# NP-hard does not mean hard

When NP-hardness pops up on the internet, say because some silly blogger wants to write about video games, it’s often tempting to conclude that the problem being proved NP-hard is actually very hard!

“Scientists proved Super Mario is NP-hard? I always knew there was a reason I wasn’t very good at it!” Sorry, these two are unrelated. NP-hardness means hard in a narrow sense this post should hopefully make clear. After that, we’ll explore what “hard” means in a mathematical sense that you can apply beyond NP-hardness to inform your work as a programmer.

When a problem is NP-hard, that simply means that the problem is sufficiently expressive that you can use the problem to express logic. By which I mean boolean formulas using AND, OR, and NOT. In the Super Mario example, the “problem” is a bundle of (1) the controls for the player (2) the allowed tiles and characters that make up a level, and (3) the goal of getting from the start to the end. Logic formulas are encoded in the creation of a level, and solving the problem (completing the level) is the same as finding conditions to make the logical formula true.

The clause gadget for the original Super Mario Brothers, encoding an OR of three variables.

In this sense, NP-hardness doesn’t make all of Super Mario hard. The levels designed to encode logical formulas are contrived, convoluted, and contorted. They abuse the rules of the game in order to cram boolean logic into it. These are worst case levels. It’s using Mario for a completely unintended purpose, not unlike hacking. And so NP-hardness is a worst case claim.

To reiterate, NP-hardness means that Super Mario has expressive power. So expressive that it can emulate other problems we believe are hard in the worst case. And, because the goal of mathematical “hardness” is to reason about the limitations of algorithms, being able to solve Super Mario in full generality implies you can solve any hard subproblem, no matter how ridiculous the level design.

The P != NP conjecture says that there’s no polynomial time algorithm to determine whether boolean logic formulas are satisfiable, and so as a consequence Super Mario (in full generality) also has no polynomial time algorithm.

That being said, in reality Super Mario levels do not encode logical formulas! If you use the knowledge that real-world Super Mario levels are designed in the way they are (to be solvable, fun), then you can solve Super Mario with algorithms. There are many examples.

In general, the difficulty of a problem for humans is unrelated to the difficulty for algorithms. Consider multiplication of integers. This is a trivial problem for computers to solve, but humans tend to struggle with it. It’s an amazing feat to be able to multiply two 7 digit numbers in less than 5 seconds, whereas computers can multiply two thousand-digit numbers in milliseconds.

Meanwhile, protein folding is known to be an NP-hard problem, but it’s been turned into a game sufficiently easy for humans to solve that players have contributed to scientific research. Indeed, even some of the most typically cited NP-hard problems, like traveling salesman, have heuristic, practical algorithmic solutions that allow one to solve them (very close to optimally) in hours on inputs as large as every city on earth.

So the mathematical notions of hardness are quite disconnected from practical notions of hardness. This is not even to mention that some NP-hard problems can be efficiently approximated to within any desired accuracy.

Let’s dig into the math a bit more. “Hardness” is a family of ideas about comparisons between problems based on reusability of algorithmic solutions. Loosely speaking, a problem $R$ is hard with respect to a class of problems $C$ if an algorithm solving $R$ can be easily transformed into an algorithm solving any problem in $C$. You have to say what kinds of transformations are allowed, and the transformation can be different for different target problems in $C$, but that’s the basic idea.

In the Super Mario example, if you want to solve logical formulas, you can transform a hypothetically perfect mario-level-playing algorithm into a logic solver by encoding the formula as a level and running the mario-level-playing algorithm on it as a black box. Add an if statement to the end to translate “level can/can’t be finished” to “formula can/can’t be satisfied,” and the transformation is complete. It’s important for NP-hardness that the transformation only takes polynomial time. Other kinds of hardness might admit more or restrict to fewer resources.

And so this is what makes Mario NP-hard, because boolean logic satisfiability is NP-hard. Any problem in NP can be solved by a boolean logic solver, and hence also by a mario-level-player. The fact that boolean logic solving is NP-hard is a difficult theorem to prove. But if we assume it’s true, you can compose the transformations to get from any NP problem to Super Mario.

As a simple example of a different kind of hardness, you can let $C$ be the class of problems solvable using only a finite amount of memory (independent of the input). You have probably heard of this class of problems by another name, but I’ll keep you guessing until the end of the post. A $C$-hard problem $R$ is one for which an algorithmic solution can be repurposed to solve any finite-memory-solvable problem.

