# Tag Info

27

Far from it. Indeed, any countable distributive lattice embeds as a sub-partial-order of $\leq_p$, even if we only consider those degrees in between two given fixed languages (K. Ambos-Spies, Sublattices of the polynomial time degrees, Inform. & Control 65(1):63-84, 1985).

20

Predecessor versions of this paper have been around for more than 15 years. I remember that there were counter-examples to the first versions, then first revisions, counter-examples to the first revisions, second revisions, new counter-examples, further revisions, further counter-examples, and so on. It would be much better, if the authors were able to ...

15

A whole lot of fun things happen. Most of the ones I know of start with the IKW paper. There, the collapse $\textrm{NEXP} = \textrm{MA}$ is shown, and (I think) is the strongest literal collapse of complexity classes that we know of. There are other sorts of "collapses" though that I think should be pointed out. Most importantly, I think, is the "universal ...

15

To be clear, it's not meant to be formalizable. It's not a theorem, it's an observation about the world -- it's okay if "natural" is subjective here. For analogy, if someone says "differentiation is mechanics while integration is art", they're not inviting you to formalize "mechanics" and "art" and prove the statement, they're trying to convey a general ...

14

I believe the strongest is that $NEXP = MA$. This was proved by Impagliazzo Kabanets and Wigderson. See https://scholar.google.com/scholar?cluster=17275091615053693892&hl=en&as_sdt=0,5&sciodt=0,5 I'd be also interested to know of any stronger collapses than this. Edit (8/24): OK, I thought of some potentially stronger collapse, which ...

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 $\mathsf{coAM}$ ...

