Question

Which of the following languages are regular. Prove (by providing a regular expression) or disprove.

a. L1 = {aⁱ b^j c^k d^l | (i + j)mod 2 = (k + l)mod 2 , i, j, k, l ≥ 0}

b. L2 = {aⁱ b^j c^k d^l | (i + j) = (k + l), i, j, k, l ≥ 0}

Answer 1

1. a, b, c, d * prove above language a) La & a bicka (Citj) mod2 = (KHL) mod2, i, j, k, 120} are infinitely produced strings. -> count and comparision of infinitely produced strings need to do here because citj mod 2 = (x+1) moda verify. -> count and companiion of infinitely produced strings not possible regular language so it is not regular. is not regular by using pumping lemma. Ster!!- assume that language regular language. step?!- Consider string generated by language L. iz 2, 1-1, Kezidel str=a²b²c²d' aabed where str EL. Stef3!- Now split the string into 3 parts. а| a albccd al y z Ll as da, ya, z = bccd. of pumping lemma Stepy check 3 Cases NOW Сам сі) " lylo lalo 1 [True) case (ii) :- sxyl s pamping length. laal 26 2 True

Care & Check xyk z el for different values of K. Kal =) ) a call bred a aa bccd El a can bred K=2 =) aaabeed &L Reason - i=3, j = 1, K22, d = 1 (it j) mod2 - 4 moda (KHL) moda =3moda b (itj) moda (kH) Moda care linis fail) Conclusion ! - Since Case di ii ) fail our assumption that langelaze hias regular false. 22 is not regular 22 = {a bick de 1 citj) = (x+1), i, j, k, l=0} infinitely produced strings a need to do count and comparision of infinitely produced string but it is not possible in regular language. * prove this it is not regular by using pumping lemma. step! :- Assume that language to as regular language. Step? : - Consider string generated by language L2 - Here also a, b, c, d are Lz.

in 2, j = 1, K22 dal a str=a²b²c²d zaabed where strela Stelp!- Nowo split the string into 3 parts. a albccd al y z da.y=0,2 = bccd. Slepy: - NOW check 3 cases of pumping lemma care ûį 141 > 0 lalo True length, case (11) :- styl s pamping Iaal 6 2 Frue Case (ii) : Check xyk z Eh for different values of Kil =) a callbeed = aa b ccd El k=2 =) a ca2 bred i acabeed &L Reason i=3, j = 1, K22, dal

itj= 3+1 = 4 KAL - 2+1=3 (i+) + (K+1) Caselii) fail. conclusion - since Caselii) fail our assumption that language L2 as regular false. :ok2 is not regular