What are the implications of solving the P versus NP problem? What practical effects would that have? Not trying to belittle the problem, just curious as an outsider.
We basically operate under the assumption that P!=NP currently. Validation that this is true doesn't really change much. I can't speak to how this may help a academic researcher in CompSci, but it probably doesn't change much for most programmers.
The problem of finding the optimum path is not in NP, if I give you a candidate solution you can't easily check if it's the global optimum.
What is in NP is the decision problem, finding a path that is better than a given bound. If I hand you a candidate solution, you just have to compare the sum of the distances to the bound to check it.
edit: The wikipedia mentions that the TSP problem is NP-hard and explicitly says that the decision version of this problem is NP-complete. My assumption is that if the optimization version was proven to be NP-hard there would be no need to explicitly mention the decision version.
I can think of a way to use the decision version to find a solution to the optimization version but i feel like it must be flawed:
First we do a binary search on `L` (the length of the tour- input to the decision version) so we can find the optimal L within a factor of epsilon (maybe this epsilon is the flaw? But I don't think that is the case.) Now we pick an edge and increase its weight to infinity. Now we do the binary search again on the new graph. If the value of the optimal solution has changed it means that the edge must be in the optimal optimization solution. By doing the same process on all the edges we can find the optimal solution.
As I said there must be a flaw in the above algorithm but I can't find it.
Hence, you can use bisection to compute the actual optimum, not involving epsilon at all.
I don't think there is anything wrong with optimization been reducible to decision - it's quite common method both in theory and in practice.
note: I am not implying that the above source is reputable. But it does hint that the solution to this problem probably is not this trivial.
Here is why I think the algorithm is correct:
At each step the edge that we are considering is either contained in all the optimal solutions or only some of them. If the edge is contained in all the solutions, increasing its weight to infinity would change the optimal solution and we pick that edge in our solution. Otherwise (if the edge is contained in only some of the solutions) increasing the weight would not change the solution because there is another optimal solution that does not contain that edge so we do not pick that edge.
So we can prove this theorem: At every step of the algorithm if an edge is picked, it is contained in all the optimal solutions.
So the algorithm does not pick any extra edges. Now we have to prove that it includes all the necessary edges. But that is easy because each time that we choose not to including an edge, we are sure that there is an optimal solution in the remaining graph so we are never left with a graph with no optimal solution.
I think you assumed that I meant we change the edge weights from infinity back to their original value at each step but that is not what I meant.
I interpreted this as meaning you were selecting the falsifying removals' edges, instead of removing the non-falsifying ones. You've got it.
It's interesting because it mix two interesting topics, that are well known in the popular science forums, but are very technical and most people don't want to read all technical the details of both.
Relevant xkcd: https://xkcd.com/1240/
If someone could prove P=NP but no one could find an algorithm. That would be incredibly funny in some sense. Like a huge joke played on us by the universe.
However, the reason why I chose to single out crypto specifically is because it has the most to lose if that algorithm exists. Our current methods of encryption become unsafe regardless of whether the algorithm is known or not. I don't think you can claim that your encryption is secure if there is an algorithm that can crack it in polynomial time, regardless of whether the algorithm is known or not.
This is 100% true for all practical purposes. But there is an explicit algorithm for NP-complete problems that runs in polynomial time iff P=NP. The Wikipedia page has it written down. https://en.m.wikipedia.org/wiki/P_versus_NP_problem
Could you back that up with some citations? This doesn't ring true. But my pure CS has withered a bit...
However, while P=NP, the algorithm (oracle) resides on the other side of the event horizon. This is called the MAD paradox.
https://nerdynotmad.com/p-equals-np/
The submitted proof will be found to be incorrect.
Sure the race would be on to improve that, but in the meanwhile no difficult problems would become solvable.
I especially liked this bit:
David Johnson famously once said, For any instance {G = (V, E)} that one could fit into the known universe, one would easily prefer {|V |^{70}} to even constant time, if that constant had to be one of Robertson and Seymour’s.
I was curious so I tracked down this:
Johnson estimated that the hidden constant is “somewhat larger” than 2 ⇑ (2 ⇑ (2 ⇑ (h/2)) + 3), where 2 ⇑ t denotes an exponential tower of t 2s (2 ⇑ 0 = 1 and 2 ⇑ t = 2^2⇑(t−1)) and h is the number of vertices in H.
If NPC problems where P in n^{10^100}, wouldn't we expect a wealth of problems between there and the myriad at n^2 or so?
It's easier to get people to devote resources to a hard but solvable problem than to one which may not even be solvable.
What does an O(n^10,000) algorithm do? What understandable problem yields a solution that behaves that way?
