# On the Computational Complexity of MapReduce

I recently wrapped up a fun paper with my coauthors Ben Fish, Adam Lelkes, Lev Reyzin, and Gyorgy Turan in which we analyzed the computational complexity of a model of the popular MapReduce framework. Check out the preprint on the arXiv.

As usual I’ll give a less formal discussion of the research here, and because the paper is a bit more technically involved than my previous work I’ll be omitting some of the more pedantic details. Our project started after Ben Moseley gave an excellent talk at UI Chicago. He presented a theoretical model of MapReduce introduced by Howard Karloff et al. in 2010, and discussed his own results on solving graph problems in this model, such as graph connectivity. You can read Karloff’s original paper here, but we’ll outline his model below.

Basically, the vast majority of the work on MapReduce has been algorithmic. What I mean by that is researchers have been finding more and cleverer algorithms to solve problems in MapReduce. They have covered a huge amount of work, implementing machine learning algorithms, algorithms for graph problems, and many others. In Moseley’s talk, he posed a question that caught our eye:

Is there a constant-round MapReduce algorithm which determines whether a graph is connected?

After we describe the model below it’ll be clear what we mean by “solve” and what we mean by “constant-round,” but the conjecture is that this is impossible, particularly for the case of sparse graphs. We know we can solve it in a logarithmic number of rounds, but anything better is open.

In any case, we started thinking about this problem and didn’t make much progress. To the best of my knowledge it’s still wide open. But along the way we got into a whole nest of more general questions about the power of MapReduce. Specifically, Karloff proved a theorem relating MapReduce to a very particular class of circuits. What I mean is he proved a theorem that says “anything that can be solved in MapReduce with so many rounds and so much space can be solved by circuits that are yae big and yae complicated, and vice versa.

But this question is so specific! We wanted to know: is MapReduce as powerful as polynomial time, our classical notion of efficiency (does it equal P)? Can it capture all computations requiring logarithmic space (does it contain L)? MapReduce seems to be somewhere in between, but it’s exact relationship to these classes is unknown. And as we’ll see in a moment the theoretical model uses a novel communication model, and processors that never get to see the entire input. So this led us to a host of natural complexity questions:

1. What computations are possible in a model of parallel computation where no processor has enough space to store even one thousandth of the input?
2. What computations are possible in a model of parallel computation where processor’s can’t request or send specific information from/to other processors?
3. How the hell do you prove that something can’t be done under constraints of this kind?
4. How do you measure the increase of power provided by giving MapReduce additional rounds or additional time?

These questions are in the domain of complexity theory, and so it makes sense to try to apply the standard tools of complexity theory to answer them. Our paper does this, laying some brick for future efforts to study MapReduce from a complexity perspective.

In particular, one of our theorems is the following:

Theorem: Any problem requiring sublogarithmic space $o(\log n)$ can be solved in MapReduce in two rounds.

This theorem is nice for two reasons. First is it’s constructive. If you give me a problem and I know it classically takes less than logarithmic space, then this gives an automatic algorithm to implement it in MapReduce, and if you were so inclined you could even automate the process of translating a classical algorithm to a MapReduce algorithm (it’s not pretty, but it’s possible).

Our other results are a bit more esoteric. We relate the questions about MapReduce to old, deep questions about computations that have simultaneous space and time bounds. In the end we give a (conditional, nonconstructive) answer to question 4 above, which I’ll sketch without getting too deep in the details.

So let’s start with the model of Karloff et al., which they named “MRC” for MapReduce Class.

## The MRC Model of Karloff et al.

MapReduce is a programming paradigm which is intended to make algorithm design for distributed computing easier. Specifically, when you’re writing massively distributed algorithms by hand, you spend a lot of time and effort dealing with really low-level questions. Like, what do I do if a processor craps out in the middle of my computation? Or, what’s the most efficient way to broadcast a message to all the processors, considering the specific topology of my network layout means the message will have to be forwarded? Note these are questions that have nothing to do with the actual problem you’re trying to solve.

So the designers of MapReduce took a step back, analyzed the class of problems they were most often trying to solve (turns out, searching, sorting, and counting), and designed their framework to handle all of the low-level stuff automatically. The catch is that the algorithm designer now has to fit their program into the MapReduce paradigm, which might be hard.

In designing a MapReduce algorithm, one has to implement two functions: a mapper and a reducer. The mapper takes a list of key-value pairs, and applies some operation to each element independently (just like the map function in most functional programming languages). The reducer takes a single key and a list of values for that key, and outputs a new list of values with the same key. And then the MapReduce protocol iteratively applies mappers and reducers in rounds to the input data. There are a lot of pictures of this protocol on the internet. Here’s one

Image source: ibm.com

An interesting part of this protocol is that the reducers never get to communicate with each other, except indirectly through the mappers in the next round. MapReduce handles the implied grouping and message passing, as well as engineering issues like fault-tolerance and load balancing. In this sense, the mappers and reducers are ignorant cogs in the machine, so it’s interesting to see how powerful MapReduce is.

The model of Karloff, Suri, and Vassilvitskii is a relatively straightforward translation of this protocol to mathematics. They make a few minor qualifications, though, the most important of which is that they encode a bound on the total space used. In particular, if the input has size $n$, they impose that there is some $\varepsilon > 0$ for which every reducer uses space $O(n^{1 - \varepsilon})$. Moreover, the number of reducers in any round is also bounded by $O(n^{1 - \varepsilon})$.

We can formalize all of this with a few easy definitions.

