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p q r sa
b
a
b
a
b
a
b
w1 w2 w3 w4 w5 w6 w7 w8 w9a a b a a b b a b
p q r s r q q q r sq0 q1 q2 q3 q4 q5 q6 q7 q8 q9
So far we have seen that for allw L
|w| p , x,y,z s.t.w = xyz withxyz L
Let us enlist a few of the decompositions
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w1 w2 w3 w4 w5 w6 w7 w8 w9
a a b a a b b a b
p q r s r q q q r sq0 q1 q2 q3 q4 q5 q6 q7 q8 q9
aax
bay
abbabz
q2 =q4
a
x
abaa
y
bbab
z
q1 =q5
ax
abaaby
babz
q1 =q6
ax
abaabby
abz
q1 =q7
aa
x
baabba
y
b
z
q2 =q8
aabx
aabbaby
z
q3 =q9
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Pumping Lemma
For every regular languageLthere existssome constantpsuch that,
for every stringw L with|w| p, there existx,y,z with
w=
xyz,|y
| 1,
|xy
| p, and for alli
N,xyiz L.
Proof
LetLbe a regular language.
LetM = (R,, , r0, F) be a DFArecognizingL
Let
p (1)
be the number of states inM
Considerw L with|w| p
That is
w = w1w2 wn s.t. n p
Herewj ,j = 1, 2, 3, . . . ,n
Let
q0q1q3 qn (2)
thatMenters while processingw.
Here
q0 =r0
andqj = (qj1, wj) for j= 1, 2, 3, . . . ,n
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The length of sequence (2) is at leastp + 1
Among the firstp + 1 elements in thesequence, two must be the
same, by thepigeonhole principle.
We call the first of theseqkand the secondql.
That isqk=ql with 0 k< l p (3)
Let
x=
w1w2
wky = wk+1wk+2 wl
z =wl+1wl+2 wn
Obviouslyw = xyz.
From (3) we have
kl = y = |y|> 0 = |y| 1 (4)
Furthermore, from the decomposition ofwwe have|xy| =l.
And from (3) it can be inferred thatxy p (5)
AsxtakesMfromq0to qk
ytakesMfromqktoq
k(=q
l)
and ztakesMfromqkto qn F
Therefore,Mmust accept
xyiz for i 0
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That isxyiz L , i N (6)
From (1), (3), (4), and (6) we have establishedthe truth of the
pumping lemma.
Succinctly the pumping lemma states
Lis regular =p w , |w| p, x,y,z , |xy| p , |y| 1 , i , xyiz
L
(7)
The contrapositive of (7)
n w , |w| n, x,y,z , |xy| n , |y| 1 , i , xyiz L
= Lis not regular
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