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Consider the grammar given below T-> Ta|epsilon ,E-> T|E;T , S-> E#. ?
insu_prince
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May 20, 2022
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The above grammar is
slr(1) ,lr(0) ?
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The LL(1) parsing table for the above grammar is Nonterminal a b e i t ... my question is WHY we always choose S' ->eS production to remain in this parse table why not S' -> epsilon? Answer this..???
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