I guess first one is NO. Second one seems no too as we get {0,1,00,11,000,111,...}. Not all of them contains even number of 0s and odd number of 1s.
Am I on the right track? If not, then give me a detailed explanation.
Thanks.
For question 1: try to come up with a proof that the language is not regular.
Hint: Let your language be $L$ and suppose towards a contradiction that it is regular. Use the closure properties of regular languages (in particular, regular languages are closed under intersection) to show that there is some regular language $L'$ such that $L \cap L'$ is a well-known non regular language.
Question 2 is not asking you whether all words in $\{0,1\}^*$ contain an even number of $0$s and an odd number of $1$s (that's clearly false). Rather, it wants you to consider the language defined as the set of all possible words (i.e., words from $\{0,1\}^*$) that also satisfy some additional property. This additional property is having both an even number of $0$s and an odd number of $1$s.
Hint: Try to design a DFA $D$ for this language. Suppose you are reading a word from left-to-right and you want to determine whether it satisfies the above property. What do you need to remember? Encode this information in the states of $D$.
The set of all words with the same number of 0’s and 1’s.
Isn't regular.
Proof (by contradiction):
Let $\ell=\{w\in\Sigma^*\mid \#w_a=\#w_b\}$, and suppose $\ell$ is regular, and let $M=\{a^nb^n\mid n\geq 0\}$ that $M\subseteq \ell$.
It's clear that $M$ isn't regular, on the other hand, we know that intersection of two regular languages are regular (by construction method on fsm), so $$\ell\cap a^*b^*=M$$ But $M$ isn't regular and it contradict with the fact that intersection of two regular languages are regular, hence, $\ell$ isn't regular.
The set of all words contained in {0,1}* that have an even number of 0’s and an odd number of 1’s.
