FAV-ZCU/KIV TI/Cvičení/Cviceni05.md

Př. 1: Vytvořte gramatiku, která bude nad abecedou \{0, 1\} generovat řetězec obsahující lichý počet 0 a sudý počet 1.

• \{0, 1\} \quad w \dots \text{lichý počet } 0 \text{, sudý počet } 1
• 4 stavy

Př. 2:

• S \to abA \mid bS \mid aa \mid A
• A \to abA
• B \to aS \mid baA \mid a

a) Najděte G' typu G3R takovou, že L(G') = L(G).

• S \to bS | aS_{1} | aS_{2} \mid aS | bB_{1} | aB_{2} \mid aA_{1}
• A \to aA_{1} \mid aS | bB_{1} | aB_{2}
• B \to aS | bB_{1} | aB_{2}

• S_{1} \to bA
• B_{1} \to aA
• A_{1} \to bA
• S_{2} \to aS_{3}
• S_{3} \to e
• B_{2} \to e

b) Vytvořte tabulku popisující nedeterministický konečný automat A takový, že platí L(A) = L(G') = L(G).

	a	b
\to S	\{S, S_{1}, S_{2}, A_{1}, B_{2}\}	\{S, B_{1}\}
S_{1}	-	\{A\}
S_{2}	\{S_{3}\}	-
\leftarrow S_{3}	-	-
A	\{S, A_{1}, B_{2}\}	\{B_{1}\}
A_{1}	-	\{A\}
B	\{S, B_{2}\}	\{B_{1}\}
B_{1}	\{A\}	-
\leftarrow B_{2}	-	-

c) Vytvořte tabulku popisující deterministický konečný automat A' takový, že platí L(A') = L(G') = L(G).

	a	b
\to S (0)	\{S, S_{1}, S_{2}, A_{1}, B_{2}\} (1)	\{S, B_{1}\} (2)
\leftarrow \{S, S_{1}, S_{2}, A_{1}, B_{2}\} (1)	\{S, S_{1}, S_{2}, S_{3}, A_{1}, B_{2}\} (3)	\{S, A, B_{1}\} (4)
\{S, B_{1}\} (2)	\{S, S_{1}, S_{2}, A, A_{1}, B_{2}\} (5)	\{S, B_{1}\} (2)
\leftarrow\{S, S_{1}, S_{2}, S_{3}, A_{1}, B_{2}\} (3)	\{S, S_{1}, S_{2}, S_{3}, A_{1}, B_{2}\} (3)	\{S, A, B_{1}\} (4)
\{S, A, B_{1}\} (4)	\{S, S_{1}, S_{2}, A, A_{1}, B_{2}\} (5)	\{S, B_{1}\} (2)
\leftarrow\{S, S_{1}, S_{2}, A, A_{1}, B_{2}\} (5)	\{S, S_{1}, S_{2}, S_{3}, A_{1}, B_{2}\} (3)	\{S, A, B_{1}\} (4)

Př. 3: Sestrojte NKA A, kde platí L(A) = L(G_{1})^R \cup L(G_{2}).

• G_{1}
  • S \to aS | bbA
  • A \to aaA | B
  • B \to bbB | e
• G_{2}
  • S \to Aba | Ab | B
  • A \to Aaa | B
  • B \to Bbb | e
• G3L -> reverze -> G3P -> NKA -> reverze -> NKA

Plán

1. A_{1} \qquad L(A_{1}) = L(G_{1})
2. A_{1}^R \qquad L(A_{1}^R) = L(A_{1})^R = L(G_{1})^R
3. G_{2}^R \qquad L(G_{2}^R) = L(G_{2})^R
4. A_{2}^R \qquad L(A_{2}^R) = L(G_{2}^R) = L(G_{2})^R
5. A_{2} \qquad A_{2} = (A_{2}^R)^R \quad L(A_{2}) = \dots = L(G_{2})
6. A \qquad L(A) = L(A_{1}^R) \cup L(A_{2}) = L(G_{1})^R \cup L(G_{2})

G_{2}^R

• S \to abA | bA | B
• A \to aaA | B
• B \to bbB | e