12

Due to popular demand, I’m converting my comment to an answer. A simple padding argument shows that for every constant $\epsilon>0$, there exist EXP-complete problems in $\mathrm{DTIME}(2^{n^\epsilon})$. Indeed, fix an arbitrary EXP-complete problem $L$, and assume that it is computable in time $2^{n^c}$. Let $d>c/\epsilon$, and consider the problem $$... 11 For bipartite graphs, vertex cover is polynomially solvable by routine techniques from matching theory. For k-partite graphs with k\ge3, we observe the following: Vertex cover is NP-complete on cubic graphs By Brooks' theorem, every cubic graph (except K_4) is 3-colorable and hence 3-partite. 11 Emil Jerabek's comment is a nice summary, but I wanted to point out that there are other classes with clearer definitions that capture more-or-less the same concept, and to clarify the relation between all these things. [Warning: while I believe I've gotten the definitions right, some of the things below reflect my personal preferences - I've tried to be ... 11 What you are actually asking is for the performance of the Schönhage–Strassen algorithm in the unit cost RAM (rather than its bit complexity). This is covered in Fürer's paper How Fast Can We Multiply Large Integers on an Actual Computer?, likely written with similar motivation to yours. 9 \let\mr\mathrmThere are several results in the literature stating that a certain class C satisfies C\nsubseteq\mr{SIZE}(n^k) for any k, and usually it is straightforward to pad them to show that any barely superpolynomially expanded version of C is not in \mr{P/poly}. Let me say that f\colon\mathbb N\to\mathbb N is a superpolynomial bound if ... 8 \let\mr\mathrmLet me denote the problem as S. Gottlob and Fermüller state that if S is solvable in \mr{FP}^{\mr{SAT}[\log n]}, then \mr{P}=\mr{NP}. However, the argument actually shows more generally that if S is in \mr{FP}^{\mr{SAT}[q(n)]}, then SAT is solvable in time \mr{poly}(n2^{q(n)}). In particular, assuming the exponential time ... 7 The state of the art for the theoretical complexity of go is well summed up on Wikipedia, with relevant references. The main remaining open problem is for rules using a superko, i.e. repeating any past position is forbidden. This is the rule used for instance in China and western countries. It is simple to state, but could bring some difficulties to ... 6 EDIT: Now that I have fresh eyes in the morning, I see that I have thoroughly misread the question. The answer below applies to “if f(n)\ne O(g(n)), then f(n)=\Omega(g(n))”. As noted in comments above, the question as actually stated has a trivially true answer for all functions, that is, f(n)=O(g(n)) is equivalent to g(n)=\Omega(f(n)) immediately ... 6 The answer to (1) is yes (regardless of the properties D.W. asked for in the comments), depending on how R is given: First, note that since R is finite, the abelian group (R,+) is of the form \mathbb{Z}_{p_1^{k_1}} \oplus \dotsb \oplus \mathbb{Z}_{p_\ell^{k_\ell}}, where \ell \leq \log_2|R|. Now, if R is only given to you by generators and ... 6 In terms of these complexity classes as a whole, not much is known that distinguishes characteristic 2 from other characteristics. The most frequently arising difference is that the permanent is easy in characteristic 2 (so not VNP-complete unless VP=VNP). (In contrast, the Hamiltonian Cycle polynomial is VNP-complete in all characteristics.) The ... 6 This is usually called the (constructive) membership problem (rather than a "factorization" problem). The membership problem is to decide whether C \in \langle A,B \rangle; the constructive membership problem is to actually find a word (if any) in A,B that equals C. Its complexity may depend on whether you want to allow A^{-1}, B^{-1} in your word (... 6 In the following paper my colleague Uli Schöpp presents a formal verification (in Coq) of a nontrivial result by Cook and Rackoff on the computational power of graph automata. https://scholar.google.at/scholar?oi=bibs&cluster=4944920843669159892&btnI=1&hl=de (Schöpp, U. (2008). A formalised lower bound on undirected graph reachability. In Logic ... 6 It seems it would depend on your particular model, in particular what information you have access to. From what I infer, you are thinking of the following model: you have a memory m, for instance of size O(\log n). at each step, you read a new letter a\in\Sigma of your stream, and you are allowed to modify your memory m you then have to say whether ... 5 If I understood it well, (1) is also NP-complete, a possible reduction is from SUBSET SUM: Given a set of m positive integers A = \{a_1, ..., a_m\}, and a positive integer B, is there a subset of A' \subseteq A a such that \sum_{a_i \in A'} a_i You simply pick n = (m+1) and build your set in this way: add  a_1, a_2, , ..., a_m add m zeros ... 5 I think of the following argument: if we can check whether two sequences have equal Kolmogorov complexity we can write a program that enumerates all sequences of length \le N and divides them into equivalence classes. We know that K(x) \le |x|. So, we have at most N equivalence classes of sequences. One of this classes should be of size at least 2^{... 5 Sorry if this answer doesn't tell anything nontrivial, but you don't seem to imply these results in the questionm. Consider first the problem of computing a modular exponentiation  a^r \mod m . You say above that you can compute this by repeated squaring modulo  m , and that this needs  O(\log r)  multiplications. This is true, and it's certainly ... 5 My intuition is that an NP-complete language of this type would cause a collapse of the polynomial hierarchy much like the one in the Karp–Lipton theorem. More specifically, if you go up to the second level of the polynomial hierarchy, you can use the power of the hierarchy to guess the equivalence between a given group element and some representative of an ... 5 This problem is FP^{NP}-complete, as shown here. It means that the lexicographical leader of the orbit is built in deterministic polynomial time with access to a NP-oracle. 5 This problem is NP-hard. Although it may be possible to find some canonical form for string isomorphism, say, in quasi-poly time, without upsetting our current guesses as to how the complexity world looks, finding the lexicographically least isomorphic string is NP-hard. This is precisely the content of Proposition 3.1 here. In fact, they show it remains NP-... 5 In "Branching Programs and Binary Decision Diagrams" by Ingo Wegener [1] (very good, complete reference to check this kind of fact on branching programs), Section 5.7 deals with how you can transform a given \pi_1-OBDD F_1 into an equivalent \pi_2-OBDD F_2 by using syntactic rules. If you have a bound B on the representation of F_1 by \pi_2-... 5 The problem (let me call it s-t-prime-connectivity) is in P; more precisely, it is NL-complete. NL-hardness is clear: we can reduce plain s-t-connectivity to s-t-prime-connectivity just by making sure that every vertex has a self-loop. Then, if there is any path from s to t, there are paths of arbitrary lengths \ge n, and in particular, of ... 5 In 1998, Michel X. Goemans gave an ICM talk, in which, he addressed this issue:"Semidefinite programs can be solved(or more precisely, approximated) in polynomial-time within any specific accuracy either by the ellipsoid algorithm or more efficiently through interior-point algorithms...The above algorithms produce a strictly feasible solution(or slightly ... 4 The obvious N2Exp problem is of course the word acceptance problem for 2exp time bounded nondeterministic Turing machines. Using this might be as hard/easy as 2exp tiling, because the simulation of such a Turing machine computation in essence also requires you to define a double-exponentially large grid (2exp many configurations of memory tapes of length ... 4 First, if k(n) and T(n) are non-negative functions satisfying$$T(n)=3T(n-1)-T(n-2)+T(n-k(n))+3^{k(n)}\tag{$*$} for all sufficiently large $n$, it is easy to see that $T(n)$ cannot have a finite limit, and in particular, it cannot be decreasing. But if $T(n_0+1)\ge T(n_0)$, then $T(n+1)>T(n)$ for all $n>n_0$ by induction on $n$. Thus, $T$ is ...

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