Is this an LL(1) grammar? How to solve First – Follow conflict?

im trying to check if this grammar is LL(1).

S -> L = R
L -> * L | id
R -> L | R + R | num

As you can see there is a Left recursion on R production. So i remove that and what i get is:

S -> L = R
L -> * L | id
R -> L R’ | num R’
R’ -> + R R’ | ε

Now the problem that i have is that First and Follow set of R’ rule have a common non-terminal (“+”) and also FIRST(R) and FOLLOW(R’) has a common non-terminal.
So i wonder how to create the parsing table if there’s this conflict. My question is: is there a way to solve this problem or simply this isn’t an LL(1) grammar?


Define nullable symbols and the first sentence of a grammar

I am practicing for an upcoming exam and am stumbled upon by a review problem. The problem gives the following grammar:

$$ S rightarrow AB $$$
$$ A rightarrow epsilon | a | (T) $$
$$ T rightarrow T, S | S $$
$$ B rightarrow b $$

As far as I can tell, the only nullable symbol is $ A $. It is the only non-terminal whose production contains the zero symbol $ epsilon $. I do not think so $ S $, that includes $ A $ is a nullable symbol in its production, since the same production also contains $ B $which is not a nullable symbol, and both $ A $ and $ B $ should be nullable for zero $ S $ also be nullable. Is $ A $ really the only nullable symbol in this grammar, or am I misinformed?

As far as the first sentence is concerned, I honestly only have problems following my professor's notes to create the first sentence. Could someone help here or point me to a good resource for it?

Thank you all.

Language of context sensitive grammar – Computer Science Stack Exchange

I have the following context sensitive grammar:

begin {align *}
& S to xSy mid a mid b \
& Xa to aa \
& Xb to bb \
& Y to a
end {align *}

I know what it does because it always ends $ a $ and is preceded by 3 $ a $s or 3 $ b $s. I'm just not sure how to write that in sentence notation and would appreciate any help. Would it be something like that?
$$ L = {a ^ n, b ^ m mid n ge 1, 0 <m le 3 } $$

Automata – context-free grammar for words of odd length

I have to write a CFG that has length strings, is divided by a b in the middle, and which are a before the b It's a language about {a, b}

I wrote the following grammar:

 S-> AbC
 A -> BaBaBA | epsilon
 B -> bB | epsilon
 C -> aC | bC | a | b

This grammar should ensure the even number of a and be divided by b. How do I make sure the length is odd?
Does this grammar ensure that b is in the middle?
Any help is appreciated!

What is the best way to improve your English grammar skills?

All valuable contributions. Thanks a lot.

I don't think I can do good grammar when I watch or read films and so on. It will improve word flow, but for grammar it seems like I should go the dry way.

When I write in my mother tongue, the words flow like a river, smoothly, but when I write in English, I get blocks. Heavy. Sometimes I write in my mother tongue and then convert to English, but it seems to be an intellectual exercise. Needs a lot of coffee.

Learning grammar in English is like learning a programming language. Hard work for a perfectionist writer. But I should make it. I like to give my readers only decent pieces.

regular languages ​​- Let 𝐺 = (𝑉, Σ, 𝑆, 𝑃) be a CFG. How is it possible to derive a grammar that generates 𝐿 (𝐺) +?

To let $ 𝐺 = (𝑉, Σ, 𝑆, 𝑃) $ be a CFG. How is it possible to derive a grammar that generates $ 𝐿 (𝐺) ^ + $?

I have three possible answers:

  1. $ 𝐴 → 𝑆𝐵 $ e $ 𝐵 → 𝑆𝐵 | 𝜀 $ for new variables $ 𝐴 $ e $ 𝐵 $.
  2. $ 𝐴 → 𝑆𝐴 | 𝜀 $ for a new variable $ 𝐴 $.
  3. $ 𝐴 → 𝐵𝐵 $ e $ 𝐵 → 𝑆 | 𝜀 $ per new variable $ 𝐴 $ e $ 𝐵 $.

My ideas are either Answer 1 or Answer 2, but I can't understand which one is the right answer.

Thank you in advance!