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

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Read the following grammar:

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

This grammar is NOT:

  1. LALR(1)
  2. SLR(1)
  3. LR(1)
  4. None of the above
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2 Answers

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

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$[Closure]$ 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

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