# Zero-One Laws for Random Graphs

Last time we saw a number of properties of graphs, such as connectivity, where the probability that an Erdős–Rényi random graph $G(n,p)$ satisfies the property is asymptotically either zero or one. And this zero or one depends on whether the parameter $p$ is above or below a universal threshold (that depends only on $n$ and the property in question).

To remind the reader, the Erdős–Rényi random “graph” $G(n,p)$ is a distribution over graphs that you draw from by including each edge independently with probability $p$. Last time we saw that the existence of an isolated vertex has a sharp threshold at $(\log n) / n$, meaning if $p$ is asymptotically smaller than the threshold there will certainly be isolated vertices, and if $p$ is larger there will certainly be no isolated vertices. We also gave a laundry list of other properties with such thresholds.

One might want to study this phenomenon in general. Even if we might not be able to find all the thresholds we want for a given property, can we classify which properties have thresholds and which do not?

The answer turns out to be mostly yes! For large classes of properties, there are proofs that say things like, “either this property holds with probability tending to one, or it holds with probability tending to zero.” These are called “zero-one laws,” and they’re sort of meta theorems. We’ll see one such theorem in this post relating to constant edge-probabilities in random graphs, and we’ll remark on another at the end.

## Sentences about graphs in first order logic

A zero-one law generally works by defining a class of properties, and then applying a generic first/second moment-type argument to every property in the class.

So first we define what kinds of properties we’ll discuss. We’ll pick a large class: anything that can be expressed in first-order logic in the language of graphs. That is, any finite logical statement that uses existential and universal quantifiers over variables, and whose only relation (test) is whether an edge exists between two vertices. We’ll call this test $e(x,y)$. So you write some sentence $P$ in this language, and you take a graph $G$, and you can ask $P(G) = 1$, whether the graph satisfies the sentence.

This seems like a really large class of properties, and it is, but let’s think carefully about what kinds of properties can be expressed this way. Clearly the existence of a triangle can be written this way, it’s just the sentence

$\exists x,y,z : e(x,y) \wedge e(y,z) \wedge e(x,z)$

I’m using $\wedge$ for AND, and $\vee$ for OR, and $\neg$ for NOT. Similarly, one can express the existence of a clique of size $k$, or the existence of an independent set of size $k$, or a path of a fixed length, or whether there is a vertex of maximal degree $n-1$.

Here’s a question: can we write a formula which will be true for a graph if and only if it’s connected? Well such a formula seems like it would have to know about how many vertices there are in the graph, so it could say something like “for all $x,y$ there is a path from $x$ to $y$.” It seems like you’d need a family of such formulas that grows with $n$ to make anything work. But this isn’t a proof; the question remains whether there is some other tricky way to encode connectivity.

But as it turns out, connectivity is not a formula you can express in propositional logic. We won’t prove it here, but we will note at the end of the article that connectivity is in a different class of properties that you can prove has a similar zero-one law.

## The zero-one law for first order logic

So the theorem about first-order expressible sentences is as follows.

Theorem: Let $P$ be a property of graphs that can be expressed in the first order language of graphs (with the $e(x,y)$ relation). Then for any constant $p$, the probability that $P$ holds in $G(n,p)$ has a limit of zero or one as $n \to \infty$.

Proof. We’ll prove the simpler case of $p=1/2$, but the general case is analogous. Given such a graph $G$ drawn from $G(n,p)$, what we’ll do is define a countably infinite family of propositional formulas $\varphi_{k,l}$, and argue that they form a sort of “basis” for all first-order sentences about graphs.

First let’s describe the $\varphi_{k,l}$. For any $k,l \in \mathbb{N}$, the sentence will assert that for every set of $k$ vertices and every set of $l$ vertices, there is some other vertex connected to the first $k$ but not the last $l$.

$\displaystyle \varphi_{k,l} : \forall x_1, \dots, x_k, y_1, \dots, y_l \exists z : \\ e(z,x_1) \wedge \dots \wedge e(z,x_k) \wedge \neg e(z,y_1) \wedge \dots \wedge \neg e(z,y_l)$.

In other words, these formulas encapsulate every possible incidence pattern for a single vertex. It is a strange set of formulas, but they have a very nice property we’re about to get to. So for a fixed $\varphi_{k,l}$, what is the probability that it’s false on $n$ vertices? We want to give an upper bound and hence show that the formula is true with probability approaching 1. That is, we want to show that all the $\varphi_{k,l}$ are true with probability tending to 1.

Computing the probability: we have $\binom{n}{k} \binom{n-k}{l}$ possibilities to choose these sets, and the probability that some other fixed vertex $z$ has the good connections is $2^{-(k+l)}$ so the probability $z$ is not good is $1 - 2^{-(k+l)}$, and taking a product over all choices of $z$ gives the probability that there is some bad vertex $z$ with an exponent of $(n - (k + l))$. Combining all this together gives an upper bound of $\varphi_{k,l}$ being false of:

$\displaystyle \binom{n}{k}\binom{n-k}{l} (1-2^{-k-1})^{n-k-l}$

And $k, l$ are constant, so the left two terms are polynomials while the rightmost term is an exponentially small function, and this implies that the whole expression tends to zero, as desired.

Break from proof.

