An Un-PAC-learnable Problem

In a previous post we introduced a learning model called Probably Approximately Correct (PAC). We saw an example of a concept class that was easy to learn: intervals on the real line (and more generally, if you did the exercise, axis-aligned rectangles in a fixed dimension).

But PAC learning wouldn’t be an interesting model if every concept class was PAC-learnable. So as a technical aside in our study of learning theory, this post presents the standard example of a problem that isn’t learnable in the PAC model. Afterward we’ll see that allowing the learner to be more expressive can be helpful, and by doing so we can make this unlearnable problem learnable.

3-Term DNF Formulas

Readers of this blog will probably have encountered a boolean formula before. A boolean formula is just a syntactic way to describe some condition (like, exactly one of these two things has to be true) using variables and logical connectives. The best way to recall it is by example: the following boolean formula encodes the “exclusive or” of two variables.

\displaystyle (x \wedge \overline{y}) \vee (\overline{x} \wedge y)

The wedge \wedge denotes a logical AND and the vee \vee denotes a logical OR. A bar above a variable represents a negation of a variable. (Please don’t ask me why the official technical way to write AND and OR is in all caps, I feel like I’m yelling math at people.)

In general a boolean formula has literals, which we can always denote by an x_i or the negation \overline{x_i}, and connectives \wedge and \vee, and parentheses to denote order. It’s a simple fact that any logical formula can be encoded using just these tools, but rather than try to learn general boolean formulas we look at formulas in a special form.

Definition: A formula is in three-term disjunctive normal form (DNF) if it has the form C_1 \vee C_2 \vee C_3 where each $C_i$ is an AND of some number of literals.

Readers who enjoyed our P vs NP primer will recall a related form of formulas: the 3-CNF form, where the “three” meant that each clause had exactly three literals and the “C” means the clauses are connected with ANDs. This is a sort of dual normal form: there are only three clauses, each clause can have any number of variables, and the roles of AND and OR are switched. In fact, if you just distribute the \vee‘s in a 3-term DNF formula using DeMorgan’s rules, you’ll get an equivalent 3-CNF formula. The restriction of our hypotheses to 3-term DNFs will be the crux of the difficulty: it’s not that we can’t learn DNF formulas, we just can’t learn them if we are forced to express our hypothesis as a 3-term DNF as well.

The way we’ll prove that 3-term DNF formulas “can’t be learned” in the PAC model is by an NP-hardness reduction. That is, we’ll show that if we could learn 3-term DNFs in the PAC model, then we’d be able to efficiently solve NP-hard problems with high probability. The official conjecture we’d be violating is that RP is different from NP. RP is the class of problems that you can solve in polynomial time with randomness if you can never have false positives, and the probability of a false negative is at most 1/2. Our “RP” algorithm will be a PAC-learning algorithm.

The NP-complete problem we’ll reduce from is graph 3-coloring. So if you give me a graph, I’ll produce an instance of the 3-term DNF PAC-learning problem in such a way that finding a hypothesis with low error corresponds to a valid 3-coloring of the graph. Since PAC-learning ensures that you are highly likely to find a low-error hypothesis, the existence of a PAC-learning algorithm will constitute an RP algorithm to solve this NP-complete problem.

In more detail, an “instance” of the 3-term DNF problem comes in the form of a distribution over some set of labeled examples. In this case the “set” is the set of all possible truth assignments to the variables, where we fix the number of variables to suit our needs, along with a choice of a target 3-term DNF to be learned. Then you’d have to define the distribution over these examples.

But we’ll actually do something a bit slicker. We’ll take our graph G, we’ll construct a set S_G of labeled truth assignments, and we’ll define the distribution D to be the uniform distribution over those truth assignments used in S_G. Then, if there happens to be a 3-term DNF that coincidentally labels the truth assignments in S_G exactly how we labeled them, and we set the allowed error \varepsilon to be small enough, a PAC-learning algorithm will find a consistent hypothesis (and it will correspond to a valid 3-coloring of G). Otherwise, no algorithm would be able to come up with a low-error hypothesis, so if our purported learning algorithm outputs a bad hypothesis we’d be certain (with high probability) that it was not bad luck but that the examples are not consistent with any 3-term DNF (and hence there is no valid 3-coloring of G).

This general outline has nothing to do with graphs, and so you may have guessed that the technique is commonly used to prove learning problems are hard: come up with a set of labeled examples, and a purported PAC-learning algorithm would have to come up with a hypothesis consistent with all the examples, which translates back to a solution to your NP-hard problem.

The Reduction

Now we can describe the reduction from graphs to labeled examples. The intuition is simple: each term in the 3-term DNF should correspond to a color class, and so any two adjacent vertices should correspond to an example that cannot be true. The clauses will correspond to…

For a graph G with n nodes v_1, \dots, v_n and a set of m undirected edges E, we construct a set of examples with positive labels S^+ and one with negative examples S^-. The examples are truth assignments to n variables, which we label x_1, \dots, x_n, and we identify a truth assignment to the \left \{ 0,1 \right \}-valued vector (x_1, x_2, \dots, x_n) in the usual way (true is 1, false is 0).

