Consider the following context-free grammar 𝐺. 𝑆→𝑎𝑏𝑎𝐴𝐵𝐴𝑏𝑏𝑎…
2026
Consider the following context-free grammar 𝐺.
𝑆→𝑎𝑏𝑎𝐴𝐵𝐴𝑏𝑏𝑎
𝐴→𝑎𝑎𝐵𝐵𝐴𝑏 | 𝑏𝐵𝑎𝑏𝑎𝑎
𝐵→𝑎𝐵𝑏 | 𝑎𝑏
In the above grammar, 𝑆 is the start symbol, 𝑎 and 𝑏 are terminal symbols, and 𝐴 and B are non-terminal symbols.
Let 𝐿(𝐺) be the language generated by the grammar 𝐺. For a string 𝑠∈𝐿(𝐺), let n1(𝑠) be the number of 𝑎’s in 𝑠 and 𝑛2(𝑠) be the number of 𝑏’s in 𝑠.
Which of the following statements is/are true?
- A.
There is a string 𝑠∈𝐿(𝐺) such that 𝑛1(𝑠)<𝑛2(𝑠)
- B.
For every string 𝑠∈𝐿(𝐺), 𝑛1(𝑠)≥𝑛2(𝑠)
- C.
There is a string 𝑠∈𝐿(𝐺) such that 𝑛1(𝑠)>2𝑛2(𝑠)
- D.
For every string 𝑠∈𝐿(𝐺), 𝑛1(𝑠)≤2𝑛2(𝑠)
Attempted by 13 students.
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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 → 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.