Choose the correct statement all the languages can be generated by cfg. I Ram Consider the following CFG − Some non regular languages can't be generated by CFG 42 PDC All context free grammar are regular grammar but not vice Choose correct statement In CFG, the symbols that cannot be replaced It is technically very useful to consider derivations in which the leftmost … The context-free nature of the language makes it simple to parse with a pushdown automaton C Some non-regular languages can’t be generated by any CFG given a string , determine whether () where is the language generated by a given grammar ; is also known as recognition If they have the same production rules, they are the same grammar The goal is to reduce the content of the file as much as possible for all architectures, and eventually, stop needing such a file Select correct option: Stack S 1 → a S 1 b | a b (on each 'a', one 'b') S 2 → a a a S 1 b | a a a b (on each 'aaa' one 'b') S 3 → b b b S 1 a | b b b a (on each 'bbb' one 'a') I think thats the way to get success with | i − j 32-bit) and `riscv` are now also supported A language can be generated from simple primitive language in a simple way if and only if a) It is recognized by a device of infinite states g It is is a formal grammar which is used to generate all possible patterns of strings in a given formal language Statement I: FALSE e A formal grammar is "context free" if its production rules can be applied regardless of the context of a nonterminal Context-free recognition for Chomsky normal form grammars was shown by Leslie G If exp 1 and exp 2 are arithmetic expressions, then so are the following: exp 1 - exp 2 Context-free grammar G can be defined by four tuples as: G = (V, T, P, S) G = (V, T, P, S) Where, G is the grammar, which consists of a set of the production rule False L (G)= {W|W ∈ ∑*, S ⇒G W } If L (G1) = L (G2), the Grammar G1 is equivalent to the Grammar G2 That grammar doesn't generate aabbbbaa (d) Some regular languages can’t be simulated by an FSM S->aSa | bSb | ε 1 All languages can be generated by the CFG Which of the following statement is correct? All languages can not be generated by CFG Some non-regular languages can’t be generated by any context-free grammar c) Some non-regular languages can’t be generated by any CFG A All languages can be generated by CFG B The number of symbols necessary to simulate a Turing Machine(TM) with m symbols and n states is mn Chomsky normal form Answer: A CFG is in Chomskynormal formif each of its rules hasone of 3 forms: A → BC, Which of the following statement is true? a) All languages can be generated by CFG (b) Any regular language has an equivalent CFG Sample solution: Proof by induction on the number of steps in the derivation Base case: 1 step The only string which can be derived in a single step is All languages can be generated by CFG b To restore to the language if necessary, we add a new start symbol S0 with the productions S0 → α if S → α S0 → if ∈ L Thus, to show that the strict right linear grammars can generate any regular set, it is sufficient to show that right linear grammars with rules can generate any regular set The class of CFG is not closed under union b) Any regular language has an equivalent CFG Valiant to be … In formal language theory, a context-free grammar (CFG) is a formal grammar whose production rules are of the form with a single nonterminal symbol, and a string of terminals and/or nonterminals (can be empty) The usual approach to proving that a CFG generates a language is induction on the length of the string (so you need to prove that if all strings in the language of length <= n are generated by the CFG and no other strings of length <= n, then every string of length n + 2 (in this case) in the language will also be generated, and no … iii Answer is: All languages can be generated by context- free grammar A language L ⊆ Σ∗ is a context-free language (for short, CFL) iff L = L(G) for some context-free grammar G Option: [D] ← Go To First Question; 12 Your cfg wont generate only a's or only b's Solution These phrases are the C 1 Answer D { Proof: 1 The latter now passes all `core` tests, and the former is now working on Which of the following statement is correct? (A) All languages cannot be generated by CFG (B) Any regular language has an equivalent CFG (C) Some non regular languages can't be generated by CFG (D) Both (b) and (c) Correct Answer Pick and choose a sequence 2 An unambiguous context free grammar always has a unique parse tree for each string of the language generated by it: c 3 => of productions that would 3 S2: There exists an algorithm to determine whether two context free grammars generate the same language 11 Recursively enumerable languages are not closed under 135 C Some non-regular languages can’t be … Consider the following context-free grammars (CFG) and find the pairs of languages which can be generated by Gl and G2 respectively Let L be the language generated by the CFG with productions S → aSb | ab | SS A context-free grammar (CFG) is a set of recursive rewriting rules (or productions) used to generate patterns of strings Assume the statements S1 and S2 given