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Do we measure the NP-hardness in terms of average-case hardness or worst-case hardness?
I've found this here:
"However, NP-completeness is defined in terms of worst-case complexity".
Does it remain true to NP-hardness?
I don't know what the term "worst-case complexity" means. What is the difference between worst-case complexity and worst-case problems?
An interesting nuance here is that NP-hardness, by itself, doesn’t speak about worst-case or average case complexity. Rather, the formal definition of NP-hardness purely says that there’s a polynomial-time reduction from every problem in NP to any NP-hard problem. That reduction means that any instance of any problem in NP could be solved by applying the reduction and then solving the NP-hard problem. But because this applies to “any instance” and the specific transform done by the reduction isn’t specified, that definition by itself doesn’t say anything about average-case complexity.
We can artificially construct NP-hard problems that are extremely easy to solve on average. Here’s an example. Take an NP-hard problem - say, the problem of checking whether a graph is 3-colorable. We can solve this in time (roughly) O(3n) by simply trying all possible colorings. (The actual time complexity is a bit higher because we need to check edges in each step, but let’s ignore that for now). Now, we’ll invent a contrived problem of the following form:
Given a string of 0s, 1s, and 2s, determine whether
The first half of the string contains a 1 or a 2, or
Whether it doesn’t and the back half of the string is a base-3 encoding of a graph that’s 3-colorable.
This problem is NP-hard because we can reduce graph 3-colorability to it by just prepending a bunch of 0s to any input instance of 3-colorability. But on average it’s very easy to solve this problem. The probability that a string’s first half is all 0s is 1 / 3n/2, where n is the length of the string. This means that even if it takes O(3n/2) time to check the coloring of the graph encoded in the back half of a suitable string, mathematically the average amount of work required to solve this problem is O(1). (I’m aware I’m conflating the meaning of n as “the number of nodes in a graph” and “how long the string is,” but the math still checks out here.)
What’s worrisome is that we still don’t have a very well-developed theory of average-case complexity for NP-hard problems. Some NP-hard problems, like the one above, are very easy on average. But others like SAT, graph coloring, etc. are mysteries to us, where we legitimately don’t know how hard they are for random instances. It’s entirely possible, for example, that P ≠ NP and yet the average-case hardness of individual NP-hard problems are not known.
I'm studying about the NP class and one of the slides mentions:
It seems that verifying that something is not present is more difficult than verifying that it is present.
______ _________
Hence, CLIQUE (complement) and SubsetSUM (complement) are not obviously members of NP.
Was it ever proved, whether the complement of CLIQUE is an element of NP?
Also, do you have the proof?
This is an open problem, actually! The complexity class co-NP consists of the complements of all problems in NP. It's unknown whether NP = co-NP right now, and many people suspect the answer is no.
Just as CLIQUE is NP-complete, the complement of CLIQUE is co-NP-complete. (More generally, the complement of any NP-complete problem is co-NP-complete). There's a theorem that if any co-NP-complete problem is in NP, then co-NP = NP,which would be a huge theoretical breakthrough.
If you're interested in learning more about this, check out the Wikipedia article on co-NP and look around online for more resources.
The general intuition here: it's easy to prove a graph contains an N-clique: just show me the clique. It's hard to prove a graph that doesn't have an N-clique in fact doesn't have an N-clique. What property of the graph are you going to exploit to do that?
Sure, for some families of graphs you can -- for example, graphs with sufficiently few edges can't possibly have such a clique. It's entirely possible that all graphs can have similar proofs built around them, although it'd be surprising -- almost as surprising as P=NP.
This is why the complement of languages in NP are not, in general, obviously in NP -- in fact, we have the term "co-NP" (as in "the complement is in NP") to refer to languages like !CLIQUE.
One common approach to make progress in complexity theory, where we haven't made progress against the hard questions, is to show that some specific hard-to-prove result would imply something surprising. Showing that NP=co-NP is a common target of these proofs -- for example, any hard problem in both NP and co-NP probably isn't complete for either, because if it were it is complete for both and thus both are equal, so somehow you have those crazy general graph proofs.
