I would like to quote from Wikipedia
In mathematics, the minimum k-cut, is a combinatorial optimization
problem that requires finding a set of edges whose removal would
partition the graph to k connected components.
It is said to be the minimum cut if the set of edges is minimal.
For a k = 2, It would mean Finding the set of edges whose removal would Disconnect the graph into 2 connected components.
However, The same article of Wikipedia says that:
For a fixed k, the problem is polynomial time solvable in O(|V|^(k^2))
My question is Does this mean that minimum 2-cut is a problem that belongs to complexity class P?
The min-cut problem is solvable in polynomial time and thus yes it is true that it belongs to complexity class P. Another article related to this particular problem is the Max-flow min-cut theorem.
First of all, the time complexity an algorithm should be evaluated by expressing the number of steps the algorithm requires to finish as a function of the length of the input (see Time complexity). More or less formally, if you vary the length of the input, how would the number of steps required by the algorithm to finish vary?
Second of all, the time complexity of an algorithm is not exactly the same thing as to what complexity class does the problem the algorithm solves belong to. For one problem there can be multiple algorithms to solve it. The primality test problem (i.e. testing if a number is a prime or not) is in P, but some (most) of the algorithms used in practice are actually not polynomial.
Third of all, in the case of most algorithms you'll find on the Internet evaluating the time complexity is not done by definition (i.e. not as a function of the length of the input, at least not expressed directly as such). Lets take the good old naive primality test algorithm (the one in which you take n as input and you check for division by 2,3...n-1). How many steps does this algo take? One way to put it is O(n) steps. This is correct. So is this algorithm polynomial? Well, it is linear in n, so it is polynomial in n. But, if you take a look at what time complexity means, the algorithm is actually exponential. First, what is the length of the input to your problem? Well, if you provide the input n as an array of bits (the usual in practice) then the length of the input is, roughly said, L = log n. Your algorithm thus takes O(n)=O(2^log n)=O(2^L) steps, so exponential in L. So the naive primality test is in the same time linear in n, but exponential in the length of the input L. Both correct. Btw, the AKS primality test algorithm is polynomial in the size of input (thus, the primality test problem is in P).
Fourth of all, what is P in the first place? Well, it is a class of problems that contains all decision problems that can be solved in polynomial time. What is a decision problem? A problem that can be answered with yes or no. Check these two Wikipedia pages for more details: P (complexity) and decision problems.
Coming back to your question, the answer is no (but pretty close to yes :p). The minimum 2-cut problem is in P if formulated as a decision problem (your formulation requires an answer that is not just a yes-or-no). In the same time the algorithm that solves the problem in O(|V|^4) steps is a polynomial algorithm in the size of the input. Why? Well, the input to the problem is the graph (i.e. vertices, edges and weights), to keep it simple lets assume we use an adjacency/weights matrix (i.e. the length of the input is at least quadratic in |V|). So solving the problem in O(|V|^4) steps means polynomial in the size of the input. The algorithm that accomplishes this is a proof that the minimum 2-cut problem (if formulated as decision problem) is in P.
A class related to P is FP and your problem (as you formulated it) belongs to this class.
Related
I am actually looking for description what NP alogrithm actually means and what kind of algo/problem can be classified as NP problem
I have read many resources on net . I liked
https://www.quora.com/What-are-P-NP-NP-complete-and-NP-hard
What are the differences between NP, NP-Complete and NP-Hard?
Non deterministic Turing machine
What are NP problems?
What are NP and NP-complete problems?
Polynomial problem :-
If the running time is some polynomial function of the size of the input**, for instance if the algorithm runs in linear time or quadratic time or cubic time, then we say the algorithm runs in polynomial time . Example can be binary search
Now I do understand Polynomial problem . But not able to contrast it with NP.
NP(nondeterministic polynomial Problem):-
Now there are a lot of programs that don't (necessarily) run in polynomial time on a regular computer, but do run in polynomial time on a nondeterministic Turing machine. These programs solve problems in NP, which stands for nondeterministic polynomial time.
