Computer Science Project

Vedic Mathematics Teacher

(With illustration)

Please review this project.

iamashishchaudhary@outlook.com

By- Ashish Chaudhary

XII A

Computer Science

Army Public School, Meerut Cantt.

Board Roll No. 5716336

Contents:1.) Acknowledgement2.) System Requirements

3.) Introduction To Vedic Mathematics4.) Advantages of Vedic Mathematics5.) Use of Vedic Mathematics in the field of Computers6.) Vedic Mathematics in schools and colleges 7.) Researches on Vedic Mathematics8.) The Vedic Mathematics - Sutras9.) Basics of Vedic Mathematics10.) Concepts of C++ used11.) The Code12.) References

Acknowledgement

I would like to express my special thanks of gratitude to my teacher Mr. Sachin Chahal as well as my school Army Public School, Meerut Cantt. who gave me the golden opportunity to do this project on the topic Vedic Mathematics Teacher With illustration , which also helped me a lot in remodeling the way I calculate and also I came to know about so many new things.

I am really thankful to them.Secondly i would also like to thank my parents and friends who helped me a lot in finishing this project within the limited time.

THANKS AGAIN TO ALL WHO HELPED ME.

Certificate

This is to certify that Ashish Chaudhary (Board Roll No. 5716336) of class XII A, Army School Meerut Cantonment has successfully completed his project work on the topic

“Veda Mathematics Teacher ( With Illustration).”

Mr. Sachin Chahal

System Requirements:

Minimum System Requirements for running the program:

• Intel Pentium II/300 MHz or better• Microsoft Windows Server 2003 (SP1)Microsoft Windows XP Professional (SP2)Windows 2000 Professional (SP4)Windows 2000 Server (SP4)

• 128 MB RAM or better• CD-ROM or DVD-ROM drive• SVGA or higher-resolution monitor (XGA recommended)

The following C++ program has been constructed via a free software Turbo C++ for windows 7 and has been compiled on Windows 8 Professional 32-bit (6.1, Build 7601) Service Pack 1.

The program has been successfully compiled with the following specifications

of the machine used:

------------------

System Information

------------------

Time of this report: 1/11/2013, 18:36:32

Machine name: GAMING-PC

Operating System: Windows 8 Professional 32-bit (6.1, Build 7601) Service Pack 1

(7601.win8sp1_rtm.101119-1850)

Language: English (Regional Setting: English)

System Manufacturer: INTEL_

System Model: DG41RQ__

BIOS: BIOS Date: 02/08/08 17:50:03 Ver: 08.00.10

Processor: Intel(R) Core(TM)2 Quad CPU E7500 @ 2.93GHz (4 CPUs), ~2.9GHz

Memory: 6144MB RAM

Available OS Memory: 6144MB RAM

Page File: 1287MB used, 2800MB available

Windows Dir: C:\Windows

DirectX Version: DirectX 11

DX Setup Parameters: Not found

User DPI Setting: Using System DPI

System DPI Setting: 96 DPI (100 percent)

DWM DPI Scaling: Enabled

DxDiag Version: 6.01.7601.17514 32bit Unicode

---------------

Display Devices

---------------

Card name: AMD Radeon HD 6990

Manufacturer: Advanced Micro Devices, Inc.

Chip type: ATI display adapter (0x954F)

DAC type: Internal DAC(400MHz)

Device Key: Enum\PCI\VEN_1002&DEV_954F&SUBSYS_03541043&REV_00

Display Memory: 2803 MB

Dedicated Memory: 2038 MB

Shared Memory: 765 MB

Current Mode: 1980 x 1080 (32 bit) (120Hz)

Monitor Name: Generic PnP Monitor

Monitor Model: W2043

Monitor Id: GSM4E9E

Native Mode: 1980 x 1080(p) (120.000Hz)

Output Type: DVI

Driver Name: aticfx32.dll,aticfx32.dll,atiumdag.dll,atidxx32.dll,atiumdva.cap

Driver File Version: 8.17.0010.1129 (English)

