Friday 1 November 2013

DMDW

 
 All the material is licensed under Creative Commons Attribution 3.0 Unported (CC-BY 3.0) and you are free to use them under that license.

Slides

Class GDrive Alternative
Trailer english en
1 Getting started with Weka en en
2 Evaluation en en
3 Simple classifiers en en
4 More classifiers en en
5 Putting it all together en en

Videos

Class Lesson YouTube Youku GDrive
Trailer
english en zurich (no-captions) en zh
1 Getting started with Weka 1 Introduction en en zh (no-captions) en zh


2 Exploring the Explorer en en zh (no-captions) en zh


3 Exploring datasets en en zh (no-captions) en zh


4 Building a classifier en en zh (no-captions) en zh


5 Using a filter en en zh (no-captions) en zh


6 Visualizing your data en en zh (no-captions) en zh


Q Questions answered en

2 Evaluation 1 Be a classifier! en en zh (no-captions) en zh


2 Training and testing en en zh (no-captions) en zh


3 Repeated training and testing en en zh (no-captions) en zh


4 Baseline accuracy en en zh (no-captions) en zh


5 Cross-validation en en zh (no-captions) en zh


6 Cross-validation results en en zh (no-captions) en zh


Q Questions answered en

3 Simple classifiers 1 Simplicity first en en zh (no-captions) en zh


2 Overfitting en en zh (no-captions) en zh


3 Using probabilities en en zh (no-captions) en zh


4 Decision trees en en zh (no-captions) en zh


5 Pruning decision trees en en zh (no-captions) en zh


6 Nearest neighbor en en zh (no-captions) en zh


Q Questions answered en

4 More classifiers 1 Classification boundaries en en zh (no-captions) en zh


2 Linear regression en en zh (no-captions) en zh


3 Classification by regression en en zh (no-captions) en zh


4 Logistic regression en en zh (no-captions) en zh


5 Support vector machines en en zh (no-captions) en zh


6 Ensemble learning en en zh (no-captions) en zh


Q Questions answered en

5 Putting it all together 1 The data mining process en en zh (no-captions) en zh


2 Pitfalls and pratfalls en en zh (no-captions) en zh


3 Data mining and ethics en en zh (no-captions) en zh


4 Summary en en zh (no-captions) en zh


Q Questions answered en

Subtitles

Class Lesson GDrive Alternative
Trailer
english zh en zh
1 Getting started with Weka 1 Introduction en zh en zh


2 Exploring the Explorer en zh en zh


3 Exploring datasets en zh en zh


4 Building a classifier en zh en zh


5 Using a filter en zh en zh


6 Visualizing your data en zh en zh


Q Questions answered en en
2 Evaluation 1 Be a classifier! en zh en zh


2 Training and testing en zh en zh


3 Repeated training and testing en zh en zh


4 Baseline accuracy en zh en zh


5 Cross-validation en zh en zh


6 Cross-validation results en zh en zh


Q Questions answered en en
3 Simple classifiers 1 Simplicity first en zh en zh


2 Overfitting en zh en zh


3 Using probabilities en zh en zh


4 Decision trees en zh en zh


5 Pruning decision trees en zh en zh


6 Nearest neighbor en zh en zh


Q Questions answered en en
4 More classifiers 1 Classification boundaries en zh en zh


2 Linear regression en zh en zh


3 Classification by regression en zh en zh


4 Logistic regression en zh en zh


5 Support vector machines en zh en zh


6 Ensemble learning en zh en zh


Q Questions answered en en
5 Putting it all together 1 The data mining process en zh en zh


2 Pitfalls and pratfalls en zh en zh


3 Data mining and ethics en zh en zh


4 Summary en zh en zh

Transcripts

Class Lesson GDrive Alternative
1 Getting started with Weka 1 Introduction en en


2 Exploring the Explorer en en


3 Exploring datasets en en


4 Building a classifier en en


5 Using a filter en en


6 Visualizing your data en en


Q Questions answered en en
2 Evaluation 1 Be a classifier! en en


2 Training and testing en en


3 Repeated training and testing en en


4 Baseline accuracy en en


5 Cross-validation en en


6 Cross-validation results en en


Q Questions answered en en
3 Simple classifiers 1 Simplicity first en en


2 Overfitting en en


3 Using probabilities en en


4 Decision trees en en


5 Pruning decision trees en en


6 Nearest neighbor en en


Q Questions answered en en
4 More classifiers 1 Classification boundaries en en


2 Linear regression en en


3 Classification by regression en en


4 Logistic regression en en


5 Support vector machines en en


6 Ensemble learning en en


Q Questions answered en en
5 Putting it all together 1 The data mining process en en


2 Pitfalls and pratfalls en en


3 Data mining and ethics en en


4 Summary en en

Music

Artist Title GDrive
Woodside Clarinets Divertimento No. 2 Movement 1 - Allegro mp3

Divertimento No. 2 Movement 2 - Menuetto mp3

Divertimento No. 2 Movement 3 - Larghetto mp3

Divertimento No. 2 Movement 4 - Menuetto mp3
Teresa Connors Opening mp3

Incidental mp3

Closing mp3
Data Mining with Weka is brought to you by the Department of Computer Science at the University of Waikato, New Zealand.

DWDM

Data Mining:
Practical Machine Learning Tools and Techniques

We have written a companion book for the Weka software, now into its third edition, that describes the machine learning techniques that it implements and how to use them. It is structured into three parts. The first part is an introduction to data mining using basic machine learning techniques, the second part describes more advanced machine learning methods, and the third part is a user guide for Weka. The third edition was published in January 2011 by Morgan Kaufmann Publishers (ISBN: 978-0-12-374856-0). Mark Hall has joined Ian Witten and Eibe Frank as co-author for this edition, which has expanded to 629 pages.
"If you have data that you want to analyze and understand, this book and the associated Weka toolkit are an excellent way to start."
-Jim Gray, Microsoft Research
"The authors provide enough theory to enable practical application, and it is this practical focus that separates this book from most, if not all, other books on this subject."
-Dorian Pyle, Director of Modeling at Numetrics
"This book would be a strong contender for a technical data mining course. It is one of the best of its kind."
-Herb Edelstein, Principal, Data Mining Consultant, Two Crows Consulting
"It is certainly one of my favourite data mining books in my library."
-Tom Breur, Principal, XLNT Consulting, Tiburg, Netherlands
Features
  • Explains how data mining algorithms work.
  • Helps you select appropriate approaches to particular problems and to compare and evaluate the results of different techniques.
  • Covers performance improvement techniques, including input preprocessing and combining output from different methods.
  • Shows you how to use the Weka machine learning workbench.
Translations The book has been translated into German (first edition) and Chinese (second edition).
Errata Click here to get to a list of errata.
Teaching material Slides for Chapters 1-5 of the 3rd edition can be found here. WILL UPLOAD SOON
Slides for Chapters 6-8 of the 3rd edition can be found here. WILL UPLOAD SOON
These archives contain .pdf files as well as .odp files in Open Document Format that were generated using OpenOffice 2.0. Note that there are several free office programs now that can read .odp files. There is also a plug-in for Word made by Sun for reading this format. Corresponding information is on this Wikipedia page.
Reviews of the first edition Review by J. Geller (SIGMOD Record, Vol. 32:2, March 2002).
Review by E. Davis (AI Journal, Vol. 131:1-2, September 2001).
Review by P.A. Flach (AI Journal, Vol. 131:1-2, September 2001).