I am not able to to understand/think of example that does not run in polynomial time on a regular computer. Per mine current understanding, Every problem/algo can be solved
in some polynomial function of time which can or can't be proportional to time. I know i am missing something here but really could not grasp this concept. Could someone
give example of problem which can not be solved in polynomial time on regular computer but can be verified in polynomial time ?
One of the example given at second link mentioned above is Integer factorization is in NP. This is the problem that given integers n and m, is there an integer f with 1 < f < m, such that f divides n (f is a small factor of n)? why this can't be solved in some polynomial time on regular computer ? we can check for all number from 1 to n if they divide n or not. Right ?
Also where verification part come here(i mean if it can be solved in polynomial time but then how the problem solution can be verified in polynomial time)?
Your question touches several points.
First, in the sense relevant to your question, the size of a problem is defined to be the size of the representation of the problem. So, for example, when you write about the problem of a divisor of n. What is the representation of n? It is a series of characters of length q (I don't want to be more specific than that). In general, n is exponential in q. So when you talk about a simple loop from 1 to n, you're talking about something that is exponential in the size of the input. For example, the number "999999999999999" represents the number 999999999999999. That is quite a large number, but it is represented by 12 characters here.
Second, while there is more than a single way to define the class NP, perhaps the simplest one for decision problems (which is the type you raise in your question, namely is something true or not) is that if the answer is true, then there is an "certificate" that can be verified in polynomial time. For example, consider the Hamilton Path Problem. This is (probably) a hard problem to solve, but, if you are given a hamilton path as an answer, it is very easy to verify that it is so; specifically, it can be done in polynomial time. For the Hamilton Path Problem, the path is a polynomial-time verifiable certificate, and therefore this problem is NP.
It's probably worth noting how the idea of "checking a solution in polynomial time" relates to a nondeterministic Turing Machine solving a problem: in a normal (deterministic) Turing Machine, there is a well-defined set of instructions telling the machine exactly what to do in any situation ("if you're in state 3 and see an 'a', move left, if you're in state 7 and see a 'c', overwrite it with a 'b', etc.") whereas in a nondeterministic Turing Machine there is more than one option for what to do in some situations ("if you're in state 3 and see an 'a', either move right or overwrite it with a 'b'"). In terms of algorithms, this lets us "guess" solutions in the sense that if we can encode a problem into a language on an alphabet* then we can use a nondeterministic Turing Machine to generate strings on this alphabet, and then use a standard (deterministic) Turing Machine to ensure that it is correct. If we assume that we always make the right guess, then the runtime of our algorithm is simply the runtime of the deterministic checking part, which for NP problems runs in polynomial time. This is what it means for a problem to be 'decidable in polynomial time on a nondeterministic Turing Machine', and why it is often simply phrased as 'checking a solution/ certificate in polynomial time'.
*
Example: The Hamiltonian Path problem could be encoded as follows:
Label the vertices of the graph 1 through n, where n is the number of vertices. Our alphabet is then the numbers 1 through n, and our language consists of all words such that
a) every integer from 1 to n appears exactly once
and
b) for every consecutive pair of integers in a word, the vertices with those labels are connected
Polynomial Time :- Problem which can be solved in polynomial time of input size is called polynomial problem. In plain simple words :- Here Solution to problem is fast. For
example sorting, binary search
Non deterministic polynomial :- Theoretically the problems which can be verified in polynomial time irrespective of actual solution time complexity (which can be polynomial or not polynomial). So some problem which are P can also be NP.
But Informally people while conversation/posts use the NP term in below sense
Problem which can not be solved in polynomial time of input size is called polynomial problem. In plain simple words :- Here Solution to problem is not fast. You may have to try different permutation/combination or guessing work. But Verification part is fast and can be done in polynomial time. Like
input some numbers X and divide the numbers into two groups that difference in their sum is minimum
I really liked the Alex Flint answer at https://www.quora.com/What-are-P-NP-NP-complete-and-NP-hard .Above is just gist of that.
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.
When learning for exam in Algorithms and Data Structures i have stumbled upon a question, what does it mean if an algorithm has pseudo polynomial time efficiency( analysis)
Did a lot of searching but turned empty handed
It means that the algorithm is polynomial with respect to the size of the input, but the input actually grows exponentially.
