1

Lecture 18

• Closure Properties of Language class LFSA– Remember ideas used in solvable languages

unit– Set complement– Set intersection, union, difference, symmetric

difference

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LFSA is closed under set complement

• If L is in LFSA, then Lc is in LFSA• Proof

– Let L be an arbitrary language in LFSA

– Let M be the FSA such that L(M) = L• M exists by definition of L in LFSA

– Construct FSA M’ from M

– Argue L(M’) = Lc

– There exists an FSA M’ such that L(M’) = Lc

– Lc is in LFSA

3

Visualization

•Let L be an arbitrary language in LFSA•Let M be the FSA such that L(M) = L

•M exists by definition of L in LFSA

•Construct FSA M’ from M•Argue L(M’) = Lc

•Lc is in LFSA

LcL

LFSA

FSA’s

MM’

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Construct FSA M’ from M

• What did we do when we proved that REC, the set of solvable languages, is closed under set complement?

• Construct program P’ from program P– P’ is basically the same program as program P

– The only difference comes at the end

– If P is about to return no, P’ returns yes instead

– If P is about to return yes, P’ returns no instead

• Can we translate this to the FSA setting?

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Construct FSA M’ from M

• M = (Q, , q0, A, )• M’ = (Q’, ’, q’, A’, ’)

– M’ should say yes when M says no– M’ should say no when M says yes– How?

• Q’ = Q• ’ = • q’ = q0

• ’ = • A’ = Q-A

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Example

1

2

3

a

a

a

b

b

b

FSA M

1

2

3

a

a

a

b

b

b

FSA M’

Q’ = Q’ = q’ = q0

’ = A’ = Q-A

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Construction is an algorithm

• Set Complement Construction– Algorithm Specification

• Input: FSA M

• Output: FSA M’ such that L(M’) = L(M)c

– Comments• This algorithm can be in any computational model.

– It does not have to be (and typically is not) an FSA

• These set closure constructions are useful. – More on this later

Construction Algorithm

FSA M

FSA M’

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Construction is an algorithm

• Your algorithm must give a complete specification of M’ in terms of M– Example:

• Let input FSA M = (Q, , q0, A, )

• Output FSA M’ = (Q’, ’, q’, A’, ’) where – Q’ = Q– ’ = – q’ = q0

– ’ = – A’ = Q-A

• When describing such constructions, I will often only focus on the critical component, in this case A’ = Q-A, but you must specify all components of M’ in terms of M.

Construction Algorithm

FSA M

FSA M’

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LFSA closed under Set Intersection Operation

(also set union, set difference, and symmetric difference)

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LFSA closed under set intersection operation

• Let L1 and L2 be arbitrary languages in LFSA

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

– M1 and M2 exist by definition of L1 and L2 in LFSA

• Construct FSA M3 from FSA’s M1 and M2

• Argue L(M3) = L1 intersect L2

• There exists FSA M3 s.t. L(M3) = L1 intersect L2

• L1 intersect L2 is in LFSA

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Visualization

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

•M1 and M2 exist by definition of L1 and L2 in LFSA

•Construct FSA M3 from FSA’s M1 and M2

•Argue L(M3) = L1 intersect L2

•There exists FSA M3 s.t. L(M3) = L1 intersect L2

•L1 intersect L2 is in LFSA

L1 intersect L2

L1

L2

LFSA

M3

M1

M2

FSA’s

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Algorithm Specification

• Input– Two FSA’s M1 and M2

• Output– FSA M3 such that L(M3) = L(M1) intersection L(M2)

FSA M1

FSA M2

FSA M3Alg

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Use Old Ideas

• Key concept: Try ideas from previous closure property proofs

• Example– How did the algorithm for the intersection of two

solvable languages is a solvable language work?• Run both input programs on the input string

• Say yes only if both say yes

– Try to create an FSA M3 that simultaneously runs M1 and M2 on the input string

FSA M1

FSA M2

FSA M3Alg

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1

Run M1 and M2 Simultaneously

0 20

0

0

0

1

1

1

1

M1

A B0

1 0,1

M2

,A 0,A 1,A 2,A

,B 0,B 1,B 2,B

M3

What happens when M1 and M2 run on input string 11010?

