{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

class14 - More Applications of The Pumping Lemma 1 The...

Info iconThis preview shows pages 1–14. Sign up to view the full content.

View Full Document Right Arrow Icon
1 More Applications of The Pumping Lemma
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 The Pumping Lemma: there exists an integer such that m for any string m w L w | | , we can write For infinite context-free language L uvxyz w = with lengths 1 | | and | | vy m vxy and it must be: 0 all for , i L z xy uv i i
Background image of page 2
3 Context-free languages } 0 : { n b a n n Non-context free languages } 0 : { n c b a n n n }*} , { : { b a w ww R }} , { : { b a v vv
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4 Theorem: The language }*} , { : { b a v vv L = is not context free Proof: Use the Pumping Lemma for context-free languages
Background image of page 4
5 Assume for contradiction that is context-free Since is context-free and infinite we can apply the pumping lemma L L }*} , { : { b a v vv L =
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
6 Pumping Lemma gives a magic number such that: m Pick any string of with length at least m we pick: L b a b a m m m m L }*} , { : { b a v vv L =
Background image of page 6
7 We can write: with lengths and m vxy | | 1 | | vy uvxyz b a b a m m m m = Pumping Lemma says: L z xy uv i i for all 0 i }*} , { : { b a v vv L =
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
8 We examine all the possible locations of string in vxy m vxy | | 1 | | vy uvxyz b a b a m m m m = m m m m b a b a }*} , { : { b a v vv L =
Background image of page 8
9 Case 1: vxy is within the first m a b b a a b b a a ...... ...... ...... ...... v m m m u z m vxy | | 1 | | vy uvxyz b a b a m m m m = x y m 1 k a v = 2 k a y = 1 2 1 + k k }*} , { : { b a v vv L =
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
10 b b a a b b a a ...... ...... ...... ...... .......... 2 v 2 1 k k m + + m m u z m vxy | | 1 | | vy uvxyz b a b a m m m m = x 2 y m Case 1: vxy is within the first m a 1 k a v = 2 k a y = 1 2 1 + k k }*} , { : { b a v vv L =
Background image of page 10
11 m vxy | | 1 | | vy uvxyz b a b a m m m m = Case 1: vxy is within the first m a 1 2 1 + k k L z xy uv b a b a m m m k k m = + + 2 2 2 1 }*} , { : { b a v vv L =
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
12 m vxy | | 1 | | vy uvxyz b a b a m m m m = Case 1: vxy is within the first m a L z xy uv 2 2 Contradiction!!! L z xy uv b a b a m m m k k m = + + 2 2 2 1 However, from Pumping Lemma: }*} , { : { b a v vv L =
Background image of page 12
13 is in the first is in the first Case 2: b b a a b b a a ...... ...... ...... ......
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 14
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}