For example take the subset sum problem. We have a set S of n integers and we want to find a subset which sums up to t.
To solve this problem you can just check the sum of each subset, so it is O(P) where P is the number of subsets. However in fact the number of subsets is 2^n so the algorithm has exponential complexity.
I hope this introduction helps to understand the wikipedia's article about it http://en.wikipedia.org/wiki/Pseudo-polynomial_time :)
Some problems that are NP-hard are also fixed-parameter tractable, or FPT. Wikipedia describes a problem as fixed-parameter tractable if there's an algorithm that solves it in time f(k) · |x|O(1).
What does this mean? Why is this concept useful?
To 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!
I believe that the explanation of #templatetypedef was already quite comprehensive of the generality of FPT.
I would like to add that in practice, it appears quite often that the class of problem one is trying to solve is FPT, such as above examples.
In the case of problems expressed as set of constraints (e.g. SAT, CSP, ILP, etc.) a very common parameter is treewidth, which basically explicits how much your problem is organized as a tree.
This allows to split ones problem into a tree of subproblems which can then be solved more individually using dynamic programming.
In such case, many problems are linear-time fixed-parameter tractable, that is the complexity grows linearly with the number of components (i.e. the size of the system) by exponentially in the size of its biggest component.
Although the use of explicit techniques is possible to solve sub-problems is possible, in order to scale-up to more reasonnable instances, using symbolic representations is recomended.
Quoted from wikipedia, the P vs NP problem, regarding the time complexity of algorithms "... asks whether every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer."
I am hoping that somebody can clarify what the difference between "verifying the problem" and "solving the problem" is here.
I am hoping that somebody can clarify what the difference between "verifying the problem" and "solving the problem" is here.
It's not "verifying the problem", but "verifying the solution". For example you can check in polynomial time whether a given set is valid for SAT. The actual generation of such set is NP hard. The section Verifier-based definition in the Wikipedia article NP (complexity) could help you a little bit:
Verifier-based definition
In order to explain the verifier-based definition of NP, let us consider the subset sum problem: Assume that we are given some integers, such as {−7, −3, −2, 5, 8}, and we wish to know whether some of these integers sum up to zero. In this example, the answer is "yes", since the subset of integers {−3, −2, 5} corresponds to the sum (−3) + (−2) + 5 = 0. The task of deciding whether such a subset with sum zero exists is called the subset sum problem.
As the number of integers that we feed into the algorithm becomes larger, the number of subsets grows exponentially, and in fact the subset sum problem is NP-complete. However, notice that, if we are given a particular subset (often called a certificate), we can easily check or verify whether the subset sum is zero, by just summing up the integers of the subset. So if the sum is indeed zero, that particular subset is the proof or witness for the fact that the answer is "yes". An algorithm that verifies whether a given subset has sum zero is called verifier. A problem is said to be in NP if and only if there exists a verifier for the problem that executes in polynomial time. In case of the subset sum problem, the verifier needs only polynomial time, for which reason the subset sum problem is in NP.
If you're more into graph theory the Hamilton cycle is a NP-complete problem. It's trivial to check whether a given solution is a Hamilton cycle (linear complexity, traverse the solution path), but if P != NP than there exists no algorithm with polynomial runtime which solves the problem.
Maybe the term "quick" is misleading in this context. An algorithm is quick in this regard if and only if it's worst-case runtime is bounded by a polynomial function, such as O(n) or O(n log n). The creation of all permutations for a given range with the length n is not bounded, as you have n! different permutations. This means that the problem could be solved in n 100 log n, which will take a very long time, but this is still considered fast. On the other hand, one of the first algorithms for TSP was O(n!) and another one was O(n2 2n). And compared to polynomial functions these things grow really, really fast.
The RSA encryption uses prime numbers like following: two big prime numbers P and Q (200-400 digits each) are multiplied to form the public key N. N=P*Q
To break the encryption one needs to figure out P and Q given the N.
While finding P and Q is very difficult and can take years, verifing the solution is just about multiplying P by Q and comparing with N.
So solving the problem is very hard while it takes nothing to verifying the solution.
P.S. The example is only part of the RSA simplified for this question. The real RSA is much more complex.