Driver Version: 8.970.100.3000

DDI Version: 10.1

Driver Model: WDDM 1.1

Driver Attributes: Final Retail

Driver Date/Size: 7/4/2012 11:57:08, 918528 bytes

WHQL Logo'd: No

WHQL Date Stamp: None

Device Identifier: {D7B71EE2-D60F-11CF-9A77-5E23BEC2C535}

Vendor ID: 0x1002

Device ID: 0x954F

SubSys ID: 0x03541043

Revision ID: 0x0000

Driver Strong Name: oem5.inf:ATI.Mfg.NTx86.6.1:ati2mtag_R7X:8.970.100.3000:pci\

ven_1002&dev_954f

Rank Of Driver: 00E62001

-------------

Sound Devices

-------------

Description: Speakers (Realtek High Definition Audio)

Default Sound Playback: Yes

Default Voice Playback: Yes

Hardware ID: HDAUDIO\FUNC_01&VEN_10EC&DEV_0662&SUBSYS_8086D613&REV_1001

Manufacturer ID: 1

Product ID: 100

Type: WDM

Driver Name: RTKVHDA.sys

Driver Version: 6.00.0001.6662 (English)

Driver Attributes: Final Retail

WHQL Logo'd: Yes

Date and Size: 6/19/2012 16:54:20, 3240400 bytes

Other Files:

Driver Provider: Realtek Semiconductor Corp.

HW Accel Level: Basic

Cap Flags: 0xF1F

Min/Max Sample Rate: 100, 200000

 

Introduction To Vedic Mathematics

About a hundred years ago the scholars in the west discovered the IndianVedas: ancient texts in their millions containing some of the most profoundknowledge. In fact the Sanskrit word ‘Veda’ means ‘knowledge’. They foundVedic texts on medicine, architecture, astronomy, ethics etc and according to the Indian tradition all knowledge is contained in these Vedas.

The age of these Vedic texts is disputed, tradition telling us that oginally the

Vedas were carried on by word of mouth, by pandits who recited the textsexactly and passed them on to their children.

However certain writings that were entitled Ganita Sutras, which meansmathematics, were not found to contain any mathematics and the scholarsrejected them saying they were nonsense.

Now a brilliant eastern scholar, Sri Bharata Krsna Tirthaji (1884-1960), heard what was being said about the Ganita texts and determined to uncover the mathematics in them. Between 1911 and 1918 he studied the texts and eventually was able to reconstruct the system that had been lost long ago.

He discovered that all of mathematics is governed by just sixteen sutras orword-formula (and some sub-Sutras), and he showed some really amazingexamples of their power and versatility. He wrote sixteen books (one on each sutra) but these have been lost. In 1958 when the loss was confirmed he wrote an introductory book, which is currently available (Reference 1).

Recent attempts to reconstruct the system from this book have led to anupsurge of interest and the Vedic system which being taken up by researchers, teachers and others all over the world.

Advantages of Vedic Mathematics:We are living in the age of tredemdous amount of competitions and Vedic Mathematics methods come to us as a boon for all the competitions. Present maths,a scary subject  requires higher amount of effort in learning. Maths can be learnt and mastered with miminum efforts in a very short span of time and can be translated into a playful and a blissful subject with the help of Vedic Maths.

The several advantages of Vedic Maths are:

It reduces the burden of remembering large amount of stuff because it requires you to learn tables upto 9 only.

It enables faster calculations when compared to the conventional method. Thus, the time that gets saved in the process can be used to answer more questions

It acts as a tool for reducing finger counting and scratch work. It plays an important role in increasing concentration as well as improving

confidence. It is very  simple, direct, totally unconventional, original and straight

forward. It encourages mental calculations. It enriches our understanding of maths and enables us to see links and

continuity between different branches of maths. Vedic Maths system also gives us a set of checking procedures for

independent crosschecking of whatever we do. It keeps the mind alert and lively because of th element of choice and

flexibility at each. Holistic development of the human brain takes place through Vedic

Mathematics along with multidimensional thinking. Vedic Mathematics system to quite an extent also helps us in developing

our spiritual part of personality. It can introduce creativity in intelligent and smart students, while helping

the slow-learners grasp the basic concepts of mathematics. More and more use of Vedic math can without any doubts generate interest in a subject that is generally dreaded by children.

