# 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 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.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:
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.

# Want to make a great puzzle game? Get inspired by theoretical computer science.

Two years ago, Erik Demaine and three other researchers published a fun paper to the arXiv proving that most incarnations of classic nintendo games are NP-hard. This includes almost every Super Mario Brothers, Donkey Kong, and Pokemon title. Back then I wrote a blog post summarizing the technical aspects of their work, and even gave a talk on it to a room full of curious undergraduate math majors.

But while bad tech-writers tend to interpret NP-hard as “really really hard,” the truth is more complicated. It’s really a statement about computational complexity, which has a precise mathematical formulation. Sparing the reader any technical details, here’s what NP-hard implies for practical purposes:

You should abandon hope of designing an algorithm that can solve any instance of your NP-hard problem, but many NP-hard problems have efficient practical “good-enough” solutions.

The very definition of NP-hard means that NP-hard problems need only be hard in the worst case. For illustration, the fact that Pokemon is NP-hard boils down to whether you can navigate a vastly complicated maze of trainers, some of whom are guaranteed to defeat you. It has little to do with the difficulty of the game Pokemon itself, and everything to do with whether you can stretch some subset of the game’s rules to create a really bad worst-case scenario.

So NP-hardness has very little to do with human playability, and it turns out that in practice there are plenty of good algorithms for winning at Super Mario Brothers. They work really well at beating levels designed for humans to play, but we are highly confident that they would fail to win in the worst-case levels we can cook up. Why don’t we know it for a fact? Well that’s the $P \ne NP$ conjecture.

Since Demaine’s paper (and for a while before it) a lot of popular games have been inspected under the computational complexity lens. Recently, Candy Crush Saga was proven to be NP-hard, but the list doesn’t stop with bad mobile apps. This paper of Viglietta shows that Pac-man, Tron, Doom, Starcraft, and many other famous games all contain NP-hard rule-sets. Games like Tetris are even known to have strong hardness-of-approximation bounds. Many board games have also been studied under this lens, when you generalize them to an $n \times n$ sized board. Chess and checkers are both what’s called EXP-complete. A simplified version of Go fits into a category called PSPACE-complete, but with the general ruleset it’s believed to be EXP-complete [1]. Here’s a list of some more classic games and their complexity status.

So we have this weird contrast: lots of NP-hard (and worse!) games have efficient algorithms that play them very well (checkers is “solved,” for example), but in the worst case we believe there is no efficient algorithm that will play these games perfectly. We could ask, “We can still write algorithms to play these games well, so what’s the point of studying their computational complexity?”

I agree with the implication behind the question: it really is just pointless fun. The mathematics involved is the very kind of nuanced manipulations that hackers enjoy: using the rules of a game to craft bizarre gadgets which, if the player is to surpass them, they must implicitly solve some mathematical problem which is already known to be hard.

But we could also turn the question right back around. Since all of these great games have really hard computational hardness properties, could we use theoretical computer science, and to a broader extent mathematics, to design great games? I claim the answer is yes.

[1] EXP is the class of problems solvable in exponential time (where the exponent is the size of the problem instance, say $n$ for a game played on an $n \times n$ board), so we’re saying that a perfect Chess or Checkers solver could be used to solve any problem that can be solved in exponential time. PSPACE is strictly smaller (we think; this is open): it’s the class of all problems solvable if you are allowed as much time as you want, but only a polynomial amount of space to write down your computations.

## A Case Study: Greedy Spiders

Greedy spiders is a game designed by the game design company Blyts. In it, you’re tasked with protecting a set of helplessly trapped flies from the jaws of a hungry spider.

A screenshot from Greedy Spiders. Click to enlarge.

In the game the spider always moves in discrete amounts (between the intersections of the strands of spiderweb) toward the closest fly. The main tool you have at your disposal is the ability to destroy a strand of the web, thus prohibiting the spider from using it. The game proceeds in rounds: you cut one strand, the spider picks a move, you cut another, the spider moves, and so on until the flies are no longer reachable or the spider devours a victim.

Aside from being totally fun, this game is obviously mathematical. For the reader who is familiar with graph theory, there’s a nice formalization of this problem.