Table of Contents for the 3rd Edition:
Sections and chapters with new material are marked in red.
Preface

Part I: Practical Machine Learning Tools and Techniques

1. What’s it all about?
1.1 Data Mining and Machine Learning
1.2 Simple Examples: The Weather Problem and Others
1.3 Fielded Applications
1.4 Machine Learning and Statistics
1.5 Generalization as Search
1.6 Data Mining and Ethics
1.7 Further Reading

2. Input: Concepts, instances, attributes
2.1 What’s a Concept?
2.2 What’s in an Example?
2.3 What’s in an Attribute?
2.4 Preparing the Input
2.5 Further Reading

3. Output: Knowledge representation
3.1 Tables
3.2 Linear Models
3.3 Trees
3.4 Rules
3.5 Instance-Based Representation
3.6 Clusters
3.7 Further Reading

4. Algorithms: The basic methods
4.1 Inferring Rudimentary Rules
4.2 Statistical Modeling
4.3 Divide-and-Conquer: Constructing Decision Trees
4.4 Covering Algorithms: Constructing Rules
4.5 Mining Association Rules
4.6 Linear Models
4.7 Instance-Based Learning
4.8 Clustering
4.9 Multi-Instance Learning
4.10 Further Reading
4.11 Weka Implementations

5. Credibility: Evaluating what’s been learned
5.1 Training and Testing
5.2 Predicting Performance
5.3 Cross-Validation
5.4 Other Estimates
5.5 Comparing Data Mining Schemes
5.6 Predicting Probabilities
5.7 Counting the Cost
5.8 Evaluating Numeric Prediction
5.9 The Minimum Description Length Principle
5.10 Applying MDL to Clustering
5.11 Further Reading

Part II: Advanced Data Mining

6. Implementations: Real machine learning schemes
6.1 Decision Trees
6.2 Classification Rules
6.3 Association Rules
6.4 Extending Linear Models
6.5 Instance-Based Learning
6.6 Numeric Prediction with Local Linear Models
6.7 Bayesian Networks
6.8 Clustering
6.9 Semisupervised Learning
6.10 Multi-Instance Learning
6.11 Weka Implementations

7. Data Transformations
7.1 Attribute Selection
7.2 Discretizing Numeric Attributes
7.3 Projections
7.4 Sampling
7.5 Cleansing
7.6 Transforming Multiple Classes to Binary Ones
7.7 Calibrating Class Probabilities
7.8 Further Reading
7.9 Weka Implementations

8. Ensemble Learning
8.1 Combining Multiple Models
8.2 Bagging
8.3 Randomization
8.4 Boosting
8.5 Additive Regression
8.6 Interpretable Ensembles
8.7 Stacking
8.8 Further Reading
8.9 Weka Implementations

9. Moving on: Applications and Beyond
9.1 Applying Data Mining
9.2 Learning from Massive Datasets
9.3 Data Stream Learning
9.4 Incorporating Domain Knowledge
9.5 Text Mining
9.6 Web Mining
9.7 Adversarial Situations
9.8 Ubiquitous Data Mining
9.9 Further Reading

Part III: The Weka Data Mining Workbench

10. Introduction to Weka
10.1 What’s in Weka?
10.2 How Do You Use It?
10.3 What Else Can You Do?

11. The Explorer
11.1 Getting Started
11.2 Exploring the Explorer
11.3 Filtering Algorithms
11.4 Learning Algorithms
11.5 Meta-Learning Algorithms
11.6 Clustering Algorithms
11.7 Association-Rule Learners
11.8 Attribute Selection

12. The Knowledge Flow Interface
12.1 Getting Started
12.2 Knowledge Flow Components
12.3 Configuring and Connecting the Components
12.4 Incremental Learning

13. The Experimenter
13.1 Getting Started
13.2 Simple Setup
13.3 Advanced Setup
13.4 The Analyze Panel
13.5 Distributing Processing over Several Machines

14. The Command-Line Interface
14.1 Getting Started
14.2 The Structure of Weka
14.3 Command-Line Options

15. Embedded Machine Learning
15.1 A Simple Data Mining Application

16. Writing New Learning Schemes
16.1 An Example Classifier
16.2 Conventions for Implementing Classifiers

17. Tutorial Excercises for the Weka Explorer
17.1 Introduction to the Explorer Interface
17.2 Nearest-Neighbor Learning and Decision Trees
17.3 Classification Boundaries
17.4 Preprocessing and Parameter Tuning
17.5 Document Classification
17.6 Mining Association Rules

References
Index

Saturday 19 October 2013

Non-Muslims please its a request atleast now revert to Islam

Signs of QAYAMAT or Judgement Day..!!

Though Allah only knows when the Last Hour before the Day of Resurrection will come, He has given us 'clues' about its approach through information in the Qur'an and the hadith of the Prophet.This article does not cover all the signs, but most of the important ones are here. All these signs are, of course, based on hadith and/or Qur'an, mainly the former.

The signs are divided into PAST,PRESENT and FUTURE

Past

1. Splitting of the Moon.
2. Death of the Prophet Muhammad, may Allah bless him and grant him peace.
3. A form of death which will kill thousands of Muslims. (Understood to refer to the plague of Amwas during the caliphate of ‘Umar ibn al-Khattab.)
4. A major fighting in Madinah (understood to refer to the battle of al-Harrah during the caliphate of Yazid, 63 AH).
5. The Muslim conquest of Jerusalem.
6. The Muslim conquest of Constantinople.
7. Two large groups of Muslims will fight in war.
8. A war between the Muslims and a reddish people with small eyes, wearing sandals made of hair (understood to refer tothe Mongol Tatar invasion of the Islamic lands.)
9. A peace agreement between the Muslims and non-Muslims from the yellow race (Chinese, Mongols, etc.)
10. Thirty impostors (dajjal) will appear, each thinking he is a prophet.

Present

11. Naked, destitute, barefoot shepherds will compete in building tall buildings. [The modern ex-nomads of Saudi Arabia are an excellent example of this.]