The positive examples S^+ are simple: for each v_i add a truth assignment x_i = T, x_j = F for j \neq i. I.e., the binary vector is (1, \dots, 1,0,1, \dots, 1), and the zero is in the i-th position.

The negative examples S^- come from the edges. For each edge (v_i, v_j) \in E, we add the example with a zero in the i-th and j-th components and ones everywhere else. Here is an example graph and the corresponding positive and negative examples:

PAC-reduction

Claim: G is 3-colorable if and only if the corresponding examples are consistent with some 3-term DNF formula \varphi.

Again, consistent just means that \varphi is satisfied by every truth assignment in S^+ and unsatisfied by every example in S^-. Since we chose our distribution to be uniform over S^+ \cup S^-, we don’t care what \varphi does elsewhere.

Indeed, if G is three-colorable we can fix some valid 3-coloring with colors red, blue, and yellow. We can construct a 3-term DNF that does what we need. Let T_R be the AND of all the literals x_i for which vertex v_i is not red. For each such i, the corresponding example in S^+ will satisfy T_R, because we put a zero in the i-th position and ones everywhere else. Similarly, no example in S^- will make T_R true because to do so both vertices in the corresponding edge would have to be red.

To drive this last point home say there are three vertices and your edge is (v_1,v_2). Then the corresponding negative example is (0,0,1). Unless both v_1 and v_2 are colored red, one of x_1, x_2 will have to be ANDed as part of T_R. But the example has a zero for both x_1 and x_2, so T_R would not be satisfied.

Doing the same thing for blue and yellow, and OR them together to get T_R \vee T_B \vee T_Y. Since the case is symmetrically the same for the other colors, we a consistent 3-term DNF.

On the other hand, say there is a consistent 3-term DNF \varphi. We need to construct a three coloring of G. It goes in largely the same way: label the clauses \varphi = T_R \vee T_B \vee T_Y for Red, Blue, and Yellow, and then color a vertex v_i the color of the clause that is satisfied by the corresponding example in S^+. There must be some clause that does this because \varphi is consistent with S^+, and if there are multiple you can pick a valid color arbitrarily. Now we argue why no edge can be monochromatic. Suppose there were such an edge (v_i, v_j), and both v_i and v_j are colored, say, blue. Look at the clause T_B: since v_i and v_j are both blue, the positive examples corresponding to those vertices  (with a 0 in the single index and 1′s everywhere else) both make T_B true. Since those two positive examples differ in both their i-th and j-th positions, T_B can’t have any of the literals x_i, \overline{x_i}, x_j, \overline{x_j}. But then the negative example for the edge would satisfy T_B because it has 1′s everywhere except i,j! This means that the formula doesn’t consistently classify the negative examples, a contradiction. This proves the Claim.

Now we just need to show a few more details to finish the proof. In particular, we need to observe that the number of examples we generate is polynomial in the size of the graph G; that the learning algorithm would still run in polynomial time in the size of the input graph (indeed, this depends on our choice of the learning parameters); and that we only need to pick \delta < 1/2 and \varepsilon \leq 1/(2|S^+ \cup S^-|) in order to enforce that an efficient PAC-learner would generate a hypothesis consistent with all the examples. Indeed, if a hypothesis errs on even one example, it will have error at least 1 / |S^+ \cup S^-|, which is too big.

Everything’s not Lost

This might seem a bit depressing for PAC-learning, that we can’t even hope to learn 3-term DNF formulas. But we will give a sketch of why this is not a problem with PAC but a problem with DNFs.

In particular, the difficulty comes in forcing a PAC-learning algorithm to express its hypothesis as a 3-term DNF, as opposed to what we might argue is a more natural representation. As we observed, distributing the ORs in a 3-term DNF produces a 3-CNF formula (an AND of clauses where each clause is an OR of exactly three literals). Indeed, one can PAC-learn 3-CNF formulas efficiently, and it suffices to show that one can learn formulas which are just ANDs of literals. Then you can blow up the number of variables only polynomially larger to get 3-CNFs. ANDs of literals are just called “conjunctions,” so the problem is to PAC-learn conjunctions. The idea that works is the same one as in our first post on PAC where we tried to learn intervals: just pick the “smallest” hypothesis that is consistent with all the examples you’ve seen so far. We leave a formal proof as an (involved) exercise to the reader.

The important thing to note is that a concept class C (the thing we’re trying to learn) might be hard to learn if you’re constrained to work within C. If you’re allowed more expressive hypotheses (in this case, arbitrary boolean formulas), then learning C suddenly becomes tractable. This compels us to add an additional caveat to the PAC definition from our first post.