FSA M1

FSA M2

FSA M3Alg

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Construction

• Input – FSA M1 = (Q1, , q1, , A1)

– FSA M2 = (Q2, , q2, , A2)

• Output– FSA M3 = (Q3, , q3, , A3)

– What is Q3?

• Q3 = Q1 X Q2 where X is cartesian product

• In this case, Q3 = {(,A), (,B), (0,A), (0,B), (1,A), (1,B), (2,A), (2,B)}

– What is 3?

• 3 = 1 = 2

• In this case, 3 = {0,1}

1 0 20

000

1

11

1

M1

A B0

1 0,1

M2

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Construction

• Input – FSA M1 = (Q1, , q1, , A1)

– FSA M2 = (Q2, , q2, , A2)

• Output– FSA M3 = (Q3, , q3, , A3)

– What is q3?

• q3 = (q1, q2)

• In this case, q3 = (,A)

– What is A3?

• A3 = {(p, q) | p in A1 and q in A2}

• In this case, A3 = {(0,B)}

1 0 20

000

1

11

1

M1

A B0

1 0,1

M2

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Construction

• Input – FSA M1 = (Q1, , q1, , A1)

– FSA M2 = (Q2, , q2, , A2)

• Output– FSA M3 = (Q3, , q3, , A3)

– What is 3?

• For all p in Q1, q in Q2, a in , 3((p,q),a) = (1(p,a),2(q,a))

• In this case,– 3((0,A),0) = (1(0,0),2(A,0))

– = (0,B)

– 3((0,A),1) = (1(0,1),2(A,1))

– = (1,A)

1 0 20

000

1

11

1

M1

A B0

1 0,1

M2

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Example Summary

1 0 200

01

1

1

M1

A B0

1 0,1

M2

,A 0,A 1,A 2,A

,B 0,B 1,B 2,B

M3

0 1

0

1

0

1

0

1

0

1

0

1 0

1

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Observation

• Input – FSA M1 = (Q1, , q1, , A1)

– FSA M2 = (Q2, , q2, , A2)

• Output– FSA M3 = (Q3, , q3, , A3)

– What is A3?

• A3 = {(p, q) | p in A1 and q in A2}

• What if operation were different?– Set union, set difference, symmetric difference

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Observation continued

• Input – FSA M1 = (Q1, , q1, , A1)

– FSA M2 = (Q2, , q2, , A2)

• Output– FSA M3 = (Q3, , q3, , A3)

– What is A3?

• Set intersection: A3 = {(p, q) | p in A1 and q in A2}

• Set union: A3 = {(p, q) | p in A1 or q in A2}

• Set difference: A3 = {(p, q) | p in A1 and q not in A2}

• Symmetric difference: A3 = {(p, q) | (p in A1 and q not in A2) or (p not in A1 and q in A2) }

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Observation conclusion

• LFSA is closed under– set intersection

– set union

– set difference

– symmetric difference

• The constructions used to prove these closure properties are essentially identical

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Comments• You should be able to execute this algorithm

– Convert two FSA’s into a third FSA with the correct properties.

• You should understand the idea behind this algorithm– The third FSA essentially runs both input FSA’s simultaneously on

any input string

– How we set A3 depending on the specific set operation

• You should understand the importance of this algorithm– Design tool

• Suppose you need to build an FSA to accept some language L3

• You observe L3 = L1 intersection L2

• You already have or can easily build FSA’s to accept L1 and L2

• Use this algorithm on those FSA’s to constrct an FSA to accept L3

• You should be able to construct new algorithms for new closure property proofs

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Comparison

L1 intersect L2

L1

L2

LFSA

M3

M1

M2

FSA’s

LFSAREC

FSA’s C++ Programs

L L

M P