Search Results

You could start from the fact that the language $HP$ of the Halting problem is r.e. If its complement $\overline{HP}$ were r.e. too then that would mean that $HP$ is recursive (in $R$) which is imposs …

Yes there is. $w \in \{ a, b\}$ means that $w$ may be either $a$ or $b$, while $w \in \{ a, b\}^*$ means that $w$ may be a string over $a$ and $b$ including the empty string $\epsilon$. In particular …

This is true for any recursive language $A$ different from $\emptyset$ and $\Sigma^*$ for some finite alphabet $\Sigma$. Your definition is correct. In general you could define $f$ as following. Let $ …

Consider an alphabet $\Sigma = \{0,1\}$ and the set of all infinite length strings over $\Sigma$. Then this set is of course infinite, but uncountable which can be easily proved by the diagonalization …

The empty language is recursive. The TM accepting this language just returns $0$/REJECT on any input. Similarly, the language $\Sigma^*$ is also recursive. Just return $1$/ACCEPT on any input. Basi …

You can use Thompson's construction. It constructs NFA with $\epsilon$ moves. But remove the forward $\epsilon$-transition from the new start state into the new final state since you are interested in …

The Rice's theorem can also be stated in terms of index sets of TMs. Lets start with a basic definition of a property of a language. A property of a language is a set of languages. For example $$P_{ …

This trivially follows from the definition of the Kleene plus operation. $A^+$ is known as the Kleene plus operation on the set $A$. This operation omits the $A^0=\{\epsilon\}$ in the the Kleene star …

This language is regular. Let $M$ be a DFA accepting $L$ and assume it has a single initial and final state. We can construct a NFA accepting $L_2$. We create two copies of $M$: $M_1$ and $M_2$. Let u …

Use De Morgan rule $A\cup B = \overline{\overline{A} \cap \overline{B}}$. Also you could prove it using DFA for $L_1$ and $L_2$ without using complement and intersection. The latter is systematic an …

My answer is just a hint. The following pattern should lead you to the right answer: These strings belong to the language: $w = 11 \Rightarrow 1^1\epsilon 1^1, n=1$ $w = 1011 \Rightarrow 1^1(01) 1^1, …

Take $r = (r_1 + r_2)^*$. We have to prove $$(r_1 + r_2)^* = r_1(r_1 + r_2)^* + r_2$$ First note that both $(r_1 + r_2)^*$ and $r_1(r_1 + r_2)^* + r_2$ contain $\epsilon$ since we are given that $\e …

The fact that is a proper subset does not inherit the global properties in general is common in mathematics and computer science. A proper subset does not have to inherit the global properties of its …

The problem asks to draw a deterministic finite automaton (DFA), not a pushdown automaton or a Turing machine. If you cannot directly come up with a DFA then you can draw a NFA (or NFA with epsilon mo …

Try the following grammar (which can be transformed into a regular grammar): $S \rightarrow A \mid E \mid I \mid O\mid U$ $A \rightarrow aA \mid bA \mid cA \mid \dots \mid zA \mid \epsilon$ (all con …