Thus, Vedic maths is considered to be more than the blessing of the Veda for the entire humanity.

A large number of people in the field of education, especially maths professors consider Vedic Maths to be a new and a better approach in studying and understanding Maths. Even students studying at the Indian Institute of Technology are said to be making use of this technique for quicker calculations. Several other institutes are also emphasizing the significance of Vedic Maths.

Use of Vedic Mathematics in the field of Computers:

Vedic mathematics is used by several researchers in the field of Digital signal processing, Chip designing, Discrete Fourier Transform , High speed low power VLSI arithmetic and algorithm, RSA encryption system . Most of the researchers have used the vedic mathematics method such as multiplication, division, squares and cubes in above mention fields.

VLSI Implementation of RSA encryption

H. Thapliyal implementated RSA encryption or decryption algorithm using the algorithm of ancient

Indian vedic mathematics which is being modified to improve the performance .The most significant aspect is the development of divisor architecture based on straight division algorithm of Vedic mathematics and embedding it in RSA encryption or decryption circuitry for improved efficiency .They proved that RSA circuitry implemented using vedic division and multiplication is efficient in terms of area and speed compared to its implementation using conventional multiplication and division architecture.

Multiplier and square architecture

Time, area and power efficient multiplier and square architecture based on ancient Indian vedic mathematics is developed by H. Thapliyal. Developed algorithm was for low power and high speed applications. It is based on generating partial product and their sums in one step. The design implemented is described at both gate andhigh level using verilog Hardware Description Language. The design code is tested using “Veriwelsimulator. Their work relates to the field of mathematical coprocessor in computers and more specifically to improve speed and power over the conventional co-processor. The Vedic multiplier and square architecture developed by them were faster than array multiplier and booth multiplier in FPGA implementation.

Digital Signal Processing

Digital Signal processing is a technology that is present in almost every engineering discipline. It is also the fastest growing technology of the century and hence it posses tremendous challenges to the engineering community. Faster addition and multiplication are of extreme importance in DSP for Convolution, DFT and Digital filters .The core computing process is always a multiplication routine and hence DSP

engineers are constantly looking for newalgorithms and hardware to implement them. The methods in Vedic Mathematics are complementary directly and easy. Mr. Mangesh Karad and Mr. Chidgupkar highlight the use of multiplication process based on vedic algorithms and implemented on 8085 and 8086 microprocessors . Use of Vedic algorithms shows appreciable saving of processing time.

ALU Design

The ever increasing demands in enhancing the ability of processors to handle the complex and challenging processes has resulted in the integration of number of processor cores into one chip still the load on processor is not less. This load is reduced by supplementing the main coprocessor which is designed to work on specific types of functions like numeric computations, signal processing, graphics etc. The speed of ALU depends on the multipliers. In algorithm and structure levels, numerous multiplication techniques have been developed to enhance the efficiency of multiplier which concentrates on reducing the partial product and their additions. But in this case principle behind the multiplication remains same. Use of Vedic mathematics for multiplication strikes difference in actual process. M. Ramalatha used Urdhva tiryakbhyam Sutra of Vedic

mathematics to build a high speed power efficient multiplier in the coprocessor. Using Vedic techniques various arithmetic modules can be designed and integrated into a Vedic ALU, which is compatible for co-processor. The advantages of Vedic multipliers are increase in speed, decrease in delay, decrease in power consumption and decrease in area occupancy. This Vedic coprocessor is more efficient than the conventional one.

Performance Analyses of Vedic Algorithms

Various parameters are recommended byresearchers to evaluate the performance of Vedic Maths

algorithm. Researchers suggested many parameters few of them are: Time, Delay, Power and Number of slices. The following table shows the comparison of Delay (ns) factor for multiplication implemented in different algorithms between Conventionaland Vedic way.

