part 1. (a) a*b(a U ba*b)* or (a U ba*b)*ba*
Part (b) One of the possible solution is
((aUb)a*b(ba*b)*a)*((aUb)a*b(ba*b)*)*
2. Part a: You can either use Myhill-Nerode or pumping lemma to prove.
For pumping lemma, the idea is that S(n) can be written as n(n+1)/2.
If we choose string s = xyz = 0^{S(2p)}, then |s| = 2p^{2} + p. If we
pump the string to xyyz, then
S(2p) = 2p^2 + p < |xyyz| <= 2p^2 + 2p < S(2p+1) = 2p^2 + 3p + 1.
Therefore, xyyz is not in the language.
For part b, you can use the similar idea, but instead of sum of the
first n-th natural numbers, it's n-factorial. You can pick string s to
be 0^{P(p)}, |s| = p! > p, then if you pump it to xyyz, then
p! < |xyyz| <= p! + p < (p+1)p!
3. Part (a): This one is very similar to the proof for language of
O^{n}1^{n}. I won't go to the detail about it. Choose s = 0^{p}10^{p}
to avoid proving for multiple ways of breaking down s to x,y,z
components.
Part (c): Let W be the language in question. If W is regular, then W'
(complement), which is basically the language of palindromes, must also
be regular under the closure of class under complement. We have showned
in class that W' is non-regular. Therefore, we've reached a
contradiction, and thus W cannot be regular.
Part (d): If you choose to prove by pumping lemma, then it will not be
possible to prove by choosing string s = 0^{p}10^{p} or any string with
similar format. The reason being once that, you pump this string, up or
down, then it will still be in the language of L = {wtw}. Because,
then, the extra 0's will become part of t.
To illustrate, consider a string
s = 0^p 11 0^p.
For simplicity, we can think of w = 0^p, and t = 11.
Then, if we pump the string up to xyyz, we get
xyyz = 0^p 0^k 11 0^p, where 1 <= k <= p.
This is not the same as part (a). Here, you can think of w = 0^p, and t
becomes 0^k 11. Thus, xyyz is still in the language.
So how to make this work? You can choose string s = 0^{p}1010^{p}1, and
then pump it UP.
4. (a) You can use Myhill-Nerode thm, or use the closure properties to
show that F is not regular.
Important: For people who use the closure properties.
Please remember that we want to prove that F is not regular.
Let # represent a regular operation.
Let languages A and B be reg...