Question

The pumping lemma for regular languages is Theorem 1.70 on page 78 of the required text. Definition: w is a string if and only if there exists an alphabet such that w is a string over that alphabet. Note: For every alphabet, the empty string is a string over that alphabet. Notation: For any symbol o, gº denotes the empty string, and for every positive integer k, ok denotes the string of length k over the alphabet {o}. 1) (20%] Using the pumping lemma for regular languages, prove that the language { 0j12j | j is a non negative integer } is not regular. 2) (20%] Using the pumping lemma for regular languages, prove that the language { alb j is a non negative integer } is not regular. 3) [20%] Using the pumping lemma for regular languages, prove that the language { 01132 | j is a non negative integer } is not regular.

Answer 1

pumping Lemma - *) If A is a regular Language, then A has a Pumping length p such that any string 'g' where 18129 may be divided into 3 pasos G=XYZ such that The following conditions must te the Дхуі 2 ел Рех елеу j20 2) 140 3) IXY) SP To prove that a longuage is not regular using Pumping Lemma follow the below steps: cwe poove using contradico op 1. Assume that A is regular 2. It has to have a pumping Length (Say D 3 All Songs Longer than p can be Pumped ISIZP 4 wow find a sting &' n A such that 18/21 5. onide s into xyz 6- show that xy/Z&A for some i 7. Then consides all ways that I can te divided into Xyz

L=800, au : j 2012, 3..3 To Assume that Lig Regular 2. Pumping length according to condition.. I will take s=2 L=30°223 - S0013 3. Divide Language into 3 Parts X Y Z L=EQOLL? X-00 Y Z z =1 Soung lengonan) -6 case is IxYILM 556 -ruiz 120 c m case 3:-1=xyz 120 lojas - I will take i L2x422 =O0C11D27 200 MAH X foulea & our assumption is useng because i's must te it right now 7i's It is not even - xyiz&L 3 j20,123 - - 3 not jeglas even

2) 2=3 a 35 j2 0,1,2,3 1-3 To Assume that t is peglar 2. Pumping Lengon accosding to according to condition I will take j=2 L=522 323 = qaabbbbb63 3. Divide Language into 3 Posts XYZ L={aabbbbbby xza96 4=bb zoo66 string Lengonan) =8 case 1:- 1x4) 6585 casea -- lyzl 221 cases. L=xglz izo . I will take ia arte 2-x922 =dabc666666 saabbbbbbbbe x fouled 4 ous assumption is useng no of bs most te multiply of 3 but right 865. 8 is not multiply of 3 . xyiz EL L={auzi: 120,1,2,3 -- 3 not rester

3 L=Loj 132: 320,1,2,3- to Assume that t is Reglar 3 2. Puning length accol ength according to condition - I will take j-2 ? L={ 02, 223 = 2001111} 3. Divide Language into 3 pasos XYZ L=400! 3 x200 Z=11 Stang Length an)=6 casel : - 1x 4 cm 456V X Y Z 221 v Po cose 3 :- = xyiz izo i will take i-a L=XYZ =O0CMR11 =0010111 x fouled *ous assumption is useng because is must be sense of the Some nuntes like R23 right now 61's It is not a sense of any number - XyiZEL L=800,32 = 120,123 -3 Š not seguor
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