How can I prove that the language $\{ab^kab^kab^k\subset \{a,b\}^* | k \geq 0\}$ is non context-free? I've tried applying the pumping lemma but can't write a proof without considering multiple different cases of possible subwords.
2025-01-13 08:01:13.1736755273
Proving this language is non context-free
65 Views Asked by J. Young https://math.techqa.club/user/j-young/detail At
1
There are 1 best solutions below
Related Questions in AUTOMATA
- If $L_1$ and $L_2$ are regular then $L_1 \cup\;L_2\; = L$ is regular. Is the converse true?
- Can $L$ be regular language if it is a union of infinitely many regular languages $L_1,L_2,L_3,...$ over the same alphabet?
- Is C* regular if C is a language with strings of prime length?
- Union of two Non deterministic Finite automata
- number of NFAs given $a$ states
- For DFAs, NFAs how can you show $L(D) = \overline {L(D')}$ or $L(N) = \overline {L(N')}$?
- Construct DFA from context-free grammar
- Probability Assessment of Interactive Markov Chain (IMC)
- Is $L = \left \{ a^m b^mca^nb^m \mid m,n \geq 0 \right \}$ context free language?
- Find languages L1 and L2, neither of which contains the other, such that (L1* ∪ L2*) = (L1 ∪ L2)*.
Related Questions in CONTEXT-FREE-GRAMMAR
- How would I find the Context Free Grammar for the complement of L = {a^n | n>= 0}? The alphabet is {a}.
- Construct DFA from context-free grammar
- Is $L_1$ context free language?
- Is $L = \left \{ a^m b^mca^nb^m \mid m,n \geq 0 \right \}$ context free language?
- Suppose that CFG G defines $T^*$. Does G defines every language $L$ over the alphabet T?
- What is meant by S -> SS in a CFG?
- Context free grammar for these langauges?
- How to draw DPDA for language $L = \{a^ncb^{2n} | n \geq1\}$over the alphabet $\Sigma =\{a,b,c\} ?$
- What happens when all productions in a grammar are useless?
- When eliminating the lambda expressions from this grammar, are we missing a production?
Related Questions in PUMPING-LEMMA
- Understanding the proof that uses Pumping Lemma that the language $C =\{w \mid w$ has an equal number of $0$'s and $1$'s$\}$ is not regular
- Prove that $\{w \mid \text{ w has even length and the first half of w has more 0s than the second half of w} \}$ is not regular?
- Prove that the class of Turing Decidable Languages is strictly larger than class of Context Free Languages
- Prove that a language is not regular
- Pumping lemma for context free. How do I define the string 'w' and define cases?
- Using the Pumping Lemma to show that the language of all strings of even length having no $0s$ in their second half is not regular
- General question about pumping lemma statement for regular languages
- Proving that the language $\{w\in \{a,b\}^* \big|\#_a(w)< \#_b (w)\}$ is non regular using the pumping lemma
- Proving that $\mathscr L=\{0^n \big|\text{n is the square of a natural number }\}$ is non regular using the pumping lemma
- Proving that the language $\mathscr L$ is non regular using the pumping lemma
Trending Questions
- Induction on the number of equations
- How to convince a math teacher of this simple and obvious fact?
- Refuting the Anti-Cantor Cranks
- Find $E[XY|Y+Z=1 ]$
- Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space
- Why does this innovative method of subtraction from a third grader always work?
- What are the Implications of having VΩ as a model for a theory?
- How do we know that the number $1$ is not equal to the number $-1$?
- Defining a Galois Field based on primitive element versus polynomial?
- Is computer science a branch of mathematics?
- Can't find the relationship between two columns of numbers. Please Help
- Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
- Identification of a quadrilateral as a trapezoid, rectangle, or square
- A community project: prove (or disprove) that $\sum_{n\geq 1}\frac{\sin(2^n)}{n}$ is convergent
- Alternative way of expressing a quantied statement with "Some"
Popular # Hahtags
real-analysis
calculus
linear-algebra
probability
abstract-algebra
integration
sequences-and-series
combinatorics
general-topology
matrices
functional-analysis
complex-analysis
geometry
group-theory
algebra-precalculus
probability-theory
ordinary-differential-equations
limits
analysis
number-theory
measure-theory
elementary-number-theory
statistics
multivariable-calculus
functions
derivatives
discrete-mathematics
differential-geometry
inequality
trigonometry
Popular Questions
- How many squares actually ARE in this picture? Is this a trick question with no right answer?
- What is the difference between independent and mutually exclusive events?
- Visually stunning math concepts which are easy to explain
- taylor series of $\ln(1+x)$?
- Determine if vectors are linearly independent
- What does it mean to have a determinant equal to zero?
- How to find mean and median from histogram
- Difference between "≈", "≃", and "≅"
- Easy way of memorizing values of sine, cosine, and tangent
- How to calculate the intersection of two planes?
- What does "∈" mean?
- If you roll a fair six sided die twice, what's the probability that you get the same number both times?
- Probability of getting exactly 2 heads in 3 coins tossed with order not important?
- Fourier transform for dummies
- Limit of $(1+ x/n)^n$ when $n$ tends to infinity
Assume that the language is context-free. Then the pumping lemma says (since the language is clearly infinite) that there is a string in the language that splits into $uvwxy$ such that $uv^nwx^ny$ is in the language for every $n\ge 0$ and $v, x$ are not both empty. In particular, both $uvwxy$ and $uwy$ are in the language.
Now neither $v$ nor $x$ can contain an $a$, because then $uvwxy$ and $uwy$ would have different numbers of $a$s, but every word in the language has exactly 3 $a$s. So both $v$ and $x$ must consist entirely of $b$s.
However, then going from $uvwxy$ ot $uwy$ would correspond to taking one or more $b$s away from at most two of the $b^k$ groups in $uvwxy$, so $uwy$ cannot be in the language after all.