Definition: A concept class \mathsf{C} over a set X is efficiently PAC-learnable using the hypothesis class \mathsf{H} if there exists an algorithm A(\varepsilon, \delta) with access to a query function for \mathsf{C} and runtime O(\text{poly}(1/\varepsilon, 1/\delta)), such that for all c \in \mathsf{C}, all distributions D over X, and all 0 < \delta , \varepsilon < 1/2, the probability that A produces a hypothesis h \in \mathsf{H} with error at most \varepsilon is at least 1-\delta.

And with that we’ll end this extended side note. The next post in this series will introduce and analyze a fascinating notion of dimension for concept classes, the Vapnik-Chervonenkis dimension.

Until then!

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Stable Marriages and Designing Markets

Here is a fun puzzle. Suppose we have a group of 10 men and 10 women, and each of the men has sorted the women in order of their preference for marriage (that is, a man prefers to marry a woman earlier in his list over a woman later in the list). Likewise, each of the women has sorted the men in order of marriageability. We might ask if there is any way that we, the omniscient cupids of love, can decide who should marry to make everyone happy.

Of course, the word happy is entirely imprecise. The mathematician balks at the prospect of leaving such terms undefined! In this case, it’s quite obvious that not everyone will get their first pick. Indeed, if even two women prefer the same man someone will have to settle for less than their top choice. So if we define happiness in this naive way, the problem is obviously not solvable in general.

Now what if instead of aiming for each individual’s maximum happiness we instead shoot for mutual contentedness? That is, what if “happiness” here means that nobody will ever have an incentive to cheat on their spouse? It turns out that for a mathematical version of this condition, we can always find a suitable set of marriages! These mathematical formalisms include some assumptions, such as that preferences never change and that no new individuals are added to the population. But it is nevertheless an impressive theorem that we can achieve stability no matter what everyone’s preferences are. In this post we’ll give the classical algorithm which constructs so-called “stable marriages,” and we’ll prove its correctness. Then we’ll see a slight generalization of the algorithm, in which the marriages are “polygamous,” and we’ll apply it to the problem of assigning students to internships.

As usual, all of the code used in this post is available for download at this blog’s Github page.

Historical Notes

The original algorithm for computing stable marriages was discovered by Lloyd Shapley and David Gale in the early 1960′s. Shapely and Alvin Roth went on to dedicate much of their career to designing markets and applying the stable marriage problem and its generalizations to such problems. In 2012 they jointly received the Nobel prize in economics for their work on this problem. If you want to know more about what “market design” means and why it’s needed (and you have an hour to spare), consider watching the talk below by Alvin Roth at the Simons Institute’s 2013 Symposium on the Visions of the Theory of Computing. Roth spends most of his time discussing the state of one particular economy, medical students and residence positions at hospitals, which he was asked to redesign. It’s quite a fascinating tale, although some of the deeper remarks assume knowledge of the algorithm we cover in this post.

Alvin Roth went on to apply the ideas presented in the video to economic systems in Boston and New York City public schools, kidney exchanges, and others. They all had the same sort of structure: both parties have preferences and stability makes sense. So he actually imposed the protocol we’re about to describe in order to guarantee that the process terminates to a stable arrangement (and automating it saves everyone involved a lot of time, stress, and money! Watch the video above for more on that).

The Monogamous Stable Marriage Algorithm

Let’s formally set up the problem. Let X = \left \{ 1, 2, \dots, n \right \} be a set of n suitors and Y = \left \{ 1,2,\dots ,n \right \} be a set of n “suited.” Let \textup{pref}_{X \to Y}: X \to S_n be a list of preferences for the suitors. In words, \textup{pref}_{X \to Y} accepts as input a suitor, and produces as output an ordering on the suited members of Y. We denote the output set as S_n, which the group theory folks will recognize as the permutation group on 1, \dots, n. Likewise, there is a function \textup{pref}_{Y \to X}: Y \to S_n describing the preferences of each of the suited.

An example will help clarify these stuffy definitions. If X = \left \{ 1, 2, 3 \right \} and Y = \left \{ 1, 2, 3 \right \}, then to say that

\textup{pref}_{X \to Y}(2) = (3, 1, 2)

is to say that the second suitor prefers the third member of Y the most, and then the first member of Y, and then the second. The programmer might imagine that the datum of the problem consists of two dictionaries (one for X and one for Y) whose keys are integers and whose values are lists of integers which contain 1 through n in some order.

A solution to the problem, then, is a way to match (or marry) suitors with suited. Specifically, a matching is a bijection m: X \to Y, so that x is matched with m(x). The reason we use a bijection is because the marriages are monogamous: only one suitor can be matched with one suited and vice versa. Later we’ll see this condition dropped so we can apply it to a more realistic problem of institutions (suited) which can accommodate many applicants (suitors). Because suitor and suited are awkward to say, we’ll use the familiar, antiquated, and politically incorrect terms “men and women.”

