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Properties of Regular Languages

Properties of Regular Languages. Closure property : New recognizers for languages that are constructed from other languages by certain operations can be built. Decision Property: This property gives algorithms for answering important questions about automata. Proving a language to be Regular.

linda-welch
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Properties of Regular Languages

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  1. Properties of Regular Languages • Closure property : New recognizers for languages that are constructed from other languages by certain operations can be built. • Decision Property: This property gives algorithms for answering important questions about automata.

  2. Proving a language to be Regular • A language L is regular if there exists a DFA which accepts the strings of L • Example: Is L={01,0011,000111,…}regular?

  3. Try constructing a DFA for given L={0n1n|n>=1} 0 q0 1 q1 • Think of all strings 00001, 0001 etc which are not in L. These are also accepted by the DFA • Try any other DFA

  4. L={0n1n|n>=1} is not regular • The language contains strings containing equal number of zeroes followed by equal number of ones. • If L is regular, then L should be the language of some DFA • Before learning the proof, we understand an important principle.

  5. Pigeonhole Principle • If there are more pigeons than the pigeonholes, then there must be at least one hole that has more than one pigeon.

  6. Pigeonhole Principle • The Pigeonhole Principle depends on the number of pigeonholes being finite. • Therefore this principle works for finite state automata with the states as pigeonholes and the symbols in a string w as pigeons • This principle does not work for other kinds of automata that have an infinite number of states

  7. Pigeonhole Principle • If we put n objects into m boxes, and if n>m, then at least one box must have more than one item in it

  8. DFA and Pigeonhole Principle q2 q3 q4 q0 0 q1 1 1 0 q4 • Total states: 5 • Minimum size of acceptable input string: 1(direct) • Minimum size of acceptable input string that passes through all the states : 4 • What if a string is of size 5? Say 01110

  9. DFA and Pigeonhole Principle 1 q2 q3 q4 q0 0 q1 1 1 0 q4 • What if a string is of size 5? Say 01110 • Certainly there is an arc that repeats at least one state. • This also means that the DFA also accepts even bigger strings e.g. 01111110, 011110

  10. Proving another language to be regular using Pigeonhole Principle • This gives us an idea of extending our knowledge about the regular language that is accepted by the DFA • It states that now any language which is having a regular expression 0111*0 is also regular as there is a DFA that accepts all the strings in this language

  11. Example • We had considered 5 states and a string of 5 inputs say 01110, and observed that there is certainly a loop at 4th state. • In a transition graph with n vertices (states), any walk of length n or longer must repeat some vertex, that is contain a cycle.

  12. Pumping Lemma • Let L be a regular language . Then there exists a constant n (which depends on L) such that every string w in L such that |w|>=n, we can break w into three strings, w=xyz, such that • y≠є • |xy|<=n • For all k>=0, the string xykz is also in L

  13. Use Pumping Lemma to understand previous example • Number of states :5 • String 01110 can be written as xyz where x=011 y=1 z=0 • We can now easily see that the string 01111110 is of the form 011140 for k=4 and is accepted by the DFA, hence is in L

  14. Pumping Lemma • This way we have seen that if any string in a language is of larger size than the total number of states, then we can always find a nonempty string y that can be ‘pumped’, that is repeating y any number of times, or deleting it keeps the resulting string in the language L

  15. Proof of Pumping Lemma • Let L be regular. Then L=L(A) for some DFA A. • Let A has n states • Consider any string of length n or more, say m w=a1a2….. anan+1an+2 ….. Am where m>=n

  16. Proof of Pumping Lemma • For i=0,1,2,……n define state pi to be d(q0,a1a2….. ai) where d is the transition function of A. then p0 =q0 • By the pigeonhole principle, it is not possible for the n+1 different pi for i=0,1,2,….n to be distinct since there are only n states

