Login
Remember
Register
Certificationglobe
Questions
Unanswered
Users
Ask a Question
Ask a Question
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?
0
votes
asked
Feb 14
by
Mark Pearson
(
195k
points)
Que.
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?
a.
Removing left recursion alone
b.
Factoring the grammar alone
c.
Removing left recursion and factoring the grammar
d.
None of these
Please
log in
or
register
to answer this question.
1
Answer
0
votes
answered
Feb 14
by
Raelynn
(
407k
points)
Answer: None of these
Please
log in
or
register
to add a comment.
10.2k
questions
10.2k
answers
0
comments
3
users
Related questions
0
votes
1
answer
Consider the following expression grammar. The semantic rules for expression calculation are stated next to each grammar production. E → number E.val = number. val | E '+' E E(1).val = E(2).val + E(3).val | E '×' E E(1).val = E(2).val × E(3).val Assume the conflicts in Part (a) of this question are resolved and an LALR(1) parser is generated for parsing arithmetic expressions as per the given grammar. Consider an expression 3 × 2 + 1. What precedence and associativity properties does the generated parser realize?
asked
Feb 1
by
John Adam
(
11.0k
points)
0
votes
1
answer
The following Context-Free Grammar (CFG) : S → aB | bA A → a | as | bAA B → b | bs | aBB will generate?
asked
Jan 14
by
Mark Pearson
(
195k
points)
0
votes
1
answer
Consider the following SQL query: select distinct al, a2,........., an from r1, r2,........, rm where P For an arbitrary predicate P, this query is equivalent to which of the following relational algebra expressions ?
asked
Jan 25
by
Mark Pearson
(
195k
points)
0
votes
1
answer
What is the min and max number of tables required to convert an ER diagram with 2 entities and 1 relationship between them with partial participation constraints of both entities?
asked
Feb 5
by
Mark Pearson
(
195k
points)
0
votes
1
answer
The Greibach normal form grammar for the language L = {an bn+1 | n ≥ 0 } is a. S → aSB, B →bB I λ b. S → aSB, B →bB I b c. S → aSB I b, B→b d. S → aSB I b?
asked
Feb 5
by
Mark Pearson
(
195k
points)
...