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This is only possible if there are many admissible outputs for a given input. I.e., when the relation $R$ is not a function because it violates uniqueness.
For instance, consider this problem:
Given $n \in \mathbb{N}$ (represented in unary) and a TM $M$, produce another TM $N$ such that $L(M)=L(N)$ and $\# N > n$ (where $\# N$ stands for the encoding (...
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There are two cases:
$P = NP$ non-constructively: this means we have derived a contradiction from the assumption that $P \neq NP$, and thus can conclude that $P = NP$ by the law of the excluded middle. In this case, we have no idea what an algorithm to solve graph coloring in polynomial time looks like, or any other problem. We know one exists, because we ...
36
NP is the class of problems where you can verify "yes" instances. No guarantee is given that you can verify "no" instances.
The class of problems where you can verify "no" instances in polynomial time is co-NP. Any language in co-NP is the complement of some language in NP, and vice-versa. Examples include things like non-3-colourability. The problem you ...
27
Generally speaking, the following is true for any algorithm:
Suppose $A$ is an algorithm that runs in $f(n)$ time. Then $A$ could not take more than $f(n)$ space, since writing $f(n)$ bits requires $f(n)$ time.
Suppose $A$ is an algorithm that requires $f(n)$ space. Then in $2^{f(n)}$ time, $A$ can visit each of its different states, therefore can gain ...
24
Proof complexity only makes sense when there is a sequence of statements depending on a parameter $n$. For example, the proposition $\mathsf{PHP}_n$ states (informally) that there is no bijection $[n+1] \to [n]$. This sequence of propositions is hard for certain propositional proof systems.
The statement $\mathsf{P} \neq \mathsf{NP}$ is a single statement, ...
23
Your question doesn't make sense:
The problem is NP-complete (proven) for all input data (without exception).
This is not a thing. NP-completeness is a property of entire sets, not of specific inputs. It's fairly trivial to show that, if you choose a specific input, any problem is $O(1)$ on that input: you just output yes or no, depending which is ...
22
No. This direction is unlikely to be useful, for two reasons:
Most computer scientists believe that P $\ne$ NP. Assuming P $\ne$ NP, this means there does not exist any polynomial-time algorithm to solve any NP-complete problem. If you want your neural network to solve the problem in a reasonable amount of time, then it can't be too large, and thus the ...
20
So your problem is as follows:
Input: integers $\ell,u$
Question: does there exist a prime in $[\ell,u]$?
As far as I know, it is not known whether that problem is in P or not.
Here's what I do know:
Primality testing (given a single number, test whether it is prime) is in P, so if the range is small enough, you can exhaustively test each number in the ...
19
There are at least four such $NP$-complete problems listed in the appendix of Garey and Johnson's COMPUTERS AND INTRACTABILITY: A Guide to the Theory of NP-Completeness.
[AN6] NON-DIVISIBILITY OF A PRODUCT POLYNOMIAL
INSTANCE: Sequences $A_i = \langle (a_i[1],b_i[1]), ..., (a_i[k],b_i[k]) \rangle,\ 1 \leqslant i \leqslant m,$ of pairs of integers, with each ...
19
The Post correspondence problem is undecidable, and in particular it is not in NP. The reason that your idea doesn't work is that the witness is not necessarily of polynomial size (in fact, you just proved it). That is, for your certifier to prove that the Post correspondence problem is in NP, it needs to run in polynomial time (in terms of the size of the ...
17
First of all, the question you are asking is open, since an affirmative answer shows that $\sf NP = coNP$. In fact it is one of the most prominent open problems in computer science.
If $\sf P= NP$, then the class $\sf NP$ is closed under complement since $\sf P$ is. If on the other hand $\sf P \not = NP$ then we cannot say whether $\sf NP = coNP$ or not. ...
17
The problem isn't really well-posed. For any particular instance, there is a single solution, say $S$. Consequently, we can imagine an algorithm that has the answer $S$ hardcoded in: no matter what input you give it, all it does is just print $S$. This answer counts as a deterministic polynomial-time algorithm that solves this particular instance $I$.
...
16
Your language is in P. Suppose that the matrix is $n\times n$ and that the words have total length $\ell$. Each word can start at at most $n^2$ positions and be written in $O(1)$ many orientations, for a total of $O(n^2)$ possible placements. Checking each one costs at most $O(m)$, where $m$ is the length of the word. In total, we obtain an algorithm whose ...
15
One reason that we see different approximation complexities for NP-complete problems is that the necessary conditions for NP-complete constitute a very coarse grained measure of a problem's complexity. You may be familiar with the basic definition of a problem $\Pi$ being NP-complete:
$\Pi$ is in NP, and
For every other problem $\Xi$ in NP, we can turn an ...
15
If a finite problem is NP-complete then P=NP, since every finite problem has a polynomial time algorithm (even a constant time algorithm).
When we say that Sudoku is NP-complete, we mean that a generalized version of Sudoku played on an $n^2 \times n^2$ board is NP-complete.
Finally, the 4-clique problem, while not a finite problem (the input graph has ...
