Tutorial Exercise (Recurrences)
CS3381 Design and Analysis of Algorithms
Helena Wong, 2001
1. Describe how Merge-Sort follows the Divide-and-Conquer paradigm.
2. To prove the correctness of the Merge(A,p,q,r) procedure. Let’s state the loop invariant as
“At the start of each iteration of the for loop of lines 12-17, the subarray A[p..k-1] contains
the k-p smallest elements of L[1..n1+1] and R[1..n2+1], in sorted order. Moreover, L[I] and
R[j] are the smallest elements of their arrays that have not been copied back to A.”
Complete the prove by showing the 3 properties of the invariant.
3. Use a recursion tree to determine a good asymptotic upper bound on the recurrence:
T(n) =
3T(n/2)+n
1
if n>1
if n=1
Level 0
=> n
n
n
T(n)
T(n/2)
T(n/2)
Level 1
T(n/2)
n/2
n/2
n/2
=> (3/2)n
Level 2 T(n/4) T(n/4) T(n/4)
2
T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) => (3/2) n
=> (3/2)(lg n)-1n
Level (lg n)-1
Level lg n
1 1 1 1
1
1 1 1
1
1 1 1
3lg n
T(n)=n+1.51n+1.52n+..+1.5(lg n – 1)n + 3lg n
can be verified using T(4)=3T(2)+4=3[3T(1)+2]+4=3*5+4=19
T(n)=n(1+1.5+1.52+1.5(lg n – 1))+3lg n =n *(1.5lg n – 1) / 0.5 + 3 lg n = 2n*1.5lg n - 2n + 3lg n = 2n*nlg1.5 – 2n + nlg3
= 2n1+lg1.5-2n+nlg3=2nlg3-2n+nlg3=3nlg3-2n => Answer: O(nlg3)
1
=> 3lg n * 1
1+x+x2+..+xk
= (xk+1-1)/(x-1)
alogbC=clogba,
logba=logca / logcb
Tutorial Exercise (Recurrences)
CS3381 Design and Analysis of Algorithms
Helena Wong, 2001
4. Given T(n)=T(n/2)+T(n/2)+1.
We guess that it is O(n). However, it is difficult to prove that “T(n)  cn for some positive constant
c”, because by substitution (T(n/2) and T(n/2) are true), T(n)  c(n/2)+c(n/2)+1=cn+1, that
does not imply T(n)  cn.
Try to prove that T(n) is O(n) by showing T(n)  cn -b, for some positive constants c and b.
Note: Normally, if not specified, we assume T(1)=1.
When n=1, T(1)
=1
cn-b when c=2 and b=1
Assume that “T(n)  cn –b, for some +ve c,b” is true for T(n/2) and T(n/2),
ie. T(n/2)  cn/2 -b and T(n/2)  cn/2 -b for some +ve c and b.
T(n)
=T(n/2)+T(n/2)+1
 (cn/2-b)+(cn/2-b)+1
= cn-2b+1
 cn-b as long as b>=1
Conclusion: taking c as 2 and b as 1,
we have seen T(n)  cn –b is true when n=1, and
we have proven that it is true for T(n) by assuming T(n/2) and T(n/2) are true.
 “T(n)  cn –b, for some +ve c,b” is true
 “T(n)  cn, for some +ve c,b” is true
 T(n) is O(n).
5. Use the master method to give tight asymptotic bounds for the following recurrences:
a. T(n) = 4T(n/2)+n
b. T(n) = 4T(n/2)+n2
c. T(n) = 4T(n/2)+n3
Master Theorem For T(n) = a T(n/b) + f(n) where a >=1 and b > 1.

If f(n) = O(nlogba-)
If f(n) = (nlogba),
for some constant
>0,
then T(n) = (nlogbalg n)
If f(n) = (nlogba+)
for some constant
then T(n) = (nlogba)
>0,
and if a f (n/b) <= c f(n) for some constant c <1 and all sufficiently large n,
then T(n) = (f(n))
a=4, b=2, => nlogba=n2
(a) f(n)=n, => f(n)=nlogba-1=> f(n)=O(n logba-) where =1>0
=> case 1 => T(n)=(nlogba)= (n2)
(b) f(n)=n2, => f(n)=nlogba=> f(n)=  (n logba)
=> case 2 => T(n)=(nlogbalg n)= (n2lg n)
(c) f(n)=n3, => f(n)=nlogba+1=> f(n)=  (n logba+) where =1>0
a f(n/b)=4*f(n/2)=4*(n/2)3=n3/2=0.5*f(n)=c*f(n)
where c=0.5<1
=> case 3 => T(n)=(f(n)) = (n3)