Equivalence with FA * Any Regex can be converted to FA and vice versa, because: * Regex and FA are equivalent…

Equivalence with FA* Any Regex can be converted to FA and vice versa, because:* Regex and FA are equivalent in their descriptive power** Regular language is one that is recognized by some FA, remember?Hence: (Lemma) if a language is described by a RegEx, then it is a regular language.See Proof: P.67, example 1.56

Lemma: If a language is regular, then it is descried by a RegEx.

Because A is regular, it is accepted by a DFA. We describe a procedure for converting DFAs into equivalent RegEx.

Breaking the procedure into 2 parts, using a new type of FA called Generalized Nondeterministic Finite Automaton (GNFA)

Hence, 1st convert DFA to GNFA, then GNFA to RegEx

GNFA is NFA, but Regex could be used as labels on arrows. Figure 1.61 P.70 shows an example of GNFA, which is on the special form that it should ALWAYS have.

To convert a NFA to GNFA (and keep the special form)

Add a new start state with ε arrow to the old start Add a new accept state with ε arrow coming from

the old accept state. If any arrow has more than one label, or there are

more than one arrow between two states on the same direction, replace with one arrow and “union” the labels

If two states have no arrows between them, add an arrow with Φ label

Convert the GNFA to RegEx In GNFA, the start and accept states are must,

and they're different, hence no. of states (k) of any GNFA is always >=2.

If k>2, then GNFA is constructed by reducing the number to k=2 (an arrow form start to accept states) whose label is the RegEx.

Example, converting a 3 state DFA to RegEx.

3k DFA > 5k GNFA > 4k GNFA > 3k GNFA > 2k GNFA > RegEx

Convert the GNFA to RegExThe Challenge in reducing the states if

greater than 2, by Ripping the a state out of the machine (other than the start, accept states), we call it qrip , we repair the machine by altering the removed qrip with labeled arrow. The label should cover the absence of qrip .The new label form start to accept states is the RegEx.

Figure 1.63 P. 72

Formal definition of GNFA Just like the DFA, it's a 5-tuple with same parameters

except for δ, which can be given as:

δ: (Q - {qaccept }) X (Q – {qstart}) → R

R is the collection of all RegEx over the alphabet ∑, The domain of the transition function is (Q - {qaccept }) X (Q – {qstart}) simply because in GNFA, an arrow connects every state to every other state, except no arrows are coming from qaccept or going to qstart.Read: P.73 - 76

Nonregular Languages Some languages cannot be recognized by any

FA (a limitation of FA) Assume the language B = {0n1n | n>=0}. No DFA

can recognize it because the resulting number of 0s is unlimited, and therefore no DFA can remember this number of possibilities.

But just because the language appears to require unbounded memory doesn't mean it's nonregular. So we need a method to prove the nonregularity

Nonregular LanguagesAssume we have 2 languages C and D, over

the alphabet ∑ = {0,1}, where:

C = { w | w has an equal no. of 0s and 1s}, andD = { w | w has an equal number of occurrences of

01 and 10 as substrings} Which one is nonregular? See problem 1.48>> Difficult? We need a mathematical proof for

certainty.

The pumping lemma for regular languages

Theorem (Pumping Lemma): “All regular languages have a special property, if a language doesn't have that property, it is not regular”.

The property states that all strings in the language can be “pumped” if they are at least as long as a certain special value called the “pumping length”. In other words:

Such string contains a section that can be repeated any number of times with the resulting string remaining in the language

Pumping lemma: if A is a regular language, then there is a number p (the pumping length) where, if s is any string in A of length at least p, then s may be divided into three pieces, s=xyz, satisfying the conditions:

1- for each i>=0, xyiz A∈2- |y| >0, and

3- |xy| <= p.

Figure 1.71 P.78

Fall 2006 Costas Busch - RPI 12

Regular languages

ba* acb *

...etc*)( bacb

Non-regular languages }0:{ nba nn

}*},{:{ bavvvR


How can we prove that a language Lis not regular?

Prove that there is no DFA or NFA or RE that accepts L

Difficulty: this is not easy to prove(since there is an infinite number of them)

Solution: use the Pumping Lemma !!!


The Pigeonhole Principle

pigeons

pigeonholes

4

3


A pigeonhole mustcontain at least two pigeons


...........

...........

pigeons

pigeonholes

n

m mn



...........

pigeonspigeonholesnm

mn There is a pigeonhole with at least 2 pigeons



and

DFAs


Consider a DFA with states 4

1q 2q 3qa

b

4q

b

a b

b

a a


Consider the walk of a “long’’ string:

1q 2q 3qa

b

4q

b

b

b

a a

a

aaaab

1q 2q 3q 2q 3q 4qa a a a b

A state is repeated in the walk of

(length at least 4)

aaaab


aaaab

1q 2q 3q 2q 3q 4qa a a a b

1q 2q 3q 4q

Pigeons:

Nests:(Automaton states)

Are more than

Walk of

The state is repeated as a result of the pigeonhole principle

(walk states)

Repeatedstate


Consider the walk of a “long’’ string:

