My Argument/Answer is if y is aregular set then there exits a DFA which accepts y. In the L1 there is a condition that y=x^n, that x will belong to L1,as y is accepted by DFA. So is x^n and so is x so L1 is regular. Now L2 --> here the condition is x=y^n. Here y is accepted by DFA so is y^n so which is equal to x so x can be accepted by DFA. This makes both L1,L2 regular
Is my argument right?
This question seems like it's poorly-posed. For example, if we take A = {a}, then L1 is the language {a} and L2 is the language a*, both of which are regular. If we pick A = a*b, then L1 = a *b (which is regular) and L2 = { (anb)m | m, n ≥ 0 }, which is not regular (using the pumping lemma). In other words, the answer depends on the choice of A.
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I need to decide whether the language L = {wxw| x, w ∈ {a,b}* , w≠ε} is regular or not.
I know that L2 = {wxwR| x, w ∈ {a,b}* , w≠ε} is regular, since you can make sure that the word start and ends in the same letter, but it doesn't seem to work without the reversing, (for example w = 10, x = ε)
How do I prove it though?
It is not regular. Use Myhill-Nerode. Consider the prefix a^n b. The shortest string ending with b that can be appended to this to get a string in the language is a^n b. This means every such prefix is distinguishable and there's no DFA.
This is weird but by pumping lemma, say
Let L be a regular language. There exists a constant n such that for every string w in L such that |w| >= n, we can break w in to xyz such that xy*z is also in L.
This lemma is strong because it argues for all regular languages. But what if the regular language L = a? There is only one word (a) in it. How the pumping lemma works for this case?
If n = 2 then it is vacuously true that any w in L with |w| >= n satisfies the conclusion of the pumping lemma. No words in L are long enough to serve as counterexamples. More generally, if L is any finite language then L satisfies the pumping lemma: just take n to be greater than the length of the longest word in L.
Ok, I know that this isn't a programming question but it is a computing question so it is relevant.
Basically, how can I use the pumping lemma to prove that this language is not regular?
{w in {0,1}* | if the length of w is odd then the middle symbol is 0}
Please answer this as simple as possible as whilst I know about models of computation, I am relatively new to it.
Thank you very much in advance!
According to the pumping lemma, if that language is regular then there must exist a number p such that for all strings longer than p in the language, we can decompose that string into x + y + z, where each of x, y, and z are strings and |y| >= 1, |x + y| <= p, and x + (y * i) + z is in the language for all non-negative integers i.
Now observe that for every non-negative integer i, the string "1" * i + "0" + "1" * i is in the language. (That is, the string of i 1s followed by a single 0 and then i more 1s)
Specifically, the string S consisting of p 1s followed by a 0 and then p more 1s is in the language. Since this string has length 2 p + 1, this string is long enough that it can be broken into three strings x, y, and z as in the pumping lemma. Since |x + y| <= p, it must be that x and y are all 1s, and the only 0 character in S is in z. Now consider the string S' = x + y + y + y + z. Since we added 2*|y| characters to it, S' must also have an odd length. But we added some number of 1 characters to the left of the only 0 in S, and didn't add any 1 characters to the right of the 0. So S' doesn't have a 0 as its middle character, and therefore S' isn't in the language.
Therefore, we've shown that the language can't be pumped as the pumping lemma requires. Therefore, the language is not regular.
Given an alphabet of {a, b} where Na denotes the number of occurrences of a, and Nb the number of occurrences of b:
L1 = {xy | Na(x) = Nb(y)}
L2 = {w | Na(w) and Nb(w) are even number}
Wouldn't a single DFA with four states and using mod be able to accept both languages?
No, because both languages are different so you can't draw single DFA for both languages.
An automaton uniquely defined a language, but yes of-course for a language more than one automata are possible called 'equivalent automata'.
Language L1 = A = {xy | Na(x) = Nb(y)} is a regular language. Regular expression for this language is:
(a + b)*a(a + b)*b(a + b)* + ^
To understand this language and regular expression read: "Show that the following set over {a, b} is regular".
Language L2 = A = {w | Na(w) and Nb(w) are even number} is also a regular language. Regular expression for this language is:
((a + b(aa)*ab)(bb)*(ba(aa)*ab(bb)*)*a + (b + a(bb)*ba)(aa)*(ab(bb)*ba(aa)*)*b)*
To understand this language and regular expression read: "Need Regular Expression for Finite Automata".
But both languages are not equal because there are some strings in language L1 those are not belongs to language L2 e.g. ab is a string in L1 but doesn't not consist of even number of a and b hence doesn't belongs to language L2.
Note: Language L2 is either not a subset of language L1, because in L2 a strings of even length and single symbol is possible like aa, aaaa, bb, bbbb but these strings are not member in L1.
Both languages are different hence single DFA is not possible for both languages.
Both the languages L1 = {xy | Na(x) = Nb(y)} and
L2 = {w | Na(w) and Nb(w) are even number}are different so we cannot draw a single DFA for both languages.
