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November 1st, 2015, 03:34 AM   #1
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Automata Question

Can you prove that this RE : 0^n1^n can't make a regular Language ?

With Pumping Lemma !
Saman Q is offline  
 
November 7th, 2015, 11:26 PM   #2
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suppose RE is regular,
by pumping lemma, there exists $\displaystyle p\geq 1$ such that for all $\displaystyle w \in RE, |w|\geq p \implies$ there exist a decomposition w=xyz such that $\displaystyle |xy|\leq p , |y|\geq 1 \text{ and }xy^iz \in RE \text{ for all i>0 }$

We can challenge whoever claim that it is a regular language to choose a p.
Given this p, we have the freedom to choose any arbitrary w with length $\displaystyle |w|\geq p$

in this case,
given any p, we choose $\displaystyle w=0^p1^p$

this forces $\displaystyle xy=0^k, \text{ for some }k\leq p \text{ since }|xy|\leq p$
and hence, y =$\displaystyle 0^m,\text{ for some }m\geq 1$

now, consider $\displaystyle xy^2z=0^{p+m}1^p \notin RE$ hence a contradiction which proofs that RE cannot be a regular language
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November 8th, 2015, 01:42 AM   #3
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Actually this isn't a good problem to start off with as you can practically choose w to be anything but your proof will be longer to consider the various cases.

For your understanding,
let's consider a separate problem where the choice of w is crucial.
I would like to encourage you to think of proving a language to be irregular as a game between you and someone who wants to claim that it is regular.

step 1: you challenge your opponent to select a number p such that pumping lemma holds.
step 2: you carefully select a w (so that you can win the game)
step 3: your opponent finds a prefix xy of w (w=xyz) and denote which part of the prefix is the part where we can pump. His choice is restricted by the fact that the length of xy |xy|<=p
step 4: now you find a k such that $\displaystyle xy^kz$ is not in the language, usually k=2 works.

as a demonstration, i shall define another language
$\displaystyle L=\{0^m1^n|n>m\}$
and prove it is not regular.

step 1: let him select a p for which he thinks his pumping lemma will hold.
step 2: you select w to be $\displaystyle 0^p1^{p+1}$ check that |w|=2p >=p
step 3: you will find that as usual, no matter how your opponent choose, the prefix must be of the form $\displaystyle 0^i$ and therefore his y must also be of the form $\displaystyle 0^j$ and $\displaystyle z=0^{p-i}1^{p+1}$
step 4: $\displaystyle xy^2z=0^{i-j}0^{2j}0^{p-i}1^{p+1}=0^{p+j}1^{p+1}$
but p+j>=p+1 means that $\displaystyle xy^2z \notin L$
which is well and good and you win.

suppose instead at step 2 we are careless and choose w to be $\displaystyle 0^{p-1}1^{p}$
now his turn in step 3, he can actually choose $\displaystyle xy=0^{p-1}1, |xy|=p,y=1$
now, no matter what k you choose,$\displaystyle xy^kz=0^{p-1}1^{p+k}\in L$. He'd be happy he won the game, but this doesn't mean that L is actually regular, he is just lucky that you have not chosen the right w. So the idea behind using pumping lemma is to choose the w to force the choice by your opponent so that you win no matter what he choose.

Now as an exercise, go back to the original question and see why I said that any choice of w with length |w|>=p works.

Last edited by changcw83; November 8th, 2015 at 01:50 AM.
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