2024 Every language in np is decidable

Every language in np is decidable

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August undefined, 2024

WebSep 14, 2010 · The halting problem is a classical example of NP-hard but not in NP problem; it can't be in NP since it's not even decidable, and it's NP-hard since given any NP-language L and an NP machine M for it, then the reduction from L to halting problem goes like this: Reduce the input x to the input ( M ′, x), where M ′ is a machine that runs … WebDefinition 1 The class P/poly consists of every language that is decidable by a circuit family of size O(nc) for some c>0. Theorem 1 P⊆ P/poly. Proof: We show that every language that is decidable in t(n) time has circuits of size O(t(n)2). (A more careful argument can actually yield circuits of size O(t(n)logt(n)). The

[Solved] Consider the following statements. I. The

WebDe nition 2 Analogous to P and NP, we have that for every O f0;1g, PO is the set containing all languages decidable by a polynomial-time deterministic oracle Turing Machine with oracle access to O. It follows that NPOis the set containing every language that can be decided by a polynomial-time nondeterministic Turing machine with oracle access ... WebA decision problem is a problem that can be posed as a yes or no. If we can decide a problem, then its complement is also decidable. So, given statement is true. NP class … bush hall london

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Web1 is decidable. This language is ambiguous. 4.2 Some non-algebraic languages decided in -CL-PRec Languages a n 1 a n 2 a n 1. Since intersection of decidable languages is decidable, the language a n 1 a n 2 a n 1 = a n 1 a n 2 a m 1 \ a n 1 a m 2 a m 1 is decidable. This language is not alge-braic. Similarly, it is possible to prove that the ... WebThe polynomial-time reduction can increase the instance size, so all you get is that every language in NP is decidable in time $2^{O(\sqrt{f(n)})}$ where f(n) bounds the size of a reduction of that problem to SAT. Share. Cite. Follow answered Dec … WebIt is also easy to see that the halting problem is not in NP since all problems in NP are decidable in a finite number of operations, while the halting problem, in general, is undecidable. ... A language L is NP-hard if and only if L' ≤ p L for every language L' ∈ NP. An NP-hard language may or may not be in NP itself. A language L is NP ... handheld streamer for wedding send off

Relationship between context-free/decidable languages and NP

undecidability - Do all decidable problems lie in the class …

WebApr 8, 2024 · Partially decidable or Semi-Decidable Language -– A decision problem P is said to be semi-decidable (i.e., have a semi-algorithm) if the language L of all yes instances to P is RE. A language ‘L’ is partially decidable if ‘L’ is a RE but not REC language. Recursive language (REC) – A language ‘L’ is said to be recursive if there ... WebYou can check, however, that $\emptyset$ and $\{0,1\}^*$ are two examples of languages not reducible to their complement. Indeed, these are the only examples, which you can try to prove. Share bush hall music venueWebIn fact, there do exist languages for which it is not known whether they are in P or not, which means that we do not have a polynomial-time algorithm to decide those languages. However, we do know that the decision problem associated with P is decidable, even if we cannot always construct a Turing machine that decides every language in P. bush hall ohio university

"WebJan 9, 2024 · Every language that is in NP is by definition decidable. Becase if a language L is in NP, than there is a nondeterministic Turing Machine that decides it in polynomial time, and thus L is decidable. " - Every language in np is decidable

Every language in np is decidable

Decidability and Undecidability - Stanford University

WebPSPACE is the set of languages decidable in polynomial space. EXPTIME is the set of languages decidable in exponential time. It is known that P ⊆ PSPACE ⊆ EXPTIME. ... concept to prove P = NP it was necessary to prove that every language in the P-NP gap is actually in P. However, there are many of problems in this gap, and in fact there ... WebThe polynomial-time reduction can increase the instance size, so all you get is that every language in NP is decidable in time $2^{O(\sqrt{f(n)})}$ where f(n) bounds the size of a …

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WebAnswer (1 of 4): As others have mentioned, the answer is obviously No if P=NP, and not-obviously Yes if P\neqNP, thanks to Ladner’s Theorem. What is perhaps more … WebA crucial property of f is that it's computable. That is, given n we can compute f ( n). It is easy to see that functions such as polynomials and exponentials are computable. Since V is a verifier for B, then we have can write the following: B = { x: ∃ y such that ( x, y) ∈ L ( V) } where V has the property that given input ( x, y), V halts ...

WebExplore 75 Papers presented at Symposium on Theoretical Aspects of Computer Science in 1993. Symposium on Theoretical Aspects of Computer Science is an academic conference. The conference publishes majorly in the area(s): Time complexity & Upper and lower bounds. Over the lifetime, 2012 publication(s) have been published by the conference …

WebDecidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). Run M on w. Although it might take a staggeringly long time, M will eventually accept or reject w. The set R is the set of all decidable languages. L ∈ R iff L is decidable Web(i)Every language reduces to itself. True. By trivial reduction. (j)Every language reduces to its complement. False. A TM can not reduce to its complement. 2.(10 points) Determine whether the following languages are decidable, recognizable, or undecidable. Brie y justify your answer for each statement. (a) L 1 = fhD;wi: Dis a DFA and w62L(D)g ...

WebIf P=NP, the proof does not work and the statement is false: if P=NP, then any language L that is nonempty and nonfull is NP-hard. Proof: suppose u is any word in L and v is any …

WebIt remains unknown whether NP=EXP, but it is known that P⊊EXP. Prove that NP ⊆ EXP. In other words, any efficiently verifiable language is decidable, in exponential time. Hint: For a; Question: Let EXP be the class of all languages that are decidable in exponential time, i.e., in time O(2nk) for some constant k (where n is the input size ... bush hall nearest undergroundWebMar 13, 2024 · Algorithms and Theory of Computation Handbook, CRC Press LLC, 1999, "NP-complete language", in Dictionary of Algorithms and Data Structures [online], Paul … handheld stream flow meterWebThe complement of every Turing decidable language is Turing decidable. II. There exists some language which is in NP but is not Turing decidable. III. If L is a language in NP, L is Turing decidable. Which of the above statements is/are true? This question was previously asked in. GATE CS 2015 Official Paper: Shift 2 Attempt Online. hand held storage containersWebPNP = the class of languages decidable by some polynomial-time oracle TM MB for some B in NP Informally, PNP is the class of problems you can solve in polynomial time, if SAT solvers work A problem in PNP that looks harder than SAT or TAUT: FIRST-SAT = { ( , i) ∊SAT and the i-th bit of the lexicographically first SAT assignment of is 1} ... handheld string trimmer partsWebFeb 20, 2014 · A decidable language L is NP-complete if: L is in NP, and; L' can be reduced to L in polynomial time for every L' in NP. If a language L satisfies property 2, but not necessarily property 1, we say that L is NP … hand held stretch film dispenserWebBut not all decidable Languages are in NP, because NP only contains decision problems. ... Every CFL is generated by a CFG in Chomsky normal form; Every regular language … hand held stretch wrapperWebBut not all decidable Languages are in NP, because NP only contains decision problems. ... Every CFL is generated by a CFG in Chomsky normal form; Every regular language is (also) context-free. That answers immediately all those questions: What is the relationship between NP and context-free/regular languages? Are there context-free languages ... bush hall shepherds bush capacity