Now if we’re given a monogamous matching m, a pair x \in X, y \in Y is called unstable for m if both x,y prefer each other over their partners assigned by m. That is, (x,y) is unstable for m if y appears before m(y) in the preference list for x, \textup{pref}_{X \to Y}(x), and likewise x appears before m^{-1}(y) in \textup{pref}_{Y \to X}(y).

Another example to clarify: again let X = Y = \left \{ 1,2,3 \right \} and suppose for simplicity that our matching m pairs m(i) = i. If man 2 has the preference list (3,2,1) and woman 3 has the preference list (2,1,3), then 2 and 3 together form an unstable pair for m, because they would rather be with each other over their current partners. That is, they have a mutual incentive to cheat on their spouses. We say that the matching is unstable or admits an unstable pair if there are any unstable pairs for it, and we call the entire matching stable if it doesn’t admit any unstable pairs.

Unlike real life, mathematically unstable marriages need not have constant arguments.

Unlike real life, mathematically unstable marriages need not feature constant arguments.

So the question at hand is: is there an algorithm which, given access to to the two sets of preferences, can efficiently produce a stable matching? We can also wonder whether a stable matching is guaranteed to exist, and the answer is yes. In fact, we’ll prove this and produce an efficient algorithm in one fell swoop.

The central concept of the algorithm is called deferred acceptance. The gist is like this. The algorithm operates in rounds. During each round, each man will “propose” to a woman, and each woman will pick the best proposal available. But the women will not commit to their pick. They instead reject all other suitors, who go on to propose to their second choices in the next round. At that stage each woman (who now may have a more preferred suitor than in the first round) may replace her old pick with a new one. The process continues in this manner until each man is paired with a woman. In this way, each of the women defers accepting any proposal until the end of the round, progressively increasing the quality of her choice. Likewise, the men progressively propose less preferred matches as the rounds progress.

It’s easy to argue such a process must eventually converge. Indeed, the contrary means there’s some sort of cycle in the order of proposals, but each man proposes to only strictly less preferred women than any previous round, and the women can only strictly increase the quality of their held pick. Mathematically, we’re using an important tool called monotonicity. That some quantity can only increase or decrease as time goes on, and since the quantity is bounded, we must eventually reach a local maximum. From there, we can prove that any local maximum satisfies the property we want (here, that the matching is stable), and we win. Indeed, supposing to the contrary that we have a pair (x,y) which is unstable for the matching m produced at the end of this process, then it must have been the case that x proposed to y in some earlier round. But y has as her final match some other suitor x' = m^{-1}(y) whom she prefers less than x. Though she may have never picked x at any point in the algorithm, she can only end up with the worse choice x' if at some point y chose a suitor that was less preferred than the suitor she already had. Since her choices are monotonic this cannot happen, so no unstable pairs can exist.

Rather than mathematically implement the algorithm in pseudocode, let’s produce the entire algorithm in Python to make the ideas completely concrete.

Python Implementation

We start off with some simple data definitions for the two parties which, in the renewed interest of generality, refer to as Suitor and Suited.

class Suitor(object):
   def __init__(self, id, prefList):
      self.prefList = prefList
      self.rejections = 0 # num rejections is also the index of the next option
      self.id = id

   def preference(self):
      return self.prefList[self.rejections]

   def __repr__(self):
      return repr(self.id)

A Suitor is simple enough: he has an id representing his “index” in the set of Suitors, and a preference list prefList which in its i-th position contains the Suitor’s i-th most preferred Suited. This is identical to our mathematical representation from earlier, where a list like (2,3,1) means that the Suitor prefers the second Suited most and the first Suited least. Knowing the algorithm ahead of time, we add an additional piece of data: the number of rejections the Suitor has seen so far. This will double as the index of the Suited that the Suitor is currently proposing to. Indeed, the preference function provides a thin layer of indirection allowing us to ignore the underlying representation, so long as one updates the number of rejections appropriately.

Now for the Suited.

class Suited(object):
   def __init__(self, id, prefList):
      self.prefList = prefList
      self.held = None
      self.currentSuitors = set()
      self.id = id

   def __repr__(self):
      return repr(self.id)

A Suited likewise has a list of preferences and an id, but in addition she has a held attribute for the currently held Suitor, and a list currentSuitors of Suitors that are currently proposing to her. Hence we can define a reject method which accepts no inputs, and returns a list of rejected suitors, while updating the woman’s state to hold onto her most preferred suitor.

   def reject(self):
      if len(self.currentSuitors) == 0:
         return set()

      if self.held is not None:
         self.currentSuitors.add(self.held)

      self.held = min(self.currentSuitors, key=lambda suitor: self.prefList.index(suitor.id))
      rejected = self.currentSuitors - set([self.held])
      self.currentSuitors = set()

      return rejected

The call to min does all the work: finding the Suitor that appears first in her preference list. The rest is bookkeeping. Now the algorithm for finding a stable marriage, following the deferred acceptance algorithm, is simple.