Conclusion

Vedic mathematics can be used inimplementation of fast algorithms in various fields of engineering. The performance analysis of Vedic algorithms is done on the basis of delay. This shows efficiency of Vedic multiplier interms of high speed and less complexity.

VEDIC MATHS IN SCHOOLS AND COLLEGES: 

A few years ago several schools began to teach the Vedic system. The response received was so huge that this remarkable system is today in many schools and institutions in India as well as abroad. It is not only taught in schools but also to a lot of MBA and Economic students. There are obviously many benefits of making use of a flexible, distinguished and proficient mental system like Vedic math. Pupils can come out of the imprisonment of the ‘only one correct’ way, and make their own distinct methods under the Vedic system.

Ever since the year 1999, a forum which has been initiated in Delhi called International Research Foundation for Vedic Mathematics and Indian Heritage, has arranged several lectures on Vedic maths in various schools and colleges in Delhi.

RESEARCHES ON VEDIC MATHS: 

People like Aryabhatta, who laid the foundation of Algebra, Baudhayan, the great geometer and the saint duo, Medhatithi and Madhyatithi, who formulated the basic structure for numerals, have conducted large number of researches.

Several research programmes have currently been undertaken in many areas. These researches have been initiated in order to try and find the effects of learning Vedic maths on children, to develop more commanding and simple applications of the Vedic sutras in different fields such as geometry, calculus, and computing.

THE VEDIC MATHEMATICS – SUTRAS

Sutras: The Natural Formula

In Sanskrit, the ancient language of India, the word Sutra means "Thread of Knowledge". In English we have the word Suture from Sutra and Suture is the thread used for stitching wounds together. So each rule is a thread of knowledge and the whole subject is based on these sutras. The system is based on 16 Vedic Main-Sutras together with 13 Sub-Sutras which are actually word formulae describing natural ways of solving a whole range of mathematical problems. These are 16 one-line formulae originally written in Sanskrit,which can be easily memorized, enable one to solve long mathematical problems quickly.

• The Sutras (aphorisms) apply to and cover each and every part of each andevery chapter of each and every branch of mathematics (including Arithmetic,Algebra, Geometry – plane and solid, Trigonometry – plane and spherical,Conics – geometrical and analytical, Astronomy, Calculus – differential andintegral etc.) In fact, there is no part of mathematics, pure or applied, that isbeyond their jurisdiction.

• The Sutras are easy to understand, easy to apply and easy to remember, and the whole work can be truthfully summarized in one word ‘Mental’!

Basics Of Vedic Mathematics:1GIVE AND TAKE

A Addition and subtraction always seem to come together. For example, if you take some money out of your pocket it is now in your hand but is no longer in your pocket.So it is no surprise that there is a Vedic Sutra for this:By Addition and By SubtractionWe naturally use this when, say, we add 55 and 19.Most people will add 20 and take 1 away, so 55 + 19 = 74.Similarly for 198 + 64 we could add 200 to 64 and take off 2 after.So 198 + 64 = 262.

In the same way we can subtract easily:For 44 – 19 we take 20 from 44 and add 1 back on.So 44 – 19 = 25.And for 333 – 198 we can take 200 from 333 and add 2 back.We get 333 – 198 = 135.

This is what many people would naturally do.

B For this we use the Sutra:

All from 9 and the Last from 10

To take 357 from 1000 for example,we apply All from 9 and the Last from 10 to the figures of 357.So 1000 – 357 = 643.We take the 3 from 9 to get 6,take the 5 from 9 to get 4,and take the 7 from 10 to get 3.Similarly 10,000 – 7621 = 2379,we take 7, 6 and 2 from 9 and the 1 from 10.

If the number of figures in the number being subtracted is less than the number of zeros in the base number, we just add as many zeros as needed to make them the same: For example in 10,000 – 643 there are four zeros and only three figures in 643.So we write 10,000 – 0643 so that 0643 has four figures.Then 10,000 – 0643 = 9357.(We take 0, 6, 4 from 9 and 3 from 10)Similarly, 10,000 – 77 = 10,000 – 0077 = 9923.

