Below question was asked in GATE 2008 paper :
If L and L' (L complement) are Recursively enumerable then L is ?
a) Regular
b) CFL
c) CSL
d) Recursive
Correct option was option (d) and I accept that it's true. But my question is why can't it be regular or CSL ?
Because I think if we consider L is regular, then L' is also regular (As Regular languages are closed under complementation). And now as L' is regular so according to 'Chomsky hierarchy' L' is also Recursively enumerable. As even L after being regular, it fits into the question statement then why option (a) is not a correct option ? Same goes for CSL, so why option (c) is also not a correct option?
A quick review of language classes -- we know that these 5 language classes are all (strict) subsets of each other:
regular ⊂ CFL ⊂ CSL ⊂ recursive ⊂ recursively enumerable
The question is asking, if we know a language L is recursively enumerable AND we know it's complement L' is also recursively enumerable, what can we say for certain which smaller class L is in?
The answer is equivalent to saying that if a language L is recursively enumerable and NOT recursive, then L' is NOT recursively enumerable. That statement is true, but the equivalent statement for any of the other language classes is not.
If L and L' are both recursively enumerable, then
a) L may be regular (indeed, if L is regular, then L' is regular as well, and all regular languages are recursively enumerable)... but there are non-regular languages whose complements are recursively enumerable
b) L may be a CFL (there are CFLs whose complements are also CFLs, as well as CFLs whose complements are not CFLs)... but there are non-context-free languages whose complements are recursively enumerable
c) L may be a CSL (there are CSLs whose complements are CSLs) ... but there are non-context-sensitive languages whose complements are recursively enumerable
d) L must be recursive because, by virtue of both L and L' being recursively enumerable, we have an effectively computable procedure for deciding whether or not any given string is in L: begin enumerating strings in each language, interleaving the enumerations (so you give the next string in L, then the next string in L', then back to L, etc.). Continuing this process will eventually find the target string either in L or L', at which point you can return true (if it was enumerated in L) or false (if enumerated in L').
Therefore, while it's true that L could be regular, CFL or CSL, it is also true that it might not be any of those; but it must absolutely be recursive. Therefore, that is the "best" answer and the only one that is generally correct in all cases.
Related
When I asked this question, one of the answers, now deleted, was suggesting that the type Either corresponds to XOR, rather than OR, in the Curry-Howard correspondence, because it cannot be Left and Right at the same time.
Where is the truth?
If you have a value of type P and a value of type Q (that is, you have both a proof of P and a proof of Q), then you are still able to provide a value of type Either P Q.
Consider
x :: P
y :: Q
...
z :: Either P Q
z = Left x -- Another possible proof would be `Right y`
While Either does not have a specific case that explicitly represents this situation (unlike These), it does not do anything to exclude it (as in exclusive OR).
This third case where both have proofs is a bit different than the other two cases where only one has a proof, which reflects the fact that "not excluding" something is a bit different than "including" something in intuitionistic logic, since Either does not provide a particular witness for this fact. However Either is not an XOR in the way that XOR would typically work since, as I said, it does not exclude the case where both parts have proofs. What Daniel Wagner proposes in this answer, on the other hand, is much closer to an XOR.
Either is kind of like an exclusive OR in terms of what its possible witnesses are. On the other hand, it is like an inclusive OR when you consider whether or not you can actually create a witness in four possible scenarios: having a proof of P and a refutation of Q, having a proof of Q and a refutation of P, having a proof of both or having a refutation of both.[1] While you can construct a value of type Either P Q when you have a proof of both P and Q (similar to an inclusive OR), you cannot distinguish this situation from the situation where only P has a proof or only Q has a proof using only a value of type Either P Q (kind of similar to an exclusive OR). Daniel Wagner's solution, on the other hand, is similar to exclusive OR on both construction and deconstruction.
