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

Every derivation from this grammar has to end at one of the base rules, \(S \to a\) or \(S \to b\), which are single characters, so length 1.
Every other rule, \(S \to aSa\) or \(S \to bSb\), just wraps whatever string \(S\) already produced with a matching symbol on both sides, so it adds exactly one symbol to the front and the same symbol to the back. That keeps the string symmetric (a palindrome) and adds 2 to its length each time, so the length always stays odd since it starts odd (at 1).
Because we can never reach an even length or break the mirror symmetry, the exact language is all odd-length palindromes over {a,b}.
So the correct answer is all odd length palindromes.

Answer 2

A cleaner way to see this is to look at what sits at the exact center of any string this grammar produces:

1. All palindromes: This would require the grammar to also be able to produce strings with an even-length "empty center" (two symbols meeting in the middle with nothing between them, like "aa"), but there is no production that lets \(S\) vanish or become empty, so a center always exists as a real single symbol, ruling out even-length palindromes and hence "all palindromes" as too broad.
2. Odd length palindromes: Because \(S\) always eventually resolves to a single central symbol (\(a\) or \(b\)), and every wrapping step adds a mirrored pair around that fixed center, every generated string has one clear middle character with a perfectly symmetric string of equal-length arms on either side, which is exactly the definition of an odd-length palindrome.
3. Strings that begin and end with the same symbol: This only constrains the two endpoints, but says nothing about the symbols in between needing to mirror each other, whereas this grammar forces every pair of positions equidistant from the center to match, a much stronger condition than just the endpoints agreeing.
4. Even length palindromes: These have two middle positions instead of one true center, but this grammar's recursion always leaves exactly one central symbol from a base rule, so it structurally cannot produce a two-symbol center.

The existence of a single, fixed central symbol surrounded by mirrored pairs is the signature of odd-length palindromes specifically.

Therefore, the correct answer is All odd length palindromes.