Formal Language and Automata Theory: Quiz 1
2013/10/30
Name:_________________ Reg.No.:___Solution________
Score:_______________
1. [10pts] Let A= {01,1,101} and B = {0,10,110}.
1.1 Which of the following strings are in AB ?
(a) 1010
(b) 1100
(c) 0101
(d) 110 (e) 10101
1.2 List all strings x in A3 with |x| < 5.
ANS:
111, 0111, 1011, 1101
2. [15pts] Which of the following statements are correct? Let A,B and C are arbitrary languages.
(a) If A  B then CA  CB
(b) A (B  C) = AB  AC.
(c) (A*)* = A*
(d) If A2  A then A+ = A.
(e) {x,y,z}*  {xyz}*, where x,y and z are arbitrary strings.
3. [25pts] Given a DFA M shown below as a state transition diagram :
M
a
b
1F
1
2
2
3
4
3
2
1
4F
3
4
(a) Fill the above table to get a complete state transition table for the DFA M.
(b) Let M = (Q, , , s, F). Then what are the contents of Q, , , S and F, respectively.
 Q = _{1,2,3,4}__________
 s =__1_________________
 = __{a,b}_______________
F =__{1,4}_______________
  = { (1,a,1),(2,a,3),(3,a, 2),(4,a,3),
(1,b,)(2,b,4),(3,b,1),(4,b,4)
}
Note: Use triple (p,a,q) to represent the case that (p,a) = q.
(c) For each input given in the table below, list the sequence of states traversed while running
the machine with the input and determine if the input is accepted.
1
input
Sequence of states traversed
accepted?
q1
yes
abba
11243
no
baab
12324
yes

4. [35pts] Let = {a,b}. Construct a DFA for each of the following languages. Use state transition
diagram for your answer. Note: #c(x) denotes the number of occurrences of symbol c in x.
(a)
L1 = { x {a,b}* | x starts with a and ends with b }.
a
b
b
a
a
b
a,b
(b)
L2 = { x{0,1}* | x contains the substring 001
0,1
1
0
0
1
0
1
(c)
}
L3 = { x{0,1}* | x ends with 1001 }.
0
1
1
1
0
1
0 0
1
0
(d) [5pts] Given two arbitrary binary number x= x1x2…xn and y = y1y2…yn of the same length, we
want to determine if x > y. Let 2 = { (0,0),(0,1),(1,0),(1,1) }, and for the pair (x,y) of binary
number x and y of the same length, let e(x,y)= (x1,y1) (x2,y2) …(xn,yn) be a string over 2.
Design a DFA M such that L(M) = { e(x,y) | x and y are binary numbers of the same length and
x > y.}
ANS: M can be given as follows:
(0,0)
(1,1)
(1,0)
(0,1)
1
1
1
2
3
2F
2
2
2
2
3
3
3
3
3
2
5. [25pts]
(a) Find a DFA M1 with two states: p,q such that L(M1) = { x | the length of x is odd }.
(b) Find a DFA M2 with three states:1,2,3 such that L(M2) = { x | x ends with bb }
Note: remember to mark starting and final states!
M1
a
b
M2
a
b
p
q
q
1
1
2
qF
p
p
2
1
3
3F
1
3
(c) Use the product construction to generate a DFA M3 from M1 and M2 with the set of states
{p,q}x{1,2,3} such that L(M3) = { x | The length of x is odd and x does not ends with bb. }. Give your
answer using state transition table. [15pts]
M3
a
b
p1
q1
q2
q1F
p1
p2
p2
q1
q3
q2F
p1
p3
p3
q1
q3
q3
p
p3
6. [25pts] For a given alphabet , the regular expression over  is defined inductively as follows:
Basis: 1. The symbol e, , and every symbol a   are regular expressions.
Recursion: 2. If  and  are regular expressions, then so are (), (), and *.
Given a regular expression , the language L() denoted by  is defined inductively as follows:
Basis: L(e) = {}, L()= {} and L(a) = {a} for every symbol a  .
Recursion: L(()) = L()  L(), L(()) = L()  L(), L(*)= L()*.
For any string w = x1x2 · · ·xn, the reverse of w, written rev(w), is the string xn · · ·x2x1 ,i.e.,w
written in reverse order. For any language A  *, let rev(A) = {rev(w) | w  A}. We now want you
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to construct a mapping R from regular expressions to regular expressions such that if A = L(r) for
some regular expression r, then rev(A) = L( R(r) ). The mapping R should be defined recursively
as follows:
(a) Basis: R(e) = _e___, R()= ___ and R(a) = ___a_______ for every a  .
(b) Recursion: R(()) = (R() + R()), R(()) = _(R()  R())_, R(*)= _R()*__.
(c) Show the correctness of your answer by proving that : for all regular expressions ,
rev(L()) = L( R()). Hint: Use induction on the structure of . [13pts]
pf: The proof is by induction on the structure of .
Basic cases:  = e,  , or a  . Then
rev(L(e)) = rev({}) = {} = L(e) = L(R(e)),
rev(L()) = rev({}) = {} = L() = L(R()), and
rev(L(a)) = rev({a}) = {a} = L(a) = L(R(a)).
Ind. cases:  = +,  or * for some regular expressions . Then
rev(L(+)) = rev(L() U L()) = rev(L()) U rev(L())
= L(R()) U L(R()) -- by ind.hyp.
= L(R() + R()) = L(R()).
rev(L()) = rev(L()L()) = rev(L()) rev(L())
= L(R())  L(R()) -- by ind.hyp.
= L(R()R()) = L(R()).
rev(L(*)) = rev(L()*) = (rev(L()))*
= (L(R()))* -- by ind.hyp.
= L(R()*)
= L(R(*)) .
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