CS计算机代考程序代写 algorithm FIT2014 Theory of Computation Lecture 8 Kleene’s Theorem. I. Regexp -3mu NFA -3mu FA

FIT2014 Theory of Computation Lecture 8 Kleene’s Theorem. I. Regexp -3mu NFA -3mu FA

Monash University
Faculty of Information Technology

FIT2014 Theory of Computation

Lecture 8
Kleene’s Theorem. I.

Regexp −→ NFA −→ FA

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Overview

I Questions

I Kleene’s Theorem

I Convert Regular Expressions to NFA

I Convert NFA to FA

I Next lecture:
Convert FA to Regular Expression

Kleene (1909–1994)
https://mathshistory.st-andrews.

ac.uk/Biographies/Kleene/

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https://mathshistory.st-andrews.ac.uk/Biographies/Kleene/
https://mathshistory.st-andrews.ac.uk/Biographies/Kleene/

Questions
I Can every language which is represented by a regular expression

be described by a finite automaton?

I Can every language which is described by a finite automaton
be represented by a regular expression?

I Can every language be represented by a regular expression or a finite automaton?

{ all languages } representable by
regexp

?

recognised by
FA

?
?

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Kleene’s Theorem

Theorem.
Any language which can be defined by

I Regular Expressions

I Finite Automata

I Nondeterministic Finite Automata (NFA)

I Generalized Nondeterministic Finite Automata (GNFA)

can be defined by any of the other methods.

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Kleene’s Theorem

Regular Expression NFA

Finite AutomatonGNFA

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Converting Regular Expression to NFA

Start with:

Regular Expression

Apply the following rules until all edges are labelled with a letter or ε:

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Converting Regular Expression to NFA

∅ (R)

R

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Converting Regular Expression to NFA

RS

R S

R ∪ S

R

S

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Converting Regular Expression to NFA

R∗

ε ε

R

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Converting Regular Expression to NFA

The two ε transitions
are necessary,
in general.

Here, you can’t
match PR∗Q
or SR∗T ,
in general.

So the R loop
cannot be at
left node or
right node.

P

Q

S

T

P

Q

S

T

R∗

ε ε

R

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Converting Regular Expression to NFA. Example: a((ab)∗ ∪ (ba))b∗

a((ab)∗ ∪ (ba))b∗

a (ab)
∗ ∪ (ba) b∗

a

(ab)∗

ba

b∗

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a

(ab)∗

b a
b∗

a
ε ε

ab

b a
b∗

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a
ε ε

ab

b a
ε ε

b

a
ε ε

b a
ε ε

ba b

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Converting a NFA to a FA

Complexity?

How reversible is this construction?

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Kleene’s Theorem

Regular Expression NFA

Finite AutomatonGNFA

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Converting a NFA to a FA

In a FA:

I Any string w traces a unique path, starting from the Start State and ending at
some unique state, which we’ll call endState(w).

I The string w is accepted if endState(w) is a Final State, otherwise it is rejected.

I endState(ε) = Start State.

In a NFA:

I Any string w traces a set of paths, starting from the Start State and ending at
some set of states, which we’ll call endStates(w).

I The set might have zero, one or more members.

I The string w is accepted if endStates(w) contains a Final State, otherwise it is
rejected.

I endStates(ε) = { Start State } if there are no ε transitions.

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Converting a NFA to a FA

1 2 3

a,b

b a,b

endStates(ab) = {1, 2}
endStates(aba) = {1, 3}

In general, if w is a string and x is a single letter, then

endStates(wx) = {q : for some state p ∈ endStates(w), there is a transition p x−→ q}
. . . provided there are no empty transitions.

This suggests part of a method for constructing endStates(w) for all strings w .
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Converting a NFA to a FA
Idea:

sets of states in the NFA −→ states in the FA.
Informally (and assuming no empty transitions, for the time being):

Start with the one-element set { Start State }.
I This is endStates(ε).
I It’s the set of NFA-states we can possibly be in at the very start.

Construct endStates(a), the set of all states we could then get to by reading a single a.
Construct endStates(b), the set of all states we could then get to by reading a single b.

For each set of states, X , that we construct:
I find the set of states we can get to from X , by reading a single a.
I find the set of states we can get to from X , by reading a single b.

