Question

1. L is the set of strings over {a, b) that begin with a and do not contain the substring bb. a. Show L is regular by giving a regular expression that denotes the language. b. Show L is regular by giving a DETERMINISTIC finite automaton that recognizes the language.

Answer 1

A regular language over alphabet A is a language constructed by the following rules . Ø and A} are regular languages · lat is a regular language for all a E A. If M and V are regular language, then so are MUN, MN, and M* Example, Let A = ta, bt. Then the following languages are a sampling of the regular languages over A: A regular expression over alphabet A is an expression constructed by the following rules . Ø and Л are regular expressions . a is a regular expression for all a E A. If R and S are regular expressions, then so are (R), R+ S, R-S, andR* The hierarchy in the absence of parenthesæs is, * (doit first), , Juxtaposition will be used in place of (do it last) Example. Let A- fa, bt. Then the following expressions are a sampling of the regular expressions ver A 0, A, a, b, ab, a+ ab, (a+ b)*.

Regular expressions represent regular languages Regular expressons represent regular languages by the following correspondence, where LIR) denotes the regular language of the regular expression R. L(a)- fa} for all a E A ぶample. The regular expression αο + α represents the rollowing regular language -lab, Л, a, da, daa, } represents the following regular language , an, Example. The regular expression a+b) し((a + b) *) = (L(a+ b) *-{a, 하*, the set of all possible strings over {α하 Back to the Problem: Suppose the input strings to a program must be strings over the alphabet fa, b} that contain exactly one substring bb. In other words, the strings must be of the form xbby, where x and y are strings over fa, b} that do not contain bb, x does not end in b, and y does not begin with b. How can we descrnibe the set of inputs formally? Solution let x = (a + ba) * and y= (a + ab) *

(a) The language { w E Σ* Maul is odd } Answer: (aU b)((a Ub)(aUb)) (b) The language fw w has an odd number of a's Answer: b*a (ab aU b)* c) The language {w contains at least two a's, or exactly two b's Answer: b ab*a(aU b)* U a ba ba* (d) The language w E w ends in a double letter . (A string contains a double letter if it contains aa or bb as a substring.) Answer: (aU b)*(aa Ubb) (e) The language w E* w does not end in a double letter H Answer: E U a U b U (aUb)*(ab U ba) (f) The language {w E Σ* | w contains exactly one double letter). For example, baaba has exactly one double letter, but baaaba has two double letters. Answer: (EU b)(ab)*aa(ba)*(e U b) U (EU a) (ba) bb(ab)*(E U a)