Definition: An input to a MapReduce algorithm is a list of key-value pairs $\langle k_i,v_i \rangle_{i=1}^N$ of total size $n = \sum_{i=1}^N |k_i| + |v_i|$.

For binary languages (e.g., you give me a binary string $x$ and you want to know if there are an odd number of 1’s), we encode a string $x \in \{ 0,1 \}^m$ as the list $\langle x \rangle := \langle i, x_i \rangle_{i=1}^n$. This won’t affect any of our space bounds, because the total blowup from $m = |x|$ to $n = |\langle x \rangle |$ is logarithmic.

Definition: A mapper $\mu$ is a Turing machine which accepts as input a single key-value pair $\langle k, v \rangle$ and produces as output a list of key-value pairs $\langle k_1', v'_1 \rangle, \dots, \langle k'_s, v'_s \rangle$.

Definition: reducer $\rho$ is a Turing machine which accepts as input a key $k$ and a list of values $v_1, \dots, v_m$ and produces as output the same key and a new list of values $v'_1, \dots, v'_M$.

Now we can describe the MapReduce protocol, i.e. what a program is and how to run it. I copied this from our paper because the notation is the same so far.

The MRC protocol

All we’ve done here is just describe the MapReduce protocol in mathematics. It’s messier than it is complicated. The last thing we need is the space bounds.

Definition: A language $L$ is in $\textup{MRC}[f(n), g(n)]$ if there is a constant $0 < c < 1$ and a sequence of mappers and reducers $\mu_1, \rho_1, \mu_2, \rho_2, \dots$ such that for all $x \in \{ 0,1 \}^n$ the following is satisfied:

1. Letting $R = O(f(n))$ and $M = (\mu_1, \rho_1, \dots, \mu_R, \rho_R)$, $M$ accepts $x$ if and only if $x \in L$.
2. For all $1 \leq r \leq R$, $\mu_r, \rho_r$ use $O(n^c)$ space and $O(g(n))$ time.
3. Each $\mu_r$ outputs $O(n^c)$ keys in round $r$.

To be clear, the first argument to $\textup{MRC}[f(n), g(n)]$ is the round bound, and the second argument is the time bound. The last point implies the bound on the number of processors. Since we are often interested in a logarithmic number of rounds, we can define

$\displaystyle \textup{MRC}^i = \textup{MRC}[\log^i(n), \textup{poly}(n)]$

So we can restate the question from the beginning of the post as,

Is graph connectivity in $\textup{MRC}^0$?

Here there’s a bit of an issue with representation. You have to show that if you’re just given a binary string representing a graph, that you can translate that into a reasonable key-value description of a graph in a constant number of rounds. This is possible, and gives evidence that the key-value representation is without loss of generality for all intents and purposes.

## A Pedantic Aside

If our goal is to compare MRC with classes like polynomial time (P) and logarithmic space (L), then the definition above has a problem. Specifically the definition above allows one to have an infinite list of reducers, where each one is potentially different, and the actual machine that is used depends on the input size. In formal terminology, MRC as defined above is a nonuniform model of computation.

Indeed, we give a simple proof that this is a problem by showing any unary language is in $\textup{MRC}^1$ (which is where many of the MRC algorithms in recent years have been). For those who don’t know, this is a problem because you can encode versions of the Halting problem as a unary language, and the Halting problem is undecidable.

While this level of pedantry might induce some eye-rolling, it is necessary in order to make statements like “MRC is contained in P.” It’s simply not true for the nonuniform model above. To fix this problem we refined the MRC model to a uniform version. The details are not trivial, but also not that interesting. Check out the paper if you want to know exactly how we do it. It doesn’t have that much of an effect on the resulting analysis. The only important detail is that now we’re representing the entire MRC machine as a single Turing machine. So now, unlike before, any MRC machine can be written down as a finite string, and there are infinitely many strings representing a given MRC machine. We call our model MRC, and Karloff’s model nonuniform MRC.

You can also make randomized versions of MRC, but we’ll stick to the deterministic version here.

## Sublogarithmic Space

Now we can sketch the proof that sublogarithmic space is in $\textup{MRC}^0$. In fact, the proof is simpler for regular languages (equivalent to constant space algorithms) being in $\textup{MRC}^0$. The idea is as follows:

A regular language is one that can be decided by a DFA, say $G$ (a graph representing state transitions as you read a stream of bits). And the DFA is independent of the input size, so every mapper and reducer can have it encoded into them. Now what we’ll do is split the input string up in contiguous chunks of size $O(\sqrt{n})$ among the processors (mappers can do this just fine). Now comes the trick. We have each reducer compute, for each possible state $s$ in $G$, what the ending state would be if the DFA had started in state $s$ after processing their chunk of the input. So the output of reducer $j$ would be an encoding of a table of the form:

$\displaystyle s_1 \to T_j(s_1) \\ s_2 \to T_j(s_2) \\ \vdots \\ s_{|S|} \to T_j(s_{|S|})$

And the result would be a lookup table of intermediate computation results. So each reducer outputs their part of the table (which has constant size). Since there are only $O(\sqrt{n})$ reducers, all of the table pieces can fit on a single reducer in the second round, and this reducer can just chain the functions together from the starting state and output the answer.

The proof for sublogarithmic space has the same structure, but is a bit more complicated because one has to account for the tape of a Turing machine which has size $o(\log n)$. In this case, you split up the tape of the Turing machine among the processors as well. And then you have to keep track of a lot more information. In particular, each entry of your function has to look like

“if my current state is A and my tape contents are B and the tape head starts on side C of my chunk of the input”

then

“the next time the tape head would leave my chunk, it will do so on side C’ and my state will be A’ and my tape contents will be B’.”

