15159222.ppt

Dete rministic Finite
Autom ata (DFA)
• DFA can be defined by quintuples
( 𝑄 , Σ , 𝛿 , 𝑞 0 , 𝐹 )
– 𝑄 is finite set of states
– Σ is finite set of alphabets
– 𝛿 is transition function
– 𝑞 0 is initial state
– 𝐹 finite set of final states
1

Example of DFA
𝑄 = 𝐴, 𝐵, 𝐶
Σ = 𝑎, 𝑏
𝛿: 𝑄 Χ Σ → 𝑄
𝑞0 = 𝐴
𝐹 = {𝐵}
2

Transition Function
𝛿: 𝑄 Χ Σ → 𝑄
{A, B, C} X {a, b} {A, B, C}
(A, a)
(A, b)
(B, a)
(B, b)
(C, a)
(C, b)
A
B
C
For every input on a
state there is
exactly one
transition.
3
A
B
C

Question
• Why this FA is called Deterministic Finite
Automata?
– Every state on seeing any input it is always going
to go to only one state.
4

DFA
• Construct a DFA, that accepts set of all strings
over {a, b} of length 2.
• Ans:
– Σ = 𝑎, 𝑏
– 𝐿 = {𝑎𝑎, 𝑎𝑏, 𝑏𝑎, 𝑏𝑏}
B C
A
a a
5

DFA
• Is this a complete DFA?
• No
• Now is this a complete DFA?
• No
• After reaching sate ‘D’ we can’t go back to the final state, that is
why this state is called dead state/trap state.
B C
A
a a
B C
A
B C
A
a, b a, b
a, b a, b
D
a, b
a, b
6

Acceptance
• Acceptance of a string by a FA:
– Scan the entire string and if we reach a final state
from initial state.
• Acceptance of a language by a FA:
– If all the strings in the language are accepted by
the FA and all the strings those are not in the
language must reject by the FA.
7

DFA
• Construct a DFA that accept all strings over the
alphabets {a, b} where the string length is at least 2.
– 𝝎 ∈ 𝒂, 𝒃 and 𝝎 ≥ 𝟐
– 𝜮 = 𝒂, 𝒃
– 𝑳 = {𝒂𝒂, 𝒂𝒃, 𝒃𝒂, 𝒃𝒃, 𝒂𝒂𝒂, … … 𝒃𝒃𝒃, … … … }
– Up to this is a DFA of string length 2. This is not the
required DFA.
– Now this is required DFA.
A B
a,b
C
a,b
a, b
8

DFA
alphabets {a, b} where the string length is at most
two.
– 𝝎 ∈ 𝒂, 𝒃 and 𝝎 ≤ 𝟐
– 𝜮 = 𝒂, 𝒃
– 𝑳 = {𝝐, 𝒂, 𝒃, 𝒂𝒂, 𝒂𝒃, 𝒃𝒂, 𝒃𝒃}
– To accept null string (𝝐) the only way to make initial
state to both initial state and final state.
A B
a,b
C
a,b
D
a, b
a, b
9

Minimum numbe r of state s
𝝎 = 𝟐 𝝎 ≥ 𝟐 𝝎 ≤ 𝟐
𝝎 = 𝒏
n + 2
𝝎 ≥ 𝒏
n + 1
𝝎 ≤ 𝒏
n + 2
10

DFA
• Construct a DFA that accept all strings over
the alphabets {a, b} where the string length is
divisible by two.
• 𝑳 = {𝜖, 𝑎𝑎, 𝑎𝑏, 𝑏𝑎, 𝑏𝑏, 𝑎𝑎𝑎𝑎 … 𝑏𝑏𝑏𝑏 … … … }
• This is not a finite automata
A B
a, b
C
a, b
D
a, b
E
a, b
Length 0 Length 1 Length 2 Length 3 Length 4
11

DFA
divisible by two.
• This is not minimal
A B
a, b
C
a, b
Length 0 Length 1 Length 2
A B
a, b
Length
Even
Length
Odd
a, b
a, b
12

DFA
alphabets {a, b} where the string length is not
divisible by two.
• We will get 0 or 1 after diving the length of a string
by two. That is why we need two state.
• If string length is 3 then we will get 0 or 1 or 2 as
remainder after dividing any length by 3. So we will
be requiring three state in that case.
A B
a, b
Length
Even
Length
Odd
a, b
13

