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0 votes

Please suggest me the approach to solve this kind of questions

Options are-

- Max(B1.val , value(n) ) , Min(C1.val , Value(n) )
- Min(B1.val , value(n )) , Min(C1.val , value(n) )
- Max(B1.val , value(n)) , Max(C1.val, value(n) )
- Min(B1.val , value(n) ) , Max(C1.val , value(n)

0

should be D only.

But in the original SDT,

if it is__ not L-attributed,__ then **S -> A sign** // A.sign = Sign.sign, S.value = A.value

if it is __L-attributed__, then **S -> sign A** // A.sign = Sign.sign, S.value = A.value

0

it doesn't mean, we are evaluating using only bottom-up parsers...

for using in bottom up parsers we redefine it

for using in bottom up parsers we redefine it

0

@srestha The description of SDT says if the sign is $1\,i.e\,'+'$ we will take minimum of the two values, in $(C)$ we are taking maximum in $'+'$ and in $(D)$ we take minimum at position $B1$ as $B1$ came when the sign was $'+'$

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