CSE 311 Lecture 26: Limitations of DFAs, NFAs, and Regular ...CSE 311 Lecture 26: Limitations of DFAs, NFAs, and Regular Expressions Emina Torlak and Kevin Zatloukal 1

CSE 311 Lecture 26: Limitations of DFAs, NFAs,and Regular ExpressionsEmina Torlak and Kevin Zatloukal

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture26.html?print-pdf#/

TopicsFrom NFAs to DFAs

A quick wrap-up of .DFAs NFAs regular expressions

They are all the same and represent regular languages.Languages and representations

Regular, context-free, and other languages.Proving irregularity

A proof template for showing that a language is not regular.

Lecture 25≡ ≡

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture25.html


From NFAs to DFAsA quick wrap-up of .Lecture 25

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NFAs and DFAs

Every DFA is an NFA.A DFA is an NFA that satisfies more constraints.

4


NFAs and DFAs


TheoremFor every NFA there is a DFA that recognizes exactly the same language.

4


NFAs and DFAs



Proof (and algorithm) idea:The DFA constructed for an NFA keeps track of all the states that a prefixof an input string can reach in the NFA.

4


NFAs and DFAs



Proof (and algorithm) idea:The DFA constructed for an NFA keeps track of all the states that a prefixof an input string can reach in the NFA.So there will be one state in the DFA for each subset of the states of theNFA that can be reached by some string.

4


NFAs and DFAs



Proof (and algorithm) idea:The DFA constructed for an NFA keeps track of all the states that a prefixof an input string can reach in the NFA.So there will be one state in the DFA for each subset of the states of theNFA that can be reached by some string.We’ll see how to construct the start state, remaining states andtransitions, and the final states of the DFA.

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NFAs to DFAs: the start state

The start state of the DFA represents the following set of states in the NFA:All states reachable from the start state of the NFA using only edges.ε

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The start state of the DFA represents the following set of states in the NFA:All states reachable from the start state of the NFA using only edges.

NFA

a bε

eε

fε

c

1

ε

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The start state of the DFA represents the following set of states in the NFA:All states reachable from the start state of the NFA using only edges.

NFA

a bε

eε

fε

c

1

DFA

a, b, e, f

ε

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NFAs to DFAs: states and transitions

Repeat until fixed point:

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Repeat until fixed point:Let be a state of the DFA corresponding to a set of the NFA states.DQ Q

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Repeat until fixed point:Let be a state of the DFA corresponding to a set of the NFA states.Let be a symbol for which has no outgoing edge.

DQ Qa ∈ Σ DQ

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Repeat until fixed point:Let be a state of the DFA corresponding to a set of the NFA states.Let be a symbol for which has no outgoing edge.Let be the (possibly empty) set of NFA states reachable from some statein by following one edge and zero or more edges.

DQ Qa ∈ Σ DQT

Q a ε

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Repeat until fixed point:Let be a state of the DFA corresponding to a set of the NFA states.Let be a symbol for which has no outgoing edge.Let be the (possibly empty) set of NFA states reachable from some statein by following one edge and zero or more edges.Add a state to the DFA, if not included, that represents the set .

DQ Qa ∈ Σ DQT

Q a εDT T

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Repeat until fixed point:Let be a state of the DFA corresponding to a set of the NFA states.Let be a symbol for which has no outgoing edge.Let be the (possibly empty) set of NFA states reachable from some statein by following one edge and zero or more edges.Add a state to the DFA, if not included, that represents the set .Add an edge labeled from to .

DQ Qa ∈ Σ DQT

Q a εDT T

a DQ DT

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NFA

b

eε

fεd1

1

c1

1gε

DFA

b, e, f

DQ Qa ∈ Σ DQT

Q a εDT T

a DQ DT

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NFA

b

eε

fεd1

1

c1

1gε

DFA

b, e, f c, d, e, g1

DQ Qa ∈ Σ DQT

Q a εDT T

a DQ DT

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NFAs to DFAs: final states

The final states of the DFA:Every DFA state that represents a set of NFA states containing a final state.