This even generalizes -- you can start talking about what happens if your NTMs (or certificate checkers) have an oracle for an NP complete language like CLIQUE. Obviously both CLIQUE and !CLIQUE is in P^CLIQUE, but now there are (probably) new languages in NP^CLIQUE and co-NP^CLIQUE, and you can keep going further until you have an entire hierarchy of complexity classes -- the "polynomial hierarchy". This hierarchy intuitively goes on forever, but may well collapse at some point or even not exist at all (if P=NP).
The polynomial hierarchy makes this general argument technique more powerful -- showing that some result would make the polynomial hierarchy collapse to the 2nd or 3rd level would make that result pretty surprising. Even showing it collapses at all would be somewhat surprising.
There are a lot of real-world problems that turn out to be NP-hard. If we assume that P ≠ NP, there aren't any polynomial-time algorithms for these problems.
If you have to solve one of these problems, is there any hope that you'll be able to do so efficiently? Or are you just out of luck?
If a problem is NP-hard, under the assumption that P ≠ NP there is no algorithm that is
deterministic,
exactly correct on all inputs all the time, and
efficient on all possible inputs.
If you absolutely need all of the above guarantees, then you're pretty much out of luck. However, if you're willing to settle for a solution to the problem that relaxes some of these constraints, then there very well still might be hope! Here are a few options to consider.
Option One: Approximation Algorithms
If a problem is NP-hard and P ≠ NP, it means that there's is no algorithm that will always efficiently produce the exactly correct answer on all inputs. But what if you don't need the exact answer? What if you just need answers that are close to correct? In some cases, you may be able to combat NP-hardness by using an approximation algorithm.
For example, a canonical example of an NP-hard problem is the traveling salesman problem. In this problem, you're given as input a complete graph representing a transportation network. Each edge in the graph has an associated weight. The goal is to find a cycle that goes through every node in the graph exactly once and which has minimum total weight. In the case where the edge weights satisfy the triangle inequality (that is, the best route from point A to point B is always to follow the direct link from A to B), then you can get back a cycle whose cost is at most 3/2 optimal by using the Christofides algorithm.
As another example, the 0/1 knapsack problem is known to be NP-hard. In this problem, you're given a bag and a collection of objects with different weights and values. The goal is to pack the maximum value of objects into the bag without exceeding the bag's weight limit. Even though computing an exact answer requires exponential time in the worst case, it's possible to approximate the correct answer to an arbitrary degree of precision in polynomial time. (The algorithm that does this is called a fully polynomial-time approximation scheme or FPTAS).
Unfortunately, we do have some theoretical limits on the approximability of certain NP-hard problems. The Christofides algorithm mentioned earlier gives a 3/2 approximation to TSP where the edges obey the triangle inequality, but interestingly enough it's possible to show that if P ≠ NP, there is no polynomial-time approximation algorithm for TSP that can get within any constant factor of optimal. Usually, you need to do some research to learn more about which problems can be well-approximated and which ones can't, since many NP-hard problems can be approximated well and many can't. There doesn't seem to be a unified theme.
Option Two: Heuristics
In many NP-hard problems, standard approaches like greedy algortihms won't always produce the right answer, but often do reasonably well on "reasonable" inputs. In many cases, it's reasonable to attack NP-hard problems with heuristics. The exact definition of a heuristic varies from context to context, but typically a heuristic is either an approach to a problem that "often" gives back good answers at the cost of sometimes giving back wrong answers, or is a useful rule of thumb that helps speed up searches even if it might not always guide the search the right way.
As an example of the first type of heuristic, let's look at the graph-coloring problem. This NP-hard problem asks, given a graph, to find the minimum number of colors necessary to paint the nodes in the graph such that no edge's endpoints are the same color. This turns out to be a particularly tough problem to solve with many other approaches (the best known approximation algorithms have terrible bounds, and it's not suspected to have a parameterized efficient algorithm). However, there are many heuristics for graph coloring that do quite well in practice. Many greedy coloring heuristics exist for assigning colors to nodes in a reasonable order, and these heuristics often do quite well in practice. Unfortunately, sometimes these heuristics give terrible answers back, but provided that the graph isn't pathologically constructed the heuristics often work just fine.
