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Consider the following grammar:
S → Aa | bAc | dc | bda
A → e
Which of the following is correct about the above grammar?
  • a)
    LALR(1) but not SLR(1)
  • b)
    SLR(1) and CLR(1)
  • c)
    LL(1) and LALR(1)
  • d)
    Only CLR(1)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider the following grammar:S → Aa | bAc | dc | bdaA → eWhich of t...
There is no conflict in any of the state so it SLR, LALR(1), CLR(1).
For LL(1)
First(S) = First (AA) ∩ First (bAC) ∩ First (dC) ∩ First (bda)
= {e} ∩ {b} ∩ {d} ∩ {b} ∩ Φ
Not LL(1).
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Most Upvoted Answer
Consider the following grammar:S → Aa | bAc | dc | bdaA → eWhich of t...
SLR(1) and CLR(1)

To determine if the given grammar is SLR(1) and CLR(1), let's analyze the grammar based on the properties of these parsing techniques.

SLR(1)
SLR(1) parsing requires that the grammar be unambiguous and have no shift-reduce or reduce-reduce conflicts. It also requires that for each state in the parsing table, there should be at most one action for each terminal symbol and at most one goto action for each non-terminal symbol.

Let's construct the SLR(1) parsing table for the given grammar:

Grammar:
S → Aa | bAc | dc | b
A → e

First and Follow Sets:
First(S) = {b, d}
First(A) = {e}
Follow(S) = {$, a}
Follow(A) = {a}

Canonical LR(0) Items:
S' → .S
S → .Aa
S → .bAc
S → .dc
S → .b
A → .e

LR(0) Closure:
Closure({S' → .S}) = {S' → .S}
Closure({S → .Aa}) = {S → .Aa, A → .e}
Closure({S → .bAc}) = {S → .bAc}
Closure({S → .dc}) = {S → .dc}
Closure({S → .b}) = {S → .b}
Closure({A → .e}) = {A → .e}

LR(0) Transitions:
Closure({S' → .S})
- On 'S': Go to {S' → S.}
Closure({S → .Aa, A → .e})
- On 'a': Go to {S → A.a, A → .e}
Closure({S → .bAc})
- On 'b': Go to {S → b.Ac}
Closure({S → .dc})
- On 'd': Go to {S → d.c}
Closure({S → .b})
- On 'b': Go to {S → b.}
Closure({A → .e})
- No transitions

SLR(1) Parsing Table:

State | a | b | c | d | $ |
------|---|---|---|---|---|
0 | | s1| | s3| |
1 | | | | | acc |
2 | | r5| s4| | r5|
3 | | | | | r4|
4 | | | | | r3|

s -> shift
r -> reduce
acc -> accept

SLR(1) Analysis:
- There are no shift-reduce or reduce-reduce conflicts in the parsing table.
- The table has at most one action for each terminal symbol and at most one goto action for each non-terminal symbol.

Therefore, the given grammar is SLR(1).

CLR(1)
CLR(1) parsing requires that the grammar be unambiguous and have no shift-reduce or reduce-reduce conflicts. It also requires that for each state in the parsing table
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Consider the following grammar:S → Aa | bAc | dc | bdaA → eWhich of the following is correct about the above grammar?a)LALR(1) but not SLR(1)b)SLR(1) and CLR(1)c)LL(1) and LALR(1)d)Only CLR(1)Correct answer is option 'B'. Can you explain this answer?
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