Lecture-8-Pumpdndndndndnddndning Lemma.pdf
- 2. Finite Automata 2 How did we prove that a language is Regular? If we can draw a finite automaton then it will be Regular Finite means that it contains a finite number of states every finite automata is Regular Example of infinite language: L={w|w contains an even number of symbols} L={ɛ,00,01,11,10,0000,…} 0,1 0,1 Repeated states An infinite language has to be accepted by a machine that is called a finite automata, then there should be a loop or repeated states and inside the loop there should be a specific pattern.
- 3. Infinite Languages 3 Each state represents a memory, its memorizing a specific property. All finite languages are Regular but not necessarily all the infinite languages are not Regular Suppose we have L={ an | n>=1} L={ɛ, a, aa, aaa, aaaa, …} a a Repeated states
- 4. Infinite Languages 4 Let L={anbm | n,m>=1} The property of this language is that all a’s should come before all b’s a a Repeated states b b
- 5. Infinite Languages 5 What if we have L={anbn | n>=1} Here the count factor is important as I can have the number of a’s less than or greater than the number of b’s Is it a Regular language? What if L={anbn | n<=100999} It becomes a regular language because it is finite and every finite language is Regular. How to prove that a specific language is Regular/ not Regular.
- 6. Pumping Lemma 6 How can we prove that a language L is not regular? Prove that there is no DFA or NFA or RE that accepts L Difficulty: this is not easy to prove (since there is an infinite number of them) Solution: use the Pumping Lemma !!!
- 7. Pumping Lemma 7 • In the theory of formal languages, the pumping lemma for regular languages is a lemma that describes an essential property of all regular languages. • Informally, it says that all sufficiently long strings in a regular language may be pumped—that is, have a middle section of the string repeated an arbitrary number of times—to produce a new string that is also part of the language.
- 8. Pumping Lemma 8 • Specifically, the pumping lemma says that for any regular language L • There exists a constant P such that any string w in L with length at least P can be split into three substrings x,y and z (w=xyz, with y > 0, such that the strings xz, xyz, xyyz, xyyyz, … constructed by repeating y zero or more times are still in L • This process of repetition is known as "pumping". • Moreover, the pumping lemma guarantees that the length of xy will be at most p, imposing a limit on the ways in which w may be split.
- 9. Pumping Lemma (For Regular Languages) 9 • Pumping Lemma is used to prove that a Language is NOT REGULAR • It cannot be used to prove that a Language is Regular • If A is a Regular Language, then A has a Pumping Length 'P' such that any string 'S' may be divided into 3 parts S = xyz such that the following conditions must be true: 1. x yi z ∈ A for every i > 0 2. |y| > 0 3. |xy| < P
- 10. Pumping Lemma (For non Regular Languages) 10 To prove that a language is not Regular using PUMPING LEMMA, follow the below steps: (We prove using Contradiction) • Assume that A is Regular • It has to have a Pumping Length (say P) • All strings longer than P can be pumped |S|>= P • Now find a string 'S' in A such that ISI>=P • Divide S into x y z • Show that xyiz ∉ A for some i • Then consider all ways that S can be divided into x y z • Show that none of these can satisfy all the 3 pumping conditions at the same time • S cannot be Pumped == CONTRADICTION
- 11. EXAMPLE 1 11 Pumping Lemma (For Regular Languages) Using Pumping Lemma prove that the language A = {an bn | n >= 0} is Not Regular Proof: Assume that A is Regular Pumping length = P S = ap bp Assume p = 7 Then, S = aaaaaaabbbbbbb X Y Z
- 12. EXAMPLE 1 12 Case 1: the Y is in ‘a’ part Z Y X Case 2: the Y is in ‘b’ part Z Y X Case 3: the Y is in the ‘a’ and in the ‘b’ part Z Y X xyiz => xy2z aa aaaa aaaa abbbbbbb xyiz => xy2z aaaaaaabb bbbb bbbb b xyiz => xy2z aaaaa aabb aabb bbbbb #a’s ≠ #b’s 7 11 #a’s ≠ #b’s 11 7 #a’s ≠ #b’s 11 11 Still doesn’t satisfy anbn √ √ √ √ X X X X X
- 13. EXAMPLE 2 13 Pumping Lemma (For Regular Languages) - Using pumping lemma prove that the language A= {yy|y ∈ {0,1}*} is Not Regular X Y Z X Y Z xyiz = xy2z 00 0000 0000 0100000001 A is not Regular X √ X Proof: Assume that A is Regular Then it must have a pumping length = P S = 0p 1 0p 1
- 14. Closure Property 14 101010 is an accepted string in this language 000111 is also accepted string 0*1* L1 = { w ∈ ∑* | w contains zero or more 0’s followed by zero or more 1’s} This language is Regular L2 = {0n 1n | n >= 0} This language is not Regular L3 = { w ∈ ∑* | w has an equal number of 0’s and 1’s} all strings in L2 are contained in L3 all strings in L2 are contained in this language L1 ∩ L3 = L2 L1 is Regular L2 is not Regular (proved) Based on Closure property the intersection of 2 Regular languages is regular L3 is not Regular