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THEOREM 1.39 Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let Every nondeterministic finite automaton has an equivalent deterministic finite automaton. PROOF Let N = (Q, E, 6, qo, F) be the NFA recognizing some language A. Ne construct a DFA M = (Q', £, 8', qo', F") recognizing A. Before doing the full onstruction, let's first consider the easier case wherein N has no e arrows. Later E(R) = {q\ q can be reached from R by traveling along 0 or more e arrows}. Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing 6(r, a) by E(8(r, a)) achieves this effect. Thus ve take the e arrows into account. 1. Q' = P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a € E, let 8'(R, a) = {q € Q] q € 6(r, a) for some r € R}. If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is 8'(R,a) = {q € Qlq € E(6(r, a)) for some r € R}. Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing 4o' to be E({q}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. 8 (R,a) = U 6(r,a).* rER

THEOREM 1.39 Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let Every nondeterministic finite automaton has an equivalent deterministic finite automaton. PROOF Let N = (Q, E, 6, qo, F) be the NFA recognizing some language A. Ne construct a DFA M = (Q', £, 8', qo', F") recognizing A. Before doing the full onstruction, let's first consider the easier case wherein N has no e arrows. Later E(R) = {q\ q can be reached from R by traveling along 0 or more e arrows}. Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing 6(r, a) by E(8(r, a)) achieves this effect. Thus ve take the e arrows into account. 1. Q' = P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a € E, let 8'(R, a) = {q € Q] q € 6(r, a) for some r € R}. If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is 8'(R,a) = {q € Qlq € E(6(r, a)) for some r € R}. Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing 4o' to be E({q}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. 8 (R,a) = U 6(r,a).* rER

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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THEOREM 1.39 Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let Every nondeterministic finite automaton has an equivalent deterministie finite automaton. PROOF Let N = (Q.E. 8. qo, F) be the NFA recognizing some language A. We construct a DFA M = (Q', £,8', qo', F') recognizing A. Before doing the full construction, let's first consider the easier case wherein N has no e arrows. Later we take the e arrows into account. E(R) = {ql q can be reached from R by traveling along 0 or more e arrows}. Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing 8(r, a) by E(8(r, a)) achieves this effect. Thus 1. Q' = P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a e £, let 8'(R, a) = {q € QI q € d(r, a) for some r e R}. If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is 8'(R, a) = {q € Q]q € E(5(r, a)) for some r E R}. Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing qo' to be E({go}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. 8 (R, a) = U 6(r,a).* rER 3. qo' = {q0}. M starts in the state corresponding to the collection containing just the Theorem 1.39 states that every NFA can be converted into an equivalent DFA. Thus nondeterministic finite automata give an alternative way of characterizing the regular languages. We state this fact as a corollary of Theorem 1.39. start state of N. 4. F' = {R€ Q'| R contains an accept state of N}. The machine M accepts if one of the possible states that N could be in at this point is an accept state.
Transcribed Image Text:THEOREM 1.39 Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let Every nondeterministic finite automaton has an equivalent deterministie finite automaton. PROOF Let N = (Q.E. 8. qo, F) be the NFA recognizing some language A. We construct a DFA M = (Q', £,8', qo', F') recognizing A. Before doing the full construction, let's first consider the easier case wherein N has no e arrows. Later we take the e arrows into account. E(R) = {ql q can be reached from R by traveling along 0 or more e arrows}. Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing 8(r, a) by E(8(r, a)) achieves this effect. Thus 1. Q' = P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a e £, let 8'(R, a) = {q € QI q € d(r, a) for some r e R}. If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is 8'(R, a) = {q € Q]q € E(5(r, a)) for some r E R}. Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing qo' to be E({go}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. 8 (R, a) = U 6(r,a).* rER 3. qo' = {q0}. M starts in the state corresponding to the collection containing just the Theorem 1.39 states that every NFA can be converted into an equivalent DFA. Thus nondeterministic finite automata give an alternative way of characterizing the regular languages. We state this fact as a corollary of Theorem 1.39. start state of N. 4. F' = {R€ Q'| R contains an accept state of N}. The machine M accepts if one of the possible states that N could be in at this point is an accept state.
Use the construction given in Theorem 1.39 to convert the following two nonde- terministic finite automata to equivalent deterministic finite automata. 2. b a,b 2 3
Transcribed Image Text:Use the construction given in Theorem 1.39 to convert the following two nonde- terministic finite automata to equivalent deterministic finite automata. 2. b a,b 2 3
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