Non-deterministic and deterministic construction based on language

2. • Draw a DFA for the language accepting strings ending with ’01’ over input alphabets ∑ = {0, 1}
• Draw a DFA for the language accepting strings ending with ‘abb’ over input alphabets ∑ = {a, b}
• Draw a DFA for the language accepting strings ending with ‘abba’ over input alphabets ∑ = {a, b}
minimum length of string = 4 no. states required in the DFA = 4 + 1 = 5.

3. • Design a FA that accepts language containing strings an 101’ as substring.
• automata that accepts language containing strings which have ‘Design FA with
∑ = {0, 1} accepts the set of all strings with three consecutive 0’s.
• Design a FA with ∑ = {0, 1} accepts those string which starts with 1 and ends
with 0.

4. • Draw a DFA for the language accepting strings starting with ‘ab’ over
input alphabets ∑ = {a, b}
• Draw a DFA for the language accepting strings starting with ‘101’ over
input alphabets ∑ = {0, 1}
• Construct a DFA that accepts a language L over input alphabets ∑ =
{a, b} such that L is the set of all strings starting with ‘aa’ or ‘bb’

5. • Draw a DFA for the language accepting strings not having substring
‘101’ over input alphabets ∑ = {0, 1}
• Construct a DFA machine over input alphabet ∑ = {0, 1}, that
accepts:
1.Odd number of 0’s or even number of 1’s
2.Odd number of 1’s and even number of 0’s
3.Either odd number of 0’s or even number of 1’s
but not the both together

6. NFA CONSTRUCTION
δ: Q x ∑ →2Q
1. NFA with ∑ = {0, 1} accepts all strings with 01.
2. NFA with ∑ = {0, 1} and accept all string of length atleast 2.
3. Design an NFA with ∑ = {0, 1} accepts all string ending with 01.
4. Design an NFA with ∑ = {0, 1} in which double '1' is followed by double ‘0’
1. Design an NFA with ∑ = {0, 1} accepts all string in which the third symbol from the right end is always 0.

7. Find the language of the automata:
State a b
q0 [q0,q1] [q0]
q0, q1 [q0,q1] [q0,q2]
q0, q2 [q0,q1] [q0]
State a b
q0 {q0,q1} {q0}
q1 ∅ {q2}
q2 ∅ ∅
Transition Function:
δ ( q0, a )= [q0,q1]
δ ( q0, b)= [q0]
δ ( [q0, q1 ], a )= [q0,q1] δ ( { q0, q2 }, a )= [q0,q1]
δ ( [q0, q1 ], b )= [q0,q2] δ ( { q0, q2 }, b )=[q0]
Transition Function:
δ ( q0, a )= {q0,q1}
δ ( q0, b)={q0}
δ ( q1 , a )= ∅ δ ( q2, a )=∅
δ ( q1 , b )={q2} δ ( q2 , b )=∅
Transition table

8. NFA - ∈
Epsilon (∈) – closure : Epsilon closure for a given state X is a set
of states which can be reached from the states X with only (null) or ε
moves including the state X itself.

9. Example:2
∈ closure(1) : {1,2,3,4,8 }
∈ closure(2) : {2,3,5 }
∈ closure(3) : { 3}
∈ closure(4) : {4,2,3,5,7,8 }
∈ closure(5) : {5, }
∈ closure(6) : {6,7,8,2,3,5 }
∈ closure(7) : { 7,8,2,3,5}
∈ closure(8) : { 8}

10. Conversion problem: