# Show this language is non-regular using pumping lemma: B = {ww | w ∈ {a,b,c,…,z)*} [duplicate]

The question I'm working from is:

Prove whether or not a finite automation exists that recognises the following language:

B = {ww | w ∈ {a,b,c,...,z)*}

EDIT

So I believe this is a non-regular language. My understanding of pumping lemma is not great but this was my solution:

S = apbapb

Where S is a valid string in the language and p is the pumping length.

S = aaaabaaaab for example when p = 4

S = xyz // s can be split into xyz components

| x y | <= p

SO y must be all a's before the first b e.g. a | aaa | baaaab

xy2z = aaaaaaabaaaab

xy2z is not in B

Therefore B is not regular

Apparently though this is wrong, please could someone explain why / how to obtain the right answer?

• the task can indeed be solved by finding a regular expression for $B$ but maybe it can be solved by proving that $B$ is not regular – phs Jan 7 '16 at 19:46
• I don't see a question here. What are you asking? If you want us to solve the exercise for you, that's off-topic. – David Richerby Jan 7 '16 at 19:50
• We are trying to hint that perhaps the language isn't regular after all. – Yuval Filmus Jan 7 '16 at 20:34
• Welcome to Computer Science! Your question is a very basic one. Let me direct you towards our reference questions which cover your problem in detail, especially cs.stackexchange.com/questions/1031/…. – D.W. Jan 7 '16 at 22:51
• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – D.W. Jan 7 '16 at 22:52

Your proof using the pumping lemma is wrong. You choose a pumping lemma constant $p=4$ but what happens if $p=5$ works? The pumping lemma tell us that there exists a constant $p$. Now you have to try all the remaining possible values of $p$.
• For example. Consider the word $\sigma = a^pba^pb$ with $|\sigma|=2p+2 \geq p$. You can say that $xy$ are only letter a's because $|xy|\leq p$. I really can't be more precise without solving the exercise. – Renato Sanhueza Jan 7 '16 at 23:00