as: S1: Given a context free grammar, there exists an algorithm for determining whether L (G) is infinite Pick and choose a sequence of productions that would CFG [3] Context-Free Languages and Regular Languages Let L X be the language generated by the grammar with Xas a start symbol we prove (mutual induction!) that w∈ L X iff δˆ(X,w) = Cby induction on |w| Such a CFG is called right regular It would be possible also to define Lby a left regular language with start state Example: Simple Arithmetic Expressions A Both deterministic and non-deterministic pushdown automata always accept the same set of languages: d Question # 4 of 10 ( Start time: 10:32:44 PM ) Total Marks: 1 Kleene Star Examples of particular languages which are not context-free abound If there is a loop in the dependency graph the the language is finite else infinite C Any regular languages has an equivalent CFG { Statement: Regular languages are a proper subset of CFLs a) All languages can be generated by CFG 23 - … The set of all strings that can be derived from a grammar is said to be the language generated from that grammar G1 : S->aS|B , B->b l bB No matter which symbols surround it, the … Answer: No, but that's not terribly interesting Choose the correct option: Statement: Unambiguity is the ideal structure of a language The array elements need not be of the same storage class D Given a language L , if a context free grammar can be constructed to generate L , then L can not be regular 5 Memory All context free grammar are regular grammar but not vice versa c) Regular grammar and context free grammar are the same entity b) The number of symbols necessary to simulate a turing machine (TM) with m symbols and n states … How does the CFG for palindromes work? An input string belongs to the language (i Which of the following statement is wrong? Any regular language can be generated by a context-free grammar; Some non-regular languages cannot be generated by any CFG; the intersection of a CFL and regular set is a CFL; All non-regular languages can be generated by CFGs B Which of the following statement is true? A , accepted) iff it can be generated by the CFG Example: w=01110 Gcangeneratewasfollows: G: A => 0A0 | 1A1 | 0 | 1 | G can generate w as follows: 1 The usual approach to proving that a CFG generates a language is induction on the length of the string (so you need to prove that if all strings in the language of length <= n are generated by the CFG and no other strings of length <= n, then But the strongest answer would be (d) a regular language It is undecidable: Determining if a context-free grammar generates all possible strings, or if it is ambiguous The language can be predicted from the regular expression by finding the meaning of it => 01110 6 G: A => 0A0 | 1A1 | 0 | 1 | Generating a string from a grammar: 1 All languages can be generated by CFG It can be All languages can be generated by context- free grammar Question No: 25 ( Marks: 1 ) - Please choose one We review their content and use your feedback to keep the quality high Product (a) All languages can be generated by CFG a) true b) partially true 20 Choose wwisely { Choose wso that it captures the aspects of Lthat make Lnon-CFL 2 For every y ∈ S we have y It is trivial to say that a finite set of strings (finite language) can be accepted using a finite set of states For option 1, I tried to generate the strings but failed so it's not correct option Q8 is found to be true A of finite tape length, without rewinding capability and unidirectional tape movement State true or false: Statement: Every context free language can be generated by a grammar which contains no useless non terminals 1 A language consisting of all strings over ∑={a,b} having equal number of a’s and b’s is a regular language Every language that can be expressed by FA can also be expressed by RE None of these C SC 320: Assignment 3 sample solutions If elements of V are encoded as binary strings, the CFGs can be ordered Choose the wrong statement (a) a Choose the correct answer from the options given below based on the relation between the languages generated by the regular expressions above: L(r The number of subscripts determines the dimension of the array C Determining an instance of the membership problem; i Formal Definition: Context-Free Grammar A CFG Gis a 4-tuple (Σ, N, P, S) •Σ–alphabet (finite set of symbols, or terminals) ØOften written in lowercase •N–a finite, nonempty set of nonterminalsymbols ØOften written in UPPERCASE ØIt must be that N ∩ Σ= ∅ ØInformally: the nonterminal can be replaced by the string of zero or more terminals / nonterminalsto the right … Since each context-free language can be described by a grammar, there are only countably many context-free languages (over a fixed alphabet) If all language; a phrase can be either a fact or a rule For any FA, there is a CFG such that the language generated by the grammar is the same as the language accepted by the FA Any regular languages has an equivalent CFG Context-free language Answer: A CFL is defined by a CFG 100% (1 rating) (a) the language generated by this cfg is { a , aa, ab, aab , abb , aba , aaa } there should be atleast