Lecture Note for 資管系 Algorithms
國立中山大學資訊管理系
黃三益
2001/9/25
Chapter 1, 2 Algorithm Analysis
Analyze the insertion sort algorithm
INSERTION-SORT(A)
Cost
times
For j2 to length[A]
c1
n
c2
n 1
/* insert A[j] into the sorted 0
sequence A[1..j-1] */
n 1
c4
n 1
keyA[j]
ij
While i>0 and A[i]> key c5
A[i+1] A[i]
c6
i i-1
c7
n

n

n
t
j 2 j
j 2
j 2
(t j  1)
(t j  1)
n 1
c8
A[i+1] key

T (n)  c1n  c2 (n  1)  c4 (n  1)  c5  j  2 t j  c6  j  2 (t j  1)  c7  j  2 (t j  1)  c8 (n  1)
n
n
n

In the best case, where the input array is already sorted, tj =1 for j=2, 3, …, n.
Therefore the running time is
T (n)  c1n  c2 (n  1)  c4 (n  1)  c5 (n  1)  c8 (n  1)
 (c1  c2  c4  c5  c8 )n  (c2  c4  c5  c8 )
We call T(n) a linear function of n.

In the worst case, where the input array is listed in descending order, tj =j for j=2,
3, …, n.

n
j 2
j
n
n(n  1)
n(n  1)
 1,  j  2 j  1 
. Thus, the running
2
2
time becomes
T (n)  c1n  c2 (n  1)  c4 (n  1)  c5 (
n(n  1)
n(n  1)
n(n  1)
 1)  c6 (
)  c7 (
)  c8 (n  1)
2
2
2
We call T(n) a quadratic function of n.

In most cases, only the worst case is considered. Sometimes, we also consider

the average-case or the expected running time of an algorithm. For the average
case analysis, it is commonly assumed that all inputs of a given size are equally
likely.
Q: What is the average running time of INSERTION-SORT?
Merge Sort


It uses the divide-and-conquer approach:

Divide: Dividing the n-element sequence into two subsequences of n/2


elements each.
Conquer: Sort the two subsequences recursively using merge sort.
Combine: Merge the two sorted subsequences to produce the sorted answer.
The pseudo-code is as follows:
MERGE-SORT(A, p, r)
If p<r
Then q(p+r)/2
MERGE-SORT(A, p, q)
MERGE-SORT(A, q+1, r)
MERGE(A, p, q, r)

The running time T(n) is given below:
1
if n  1

T ( n)  
2T (n / 2)  n if n  1

Q: Show that if n=2k, T(n)=nlgn.
-notation





(g(n)) is a set of functions as defined below:
(g(n))={f(n): there exist positive constants c1, c2, and n0 such that 0c1g(n)
f(n) c2g(n) for all nn0}.
It is obvious (n2) includes many quadratic functions as such 5n2, 0.5n2-3n, and
0.5n2+9n+3. [Q: what are the values for c1, c2, and n0 in each case?]
Strictly speaking, we should denote 2n2+9n+3(n2). However, in convention,
equality (=) is abused to denote set membership. That is, we usually write
2n2+9n+3=(n2).
We say that g(n) is asymptotically tight bound for f(n) if f(n) = (g(n)).
More precisely, the running time T(n) of merge-sort should be written as
(1)
if n  1

T ( n)  
2T (n / 2)  (n) if n  1
O-notation



O-notation deals with only an asymptotic upper bound. It is formally defined as
below:
O(g(n))={f(n): there exist positive constants c and n0 such that 0f(n) cg(n) for
all nn0}.
Similarly, to denote that a function f(n) is a member of O(g(n)), we write f(n) =
O(g(n)).
O-notation is mainly used to analyze the worst-case running time of an
algorithm.
-notation


In contrast to O-notation, -notation provides an asymptotic lower bound. It is
formally defined as below:
 (g(n))={f(n): there exist positive constants c and n0 such that 0cg(n) f(n) for
all nn0}.
O-notation is mainly used to analyze the best-case running time of an algorithm.
o-notation


o-notation denote an upper bound that is not asymptotically tight. That is
o(g(n))={f(n): for any positive constant c > 0, there exists a constant n0>0 such
that 0f(n) <cg(n) for all nn0}.
That is
f ( n)
lim
 0.
n   g ( n)
-notation

By analogy, -notation is used to denote a lower bound that is not asymptotically
tight. It is formally defined as below:
 (g(n))={f(n): for any positive constant c > 0, there exists a constant n0>0

such that 0 cg(n) <f(n) for all nn0}.
That is
f ( n)
lim
 .
n   g ( n)
Exercises

2.1-1 and 2.1-4
Other common functions




(monotonically increasing): A function f(n) is monotonically increasing if mn
implies f(m) f(n).
(strictly increasing): A function f(n) is strictly increasing if m<n implies
f(m)<f(n).
(monotonically decreasing): A function f(n) is monotonically decreasing if mn
implies f(m)f(n).
(strictly decreasing): A function f(n) is strictly increasing if m<n implies
f(m)>f(n).

Floors and ceilings
x  1  x  x  x  x  1 .

A polynomial in n of degree d is a function of the form:
d
p (n)   ai ni , where
i 0
a0, a1, …, ad are called coefficients.
1 m n
, (a )  a mn  (a n ) m , a m a n  a m  n
a

a 0  1, a 1 

nb
lim
 0 . That is, any exponential function grows faster than any polynomial.
n a n

ex  1  x 

x 2 x3
xi
[How is it computed? Hint: use Taylor’s
  
2! 3!
i  0 i!
expansion equation]



1  x  e x  1  x  x 2 [Can you show it?]
x
lim (1  ) n  e x [Can you show it? Hint: use Taylor’s expansion equation]
n
n 
Logarithms
lg n  log 2 n
ln n  log e n
lg k n  (lg n) k
lg lg n  lg(lg n)
a  b logb a
log c (ab)  log c a  log c b
log b a n  n log b a
log b a 
log c a
log c b
log b (1 / a )   log b a
log b a 
1
log a b
a logb n  n logb a
ln( 1  x)  x 
x 2 x3 x 4 x5
     , when |x|<1. [Can you prove it?]
2
3
4
5
x
 ln( 1  x)  x
1 x

Factorials

The following is Stirling’s approximation:
e
1
n! 2n ( ) n (1  ( ))
n
n
n! o(n n )
n!  (2n )
lg( n!)  (n lg n)

Fibonacci numbers:
F0=0, F1=1, Fi=Fi-1+Fi-2.

i  i
Fi 
, where
5

1 5
.
2
The Fibonacci numbers grow exponentially. (Why?)



1 5
(golden ratio),
2
Exercises

2.2-5.