As complicated as it sounds, the size of one of these tables for one reducer is still less than $n^c$ for some $c < 1/2$. And so we can do the same trick of chaining the functions together to get the final answer. Note that this time the chaining will be adaptive, because whether the tape head leaves the left or right side will determine which part of the table you look at next. And moreover, we know the chaining process will stop in polynomial time because we can always pick a Turing machine to begin with that will halt in polynomial time (i.e., we know that L is contained in P).

The size calculations are also just large enough to stop us from doing the same trick with logarithmic space. So that gives the obvious question: is $\textup{L} \subset \textup{MRC}^0$? The second part of our paper shows that even weaker containments are probably very hard to prove, and they relate questions about MRC to the problem of L vs P.

## Hierarchy Theorems

This part of the paper is where we go much deeper into complexity theory than I imagine most of my readers are comfortable with. Our main theorem looks like this:

Theorem: Assume some conjecture is true. Then for every $a, b > 0$, there is are bigger $c > a, d > b$ such that the following hold:

$\displaystyle \textup{MRC}[n^a, n^b] \subsetneq \textup{MRC}[n^c, n^b],$
$\displaystyle \textup{MRC}[n^a, n^b] \subsetneq \textup{MRC}[n^a, n^d].$

This is a kind of hierarchy theorem that one often likes to prove in complexity theory. It says that if you give MRC enough extra rounds or time, then you’ll get strictly more computational power.

The conjecture we depend on is called the exponential time hypothesis (ETH), and it says that the 3-SAT problem can’t be solved in $2^{cn}$ time for any $0 < c < 1$. Actually, our theorem depends on a weaker conjecture (implied by ETH), but it’s easier to understand our theorems in the context of the ETH. Because 3-SAT is this interesting problem: we believe it’s time-intensive, and yet it’s space efficient because we can solve it in linear space given $2^n$ time. This time/space tradeoff is one of the oldest questions in all of computer science, but we still don’t have a lot of answers.

For instance, define $\textup{TISP}(f(n), g(n))$ to the the class of languages that can be decided by Turing machines that have simultaneous bounds of $O(f(n))$ time and $O(g(n))$ space. For example, we just said that $\textup{3-SAT} \in \textup{TISP}(2^n, n)$, and there is a famous theorem that says that SAT is not in $\textup{TISP}(n^{1.1} n^{0.1})$; apparently this is the best we know. So clearly there are some very basic unsolved problems about TISP. An important one that we address in our paper is whether there are hierarchies in TISP for a fixed amount of space. This is the key ingredient in proving our hierarchy theorem for MRC. In particular here is an open problem:

Conjecture: For every two integers $0 < a < b$, the classes $\textup{TISP}(n^a, n)$ and $\textup{TISP}(n^b, n)$ are different.

We know this is true of time and space separately, i.e., that $\textup{TIME}(n^a) \subsetneq \textup{TIME}(n^b)$ and $\textup{SPACE}(n^a) \subsetneq \textup{SPACE}(n^b)$. but for TISP we only know that you get more power if you increase both parameters.

So we prove a theorem that address this, but falls short in two respects: it depends on a conjecture like ETH, and it’s not for every $a, b$.

Theorem: Suppose ETH holds, then for every $a > 0$ there is a $b > a$ for which $\textup{TIME}(n^a) \not \subset \textup{TISP}(n^b, n)$.

In words, this suggests that there are problems that need both $n^2$ time and $n^2$ space, but can be solved with linear space if you allow enough extra time. Since $\textup{TISP}(n^a, n) \subset \textup{TIME}(n^a)$, this gives us the hierarchy we wanted.

To prove this we take 3-SAT and give it exponential padding so that it becomes easy enough to do in polynomial TISP (and it still takes linear space, in fact sublinear), but not so easy that you can do it in $n^a$ time. It takes some more work to get from this TISP hierarchy to our MRC hierarchy, but the details are a bit much for this blog. One thing I’d like to point out is that we prove that statements that are just about MRC have implications beyond MapReduce. In particular, the last corollary of our paper is the following.

Corollary: If $\textup{MRC}[\textup{poly}(n), 1] \subsetneq \textup{MRC}[\textup{poly}(n), \textup{poly}(n)]$, then L is different from P.

In other words, if you’re afforded a polynomial number of rounds in MRC, then showing that constant time per round is weaker than polynomial time is equivalently hard to separating L from P. The theorem is true because, as it turns out, $\textup{L} \subset \textup{MRC}[textup{poly}(n), 1]$, by simulating one step of a TM across polynomially many rounds. The proof is actually not that complicated (and doesn’t depend on ETH at all), and it’s a nice demonstration that studying MRC can have implications beyond parallel computing.

The other side of the coin is also interesting. Our first theorem implies the natural question of whether $\textup{L} \subset \textup{MRC}^0$. We’d like to say that this would imply the separation of L from P, but we don’t quite get that. In particular, we know that

$\displaystyle \textup{MRC}[1, n] \subsetneq \textup{MRC}[n, n] \subset \textup{MRC}[1, n^2] \subsetneq \textup{MRC}[n^2, n^2] \subset \dots$

But at the same time we could live in a world where

$\displaystyle \textup{MRC}[1, \textup{poly}(n)] = \textup{MRC}[\textup{poly}(n), \textup{poly}(n)]$

It seems highly unlikely, but to the best of our knowledge none of our techniques prove this is not the case. If we could rule this out, then we could say that $\textup{L} \subset \textup{MRC}^0$ implies the separation of L and P. And note this would not depend on any conjectures.