  17. Proof of Pumping Lemma • This indicates a repetition of some state and we can find two integers i and j, with 0<=i<j<=n such that pi=pj. • We can break w=xyz as follows • x=a1a2….. ai • y= ai+1ai+2….. aj • z= aj+1aj+2….. am

  18. Proof of Pumping Lemma y=ai+1ai+2….. aj x=a1a2….. ai pi p0 z=aj+1aj+2….. am

  19. L={0n1n|n>=1} is not regular: Proving by the Pumping Lemma • Let L be regular • So Pumping lemma must hold (as L is infinite and a DFA has finite number of states) • Let the number of states in the DFA be ‘m’ • let us consider a string 0m1m

  20. 0m is accepted by the DFA • Consider a string0m 1m, which means there is a y such that 0m 1mcan be written as xyz (Pumping Lemma) 0 q0 q1 0 q2 0 0 0 0 qm qm • Let x= 00000..000(m times) • y=1 • z=111 …..1111(m-1 times) • But here |xy|>m (states) Contradiction 1 Hence L is not regular

  21. Closure Properties of Regular Languages • Closure properties express the idea that any language L formed from any regular language by certain operations, then L is also regular.

  22. Closure Properties • Union (of two regular language is regular) • Intersection • Complement • Difference • Reversal • Star closure • Concatenation • Homomorphism • Inverse homomorphism

  23. Closure under Union • Let L and M are regular • Then these languages have regular expressions R and S i.e. L=L(R) and M=L(S) • LUM = L(R) U L(S) = L(R+S) • Since R and S are the regular expressions then R+S is also regular • This implies LUM is also regular (Union is Closed)

  24. Closure under complementation • If L is regular then to prove that its complement Ľ (L bar), which is defined as S*- L, is also regular. • If there exists any DFA that accepts Ľ, then we say that the complement of L is also regular. • For this we learn to construct such DFA

  25. Constructing a DFA to accept L complement • Since L is regular, there exists a DFA that accepts it. • Let L=L(R ). • Convert the regular expression R to an є-NFA. • Convert the є-NFA to a DFA by the subset construction.

  26. Constructing a DFA to accept L complement • Complement the accepting states of that DFA i.e. now the accepting states of this new DFA are the states other than the accepting states of previous one (i.e. Q-F) • Turn the complement DFA into a regular expression

  27. Proving that the complement of a regular language is also regular • Let L=L(A) for some DFA A • Where A= (Q, S,d, q0,F) • Then Ľ=L(B) where B=(Q, S,d, q0,Q-F) • Then A and B are same other than the fact that their accepting states are different • Then any string w is in L(B) if and only if d*(q0,w) is in Q-F which occurs only if w is not in L(A). Hence Ľ is regular.

  28. Example :Class assignment • Prove that L(A) = (0+1)*01 is regular. • Prove that the language consisting of all strings that do not end in 01 is regular. • If this language is regular, draw a DFA accepting the language.

  29. DFA accepting language that does not end with 01 1 0 {q0 ,q2 } {q0 ,q1 } {q0} 0 {q0 ,q1 } 1 0 1

  30. Closure under Intersection • Intersection of two regular languages L and M is written as L ∩ M • Demorgan’s Law __ __ ________ _______ _ ______ L ∩ M = L U M M L L ∩ M

  31. Closure under Intersection • If L is regular, then complement of L is also regular. • If M is regular, then complement of M is also regular. • Also union of two regular languages is proved to be regular • Hence intersection of two regular languages is also regular.