## A bit of model theory

So what we’ve proved so far is that the probability of every formula of the form $\varphi_{k,l}$ being satisfied in $G(n,1/2)$ tends to 1.

Now look at the set of all such formulas

$\displaystyle \Phi = \{ \varphi_{k,l} : k,l \in \mathbb{N} \}$

We ask: is there any graph which satisfies all of these formulas? Certainly it cannot be finite, because a finite graph would not be able to satisfy formulas with sufficiently large values of $l, k > n$. But indeed, there is a countably infinite graph that works. It’s called the Rado graph, pictured below.

The Rado graph has some really interesting properties, such as that it contains every finite and countably infinite graph as induced subgraphs. Basically this means, as far as countably infinite graphs go, it’s the big momma of all graphs. It’s the graph in a very concrete sense of the word. It satisfies all of the formulas in $\Phi$, and in fact it’s uniquely determined by this, meaning that if any other countably infinite graph satisfies all the formulas in $\Phi$, then that graph is isomorphic to the Rado graph.

But for our purposes (proving a zero-one law), there’s a better perspective than graph theory on this object. In the logic perspective, the set $\Phi$ is called a theory, meaning a set of statements that you consider “axioms” in some logical system. And we’re asking whether there any model realizing the theory. That is, is there some logical system with a semantic interpretation (some mathematical object based on numbers, or sets, or whatever) that satisfies all the axioms?

A good analogy comes from the rational numbers, because they satisfy a similar property among all ordered sets. In fact, the rational numbers are the unique countable, ordered set with the property that it has no biggest/smallest element and is dense. That is, in the ordering there is always another element between any two elements you want. So the theorem says if you have two countable sets with these properties, then they are actually isomorphic as ordered sets, and they are isomorphic to the rational numbers.

So, while we won’t prove that the Rado graph is a model for our theory $\Phi$, we will use that fact to great benefit. One consequence of having a theory with a model is that the theory is consistent, meaning it can’t imply any contradictions. Another fact is that this theory $\Phi$ is complete. Completeness means that any formula or it’s negation is logically implied by the theory. Note these are syntactical implications (using standard rules of propositional logic), and have nothing to do with the model interpreting the theory.

The proof that $\Phi$ is complete actually follows from the uniqueness of the Rado graph as the only countable model of $\Phi$. Suppose the contrary, that $\Phi$ is not consistent, then there has to be some formula $\psi$ that is not provable, and it’s negation is also not provable, by starting from $\Phi$. Now extend $\Phi$ in two ways: by adding $\psi$ and by adding $\neg \psi$. Both of the new theories are still countable, and by a theorem from logic this means they both still have countable models. But both of these new models are also countable models of $\Phi$, so they have to both be the Rado graph. But this is very embarrassing for them, because we assumed they disagree on the truth of $\psi$.

So now we can go ahead and prove the zero-one law theorem.

Given an arbitrary property $\varphi \not \in \Psi$. Now either $\varphi$ or it’s negation can be derived from $\Phi$. Without loss of generality suppose it’s $\varphi$. Take all the formulas from the theory you need to derive $\varphi$, and note that since it is a proof in propositional logic you will only finitely many such $\varphi_{k,l}$. Now look at the probabilities of the $\varphi_{k,l}$: they are all true with probability tending to 1, so the implied statement of the proof of $\varphi$ (i.e., $\varphi$ itself) must also hold with probability tending to 1. And we’re done!

$\square$

If you don’t like model theory, there is another “purely combinatorial” proof of the zero-one law using something called Ehrenfeucht–Fraïssé games. It is a bit longer, though.

## Other zero-one laws

One might naturally ask two questions: what if your probability is not constant, and what other kinds of properties have zero-one laws? Both great questions.

For the first, there are some extra theorems. I’ll just describe one that has always seemed very strange to me. If your probability is of the form $p = n^{-\alpha}$ but $\alpha$ is irrational, then the zero-one law still holds! This is a theorem of Baldwin-Shelah-Spencer, and it really makes you wonder why irrational numbers would be so well behaved while rational numbers are not :)

For the second question, there is another theorem about monotone properties of graphs. Monotone properties come in two flavors, so called “increasing” and “decreasing.” I’ll describe increasing monotone properties and the decreasing counterpart should be obvious. A property is called monotone increasing if adding edges can never destroy the property. That is, with an empty graph you don’t have the property (or maybe you do), and as you start adding edges eventually you suddenly get the property, but then adding more edges can’t cause you to lose the property again. Good examples of this include connectivity, or the existence of a triangle.

So the theorem is that there is an identical zero-one law for monotone properties. Great!

It’s not so often that you get to see these neat applications of logic and model theory to graph theory and (by extension) computer science. But when you do get to apply them they seem very powerful and mysterious. I think it’s a good thing.

Until next time!

# The Giant Component and Explosive Percolation

Last time we left off with a tantalizing conjecture: a random graph with edge probability $p = 5/n$ is almost surely a connected graph. We arrived at that conjecture from some ad-hoc data analysis, so let’s go back and treat it with some more rigorous mathematical techniques. As we do, we’ll discover some very interesting “threshold theorems” that essentially say a random graph will either certainly have a property, or it will certainly not have it.

The phase transition we empirically observed from last time.