Now suppose with don’t have a base number, but a multiple of it: like 200,3000 etc.Suppose you had 3000 – 444.We get 3000 – 444 = 2556.The 444 will come off one of the three thousands, so the initial 3 becomes 2.Then applying “All from 9 and the Last from 10” to the 444 we put down 556.Similarly 700 – 52 = 648.The 7 reduces to 6, and applying the sutra to 52 gives 48.

There is one thing to be careful of though.Suppose you had 7000 – 430.When you apply the Sutra to 430 you just apply it to the 43, not to 430.So you get 7000 – 430 = 6570.(7 becomes 6, the Sutra converts 43 to 57 and the zero is carried over).

C Amazingly, this simple method can handle all subtractions!Suppose we had: 444 – 286.Subtracting in each column we get 4 – 2 = 2, 4 – 8 = –4, 4 – 6 = –2.Since these negative answers can be written like this :2 (42) and 2 (42) is easily converted into 158 because this is justthe kind of sum you have just done:2 (42) means 200 – 42 = 158.

Similarly

6767-1908= 5(2)6(1)

Here to convert 5(2)6(1) we split it into 5(2) / 6(1)

and 5(2)=48 (because 50-2) also 6(1)=59 (because 60-1=59).

2MIGHTY MULTIPLICATIONS

In the Vedic system there are general methods that always work (see Chapter

3 for general multiplication) and there are special methods which can beapplied in special cases. Here we look at a very powerful special method that works when the numbers are near a base number (like 10, 100, 1000 etc.) or a multiple of a basenumber.

A Numbers below a base number.88 × 97 = 8536We notice here that both numbers being multiplied are close to 100:88 is 12 below 100, and 97 is 3 below it.

So we write 88 = 88(3) or 97= 97(12) which is equal to 88-3=85 or 97-12=85 (whichever you like).

Then we multiply the deficiencies ie 12 and 3 to get 12 × 3= 36.

So the answer (in two parts) becomes 85 / 36 = 8536.

Similarly, 93 × 96 = 8928

Deficiencies are 7 and 4 so

93 =93(4) and 96= 96(7)

Which equals 93-4=89 or 96-7=89.

Then 7 × 4= 28.

Therefore the answer is 89 / 28 =8928.

89 × 89= 7921

Here the first part is 89(11)=89-11=78 and multiplying the deficiencies gives 121.

So we write 21 out of the 121 and carry 1 to 78 to become 79.

This gives the answer is 79 / 21 =7921.

876 × 998 = 874 / 248

Same rule here, but the base number is 1000 so the deficiencies are 1000-876= 124 by All from 9 and the Last from 10 sutra and 1000-998=2.

So the first part is 876-2=998-124=874 and the second part is 124×2=248.

So the answer is 874 / 248 = 874248.

D But this is not all: if the numbers are above the base number the methodstill works:103 × 104 = 10712

Here the excess is 103-100=3 and 104-100=4.

So the first part in this case becomes 103+4=104+3=107 and the second part becomes 4×3=12.

This gives the answer 10712.

E But suppose one number is above the base and the other number isbelow it.Can the method still be used?The answer is yes, most certainly.

124 × 98 = 122(48)= 12152

Here the base number is 100 and the deficiency is 100-98=2 also the efficiency is 124-100=24.

Therefore the first part is 124-2 or 98+24= 122 (whichever you like), the second part is 24×minus 2 =(48).

So the answer is 122(48)= 12200-48. Now applying All from 9 and the Last from 10 sutra 122 becomes 121 and 48 gives 52 therefore the answer becomes 12152.

F Let’s look at one more extension of this method: where the numbers arenot near a base number, but are near a multiple of a base number, like 30,200, 6000.

In fact we use a sub-sutra called Proportionately.

213 × 203 = 2 × 216/39 = 43239

Here excess is 213-200=13 and 203-200=3.

We observe here that the numbers are not near any of bases used before:10, 100, 1000 etc. But they are close to 200, and are 13 and 3 above 200.

