Review Languages and Grammars Alphabets, strings, languages - PDF document

Review • Languages and Grammars – Alphabets, strings, languages • Regular Languages CS 301 - Lecture 7 – Deterministic Finite Automata – Nondeterministic Finite Automata More Pumping Lemma – Equivalence of NFA and DFA – Minimizing a DFA Fall 2008 – Regular Expressions – Regular Grammars – Properties of Regular Languages – Pumping lemma for Regular Languages • Today: – More pumping lemma for regular languages The Pumping Lemma: n n Theorem: The language L { a b : n 0 } = ≥ • Given a infinite regular language L is not regular m • there exists an integer | w ≥ | m • for any string with length w ∈ L w = x y z • we can write Proof: Use the Pumping Lemma | x y | m | y | 1 • with and ≤ ≥ i i 0 , 1 , 2 , ... x y z L • such that: = ∈ 1

n n n n L { a b : n 0 } L { a b : n 0 } = ≥ = ≥ m Let be the integer in the Pumping Lemma Assume for contradiction L that is a regular language Pick a string such that: w w ∈ L | w ≥ | m length L Since is infinite we can apply the Pumping Lemma m b m w = a We pick m m m b m k a b x y z x y z a y a , k 1 Write: = = = ≥ From the Pumping Lemma i | x y | m , | y | 1 it must be that length ≤ ≥ x y z L From the Pumping Lemma: ∈ m m i 0 , 1 , 2 , ... = m m xyz a b a ... aa ... aa ... ab ... b = = x y z 2 x y z L Thus: ∈ k y a , k 1 Thus: = ≥ 2

m b m k m k m + x y z a y a , k 1 a b L = = ≥ ∈ 2 x y z L From the Pumping Lemma: ∈ n n BUT: L { a b : n 0 } = ≥ m m + k 2 xy z a ... aa ... aa ... aa ... ab ... b L = ∈ m k m + a b L ∉ x y y z CONTRADICTION!!! m k m + a b L Thus: ∈ L Therefore: Our assumption that n n Non-regular languages { a b : n 0 } ≥ is a regular language is not true Regular languages Conclusion: L is not a regular language 3

More Applications Non-regular languages R L { vv : v *} = ∈ Σ of Regular languages the Pumping Lemma R L { vv : v *} = ∈ Σ Theorem: The language R L { vv : v *} { a , b } Σ = = ∈ Σ Assume for contradiction is not regular L that is a regular language Proof: Use the Pumping Lemma Since is infinite L we can apply the Pumping Lemma 4

m m m m a b b a x y z Write R = L { vv : v *} = ∈ Σ m Let be the integer in the Pumping Lemma From the Pumping Lemma | x y | m , | y | 1 it must be that length ≤ ≥ Pick a string such that: w w ∈ L and m m m m | w ≥ | m length xyz a ... aa ... a ... ab ... bb ... ba ... a = x y z m m m m w = a b b a We pick k y a , k 1 Thus: = ≥ m m m m k m m m m k x y z a b b a y a , k 1 x y z a b b a y a , k 1 = = ≥ = = ≥ 2 x y z L From the Pumping Lemma: ∈ i m m m From the Pumping Lemma: x y z L m + k ∈ 2 xy z = a ... aa ... aa ... a ... ab ... bb ... ba ... a ?¸ L i 0 , 1 , 2 , ... = x y y z 2 x y z L Thus: ∈ m k m m m + a b b a L Thus: ∈ 5

m k m m m + a b b a L k 1 ∈ ≥ L Therefore: Our assumption that is a regular language is not true R BUT: L { vv : v *} = ∈ Σ m k m m m + a b b a L Conclusion: L is not a regular language ∉ CONTRADICTION!!! Non-regular languages Theorem: The language n l n l + L { a b c : n , l 0 } = ≥ n l n l + L { a b c : n , l 0 } = ≥ is not regular Regular languages Proof: Use the Pumping Lemma 6

