Prove that the language defined over {0,1} where the number of 0s of x=number of 1s of x is not regular.
Myhill-Nerode says that there are as many equivalence classes over the indistinguishability relation w.r.t. a regular language as there are states in a minimal DFA for that language.
Two strings w and x are indistinguishable w.r.t. a regular language L if for any y such that wy is in L, xy is also in L; and if for any z such that xz is in L, wz is in L also. In other words, w and x are indistinguishable if they can have exactly the same set of strings concatenated to them to get some string in L.
For your language, we can show the number of equivalence classes under this relation is infinite. Because no DFA has infinitely many states, this is a contradiction; by showing this we show the language is not regular.
In your case, we can show that 0^n is distinguishable from 0^m, for m > n, since the shortest string that can be appended to 0^n to get a string in your language is 1^n, whereas the shortest string that can be appended to 0^m to get a string in your language is 1^m, and m > n. Thus, each string 0^n is distinguishable from every other string 0^m, n != m, so there are infinitely many equivalence classes. As we said earlier, this means the language cannot be regular.
If we had found that a finite number of equivalence classes existed and named them, we'd have also found a minimal DFA for the language. Simply add one state per equivalence class and figure out the transitions.
Related
I'm having trouble understanding when the empty string (epsilon) is a subset or element of an alphabet? My understanding was that epsilon was only part of a language, but my TA in the class said it was an element of all alphabets so now I am confused.
e.g. would {a,b,c} contain epsilon as an element?
e.g. would {} contain epsilon as an element?
e.g. is {eps} a subset of all alphabets or languages?
This question is probably better suited for cs.stackexchange but I will try to help you according to my understanding, please do correct me if necessary.
In general, your own intuition seems quite correct to me. ϵ is not automatically part of every alphabet. It is the empty string of characters.
However, this means that ϵ is a string over any alphabet, even your alphabet {a, b, c}.
So to answer your three examples:
No, it does not. If {a, b, c} is an alphabet, it is a set of symbols, and ϵ is a string. However, ϵ is definitely part of some languages defined over this alphabet.
No, {} is the empty set, and it contains nothing, not even ϵ.
{ϵ} is the set containing only ϵ. ϵ is a string, not a symbol, so it is not a subset of all alphabets (however, it seems there are cases where some alphabets are defined to contain ϵ but that is a different confusing story). It is also not a subset of all languages, because consider the language L = {aa, ab, ba, bb}. The empty string ∈ is not one of these elements.
The analogue to set theory might cause some additional confusion. Notice that the empty set is a subset of every set. The empty string is not a subset of every language, but rather it is a substring of any string.
I have been asked to prove if the following set is decidible, semi-decidible or not semi-decidible:
In other words, it is the set of inputs such that exists a Turing Machine encoded with the natural y with input p that returns its input.
Consider the set K as the set of naturals such that the Turing machine encoded with x and input x stops. This is demonstrated to be a non-decidible set.
I think that what I need is to find a reduction of K to L, but I don't know how to prove that L is decidible, semi-decidible or not semi-decidible.
L may not look decidable at first glance, because there is this nasty unbounded quantifier included, which seems to make necessary a possibly infinite search when you look for a y satisfying the condition for a specific p.
However, the answer is much simpler: There is a turing machine M which always returns its input, i.e. M(p) = p holds for all p in the considered language. Let y be a code of M. Then you can use this same y for all p, showing that L contains all words of the language. Hence L is of course decidable.
In fact, this is an example to demonstrate the principle of extensionality (if two sets have the same elements and one is decidable, then the other is decidable too, even if it doesn't look so).
I encountered this doubt in an online intro-logic open course offered by Stanford Uni.
Under the section 9.4 of this textbook here: http://logic.stanford.edu/intrologic/secondary/notes/chapter_09.html
It says:
The axioms shown here define the same relation in terms of 0 and s.(where the functional constant letter s below represents the successor function, e.g. s(0)=1, s(1)=2, s(2)=3 )
∀x.same(x,x)
∀x.(¬same(0,s(x)) ∧ ¬same(s(x),0))
∀x.∀y.(¬same(x,y) ⇒ ¬same(s(x),s(y)))
As my understanding, :
The first sentence says two identical numbers are same. The second and third sentences are used to define what is not same.
The second says no successor of any number is same to 0.