  32. Special DFA construct for intersection of two regular languages • Let L and M be the languages of automata AL=(QL, S,dL, qL,FL) and AM=(QM, S,dM, qM,FM) • Construct an automaton A = (QL XQM, S,d, qLXqM,FLXFM) Where the states are the pairs of type (p,q) such that p is in QL and q is in QM

  33. Special DFA construct for intersection of two regular languages • The alphabet S is assumed to be the same. • The transition function d = dLXdM such that d((p,q),a)= an intermediate state of A which is a pair of states obtained by transition of state p (in AL) and of state q (in AM) • d((p,q),a)=(dL(p,a) ,dM (q,a))

  34. Special DFA construct for intersection of two regular languages • The start state of the new DFA =(qL ,qM) • Final states of the new DFA = cross products of FL and FM • Now to prove that L ∩ M is regular • Prove that there is a DFA that accepts all strings of L ∩ M • Letus assume that the newly constructed DFAis the required one

  35. Special DFA construct for intersection of two regular languages • Let a stringw ∈L ∩ M • This meansw ∈ L andw ∈M • Since L and M are regular, w is accepted by the corresponding automata i.e. dL(qL,w)∈FL anddM (qM,w) ∈ FM • This indicates that • d(( qL, qM),w)∈(FL,FM) • HenceL ∩ M is Regular

  36. Closure under Difference • L-M = set of all strings that are in L and not in M • i.e.L-M =L ∩ M • Since Complement of M is regular as M is regular, Intersection of L and Mis also regular • Hence the difference of two regular languages is regular

  37. p q Example • Let L=L(A) and M= L(B) where automata A and B are described by the following transition diagrams 0,1 1 q 0 0 0,1 q r s 1

  38. Reversal of a regular language is regular • Reversal of a string w = a1a2……an is the string anan-1……a1 denoted as wR. • Reversal of string 110101 is 101011 • Reversal of string abaaa is aaaba • Reversal of a language L is the language consisting of the reversals of all its strings • LR= {wR | w∈ L}

  39. Example : reversal of a language • Let L= {001,10,111} then LR = {100,01,111}

  40. Proving reversal of a regular language is regular: through DFA construct • Construct an automaton forLR • Reverse all arcs in the transition diagram for A • Make the start state of A the only accepting state for the new automaton • Create a new start state P0 with transitions on єto all the accepting states of A

  41. Example : L=1*0(0+1)* • L={100,1100,…101,1101,…} • LR = {001,0011,….101,1011,…} • First we draw a DFA for this language L 1 0,1 q p 0 q

  42. Example : drawing a new automaton for LR= (0+1)*01* • Reverse the arcs and make p as accepting state • Introduce a new start state p0 and draw an arc with symbol єfrom p0 to the final state of A 1 0,1 p є p 0 q p0

  43. Homomorphism • A string homomorphism is a function on strings that works by substituting a particular string for each symbol. • Example : Let S= {0,1}; w=0011 then function h(0)=ab h(1)=є • Then h(w)= h(0)h(0)h(1)h(1)=ababєє • h(w)=abab

  44. Homomorphism to a language • Homomorphism to a language is applied by applying it to each of the strings in the language • h(L) = {h(w) | w is in L} • Let L= 10*1 i.e. L={11,101,1001,…} • Then h(L) = {єє, єabє, єababє,…} or (ab)*

  45. If L is regular then h(L) is also regular • Let L=L( R) for regular expressionR • Let E be an expression over S • Then let h(E) be an expression obtained by replacing each symbol a of S in E by h(a) Then h(R) defines the language h(L).

  46. Equivalence and Minimization of Automata • Testing whether two descriptors for regular languages are equivalent, in the sense that they define the same languages. • This gives a method to minimize a DFA

  47. Definition: Equivalent states • Two states p and q are said to be equivalent if for all input strings w, d(p,w) is an accepting state if and only if d(q,w) is an accepting state • If two states are not equivalent then we say they are distinguishable (i.e. if they are not equivalent for at least one w)

  48. Example 1 0 0 1 C A B 0 D C 1 0 1 1 0 1 1 0 H E F G 0 1 0

  49. Table filling algorithm B x C x x D x x x E x x x F x x x x G x x x x x x H x x x x x x A B C D E F G

  50. Minimum state DFA equivalent to given DFA G 1 D,F 1 1 A,E 0 0 0 0 1 C B,H C 1 0

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