1.21 b
thus, the regular expression is:
E U (a U b) a* b ((b U (a(a U b))) a * b)*
(E U a)
Some good regular expressions to test to see if your regular expressions work are: the
empty string, "abbb," and "bbbb."
1.30
In this problem you need to merely explain the error in the "false proof."
In the proof a reference is made to example 1.73, however, the logic that for S=0
p
1
p
cannot be pumped is flawed. In cases 2 and 3 of example 1.73 the pumping of 0
p
1
p
is
possible, as a contradiction does not arise. This is because although there are more ones
or zeros, the strings generated are still in the language (0*1*). The pumping lemma
states only that a subdivison allowing pumping exists, NOT that all possible sub
divisions can be pumped.
1.46 a
There are multiple possible proofs, here is one using the pumping lemma.
Suppose L were regular. Let p be the pumping length given by the pumping lemma.
Consider the string w
p
= 0
p
10
p
Clearly, w
p
 >= p and w
p
is in L. So, by the pumping lemma, there must be some choice of
x,y,z satisfying the conditions of the pumping lemma.
But, consider
any
choice of x,y,z for which w
p
= xyz, xy <= p, and y >= 1.
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Because
xy <= p, we know that x and y must be 0s, (otherwise xy>p ) . This allows us to
define y = 0
j
and we know that j>0 from point 2 (y>=1) of the pumping lemma. It follows
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 Spring '11
 sm
 Formal language, Formal languages, Regular expression, Regular language, Kleene star

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