Question 11791 – Compiler-Design

Consider the grammar
S→AS/b A→SA / a then Closure (S’→.S,$) is:

Correct Answer: A

Question 343 Explanation:

The given grammar is
S → AS|b
A → SA|a
Now closure for S →.S, $,
S’ → .S, $
S → .AS|.b, $
A → .SA|.a, a|b
S → .AS|.b, a|b
⇓
S → .S, $
S → .AS|.b, $|a|b
A → .SA|.a, a|b

S¹→.S,$
S→.AS,$ / a/b
S→.b,$/a/b
A→.SA, a/b
A→.a,a /b

S¹→.S,$
S→.AS,$ s/ b
S→.b,$ / b

S¹→.S,$
S→.AS,$ / a/b
S→.AS,$ / a/b
S→.b,$ / a/b

S¹→.S,$
S→.AS,$
S→.b,$

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