15
Actually you're wrong. In the example you give, you take a list of integers, but even with binary representation for the numbers in the list, the input length is still linear in the number of elements of that list, so the algorithm is polynomial ( not pseudo ). Complexity is always measured in respect to the total input length, not some arbitrary number ( b )...
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Typically, we use $\alpha < 1$ for maximization problems, and $\alpha > 1$ for minimization problems, where $\alpha$ is the approximation guarantee. So, a $2$-approximation algorithm returns a solution whose cost is at most twice the optimal. But as always, to be absolutely sure, go back to the definitions of the text you are reading (if a definition ...
14
One way to consider the difference between decision version and optimization version is by considering different optimization versions of the same decision version. Take for example the MAX-CLIQUE problem, which is very hard to approximate in terms of the usual parameter – the size of the clique. If we change the optimization parameter to the logarithm of ...
14
Yes, every NP problem has an exponential-time algorithm. One definition of NP is the "succinct certificates" definition: a language $L$ is in NP if, and only if, there is a relation $R$ on strings such that:
there is a polynomial $p$ such that, whenever $(x,y)\in R$, $|y|\leq p(|x|)$,
$x\in L$ if, and only if, $(x,y)\in R$ for some $y$, and
there ...
14
It is known that any proof of super-polynomial circuit lower bounds (which are slightly stronger statements than $P\neq NP$) require super-polynomial, even exponential size proofs in weak proof systems like resolution. Generalizing this to stronger proof systems is a well known open problem.
See section 5 of this survey by A. Razborov where these things ...
13
We do know all problems in NP. Each problem in NP is given by a non-deterministic Turing machine running in polynomial time. Steve Cook (and, independently, Leonid Levin) proved that SAT is NP-complete by encoding the statement "Machine $M$ accepts $x$ given non-deterministic choices $y$" as a logical formula for every non-deterministic Turing machine; for a ...
13
First, let's refresh the proof that the definition given is indeed equivalent to the standard definition of NP, i.e., the class of languages accepted by polynomial-time nondeterministic Turing machines.
Suppose $L$ is accepted by a polynomial-time Turing machine $M$. There is some $c$ such that $M$ runs in time $|x|^c$ for all inputs $x$. The verification ...
13
Your intuition about "relative hardness" is correct, the underlying mathematics is why III. is true. However your justification about I. is a little off (not wrong, but the reasoning is possibly not accurate).
It might help to think about reductions in these terms (everything I'll talk about here will be polynomial time, so I will leave that out just so I ...
13
The fact that P ≠ NP does not preclude the possibility that NP = co-NP, in which case NP ∩ co-NP = NP. So to further the discussion, let us assume that NP ≠ co-NP. In that case, Corollary 9 in Schöning's A uniform approach to obtain diagonal sets in complexity classes shows that there exists some language in NP – co-NP which is NP-intermediate. So NPI ...
13
In fact, you don't need the P$\,\neq\,$NP hypothesis, since there are even infinite constant-time decidable subsets of NP-complete problems. For any NP-complete language $L\subseteq\{0,1\}^*$, let $L' = \{0w\mid w\in L\}\cup\{1w\mid w\in\{0,1\}^*\}$. $L'$ is still NP-complete (trivial reduction from $L$), but it contains the infinite constant-...
12
Yes, the class is called UP (the U standing for "unambiguous"). David points out in the comments that another answer is US.
UP: If $x \in L$, then there is exactly one "proof" ("witness", "certificate", "accepting path"). If $x \not\in L$, there are exactly zero "proofs".
US: If $x \in L$, then there is exactly one "proof". If $x \not\in L$, there may be ...
12
A concrete example is repeated squaring. Squaring an integer of length $n$ takes time $O(n^2)$ (using the naive algorithm; you can do much better with more complicated algorithms). If you square an $n$-bit integer $i$ times in a row, its length becomes $2^i n$. So if you square it $n$ times in a row, the complexity is $O(4^n n^2)$, which is exponential.
The ...
12
Since L is in P, you can answer the word problem in polynomial time. To show that L is in NP as well, we need to provide a polynomial checking relation $R$ such that
$$ w\in L \Leftrightarrow \exists y.(|y|\le |w^k| \text{ and } R(w,y))$$
Now Prof. Cook says to take a very simple $R$. For every $w$ in $L$, no matter what $y\in \Sigma^*$ you take, $R(w,y)$ ...
12
Actually, your version is correct and Wikipedia's is wrong! (Except that it has a tiny disclaimer at the bottom.)
If $\mathrm{P}=\mathrm{NP}$, Wikipedia claims that every problem in $\mathrm{P}$ is $\mathrm{NP}$-complete. However, this is not true: in fact, every problem in $\mathrm{P}$ would be $\mathrm{NP}$-complete, except for the trivial languages $\...
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You can prove it by contradiction:
Suppose that $P \neq NP$ and there is a polynomial-time reduction from 3-SAT to 2-SAT; then 2-SAT is NP-complete, but 2-SAT is also solvable in polynomial time, so
for all decision problems $A \in NP$ you can decide $x \in A$ reducing $x$ to the corresponding 2-SAT instance in polynomial time and solve it in polynomial ...
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