We have two divide$-$&$-$conquer algorithms, let's call them A and B$,$ where A calls B$.$

Their time complexities, TA$($n$)$

and TB$($n$)$

$,$ are expressed by the following recurrence relations

TA$($n$)=4$TA$($n$/2)+$ TB$($n$)+\backslash$Theta $($n$)$

TB$($n$)=6$TB$($n$/3)+\backslash$Theta $($n$2$logn$)$

$.$

Then

Question $21$Answer

a$.$

TA$($n$)=\backslash$Theta $($n$2$log$3$n$)$

b$.$

TA$($n$)=\backslash$Theta $($n$2$log$2$n$)$

c$.$

TA$($n$)=\backslash$Theta $($n$2$logn$)$

d$.$

TB$($n$)=\backslash$Theta $($n$2$log$2$n$)$