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I don't really have references for these results--they aren't hard to prove once you understand Ladner's theorem.
No, for any NP-incomplete set A there is another set B strictly between A and SAT.
These equivalence classes are known as polynomial-many-one degrees. You can embed any finite poset into the degrees below NP. In particular the degrees are not ...
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The class ${\cal C}$ you are proposing is probably not $NP$. (If ${\cal C} = NP$, then every $NP$ language would have linear-size witnesses, which would imply that every $NP \subseteq TIME[2^{O(n)}]$ and $NP \neq EXP$, among other things).
It is very natural to consider such classes; they arise in several settings. In this paper, Rahul Santhanam (...
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I asked this question a few years ago and Boaz Barak positively answered it.
The statement is equivalent to the existence of an NP-complete language $L$ where $|L_n|$ is polynomial-time computable.
Consider Boolean formulas and SAT.
Using padding and slightly modifying the encoding of formulas
we can make sure that
$\varphi$ and $\lnot \varphi$ have the ...
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Here is a simple reduction for the TSP problem to the metric TSP problem:
For the given TSP instance with $n$ cities, let $D(i,j) \geq 0$ denote the distance between $i$ and $j$. Now let $M = \max_{i,j} D(i,j)$. Define the metric TSP instance by the distances $D'(i,j) := D(i,j)+M$. To see that this gives a metric TSP instance, let $i,j,k$ be arbitrary. Then ...
answered Oct 22 '12 at 15:02
Kristoffer Arnsfelt Hansen
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No.
Turing-recognizable undecidable languages can be unary (define $x \not\in L$ unless $x = 0000\ldots 0$, so the only difficult strings are composed solely of 0's). Mahaney's theorem says that no unary language can be NP-complete unless P=NP.
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Integer Programming.
Showing that if there is an integer solution then there is a polynomial size integer solution is quite involved. See
Christos Papadimitriou, "On the Complexity of Integer Programming", JACM, 1981.
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The most natural restriction on the proof DAG is that it be a tree – that is, any "lemma" (intermediate conclusion) is not used more than once. This property is called being "tree-like". General resolution is exponentially more powerful than tree-like resolution, as shown for example by Ben-Sasson, Impagliazzo and Wigderson. The concept has also been ...
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Just an extended comment to better explain ARi's comment (I was writing it while I saw it).
It is sufficient to use a "large gap" approach similar to the one used in Lardner's theorem; for example:
$A_1 = \{ x \mid x \in SAT \land f(|x|) \text { is even}\} \cup \{x \mid f(|x|) \text{ is odd} \}$
$A_2 = \{ x \mid x \in SAT \land f(|x|) \text { is odd}\} \...
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Q1. This is a notorious open problem. It is known to be in the fourth level of the counting hierarchy, due to [ABKM]. Not known to be in NP. The problem is not really in computing square roots as claimed in the lecture notes: $n$ bits of a square root of an integer can be computed in time polynomial in $n$ and the bitsize of the integer. The problem is, ...
17
While the problem "is the crossing number of a graph at most $k$?" is trivially in NP,
the NP-membership of the related problems for the rectilinear crossing number and
the pair crossing number are highly not obvious;
cf. Bienstock, Some probably hard crossing number problems, Discrete Comput. Geometry 6 (1991) 443-459, and
Schaefer et al., Recognizing ...
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No, $\mathrm{NP}\subseteq\mathrm{BQP}$ is not known to imply $\mathrm P=\mathrm{NP}$. Even the stronger assumption $\mathrm{NP}\subseteq\mathrm{BPP}$ is not known to yield a deeper collapse than $\mathrm{NP}=\mathrm{RP}$ and $\mathrm{PH}=\mathrm{ZPP^{RP}}=\mathrm{BPP}$; in particular, it is not even known to imply $\mathrm{NP}=\mathrm{coNP}$. (However, all ...
answered Jul 15 '15 at 14:09
Emil Jeřábek supports Monica
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If you assume that $P=^?NP$ is provable in PA (or ZFC), a trivial example is the following:
Input: N (integer in binary format)
For I = 1 to N do
begin
if I is a valid encoding of a proof of P = NP in PA (or ZFC)
then halt and accept
End
Reject
Another - less trivial - example that relies on no assumption is the following:
Input: x (boolean ...
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Adachi, Iwata, and Kasai in a 1984 JACM paper show by reduction that the Cat and $k$-Mice game has an $n^{\Omega(k)}$ time lower bound. The problem is in P for each $k$. The problem is played on a directed graph. The moves consist of the cat and then one of the $k$ mice alternating steps. The mice win if they can land on a designated cheese node ...
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You may wish to look at cost semantics for functional languages. These are various computational complexity measures for functional languages that do not pass through any kind of Turing machine, RAM machine, etc. A good place to start looking is this Lambda the Ultimate post, which has some good further references.
Section 7.4 of Bob Harper's Practical ...
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I can only give a partial answer to this question.
