7/28/2019 TA ZC142-L7

1/45

TA ZC142 Computer ProgrammingLecture 7Date: 30/01/2013

7/28/2019 TA ZC142-L7

2/45

Last Lecture

Character Arrays (Strings)

7/28/2019 TA ZC142-L7

3/45

Array of strings

Programming Exercise

Efficiency and Complexity

Resources and Measurements

Complexity

Time Complexity

Space Complexity

Todays Agenda

7/28/2019 TA ZC142-L7

4/45

Arrays of Strings

Declaration:

char name[5][30];

Five strings each contains maximum thirty characters.

Initialization: char[5][10]={One,Two,Three,Four,Five};

Other valid declarations: char[][]={One,Two,Three,Four,Five};

char[5][]={One,Two,Three,Four,Five};

7/28/2019 TA ZC142-L7

5/45

String Arrays: Reading and Displaying

void main()

{ char name[5][30];

printf(\n Enter five strings);

/* Reading strings */

for(i=0;i

7/28/2019 TA ZC142-L7

6/45

Programming on Array of

String

7/28/2019 TA ZC142-L7

7/45

Write a C program to read the full name ofthe customers and their telephone numbers.

Full name includes Surname followed by firstname and then the middle name.

Sort the customer names in alphabetical

order with surname first followed by comma(,) and initials of the first and middle names.Display this sorted customer list along with

their telephone no. E.g Ghokle Suraj Ram should be written as

Ghokle,S.R

7/28/2019 TA ZC142-L7

8/45

OUTPUT

asma@atlas:~/cp1$ ./a.out

Enter names and telephone numberssharma anil sudhir 1234

verma mona abhay 9876

garg sona abhi 7654

modak tinu pintu 9870shree shits sumit 8765

Customer list in alphabetical order

garg,s.a 7654modak,t.p 9870

sharma,a.s 1234

shree,s.s 8765

verma,m.a 9876

7/28/2019 TA ZC142-L7

9/45

Declarations

charfname[20][10],mname[20][10],sname[20][10];

charname[20][15],telephone[20][10],temp[20];

7/28/2019 TA ZC142-L7

10/45

for(i=0;i

7/28/2019 TA ZC142-L7

11/45

Code for Sorting list in alphabetical order along

with telephone numbers

for(i=1;i

7/28/2019 TA ZC142-L7

12/45

Print the list

for(i=0;i

7/28/2019 TA ZC142-L7

13/45

Exercise : Array of Strings

Problem Statement: Write a C program thatwill read and store the details of a list ofstudents in the format

ID NAME MARKS

And produce the following output

1. Alphabetical list of Names, IDs and Marks.

2. List sorted on IDs3. List sorted on Marks

7/28/2019 TA ZC142-L7

14/45

Exercise

1. Write a C program which will read aline of text and rewrite it in the

alphabetical order.

2. Write a C program to replace a

particular word by another word in agiven string. Both the words areprovided by the user at run time.

7/28/2019 TA ZC142-L7

15/45

Efficiency of algorithm needs algorithm

analysis.

Analyzing algorithms means to predict theresources that the algorithm requires.

Efficiency

7/28/2019 TA ZC142-L7

16/45

Resources

CPU Time

Storage

Complexity

Time Complexity

Space Complexity

Complexity of Algorithm

7/28/2019 TA ZC142-L7

17/45

Running Time of Algorithm is affected by

Hardware

CPU, Clock Rate, memory, disk etc

Software

Operating system, programming language,

compilers etc

Running time of an algorithm is always

proportional to the input size.

Time Complexity

7/28/2019 TA ZC142-L7

18/45

Efficiency of algorithms must be

compared on their running time in the

same h/w and s/w environment. So implementation of the algorithm is

needed to study time complexity.

But it is always not possible.

Time Complexity

7/28/2019 TA ZC142-L7

19/45

Based on algorithm (high level description

of problem)

It takes into account all possible inputs

It also allows us to evaluate relative

efficiency of two algorithms

Analytical Framework

7/28/2019 TA ZC142-L7

20/45

Primitive Operations

Assigning a value to a variable

Calling a function

Performing an arithmetic operation

Comparing two numbers

Indexing into an array

Returning value from function

Computational Model

7/28/2019 TA ZC142-L7

21/45

Each Primitive operation is an instruction

One instruction takes one unit execution time.

This approach assumes that all primitive

operations take fairly similar execution time.

Primitive operations does not depend on input

size.

Run time of an algorithm is estimated based on

number of primitive operations.

