Q. Given a language πΏ, define πΏπ as follows:
πΏ0 = {π}
πΏπ = πΏπβ1 β πΏ πππ πππ π > 0
The order of a language L is defined as the smallest k such that πΏπ =Β Β Β πΏπ+1. Consider the language L1 (over alphabet 0) accepted by the following automaton.
The order of L1 is
Ans: 2
Sol:
L1Β = Ξ΅ + 0(00)*
L0Β = Ξ΅
L1Β = Ξ΅ . (Ξ΅ + 0(00)*)
Β Β = Ξ΅ + 0(00)* = L1
L2Β = L1Β .Β L1Β =Β L1Β .Β L1
Β Β = (Ξ΅ + 0(00)*) (Ξ΅ + 0(00)*)
Β Β = Ξ΅ + 0(00)* + 0(00)* + 0(00)* . 0(00)*
Β Β = Ξ΅ + 0(00)* + 0(00)* 0(00)* =Β 0*
Given automaton contains epsilon and even number of 0βs and odd numbers of 0βs.
L3Β = L2Β .Β L1
Β Β = {0*} . {Ξ΅ + 0(00)*} = 0* 0(00)* = 0*
So, L2Β = L3Β = L2+1
So, smallest value ofΒ k = 2