## Open Problems

Our paper has a list of open problems at the end. My favorite is essentially: how do we prove better lower bounds in MRC? In particular, it would be great if we could show that some problems need a lot of rounds simply because the communication model is too restrictive, and nobody has true random access to the entire input. For example, this is why we think graph connectivity needs a logarithmic number of rounds. But as of now nobody really knows how to prove it, and it seems like we’ll need some new and novel techniques in order to do it. I only have the wisps of ideas in that regard, and it will be fun to see which ones pan out.

Until next time!

# Occam’s Razor and PAC-learning

So far our discussion of learning theory has been seeing the definition of PAC-learningtinkering with it, and seeing simple examples of learnable concept classes. We’ve said that our real interest is in proving big theorems about what big classes of problems can and can’t be learned. One major tool for doing this with PAC is the concept of VC-dimension, but to set the stage we’re going to prove a simpler theorem that gives a nice picture of PAC-learning when your hypothesis class is small. In short, the theorem we’ll prove says that if you have a finite set of hypotheses to work with, and you can always find a hypothesis that’s consistent with the data you’ve seen, then you can learn efficiently. It’s obvious, but we want to quantify exactly how much data you need to ensure low error. This will also give us some concrete mathematical justification for philosophical claims about simplicity, and the theorems won’t change much when we generalize to VC-dimension in a future post.

## The Chernoff bound

One tool we will need in this post, which shows up all across learning theory, is the Chernoff-Hoeffding bound. We covered this famous inequality in detail previously on this blog, but the part of that post we need is the following theorem that says, informally, that if you average a bunch of bounded random variables, then the probability this average random variable deviates from its expectation is exponentially small in the amount of deviation. Here’s the slightly simplified version we’ll use:

Theorem: Let $X_1, \dots, X_m$ be independent random variables whose values are in the range $[0,1]$. Call $\mu_i = \mathbf{E}[X_i]$, $X = \sum_i X_i$, and $\mu = \mathbf{E}[X] = \sum_i \mu_i$. Then for all $t > 0$,

$\displaystyle \Pr(|X-\mu| > t) \leq 2e^{-2t^2 / m}$

One nice thing about the Chernoff bound is that it doesn’t matter how the variables are distributed. This is important because in PAC we need guarantees that hold for any distribution generating data. Indeed, in our case the random variables above will be individual examples drawn from the distribution generating the data. We’ll be estimating the probability that our hypothesis has error deviating more than $\varepsilon$, and we’ll want to bound this by $\delta$, as in the definition of PAC-learning. Since the amount of deviation (error) and the number of samples ($m$) both occur in the exponent, the trick is in balancing the two values to get what we want.

## Realizability and finite hypothesis classes

Let’s recall the PAC model once more. We have a distribution $D$ generating labeled examples $(x, c(x))$, where $c$ is an unknown function coming from some concept class $C$. Our algorithm can draw a polynomial number of these examples, and it must produce a hypothesis $h$ from some hypothesis class $H$ (which may or may not contain $c$). The guarantee we need is that, for any $\delta, \varepsilon > 0$, the algorithm produces a hypothesis whose error on $D$ is at most $\varepsilon$, and this event happens with probability at least $1-\delta$. All of these probabilities are taken over the randomness in the algorithm’s choices and the distribution $D$, and it has to work no matter what the distribution $D$ is.

Let’s introduce some simplifications. First, we’ll assume that the hypothesis and concept classes $H$ and $C$ are finite. Second, we’ll assume that $C \subset H$, so that you can actually hope to find a hypothesis of zero error. This is called realizability. Later we’ll relax these first two assumptions, but they make the analysis a bit cleaner. Finally, we’ll assume that we have an algorithm which, when given labeled examples, can find in polynomial time a hypothesis $h \in H$ that is consistent with every example.

These assumptions give a trivial learning algorithm: draw a bunch of examples and output any consistent hypothesis. The question is, how many examples do we need to guarantee that the hypothesis we find has the prescribed generalization error? It will certainly grow with $1 / \varepsilon$, but we need to ensure it will only grow polynomially fast in this parameter. Indeed, realizability is such a strong assumption that we can prove a polynomial bound using even more basic probability theory than the Chernoff bound.

Theorem: A algorithm that efficiently finds a consistent hypothesis will PAC-learn any finite concept class provided it has at least $m$ samples, where

$\displaystyle m \geq \frac{1}{\varepsilon} \left ( \log |H| + \log \left ( \frac{1}{\delta} \right ) \right )$

Proof. All we need to do is bound the probability that a bad hypothesis (one with error more than $\varepsilon$) is consistent with the given data. Now fix $D, c, \delta, \varepsilon$, and draw $m$ examples and let $h$ be any hypothesis that is consistent with the drawn examples. Suppose that the bad thing happens, that $\Pr_D(h(x) \neq c(x)) > \varepsilon$.

Because the examples are all drawn independently from $D$, the chance that all $m$ examples are consistent with $h$ is

$\displaystyle (1 - \Pr_{x \sim D}(h(x) \neq c(x)))^m < (1 - \varepsilon)^m$

What we’re saying here is, the probability that a specific bad hypothesis is actually consistent with your drawn examples is exponentially small in the error tolerance. So if we apply the union bound, the probability that some hypothesis you could produce is bad is at most $(1 - \varepsilon)^m S$, where $S$ is the number of hypotheses the algorithm might produce.

