10. Context-Free Languages & Lexers

The start symbol of the PDA is S.
The PDA will only have one state q and it will accept via empty stack.
For each nonterminal/rule $ A \leftarrow \gamma $ add a transition q —$(\epsilon, A/ \gamma)$–➤ q and for each terminal c add q —$(c, c/ \epsilon)$–➤ q

Thus, our PDA has the following transitions looping on state q:

• $ \epsilon, S/aS $
• $ \epsilon, S/aSb $
• $ \epsilon, S/\epsilon $
• $ a, a/\epsilon $
• $ b, b/\epsilon $

Input: aaabb
(aaabb, q, S) ⇒ (aaabb, q, aSb) ⇒ (aabb, q, Sb) ⇒ (aabb, q, aSbb) ⇒ (abb, q, Sbb) ⇒ (abb, q, aSbb) ⇒ (bb, q, Sbb) ⇒ (bb, q, bb) ⇒ (b, q, b) ⇒ ($\epsilon$, q, $\epsilon$)

Is the grammar ambiguuous? yes, because there exist 2 different left-derivations for word aaabb
S ⇒ aSb ⇒ aaSbb ⇒ aaaSbb ⇒ aaabb
S ⇒ aS ⇒ aaSb ⇒ aaaSbb ⇒ aaabb

The accepted language is $ L(G) = \{a^{m}b^{n} | m \ge n \ge 0\} $

Repaired grammar:
$ S \leftarrow aS | A \\ A \leftarrow aAb | \epsilon $

The PDA has the following transitions looping on state q:

• $ \epsilon, S/ABC $
• $ \epsilon, A/aA $
• $ \epsilon, A/\epsilon $
• $ \epsilon, B/bbB $
• $ \epsilon, B/b $
• $ \epsilon, C/cC $
• $ \epsilon, C/c $
• $ a, a/\epsilon $
• $ b, b/\epsilon $
• $ c, c/\epsilon $

Input: aaabbbbbccc
(aaabbbbbccc, q, S) ⇒ (aaabbbbbccc, q, ABC) ⇒ (aaabbbbbccc, q, aABC) ⇒ (aabbbbbccc, q, ABC) ⇒ ⇒ (aabbbbbccc, q, aABC) ⇒ (abbbbbccc, q, ABC) ⇒ (aabbbbbccc, q, ABC) ⇒ (aabbbbbccc, q, aABC) ⇒ ⇒ (abbbbbccc, q, ABC) ⇒ ⇒ (abbbbbccc, q, aABC) ⇒ (bbbbbccc, q, ABC) ⇒ (bbbbbccc, q, BC) ⇒ (bbbbbccc, q, bbBC) ⇒ (bbbbccc, q, bBC) ⇒ (bbbccc, q, BC) ⇒ (bbbccc, q, bbBC) ⇒ (bbccc, q, bBC) ⇒ (bccc, q, BC) ⇒ (bccc, q, bC) ⇒ (ccc, q, C) ⇒ (ccc, q, cC) ⇒ (cc, q, C) ⇒ (cc, q, cC) ⇒ (c, q, C) ⇒ (c, q, c) ⇒ ($\epsilon$, q, $\epsilon$)

Is the grammar ambiguuous? no

The accepted language is $ L(G) = \{a^{m}b^{2n + 1}c^{p+1} | m,n,p \ge 0\} $