NFA

c

a b

e

DFA

a, b, c, e

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Example: NFA to DFA conversion

NFA

a b

ε

c1

0

0, 1

0

DFA

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

8



NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

1

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

1

0

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

1

0

1

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

1

0

1

0

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

1

0

1

0∅

1

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NFA

a b

ε

c1

0

0, 1

0

DFA

a, b

0

c1

b, c0

b1

a, b, c0

1

0

1

0∅

1

0, 1

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Exponential blow-up in simulating nondeterminism

In general the DFA might need a state for every subset of states of the NFA.Power set of the set of states of the NFA.-state NFA yields DFA with up to states.

We saw an example of this worst case outcome.n 2n

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Exponential blow-up in simulating nondeterminism

In general the DFA might need a state for every subset of states of the NFA.Power set of the set of states of the NFA.-state NFA yields DFA with up to states.

We saw an example of this worst case outcome.n 2n

The famous “P=NP?” question asks whether a similar blow-up is alwaysnecessary to get rid of nondeterminism for polynomial-time algorithms.

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DFAs NFAs regular expressionsThey are all the same and represent regular languages.

≡ ≡

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Equivalence of DFAs, NFAs, and regular expressions

We have shown how to build an optimal DFA for every regular expression.Build an NFA.Convert the NFA to a DFA using the subset construction.Minimize the resulting DFA.

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TheoremA language is recognized by a DFA (or NFA) if and only if it has a regularexpression.

You need to know this fact but we won’t ask you anything about the “onlyif” direction from DFAs/NFAs to regular expressions.

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TheoremA language is recognized by a DFA (or NFA) if and only if it has a regularexpression.

Languages represented by NFAs, DFAs, and regular expressions arecalled regular languages.

You need to know this fact but we won’t ask you anything about the “onlyif” direction from DFAs/NFAs to regular expressions.

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Languages and representationsRegular, context-free, and other languages.

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A hierarchy of languages and representations

Finite{010,11, 21}

Regular0*1* DFA

NFARegex

Context-FreeCFG

All

S → 0S1 | ε

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Finite{010,11, 21}

Regular0*1* DFA

NFARegex

Context-FreeCFG

All

S → 0S1 | ε

In HW8, you’ll (almost :) prove that regular languages are context-free. Howdo we prove that all finite languages are regular?

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Finite{010,11, 21}

Regular0*1* DFA

NFARegex

Context-FreeCFG

All

S → 0S1 | ε

In HW8, you’ll (almost :) prove that regular languages are context-free. Howdo we prove that all finite languages are regular? By showing how toconstruct a DFA/NFA/RegEx for every finite language!

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Converting a finite language to a regular expression

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Converting a finite language to a regular expression

Convert each string in the language to a regular expression.This is just each string itself.

Then put these regular expressions together using the operator.The resulting regular expression accepts exactly the strings in .

Example

L

∪L

{010, 11, 21} ⟶ 010 ∪ 11 ∪ 21

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Back to languages and representations …

Finite{010,11, 21}

Regular0*1* DFA

NFARegex

Context-FreeCFG

All

S → 0S1 | ε

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Back to languages and representations …

Finite{010,11, 21}

Regular0*1* DFA

NFARegex

Context-FreeCFG

All

S → 0S1 | ε

We saw in that the language of all binary palindromes can berepresented by a CFG. We also said that can’t be represented by anyregular expression. How would you prove that?

Lecture 21 BB

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Why isn’t (binary palindromes) a regular language?

If were regular, we could express it as a DFA/NFA/RegEx.Let’s choose a DFA as our hypothetical representation.

BB

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Now, recall that consists of infinitely many strings where is the reverse of and .

BB

B wvw⎯ ⎯⎯⎯

w⎯ ⎯⎯⎯ w v ∈ {ε, “0”, “1”}

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What would a DFA need to keep track of to decide ?

BB

B wvw⎯ ⎯⎯⎯

w⎯ ⎯⎯⎯ w v ∈ {ε, “0”, “1”}B

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What would a DFA need to keep track of to decide ?It would need to keep track of in order to check against it.

BB

B wvw⎯ ⎯⎯⎯

w⎯ ⎯⎯⎯ w v ∈ {ε, “0”, “1”}B

w w⎯ ⎯⎯⎯

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What would a DFA need to keep track of to decide ?It would need to keep track of in order to check against it.But there are infinitely many possible ’s and finitely many DFA states!

