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Will you please provide me the UGC NET computer science solved Question papers??? As per your request here I am sharing the UGC NET computer science solved Question papers: 1. The context-free languages are closed for : (i) Intersection (ii) Union (iii) Complementation (iv) Kleene Star then (A) (i) and (iv) (B) (i) and (iii) (C ) (ii) and (iv) (D) (ii) and (iii) Ans:-C Explanation:- Context-free languages are closed under union,intersection and star-closure. Context-free languages are not closed under intersection and complementation. June - 2005 2. Which of the following is not true ? (A) Power of deterministic automata is equivalent to power of non-deterministic automata. (B) Power of deterministic pushdown automata is equivalent to power of non-deterministic pushdown automata. (C) Power of deterministic turing machine is equivalent to power of non-deterministic turing machine. (D) All the above Ans:-B Explanation:- Deterministic and nondeterministic finite automata have equivalent computational capabilities. Deterministic and nondeterministic Turing machines have the same computational capabilities. However, the computational capabilities of deterministic push-down automata are less than those of nondeterministic push-down automata. So, the answer is B. 3. Identify the language which is not context - free. (A) L={wwR|w is a member of {0,1}*} (B) L={anbn|n>=0} (C )L={ww|w is a member of {0,1}*} (D) L={anbmcmdn|n,m>=0} Ans:-B December 2005 4. Which sentence can be generated by S → d/bA, A → d/ccA : (A) bccddd (B) aabccd (C) ababccd (D) abbbd Ans:-A Explanation:- The most closest answer seems to be A only. The terminals in the grammar are d,b and c. There is no terminal 'a' at all. So the options B,C and D are ruled out. option A can also be derived at something like this. S->bA ->bccA ->bccccA ->bccccd So the sentence generated should be bccccd. If there are any other explanations for the same please post them. 5. Regular expression a+b denotes the set : (A) {a} (B) {Epsilon, a, b} (C) {a, b} (D) None of these Ans:-C June 2006 6. Which of the following strings is in the language defined by grammar S → OA, A → 1A/0A/1 (A) 01100 (B) 00101 (C ) 10011 (D) 11111 Ans:-B Explanation:- Non terminals in the above grammar are S, and A. Terminals are 0 and 1. The start symbol of the grammar is S. S->0A. So rule out strings beginning with 1. so that leaves us with two options A and B. S->0A ->00A ->001A ->0010A ->00101 So the option B is correct. If you try to derive the string for option A you would not be able to get it. So, the correct answer is option A. 7. The logic of pumping lemma is a good example of : (A) pigeon hole principle (B) recursion (C) divide and conquer technique (D) iteration Ans:-A December - 2006 8. Which of the regular expressions corresponds to this grammar ? S → AB/AS, A → a/aA, B → b (A) aa*b+ (B) aa*b (C ) (ab)* (D) a(ab)* Ans:-B JUNE 2007 9. The regular expression given below describes : r=(1+01)*(0+λ) (A) Set of all string not containing '11' (B) Set of all string not containing '00' (C) Set of all string containing '01' (D) Set of all string ending in '0' Ans:-B Explanation:- The meaning of (1+01)* means that the set of strings of 1 and 01 of any length including the NULL string. (0+λ) means either 0 or λ. So the full regular expression stands for the set of strings of 1's and 01's of any length ending with 0 or λ. So we cannot say the string will only end in '0'. It could end in λ also. The string can begin with 1 or 01 and multiple times it can get repeated.But the string '11' cannot be together. So the option could be B. Posted by UGC NET SOLVED at 00:55 5 comments: Wednesday, 15 May 2013 JUNE 2012 - PAPER III 21. A* algorithm uses f' = g + h' to estimate the cost of getting from the initial state to the goal state, where g is a measure of the cost of getting from initial state to the current node and the function h' is an estimate of the cost of getting from the current node to the goal state. To find