Foundations of Algorithms, Fourth EditionRichard Neapolitan, Kumarss Naimipour

Updated by Richard P. Simpson

Chapter 1Algorithms: Efficiency, Analysis, and Order

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?

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.

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).

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 Svoid seqsearch(int n, const int S[], int x, index& location){ location = 1; while(location<=n && S[location]!=x) location++; if (location >n)location=0;}

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]; }}

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

Recursive FibonacciSee Wolfram MathWorld Discussion

1 2 3 5 8 13 21 34 55 89 . . .f(0) = 0, f(1)=1, f(n)=f(n-1)+f(n-2)

Recursive Solution

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

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

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) 0 1 2 3 4 5 6 7 8 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 }

0 1 1 2 3 5 8

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.

Types of complexity

• Worst Case ( our main focus)• Best Case• Average Case• Every Case (i.e. Best = Worst)

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)

) = See the analysis for the possibility that x is not in the array. p22

Complexity Classesrecall 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.

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)

Figure 1.3: Growth rates of some common complexity

functions.

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 itemsHence Ɵ((2n)2)=Ɵ(4n2) = 4Ɵ(n2)

Four times as long!

What about cubics?

Big O (memorize)This is not a suggestion

DefinitionFor 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)

Showing that We need to find a c and a N that will make the following inequality true

What would a good choice for c be?

We can solve this or just guess.A solution : c=3 and c=6 works.

Big O , Big Ω, Big θ

Θ( n2 ) All quadratics

Greater than quadratics

N3,2n

Less than quadraticsnlgn,n,lg

O(n2)

Ω(n2)

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

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

Figure 1.6: The sets O (n2), Ω (n2)and Θ (n2). Some exemplary members are shown.

Another way of looking at it

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 > 1logbx = 0 if x = 1logbx < 0 if 0 < x < 1

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

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

The Little o Theorem: If log(f)o(log(g)) andlim 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)=nlog22and log( nn) = nlog2n.

Hence

Implies that 2n o(nn)

2loglog

2loglog limlim nn

nnnn

Theorem: )(lg non

nn

nn

nn 2lnlnlimlglim

nn

n

lnlim2ln1

)2(1/1lim

2ln1

nn

n

L’HospitalsRule

Homework

• 1.1 problem 7• 1.3 problem 14• 1.4 problem 15, 19