Question

Consider the grammar $S \rightarrow aSa \mid bSb \mid a \mid b$. Which one of the following options correctly characterizes the language generated by the given grammar over the alphabet {a,b} 
 

• A. All palindromes
• B. All odd length palindromes
• C. Strings that begin and end with the same symbol
• D. All even length palindromes

Hint:

Palindromic grammars often use symmetric productions like $aSa$ or $bSb$ to preserve symmetry.

Answer 1

The Correct Option is B

Step 1: Understanding the base productions.
The productions $S \rightarrow a$ and $S \rightarrow b$ generate strings of length $1$, which are trivially palindromes and have odd length.
Step 2: Understanding recursive productions.
The productions $S \rightarrow aSa$ and $S \rightarrow bSb$ add the same symbol to both ends of the string, preserving the palindrome property.
Step 3: Analyzing string length.
Each recursive step increases the string length by $2$. Since the base strings have odd length, all derived strings will also have odd length.
Step 4: Final conclusion.
Therefore, the grammar generates exactly all odd-length palindromes over the alphabet $\{a,b\}$.

Answer 2

The grammar is \(S \rightarrow aSa \mid bSb \mid a \mid b\). Let's generate a few strings and check each characterization against them: from \(S \rightarrow a\) we get "a" (length 1); from \(S \rightarrow aSa\) applied to "a" we get "aaa" (length 3); from \(S \rightarrow bSb\) applied to "a" we get "bab" (length 3); applying \(aSa\) to "bab" gives "ababa" (length 5).

1. All palindromes: Every string this grammar can generate does happen to read the same forwards and backwards, but this option would also have to include even-length palindromes like "abba" or "aa". The grammar can never produce those, since it only bottoms out at the single-symbol productions \(a\) or \(b\), never at an empty string or a doubled pair, so the generated language is a strict subset of all palindromes, not all of them.
2. Odd-length palindromes: Every derivation starts from \(a\) or \(b\) (length 1, odd), and each further step wraps the string in a matching pair of symbols, adding exactly 2 to the length each time. Odd plus an even number stays odd, and the matching outer symbols keep the palindrome property intact at every step. Strings like "a", "aaa", "bab", "ababa" all fit this exactly.
3. Strings that begin and end with the same symbol: This is true of every string the grammar produces, but it's a weaker property than what the grammar actually enforces, the grammar structurally forces full symmetry between every mirrored pair of positions, not just matching endpoints.
4. Even-length palindromes: Since every derivation bottoms out at a length-1 base case and each recursive step adds 2 to the length, no string generated by this grammar can ever have even length.

The base cases fix an odd starting length, and the recursive rule preserves both the palindrome symmetry and the oddness of the length at every step.

Therefore, the correct answer is All odd length palindromes.