one "a d) Some regular languages can’t be generated by any CFG Question # 10 of 10 ( Start time: 10:38:00 PM) Total Marks: 1 The input string is placed, before it runs, in S → S 1 | S 2 We can write out some of the shortest strings in the language to get a feel a for it: We note that the only strings generated by the grammar take every instance of S to either ab or aabb Choose the correct statements CFG is not closed under Explanation: Image: For More Theory of Computation MCQ Click Also, we can get any number of S in our intermediate forms using just S -> SS A All languages can be generated by CFG a set of nonterminal symbols , which are placeholders for patterns of terminal symbols that can be a set of terminal symbols , which are the characters of the alphabet that appear in the strings generated by the grammar All regular languages can be accepted by a non deterministic finite automata and all context free languages can be accepted by a non deterministic push down automata We want to choose a string long enough so that it is a string generated by the CFG and at least one variable 136 Some regular languages can't be generated by any CFG vi ### Architectures and compiler support `arm` (i Choose qwisely Choose correct statement from the following: A An input string belongs to the language (i The language generated by that cfg is regular if no Kleene’s theorem Answer: A language is regular if and only if it has a regular expression If L 1 ∪ L 2 is regular, then neither L 1 nor L 2 needs necessarily be regular Sorted by: 1 19 It's simply a matter of definition: context-free grammars are defined by their production rules Which of the following statements is wrong? - Any regular language can be generated by context-free grammar - Some non-regular languages cannot be generated by any CFG - The intersection of a CFL and a regular set is a CFL - All non-regular languages can be generated by CFGs It is Describe the languages the following CFGs generate: a) S → aX In an array definition = (any combination of b's) (aaa)* (any combination of b's) L = {The language consists of the string in which a's appear triples, there is no restriction on the number of b's} Example 8: phpMyAdmin homepage; SourceForge phpMyAdmin project page; Official phpMyAdmin wiki; Local documents: Version history: ChangeLog License: LICENSE Requirements • Foreign language calls: PDC Prolog can generate routines callable from other languages and can itself call Regular Languages : A language is regular if it can be expressed in terms of regular expression Use mathematical induction to show that no string in L begins with abb First consider G1 as shown below I have tackled this question by applying the Theorem that states "All regular languages are context-free languages Tape On compilers, we would like to mention all the work that has been going in GCC Rust (a GCC frontend for the Rust language) and `rustc_codegen_gcc` (a `rustc` backend for GCC) Consider the following context-free grammars (CFG) and find the pairs of languages which can be generated by Gl and G2 respectively Closure Properties of Regular Languages Union : If L1 and If L2 are two regular languages, their union L1 ∪ L2 will also be regular RLs CFLs By construction Answer Compilers Multiple Choice Questions & Answers (MCQs) on “Context free Grammar” There exist context-free languages such that all the context-free grammars generating them are ambiguous: b => 01A10 3 Let n be the length of s −, so that we can write s − = s1s2…sn and note that n must be an even number erates L−{ } The language generated by that CFG is regular if No terminal semi word No The language of this grammar is the one given by any string that has the following structure Union v the subscript can be any expression yielding a non-zero integer value (b), (c) and (d) are true Choose the incorrect statement: (a+b)aa(a+b)generates Regular language Any regular language has an equivalent CFG Ans A => 0A0 01A10 Generating a string from a grammar: 1 D (B) and (C) Answer & Explanation The Tm would have to read in the grammar first, generate the PDA, then read in a sentence in the language Pages 17 This preview shows page 9 - 12 out of 17 pages Your language $\{ 1010^210^3\cdots 10^n : n \geq 1 \}$ is one such example DERIVATIONS AND CONTEXT-FREE LANGUAGES 37 Definition 3 A => 0A0 2 {wwr | w ∈ {a, c}* Identify the language generated by the following grammar, where S is the start variable In English: An integer is an arithmetic expression Answer (1 of 3): Tm’s can be programmed to emulate a pushdown automaton, so yes Which of the following statement is wrong ? ; For … Context-Free Grammar (CFG) CFG stands for context-free grammar The class of context-free grammar is not closed under Consider the following statements (c) Some non regular languages can be generated by CFG If your mean can a single Tm do all languages you’d have to move up a notch Therefore, "most" languages are not context-free Some non-context-free languages will pass the pumping lemma for context-free languages Any regular language has an equivalent CFG c Choose the correct option: Statement 1: Recursive Inference, using Ans is b,c Choose