1. 2-Way Finite Automata: An Example
2DFA accepting {x{a,b}* | #a(x) is even, #b(x) is a multiple of 3}, where #c(z) is the # of c’s in z.
qeven is the initial state.
Phase1: Check the # of a’s:
(qeven,a)=(qodd,+1), (qodd,a)=(qeven,+1),
(q,c)=(q,+1) if q{qeven,qodd},c{¢,b};
If Phase1 was ok, then (qeven,$)=(q0,-1);
Phase2: Check the # of b’s:
(q0,b)=(q1,-1), (q1,b)=(q2,-1),
(q2,b)=(q0,-1), (qi,a)=(qi,-1) for i=0,1,2;
Accept if Phase 2 was ok:
(q0,¢)=(qaccept,0);
(qeven,¢aba$,1)(qeven,¢aba$,2)(qodd,¢aba, 3)
(qodd,¢aba$,4)(qeven,¢aba$,5)(q0,¢aba$,4)
(q0,¢aba$,3)(q1,¢aba$,2)(q1,¢aba$,1)No more
aba is not accepted by the above 2DFA.
•  L ¢
003-03

2. There is any kind of finite automaton …
A DFA that accepts the language on the previous slide  DFAs/NFAs that simulates the moves of two DFAs/NFAs
Finite automata with output
Finite automata with multi input heads  Strictly more powerful than ordinary FAs
Finite automata on arbitrary labeled digraphs; especially tree automata
Parallelization in terms of a number of finite automata  cellular automata
003-03

3. Quiz #01 (5 points)
Let M be an -NFA defined by the state transition table given below. Let ={a,b}.
Draw the state transition graph (STG) of M.
Determine *(q0, a1b2).
What is X={x* | (q3,x) * (q3,)}?
What is L(M)?
Convert M to an equivalent DFA (Draw the STG for it). Use Algorithm -NFAtoDFA
a b 
 q {q1} {q1} {q1,q4}
q1 {q2,q3}  {q2}
q  {q0} 
q  {q4} 
q  {q3} 
q5 {q3,q4} {q3} 
•  L ¢
003-03

4. Solution for quiz #01 (1) Shown below (left).
(2) *(q0, a1b2)={q0,q1,q2,q3,q4}.
(3) X=(bb)*={b2n | n0}. The set of words consisting of even number of b’s.
(4) L(M)={a,b}+. The set of all words over {a,b} except for the empty word.
(5) Shown below (right).
a,b, q0 q1
a, a
{q0} b
a,b q2 q3  b
{q0,q1,q2,q3,q4}
a,b b q4 q5 a
a,b
003-03