## Big components

Recalling the basic definition: an Erdős-Rényi (ER) random graph with $n$ vertices and edge probability $p$ is a probability distribution over all graphs on $n$ vertices. Generatively, you draw from an ER distribution by flipping a $p$-biased coin for each pair of vertices, and adding the edge if you flip heads. We call the random event of drawing a graph from this distribution a “random graph” even though it’s not a graph, and we denote an ER random graph by $G(n,p)$. When $p = 1/2$, the distribution $G(n,1/2)$ is the uniform distribution over all graphs on $n$ vertices.

Now let’s get to some theorems. The main tools we’ll use are called the first and second moment method. Let’s illustrate them by example.

### The first moment method

Say we want to know what values of $p$ are likely to produce graphs with isolated vertices (vertices with no neighbors), and which are not. Of course, the value of $p$ will depend on $n \to \infty$ in general, but we can already see by example that if $p = 1/2$ then the probability of a fixed vertex being isolated is $2^{-n} \to 0$. We can use the union bound (sum this value over all vertices) to show that the probability of any vertex being isolated is at most $n2^{-n}$ which also tends to zero very quickly. This is not the first moment method, I’m just making the point that all of our results will be interpreted asymptotically as $n \to \infty$.

So now we can ask: what is the expected number of isolated vertices? If I call $X$ the random variable that counts the expected number of isolated vertices, then I’m asking about $\mathbb{E}[X]$. Really what I’m doing is interpreting $X$ as a random variable depending on $n, p(n)$, and asking about the evolution of $\mathbb{E}[X]$ as $n \to \infty$.

Now the first moment method states, somewhat obviously, that if the expectation tends to zero then the value of $X$ itself also tends to zero. Indeed, this follows from Markov’s inequality, which states that the probability that $X \geq a$ is bounded by $\mathbb{E}[X]/a$. In symbols,

$\displaystyle \Pr[X \geq a] \leq \frac{\mathbb{E}[X]}{a}$.

In our case $X$ is counting something (it’s integer valued), so asking whether $X > 0$ is equivalent to asking whether $X \geq 1$. The upper bound on the probability of $X$ being strictly positive is then just $\mathbb{E}[X]$.

So let’s find out when the expected number of isolated vertices goes to zero. We’ll use the wondrous linearity of expectation to split $X$ into a sum of counts for each vertex. That is, if $X_i$ is 1 when vertex $i$ is isolated and 0 otherwise (this is called an indicator variable), then $X = \sum_{i=1}^n X_i$ and linearity of expectation gives

$\displaystyle \mathbb{E}[X] = \mathbb{E}[\sum_{i=1}^n X_i] = \sum_{i=1}^n \mathbb{E}[X_i]$

Now the expectation of an indicator random variable is just the probability that the event occurs (it’s trivial to check). It’s easy to compute the probability that a vertex is isolated: it’s $(1-p)^n$. So the sum above works out to be $n(1-p)^n$. It should really be $n(1-p)^{n-1}$ but the extra factor of $(1-p)$ doesn’t change anything. The question is what’s the “smallest” way to set $p$ as a function of $n$ in order to make the above thing go to zero? Using the fact that $(1-x) < e^{-x}$ for all $x > 0$, we get

$n(1-p)^n < ne^{-pn}$

And setting $p = (\log n) / n$ simplifies the right hand side to $ne^{- \log n} = n / n = 1$. This is almost what we want, so let’s set $p$ to be anything that grows asymptotically faster than $(\log n) / n$. The notation for this is $\omega((\log n) / n)$. Then using some slick asymptotic notation we can prove that the RHS of the inequality above goes to zero, and so the LHS must as well. Back to the big picture: we just showed that the expectation of $X$ (the expected number of isolated vertices) goes to zero, and so by the first moment method the value of $X$ (the actual number of isolated vertices) has to go to zero with probability tending to 1.

Some quick interpretations: when $p = (\log n) / n$ each vertex has $\log n$ neighbors in expectation. Moreover, having no isolated vertices is just a little bit short of the entire graph being connected (our ultimate goal is to figure out exactly when this happens). But already we can see that our conjecture from the beginning is probably false: we aren’t able to use this same method to show that when $p = c/n$ for some constant $c$ rules out isolated vertices as $n \to \infty$. We just got lucky in our data analysis that 5 is about the natural log of 100 (which is 4.6).

### The second moment method

Now what about the other side of the coin? If $p$ is asymptotically less than $(\log n) / n$ do we necessarily get isolated vertices? That would really put our conjecture to rest. In this case the answer is yes, but it might not be in general. Let’s discuss.

We said that in general if $\mathbb{E}[X] \to 0$ then the value of $X$ has to go to zero too (that’s the first moment method). The flip side of this is: if $\mathbb{E}[X] \to \infty$ does necessarily the value of $X$ also tend to infinity? The answer is not always yes. Here is a gruesome example I originally heard from a book: say $X$ is the number of people that will die in the next decade due to an asteroid hitting the earth. The probability that the event happens is quite small, but if it does happen then the number of people that will die is quite large. It is perfectly reasonable for this to drag up the expectation (as the world population grows every decade), but at least we hope a growing population doesn’t by itself increase the value of $X$.

Mathematics is on our side here. We’re asking under what conditions on $\mathbb{E}[X]$ does the following implication hold: $\mathbb{E}[X] \to \infty$ implies $\Pr[X > 0] \to 1$.

