1-Pumping Lemma (1)

8/8/2019 1-Pumping Lemma (1)

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The Pumping Lemma forRegular Languages

The Pumping Lemma forRegular Languages p.1/3


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Nonregular languages

Consider the language

.



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.

If we attempt to find a DFA that recognizes

we discover thatsuch a machine needs to remember how many

s have been

seen so far as it reads the input



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.



s have been


Because the number of

s isnt limited, the machine needs tokeep track of an unlimited number of possibilities



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.



s have been


Because the number of

s isnt limited, the machine needs tokeep track of an unlimited number of possibilities

This cannot be done with any finite number of states



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Intuition may fail us

Just because a language appears to require

unbounded memory to be recognized, it doesnt meanthat it is necessarily so



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Example:



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Example:

has an equal number of 0s and 1s



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Example:


not regular



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Example:


not regular

has equal no of 01 and 10 substrings



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Example:


not regular

has equal no of 01 and 10 substrings

regular



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Language nonregularity

The technique for proving nonregularity of some

language is provided by a theorem about regularlanguages called pumping lemma



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Pumping lemma states that all regular languages have

a special property



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Pumping lemma states that all regular languages have

a special property

If we can show that a language

does not have this

property we are guaranteed that

is not regular.



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Observation

Pumping lemma states that all regular languages have a

special property.



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Observation


special property.Pumping lemma does not state that only regular languages

have this property. Hence, the property used to prove that

a language

is not regular does not ensure that language

is

regular.



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Observation


special property.Pumping lemma does not state that only regular languages

have this property. Hence, the property used to prove that

a language

is not regular does not ensure that language

is

regular.

Consequence: A language may not be regular and still have

strings that have all the properties of regular languages.



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Pumping property

All strings in the language can be pumped" if they are at

least as long as a certain value, called the pumping length



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Pumping property

All strings in the language can be pumped" if they are at

least as long as a certain value, called the pumping length

Meaning: each such string in the language contains a sec-

tion that can be repeated any number of times with the re-sulting string remaining in the language.



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Theorem 1.70

Pumping Lemma: If

is a regular language, then there is

a pumping length

such that:



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Theorem 1.70

Pumping Lemma: If


a pumping length

such that:

If is any string in

of length at least ,



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Theorem 1.70

Pumping Lemma: If


a pumping length

such that:

If is any string in


Then may be divided into three pieces, ,

satisfying the following conditions:



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Theorem 1.70

Pumping Lemma: If


a pumping length

such that:

If is any string in




1. for each

,



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Theorem 1.70

Pumping Lemma: If


a pumping length

such that:

If is any string in




1. for each

,

2.



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Theorem 1.70

Pumping Lemma: If


a pumping length

such that:

If is any string in




1. for each

,

2.

3.



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Interpretation

Recall that

represents the length of string and

means that

may be concatenated

times, and



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Interpretation

Recall that


means that

may be concatenated

times, and

When , either or may be , but



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Interpretation

Recall that


means that

may be concatenated

times, and

When , either or may be , but

Without condition

theorem would be trivially true



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Proof idea

Let

be a DFA that recognizes



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Proof idea

Let


Assign a pumping length

to be the number of states of



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Proof idea

Let


Assign a pumping length

to be the number of states of

Show that any string

,

may be broken into three

pieces satisfying the pumping lemmas conditions



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More ideas

If

and

, consider a sequence of states that

goes through to accept

, example:



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More ideas

If

and



, example:

Since accepts , must be fi nal;if

then the

length of is



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More ideas

If

and



, example:


then the

length of is

Because

and

it result that

.



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More ideas

If

and



, example:


then the

length of is

Because

and

it result that

.

By pigeonhole principle:



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More ideas

If

and



, example:


then the

length of is

Because

and

it result that

.


- If p pigeons are placed into fewer than p holes, some holes musthold more than one pigeon



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More ideas

If

and



, example:


then the

length of is

Because

and

it result that

.


- If p pigeons are placed into fewer than p holes, some holes musthold more than one pigeon

the sequence must contain a repeated

state, see Figure 1



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Recognition sequence

Figure 1: State repeats when reads



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More ideas, continuation

Divide in to the three pieces: , , and



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Piece

is the part of

appearing before



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Piece is the part of appearing before

Piece is the part of between two appearances of



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Piece is the part of after the 2nd appearance of



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In other words:


M id i i


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In other words:

takes

from to ,


M id i i


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In other words:

takes

from to ,

takes

from to ,


M id ti ti


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In other words:

takes

from to ,

takes

from to ,

takes

from to


N t


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Note

The division specifi ed above satisfi es the 3 conditions


Ob ti


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Observations

Suppose that we run on


Ob ti


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Observations


Condition 1: it is obvious that

accepts , , and in

general

for all

. For

,

which is also

accepted because takes

to


Observations


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Observations



accepts , , and in

general

for all

. For

,

which is also


to

Condition 2: Since

, state is repeated. Then because is

the part between two successive occurrences of ,

.