We have to be careful: if the transformation between solutions allows us polynomial time (in the size of the input) like it did for NP-hardness, then we might have enough time in the transformation alone to solve the entire problem, removing the need for a solution to $R$ in the first place! For this reason, we have to limit the amount of work that can be done in the transformation. We get a choice here that influences how interesting or useful the definition of hardness is, but let’s just pick one and say that the transformation can only use finite time (independent of the input).

To be fair, I actually don’t know if there are any hard problems with respect to this definition. There probably are, but chances are good that they are not members of $C$, and that’s where the definition of hardness gets really interesting. If you have a problem in $C$ which is also $C$-hard, it’s called complete for $C$. And once you’ve found a complete problem, from a theoretical perspective you’re a winner. You’ve found a problem which epitomizes the difficulty of solving problems in $C$. And so it’s a central aim of researchers studying a complexity class to find complete problems. As they say in the business, “ABC: always be completing.”

As a more concrete and interesting example, the class $P$ of all polynomial-time solvable problems has a complete problem. Here the transformations are a bit up in the air. They could either be logarithmic-space computations, or what’s called NC, which can be thought of as poly-logarithmic time (very fast) parallel computations. I only mention NC because it allows you to say “P-complete problems are hard to parallelize.”

Regardless of the choice, there are a number of very useful problems known to be P-complete. The first is the Circuit Value Problem, given a circuit (described by its gates and wires using any reasonable encoding) and an input to the circuit, what is the output?

Others include linear programming (optimize this linear function with respect to linear constraints), data compression (does the compressed version of a string $s$ using Lempel–Ziv–Welch contain a string $t$?), and type inference for partial types. There are many more in this compendium of Greenlaw et al. Each one is expressive enough to encode any instance of the other, and any instance of any problem in P. It’s quite curious to think that gzip can solve linear programs, but that’s surely no curiouser than super mario levels encoding boolean logic.

Just as with NP-hardness, when a problem is P-hard that doesn’t automatically mean it’s easy or hard for humans, or that typical instances can’t be easily parallelized. P-hardness is also a worst case guarantee.

Studying P-completeness is helpful in the same way NP-completeness is helpful. Completeness informs you about whether you should hope to find a perfect solution or be content with approximations and heuristics (or incorporate problem context to make it easier). Knowing a problem is P-complete means you should not expect perfect efficient parallel algorithms, or perfect efficient algorithms that use severely limited space. Knowing a problem is NP-hard means you should not expect a perfect polynomial time solution. In other words, if you are forced to work with those restrictions, the game becomes one of tradeoffs. Hardness and completeness focus and expedite your work, and clarify a principled decision making process.

Until next time!

P.S. The class of problems solvable in a finite amount of memory is just the class of regular languages. The “finite memory” is the finite state machine used to solve them.

# A problem that is not (properly) PAC-learnable

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).

One of the primary goals of studying models of learning is to figure out what is learnable and what is not learnable in the various models. 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 we presented last time. 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.

Addendum: This post is dishonest in the following sense. The original definition I presented of PAC-learning is not considered the “standard” version, precisely because it forces the learning algorithm to produce hypotheses from the concept class it’s trying to learn. As this post shows, that prohibits us from learning concept classes that should be easy to learn. So to quell any misconceptions, we’re not saying that 3-term DNF formulas (defined below) are not PAC-learnable, just that they’re not PAC-learnable under the definition we gave in the previous post. In other words, we’ve set up a straw man (or, done some good mathematics) in order to illustrate why we need to add the extra bit about hypothesis classes to the definition at the end of this post.

## 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:

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 mostly 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!

# 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!

# How to Conquer Tensorphobia

A professor at Stanford once said,

If you really want to impress your friends and confound your enemies, you can invoke tensor products… People run in terror from the $\otimes$ symbol.

He was explaining some aspects of multidimensional Fourier transforms, but this comment is only half in jest; people get confused by tensor products. It’s often for good reason. People who really understand tensors feel obligated to explain it using abstract language (specifically, universal properties). And the people who explain it in elementary terms don’t really understand tensors.

This post is an attempt to bridge the gap between the elementary and advanced understandings of tensors. We’ll start with the elementary (axiomatic) approach, just to get a good feel for the objects we’re working with and their essential properties. Then we’ll transition to the “universal” mode of thought, with the express purpose of enlightening us as to why the properties are both necessary and natural.