a path involving the fewest number of steps, we should set (A) g=1 (B) g=0 (C) h'=0 (D) h'=1 Ans:-A 22. The transform which possesses the highest ‘energy compaction’ property is (A) Slant transform (B) Cosine transform (C) Fourier transform (D) Karhunen-Loeve transform Ans:-D 23. Which one of the following prolog programs correctly implement “if G succeeds then execute goal P else execute goal θ ?” (A) if-else (G, P, θ) :- !, call(G), call(P). if-else (G, P, θ) :- call(θ). (B) if-else (G, P, θ) :- call(G), !, call(P). if-else (G, P, θ) :- call(θ). (C) if-else (G, P, θ) :- call(G), call(P), !. if-else (G, P, θ) :- call(θ). (D) All of the above Ans:-B Explanation:- The syntax of If--then---else in prolog goes the following way… (A->B;C) :- call(A), !, call(B) (A->B;C) :- call(C ) So according to the above syntax option B is correct. 24. The _______ memory allocation function modifies the previous allocated space. (A) calloc( ) (B) free() (C) malloc( ) (D) realloc() Ans:-D 25. Which is not the correct statement(s) ? (i) Every context sensitive language is recursive. (ii) There is a recursive language that is not context sensitive. (A) (i) is true, (ii) is false. (B) (i) is true and (ii) is true. (C) (i) is false, (ii) is false. (D) (i) is false and (ii) is true. Ans:-B 26. The mechanism that binds code and data together and keeps them secure from outside world is known as (A) Abstraction (B) Inheritance (C) Encapsulation (D) Polymorphism Ans:-C 27. Identify the addressing modes of below instructions and match them : (a) ADI (1) Immediate addressing (b) STA (2) Direct addressing (C )CMA (3) Implied addressing (d) SUB (4) Register addressing (A) a-1,b-2,c-3,d-4 (B) a-2,b-1,c-4,d-3 (C ) a-3,b-2,c-1,d-4 (D) a-4,b-3,c-2,d-1 Ans:- A Explanation:- The instruction ADI adds some content to the accumulator. It is an immediate addressing mode instruction. The instruction STA stores the contents of the accumulator in the particular memory location specified as operand. CMA instruction takes complement of the contents of the accumulator. SUB instruction subtracts the contents of the register to the contents of the accumulator. . So the option is A. 28. Which one of the following is not a Greibach Normal form grammar ? (i) S → a | bA | aA | bB A→a B→b (ii) S→a|aA|AB A→a B→b (iii) S→a|A|aA A→a (A) (i) and (ii) (B) (i) and (iii) (C) (ii) and (iii) (D) (i), (ii) and (iii) Ans:-C Explanation:- Restriction for GNF:- The first symbol on the right hand side of the production must be a terminal. It can be followed by zero or more variables. In grammar (ii) of the question, S->AB is a production. AB are two non-terminals and it can be in GNF. In grammar (iii) S->A is given and it is a unit production and that is not allowed in GNF. So the grammar which is not in GNF is (ii) and (iii). So the option is C. 29. Which of the following IP address class is a multicast address ? (A) Class A (B) Class B (C) Class C (D) Class D Ans:-D 30. While unit testing a module, it is found that for a set of test data, maximum 90% of the code alone were tested with a probability of success 0.9. The reliability of the module is (A) atleast greater than 0.9 (B) equal to 0.9 (C) atmost 0.81 (D) atleast 1/0.81 Ans:-C Posted by UGC NET SOLVED at 04:38 1 comment: Wednesday, 20 March 2013 JUNE 2012 - PAPER III 18. Consider a schema R(A, B, C, D) and functional dependencies A → B and C → D. Then the decomposition R1(A, B) and R2(C, D) is (A) Dependency preserving but not lossless join (B) Dependency preserving and lossless join (C) Lossless Join but not dependency preserving (D) Lossless Join Ans:-A Explanation:- I have given the explanation in question no. 17 of December 2010 UGC paper. I am repeating it once again here for dependency preservation and lossless join. First of all let us consider the dependency preservation and how to understand it. Definition of Dependency preservation decomposition:- Each FD specified in F either appears directly in one of the relations in the decomposition, or be inferred from FDs that appear