With the first moment method we used Markov’s inequality (a statement about expectation, also called the first moment). With the second moment method we’ll use a statement about the second moment (variances), and the most common is Chebyshev’s inequality. Chebyshev’s inequality states that the probability $X$ deviates from its expectation by more than $c$ is bounded by $\textup{Var}[X] / c^2$. In symbols, for all $c > 0$ we have

$\displaystyle \Pr[|X - \mathbb{E}[X]| \geq c] \leq \frac{\textup{Var}[X]}{c^2}$

Now the opposite of $X > 0$, written in terms of deviation from expectation, is $|X - \mathbb{E}[X]| \geq \mathbb{E}[X]$. In words, in order for any number $a$ to be zero, it has to have a distance of at least $b$ from any number $b$. It’s such a stupidly simple statement it’s almost confusing. So then we’re saying that

$\displaystyle \Pr[X = 0] \leq \frac{\textup{Var}[X]}{\mathbb{E}[X]^2}$.

In order to make this probability go to zero, it’s enough to have $\textup{Var}[X] = o(\mathbb{E}[X]^2)$. Again, the little-o means “grows asymptotically slower than.” So the numerator of the fraction on the RHS will grow asymptotically slower than the denominator, meaning the whole fraction tends to zero. This condition and its implication are together called the “second moment method.”

Great! So we just need to compute $\textup{Var}[X]$ and check what conditions on $p$ make it fit the theorem. Recall that $\textup{Var}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$, and we want to upper bound this in terms of $\mathbb{E}[X]^2$. Let’s compute $\mathbb{E}[X]^2$ first.

$\displaystyle \mathbb{E}[X]^2 = n^2(1-p)^{2n}$

Now the variance.

$\displaystyle \textup{Var}[X] = \mathbb{E}[X^2] - n^2(1-p)^{2n}$

Expanding $X$ as a sum of indicator variables $X_i$ for each vertex, we can split the square into a sum over pairs. Note that $X_i^2 = X_i$ since they are 0-1 valued indicator variables, and $X_iX_j$ is the indicator variable for both events happening simultaneously.

\displaystyle \begin{aligned} \mathbb{E}[X^2] &= \mathbb{E}[\sum_{i,j} X_{i,j}] \\ &=\mathbb{E} \left [ \sum_i X_i^2 + \sum_{i \neq j} X_iX_j \right ] \\ &= \sum_i \mathbb{E}[X_i^2] + \sum_{i \neq j} \mathbb{E}[X_iX_j] \end{aligned}

By what we said about indicators, the last line is just

$\displaystyle \sum_i \Pr[i \textup{ is isolated}] + \sum_{i \neq j} \Pr[i,j \textup{ are both isolated}]$

And we can compute each of these pieces quite easily. They are (asymptotically ignoring some constants):

$\displaystyle n(1-p)^n + n^2(1-p)(1-p)^{2n-4}$

Now combining the two terms together (subtracting off the square of the expectation),

\displaystyle \begin{aligned} \textup{Var}[X] &\leq n(1-p)^n + n^2(1-p)^{-3}(1-p)^{2n} - n^2(1-p)^{2n} \\ &= n(1-p)^n + n^2(1-p)^{2n} \left ( (1-p)^{-3} - 1 \right ) \end{aligned}

Now we divide by $\mathbb{E}[X]^2$ to get $n^{-1}(1-p)^{-n} + (1-p)^{-3} - 1$. Since we’re trying to see if $p = (\log n) / n$ is a sharp threshold, the natural choice is to let $p = o((\log n) / n)$. Indeed, using the $1-x < e^{-x}$ upper bound and plugging in the little-o bounds the whole quantity by

$\displaystyle \frac{1}{n}e^{o(\log n)} + o(n^{1/n}) - 1 = o(1)$

i.e., the whole thing tends to zero, as desired.

## Other thresholds

So we just showed that the property of having no isolated vertices in a random graph has a sharp threshold at $p = (\log n) / n$. Meaning at any larger probability the graph is almost surely devoid of isolated vertices, and at any lower probability the graph almost surely has some isolated vertices.

This might seem like a miracle theorem, but there turns out to be similar theorems for lots of properties. Most of them you can also prove using basically the same method we’ve been using here. I’ll list some below. Also note they are all sharp, two-sided thresholds in the same way that the isolated vertex boundary is.

• The existence of a component of size $\omega(\log (n))$ has a threshold of $1/n$.
• $p = c/n$ for any $c > 0$ is a threshold for the existence of a giant component of linear size $\Theta(n)$. Moreover, above this threshold no other components will have size $\omega(\log n)$.
• In addition to $(\log n) / n$ being a threshold for having no isolated vertices, it is also a threshold for connectivity.
• $p = (\log n + \log \log n + c(n)) / n$ is a sharp threshold for the existence of Hamiltonian cycles in the following sense: if $c(n) = \omega(1)$ then there will be a Hamilton cycle almost surely, if $c(n) \to -\infty$ there will be no Hamiltonian cycle almost surely, and if $c(n) \to c$ the probability of a Hamiltonian cycle is $e^{-e^{-c}}$. This was proved by Kolmos and Szemeredi in 1983. Moreover, there is an efficient algorithm to find Hamiltonian cycles in these random graphs when they exist with high probability.

