What regular language intersects with 1*0* gives 1n0n

Question

I'm reading a book on automata theory, and the book gives an example that a language with equal number of 0s and 1s intersects with 1*0* would result 1n0n, where n > 0

So my question is, how can I find some regular languages that when intersected with 1*0*, would also results in 1n0n. Is there a way to think about that?

update: Thanks for the answers! I guess what I'm trying to find is some regular languages, so the ones like 1n0n wouldn't work ;) Is it possible? Any ideas?

Answer 1

N.B. The language with an equal and unbounded number of 0s and 1s is not a regular language.

As for your question, I don't think there are any more restrictions you can add to some ones followed by some zeros to get n ones followed by n zeros other than the two you have given.

There are an infinite number of trivially-constructed languages that satisfy the conditions: A1nB0nC where A, B and C are any expressions that can match zero width.

Answer 2

Just think of the question as: "What languages, when intersected with 1n0m, give the language 1n0n?" Basically, anything that adds the constraint that n=m.

One example is anbn, where a!=b. Another one is L = { 1n0n1m0m | n!=m, n >= 0, m >= 0 }.

Also, as OrangeDog pointed out, 1n0n is not regular, and since regular languages are closed under intersection, it follows that any language whose intersection with 1*0* gives 1n0n is not regular.