The Greedy Spiders Problem: You are given a graph $G_0 = (V, E_0)$ and two sets $S_0, F \subset V$ denoting the locations of the spiders and flies, respectively. There is a fixed algorithm $A$ that the spiders use to move. An instance of the game proceeds in rounds, and at the beginning of each round we call the current graph $G_i = (V, E_i)$ and the current location of the spiders $S_i$. Each round has two steps:

1. You pick an edge $e \in E_i$ to delete, forming the new graph $G_{i+1} = (V, E_i)$.
2. The spiders jointly compute their next move according to $A$, and each spider moves to an adjacent vertex. Thus $S_i$ becomes $S_{i+1}$.

Your task is to decide whether there is a sequence of edge deletions which keeps $S_t$ and $F$ disjoint for all $t \geq 0$. In other words, we want to find a sequence of edge deletions that disconnects the part of the graph containing the spiders from the part of the graph containing the flies.

This is a slightly generalized version of Greedy Spiders proper, but there are some interesting things to note. Perhaps the most obvious question is about the algorithm $A$. Depending on your tastes you could make it adversarial, devising the smartest possible move at every step of the way. This is just as hard as asking if there is any algorithm that the spiders can use to win. To make it easier, $A$ could be an algorithm represented by a small circuit to which the player has access, or, as it truly is in the Greedy Spiders game, it could be the greedy algorithm (and the player can exploit this).

Though I haven’t heard of the Greedy Spiders problem in the literature by any other name, it seems quite likely that it would arise naturally. One can imagine the spiders as enemies traversing a network (a city, or a virus in a computer network), and your job is to hinder their movement toward high-value targets. Perhaps people in the defense industry could use a reasonable approximation algorithm for this problem. I have little doubt that this game is NP-hard [2], but the purpose of this article is not to prove new complexity results. The point is that this natural theoretical problem is a really fun game to play! And the game designer’s job is to do what game designers love to do: add features and design levels that are fun to play.

Indeed the Greedy Spiders folks did just that: their game features some 70-odd levels, many with multiple spiders and additional tools for the player. Some examples of new tools are: the ability to delete a vertex of the graph and the ability to place a ‘decoy-fly’ which is (to the greedy-algorithm-following spiders) indistinguishable from a real fly. They player is usually given only one or two uses of these tools per level, but one can imagine that the puzzles become a lot richer.

[2]: In the adversarial case it smells like it’s PSPACE-complete, being very close to known PSPACE-hard problems like Cops and Robbers and Generalized Geography

## Examples

I can point to a number of interesting problems that I can imagine turning into successful games, and I will in a moment, but before I want to make it clear that I don’t propose game developers study theoretical computer science just to turn our problems into games verbatim. No, I imagine that the wealth of problems in computer science can serve as inspiration, as a spring board into a world of interesting gameplay mechanics and puzzles. The bonus for game designers is that adding features usually makes problems harder and more interesting, and you don’t need to know anything about proofs or the details of the reductions to understand the problems themselves (you just need familiarity with the basic objects of consideration, sets, graphs, etc).

For a tangential motivation, I imagine that students would be much more willing to do math problems if they were based on ideas coming from really fun games. Indeed, people have even turned the stunningly boring chore of drawing an accurate graph of a function into a game that kids seem to enjoy. I could further imagine a game that teaches programming by first having a student play a game (based on a hard computational problem) and then write simple programs that seek to do well. Continuing with the spiders example they could play as the defender, and then switch to the role of the spider by writing the algorithm the spiders follow.

But enough rambling! Here is a short list of theoretical computer science problems for which I see game potential. None of them have, to my knowledge, been turned into games, but the common features among them all are the huge potential for creative extensions and interesting level design.

### Graph Coloring

Graph coloring is one of the oldest NP-complete problems known. Given a graph $G$ and a set of colors $\{ 1, 2, \dots, k \}$, one seeks to choose colors for the vertices of $G$ so that no edge connects two vertices of the same color.

Now coloring a given graph would be a lame game, so let’s spice it up. Instead of one player trying to color a graph, have two players. They’re given a $k$-colorable graph (say, $k$ is 3), and they take turns coloring the vertices. The first player’s goal is to arrive at a correct coloring, while the second player tries to force the first player to violate the coloring condition (that no adjacent vertices are the same color). No player is allowed to break the coloring if they have an option. Now change the colors to jewels or vegetables or something, and you have yourself an award-winning game! (Or maybe: Epic Cartographer Battles of History)

An additional modification: give the two players a graph that can’t be colored with $k$ colors, and the first player to color a monochromatic edge is the loser. Add additional move types (contracting edges or deleting vertices, etc) to taste.