DFA
divisible by three.
– 𝜔 𝑚𝑜𝑑 3 = 0
– 𝐿 = {𝜖, 𝑎𝑎𝑎, … … … , 𝑏𝑏𝑏, 𝑎𝑎𝑎𝑎𝑎𝑎, … … … , 𝑏𝑏𝑏𝑏𝑏𝑏, … … … … … … }
A B
a, b
C
a, b
Length
0,3,6,…
Length
1,4,7,…
Length
2,5,8,…
a, b
14

DFA
• Construct a DFA where 𝜔 ≅ 1 𝑚𝑜𝑑 3
• 𝜔 ≅ 1 𝑚𝑜𝑑 3 means 𝜔 𝑚𝑜𝑑 3 = 1
• For 𝜔 𝑚𝑜𝑑 𝑛 = 0 𝑜𝑟 𝜔 ≅ 𝑚 𝑚𝑜𝑑 n (m is any number)
in general for minimal DFA we need 𝑛 number of states.
A
a, b
C
a, b
Length
0,3,6,…
Length
1,4,7,…
Length
2,5,8,…
a, b
B
15

DFA
the alphabets {a, b} where number of ‘a’ in
the string is two.
– 𝑛𝑎 𝜔 = 2
– 𝐿 = 𝑎𝑎, 𝑏𝑎𝑎, 𝑎𝑏𝑎, 𝑎𝑎𝑏, 𝑏𝑏𝑎𝑎, … … …
B
a
C
a
0 ‘a’s 1 ‘a’s 2 ‘a’s
A D
b b b
a
3 ‘a’s
a,b
16

DFA
the string at least two.
– 𝑛𝑎 𝜔 ≥ 2
B
a
C
a
0 ‘a’s 1 ‘a’s 2 ‘a’s
A
b b a, b
17

DFA
the string at most two.
– 𝑛𝑎 𝜔 ≤ 2
b
B
a
C
a
0 ‘a’s 1 ‘a’s 2 ‘a’s
A
b b
D
3 ‘a’s
a,b
a
18

DFA
the string is even.
– 𝑛𝑎 𝜔 𝑚𝑜𝑑 2 = 0
– 𝑛𝑎 𝜔 ≅ 0 𝑚𝑜𝑑 2
A B
a
Length
Even
Length
Odd
a
b
b
19

DFA
the string is odd.
– 𝑛𝑎 𝜔 𝑚𝑜𝑑 2 = 1
– 𝑛𝑎 𝜔 ≅ 1 𝑚𝑜𝑑 2
b
A B
a
Length
Even
Length
Odd
a
b
20

DFA
the string is divisible by 3.
– 𝑛𝑎 𝜔 𝑚𝑜𝑑 3 = 0
– 𝑛𝑎 𝜔 ≅ 0 𝑚𝑜𝑑 3
A B
a
a
b
b
C
a
b
21

DFA
• 𝑖𝑓𝑛𝑎 𝜔 ≅ 𝑘 𝑚𝑜𝑑 n , how to draw DFA??
– 𝑛 number of states will be there
• If k=0 𝑞0 will be final state
• So on
22

Assignment
1. Construct a DFA that accept all strings over the
alphabets {a, b} where number of ‘b’ in the string is
divisible by 5.
alphabets {a, b} such that 𝑛𝑎 𝜔 ≅ 2 𝑚𝑜𝑑 5
alphabets {a, b} such that 𝑛𝑏 𝜔 ≅ 1 𝑚𝑜𝑑 3
23

DFA
the alphabets {a, b} where number of ‘a’ and
number of ‘b’ in the string is even.
– 𝑛𝑎 𝜔 𝑚𝑜𝑑 2 = 0 & 𝑛𝑏 𝜔 𝑚𝑜𝑑 2 = 0
– 𝑛𝑎 𝜔 ≅ 0 𝑚𝑜𝑑 2 & 𝑛𝑏 𝜔 ≅ 0 𝑚𝑜𝑑 2
𝑛𝑎 𝜔 𝑛𝑏 𝜔
E E
O E
E O
O O
Four possible
states
representing this
situation
24

DFA
ee eo
oe oo
1. String 𝜖 state ee
2. String ‘a’ state oe
3. String ‘b’ state eo
4. String ‘aa’ state ee
5. String ‘ab’ state oo
6. String ‘ba’ state oo
7. String ‘bb’ state ee
8. String ‘aba’ or ‘baa’ state eo
9. String ‘bab’ or ‘abb’ state oe
a
b
a
b
a
b
a
b
25

