Compiler Design #Made_Easy_Test_Series

Question

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 _____________.

Please explain your answer.

Comments

Yes it will be 3as you shown

---

Yes I'm also getting total 5

---

Me too $5$.

---

then it's Wrong Answer by ME, error in almost every test.

---

no it will be {$} for X->.Y and also for X->X+Y

---

yes you are right + will come in lookahead. Solution is wrong.

---

Hey adarsh in same state again we have to do closure of stmt X->.X+Y{$}
so we get

X->.X+Y{+} and X->.Y{+}
and after combine we
X->.X+Y{$,+} and X->.Y{$,+}
Explain me if I am wrong

---

You are right @Ravi kumar singh. As it is left recursive we have to find its closure again which will add + to lookahead

---

i missd this part thanks ravi u r right.thanks

---

@Hemanth_13 Total number of look ahead should be 2+3 = 5 right ?? How is it 3 ?

---

Sorry I just considered for Z->.id  ,$/+/* only.

---

check this https://gateoverflow.in/269254/made-easy-test-series#c274758   @Ravi kumar singh

---

So, final ans $5$

right?

---

Yes it should be 5.

---

I am getting only 4

X-->.Y {dollar,+}
 
Z-->.id {dollar,*}

how you all got 5. I think I am missing something. please explain.

---

X->.Y ,$/+

Z->.id,$/+/*  @Gupta731

---

Yes ans is 5 thanks I also missed 1 lookahead got it now.

---

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?

---

but hemanth it is asking for particular items right? so it will still different for both the items .isnt so?

---

@adarsh_1997 Yes you are right. After seeing the answer I just thought in what way the answer could be 3.

---

It is a duplicate question

Answer is 5

X-->.Y{$$,+}

Z--> .id{$,+,*}

---

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

---

The answer is 5. They have provided wrong answer.

---

Yeah it should be $5$. 2 for $X \rightarrow .Y$ and 3 for $Z \rightarrow .id$.

---

yes that's what i've got but i don't know why they incorrect the 5.

---

@ghostman23111   madeeasy have 1 solution incorrect for almost every test. so better trust your intuition and judgement.

---

2 +3 = 5

for X->.Y +/\$

for Z->.id */+/\$

---

This question is from made easy test na??. But I think they have given wrong answer

---

Yes, it is from ME test series.

Are you sure about the answer?

---

@VikramRB Yeah! I am sure.

 

---

Thanks for the answer Kunal.

---

https://gateoverflow.in/283612/lr-1

---

why $5$ lookaheads?

here lookaheads are $,+, *

what other two?

---

@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

---

@adarsh_1997 here look ahead will be only 3

$,*,+

check this https://gateoverflow.in/184648/m-e-test

---

please provide the detail solution, unable to find the look ahead.

---

$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 $* , + ,\$ $

Answer 1

yes it will have 5 .