Question

Write the recurrence equation for the Worst Case complexity function for Ternary
Search (counting 4 comparisons per recursive call, and assuming W(1) = 1) and solve this
equation. You may assume that n is a power of 3.

Answer 1

The recurrence for normal binary search is T₂(n) = T₂(n=2)+1. For ternary search, we make two comparisons on elements that partition the list into three sections with roughly n=3 elements and recurse on the appropriate partition. Thus analogously, the recurrence for the number of comparisons made by this ternary search is T₃(n) = T₃(n/3) + 2.

However, just as for binary search the second case of the Master Theorem applies. We therefore conclude that T₃(n) 2 € Θ(log(n)). We now consider a slightly modified take on a ternary search in which only one comparison is made which creates two partitions, one of roughly n=3 elements and the other of 2n=3.

Here the worst case arises when the recursive call is on the larger 2n=3-element partition. Thus the recurrence corresponding to this worst-case number of comparisons is

                                   T3w(n) = T3w(2n=3) + 1

But again the second case of the Master Theorem applies to place T3w € Θ (log n). The best case is exactly the same. It is interesting to note that we will get the same results for general k-ary search as n approaches infinity.

The recurrence becomes one of T_k(n) = T_k(n/k) + k – 1 depending on whether we are talking about part (a) or part (b). For each recurrence case, two of the Master Theorem applies.