Cross Product Method
• 𝑛𝑎 𝜔 𝑚𝑜𝑑 2 = 0 & 𝑛𝑏 𝜔 𝑚𝑜𝑑 2 = 0
𝐴, 𝐵 × 𝐶, 𝐷 = 𝐴𝐶, 𝐴𝐷, 𝐵𝐶, 𝐵𝐷
A B
a
a
b
b
C D
b
b
a
a
b
a
b
a
b
a
b
a
A B
C C
a
a
b
b
A A
C D
AC BC
AD BD
ee oe
oo
eo
Observation: ‘a’s are counting
in horizontal way and ‘b’s are
counting in vertical way. 26

alphabets {a, b} where number of ‘a’ is divisible by 3
and number of ‘b’ in the string is divisible by 2.
b
a
a
b
00 10
01 11
20
b
21
a
a
a
a
b
b
b
27

• Construct a DFA that accept all strings over the alphabets {a,
b} where number of ‘a’ and number of ‘b’ in the string is
divisible by 3.
– 𝑛𝑎 𝜔 𝑚𝑜𝑑 3 = 0 & 𝑛𝑏 𝜔 𝑚𝑜𝑑 3 = 0
– 𝑛𝑎 𝜔 ≅ 0 𝑚𝑜𝑑 3 & 𝑛𝑏 𝜔 ≅ 0 𝑚𝑜𝑑 3
b
a
a
b
A B
D E
C
b
F
a
a
a
a
b
a
b
G H
b
I
a
a
b
b
b
28

DFA
• Construct a DFA that accept all strings over the alphabets {a,
b} such that 𝑛𝑎 𝜔 𝑚𝑜𝑑 3 ≥ 𝑛𝑏 𝜔 𝑚𝑜𝑑 2
b
a
a
b
00 10
01 11
20
b
21
a
a
a
a
b
b
b
No. of ‘a’s mod 3 >
no. of ‘b’s mod 2
this will be a final
state
state
No. of ‘a’s mod 3 =
state
state
No. of ‘a’s mod 3 =
So this will be a
final state
No. of ‘a’s mod 3 <
this will not be a
final state
29

DFA
• Construct a minimal DFA which accepts all
strings over {0,1}, which when interpreted as
binary number is divisible by 2.
– Two remainders are possible, so two states
𝒒𝟎 𝒒𝟏
1
0
1
0
30

DFA
binary number is divisible by 3.
– Three remainders are possible, so three states
𝒒𝟎 𝒒𝟏
1
1
0
0
𝒒𝟐
0
1
If the question is divisible by ‘n’ then number of state is n.
31

Transition Table
0 1
𝑞0 𝑞0 𝑞1
𝑞1 𝑞2 𝑞0
𝑞2 𝑞1 𝑞2
q0 q1
q2 q0
q1 q2
32

Transition Table
binary number is divisible by 4
0 1
𝑞0 𝑞0 𝑞1
𝑞1 𝑞2 𝑞3
𝑞2 𝑞0 𝑞1
𝑞3 𝑞2 𝑞3
33

DFA
binary number is ≅ 1 𝑚𝑜𝑑 3.
𝒒𝟎 𝒒𝟏
1
1
0
0
𝒒𝟏
0
1
34

DFA
binary number is ≅ 2 𝑚𝑜𝑑 3.
𝒒𝟎 𝒒𝟏
1
1
0
0
𝒒𝟏
0
1
35

Transition Table
strings over {0,1,2}, which when interpreted as
ternary number is divisible by 4
0 1 2
𝑞0 𝑞0 𝑞1 𝑞2
𝑞1 𝑞3 𝑞0 𝑞1
𝑞2 𝑞2 𝑞3 𝑞0
𝑞3 𝑞1 𝑞2 𝑞3
36

DFA
• Construct a minimal DFA which accepts set of
all strings over {a, b} where each string starts
with an ‘a’.
all strings over {a, b} where each string
contains an ‘a’.
all strings over {a, b} where each string ends
with an ‘a’.
37

Assignment
1. Construct a DFA that accept all strings over the alphabets {a, b} where number of
‘a’ and number of ‘b’ is odd in the string.
‘a’ is even and number of ‘b’ is odd in the string .
‘a’ is odd and number of ‘b’ is even in the string .
‘a’ is multiple of three and number of ‘b’ is multiple of two in the string.
5. Construct a DFA that accept all strings over the alphabets {a, b} such that
𝑛𝑎 𝜔 𝑚𝑜𝑑 3 = 𝑛𝑏 𝜔 𝑚𝑜𝑑 2
𝑛𝑎 𝜔 𝑚𝑜𝑑 3 ≤ 𝑛𝑏 𝜔 𝑚𝑜𝑑 2
𝑛𝑎 𝜔 𝑚𝑜𝑑 3 = 1 and 𝑛𝑏 𝜔 𝑚𝑜𝑑 3 = 2
𝑛𝑎 𝜔 𝑚𝑜𝑑 3 > 𝑛𝑏 𝜔 𝑚𝑜𝑑 3
9. Construct a minimal DFA which accepts all strings over Octal number system,
which when interpreted as binary number ≅ 1 𝑚𝑜𝑑 5
38