Observations


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Observations



accepts , , and in

general

for all

. For

,

which is also


to

Condition 2: Since

, state is repeated. Then because is

the part between two successive occurrences of ,

.

Condition 3: makes sure that is the first repetition in the

sequence. Then by pigeonhole principle, the first

states in

the sequence must contain a repetition. Therefore,


Pumping lemmas proof


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Let

be a DFA that has states and

recognizes

. Let be a string over

of

length

. Let be the sequence of states

while processing , i.e.,

,




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Pumping lemma s proof

Let


recognizes


of

length



,

and among the first

elements in

two must be the same state, say .




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Let


recognizes


of

length



,

and among the first

elements in


Because occurs among the first

places in the sequence

starting at

, we have




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Let


recognizes


of

length



,

and among the first

elements in


Because occurs among the first

places in the sequence

starting at

, we have

Now let , , .


Note


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Note

As takes

from to , takes

from to , and takes

from to , which is an accept state,

must accept

,

for


Note


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Note

As takes

from to , takes

from to , and takes


must accept

,

for

We know that

, so

;



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Note


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Note

As takes

from to , takes

from to , and takes


must accept

,

for

We know that

, so

;

We also know that

, so

Thus, all conditions are satisfi ed and lemma is proven


Before using lemma


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Before using lemma

Note: To use this lemma we must also ensure that if the

property stated by the pumping lemma is true then the

language is regular.


Before using lemma


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Before using lemma




Proof: assuming that each element of language

satisfi es

the three conditions stated in pumping lemma we caneasily construct a FA that recognizes

, that is,

is regular.


Before using lemma


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Before using lemma




Proof: assuming that each element of language

satisfi es

the three conditions stated in pumping lemma we caneasily construct a FA that recognizes

, that is,

is regular.

Note: if only some elements of

satisfy the three conditions it does not

mean that

is regular.


Using pumping lemma (PL)


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Proving that a language

is not regular using PL:




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g p p g ( )



1. Assume that

is regular in order to obtain a contradiction




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g p p g ( )



1. Assume that


2. The pumping lemma guarantees the existence of a pumping

length s.t. all strings of length or greater in

can be pumped




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g p p g ( )



1. Assume that


2. The pumping lemma guarantees the existence of a pumping

length s.t. all strings of length or greater in

can be pumped

3. Find

,

, that cannot be pumped: demonstrate that

cannot be pumped by considering all ways of dividing into , , ,

showing that for each division one of the pumping lemma

conditions, (1)

, (2)

, (3)

, fails.


Observations


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The existence of contradicts pumping lemma, hence

cannot be regular


Observations


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The existence of contradicts pumping lemma, hence

cannot be regular

Finding sometimes takes a bit of creative thinking.

Experimentation is suggested


Applications


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pp

Example 1: prove that

is not regular


Applications


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Example 1: prove that

is not regular

Assume that

is regular and let

be the pumping length of

. Choose

; obviously

. By pumping

lemma such that for any

,


Example, continuation


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Consider the cases:




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Consider the cases:

1. consists of

s only. In this case has more

s than

s and

so it is not in

, violating condition 1




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Consider the cases:

1. consists of


s than

s and

so it is not in


2. consists of

s only. This leads to the same contradiction




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Consider the cases:

1. consists of


s than

s and

so it is not in


2. consists of


3. consists of

s and

s. In this case may have the same

number of

s and

s but they are out of order with some

s before

some

s hence it cannot be in

either




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Consider the cases:

1. consists of


s than

s and

so it is not in


2. consists of


3. consists of

s and

s. In this case may have the same

number of

s and

s but they are out of order with some

s before

some

s hence it cannot be in

either

The contradiction is unavoidable if we make the assumptionthat

is regular so

is not regular


Example 2


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Prove that


is not regular


Example 2


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Prove that


is not regular

Proof: assume that

is regular and is its pumping length.

Let

with

. Then pumping lemma guarantees

that , where

for any

.


Note


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If we take the division ,

it seems thatindeed, no contradiction occurs. However:


Note


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Condition 3 states that

, and in our case

and

. Hence,

cannot be pumped


Note


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, and in our case

and

. Hence,

cannot be pumped

If

then must consists of only

s, so

because

there are more 1-s than 0-s.


Note


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, and in our case

and

. Hence,

cannot be pumped

If

then must consists of only

s, so

because

there are more 1-s than 0-s.

This gives us the desired contradiction


Other selections


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Selecting

leads us to trouble because this string

can be pumped by the division: ,

,

.

Then

for any


An alternative method


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Use the fact that

is nonregular.




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Use the fact that

is nonregular.

If

were regular then

would also be regular because

is regular and

of regular languages is a regular language.




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Use the fact that

is nonregular.

If

were regular then


is regular and


But

which is not regular.




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Use the fact that

is nonregular.

If

were regular then


is regular and


But

which is not regular.

Hence,

is not regular either.


Example 3


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Show that

is nonregular using

pumping lemma


Example 3


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Show that

is nonregular using

pumping lemma

Proof: Assume that

is regular and is its pumping length.

