∑ = {0, 1}
Answer:
In that case, we try to develop an automata when substring ‘001’ then make it complement. Containing substring ‘001’.
L = {001, 0001, 00011, ……}
RE = (0 + 1)* 001 (0 + 1)*
Now this machine for “Not containing substring 001”
RE = (1 + 01)* + (1 + 01)*0 + (1 + 01)*000*
= (1 + 01)* [λ + 0 + 000*]
= (1 + 01)* [λ + 0(λ + 00*)]
= (1 + 01)* [λ + 00*]
= (1 + 01)*0*