Exercise 4.3-6*
Show that the solution to T(n) = 2T( n / 2  17 ) + n is O( n lg n )
Solution:
Assume T(n) = 1 if n ≤ 34
Let n0 = 35.
Base case: T(35) = 1 + 35 ≤ c 34 lg (34) hold.
To show that T(n) = O(n lg n), it is enough to show that T(n) < cn lg n
First assume (inductive hypothesis) that T(n) < cn lg n, for all n ≤ 35, 36, …, k
Therefore, T( n / 2 ) ≤ c( n / 2  17 ) lg( n / 2 + 17)
Then show that the inductive hypothesis is true also for n = k + 1:
T(n) = 2T( n / 2  17 ) + n
≤ 2c( n / 2  17 ) lg( n / 2 + 17) + n
≤ c(n + 34) lg( /2 + 17) + n
≤ c(n + 34) lg n + n
= cn lg (n) + 34 lg (n) + n
= O(n lgn)
Exercise 4.3-8
Using the master method in Section 4.5, you can show that the solution to the
recurrence T(n) = 4T(n/2)+n is T(n)=Θ(n2). Show that a substitution proof with the
assumption T(n) ≤ cn2 fails. Then show how to subtract off a lower-order term to make
the substitution proof work.
Solution:
If we guess T(n) ≤ cn2
T(n) ≤ 4c(n/2)2 +n
≤ cn2+n => This is not equal to cn2
Now we guess T(n)≤cn2−n
T(n) ≤ 4(c(n/2)2−n/2)+n
≤ cn2−2n+n
≤ cn2−n => exact equal to our assumption.
4.4-6
Argue that the solution to the recurrence T(n) = T(n/3) + T(2n/3) + cn, where c is a
constant, is Ω(n lg n) by appealing to the recursion tree.
Solution:
Note that each layer of the recursion tree totally cost cn.
Layer 0: cn
Layer 1: c(n/3) + c(2n/3) = cn
Layer 2: c(n/9) + c(2n/9) + c(2n/9) + c(4n/9) = cn
Because the recursion tree is unbalanced, the depth varies. To show T(n) = Ω(n lg n) is
to find a lower bound, so we need to consider the shortest path first. It is trivial that the
shortest path from root to a leaf is cn→ /3 → /9 → /27 →…→1. The height k
is given by
(1/3) ≤ 1, meaning cn ≤ 3 and k ≥ lg3 cn. Then we have T(n)≥
cn lg3 cn =cn( lg n +lg c)/lg 3≥(c/lg 3) n lg n, which implies T(n) =Ω(n lg n).