As an example of the second type of heuristic, it's helpful to look at SAT solvers. SAT, the Boolean satisfiability problem, was the first problem proven to be NP-hard. The problem asks, given a propositional formula (often written in conjunctive normal form), to determine whether there is a way to assign values to the variables such that the overall formula evaluates to true. Modern SAT solvers are getting quite good at solving SAT in many cases by using heuristics to guide their search over possible variable assignments. One famous SAT-solving algorithm, DPLL, essentially tries all possible assignments to see if the formula is satisfiable, using heuristics to speed up the search. For example, if it finds that a variable is either always true or always false, DPLL will try assigning that variable its forced value before trying other variables. DPLL also finds unit clauses (clauses with just one literal) and sets those variables' values before trying other variables. The net effect of these heuristics is that DPLL ends up being very fast in practice, even though it's known to have exponential worst-case behavior.
Option Three: Pseudopolynomial-Time Algorithms
If P ≠ NP, then no NP-hard problem can be solved in polynomial time. However, in some cases, the definition of "polynomial time" doesn't necessarily match the standard intuition of polynomial time. Formally speaking, polynomial time means polynomial in the number of bits necessary to specify the input, which doesn't always sync up with what we consider the input to be.
As an example, consider the set partition problem. In this problem, you're given a set of numbers and need to determine whether there's a way to split the set into two smaller sets, each of which has the same sum. The naive solution to this problem runs in time O(2n) and works by just brute-force testing all subsets. With dynamic programming, though, it's possible to solve this problem in time O(nN), where n is the number of elements in the set and N is the maximum value in the set. Technically speaking, the runtime O(nN) is not polynomial time because the numeric value N is written out in only log2 N bits, but assuming that the numeric value of N isn't too large, this is a perfectly reasonable runtime.
This algorithm is called a pseudopolynomial-time algorithm because the runtime O(nN) "looks" like a polynomial, but technically speaking is exponential in the size of the input. Many NP-hard problems, especially ones involving numeric values, admit pseudopolynomial-time algorithms and are therefore easy to solve assuming that the numeric values aren't too large.
For more information on pseudopolynomial time, check out this earlier Stack Overflow question about pseudopolynomial time.
Option Four: Randomized Algorithms
If a problem is NP-hard and P ≠ NP, then there is no deterministic algorithm that can solve that problem in worst-case polynomial time. But what happens if we allow for algorithms that introduce randomness? If we're willing to settle for an algorithm that gives a good answer on expectation, then we can often get relatively good answers to NP-hard problems in not much time.
As an example, consider the maximum cut problem. In this problem, you're given an undirected graph and want to find a way to split the nodes in the graph into two nonempty groups A and B with the maximum number of edges running between the groups. This has some interesting applications in computational physics (unfortunately, I don't understand them at all, but you can peruse this paper for some details about this). This problem is known to be NP-hard, but there's a simple randomized approximation algorithm for it. If you just toss each node into one of the two groups completely at random, you end up with a cut that, on expectation, is within 50% of the optimal solution.
Returning to SAT, many modern SAT solvers use some degree of randomness to guide the search for a satisfying assignment. The WalkSAT and GSAT algorithms, for example, work by picking a random clause that isn't currently satisfied and trying to satisfy it by flipping some variable's truth value. This often guides the search toward a satisfying assignment, causing these algorithms to work well in practice.
It turns out there's a lot of open theoretical problems about the ability to solve NP-hard problems using randomized algorithms. If you're curious, check out the complexity class BPP and the open problem of its relation to NP.
Option Five: Parameterized Algorithms
Some NP-hard problems take in multiple different inputs. For example, the long path problem takes as input a graph and a length k, then asks whether there's a simple path of length k in the graph. The subset sum problem takes in as input a set of numbers and a target number k, then asks whether there's a subset of the numbers that dds up to exactly k.
Interestingly, in the case of the long path problem, there's an algorithm (the color-coding algorithm) whose runtime is O((n3 log n) · bk), where n is the number of nodes, k is the length of the requested path, and b is some constant. This runtime is exponential in k, but is only polynomial in n, the number of nodes. This means that if k is fixed and known in advance, the runtime of the algorithm as a function of the number of nodes is only O(n3 log n), which is quite a nice polynomial. Similarly, in the case of the subset sum problem, there's a dynamic programming algorithm whose runtime is O(nW), where n is the number of elements of the set and W is the maximum weight of those elements. If W is fixed in advance as some constant, then this algorithm will run in time O(n), meaning that it will be possible to exactly solve subset sum in linear time.
