Problem $2.$ In this module the number of bits in the connections carrying parentheses counts is hardcoded to $8.$ Though the hardware is correct for n $<mtext></mtext><mn>2</mn><mn>5</mn><mn>6</mn>$ it is more costly and slower than it needs to be$.$ But for this problem it$$s good enough. Show the hardware that will be inferred for this module for n $>1($the non$-$base case$).$ That is$,$ don$$t show the hardware computing left$\_$out$\_$n$\_$unmat$\_$close and right$\_$out$\_$n$\_$unmat$\_$open.

a$.$ Show the inferred hardware at one level for n $>1.$

module pmatch$\_$mark$\_$big #$($ int n $-5,$ wn $-8)$

$($ output logic $[$wn$-1$:$0]$ left$\_$out$\_$n$\_$unnat$\_$close, right$\_$out$\_$n$\_$unnat$\_$open,

output uwire $[3$:$0]$ str$\_$marked $[0$:n$-1],$

input uwire $[$wn$-1$:$0]$ left$\_$in$\_$n$\_$unnat$\_$open, right$\_$in$\_$n$\_$unmat$\_$close,

input uwire $[3$:$0]$ str $[0$:n$-1])$;

if $($ n $==1)$ begin

assign left$\_$out$\_$n$\_$unmat$\_$close $=$ str$[0]==$ Char$\_$Close;

assign right$\_$out$\_$n$\_$unmat$\_$open $=$ str$[0]==$ Char$\_$Open;

assign str$\_$marked$[0]=$

str$[0]==$ Char$\_$Close $$ left$\_$in$\_$n$\_$unnat$\_$open $?$ Char$\_$Close$\_$Okay :

str$[0]=-$ Char$\_$Open $=0?$ lt$\_$close : lt$\_$close $-$ delta;

right$\_$out$\_$n$\_$unmat$\_$open $=$ delta rt$\_$open : rt$\_$open $+$ delta;

end

end

endmodule