It is also worth mentioning that These more explicitly represents the possibility of both having proofs. These is similar to inclusive OR on both construction and deconstruction. However, it's also worth noting that there is nothing preventing you from using an "incorrect" constructor when you have a proof of both P and Q. You could extend These to be even more representative of an inclusive OR in this regard with a bit of additional complexity:
data IOR a b
= OnlyFirst a (Not b)
| OnlySecond (Not a) b
| Both a b
type Not a = a -> Void
The potential "wrong constructor" issue of These (and the lack of a "both" witness in Either) doesn't really matter if you are only interested in a proof irrelevant logical system (meaning that there is no way to distinguish between any two proofs of the same proposition), but it might matter in cases where you want more computational relevance in the logic.[2]
In the practical situation of writing computer programs that are actually meant to be executed, computational relevance is often extremely important. Even though 0 and 23 are both proofs that the Int type is inhabited, we certainly like to distinguish between the two values in programs, in general!
Regarding "construction" and "destruction"
Essentially, I just mean "creating values of a type" by construction and "pattern matching" by destruction (sometimes people use the words "introduction" and "elimination" here, particularly in the context of logic).
In the case of Daniel Wagner's solutions:
Construction: When you construct a value of type Xor A B, you must provide a proof of exactly one of A or B and a refutation of the other one. This is similar to exclusive or. It is not possible to construct a value of this unless you have a refutation of either A or B and a proof of the other one. A particularly significant fact is that you cannot construct a value of this type if you have a proof of both A and B and you don't have a refutation of either of them (unlike inclusive OR).
Destruction: When you pattern match on a value of type Xor A B, you always have a proof of one of the types and a refutation of the other. It will never give you a proof of both of them. This follows from its definition.
In the case of IOR:
Construction: When you create a value of type IOR A B, you must do exactly one of the following: (1) provide only a proof of A and a refutation of B, (2) provide a proof of B and a refutation of B, (3) provide a proof of both A and B. This is like inclusive OR. These three possibilities correspond exactly to each of the three constructors of IOR, with no overlap. Note that, unlike the situation with These, you cannot use the "incorrect constructor" in the case where you have a proof of both A and B: the only way to make a value of type IOR A B in this case is to use Both (since you would otherwise need to provide a refutation of either A or B).
Destruction: Since the three possible situations where you have a proof of at least one of A and B are exactly represented by IOR, with a separate constructor for each (and no overlap between the constructors), you will always know exactly which of A and B are true and which is false (if applicable) by pattern matching on it.
Pattern matching on IOR
Pattern matching on IOR works exactly like pattern matching on any other algebraic datatype. Here is an example:
x :: IOR Char Int
x = Both 'c' 3
y :: IOR Char Void
y = OnlyFirst 'a' (\v -> v)
f :: Not p -> IOR p Int
f np = OnlySecond np 7
z :: IOR Void Int
z = f notVoid
g :: IOR p Int -> Int
g w =
case w of
OnlyFirst p q -> -1
OnlySecond p q -> q
Both p q -> q
-- We can show that the proposition represented by "Void" is indeed false:
notVoid :: Not Void
notVoid = \v -> v
Then a sample GHCi session, with the above code loaded:
ghci> g x
3
ghci> g z
7
[1]This gets a bit more complex when you consider that some statements are undecidable and therefore you cannot construct a proof or a refutation for them.
[2]Homotopy type theory would be one example of a proof relevant system, but this is reaching the limit of my knowledge as of now.
The confusion stems from the Boolean truth-table exposition of logic. In particular, when both arguments are True, OR is True, whereas XOR is False. Logically it means that to prove OR it's enough to provide the proof of one of the arguments; but it's okay if the other one is True as well--we just don't care.
In Curry-Howard interpretation, if somebody gives you an element of Either a b, and you were able to extract the value of a from it, you still know nothing about b. It could be inhabited or not.
On the other hand, to prove XOR, you not only need the proof of one argument, you must also provide the proof of the falsehood of the other argument.
