PHYS771 Lecture 6: P, NP, and Friends

We've seen that if we want to make progress in complexity, then we need to talk about asymptotics: not which problems can be solved in 10,000 steps, but for which problems can instances of size n be solved in cn2 steps as n goes to infinity? We met TIME(f(n)), the class of all problems solvable in O(f(n)) steps, and SPACE(f(n)), the class of all problems solvable using O(f(n)) bits of memory.

But if we really want to make progress, then it's useful to take an even coarser-grained view: one where we distinguish between polynomial and exponential time, but not between O(n2) and O(n3) time. From this remove, we think of any polynomial bound as "fast," and any exponential bound as "slow."

Now, I realize people will immediately object: what if a problem is solvable in polynomial time, but the polynomial is n50000? Or what if a problem takes exponential time, but the exponential is 1.00000001n? My answer is pragmatic: if cases like that regularly arose in practice, then it would've turned out that we were using the wrong abstraction. But so far, it seems like we're using the right abstraction. Of the big problems solvable in polynomial time -- matching, linear programming, primality testing, etc. -- most of them really do have practical algorithms. And of the big problems that we think take exponential time -- theorem-proving, circuit minimization, etc. -- most of them really don't have practical algorithms. So, that's the empirical skeleton holding up our fat and muscle.

Petting Zoo

It's now time to meet the most basic complexity classes -- the sheep and goats of the Complexity Zoo.

• P is the class of problems solvable by a Turing machine in polynomial time. In other words, P is the union, over all positive integers k, of TIME(nk). (Note that by "problem," we'll always mean decision problem: a problem where the inputs are n-bit strings and the outputs are either yes or no.)
• PSPACE is the class of problems solvable in polynomial space (but unlimited time). In other words, it's the union over all integers k of SPACE(nk).
• EXP is the class of problems solvable in exponential time. In other words, it's the union over all integers k of TIME(2n^k).

Certainly P is contained in PSPACE. I claim that PSPACE is contained in EXP. Why?

Right: a machine with nk bits of memory can only go through 2n^k different configurations, before it either halts or else gets stuck in an infinite loop.

Now, NP is the class of problems for which, if the answer is yes, then there's a polynomial-size proof of that fact that you can check in polynomial time. (The NP stands for "Nondeterministic Polynomial," in case you were wondering.) I could get more technical, but it's easiest to give an example: say I give you a 10,000-digit number, and I ask whether it has a divisor ending in 3. Well, answering that question might take a Long, Long TimeTM. But if your grad student finds such a divisor for you, then you can easily check that it works: you don't need to trust your student (always a plus).

I claim that NP is contained in PSPACE. Why?

Right: in polynomial space, you can loop over all possible nk-bit proofs and check them one by one. If the answer is "yes," then one of the proofs will work, while if the answer is "no," then none of them will work.

Certainly P is contained in NP: if you can answer a question yourself, then someone else can convince you that the answer is yes (if it is yes) without even telling you anything.

One interesting little puzzle, which I'll leave as a homework exercise for Tuesday, is whether P equals NP. In other words, if you can recognize an answer efficiently, can you also find one efficiently?

Oh, alright, you don't have to solve this one by Tuesday. You can have till Thursday.

Look, this question of whether P=NP, what can I say? People like to describe it as "probably the central unsolved problem of theoretical computer science." That's a comical understatement. P vs. NP is one of the deepest questions that human beings have ever asked.

And not only that: it's one of the seven million-dollar prize problems of the Clay Math Institute! What an honor! Imagine: our mathematician friends have decided that P vs. NP is as important as the Hodge Conjecture, or even Navier-Stokes existence and smoothness! (Apparently they weren't going to include it, until they asked around to make sure it was important enough.)

Dude. One way to measure P vs. NP's importance is this. If NP problems were feasible, then mathematical creativity could be automated. The ability to check a proof would entail the ability to find one. Every Apple II, every Commodore, would have the reasoning power of Archimedes or Gauss. So by just programming your computer and letting it run, presumably you could immediately solve not only P vs. NP, but also the other six Clay problems. (Or five, now that Poincar� is down.)

But if that's the case, then why isn't it obvious that P doesn't equal NP? Surely God wouldn't be so benign as to grant us these extravagant powers! Surely our physicist-intuition tells us that brute-force search is unavoidable! (Leonid Levin told me that Feynman -- the king (or maybe court jester) of physicist-intuition -- had trouble even being convinced that P vs. NP was an open problem!)