12. The slave-woman will give birth to her master or mistress.

13. A trial (fitnah) which will enter every Arab household.

14. Knowledge will be taken away (by the death of people of knowledge), and ignorance will prevail.

"Knowledge of Islam will be taken away while ignorance will increase. This removal of knowledge will be because old scholars will die and fewer and fewer new scholars will replace them. The leaders of the Muslims will be chosen from ignorant people, and they will rule according to their whims." [Today, studying Islam formally is considered to be an inferior career in most of the Muslim world. The leaders of the Muslim world are more known for their political or military prowess (oppressive ability?) than Islamic knowledge.]
15. Wine (intoxicants, alcohol) will be drunk in great quantities.
16. Illegal s3x will become widespread.
17. Earthquakes will increase.
18. Time will pass more quickly.
19. Tribulations (fitan) will prevail.
20. Bloodshed will increase.
21. A man will pass by the grave of another and wish he was in the latter’s place.
22. Trustworthiness will be lost, i.e. when authority is given to those who do not deserve it.
23. People will gather for prayer, but will be unable to find an imam to lead them.
Future
24. The number of men will decrease, whilst the number of women will increase, until for every man there are 50 women.
25. The Euphrates will reveal a treasure of gold, and
many will die fighting over it, each one hoping to be the one who gains the treasure.
26. The Romans (Europeans) will come to a place called A’maq or Wabiq, and an army of the best people will go forth from Madinah to face them.
27. The Muslim conquest of Rome.
28. The Mahdi (guided one) will appear, and be the Imam of the Muslims.
29. Jesus Christ(Prophet Isaa) will descend in Damascus, and pray behind the Mahdi.
30. Jesus will break the cross and kill the swine, i.e. destroy the false christianity.
31. The Antichrist (al-masih al-dajjal, the false christ) will appear, with all his tools of deception, and be an immense trial. He will be followed by 70,000 Jews from Isfahan (present-day Iran).
32. The appearance of Ya’juj and Ma’juj (Gog and Magog), and the associated tribulations.
("Gog and Magog (two hidden tribes of people) will break free of the dam that holds them back, and ravage the earth. They will drink all the water, and kill people until Allah sends against them a worm which will wipe them out." [This is roughly before/during the time of Isa, although it is not entirely clear.] )
33. The emergence of the Beast from the Earth, carrying the Staff of Moses and the Seal of Solomon, who will speak to the people, telling them they did not believe with certainty in the Divine Signs.
34. A major war between the Muslims (including Jews and Christians who truly believe in Jesus after his return) led by the Imam Mahdi, and the Jews plus other non-Muslims led by the Antichrist.
35. Jesus will kill the Antichrist at the gate of Ludd (Lod in present-day Israel, site of an airport and a major Israeli military base).
36. A time of great peace and serenity during and after the remaining lifetime of Jesus.
37. Wealth will come so abundant that it will become difficult to find someone to accept charity.
38. Arabia will become a land of gardens and rivers. [ it has started, many hotels and big construction projects are being made there, example, Hotel Burj-al-Arab]
39. Society will then decay.
40. The buttocks of the women of the tribe of Daws will again sway in circumambulation (tawaf) around the idol Dhul-Khulsah.
41. A great fire in the Hijaz, seen by the inhabitants of Busra.
42. Three major armies will sink into the earth: one in the east, one in the west, one in Arabia.
43. An Abyssinian leader with thin shins will destroy the Ka’bah.
44. The huge cloud of smoke.
45. The sun will rise from the west (its place of setting).
46. A gentle wind which will take the souls of the believers.
47. There is no-one left on the earth saying, "Allah, Allah" or "There is no god except Allah."
48. Eventually the Day of Judgment is established upon the worst of the people, who copulate like donkeys in public.
49. The blowing in the Trumpet by the Angel Israfil, upon which everyone will faint except as Allah wills.
50. The second blowing in the Trumpet, upon which everyone will be resurrected.
The Prophet Muhammad (PBUH) also said:
1. Alcohol will be called Sherbat, and will be considered Halal 2. Interest will be called Trade, and will be considered Halal 3. Bribes will be called Gifts, and will be considered Halal
4. Women will be naked in spite of wearing dresses. This Hadith has baffled the Ulema for a very long time-until now: The 3 kind of naked women are (i) Those who wear see-through dresses (ii) Those who wear tight dresses and (iii) those whose dresses are so short,that they expose the body
5. Women will have hair, like the hump of a camel. (This hadith too, had baffled the Ulema for quite a long time till they saw current hairstyles. I believe it came into fashion about 5 years ago.)
This list is far from complete; there are many other signs. A couple more of interest are: Constantinople (Istanbul) will fall before Rome to Islam (note that this means Rome will indeed fall), and the Muslim political system will go from Prophetic leadership to a just Khilafah to a 'biting' authority to an oppressive authority and finally back to a just Khilafah. It seems that today we are in the next to last stage of this chain; now is a good time to actively work for the reestablishment of the Muslim state and Khilafah.
Finally, the Prophet has said that when the Signs start to come, they will come FAST - like a string of beads falling one after the other.
And Allah Knows Best.
Have fun praying and getting ready for the day of judgement coz that day will NOT be fun :) P.S.: dont forget to make dua for me. !!

Friday 4 October 2013

Email

How Email Works

In this diagram, the sender is a human being using their company account to send an email to someone at a different company.

Step A: Sender creates and sends an email

The originating sender creates an email in their Mail User Agent (MUA) and clicks 'Send'. The MUA is the application the originating sender uses to compose and read email, such as Eudora, Outlook, etc.

Step B: Sender's MDA/MTA routes the email

The sender's MUA transfers the email to a Mail Delivery Agent (MDA). Frequently, the sender's MTA also handles the responsibilities of an MDA. Several of the most common MTAs do this, including sendmail and qmail (which Kavi uses).
The MDA/MTA accepts the email, then routes it to local mailboxes or forwards it if it isn't locally addressed.
In our diagram, an MDA forwards the email to an MTA and it enters the first of a series of "network clouds," labeled as a "Company Network" cloud.

Step C: Network Cloud

An email can encounter a network cloud within a large company or ISP, or the largest network cloud in existence: the Internet. The network cloud may encompass a multitude of mail servers, DNS servers, routers, lions, tigers, bears (wolves!) and other devices and services too numerous to mention. These are prone to be slow when processing an unusually heavy load, temporarily unable to receive an email when taken down for maintenance, and sometimes may not have identified themselves properly to the Internet through the Domain Name System (DNS) so that other MTAs in the network cloud are unable to deliver mail as addressed. These devices may be protected by firewalls, spam filters and malware detection software that may bounce or even delete an email. When an email is deleted by this kind of software, it tends to fail silently, so the sender is given no information about where or when the delivery failure occurred.
Email service providers and other companies that process a large volume of email often have their own, private network clouds. These organizations commonly have multiple mail servers, and route all email through a central gateway server (i.e., mail hub) that redistributes mail to whichever MTA is available. Email on these secondary MTAs must usually wait for the primary MTA (i.e., the designated host for that domain) to become available, at which time the secondary mail server will transfer its messages to the primary MTA.

Step D: Email Queue

The email in the diagram is addressed to someone at another company, so it enters an email queue with other outgoing email messages. If there is a high volume of mail in the queue—either because there are many messages or the messages are unusually large, or both—the message will be delayed in the queue until the MTA processes the messages ahead of it.

Step E: MTA to MTA Transfer

When transferring an email, the sending MTA handles all aspects of mail delivery until the message has been either accepted or rejected by the receiving MTA.
As the email clears the queue, it enters the Internet network cloud, where it is routed along a host-to-host chain of servers. Each MTA in the Internet network cloud needs to "stop and ask directions" from the Domain Name System (DNS) in order to identify the next MTA in the delivery chain. The exact route depends partly on server availability and mostly on which MTA can be found to accept email for the domain specified in the address. Most email takes a path that is dependent on server availability, so a pair of messages originating from the same host and addressed to the same receiving host could take different paths. These days, it's mostly spammers that specify any part of the path, deliberately routing their message through a series of relay servers in an attempt to obscure the true origin of the message.
To find the recipient's IP address and mailbox, the MTA must drill down through the Domain Name System (DNS), which consists of a set of servers distributed across the Internet. Beginning with the root nameservers at the top-level domain (.tld), then domain nameservers that handle requests for domains within that .tld, and eventually to nameservers that know about the local domain.
DNS resolution and transfer process
  • There are 13 root servers serving the top-level domains (e.g., .org, .com, .edu, .gov, .net, etc.). These root servers refer requests for a given domain to the root name servers that handle requests for that tld. In practice, this step is seldom necessary.
  • The MTA can bypass this step because it has already knows which domain name servers handle requests for these .tlds. It asks the appropriate DNS server which Mail Exchange (MX) servers have knowledge of the subdomain or local host in the email address. The DNS server responds with an MX record: a prioritized list of MX servers for this domain.
    An MX server is really an MTA wearing a different hat, just like a person who holds two jobs with different job titles (or three, if the MTA also handles the responsibilities of an MDA). To the DNS server, the server that accepts messages is an MX server. When is transferring messages, it is called an MTA.
  • The MTA contacts the MX servers on the MX record in order of priority until it finds the designated host for that address domain.
  • The sending MTA asks if the host accepts messages for the recipient's username at that domain (i.e., username@domain.tld) and transfers the message.