A crude upper bound on the number of hypotheses you could produce is just the total number of hypotheses, $|H|$. Even cruder, let’s use the inequality $(1 - x) < e^{-x}$ to give the bound

$\displaystyle (1 - \varepsilon)^m |H| < e^{-\varepsilon m} |H|$

Now we want to make sure that this probability, the probability of choosing a high-error (yet consistent) hypothesis, is at most $\delta$. So we can set the above quantity less than $\delta$ and solve for $m$:

$\displaystyle e^{-\varepsilon m} |H| \leq \delta$

Taking logs and solving for $m$ gives the desired bound.

$\square$

An obvious objection is: what if you aren’t working with a hypothesis class where you can guarantee that you’ll find a consistent hypothesis? Well, in that case we’ll need to inspect the definition of PAC again and reevaluate our measures of error. It turns out we’ll get a similar theorem as above, but with the stipulation that we’re only achieving error within epsilon of the error of the best available hypothesis.

But before we go on, this theorem has some deep philosophical interpretations. In particular, suppose that, before drawing your data, you could choose to work with one of two finite hypothesis classes $H_1, H_2$, with $|H_1| > |H_2|$. If you can find a consistent hypothesis no matter which hypothesis class you use, then this theorem says that your generalization guarantees are much stronger if you start with the smaller hypothesis class.

In other words, all else being equal, the smaller set of hypotheses is better. For this reason, the theorem is sometimes called the “Occam’s Razor” theorem. We’ll see a generalization of this theorem in the next section.

## Unrealizability and an extra epsilon

Now suppose that $H$ doesn’t contain any hypotheses with error less than $\varepsilon$. What can we hope to do in this case? One thing is that we can hope to find a hypothesis whose error is within $\varepsilon$ of the minimal error of any hypothesis in $H$. Moreover, we might not have any consistent hypotheses for some data samples! So rather than require an algorithm to produce an $h \in H$ that is perfectly consistent with the data, we just need it to produce a hypothesis that has minimal empirical error, in the sense that it is as close to consistent as the best hypothesis of $h$ on the data you happened to draw. It seems like such a strategy would find you a hypothesis that’s close to the best one in $H$, but we need to prove it and determine how many samples we need to draw to succeed.

So let’s make some definitions to codify this. For a given hypothesis, call $\textup{err}(h)$ the true error of $h$ on the distribution $D$. Our assumption is that there may be no hypotheses in $H$ with $\textup{err}(h) = 0$. Next we’ll call the empirical error $\hat{\textup{err}}(h)$.

Definition: We say a concept class $C$ is agnostically learnable using the hypothesis class $H$ if for all $c \in C$ and all distributions $D$ (and all $\varepsilon, \delta > 0$), there is a learning algorithm $A$ which produces a hypothesis $h$ that with probability at least $1 - \delta$ satisfies

$\displaystyle \text{err}(h) \leq \min_{h' \in H} \text{err}(h') + \varepsilon$

and everything runs in the same sort of polynomial time as for vanilla PAC-learning. This is called the agnostic setting or the unrealizable setting, in the sense that we may not be able to find a hypothesis with perfect empirical error.

We seek to prove that all concept classes are agnostically learnable with a finite hypothesis class, provided you have an algorithm that can minimize empirical error. But actually we’ll prove something stronger.

Theorem: Let $H$ be a finite hypothesis class and $m$ the number of samples drawn. Then for any $\delta > 0$, with probability $1-\delta$ the following holds:

$\displaystyle \forall h \in H, \hat{\text{err}}(h) \leq \text{err}(h) + \sqrt{\frac{\log |H| + \log(2 / \delta)}{2m}}$

In other words, we can precisely quantify how the empirical error converges to the true error as the number of samples grows. But this holds for all hypotheses in $H$, so this provides a uniform bound of the difference between true and empirical error for the entire hypothesis class.

Proving this requires the Chernoff bound. Fix a single hypothesis $h \in H$. If you draw an example $x$, call $Z$ the random variable which is 1 when $h(x) \neq c(x)$, and 0 otherwise. So if you draw $m$ samples and call the $i$-th variable $Z_i$, the empirical error of the hypothesis is $\frac{1}{m}\sum_i Z_i$. Moreover, the actual error is the expectation of this random variable since $\mathbf{E}[1/m \sum_i Z_i] = Z$.

So what we’re asking is the probability that the empirical error deviates from the true error by a lot. Let’s call “a lot” some parameter $\varepsilon/2 > 0$ (the reason for dividing by two will become clear in the corollary to the theorem). Then plugging things into the Chernoff-Hoeffding bound gives a bound on the probability of the “bad event,” that the empirical error deviates too much.

$\displaystyle \Pr[|\hat{\text{err}}(h) - \text{err}(h)| > \varepsilon / 2] < 2e^{-\frac{\varepsilon^2m}{2}}$

Now to get a bound on the probability that some hypothesis is bad, we apply the union bound and use the fact that $|H|$ is finite to get

$\displaystyle \Pr[|\hat{\text{err}}(h) - \text{err}(h)| > \varepsilon / 2] < 2|H|e^{-\frac{\varepsilon^2m}{2}}$

Now say we want to bound this probability by $\delta$. We set $2|H|e^{-\varepsilon^2m/2} \leq \delta$, solve for $m$, and get

$\displaystyle m \geq \frac{2}{\varepsilon^2}\left ( \log |H| + \log \frac{2}{\delta} \right )$

This gives us a concrete quantification of the tradeoff between $m, \varepsilon, \delta,$ and $|H|$. Indeed, if we pick $m$ to be this large, then solving for $\varepsilon / 2$ gives the exact inequality from the theorem.