BB

B wvw⎯ ⎯⎯⎯

w⎯ ⎯⎯⎯ w v ∈ {ε, “0”, “1”}B

w w⎯ ⎯⎯⎯

w

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What would a DFA need to keep track of to decide ?It would need to keep track of in order to check against it.But there are infinitely many possible ’s and finitely many DFA states!

BB

B wvw⎯ ⎯⎯⎯

w⎯ ⎯⎯⎯ w v ∈ {ε, “0”, “1”}B

w w⎯ ⎯⎯⎯

w

This is the intuition for why is not regular. Let’s see how to turn thisintuition into a formal proof.

B

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A strategy for proving that is not regular

Proof by contradiction:Assume that is regular.Therefore, there is a DFA that recognizes .Show that accepts or rejects a string it shouldn’t.

B

BM B

M

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Key Idea 1If two string prefixes collide by reaching the samestate, a DFA can no longer distinguish their suffixes.

0a10b1

?

B

BM B

M

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0a10b1

?

Key Idea 2The machine has finitely many states, and sincethe strings in have infinitely many distinct prefixes,two of them must collide!

B

BM B

M

MB

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0a10b1

?

Key Idea 2The machine has finitely many states, and sincethe strings in have infinitely many distinct prefixes,two of them must collide!

B

BM B

M

MB

We choose an infinite set of prefixes. This choice mustensure that for every pair ofprefixes in ,there is a suffix such thatone of of is in butnot the other.

S

≠ ∈ Ssa sbt

t, tsa sb B

S= {1, 01, 001, 0001, …}= { 1 : n ≥ 0}0n

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Proving that is not regularSuppose that some DFA recognizes . We show accepts or rejectsa string it shouldn’t. Consider the set .

BM B M

S = { 1 : n ≥ 0}0n

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Since there are finitely many states in and infinitely many strings in , there exist strings and with that end in the

same state of .

BM B M

S = { 1 : n ≥ 0}0n

MS 1 ∈ S0a 1 ∈ S0b a ≠ b

M

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same state of .

BM B M

S = { 1 : n ≥ 0}0n

MS 1 ∈ S0a 1 ∈ S0b a ≠ b

MImportant: We don’t get to choose

anda! We just know they exist.b

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same state of .

Now, consider appending to both strings. 0a10b1

0a

BM B M

S = { 1 : n ≥ 0}0n

MS 1 ∈ S0a 1 ∈ S0b a ≠ b

M0a

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same state of .

Now, consider appending to both strings. 0a10b1

0a

Since and end in the same state, so do and , call it. But then must make a mistake: needs to be an accept state since

, but then would accept , which is an error.

BM B M

S = { 1 : n ≥ 0}0n

MS 1 ∈ S0a 1 ∈ S0b a ≠ b

M0a

10a 10b 0a10a 0b10a

q M q∈ B0a10a M ∉ B0b10a ◻

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Proving irregularityA proof template for showing that a language is not regular.

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A template for proving that a language is not regular

ɠ Suppose for contradiction that some DFA recognizes .

L

ML

20




ɡ Consider the set { }.

L

ML

S = …

must be infinite, andfor every pair ofprefixes ,there is a suffix suchthat one of of is in but not theother.

S

≠ ∈ Ssa sbtt, tsa sb

L

20





ɢ Since is infinite and has finitely many states,there must be two strings such that

and both end in the same state of .

L

ML

S = …S M

,sa sb ∈ S≠sa sb M


We don’t get to choose and !

S


L

sa sb

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ɣ Consider appending to both and .

L

ML

S = …S M


t sa sb



should be an acceptsuffix for but not .

S


L

sa sb

tsa sb

20








ɤ Since and end in the same state of , then and also end in the same state of . Since

and , does not recognize .

L

ML

S = …S M


t sa sb

sa sb Mtsa tsb q Mt ∈ Lsa t ∉ Lsb M L




S


L

sa sb

tsa sb

20








ɤ Since and end in the same state of , then and also end in the same state of . Since

and , does not recognize .

ɥ Since was arbitrary, no DFA recognizes .

L

ML

S = …S M


t sa sb

sa sb Mtsa tsb q Mt ∈ Lsa t ∉ Lsb M L

M L




S


L

sa sb

tsa sb

20


Example: prove that is not regularSuppose for contradiction that some DFA recognizes .