Step F: Firewalls, Spam and Virus Filters

The transfer process described in the last step is somewhat simplified. An email may be transferred to more than one MTA within a network cloud and is likely to be passed to at least one firewall before it reaches it's destination.
An email encountering a firewall may be tested by spam and virus filters before it is allowed to pass inside the firewall. These filters test to see if the message qualifies as spam or malware. If the message contains malware, the file is usually quarantined and the sender is notified. If the message is identified as spam, it will probably be deleted without notifying the sender.
Spam is difficult to detect because it can assume so many different forms, so spam filters test on a broad set of criteria and tend to misclassify a significant number of messages as spam, particularly messages from mailing lists. When an email from a list or other automated source seems to have vanished somewhere in the network cloud, the culprit is usually a spam filter at the receiver's ISP or company. This explained in greater detail in Virus Scanning and Spam Blocking.

Delivery

In the diagram, the email makes it past the hazards of the spam trap...er...filter, and is accepted for delivery by the receiver's MTA. The MTA calls a local MDA to deliver the mail to the correct mailbox, where it will sit until it is retrieved by the recipient's MUA.

RFCs

Documents that define email standards are called "Request For Comments (RFCs)", and are available on the Internet through the Internet Engineering Task Force (IETF) website. There are many RFCs and they form a somewhat complex, interlocking set of standards, but they are a font of information for anyone interested in gaining a deeper understanding of email.

Top 30 "C" programs asked in interview

Programs :1. Write a program to find factorial of the given number...
2. Write a program to check whether the given number is even or odd.
3. Write a program to swap two numbers using a temporary variable.
4. Write a program to swap two numbers without using a temporary variable.
5. Write a program to swap two numbers using bitwise operators.
6. Write a program to find the greatest of three numbers.
7. Write a program to find the greatest among ten numbers.
8. Write a program to check whether the given number is a prime.
9. Write a program to check whether the given number is a palindrome c number.
10.Write a program to check whether the given string is a palindrome .
11.Write a program to generate the Fibonacci series.
12.Write a program to print"Hello World"without using semicolon anywhere in the code.
13.Write a program to print a semicolon without using a semicolon anywhere in the code.
14.Write a program to compare two strings without using strcmp() function.
15.Write a program to concatenat e two strings without using strcat() function.
16.Write a program to delete a specified line from a text file.
17.Write a program to replace a specified line in a text file.
18.Write a program to find the number of lines in a text file..
19.Write a C program which asks the user for a number between 1 to 9 and shows the number. If the user
inputs a number out of the specified range, the program should show an error and prompt the user for a
valid input.
20.Write a program to display the multiplica tion table of a given number..
21.WAP to check a string is Caliondrom e or not. // Maventic question.
22.WAP to print DONE,witho ut using any loop. // asked to my frnd in any company.
23.WAP to print DONE,witho ut using any loop and any conditonal clause or operators. // asked to me as a cross question of 22th question by the person i asked 22th ques.
24. WAP to find out the longest word in a string.
25.Prog of WORLD MAP. // this code was written by someone,i forgot his name,he won award for this code as short and best c code. JUST FOR FUN //
26.WAP to print the triangle of letters in increasing order of lines..
27.WAP to print'xay'in place of every'a'in a string.// DOC Update on 24-jan-12.
28.Count the Total Number of 7 comming between 1 to 100.
/* I made this code in a way that u can give Upper limit i.e. 100,Lower limit i.e. 1 and the specific number u wants to count in between i.e. 7 */ // asked by: Vishwa Pratap Rana..
29. Code for duplicate' s removal,by Amit Aru.. // Similar question was asked in Maventic 2nd round to me,,
30. WAP to find out if a given number is a power series of 2 or not,withou t any loop and without using % modulo operator..

ANSWERS

1. Write a program to find factorial of the given number.
Recursion: A function is called'recursive 'if a statement within the body of a function calls the same function. It
is also called'circular definition '. Recursion is thus a process of defining something in terms of itself.
Program: To calculate the factorial value using recursion.
#include<stdio.h>
int fact(int n);
int main(){
int x, i;
printf("En ter a value for x: \n");
scanf("%d" ,&x);
i = fact(x);
printf("\n Factorial of %d is %d", x, i);
return 0;
}int fact(int n){
/* n=0 indicates a terminatin g condition */
if (n
return (1);
}else{
/* function calling itself */
return (n * fact(n - 1));
/*n*fact(n -1) is a recursive expression */
}
}
Output:
Enter a value for x:
4
Factorial of 4 is 24
Explanatio n:
fact(n) = n * fact(n-1)
If n=4
fact(4) = 4 * fact(3) there is a call to fact(3)
fact(3) = 3 * fact(2)
fact(2) = 2 * fact(1)
fact(1) = 1 * fact(0)
fact(0) = 1
fact(1) = 1 * 1 = 1
fact(2) = 2 * 1 = 2
fact(3) = 3 * 2 = 6
Thus fact(4) = 4 * 6 = 24
Terminatin g condition( n
infinite loop.

2. Write a program to check whether the given number is even or odd.
Program:
#include
int main(){
int a;
printf("En ter a: \n");
scanf("%d" ,&a);
/* logic */
if (a % 2 == 0){
printf("Th e given number is EVEN\n");
}
else{
printf("Th e given number is ODD\n");
}
return 0;
}
Output:
Enter a: 2
The given number is EVEN
Explanatio n with examples:
Example 1: If entered number is an even number
Let value of'a'entered is 4
if(a%2==0) then a is an even number, else odd.
i.e. if(4%2==0) then 4 is an even number, else odd.
To check whether 4 is even or odd, we need to calculate (4%2).
/* % (modulus) implies remainder value. */
/* Therefore if the remainder obtained when 4 is divided by 2 is 0, then 4 is even. */
4%2==0 is true
Thus 4 is an even number.
Example 2: If entered number is an odd number.
Let value of'a'entered is 7
if(a%2==0) then a is an even number, else odd.
i.e. if(7%2==0) then 4 is an even number, else odd.
To check whether 7 is even or odd, we need to calculate (7%2).
7%2==0 is false /* 7%2==1 condition fails and else part is executed */
Thus 7 is an odd number.