## Explosive Percolation

So now we know that as the probability of an edge increases, at some point the graph will spontaneously become connected; at some time that is roughly $\log(n)$ before, the so-called “giant component” will emerge and quickly engulf the entire graph.

Here’s a different perspective on this situation originally set forth by Achlioptas, D’Souza, and Spencer in 2009. It has since become called an “Achlioptas process.”

The idea is that you are watching a random graph grow. Rather than think about random graphs as having a probability above or below some threshold, you can think of it as the number of edges growing (so the thresholds will all be multiplied by $n$). Then you can imagine that you start with an empty graph, and at every time step someone is adding a new random edge to your graph. Fine, eventually you’ll get so many edges that a giant component emerges and you can measure when that happens.

But now imagine that instead of being given a single random new edge, you are given a choice. Say God presents you with two random edges, and you must pick which to add to your graph. Obviously you will eventually still get a giant component, but the question is how long can you prevent it from occurring? That is, how far back can we push the threshold for connectedness by cleverly selecting the new edge?

What Achlioptas and company conjectured was that you can push it back (some), but that when you push it back as far as it can go, the threshold becomes discontinuous. That is, they believed there was a constant $\delta \geq 1/2$ such that the size of the largest component jumps from $o(n)$ to $\delta n$ in $o(n)$ steps.

This turned out to be false, and Riordan and Warnke proved it. Nevertheless, the idea has been interpreted in an interesting light. People have claimed it is a useful model of disaster in the following sense. If you imagine that an edge between two vertices is a “crisis” relating two entities. Then in every step God presents you with two crises and you only have the resources to fix one. The idea is that when the entire graph is connected, you have this one big disaster where all the problems are interacting with each other. The percolation process describes how long you can “survive” while avoiding the big disaster.

There are critiques of this interpretation, though, mainly about how simplistic it is. In particular, an Achlioptas process models a crisis as an exogenous force when in reality problems are usually endogenous. You don’t expect a meteor to hit the Earth, but you do expect humans to have an impact on the environment. Also, not everybody in the network is trying to avoid errors. Some companies thrive in economic downturns by managing your toxic assets, for example. So one could reasonably argue that Achlioptas processes aren’t complex enough to model the realistic types of disasters we face.

Either way, I find it fantastic that something like a random graph (which for decades was securely in pure combinatorics away from applications) is spurring such discussion.

Next time, we’ll take one more dive into the theory of Erdős-Rényi random graphs to prove a very “meta” theorem about sharp thresholds. Then we’ll turn our attention to other models of random graphs, hopefully more realistic ones :)

Until then!

# 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!

# When Greedy Algorithms are Perfect: the Matroid

Greedy algorithms are by far one of the easiest and most well-understood algorithmic techniques. There is a wealth of variations, but at its core the greedy algorithm optimizes something using the natural rule, “pick what looks best” at any step. So a greedy routing algorithm would say to a routing problem: “You want to visit all these locations with minimum travel time? Let’s start by going to the closest one. And from there to the next closest one. And so on.”

Because greedy algorithms are so simple, researchers have naturally made a big effort to understand their performance. Under what conditions will they actually solve the problem we’re trying to solve, or at least get close? In a previous post we gave some easy-to-state conditions under which greedy gives a good approximation, but the obvious question remains: can we characterize when greedy algorithms give an optimal solution to a problem?

The answer is yes, and the framework that enables us to do this is called a matroid. That is, if we can phrase the problem we’re trying to solve as a matroid, then the greedy algorithm is guaranteed to be optimal. Let’s start with an example when greedy is provably optimal: the minimum spanning tree problem. Throughout the article we’ll assume the reader is familiar with the very basics of linear algebra and graph theory (though we’ll remind ourselves what a minimum spanning tree is shortly). For a refresher, this blog has primers on both subjects. But first, some history.

## History

Matroids were first introduced by Hassler Whitney in 1935, and independently discovered a little later by B.L. van der Waerden (a big name in combinatorics). They were both interested in devising a general description of “independence,” the properties of which are strikingly similar when specified in linear algebra and graph theory. Since then the study of matroids has blossomed into a large and beautiful theory, one part of which is the characterization of the greedy algorithm: greedy is optimal on a problem if and only if the problem can be represented as a matroid. Mathematicians have also characterized which matroids can be modeled as spanning trees of graphs (we will see this momentarily). As such, matroids have become a standard topic in the theory and practice of algorithms.

## Minimum Spanning Trees

It is often natural in an undirected graph $G = (V,E)$ to find a connected subset of edges that touch every vertex. As an example, if you’re working on a power network you might want to identify a “backbone” of the network so that you can use the backbone to cheaply travel from any node to any other node. Similarly, in a routing network (like the internet) it costs a lot of money to lay down cable, it’s in the interest of the internet service providers to design analogous backbones into their infrastructure.

A minimal subset of edges in a backbone like this is guaranteed to form a tree. This is simply because if you have a cycle in your subgraph then removing any edge on that cycle doesn’t break connectivity or the fact that you can get from any vertex to any other (and trees are the maximal subgraphs without cycles). As such, these “backbones” are called spanning trees. “Span” here means that you can get from any vertex to any other vertex, and it suggests the connection to linear algebra that we’ll describe later, and it’s a simple property of a tree that there is a unique path between any two vertices in the tree.

