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NTA NET Nov 2020 Computer Science Question Paper (P1 + P2)

NTA NET Dec 2019 Computer Science Question Paper (P1 + P2)

NTA NET June 2019 Computer Science Question Paper (P1 + P2)

NTA NET Dec 2018 Computer Science Question Paper (P1 + P2)

Q1. A grammar G is LL(1) if and only if the following conditions hold for two distinct productions A → α | β
I. First (α) ∩ First (β) ≠ {a} where a is some terminal symbol of the grammar.
II. First (α) ∩ First (β) ≠ λ
III. First (α) ∩ Follow(A) = φ if λ є First (β)
(A) I and II (B) I and III
(C) II and III (D) I, II and III

Ans: D
assume
S—> aX |aYb first(aX)= a first(aYb)= a
so there will 2 entries for S in terminal ‘a’ which violates the rule of LL(1) grammar because for every variable and for every terminal there can be at most 1 production .
hence ,their intersection should be ∅
first one statement is correct
second one is true with similar reasoning as for first. now third assume example
1) S–>aSA | ∈
2) A–> C | ∈

now take 2) production first(c)=c
follow(A) = follow(S) = => first(A)= c, ∈ ,$ now the first(c)∩ follow(A)= c means there will be again 2 entries for A in c place that means again violation of LL(1) rules. so first(c)∩ follow(A)= ∅ is true.
hence answer is (D)

Q2. Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar ?
(A) Removing left recursion alone.
(B) Removing the grammar alone
(C) Removing left recursion and factoring the grammar
(D) None of the above

Ans: D The grammar not only need to be free from left recursion and left factoring but also it need to be free from ambiguity

Q3. A shift reduce parser suffers from
(A) shift reduce conflict only
(B) reduce reduce conflict only
(C) both shift reduce conflict and reduce reduce conflict
(D) shift handle and reduce handle conflicts

Ans: C

Both shift reduce conflict and reduce reduce conflict A shift-reduce parser scans and parses the input text in one forward pass over the text, without backing up.

Compiler Questions of UGC NET CS Paper – II December – 2014

Q1. Shift-Reduce parsers perform the following :
(A) Shift step that advances in the input stream by K(K>1) symbols and Reduce step that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.
(B) Shift step that advances in the input stream by one symbol and Reduce step that applies a completed grammar rule to some recent
parse trees, joining them together as one tree with a new root symbol.
(C) Shift step that advances in the input stream by K(K = 2) symbols and Reduce step that applies a completed grammar rule to form a single tree.
(D) Shift step that does not advance in the input stream and Reduce step that applies a completed grammar rule to form a single tree.

Ans: B

Q2. Which of the following is true ?
(A) Canonical LR parser is LR (1) parser with single look ahead terminal
(B) All LR(K) parsers with K > 1 can be transformed into LR(1) parsers.
(C) Both (A) and (B)
(D) None of the above

Ans: C A canonical LR parser or LR(1) parser is an LR(k) parser for k=1, i.e. with a single look-ahead terminal. The special attribute of this parser is that all LR(k) parser’s with k>1 can be transformed into a LR(1) parser.

Q3. In a two-pass assembler, symbol table is
(A) Generated in first pass
(B) Generated in second pass
(C) Not generated at all
(D) Generated and used only in second pass

Ans: A
In first pass it does following things It allocates space for the literals.
It computes the total length of the program.
It builds the symbol table for the symbols and their values

Q4. Debugger is a program that
(A) allows to examine and modify the contents of registers
(B) does not allow execution of a segment of program
(C) allows to set breakpoints, execute a segment of program and display contents of
register
(D) All of the above

Ans: C
Debugger allows to set breakpoints, execute a segment of program and display
contents of register. One of the Debugger name is GDB used under Unix.

Here I am going to post PREVIOUS YEAR SOLVED QUESTIONS OF TOC AND COMPILER 2005 .

There are 2 Questions from TOC and 3 from COMPILER IN YEAR 2005 (D2)

Q. 1 Which sentence can be generated by

S -> d | bA, A -> d | ccA

a) bccccd

b) aabccd

c) ababccd

d) abbbd

Ans-  A -> d|ccA 

Explanation:  A -> (cc)*d

Then S -> d| bA it give

S -> d| b(cc)*d

L= b( c c )* d+d

Now finally it become

bccccd ,which is generated by the language.