The usual procedure gives us 216/39 (213+3=216, 13×3=39).Now since our base is 200 which is 100×2 we multiply only the left-handpart of the answer by 2 to get 43239.

29 × 28 = 3×27/2 = 812

Deficiency is 30-29=1 and 30-28=2.

So 29-2=28-1= 27 and 3 × 27=81 and 2 × 1 =2, therefore the answer becomes 812.

3VERTICAL AND CROSSWISE PATTERNS_____________________________________________________The Vedic Sutra Vertically and Crosswise enables us to multiply any twonumbers together in one line (from left to right or from right to left) no matterhow big the numbers are.A Let us start with multiplying 2-figure numbers.21 × 23 = 483we put the numbers one below the other and multiply:A. vertically on the right,B. then crosswise and add,

C. and then vertically on the left.

C. multiply vertically in the right-handcolumn: 1 × 3 = 3,so 3 is the last figure of the answer.D. multiply crosswise and add:2 × 3 = 6,1 × 2 = 2, 6 + 2 = 8,so 8 is the middle figure of the answer.E. multiply vertically in the left-handcolumn: 2 × 2 = 4,4 is the first figure of the answer.

So the answer is 483.

The previous examples involved no carry figures, so let us consider this next.

23×41=943

The 3 steps give us: 3×1 = 3,2×1 + 3×4 = 14, so we carry 1 to the left as shown.2×4 = 8, and 8 + the carried 1 = 9.

So 23 × 41 = 943.

Concepts of C++ used:

a. Switch case: switch ( expression )case constant-expression : statement[default : statement]

If a matching expression is found, control is not impeded by subsequent case or default labels. The break statement is used to stop execution and transfer control to the statement after the switch statement.

Without a break statement, every statement from the matched case label to the end of the switch, including the default, is executed. For example:

// switch_statement1.cpp#include <stdio.h>

int main() { char *buffer = "Any character stream"; int capa, lettera, nota; char c; capa = lettera = nota = 0;

while ( c = *buffer++ ) // Walks buffer until NULL { switch ( c ) { case 'A': capa++; break; case 'a': lettera++; break; default: nota++; } } printf_s( "\nUppercase a: %d\nLowercase a: %d\nTotal: %d\n", capa, lettera, (capa + lettera + nota) );}In the above example, capa is incremented if c is an uppercase A. The break statement after capa++ terminates execution of the switch statement body and control passes to the while loop. Without the break statement, lettera and nota would also be incremented. A similar purpose is served by the break statement for case 'a'. If c is a lowercase a, lettera is incremented and the break statement terminates the switch statement body. If c is not an a or A, the default statement is executed.

A switch statement can be nested. In such cases, case or default labels associate with the closest switch statement that encloses them.

////////////////////////////////////////////////////////////////////////

b. Classes:

A class is an expanded concept of a data structure: instead of holding only data, it can hold both data and functions.

An object is an instantiation of a class. In terms of variables, a class would be the type, and an object would be the variable.

Classes are generally declared using the keyword class, with the following format:

class class_name { access_specifier_1: member1; access_specifier_2: member2; ... } object_names;

Where class_name is a valid identifier for the class, object_names is an optional list of names for objects of this class. The body of the declaration can contain members, that can be either data or function declarations, and optionally access specifiers.

All is very similar to the declaration on data structures, except that we can now include also functions and members, but also this new thing called access specifier. An access specifier is one of the following three keywords:private, public or protected. These specifiers modify the access rights that the members following them acquire:

private members of a class are accessible only from within other members of the same class or from theirfriends.

protected members are accessible from members of their same class and from their friends, but also from members of their derived classes.

Finally, public members are accessible from anywhere where the object is visible.

By default, all members of a class declared with the class keyword have private access for all its members. Therefore, any member that is declared before one other class specifier automatically has private access.

For example: 

class CRectangle { int x, y; public: void set_values (int,int); int area (void); } rect;

Declares a class (i.e., a type) called CRectangle and an object (i.e., a variable) of this class called rect. This class contains four members: two data members of type int (member x and member y) with private access (because private is the default access level) and two member functions with public access: set_values() and area(), of which for now we have only included their declaration, not their definition.

