over the alphabet {0,1}
a) L = { w $\in$ {0,1}$^*$ | |w|$_{0} \leq$ 2 }
b) L = { w $\in$ {0,1}$^*$ | w ends with 0}
over the alphabet {a,b,c}
L = { w $\in$ {a,b,c}$^*$ | |w|$_{ab}$ = 0 mod 2 and |w|$_{c} \geq$ 1 }
My work so far:
So I solved a) and b). But c) is really a problem. I've tried so many things but nothing worked. I will upload now my work a),b) and c). And like I said: I know that my c) isnt correct. Please help me and thank you in advance.
Remark: |w|$_{a}$ counts the letter a in w.
For c) consider the DFAs for the two following languages : $\left\{w\in\{a,b,c\}: |a|_{ab} \equiv 0 \mod 2\right\}$ and $\left\{w\in\{a,b,c\}: |a|_{c} \ge 1\right\}$.
Then consider combining the DFAs into their cartesian product. The states are pairs $z=(x,y)$ where $x$ is a state of the first DFA and y a state of the second DFA. A transition from $z_1=(x_1,y_1)$ to $z_2$ on character $\xi$ occurs if there was a transition from $x_1$ to $x_2$ on $\xi$ and a transition from $y_1$ to $y_2$ on $\xi$. Trim out unreachable states, possibly simplify if needed and voilà.
This little operation, the cartesian product, is actually a way to follow two DFAs at the same time. While parsing a string, instead of registering one state, you register two states, one for the left DFA and one for the right DFA. But you can view the pair of states as a single state in itself and simplify that DFA as usual.