We all know that (a + b)* is a regular language for containing only symbols a and b.
But (a + b)* is a string of infinite length and it is regular as we can build a finite automata, so it should be finite.
Can anyone please explain this?
Finite automaton can be constructed for any regular language, and regular language can be a finite or an infinite set. Of-course there are infinite sets those are not regular sets. Check the Venn diagram below:
Notes:
1. every finite set is a regular set.
2. any dfa for an infinite set will always contains loop (or dfa without loop is not possible for infinite set).
3. every non-regular language is an infinite set.
The word "finite" in finite automata significance the presence of 'finite amount of memory' in automata for the class of regular languages, hence only 'finite' (or says bounded) amount of information can be stored at any instance of time while processing a string of language.
In finite automata, memory is present in the form of states only (whereas in the other class of automata like Pda, Turing Machines external memory are used to store unbounded information). You can think a finite automata as a CPU without explicit memory; that can only store recent results in its registers.
So, we can define "regular language" as — a class of languages for which only bounded (finite) information is required to stored at any instance of time while processing language strings.
Further read (for infinite languages):
What is regular language: What is basically a regular language? And Why a*b* is regular? But language { anbn | n > 0 } is not a regular language
To understand how states are uses as memory element read this answer: How to write regular expression for a DFA
And difference between automate for finite ans infinite regular language: To make sure: Pumping lemma for infinite regular languages only?
Each word in the language (a+b) is of finite length. The same way as there are infinitely many integers, but each of them finite.
Yes, the language itself is an infinite set. Most languages are. But a finite automaton (NB: automata is plural) works just fine for them, provided each word is of finite length.
As an aside: This type of question probably should go to cs.stackexchange.com.
But (a + b)* is a string of infinite length
No, (a + b)* is a finite way to express an infinite set (language) of finite strings.
1. A regular expression describes the string generated by some language. Applying that regular expression gives you all the strings that can be described by that language.
2. When you convert that regular expression to a finite automaton (automata with finite states) , it means that those same strings can also be generated by traversing from state-to-state on that automaton. Now, intuitively, each state here represents a group of strings belonging to that language. It says, after having "absorbed" some input, the string is now in state X.
Example:
If you want a regex to accept strings with even numbers of 0 , then you'll have one state (group) which indicates that even number of 0 has been observed in the input so far. And another state (group) for odd numbers --> this state would be your non-accepting state in the FA.
As shown here, you just needed 2 (finite) states to generate an infinite number of strings, because of the grouping of odd and even we did.
And that is why it is regular.
It just means there exists a finite regular expression for the specified language and is no where related to no of strings generated from the expression.
For many regular languages we can generate infinite number of strings which follow that language but to that language is regular to prove that we need a regular expression which must be finite.
So here the expression (a+b)* is finite way of expressing 0-n number of a's or b's or combination of that but n can take any value which results in infinite no. of strings.
Related
There is this proof that I thought of that I am not quite sure if it's valid or not.
Suppose you had to prove the nonregularity of the following language:
A = { 0^n 1^n 2^n | n>= 0 }
The proof I devised picks a string that belongs in the language, such as 012, and show that it doesn't matter how it's divided, the pumping lemma is not wholly satisfied(I could post the entire proof, but the post is verbose as it is). According to my professor however, this proof cannot be accepted.
He did not explain why, and I don't see how such a proof would be insufficent to demonstrate that a language is not regular. If a string clearly belonging to an assumed regular language does not satisfy the pumping lemma, the language clearly has strings that are not regular as part of it set of strings, therefore the language is not regular.
I believe the reason my professor rejected this proof is because in the majority of problems the pumping length P cannot be correctly guessed. At the same time I do not see how my proof could be proven wrong with a counterexample.
You can only choose p (the pumping length) to be a specific number if the language is regular and p actually exists. The fact itself, that you pick an exact number, means that p exists, which is the thing to be actually proven.
Suppose that p exists. Lets choose a word, that is long enough: w=0^{p}1^{p}2^{p}. According to the pumping lemma there must exist a decomposition of each string in language A as w=xyz with |xy|≤p and |y|≥1 such that xy^{i}z in language A for every i≥0. To satisfy |xy|≤p choose x to be empty, y=0^{p}, and as a consequence z=1^{p}2^{p}. From the lemma, |y|≥1 so |xy|≥1. The strings with i≠1 (in xy^{i}z) are not in the language A. The language is thus not regular, and p does not exist.
If p existed then a finite state automaton could be constructed for this language. But no such automaton exists, because it would need memory to remember the number of 0 to later match the same number of 1 and 2. If n was a finite number, then you could construct, a probably large, automaton, but for infinite n no finite automaton can be constructed.
This language is not even context-free, because there is no push-down automaton that can be constructed for it. It is context-sensitive.