Well, we certainly believe PNP. Indeed, we don't even believe there's a general way to solve NP problems that's dramatically better than brute-force search through every possibility. But if you want to understand why it's so hard to prove these things, let me tell you something.

Let's say you're given an N-digit number, but instead of factoring it, you just want to know if it's prime or composite.

Or let's say you're given a list of freshmen, together with which ones are willing to room with which other ones, and you want to pair off as many willing roommates as you can.

Or let's say you're given two DNA sequences, and you want to know how many insertions and deletions are needed to convert the one sequence to the other.

Surely these are fine examples of the sort of exponentially-hard NP problem we were talking about! Surely they, too, require brute-force search!

Except they don't. As it turns out, all of these problems have clever polynomial-time algorithms. The central challenge any PNP proof will have to overcome is to separate the NP problems that really are hard from the ones that merely look hard. I'm not just making a philosophical point. While there have been dozens of purported PNP proofs over the years, almost all of them could be rejected immediately for the simple reason that, if they worked, then they would rule out polynomial-time algorithms that we already know to exist.

So to summarize, there are problems like primality testing and pairing off roommates, for which computer scientists (often after decades of work) have been able to devise polynomial-time algorithms. But then there are other problems, like proving theorems, for which we don't know of any algorithm fundamentally better than brute-force search. But is that all we can say -- that we have a bunch of these NP problems, and some of them seem easy and others seem hard?

As it turns out, we can say something much more interesting than that. We can say that almost all of the "hard" problems are the same "hard" problem in different guises -- in the sense that, if we had a polynomial-time algorithm for any one of them, then we'd also have polynomial-time algorithms for all the rest. This is the upshot of the theory of NP-completeness, which was created in the early 1970's by Cook, Karp, and Levin.

The way it goes is, we define a problem B to be "NP-hard" if any NP problem can be efficiently reduced to B. What the hell does that mean? It means that, if we had an oracle to immediately solve problem B, then we could solve any NP problem in polynomial time.

That gives one notion of reduction, which is called Cook reduction. There's also a weaker notion of reduction, which is called Karp reduction. In a Karp reduction from problem A to problem B, we insist that there should be a polynomial-time algorithm that transforms any instance of A to an instance of B having the same answer.

What's the difference between Cook and Karp?

Right: with a Cook reduction, in solving problem A we get to call the oracle for problem B more than once. We can even call the oracle adaptively -- that is, in ways that depend on the outcomes of the previous calls. A Karp reduction is weaker in that we don't allow ourselves these liberties. Perhaps surprisingly, almost every reduction we know of is a Karp reduction -- the full power of Cook reductions is rarely needed in practice.

Now, we say a problem is NP-complete if it's both NP-hard and in NP. In other words, NP-complete problems are the "hardest" problems in NP: the problems that single-handedly capture the difficulty of every other NP problem. As a first question, is it obvious that NP-complete problems even exist?

I claim that it is obvious. Why?

Well, consider the following problem, called DUH: we're given a polynomial-time Turing machine M, and we want to know if there exists an nk-bit input string that causes M to accept. I claim that any instance of any NP problem can be converted, in polynomial time, into a DUH instance having the same answer. Why? Well, DUH! Because that's what it means for a problem to be in NP!

The discovery of Cook, Karp, and Levin was not that there exist NP-complete problems -- that's obvious -- but rather that many natural problems are NP-complete.

The king of these natural NP-complete problems is called 3-Satisfiability, or 3SAT. (How do I know it's the king? Because it appeared on the TV show "NUMB3RS.") Here we're given n Boolean variables, x1,...,xn, as well as a set of logical constraints called clauses that relate at most three variables each:

x2 or x5 or not(x6)
not(x2) or x4
not(x4) or not(x5) or x6
...

Then the question is whether there's some way to set the variables x1,...,xn to TRUE or FALSE, in such a way that every clause is "satisfied" (that is, every clause evaluates to TRUE).

It's obvious that 3SAT is in NP. Why? Right: Because if someone gives you a setting of x1,...,xn that works, it's easy to check that it works!

Our goal is to prove that 3SAT is NP-complete. What will that take? Well, we need to show that if we had an oracle for 3SAT, then we could use it to solve not only 3SAT in polynomial time, but any NP problem whatsoever. That seems like a tall order! Yet in retrospect, you'll see that it's almost a triviality.