For Language L1 :
A = {xy | Na(x) = Nb(y)} is a regular language.
Regular expression for this language is:
(a + b)*a(a + b)b(a + b)
Language L2 :
A = {w | Na(w) and Nb(w) are even number} is also a regular language. Regular expression for this language is:
((a + b(aa)ab)(bb)(ba(aa)ab(bb))*a + (b + a(bb)ba)(aa)(ab(bb)ba(aa))b)
Both languages are not equal because there are some strings in language L1 which doesnt belong to language L2. ab is a string in L1 but doesn't not consist of even number of a and b hence doesn't belongs to language L2.
As Both languages are different,single DFA cannot be constructed that accepts both the languages.
I'm learning the difference between the lemmata in the question. Every reference I can find uses the example:
{(a^i)(b^j)(c^k)(d^l) : i = 0 or j = k = l}
to show the difference between the two. I can find an example using the regular lemma to "disprove" it.
Select w = uvxyz, s.t. |vy| > 0, |vxy| <= p.
Suppose w contains an equal number of b's, c's, d's.
I selected:
u,v,x = ε
y = (the string of a's)
z = (the rest of the string w)
Pumping y will just add to the number of a's, and if |b|=|c|=|d| at first, it still will now.
(Similar argument for if w has no a's. Then just pump whatever you want.)
My question is, how does Ogden's lemma change this strategy? What does "marking" do?
Thanks!
One important stumbling issue here is that "being able to pump" does not imply context free, rather "not being able to pump" shows it is not context free. Similarly, being grey does not imply you're an elephant, but being an elephant does imply you're grey...
Grammar context free => Pumping Lemma is definitely satisfied
Grammar not context free => Pumping Lemma *may* be satisfied
Pumping Lemma satisfied => Grammar *may* be context free
Pumping Lemma not satisfied => Grammar definitely not context free
# (we can write exactly the same for Ogden's Lemma)
# Here "=>" should be read as implies
That is to say, in order to demonstrate that a language is not context free we must show it fails(!) to satisfy one of these lemmata. (Even if it satisfies both we haven't proved it is context free.)
Below is a sketch proof that L = { a^i b^j c^k d^l where i = 0 or j = k = l} is not context free (although it satisfies The Pumping Lemma, it doesn't satisfy Ogden's Lemma):
Pumping lemma for context free grammars:
If a language L is context-free, then there exists some integer p ≥ 1 such that any string s in L with |s| ≥ p (where p is a pumping length) can be written as
s = uvxyz
with substrings u, v, x, y and z, such that:
1. |vxy| ≤ p,
2. |vy| ≥ 1, and
3. u v^n x y^n z is in L for every natural number n.
In our example:
For any s in L (with |s|>=p):
If s contains as then choose v=a, x=epsilon, y=epsilon (and we have no contradiction to the language being context-free).
If s contains no as (w=b^j c^k d^l and one of j, k or l is non-zero, since |s|>=1) then choose v=b (if j>0, v=c elif k>0, else v=c), x=epsilon, y=epsilon (and we have no contradiction to the language being context-free).
(So unfortunately: using the Pumping Lemma we are unable to prove anything about L!
Note: the above was essentially the argument you gave in the question.)
Ogden's Lemma:
If a language L is context-free, then there exists some number p > 0 (where p may or may not be a pumping length) such that for any string w of length at least p in L and every way of "marking" p or more of the positions in w, w can be written as
w = uxyzv
with strings u, x, y, z, and v such that:
1. xz has at least one marked position,
2. xyz has at most p marked positions, and
3. u x^n y z^n v is in L for every n ≥ 0.
Note: this marking is the key part of Ogden's Lemma, it says: "not only can every element be "pumped", but it can be pumped using any p marked positions".
In our example:
Let w = a b^p c^p d^p and mark the positions of the bs (of which there are p, so w satisfies the requirements of Ogden's Lemma), and let u,x,y,z,v be a decomposition satisfying the conditions from Ogden's lemma (z=uxyzv).
If x or z contain multiple symbols, then u x^2 y z^2 w is not in L, because there will be symbols in the wrong order (consider (bc)^2 = bcbc).
Either x or z must contain a b (by Lemma condition 1.)
This leaves us with five cases to check (for i,j>0):
x=epsilon, z=b^i
x=a, z=b^i
x=b^i, z=c^j
x=b^i, z=d^j
x=b^i, z=epsilon
in every case (by comparing the number of bs, cs and ds) we can see that u x^2 v y^2 z is not in L (and we have a contradiction (!) to the language being context-free, that is, we've proved that L is not context free).
.
To summarise, L is not context-free, but this cannot be demonstrated using The Pumping Lemma (but can by Ogden's Lemma) and thus we can say that:
Ogden's lemma is a second, stronger pumping lemma for context-free languages.
I'm not too sure about how to use Ogden's lemma here but your "proof" is wrong. When using the pumping lemma to prove that a language is not context free you cannot choose the splitting into uvxyz. The splitting is chosen "for you" and you have to show that the lemma is not fulfilled for any uvxyz.