3. Write a program to swap two numbers using a temporary variable.
Swapping interchang es the values of two given variables.
Logic:
step1: temp=x;
step2: x=y;
step3: y=temp;
Example:
if x=5 and y=8, consider a temporary variable temp.
step1: temp=x=5;
step2: x=y=8;
step3: y=temp=5;
Thus the values of the variables x and y are interchang ed.
Program:
#include
int main(){
int a, b, temp;
printf("En ter the value of a and b: \n");
scanf("%d %d",&a,&b);
printf("Be fore swapping a=%d, b=%d \n", a, b);
/*Swapping logic */
temp = a;
a = b;
b = temp;
printf("Af ter swapping a=%d, b=%d", a, b);
return 0;
}
Output:
Enter the values of a and b: 2 3
Before swapping a=2, b=3
After swapping a=3, b=2
4. Write a program to swap two numbers without using a temporary variable.
Swapping interchang es the values of two given variables.
Logic:
step1: x=x+y;
step2: y=x-y;
step3: x=x-y;
Example:
if x=7 and y=4
step1: x=7+4=11;
step2: y=11-4=7;
step3: x=11-7=4;
Thus the values of the variables x and y are interchang ed.
Program:
#include
int main(){
int a, b;
printf("En ter values of a and b: \n");
scanf("%d %d",&a,&b);
printf("Be fore swapping a=%d, b=%d\n", a,b);
/*Swapping logic */
a = a + b;
b = a - b;
a = a - b;
printf("Af ter swapping a=%d b=%d\n", a, b);
return 0;
}
Output:
Enter values of a and b: 2 3
Before swapping a=2, b=3
The values after swapping are a=3 b=2

5. Write a program to swap two numbers using bitwise operators.
Program:
#include
int main(){
int i = 65;
int k = 120;
printf("\n value of i=%d k=%d before swapping", i, k);
i = i ^ k;
k = i ^ k;
i = i ^ k;
printf("\n value of i=%d k=%d after swapping", i, k);
return 0;
}
Explanatio n:
i = 65; binary equivalent of 65 is 0100 0001
k = 120; binary equivalent of 120 is 0111 1000
i = i^k;
i...0100 0001
k...0111 1000
---------
val of i = 0011 1001
---------
k = i^k
i...0011 1001
k...0111 1000
---------
val of k = 0100 0001 binary equivalent of this is 65
---------( that is the initial value of i)
i = i^k
i...0011 1001
k...0100 0001
---------
val of i = 0111 1000 binary equivalent of this is 120
--------- (that is the initial value of k)

6. Write a program to find the greatest of three numbers.
Program:
#include
int main(){
int a, b, c;
printf("En ter a,b,c: \n");
scanf("%d %d %d",&a,&b,&c);
if (a>b&&a>c){
printf("a is Greater than b and c");
}
else if (b>a&&b>c){
printf("b is Greater than a and c");
}
else if (c>a&&c>b){
printf("c is Greater than a and b");
}
else{
printf("al l are equal or any two values are equal");
}
return 0;
}
Output:
Enter a,b,c: 3 5 8
c is Greater than a and b
Explanatio n with examples:
Consider three numbers a=5,b=4,c= 8
if(a>b&&a>c) then a is greater than b and c
now check this condition for the three numbers 5,4,8 i.e.
if(5>4&&5>8) /* 5>4 is true but 5>8 fails */
so the control shifts to else if condition
else if(b>a&&b>c) then b is greater than a and c
now checking this condition for 5,4,8 i.e.
else if(4>5&&4>8) / * both the conditions fail */
now the control shifts to the next else if condition
else if(c>a&&c>b) then c is greater than a and b
now checking this condition for 5,4,8 i.e.
else if(8>5&&8>4) / * both conditions are satisfied */
Thus c is greater than a and b.

7. Write a program to find the greatest among ten numbers.
Program:
#include
int main(){
int a[10];
int i;
int greatest;
printf("En ter ten values:");
//Store 10 numbers in an array
for (i = 0; i<10; i++){
scanf("%d" ,&a[i]);
}
//Assume that a[0] is greatest
greatest = a[0];
for (i = 0; i<10; i++){
if (a[i]>greatest){
greatest = a[i];
}
}
printf("\n Greatest of ten numbers is %d", greatest);
return 0;
}
Output:
Enter ten values: 2 53 65 3 88 8 14 5 77 64 Greatest of ten numbers is 88
Explanatio n with example:
Entered values are 2, 53, 65, 3, 88, 8, 14, 5, 77, 64
They are stored in an array of size 10. let a[] be an array holding these values.
/* how the greatest among ten numbers is found */
Let us consider a variable'greatest' . At the beginning of the loop, variable'greatest' is assinged with the value of
first element in the array greatest=a [0]. Here variable'greatest' is assigned 2 as a[0]=2.
Below loop is executed until end of the array'a[]';.
for(i=0; i
{
if(a[i]>gr eatest)
{
greatest= a[i];
}
}
For each value of'i', value of a[i] is compared with value of variable'greatest' . If any value greater than the value
of'greatest' is encountere d, it would be replaced by a[i]. After completion of'for'loop, the value of variable
'greatest' holds the greatest number in the array. In this case 88 is the greatest of all the numbers.
8. Write a program to check whether the given number is a prime.
A prime number is a natural number that has only one and itself as factors. Examples: 2, 3, 13 are prime
numbers.
Program:
#include
main(){
int n, i, c = 0;
printf("En ter any number n: \n");
scanf("%d" ,&n);
/*logic*/
for (i = 1; i
if (n % i == 0){
c++;
}
}
if (c == 2){
printf("n is a Prime number");
}
else{
printf("n is not a Prime number");
}
return 0;
}
Output:
Enter any number n: 7
n is Prime
Explanatio n with examples:
consider a number n=5
for(i=0;i
i.e. for(i=0;i
1st iteration: i=1;i
here i is incremente d i.e. i value for next iteration is 2
now if(n%i==0) then c is incremente d
i.e.if(5%1 ==0)then c is incremente d, here 5%1=0 thus c is incremente d.
now c=1;
2nd iteration: i=2;i
here i is incremente d i.e. i value for next iteration is 3
now if(n%i==0) then c is incremente d
i.e.if(5%2 ==0) then c is incremente d, but 5%2!=0 and so c is not incremente d, c remains 1
c=1;
3rd iteration: i=3;i
here i is incremente d i.e. i value for next iteration is 4
now if(n%i==0) then c is incremente d
i.e.if(5%3 ==0) then c ic incremente d, but 5%3!=0 and so c is not incremente d, c remains 1
c=1;
4th iteration: i=4;i
here i is incremente d i.e. i value for next iteration is 5
now if(n%i==0) then c is incremente d
i.e. if(5%4==0) then c is incremente d, but 5%4!=0 and so c is not incremente d, c remains 1
c=1;
5th iteration: i=5;i
here i is incremente d i.e. i value for next iteration is 6
now if(n%i==0) then c is incremente d
i.e. if(5%5==0) then c is incremente d, 5%5=0 and so c is incremente d.
i.e. c=2
6th iteration: i=6;i
here i value is 6 and 6
now if(c==2) then n is a prime number
we have c=2 from the 5th iteration and thus n=5 is a Prime number.