# monogamousStableMarriage: [Suitor], [Suited] -> {Suitor -> Suited}
# construct a stable (monogamous) marriage between suitors and suiteds
def monogamousStableMarriage(suitors, suiteds):
   unassigned = set(suitors)

   while len(unassigned) > 0:
      for suitor in unassigned:
         suiteds[suitor.preference()].currentSuitors.add(suitor)
      unassigned = set()

      for suited in suiteds:
         unassigned |= suited.reject()

      for suitor in unassigned:
         suitor.rejections += 1

   return dict([(suited.held, suited) for suited in suiteds])

All the Suitors are unassigned to begin with. Each iteration of the loop corresponds to a round of the algorithm: the Suitors are added to the currentSuitors list of their next most preferred Suited. Then the Suiteds “simultaneously” reject some Suitors, whose rejection counts are upped by one and returned to the pool of unassigned Suitors. Once every Suited has held onto a Suitor we’re done.

Given a matching, we can define a function that verifies by brute force that the marriage is stable.

# verifyStable: [Suitor], [Suited], {Suitor -> Suited} -> bool
# check that the assignment of suitors to suited is a stable marriage
def verifyStable(suitors, suiteds, marriage):
   import itertools
   suitedToSuitor = dict((v,k) for (k,v) in marriage.items())
   precedes = lambda L, item1, item2: L.index(item1) < L.index(item2)

   def suitorPrefers(suitor, suited):
      return precedes(suitor.prefList, suited.id, marriage[suitor].id)

   def suitedPrefers(suited, suitor):
      return precedes(suited.prefList, suitor.id, suitedToSuitor[suited].id)

   for (suitor, suited) in itertools.product(suitors, suiteds):
      if suited != marriage[suitor] and suitorPrefers(suitor, suited) and suitedPrefers(suited, suitor):
         return False, (suitor.id, suited.id)

   return

Indeed, we can test the algorithm on an instance of the problem.

>>> suitors = [Suitor(0, [3,5,4,2,1,0]), Suitor(1, [2,3,1,0,4,5]),
...            Suitor(2, [5,2,1,0,3,4]), Suitor(3, [0,1,2,3,4,5]),
...            Suitor(4, [4,5,1,2,0,3]), Suitor(5, [0,1,2,3,4,5])]
>>> suiteds = [Suited(0, [3,5,4,2,1,0]), Suited(1, [2,3,1,0,4,5]),
...            Suited(2, [5,2,1,0,3,4]), Suited(3, [0,1,2,3,4,5]),
...            Suited(4, [4,5,1,2,0,3]), Suited(5, [0,1,2,3,4,5])]
>>> marriage = monogamousStableMarriage(suitors, suiteds)
{3: 0, 4: 4, 5: 1, 1: 2, 2: 5, 0: 3}
>>> verifyStable(suitors, suiteds, marriage)
True

We encourage the reader to check this by hand (this one only took two rounds). Even better, answer the question of whether the algorithm could ever require n steps to converge for 2n individuals, where you get to pick the preference list to try to make this scenario happen.

Stable Marriages with Capacity

We can extend this algorithm to work for “polygamous” marriages in which one Suited can accept multiple Suitors. In fact, the two problems are entirely the same! Just imagine duplicating a Suited with large capacity into many Suiteds with capacity of 1. This particular reduction is not very efficient, but it allows us to see that the same proof of convergence and correctness applies. We can then modify our classes and algorithm to account for it, so that (for example) instead of a Suited “holding” a single Suitor, she holds a set of Suitors. We encourage the reader to try extending our code above to the polygamous case as an exercise, and we’ve provided the solution in the code repository for this post on this blog’s Github page.

Ways to Make it Harder

When you study algorithmic graph problems as much as I do, you start to get disheartened. It seems like every problem is NP-hard or worse. So when we get a situation like this, a nice, efficient algorithm with very real consequences and interpretations, you start to get very excited. In between our heaves of excitement, we imagine all the other versions of this problem that we could solve and Nobel prizes we could win. Unfortunately the landscape is bleaker than that, and most extensions of stable marriage problems are NP-complete.

For example, what if we allow ties? That is, one man can be equally happy with two women. This is NP-complete. However, it turns out his extension can be formulated as an integer programming problem, and standard optimization techniques can be used to approximate a solution.

What if, thinking about the problem in terms of medical students and residencies, we allow people to pick their preferences as couples? Some med students are married, after all, and prefer to be close to their spouse even if it means they have a less preferred residency. NP-hard again. See page 53 (pdf page 71) of these notes for a more detailed investigation. The problem is essentially that there is not always a stable matching, and so even determining whether there is one is NP-complete.

So there are a lot of ways to enrich the problem, and there’s an interesting line between tractable and hard in the worst case. As a (relatively difficult) exercise, try to solve the “roommates” version of the problem, where there is no male/female distinction (anyone can be matched with anyone). It turns out to have a tractable solution, and the algorithm is similar to the one outlined in this post.

