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.