But above all, we intend to be sufficiently friendly so as to not make anybody run in fear. This means lots of examples and preferring words over symbols. Unfortunately, we simply can’t get by without the reader knowing the very basics of linear algebra (the content of our first two primers on linear algebra (1) (2), though the only important part of the second is the definition of an inner product).

So let’s begin.

## Tensors as a Bunch of Axioms

Before we get into the thick of things I should clarify some basic terminology. Tensors are just vectors in a special vector space. We’ll see that such a vector space comes about by combining two smaller vector spaces via a tensor product. So the tensor product is an operation combining vector spaces, and tensors are the elements of the resulting vector space.

Now the use of the word product is quite suggestive, and it may lead one to think that a tensor product is similar or related to the usual direct product of vector spaces. In fact they are related (in very precise sense), but they are far from similar. If you were pressed, however, you could start with the direct product of two vector spaces and take a mathematical machete to it until it’s so disfigured that you have to give it a new name (the tensor product).

With that image in mind let’s see how that is done. For the sake of generality we’ll talk about two arbitrary finite-dimensional vector spaces $V, W$ of dimensions $n, m$. Recall that the direct product  $V \times W$ is the vector space of pairs $(v,w)$ where $v$ comes from $V$ and $w$ from $W$. Recall that addition in this vector space is defined componentwise ($(v_1,w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2$)) and scalar multiplication scales both components $\lambda (v,w) = (\lambda v, \lambda w)$.

To get the tensor product space $V \otimes W$, we make the following modifications. First, we redefine what it means to do scalar multiplication. In this brave new tensor world, scalar multiplication of the whole vector-pair is declared to be the same as scalar multiplication of any component you want. In symbols,

$\displaystyle \lambda (v, w) = (\lambda v, w) = (v, \lambda w)$

for all choices of scalars $\lambda$ and vectors $v, w$. Second, we change the addition operation so that it only works if one of the two components are the same. In symbols, we declare that

$(v, w) + (v', w) = (v + v', w)$

only works because $w$ is the same in both pieces, and with the same rule applying if we switch the positions of $v,w$ above. All other additions are simply declared to be new vectors. I.e. $(x,y) + (z,w)$ is simply itself. It’s a valid addition — we need to be able to add stuff to be a vector space — but you just can’t combine it any further unless you can use the scalar multiplication to factor out some things so that $y=w$ or $x=z$. To say it still one more time, a general element of the tensor $V \otimes W$ is a sum of these pairs that can or can’t be combined by addition (in general things can’t always be combined).

Finally, we rename the pair $(v,w)$ to $v \otimes w$, to distinguish it from the old vector space $V \times W$ that we’ve totally butchered and reanimated, and we call the tensor product space as a whole $V \otimes W$. Those familiar with this kind of abstract algebra will recognize quotient spaces at work here, but we won’t use that language except to note that we cover quotients and free spaces elsewhere on this blog, and that’s the formality we’re ignoring.

As an example, say we’re taking the tensor product of two copies of $\mathbb{R}$. This means that our space $\mathbb{R} \otimes \mathbb{R}$ is comprised of vectors like $3 \otimes 5$, and moreover that the following operations are completely legitimate.

$3 \otimes 5 + 1 \otimes (-5) = 3 \otimes 5 + (-1) \otimes 5 = 2 \otimes 5$

$6 \otimes 1 + 3\pi \otimes \pi = 3 \otimes 2 + 3 \otimes \pi^2 = 3 \otimes (2 + \pi^2)$

Cool. This seemingly innocuous change clearly has huge implications on the structure of the space. We’ll get to specifics about how different tensors are from regular products later in this post, but for now we haven’t even proved this thing is a vector space. It might not be obvious, but if you go and do the formalities and write the thing as a quotient of a free vector space (as we mentioned we wouldn’t do) then you know that quotients of vector spaces are again vector spaces. So we get that one for free. But even without that it should be pretty obvious: we’re essentially just declaring that all the axioms of a vector space hold when we want them to. So if you were wondering whether

$\lambda (a \otimes b + c \otimes d) = \lambda(a \otimes b) + \lambda(c \otimes d)$

The answer is yes, by force of will.

