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$S\rightarrow aA|bAc|dc$

$A\rightarrow d$

Number of states in $CLR\left ( 1 \right )$ parser construction _______________

Is

$S\rightarrow d.c|$A\rightarrow d.,a$will be in$1$state or in$2$different states?? asked | 196 views 0 Check the 3rd for loop,$First(\beta a)$Now if$\beta =\epsilon$Then lookahead will be$a$, i.e. lookahead for$A\rightarrow \alpha B\beta$0 Then lookahead will be a both a and b 0 yes, merging is a shortcut for lookahead right?? 0 this line of ur is not correct We don't need to find lookahead1 as we have it already found that in$I_{0}$However , I understood it 0 by merging you mean$A \rightarrow a.A , a $and$A \rightarrow a.A ,  b$can be writen as$A \rightarrow a.A , a | b$? 0 that line is correct coz that production is not added in that state and it was added in$I_0$0 No suppose for$A \rightarrow a.A , b $here lookahead just merged with prev. grammar, when A has no terminal after it. I mean this$A \rightarrow a.A , b $in next level becomes$A \rightarrow a.Ab $And Next we got look ahead for A will be$\left \{ b \right \}$Check the 3rd for loop of the algo, u will get it 0 your basics are not clear lookaheads don't merge You should read from page 260 0 that line is correct coz that production is not added in that state yes, because it shouldnot add that state Check the rule lookahead1 is lookahead to that$A$in red color${\color{Red} A}\rightarrow a.A,a|b$And that$A$comes from prev. state And in prev state we got$A\rightarrow .aA$which again has no follow set So, go to prev production again, i.e.$S\rightarrow .AA$And from here we got value of$Follow\left ( A \right )=\left \{ a,b \right \}$Ok?? 0 your basics are not clear I just looking for a shortcut : | check the 3rd for loop for(each terminal$b$in$first\left ( \beta a \right )$) What is$a\$ here??