### Art Gallery Problem

Given a layout of a museum, the art gallery problem is the problem of choosing the minimal number of cameras so as to cover the whole museum.

This is a classic problem in computational geometry, and is well-known to be NP-hard. In some variants (like the one pictured above) the cameras are restricted to being placed at corners. Again, this is the kind of game that would be fun with multiple players. Rather than have perfect 360-degree cameras, you could have an angular slice of vision per camera. Then one player chooses where to place the cameras (getting exponentially more points for using fewer cameras), and the opponent must traverse from one part of the museum to the other avoiding the cameras. Make the thief a chubby pig stealing eggs from birds and you have yourself a franchise.

For more spice, allow the thief some special tactics like breaking through walls and the ability to disable a single camera.

This idea has of course been the basis of many single-player stealth games (where the guards/cameras are fixed by the level designer), but I haven’t seen it done as a multiplayer game. This also brings to mind variants like the recent Nothing to Hide, which counterintuitively pits you as both the camera placer and the hero: you have to place cameras in such a way that you’re always in vision as you move about to solve puzzles. Needless to say, this fruit still has plenty of juice for the squeezing.

### Pancake Sorting

Pancake sorting is the problem of sorting a list of integers into ascending order by using only the operation of a “pancake flip.”

Just like it sounds, a pancake flip involves choosing an index in the list and flipping the prefix of the list (or suffix, depending on your orientation) like a spatula flips a stack of pancakes. Now I think sorting integers is boring (and it’s not NP-hard!), but when you forget about numbers and that one special configuration (ascending sorted order), things get more interesting. Instead, have the pancakes be letters and have the goal be to use pancake flips to arrive at a real English word. That is, you don’t know the goal word ahead of time, so it’s the anagram problem plus finding an efficient pancake flip to get there. Have a player’s score be based on the number of flips before a word is found, and make it timed to add extra pressure, and you have yourself a classic!

The level design then becomes finding good word scrambles with multiple reasonable paths one could follow to get valid words. My mother would probably play this game!

### Bin Packing

Young Mikio is making sushi for his family! He’s got a table full of ingredients of various sizes, but there is a limit to how much he can fit into each roll. His family members have different tastes, and so his goal is to make everyone as happy as possible with his culinary skills and the options available to him.

Another name for this problem is bin packing. There are a collection of indivisible objects of various sizes and values, and a set of bins to pack them in. Your goal is to find the packing that doesn’t exceed the maximum in any bin and maximizes the total value of the packed goods.

I thought of sushi because I recently played a ridiculously cute game about sushi (thanks to my awesome friend Yen over at Baking And Math), but I can imagine other themes that suggest natural modifications of the problem. The objects being packed could be two-dimensional, there could be bonuses for satisfying certain family members (or penalties for not doing so!), or there could be a super knife that is able to divide one object in half.

I could continue this list for quite a while, but perhaps I should keep my best ideas to myself in case any game companies want to hire me as a consultant. :)

Do you know of games that are based on any of these ideas? Do you have ideas for features or variations of the game ideas listed above? Do you have other ideas for how to turn computational problems into games? I’d love to hear about it in the comments.

Until next time!

# 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)

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

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

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!

# Lagrangians for the Amnesiac

For a while I’ve been meaning to do some more advanced posts on optimization problems of all flavors. One technique that comes up over and over again is Lagrange multipliers, so this post is going to be a leisurely reminder of that technique. I often forget how to do these basic calculus-type things, so it’s good practice.

We will assume something about the reader’s knowledge, but it’s a short list: know how to operate with vectors and the dot product, know how to take a partial derivative, and know that in single-variable calculus the local maxima and minima of a differentiable function $f(x)$ occur when the derivative $f'(x)$ vanishes. All of the functions we’ll work with in this post will have infinitely many derivatives (i.e. smooth). So things will be nice.

The gradient of a multivariable function is the natural extension of the derivative of a single-variable function. If $f(x_1, \dots, x_n)$ is a differentiable function, the data of the gradient of $f$ consists of all of the partial derivatives $\partial f / \partial x_i$. It’s usually written as a vector

$\displaystyle \nabla f = \left ( \frac{\partial f}{\partial x_1}, \dots, \frac{\partial f}{\partial x_n} \right )$

To make things easier for ourselves, we’ll just call $f$ a function $f(x)$ and understand $x$ to be a vector in $\mathbb{R}^n$.