So just to recall, the axioms of a tensor space $V \otimes W$ are

1. The “basic” vectors are $v \otimes w$ for $v \in V, w \in W$, and they’re used to build up all other vectors.
2. Addition is symbolic, unless one of the components is the same in both addends, in which case $(v_1, w) + (v_2, w) = (v_1+ v_2, w)$ and $(v, w_1) + (v,w_2) = (v, w_1 + w_2)$.
3. You can freely move scalar multiples around the components of $v \otimes w$.
4. The rest of the vector space axioms (distributivity, additive inverses, etc) are assumed with extreme prejudice.

Naturally, one can extend this definition to $n$-fold tensor products, like $V_1 \otimes V_2 \otimes \dots \otimes V_d$. Here we write the vectors as sums of things like $v_1 \otimes \dots \otimes v_d$, and we enforce that addition can only be combined if all but one coordinates are the same in the addends, and scalar multiples move around to all coordinates equally freely.

## So where does it come from?!

By now we have this definition and we can play with tensors, but any sane mathematically minded person would protest, “What the hell would cause anyone to come up with such a definition? I thought mathematics was supposed to be elegant!”

It’s an understandable position, but let me now try to convince you that tensor products are very natural. The main intrinsic motivation for the rest of this section will be this:

We have all these interesting mathematical objects, but over the years we have discovered that the maps between objects are the truly interesting things.

A fair warning: although we’ll maintain a gradual pace and informal language in what follows, by the end of this section you’ll be reading more or less mature 20th-century mathematics. It’s quite alright to stop with the elementary understanding (and skip to the last section for some cool notes about computing), but we trust that the intrepid readers will push on.

So with that understanding we turn to multilinear maps. Of course, the first substantive thing we study in linear algebra is the notion of a linear map between vector spaces. That is, a map $f: V \to W$ that factors through addition and scalar multiplication (i.e. $f(v + v') = f(v) + f(v')$ and $f(\lambda v) = \lambda f(v)$).

But it turns out that lots of maps we work with have much stronger properties worth studying. For example, if we think of matrix multiplication as an operation, call it $m$, then $m$ takes in two matrices and spits out their product

$m(A,B) = AB$

Now what would be an appropriate notion of linearity for this map? Certainly it is linear in the first coordinate, because if we fix $B$ then

$m(A+C, B) = (A+C)B = AB + CB = m(A,B) + m(C,B)$

And for the same reason it’s linear in the second coordinate. But it is most definitely not linear in both coordinates simultaneously. In other words,

$m(A+B, C+D) = (A+B)(C+D) = AC + AD + BC + BD \neq AC + BD = m(A,C) + m(B,D)$

In fact, there is only one function that satisfies both “linearity in its two coordinates separately” and also “linearity in both coordinates simultaneously,” and it’s the zero map! (Try to prove this as an exercise.) So the strongest kind of linearity we could reasonably impose is that $m$ is linear in each coordinate when all else is fixed. Note that this property allows us to shift around scalar multiples, too. For example,

$\displaystyle m(\lambda A, B) = \lambda AB = A (\lambda B) = m(A, \lambda B) = \lambda m(A,B)$

Starting to see the wispy strands of a connection to tensors? Good, but hold it in for a bit longer. This single-coordinate-wise-linear property is called bilinearity when we only have two coordinates, and multilinearity when we have more.

Here are some examples of nice multilinear maps that show up everywhere:

• If $V$ is an inner product space over $\mathbb{R}$, then the inner product is bilinear.
• The determinant of a matrix is a multilinear map if we view the columns of the matrix as vector arguments.
• The cross product of vectors in $\mathbb{R}^3$ is bilinear.

There are many other examples, but you should have at least passing familiarity with these notions, and it’s enough to convince us that multilinearity is worth studying abstractly.

And so what tensors do is give a sort of classification of multilinear maps. The idea is that every multilinear map $f$ from a product vector space $U_1 \times \dots \times U_d$ to any vector space $Y$ can be written first as a multilinear map to the tensor space

$\displaystyle \alpha : U_1 \times \dots \times U_d \to U_1 \otimes \dots \otimes U_d$

Followed by a linear map to $Y$,

$\displaystyle \hat{f} : U_1 \otimes \dots \otimes U_d \to Y$

And the important part is that $\alpha$ doesn’t depend on the original $f$ (but $\hat{f}$ does). One usually draws this as a single diagram:

And to say this diagram commutes is to say that all possible ways to get from one point to another are equivalent (the compositions of the corresponding maps you follow are equal, i.e. $f = \hat{f} \alpha$).

