2024 Every np language is decidable

Every np language is decidable

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August undefined, 2024

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 … 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

Relationship between context-free/decidable languages and NP

WebFeb 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 … 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 … is swimming a verb

undecidability - Do all decidable problems lie in the class …

WebModified 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 … Web“Turing recognizable” vs. “Decidable” L(M) –“language recognized by M” is set of strings M accepts Language is Turing recognizable if some Turing machine recognizes it •Also called “recursively enumerable” Machine that halts on all inputs is a decider. A decider that recognizes language L is said to decide language L 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 is swimming a team or individual sport

“Turing recognizable” vs. “Decidable” - Fordham University

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 ... WebMar 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 ... is swimming a physical activityWebAnswer (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 … if that\u0027s the way you want it lyrics

"WebThe 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. " - Every np language is decidable

Every np language is decidable

undecidability - Do all decidable problems lie in the class NP ...

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 … 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 …

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WebModified 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 ... 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 ...

WebFeb 19, 2014 · $\begingroup$ A good pool for examples of nontrivial membership in NP may come from problems for which it has been open for some time whether they were even decidable. Two problems from the top of my hat : string graph recognition and unknot recognition (and the more general knot genus). In both cases though, there does exist a … 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.

WebJul 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 ... WebAug 27, 2014 · 1. Nope, not all languages in in P or NP. Here are a few ways to see this: There are uncountably many languages, but only countably many languages in P and …

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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 … is swimming a team sportWebDec 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. if that\u0027s what it is 13Web(xxi) T Every NP language is decidable. 1 (xxii) T The intersection of two NP languages must be NP. ... T If L is RE and also co-RE, then L must be decidable. (lvii) T Every language is enumerable. (lviii) F If a language L is undecidable, then there can be no machine that enumerates L. if that\u0027s what it takes albumWebEvery t(n ) nondeterministic TM has an equivalent 2O (t n )) ... directed path containing each node once Decidable ) complement is decidable I Finding: hard (NP-complete, will revisit) I Checking: easy (traverse path, mark all graph nodes) Complexity: polynomial O (n 2) ... A language is in NP i it is decided by some is swimming a verb or adjectiveWebA 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 … is swimming bad for skinWebJan 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. if that\u0027s what it is lyricsWebSep 25, 2012 · L is in NP if and only if there is a polynomial p and a PTIME language L' such that. x in L if and only if there exists y of length p ( x ) such that (x,y) is in L'. To … is swimming bad for your back