CS103
HO#33
SlidesContextFree Languages
4/29/11
1
Show that L = { w  0
n
where n is prime } is not regular.
Suppose L is regular, and that p is the pumping length guaranteed to
exist by the lemma.
We will find a string s such that s
L and s > p
but s does not satisfy the conditions of the lemma.
Show that L = { w  0
n
where n is prime } is not regular.
Suppose L is regular, and that p is the pumping length guaranteed to
exist by the lemma.
We will find a string s such that s
L and s > p
but s does not satisfy the conditions of the lemma.
Let s = 0
k
, where k is the smallest prime > p + 1.
s
L, and s > p.
According to the lemma, there must exist strings x, y, and z such that
(s = xyz
y > 0
xy
p) and xy
i
z
L for all i
0.
That is,
for all i
0, x + z + iy must be prime.
We will show that for any
x, y, z satisfying (….), there is an i where
the last condition fails.
Suppose
(s = xyz
y > 0
xy
p)
and let
i = x + z.
Then
x + z + iy = x + z + (x + z)y =
(x + z) (1 + y)
x + z > 1
since s = xyz, s > p + 1, and y
p.
1 + y > 1
since y > 0.
But that means (x + z)(1 + y) is not prime.
So for at least one value of i, xy
i
z
L, and thus L is not regular.
Closure Properties of Regular Languages
Regular languages are closed under:
Union
Intersection
Concatenation
Difference
Complement
Reversal
Closure (star)
(You may use these facts in your proofs.)
Using Closure Properties to Show Nonregularity
Consider D = { w
{0, 1}*
 the number of 0's in w is the same as the number of 1's }
Suppose D is regular.
Then
D
0*1*
is regular since regular languages are closed under intersection.
But
D
0*1* = { 0
n
1
n
 n
0 }
, which we have shown to be nonregular.
So D is not regular.
Regular Grammars
A
1
0
1
0
B
Instead of accepting its language, we can think of the machine
as
generating
it.
That is, every transition on a path to the accepting
state generates a string containing the symbols on the arcs of the
path.
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 Spring '11
 PLUMMER
 Formal language, Formal languages, Contextfree grammar, regular languages, contextfree languages

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