Until next time!

PS. I originally wrote this post about a year ago when I was contacted by someone in industry who agreed to provide some (anonymized) data listing the preferences of companies and interns applying to work at those companies. Not having heard from them for almost a year, I figure it’s a waste to let this finished post collect dust at the risk of not having an interesting data set. But if you, dear reader, have any data you’d like to provide that fits into the framework of stable marriages, I’d love to feature your company/service on my blog (and solve the matching problem) in exchange for the data. The only caveat is that the data would have to be public, so you would have to anonymize it.

Where Math ∩ Programming is Headed

This week is Spring break at UI Chicago. While I’ll be spending most of it working, it does give me some downtime to reflect. We’ve come pretty far, dear reader, in these almost three years. I learned, you learned. We all laughed. My blog has become my infinite source of entertainment and an invaluable tool for synthesizing my knowledge.

But the more I write the more ideas I have for articles, and this has been accelerating. I’m currently sitting on 55 unfinished drafts ranging from just a title and an idea to an almost-completed post. A lot of these ideas have long chains of dependencies (I can’t help myself but write on all the background math I can stomach before I do the applications). So one day I decided to draw up a little dependency graph to map out my coarse future plans. Here it is:

A map of most of my current plans for blog posts and series, and their relationships to one another. Click to enlarge.

A map of most of my current plans for blog posts and series, and their relationships to one another. Click to enlarge.

Now all you elliptic curve fanatics can rest assured I’ll continue working that series to completion before starting on any of these big projects. This map basically gives a rough picture of things I’ve read about, studied, and been interested in over the past two years that haven’t already made it onto this blog. Some of the nodes represent passed milestones in my intellectual career, while others represent topics yet to be fully understood. Note very few specific applications are listed here (e.g., what might I use SVM to classify?), but I do have ideas for a lot of them. And note that these are very long term plans, some of which are likely never to come to fruition.

So nows your chance to speak up. What do you want to read about? What do you think is missing?

On Coloring Resilient Graphs

I’m pleased to announce that another paper of mine is finished. This one is submitted to ICALP, which is being held in Copenhagen this year (this whole research thing is exciting!). This is joint work with my advisor, Lev Reyzin. As with my first paper, I’d like to explain things here on my blog a bit more informally than a scholarly article allows.

A Recent History of Graph Coloring

One of the first important things you learn when you study graphs is that coloring graphs is hard. Remember that coloring a graph with k colors means that you assign each vertex a color (a number in \left \{ 1, 2, \dots, k \right \}) so that no vertex is adjacent to a vertex of the same color (no edge is monochromatic). In fact, even deciding whether a graph can be colored with just 3 colors (not to mention finding such a coloring) has no known polynomial time algorithm. It’s what’s called NP-hard, which means that almost everyone believes it’s hopeless to solve efficiently in the worst case.

One might think that there’s some sort of gradient to this problem, that as the graphs get more “complicated” it becomes algorithmically harder to figure out how colorable they are. There are some notions of “simplicity” and “complexity” for graphs, but they hardly fall on a gradient. Just to give the reader an idea, here are some ways to make graph coloring easy:

  • Make sure your graph is planar. Then deciding 4-colorability is easy because the answer is always yes.
  • Make sure your graph is triangle-free and planar. Then finding a 3-coloring is easy.
  • Make sure your graph is perfect (which again requires knowledge about how colorable it is).
  • Make sure your graph has tree-width or clique-width bounded by a constant.
  • Make sure your graph doesn’t have a certain kind of induced subgraph (such as having no induced paths of length 4 or 5).

Let me emphasize that these results are very difficult and tricky to compare. The properties are inherently discrete (either perfect or imperfect, planar or not planar). The fact that the world has not yet agreed upon a universal measure of complexity for graphs (or at least one that makes graph coloring easy to understand) is not a criticism of the chef but a testament to the challenge and intrigue of the dish.

Coloring general graphs is much bleaker, where the focus has turned to approximations. You can’t “approximate” the answer to whether a graph is colorable, so now the key here is that we are actually trying to find an approximate coloring. In particular, if you’re given some graph G and you don’t know the minimum number of colors needed to color it (say it’s \chi(G), this is called the chromatic number), can you easily color it with what turns out to be, say, 2 \chi(G) colors?

Garey and Johnson (the gods of NP-hardness) proved this problem is again hard. In fact, they proved that you can’t do better than twice the number of colors. This might not seem so bad in practice, but the story gets worse. This lower bound was improved by Zuckerman, building on the work of Håstad, to depend on the size of the graph! That is, unless P=NP, all efficient algorithms will use asymptotically more than \chi(G) n^{1 - \varepsilon} colors for any \varepsilon > 0 in the worst case, where n is the number of vertices of G. So the best you can hope for is being off by something like a multiplicative factor of n / \log n. You can actually achieve this (it’s nontrivial and takes a lot of work), but it carries that aura of pity for the hopeful graph colorer.

