6
In computer science, "automaton" refers to some kind of finite state machine. This is a basic and fundamental model of computation, and automata are widely used in implementing simple electronic devices and in writing parsers, e.g., for programming languages.
5
There is no known natural example of such a pair, and indeed there are various results in computability theory suggesting that such a pair does not exist. So to whip up an example one has to do some work.
The simplest approach (indeed, the easiest I know) is via mutual diagonalization: we build inductively a pair of increasing sequences of infinite binary ...
5
The automaton is mainly used as a simple model of computation to check input strings on some defined conditions by reading the string and giving out whether the string is accepted in a defined language or not.
There are a lot of examples. A really crucial for example in terms of computing are the RegEx-expressions, if you heard of that. There are some ...
3
I think you have to be more careful in the inductive steps, but it seems you basically have the argumentation ready. According to me proper induction would be
If $R_1$ and $R_2$ can be written as sum of products, then so can
$R_1\cdot R_2$, $R_1+R_2$ and $R_1^*$.
The proof could be completed as:
If $R_1$ and $R_2$ can be represented as sum of products ...
1
Yes, as long as your alphabet is defined as a set of different couples of binary digits
1
Call $(0+11^*0)$ as A and $(1+01^*0)$ as B. Then what you have on left-hand side is just: $$ R = A + AB^*B$$
Using Arden's Theorem we know that in this case: $$R = AB^*$$
Then what we have simplified is: $$(11^*0+0)(1+01^*0)^*$$
Another simplification using Arden's Theorem would give us:
$$(11^*+ \epsilon)0(1+01^*0)^*$$
Which is finally equivalent to: $$1^*...
1
The two regular expressions describe the same language.
1
Imagine an NFA with a single state, which is final/accepting. It has no edges.
This NFA accepts the language $\{\varepsilon\}$—that is, only the empty string. If it is given a non-empty string, it looks for appropriate edges leading away from the starting state, finds none, and fails.
1
any location except i/p part means
(total unbounded tape area - tape area that includes i/p string)
= (infinity length - finite length) as we know i/p string length is always finite
= infinity length
in the definition of linear bounded automata it is stated that the tape can be used as a function of the input string length.but here the portion that can ...
Only top voted, non community-wiki answers of a minimum length are eligible
