Assignment_4_Answer_Key

Assignment_4_Answer_Key - The alphabet is the same as the...

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Assignment 4 Answer Key 1.7.b 1.7.c 1.9.b 1.10a 1.16a
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1.19a 1.18a 1 (0 U 1)* 0 1.18b 0* 1 0* 1 0* 1 0* (10*)* 1.18c (0 U 1)* 0101 (0 U 1)* 1.18d (0 U 1) (0 U 1) 0 (0 U 1)* 1.18e ( (0 U (10 U 11)) ((0 U 1)(0 U 1))* ) 1.31 Since A is regular, we can assume there is a DFA that recognizes A. To prove that A R is regular it is sufficient to prove that there exists a DFA which recognizes A R . The following is a constructions of such a DFA. From the DFA for A we have (Q, , , q, F) To construct a DFA for A R , we define the following: (Q U {q0}, , ', q0, {q}) The states include all the original states from the DFA for A plus a new state q0.
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Unformatted text preview: The alphabet is the same as the original DFA for A. The starting states is the new state q0. The accepting state for the DFA for A R is the original starting state for the DFA for A, q. ' has all the arrows from the original transition arrows, except they are all reversed. In addition, there are transitions from q0 to all of the original accepting states of the DFA for A, F. This ensures that there is only one starting state for the DFA for A R ....
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This note was uploaded on 02/07/2011 for the course CS 501 taught by Professor Sm during the Spring '11 term at Indiana.

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Assignment_4_Answer_Key - The alphabet is the same as the...

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