Computational Model

7/28/2019 TA ZC142-L7

22/45

Example 1 (Y and Z are input)

X = 3 * Y + Z;

X = 2 + X;// 2 units of time and 1 unit of storage

Complexity Example [1]

7/28/2019 TA ZC142-L7

23/45

Example 2 (a and N are input)

j = 0;

while (j < N) do

a[j] = a[j] * a[j];

b[j] = a[j] + j;

j = j + 1;endwhile;

// 4N + 1 units of time and 2N+1 units of

storage

Complexity Example [2]

7/28/2019 TA ZC142-L7

24/45

Example 1 (Y and Z are input)

X = 3 * Y + Z;

X = 2 + X;

// Constant Unit of time and Constant Unit of

storage

Order of complexity [1]

7/28/2019 TA ZC142-L7

25/45

Example 2 (a and N are input)

j = 0;

while (j < N) do

a[j] = a[j] * a[j];

b[j] = a[j] + j;

j = j + 1;endwhile;

// time units prop. to N and storage prop. to N

Order of complexity [2]

7/28/2019 TA ZC142-L7

26/45

Example 2 (a and N are input)j = 0;

while (j < N) do

k = 0;while (k < N) do

a[k] = a[j] + a[k];

k = k + 1;

endwhile;

b[j] = a[j] + j;j = j + 1;

endwhile;

// time prop. to N2 and storage prop. to N

Order of complexity [3]

7/28/2019 TA ZC142-L7

27/45

What if the algorithm is more

complicated?

How this analytical approach will work forlarge input size?

Each statement may have small number of

primitive operations.

So there are limitations of analytical

approach.

7/28/2019 TA ZC142-L7

28/45

Allows to characterize main factors

affecting an algos running time without

going in much details of the algorithm. Run time is proportional to the input size

of the algorithm.

Asymptotic Notation

7/28/2019 TA ZC142-L7

29/45

Order Notation

log2N N N2 N3 2N

1 2 4 8 4

3.3 10 102 103 >103

6.6 100 104 106 >1025

9.9 1000 106 109 >10250

13.2 10000 108 1012 >102500

7/28/2019 TA ZC142-L7

30/45

Purpose

Capture proportionality

Machine independent measurement

Asymptotic growth (I.e. large values of input

size N)

Order Notation

7/28/2019 TA ZC142-L7

31/45

f(n)function which characterizes the

running time of algorithm.

Notation is f(n) 0, n >= n0 where n0 >= 1

f(n) is order of g(n)

f(n) is O(g(n))

Asymptotic Notation

7/28/2019 TA ZC142-L7

32/45

Motivation for Order

Notationlog2N/10 N/10

N2/10 N3/102N/10

0.1 0.2 0.4 0.8 0.4

0.33 1 10 102 >102

0.66 10 103 105 >1024

0.99 100 105

108

>10249

1.32 1000 107 1011 >102499

Speed factor of 10

7/28/2019 TA ZC142-L7

33/45

Motivation for Order Notation

log2N/104 N/104

N2/104 N3/104 2N/104

0.0001 0.0002 0.0004 0.0008 0.0004

0.00033 0.0001 0.01 0.1 >0.1

0.00066 0.001 1 102 >1021

0.00099 0.01 102 105 >10246

0.00132 0.1 104 108 >102496

Speed factor of 104

7/28/2019 TA ZC142-L7

34/45

Examples

(17*N + 5) is O(N)

(5*N3

+ 10*N2

+ 3) is O(N3

) (C1*Nk+ C2*Nk-1 + + Ck*N + C)

is O(Nk)

(2N + 4*N3 + 16) is O(2N)

(5*N*log(N) + 3*N) is O(N*log(N))

1789 is O(1)

Order Notation

7/28/2019 TA ZC142-L7

35/45

function search(X, A, N)

j = 0;

while (j < N)

if (A[j] == X) return j;

j++;

endwhile;

return Not_found;

Linear Search - Complexity

7/28/2019 TA ZC142-L7

36/45

low = 1; high = N;

while (low

7/28/2019 TA ZC142-L7

37/45

Let key=7; BinarySearch(A, 0,16, 7)

Find the index of the middle element:

A[mid]=A[(0+16)/2]=A[8]=45 A[mid]==7?

Binary Search

7/28/2019 TA ZC142-L7

38/45

7 < 45, search left sublist:A[0..7]

Begin BinarySearch(A, 0, 7, 7)

Find the index of the middle element:A[mid]=A[(0+7)/2]=A[3]=8

A[mid]==7?

7/28/2019 TA ZC142-L7

39/45

7 < A[mid]=8, perform

BinarySearch(A, 0, 2, 7)

mid=1, so we compare 7 with A[1]= 6.

7/28/2019 TA ZC142-L7

40/45

7 >A[mid]=6, look at sublist A[2..2]

The value of element 2 matches the search

key, so our search is successful and we

return the index 2.

7/28/2019 TA ZC142-L7

41/45

Sorting Algorithm

int main()

7/28/2019 TA ZC142-L7

42/45

int main(){

int a[10],t,i,j,n=5;for(i=0;i

7/28/2019 TA ZC142-L7

43/45

int main(){

int a[10],n=5,i,j,x;

for(i=0;i=i+1;j--)a[j] = a[j-1];

a[j] = x; n++;for(i=0;i

7/28/2019 TA ZC142-L7

44/45

int main(){

int a[10],n=5,i,j,x;

for(i=0;i

7/28/2019 TA ZC142-L7

45/45

Any Doubts ?