$\square$

Now we know that if we pick enough samples (polynomially many in all the parameters), and our algorithm can find a hypothesis $h$ of minimal empirical error, then we get the following corollary:

Corollary: For any $\varepsilon, \delta > 0$, the algorithm that draws $m \geq \frac{2}{\varepsilon^2}(\log |H| + \log(2/ \delta))$ examples and finds any hypothesis of minimal empirical error will, with probability at least $1-\delta$, produce a hypothesis that is within $\varepsilon$ of the best hypothesis in $H$.

Proof. By the previous theorem, with the desired probability, for all $h \in H$ we have $|\hat{\text{err}}(h) - \text{err}(h)| < \varepsilon/2$. Call $g = \min_{h' \in H} \text{err}(h')$. Then because the empirical error of $h$ is also minimal, we have $|\hat{\text{err}}(g) - \text{err}(h)| < \varepsilon / 2$. And using the previous theorem again and the triangle inequality, we get $|\text{err}(g) - \text{err}(h)| < 2 \varepsilon / 2 = \varepsilon$. In words, the true error of the algorithm’s hypothesis is close to the error of the best hypothesis, as desired.

$\square$

## Next time

Both of these theorems tell us something about the generalization guarantees for learning with hypothesis classes of a certain size. But this isn’t exactly the most reasonable measure of the “complexity” of a family of hypotheses. For example, one could have a hypothesis class with a billion intervals on $\mathbb{R}$ (say you’re trying to learn intervals, or thresholds, or something easy), and the guarantees we proved in this post are nowhere near optimal.

So the question is: say you have a potentially infinite class of hypotheses, but the hypotheses are all “simple” in some way. First, what is the right notion of simplicity? And second, how can you get guarantees based on that analogous to these? We’ll discuss this next time when we define the VC-dimension.

Until then!

# Parameterizing the Vertex Cover Problem

I’m presenting a paper later this week at the Matheamtical Foundations of Computer Science 2014 in Budapest, Hungary. This conference is an interesting mix of logic and algorithms that aims to bring together researchers from these areas to discuss their work. And right away the first session on the first day focused on an area I know is important but have little experience with: fixed parameter complexity. From what I understand it’s not that popular of a topic at major theory conferences in the US (there appears to be only one paper on it at this year’s FOCS conference), but the basic ideas are worth knowing.

The basic idea is pretty simple: some hard computational problems become easier (read, polynomial-time solvable) if you fix some parameters involved to constants. Preferably small constants. For example, finding cliques of size $k$ in a graph is NP-hard if $k$ is a parameter, but if you fix $k$ to a constant then you can check all possible subsets of size $k$ in $O(n^k)$ time. This is kind of a silly example because there are much faster ways to find triangles than checking all $O(n^3)$ subsets of vertices, but part of the point of fixed-parameter complexity is to find the fastest algorithms in these fixed-parameter settings. Since in practice parameters are often small [citation needed], this analysis can provide useful practical algorithmic alternatives to heuristics or approximate solutions.

One important tool in the theory of fixed-parameter tractability is the idea of a kernel. I think it’s an unfortunate term because it’s massively overloaded in mathematics, but the idea is to take a problem instance with the parameter $k$, and carve out “easy” regions of the instance (often reducing $k$ as you go) until the runtime of the trivial brute force algorithm only depends on $k$ and not on the size of the input. The point is that the solution you get on this “carved out” instance is either the same as the original, or can be extended back to the original with little extra work. There is a more formal definition we’ll state, but there is a canonical example that gives a great illustration.

Consider the vertex cover problem. That is, you give me a graph $G = (V,E)$ and a number $k$ and I have to determine if there is a subset of $\leq k$ vertices of $G$ that touch all of the edges in $E$. This problem is fixed-parameter tractable because, as with $k$-clique one can just check all subsets of size $k$. The kernel approach we’ll show now is much smarter.

What you do is the following. As long as your graph has a vertex of degree $> k$, you remove it and reduce $k$ by 1. This is because a vertex of degree $> k$ will always be chosen for a vertex cover. If it’s not, then you need to include all of its neighbors to cover its edges, but there are $> k$ neighbors and your vertex cover is constrained by size $k$. And so you can automatically put this high-degree vertex in your cover, and use induction on the smaller graph.

Once you can’t remove any more vertices there are two cases. In the case that there are more than $k^2$ edges, you output that there is no vertex cover. Indeed, if you only get $k$ vertices in your cover and you removed all vertices of degree $> k$, then each can cover at most $k$ edges, giving a total of at most $k^2$. Otherwise, if there are at most $k^2$ edges, then you can remove all the isolated vertices and show that there are only $\leq 2k^2$ vertices left. This is because each edge touches only two vertices, so in the worst case they’re all distinct. This smaller subgraph is called a kernel of the vertex cover, and the fact that its size depends only on $k$ is the key. So you can look at all $2^{2k^2} = O(1)$ subsets to determine if there’s a cover of the size you want. If you find a cover of the kernel, you add back in all the large-degree vertices you deleted and you’re done.

Now, even for small $k$ this is a pretty bad algorithm ($k=5$ gives $2^{50}$ subsets to inspect), but with more detailed analysis you can do significantly better. In particular, the best known bound reduces vertex cover to a kernel of size $2k - c \log(k)$ vertices for any constant $c$ you specify. Getting $\log(k)$ vertices is known to imply P = NP, and with more detailed complexity assumptions it’s even hard to get a graph with fewer than $O(k^{2-\varepsilon})$ edges for any $\varepsilon > 0$. These are all relatively recent results whose associated papers I have not read.

