COSC 600: Advanced Data Structure and Algorithm
Lecture Note - 1 (Chapter 1)
In order to analyze a given algorithm, we need to have some mathematical
background. Some of formulas that aid in analysis are as follows
1) Exponents
Definition
The base X raised to power N is equal to product of X multiplied N many times. The
power N in this case is exponent
X N = X · X · X ·....· X
Exponent Rules
X A.X B = X A+B for example 25.2 7 = 25+7 = 212
(X A )B = X A.B for example (25 )3 = 215
X N + X N = X N+1 for example 25 + 25 = 26
2) Logarithms
Definition
X A = B if and only if log X B = A
Theorem 1.1
log A B =
logC B
logC A
where A, B, C > 0, A ¹1
Theorem 1.2
log AB = log A + log B where A, B > 0
3) Series
Let A1, A2 , A3,........., AN be a sequence then the sum of the sequence is called series
denoted as
N
åA = A + A
i
1
2
+ A3 +......... + AN
i=1
The easiest formula to remember is
N
å2
= (21 + 2 2 + 23 + 2 4 +..... + 2 N ) = 2 N+1 -1
i
i=1
N
if
å2
i
= (2 0 + 21 + 2 2 +..... + 2 N ) = 2 N+1
i=0
The companion formula is
N
åA
i
= (A1 + A 2 + A3 + A 4 +..... + A N ) =
i=1
if 0 < A < 1,
A N+1 -1
1
=
A -1 1- A
N
å Ai = lim
lim
N®¥
N®¥
i=1
A N+1 -1
where A ¹1 Þ
A -1
Important formulas used for analysis
N
N(N +1) N 2
1) åi =1+ 2 + 3.... + N =
»
2
2
i=1
2)
N(N +1)(2N +1) N 3
»
åi =1 + 2 + 3 +.... + N =
6
3
i=1
3)
N(N +1) 2 N 4
åi =1 + 2 + 3 +.... + N = ( 2 ) » 4
i=1
N
2
2
2
2
2
N
3
3
3
3
3
N
4)
N k+1
åi =1 + 2 + 3 +.... + N » | k +1| where k ¹ -1
i=1
k
k
N-1
for example
åi =
i=1
for example
k
k
k
(N -1)(N -1+1) N(N -1)
=
2
2
N
N
N
N
N
i=1
i=1
i=1
i=1
i=1
å(2i -1) = 1+ 3+ 5 +.... + (2N -1) = å 2i - å1 = 2åi - å1 = 2 ·
if N = 100
100
å(2i -1) = (1+ 3+ 5 +.... +199) = 10, 000
i=1
N(N +1) - N
= N2
2
Rules
Rule 1
N
N
N
N
i=1
i=1
i=1
i=1
å f (N ) = f (N )· å1 = N · f (N ) ¹ å f (i) ¹ f (i)å1 = Nf (i)
Rule 2
N
N 0 -1
N 0 -1
i=N 0
i=1
i=1
å f (i) = å f (i) - å f (i)
100
for example
å i = (10 +11+12 +.... +100) = (1+ 2 +.... +10 +11+....100) - (1+ 2 +.... + 9)
i=10
100
9
i=1
i=1
= åi - åi =
N
100.101 10.11
2
2
å i = (1+ 2 + 2
1
1
1
2
i=1
+
1
1
+.... + N ) » log e N
3
2
2
for example
2 + 7 +12 +17 + 22 +.... + (5N + 2)
N
= å (5i + 2)
i=0
N
= å (5i + 2) + 2
i=1
N
N
i=1
i=1
N+1
= å (5i - 3)
i=1
N+1
N+1
i=1
i=1
= 5å i - 3å1
5(N +1)(N + 2)
- 3(N +1)
2
= 5å i + 2å1 + 2
=
5N(N +1)
+ 2N + 2
2
5N 2 + 5N + 4N + 4
=
2
2
5N + 9N + 4
=
2
5N 2 +15N +10
- 3N - 3
2
5N 2 +15N +10 - 6N - 6
=
2
2
5N + 9N + 4
=
2
=
=
4) Induction Method
Principle of Mathematical Induction
Let P be a property of positive integers such that
1) Base Case: P (1) is true, and
2) Inductive Step: if P (n) is true, then P (n+1) is true.
Then P (n) is true for all positive integers.
The premise P (n) in the inductive step is called Induction Hypothesis.