A result by Lenstra (later improved by Kannan, and Frank and Tardos) states that ILP with $k$ variables can be solved in time $k^{O(k)}$ (times a polynomial in the size of the ILP). Therefore, ILP is in P when the number of variables is $O(\log n/\log\log n)$. I am not sure if a $2^{O(k)}$ algorithm is ...
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Actually, acceptance of nondeterministic Turing machines in time $t$ is $O(t \log t)$-time reducible to SAT (the construction is via oblivious simulation, see Arora-Barak), so typically any time a nondeterministic machine is appreciably faster than a deterministic one, we'll see at least some speedup with a SAT oracle.
To be more concrete, primality testing ...
14
At the request of Andrej and PhD, I am turning my comment into an answer, with apologies for self-advertising.
I recently wrote a paper in which I look at how to prove the Cook-Levin theorem ($\mathsf{NP}$-completeness of SAT) using a functional language (a variant of the λ-calculus) instead of Turing machines. A summary:
the key notion is that of affine ...
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My favourite example is a classic 1977 result of Ashok Chandra and Philip Merlin. They showed that the query containment problem was decidable for conjunctive queries. The conjunctive query containment problem turns out to be equivalent to deciding whether there is a homomorphism between the two input queries. This rephrases a semantics problem, involving ...
13
The subtlety comes in where we introduce the notion of "harder". The reduction showing that SAT can be reduced to Hamiltonian Cycle shows that the latter is "harder" up to polynomial factors. In doing the reduction, we may very well increase the variable $n$ in question substantially (by some polynomial). And the canonical reductions do blow it up quite ...
12
The classic result I know of is due to Paul, Pippenger, Szemeredi and Trotter (1983) and separates deterministic from non-deterministic linear time.
Then, there is the more recent result by Fortnow,Lipton, van Melkebeek and Viglas (2004) that was already mentioned. The uniqueness of this result is that it is a time-space tradeoff result, bounding space as ...
12
For NP, this seems hard to construct. In particular, if you can also sample (nearly) uniform elements from your group - which is true for many natural ways of constructing groups - then if an NP-complete language has a poly-time group action with few orbits, PH collapses. For, with this additional assumption about sampleability, the standard $\mathsf{coAM}$ ...
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It might be conjectured that the Learning Parity with Noise Problem (LPN) at constant error rate requires time $2^{n^{1-o(1)}}$. The fastest known algorithm (Blum-Kalai-Wasserman) uses time $2^{O(n/\log n)}$.
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This is a very interesting question.
First, a clarifying remark. Note that "upper bound on the number of witnesses" is not a property of a computational problem per se, but of a particular verifier used to decide an $NP$ problem, just as an "upper bound on number of states" would not be a property of a problem but of a Turing machine deciding it. So saying ...
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You can take a look to:
Peter Golbus, Robert W. McGrail, Tomasz Przytycki, Mary Sharac, and Aleksandar Chakarov. 2009. Tricolorable torus knots are NP-complete. In Proceedings of the 47th Annual Southeast Regional Conference (ACM-SE 47). ACM, New York, NY, USA, , Article 42 , 6 pages.
Abstract: This work presents a method for associating a class of ...
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“Berman-Hartmanis Conjecture Separates NP From All Super-Poly. DTIME Classes” — Worthy of arXiv.org?
I'm glad you are interested in complexity but there are some issues in your paper. Your techniques relativize and there is an oracle relative to which the Berman-Hartmanis conjecture is true and NP = EXP.
The main issue is that you can't do self-reference for time-bounded machines since you can't simulate and stay within the time bound.
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It's not quite the same as "every algorithm", but in SODA'04 Achlioptas Beame and Molloy suggested that every backtracking algorithm should require exponential time on random 3SAT instances with $n$ variables and $cn$ clauses, with $c$ chosen within a range of values near the satisfiability threshold.
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You might be interested in Scott Aaronson's talk "Has There Been Progress on the P vs. NP Question?" (earlier version)
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How about the edge coloring number in a dense graph (aka Chromatic index)? You are given the adjacency matrix of an $n$ vertex graph ($n^2$ bit input), but the natural witness describing the coloring has size $n^2\log n$. Of course, there might be shorter proofs for class 1 graphs in Vizing's theorem.
See also this possibly related question
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First, this result is listed in the complexity zoo: https://complexityzoo.uwaterloo.ca/Complexity_Zoo:N#npiconp. Alternatively, it's possible to prove without much trouble (which I do below).
We want to show that $P^{NP \cap coNP} = NP \cap coNP$. Clearly, one direction is obviously true: $NP \cap coNP \subseteq P^{NP \cap coNP}$. To prove the other ...
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I think the $k$-clique cover problem, which asks whether there exists a partition of the vertices in $k$ sets such that each set induces a clique, has the desired properties.
Clearly, taking induced subgraphs can't make the minimum size of such partition increase. On the other hand, when you take the disjoint union of two graphs, you have to take the union ...
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