Chapter 1: Q42P (page 89)
For languages , let the shuffle of be the language
Show that the class of regular languages is closed under shuffle.
Short Answer
The class of regular languages is closed under shuffle.
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Chapter 1: Q42P (page 89)
For languages , let the shuffle of be the language
Show that the class of regular languages is closed under shuffle.
The class of regular languages is closed under shuffle.
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a. Let be an infinite regular language. Prove that can be split into two infinite disjoint regular subsets.
b. Let be two languages. Write and contains infinitely many strings that are not in . Show that if and are two regular languages where , then we can find a regular language where .
The pumping lemma says that every regular language has a pumping length P , such that every string in the language can be pumped if it has length p or more. If P is a pumping length for language A, so is any length The minimum pumping length for A is the smallest p that is a pumping length for A . For example, if , the minimum pumping length is 2.The reason is that the string is in A and has length 1 yet s cannot be pumped; but any string A in of length 2 or more contains a 1 and hence can be pumped by dividing it so that is the rest. For each of the following languages, give the minimum pumping length and justify your answer.
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Let . Let . Show that is a CFL.
Let be a DFA and let be a state of Mcalled its 鈥渉ome鈥. A synchronizing sequence for M and h is a string s鈭埼b垪where (Here we have extended to strings, so that equals the state where M ends up when M starts at state q and reads input s .) Say that M is synchronizable if it has a synchronizing sequence for some state h . Prove that if M is a state synchronizable DFA, then it has a synchronizing sequence of length at most . Can you improve upon this bound?
Question: Let and let
Thus because 101 contains a single 01 and a single 10, but because 1010 contains two 10 s and one .01 Show that D is a regular language.
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