Both of these algorithms are examples of parameterized algorithms, algorithms for solving NP-hard problems that split the hardness of the problem into two pieces - a "hard" piece that depends on some input parameter to the problem, and an "easy" piece that scales gracefully with the size of the input. These algorithms can be useful for finding exact solutions to NP-hard problems when the parameter in question is small. The color-coding algorithm mentioned above, for example, has proven quite useful in practice in computational biology.
However, some problems are conjectured to not have any nice parameterized algorithms. Graph coloring, for example, is suspected to not have any efficient parameterized algorithms. In the cases where parameterized algorithms exist, they're often quite efficient, but you can't rely on them for all problems.
For more information on parameterized algorithms, check out this earlier Stack Overflow question.
Option Six: Fast Exponential-Time Algorithms
Exponential-time algorithms don't scale well - their runtimes approach the lifetime of the universe for inputs as small as 100 or 200 elements.
What if you need to solve an NP-hard problem, but you know the input is reasonably small - say, perhaps its size is somewhere between 50 and 70. Standard exponential-time algorithms are probably not going to be fast enough to solve these problems. What if you really do need an exact solution to the problem and the other approaches here won't cut it?
In some cases, there are "optimized" exponential-time algorithms for NP-hard problems. These are algorithms whose runtime is exponential, but not as bad an exponential as the naive solution. For example, a simple exponential-time algorithm for the 3-coloring problem (given a graph, determine if you can color the nodes one of three colors each so that no edge's endpoints are the same color) might work checking each possible way of coloring the nodes in the graph, testing if any of them are 3-colorings. There are 3n possible ways to do this, so in the worst case the runtime of this algorithm will be O(3n · poly(n)) for some small polynomial poly(n). However, using more clever tricks and techniques, it's possible to develop an algorithm for 3-colorability that runs in time O(1.3289n). This is still an exponential-time algorithm, but it's a much faster exponential-time algorithm. For example, 319 is about 109, so if a computer can do one billion operations per second, it can use our initial brute-force algorithm to (roughly speaking) solve 3-colorability in graphs with up to 19 nodes in one second. Using the O((1.3289n)-time exponential algorithm, we could solve instances of up to about 73 nodes in about a second. That's a huge improvement - we've grown the size we can handle in one second by more than a factor of three!
As another famous example, consider the traveling salesman problem. There's an obvious O(n! · poly(n))-time solution to TSP that works by enumerating all permutations of the nodes and testing the paths resulting from those permutations. However, by using a dynamic programming algorithm similar to that used by the color-coding algorithm, it's possible to improve the runtime to "only" O(n2 2n). Given that 13! is about one billion, the naive solution would let you solve TSP for 13-node graphs in roughly a second. For comparison, the DP solution lets you solve TSP on 28-node graphs in about one second.
These fast exponential-time algorithms are often useful for boosting the size of the inputs that can be exactly solved in practice. Of course, they still run in exponential time, so these approaches are typically not useful for solving very large problem instances.
Option Seven: Solve an Easy Special Case
Many problems that are NP-hard in general have restricted special cases that are known to be solvable efficiently. For example, while in general it’s NP-hard to determine whether a graph has a k-coloring, in the specific case of k = 2 this is equivalent to checking whether a graph is bipartite, which can be checked in linear time using a modified depth-first search. Boolean satisfiability is, generally speaking, NP-hard, but it can be solved in polynomial time if you have an input formula with at most two literals per clause, or where the formula is formed from clauses using XOR rather than inclusive-OR, etc. Finding the largest independent set in a graph is generally speaking NP-hard, but if the graph is bipartite this can be done efficiently due to König’s theorem.
As a result, if you find yourself needing to solve what might initially appear to be an NP-hard problem, first check whether the inputs you actually need to solve that problem on have some additional restricted structure. If so, you might be able to find an algorithm that applies to your special case and runs much faster than a solver for the problem in its full generality.
Conclusion
If you need to solve an NP-hard problem, don't despair! There are lots of great options available that might make your intractable problem a lot more approachable. No one of the above techniques works in all cases, but by using some combination of these approaches, it's usually possible to make progress even when confronted with NP-hardness.
Would it be an polynomial time algorithm to a specific NP-complete problem, or just abstract reasonings that demonstrate solutions to NP-complete problems exist?
