For example, if I have a non-context free language of B, is there such a context free language A such that A is a subset of B? I have been thinking of examples but unable to think of any valid ones.
I thought I got it when I said that A = {A = {w | w is of even length and w ∈ {a, b, c}}, which is context-free, and B = {ww | w ∈ {a,b,c}}, which is not context free. However, I realized that there are some strings A can produce that B cannot, and therefore, A is not a subset of B.
Does anyone know of any examples that could be valid in my situation?
Any finite set of strings is a context-free language. (Indeed, it is a regular language.) So any finite subset of a language is context-free, regardless of what the language is.
Another trivial case is the language L = L1 ∪ L2 where the alphabet of L1 is Σ1 and the alphabet of L2 is Σ2 and Σ1 ∩ Σ2 = ∅. Now L is context-free only if both L1 and L2 are context-free. (This is not the case if the alphabets are not disjoint.) So if exactly one of L1 and L2 is context-free, then it is a subset of the non-context-free language L.
If neither of those are interesting enough for you, then the language { a* } (where a is a symbol) is a subset of { ww | w ∈ {a, b}* }. Another subset of the same classic non-context-free language is { ww | w ∈ {a, b}* ∧ w = wR } (that is, the language of all duplicated even-length palindromes) which is context-free because it is exactly the same as { wwR | w ∈ {a, b}* ∧ w = wR }, as a result of the second condition.
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For DFA, the alphabet Σ = {a,b} the language this DFA will recognise is {w|w must contain a}. For this language that the DFA recognises, what would be the compliment of this language or how would it be written please?
I can't work out if b should also be in the compliment of the language or not. a I know should be excluded from the compliment as it is in the original L or language.
I would have posted an image of the alphabet but I don't have enough points or reputation to.
I am unable to determine the complement of the language that the DFA recognises
The complement of the language {w in {a, b}* | w must contain a} is the language {w in {a, b}* | w contains no a}. Strings in this language can be empty or consist of any number of b; the regular expression for this is b* and there is a 2-state minimal DFA that loops on the accepting initial state in case of b, and goes to a non-accepting dead state in case of a, where both transitions from the dead state are loops back to the dead state itself.
I have a vital misunderstanding of the pumping lemma. In the following example I show an example of using it on a regular language to come to incorrect conclusions. What am I doing wrong?
L={a*b*}, assume the language is regular so by the pumping lemma there exists some n, and σ = αβγ and σ' = αβ^kγ ∈ L for all non negative k.
σ = aaabbb
α = aa
β = ab
γ = bb
then σ'= αβ^2γ for k=2, σ' =aaababbb
σ'∉ L, a contradiction, thus L is not regular.
L as described I know to be a regular language so I would expect to find ∈ L. This is due to my choice of β spanning across two characters but there is nothing I can find in the pumping lemma which forbids this.
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've been watching lectures from Coderisland on YouTube about finite state machines, DFAs and NFAs, and in one discussion he talks about how to use the pumping lemma to show how a language is not regular. I don't know quite how to apply the lemma and want to understand if I'm doing it right. If I had something like:
w = {anbk, n =/= k}
am I correct in that I can say that:
h = {anbn + r, r > 0} is a subset of w, and thus if I show by the lemma that h is not regular, that w must not be regular since h is a subset of w.
The way I would show this is as follows:
h = xyz
|xy| <= n
x = an-r
y = ar
z = bn + r
xyz = an-rarbn + r
xy2z = an-ra2rbn + r = an + rbn + r
Thus h cannot be regular since an + rbn + r is not of the form {anbn + r, r > 0}, and since h is not regular w must not be regular, since h is an element of w.
Have I applied it correctly? I understand how to apply it for an easy language like {anbn}, because I can apply the lemma directly to this language, but the only way I could think of for my language was to create a subset that belongs to my language, and apply the lemma to that.
If I haven't applied it correctly, is there a way to show that my language is not regular (or regular), using another lemma, or perhaps with closure properties?
This is a really awesome topic, even if I don't understand the pumping lemma fully, I'm excited to explore it further!
There are two mistakes in your proof.
First, is the mistake in this statement of yours:
if I show by the lemma that h is not regular, that w must not be regular since h is a subset of w.
Because consider the language L = {anbm | n,m natural numbers}
h is a subset of L, but clearly L is regular, as it can be represented by the regular expression a*b*.
But you're so close to the solution, in which actually you don't need to consider h. You should instead choose a string in w so that however you apply Pumping Lemma to it, you will always get a string which is not in w.
Now this is your second mistake in step 3 of your pumping lemma. In pumping lemma, we are to prove that "there is an element of that language, such that however you apply pumping lemma to that string, you will always get a string that is outside the language".
In your proof, you deliberately pick x = an-r, without explaining why it must be the case. There might be the case that actually x = an-2 for example. In this case, fortunately, you still have the same conclusion that it doesn't satisfy pumping lemma (and hence it's not regular) since by considering xyr+1z you will definitely have more a's than b's.
One correct way to prove your problem is applying pumping lemma to it directly (there are other methods such as using complement, or intersecting it with a regular language and prove that the intersection is not regular), but for the purpose of explaining pumping lemma, I will show you how to apply pumping lemma for this language.
So, the problem states that we have w = {anbk, n=/=k} and we are to prove that it's not regular.
Now consider the string s = anbn!+n (that is, n number of a's followed by n factorial plus n number of b's). By pumping lemma, if w is regular, there should be xyz such that s = xyz, and that |xy|<=n.
Since |xy|<=n and that we have an at the beginning of s, then both x and y must contain only a's. Let y = am. We know that 1<=m<=n.
Now we have the difference between the number of a's and b's is (n!+n)-n = n!. This number is divisible by any number in the range 1 to n, inclusive. So we have n!/m is a whole number. Let q = n!/m.
Considering the string xyq+1z, we have the number of a's as (n-m)+(q+1)*m = n-m+qm+m = n+qm = n+n!, which is the same as the number of b's. So xyq+1z is not in the language w since number of a's is the same as the number of b's. Therefore the language w doesn't satisfy the pumping lemma.
Therefore w is not regular.
To address the question in your comment, prove using complement.
Suppose w is regular. Then the complement w' should also be regular.
But the complement w' = {anbn, n natural number} is not regular (you showed that already, right?).
Therefore w is not regular.