In fuzzy words, the tensor product is like the gatekeeper of all multilinear maps, and $\alpha$ is the gate. Yet another way to say this is that $\alpha$ is the most general possible multilinear map that can be constructed from $U_1 \times \dots \times U_d$. Moreover, the tensor product itself is uniquely defined by having a “most-general” $\alpha$ (up to isomorphism). This notion is often referred to by mathematicians as the “universal property” of the tensor product. And they might say something like “the tensor product is initial with respect to multilinear mappings from the standard product.” We discuss language like this in detail in this blog’s series on category theory, but it’s essentially a super-compact (and almost too vague) way of saying what the diagram says.

Let’s explore this definition when we specialize to a tensor of two vector spaces, and it will give us a good understanding of $\alpha$ (which is really incredibly simple, but people like to muck it up with choices of coordinates and summations). So fix $V, W$ as vector spaces and look at the diagram

What is $\alpha$ in this case? Well it just sends $(v,w) \mapsto v \otimes w$. Is this map multilinear? Well if we fix $w$ then

$\displaystyle \alpha(v_1 + v_2, w) = (v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w = \alpha(v_1, w) + \alpha (v_2, w)$

and

$\displaystyle \alpha(\lambda v, w) = (\lambda v) \otimes w = (\lambda) (v \otimes w) = \lambda \alpha(v,w)$

And our familiarity with tensors now tells us that the other side holds too. Actually, rather than say this is a result of our “familiarity with tensors,” the truth is that this is how we know that we need to define the properties of tensors as we did. It’s all because we designed tensors to be the gatekeepers of multilinear maps!

So now let’s prove that all maps $f : V \times W \to Y$ can be decomposed into an $\alpha$ part and a $\hat{f}$ part. To do this we need to know what data uniquely defines a multilinear map. For usual linear maps, all we had to do was define the effect of the map on each element of a basis (the rest was uniquely determined by the linearity property). We know what a basis of $V \times W$ is, it’s just the union of the bases of the pieces. Say that $V$ has a basis $v_1, \dots, v_n$ and $W$ has $w_1, \dots, w_m$, then a basis for the product is just $((v_1, 0), \dots, (v_n,0), (0,w_1), \dots, (0,w_m))$.

But multilinear maps are more nuanced, because they have two arguments. In order to say “what they do on a basis” we really need to know how they act on all possible pairs of basis elements. For how else could we determine $f(v_1 + v_2, w_1)$? If there are $n$ of the $v_i$‘s and $m$ of the $w_i$‘s, then there are $nm$ such pairs $f(v_i, w_j)$.

Uncoincidentally, as $V \otimes W$ is a vector space, its basis can also be constructed in terms of the bases of $V$ and $W$. You simply take all possible tensors $v_i \otimes w_j$. Since every $v \in V, w \in W$ can be written in terms of their bases, it’s clear than any tensor $\sum_{k} a_k \otimes b_k$ can also be written in terms of the basis tensors $v_i \otimes w_j$ (by simply expanding each $a_k, b_k$ in terms of their respective bases, and getting a larger sum of more basic tensors).

Just to drive this point home, if $(e_1, e_2, e_3)$ is a basis for $\mathbb{R}^3$, and $(g_1, g_2)$ a basis for $\mathbb{R}^2$, then the tensor space $\mathbb{R}^3 \otimes \mathbb{R}^2$ has basis

$(e_1 \otimes g_1, e_1 \otimes g_2, e_2 \otimes g_1, e_2 \otimes g_2, e_3 \otimes g_1, e_3 \otimes g_2)$

It’s a theorem that finite-dimensional vector spaces of equal dimension are isomorphic, so the length of this basis (6) tells us that $\mathbb{R}^3 \otimes \mathbb{R}^2 \cong \mathbb{R}^6$.

So fine, back to decomposing $f$. All we have left to do is use the data given by $f$ (the effect on pairs of basis elements) to define $\hat{f} : V \otimes W \to Y$. The definition is rather straightforward, as we have already made the suggestive move of showing that the basis for the tensor space ($v_i \otimes w_j$) and the definition of $f(v_i, w_j)$ are essentially the same.