9. Write a program to check whether the given number is a palindromi c number.
If a number, which when read in both forward and backward way is same, then such a number is called a
palindrome number.
Program:
#include
int main(){
int n, n1, rev = 0, rem;
printf("En ter any number: \n");
scanf("%d" ,&n);
n1 = n;
/* logic */
while (n>0){
rem = n % 10;
rev = rev * 10 + rem;
n = n / 10;
}
if (n1 == rev){
printf("Gi ven number is a palindromi c number");
}
else{
printf("Gi ven number is not a palindromi c number");
}
return 0;
}
Output:
Enter any number: 121
Given number is a palindrome
Explanatio n with an example:
Consider a number n=121, reverse=0, remainder;
number=121
now the while loop is executed /* the condition (n>0) is satisfied */
/* calculate remainder */
remainder of 121 divided by 10=(121%10 )=1;
now reverse=(r everse*10) +remainder
=(0*10)+1 / * we have initialized reverse=0 */
=1
number=num ber/10
=121/10
=12
now the number is 12, greater than 0. The above process is repeated for number=12.
remainder= 12%10=2;
reverse=(1 *10)+2=12;
number=12/ 10=1;
now the number is 1, greater than 0. The above process is repeated for number=1.
remainder= 1%10=1;
reverse=(1 2*10)+1=12 1;
number=1/ 10 / * the condition n>0 is not satisfied,co ntrol leaves the while loop */
Program stops here. The given number=121 equals the reverse of the number. Thus the given number is a
palindrome number.

10.Write a program to check whether the given string is a palindrome .
Palindrome is a string, which when read in both forward and backward way is same.
Example: radar, madam, pop, lol, rubber, etc.,
Program:
#include
#include
int main(){
char string1[20 ];
int i, length;
int flag = 0;
printf("En ter a string: \n");
scanf("%s" , string1);
length = strlen(str ing1);
for(i=0;i<length ;i++){
if(string1 [i] != string1[le ngth-i-1]) {
flag = 1;
break;
}
}
if (flag){
printf("%s is not a palindrome \n", string1);
}
else{
printf("%s is a palindrome \n", string1);
}
return 0;
}
Output:
Enter a string: radar
"radar"is a palindrome
Explanatio n with example:
To check if a string is a palindrome or not, a string needs to be compared with the reverse of itself.
Consider a palindrome string:"radar",
---------- ---------- -------
index: 0 1 2 3 4
value: r a d a r
---------- ---------- -------
To compare it with the reverse of itself, the following logic is used:
0th character in the char array, string1 is same as 4th character in the same string.
1st character is same as 3rd character.
2nd character is same as 2nd character.
. . . .
ith character is same as'length-i- 1'th character.
If any one of the above condition fails, flag is set to true(1), which implies that the string is not a palindrome .
By default, the value of flag is false(0). Hence, if all the conditions are satisfied, the string is a palindrome .

11.Write a program to generate the Fibonacci series.
Fibonacci series: Any number in the series is obtained by adding the previous two numbers of the series.
Let f(n) be n'th term.
f(0)=0;
f(1)=1;
f(n)=f(n-1 )+f(n-2); (for n>=2)
Series is as follows
011
(1+0)
2 (1+1)
3 (1+2)
5 (2+3)
8 (3+5)
13 (5+8)
21 (8+13)
34 (13+21)
...and so on
Program: to generate Fibonacci Series(10 terms)
#include
int main(){
//array fib stores numbers of fibonacci series
int i, fib[25];
// initialized first element to 0
fib[0] = 0;
// initialized second element to 1
fib[1] = 1;
//loop to generate ten elements
for (i = 2; i<10; i++){
//i'th element of series is equal to the sum of i-1'th element and i-2'th element.
fib[i] = fib[i - 1] + fib[i - 2];
}
printf("Th e fibonacci series is as follows \n");
//print all numbers in the series
for (i = 0; i<10; i++){
printf("%d \n", fib[i]);
}
return 0;
}
Output:
The fibonacci series is as follows
01123581
3
21
34
Explanatio n:
The first two elements are initialize d to 0, 1 respective ly. Other elements in the series are generated by looping
and adding previous two numbes. These numbers are stored in an array and ten elements of the series are
printed as output.

12.Write a program to print"Hello World"without using semicolon anywhere in the code.
Generally when we use printf("") statement, we have to use a semicolon at the end. If printf is used inside an if
Condition, semicolon can be avoided.
Program: Program to print something without using semicolon (;)
#include
int main(){
//printf returns the length of string being printed
if (printf("H ello World\n")) //prints Hello World and returns 11
{
//do nothing
}
return 0;
}
Output:
Hello World
Explanatio n:
The if statement checks for condition whether the return value of printf("He llo World") is greater than 0. printf
function returns the length of the string printed. Hence the statement if (printf("H ello World")) prints the string
"Hello World".

13.Write a program to print a semicolon without using a semicolon anywhere in the code.
Generally when use printf("") statement we have to use semicolon at the end.
If we want to print a semicolon, we use the statement: printf(";" );
In above statement, we are using two semicolons . The task of printing a semicolon without using semicolon anywhere in the code can be accomplish ed by using the ascii value of';'which is equal to 59.
Program: Program to print a semicolon without using semicolon in the code.
#include
int main(void) {
//prints the character with ascii value 59, i.e., semicolon
if (printf("% c\n", 59)){
//prints semicolon
}
return 0;
}
Output:
;
Explanatio n:
If statement checks whether return value of printf function is greater than zero or not. The return value of function
call printf("%c ",59) is 1. As printf returns the length of the string printed. printf("%c ",59) prints ascii value that
correspond s to 59, that is semicolon( .

14.Write a program to compare two strings without using strcmp() function.
strcmp() function compares two strings lexicograp hically. strcmp is declared in stdio.h
Case 1: when the strings are equal, it returns zero.
Case 2: when the strings are unequal, it returns the difference between ascii values of the characters that differ.
a) When string1 is greater than string2, it returns positive value.
b) When string1 is lesser than string2, it returns negative value.
Syntax:
int strcmp (const char *s1, const char *s2);
Program: to compare two strings.
#include
#include
int cmpstr(cha r s1[10], char s2[10]);
int main(){
char arr1[10] ="Nodalo";
char arr2[10] ="nodalo";
printf("%d", cmpstr(arr 1, arr2));
// cmpstr() is equivalent of strcmp()
return 0;
}/
/s1, s2 are strings to be compared
int cmpstr(cha r s1[10], char s2[10]){
//strlen function returns the length of argument string passed
int i = strlen(s1) ;
int k = strlen(s2) ;
int bigger;
if (i<k){
bigger = k;
}
else if (i>k){
bigger = i;
}
else{
bigger = i;
}
//loops'bigger'times
for (i = 0; i<bigger; i++){
// if ascii values of characters s1[i], s2[i] are equal do nothing
if (s1[i] == s2[i]){
}
//else return the ascii difference
else{
return (s1[i] - s2[i]);
}
}
//return 0 when both strings are same
//This statement is executed only when both strings are equal
return (0);
}
Output:
-32
Explanatio n:
cmpstr() is a function that illustrate s C standard function strcmp(). Strings to be compared are sent as arguments
to cmpstr().
Each character in string1 is compared to its correspond ing character in string2. Once the loop encounters a
differing character in the strings, it would return the ascii difference of the differing characters and exit.