This is a grammar and I wan to check if this language is regular or not.
L → ε | aLcLc | LL
For example the result of this grammar is:
acc, accacc ..., aacccc, acaccc, accacc, aaacccccc, ...
I know that is not a regular language but how to prove it? Is building an automata the right way to prove it? What is the resulting automata. I don't see pattern to use it for build the automata.
Thank you for any help!
First, let me quickly demonstrate that you cannot deduce the language of a grammar is irregular based solely on the grammar's being irregular. To see this, consider the unrestricted grammar:
S -> SSaSS | aS | e
SaS -> aSa
aaS -> SSa
This is clearly not a regular grammar but you should be able to verify it generates the infinite regular language of all strings of a.
That said, how should we proceed? We will need to figure out what language your grammar generates, and then argue that particular language cannot be regular. We notice that the only rule that introduces terminal symbols always introduces twice as many c as it does a. Furthermore, it's not hard to see the language must be infinite. We can use the Myhill-Nerode theorem to show that these observations imply the language must be irregular.
Consider the prefix a^n of a hypothetical string in the language of this grammar. The shortest string which can be appended to the end of this prefix to give us a string generated by this grammar is c^(2n). No shorter string will work, and that string always works. Imagine now that we were looking at a correct deterministic finite automaton for the language of the grammar. Then, whatever state processing the prefix a^n left us in, we'd need the shortest path from there to an accepting state in the automaton to have length 2n. But a DFA must have finitely many states, and n is an arbitrary natural number. Our DFA cannot work for all possible n (it would need to have arbitrarily many states). This is a contradiction, so there can be no correct DFA for the language of the grammar. Since all regular languages have DFAs, that means the language of this grammar cannot be regular.
I want to understand if RE a∗ba∗ab∗ is same as the following the finite automata. The Part where I am confused is that, From state 3 to state 4 , there is a b , which means that the language needs to have a b at the end , while the RE just has b* which means 0 or more b . If not what is correct finite automata for this RE ?
Indeed, the regular expression a*ba*ab* is not equivalent to the DFA shown for exactly the reason you stated in your question.
Thompson's algorithm is a standard way of systematically converting a regular expression into an NFA. (If you need the finite automaton to be deterministic, you can then run the subset construction.)
How does the pumping length of a regular language relate to the pumping length of a related language. For example, if A :< B :< C are all regular languages and k is the pumping length of B, do we know anything about the pumping lengths of A and C?
One might be inclined naively to think that a sublanguage has a smaller (<=) pumping length when we look at finite languages. {a,ab,abc} :< {a,ab,abc,abcd} have respective pumping lengths 4 <= 5. Taking an element out of a set can't make its longest word even longer.
On the other hand if you look at the state machine formed by the synchronized product of two languages, the intersection language and the union language have the same state machine structure, but differ in that the set of final states of the intersection is a subset of the set of final states of the union. Having more final states, could make it more probable to find a shorter path through the state machine. But on the contrary having fewer final states makes it more likely that the state machine has non-co-accessible states, and is thus reducible.
Note first that all languages over an alphabet are related to all other languages over that alphabet by some relation, however contrived. We really do need to limit discussion, as you have suggested, to something like subset in order to meaningfully scope the question.
Now, you've already correctly noted that in the general case, the subset relation doesn't have a clear bearing on the pumping lengths of relative languages. You can easily take a* and compare to {a^n} and show that a minimal DFA for a* is almost always simpler than one for {a^n}.
Let us further restrict ourselves to languages that differ by a finite number of entries; that is, L \ R is finite and R \ L is finite. This is an indicator of similarity different from subset; but if we require that, w.l.o.g., that R \ L be empty, then we recover a restricted version of subset. We might now ask the same question given this modified version: for languages that differ in a finite number of entries, does the subset relation tell us anything?
The answer is still no. Consider L = a* and R = a* \ A, where A is a finite non-empty subset of a*. Even still, L takes one state and R takes potentially many more.
Restricting ourselves to finite sets only, as you suggest, does let us deduce what you propose: that a minimal automaton for R will have no more states than the one for L. Why is this? We must have n+1 states to accept any string of length n, and we must have a dead state to accept strings not in the language (of which there will be infinitely many). A minimal DFA will have no loops at all (except centering on the dead state) if the language is finite, since otherwise you'd be able to get infinitely many strings.
Your observation about taking the Cartesian product is correct, in that the result of applying the construction gives a structurally identical DFA for any set operation (union, difference, intersection, etc.); however, these DFAs are not guaranteed to be minimal and the point for finite languages still holds, namely, the DFA for intersection will have no more states than the one for union, before and after DFA minimization of both machines.
