1

Module 21

• Closure Properties for LFSA using NFA’s– From now on, when I say NFA, I mean any

NFA including an NFA- unless I add a specific restriction

– union (second proof)– concatenation– Kleene closure

2

LFSA closed under set union(again)

3

LFSA closed under set union

• Let L1 and L2 be arbitrary languages in LFSA

• Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2

– M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA

• Construct NFA M3 from NFA’s M1 and M2

• Argue L(M3) = L1 union L2

• There exists NFA M3 s.t. L(M3) = L1 union L2

• L1 union L2 is in LFSA

4

Visualization

L1 union L2

L1

L2

LFSA

M3

M1

M2

NFA’s

•Let L1 and L2 be arbitrary languages in LFSA•Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2

•M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA

•Construct NFA M3 from NFA’s M1 and M2

•Argue L(M3) = L1 union L2

•There exists NFA M3 s.t. L(M3) = L1 union L2

•L1 union L2 is in LFSA

5

Algorithm Specification

• Input– Two NFA’s M1 and M2

• Output– NFA M3 such that L(M3) = ?

NFA M1

NFA M2

NFA M3A

6

Use -transitionNFA M1

NFA M2

NFA M3A

a

M1

a,b a,b

a,b

M2

a

a,b a,b

a,b

M3

7

General Case *NFA M1

NFA M2

NFA M3A

M1

M2

M3

8

Construction *

• Input – NFA M1 = (Q1, , q1, , A1)

– NFA M2 = (Q2, , q2, , A2)

• Output– NFA M3 = (Q3, , q3, , A3)

– What is Q3?

• Q3 =

– What is 3?

• 3 = 1 = 2

– What is q3?

• q3 =

NFA M1

NFA M2

NFA M3A

9

Construction

• Input – NFA M1 = (Q1, , q1, , A1)

– NFA M2 = (Q2, , q2, , A2)

• Output– NFA M3 = (Q3, , q3, , A3)

– What is A3?

• A3 =

– What is 3?

• 3 =

NFA M1

NFA M2

NFA M3A

10

Comments

• You should be able to execute this algorithm

• You should understand the idea behind this algorithm

• You should understand how this algorithm can be used to simplify design

• You should be able to design new algorithms for new closure properties

• You should understand how this helps prove result that regular languages and LFSA are identical– In particular, you should understand how this is used to construct an

NFA M from a regular expression r s.t. L(M) = L(r)

– To be seen later

11

LFSA closed under set concatenation

12

LFSA closed under set concatenation

• Let L1 and L2 be arbitrary languages in LFSA

• Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2

– M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA

• Construct NFA M3 from NFA’s M1 and M2

• Argue L(M3) = L1 concatenate L2

• There exists NFA M3 s.t. L(M3) = L1 concatenate L2

• L1 concatenate L2 is in LFSA

13

Visualization

L1 concatenate L2

L1

L2

LFSA

M3

M1

M2

NFA’s

• Let L1 and L2 be arbitrary languages in LFSA

• Let M1 and M2 be NFA’s s.t. L(M1) = L1, L(M2) = L2

– M1 and M2 exist by definition of L1 and L2 in LFSA and the fact that every FSA is an NFA

• Construct NFA M3 from NFA’s M1 and M2

• Argue L(M3) = L1 concatenate L2

• There exists NFA M3 s.t. L(M3) = L1 concatenate L2

• L1 concatenate L2 is in LFSA

14

Algorithm Specification

• Input– Two NFA’s M1 and M2

• Output– NFA M3 such that L(M3) =

NFA M1

NFA M2

NFA M3A

15

Use -transitionNFA M1

NFA M2

NFA M3A

a

M1

a,b a,b

a,b

M2

a

a,b a,b

a,b

M3

16

General CaseNFA M1

NFA M2

NFA M3A

M1

M2

M3

17

Construction

• Input – NFA M1 = (Q1, , q1, , A1)

– NFA M2 = (Q2, , q2, , A2)

• Output– NFA M3 = (Q3, , q3, , A3)

– What is Q3?

• Q3 =

– What is 3?

• 3 = 1 = 2

– What is q3?

• q3 =

NFA M1

NFA M2

NFA M3A

18

Construction

• Input – NFA M1 = (Q1, , q1, , A1)

– NFA M2 = (Q2, , q2, , A2)

• Output– NFA M3 = (Q3, , q3, , A3)

– What is A3?

• A3 =

– What is 3?

• 3 =

NFA M1

NFA M2

NFA M3A

19

Comments

• You should be able to execute this algorithm

• You should understand the idea behind this algorithm

• You should understand how this algorithm can be used to simplify design

• You should be able to design new algorithms for new closure properties

• You should understand how this helps prove result that regular languages and LFSA are identical– In particular, you should understand how this is used to construct an

NFA M from a regular expression r s.t. L(M) = L(r)

– To be seen later

20

LFSA closed under Kleene Closure

21

LFSA closed under Kleene Closure

• Let L be arbitrary language in LFSA

• Let M1 be an NFA s.t. L(M1) = L

– M1 exists by definition of L1 in LFSA and the fact that every FSA is an NFA

• Construct NFA M2 from NFA M1

• Argue L(M2) = L1*

• There exists NFA M2 s.t. L(M2) = L1*

• L1* is in LFSA

22

Visualization

L1*

L1

LFSA

NFA’s

• Let L be arbitrary language in LFSA

• Let M1 be an NFA s.t. L(M1) = L

– M1 exists by definition of L1 in LFSA and the fact that every FSA is an NFA

• Construct NFA M2 from NFA M1

• Argue L(M2) = L1*

• There exists NFA M2 s.t. L(M2) = L1

*

• L1* is in LFSA

M1 M2

23

Algorithm Specification

• Input– NFA M1

• Output– NFA M2 such that L(M2) =

NFA M1 NFA M2A

24

Use -transition NFA M1 NFA M2A

a

M1a

M2

25

General Case * NFA M1 NFA M2A

M1

M2

26

Construction

• Input – NFA M1 = (Q1, , q1, , A1)

• Output– NFA M2 = (Q2, , q2, , A2)

– What is Q2?

• Q2 =

– What is 2?

• 2 = 1

– What is q2?

• q2 =

NFA M1 NFA M2A

27

Construction

• Input – NFA M1 = (Q1, , q1, , A1)

• Output– NFA M2 = (Q2, , q2, , A2)

– What is A2?

• A2 =

– What is 2?

• 2 =

NFA M1 NFA M2A

28

Comments

• You should be able to execute this algorithm • You should understand the idea behind this algorithm

– Why do we need to make an extra state p?

• You should understand how this algorithm can be used to simplify design

• You should be able to design new algorithms for new closure properties

• You should understand how this helps prove result that regular languages and LFSA are identical– In particular, you should understand how this is used to construct an NFA

M from a regular expression r s.t. L(M) = L(r)– To be seen later