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Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer?.

Solutions for Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE). Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.

Here you can find the meaning of Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider two languages L1 and L2 each on the alphabet ∑. Let f : ∑ → ∑ be a polynomial time computable bijection such that (∀ x) [x ∈ L1 iff f(x) ∈ L2]. Further, let f-1be also polynomial time computable. Which of the following CANNOT be true?a)L1 ∈ P and L2 is finiteb)L1 ∈ NP and L2 ∈ Pc)L1 is undecidable and L2 is decidabled)L1 is recursively enumerable and L2 is recursiveCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.