We can also think of $\nabla f$ as a function which takes in vectors and spits out vectors, by plugging in the input vector into each $\partial f / \partial x_i$. And the reason we do this is because it lets us describe the derivative of $f$ at a point as a linear map based on the gradient. That is, if we want to know how fast $f$ is growing along a particular vector $v$ and at a particular point $(x, f(x))$, we can just take a dot product of $v$ with $\nabla f(x)$. I like to call dot products inner products, and use the notation $\left \langle \nabla f(x), v \right \rangle$. Here $v$ is a vector in $\mathbb{R}^n$ which we think of as “tangent vectors” to the surface defined by $f$. And if we scale $v$ bigger or smaller, the value of the derivative scales with it (of course, because the derivative is a linear map!). Usually we use unit vectors to represent directions, but there’s no reason we have to. Calculus textbooks often require this to define a “directional derivative,” but perhaps it is better to understand the linear algebra over memorizing these arbitrary choices.

For example, let $f(x,y,z) = xyz$. Then $\nabla f = (yz, xz, xy)$, and $\nabla f(1,2,1) = (2, 1, 2)$. Now if we pick a vector to go along, say, $v = (0,-1,1)$, we get the derivative of $f$ along $v$ is $\left \langle (2,1,2), (0,-1,1) \right \rangle = 1$.

As importantly as computing derivatives is finding where the derivative is zero, and the geometry of the gradient can help us here. Specifically, if we think of our function $f$ as a surface sitting in $\mathbb{R}^{n+1}$ (as in the picture below), it’s not hard to see that the gradient vector points in the direction of steepest ascent of $f$. How do we know this? Well if you fix a point $(x_1, \dots, x_n)$ and you’re forced to use a vector $v$ of the same magnitude as $\nabla f(x)$, how can you maximize the inner product $\left \langle \nabla f(x), v \right \rangle$? Well, you just pick $v$ to be equal to $\nabla f(x)$, of course! This will turn the dot product into the square norm of $\nabla f(x)$.

The gradient points in the direction of steepest ascent. (image source)

More generally, the operation of an inner product $\left \langle -, v \right \rangle$ is geometrically the size of the projection of the argument onto $v$ (scaled by the size of $v$), and projections of a vector $w$ onto different directions than $w$ can only be smaller in magnitude than $w$. Another way to see this is to know the “alternative” formula for the dot product

$\displaystyle \left \langle v,w \right \rangle = \left \| v \right \| \left \| w \right \| \cos(\theta)$

where $\theta$ is the angle between the vectors (in $\mathbb{R}^n$). We might not know how to get that angle, and in this post we don’t care, but we do know that $\cos(\theta)$ is between -1 and 1. And so if $v$ is fixed and we can’t change the norm of $w$ but only its direction, we will maximize the dot product when the two vectors point in the same direction, when $\theta$ is zero.

All of this is just to say that the gradient at a point can be interpreted as having a specific direction. It’s the direction of steepest ascent of the surface $f$, and it’s size tells you how steep $f$ is at that point. The opposite direction is the direction of steepest descent, and the orthogonal directions (when $\theta = \pi /2$) have derivative zero.

Now what happens if we’re at a local minimum or maximum? Well it’s necessary that $f$ is flat, and so by our discussion above the derivatives in all directions must be zero. It’s a basic linear algebra proof to show that this means the gradient is the zero vector. You can prove this by asking what sorts of vectors $w$ have a dot product of zero with all other vectors $v$?

Now once we have a local max or a local min, how do we tell which? The answer is actually a bit complicated, and it requires you to inspect the eigenvalues of the Hessian of $f$. We won’t dally on eigenvalues except to explain the idea in brief: for an $n$ variable function $f$ the Hessian of $f$ at $x$ is an $n$-by-$n$ matrix where the $i,j$ entry is the value of $(\partial f / \partial x_i \partial x_j )(x)$. It just so turns out that if this matrix has only positive eigenvalues, then $x$ is a local minimum. If the eigenvalues are all negative, it’s a local max. If some are negative and some are positive, then it’s a saddle point. And if zero is an eigenvalue then we’re screwed and can’t conclude anything without more work.

