Stack Overflow
Stack Overflow

Questions tagged [pumping-lemma]

A lemma mostly used to prove that a language is not regular/context-free.

From wikipedia,

In the theory of formal languages in computability theory, a pumping lemma or pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language. The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.

128 questions
0
votes
1 answer

{w is in {a,b}*|w contains b only in some odds position}

I need the DFA and the regular expression for this language. I think the DFA is this but the Regular expression I get is this ((aUb)a)* and I think it's not correct.
Matioski
  • 117
  • 1
  • 2
  • 11
0
votes
1 answer

Unable to prove this language is not context-free using Pumping Lemma

I'm trying to prove that the language { w ϵ {a, b, c}* | n_a(w) < n_b(w) and n_a(w) < n_c(w) } is not a CFL using Pumping Lemma. The symbol "n_a" represents "number of 'a'". For pumping lemma, z = u(v^i)x(w^i)y, |vxw| <= m, |vw| >= 1. I've chosen to…
Ioannis Georgantas
  • 47
  • 3
0
votes
1 answer

Proving nonregularity

Suppose I have a language L = {wxwR} where wR is the reverse of w, w and x has minimum length of 1, w can consist of either 0's or 1's, while x can only consist of 1's. How do I prove that this language is not regular? Is there any other way than…
Darren B
  • 31
  • 5
0
votes
1 answer

Pumping Lemma Assistance

I recently had an assignment where I was asked to use pumping lemma to show that a language was not regular, and unfortunately got the wrong answer. The language to prove is non-regular is as follows: L = {ai bj ck: i = j or j = k} The definition…
Ben Rogers
  • 45
  • 6
0
votes
1 answer

Why the following language satisfies the pumping lemma for cfl

L = {a^n b c^n| i is greater than 1 and less than 100 , n is greater than 1} I think i misunderstood the pumping lemma for cfl. why cant i pick a word z = a^n c b^n then break it apart to u= a^s v = a^n-s w=epsilon x=b ,y= b^n then pump it with i=0…
gil
  • 2,388
  • 1
  • 21
  • 29
0
votes
1 answer

Is any regular language L has infinite words?

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…
yuanqili
  • 105
  • 1
  • 4
0
votes
1 answer

Proving language is context-free with pumping lemma

I've got a test coming up in using the pumping lemma to prove whether or not a language is context free. I'm trying to work through some practice problems and things aren't going so great... The practice problem is: For a) through j), prove whether…
user1072692
  • 687
  • 1
  • 8
  • 19
0
votes
1 answer

Why pumping lemma for CFG doesn't work

Language: {(a^i)(b^j)(c^k)(d^l) : i = 0 or j = k = l} We take word w = a^0 b^n c^n d^n Which obviously belongs to the language because j = k = l w = uvxyz |vxy| <= n |vy| > 1 and now v and y can be: just a single character and if we pump…
signingPls
  • 3
  • 1
0
votes
1 answer

Minimum Pumping Length

What is the minimum pumping length of following language L=10 (11* 0)* 0 I have read a statement like this s = xyz = 10100 where x=10,y=10 and z=0 such that xyiz∈L (that is 10(1∊0)*0 ) Well it looks Minimum Pumping length is 5 , But it is not,…
Akhil Nadh PC
  • 574
  • 1
  • 7
  • 24
0
votes
4 answers

How a^n b^n where n>=1 is not regular?

This is the simple finite automata I tried, what am I doing it wrong?
user
  • 53
  • 2
  • 7
0
votes
1 answer

Proof that a regular expression is not a regular language using pumping lemma

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…
HJGBAUM
  • 297
  • 5
  • 16
0
votes
0 answers

Regular pumping lemma proving L={Am+n Bm C2n} is not regular

I wanted to prove that this language is not regular L={Am+n Bm C2n} so this is what i did : Let m be the critical length of L and length |W|>/m and i picked W=A3m Bm C2m then from the pumping lemma i got: W=A3m Bm C2m with lengths |YZ|<\m,…
Jacob13000x
  • 3
  • 5
0
votes
2 answers

pumping lemma for very simple regular expression

Pumping lemma definition (from wiki) Let L be a regular language. Then there exists an integer p ≥ 1 depending only on L such that every string w in L of length at least p (p is called the "pumping length"[4]) can be written as w = xyz (i.e., w can…
YeonJae Rho
  • 1
0
votes
1 answer

Is the following language context free grammar?

For n>=0, is the given grammar (a^na^na^n) context free? I tried using pumping lemma, and the result was, no it is not context free.
Patrick Dolan
  • 21
  • 2
0
votes
2 answers

Proof the language is regular or not regular using Pumping lemma?

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…
Rahim Khalid
  • 469
  • 6
  • 15
1 2 3
…
8 9