Steps for proof by Induction Method
1) Base condition: prove the theorem for base condition. For example N=1
2) Assume that the given theorem is true for N=k case
3) Using assumption, prove the theorem is true for N=k+1 case step by step
Example I
N
For example, proof of åi =
i=1
Proof
N(N +1)
for any positive integer N
2
Base condition: For N=1
1
åi = 1 = L.H.S
i=1
1(1+1)
=1 = R.H.S
2
k
N(N +1)
k(k +1)
Assume that åi =
for N = k Þ åi =
2
2
i=0
i=1
N
For N=k+1 case,
k+1
åi = (1+ 2 +.... + k + (k +1))
i=1
k(k +1)
+ (k +1)
2
k
= (k +1)( +1)
2
k+2
= (k +1)(
)
2
(k +1)(k + 2)
=
2
=
Þ Theorem is true for n=k+1 case
Þ Theorem is true for any positive integer N
Example II
For example, proof of Fibonacci number
Let Fi be Fibonacci number
(F0 =1, F1 =1, F2 = 2, F3 = 3, F4 = 5,...., Fi = Fi-1 + Fi-2 ) Prove Fi < ( 5 3 )i for i ³1
Proof
Base condition: For i=1 case, F1 =1 < 5
3
is true
Assume that Fk < ( 5 )k for any k
3
For i=k+1 case,
Fk+1 = Fk-1 + Fk < ( 5 )k-1 + ( 5 )k
3
3
k-1
= ( 5 ) (1+ 5 )
3
3
k-1
= (5 ) (8 )
3
3
9 8
= ( 5 )k+1 ( ´ )
3
25 3
k+1
= ( 5 ) ( 3 )2 .( 8 )
3
5
3
5
28
5
= ( )k+1 ( ) < ( )k+1
3
25
3
i = k +1 case is true
5) Proof by Counterexample
This method of proof involves giving an example that fails the stated theorem, and
hence, disproves the theorem.
For example, let Fi is Fibonacci number
Fk £ k 2 is false
F0 =1, F1 =1, F2 = 2, F3 = 3,.....
Proof:
F11 =144 £121 is not true
6) Proof by Contradiction
This method of proof begins with the assumption that the stated theorem is false,
and during the process shows that the assumption was wrong and, hence proves the
stated theorem.
For example,
Theorem: There is an infinite number of prime. e.g. 2,3,5,7,13,17,19,…
Proof:
Assume that there is finite number of prime
Let, Pk be the largest prime
Let, N = P1 ·P2 ·P3 ·........·Pk +1
None of P1, P2 , P3........, Pk divide N exactly ( there will be a remainder of 1)
Þ N is a prime number
Þ N> Pk ( Pk is not the largest prime which is contradiction to initial assumption that
Pk is the largest prime)
Þ Theorem is true i.e. there is an infinite number of prime
7) Recursion
Any function that is defined in terms of itself is called recursive function.
For example, Divide & Conquer
Like Merge Sort for traversing an array, whole array is divided in half, and, then
each half is further divided into half, until search item is found.
Rules of recursion:
1. Base cases-must always have some base case that can be solved without
recursion
2. Making progress-must always to be a case that makes a progress toward a
base case
Example I
For example, consider following program
Public static int f (int n)
{
if (n == 0)
return 0;
else
return f (n/3 +1) + n-1;
}
If we pass value of n as 10, the program will return f (4)+9, here we do not know
what is f (4) so, the program gets called again with value of n= 4. The program now
returns f (2)+3, the program gets called again with value of n=2 to yield f (1)+2. The
program gets called again with n=1 to yield f (1)+0. The program gets called again
and again with n=1. This is bad example. To make this example good one we can
change if (n==0) to if (n==1).
F (10)  F (4) +9  F (2) +3  F (1) +2  F (1) + 0
Example II
Consider another example, Hanoi Tower Problem. This problem involves moving
different size disk from source to destination tower with temporary tower in
between following condition that bigger size disk cannot be placed over smaller as it
will crush smaller disk. In this case, the number of movements needed to move disks
from source, to destination along with movements to and from temporary tower
increases with increase in number of disks and is given by formula N (number of
disk movements)= 2 k -1(k is number of disc). A solution to the problem is as follows
T (n, a, b, c) //n= number of disc, a=source tower, b=temporary tower, c=
destination tower
{
if (n=1)
{
move disk from source to destination
}
else
{
}
}
T (n-1, A, C, B)
Move nth disc from source A to destination C
T (n-1, B, A, C)
References
1) LERMA, M. (2003, February 8). MATHEMATICAL INDUCTION. Retrieved
September 6, 2014, from
http://www.math.northwestern.edu/~mlerma/problem_solving/results/in
duction.pdf