15.Write a program to concatenat e two strings without using strcat() function.
strcat(str ing1,strin g2) is a C standard function declared in the header file string.h
The strcat() function concatenat es string2, string1 and returns string1.
Program: Program to concatenat e two strings
#include
#include
char *strct(cha r *c1, char *c2);
char *strct(cha r *c1, char *c2){
//strlen function returns length of argument string
int i = strlen(c1) ;
int k = 0;
// loops until null is encountered and appends string c2 to c1
while (c2[k] !='\0'){
c1[i + k] = c2[k];
k++;
}
return c1;
}
int main(){
char string1[15 ] ="first";
char string2[15 ] ="second";
char *finalstr;
printf("Be fore concatenat ion:"
"\n string1 = %s \n string2 = %s", string1, string2);
// addresses of string1, string2 are passed to strct()
finalstr = strcat(str ing1, string2);
printf("\n After concatenat ion:");
//prints the contents of string whose address is in finalstr
printf("\n finalstr = %s", finalstr);
//prints the contents of string1
printf("\n string1 = %s", string1);
//prints the contents of string2
printf("\n string2 = %s", string2);
return 0;
}
Output:
Before concatenat ion:
string1 = first
string2 = second
After concatenat ion:
finalstr = firstsecon d
string1 = firstsecon d
string2 = second
Explanatio n:
string2 is appended at the end of string1 and contents of string2 are unchanged.
In strct() function, using a for loop, all the characters of string'c2'are copied at the end of c1. return (c1) is
equivalent to return&c1[0] and it returns the base address of'c1'.'finalstr' stores that address returned by the
function strct().

16.Write a program to delete a specified line from a text file.
In this program, user is asked for a filename he needs to change. User is also asked for the line number that is
to be deleted. The filename is stored in'filename' . The file is opened and all the data is transferre d to another file
except that one line the user specifies to delete.
Program: Program to delete a specific line.
#include
int main(){
FILE *fp1, *fp2;
// consider 40 character string to store filename
char filename[4 0];
char c;
int del_line, temp = 1;
//asks user for file name
printf("En ter file name:");
// receives file name from user and stores in'filename'
scanf("%s" , filename);
//open file in read mode
fp1 = fopen(file name,"r");
c = getc(fp1);
//until the last character of file is obtained
while (c != EOF)
{
printf("%c ", c);
//print current character and read next character
c = getc(fp1);
}
//rewind
rewind(fp1 );
printf("\n Enter line number of the line to be deleted:") ;
//accept number from user.
scanf("%d" ,&del_line) ;
//open new file in write mode
fp2 = fopen("cop y.c","w");
c = getc(fp1);
while (c != EOF){
c = getc(fp1);
if (c =='\n')
temp++;
//except the line to be deleted
if (temp != del_line)
{
//copy all lines in file copy.c
putc(c, fp2);
}
}
//close both the files.
fclose(fp1 );
fclose(fp2 );
//remove original file
remove(fil ename);
//rename the file copy.c to original name
rename("co py.c", filename);
printf("\n The contents of file after being modified are as follows:\n ");
fp1 = fopen(file name,"r");
c = getc(fp1);
while (c != EOF){
printf("%c ", c);
c = getc(fp1);
}
fclose(fp1 );
return 0;
}
Output:
Enter file name:abc.t xt
hi.
Hello
how are you?
I am fine
hope the same
Enter line number of the line to be deleted:4
The contents of file after being modified are as follows:
hi.
hello
how are you?
hope the same
Explanatio n:
In this program, user is asked for a filename that needs to be modified. Entered file name is stored in a char
array'filename' . This file is opened in read mode using file pointer'fp1'. Character'c'is used to read characters
from the file and print them to the output. User is asked for the line number in the file to be deleted. The file
pointer is rewinded back and all the lines of the file except for the line to be deleted are copied into another file
"copy.c". Now"copy.c"is renamed to the original filename. The original file is opened in read mode and the
modified contents of the file are displayed on the screen.

17.Write a program to replace a specified line in a text file.
Program: Program to replace a specified line in a text file.
#include
int main(void) {
FILE *fp1, *fp2;
// 'filename'i s a 40 character string to store filename
char filename[4 0];
char c;
int del_line, temp = 1;
//asks user for file name
printf("En ter file name:");
// receives file name from user and stores in'filename'
scanf("%s" , filename);
fp1 = fopen(file name,"r");
//open file in read mode
c = getc(fp1);
//print the contents of file .
while (c != EOF){
printf("%c ", c);
c = getc(fp1);
}
//ask user for line number to be deleted.
printf("\n Enter line number to be deleted and replaced") ;
scanf("%d" ,&del_line) ;
//take fp1 to start point.
rewind(fp1 );
//open copy.c in write mode
fp2 = fopen("cop y.c","w");
c = getc(fp1);
while (c != EOF){
if (c =='\n'){
temp++;
}
// till the line to be deleted comes,copy the content from one file to other
if (temp != del_line){
putc(c, fp2);
}
else //when the line to be deleted comes
{
while ((c = getc(fp1)) !='\n'){
}
//read and skip the line ask for new text
printf("En ter new text");
//flush the input stream
fflush(std in);
putc('\n', fp2);
//put'\n'in new file
while ((c = getchar()) !='\n')
putc(c, fp2);
//take the data from user and place it in new file
fputs("\n" , fp2);
temp++;
}
// continue this till EOF is encountere d
c = getc(fp1);
}
//close both files
fclose(fp1 );
fclose(fp2 );
//remove original file
remove(fil ename);
//rename new file with old name opens the file in read mode
rename("co py.c", filename);
fp1 = fopen(file name,"r");
//reads the character from file
c = getc(fp1);
// until last character of file is encountered
while (c != EOF){
printf("%c ", c);
// all characters are printed
c = getc(fp1);
}
//close the file pointer
fclose(fp1 );
return 0;
}
Output:
Enter file name:abc.t xt
hi.
hello
how are you?
hope the same
Enter line number of the line to be deleted and replaced:4
Enter new text: sayonara see you soon
hi.
hello
how are you?
sayonara see you soon
Explanatio n:
In this program, the user is asked to type the name of the file. The File by name entered by user is opened in
read mode. The line number of the line to be replaced is asked as input. Next the data to be replaced is asked. A
new file is opened in write mode named"copy.c". Now the contents of original file are transferre d into new file
and the line to be modified is deleted. New data is stored in its place and remaining lines of the original file are
also transferre d. The copied file with modified contents is replaced with the original file's name. Both the file
pointers are closed and the original file is again opened in read mode and the contents of the original file is
printed as output.

18.Write a program to find the number of lines in a text file.
Number of lines in a file can be determined by counting the number of new line characters present.
Program: Program to count number of lines in a file.
#include
int main()
/* Ask for a filename and count number of lines in the file*/
{
//a pointer to a FILE structure
FILE *fp;
int no_lines = 0;
// consider 40 character string to store filename
char filename[4 0], sample_chr ;
//asks user for file name
printf("En ter file name:");
// receives file name from user and stores in a string named'filename'
scanf("%s" , filename);
//open file in read mode
fp = fopen(file name,"r");
//get character from file and store in sample_chr
sample_chr = getc(fp);
while (sample_ch r != EOF){
// Count whenever sample_chr is'\n'(new line) is encountere d
if (sample_ch r =='\n')
{
// increment variable'no_lines' by 1
no_lines=n o_lines+1;
}
//take next character from file.
sample_chr = getc(fp);
}
fclose(fp) ; //close file.
printf("Th ere are %d lines in %s \n", no_lines, filename);
return 0;
}
Output:
Enter file name:abc.t xt
There are 4 lines in abc.txt
Explanatio n:
In this program, name of the file to be read is taken as input. A file by the given name is opened in read-mode
using a File pointer'fp'. Characters from the file are read into a char variable'sample_ch r'with the help of getc
function. If a new line character( '\n') is encountere d, the integer variable'no_lines' is incremente d. If the
character read into'sample_ch ar'is not a new line character, next character is read from the file. This process is
continued until the last character of the file(EOF) is encountere d. The file pointer is then closed and the total
number of lines is shown as output.


