# 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.

# Determinism and Finite Automata – A Primer

This half of the theory of computing primer will cover the various finite automata, including deterministic, nondeterministic, and pushdown automata. We devote the second half [upcoming] entirely to Turing machines and the halting problem, but to facilitate the discussion of Turing machines we rely on the intuition and notation developed here.

## Defining Computation

The first step in studying the sorts of possible computations (and more interestingly, those things which cannot be computed) is to define exactly what we mean by a “computation.” At a high level, this is easy: a computation is simply a function. Given some input, produce the appropriate output.

Unfortunately this is much too general. For instance, we could define almost anything we want in terms of functions. Let $f$ be the function which accepts as input the date of California Super Lotto drawings, and returns the set of winning numbers for that date. Of course (assuming the winning numbers are chosen randomly), this is ridiculous and totally uninformative. We cannot learn anything interesting about computations if they are performed behind an opaque veil. Instead, we need to rigorously define the innards of our model for computation, and then see what kinds of things it can do.

Because our inputs may be long and complicated, our model should break an input into pieces and work on the pieces in sequence. This motivates the following two definitions:

Definition: An alphabet is a finite set, usually denoted $\Sigma$.

Definition: $\Sigma^*$ is the set of finite strings whose characters come from $\Sigma$. We include the empty string $\varepsilon$ as the string of zero length. This operation is called the Kleene star.

We will henceforth let inputs to our model be elements of $\Sigma^*$ for some fixed alphabet $\Sigma$, and operate on each character one at a time.

Since $\Sigma$ is finite, we may assign to each element a unique finite string of ones and zeros (to keep things simple, let each string have length equal to ceiling of $\log |\Sigma|)$. Then, we can simulate any alphabet by the appropriate subset of $\left \{0,1 \right \}^*$. So without loss of generality, we may always assume our input comes from $\left \{ 0,1 \right \}^*$, and thus we will henceforth have $\Sigma = \left \{ 0,1 \right \}$.

Now we need a clear objective for our model, and we will phrase it again in terms of $\Sigma^*$. Specifically, we pick a subset of $A \subset \Sigma^*$, and call it a language. Our computational model will accept an input, and output “accept” or “reject.” We say that our model recognizes $A$ if it accepts and only accepts those strings in $A$.

Notice that this is a bit different from our colloquial idea of “computation.” Specifically, all our model will do is determine presence of an element in a set which has a potentially complicated construction. While it at first seems that this does not include things like “adding two numbers,” it turns out that we may phrase such objectives in terms of acceptance and rejection. Let $A = \left \{ axbxc | a,b,c \in \mathbb{N}, a+b=c \right \}$, where $x$ is some fixed separator. If we fix $a$ and $b$, then we may run our computational model on successively higher values of $c$ (ordered by absolute value, or restricting everything to be positive) until finding acceptance.

Colloquially, we have defined “computation” as recognizing all inputs which satisfy a qestion. Moreover, we will see that very many complex problems can be recognized as recognition problems, including things that are logically impossible to compute within our model.

The reason we have been heretofore vague in naming our model is that we will actually define four different models of progressively stronger computational ability, and this framework will hold for all of them. So here our first try.

## Deterministic Finite Automata

This definition comes from the intuitive idea that a computation can be carried out via a set of states and transitions between those states. A very simple example is a light switch, which has the states ‘on’ and ‘off’, and which can accept as input ‘switch’ or ‘do nothing’. This model is rigorized here:

Definition: A deterministic finite automaton, abbreviated DFA, is a five-tuple $D = (S, \Sigma, \tau, s_0, F)$, where:

• $S$ is a set of states, with initial state $s_0$.
• $\Sigma$ is our working alphabet, and inputs come from $\Sigma^*$.
• $\tau$ is a transition function $S \times \Sigma \to S$, which maps the current state and next input character to the next appropriate state.
• $F \subset S$ is a set of final states. If the last input character lands us in any state $f \in F$, we accept.

