I stuck on this question for a long time and cannot figure out how to prove it?

# Tag: context

## a^ib^jc^k, i < j < k is a context-sensitive language, how can prove it as a context sensitive

I am thinking this question for a long time, that a^ib^jc^k, i < j < k is a context-sensitive language, how we can prove it as a context sensitive or which grammar can generate such a language. Thanks for all of your help

## What does the ‘ (prime) mean in this context?

I’m trying to check a simple calculation regarding the derivative of a matrix exponential. By hand I get $ $ \frac{d}{d\theta_1}e^{\theta_1 \mathbf{W_1}+\theta_2\mathbf{W_2}} = \mathbf{W_1}e^{\theta_1 \mathbf{W_1}+\theta_2\mathbf{W_2}}. $ $ Checking this with Mathematica, I have

`Assuming [{\[Theta]1 \[Element] Reals, {W1, W2} \[Element] Matrices[{d, d}, Reals, Symmetric]}, D[MatrixExp[ (\[Theta]1 W1 + \[Theta]2 W2)], \[Theta]1]] `

which returns

`(* W1 Derivative[1][MatrixExp][W1 \[Theta]1 + W2 \[Theta]2] *) `

Why is there a derivatie here? And what’s the `[1]`

?

## determining decidability with intersection of context free languages

I am trying to solve this problem: Given context-free languages A, B, and C find if the language $ (A\cap B)\cup (B\cap C)$ is empty. Is this problem decidable.

I know the CFG is not closed under intersection so we do not know that $ A\cap B$ is also CFG. If it was CFG, I know how to prove decidability. Since we can’t determine about $ A\cap B$ , is there a way to prove whether or not this is decidable?

## Context free languages invariant by “shuffling” right hand side

Given a grammar for a Context Free language $ L$ , we can augment it by "shuffling" the right hand side of each production, e.g.:

$ A \to BCD$ is expanded to $ A \to BCD \; | \; BDC \; | \; CBD \; | CDB \; | \; DBC \; | \; DCB$

It may happen that the resulting language $ L’$ is equal to $ L$

For example:

`Source Shuffled S -> XA | YB S -> XA | AX | YB | BY A -> YS | SY A -> YS | SY B -> XS | SX B -> XS | SX X -> 1 X -> 1 Y -> 0 Y -> 0 `

Is there a name for such class of CF languages ($ L = \text{shuffled}(L)$ ?

## If A is context free then A* is regular

I am currently studying for my exam and I am having trouble to solve this question :

If A is context free then A* is regular . (Wrong or right)

I think its wrong because if A is context free it means that A can be a non regular languague. And the non regular languagues are not closed under the kleene star operation(At least I think so). I am not sure how write this in a more formal way.

Maybe like this ? : Let A=$ \{a^nb^n|n \in \mathbb{N}\}$ . Then we know that A is non regular and context free. Then A*= I am not sure about this :/

## Help with context free grammar excercise

So, I have an exercise in which I have to write a context free grammar for this language:

$ $ L = \{x \in L(a^∗b^∗c^∗) : |x|_a > |x|_c; |x|_b > 0; |x|_c ≥ 0\}$ $

meaning every string with any number of $ a$ ‘s, $ b$ ‘s and $ c$ ‘s in that order, with the amount of $ a$ ‘s greater than the amount of $ c$ ‘s and the amount of $ b$ ‘s greater than zero.

I am having trouble figuring out the rule that makes sure there are more $ a$ s than $ c$ s.

I have: $ $ \begin{align}S&\to aABC | ab\ A&\to aA | a\ B&\to bB | b\ C&\to cC | c\ \end{align}$ $ I know this is wrong because I should be adding an $ a$ every time I add a $ c$ , but I don’t know how to write that.

## How to interpret this context free language?

$ S -> aAA$

$ A -> aS | bS | a$

Trivial thing:

starting and ending with a

Atleast 3 a’s are definitely present

(These are very layman observations…but seriously I am unable to figure out what exactly this language is all about. …) My attempt:

What I am able to generate

1)aaa

2)aAA

a bS A

a ba AA A

a ba ***

This suggests after b there should be atleast 1 a

*(Coz those *** have all

- a’s

or

- if bS used then again a ‘ba**’

Or

- if aS used length also increases by atleast 3.)*

I know this is not a good analysis..and so I am not expecting any answer but would definitely expect some comments about some intuition or idea..plz any help would be much regarded..

(Although answers are always welcome 😉 )

## Are there context free grammars for all restricted Dyck paths?

A *Dyck path* is a finite list of $ 1$ ‘s and $ -1$ ‘s whose partial sums are nonnegative and whose total sum is $ 0$ . For example, `[1, 1, -1, -1]`

is a Dyck path. Rather than use $ 1$ and $ -1$ , it is common to use "U" and "D" for $ 1$ and $ -1$ , respectively, so we might write `UUDD`

instead. (These might be more familiar as *Dyck words*.)

It is well-known that Dyck paths have a standard "grammar." The "Dyck grammar" for a path $ P$ is

$ $ P = \epsilon \quad | \quad U P D P.$ $

This grammar is very useful because it lets us quickly compute the generating function which enumerates the number of Dyck paths of given lengths.

I am interested not in *all* Dyck paths, but *restricted sets* of Dyck paths. For example, consider the Dyck paths which avoid "peaks" (the bigram `UD`

) at positive even heights; `UUUDDD`

is such a path, while `UUDD`

is not. If we could devise an unambiguous grammar for such paths, then we could write down an equation for the generating function which counts them. (There is such a grammar. It requires two production rules – one for the paths which avoid peaks at even heights, and one for paths which avoid peaks at odd heights.)

This idea leads to a natural question:

Given a (possibly infinite) set of positive integers $ A$ , does there exist a finite, unambiguous context-free grammar which generates precisely the Dyck paths whose peaks avoid $ A$ ?

The answer (I think) is "yes" when $ A$ is finite or an arithmetic progression. I suspect that the answer is "no" in every other case, but I have no idea how to begin proving this.

For example, here is a special case that I cannot decide:

Is the set of Dyck paths which avoid peaks at positive squares a context-free language?

## closure of Context free grammer to homomorphism using PDA

I was looking online, on sipser book, and on lecture notes and I can’t find a proof to closure of context free languages to homomorphism that using PDA instead of CFG. I’m not looking for a full and formal proof, I just want to understand the general idea.