Even with these hardness results, there are two reasons why this kind of analysis is useful. The first is that it gives us a clearer picture of the complexity of these problems. In particular, the reduction we showed for vertex cover gives a time $O(2^{2k^2} + n + m)$-time algorithm, which you can then compare directly to the trivial $O(n^k)$ time brute force algorithm and measure the difference. Indeed, if $k = o(\sqrt{(k/2) log(n)})$ then the kernelized approach is faster.

The second reason is that the kernel approach usually results in simple and quick checks for negative answers to a problem. In particular, if you want to check for $k$-sized set covers in a graph in the real world, this analysis shows that the first thing you should do is check if the kernel has size $> k^2$. If so, you can immediately give a “no” answer. So useful kernels can provide insight into the structure of a problem that can be turned into heuristic tools even when it doesn’t help you solve the problem exactly.

So now let’s just see the prevailing definition of a “kernelization” of a problem. This comes from the text of Downey and Fellows.

Definition: kernelization of a parameterized problem $L$ (formally, a language where each string $x$ is paired with a positive integer $k$) is a $\textup{poly}(|x|, k)$-time algorithm that converts instances $(x,k)$ into instances $(x', k')$ with the following three properties.

• $(x,k)$ is a yes instance of $L$ if and only if $(x', k')$ is.
• $|x'| \leq f(k)$ for some computable function $f: \mathbb{N} \to \mathbb{N}$.
• $k' \leq g(k)$ for some computable function $g: \mathbb{N} \to \mathbb{N}$.

The output $(x', k')$ is called a kernel, and the problem is said to admit a polynomial kernel if $f(k) = O(k^c)$ for some constant $c$.

So we showed that vertex cover admits a polynomial kernel (in fact, a quadratic one).

Now the nice theorem is that a problem is fixed-parameter tractable if and only if it admits a polynomial kernel. Finding a kernel is conceptually easier because, like in vertex cover, it allows you to introduce additional assumptions on the structure of the instances you’re working with. But more importantly from a theoretical standpoint, measuring the size and complexity of kernels for NP-hard problems gives us a way to discriminate among problems within NP. That and the chance to get some more practical tools for NP-hard problems makes parameterized complexity more interesting than it sounds at first.

Until next time!

# An Update on “Coloring Resilient Graphs”

A while back I announced a preprint of a paper on coloring graphs with certain resilience properties. I’m pleased to announce that it’s been accepted to the Mathematical Foundations of Computer Science 2014, which is being held in Budapest this year. Since we first published the preprint we’ve actually proved some additional results about resilience, and so I’ll expand some of the details here. I think it makes for a nicer overall picture, and in my opinion it gives a little more justification that resilient coloring is interesting, at least in contrast to other resilience problems.

## Resilient SAT

Recall that a “resilient” yes-instance of a combinatorial problem is one which remains a yes-instance when you add or remove some constraints. The way we formalized this for SAT was by fixing variables to arbitrary values. Then the question is how resilient does an instance need to be in order to actually find a certificate for it? In more detail,

Definition: $r$-resilient $k$-SAT formulas are satisfiable formulas in $k$-CNF form (conjunctions of clauses, where each clause is a disjunction of three literals) such that for all choices of $r$ variables, every way to fix those variables yields a satisfiable formula.

For example, the following 3-CNF formula is 1-resilient:

$\displaystyle (a \vee b \vee c) \wedge (a \vee \overline{b} \vee \overline{c}) \wedge (\overline{a} \vee \overline{b} \vee c)$

The idea is that resilience may impose enough structure on a SAT formula that it becomes easy to tell if it’s satisfiable at all. Unfortunately for SAT (though this is definitely not the case for coloring), there are only two possibilities. Either the instances are so resilient that they never existed in the first place (they’re vacuously trivial), or the instances are NP-hard. The first case is easy: there are no $k$-resilient $k$-SAT formulas. Indeed, if you’re allowed to fix $k$ variables to arbitrary values, then you can just pick a clause and set all its variables to false. So no formula can ever remain satisfiable under that condition.

The second case is when the resilience is strictly less than the clause size, i.e. $r$-resilient $k$-SAT for $0 \leq r < k$. In this case the problem of finding a satisfying assignment is NP-hard. We’ll show this via a sequence of reductions which start at 3-SAT, and they’ll involve two steps: increasing the clause size and resilience, and decreasing the clause size and resilience. The trick is in balancing which parts are increased and decreased. I call the first step the “blowing up” lemma, and the second part the “shrinking down” lemma.

## Blowing Up and Shrinking Down

Here’s the intuition behind the blowing up lemma. If you give me a regular (unresilient) 3-SAT formula $\varphi$, what I can do is make a copy of $\varphi$ with a new set of variables and OR the two things together. Call this $\varphi^1 \vee \varphi^2$. This is clearly logically equivalent to the original formula; if you give me a satisfying assignment for the ORed thing, I can just see which of the two clauses are satisfied and use that sub-assignment for $\varphi$, and conversely if you can satisfy $\varphi$ it doesn’t matter what truth values you choose for the new set of variables. And further you can transform the ORed formula into a 6-SAT formula in polynomial time. Just apply deMorgan’s rules for distributing OR across AND.

Now the choice of a new set of variables allows us to give some resilient. If you fix one variable to the value of your choice, I can always just work with the other set of variables. Your manipulation doesn’t change the satisfiability of the ORed formula, because I’ve added all of this redundancy. So we took a 3-SAT formula and turned it into a 1-resilient 6-SAT formula.