The proof has two steps. Step 1 is to show that, if we could solve 3SAT, then we could solve a more "general" problem called CircuitSAT. Step 2 is to show that, if we could solve CircuitSAT, then we could solve any NP problem.

In CircuitSAT, we're given a Boolean circuit and ... wait, do we have any engineers in this class? We do? Then listen up: in computer science, a "circuit" never has loops! Nor does it have resistors or diodes or anything weird like that. For us, a circuit is just an object where you start with n Boolean variables x1,...,xn, and then you can repeatedly define a new variable that's equal to the AND, OR, or NOT of variables that you've previously defined. Like so:

xn+1 := x3 or xn
xn+2 := not(xn+1)
xn+3 := x1 and xn+2
...

We designate the last variable in the list as the circuit's "output." Then the goal, in CircuitSAT, is to decide whether there's a setting of x1,...,xn such that the output is TRUE.

I claim that, if we could solve 3SAT, then we could also solve CircuitSAT. Why?

Well, all we need to do is notice that every CircuitSAT instance is really a 3SAT instance in disguise! Every time we compute an AND, OR, or NOT, we're relating one new variable to one or two old variables. And any such relationship can be expressed by a set of clauses involving at most three variables each. So for example,

xn+1 := x3 or xn

becomes

xn+1 or not(x3)
xn+1 or not(xn)
not(xn+1) or x3 or xn.

So, that was Step 1. Step 2 is to show that, if we can solve CircuitSAT, then we can solve any NP problem.

Alright, so consider some instance of an NP problem. Then by the definition of NP, there's a polynomial-time Turing machine M such that the answer is "yes" if and only if there's a polynomial-size witness string w that causes M to accept.

Now, given this Turing machine M, our goal is to create a circuit that "mimics" M. In other words, we want there to exist a setting of the circuit's input variables that makes it evaluate to TRUE, if and only if there exists a string w that causes M to accept.

How do we achieve that? Simple: by defining a whole shitload of variables! We'll have a variable that equals TRUE if and only if the 37th bit of M's tape is set to '1' at the 42nd time step. We'll have another variable that equals TRUE if and only if the 14th bit is set to '1' at the 52nd time step. We'll have another variable that equals TRUE if and only if M's tape head is in the 15th internal state and the 74th tape position at the 33rd time step. Well, you get the idea.

Then, having written down this shitload of variables, we write down a shitload of logical relations between them. If the 17th bit of the tape is '0' at the 22nd time step, and the tape head is nowhere near the 17th bit at that time, then the 17th bit will still be '0' at the 23rd time step. If the tape head is in internal state 5 at the 44th time step, and it's reading a '1' at that time step, and internal state 5 transitions to internal state 7 on reading a '1', then the tape head will be in internal state 7 at the 45th time step. And so on, and so on. The only variables that are left unrestricted are the ones that constitute the string w at the first time step.

The key point is that, while this is a very large shitload of variables and relations, it's still only a polynomial shitload. We therefore get a polynomially-large CircuitSAT instance, which is satisfiable if and only if there exists a w that causes M to accept.

[Devin, our resident engineer, is laughing hysterically]

Scott: What?

Devin: Nothing! That is just so simple in the basic idea, and so messy in the details...

Scott: Exactly. You've understood completely.

We've just proved the celebrated Cook-Levin Theorem: that 3SAT is NP-complete. This theorem can be thought of as the "initial infection" of the NP-completeness virus. Since then, the virus has spread to thousands of other problems. What I mean is this: if you want to prove that your favorite problem is NP-complete, all you have to do is prove it's as hard as some other problem that's already been proved NP-complete. (Well, you also have to prove that it's in NP, but that's usually trivial.) So there's a rich-get-richer effect: the more problems that have already been proved NP-complete, the easier it is to induct a new problem into the club. Indeed, proving problems NP-complete had become so routine by the 80's or 90's, and people had gotten so good at it, that (with rare exceptions) STOC and FOCS stopped publishing yet more NP-completeness proofs.

I'll just give you a tiny sampling of some of the earliest problems that were proved NP-complete:

• Map Colorability: Given a map, can you color every country red, green, or blue, in such a way that no two neighboring countries are colored the same? (Interestingly, if you're only allowed to use two colors, then it's easy to decide whether or not such a coloring is possible -- why? On the other hand, if you're allowed four colors, then it always is possible, at least for maps drawn in the plane -- that's the famous Four-Color Theorem. So then the problem is again easy. Only with three colors is the problem NP-complete.)
• Clique: Given a set of N high-school students, together with which ones will sit at a cafeteria table with which other ones, is there a "clique" of N/3 students who will all sit at a table with each other?
• Packing: Given a set of boxes of specified dimensions, can you fit them into the trunk of your car?