Rigorously, let our input word $w \in \Sigma^*$ have length $n$ and characters indexed $w_k$. Call the sequence of states in the computation $s_k$, where $s_{k+1} = \tau(s_k, w_k)$. Then our DFA accepts $w$ if and only if $s_n \in F$. We will use this notation throughout our discussion of finite automata. We call the set of languages which a DFA can recognize the regular languages. We will soon see that this class is rather small.

Of course, we have some examples of DFAs. The simplest possible DFA has only one state, $s$, and the transition function $\tau(s,a) = s$ for all $a \in \Sigma$. Depending on whether $F$ is empty, this DFA will accept or reject all of $\Sigma^*$. We may visualize this DFA as a directed graph:

A trivial DFA, with one state.

Indeed, this visualization provides a convenient way for us to write out the transition function. We simply draw an edge originating at each state for each element of our alphabet. When two or more inputs have the same behavior, we label one edge with multiple input symbols, as above.

Here is a more interesting example: let $A$ be the set of binary strings which have an even number of zeros. We design a DFA to accept this as follows:

A DFA which accepts binary numbers with evenly many zeros.

The shaded state, $s_0$ is the only final state. It is obvious from tracing the diagram that this DFA accepts precisely the set of strings with evenly many zeros. Let’s try something harder:

Let $A = \left \{ 0^n1^n | n \in \mathbb{N} \right \}$. This is a simple enough language, so let us develop a DFA to recognize it. We can easily construct a DFA which recognizes $0^n1^n$ for a fixed $n$, and if we have two DFAs we can construct a DFA which recognizes their union quite easily (do so as an exercise!). However, due to the restriction that $S$ is finite, we cannot connect these pieces together indefinitely! While we might imagine some cleverly designed DFA exists to recognize $A$, we find it more likely that no such DFA exists.

Indeed, it has been proven that no such DFA exists. The key to the proof lies in an esoteric lemma called the Pumping Lemma. Being a notational beast of a lemma, we will not state it rigorously. Colloquially, the lemma says that if $A$ is a regular language, then any sufficiently large word $w \in A$ can be split up into three pieces $xyz$, such that the middle piece may be repeated arbitrarily many times, as $xyyyy \dots yz$, with the resulting word still being in $A$. Clearly this breaks the “balance” restraint of $A$ no matter where the splitting is done, and so $A$ cannot be regular.

Before we increase our model’s expressive power, let us make an “enhancement” to the DFA.

## Nondeterministic Finite Automata

Following the colloquial definition of nondeterminism, we can design our computational system to make state transitions “branch out.” This will be made clear by a slight modification of the DFA.

Definition: A nondeterministic finite automaton, abbreviated NFA, is a DFA with the following two modifications:

• Instead of having a single state $s_k$ at each step, we allow a set of possible states, which we denote $S_k \subset S$.
• Our initial state $s_0$ is replaced with an initial set of states $S_0 \subset S$.
• We include $\varepsilon$ in $\Sigma$, allowing for immediate transitions that require no input.
• The transition function $\tau$ now has signature $S \times \Sigma \to P(S)$, where $P(S)$ denotes the power set of $S$, the set of all subsets of $S$.
• At each step (with input character $w_k$), we map $\tau(-,w_k)$ over $S_k$ to get $S_{k+1}$, i.e. $S_{k+1} = \left \{ \tau(s, w_k) | s \in S_k \right \}$
• The NFA accepts if any state in $S_n$ is final.

In this way, our machine can get a character as input, and proceed to be in two separate states at the same time. If any of the branches of computation accept, then they all accept.

Extending our example from DFAs, here is an NFA which accepts the language of binary strings which have either an even number of zeros or an even number of ones (or both).

An NFA recognizing strings containing evenly many zeros or evenly many ones

Here, $S_0 = F = \left \{ s_0, s_2 \right \}$, and by tracing (a little more carefully this time), we can see that it accepts what we expect.

Now, with the added strength of nondeterminism, we might expect that NFAs can compute some things that DFAs cannot. Amazingly, this is not the case. We point the reader to a more detailed description, but trust that our quick explanation will give the necessary intuition to see the equivalence.