The idea generalizes to the blowing up lemma, which says that you can measure the effects of a blowup no matter what you start with. More formally, if $s$ is the number of copies of variables you make, $k$ is the clause size of the starting formula $\varphi$, and $r$ is the resilience of $\varphi$, then blowing up gives you an $[(r+1)s - 1]$-resilient $(sk)$-SAT formula. The argument is almost identical to the example above the resilience is more general. Specifically, if you fix fewer than $(r+1)s$ variables, then the pigeonhole principle guarantees that one of the $s$ copies of variables has at most $r$ fixed values, and we can just work with that set of variables (i.e., this small part of the big ORed formula is satisfiable if $\varphi$ was $r$-resilient).

The shrinking down lemma is another trick that is similar to the reduction from $k$-SAT to 3-SAT. There you take a clause like $v \vee w \vee x \vee y \vee z$ and add new variables $z_i$ to break up the clause in to clauses of size 3 as follows:

$\displaystyle (v \vee w \vee z_1) \wedge (\neg z_1 \vee x \vee z_2) \wedge (\neg z_2 \vee y \vee z)$

These are equivalent because your choice of truth values for the $z_i$ tell me which of these sub-clauses to look for a true literal of the old variables. I.e. if you choose $z_1 = T, z_2 = F$ then you have to pick either $y$ or $z$ to be true. And it’s clear that if you’re willing to double the number of variables (a linear blowup) you can always get a $k$-clause down to an AND of 3-clauses.

So the shrinking down reduction does the same thing, except we only split clauses in half. For a clause $C$, call $C[:k/2]$ the first half of a clause and $C[k/2:]$ the second half (you can see how my Python training corrupts my notation preference). Then to shrink a clause $C_i$ down from size $k$ to size $\lceil k/2 \rceil + 1$ (1 for the new variable), add a variable $z_i$ and break $C_i$ into

$\displaystyle (C_i[:k/2] \vee z_i) \wedge (\neg z_i \vee C[k/2:])$

and just AND these together for all clauses. Call the original formula $\varphi$ and the transformed one $\psi$. The formulas are logically equivalent for the same reason that the $k$-to-3-SAT reduction works, and it’s already in the right CNF form. So resilience is all we have to measure. The claim is that the resilience is $q = \min(r, \lfloor k/2 \rfloor)$, where $r$ is the resilience of $\varphi$.

The reason for this is that if all the fixed variables are old variables (not $z_i$), then nothing changes and the resilience of the original $\phi$ keeps us safe. And each $z_i$ we fix has no effect except to force us to satisfy a variable in one of the two halves. So there is this implication that if you fix a $z_i$ you have to also fix a regular variable. Because we can’t guarantee anything if we fix more than $r$ regular variables, we’d have to stop before fixing $r$ of the $z_i$. And because these new clauses have size $k/2 + 1$, we can’t do this more than $k/2$ times or else we risk ruining an entire clause. So this give the definition of $q$. So this proves the shrinking down lemma.

## Resilient SAT is always hard

The blowing up and shrinking down lemmas can be used to show that $r$-resilient $k$-SAT is NP-hard for all $r < k$. What we do is reduce from 3-SAT to an $r$-resilient $k$-SAT instance in such a way that the 3-SAT formula is satisfiable if and only if the transformed formula is resiliently satisfiable.

What makes these two lemmas work together is that shrinking down shrinks the clause size just barely less than the resilience, and blowing up increases resilience just barely more than it increases clause size. So we can combine these together to climb from 3-SAT up to some high resilience and satisfiability, and then iteratively shrink down until we hit our target.

One might worry that it will take an exponential number of reductions (or a few reductions of exponential size) to get from 3-SAT to the $(r,k)$ of our choice, but we have a construction that does it in at most four steps, with only a linear initial blowup from 3-SAT to $r$-resilient $3(r+1)$-SAT. Then, to deal with the odd ceilings and floors in the shrinking down lemma, you have to find a suitable larger $k$ to reduce to (by padding with useless variables, which cannot make the problem easier). And you choose this $k$ so that you only need at most two applications of shrinking down to get to $(k-1)$-resilient $k$-SAT. Our preprint has the gory details (which has an inelegant part that is not worth writing here), but in the end you show that $(k-1)$-resilient $k$-SAT is hard, and since that’s the maximal amount of resilience before the problem becomes vacuously trivial, all smaller resilience values are also hard.

## So how does this relate to coloring?

I’m happy about this result not just because it answers an open question I’m honestly curious about, but also because it shows that resilient coloring is more interesting. Basically this proves that satisfiability is so hard that no amount of resilience can make it easier in the worst case. But coloring has a gradient of difficulty. Once you get to order $k^2$ resilience for $k$-colorable graphs, the coloring problem can be solved efficiently by a greedy algorithm (and it’s not a vacuously empty class of graphs). Another thing on the side is that we use the hardness of resilient SAT to get the hardness results we have for coloring.

If you really want to stretch the implications, you might argue that this says something like “coloring is somewhat easier than SAT,” because we found a quantifiable axis along which SAT remains difficult while coloring crumbles. The caveat is that fixing colors of vertices is not exactly comparable to fixing values of truth assignments (since we are fixing lots of instances by fixing a variable), but at least it’s something concrete.

Coloring is still mostly open, and recently I’ve been going to talks where people are discussing startlingly similar ideas for things like Hamiltonian cycles. So that makes me happy.

Until next time!