isha mudgal * passed the exams or passed away from the world?? * 😂😂

HAHAHA! Goteem. All of her classmates and 2 generations of her seniors have all passed away on their seats watching these lectures.

@@kuelf123 HAHAHAHAH

Except you?

Really me too😂😂😂

@porus how was your result?

@@tarunreddy9219 good enough to get into One of the Old Nits...if you are preperaing now...focus and work hard...hope you will get one of the iits...best of luck

@porus Thanks brother.Hoping for the best

@@PratikShende91 What are you upto now days. 9 years have gone by when you posted that comment

you are wrong because first(s)->{a,b} and first(A)->{€} first(B)->{€} and follow{S}->{$} Follow(A)->{a,b} and Follow(B)->{a,b}

Even I had the same doubt but found out later that each state as a whole is considered while merging, not any single lr(1) item. If you observe carefully, other than the those two items, both of the states contain various other lr(1) items. So, they don't qualify the criteria for merging and hence, shall not be taken into account.

first one is correct

i love you

Because while taking the closure, we encounter A at (S -> .Aa, $) and the first of a$ is a. Similarly, we encounter B at (S -> .Bc, $) and the first of c$ is c. Therefore we have A -> .d, a and B -> .d, c in the first state.

@@GauravJain108 thankyou

If A-> B|C, then first(B) and first(C) should be disjoint which implies that if the grammar has common prefixes in the production i.e. left factors, it can't be LL(1).

Transitions are done only for the items which are on the RHS of dot (.). For the first state, a and b are present on look ahead and not on the RHS of dot (.). Hope this helps!

2.6 :)

exactly

Because both follow of A and B has {a,b} and therefore.....A->. and B->. ; therefore in columns of a and b; there will be RR conflict.

The final items in state I(zero) are :(i) A -> . , a (Look-ahead is 'a') and (ii) B -> . , b (Look-ahead is 'b'). So reduce move (i) would be placed in the I(zero) state row, under the 'a' column and reduce move (iI) would be placed in the I(zero) state row, under the 'b' column. So, no conflicts would arise.

YES..

It is simply a. Refer to: goo.gl/p98ZK3

+Gate Lectures by Ravindrababu Ravula sir, in your last example why did u consider lookahead of A->.d , a and for B-> .d, c and why not A-> .d,a/c and B-> .d, c/d because if we consider the other production c and d are followed by A and B respectively...so shouldn't we consider the 2nd production first(c,$) and 4th production (a,$) while calculating the lookahead for A and B.?

+Gate Lectures by Ravindrababu Ravula sir, in your last example why did u consider lookahead of A->.d , a and for B-> .d, c and why not A-> .d,a/c and B-> .d, c/d because if we consider the other production c and d are followed by A and B respectively...so shouldn't we consider the 2nd production first(c,$) and 4th production (a,$) while calculating the lookahead for A and B.

+Eshita Nimje because the production A->.d was instantiated by the production S->.Aa,$ And not by the production S->.bAc,$. So we only consider the follow of A in the first production. To make it clear if the production was S->>.Aa,$ /.Ac,$instead then you could have written A->.d,a/c since it was instantiated by both the production so you have to consider the follow of A from both the production.

+Gate Lectures by Ravindrababu Ravula Did u want website for lectures Delivery?? I can build website for you.

Gate Lectures by Ravindrababu Ravula