Notes:Possible error 2016 compiler design q2b

Question

File:Cs3250-2016-with-solutions.pdf - question 2 part b - which can be found on the front page

The same as mine, but with a transition via $y$ from my $A_5$ to $G_2$ , see bottom of page 5 in File:Cs3250-2016-with-solutions.pdf

#	Production	DFA	Comment
0	$S::\eq\ A\#$	$\begin{xy}\xymatrix{& & & ++[o][F=]{C_0} \\ \text{start} \ar[r] & ++[o][F=]{A_5} \ar[dr]^B \ar[ddr]_x \ar[r]^A \ar & ++[o][F-]{B} \ar[ur]^{\#} \ar[r]^z \ar[dr]^x & ++[o][F=]{D_1} \\ & & ++[o][F-]{F} \ar[dr]_y & ++[o][F=]{E_3} \\ & & ++[o][F-]{H} \ar[dr]_y & ++[o][F=]{G_2} \\ & & & *++[o][F=]{I_4} }\end{xy}$	Notice that if we just have the input " $y$ " we are stuck on $A_5$ , this is accepting so we "construct" (reduce $\epsilon$ - nothing - to $B$ ) $B$ from an $\epsilon$ and go back to the start. We now have a $B$ on the top of the stack, we shift this and end up at state $F$ , we shift the next top-of-the-stack (the $y$ from before) ending up at $G_2$ , this is an accepting state, so: We reduce the $y$ and $B$ on the top of the stack to an $A$ - as per production $2$ Go back to the start (ie $A_5$ ) We then shift the $A$ and shit the $\#$ (assuming this means "end of input") then reduce both to the start symbol, which is the accept-the-parse of accepting states, we're done. There is some abuse of terminology here, but you see what I mean, and we'd usually use numbers for the states. Regardless we reduce to "logically" get a $B$ from $\epsilon$ , shift the $y$ over that, reduce them both to get an $A$ , shift the $\#$ over that, reduce them both to get an $S$ , we're done.
1	$A::\eq\ Az$
2	$A::\eq\ By$
3	$B::\eq\ Ax$
4	$B::\eq\ xy$
5	$B::\eq\ \epsilon$