in the theorem above. There is an extra technical condition one can add to strong connectivity, called aperiodicity, which allows one to beef up the theorem so that $x_t$ itself converges to the stationary distribution. Rigorously, aperiodicity is the property that, regardless of where you start your random walk, after some sufficiently large number of steps $n$ the random walk has a positive probability of being at every vertex at every subsequent step. As an example of a graph where aperiodicity fails: an undirected cycle on an even number of vertices. In that case there will only be a positive probability of being at certain vertices every other step, and averaging those two long term sequences gives the actual stationary distribution.

Image source: Wikipedia

One way to guarantee that your Markov chain is aperiodic is to ensure there is a positive probability of staying at any vertex. I.e., that your graph has a self-loop. This is what we’ll do in the next section.

## Constructing a graph to walk on

Recall that the problem we’re trying to solve is to draw from a distribution over a finite set $X$ with probability function $p(x)$. The MCMC method is to construct a Markov chain whose stationary distribution is exactly $p$, even when you just have black-box access to evaluating $p$. That is, you (implicitly) pick a graph $G$ and (implicitly) choose transition probabilities for the edges to make the stationary distribution $p$. Then you take a long enough random walk on $G$ and output the $x$ corresponding to whatever state you land on.

The easy part is coming up with a graph that has the right stationary distribution (in fact, “most” graphs will work). The hard part is to come up with a graph where you can prove that the convergence of a random walk to the stationary distribution is fast in comparison to the size of $X$. Such a proof is beyond the scope of this post, but the “right” choice of a graph is not hard to understand.

The one we’ll pick for this post is called the Metropolis-Hastings algorithm. The input is your black-box access to $p(x)$, and the output is a set of rules that implicitly define a random walk on a graph whose vertex set is $X$.

It works as follows: you pick some way to put $X$ on a lattice, so that each state corresponds to some vector in $\{ 0,1, \dots, n\}^d$. Then you add (two-way directed) edges to all neighboring lattice points. For $n=5, d=2$ it would look like this:

And for $d=3, n \in \{2,3\}$ it would look like this:

You have to be careful here to ensure the vertices you choose for $X$ are not disconnected, but in many applications $X$ is naturally already a lattice.

Now we have to describe the transition probabilities. Let $r$ be the maximum degree of a vertex in this lattice ($r=2d$). Suppose we’re at vertex $i$ and we want to know where to go next. We do the following:

1. Pick neighbor $j$ with probability $1/r$ (there is some chance to stay at $i$).
2. If you picked neighbor $j$ and $p(j) \geq p(i)$ then deterministically go to $j$.
3. Otherwise, $p(j) < p(i)$, and you go to $j$ with probability $p(j) / p(i)$.

We can state the probability weight $p_{i,j}$ on edge $(i,j)$ more compactly as

$\displaystyle p_{i,j} = \frac1r \min(1, p(j) / p(i)) \\ p_{i,i} = 1 - \sum_{(i,j) \in E(G); j \neq i} p_{i,j}$

It is easy to check that this is indeed a probability distribution for each vertex $i$. So we just have to show that $p(x)$ is the stationary distribution for this random walk.

Here’s a fact to do that: if a probability distribution $v$ with entries $v(x)$ for each $x \in X$ has the property that $v(x)p_{x,y} = v(y)p_{y,x}$ for all $x,y \in X$, the $v$ is the stationary distribution. To prove it, fix $x$ and take the sum of both sides of that equation over all $y$. The result is exactly the equation $v(x) = \sum_{y} v(y)p_{y,x}$, which is the same as $v = Av$. Since the stationary distribution is the unique vector satisfying this equation, $v$ has to be it.

Doing this with out chosen $p(i)$ is easy, since $p(i)p_{i,j}$ and $p(i)p_{j,i}$ are both equal to $\frac1r \min(p(i), p(j))$ by applying a tiny bit of algebra to the definition. So we’re done! One can just randomly walk according to these probabilities and get a sample.

## Last words

The last thing I want to say about MCMC is to show that you can estimate the expected value of a function $\mathbb{E}(f)$ simultaneously while random-walking through your Metropolis-Hastings graph (or any graph whose stationary distribution is $p(x)$). By definition the expected value of $f$ is $\sum_x f(x) p(x)$.

Now what we can do is compute the average value of $f(x)$ just among those states we’ve visited during our random walk. With a little bit of extra work you can show that this quantity will converge to the true expected value of $f$ at about the same time that the random walk converges to the stationary distribution. (Here the “about” means we’re off by a constant factor depending on $f$). In order to prove this you need some extra tools I’m too lazy to write about in this post, but the point is that it works.

The reason I did not start by describing MCMC in terms of estimating the expected value of a function is because the core problem is a sampling problem. Moreover, there are many applications of MCMC that need nothing more than a sample. For example, MCMC can be used to estimate the volume of an arbitrary (maybe high dimensional) convex set. See these lecture notes of Alistair Sinclair for more.

If demand is popular enough, I could implement the Metropolis-Hastings algorithm in code (it wouldn’t be industry-strength, but perhaps illuminating? I’m not so sure…).

Until next time!

# 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.