Post on 28-Dec-2015
description
The Pumping LemmaComputation Theory
M Ayzed Mirza13-NTU-4005
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The Pumping Lemma
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Take an infinite regular languageL
There exists a DFA that accepts L
mstates
(contains an infinite number of strings)
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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
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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
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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
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zyxw
q... ...
x
y
z
In DFA:
...
...
1 ki1ij
1j
contains only first occurrence of q
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Observation: myx ||length numberof statesof DFA
Since, in no state is repeated (except q)
xy
Unique States
q...
x
y
...
1 i1ij
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Observation: 1|| ylength
Since there is at least one transition in loop
q
y
...
1ij
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We do not care about the form of string z
q...
x
y
z
...
zmay actually overlap with the paths of and x y
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The string is accepted
zxAdditional string:
q... ...
x z
...
Do not follow loopy
...
1 ki1ij
1j
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The string is accepted
zyyx
q... ... ...
x z
Follow loop2 times
Additional string:
y
...
1 ki1ij
1j
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The string is accepted
zyyyx
q... ... ...
x z
Follow loop3 times
Additional string:
y
...
1 ki1ij
1j
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The string is accepted
zyx iIn General:
...,2,1,0i
q... ... ...
x z
Follow loop timesi
y
...
1 ki1ij
1j
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Lzyx i Therefore: ...,2,1,0i
Language accepted by the DFA
q... ... ...
x z
y
...
1 ki1ij
1j
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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)
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Applications
of
the Pumping Lemma
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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)
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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
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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
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Note: It suffices to show that only one stringgives a contradiction
Lw
You don’t need to obtaincontradiction for every Lw
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Theorem: The language }0:{ nbaL nn
is not regular
Proof: Use the Pumping Lemma
Example of Pumping Lemma application
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Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
}0:{ nbaL nn
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Let be the critical length for
Pick a string such that: w Lw
mw ||and length
mmbawWe pick
m
}0:{ nbaL nn
L
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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
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From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmbazyx
Lzyx 2
mkay k 1,
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From the Pumping Lemma:
Lbabaaaaaaazxy ...............2
x y z
km m
Thus:
Lzyx 2
mmbazyx
y
Lba mkm
mkay k 1,
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Lba mkm
}0:{ nbaL nnBUT:
Lba mkm
CONTRADICTION!!!
1≥k
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Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF
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More Applications
of
the Pumping Lemma
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The Pumping Lemma:• Given a infinite regular language L
• there exists an integer (critical length)m
• for any string with length Lw mw ||
• we can write zyxw
• with andmyx || 1|| y
• such that: Lzyx i ...,2,1,0i
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Regular languages
Non-regular languages *}:{ vvvL R
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Theorem:The language
is not regular
Proof: Use the Pumping Lemma
*}:{ vvvL R },{ ba
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Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
*}:{ vvvL R
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mmmm abbaw We pick
Let be the critical length for
Pick a string such that: w Lw
mw ||length
m
and
*}:{ vvvL R
L
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we can write: zyxabbaw mmmm
with lengths:
From the Pumping Lemma:
ababbabaaaaxyz ..................
x y z
m m m m
1||,|| ymyx
mkay k 1,Thus:
w
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From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus: Lzyx 2
mmmm abbazyx mkay k 1,
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From the Pumping Lemma:
Lababbabaaaaaazxy ∈.....................=2
x y z
km + m m m
y
Lzyx 2
Thus:
mmmm abbazyx
Labba mmmkm
mkay k 1,
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Labba mmmkm
Labba mmmkm
BUT:
CONTRADICTION!!!
1k
*}:{ vvvL R
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Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF