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- pumping lemma: ww^R not regular - Mathematics Stack Exchange
I'm trying to prove that L = {wwR: w ∈ {a, b}∗} L = {w w R: w ∈ {a, b} ∗} (wR w R is the reverse of w w) is not regular using the pumping lemma Let p p be the pumping length and s =apbbap s = a p b b a p x = ϵ x = ϵ, y = ap y = a p, z = bbap s = ϵapbbap =apbbap z = b b a p s = ϵ a p b b a p = a p b b a p 1) Was s s properly divided? Let's see: |xy| ≤ p | x y | ≤ p |y|> 0 | y
- Context free grammar: Meaning of notation ww^R
A common example in CFG is the palindrome example These examples often contain the wwR w w R notation for the string An example from my class could be: Strings wwR w w R over the alphabet Σ = {0, 1} Σ = {0, 1} (a subset of palindromes over Σ Σ), or
- Is it possible to make a PDA for - Mathematics Stack Exchange
I know it's easy to make a PDA for {wwR: w ∈ {0, 1}∗} {w w R: w ∈ {0, 1} ∗} where wR w R is the reverse of w w, but I can't think of a PDA that recognises the language L L
- regular expressions - Isnt $L=\ {ww|w \in \ {0,1\}^*\}$ a Non . . .
The problem is that L = {w w: w ∈ {0, 1} ∗} isn’t context-free at all You can use the pumping lemma to prove this Suppose that L is context-free, and let p be
- formal languages - Pumping lemma, L= {WW^R | W can be {1 . . .
im trying to find out, if L is regular or not using pumping lemma I have L= {WW^R | W can be {1}+} So possible strings would be 11, 1111, 111111 In every cases i have googled so far are example
- Proving $ww^Ru$ is not a regular language with Pumping Lemma
I would suggest wwR w w R with w = ababbabbbabbbb w = a b a b b a b b b a b b b b i e increasing sequences of b b 's separated by an a a I think it's easy to find a contradiction to the pumping lemma using this string
- Prove that $\\{ww^R\\#ww^R\\}$ is not context free
I need to prove that L = {wwR#wwR |w is in {a, b}∗} L = {w w R # w w R | w is in {a, b} ∗} is not context free I have tried using the pumping lemma for this For w =apbpbpap#apbpbpap w = a p b p b p a p # a p b p b p a p I have two cases for w = uvxyw w = u v x y w: (1) v v and y y are both before the # # or both after the # # Obviously w2 w 2 is not in L L, contradiction (2) v v is
- computer science - two tape Turing machine that accepts L = {ww | w . . .
in a linear time complexity The input will be on Tape #1 # 1 and the marker will point to the first element of the tape basically my idea was to copy the first string to the other going to the middle in the first tape, going to the start in the second tape and comparing the 1st half of the word with the second but the problem is that there is no STAY operation for the marker, only Left or
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