So, with Curry-Howard interpretation, if somebody gives you an element of Xor a b and you were able to extract the value of a from it, you would conclude that b is uninhabited (that is, isomorphic to Void). Conversely, if you were able to extract the value of b, then you'd know that a was uninhabited.
The proof of falsehood of a is a function a->Void. Such a function would be able to produce a value of Void, given a value of a, which is clearly impossible. So there can be no values of a. (There is only one function that returns Void, and that's the identity on Void.)
Perhaps try replacing “proof” in the Curry-Howard isomorphism with “evidence”.
Below I will use italics for propositions and proofs (which I will also call evidence), the mathematical side of the isomorphism, and I will use code for types and values.
The question is: suppose I know the type for [values corresponding to] evidence that P is true (I will call this type P), and I know the type for evidence that Q is true (I call this type Q), then what is the type for evidence of the proposition R = P OR Q?
Well there are two ways to prove R: we can prove P, or we can prove Q. We could prove both but that would be more work than necessary.
Now ask what the type should be? It is the type for things which are either evidence of P or evidence of Q. I.e. values which are either things of type P or things of type Q. The type Either P Q contains precisely those values.
What if you have evidence of P AND Q? Well this is just a value of type (P, Q), and we can write a simple function:
f :: (p,q) -> Either p q
f (a,b) = Left a
And this gives us a way to prove P OR Q if we can prove P AND Q. Therefore Either cannot correspond to xor.
What is the type for P XOR Q?
At this point I will say that negations are a bit annoying in this sort of constructive logic.
Let’s convert the question to things we understand, and a simpler thing we don’t:
P XOR Q = (P AND (NOT Q)) OR (Q AND (NOT P))
Ask now: what is the type for evidence of NOT P?
I don’t have an intuitive explanation for why this is the simplest type but if NOT P were true then evidence of P being true would be a contradiction, which we say as proving FALSE, the unprovable thing (aka BOTTOM or BOT). That is, NOT P may be written in simpler terms as: P IMPLIES FALSE. The type for FALSE is called Void (in haskell). It is a type which no values inhabit because there are no proofs of it. Therefore if you could construct a value of that type you would have problems. IMPLIES corresponds to functions and so the type corresponding to NOT P is P -> Void.
We put this with what we know and get the following equivalence in the language of propositions:
P XOR Q = (P AND (NOT Q)) OR (Q AND (NOT P)) = (P AND (Q IMPLIES FALSE)) OR ((P IMPLIES FALSE) AND Q)
The type is then:
type Xor p q = Either (p, q -> Void) (p -> Void, q)
Suppose L is an infinite regular language. Does it follow that there exists a finite language S such that L = SS* ? Prove or disprove by finding a counterexample.
What i have tried:
Intuitively this should be true. Any infinite language can be represented by a finite language S if S has the same alphabets as L e.g if L is the infinite language over the alphabet {a, b}* then S = {a, b} works, so essentially S contains just one occurrence of all the alphabets in L. Is this correct or am i missing something fundamental? or is this just not valid at all?
Any help would be appreciated!
My intuition on this is that it's not true. Let's take the example of the language of all odd-length strings over {a, b}. This is trivially regular, and trivially infinite. However, any finite subset of this language will have odd-length strings, and any infinite suffix would have to have an even length, so there is no reasonable construction of L = SS* for some finite language S.
I'll leave turning this intuition into a formal proof to the reader.
Here's an alternative counterexample that might be easier. Consider the infinite regular language ab*. Now suppose L = SS* for some S. Now, either S contains a string with a in it, or it does not. If it does, then L = SS* contains strings with multiple a's, so it cannot be the language ab*. If S does not contain a, then L = SS* contains no strings with any a's at all, and can't be the language ab*. In either case, L is not ab*, a contradiction. So L cannot be written as SS* for any S.