n l n l n l n l + + L { a b c : n , l 0 } L { a b c : n , l 0 } = ≥ = ≥ m Let be the integer in the Pumping Lemma Assume for contradiction L that is a regular language Pick a string such that: w w ∈ L and | w ≥ | m length L Since is infinite we can apply the Pumping Lemma m m 2 m We pick w a b c = m m 2 m m m 2 m k a b c x y z x y z a b c y a , k 1 Write = = = ≥ From the Pumping Lemma | x y | m , | y | 1 it must be that length ≤ ≥ i From the Pumping Lemma: x y z L ∈ m m 2 m i 0 , 1 , 2 , ... = xyz a ... aa ... aa ... ab ... bc ... cc ... c = y x z Thus: k Thus: y a , k 1 = ≥ 7

m m 2 m k m k m 2 m − k 1 x y z a b c y a , k 1 a b c L ≥ = = ≥ ∈ From the Pumping Lemma: xz ∈ L n l n l BUT: + L { a b c : n , l 0 } = ≥ m − k m 2 m xz a ... aa ... ab ... bc ... cc ... c L = ∈ m k m 2 m − a b c L ∉ x z CONTRADICTION!!! m k m 2 m − a b c L Thus: ∈ L Therefore: Our assumption that n ! Non-regular languages L { a : n 0 } = ≥ is a regular language is not true Regular languages Conclusion: L is not a regular language 8

n ! L { a : n 0 } = ≥ n ! Theorem: The language L { a : n 0 } = ≥ is not regular Assume for contradiction L that is a regular language n ! 1 2  ( n 1 ) n = ⋅ − ⋅ L Since is infinite Proof: Use the Pumping Lemma we can apply the Pumping Lemma n ! a m = ! L { a : n 0 } x y z Write = ≥ m Let be the integer in the Pumping Lemma From the Pumping Lemma | x y | m , | y | 1 it must be that length ≤ ≥ w w ∈ L Pick a string such that: m m − ! m | w ≥ | m length m ! = xyz a a ... aa ... aa ... aa ... aa ... a = x y z m ! w = a We pick k y a , 1 k m Thus: = ≤ ≤ 9

m ! k m ! k x y z a y a , 1 k m x y z a y a , 1 k m = = ≤ ≤ = = ≤ ≤ 2 x y z L From the Pumping Lemma: ∈ i From the Pumping Lemma: x y z L ∈ m − ! m m + k 2 i 0 , 1 , 2 , ... = xy z a ... aa ... aa ... aa ... aa ... aa ... a L = ∈ x y y z 2 Thus: x y z L ∈ m ! k + Thus: a L ∈ m + ! k m ! m m ! k However: for m 1 + ≤ + > a L 1 k ≤ m ∈ ≤ m ! m ! ≤ + m ! m m ! < + n ! Since: L { a : n 0 } = ≥ m ! ( m 1 ) = + ( m 1 )! = + p There must exist such that: m ! k ( m 1 )! + < + m ! k p ! + = for any p m ! k p ! + ≠ 10

m ! k 1 k ≤ m + a L ≤ ∈ L Therefore: Our assumption that is a regular language is not true n ! BUT: L { a : n 0 } = ≥ m ! k + a L Conclusion: L is not a regular language ∉ CONTRADICTION!!! What’s Next • Read – Linz Chapter 1, 2.1, 2.2, 2.3, (skip 2.4) 3, 4, 5.1-5.3 – JFLAP Startup, Chapter 1, 2.1, 3, 4, 6.1 • Next Lecture Topics from Chapter 5.1-5.3 – Context Free Grammars • Quiz 1 in Recitation on Wednesday 9/17 – Covers Linz 1.1, 1.2, 2.1, 2.2, 2.3 and JFLAP 1, 2.1 – Closed book, but you may bring one sheet of 8.5 x 11 inch paper with any notes you like. – Quiz will take the full hour • Homework – Homework Due Thursday 11