An example of a spanning tree

When your edges $e \in E$ have nonnegative weights $w_e \in \mathbb{R}^{\geq 0}$, we can further ask to find a minimum cost spanning tree. The cost of a spanning tree $T$ is just the sum of its edges, and it’s important enough of a definition to offset.

Definition: A minimum spanning tree $T$ of a weighted graph $G$ (with weights $w_e \geq 0$ for $e \in E$) is a spanning tree which minimizes the quantity

$w(T) = \sum_{e \in T} w_e$

There are a lot of algorithms to find minimal spanning trees, but one that will lead us to matroids is Kruskal’s algorithm. It’s quite simple. We’ll maintain a forest $F$ in $G$, which is just a subgraph consisting of a bunch of trees that may or may not be connected. At the beginning $F$ is just all the vertices with no edges. And then at each step we add to $F$ the edge $e$ whose weight is smallest and also does not introduce any cycles into $F$. If the input graph $G$ is connected then this will always produce a minimal spanning tree.

Theorem: Kruskal’s algorithm produces a minimal spanning tree of a connected graph.

Proof. Call $F_t$ the forest produced at step $t$ of the algorithm. Then $F_0$ is the set of all vertices of $G$ and $F_{n-1}$ is the final forest output by Kruskal’s (as a quick exercise, prove all spanning trees on $n$ vertices have $n-1$ edges, so we will stop after $n-1$ rounds). It’s clear that $F_{n-1}$ is a tree because the algorithm guarantees no $F_i$ will have a cycle. And any tree with $n-1$ edges is necessarily a spanning tree, because if some vertex were left out then there would be $n-1$ edges on a subgraph of $n-1$ vertices, necessarily causing a cycle somewhere in that subgraph.

Now we’ll prove that $F_{n-1}$ has minimal cost. We’ll prove this in a similar manner to the general proof for matroids. Indeed, say you had a tree $T$ whose cost is strictly less than that of $F_{n-1}$ (we can also suppose that $T$ is minimal, but this is not necessary). Pick the minimal weight edge $e \in T$ that is not in $F_{n-1}$. Adding $e$ to $F_{n-1}$ introduces a unique cycle $C$ in $F_{n-1}$. This cycle has some strange properties. First, $e$ has the highest cost of any edge on $C$. For otherwise, Kruskal’s algorithm would have chosen it before the heavier weight edges. Second, there is another edge in $C$ that’s not in $T$ (because $T$ was a tree it can’t have the entire cycle). Call such an edge $e'$. Now we can remove $e'$ from $F_{n-1}$ and add $e$. This can only increase the total cost of $F_{n-1}$, but this transformation produces a tree with one more edge in common with $T$ than before. This contradicts that $T$ had strictly lower weight than $F_{n-1}$, because repeating the process we described would eventually transform $F_{n-1}$ into $T$ exactly, while only increasing the total cost.

$\square$

Just to recap, we defined sets of edges to be “good” if they did not contain a cycle, and a spanning tree is a maximal set of edges with this property. In this scenario, the greedy algorithm performed optimally at finding a spanning tree with minimal total cost.

## Columns of Matrices

Now let’s consider a different kind of problem. Say I give you a matrix like this one:

$\displaystyle A = \begin{pmatrix} 2 & 0 & 1 & -1 & 0 \\ 0 & -4 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 7 \end{pmatrix}$

In the standard interpretation of linear algebra, this matrix represents a linear function $f$ from one vector space $V$ to another $W$, with the basis $(v_1, \dots, v_5)$ of $V$ being represented by columns and the basis $(w_1, w_2, w_3)$ of $W$ being represented by the rows. Column $j$ tells you how to write $f(v_j)$ as a linear combination of the $w_i$, and in so doing uniquely defines $f$.

Now one thing we want to calculate is the rank of this matrix. That is, what is the dimension of the image of $V$ under $f$? By linear algebraic arguments we know that this is equivalent to asking “how many linearly independent columns of $A$ can we find”? An interesting consequence is that if you have two sets of columns that are both linearly independent and maximally so (adding any other column to either set would necessarily introduce a dependence in that set), then these two sets have the same size. This is part of why the rank of a matrix is well-defined.

If we were to give the columns of $A$ costs, then we could ask about finding the minimal-cost maximally-independent column set. It sounds like a mouthful, but it’s exactly the same idea as with spanning trees: we want a set of vectors that spans the whole column space of $A$, but contains no “cycles” (linearly dependent combinations), and we want the cheapest such set.

So we have two kinds of “independence systems” that seem to be related. One interesting question we can ask is whether these kinds of independence systems are “the same” in a reasonable way. Hardcore readers of this blog may see the connection quite quickly. For any graph $G = (V,E)$, there is a natural linear map from $E$ to $V$, so that a linear dependence among the columns (edges) corresponds to a cycle in $G$. This map is called the incidence matrix by combinatorialists and the first boundary map by topologists.