I have this exercise for homework:
Say we have a language L. we know that the language pref(L) (all the prefixes of L, including all the words in L itself) is a regular language. Does this imply that the language L is regular as well?
I took the NFA of pref(L) and divided it (via 2 epsilon transitions from q0) to 2 separate NFA's, as 1 defines L and the other defines pref(L)\L.
What I actually got is a NFA for L, which means it is regular.
I am not sure this is the way or if it legal. I'd be glad for another lead.
Thanks in advance,
Yaron.
It is not necessarily the case that if pref(L) is regular, then L is regular as well.
As an example, let Σ = {a} be a unary alphabet. I'm going to claim that if L is any infinite language over Σ, then pref(L) = Σ*. To see this, first note that pref(L) ⊆ Σ* because every pref(L) is a language over Σ. Now, consider any string in Σ*, which must have the form an. If L is an infinite language over Σ, it must contain at least one string of the form am where m ≥ n. Then an would be a prefix of am, so an ∈ pref(L). This shows that Σ* ⊆ pref(L) and that pref(L) ⊆ Σ*, so in this case Σ* = pref(L).
Now, all we need to do is find a nonregular language over Σ = {a}. As an example, take the language { a2n | n ∈ N } of all strings whose length is a power of two. It's possible to prove using either the Myhill-Nerode theorem or the pumping lemma that this language is not regular. However, by the above result, we know that pref(L) is a regular language.
Hope this helps!
I've tried to look for the answer and I'm getting conflicting answers so I'm not sure. I know the reverse is true, that if L is regular then L* is regular under closure.
I imagine that if L* is regular then L is regular because the subset of L* should be regular and L is part of that subset.
If L* is regular, then L is not necessarily regular. For example, consider any nonregular language L over an alphabet Σ such that Σ ⊆ L. (That is, imagine you have a nonregular language where each individual character in the alphabet is a string in L.) In that case, L* = Σ*, since you can form any string as the concatenation of all the individual characters of Σ.
Here's one possible example. Let Σ = {a} and consider the language L = { a2n | n ∈ N }. This language is not regular, and you can prove it using either the pumping lemma for regular languages or the Myhill-Nerode theorem. However, the language L* is the language a*, which is regular. To see this, notice that since L contains the string a, the language L* contains all strings of the form an for any natural number n.
Another option: pick L to be any nonregular language over Σ, then consider the language L ∪ Σ. This is also a nonregular language (if L ∪ Σ were regular, then we could subtract out each character added in via the union, leaving a regular language at each step, to show that L is regular), and it satisfies the above requirements.
Hope this helps!
Take L = {a,b}*, which is regular, but has a non-regular subset L={a^n b^n} (this one can be proved to be non regular by pumping lemma...), so it's not the case that all subsets of a regular language are regular.
I think I know the pumping lemma and was told that Myhill-Nerode is a very elegant way to show that something is regular or not regular. But I am having a lot of trouble with it. Take this for example:
= {0k, k = 2n, n >̲ 1}
My language is the repetition of 0 to a length that's a power of 2. I want to use the Myhill-Nerode to show that this is either regular or not regular. Is it possible?
I know how to set this up to resemble other Myhill-Nerode looking proofs but I don't understand the equivalence concept that much.
I could say that I have some and where ≠ and both are of the form 2h and , I then define , and so that:
= 0j/2
= 0p/2
= 0j/2
Where = 0j/20j/2 = 0j is in my language since is of the form 2n, however = 0p/20j/2 is not guaranteed to be in my language for every p and j, since ≠
Given a languages L, two strings u, v 2 L are equivalent if for all
strings w belong to sigma * we have that u.w belongs to L iff v.w belongs to L
consider the set of {0,0^2,0^4,0^8....}, in this case for some m and n 0^m and 0^n should be mapped to the same equivalence class or else there would be infinite equivalence classes making it non-regular by Myhill-Nerode theorem. However 0^m.0^m belong to L but 0^n.0^m would not ..Hence