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Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?
- a)G is not ambiguous
- b)There exist x, y, ∈ L (G) such that xy ∉ L(G)
- c)There is a deterministic pushdown automaton that accepts L(G)
- d)We can find a deterministic finite state automaton that accepts L(G)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let G = ({S}, {a, b} R, S) be a context free grammar where the rule se...
An ambiguous grammar can be converted to unambiguous one.
Here we can get grammar in partial GNF form as S -> ab | abS | aSb | aSbS
We can convert this into GNF too but no need for PDA reasoning so, above grammar is not a ambiguous thus a definite PDA possible
Trick for this is but just deriving 3-4 strings from grammar, we can easily understand its (anbn)* above expression anbn is in CFL thus closure of DCFG is a DCFG i.e., you can get L = {ε, ab, abab, aabb, aabbab, abaabb, ababab,......}
PDA will push "a" until "b" is read, start popping "a" for the "b" read.
If "a" is read again from the tape then push only when stack is empty else terminate.
Repeat this until string is read. Remember fastest way to get answer is by elimination other options.
Here we can get grammar in partial GNF form as S -> ab | abS | aSb | aSbS
We can convert this into GNF too but no need for PDA reasoning so, above grammar is not a ambiguous thus a definite PDA possible
Trick for this is but just deriving 3-4 strings from grammar, we can easily understand its (anbn)* above expression anbn is in CFL thus closure of DCFG is a DCFG i.e., you can get L = {ε, ab, abab, aabb, aabbab, abaabb, ababab,......}
PDA will push "a" until "b" is read, start popping "a" for the "b" read.
If "a" is read again from the tape then push only when stack is empty else terminate.
Repeat this until string is read. Remember fastest way to get answer is by elimination other options.
Most Upvoted Answer
Let G = ({S}, {a, b} R, S) be a context free grammar where the rule se...
Question Analysis
The given context-free grammar G has the rule set R consisting of three production rules: S → aSb, S → SS, and S → ε. We need to determine which statement is true among the given options: a) G is not ambiguous, b) There exist x, y, L(G) such that xy ∈ L(G), c) There is a deterministic pushdown automaton that accepts L(G), and d) We can find a deterministic finite state automaton that accepts L(G).
Answer Explanation
The correct answer is option 'C' - There is a deterministic pushdown automaton that accepts L(G).
To prove this, we need to show that the given context-free grammar G can be accepted by a deterministic pushdown automaton (DPDA).
Definition: A deterministic pushdown automaton (DPDA) is a theoretical model of computation that extends a deterministic finite automaton (DFA) with a stack. It can recognize context-free languages.
Steps to Construct a DPDA for G:
1. Start with an empty stack and the initial state of the DPDA.
2. Read the input string from left to right.
3. Based on the current state and the input symbol, determine the next state and the operation to be performed on the stack.
4. Perform the stack operation (Push or Pop) and transition to the next state.
5. Repeat steps 3 and 4 until the input string is completely processed.
6. If the DPDA reaches an accepting state and the stack is empty, accept the input string. Otherwise, reject it.
DPDA Construction for G:
1. Let's consider the following DPDA configuration:
- States: q0, q1, q2, q3
- Input symbols: a, b
- Stack symbols: Z (initial stack symbol), S
- Initial state: q0
- Accepting states: q3
2. The transition function for the DPDA is as follows:
| Current State | Input Symbol | Stack Top | Next State | Stack Operation |
|:-------------:|:------------:|:---------:|:----------:|:--------------:|
| q0 | a | Z | q1 | SZ |
| q1 | a | S | q1 | SS |
| q1 | b | S | q2 | - |
| q2 | b | S | q2 | - |
| q2 | ε | Z | q3 | - |
3. The DPDA follows the following steps:
- Start in state q0 with an empty stack.
- Read 'a' from the input and push 'S' onto the stack.
- Read 'a' from the input and push 'S' onto the stack.
- Read 'b' from the input and pop 'S' from the stack.
- Read 'b' from the input and pop 'S' from the stack.
