Question

Question1. (0+1)*.1.(0+1) +(0+1) *.1.(0+1)

Question2. (((0*.1*)+1)*(0+1)*)*

Question3. (L+M*)*

Question4. (L.M*)*

Introduction to automata theory questions

Answer 1

Question1. (0+1)*.1.(0+1) +(0+1) *.1.(0+1)

• step 1 : (0+1)*.1.(0+1)   +  (0+1) *.1.(0+1)
  • use A = A.lambda
• step 2 : (0+1)*.1.(0+1).lambda   +  (0+1) *.1.(0+1).lambda
  • use A.b + A.c = A.[ b+c ]
• step 3 : (0+1)*.1.(0+1)[ lambda   + lambda ]
  • use lambda + lambda = lambda
  • use A=A.lambda
• step 4 : (0+1)*.1.(0+1)

Question2. (((0*.1*)+1)*(0+1)*)*

( ((0*.1*)+1)* (0+1)* )*

• The first underlined portion can generate all kind of strings from 0,1.
• Also the second underlined portion can generate all kind of strings from 0,1.
• So, the given regular expression refers to all the strings of 0,1.
• The simplified answer is (0+1)*, which represents the set of all the strings of 0,1

Question3. (L+M*)*

• Answer is (L+M)* , the explanation for this is similar to explanation of Question 2.

Question4. (L.M*)*

4 is in simplified form.