in some relation. Let us consider the above example for Dependency preservation Let R be a relation R(A B C D) Let there be 2 functional dependencies. FD1: A->B FD2: C->D Let the relation R be decomposed into two more relations. R1(A B ) : R2(C D) Let us first consider the relation R1(A B ). Here between A and B the functional dependency FD1 is preserved. Let us now consider the second relation R2(C D). Between C and D the FD, FD2 is preserved. So in the two relations R1 and R2, all the 2 functional dependencies are preserved. Now for the lossless join. A decomposition of R into R1 and R2 is lossless join if and only if at least one of the following dependencies is in F*: R1 ∩ R2 -> R1 R1 ∩ R2 -> R2 In the above example, R1 ∩ R2 = { null }. So, it is not a lossless join. So the answer is dependency preserving but not lossless join. So, the answer is option A. 19. The quantiser in an image-compression system is a (A) lossy element which exploits the psychovisual redundancy (B) lossless element which exploits the psychovisual redundancy (C) lossy element which exploits the statistical redundancy (D) lossless element which exploits the statistical redundancy Ans:-A 20. Data Warehouse provides (A) Transaction Responsiveness (B) Storage, Functionality Responsiveness to queries © Demand and supply Responsiveness (D) None of the above Ans:-B Posted by UGC NET SOLVED at 06:45 No comments: Tuesday, 19 March 2013 JUNE 2012 - PAPER III 11. X.25 is ________ Network. (A) Connection Oriented Network (B) Connection Less Network (C) Either Connection Oriented or Connection Less (D) Neither Connection Oriented nor Connection Less Ans:-A Explanation:- X.25 is a connection oriented protocol. 12. Which of the following can be used for clustering of data ? (A) Single layer perception (B) Multilayer perception (C) Self organizing map (D) Radial basis function Ans:-C 13. Which of the following is scheme to deal with deadlock ? (A) Time out (B) Time in (C) Both (A) & (B) (D) None of the above Ans:-A 14. If the pixels of an image are shuffled then the parameter that may change is (A) Histogram (B) Mean (C) Entropy (D) Covariance Ans:-D 15. The common property of functional language and logical programming language : (A) Both are declarative (B) Both are based on λ-calculus (C) Both are procedural (D) Both are functional Ans:-A 16. Given the following statements : (i) The power of deterministic finite state machine and non- deterministic finite state machine are same. (ii) The power of deterministic pushdown automaton and non- deterministic pushdown automaton are same. Which of the above is the correct statement(s) ? (A) Both (i) and (ii) (B) Only (i) (C) Only (ii) (D) Neither (i) nor (ii) Ans:-B Explanation:- The answer is B. But why is it so?. A very good explanation is given in the book "Theory of computation" by A.A.Puntambekar. We all know that finite machine is of two types. One is deterministic finite state machine and the other one non deterministic finite state machine. Both these machine accept regular language only. So the power of DFA = NFA. So the first statement is true. Next comes the question of pushdown automaton and the power of deterministic and non-deterministic being the same. PDA has more power than FA because PDA has a memory and so can accept large class of languages than FA. PDA accepts the language of context free grammar. The power of DPDA is less than NPDA because NPDA accepts a larger class of context free language. Turing machines can accept a more large class of language. Therefore it is the most powerful computational model. The power of deterministic and non deterministic turing machine is the same. 17. LetQ(x,y)denote “x+y=0” and let there be two quantifications given as (i) ∃y∀x Q(x, y) (ii) ∀x∃y Q(x, y) where x & y are real numbers. Then which of the following is valid ? ￼(A) (i) is true & (ii) is false. (B) (i) is false & (ii) is true. (C) (i) is false & (ii) is also false. (D) both (i) & (ii) are true. Ans:-B Last edited by Aakashd; October 17th, 2019 at 01:21 PM. |