Proving that language is regular or not regular

Question

Let $L$ be a regular language. Prove that:

1. $L_{+--}=\left\{w: \exists_u |u|=2|w| \wedge wu\in L\right\}$

2. $L_{++-}=\left\{w: \exists_u 2|u|=|w| \wedge wu\in L \right\}$

3. $L_{-+-}=\left\{w:\exists_{u,v} |u|=|w|=|v| \wedge uwv\in L\right\}$

are regular and:

1. $L_{+-+}=\left\{ uv:\exists_w |u|=|w|=|v| \wedge uwv\in L \right\}$

is not regular.

Seems very hard to me. I suppose 1-3 are similar (but I may be wrong), but I don't know how to approach. General idea is usually to modify finite state machine for $L$ to accept other language. But those constructions are often very sophisticated and I still can't come up with it alone.

Answer 1

Here is a proof that the language $L_0 = \{ w : \exists_u |u|=|w| \land uw \in L \}$ is regular. It can be modified to show that the first three on your list are regular. (Note that I changed $wu$ to $uw$.) Given a DFA for $L$, we build an NFA for $L_0$. The first thing that the NFA does is guess (take an $\epsilon$ move) a state $q$, whose intended semantics is the state that the DFA for $L$ ends up after reading $u$. It then simultaneously runs two copies of the DFA for $L$, one starting at the start state and the other starting at $q$. On reading a symbol $a$, it moves according to an arbitrary symbol on the first, and moves according to $a$ on the second. A state is accepting if the first copy is at state $q$ and the second is at an accepting state.

For the final one, consider the language $L= a^+ b^+ c^+$, and intersect $L_{+-+}$ with $a^+ c^+$.