////////////////////////////////////////////////////////////////////

c. Scope Resolution:

 ‘::’ is known as Scope Resolution Operator. C++ is a block structured language. Different program modules are written in various blocks. Same variable name can be used in different blocks. Scope of a variable extends from the point of declaration to the end of the block. A variable declared inside a block is ‘local’ variable. Blocks in C++ are often nested.

e.g.….….{           int x = 10;           ….           ….           {                  int x = 20;                  …..                  ….           }           ….}

The declaration of the inner block hides the declaration of same variable in outer block. This means, within the inner block, the variable x will refer to the data object declared therein. To access the global version of the variable, C++ provides scope resolution operator.

In the above example, x has a value of 20 but ::x has value 10.Similarly, this operator is used when a member function is defined outside the class

e.g.Class MyClass{int n1, n2;public:{void func1();                     ---------Function Declaration}};

public void MyClass::func1() ---Use of Scope Resolution Operator to write                                                          function definition outside class definition{               // Function Code }

//////////////////////////////////////////////////////////////////////////////

d. Inheritance:

Inheritance is a mechanism of reusing and extending existing classes without modifying them, thus producing hierarchical relationships between them.

Inheritance is almost like embedding an object into a class. Suppose that you declare an object x of class A in the class definition of B. As a result, class B will have access to all the public data members and member functions of class A. However, in class B, you have to access the data members and member functions of class A through objectx. The following example demonstrates this:

#include <iostream>

using namespace std;

class A {

int data;

public:

void f(int arg) { data = arg; }

int g() { return data; }

};

class B {

public:

A x;

};

int main() {

B obj;

obj.x.f(20);

cout << obj.x.g() << endl;

// cout << obj.g() << endl;

}

In the main function, object obj accesses function A::f() through its data memberB::x with the statement obj.x.f(20). Object obj accesses A::g() in a similar manner with the statement obj.x.g(). The compiler would not allow the statementobj.g() because g() is a member function of class A, not class B.

The inheritance mechanism lets you use a statement like obj.g() in the above example. In order for that statement to be legal, g() must be a member function of class B.

Inheritance lets you include the names and definitions of another class's members as part of a new class. The class whose members you want to include in your new class is called a base class. Your new class is derived from the base class. The new class contains a subobject of the type of the base class. The following example is the same as the previous example except it uses the inheritance mechanism to give class Baccess to the members of class A:

#include <iostream>

using namespace std;

class A {

int data;

public:

void f(int arg) { data = arg; }

int g() { return data; }

};

class B : public A { };

int main() {

B obj;

obj.f(20);

cout << obj.g() << endl;

}

Class A is a base class of class B. The names and definitions of the members of class Aare included in the definition of class B; class B inherits the members of class A. ClassB is derived from class A. Class B contains a subobject of type A.

You can also add new data members and member functions to the derived class. You can modify the implementation of existing member functions or data by overriding base class member functions or data in the newly derived class.

You may derive classes from other derived classes, thereby creating another level of inheritance. The following example demonstrates this:

struct A { };

struct B : A { };

struct C : B { };

Class B is a derived class of A, but is also a base class of C. The number of levels of inheritance is only limited by resources.

////////////////////////////////////////////////

e. if-else ladder:

When in any situation number of conditions arise in a sequence, then we can use ladder-ifstatement.In this, first condition is checked. If it is true then action will be taken. Otherwise furtheraction will be checked and this process will continue till the end of the condition.

The general syntax of this is :

if(condition 1){statement1}

else if(condition 2){statement 2;}

else if(condition 3){statement 3;}................................................

else if(condition n){statement-n;}else{default statement;}Statement -x

Example :

#include <iostream.h>#include <conio.h>

void main(){int x;int y;

clrscr();

cout << "\nEnter X : ;cin >> x;

cout << "\nEnter Y : ;cin >> y;

if ( x > y ){cout << "\nX is Greater Than Y";}

else if ( y > x){cout << "\nY is Greater Than X";}else{cout << "\nBoth Are Same";}getch();}

f. File Handling

So far we have done I/O of basic data types. Since the class objects are the central elements of C++ programming, it is quite natural that the language supports features for writing and reading from the disk files objects directly.