Consider

. Since

, and

satisfi esthe conditions of the pumping lemma.


Note


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Condition 3 is again crucial because without it we

could pump if we let , so


Note


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The string

exhibits the essence of the

nonregularity of

.


Note


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The string

exhibits the essence of the

nonregularity of

.

If we chose, say

we fail because this stringcan be pumped


Example 4


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Show that

is nonregular.


Example 4


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Show that

is nonregular.

Proof by contradiction: Assume that

is regular and let be its pumping length. Consider

,

.

Pumping lemma guarantees that can be split, ,

where for all

,


Searching for a contradiction


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The elements of

are strings whose lengths are perfectsquares. Looking at fi rst perfect squareswe observe that

they are: 0, 1, 4, 9, 25, 36, 49, 64, 81,




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The elements of


they are: 0, 1, 4, 9, 25, 36, 49, 64, 81,

Note the growing gap between these numbers: large members

cannot be near each other




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The elements of


they are: 0, 1, 4, 9, 25, 36, 49, 64, 81,



Consider two strings

and

which differ from each otherby a single repetition of .




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The elements of


they are: 0, 1, 4, 9, 25, 36, 49, 64, 81,



Consider two strings

and

which differ from each otherby a single repetition of .

If we chose

very large the lengths of

and

cannot be

both perfect square because they are too close to each other.


Turning this idea into a proof


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Calculate the value of

that gives us the contradiction.




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If

, calculating the difference we obtain




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If

, calculating the difference we obtain

By pumping lemma

and

are both perfect

squares. But letting

we can see that theycannot be both perfect square if

,

because they would be too close together.


Value of

for contradiction


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To calculate the value for

that leads to contradiction we

observe that:


Value of

for contradiction


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observe that:


Value of

for contradiction


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observe that:

Let

. Then


Example 5


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Sometimes pumping down" is useful when we applypumping lemma.


Example 5


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We illustrate this using pumping lemma to prove that

is not regular


Example 5


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We illustrate this using pumping lemma to prove that

is not regular

Proof: by contradiction using pumping lemma. Assume that

is

regular and its pumping length is

.




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Let

; From decomposition , from

condition 3,

it results that consists only of 0s.



f


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Let


condition 3,


Let us examine to see if it is in

. Adding an

extra-copy of increases the number of zeros. Since

contains all strings

that have more 0s than 1s, it

will still give a string in



L

F d i i f


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Let


condition 3,


Let us examine to see if it is in

. Adding an

extra-copy of increases the number of zeros. Since

contains all strings

that have more 0s than 1s, it

will still give a string in


Try something else

Si

h

id

d


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Since

even when , consider and

.


Try something else

Si

h

id

d


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Since

even when , consider and

.

This decreases the number of 0s in .


Try something else

Since

even when

consider

and


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Since even when , consider and

.


Since has just one more 0 than 1 and cannot have

more 0s than 1s,(

and

) cannot be in

.


Try something else

Since

even when

consider

and


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Since even when , consider and

.


Since has just one more 0 than 1 and cannot have

more 0s than 1s,(

and

) cannot be in

.

This is the required contradiction


Minimum pumping length

The pumping lemma says that every regular language


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has a pumping length , such that every string in the

language of length at least

can be pumped.





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can be pumped.

Hence, if is a pumping length for a regular language

so is any length

.





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can be pumped.

Hence, if is a pumping length for a regular language

so is any length

.

The minimum pumping length for

is the smallest

that is a pumping length for

.


Example

Consider

The minimum pumping length for

is 2


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Consider . The minimum pumping length for is 2.


Example

Consider

. The minimum pumping length for

is 2.


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Consider . The minimum pumping length for is 2.

Reason: the string

,

and cannot be pumped. But any

string

,

can be pumped because for where

,

,

and

. Hence, the minimum pumping length for

is 2.


Problem 1

Find the minimum pumping length for the language

.


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Find the minimum pumping length for the language .


Problem 1


.


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p p g g g g

Solution: The minimum pumping length for

is 4.

Reason:

but

cannot be pumped. Hence, 3 is not a

pumping length for

. If

and

can be pumped by

the division

,

,

,

.


Problem 2


.


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p p g g g g


Problem 2


.


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p p g g g g

Solution: The minimum pumping length of

is 1.


Problem 2


.


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Solution: The minimum pumping length of

is 1.

Reason: the minimum pumping length for

cannot be 0 because is

in the language but cannot be pumped. Every nonempty string

,

can be pumped by the division:

,

,

first character

of and the rest of .


Problem 3



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.


Problem 3



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.


is 3.


Problem 3



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.


is 3.

Reason: The pumping length cannot be 2 because the string

is in the

language and it cannot be pumped. Let

be a string in the language

of length at least 3. If is generated by

we can write is as

, , is the first symbol of , and is the rest of the string.

If is generated by

we can write it as ,

,

and

is the remainder of .


1-Pumping Lemma (1)

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