I have to check out whether my logic is on the right path.
NP-HARD: these are the hardest problems which may/may not be in NP class. If you have an efficient algorithm for these problems you have one for every problem in class NP.
NP COMPLETE: these are the hardest problems in class NP and also if you solve one of these you could solve any problem in class NP. So, NP COMPLETE problem is an NP-HARD problem.
COOK'S THEOREM: If SAT(NP-HARD) has a polynomial time algorithm then so does every problem in class NP.
Now, suppose we have to prove that CDP(clique decision problem) is NP COMPLETE.
->Step 1: Prove that CDP is in the class NP.
It is in class NP because the prover can generate a proof for yes inputs which would enable the verifier to check that it is a CDP (has a clique of size k).
->Step 2: Prove that CDP is NP HARD.
For that, we can convert the SAT to CDP by constructing a graph from clauses and supplying k.
We supply(G,k) to the clique subroutine which would verify is there a clique of size k or not. If it can figure this out in polynomial time then SAT has a polynomial time algorithm as CDP had a polynomial time algorithm and we converted SAT to CDP. So, now we proved that if there is a polynomial time algorithm for CDP then there is for the SAT. Now if we can find a polynomial time algorithm for CDP then it would imply that there is a polynomial time algorithm for SAT. This would imply that there is a polynomial time algorithm for every problem in NP by COOK'S THEOREM.
So we proved that CDP is NP COMPLETE. Once we have added CDP to NP COMPLETE class and now we come up with a new problem which we have again to prove that it is NP COMPLETE we can prove that problem to be in NP and then we could prove that if there is an efficient algorithm for given problem then that implies that there is an efficient algorithm for SAT/CDP(as we have added this to NP COMPLETE). Then as said above we can convert this problem to CDP/SAT and then prove that if there is an efficient algorithm for our problem then there is one for CDP/SAT and then by COOK'S THEOREM again we have that if there is a solution to NP-HARD problem (in this case CDP/SAT) then there is one for every problem in NP. So we again proved our problem as NP-HARD and as now it also belongs to NP as said above it is NP COMPLETE.
So we can add as many problems to the NP COMPLETE class as long as we can convert some problem which is already in NP-HARD class(in this case SAT/CDP) into our problem and we should find an efficient algorithm to our problem which would indirectly find an efficient algorithm to the NP-HARD problem and by COOK's theorem we can say that as some NP-HARD problem has an efficient algorithm we have an efficient algorithm to solve all problems in NP.
You're on the right path, but there your logic is a little incomplete.
The general structure of your proof is correct: First prove a problem is in NP, then prove the problem is NP-Hard. Those two bits of information together prove that a problem is NP-Complete.
Your proof for proving a problem is in NP is incomplete. Here are the key components to proving a problem is in NP:
Reword the problem as a decision problem that can be answered with a yes or a no.
Describe what a "certificate" would be. NOTE: a certificate is an output that can be checked to verify the answer to the decision problem. For CDP it could be a list of vertices and edges that make up the clique of size k.
Prove that this certificate can be verified in polynomial time.
Your proof for proving NP-Hard is incomplete. Here are the key components to proving a problem is NP-Hard:
Transform the input of the known NP-Hard problem into the input for the problem you are trying to prove.
Prove that this transformation can be done in polynomial time.
Transform the output of the problem your trying to prove into the output of the known NP-Hard problem.
Demonstrate how this can be done in polynomial time.
Prove that if you get an answer for the problem you are trying to prove, then you have an answer for the known problem.
Prove that if you get an answer for the known problem you have an answer for the problem you are trying to prove.
Only by meeting those 6 criterion can you say you have completely proven that a problem is NP-Hard.
Besides the specifics on that your logic is sound. Be careful when saying "efficient" if you really mean "can be solved in polynomial time".
Related
The algorithm needs to generate all possible combinations from a given list (empty set excluded).
list => [1, 2, 3]
combinations => [{1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}]
This algorithm would take O(2n) time complexity to generate all combinations. However, I'm not sure if an improved algorithm can bring this time complexity down. If an improved algorithm exists, please do share your knowledge!
In the case that it takes O(2n) which is exponential, I would like some insight regarding which class this algorithm belongs to P, NP, NP-Complete, or NP-Hard. Thanks in advance :)
P, NP, NP-complete, and NP-hard are all classes of decision problems, none of which contain problems that involve non-binary output (such as this enumeration problem).
Often people refer colloquially to problems in FNP as being in NP. This problem is not in FNP either because the length of the output string for the relation must be bounded by some polynomial function of the input length. It might be FNP-hard, but we're getting into the weeds that even a graduate CS education doesn't cover. Worth asking on the CS Stack Exchange if you care enough.
This problem is in none of them except, arguably, NP-hard.
It is not in P because there is no polynomial time algorithm to do it. You cannot generate an exponential number of things in polynomial time.
It is not in NP because there is no polynomial time algorithm to validate the answer. You cannot process an exponential number of things in polynomial time.
It is not in NP-complete because everything in NP-complete must be in NP and it is not.
The argument for it being in NP-hard goes like this. You can say anything that you want about the members of the empty set. Including that they make monkeys fly out of your nose and can solve any problem in NP in polynomial time. So if we could find a polynomial solution, we can solve any NP problem fast, and therefore it meets the definition of NP-hard. But uselessly so - we know that no polynomial solution exists.
If a polynomial time algorithm for an NP-Complete problem is found, lets say its O(n^2) hypothetically, does this imply that there is an O(n^2) solution for all NP problem? I know this would imply that there is a polynomial solution for all NP-problems, but would it necessarily be O(n^2)?
Not necessarily
A problem x that is in NP is also in NP-Complete if and only if every
other problem in NP can be quickly (ie. in polynomial time)
transformed into x.
Therefore an algorithm that solves one NP-Complete problem means we can solve any other problem in NP by transforming it to an instance of that problem and solving it. But the transformation could be any complexity as long as its polynomial we satisfy the condition.
So the answer to your question in no, an O(N^2) algorithm to solve an NP-Complete problem does not imply all NP problems can be solved in O(N^2) time, it only guarantees there exists a polynomial time algorithm to solve it.
ie O(N^2) + T(N) where T(N) is the complexity to transform the problem instance
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.
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.