But all of this Hessian business isn’t particularly important for us, because most of our applications of the Lagrangian will work with functions where we already know that there is a unique global maximum or minimum. Finding where the gradient is zero is enough. As much as this author stresses the importance of linear algebra, we simply won’t need to compute any eigenvalues for this one.

What we will need to do is look at optimizing functions which are constrained by some equality conditions. This is where Lagrangians come into play.

## Constrained by Equality

Often times we will want to find a minimum or maximum of a function $f(x)$, but we will have additional constraints. The simplest kind is an equality constraint.

For example, we might want to find the maximum of the function $f(x, y, z) = xyz$ requiring that the point $(x,y,z)$ lies on the unit circle. One could write this in a “canonical form”

maximize $xyz$
subject to $x^2 + y^2 + z^2 = 1$

Way back in the scientific revolution, Fermat discovered a technique to solve such problems that was later generalized by Lagrange. The idea is to combine these constraints into one function whose gradient provides enough information to find a maximum. Clearly such information needs to include two things: that the gradient of $xyz$ is zero, and that the constraint is satisfied.

First we rewrite the constraints as $g(x,y,z) = x^2 + y^2 + x^2 - 1 = 0$, because when we’re dealing with gradients we want things to be zero. Then we form the Lagrangian of the problem. We’ll give a precise definition in a minute, but it looks like this:

$L(x,y,z,\lambda) = xyz + \lambda(x^2 + y^2 + z^2 - 1)$

That is, we’ve added a new variable $\lambda$ and added the two functions together. Let’s see what happens when we take a gradient:

$\displaystyle \frac{\partial L}{\partial x} = yz + \lambda 2x$

$\displaystyle \frac{\partial L}{\partial y} = xz + \lambda 2y$

$\displaystyle \frac{\partial L}{\partial z} = xy + \lambda 2z$

$\displaystyle \frac{\partial L}{\partial \lambda} = x^2 + y^2 + z^2 - 1$

Now if we require the gradient to be zero, the last equation is simply the original constraint, and the first three equations say that $\nabla f (x,y,z) = \lambda \nabla g (x,y,z)$. In other words, we’re saying that the two gradients must point in the same direction for the function to provide a maximum. Solving for where these equations vanish gives some trivial solutions (one variable is $\pm 1$ and the rest zero, and $\lambda = 0$), and a solution defined by $x^2 = y^2 = z^2 = 1/3$ which is clearly the maximal of the choices.

Indeed, this will work in general, and you can see a geometric and analytic proof in these notes.

Specifically, if we have an optimization problem defined by an objective function $f(x)$ to optimize, and a set of $k$ equality constraints $g_i(x) = 0$, then we can form the Lagrangian

$\displaystyle L(x, \lambda_1, \dots, \lambda_k) = f(x) + \sum_{i=1}^k \lambda_i g_i(x)$

And then a theorem of Lagrange is that all optimal solutions $x^*$ to the problem satisfy $\nabla L(x^*, \lambda_1, \dots, \lambda_k) = 0$ for some choice of $\lambda _i$. But then you have to go solve the system and figure out which of the solutions gives you your optimum.

## Convexity

As it turns out, there are some additional constraints you can add to your problem to guarantee your system has a solution. One nice condition is that $f(x)$ is convexA function is convex if any point on a line segment between two points $(x,f(x))$ and $(y,f(y))$ has a value greater than $f$. In other words, for all $0 \leq t \leq 1$:

$\displaystyle f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)$

Some important examples of convex functions: exponentials, quadratics whose leading coefficient is positive, square norms of a vector variable, and linear functions.

Convex functions have this nice property that they have a unique local minimum value, and hence it must also be the global minimum. Why is this? Well if you have a local minimum $x$, and any other point $y$, then by virtue of being a local minimum there is some $t$ sufficiently close to 1 so that:

$\displaystyle f(x) \leq f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)$

And rearranging we get

$\displaystyle (1-t)f(x) \leq (1-t)f(y)$

So $f(x) \leq f(y)$, and since $y$ was arbitrary then $x$ is the global minimum.

This alleviates our problem of having to sort through multiple solutions, and in particular it helps us to write programs to solve optimization problems: we know that techniques like gradient descent will never converge to a false local minimum.

That’s all for now! The next question we might shadowily ask: what happens if we add inequality constraints?