The map is easy to construct: for each edge $e = (v_i,v_j)$ you add a column with a 1 in the $j$-th row and a $-1$ in the $i$-th row. Then taking a sum of edges gives you zero if and only if the edges form a cycle. So we can think of a set of edges as “independent” if they don’t contain a cycle. It’s a little bit less general than independence over $\mathbb{R}$, but you can make it exactly the same kind of independence if you change your field from real numbers to $\mathbb{Z}/2\mathbb{Z}$. We won’t do this because it will detract from our end goal (to analyze greedy algorithms in realistic settings), but for further reading this survey of Oxley assumes that perspective.

So with the recognition of how similar these notions of independence are, we are ready to define matroids.

## The Matroid

So far we’ve seen two kinds of independence: “sets of edges with no cycles” (also called forests) and “sets of linearly independent vectors.” Both of these share two trivial properties: there are always nonempty independent sets, and every subset of an independent set is independent. We will call any family of subsets with this property an independence system.

Definition: Let $X$ be a finite set. An independence system over $X$ is a family $\mathscr{I}$ of subsets of $X$ with the following two properties.

1. $\mathscr{I}$ is nonempty.
2. If $I \in \mathscr{I}$, then so is every subset of $I$.

This is too general to characterize greedy algorithms, so we need one more property shared by our examples. There are a few things we do, but here’s one nice property that turns out to be enough.

Definition: A matroid $M = (X, \mathscr{I})$ is a set $X$ and an independence system $\mathscr{I}$ over $X$ with the following property:

If $A, B$ are in $\mathscr{I}$ with $|A| = |B| + 1$, then there is an element $x \in A \setminus B$ such that $B \cup \{ a \} \in \mathscr{I}$.

In other words, this property says if I have an independent set that is not maximally independent, I can grow the set by adding some suitably-chosen element from a larger independent set. We’ll call this the extension property. For a warmup exercise, let’s prove that the extension property is equivalent to the following (assuming the other properties of a matroid):

For every subset $Y \subset X$, all maximal independent sets contained in $Y$ have equal size.

Proof. For one direction, if you have two maximal sets $A, B \subset Y \subset X$ that are not the same size (say $A$ is bigger), then you can take any subset of $A$ whose size is exactly $|B| + 1$, and use the extension property to make $B$ larger, a contradiction. For the other direction, say that I know all maximal independent sets of any $Y \subset X$ have the same size, and you give me $A, B \subset X$. I need to find an $a \in A \setminus B$ that I can add to $B$ and keep it independent. What I do is take the subset $Y = A \cup B$. Now the sizes of $A, B$ don’t change, but $B$ can’t be maximal inside $Y$ because it’s smaller than $A$ ($A$ might not be maximal either, but it’s still independent). And the only way to extend $B$ is by adding something from $A$, as desired.

$\square$

So we can use the extension property and the cardinality property interchangeably when talking about matroids. Continuing to connect matroid language to linear algebra and graph theory, the maximal independent sets of a matroid are called bases, the size of any basis is the rank of the matroid, and the minimal dependent sets are called circuits. In fact, you can characterize matroids in terms of the properties of their circuits, which are dual to the properties of bases (and hence all independent sets) in a very concrete sense.

But while you could spend all day characterizing the many kinds of matroids and comatroids out there, we are still faced with the task of seeing how the greedy algorithm performs on a matroid. That is, suppose that your matroid $M = (X, \mathscr{I})$ has a nonnegative real number $w(x)$ associated with each $x \in X$. And suppose we had a black-box function to determine if a given set $S \subset X$ is independent. Then the greedy algorithm maintains a set $B$, and at every step adds a minimum weight element that maintains the independence of $B$. If we measure the cost of a subset by the sum of the weights of its elements, then the question is whether the greedy algorithm finds a minimum weight basis of the matroid.

The answer is even better than yes. In fact, the answer is that the greedy algorithm performs perfectly if and only if the problem is a matroid! More rigorously,

Theorem: Suppose that $M = (X, \mathscr{I})$ is an independence system, and that we have a black-box algorithm to determine whether a given set is independent. Define the greedy algorithm to iteratively adds the cheapest element of $X$ that maintains independence. Then the greedy algorithm produces a maximally independent set $S$ of minimal cost for every nonnegative cost function on $X$, if and only if $M$ is a matroid.

It’s clear that the algorithm will produce a set that is maximally independent. The only question is whether what it produces has minimum weight among all maximally independent sets. We’ll break the theorem into the two directions of the “if and only if”:

Part 1: If $M$ is a matroid, then greedy works perfectly no matter the cost function.
Part 2: If greedy works perfectly for every cost function, then $M$ is a matroid.

Proof of Part 1.

Call the cost function $w : X \to \mathbb{R}^{\geq 0}$, and suppose that the greedy algorithm picks elements $B = \{ x_1, x_2, \dots, x_r \}$ (in that order). It’s easy to see that $w(x_1) \leq w(x_2) \leq \dots \leq w(x_r)$. Now if you give me any list of $r$ independent elements $y_1, y_2, \dots, y_r \in X$ that has $w(y_1) \leq \dots \leq w(y_r)$, I claim that $w(x_i) \leq w(y_i)$ for all $i$. This proves what we want, because if there were a basis of size $r$ with smaller weight, sorting its elements by weight would give a list contradicting this claim.

