Deterministic Finite Automata (DFA) - 1

q0 q1

01

0

1

strings that do not end with "1".

Build an automaton to identify strings that end with "0".

q1 q2

01

0

1

q00

1

Σ = {0,1}

Deterministic Finite Automata (DFA) - 2

What patterns does this automaton identify (over {a,b,c})?

strings that begin and end with the same letter.

aA

b B

c

C

S

a

b

c

A’b,c

B’a,c

C’a,b

b,c

a

a,c

b

a,b

c

Deterministic Finite Automata (DFA) - 3

Build an automaton to identify strings over {0,1} that are binary numbers divided by 3.

explanation: qi denotes remainder of i.

for example, if n=3k+1 (remainder 1)then adding "1" yields 2n+1 = 2(3k+1)+1 = 6k+3 = 3(2k+1) remainder 0

q0 q1 q2

01

1

0

0

1

Deterministic Finite Automata (DFA) - 4

Given 2 automata A1 and A2, that identify patterns p1 and p2 respectively,

describe an automaton that identifies strings with either p1 or p2.

solution:

states: tuples (q1,q2) for every q1 from A1 and q2 from A2

transitions: if q1 –x-> q1' and q2 –x-> q2' then (q1,q2) –x-> (q1',q2')

start state: (q1start,q2start) when q1start and q2start are A1 and A2 starting states

accepting states: (q1,q2) is accepting if q1 is accepting in A1 or q2 is accepting in A2.

Deterministic Finite Automata (DFA) - 5

Prove no finite automata can identify strings of the form anbn for unknown n.

proof: assume there is such an automaton, with N states.then for aN+1bN+1 it must repeat the same state si at least twice (pigeon hole principle).but then it must also accept strings with <N+1 a's , contradiction.

aN+1bN+1 = a a a a a … a a a b b b b b … b b b

a a … a a b b b b b … b b b

* comment: for any finite known n, there IS such an automaton…

delete

N+1 N+1

<N+1 N+1

si si