:Do you mean to say that the string $bbba\notin L$?Could you tell why we can't pump $ba$ into $bbba$.

:Yeah that's a good idea.I'm sick of drawing DFA's using jflap.

:Could you tell why we can't pump the initial part of the string $x$ instead of the middle part $y$.

:I've updated the post.

:Isn't it necessary to pump the middle section?

:Oh sorry I'm not trying to have $y$ in the beginning but trying to pump the initial part of the string as I said in the post.

@Shreeh:That's the result I'm trying to get but the path is different.I'm trying to have $x=b$,$y=a$,$z=\epsilon$ where we pump the initial part of the string i.e $x$ instead of pumping the middle part of the string so that we could make $ba$ into $bbba$ by pumping the $x$ part i.e $b$.Is this possible or is there any strict rule that you can only pump the middle part of the string i.e $y$?

:Do you mean to say that we can't pump the $x$ part i.e $b$ since it's not a loop?

:Eventhough $y=a$ is not a loop isn't $x=b$ a loop?

:Why do we make the loop as $y$ instead of $x$ which should be given preference from the dfa?

@Yeah I could see from the article that it tells us to pump the middle part.Is there any reason why we specifically pump at the middle section and not at the initial or final section of the string.

Why can't we make $z=\epsilon$ and $x$='loop' and $y$ as a string(or strings)?

:Could you show an example to get evidence about the previous comment.

:There too I can see loop in the middle part but not the initial part.I would like to get an example where we could see a loop in the initial part i.e $x$ and claiming that "this specific string won't be in $L$.

I don't understand you, pumping lemma say $xy^iz$ belongs to $L$ for all $i$. But for your choice $baaa$ definitely does not belong to $L$ as it ends at state $q_0$ which is not an accepting state.

Can you come to chat after fifteen minutes?

:Oh sure for that.

Sorry.I'm having an terrible internet connection that's why I couldn't connect with you.I'll try my best to connect.

Same problem here!

I might have to leave now since the college is going to be closed.If we can't meet now we'll meet later......

Fine

Are you in South Asia!!

I'm actually telling about the string $bbba$ not about $baaa$.Could you now help me.Yes I'm from India.

bbba can be partitioned into many way

x y z

e bbb a

e bb a

e b a

These three are valid ways of partitioning others are not.

bb ba e is not a valid partitioning

because bb(ba)^0e is bb which is not accepted.

Pay attention to Pumping Lemma

Pumping lemma says following: If L is accepted by some DFA then every accepted string w larger than some length, can be partitioned into some x, y, z so that every xy^iz will be in L.

For the DFA given by you that particular length is 4

So any accepted string of length >= 4 can be partitioned correctly into x, y, z so that the condition of pumping lemma holds.

Anything not clear till now?

Pumping lemma does not say that every partitioning will satisfy the conditions, it only says there will be at least one partition that will satisfy the conditions.

In your example there are 3 such partitions.

It seems your Internet connection is really bad. We can chat some other time.

Good Luck!