Automata Theory Questions and Answers – Eliminating Epsilon Productions

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This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Eliminating Epsilon Productions”.

1. The use of variable dependency graph is in:
a) Removal of useless variables
b) Removal of null productions
c) Removal of unit productions
d) None of the mentioned
View Answer

Answer: a
Explanation: We use the concept of dependency graph inorder to check, whether any of the variable is reachable from the starting variable or not.
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2. The variable which produces an epsilon is called:
a) empty variable
b) nullable
c) terminal
d) all of the mentioned
View Answer

Answer: b
Explanation: Any variable A for which the derivation: A->*e is possible is called Nullable.

3. Statement:
For A-> e ,A can be erased. So whenever it appears on the left side of a production, replace with another production without the A.
State true or false:
a) true
b) false
View Answer

Answer: b
Explanation: A can be erased. So whenever it appears on the right side of the production, replace with another production without the A.

4. Simplify the given grammar:
S->aXb
X->aXb | e
a) S->aXb | ab, X-> aXb | ab
b) S->X | ab, X-> aXb | ab
c) S->aXb | ab, X-> S | ab
d) None of the mentioned
View Answer

Answer: a
Explanation: As X is nullable, we replace every right hand side presence of X with e and produce the simplified result.
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5. Consider the following grammar:
A->e
B->aAbC
B->bAbA
A->bB
The number of productions added on the removal of the nullable in the given grammar:
a) 3
b) 4
c) 2
d) 0
View Answer

Answer: b
Explanation: The modified grammar aftyer the removal of nullable can be shown as:
B->aAbC| abC
B->bAbA| bbA| bAb| bb
A->bB

6. Let G=(V, T, P, S) be a CFG such that _____________. Then there exists an equivalent grammar G’ having no e productions.
a) e ∈ L(G)
b) w ∉ L(G)
c) e ∉ L(G)
d) w ∈ L(G)
View Answer

Answer: c
Explanation: Theorem: Let G = (V, T, S, P) be a CFG such that e ∉ L(G). Then there exists an equivalent grammar G’ having no e-productions.

7. For each production in P of the form:
A-> x1x2x3…xn
put into P’ that production as well as all those generated by replacing null variables with e in all possible combinations. If all x(i) are nullable,
a) A->e is put into P’
b) A->e is not put into P’
c) e is a member of G’
d) None of the mentioned
View Answer

Answer: b
Explanation: It is an exception that A->e is not put into P’ if all x(i) are nullable variables.
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8. For the given grammar G:
S->ABaC
A->BC
B->b| e
C->D| e
D-> d
Remove the e productions and generate the number of productions from S in the modified or simplified grammar.
a) 6
b) 7
c) 5
d) 8
View Answer

Answer: d
Explanation: The grammar after the removal of epsilon production can be shown as:
S->ABaC| AaC| ABa| Aa| a| aC| Ba| BaC
A->BC| B| C
B->b
C->D
D-> d

9. Consider G=({S,A,B,E}, {a,b,c},P,S), where P consists of S →AB, A →a, B →b and E →c.
Number of productions in P’ after removal of useless symbols:
a) 4
b) 3
c) 2
d) 5
View Answer

Answer: a
Explanation:
P’= S->AB, A->a, B-> b,
V’={S, A, B},
∑’={a, b}

10. Given grammar G:
S->aS| AB
A-> e
B-> e
D-> b
Reduce the grammar, removing all the e productions:
a) S->aS| AB| A| B, D-> b
b) S->aS| AB| A| B| a, D-> b
c) S->aS| AB| A| B
d) None of the mentioned
View Answer

Answer: b
Explanation: We will replace all the nullables wherever they appear in the right hand side of any production. D will not be erased as we are just removing nullable variables not completely simplifying the grammar.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn