I'm trying to prove that the complement of L= {a^i b^i c^i : i >= 1} is a context free. L complement is: {w is a word over {a,b,c}* : w not in L}.
As we know, context-free languages are closed under union. So, I'm trying to divide my language (complement of {a^i b^i c^i}) into context-free subsets in which their union must be context-free. Could anyone help me to find the subsets? Each time I try to, I end up with L*!
Thank you.
Note: In the following, I left out the constraint that L does not include the empty string, but that only requires a small adjustment.
Consider aibjck.
If i=j and j=k then you have aibici. Conversely, if i≠j or j≠k, then you have the part of the complement of aibici which consists of strings in a*b*c* (the rest of the complement is straight-forward).
In other words,
L = { aibjck | i=j } ∩ { aibjck | j=k }
and
L' = { aibjck | i≠j } ∪ { aibjck | j≠k }
It's easy to show that each subset in the above equations is context-free. As you say, context-free languages are closed under union, but they are not closed under intersection; consequently, L' is context-free even though L is not.
Related
I have understood that L={wxw^r|w,x belongs to {a,b}^* } is regular because it turns out to be the pattern of starting and ending with same symbol but I am not getting the proper explanation that how to say L={ww^rx|w,x belongs to {a,b}*} is regular language using DFA design.
Please help me in understanding this!
This is a trick question. The language L as you have specified is the language of the regular expression (a + b)*, that is, any string of a's and b's. The trick is that for any string y = s1.s2.s3...sk where si in {a, b}, we can write y = wxw^R where w is the empty string and x = y. Basically, the trick is that we can always choose w to be the empty string, and in that case we are left with L = {x | x in {a, b}^*}, clearly regular. Another way of thinking about it is this: can you find any string of a's and b's that is not in L? Is it not in L even if you take w to be the empty string?
let L = {wwRu | w, u ∈ {0,1}+}. Is L regular language ? Note that w, u cannot be empty.
I've tried to prove it is not regular language by the pumping lemma, but I failed when w = 0^p1^p, 01^p, (01)^p. Once I take y = 0^p or 1^p, xyyz will be 00.../11.../01^n0... etc.
And I cannot draw its DFA/NFA or write its regular expression to prove it is regular language.
So is L regular or not ? How can I prove it ?
The language is not regular, and we can prove it using the Myhill-Nerode theorem.
Consider the sequence of strings 01, 0101, ..., (01)^n, ...
First, notice that none of these strings are in the language. Any prefix of any of these strings which has even length is of the form (01)^2m for some m, and therefore just a shorter string in the sequence; splitting such a prefix in two either has both substrings start with 0 and end with 1, or else it has the first substring start and end with 0 and the second start and end with 1. In either case, these strings are not of the form w(w^R)u for any w or u.
Next, notice that the shortest possible string which we can append to any of these strings, to produce a string in the language, is always the reverse of itself followed by either 0 or 1. That is, to turn 01 into a string in the language, we must append 100 or 101; there are no shorter strings we can append to 01 to get a string in the language. The same holds true for 0101: 10100 and 10101 are the shortest possible strings that take 0101 to a string in L. And so on for each string of the form (01)^n.
This means that each string of the form (01)^n is distinguishable with respect to the target language w(w^R)u. The Myhill-Nerode theorem tells us that a minimal DFA for a regular language has exactly as many states as there are equivalence classes under the indistinguishability relation. Because we have infinitely many distinguishable strings with respect to our language, a minimal DFA for this language must have infinitely many states. But, a DFA cannot have infinitely many states; this is a contradiction. This means that our language cannot be regular.
The language is REGULAR:
L = 00(0+1)+ + 11(0+1)+ + 0(11)+0(0+1)+ + 1(00)+1(0+1)+
Can any one help to figure out that L = { am bn, m ≥ n + 2, m ≤ 3 } is regular or not using pumping lemma, It seems to be a bit difficult to prove.
I have tried to used pumping lemma and it shows that it is regular language but i am really confused that is this right or not.
I have tried to used pumping lemma and it shows that it is regular language but I am really confused that is this right or not.
First note, the pumping lemma can be used to proof that "certain language is not a regular language". But we can not use pumping lemma to proof that "certain language is a regular".
Yes, the pumping lemma for regular languages describes an essential property of all regular languages, and if a languages don't satisfy the conditions described by pumping lemma or say don't satisfy pumping lemma property then that language is actually not a regular language, But the converse of this is not true!! &mdash- there are "languages those satisfies pumping lemma's conditions and may still be non-regular", means:-
Pumping lemma is 'necessary but not sufficient condition' for a language to be regular.
This is something like: every engineer is math student - so if a student is engineer we can say that he knows math, but their are math students those don't know engineering. (just giving you an example to explain)
From your question, I have impression that you don't understand basically - "what is a regular language?".
Unlike pumping lemma for regular languages, we need to learn some other characteristics of regular languages - finding them in a language proofs that the language is a regular language. If you can represent a language using finite automata or/and regular expression that proofs that the language is a regular language.
Now, come to your original question:
How to proof that language { am bn, m ≥ n + 2, m ≤ 3 } is regular or not?
Because neither 'm' or 'n' can be negative (a string without occurrence of any symbol is possible eg. empty string1, but a string with negative occurrence of language symbols is not possible) and according to condition "m ≥ n + 2", 'm' is always greater than 'n' — so minimum value of 'm' (when 'n' is 0) is 2. From second condition "m ≤ 3", maximum value of 'm' is '3', and so 'm' can be either 2 or 3.
If you again notice fist condition: "m ≥ n + 2" for m = {2, 3} possible values for 'n' can be:
for m = 2, n can be 0 and that makes string 'aa'.
for m = 3, n can be 0 or 1 and that makes strings 'aaa', 'aaab'.
So, it came that the language is a finite language e. only consists of three strings. Every finite language is a regular language, check venn diagram:
(to lean more about this read: Finiteness of Regular Language)
Regular expression for your language L = { am bn, m ≥ n + 2, m ≤ 3 } would be aa(a + Λ)(b + Λ),
hence it is proofed that language is a regular language.
1 a string without occurrence of any symbol is called empty or null string in formal languages, in most of formal language books symbol Ɛ for empty string.
Λ is empty symbol.
we cannot use the pumping lemma for proving language is regular.To prove a language is regular we should design DFA for that language
I understand the reason and the proof why {a^n b^n | n >= 0} is NOT regular.
Why is {a^nb^n | n >= 0} not regular?
The solution of one of my exercises is: {a^n a^n | n >= 0} is regular. How can I prove this thesis?
Yes, Language {an an | n >= 0} is a regular language. To proof that certain language is regular, you can draw its dfa/regular expression. And you can drive do for this language as follows:
Because "anan for n >= 0" is same as "a2n for n >=0", and that is "set of all string contests of even number of symbol a" that is regular — regular expression for this is (aa)*.
Note, regular expressions is only possible for regular languages hence it is proved that {an an | n >= 0} is a regular language. and DFA would be:
I would suggest you to read this why languages like {an bn | n >= 0} are not regular.
First change the definition to the equivalent L = {a^2n | n >= 0}. Now observe that any string that belongs to L is simply a multiple of 2 as. Then change that definition to (aa)*, which is a regular expression since it only uses primitives for expressing regular languages - individual characters (a), concatenation (aa) and Kleene star (*). Now you're done.
Im looking forward to design a FA which accepts some kind of string that accept some A and B.
First a string which the number of A is five more times higher than B.
i mean L={w∈{A,B}* and (nA(W)-nB(W) mod 5=0)
And also a FA which accept different number of character in a string:
L={A^n B^m C^k | n,k>0 and m>3}
I design some FAs But they did not work perfectly on this complicated strings.
Any help on how should i design this ?
Unfortunately, your questions are confusing as the english text doesn't agree with the mathematical formula. I will try to answer to these four questions then:
A language which consists of string over {a,b} that the number of a (= #a(w))
is five times as the number of b ( #b(w)),
L = { w in {a,b}* : #a(w)>#b(w) and #a(w)=#b(w)mod5 }
This cannot be done by an NFA. The proof is simple by using the pumping lemma (P.L) with the string a^pb^5p, where p is the constant of P.L.
For the language: L={w∈{A,B}* : (nA(W)-nB(W)) mod 5=0} that you wrote,
you can do it with an DFA that consists of a cycle of 5 states.
The transitions are, if you read a go clockwise if you read b go counter-clocwise. Choose at random one state to be initial state and the same state will be the final state.
For the language L={A^n B^m C^k | n,k>0 and m>3}, it should be easy to find out
if you read L as L=A(A)* B(B)* c^4(C)*
For the language that accepts different number of character in the string (let's say over a,b). The language should be R={ w in {a,b}* : #a(w) not equal #b(w)}
This language again it cannot be recognized by an NFA. If this language was regular (recognzied by an NFA) so would be this language:
L=a*b* intersection (R complement). The language L is {a^n b^n/ n non-negative integer}.
The language L is the first example of most books when they speak about languages that are non-regular.
Hopefully, you will find this answer helpful.