In another realm of computational mathematics, matrix mjultiplication is cubic, with optimizations that can approach quadratic time with lots of effort. It's conjectured that matrix multiplication can actually be brought arbitrarily close to quadratic time, but at the expense of ever-more-complex algorithms.
It could very well be that the biggest interesting exponent in P is 3 or 4.
For example, when you have an O(n^3) algorithm and want to process 10,000 elements (which are very few in many situations), it will be, as a rough estimate, 1,000,000,000,000 times slower than processing a single element. This will be acceptable only in very specific situations. As a rule of thumb, an exponent of 2 is already unacceptably slow in many practical situations, and at least on the verge of being unacceptable in others.
You're essentially referring to the idea that could be loads of high-order poly-time algorithms, but we ignore them because those algorithms are slow, and therefore we don't use them. So in the space of all useful algorithms, we simply have a very biased sample.
I think the truth is more profound than that. There actually don't exist very many interesting* algorithms in the classes O(n^(k>3)). The real world we live in and model does not feature many interesting problems for which high-order polynomial complexity algorithms are natural solutions.
*Not sure what the right word to use here is...maybe non-trivial? The point I'm going for is to say that obviously we can invent an O(n^5) algorithm by simply nesting our loops five-deep and printing something, but that's a constructed example. I'm looking for algorithms that naturally arise as a solution to some problem.
Donald Knuth believes P = NP.
Source: http://www.informit.com/articles/article.aspx?p=2213858&WT.m... (question 17). Also cf. https://www.quora.com/Why-does-Donald-Knuth-think-that-P-NP
Let me explain why P=NP probably won't have any impact, since you see a lot of bullshit claiming lots of bad things will happen. The description of complexity classes like P and NP sweep a lot of details under the rug, and those details matter a lot of practical matters.
The more important of these matters is that complexity is based on worst-case running time, not average-case or typical-case. Often times, we can solve most instances of "hard" problems. We can factor most integers, for example--half of them are divisible by 2, and another sixth divisible by 3. If we limit are inputs to factoring products of two primes of roughly equal size, that is difficult. SAT is another example: it may be the canonical NP-complete problem, but many people think nothing of using a SAT solver (or its cousin, the SMT solver) to solve for things like "how do I find an input that can reach this program point." Except if solving that condition requires, say, finding SHA256(x) = binary digits of pi.
This is one of the main cruxes of P?=NP that doesn't come out much: it's not so much that problems are hard or easy, it's that there's this field of problem instances that seem to be intrinsically hard. Indeed, if you look at restricted versions of these NP-complete problems, you'll find that some restrictions still retain NP-complete, but a very slight reduction in those restrictions suddenly admits a very simple, fast, easy solution.
The related notion that you see people sometimes bring up and dismiss is that P and NP hide constants. This objection tends to be dismissed because most people have no familiarity with polynomial-time algorithms with massive constants. But such algorithms do exist, and they tend to crop up in combinatorial-style algorithms. Which, incidentally, is probably what a polynomial algorithm for an NP-complete problem would be.
Let me explain by analogy of a not-so-recently-solved problem that's a weaker but related notion to P?=NP, L?=SL. This question is essentially asking "could you solve every problem that's equivalent to checking for connectivity in an undirected graph using only constant memory" [1]. The answer turns out to be yes. In essence, there is a deterministic string of coin flips deciding your next vertex that will, if followed, eventually guarantee that you will hit every node in the graph that you can reach within a certain amount of time--if your graph has certain properties. And it's possible to transform every graph into one with this property by replacing every node with an expander graph that's only of size 3^2^16 at the smallest. But since the new graph is 3^2^16 * N, that large number is a constant factor that "doesn't matter".
These kinds of constants show up a lot in combinatorial algorithms. All of these algorithms basically have the property that you can solve the input fairly easy if the input has some structure, but to guarantee that the input has that structure, you need to embed it in a very large instance. I argue that, if P=NP were to hold, it would have a similar form to these algorithms. We know that there are some instances that just seem to be intrinsically harder than others, and we know that there are large classes of instances that can be easily solved with fast algorithms. If we haven't been able to generalize the fast algorithms to cover the hard instances but a "fast" (ignoring constants) algorithm must exist, then the apparent complexity has to be generated from sort of fiendish complexity-multiplying, combinatoric structure.
We already see in modern security algorithms that ciphers and hashes have to carefully choose their parameters to make sure they stay in a hard subset of their input algorithms. So even if P=NP, the hard subset is likely to remain much harder than the easy subset, and that gap is sufficient to maintain security.
[1] A very imprecise characterization.
I should say, theoretically breaking public-key. In reality the problem may remain too hard to brute force even if the P vs NP problem is solved.
This preprint "implies P not equal NP".