Q 2. Regular expression a+b denotes the set:

a) {a}

b) {€,a,b}

c) {a,b}

d) Non of these

Ans: c) {a,b}

Explanation: a+b is used to accept either a or b only in a regular expression. So we can say that c is the ans.

Q  3. A Top-down Parse generates:

a) Left most derivation

b) Right most derivation

c) Right most derivation in reverse

d) Left most derivation in reverse

Ans: a) Left most derivation

Explanation:

top down parsing -left most derivation

Q 4. A general macroprocessor is an in built function of:

a) Loader

b) Linker

c) Editor

d) Assembler

Ans: d) Assembler

Explanation: An assembler is a program that takes basic computer instructions and it converts them into a pattern of bits that the computer’s processor can use to perform it’s basic operations ,so ans is assembler.

Q 5. Which activities is not included in the first pass of two pass assembler?

a)  build the symbol table

b) Construct the Intermediate code

c) seperats memonic opcode and operand field.

d) Non of these

Ans: d) Non of these

Explanation: In Question asked not included but all the three options are included in the first Parse of two pass assembler so ans is d.

Hope you all are defenately benefited by this post and understand well. If required more detail explanation then comment in comment section ,will defenately explain that topic. And also mention is this sufficient explanation so that I can improve the pattern of post and make it more easier to you .

Thanks a lot☺️

Friends First of all I am going to give you collection of all PYQSyear wise and unit wise too ……so stay tuned with UNIFYSTUDY😊

THERE R TOTAL 5Q FROM UNIT 8 ACCORDING to new syllabus ,one is from TOC and 4 is from Compiler.

I am going to give you all questions and ans here,if you need detailed explanation of these topics then plz comment I’ll defenately create new course on detail of that particular topic, thanks a lot 🙏😊

Q 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

And: C – ii and iv

Explanation:- because context free languages are closer under

union,concatenation,kleen star,kleen plus, homomorphism, epsilon free homomorphism,substitution,inverse homomorphism,reverse,union with a regular language, Intersection with a regular language,left difference with a regular language, left quotient with a RL, right quotient with RL kleen star ,subset.

And CFL NOT closed under only

Intersection,set difference,complement, right difference with a regular language.

Q 2 . Which activity is not included in the first pass of two pass assemblers?

(a) Build the symbol table

(b)Construct the intermediate code

(c)Separate mnemonic opcode and operand fields

(d)None of these

Ans -(d) None of these

Explanation:- since all the activities are included in the first pass of two pass assemblers so there is no option which not included in first pass of two pass assemblers.

Q 3.Which activity is included in the first pass of two pass assemblers?

(a) Build the symbol table

(b) Construct the intermediate code

(c) Seprate mnemonic opcode and operand fields

(d) None of these

Ans- friends here ans is a,b,c all are correct because all are included in the first pass of two pass assemblers, so don’t get confused.

Also see both above Questions you can compare them.

If needed topic detail ,then plz comment ,sooner I’ll provide detailed explanation of these topics.

Q 4. In two pass assembler the symbol table is used to store :

(a) Label and value

(b) Only value

(c) Mnemonic

(d) Memory Location

Ans- (a)Label and value

Explanation:- dear friends As we all know that

A TWO PASS ASSEMBLER makes two passes over the input program,that is It read the program twice. On the first first pass the symbol table is constructed.On the second pass, the complete symbol table is used to allow expressions . It is used to be evaluated without problems because forward reference is used there. It’s a very important topic if needed then comment I’ll explain in detail .

Q 5. Non modifiable procedures are called

(a) Seriall usable procedures

(b) Concurrent procedures

(c) Reentrant procedures

(d) Topdown procedures

Ans – (c) dear friends non modifiable procedures are called Reentrant procedures in this a single copy of the program code can be shared by multiple users at the same period of time, so ans is c.

These all 5 are Questions from year 2004 from  unit 8 of Updated Syllabus of NTA UGC NET, you can also feel it very much beneficial for preparation of  GATE Exams and computer science STUDENTS get benefits from these questions and explanations .

Plz suggest topics which you want to understand , I’ll defenately explain and all PYQS will be here soon unit wise and year wise .

Thanks a lot : BY JYOTI KUMARI