$6.(3\mathrm{.}3\mathrm{.}26)$ Recall that a binary search exploits the ordered nature of a sorted sequence. $($a$)$ Consider the following recursive algorithm for searching a sorted sequence: TernarySearch$($xA $=($a$0$ a$1$ an $1))$ Input: An integer x and a finite, non$-$empty, sorted sequence A of n integers Output: Whether or not x is an element of A $1$: let i be n$/3$ and j be $2$n$/32$: if ai $=$ x or aj $=$ x then $3$: return T $4$: else if n $2$ then $5$: return F $6$: else if x $<$ ai then $7$: return TernarySearch$($x$($a$0$ a$18$: else if ai $<$ x $<$ aj then $9$: return TernarySearch$($x$($ai ai$+110$: else $11$: return TernarySearch$($x$($aj aj$+1$ ai $1))$ aj $1))$ an $1))$ Set up and a solve a recurrence relation giving a Big$-$O estimate for the complexity of this algorithm. $($b$)$ Compared to a binary search, would any algorithm that attempted to divide the sequence into b $>2$ subsequences offer any further improvement in time complexity?