You might be able to take the Myhill-Nerode theorem and define a notion of "close-relatedness" using that which does allow you to compare languages to determine which will have the larger minimal DFA. I am not sure about that, however, since an easy way of doing that would allow you to compare to parameterized reference languages to, for instance, easily prove any non-regular language is non-regular, which might be a big deal in mathematics in its own right (like, either it's impossible in general or it would prove P=NP, etc. to do so).
I know anbn for n > 0 is not regular by the pumping lemma but I would imagine a*b* to be regular since both a,b don't have to be the same length. Is there a proof for it being regular or not?
Answer to your question:
imagine a*b* to be regular, Is there a proof for it being regular or not?
No need to imagine, expression a*b* is called regular expression (re), and regular expressions are possible only for regular languages. If a language is not regular then regular expression is also not possible for that and if a language is regular language then we can always represent it by some regular expression.
Yes, a*b* represents a regular language.
Language description: Any number of a followed by any numbers of b (by any number I mean zero (including null ^) or more times). Some example strings are:
{^, a, b, aab, abbb, aabbb, ...}
DFA for RE a*b* will be as follows:
a- b-
|| ||
▼| ▼|
---►((Q0))---b---►((Q1))
In figure: `(())` means final state, so both `{Q0, Q1}` are final states.
You need to understand following basic concept:
What is basically a regular language? And why an infinite language `a*b*` is regular whereas languages like `{ anbn | n > 0 }` are not regular!!
A language(a set) is called regular language, if it requires only bounded(finite) amount of information to keep store at any instance of time while processing strings of the language.
So, what is 'bounded' information?
For example: Consider a fan 'on'/'off' switch. By viewing fan switch we can say whether the fan is in the on or off state (this is bounded or limited information). But we cannot tell 'how many times' a fan has been switched to on or off in the past! (to memorize this, we require a mechanism to store an 'unbounded' amount of information to count — 'how many times' e.g. the meter used in our cars/bikes).
The language { anbn | n > 0 } is not a regular language because here n is unbounded(it can be infinitely large). To verify strings in the language anbn, we need to memorize how many a symbols there are and it requires an infinite memory storage to count because the number of a symbols in the string can be infinitely large!
That means an automata is only capable of processing strings of the language anbn if it has infinite memory e.g PDA.
Whereas, a*b* is of course regular by its nature, because there is the bounded restriction ‐ that b may come after some a ( and a can't came after b). And that is why every string of this language can be easily processed (or recognized) by an automata in which we have finite memory - and finite automata is a class of automata where memory is finite. Yes, in finite automata, we have finite amount of memory in the term of states.
(Memory in finite automata is present in the form of states Q and according to automata principal: any automata can have only finite states. hence finite automata have finite memory, this is the reason the class of automata for regular languages is called finite automata. You can think of a finite automata like a CPU without memory, that has finite register to remember its internal states)
Finite State ⇒ Finite Memory ⇒ Only language can be processed for which finite memory needs to store at any instance of time while processing the string ⇒ that language is called Regular Language
Absent of external memory is limitation of finite automate ⇒ or we can say limitation of finite automata defined class of language called Regular Language.
You should read other answer "finiteness of regular language" to learn scope of regular language.
side note::
language { anbn | n > 0 } is subset of a*b*
Also a language { anbn | 10>100 n > 0 } is regular, a large set but regular because n is bounded, hence finite automata and regular expression is possible for this language.
You should also read: How to prove a language is regular?
The proof is: ((a*)(b*)) is a well-formed regular expression, hence matching a regular language. a*b* is a syntactic shortenning of the same expression.
Another proof: Regular languages are closed to concatenation. a* is a regular language. b* is a regular language, therefore their concatenation, a*b*, is also a regular expression.
You can build an automat for it:
0 ->(a) 1
0 ->(b) 2
1 ->(a) 1
1 ->(b) 2
2 ->(b) 2
2 ->(a) 3
3 ->(a,b) 3
where only 3 is not an accepting state, and prove that the language is a*b*.
To prove that a language is regular, it is sufficient to show either:
1) There exists some DFA that recognizes it. In this case, the DFA is trivial.
2) The language can be expressed as a regular expression, as mentioned in another answer. a*b* is a regular expression to recognize this language.
A regular language is a language that can be expressed with a regular expression or a deterministic or non-deterministic finite automata or state machine.
A language is a set of strings which are made up of characters from a specified alphabet, or set of symbols. Regular languages are a subset of the set of all strings.
a closure property is a statement that a certain operation on languages, when applied to languages in a class (e.g., the regular languages), produces a result that is also in that class.
this RE shows..the type of language that accepts multiple of (a) if any but before (b)
means language without containing any substring (ba)
Regular languages are not subset of context free languages. For example, ab is regular, comprising all the strings made of substring of a's followed by substring of b's. This is not subset of a^nb^n, but superset.