6. Constrict a DFA which accept number ( integer) divisible by 2 or (multiple of 2) .
Solution:
Here we have two states such as q1 , q2. where q2 is final state.
let we start
Case I: Input string = If string is even numbers such as (0,2,4,6,8) between 0 - 9.
Case II: Input string = If string is odd numbers such as (1,3,5,7,9) between 0 - 9.
Formal Definition :
Q = { q1 , q2 }
Σ = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
q1 = q1 Initial state
F = { q1 }
δ =
For CONSTRUCTION OF FINITE AUTOMATA (FA) we had seen different cases and finally construct a machine. In CASE II we get final graph.
Solution:
Here for CONSTRUCTION OF FINITE AUTOMATA (FA) input strings are combination of a and b and string having 0 - 9 number i.e 10 numbers because any number divisible by 2 the resulting number must be between 0 - 9.
For CONSTRUCTION OF FINITE AUTOMATA (FA) we will take different cases of input string and trace it.
Here we have two states such as q1 , q2. where q2 is final state.
let we start
Case I: Input string = If string is even numbers such as (0,2,4,6,8) between 0 - 9.
Case II: Input string = If string is odd numbers such as (1,3,5,7,9) between 0 - 9.
Formal Definition :
Q = { q1 , q2 }
Σ = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
q1 = q1 Initial state
F = { q1 }
δ =
For CONSTRUCTION OF FINITE AUTOMATA (FA) we had seen different cases and finally construct a machine. In CASE II we get final graph.
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