19.Write a C program which asks the user for a number between 1 to 9 and shows the number. If the
user inputs a number out of the specified range, the program should show an error and prompt
the user for a valid input.
Program: Program for accepting a number in a given range.
#include
int getnumber( );
int main(){
int input = 0;
//call a function to input number from key board
input = getnumber( );
//when input is not in the range of 1 to 9,print error message
while (!((input = 1))){
printf("[E RROR] The number you entered is out of range");
//input another number
input = getnumber( );
}
//this function is repeated until a valid input is given by user.
printf("\n The number you entered is %d", input);
return 0;
}/
/this function returns the number given by user
int getnumber( ){
int number;
//asks user for a input in given range
printf("\n Enter a number between 1 to 9 \n");
scanf("%d" ,&number);
return (number);
}
Output:
Enter a number between 1 to 9
45
[ERROR] The number you entered is out of range
Enter a number between 1 to 9
4
The number you entered is 4
Explanatio n:
getfunctio n() function accepts input from user.'while'loop checks whether the number falls within range or not
and accordingl y either prints the number(If the number falls in desired range) or shows error message(nu mber is
out of range).

20.Write a program to display the multiplica tion table of a given number.
Program: Multiplica tion table of a given number
#include
int main(){
int num, i = 1;
printf("\n Enter any Number:");
scanf("%d" ,&num);
printf("Mu ltiplicati on table of %d: \n", num);
while (i
printf("\n %d x %d = %d", num, i, num * i);
i++;
}
return 0;
}
Output:
Enter any Number:5
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4 = 20
5 x 5 = 25
5 x 6 = 30
5 x 7 = 35
5 x 8 = 40
5 x 9 = 45
5 x 10 = 50
Explanatio n:
We need to multiply the given number (i.e. the number for which we want the multiplica tion table)
with value of'i'which increments from 1 to 10.


21. .WAP to check a string is Caliondrom e or not. // Maventic question.
#include
#include
void main()
{
int i,j=0; char a[100];
clrscr();
printf("\n Enter the string to check for caliondrom e:\n");
gets(a);

if(strlen( a)%6)
{
printf("\n %s: is Not a caliondrom e..",a);
getch();
exit(0);
}
for (i=0;a[i]! ='\0'
{
if((a[i]== a[i+5])&&( a[i+1]==a[ i+4])&&(a[ i+2]==a[i+ 3]))
i=i+6;

else
{
j=1;
break;
}
}
if(j)
printf("\n %s: is Not a caliondrom e..",a);
else
printf("\n %s: is a caliondrom e..",a);
getch();
}

22.WAP to print DONE,witho ut using any loop. // asked to my frnd in any company.
#include
void main()
{
static int i=0;
printf("\n %d. DONE",i);
if(i++
main();
getch();
exit(0); / * I used exit(0) to terminate the program after 100 DONE,,i dunno why it was not terminating without using it,may be just at my system,try without it at ur sustem,it sud work */
}


23.WAP to print DONE,witho ut using any loop and any conditonal clause or operators.

/* This code is just in purpose to solve the above question,, but its not a good code in programmin g,as its terminatin g at divide error,,if anyone have a better code,let me know */

main()
{
static int i=100;
printf("%d . DONE\n",10 1-i);
main(1/ --i);
}

/* use"ctrl+f9",then"alt+f5"to see the result */

24. WAP to find out the longest word in a string.
#include
#include
#include
void main()
{
int i,max=0,co unt=0,j;
char str[100]; / * ={"INDIA IS DEMOCRATIC COUNTRY"}; u can use a string inside,in place of user input */

printf("\n Enter the string\n:" );
gets(str);

for(i=0;i
{
if(!(str[i ]==32))
{
count++;
}
else
{

if(max
{
j=i-count;
max=count;
}
count=0;
}
}
for(i=j;i
printf("%c ",str[i]);
getch();
}


25.Prog of WORLD MAP.
#include 
main(l ,a,n,d)
cha r**a;{for(d=atoi (a[1])/ 10*80- atoi(a[2]) / 5-596;n="@N KA\CLCCGZA AQBEAADAFa ISADJABBA^ \SNLGAQABD AXIMBAACTB ATAHDBAN\Z cEMMCCCCAA hEIJFAEAAA BAfHJE\TBd FLDAANEfDN BPHdBcBBBE A_AL\ H E L L O, W O R L D!"[l++-3];)f or(;n-->64 putchar(!d +++33^ l&1);print f("\n\n\n\ n\t\tFound By:\n\t\t\ t Amit Aru");getc h();}

26.WAP to print the triangle of letters in increasing order of lines.

#include
#include
void main()
{
int i,j,k;
char ch;
printf("\n Enter the number of lines wants to make the triangle \n:");
scanf("%d" ,&i);
for(j=1;j
{
ch=65;
for(k=1;k
{
printf("%c ",ch++);
}
printf("\n ");
}
getch();
}


27.WAP to print'xay'in place of every'a'in a string.

#include
#include
void main()
{
int i=0;
char str[100],x ='x',y='y' ;
printf("En ter the string\n:");
gets(str);
while(str[ i]!='\0')
{
if(str[i]= ='a')
{
printf("%c ",x);
printf("%c ",str[i++] );
printf("%c ",y);
}
else
{
printf("%c ",str[i++] );
}
}
getch();
}

28.Count the Total Number of 7 comming between 1 to 100.

/* I made this code in a way that u can give Upper limit i.e. 100,Lower limit i.e. 1 and the specific number u wants to count in between i.e. 7 */

#include
#include
void main()
{
int i,j,U=100, L=1,count= 0,r=1,n;
clrscr();
printf("\n Enter the number u wants to count\n:");
scanf("%d" ,&n);
printf("\n Enter the lower limit\n:");
scanf("%d" ,&L);
printf("\n Enter the upper limit\n:");
scanf("%d" ,&U);

for (i=L;i
{
j=i;
while(j)
{
r=j%10;
if (r==n)
{
count++;
}
j=j/10;
}
}
if(n==0&&L ==0)
count++;
printf("\n Total Number of %d between %d and %d = %d",n,L,U, count);
getch();
}


29. Code for duplicate' s removal,by Amit Aru.
#include
#include
void main()
{
int i,j,k=0,co unt[300]={ 0};
char ch,str[100 0],str1[10 00];
clrscr();
printf("\n Enter the string to remove duplicasy\ n:");
gets(str);
for (i=0;str[i ]!='\0';i+ +)
{
ch=str[i];
count['']=0; / * U can use other delimiter inplace of space''here,just put that char inside'',for ex: count['A']=0 ; if u dnt want any delimiter, just remove this line.*/

if(count[c h])
continue;
else
{
str1[k++]= ch;
count[ch]= 1;
}
}
puts(str1) ;
getch();
}

30. WAP to find out if a given number is a power series of 2 or not,withou t any loop and without using % modulo operator.

#include
#include
int pow2(float );
void main()
{
int i,flag;
clrscr();
printf("En ter the number\n") ;
scanf("%d" ,&i);
flag=pow2( i);
if(flag)
printf("\n %d is power series of 2",i);
else
printf("\n %d is not a power series of 2",i);
getch();
}

int pow2(float j)
{
static float x;
x=j/2;
if(x==2)
return 1;
if(x
return 0;
x=pow2(x);
}