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1. Use the construction in Theorem 3.1 to find an nfa that accepts the language L (a*a + ab).

1. Use the construction in Theorem 3.1 to find an nfa that accepts the language L (a*a + ab).

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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1. Use the construction in Theorem 3.1 to find an nfa that accepts the language L (a*a + ab).
Transcribed Image Text:1. Use the construction in Theorem 3.1 to find an nfa that accepts the language L (a*a + ab).
28 Let r be a regular expression. Then there exists some nondeterministic finite accepter that accepts L (r). Consequently, L (r) is a regular language. Proof: We begin with automata that accept the languages for the simple regular expressions Ø, A, and a EE. These are shown in Figure 3.1(a), (b), and (c), respectively. Assume now that we have automata M (r₁) and M (r2) that accept languages denoted by regular expressions r₁ and 72, respectively. We need not explicitly construct these automata, but may represent them schematically, as in Figure 3.2. In this scheme, the graph 90 91 tv M(r) 90 91 (a) (b) (c) FIGURE 3.1 (a) nfa accepts Ø. (b) nfa accepts {A}. (c) nfa accepts {a}. 90 ℗ 91 FIGURE 3.2 Schematic representation of an nfa accepting L (r). 3.2 Regular Expressions and Regular Languages vertex at the left represents the initial state, the one on the right the fi- nal state. In Exercise 7, Section 2.3, we claim that for every nfa there is an equivalent one with a single final state, so we lose nothing in assuming that there is only one final state. With M (ri) and M (r2) represented in this way, we then construct automata for the regular expressions r₁ + r2, T1T2, and ri. The constructions are shown in Figures 3.3 to 3.5. As indicated in the drawings, the initial and final states of the constituent machines lose their status and are replaced by new initial and final states. By stringing together coveral such stone we can build automate for arbitrary comploy 81 MacBook Air 11 Compu nework # 03/29/20 ethod: Ca a single sult in 0 ems 1, 2 blems 1,
Transcribed Image Text:28 Let r be a regular expression. Then there exists some nondeterministic finite accepter that accepts L (r). Consequently, L (r) is a regular language. Proof: We begin with automata that accept the languages for the simple regular expressions Ø, A, and a EE. These are shown in Figure 3.1(a), (b), and (c), respectively. Assume now that we have automata M (r₁) and M (r2) that accept languages denoted by regular expressions r₁ and 72, respectively. We need not explicitly construct these automata, but may represent them schematically, as in Figure 3.2. In this scheme, the graph 90 91 tv M(r) 90 91 (a) (b) (c) FIGURE 3.1 (a) nfa accepts Ø. (b) nfa accepts {A}. (c) nfa accepts {a}. 90 ℗ 91 FIGURE 3.2 Schematic representation of an nfa accepting L (r). 3.2 Regular Expressions and Regular Languages vertex at the left represents the initial state, the one on the right the fi- nal state. In Exercise 7, Section 2.3, we claim that for every nfa there is an equivalent one with a single final state, so we lose nothing in assuming that there is only one final state. With M (ri) and M (r2) represented in this way, we then construct automata for the regular expressions r₁ + r2, T1T2, and ri. The constructions are shown in Figures 3.3 to 3.5. As indicated in the drawings, the initial and final states of the constituent machines lose their status and are replaced by new initial and final states. By stringing together coveral such stone we can build automate for arbitrary comploy 81 MacBook Air 11 Compu nework # 03/29/20 ethod: Ca a single sult in 0 ems 1, 2 blems 1,
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