- Read ε (empty string) from the input and reach the accepting state q3 with
The given context-free grammar G has the rule set R consisting of three production rules: S → aSb, S → SS, and S → ε. We need to determine which statement is true among the given options: a) G is not ambiguous, b) There exist x, y, L(G) such that xy ∈ L(G), c) There is a deterministic pushdown automaton that accepts L(G), and d) We can find a deterministic finite state automaton that accepts L(G).
Answer Explanation
The correct answer is option 'C' - There is a deterministic pushdown automaton that accepts L(G).
To prove this, we need to show that the given context-free grammar G can be accepted by a deterministic pushdown automaton (DPDA).
Definition: A deterministic pushdown automaton (DPDA) is a theoretical model of computation that extends a deterministic finite automaton (DFA) with a stack. It can recognize context-free languages.
Steps to Construct a DPDA for G:
1. Start with an empty stack and the initial state of the DPDA.
2. Read the input string from left to right.
3. Based on the current state and the input symbol, determine the next state and the operation to be performed on the stack.
4. Perform the stack operation (Push or Pop) and transition to the next state.
5. Repeat steps 3 and 4 until the input string is completely processed.
6. If the DPDA reaches an accepting state and the stack is empty, accept the input string. Otherwise, reject it.
DPDA Construction for G:
1. Let's consider the following DPDA configuration:
- States: q0, q1, q2, q3
- Input symbols: a, b
- Stack symbols: Z (initial stack symbol), S
- Initial state: q0
- Accepting states: q3
2. The transition function for the DPDA is as follows:
| Current State | Input Symbol | Stack Top | Next State | Stack Operation |
|:-------------:|:------------:|:---------:|:----------:|:--------------:|
| q0 | a | Z | q1 | SZ |
| q1 | a | S | q1 | SS |
| q1 | b | S | q2 | - |
| q2 | b | S | q2 | - |
| q2 | ε | Z | q3 | - |
3. The DPDA follows the following steps:
- Start in state q0 with an empty stack.
- Read 'a' from the input and push 'S' onto the stack.
- Read 'a' from the input and push 'S' onto the stack.
- Read 'b' from the input and pop 'S' from the stack.
- Read 'b' from the input and pop 'S' from the stack.
- Read ε (empty string) from the input and reach the accepting state q3 with
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Community Answer
Let G = ({S}, {a, b} R, S) be a context free grammar where the rule se...
An ambiguous grammar can be converted to unambiguous one.
Here we can get grammar in partial GNF form as S -> ab | abS | aSb | aSbS
We can convert this into GNF too but no need for PDA reasoning so, above grammar is not a ambiguous thus a definite PDA possible
Trick for this is but just deriving 3-4 strings from grammar, we can easily understand its (anbn)* above expression anbn is in CFL thus closure of DCFG is a DCFG i.e., you can get L = {ε, ab, abab, aabb, aabbab, abaabb, ababab,......}
PDA will push "a" until "b" is read, start popping "a" for the "b" read.
If "a" is read again from the tape then push only when stack is empty else terminate.
Repeat this until string is read. Remember fastest way to get answer is by elimination other options.
Here we can get grammar in partial GNF form as S -> ab | abS | aSb | aSbS
We can convert this into GNF too but no need for PDA reasoning so, above grammar is not a ambiguous thus a definite PDA possible
Trick for this is but just deriving 3-4 strings from grammar, we can easily understand its (anbn)* above expression anbn is in CFL thus closure of DCFG is a DCFG i.e., you can get L = {ε, ab, abab, aabb, aabbab, abaabb, ababab,......}
PDA will push "a" until "b" is read, start popping "a" for the "b" read.
If "a" is read again from the tape then push only when stack is empty else terminate.
Repeat this until string is read. Remember fastest way to get answer is by elimination other options.
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Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct answer is option 'C'. Can you explain this answer?
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Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct 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 Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct 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 Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct answer is option 'C'. Can you explain this answer?.
Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct 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 Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct 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 Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε Which of the following statements is true?a)G is not ambiguousb)There exist x, y, ∈ L (G) such that xy ∉ L(G)c)There is a deterministic pushdown automaton that accepts L(G)d)We can find a deterministic finite state automaton that accepts L(G)Correct answer is option 'C'. Can you explain this answer?.
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