Consider the set

L = { : n ≥ 0}0n1n

M LS =

21



Consider the set .

L = { : n ≥ 0}0n1n

M LS = { : n ≥ 0}0n

21



Consider the set .

Since is infinite and has finitely many states, there must be twostrings with that both end in the same state of .

L = { : n ≥ 0}0n1n

M LS = { : n ≥ 0}0n

S M,0a 0b ∈ S a ≠ b M

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Consider the set .


Consider appending to both and .

L = { : n ≥ 0}0n1n

M LS = { : n ≥ 0}0n

S M,0a 0b ∈ S a ≠ b M

0a 0b

21



Consider the set .



L = { : n ≥ 0}0n1n

M LS = { : n ≥ 0}0n

S M,0a 0b ∈ S a ≠ b M

1a 0a 0b

21



Consider the set .



Since and end in the same state of , then and alsoend in the same state of . Since and , doesnot recognize .

L = { : n ≥ 0}0n1n

M LS = { : n ≥ 0}0n

S M,0a 0b ∈ S a ≠ b M

1a 0a 0b

0a 0b M 0a1a 0b1a

q M ∈ L0a1a ∉ L0b1a ML

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Consider the set .



Since and end in the same state of , then and alsoend in the same state of . Since and , doesnot recognize .

Since was arbitrary, no DFA recognizes .

L = { : n ≥ 0}0n1n

M LS = { : n ≥ 0}0n

S M,0a 0b ∈ S a ≠ b M

1a 0a 0b

0a 0b M 0a1a 0b1a

q M ∈ L0a1a ∉ L0b1a ML

M L

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Consider the set

L = { : n ≥ 0}(n)n

M LS =

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Consider the set .

L = { : n ≥ 0}(n)n

M LS = { : n ≥ 0}(n

22



Consider the set .


L = { : n ≥ 0}(n)n

M LS = { : n ≥ 0}(n

S M,(a (b ∈ S a ≠ b M

22



Consider the set .



L = { : n ≥ 0}(n)n

M LS = { : n ≥ 0}(n

S M,(a (b ∈ S a ≠ b M

(a (b

22



Consider the set .



L = { : n ≥ 0}(n)n

M LS = { : n ≥ 0}(n

S M,(a (b ∈ S a ≠ b M

)a (a (b

22



Consider the set .



Since and end in the same state of , then and also endin the same state of . Since and , does notrecognize .

L = { : n ≥ 0}(n)n

M LS = { : n ≥ 0}(n

S M,(a (b ∈ S a ≠ b M

)a (a (b

(a (b M (a)a (b)a

q M ∈ L(a)a ∉ L(b)a ML

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Consider the set .



Since and end in the same state of , then and also endin the same state of . Since and , does notrecognize .

Since was arbitrary, no DFA recognizes .

L = { : n ≥ 0}(n)n

M LS = { : n ≥ 0}(n

S M,(a (b ∈ S a ≠ b M

)a (a (b

(a (b M (a)a (b)a

q M ∈ L(a)a ∉ L(b)a ML

M L

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A fun fact about this proof methodSuppose that for a language , the set is a largest set of prefix stringswith the property that for every pair , there is some string such that one of is in but the other isn’t.

L S≠ ∈ Ssa sb t

t, tsa sb L

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If is infinite, then is not regular.

L S≠ ∈ Ssa sb t

t, tsa sb LS L

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If is infinite, then is not regular.

If is finite, then the minimal DFA for has precisely states, onereached by each member of .

L S≠ ∈ Ssa sb t

t, tsa sb LS LS L |S|

S

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SummaryDFAs NFAs regular expressions.

We’ve shown how to go from a regular expression to an NFA to a DFA.(And going from an NFA to a DFA can lead to exponential blow-up.)But we won’t show how to go from an NFA/DFA to a regular expression.

Finite regular context-free languages.To show that a language is regular, construct a DFA/NFA/RegEx for it.To show that it is not regular, use the proof method from this lecture.

≡ ≡

⊂ ⊂

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CSE 311 Lecture 26: Limitations of DFAs, NFAs, and Regular ...CSE 311 Lecture 26: Limitations of DFAs, NFAs, and Regular Expressions Emina Torlak and Kevin Zatloukal 1