https://stackoverflow.com/questions/19643939
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RussianTo begin with, under the assumption that P ≠ NP, there are no polynomial-time, exact algorithms for any NP-hard problem. Although we don't know whether P = NP or P ≠ NP, we don't have any polynomial-time algorithms for any NP-hard problems.
The idea behind fixed-parameter tractability is to take an NP-hard problem, which we don't know any polynomial-time algorithms for, and to try to separate out the complexity into two pieces - some piece that depends purely on the size of the input, and some piece that depends on some "parameter" to the problem.
As an example, consider the 0/1 knapsack problem. In this problem, you're given a list of n objects that have associated weights and values, along with some maximum weight W that you're allowed to carry. The question is to determine the maximum amount of value that you can carry. This problem is NP-hard, meaning that there's no polynomial-time algorithm that solves it. A brute-force method will take time around O(2n) by considering all possible subsets of the items, which is extremely slow for large n. However, it is possible to solve this problem in time O(nW), where n is the number of elements and W is the amount of weight you can carry. If you look at the runtime O(nW), you'll notice that it's split into two parts: a component that's linear in the number of elements (the n part) and a component that's linear in the weight (the W part). If W is any fixed constant, then the runtime of this algorithm will be O(n), which is linear-time, even though the problem in general is NP-hard. This means that if we treat W as some tunable "parameter" of the problem, for any fixed value of this parameter, the problem ends up running in polynomial time (which is "tractable," in the complexity theory sense of the word.)
As another example, consider the problem of finding long, simple paths in a graph. This problem is also NP-hard, and the naive algorithm for finding simple paths of length k in a graph takes time O(n! / (n - k)!), which for large k ends up being superexponential. However, using the technique of color-coding, it's possible to solve this problem in time O((2e)kn3 log n), where k is the length of the path to find and n is the number of nodes in the input graph. Notice that this runtime also has two "components:" one component that's a polynomial in the number of nodes in the input graph (the n3 log n part) and one component that's exponential in k (the (2e)k part). This means that for any fixed value of k, there's a polynomial-time algorithm for finding length-k paths in the graph; the runtime will be O(n3 log n).
In both of these cases, we can take a problem for which we have an exponential-time solution (or worse) and find a new solution whose runtime is some polynomial in n times some crazy-looking function of some extra "parameter." In the case of the knapsack problem, that parameter is the maximum amount of weight we can carry; in the case of finding long paths, the parameter is the length of the path to find. Generally speaking, a problem is called fixed-parameter tractable if there is some algorithm for solving the problem defined in terms of two quantities: n, the size of the input, and k, some "parameter," where the runtime is
O(p(n) · f(k))
Where p(n) is some polynomial function and f(k) is an arbitrary function in k. Intuitively, this means that the complexity of the problem scales polynomially with n (meaning that as only the problem size increases, the runtime will scale nicely), but can scale arbitrarily badly with the parameter k. This separates out the "inherent hardness" of the problem such that the "hard part" of the problem is blamed on the parameter k, while the "easy part" of the problem is charged to the size of the input.
Once you have a runtime that looks like O(p(n) · f(k)), we immediately get polynomial-time algorithms for solving the problem for any fixed k. Specifically, if k is fixed, then f(k) is some constant, so O(p(n) · f(k)) is just O(p(n)). This is a polynomial-time algorithm. Therefore, if we "fix" the parameter, we get back some "tractable" algorithm for solving the problem. This is the origin of the term fixed-parameter tractable.
(A note: Wikipedia's definition of fixed-parameter tractability says that the algorithm should have runtime f(k) · |x|O(1). Here, |x| refers to the size of the input, which I've called n here. This means that Wikipedia's definition is the same as saying that the runtime is f(k) · nO(1). As mentioned in this earlier answer, nO(1) means "some polynomial in n," and so this definition ends up being equivalent to the one I've given here).
Fixed-parameter tractability has enormous practical implications for a problem. It's common to encounter problems that are NP-hard. If you find a problem that's fixed-parameter tractable and the parameter is low, it can be significantly more efficient to use the fixed-parameter tractable algorithm than to use the normal brute-force algorithm. The color-coding example above for finding long paths in a graph, for example, has been used to great success in computational biology to find sequencing pathways in yeast cells, and the 0/1 knapsack solution is used frequently because common values of W are low enough for it to be practical.
Hope this helps!
