Whether or not is the following grammar LL(1)? Why (State the reasons.)? If thegrammar is not LL(1), then convert it to an LL(1) grammar. Next, present thetransition table and show the parsing process by parsing the string “ababbbb”according to the prepared transition table.• S->aAbB | bS• A->AbB|λ• B->Bb|a

The given grammar is not LL(1) because it does not satisfy the conditions for a grammar to be LL(1). The conditions are:

1. For each non-terminal, the productions must be disjoint. This means that for a given input symbol, there should be only one production that can be applied. In the given grammar, this condition is violated. For example, in the production S->aAbB | bS, both productions can start with 'a' or 'b', which makes it ambiguous.

2. The grammar must not be left recursive. This means that there should be no non-terminal A such that there is a derivation A =>+ Aα for some string α. In the given grammar, this condition is violated. For example, in the production A->AbB|λ, A derives A which is a left recursion.

To convert the given grammar to LL(1), we need to eliminate the ambiguity and the left recursion. Here is a possible LL(1) grammar:

1. S -> aTbB | bS
2. T -> aTb | λ
3. B -> Bb | a

The transition table for the LL(1) parsing is as follows:

1: S->aTbB 2: S->bS 3: T->aTb 4: T->λ 5: B->Bb 6: B->a

Now, let's parse the string "ababbbb" according to the transition table:

1. Initially, the stack is {S} and the input is ababbbb.
2. Apply production 1 (S->aTbB), the stack is {aTbB} and the input is ababbbb.
3. Match 'a', the stack is {TbB} and the input is babbbb.
4. Apply production 3 (T->aTb), the stack is {aTbB} and the input is abbbb.
5. Match 'a', the stack is {TbB} and the input is bbbb.
6. Apply production 4 (T->λ), the stack is {bB} and the input is bbbb.
7. Match 'b', the stack is {B} and the input is bbb.
8. Apply production 5 (B->Bb), the stack is {Bb} and the input is bbb.
9. Apply production 5 (B->Bb), the stack is {Bb} and the input is bb.
10. Apply production 5 (B->Bb), the stack is {Bb} and the input is b.
11. Apply production 5 (B->Bb), the stack is {Bb} and the input is .
12. Apply production 6 (B->a), the stack is {a} and the input is .
13. The input is not 'a', so the parsing fails.

The string "ababbbb" cannot be parsed by the LL(1) grammar.