Keep doing this, until we no longer get any new sets of states.
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Converting a NFA to a FA

1 2 3

a,b

b a,b

state a b

Start {1}
state a b

Start {1} {1} {1,2}
{1,2}

state a b

Start {1} {1} {1,2}
{1,2} {1,3} {1,2,3}
{1,3}
{1,2,3}

· · ·

state a b

Start {1} {1} {1,2}
{1,2} {1,3} {1,2,3}
{1,3} {1} {1,2}
{1,2,3}

· · · · · ·

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Converting a NFA to a FA

1 2 3

a,b

b a,b

· · ·

state a b

Start {1} {1} {1,2}
{1,2} {1,3} {1,2,3}
{1,3} {1} {1,2}
{1,2,3} {1,3} {1,2,3}

state a b

Start {1} {1} {1,2}
{1,2} {1,3} {1,2,3}

Final {1,3} {1} {1,2}
Final {1,2,3} {1,3} {1,2,3}

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Converting a NFA to a FA

1 2 3

a,b

b a,b

state a b

Start {1} {1} {1,2}
{1,2} {1,3} {1,2,3}

Final {1,3} {1} {1,2}
Final {1,2,3} {1,3} {1,2,3}

{1}

{1,2}

{1,3}

{1,2,3}

a

b

a

b

a

b

a

b

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Algorithm: Conversion of NFA without empty transitions to FA

Input: a NFA

NextSetOfStatesOfNFA := { Start State of NFA }.

Create new incomplete row in FA table, for Start State called NextSetOfStatesOfNFA.
while the FA table still has at least one incomplete row do

CurrentStateInFA := the state for the first incomplete row of the FA.
for each letter x in the alphabet do

NextSetOfStatesOfNFA :=
{q : for some NFA-state p in CurrentStateInFA, ∃ transition p x−→ q}

Write NextSetOfStatesOfNFA in table entry for row CurrentStateInFA, column x .
if NextSetOfStatesOfNFA is new then

Create new incomplete row in table, using set NextSetOfStatesOfNFA as state.

Any FA state which (as a set) contains an NFA Final State is labelled Final.
Output: the FA 22 / 28

Converting a NFA to a FA

Now suppose that the NFA might have empty transitions, q1
ε−→ q2.

These allow change of state without reading any letter of the input string.

Every time we include a new state q in NextSetOfStatesOfNFA, we also need to
include any state we can reach from it along empty transitions.

Look at all paths from q that just use ε transitions . . .

q
ε−→ q1

ε−→ q2
ε−→ · · · · · · ε−→ qi

. . . and include all states on such paths.

Modify earlier algorithm, for constructing the sets of NFA states, to take account of
empty transitions.

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Algorithm: Conversion of NFA to FA

Input: a NFA

NextSetOfStatesOfNFA := { Start State of NFA }.
for each q ∈ NextSetOfStatesOfNFA do

Add, to NextSetOfStatesOfNFA, all states reachable from q along ε-transitions.
Create new incomplete row in FA table, for Start State called NextSetOfStatesOfNFA.
while the FA table still has at least one incomplete row do

CurrentStateInFA := the state for the first incomplete row of the FA.
for each letter x in the alphabet do

NextSetOfStatesOfNFA :=
{q : for some NFA-state p in CurrentStateInFA, ∃ transition p x−→ q}

for each q ∈ NextSetOfStatesOfNFA do
Add, to NextSetOfStatesOfNFA, all states reachable from q along ε-transitions.

Write NextSetOfStatesOfNFA in table entry for row CurrentStateInFA, column x .
if NextSetOfStatesOfNFA is new then

Create new incomplete row in table, using set NextSetOfStatesOfNFA as state.

Any FA state which (as a set) contains an NFA Final State is labelled Final.
Output: the FA 24 / 28

Converting a NFA to a FA

1 2 3

a

ε

b

ε

state a b

Start {1,2,3}
state a b

Start {1,2,3} {1,2,3} {2,3}
{2,3}

state a b

Start {1,2,3} {1,2,3} {2,3}
{2,3} ∅ {2,3}

state a b

Start {1,2,3} {1,2,3} {2,3}
{2,3} ∅ {2,3}
∅ ∅ ∅

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Converting a NFA to a FA

1 2 3

a

ε

b

ε

state a b

Start {1,2,3} {1,2,3} {2,3}
{2,3} ∅ {2,3}
∅ ∅ ∅ {1,2,3} {2,3} ∅

a

b

b

a

a,b

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Converting a NFA to a FA

Complexity?

I Think about how many states the FA may have,
as a function of the number of states in the NFA.

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Revision

Today:

I Understand Kleene’s Theorem

I Be able to convert Regular Expression −→ NFA
I Be able to convert NFA −→ Finite Automaton

Next lecture:

I Be able to convert FA −→ Regular Expression

Reading:
Sipser, Ch 1, especially pp. 54–58, 66–69.

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