That is, just take $\hat{f}(v_i \otimes w_j) = f(v_i, w_j)$. Note that this is just defined on the basis elements, and so we extend to all other vectors in the tensor space by imposing linearity (defining $\hat{f}$ to split across sums of tensors as needed). Is this well defined? Well, multilinearity of $f$ forces it to be so. For if we had two equal tensors, say, $\lambda v \otimes w = v \otimes \lambda w$, then we know that $f$ has to respect their equality, because $f(\lambda v_i, w_j) = f(v_i, \lambda w_j)$, so $\hat{f}$ will take the same value on equal tensors regardless of which representative we pick (where we decide to put the $\lambda$). The same idea works for sums, so everything checks out, and $f(v,w)$ is equal to $\hat{f} \alpha$, as desired. Moreover, we didn’t make any choices in constructing $\hat{f}$. If you retrace our steps in the argument, you’ll see that everything was essentially decided for us once we fixed a choice of a basis (by our wise decisions in defining $V \otimes W$). Since the construction would be isomorphic if we changed the basis, our choice of $\hat{f}$ is unique.

There is a lot more to say about tensors, and indeed there are some other useful ways to think about tensors that we’ve completely ignored. But this discussion should make it clear why we define tensors the way we do. Hopefully it eliminates most of the mystery in tensors, although there is still a lot of mystery in trying to compute stuff using tensors. So we’ll wrap up this post with a short discussion about that.

## Computability and Stuff

It should be clear by now that plain product spaces $V \times W$ and tensor product spaces $V \otimes W$ are extremely different. In fact, they’re only related in that their underlying sets of vectors are built from pairs of vectors in $V$ and $W$. Avid readers of this blog will also know that operations involving matrices (like row reduction, eigenvalue computations, etc.) are generally efficient, or at least they run in polynomial time so they’re not crazy impractically slow for modest inputs.

On the other hand, it turns out that almost every question you might want to ask about tensors is difficult to answer computationally. As with the definition of the tensor product, this is no mere coincidence. There is something deep going on with tensors, and it has serious implications regarding quantum computing. More on that in a future post, but for now let’s just focus on one hard problem to answer for tensors.

As you know, the most general way to write an element of a tensor space $U_1 \otimes \dots \otimes U_d$ is as a sum of the basic-looking tensors.

$\sum_k \displaystyle a_{1,k} \otimes a_{2,k} \otimes \dots \otimes a_{d,k}$

where the $a_{i,d}$ are linear combinations of basis vectors in the $U_i$. But as we saw with our examples over $\mathbb{R}$, there can be lots of different ways to write a tensor. If you’re lucky, you can write the entire tensor as a one-term sum, that is just a tensor $a_1 \otimes \dots \otimes a_d$. If you can do this we call the tensor a pure tensor, or a rank 1 tensor. We then have the following natural definition and problem:

Definition: The rank of a tensor $x \in U_1 \otimes \dots \otimes U_d$ is the minimum number of terms in any representation of $x$ as a sum of pure tensors. The one exception is the zero element, which has rank zero by convention.

Problem: Given a tensor $x \in k^{n_1} \otimes k^{n_2} \otimes k^{n_3}$ where $k$ is a field, compute its rank.

Of course this isn’t possible in standard computing models unless you can represent the elements of the field (and hence the elements of the vector space in question) in a computer program. So we restrict $k$ to be either the rational numbers $\mathbb{Q}$ or a finite field $\mathbb{F}_{q}$.

Even though the problem is simple to state, it was proved in 1990 (a result of Johan Håstad) that tensor rank is hard to compute. Specifically, the theorem is that

Theorem: Computing tensor rank is NP-hard when $k = \mathbb{Q}$ and NP-complete when $k$ is a finite field.

The details are given in Håstad’s paper, but the important work that followed essentially showed that most problems involving tensors are hard to compute (many of them by reduction from computing rank). This is unfortunate, but also displays the power of tensors. In fact, tensors are so powerful that many believe understanding them better will lead to insight in some very important problems, like finding faster matrix multiplication algorithms or proving circuit lower bounds (which is closely related to P vs NP). Finding low-rank tensor approximations is also a key technique in a lot of recent machine learning and data mining algorithms.

With this in mind, the enterprising reader will probably agree that understanding tensors is both valuable and useful. In the future of this blog we’ll hope to see some of these techniques, but at the very least we’ll see the return of tensors when we delve into quantum computing.

Until next time!