Test by Bikram | Compiler Design | Test 1 | Question: 8

Question

Read the following grammar:

$S \rightarrow  Ka \mid bKc \mid dc  \mid bda$
$K \rightarrow  d$

This grammar is NOT:

• A. LALR(1)
• B. SLR(1)
• C. LR(1)
• D. None of the above

Comments

How option (b) is correct??

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see the first example

http://people.cs.vt.edu/~ryder/515/f05/homework/hw1ans.pdf

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ok..thanx.

Answer 1

Follow of $K$ contains $c$, hence it is not $SLR(1)$.

Comments

• The production $S\rightarrow d.c,K\rightarrow d.$ has SR conflict, because $follow(K)=a,c\in c$ is in shift move.
• The  has production $S\rightarrow bd.a,K\rightarrow d.$ also  has SR conflict. 

$\therefore$ given grammar is not SLR(1).

Answer 2

$[Closure]$ on $d$ would have $K\rightarrow d.$ and $S\rightarrow d.c$

So, there's one final item (reduce move/handle) and one shift move possible on c.

=> SR conflict for LR(0)

 

Check $Follow(K)$. It has c.

=> SR conflict for SLR(1), too.

 

Option B