Every np language is decidable

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August undefined, 2024

WebJan 5, 2016 · This ﬁrst step enables us to prove our main result: one can decide whether the iteration of a given regular language of MSCs is reg- ular if, and only if, the Star Problem in trace monoids is decidable too. This relationship justiﬁes the restriction to strongly connected HMSCs which de- scribe all regular ﬁnitely generated languages. 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 ...

Assuming that SAT is decidable in $2^{O(\\sqrt{n})}$, is …

Webthe nth digit of the decimal expansion of π for a given n is equal to a given digit is decidable. (xxiii) An undecidable language is necessarily NP-complete. (xxiv) Every context-free language is in the class P-time. (xxv) Every regular language is in the class NC (xxvi) Let L = {aibjck: i = jorj = k}. Then L is not generated by any ... WebMar 13, 2024 · Algorithms and Theory of Computation Handbook, CRC Press LLC, 1999, "NP-complete language", in Dictionary of Algorithms and Data Structures [online], Paul … blender create material

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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 ... Web$\begingroup$ Are you asking if every language outside NP is NP-hard? Or are you asking if there exists some language outside NP that is NP-hard? I cannot tell from the wording … frazier construction reviews

Decidability and Undecidability - Stanford University

Solved 1. True or False. T = true, F = false, and O = open, - Chegg

WebMay 10, 2016 · 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 … WebQuestion: 1. True or False. T = true, F = false, and O = open, meaning that the answer is not known science at this time. In the questions below, P and NP denote P-TIME and NP-TIME, respectively. (i) — Every language generated by an unambiguous context-free grammar is accepted by some DPDA. (ii) — The language {a"b"c" d n >0} is recursive. frazier computer softwareWebModified 8 years, 10 months ago. Viewed 3k times. 1. All decision problems (i.e.language membership problems), which are verifiable in polynomial time by a deterministic Turing machine are called NP problems. Further, these problems can be solved by a non-deterministic Turing machine in a polynomial time and in exponential time by a ... blender create menu shortcut

"WebI had the following idea: Because SAT is NP-complete, every language in NP can be reduced to it using a polynomial-time reduction. Therefore, every language in NP can be … " - Every np language is decidable

Every np language is decidable

The Class NP NP = Nondeterministic Polynomial Time

WebAug 9, 2024 · If you mean a many-one computable reduction, then since the language is in NP, the reduction itself can determine whether the instance is a Yes or a No. $\endgroup$ – Yuval Filmus Aug 10, 2024 at 7:08 WebThe question could be seen as a stronger version of the fact that every infinite Turing-recognizable language has an ... of the fact that every infinite Turing-recognizable language has an infinite decidable subset. cc.complexity-theory; np-hardness; ... of 0's). Mahaney's theorem says that no unary language can be NP-complete unless P=NP. ...

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Webevery A in NP is polynomial time reducible to B (p. 304) Theorem 7.35. If B is in NP-complete and B in P, then P = NP Any member can be converted to any other by series of polynomial time f (p. 304) Theorem 7.36. If B in NP-complete, and B≤ P C for some C in NP, then C is also NP-complete Since B is NP-complete, every language in NP is 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 …

http://web.cs.unlv.edu/larmore/Courses/CSC456/tfIans.pdf WebMore formally, an undecidable problem is a problem whose language is not a recursive set; see the article Decidable language. There are uncountably many undecidable ... (determining whether it halts for every starting configuration). Determining whether a Turing machine is a ... Algorithms from P to NP, volume 1 - Design and Efficiency ...

WebA problem is partially decidable, semidecidable, solvable, or provable if the set of inputs (or natural numbers) for which the answer is yes is a recursively enumerable set. Problems … WebAs far as I understand all languages in NP are decidable. But not all decidable Languages are in NP, because NP only contains decision problems. Are there also …

WebAlternatively, NP can be defined using deterministic Turing machines as verifiers. A language L is in NP if and only if there exist polynomials p and q, ... Whether these …

Web9. Does every Turing-recognizable undecidable language have a NP-complete subset? The question could be seen as a stronger version of the fact that every infinite Turing … frazier consultants warrentonWebBut 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 ... frazier construction companyWebDec 5, 2014 · $\begingroup$ @KimmyShao: yes, a language has to be infinite to be NP-complete. A finite language is always decidable in constant time, as a Turing machine that recognizes it simply has to check a finite number of possibilities before outputting its answer. $\endgroup$ – zarathustra. frazier church montgomeryWebMar 8, 2011 · A language is Recognizable iff there is a Turing Machine which will halt and accept only the strings in that language and for strings not in the language, the TM either rejects, or does not halt at all. Note: there is no requirement that the Turing Machine should halt for strings not in the language. A language is Decidable iff there is a ... blender create mesh at originWebDe 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 ... blender create low poly treeWebEvery language accepted by a non-deterministic machine is accepted by some deterministic machine. T. The problem of whether a given string is generated by a given context-free grammar is decidable. T. ... Every NP language is decidable. T. The intersection of two NP languages must be NP. T. frazier consultants warrenton vaWebJul 15, 2011 · 17. (1) Yes, there are decision problems that are decidable but not in NP. It is a consequence of the time hierarchy theorem that NP ⊊ NEXP, so any NEXP-hard problem is not in NP. The canonical example is this problem: Given a non-deterministic Turing machine M and an integer n written in binary, does M accept the empty string in at most n ... blender create material from texture