It seems that the a specific algoithm is much more helpful. With it, all we'll have to do to polynomially solve an NP problem is to convert it into the specific NP-complete problem for which the proof has a solution, and we are done.
P = NP: "The 3SAT problem is a classic NP complete problem. In this proof, we demonstrate an algorithm to solve it that has an asymptotic bound of (n^99 log log n). First we ..."
P != NP: "Assume there was a polynomial algorithm for the 3SAT problem. This would imply that .... which by ..... implies we can do .... and then ... and then ... which is impossible. This was all predicated on a polynomial time algorithm for 3SAT. Thus P != NP."
UPDATE: Perhaps something like this paper (for P != NP).
UPDATE 2: Here's a video of Michael Sipser sketching out a proof for P != NP
Call me pessimistic, but it will be like this:
...
∴, P ≠ NP
QED
There are some meta-results about what a P=NP or P≠NP proof can not look like. The details are quite technical, but it is known that the proof cannot be
relativizing, which kind of means that the proof must make use of the exact definition of Turing machine used, because with some modifications ("oracles", like very powerful CISC instructions added to the instruction set) P=NP, and with some other modifications, P≠NP. See also this blog post for a nice explanation of relativization.
natural, a property of several classic circuit complexity proofs,
or algebrizing, a generalization of relativizing.
It could take the form of demonstrating that assuming P ≠ NP leads to a contradiction.
It might not be connected to P and NP in a straightforward way... Many theorems now are based on P!=NP, so proving one assumed fact to be untrue would make a big difference. Even proving something like constant ratio approximation for TS should be enough IIRC. I think, existence of NPI (GI) and other sets is also based on P!=NP, so making any of them equal to P or NP might change the situation completely.
IMHO everything happens now on a very abstract level. If someone proves anything about P=/!=NP, it doesn't have to mention any of those sets or even a specific problem.
Probably it would be in the form of a reduction from an NP problem to a P problem. See the Wikipedia page on reductions.
OR
Like this proof proposed by Vinay Deolalikar.
The most straightforward way is to prove that there is a polynomial time solution to the problems in the class NP-complete. These are problems that are in NP and are reducable to one of the known np problem. That means you could give a faster algorithm to prove the original problem posed by Stephen Cook or many others which have also been shown to be NP-Complete. See Richard Karp's seminal paper and this book for more interesting problems. It has been shown that if you solve one of these problems the entire complexity class collapses. edit: I have to add that i was talking to my friend who is studying quantum computation. Although I had no clue what it means, he said that a certain proof/experiment? in the quantum world could make the entire complexity class, i mean the whole thing, moot. If anyone here knows more about this, please reply.
There have also been numerous attempts to the problem without giving a formal algorithm. You could try to count the set. Theres the Robert/Seymore proof. People have also tried to solve it using the tried and tested diagonlization proof(also used to show that there are problems that you can never solve). Razborov also showed that if there are certain one-way functions then any proof cannot give a resolution. That means that new techniques will be required in order to solve this question.
Its been 38 years since the original paper has been published and there still is no sign of a proof. Not only that but lot of problems that mathematicians had been posing before the notion of complexity classes came in has been shown to be NP. Therefor many mathematicians and computer scientists believe that some of the problems are so fundamental that a new kind of maths may be needed to solve the problem. You have to keep in mind that the best minds human race has to offer have tackled this problem without any success. I think it should be at least decades before somebody cracks the puzzle. But even if there is a polynomial time solution the constants or the exponent could be so large that it would be useless in our problems.
There is an excellent survey available which should answer most of your questions: http://www.scottaaronson.com/papers/pnp.pdf.
Certainly a descriptive proof is the most useful, but there are other categories of proof: it is possible, for example, to provide 'existence proofs' that demonstrate that it is possible to find an answer without finding (or, sometimes, even suggesting how to find) that answer.
Set N equal to the multiplicative identity. Then NP = P. QED. ;-)
It would likely look almost precisely like one of these
Good question; it could take either form. Obviously, the specific algorithm would be more helpful, yes, but there's no determining that that would be the way that a theoretical P=NP proof would occur. Given that the nature of NP-complete problems and how common they are, it would seem that more effort has been put into solving those problems than has been put into solving the theoretical reasoning side of the equation, but that's just supposition.
