retagged by
351 views

1 Answer

0 votes
0 votes
Input string ->  aaa+a*+

1st step     ->AAA+a*+

2nd step   ->AAa*+

3rd step ->AAA*+

4th step  ->AA+

5th step  ->A

answer is(c)

please correct , in case of wrong explanation.

Related questions

0 votes
0 votes
2 answers
2
sripo asked Nov 1, 2018
2,292 views
S→(XS→E]S→E)X→E)X→E]E→ϵIs this grammar CLR(1)? The answer says it is but I find a shift reduce conflict for E- epsilon with lookup symbols ),]
0 votes
0 votes
1 answer
3
sripo asked Nov 1, 2018
532 views
For given production for a LR(1) grammarB->b.C ,$|c here C is non terminalC->c. ,$|c and here c is terminal. $|c are lookup symbolsWill there be a shift reduce conflict...