Consider the following context-free grammar πΊ. πβππππ΄π΅π΄πππβ¦
2026
Consider the following context-free grammar πΊ.
πβππππ΄π΅π΄πππ
π΄βπππ΅π΅π΄π | ππ΅ππππ
π΅βππ΅π | ππ
In the above grammar, π is the start symbol, π and π are terminal symbols, and π΄ and B are non-terminal symbols.
Let πΏ(πΊ) be the language generated by the grammar πΊ. For a string π βπΏ(πΊ), let n1(π ) be the number of πβs in π and π2(π ) be the number of πβs in π .
Which of the following statements is/are true?
Answer: B. For every string π βπΏ(πΊ), π1(π )β₯π2(π ); D. For every string π βπΏ(πΊ), π1(π )β€2π2(π ) β To determine the true statements, we analyze the balance of terminal symbols 'a' and 'b' in each production rule. First, consider non-terminal B. Theβ¦
- A.
There is a string π βπΏ(πΊ) such that π1(π )<π2(π )
- B.
For every string π βπΏ(πΊ), π1(π )β₯π2(π )
- C.
There is a string π βπΏ(πΊ) such that π1(π )>2π2(π )
- D.
For every string π βπΏ(πΊ), π1(π )β€2π2(π )
Attempted by 24 students.
Show answer & explanation
Correct answer: B, D
To determine the true statements, we analyze the balance of terminal symbols 'a' and 'b' in each production rule.
First, consider non-terminal B. The productions are B β aBb | ab. In both cases, the number of 'a's equals the number of 'b's generated directly or recursively. Thus, for any string derived from B, n1 = n2.
Next, analyze non-terminal A. Production 1 (aaBBAb) adds more 'a's than 'b's and includes A recursively. Production 2 (bBabaa) adds more 'a's than 'b's and includes B. Derivation shows that for any string from A, n1 > n2.
Finally, analyze the start symbol S (ab a A B A b b a). It adds more 'a's than 'b's directly and includes two instances of A (where n1 > n2) and one B (n1 = n2). The total count confirms that for any string s in L(G), the number of 'a's is strictly greater than the number of 'b's.
Based on this analysis, the statements corresponding to options 1 and 3 are true.