Etc., etc., etc.

To reiterate: although these problems might look unrelated, they're actually the same problem in different costumes. if any one of them has an efficient solution, then all of them do, and P=NP. If any one of them doesn't have an efficient solution, then none of them do, and PNP. To prove P=NP, it's enough to show that some NP-complete problem (no matter which one) has an efficient solution. To prove PNP, it's enough to show that some NP-complete problem has no efficient solution. One for all and all for one.

So, there are the P problems, and then there are the NP-complete problems. Is there anything in between? (You should be used to this sort of "intermediate" question by now -- we saw it both in set theory and in computability theory!)

If P=NP, then NP-complete problems are P problems, so obviously the answer is no.

But what if PNP? In that case, a beautiful result called Ladner's Theorem says that there must be "intermediate" problems between P and NP-complete: in other words, problems that are in NP, but neither NP-complete nor solvable in polynomial time.

How would we create such an intermediate problem? Well, I'll give you the idea. The first step is to define an extremely slow-growing function t. Then given a 3SAT instance F of size n, the problem will be to decide whether F is satisfiable and t(n) is odd. In other words: if t(n) is odd then solve the 3SAT problem, while if t(n) is even then always output "no."

If you think about what we're doing, we're alternating long stretches of an NP-complete problem with long stretches of nothing! Intuitively, each stretch of 3SAT should kill off another polynomial-time algorithm for our problem, where we use the assumption that PNP. Likewise, each stretch of nothing should kill off another NP-completeness reduction, where we again use the assumption that PNP. This ensures that the problem is neither in P nor NP-complete. The main technical trick is to make the stretches get longer at an exponential rate. That way, given an input of size n, we can simulate the whole iterative process up to n in time polynomial in n. That ensures that the problem is still in NP.

Besides P and NP, another major complexity class is coNP: the "complement" of NP. A problem is in coNP if a "no" answer can be checked in polynomial-time. To every NP-complete problem, there's a corresponding coNP-complete problem. We've got unsatisfiability, map non-colorability, etc.

Now, why would anyone bother to define such a stupid thing? Because then we can ask a new question: does NP equal coNP? In other words: if a Boolean formula is unsatisfiable, is there at least a short proof that it's unsatisfiable, even if finding the proof would take exponential time? Once again, the answer is that we don't know.

Certainly if P=NP then NP=coNP. (Why?) On the other hand, the other direction isn't known: it could be that PNP but still NP=coNP. So if proving PNP is too easy, you can instead try to prove NPcoNP!

This seems like a good time to mention a special complexity class, a class we quantum computing people know and love: NP intersect coNP.

This is the class for which either a yes answer or a no answer has an efficiently checkable proof. As an example, consider the problem of factoring an integer into primes. Over the course of my life, I must've met at least two dozen people who "knew" that factoring is NP-complete, and therefore that Shor's algorithm -- since it lets us factor on a quantum computer -- also lets us solve NP-complete problems on a quantum computer. Often these people were supremely confident of their "knowledge."

But if you think about it for two seconds, you'll realize that factoring has profound differences from the known NP-complete problems. If I give you a Boolean formula, it might have zero satisfying truth assignments, it might have one, or it might have 10 trillion. You simply don't know, a priori. But if I give you a 5000-digit integer, you probably won't know its factorization into primes, but you'll know that has one and only one factorization. (I think some guy named Euclid proved that a while ago.) This already tells us that factoring is somehow "special": that, unlike what we believe about the NP-complete problems, factoring has some structure that algorithms could try to exploit. And indeed, algorithms do exploit it: we know of a classical algorithm called the Number Field Sieve, which factors an n-bit integer in roughly 2n^1/3 steps, compared to the ~2n/2 steps that would be needed for trying all possible divisors. (Why only ~2n/2 steps, instead of ~2n?) And of course, we know of Shor's algorithm, which factors an n-bit integer in ~n2 steps on a quantum computer: that is, in quantum polynomial time. Contrary to popular belief, we don't know of a quantum algorithm to solve NP-complete problems in polynomial time. If such an algorithm existed, it would have to be dramatically different from Shor's algorithm.