To prove the claim, suppose to the contrary that it were false, and for some $k$ we have $w(x_k) > w(y_k)$. Moreover, pick the smallest $k$ for which this is true. Note $k > 1$, and so we can look at the special sets $S = \{ x_1, \dots, x_{k-1} \}$ and $T = \{ y_1, \dots, y_k \}$. Now $|T| = |S|+1$, so by the matroid property there is some $j$ between $1$ and $r$ so that $S \cup \{ y_j \}$ is an independent set (and $y_j$ is not in $S$). But then $w(y_j) \leq w(y_k) < w(x_k)$, and so the greedy algorithm would have picked $y_j$ before it picks $x_k$ (and the strict inequality means they’re different elements). This contradicts how the greedy algorithm runs, and hence proves the claim.

Proof of Part 2.

We’ll prove this contrapositively as follows. Suppose we have our independence system and it doesn’t satisfy the last matroid condition. Then we’ll construct a special weight function that causes the greedy algorithm to fail. So let $A,B$ be independent sets with $|A| = |B| + 1$, but for every $a \in A \setminus B$ adding $a$ to $B$ never gives you an independent set.

Now what we’ll do is define our weight function so that the greedy algorithm picks the elements we want in the order we want (roughly). In particular, we’ll assign all elements of $A \cap B$ a tiny weight we’ll call $w_1$. For elements of $B - A$ we’ll use $w_2$, and for $A - B$ we’ll use $w_3$, with $w_4$ for everything else. In a more compact notation:

We need two things for this weight function to screw up the greedy algorithm. The first is that $w_1 < w_2 < w_3 < w_4$, so that greedy picks the elements in the order we want. Note that this means it’ll first pick all of $A \cap B$, and then all of $B - A$, and by assumption it won’t be able to pick anything from $A - B$, but since $B$ is assumed to be non-maximal, we have to pick at least one element from $X - (A \cup B)$ and pay $w_4$ for it.

So the second thing we want is that the cost of doing greedy is worse than picking any maximally independent set that contains $A$ (and we know that there has to be some maximal independent set containing $A$). In other words, if we call $m$ the size of a maximally independent set, we want

$\displaystyle |A \cap B| w_1 + |B-A|w_2 + (m - |B|)w_4 > |A \cap B|w_1 + |A-B|w_3 + (m-|A|)w_4$

This can be rearranged (using the fact that $|A| = |B|+1$) to

$\displaystyle w_4 > |A-B|w_3 - |B-A|w_2$

The point here is that the greedy picks too many elements of weight $w_4$, since if we were to start by taking all of $A$ (instead of all of $B$), then we could get by with one fewer. That might not be optimal, but it’s better than greedy and that’s enough for the proof.

So we just need to make $w_4$ large enough to make this inequality hold, while still maintaining $w_2 < w_3$. There are probably many ways to do this, and here’s one. Pick some $0 < \varepsilon < 1$, and set

It’s trivial that $w_1 < w_2$ and $w_3 < w_4$. For the rest we need some observations. First, the fact that $|A-B| = |B-A| + 1$ implies that $w_2 < w_3$. Second, both $|A-B|$ and $|B-A|$ are nonempty, since otherwise the second property of independence systems would contradict our assumption that augmenting $B$ with elements of $A$ breaks independence. Using this, we can divide by these quantities to get

$\displaystyle w_4 = 2 > 1 = \frac{|A-B|(1 + \varepsilon)}{|A-B|} - \frac{|B-A|\varepsilon}{|B-A|}$

This proves the claim and finishes the proof.

$\square$

As a side note, we proved everything here with respect to minimizing the sum of the weights, but one can prove an identical theorem for maximization. The only part that’s really different is picking the clever weight function in part 2. In fact, you can convert between the two by defining a new weight function that subtracts the old weights from some fixed number $N$ that is larger than any of the original weights. So these two problems really are the same thing.

This is pretty amazing! So if you can prove your problem is a matroid then you have an awesome algorithm automatically. And if you run the greedy algorithm for fun and it seems like it works all the time, then that may be hinting that your problem is a matroid. This is one of the best situations one could possibly hope for.

But as usual, there are a few caveats to consider. They are both related to efficiency. The first is the black box algorithm for determining if a set is independent. In a problem like minimum spanning tree or finding independent columns of a matrix, there are polynomial time algorithms for determining independence. These two can both be done, for example, with Gaussian elimination. But there’s nothing to stop our favorite matroid from requiring an exponential amount of time to check if a set is independent. This makes greedy all but useless, since we need to check for independence many times in every round.

Another, perhaps subtler, issue is that the size of the ground set $X$ might be exponentially larger than the rank of the matroid. In other words, at every step our greedy algorithm needs to find a new element to add to the set it’s building up. But there could be such a huge ocean of candidates, all but a few of which break independence. In practice an algorithm might be working with $X$ implicitly, so we could still hope to solve the problem if we had enough knowledge to speed up the search for a new element.

There are still other concerns. For example, a naive approach to implementing greedy takes quadratic time, since you may have to look through every element of $X$ to find the minimum-cost guy to add. What if you just have to have faster runtime than $O(n^2)$? You can still be interested in finding more efficient algorithms that still perform perfectly, and to the best of my knowledge there’s nothing that says that greedy is the only exact algorithm for your favorite matroid. And then there are models where you don’t have direct/random access to the input, and lots of other ways that you can improve on greedy. But those stories are for another time.

Until then!