Any nonconstructive proof that P=NP really is not. It would imply that the following explicit 3-SAT algorithm runs in polynomial time:
Enumerate all programs. On round i, run all programs numbered
less than i for one step. If
a program terminates with a
satisfying input to the formula, return true. If a program
terminates with a formal proof that
no such input exists, return
false.
If P=NP, then there exists a program which runs in O(poly(N)) and outputs a satisfying input to the formula, if such a formula exists.
If P=coNP, there exists a program which runs in O(poly(N)) and outputs a formal proof that no formula exists, if no formula exists.
If P=NP, then since P is closed under complement NP=coNP. So, there exists a program which runs in O(poly(N)) and does both. That program is the k'th program in the enumeration. k is O(1)! Since it runs in O(poly(N)) our brute force simulation only requires
k*O(poly(N))+O(poly(N))^2
rounds once it reaches the program in question. As such, the brute force simulation runs in polynomial time!
(Note that k is exponential in the size of the program; this approach is not really feasible, but it suggests that it would be hard to do a nonconstructive proof that P=NP, even if it were the case.)
An interesting read that is somewhat related to this
To some extent, the form such a proof needs to have depends on your philosophical point of view (= the axioms you deem to be true) - e.g., as a contructivist you would demand the construction of an actual algorithm that requires polynomial time to solve an NP-complete problem. This could be done by using reduction, but not with an indirect proof. Anyhow, it really seems to be very unlikely :)
The proof would deduce a contradiction from to the assumption that at least one element (problem) of NP isn't also an element of P.
Assuming some background in mathematics, how would you give a general overview of computational complexity theory to the naive?
I am looking for an explanation of the P = NP question. What is P? What is NP? What is a NP-Hard?
Sometimes Wikipedia is written as if the reader already understands all concepts involved.
Hoooo, doctoral comp flashback. Okay, here goes.
We start with the idea of a decision problem, a problem for which an algorithm can always answer "yes" or "no." We also need the idea of two models of computer (Turing machine, really): deterministic and non-deterministic. A deterministic computer is the regular computer we always thinking of; a non-deterministic computer is one that is just like we're used to except that is has unlimited parallelism, so that any time you come to a branch, you spawn a new "process" and examine both sides. Like Yogi Berra said, when you come to a fork in the road, you should take it.
A decision problem is in P if there is a known polynomial-time algorithm to get that answer. A decision problem is in NP if there is a known polynomial-time algorithm for a non-deterministic machine to get the answer.
Problems known to be in P are trivially in NP --- the nondeterministic machine just never troubles itself to fork another process, and acts just like a deterministic one. There are problems that are known to be neither in P nor NP; a simple example is to enumerate all the bit vectors of length n. No matter what, that takes 2n steps.
(Strictly, a decision problem is in NP if a nodeterministic machine can arrive at an answer in poly-time, and a deterministic machine can verify that the solution is correct in poly time.)
But there are some problems which are known to be in NP for which no poly-time deterministic algorithm is known; in other words, we know they're in NP, but don't know if they're in P. The traditional example is the decision-problem version of the Traveling Salesman Problem (decision-TSP): given the cities and distances, is there a route that covers all the cities, returning to the starting point, in less than x distance? It's easy in a nondeterministic machine, because every time the nondeterministic traveling salesman comes to a fork in the road, he takes it: his clones head on to the next city they haven't visited, and at the end they compare notes and see if any of the clones took less than x distance.
(Then, the exponentially many clones get to fight it out for which ones must be killed.)
It's not known whether decision-TSP is in P: there's no known poly-time solution, but there's no proof such a solution doesn't exist.
Now, one more concept: given decision problems P and Q, if an algorithm can transform a solution for P into a solution for Q in polynomial time, it's said that Q is poly-time reducible (or just reducible) to P.
A problem is NP-complete if you can prove that (1) it's in NP, and (2) show that it's poly-time reducible to a problem already known to be NP-complete. (The hard part of that was provie the first example of an NP-complete problem: that was done by Steve Cook in Cook's Theorem.)
So really, what it says is that if anyone ever finds a poly-time solution to one NP-complete problem, they've automatically got one for all the NP-complete problems; that will also mean that P=NP.
A problem is NP-hard if and only if it's "at least as" hard as an NP-complete problem. The more conventional Traveling Salesman Problem of finding the shortest route is NP-hard, not strictly NP-complete.