The key observation is that despite nondeterminism, there is still a finite number of possible states an NFA can be in. Specifically, there are at most $2^{|S|}$ possible subsets of $S$, so we can simply construct a DFA which has $2^{|S|}$ states, one corresponding to each subset of $S$, and build up a deterministic transition function that makes things work. This would be an appropriate exercise for the advanced reader, but the beginning student should see Sisper’s Introduction to the Theory of Computation, which contains a complete and thorough proof. (We hate to refer the reader to an expensive textbook, but we could only find sparse proofs of this equivalence elsewhere on the internet, and most assume a particular working set of notation. And Sisper’s book is much better than this terse post for a primer in Computing Theory. The interested reader who lacks mathematical maturity would find it very accessible.)

In other words, every NFA has a corresponding DFA which recognizes precisely the same language. On the other hand, every DFA is trivially an NFA. Thus, the class of languages which NFAs recognize is also the regular languages.

So we have discovered that regular languages, and hence the DFAs and NFAs which recognize them, are quite limited in expressive power. Now let’s truly beef up our model with some mathematical steroids.

## Pushdown Automata

Okay, so these steroids won’t be that strong. Ideally, we just want to add as little as possible to our model and still make it more expressive. This way we can better understand the nuances of computational power, thus strengthening our intuition. While “little” is a subjective notion, lots of “littler” things were tried before arriving at a model that was not equivalent in power to a DFA. We will refrain from explaining them here, except to say that DFAs are equivalent in power to the class of formal Regular Expressions (hence the name, regular languages), which most programmers are familiar with on an informal level.

Definition: A pushdown automaton, denoted PDA, is an NFA with two additional components: $\Gamma$ a set of stack symbols including $\varepsilon$, and a modified transition function $\tau$:

$\tau: S \times \Sigma \times \Gamma \to P(S) \times \Gamma^*$

Here, the left hand $\Gamma$ symbol represents the current top of the stack (which is always popped), and the right hand $\Gamma^*$ represents any modification to the stack during state transition. This modification is always a push, and can be an empty push, or a push of arbitrarily many symbols.

For brevity of a description of $\tau$, which would otherwise be infinite as there are infinitely many possible stacks, we allow $\tau$ to be a partial function, not defined for some inputs. Then a PDA automatically rejects any encountered undefined input. Alternatively, we could construct $\tau$ as a total function, if we add edges for each undefined input going to some terminal state which is inescapable and not final. For the sake of this post, we will accept a partial transition function.

Recalling our discussion of computational ability in our post on Conway’s Game of Life, we recognize that a PDA’s new power is in its ability to grow without bound. Specifically, the stack has no limit to its size, and we can remember important pieces of input which would otherwise be discarded.

This modification seems like it was made just for the $0^n1^n$ problem. Specifically, now we can just push the 0s we see on to the stack, and pop just 1’s until we see an empty stack. Here is a PDA which does just that (making heavy use of epsilon transitions):

A PDA which recognizes $0^n1^n$

Here $S_0 = \left \{ s_0 \right \}$, and $F = \left \{ s_2 \right \}$. For each state transition label, we include $w_k; a \to bc$ to mean, “upon receiving $w_k$ as input with $a$ on top of the stack, pop $a$ and replace it with $bc$, where $bc$ may be a single character or multiple characters or the empty character. The epsilon transitions allow us to move from $s_0$ to $s_1$ and from $s_1$ to $s_2$ seamlessly, but only when the stack is agreeable. As an exercise, modify the above diagram to instead recognize $0^{2n}1^n$.

Neato! It seems like we’ve got a huge boost in computational power now that we can store some information. Wrong. As great as a stack is, it’s not good enough. Here’s a language that a PDA cannot solve: $A = \left \{ 0^n1^n2^n | n \in \mathbb{N} \right \}$. This follows from an additional and more nuanced pumping lemma, and we will not dwell on it here. But look at that! This language is hardly any more complex than $0^n1^n$, and yet it stumps our PDA.

Time to up the steroid dosage. Surprisingly enough, it turns out that our solution will be equivalent to adding another stack! In doing so, we will achieve a level of computational expression so powerful that nothing we know of today can surpass it. We call this panacea of computational woes a Turing machine, and we will cover both it and the wild problems it cannot compute next time.

Until then!