The third says if two numbers are not the same, then their successors are not same. For example, if 1≠3, then 2≠4.
However, I think the third sentence should be bi-conditional because, if I'm not wrong, the definition didn't cover the instance where the number being testified are smaller than the given number,otherwise it is possible to say if 2≠4, then 1=3.
So I wondered is this an error in text book or there's something wrong of my reasoning.
There is no error in this text book. While the statement does hold in both directions, there is no need to state it as an axiom since the other direction follows from the functional property of the successor function and the three axioms listed in the textbook.
A formal proof would involve the axioms that define the successor function. Someone more accustomed to the use of automated provers or just a good student of logic might be able to complete such a formal proof.
Here is just a sketch of a proof. It uses the symbol "=" to denote term equality, i.e. u=v means u and v are syntactically identical terms written using the symbols 0 and s(). Also "u<v" means that u and v are both ground terms and u has strictly less applications of s() than v.
Suppose
∀x.∀y.(¬same(s(x),s(y)) ⇒ ¬same(x,y))
does not hold, then there exist some terms x0 and y0 such that
same(x0,y0) and ¬same(s(x0),s(y0)).
Since s(x0) is a function, it follows from ¬same(s(x0),s(y0)) and ∀x.same(x,x) that x0 and y0 are two different terms. First let us consider the case when x0 < y0, then y0 = s(...s(x0)) where there are n applications of s() and n > 0. The other case when y0 < x0 can be handled similarly.
Substituting s(...s(x0)) for y0 in same(x0,y0) we get same(x0,s(...s(x0))).
Also x0 = s(...s(0)) where there are m applications of s() for some nonnegative integer m. Using the third axiom in the direction provided we can say that if same(s(u),s(v)) then same(u,v). Thus from same(x0,s(...s(x0))) we can "strip" m applications of s() to obtain
same(0,s(...s(0))) where there are n applications of s() in the second argument. This contradicts the second axiom. Q.E.D.
I have to draw a Finite Automaton that accepts the following string
Λ, a, aabc, acba and accb
In my view a(a+b+c)* this might be it's regular expression as the string is starting from a and includes an empty string as well.
Now I didn't find the logic of drawing FA as in below image
Question 1: If the string is starting with a then in FA, We are moving from x to y while reading b
Why we don't read a here.
Question 2: Why we use loop of a,b on state y and z
The language L = {λ, a, aabc, acba, accb} is finite. Therefore, L is not equivalent to the language denoted by the Kleene closure of the regular expression a(a + b + c), which is infinite. There is a simple algorithm that generates the nondeterministic finite automaton accepting a finite language, which consists of drawing paths accepting each of the strings in the language.
It's unclear what the relationship between the two languages and the diagram in the original post is, since the automaton in the diagram accepts neither language. Assuming that the nodes are labeled with their names, and circled nodes indicate acceptance, the language accepted by the automaton in the diagram is (a + b)*. In this case, the loops are used to accept the Kleene closure of (a + b). That said, it would be useful if you could clarify the meaning of the diagram.
I am studying membership algorithms and I am working on this particular problem which says the following:
Exhibit an algorithm that, given any regular language L, determines whether or not L = L*
So, my first thought was, we have L* which is Kleene star of L and to determine if L = L*, well couldn't we just say that since L is regular, we know L* is by definition which states that the family of regular languages is closed under star-closure.
Therefore L will always be equal to L*?
I feel like there is definitely a lot more to it, there is probably something I am missing. Any help would be appreciated. Thanks again.
since L is regular, we know L* is by definition which states that the family of regular languages is closed under star-closure. Therefore L will always be equal to L*?
No. Regular(L) --> Regular(L*), but that does not mean that L == L*. Just because two languages are both regular does not mean that they are the same regular language. For instance, a* and b* are both regular languages, but this does not make them the same language.
A example of L != L* would be the language L = a*b*, and thus L* = (a*b*)*. The string abab is part of L* but not part of L.
As far as an algorithm goes, let me remind you that the concept of a regular language is one that can be parsed by a DFA - and for any given DFA, there is a single optimal reduction of that DFA.
The implication that you stated is wrong. Closedness under the Kleene star means only that L* is again regular, if L is regular.
One possibility to check whether L = L* is to compute the minimal automaton for both and then checking for equivalence.