Michael Sipser's Introduction to the Theory of Computation is a great book, and is very readable. Another great resource is Scott Aaronson's Great Ideas in Theoretical Computer Science course.
The formalism that is used is to look at decision problems (problems with a Yes/No answer, e.g. "does this graph have a Hamiltonian cycle") as "languages" -- sets of strings -- inputs for which the answer is Yes. There is a formal notion of what a "computer" is (Turing machine), and a problem is in P if there is a polynomial time algorithm for deciding that problem (given an input string, say Yes or No) on a Turing machine.
A problem is in NP if it is checkable in polynomial time, i.e. if, for inputs where the answer is Yes, there is a (polynomial-size) certificate given which you can check that the answer is Yes in polynomial time. [E.g. given a Hamiltonian cycle as certificate, you can obviously check that it is one.]
It doesn't say anything about how to find that certificate. Obviously, you can try "all possible certificates" but that can take exponential time; it is not clear whether you will always have to take more than polynomial time to decide Yes or No; this is the P vs NP question.
A problem is NP-hard if being able to solve that problem means being able to solve all problems in NP.
Also see this question:
What is an NP-complete in computer science?
But really, all these are probably only vague to you; it is worth taking the time to read e.g. Sipser's book. It is a beautiful theory.
This is a comment on Charlie's post.
A problem is NP-complete if you can prove that (1) it's in NP, and
(2) show that it's poly-time reducible to a problem already known to
be NP-complete.
There is a subtle error with the second condition. Actually, what you need to prove is that a known NP-complete problem (say Y) is polynomial-time reducible to this problem (let's call it problem X).
The reasoning behind this manner of proof is that if you could reduce an NP-Complete problem to this problem and somehow manage to solve this problem in poly-time, then you've also succeeded in finding a poly-time solution to the NP-complete problem, which would be a remarkable (if not impossible) thing, since then you'll have succeeded to resolve the long-standing P = NP problem.
Another way to look at this proof is consider it as using the the contra-positive proof technique, which essentially states that if Y --> X, then ~X --> ~Y. In other words, not being able to solve Y in polynomial time isn't possible means not being to solve X in poly-time either. On the other hand, if you could solve X in poly-time, then you could solve Y in poly-time as well. Further, you could solve all problems that reduce to Y in poly-time as well by transitivity.
I hope my explanation above is clear enough. A good source is Chapter 8 of Algorithm Design by Kleinberg and Tardos or Chapter 34 of Cormen et al.
Unfortunately, the best two books I am aware of (Garey and Johnson and Hopcroft and Ullman) both start at the level of graduate proof-oriented mathematics. This is almost certainly necessary, as the whole issue is very easy to misunderstand or mischaracterize. Jeff nearly got his ears chewed off when he attempted to approach the matter in too folksy/jokey a tone.
Perhaps the best way is to simply do a lot of hands-on work with big-O notation using lots of examples and exercises. See also this answer. Note, however, that this is not quite the same thing: individual algorithms can be described by asymptotes, but saying that a problem is of a certain complexity is a statement about every possible algorithm for it. This is why the proofs are so complicated!
I remember "Computational Complexity" from Papadimitriou (I hope I spelled the name right) as a good book
very much simplified: A problem is NP-hard if the only way to solve it is by enumerating all possible answers and checking each one.
Here are a few links on the subject:
Clay Mathematics statement of P vp NP problem
P vs NP Page
P, NP, and Mathematics
In you are familiar with the idea of set cardinality, that is the number of elements in a set, then one could view the question like P representing the cardinality of Integer numbers while NP is a mystery: Is it the same or is it larger like the cardinality of all Real numbers?
My simplified answer would be: "Computational complexity is the analysis of how much harder a problem becomes when you add more elements."
In that sentence, the word "harder" is deliberately vague because it could refer either to processing time or to memory usage.
In computer science it is not enough to be able to solve a problem. It has to be solvable in a reasonable amount of time. So while in pure mathematics you come up with an equation, in CS you have to refine that equation so you can solve a problem in reasonable time.
That is the simplest way I can think to put it, that may be too simple for your purposes.
Depending on how long you have, maybe it would be best to start at DFA, NDFA, and then show that they are equivalent. Then they understand ND vs. D, and will understand regular expressions a lot better as a nice side effect.