$L_1=\{a^ncb^n\} \cup \{a^mdb^{2m}\}$
$L_2 = \{ a^{2n}cb^{2m+1} \} \cup \{a^{2m+1}db^{2n}\}$
This is my exam question on two days ago. Option is:
$1)$ $L_1 \cap L_2$ is Regular Language.
$2)$ $L_1 \cap L_2$ is not a Context free.
$3)$ $L_1 \cap L_2$ is Non deterministic Context Free.
$4)$ $L_1 \cap L_2$ is deterministic Context Free and not Regular.
My teacher gives a short solution answer and $(4)$ is correct choice. anyone could describe me why? how we can inference this difficult question?
First realize that $L_1$ is context free and $L_2$ is regular. Can you see why? Once you realize this, just use simple closure properties (a quick wikipedia search of context free languages will help you build a closure chart). Since all four answers list intersections, you know that a context free language intersected with a regular language is in fact context free (and not regular in this case), so answer 4) is the correct answer.
EDIT: $L_1$ is context-free and you can show it by coming up with a CFG for the language. For $L_1$, a grammar could be
$S \rightarrow C|D\\ C\rightarrow c | aCb\\ D\rightarrow d | aDbb$
I assume $n\ge 0$ and $m \ge 0$.
For $L_2$, you can come up with an NFA for $L_3 = \{a^{2n}cb^{2m+1}\}$ and $L_4 = \{a^{2m+1}db^{2n}\}$ and connect them with epsilon transitions.