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Let G be any grammar with the following productions:

X → X + Y | Y

Y → Y * Z | Z

Z → (X)

Z → id

If LR(!) parser is used to parse the above grammar, then total how many look-a-heads are present for the item X → >Y and Z → .id in the initial state _____________.

reopened | 228 views
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Actually we have 3 different lookaheads as it is specified how many lookaheads the answer should  be 5 if it is specified how many different lookaheads then answer will be 3.

Am I correct?
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but hemanth it is asking for particular items right? so it will still different for both the items .isnt so?
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Yes you are right. After seeing the answer I just thought in what way the answer could be 3.

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It is a duplicate question

X-->.Y{$$,+} Z--> .id{,+,*} 0 5 will be the answer. The answer in ME test is 3 which is wrong {+,\,*} as Look Ahead for Z->.id and {\,+} for X→.Y +1 The answer is 5. They have provided wrong answer. +1 Yeah it should be 5. 2 for X \rightarrow .Y and 3 for Z \rightarrow .id. 0 yes that's what i've got but i don't know why they incorrect the 5. +1 @ghostman23111 madeeasy have 1 solution incorrect for almost every test. so better trust your intuition and judgement. +1 2 +3 = 5 for X->.Y +/\ for Z->.id */+/\ 0 This question is from made easy test na??. But I think they have given wrong answer 0 Yes, it is from ME test series. Are you sure about the answer? 0 @VikramRB Yeah! I am sure. 0 Thanks for the answer Kunal. 0 0 why 5 lookaheads? here lookaheads are ,+, * what other two? 0 @srestha mam since the production are same and carry lookahead are diff ,we can merge both of them. because if they are not merged a production will be added two times. check my answer below 0 @adarsh_1997 here look ahead will be only 3 ,*,+ check this https://gateoverflow.in/184648/m-e-test 0 please provide the detail solution, unable to find the look ahead. +1 X' \rightarrow .X, \  X \rightarrow .$$X$$+Y , \  X \rightarrow .X+Y , +  (this production is added due to X ) X \rightarrow .Y , \  X \rightarrow .Y , +  Y \rightarrow .Y*Z , + | \$$

$Y \rightarrow .Y*Z , *$

$Y \rightarrow .Z , + | \$$Y \rightarrow .Z , * Z \rightarrow .(X) , * Z \rightarrow .(X) , + | \$$$Z \rightarrow .id , + | \

$Z \rightarrow .id , *$

now $X \rightarrow .Y$ have $2$ look aheads $\$, + $and$Z \rightarrow .id $have$3$look aheads$* , + ,\