Am a bit confused about the relationship between undecidable problems and NP hard problems. Whether NP hard problems are a subset of undecidable problems, or are they just the same and equal, or is it that they are not comparable?
For me, I have been arguing with my friends that undecidable problems are a superset to the NP hard problems. There would exist some problems that are not in NP hard but are undecidable. But i am finding this argument to be weak and am confused a bit. Are there NP-complete problems that are undecidable.? is there any problem in NP hard which is decidable.??
Some discussion would be of great help! Thanks!
Undecidable = unsolvable for some inputs. No matter how much (finite) time you give your algorithm, it will always be wrong on some input.
NP-hard ~= super-polynomial running time (assuming P != NP). That's hand-wavy, but basically NP-hard means it is at least as hard as the hardest problem in NP.
There are certainly problems that are NP-hard which are not undecidable (= are decidable). Any NP-complete problem would be one of them, say SAT.
Are there undecidable problems which are not NP-hard? I don't think so, but it isn't easy to rule it out - I don't see an obvious argument that there must be a reduction from SAT to all possible undecidable problems. There could be some weird undecidable problems which aren't very useful. But the standard undecidable problems (the halting problem, say) are NP-hard.
An NP-hard is a problem that is at least as hard as any NP-complete problem.
Therefore an undecidable problem can be NP-hard. A problem is NP-hard if an oracle for it would make solving NP-complete problems easy (i.e. solvable in polynomial time). We can imagine an undecidable problem such that, given an oracle for it, NP-complete problems would be easy to solve. For example, obviously every oracle that solves the halting problem can also solve an NP-complete problem, so every Turing-complete problem is also NP-hard in the sense that a (fast) oracle for it would make solving NP-complete problems a breeze.
Therefore Turing-complete undecidable problems are a subset of NP-hard problems.
Undecidable problem e.g. Turing Halting Problem is NP-Hard only.
<---------NP Hard------>
|------------|-------------||-------------|------------|--------> Computational Difficulty
|<----P--->|
|<----------NP---------->|
|<-----------Exponential----------->|
|<---------------R (Finite Time)---------------->|
In this diagram, that small pipe shows overlapping of NP and NP-Hard and which shows NP-Completeness, i.e. set of those problems which are NP as well as NP-Hard.
Undecidable problems are NP Hard problems which do not have solution and which are not in NP.
Related
Why is NP-hard unequal to NP-complete?
My informal understanding of definitions being used:
NP - all problems that can be verified in polynomial time
NP-complete - all problems that are NP and NP-hard
NP-hard - at least as hard as the hardest problem in NP
Decision Problem - A problem that asks a question with regards to an input and outputs a bool value
Confusion:
The problem with unknown solution of P vs NP arises from the fact that we cannot prove or disprove all problems in NP can be solved in polynomial time. It feels like a similar question arises from NP-complete vs NP-hard. How do we know all problems in NP-hard cannot be verified in polynomial time and thus result in NP-hard=NP-complete?
Here is my line of reasoning
From online research the distinction seems that this has something to do with decision problems (a concept I'm entirely new to but seem simple enough). I think this means that problems in NP have complementary decision problems that ask if an input is the solution to the problem. Let's say the problem is to find an optimal solution. I believe the complementary decision problem to be "is the given input the optimal solution?"and I believe that if this decision problem is verifiable in polynomial time then the problem is NP-complete (or in NP). So this means that NP-hard problems that aren't NP-complete problems are those that either have no decision problem (which I believe is never true since any brute force solution can answer this) or a problem is NP-hard and not NP-complete if it has a decision problem that's not verifiable in polynomial time. If it is the latter then it feels like we have the same problem from P vs NP. That is, how do we confirm all decision problems in NP-hard do not have polynomial time solutions?
Sorry if the above phrasing is weird. I will try and clarify any confusion in my question.
notes
I am interested in both an intuitive explanation and a formal explanation (a proof if it's a complicated answer). The formal explanation can certainly be a link to an academic paper. I don't want anyone to invest a significant amount of time into an overly complicated proof that may be beyond the scope of my understanding (I've found complexity theory to become very quickly... complex).
If it helps for the sake of explanation I have done work on the traveling salesman problem and I am currently working on a paper for the nurse scheduling problem (I believe these are NP-hard problems).
NP-Hard includes all problems whose solutions can be used to derive solutions to problems in NP with polynomial overhead.
This includes lots of problems that aren't in NP. For instance, the halting problem - an undecidable problem - is NP-Hard, because any problem in NP can be reduced to it in polynomial time:
Reduce any problem in NP to an instance of the NP-Complete problem 3-SAT
Construct in polynomial time a TM which checks all assignments and halts iff a satisfying assignment is found.
Use a solution to the halting problem to tell whether the TM halts.
If it halts, accept; otherwise, reject.
Can someone give me an example of an undecidable problem that is not NP-hard?
I'm unable to understand the difference between the two.
Thanks very much!
