Theory of Computation and Compilers - Syntax Analysis
31. An LALR(l) parser for a grammar G can have shift-reduce (S-R) conflicts if and only if
- Option : B
- Explanation : An LALR(1) parser for a grammar G can have shift reduce S – R conflict if and only if the LR(1) parser for G has S – R conflicts.
32. The grammar S → aSa |bS|c is
- Option : C
- Explanation : t is LL(1) and LR(1)
because in S → aSa
S → bS
S → c
no mult iple ent r ies for 1 iteration, so it is LL(1)
grammer mean by LR(1) that it is Left shift
Reduced grammer and so it is also LR(1).
Hence it is both LL(1) and LR(1)
- Option : B
- Explanation : As per given question, a grammar with no epsilon and unit -production (i.e., of type A → ε and A → a). So To Reduce input string of n tokens, first Reduce n – 2 tokens using n – 2 reduce moves and then Reduce last 2 tokens using production which has. So total of n – 2 + 1 = n – 1 Reduce moves
- Option : D
- Explanation : (1) false, as merging has been done because lookahead were different. (2) false, Shift-Reduce conflict never occur at the time of merging. (3) false, In the merged set, we can’t see any Reduce-Reduce conflict (no reduction even possible, so no chances of R-R conflict ) (4) false, because goto is carried on Non-Terminals symbols, not on terminal symbols, and c is a terminal symbol.
35. A canonical set of items is given below S → L. > R Q → R. On input symbol > the set has
- Option : D
- Explanation :
Form above diagram, we can see that there is no
shift -reduce or reduce-reduce conflict .
