9. Context-Free Grammars

$ G = \{\Sigma, V, R, S \} $

S - Starting symbol

V - Set of terminals and non-terminals

R - production rules ($ R \subseteq (V \setminus \Sigma) \times V^n $)

Property: Chomsky Normal Form (CNF) - A grammar in CNF is context-free.

  1. all transitions are of the form:
    • $ X \leftarrow YZ $
    • $ X \leftarrow a $
    • $ S \leftarrow \epsilon $

S is the starting symbol.

9.1. Generating a CF language

Write CF grammar for the following languages. 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}\} $.

Solution 9.1.1

Solution 9.1.1

$ S \leftarrow aSa\ |\ bSb \ |\ a \ |\ b \ |\ \epsilon $

9.1.2. $ L = \{ a^{m} b^{m+n} c^{n} \ | \: n, m \geq 0 \} $

Solution 9.1.2

Solution 9.1.2

$ S \leftarrow AB $
$ A \leftarrow aAb\ |\ \ \epsilon $
$ B \leftarrow bBc\ |\ \ \epsilon $

9.1.3. $ L = \{w \in \{a, b\}^* | \#_a(w) = \#_b(w) \} $

Solution 9.1.3

Solution 9.1.3

$ 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) \} $

9.1.5. $ L = \{a^ib^jc^k | i = j \lor j = k \} $

Solution 9.1.5

Solution 9.1.5

$ S \leftarrow C\ |\ A $
$ C \leftarrow Cc\ |\ B $
$ B \leftarrow aBb\ |\ \ \epsilon $
$ A \leftarrow aA\ |\ D $
$ D \leftarrow bDc\ |\ \ \epsilon $

9.2. Ambiguity

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 $

Solution 9.2.1

Solution 9.2.1

$ S \leftarrow A $
$ A \leftarrow aA\ |\ B $
$ B \leftarrow bB\ |\ \epsilon $

9.2.2.

$ S \leftarrow AS\ |\ \epsilon $
$ A \leftarrow 0A1\ |\ 01\ |\ B $
$ B \leftarrow B1\ |\ \epsilon $

Solution 9.2.2

Solution 9.2.2

$ S \leftarrow AS\ |\ A $
$ A \leftarrow 0A1\ |\ B $
$ B \leftarrow B1\ |\ \epsilon $

9.2.3.

$ S \leftarrow ASB\ |\ BSA\ |\ \epsilon $
$ A \leftarrow aA\ |\ \epsilon $
$ B \leftarrow bB\ |\ \epsilon $

Solution 9.2.3

Solution 9.2.3

$ S \leftarrow aS\ |\ bS\ |\ \epsilon $

9.2.4. Write an ambiguous grammar for $ L(a^*) $.

Solution 9.2.4

Solution 9.2.4

$ S \leftarrow aS \ |\ A\ |\ \epsilon $
$ A \leftarrow aA\ |\ \epsilon $

9.3 Regular Grammars

Definition: A regular grammar is a CF grammar where the production rules follow one of these 2 patterns:

  1. all of them are of the form:
    • $ X \leftarrow aY $
    • $ X \leftarrow Y $
    • $ X \leftarrow a $
    • $ X \leftarrow \epsilon $
  2. OR all of them are of the form:
    • $ X \leftarrow Ya $
    • $ X \leftarrow Y $
    • $ X \leftarrow a $
    • $ X \leftarrow \epsilon $

9.3.1. (warmup) Find a DFA or an NFA that accepts the language generated by the following regular grammar:

$ X \leftarrow 0X\ |\ 1Y $

$ Y \leftarrow 1Y\ |\ 0\ |\ \epsilon $

Solution 9.3.1

Solution 9.3.1

Y produces $ 1^* (0 | \epsilon) $

X produces $ 0^* 1Y \rightarrow 0^* 1^+ (0 | \epsilon) $

9.4.1. Prove that the languages accepted by regular grammars are exactly the regular languages

  • prove that any regular grammar generates a regular language: start from a regular grammar G and construct an NFA $ N $ such that $ L(N) = L(G) $
  • prove that any regular language is generated by a regular grammar: start from a DFA $ A $ and construct a regular grammar such that $ L(G) = L(A) $
  • write the proofs for both definitions (the one with $ X \leftarrow aY $ and the one with $ X \leftarrow Ya $).
  • Hint: for the first definition, think top-down; for the second one, think bottom-up;
  • Hint 2: use induction in the proofs