But can we pinpoint just how factoring differs from the known NP-complete problems, in terms of complexity theory? Yes, we can. First of all, in order to make factoring a decision (yes-or-no) problem, we need to ask something like this: given a positive integer N, does N have a prime factor whose last digit is 7? I claim that this problem is not merely in NP, but in NP intersect coNP. Why? Well, suppose someone gives you the prime factorization of N. There's only one of them. So if there is a prime factor whose last digit is 7, then you can verify that, and if there's no prime factor whose last digit is 7, then you can also verify that.

You might say, "but how do I know that I really was given the prime factorization? Sure, if someone gives me a bunch of numbers, I can check that they multiply to N, but how do I know they're prime?" For this you'll have to take on faith something that I told you earlier: that if you just want to know whether a number is prime or composite, and not what its factors are, then you can do that in polynomial time. OK, so if you accept that, then the factoring problem is in NP intersect coNP.

From this we can conclude that, if factoring were NP-complete, then NP would equal coNP. (Why?) Since we don't believe NP=coNP, this gives us a strong indication (though not a proof) that, all those people I told you about notwithstanding, factoring is not NP-complete. If we accept that, then only two possibilities remain: either factoring is in P, or else factoring is one of those "intermediate" problems whose existence is guaranteed by Ladner's Theorem. Most of us incline toward the latter possibility -- though not with as much conviction as we believe PNP.

Indeed, for all we know, it could be the case that P = NP intersect coNP but still PNP. (This possibility would imply that NPcoNP.) So, if proving PNP and NPcoNP are both too easy for you, your next challenge can be to prove PNP intersect coNP!

If P, NP, and coNP aren't enough to rock your world, you can generalize these classes to a giant teetering mess that we computer scientists call the polynomial hierarchy.

Observe that you can put any NP problem instance into the form

Does there exist an n-bit string X such that A(X)=1?

Here A is a function computable in polynomial time.

Likewise, you can put any coNP problem into the form

Does A(X)=1 for every X?

But what happens if you throw in another quantifier, like so?

Does there exist an X such that for every Y, A(X,Y)=1?

For every X, does there exist a Y such that A(X,Y)=1?

Problems like these lead to two new complexity classes, which are called Σ2P and Π2P respectively. Π2P is the "complement" of Σ2P, in the same sense that coNP is the complement of NP. We can also throw in a third quantifier:

Does there exist an X such that for every Y, there exists a Z such that A(X,Y,Z)=1?

For every X, does there exist a Y such that for every Z, A(X,Y,Z)=1?

This gives us Σ3P and Π3P respectively. It should be obvious how to generalize this to ΣkP and ΠkP for any larger k. (As a side note, when k=1, we get Σ1P=NP and Π1P=coNP. Why?) Then taking the union of these classes over all positive integers k gives us the polynomial hierarchy PH.

The polynomial hierarchy really is a substantial generalization of NP and coNP -- in the sense that, even if we had an oracle for NP-complete problems, it's not at all clear how we could use it to solve (say) Σ2P problems. On the other hand, just to complicate matters further, I claim that if P=NP, then the whole polynomial hierarchy would collapse down to P! Why?

Right: if P=NP, then we could take our algorithm for solving NP-complete problems in polynomial time, and modify it to call itself as a subroutine. And that would let us "flatten PH like a steamroller": first simulating NP and coNP, then Σ2P and Π2P, and so on through the entire hierarchy.

Likewise, it's not hard to prove that if NP=coNP, then the entire polynomial hierarchy collapses down to NP (or in other words, to coNP). If Σ2P=Π2P, then the entire polynomial hierarchy collapses down to Σ2P, and so on. If you think about it, this gives us a whole infinite sequence of generalizations of the PNP conjecture, each one "harder" to prove than the last. Why do we care about these generalizations? Because often, we're trying to study conjecture BLAH, and we can't prove that BLAH is true, and we can't even prove that if BLAH were false then P would equal NP. But -- and here's the punchline -- we can prove that if BLAH were false, then the polynomial hierarchy would collapse to the second or the third level. And this gives us some sort of evidence that BLAH is true.

Welcome to complexity theory!

Since I talked about how lots of problems have non-obvious polynomial-time algorithms, I thought I should give you at least one example. So, let's do one of the simplest and most elegant in all of computer science -- the so-called Stable Marriage Problem. Have you seen this before? You haven't?

Alright, so we have N men and N women. Our goal is to marry them off. We assume for simplicity that they're all straight. (Marrying off gays and lesbians is technically harder, though also solvable in polynomial time!) We also assume, for simplicity and with much loss of generality, that everyone would rather be married than single.

