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9.1. Accepting and generating a CF language
Write a PDA as well as a CF grammar for the following languages. Specify - for each PDA, the way it accepts (by empty-stack or by final state). For each grammar, make sure it is not ambiguous. (Start with any CF grammar that accepts L. Then write another non-ambiguous grammar for the same language).
9.1.1. $ L = \{\: w \in \{A,B\}^* \ | \:w \text{ is a palindrome}\} $.
$ S \leftarrow ASA | BSB | A | B | \epsilon $
9.1.2. $ L = \{ A^{m} B^{m+n} C^{n} \ | \: n, m \geq 0 \} $
$ S \leftarrow XY $
$ X \leftarrow AXB | \epsilon $
$ Y \leftarrow BYC | \epsilon $
9.1.3. $ L = \{w \in \{a, b\}^* | \#_a(w) = \#_b(w) \} $
Try #1:
$ S \leftarrow aBS | baS | \epsilon $
ambiguuous because of: “ababab”
Try #2:
$ S \leftarrow $ aSb | bSa | SS | $ \epsilon $
ambiguuous because of: “abab”
Idea: A = rule that promises that one letter 'a' should come without a pair; B = rule that promises that one letter 'b' should come without a pair
$ S \leftarrow aBS | bAS | \epsilon $
$ A \leftarrow a | bAA $
$ B \leftarrow b | aBB $
9.1.4. $ L = \{w \in \{a, b\}^* | \#_a(w) \neq \#_b(w) \} $
$ S \leftarrow A | B $
$ A \leftarrow G |GG $
$ B \leftarrow H | HH $
$ G \leftarrow Ea $
$ H \leftarrow Eb $
$ E \leftarrow aME | bNE | \epsilon $
$ M \leftarrow b | aMM $
$ N \leftarrow a | bNN $
9.1.5. $ L = \{a^ib^jc^k | i = j \lor j = k \} $
Inherently ambiguous grammar because of the case i = j = k
$ S \leftarrow X | Y $
$ X \leftarrow ZC $
$ Z \leftarrow aZb | \epsilon $
$ C \leftarrow cC | \epsilon $
$ Y \leftarrow AT $
$ A \leftarrow aA | \epsilon $
$ T \leftarrow bTc | \epsilon $
Which of the following grammars are ambiguous? Justify each answer. Modify the grammar to remove ambiguity, wherever the case.
9.2.1.
$ S \leftarrow aA | A $
$ A \leftarrow aA | B $
$ B \leftarrow bB | \epsilon $
Ambiguuous: yes
Repaired, unambiguous grammar:
$ S \leftarrow P_1 P_2 | B $ (this matches strings like (aa*b*)*a*b*)
$ P_1 \leftarrow aAbBP_1 | \epsilon $ (this matches strings like (a+b+)*)
$ P_2 \leftarrow aAB | \epsilon $ (this matches strings like (a+b*) | $\epsilon$)
$ A \leftarrow aA | \epsilon $
$ B \leftarrow bB | \epsilon $
9.2.2.
$ S \leftarrow AS | \epsilon $
$ A \leftarrow 0A1 | 01 | B $
$ B \leftarrow B1 | \epsilon $
Ambiguuous: yes
Repaired:
$ S \leftarrow AS | \epsilon $
$ A \leftarrow A’B $
$ A’ \leftarrow 0A’1 | \epsilon $
$ B \leftarrow B1 | \epsilon $
9.2.3.
$ S \leftarrow ASB | BSA | \epsilon $
$ A \leftarrow aA | \epsilon $
$ B \leftarrow bB | \epsilon $
Ambiguuous: yes
The grammar actually generates the language {a,b}*:
$ S \Rightarrow ASB \Rightarrow aASB \Rightarrow aSB \Rightarrow aS $
$ S \Rightarrow BSA \Rightarrow bBSA \Rightarrow bSA \Rightarrow bS $
Repaired:
$ S \leftarrow aS | bS | \epsilon $
9.2.4. Write an ambiguous grammar for $ L(a^*) $.
$ S \leftarrow aS | aaS | \epsilon $
Ambiguuous: yes
