hmwk2 - Problem 2 a. L = { w | each a in w is immediately...

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Johny Nguyen October 5 th , 2010 CS 473 Homework 2 Downey
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Problem 1 1. baa L(a*b*a*b*)? Yes. Say the first a* yields the empty string. The first b* yields one b, then we will have ‘b’. Let the second a* yield ‘aa’, and the last b* yield the empty string. We will have ‘ baa’. 2. b*a* a*b* = a* b*? Yes, because both a* and b* is contained in both b*a* and a*b*. Any concatenations that have both a and b will not be included in the intersection. So only a* and b* will be the only common items in the intersection. 3. a*b* c*d* = ? No, because each set has the empty string. So the intersection will have the empty string in it as well. 4. abcd L( ( a( cd ) * b) * )? No, because no matter what, the letter ‘b’ will never appear before ‘c’ or ‘d’ based on the ordering of the concatenation.
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Unformatted text preview: Problem 2 a. L = { w | each a in w is immediately preceded and immediately followed by a b } b. L = { w | w has abab as a substring} c. L = { w | w has neither aa nor bb as a substring} d. L = { w | w has an odd number of a’s an even number of b’s} e. L = { w | w has both ab and ba as substrings} Problem 3 a. L = { xy | x and y are binary strings that each contain the same substring of length 3} b. L = {w | w is a binary string containing both the substring 010 and the substring 101 Problem 4 a. Original NFA Final DFA b. Original NFA Final DFA Problem 5 a. Original NFA Regular Expression: a*(ba*)* b. Original NFA Regular Expression: [(a ∪ b) a*(ba*)*] ∪ [(a ∪ b) a*(ba*)*a] Problem 6 a. a(abb)* ∪ b ε b. a + ∪ (ab) + c. (a ∪ b + )a + b +...
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This note was uploaded on 11/29/2010 for the course CSC 873569 taught by Professor Roberts during the Spring '10 term at ASU.

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hmwk2 - Problem 2 a. L = { w | each a in w is immediately...

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