The next avenue is to assume you know the chromatic number of your graph, and see how well you can do then. For example: if you are given the promise that a graph G is 3-colorable, can you efficiently find a coloring with 8 colors? The best would be if you could find a coloring with 4 colors, but this is already known to be NP-hard.

The best upper bounds, algorithms to find approximate colorings of 3-colorable graphs, also pitifully depend on the size of the graph. Remember I say pitiful not to insult the researchers! This decades-long line of work was extremely difficult and deserves the highest praise. It’s just frustrating that the best known algorithm to color a 3-colorable graph requires up to n^{0.2} colors. At least it bypasses the barrier of n^{1 - \varepsilon} mentioned above, so we know that knowing the chromatic number actually does help.

The lower bounds are a bit more hopeful; it’s known to be NP-hard to color a k-colorable graph using 2^{\sqrt[3]{k}} colors if k is sufficiently large. There are a handful of other linear lower bounds that work for all k \geq 3, but to my knowledge this is the best asymptotic result. The big open problem (which I doubt many people have their eye on considering how hard it seems) is to find an upper bound depending only on k. I wonder offhand whether a ridiculous bound like k^{k^k} colors would be considered progress, and I bet it would.

Our Idea: Resilience

So without big breakthroughs on the front of approximate graph coloring, we propose a new front for investigation. The idea is that we consider graphs which are not only colorable, but remain colorable under the adversarial operation of adding a few new edges. More formally,

Definition: A graph G = (V,E) is called r-resiliently k-colorable if two properties hold

  1. G is k-colorable.
  2. For any set E' of r edges disjoint from E, the graph G' = (V, E \cup E') is k-colorable.

The simplest nontrivial example of this is 1-resiliently 3-colorable graphs. That is a graph that is 3-colorable and remains 3-colorable no matter which new edge you add. And the question we ask of this example: is there a polynomial time algorithm to 3-color a 1-resiliently 3-colorable graph? We prove in our paper that this is actually NP-hard, but it’s not a trivial thing to see.

The chief benefit of thinking about resiliently colorable graphs is that it provides a clear gradient of complexity from general graphs (zero-resilient) to the empty graph (which is (\binom{k+1}{2} - 1)-resiliently k-colorable). We know that the most complex case is NP-hard, and maximally resilient graphs are trivially colorable. So finding the boundary where resilience makes things easy can shed new light on graph coloring.

Indeed, we argue in the paper that lots of important graphs have stronger resilience properties than one might expect. For example, here are the resilience properties of some famous graphs.

From left to right: the Petersen graph, 2-resiliently 3-colorable; the Dürer graph, 4-resiliently 4-colorable; the Grötzsch graph, 4-resiliently 4-colorable; and the Chvátal graph, 3-resiliently 4-colorable. These are all maximally resilient (no graph is more resilient than stated) and chromatic (no graph is colorable with fewer colors)

From left to right: the Petersen graph, 2-resiliently 3-colorable; the Dürer graph, 4-resiliently 4-colorable; the Grötzsch graph, 4-resiliently 4-colorable; and the Chvátal graph, 3-resiliently 4-colorable. These are all maximally resilient (no graph is more resilient than stated) and chromatic (no graph is colorable with fewer colors)

If I were of a mind to do applied graph theory, I would love to know about the resilience properties of graphs that occur in the wild. For example, the reader probably knows the problem of register allocation is a natural graph coloring problem. I would love to know the resilience properties of such graphs, with the dream that they might be resilient enough on average to admit efficient coloring algorithms.

Unfortunately the only way that I know how to compute resilience properties is via brute-force search, and of course this only works for small graphs and small k. If readers are interested I could post such a program (I wrote it in vanilla python), but for now I’ll just post a table I computed on the proportion of small graphs that have various levels of resilience (note this includes graphs that vacuously satisfy the definition).

Percentage of k-colorable graphs on 6 vertices which are n-resilient
k\n       1       2       3       4
  ----------------------------------------
3       58.0    22.7     5.9     1.7
4       93.3    79.3    58.0    35.3
5       99.4    98.1    94.8    89.0
6      100.0   100.0   100.0   100.0

Percentage of k-colorable graphs on 7 vertices which are n-resilient
k\n       1       2       3       4
  ----------------------------------------
3       38.1     8.2     1.2     0.3
4       86.7    62.6    35.0    14.9
5       98.7    95.6    88.5    76.2
6       99.9    99.7    99.2    98.3

Percentage of k-colorable graphs on 8 vertices which are n-resilient
k\n       1       2       3       4
  ----------------------------------------
3       21.3     2.1     0.2     0.0
4       77.6    44.2    17.0     4.5

The idea is this: if this trend continues, that only some small fraction of all 3-colorable graphs are, say, 2-resiliently 3-colorable graphs, then it should be easy to color them. Why? Because resilience imposes structure on the graphs, and that structure can hopefully be realized in a way that allows us to color easily. We don’t know how to characterize that structure yet, but we can give some structural implications for sufficiently resilient graphs.

