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For languages AandB, let the shuffle of AandBbe the language

{|=a1b1...akbk,where鈥夆赌a1...akA鈥夆赌and鈥夆赌b1...bkB,each鈥夆赌ai,bi}.

Show that the class of regular languages is closed under shuffle.

Short Answer

Expert verified

The class of regular languages is closed under shuffle.

Step by step solution

01

Introduction 

The given two languages AandB is shuffle on theAandB is as follows:

{|=a1b1...akbk,where鈥夆赌a1...akA鈥夆赌and鈥夆赌b1...bkB,each鈥夆赌ai,bi}..

Assume, DFAA=(QA,,A,SA,FA) and DFAB=(QB,,B,SB,FB)be the twoDFA s that recognizethe, Aand Brespectively. DFAPerfect-shuffle=(Q,,,S,F), and also recognizes the language is perfect shuffle on,A and B.

02

Explanation

TheDFA for perfect shuffle switches fromDFAA toDFAB after each character is read and it tracks the current states of DFAAand DFAB.

Each character should belong to DFAAor DFABi.e., ai,bi. For each character read, DFAPerfect-shufflemakes moves in the correspondingDFA (either DFAAor DFAB).

After the whole string is read, if bothDFAA andDFAB reaches to the final state, then the input string is accepted byDFAPerfect-shuffle.

q=q0

03

Simplification

The DFAPerfect-shuffleis defined as follows:

Q=QAQB{A,B}: set of all possible states of DFAAand DFABwhich should match with DFAPerfect-shuffle.

The input alphabet for DFAPerfect-shuffleis .

q=(qA,qB,A):qAandqBare the initial states forDFAA and DFABrespectively.DFAPerfect-shuffle starts withqAin DFAA,qBin DFABand the next character should be read from DFAA.

F=FAFB{A}: FAand FBare the final states forDFAA andDFAB respectively. DFAPerfect-shuffleAccepts if both DFAAand DFABreaches to the final states and the next character should be read from DFAA.

04

Transition function

The transition function is,

  1. 鈥夆赌((m,n,A),a)=(A(m,a),n,B)
  2. ((m,n,B),b)=(m,B(n,b),n,A)

Consider, the current state of DFAAism and the current state ofDFAB is n. Change the current state of AtoA(m,a)if the next character is to be read fromDFAA whenais the next character. After the character is read, read the next character from DFAB.

Change the current state ofB toB(n,b) if the next character is to be read fromDFAB when bis the next character.

The languageL is said to be regular if there exist anFA that recognizes the language L. Here, the DFAPerfect-shuffleis defined for the language perfect shuffle.

Therefore, the class of regular languages is closed under shuffle.

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Most popular questions from this chapter

a. Let Abe an infinite regular language. Prove thatA can be split into two infinite disjoint regular subsets.

b. LetBandD be two languages. Write BDifBDand Dcontains infinitely many strings that are not in B. Show that if BandD are two regular languages whereBD , then we can find a regular languageC where BCD.

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 p'pThe minimum pumping length for A is the smallest p that is a pumping length for A . For example, if A=01*, the minimum pumping length is 2.The reason is that the string s=0is 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 x=0,y=1,andzis the rest. For each of the following languages, give the minimum pumping length and justify your answer.

a).0001*b).0*1*c).0010*1*d).0*1+0+1*10*1

role="math" localid="1660797009042" e).(01)*f).g).1*01*01*h).10(11*0)*

i).1011j).*

Let ={0,1,#}. Let C={x#xR#x|x{0,1}*} . Show that C is a CFL.

Let M=(Q,,,q0,F)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(q,s)=hforeveryqQ. (Here we have extended to strings, so that(q,s) 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 k-state synchronizable DFA, then it has a synchronizing sequence of length at mostk3 . Can you improve upon this bound?

Question: Let ={0,1}and let

D={w|wcontainsanequalnumberofoccurrencesofthesubstrings01and10}.

Thus101D because 101 contains a single 01 and a single 10, but 1010Dbecause 1010 contains two 10 s and one .01 Show that D is a regular language.

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