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Which grammar rules violate the requirement of the operator grammar? A, B, C are variables and a, b, c are terminals

  1. $A \rightarrow BC$
  2. $A \rightarrow CcBb$
  3. $A \rightarrow BaC$
  4. $A \rightarrow \epsilon$

(a) 1 only 
(b) 1 and 2
(c) 1 and 3
(d) 1 and 4

in Compiler Design
edited by
option A and D both will violate

the above question is incomplete, in original question paper option is given like this:

  1. (i) Only
  2. (i) and (ii)
  3. (i) and (iii)
  4. (i) and (iv)

Hence answer is D which include both grammar. 


 dkvg1892 From where this question is taken


Original question:

3 Answers

15 votes
Best answer

A/c to Operator precedence Grammar Two production should not be Adjacent (for exp as they are given in Option A) to each other .And there should not be any Eps Production .Why two production should not be Adjacent to each other is. In general when we write any Mathematical expression in Let take the C Language we are  not going to have 2 identifier side by side that is why we should see that Production should not be side by side.

So Correct Option Must be A and D because they violate the condition .

selected by
2 votes
option A

operator grammar is racist, it does not let two non terminals come together,

in short, no production has either an empty right-hand side or two adjacent nonterminals in its right-hand side.

actually option D should also not be allowed
Option D is the answer according to answer key of ISRO.
Here both Option(A) and option(D) are clearly violating the requirement of the operator grammar.
yes option D also correct
0 votes

Option A will be right option for it.

Because in between two variable there should be exist a terminal then such grammar is called as operator grammar.

Example of operator grammar.

  1. A→CcBbA→CcBb
  2. A→BaCA→BaC
  3. A→ϵ
  4. Example of not to be operator grammar is
  5. A->BCCB

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