Pushdown Automaton: CFL and the pumping lemma

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L={a n b n c n |n0} Case 1: v and y contain only one type of alphabet symbol. By choosing i=2 we get w 2 =uv 2 xy 2 z in which the number of appearances of one or two symbols increased while the third symbol remain unchanged. So w 2 cannot contain the same number of a's, b's and c's and w 2 L. 48
L={a i b i c i |i0} Case 2: v or y contain two types of alphabet symbols. (Cannot have three because condition 2 holds). So if we choose i=2 then the order of the symbols in v 2 or y 2 was destroyed and w 2 =uv 2 xy 2 z L. Conclusion: The assumption that L is contextfree is false. and L={a n b n c n |n0} is not CFL. 49
Page 1 and 2: Unit 9 More Pushdown Automata Conte
Page 3 and 4: Example: L = {a i b i c j d j |i,j0
Page 5 and 6: CFG=PDA Theorem: A language is cont
Page 7 and 8: Different derivations for the same
Page 9 and 10: control E E E E E E input: stac
Page 11 and 12: E E E E E E control 5 E E 5
Page 13 and 14: From CFG to PDA Informally: 1. Plac
Page 15 and 16: From CFG to PDA • We define as f
Page 17 and 18: From PDA to CFG context-free gramma
Page 19 and 20: From PDA to CFG • $ is always pop
Page 21 and 22: From PDA to CFG • For any word w
Page 23 and 24: From PDA to CFG Option 1: The stack
Page 25 and 26: From PDA to CFG Let P=(Q, , , , q 0
Page 27 and 28: Example: 0,A ,$ #, q s q 0 1,A q 1
Page 29 and 30: CFG=PDA • We have shown that a la
Page 31 and 32: Non context-free Languages • Cons
Page 33 and 34: The Pumping Lemma - background •
Page 35 and 36: The Pumping Lemma - background •
Page 37 and 38: The Pumping Length • That means t
Page 39 and 40: Proof - value of p • First we fin
Page 41 and 42: Proof - condition 1 • To prove co
Page 43 and 44: Proof - condition 3 S S R R R x u v
Page 45 and 46: Usage of the lemma - Example • We
Page 47: L n n n p p p { a b c | n 0}. Let
Page 51 and 52: Intersection 1 CFL are not closed u
Page 53: Complement Theorem: CFL are not clo

Pushdown Automaton: CFL and the pumping lemma

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