CS402 Assignment No. 2 2016

	Assignment No. 02 Semester: Spring 2016 Theory of Automata – CS402	Total Marks: 20 Due Date:  14/07/2016
Objectives: Objective of this assignment is to assess the understanding of students about: Union of Regular Languages Intersection of Regular Languages Transition Table Finite Automaton Instructions: Please read the following instructions carefully before submitting assignment: You should consult the recommended books, PowerPoint slides and video lectures to clarify your concepts. You are supposed to submit your assignment in .doc format. Any other formats like scan images, PDF, zip, rar and bmp etc will not be accepted. It should be clear that your assignment will not get any credit if: ·         The assignment is submitted after due date. ·         The assignment is copied from Internet or from any other student. ·         The submitted assignment does not open or file is corrupt. Note: No assignment will be accepted after the due date through email in any case (load shedding, server down, internet malfunctioning etc.). It is recommended to upload solution file at least two days before its closing date. For any query about the assignment, contact at cs402@vu.edu.pk Question 1:                                                                                                                                Marks:10   10 Consider the following FAs corresponding to regular languages L1 and L2, defined over the alphabet Σ = {a, b}. FA1 corresponding to L1 FA2 corresponding to L2 You are required to draw a transition table for the FA corresponding to L1 ∩ L2. Question 2:                                                                                                                               Marks: 10 Given the FAs corresponding to L1 and L2 in Question 1, you are required to draw an FA corresponding to L1c ∪ L2c.
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 Solution

Answer No 1:- 

         

         

   Transition Table for FA corresponding to L₁ ∩ L₂.
 

Old State	a	b
Z1- =1,X	(1,X)=Z1	(2,Y)=Z4
Z2 =1,Y	(I,Y)=Z2	(2,X)=Z3
Z­­­3 =2,X	(1,X)=Z1	(3,Y)=Z6
­Z4 =2,Y	(1,Y)= Z2	(3,X)=Z5
Z5+ =3,X	(3,X)=Z5	(3,Y)=Z6
Z6 =3,Y	(3,Y)=Z6	(3,X)=Z5

 





 
 

            (a*bc*fa)*(bc*d)(e*gc*de)*(gc*f)

Answer No 2:

FA corresponding to L₁^c ∪ L₂^c.