GATE CSE 2026 | Set 2 | Question: 31

Question

Consider the canonical $L R(0)$ parsing of the grammar below using terminals $\{a, b, c\}$ and non-terminals $\{A, B, C, S\}$ with $S$ as the start symbol.

\[
\begin{array}{l}
S \rightarrow A C B \\
A \rightarrow a A \mid \epsilon \\
C \rightarrow c C \mid \epsilon \\
B \rightarrow b B \mid b
\end{array}
\]
Which one of the following options gives the number of shift-reduce conflicts that will occur in the $L R(0)$ ACTION table?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

....

Comments

There is "C - >. cC" item missing in goto(1,goto(I0,A)). Also also a self-loop for the same state on c.

Rest all states are fine

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why we don't make LR(0) action table here, the number of S-R conflicts in the DFA is not always the same<!--TgQPHd|||[]--> as the number of S-R conflicts in the table<!--TgQPHd|||[]-->. Why we count only sr conflict in dfa ?<!--TgQPHd|||[]-->

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Is there any shortcut to solve such questions? This is a really good method no doubt but takes time and chances of mistake are high(ek bhi operation miss hua lag gaye)

Answer 2

Ans (D) 

One item in the state has already completed its production (Reduce), while another item in the same state 

still wants to read more input (Shift). Therefore the parser doesn't know whether to Reduce or Shift.

This means Shift-Reduce conflict

 

Ex :

A → .         Finished
A → .aA    Still working              

It is shift reduce conflict.

 

 

Hence, Total SR conflicts = 5

Answer 3

finding S-R conflicts without table  

 

Answer 4

Answer: (D)

A shift-reduce conflict occurs when there is a complete item \((A \rightarrow \alpha \bullet)\) in a state and that state also has a transition on some terminal symbol.

Thus, there are total 5 shift-reduce conflicts.

Answer 5

 

 

 Correct me if you have any suggestions regarding this .