An NP-hard problem is one such that every problem in NP can be reduced to it. In fact, it is "at least as hard as" the problems in NP class. For example, TSP (Traveling Sales Person) is NP-hard. However, undecidable is a problem for which there is no algorithm that always decide correctly. For example, the question of whether a program halts at some point or not is undecidable. In fact, you may not have an algorithm that can answer this question correctly for all programs in the world. (This can be proved)
So, in brief, an undecidable problem is logically hard; no matter how strong your computers or algorithms are, they cannot be solved. But, NP-hard problems have algorithms to be solved with but those algorithms are not polynomial in time.
Assuming P != NP
The euler diagram shows a part not part of P and NP-complete. I read on wikipedia that this set is called NP-Intermediate.
Euler Diagram
I have some doubts as to how are NPI problems defined?
An NP-intermediate problem is a decision problem that
is in NP (that is, "yes" answers can be verified in polynomial time),
is not in P (that is, there is no polynomial-time algorithm for solving the problem), and
is not NP-complete.
That last criterion can be stated in a number of different ways. One way to say this is that there is no polynomial-time mapping reduction from SAT to that particular problem.
These problems are primarily of theoretical interest right now because we don't know if any NP-intermediate problems exist - if we could find one, we'd have a problem in NP that's not in P, meaning that P ≠ NP! However, they're interesting because if we can prove that P ≠ NP, then we know that there are some problems in NP that are too hard to be solved in polynomial time, but which aren't among the "hardest" of the hard problems in NP (the problems that are NP-complete).
In the event that P = NP, then there would not be any NP-intermediate problems because you couldn't have a problem in NP but not in P. If P ≠ NP, then Ladner's theorem guarantees at least one NP-intermediate problem exists, but does so by specifically constructing a problem that is highly artificial and designed solely to be NP-intermediate in that case. Right now, with a few exceptions (notably the graph isomorphism problem), all the problems we know of in NP are either squarely in P or known to be NP-complete.
I get that if you can do a polynomial time reduction from "every" problem then it proves that the problem is at least as hard as every problem in NP. Except, how do we know that we've discovered every problem in NP? Can't there exist problems that we may not have discovered or proven exist in NP but CANNOT be reduced to any np-complete problem? Or is this still an open question?
As others have correctly stated, the existence of the problem that is NP, but is not NP-complete would imply that P != NP, so finding one would bring you a million dollar and eternal glory. One famous problem that is believed to belong in this class is integer factorization. However, your original question was
Can't there exist problems that we may not have discovered or proven
exist in NP but CANNOT be reduced to any np-complete problem?
The answer is no. By definition of NP-completeness, one of two
necessary conditions for a problem A to be NP-complete is that every NP problem needs to be reducible in polynomial time to A. If you want to find out how to prove that every single NP problem can be reducible in polynomial time to some NP-complete problem, have a look at the proof of Cook-Levin theorem that states that 3-SAT problem is NP-complete. It was the first proven NP-complete problem and many other NP-complete problems are later proven to be NP-complete by finding the appropriate reduction from 3-SAT to these problems.
NP consists of all problems that could (theoretically) be solved by being able to make lucky guesses, guessing the solution and checking in polynomial time that the solution is correct. For example, the travelling salesman problem "can I visit the capitols of all 50 states of the USA with a trip of less than 9,825 miles" can be solved by guessing a trip and checking that it is not too long.
And one problem in NP is basically simulating a programmable computer circuit with various inputs and checking whether a certain output can be achieved. And that programmable computer circuit is powerful enough to solve all problems in NP.
So yes, we know all about all problems in NP.
(Then of course an NP complete problem can by definition be used to solve any problem in NP. If there is a problem that it cannot solve, that problem is not in NP).
Except, how do we know that we've discovered every problem in NP?
We don't. The set of all problems in the universe is not only infinite, but uncountable.
Can't there exist problems that we may not have discovered or proven
exist in NP but CANNOT be reduced to any np-complete problem?
We don't know that. We suspect that this is the case, but this hasn't been proven yet. If we were to find a NP problem that is not in NP-Complete, it would be proof that P =/= NP.
It is one of the great unsolved problems in CS. Many brilliant minds have been taking a go at it, but this nut has been one tough one to crack.
Wiki says that when you convert a np complte problem in poly time to A , A is np hard.
see http://en.wikipedia.org/wiki/NP-hard
But the pdf below says that when you convert a np hard problem to problem A in polynomial time , A is np - hard
http://compgeom.cs.uiuc.edu/~jeffe/teaching/algorithms/notes/21-nphard.pdf
Which one should i believe?
Both. NP-complete is a subset of NP-hard; NP-complete problems are, by definition, NP-hard. If you're going to remember only one statement, remember the latter: if a problem in NP-hard can be reduced to a problem A in polynomial time, then A is NP-hard as well.
For what it's worth, NP-hardness refers to the case in which any problem in NP can be reduced to the problem in polynomial time. NP-completeness refers to the case in which a problem is both in NP and in NP-hard.