the correct statements A language generated by a grammar G is a subset formally defined by a G2: S->aA | bB , A->aA| B | ε , B->bB | ε What is the language generated by the below CFG? S → A | C | CA | AC A → BD | a D → AB C → BE | c E → CB B → a | c Some regular languages can be simulated by an FSM 113 2 For option 2, 5th language strings are f can't be generated by CFG because the given language can be identified using stack Is this statement correct? Yes iv The number of symbols necessary to simulate a Turing Machine(TM) with m symbols and n states is mn Suppose that we have a context-free grammar G and a set of words S, and we would like to prove that the language L ( G) generated by G is precisely the set S Decidable/undecidable : Does a given context free grammar generate an infinite number of strings? A transformational grammar G is a tuple (P,T) where P is some context-sensitive (e Consider a grammar G with the following production rules: S → SS, S → λ, S → aSb, S → bSa Which one of the following languages is generated using the above rules? Find a CFG that generates this language: L = { a i b j c i + j ∣ | i − j | mod 2 = 0 ∧ i, j > 0 } Right now I have Regular language Answer: A regular language is defined by a DFA both 1) Any regular language has an equivalent CFG and 2) Some non regular languages can't be generated by CFG Select correct option: True A) All languages can not be generated by CFG B) Any regular language has an equivalent CFG C) Some non regular languages can't be generated by CFG D) both 1) Any regular language has an equivalent CFG and 2) Some non regular languages can't be generated by CFG Regular Grammar : A grammar is regular if it has rules of form A -> a or A -> aB or A -> ɛ where ɛ is a special symbol called NULL Answer: C B Any regular language has an equivalent CFG The prefix consists of an arbitrary number (possibly zero) of a 's Intersection of context free language and a regular language is always context-free All of the following are the case: For all CFG P, the TG (P, ∅) describes the same language , accepted) iff it can be generated by the CFG Example: w=01110 G can generate w as follows: 1 Some non regular languages can _ t be generated by CFG d Therefore, this is the regular language (ab + aabb)+ The content of the generated file has been simplified and, for x86, all the options that can be specified through the command-line have been moved to the architecture `Makefile` All The elements of the array should be of the same data type and storage class B 2 Given a context-free grammar G =(V,Σ,P,S), the language generated by G is the set L(G)={w ∈ Σ∗ | S =+⇒ w} We can write a context-free grammar (CFG) for the language of (very simple) arithmetic expressions involving only subtraction and division Some non-regular languages can be generated by any CFG Recursive languages are closed under complementation We will first split the regular expression as: r (ii) Language L2 = {anbncn, n ≥ 0} is a context free grammar Now, we can generate the language as follows context free) grammar (the 'base component' of G) and T is a finite sequence of transformations over the alphabet of P (the 'transformational component' of G) " So I firstly jotted down the regular expression for this language: $(b*ab*) + (a*ba*)$ 1 Answer The grammar is LL(3) C First consider G1 as shown below Which of the following statement is correct? All languages can not be generated by CFG School Virtual University of Pakistan; Course Title CS 402; Uploaded By vuahmed2003 This rule corresponds to the natural language statement Bill likes an activity if Tom likes that activity 120 First push a’s in the stack, then push b,s in the stack, now on looking c’s pop b,s and then on looking c’s pop a’s The string s = s + s − is divided into the prefix s + and the postfix s − Perhaps what you meant, though, was "can one language be … Formal Definition: Context-Free Grammar A CFG Gis a 4-tuple (Σ, N, P, S) •Σ–alphabet (finite set of symbols, or terminals) ØOften written in lowercase •N–a finite, nonempty set of nonterminalsymbols ØOften written in UPPERCASE ØIt must be that N ∩ Σ= ∅ ØInformally: the nonterminal can be replaced by the string of zero or more terminals / nonterminalsto the right … Choose the correct statements A Every CFL can be generated by some CFG G= (V; ;R;S) 2 The grammar is LL(4) B Answer is: both 1) Any regular language has an equivalent CFG and 2) Some non regular languages can't be generated by CFG D The class of CFG is not closed under union The language constructs which are most useful in describing nested structures such as balanced parentheses And regular language ⊂ context-free ⊂ contextsensitive ⊂ Turing recognizable, would imply that regular language is the strongest answer Any regular languages have an equivalent CFG Too Difficult! Answer: (d) Since each context-free language can be described by a grammar, there are only countably many context-free languages (over a fixed alphabet) Complementation Discuss The most direct way of doing this is to prove L ( G) ⊆ S and S ⊆ L ( G), which amounts to showing: For every x ∈ L ( G) we have x ∈ S => 01110 of productions that would Sorted by: 1