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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Test  >  Compiler Design  >  Test: Regular Expression - Computer Science Engineering (CSE) MCQ

Regular Expression - GATE CSE (CSE) Compiler Design Free MCQ Test


MCQ Practice Test & Solutions: Test: Regular Expression (15 Questions)

You can prepare effectively for Computer Science Engineering (CSE) Compiler Design with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Regular Expression". These 15 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Regular Expression - Question 1
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A regular expression enables a quick test to determine objects and text strings with undependable values.

Detailed Solution: Question 1

Explanation: Because it checks for all the values and determines whether the output string matches with the given string.

Test: Regular Expression - Question 2
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RE can be used only for values of type string and number.

Detailed Solution: Question 2

 No not for all values the string and numbers can we use the RE.

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Test: Regular Expression - Question 3
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You can use RE, if you expect the value of a property to change in an unpredictable way each time its run.

Detailed Solution: Question 3

Explanation: For every cycle the values does not change unpredictably because the type of grammar that it accepts is defined.

Test: Regular Expression - Question 4
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All ___________ Are automatically treated as regular expressions.

Detailed Solution: Question 4

Explanation: The programmatic description is genuinely treated as regular expression.

Test: Regular Expression - Question 5
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If a ‘/’ is used before a character that has no special meaning, ‘/’ is ignored.

Detailed Solution: Question 5

Explanation: The backslash carries no significance and it is ignored.

Test: Regular Expression - Question 6
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The regular expression denote a language comprising all possible strings of even length over the alphabet (0, 1)

Detailed Solution: Question 6

Explanation: Option A does not consider even length criteria for the question.
Option B it can so happen here that from the former bracket it takes 0 or 1 and takes null from the latter then it forms a string of odd length
Option C it gives either 1 or 0.
Hence Option D is the answer.

Test: Regular Expression - Question 7
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The RE gives none or many instances of an x or y is

Detailed Solution: Question 7

Explanation: Whether x or y is denoted by x+y and for zero or more instances it is denoted but (x+y)*.

Test: Regular Expression - Question 8
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The RE in which any number of 0′s is followed by any number of 1′s followed by any number of 2′s is

Detailed Solution: Question 8

Explanation: The order for the desired string is 012 and foe any number of 0s we write 0* for any number of 1s we denote it by 1* and similarly for 2*.Thus 0*1*2*.

Test: Regular Expression - Question 9
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The regular expression have all strings of 0′s and 1′s with no two consecutive 0′s is :

Detailed Solution: Question 9

From the former bracket we choose 0 or epsilon. Then from the latter part 1 or 10 which can be followed by 1 or 10.

Test: Regular Expression - Question 10
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The regular expression with all strings of 0′s and 1′s with at least two consecutive 0′s is:

Detailed Solution: Question 10

Explanation: The expression (0+1)*00(0+1)* is where either it initially takes 0 or 1 or 00 followed by string of combination of 0 and 1.

Test: Regular Expression - Question 11
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 Which of the following is NOT the set of regular expression R = (ab + abb)* bbab

Test: Regular Expression - Question 12
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String generated byS->aS/bA,A->d/ccA

Detailed Solution: Question 12

Explanation: S->aS (substitute S->aS)
S->aaS (substitute S->bA)
S->aabA (substitute A->ccA)
S->aabccA (substitute A->d)
S->aabccd.

Test: Regular Expression - Question 13
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Consider the productions of the grammar S->AA, A->aa, A->bb. Describe the language specified by this grammar.

Detailed Solution: Question 13

Option A is correct.

Because A → aa | bb, each occurrence of A generates either the string aa or the string bb.

Since S → AA, the string from S is the concatenation of two strings produced by A. The possible derivations are:

If both occurrences of A produce aa: S → AA → aa aa → aaaa.

If the first A produces aa and the second produces bb: S → AA → aa bb → aabb.

If the first A produces bb and the second produces aa: S → AA → bb aa → bbaa.

If both occurrences of A produce bb: S → AA → bb bb → bbbb.

Therefore the language is {aaaa, aabb, bbaa, bbbb}, which matches Option A.

Test: Regular Expression - Question 14
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If R is regular language and Q is any language (regular/ non regular), then Pref (Q in R) is _____________

Detailed Solution: Question 14

Explanation: So says the definition of Regular Grammar.

Test: Regular Expression - Question 15
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The production of the form non terminal → Λ is said to be null production.

Detailed Solution: Question 15

The productions of type ‘A -> λ’ are called λ productions ( also called lambda productions and null productions) . These productions can only be removed from those grammars that do not generate λ (an empty string). It is possible for a grammar to contain null productions and yet not produce an empty string.

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About this Computer Science Engineering (CSE) Practice Test

This test contains 15 MCQs on Regular Expression with detailed solutions for each question. It is part of the Compiler Design course on EduRev, designed to align with the current Computer Science Engineering (CSE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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