For example, a 7-resiliently 5-colorable graph can’t have any subgraphs on 6 vertices with \binom{6}{2} - 7 edges, or else we can add enough edges to get a 6-clique which isn’t 5-colorable. This gives an obvious general property about the sizes of subgraphs in resilient graphs, but as a more concrete instance let’s think about 2-resilient 3-colorable graphs G. This property says that no set of 4 vertices may have more than 4 = \binom{4}{2} - 2 edges in G. This rules out 4-cycles and non-isolated triangles, but is it enough to make 3-coloring easy? We can say that G is a triangle-free graph and a bunch of disjoint triangles, but it’s known 3-colorable non-planar triangle-free graphs can have arbitrarily large chromatic number, and so the coloring problem is hard. Moreover, 2-resilience isn’t enough to make G planar. It’s not hard to construct a non-planar counterexample, but proving it’s 2-resilient is a tedious task I relegated to my computer.

Speaking of which, the problem of how to determine whether a k-colorable graph is r-resiliently k-colorable is open. Is this problem even in NP? It certainly seems not to be, but if it had a nice characterization or even stronger necessary conditions than above, we might be able to use them to find efficient coloring algorithms.

In our paper we begin to fill in a table whose completion would characterize the NP-hardness of coloring resilient graphs

table

The known complexity of k-coloring r-resiliently k-colorable graphs

Ignoring the technical notion of 2-to-1 hardness (it’s technical), the paper accomplishes this as follows. First, we prove some relationships between cells. In particular, if a cell is NP-hard then so are all the cells to the left and below it. So our Theorem 1, that 3-coloring 1-resiliently 3-colorable graphs is NP-hard, gives us the entire black region, though more trivial arguments give all except the (3,1) cell. Also, if a cell is in P (it’s easy to k-color graphs with that resilience), then so are all cells above and to its right. We prove that k-coloring \binom{k}{2}-resiliently k-colorable graphs is easy. This is trivial: no vertex may have degree greater than k-1, and the greedy algorithm can color such graphs with k colors. So that gives us the entire light gray region.

There is one additional lower bound comes from the fact that it’s NP-hard to 2^{\sqrt[3]{k}}-color a k-colorable graph. In particular, we prove that if you have any function f(k) that makes it NP-hard to f(k)-color a k-colorable graph, then it is NP-hard to f(k)-color an (f(k) - k)-resiliently f(k)-colorable graph. The exponential lower bound hence gives us a nice linear lower bound, and so we have the following “sufficiently zoomed out” picture of the table

zoomed-out

The zoomed out version of the classification table above.

The paper contains the details of how these observations are proved, in addition to the NP-hardness proof for 1-resiliently 3-colorable graphs. This leaves the following open problems:

  • Get an unconditional, concrete linear resilience lower bound for hardness.
  • Find an algorithm that colors graphs that are less resilient than O(k^2). Even determining specific cells like (4,5) or (5,9) would likely give enough insight for this.
  • Classify the tantalizing (3,2) cell (determine if it’s hard or easy to 3-color a 2-resiliently 3-colorable graph) or even better the (4,2) cell.
  • Find a way to relate resilient coloring back to general coloring. For example, if such and such cell is hard, then you can’t approximate k-coloring to within so many colors.

But Wait, There’s More!

Though this paper focuses on graph coloring, our idea of resilience doesn’t stop there (and this is one reason I like it so much!). One can imagine a notion of resilience for almost any combinatorial problem. If you’re trying to satisfy boolean formulas, you can define resilience to mean that you fix the truth value of some variable (we do this in the paper to build up to our main NP-hardness result of 3-coloring 1-resiliently 3-colorable graphs). You can define resilient set cover to allow the removal of some sets. And any other sort of graph-based problem (Traveling salesman, max cut, etc) can be resiliencified by adding or removing edges, whichever makes the problem more constrained.

So this resilience notion is quite general, though it’s hard to define precisely in a general fashion. There is a general framework called Constraint Satisfaction Problems (CSPs), but resilience here seem too general. A CSP is literally just a bunch of objects which can be assigned some set of values, and a set of constraints (k-ary 0-1-valued functions) that need to all be true for the problem to succeed. If we were to define resilience by “adding any constraint” to a given CSP, then there’s nothing to stop us from adding the negation of an existing constraint (or even the tautologically unsatisfiable constraint!). This kind of resilience would be a vacuous definition, and even if we try to rule out these edge cases, I can imagine plenty of weird things that might happen in their stead. That doesn’t mean there isn’t a nice way to generalize resilience to CSPs, but it would probably involve some sort of “constraint class” of acceptable constraints, and I don’t know a reasonable property to impose on the constraint class to make things work.

So there’s lots of room for future work here. It’s exciting to think where it will take me.

Until then!