1q 2q 3qa

b

4q

b

bb

a a

a

aabb

1q 2q 3q 4q 4qa a b b

A state is repeated in the walk of

(length at least 4)Due to the pigeonhole principle:

aabb


aabb

1q 2q 3q 4qAutomaton States

Pigeons:


Are more than

Walk of

The state is repeated as a result of the pigeonhole principle

(walk states)1q 2q 3q 4q 4qa a b b

Repeatedstate


iq...... ......Repeated state

kw 21

1 2 k

Walk of

iq.... iq.... ....1 2 ki j1i 1j

Arbitrary DFA

DFA of states#|| wIf ,by the pigeonhole principle,a state is repeated in the walk

In General:

1q zq

zq1q

w


mDFA of states#|| w

1q 2q 1mq mq

Walk of wPigeons:


Are morethan

(walk states)iq....

iq.... ....1q

iq.... ....

zq

A state is repeated


The Pumping Lemma


Take an infinite regular language L

There exists a DFA that accepts L

mstates

(contains an infinite number of strings)


mw ||(number of states of DFA)

then, at least one state is repeated in the walk of w

q...... ......1 2 k

Take string with Lw

kw 21Walk in DFA of

Repeated state in DFA


Take to be the first state repeatedq

q....

w

There could be many states repeated

q.... ....

Second occurrence

First occurrence

Unique states

One dimensional projection of walk :

1 2 ki j1i 1j


q.... q.... ....

Second occurrence

First occurrence

1 2 ki j1i 1j

wOne dimensional projection of walk :

ix 1 jiy 1 kjz 1

xyzwWe can write


zyxw

q... ...

x

y

z

In DFA:

...

...

1 ki1ij

1j

contains only first occurrence of q


Observation: myx ||length numberof statesof DFA

Since, in no state is repeated (except q)

xy

Unique States

q...

x

y

...

1 i1ij


Observation: 1|| ylength

Since there is at least one transition in loop

q

y

...

1ij


We do not care about the form of string z

q...

x

y

z

...

zmay actually overlap with the paths of and x y


The string is accepted

zxAdditional string:

q... ...

x z

...

Do not follow loopy

...

1 ki1ij

1j



zyyx

q... ... ...

x z

Follow loop2 times

Additional string:

y

...

1 ki1ij

1j



zyyyx

q... ... ...

x z

Follow loop3 times

Additional string:

y

...

1 ki1ij

1j



zyx iIn General:...,2,1,0i

q... ... ...

x z

Follow loop timesi

y

...

1 ki1ij

1j


Lzyx i Therefore: ...,2,1,0i

Language accepted by the DFA

q... ... ...

x z

y

...

1 ki1ij

1j


In other words, we described:

The Pumping Lemma !!!


The Pumping Lemma:• Given a infinite regular language L

• there exists an integer m

• for any string with length Lw mw ||

• we can write zyxw

• with andmyx || 1|| y

• such that: Lzyx i ...,2,1,0i

(critical length)


In the book:

Critical length = Pumping lengthm p


Applications

of

the Pumping Lemma


Observation:Every language of finite size has to be regular

Therefore, every non-regular languagehas to be of infinite size (contains an infinite number of strings)

(we can easily construct an NFA that accepts every string in the language)


Suppose you want to prove thatAn infinite language is not regular

1. Assume the opposite: is regular

2. The pumping lemma should hold for

3. Use the pumping lemma to obtain a contradiction

L

L

L

4. Therefore, is not regular L


Explanation of Step 3: How to get a contradiction

2. Choose a particular string which satisfies the length condition

Lw

3. Write xyzw

4. Show that Lzxyw i for some 1i

5. This gives a contradiction, since from pumping lemma Lzxyw i

mw ||

1. Let be the critical length for m L


Note: It suffices to show that only one stringgives a contradiction

Lw

You don’t need to obtaincontradiction for every Lw


Theorem: The language }0:{ nbaL nn

is not regular

Proof: Use the Pumping Lemma

Example of Pumping Lemma application


Assume for contradictionthat is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

}0:{ nbaL nn


Let be the critical length for

Pick a string such that: w Lw

mw ||and length

mmbawWe pick

m

}0:{ nbaL nn

L


with lengths

From the Pumping Lemma:

1||,|| ymyx

babaaaaabaxyz mm ............

mkay k 1,

x y z

m m

we can write zyxbaw mm

Thus:

w


From the Pumping Lemma: Lzyx i

...,2,1,0i

Thus:

mmbazyx

Lzyx 2

mkay k 1,


From the Pumping Lemma:

Lbabaaaaaaazxy ...............2

x y z

km m

Thus:

Lzyx 2

mmbazyx

y

Lba mkm

mkay k 1,


Lba mkm

}0:{ nbaL nnBUT:

Lba mkm

CONTRADICTION!!!

1≥k


Our assumption thatis a regular language is not true

L

Conclusion:L is not a regular language

Therefore:

END OF PROOF


Regular languages

Non-regular language }0:{ nba nn

)( **baL

Read

P. 80 – 82 (examples)Solve Exercises and Problems (P. 83 – 98) Prepare for a quiz!

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Equivalence with FA * Any Regex can be converted to FA and vice versa, because: * Regex and FA are equivalent…