1. Foundations of Algorithms, Fourth Edition
Richard Neapolitan, Kumarss Naimipour
Updated by Richard P. Simpson
Chapter 1
Algorithms: Efficiency, Analysis, and Order

2. What is a problem A problem is a question to which we seek an answer.
Examples
We want to rearrange a list of numbers in numerical order. (sort)
Determine whether the number x is in a list S of n numbers.
What is the 25 Fibonacci number?

3. What is an instance of a problem?
An instance of a problem is a specific assignment of the parameters that define the problem.
For example in the case of sorting n numbers we need to be giving the n numbers in specific order and n the number of values to sort. This creates the specific case we are interested in.

4. What is an Algorithm?
In mathematics and computer science, an algorithm (from Algoritmi, the Latin form of Al-Khwārizmī) is an effective method expressed as a finite list of well-defined instructions for calculating a function
IE a step by step solution to the problem.
In computer systems, an algorithm is basically an instance of logic written in software by software developers to be effective for the intended "target" computer(s), in order for the target machines to produce output from given input (perhaps null).

5. Sequential Search Problem: Is the key x in the array S of n keys?
Inputs(parameters): integer n, array of keys indexed from 1 to n (0 to n-1 ?)
Outputs: location, 0 if not in S
void seqsearch(int n, const int S[], int x, index& location) {
location = 1;
while(location<=n && S[location]!=x) location++;
if (location >n)location=0; }

6. Matrix Multiplication
void matrixmult(int n, const int A[][],const int B[][], int c[][]); { index i,j,k; for(i=1; i<=n; i++) for(j=1, j<=n; j++){ C[i][j]= 0; for(k=1; k<=n; k++) C[i][j] = C[i][j] + C[i][k]* B[k][j]; }

7. Searching Arrays Sequential Search Binary Search
Recursive ( be able to write this !)
Non Recursive (in book)
A problem can solved using a lot of different algorithms. These may vary in efficiency and or complexity(we will discuss this later)
See table 1.1

8. Recursive Fibonacci See Wolfram MathWorld Discussion
f(0) = 0, f(1)=1, f(n)=f(n-1)+f(n-2)

9. Recursive Solution
int fib( int n) { if(n<=1) return n; else return fib(n-1) + fib(n-2); }

10. The recursive algorithm hits all these nodes!
Is this efficient??

11. Iterative Version
int fib2 ( int n) { int I; Fills array from left to right. int f[0..n]; very efficient! f[0]=0; if (n > 0) f[1]=1; n-2 n-1 n for (i=2; i<=n; i++) f[i] = f[i-1] + f[i-2]; } return f[n]; SEE Table 1.2 1 1 2 3 5 8

12. Analysis of Algorithms
Complexity Analysis
This is a measure of the amount of work done by an algorithm as a function of its input data size. IE it’s a function of n.
Efficiency
I use this term in a very specific way. If two different algorithms having the same complexity are run on the same data set the execution time will probably be different. The faster algorithm is more efficient than the other one.

13. Types of complexity Worst Case ( our main focus) Best Case
Average Case
Every Case (i.e. Best = Worst)

14. Average case complexity
Sequential Search
Suppose we have an array of n items and would like to do a sequential search for the value x. Also assume that the value x can be in any location with equal probability (ie 1/n)
𝐴 𝑛 = 𝑘=1 𝑛 (𝑘× 1 𝑛 ) = 1 𝑛 × 𝑘=1 𝑛 𝑘
= 1 𝑛 × 𝑛(𝑛+1) 2 = 𝑛+1 2
See the analysis for the possibility that x is not in the array. p22

15. Complexity Classes recall that n is the data set size!
Constant ( 1, 3, 9, 232, etc)
Linear ( n, 2n, 3n-2, 21n+100 etc)
Quadratic (n2,2n2-3,4n2-3n+23, etc)
Cubic ( n3, 4n3+3n2-2n+7, etc)
Etc
NOTE: The leading term of the polynomial is the most important term from a growth perspective.

16. Complexity Classes
The complexity class of cubic polynomials is represented by the notation Ɵ(n3)
Note that Ɵ(n3) is a set of functions.
Common complexity sets include
Ɵ(lg n) Ɵ(2n)
Ɵ(n) Ɵ(n!)
Ɵ(n lg n) etc
Ɵ(n2)

17. Figure 1.3: Growth rates of some common complexity functions.

18. Doubling the Data Size
If an algorithm is Ɵ(n2) and the data set is doubled what happens to the execution time?
Specificly assume that we have 2n items
Hence Ɵ((2n)2)=Ɵ(4n2) = 4Ɵ(n2)
Four times as long!
What about cubics?

19. Big O (memorize) This is not a suggestion
Definition For a given complexity function f(n), O(f(n)) is the set of functions g(n) for which there exists some positive real constant c and some nonnegative integer N such that for all n≥N, g(n) ≤ c × f(n)

20. Showing that 5 𝑛 2 +3𝑛−6∈𝜃( 𝑛 2 )
We need to find a c and a N that will make the following inequality true 5 𝑛 2 +3𝑛−6≤𝑐 𝑛 2 What would a good choice for c be? 5 𝑛 2 +3𝑛−6≤6 𝑛 2 3𝑛−6≤ 𝑛 2 0≤ 𝑛 2 −3𝑛+6 We can solve this or just guess. A solution : c=3 and c=6 works.

21. Greater than quadratics
Big O , Big Ω, Big θ
Greater than quadratics
N3,2n
Ω(n2)
All quadratics
Θ( n2 )
Less than quadratics
nlgn,n,lg
O(n2)

22. Figure 1.4: Illustrating "big O", Ω and Θ

23. Figure 1.5: The function n2 + 10n eventually stays beneath the function 2n2

24. Another way of looking at it
Figure 1.6: The sets O (n2), Ω (n2)and Θ (n2). Some exemplary members are shown.

25. Logarithm Rules
The logarithm to the base b of x denoted logbx is defined to that number y such that
by = x
logb(x1*x2) = logb x1 + logb x2
logb(x1/x2) = logb x1 - logb x2
logb xc = c logbx
logbx > 0 if x > 1
logbx = 0 if x = 1
logbx < 0 if 0 < x < 1

26. Additional Rules
For all real a>0, b>0 , c>0 and n logb a = logca / logc b logb (1/a) = - logb a logb a = 1/ logab a logb n = n logb a

27. Theorem: log(n!)(nlogn)
Case 1 nlogn  O(log(n!)) log(n!) = log(n*(n-1)*(n-2) * * * 3*2*1) = log(n*(n-1)*(n-2)**n/2*(n/2-1)* * 2*1 => log(n/2*n/2* * * n/2*1 *1*1* * * 1) = log(n/2)n/2 = n/2 log n/2  O(nlogn) Case 2 log(n!)  O(nlogn) log(n!) = logn + log(n-1) + log(n-2) Log(2) + log(1) < log n + log n + log n log n = nlogn

28. The Little o Theorem: If log(f)o(log(g)) and lim g(n) =inf as n goes to inf then f o(g)
Note the above theorem does not apply to big O for
log(n2) O(log n) but n2 O(n)
Application: Show that 2n  o(nn)
Taking the log of functions we have log(2n)=nlog22
and log( nn) = nlog2n.
Hence
Implies that 2n  o(nn)

29. Theorem:
L’Hospitals
Rule

30. Homework
1.1 problem 7
1.3 problem 14
1.4 problem 15, 19