DFA
all strings over {a, b} where each string start
with “ab”
– L= {ab, aba, abb, ………}
B C
A
a b
D
a, b
a
a, b
39
b

DFA
all strings over {a, b} where each string
contains “ab” as a sub string.
– L= {ab, aab, aba, bab, abb ………}
B C
A
a b
b a a,b
40

DFA
all strings over {a, b} where each string end
with “ab”
– L= {ab, aab, bab, abab ………}
B C
A
a b
b a
a
b
41

DFA
• Construct a minimal DFA which accepts set of all
strings over {a, b} where each string starts with ‘a’
and ends with ‘b’ or vice versa.
– L ={ab, ba, aab, abb, baa, bba. ……}
B C
A
a b
b
a b
a
D
a
E
b
b
42
a

DFA
• Construct a minimal DFA which accepts set of all
strings over {a, b} where each string starts and ends
with same symbol.
– L={𝜖, a, b, aa, bb, aba, bab,…………}
C
B
A
a b
b
a
b
a
E
a
D b
b
a
This DFA is complement of
previous DFA and language also.
We will get complement of a
language by removing one
language from 𝚺∗. 𝑳𝟐 = 𝚺∗ − 𝑳𝟏
L2 is complement of L1
43

Complement of DFA
A B
a
Length
Even
Length
Odd
a
b
b
A B
a
Length
Even
Length
Odd
a
b
L1 = Even number of ‘a’ L2 = Odd number of ‘a’
L3 = Starts with ‘a’ L4 = Not starts with ‘a’
A B
C
a
b
a,b
a,b
A B
C
a
b
a,b
a,b
44

Complement of DFA
• DFA can be defined by quintuples
– ( 𝑄 , Σ , 𝛿 , 𝑞 0 , 𝐹 )
• Complement of DFA can be defined by
quintuples
– ( 𝑄 , Σ , 𝛿 , 𝑞 0 , 𝑄 − 𝐹 )
45

DFA
all strings over {a, b} which accepts strings in
which every 'a' is followed by 'b'.
– L={𝜖, b, ab, bb, aba, bab, …………}
46
C
B
A a b
a
b
a
b
D
a,b

DFA
which every 'a' never followed by 'b'.
– This is not complement of previous
– L={𝜖, a, aa,…..,b, bb,……, ba, bba…………}
47
C
B
A
a b
a
a,b
b

DFA
which every 'a' is followed by ‘bb'.
48
D
B
A a b
a
b
a
b
E
a,b
C
b
a

DFA
which every 'a' never followed by ‘bb'.
49
D
B
A a b
a, b
a
b
C
b
a

DFA
• 𝐿 = 𝑎𝑛
𝑏𝑚
|𝑛, 𝑚 ≥ 1
– 𝐿 = {𝑎𝑏, 𝑎𝑎𝑏𝑏, 𝑎𝑎𝑎𝑏, 𝑎𝑏𝑏𝑏, … … … }
50
C
B
A a
b
b
a
D
a, b
b
a

DFA
• 𝐿 = 𝑎𝑛
𝑏𝑚
|𝑛, 𝑚 ≥ 0
– 𝐿 = {𝜖, 𝑎, 𝑎𝑎, … , 𝑎𝑎𝑏, 𝑎𝑏𝑏, … 𝑏, 𝑏𝑏, 𝑏𝑏𝑏 … … … }
51
C
B
A
b a
b a,b
a

DFA
• 𝐿 = 𝑎𝑛
𝑏𝑚
𝑐𝑙
|𝑛, 𝑚, 𝑙 ≥ 1
– 𝐿 = {𝑎𝑏𝑐, 𝑎𝑎𝑏𝑐, 𝑎𝑏𝑏𝑐, 𝑎𝑏𝑐𝑐, … … … }
52
D
B
A a
b, c
c
c
D
a, b, c
b
a
C
c
a
b
a, b