So, each man ranks the women, in order from his first to last choice. Each woman likewise ranks the men. There are no ties.

Obviously, not every man can marry his first-choice woman, and not every woman can marry her first-choice man. Life sucks that way.

So let's try for something weaker. Given a way of pairing off the men and women, say that it's stable if no man and woman who aren't married to each other both prefer each other to their spouses. In other words, you might despise your husband, but no man who you like better than him likes you better than his wife, so you have no incentive to leave. This is the, um, desirable property that we call "stability."

Now, given the men's and women's stated preferences, our goal as matchmakers is to find a stable way of pairing them off. Matchmaker, matchmaker, make me a match, find me a find, catch me a catch, etc.

First obvious question: does there always exist a stable pairing of men and women? What do you think? Yes? No? As it turns out, the answer is yes, but the easiest way to prove it is just to give an algorithm for finding the pairing!

So let's concentrate on the question of how to find a pairing. In total, there are N! ways of pairing off men with women. For the soon-to-be-newlyweds' sake, hopefully we won't have to search through all of them.

Fortunately we won't. In the early 1960's, Gale and Shapley invented a polynomial-time -- in fact linear-time -- algorithm to solve this problem. And the beautiful thing about this algorithm is, it's exactly what you'd come up with from reading a Victorian romance novel. Later they found out that the same algorithm had been in use since the 1950's -- not to pair off men with women, but to pair off medical-school students with hospitals to do their residencies in. Indeed, hospitals and medical schools are still using a version of the algorithm today.

But back to the men and women. If we want to pair them off by the Gale-Shapley algorithm, then as a first step, we need to break the symmetry between the sexes: which sex "proposes" to the other? This being the early 1960's, you can guess how that question was answered. The men propose to the women.

So, we loop through all the men. The first man proposes to his first-choice woman. She provisionally accepts him. Then the next man proposes to his first-choice woman. She provisionally accepts him, and so on. But what happens when a man proposes to a woman who's already provisionally accepted another man? She chooses the one she prefers, and boots the other one out! Then, the next time we come around to that man in our loop over the men, he'll propose to his second-choice woman. And if she rejects him, then the next time we come around to him he'll propose to his third-choice woman. And so on, until everyone is married off. Pretty simple, huh?

First question: why does this algorithm terminate in linear time?

Right: because each man proposes to a given woman at most once. So the total number of proposals is at most N2, which is just the amount of memory we need to write down the preference lists in the first place.

Second question: when the algorithm does terminate, why is everyone married off?

Right: because if they weren't, then there'd be some woman who'd never been proposed to, and some man who'd never proposed to her. But this is impossible. Eventually the man no one else wants will cave in, and propose to the woman no one else wants.

Third question: why is the pairing produced by this algorithm a stable one?

Right: because if it weren't, then there'd be one married couple (say Bob and Alice), and another married couple (say Charlie and Eve), such that Bob and Eve both prefer each other to their spouses. But in that case, Bob would've proposed to Eve before proposing to Alice. And if Charlie also proposed to Eve, then Eve would've made clear at the time that she preferred Bob. And this gives a contradiction.

In particular we've shown, as promised, that there exists a stable pairing: namely the pairing found by the Gale-Shapley algorithm.

Problem Set

1. We saw that 3SAT is NP-complete. By contrast, it turns out that 2SAT -- the version where we only allow two variables per clause -- is solvable in polynomial time. Explain why.

2. Recall that EXP is the class of problems solvable in exponential time. One can also define NEXP: the class of problems for which a "yes" answer can be verified in exponential time. In other words, NEXP is to EXP as NP is to P. Now, we don't know if P=NP, and we also don't know if EXP=NEXP. But we do know that if P=NP, then EXP=NEXP. Why?

3. Show that P doesn't equal SPACE(n) (the set of problems solvable in linear space). Hint: You don't need to prove that P is not in SPACE(n), or that SPACE(n) is not in P -- only that one or the other is true!

4. Show that if P=NP, then there's a polynomial-time algorithm not only to decide whether a Boolean formula has a satisfying assignment, but to find such an assignment whenever one exists.

5. [Extra credit] Give an explicit algorithm that finds a satisfying assignment whenever one exists, and that runs in polynomial time assuming P=NP. (If there's no satisfying assignment, your algorithm can behave arbitrarily.) In other words, give an algorithm for problem 4 that you could implement and run right now -- without invoking any subroutine that you've assumed to exist but can't actually describe.