#self doubt

Question

1. Consider the below SDT:

E-> E+E {print(+)}

E->a {print(a)}

What is/are the output(s) for the input string a+a+a ?

(i) a+a+a
(ii) aa+a+
(iii) aaa++
(iv) +a+aa

(a) Only (i)

(b) Only (ii) and (iii)

(c) Only (ii) and (iv)

(d) All

Solution: Option (b)

How iii can be generated

Comments

i am not getting the sdt clearly :/

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E-> E+E {print(+)}

E->a {print(a)}

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Is the input string ++ or a+a+a?

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and your input string is also not valid :/

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@minipanda yes it should be a+a+a and than we will have two parse trees.

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E-> E+E {print(+)}

-> a {print(a)} + E {print(+)}

->a {print(a)} + E+ E {print(+)} {print(+)}

-> a{print(a)} + a{print(a)} + a{print(a)}  {print(+)} {print(+)}

aaa++

 

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yes, it is right ...grammar is ambiguous.

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if u are suppose to use top down parsing then u will get second and fourth option as answer and when u use the bottom up parsing then u will get 1 and 3 as answer .