DFA
• 𝐿 = 𝑎𝑛
𝑏𝑚
𝑐𝑙
|𝑛, 𝑚, 𝑙 ≥ 0
– 𝐿 = {𝜖, 𝑎, 𝑏, 𝑐, 𝑎𝑏, 𝑎𝑐, 𝑏𝑐, 𝑎𝑏𝑐, 𝑎𝑎, 𝑏𝑏, 𝑐𝑐, … }
53
B
A b c
c
b
a
C
c
a
D
a, b, c
a, b

DFA
• 𝐿 = {𝑎3
𝑏𝑤𝑎3
|𝑤 ∈ {𝑎, 𝑏}∗
}
– 𝐿 = {𝑎3𝑏𝜖𝑎3, 𝑎3𝑏𝑎𝑎3, 𝑎3𝑏𝑏𝑎3, 𝑎3𝑏𝑎𝑎𝑎3, … … }
54
H
B
A a a
C D E F G
a b a a a
I
b
a, b
b b a
b
b
b
b
a

Operations of DFA
Different operations that we can apply on DFAs
are -
1. Union
2. Concatenation
3. Cross Product
4. Complement
5. Reversal
55

Union
• Union of two Regular language is also Regular
– L1 = starts with ‘a’ and ends with ‘b’
– L2 = starts with ‘b’ and ends with ‘a’
– 𝐿 = 𝐿1 ∪ 𝐿2 = starts with different symbol
• Let,
𝑴𝟏 = 𝑄1, Σ1, 𝛿1, 𝑞0
1, 𝐹1 DFA for L1 , 𝑴𝟐 = 𝑄2, Σ2, 𝛿2, 𝑞0
2, 𝐹2 DFA for L2
We want to construct M for 𝑳 = 𝐿1 ∪ 𝐿2
𝑴 = 𝑄, Σ, 𝛿, 𝑞0, 𝐹 DFA for L
Where,
𝑸 = 𝑄1 × 𝑄2
𝜮 = Σ1 ∪ Σ2
𝜹 𝑟1, 𝑟2 , 𝑎 = 𝛿1 𝑟1, 𝑎 , 𝛿2 𝑟2, 𝑎
𝒒𝟎= 𝑞0
1, 𝑞0
2
𝑭 = 𝑟1, 𝑟2 | 𝑟1 ∈ 𝐹1 𝑜𝑟 𝑟2 ∈ 𝐹2
56
𝑟1 ∈ 𝑄1 𝑎𝑛𝑑 𝑟2 ∈ 𝑄2

Union
57
B
A a b
b
a
C
b
D
a, b
a
B
A b a
a
b
C
a
D
a, b
b
b
B
A a b
b
a
C
a
B
b
a
a
b
C

Concatenation
• Concatenation of two Regular language is also Regular
– L1 = starts with ‘a’
– L2 = ends with ‘b’
– 𝐿 = 𝐿1 ∘ 𝐿2 = starts with ‘a’ and ends with ‘b’
𝐿1 = 𝑎, 𝑎𝑎, 𝑎𝑏, 𝑎𝑎𝑏, 𝑎𝑎𝑎, 𝑎𝑏𝑏, … …
𝐿2 = {𝑏, 𝑎𝑏, 𝑏𝑏, 𝑏𝑏𝑏, 𝑎𝑎𝑏, 𝑎𝑏𝑏, … … }
𝐿 = {𝑎𝑏, 𝑎𝑎𝑏, 𝑎𝑏𝑏, 𝑎𝑏𝑏𝑏, … … … }
• Let,
𝑴𝟏 = 𝑄1, Σ1, 𝛿1, 𝑞0
1
, 𝐹1 DFA for L1 , 𝑴𝟐 = 𝑄2, Σ2, 𝛿2, 𝑞0
2
, 𝐹2 DFA for L2
We want to construct M for 𝑳 = 𝐿1 ∘ 𝐿2
𝑴 = 𝑄, Σ, 𝛿, 𝑞0, 𝐹 DFA for L
58

Concatenation
59
A a
a, b
B
b
C
a, b
A b
b
B
a
a
B
A a b
b
a
C
b
D
a, b
a

Reversal
• L1= starts with ‘a’
– 𝐿1 = {a, aa, ab, aaa, aab, aba,……}
– 𝐿1𝑅 = {a, aa, ba, aaa, baa, aba,……}
• Steps:
– States are same as original
– Convert initial state to final state
– Convert final state to initial state
60
A a
a, b
B
b
C
a, b
A
a
a, b
B
b
C
a, b
This is not DFA it is NFA
If we reverse a DFA of a
language we will get a
reverse language, but we
may or may not get
reverse DFA. Reverse FA is
either DFA or NFA.

15159222.ppt

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