The binary input and output functions read() and write() are designed to do exactly this job.

The write() function is used to write the object of a class into the specified file and read() function is used to read the object of the class from the file. 

Both these functions take two arguments:1. address of object to be written.2. size of the object.

The address of the object must be cast to the type pointer to char.

One important point to remember is that only data members are written to the disk file and the member functions are not.

Writing an object into the file

#includeclass Person{private:char name[40];int age;public:void getData(){cout << “\n Enter name:”; cin >> name;cout << “\n Enter age:”; cin >> age;}} ; // End of the class definition

void main(){Person per ; // Define an object of Person class

per.getData(); // Enter the values to the data members of the class.

ofstream outfile(“Person.txt”); // Open the file in output mode

outfile.write((char*)&per, sizeof(per)); // Write the object into the file}

fstream object can be used for both input & output.

In the open() function we include several mode bits to specify certain aspects of the file object.

app -> To preserve whatever was in the file before. Whatever we write to the file will be appended to the existing contents.

We use in and out because we want to perform both input and output on the file.

eof() is a member function of ios class. It returns a nonzero value if EOF is encountered and a zero otherwise.

The Code:// VEDIC MATHEMATICS TEACHER BY ASHISH CHAUDHARY XII A

// CLASS XII PROJECT FOR BOARD EXAMS 2012-2013

#include<iostream.h>

#include<conio.h>

#include<stdio.h> //for gets function

#include<process.h> //for exit function

#include<fstream.h>

class io

{

public:

char name[35];

int n1,n2,k,x1,x2,y1,y2,z1,z2,l1,l2,m1,m2,k1,k2,j;

void wel();

void enter1();

void show1();

};

//function definitions of class io

void io::wel()

{

gets(name);

clrscr();

cout<<"\nWelcome "<<name;

}

void io::enter1()

{

cout<<"\nEnter the first number ";

cin>>n1;

cout<<"\nEnter the second number ";

cin>>n2;

}

void io::show1()

{

clrscr();

if(n1>=n2)

cout<<"\nThe entered numbers are "<<n1<<" and "<<n2;

else

cout<<"\nERRORENOUS CHOICE ";

}

class teacher:public io

{

public:

int c;

void choice();

void sep();

void add();

void subs();

void might();

void piece();

void give_func();

void mig_func();

void vert_func();

void mu();

};

//function definitions of class teacher

void teacher::choice()

{

cin>>c;

if(c>0 && c<6)

{

switch(c)

{

case 1:

clrscr();

cout<<"\nYou have entered piecewise automated illustration as your choice ";

piece();

break;

case 2:

clrscr();

cout<<"\nYou have entered addition as your choice ";

add();

break;

case 3:

clrscr();

cout<<"\nYou have entered substraction as your choice ";

subs();

break;

case 4:

clrscr();

cout<<"\nYou have entered mighty multiplication as your choice ";

might();

break;

}

}

else

{

cout<<"\nWrong choice, press any key then enter to exit ";

cin>>k;

exit(c);

}

}

void teacher::sep()

{

enter1();

x1=n1/100;

z1=n1%10;

y1=((n1%100)-z1)/10;

x2=n2/100;

z2=n2%10;

y2=((n2%100)-z2)/10;

l1=(x1+x2)/10;

l2=(x1+x2)%10;

m1=(y1+y2)/10;

m2=(y1+y2)%10;

k1=(z1+z2)/10;

k2=(z1+z2)%10;

}

void teacher::add()

{

sep();

cout<<"\nWe have to find "<<n1<<" + "<<n2;

cout<<"\nAdding coloumn wise :";

cout<<"\n"<<x1<<" + "<<x2<<" = "<<x1+x2;

cout<<"\n"<<y1<<" + "<<y2<<" = "<<y1+y2;

cout<<"\n"<<z1<<" + "<<z2<<" = "<<z1+z2;

m2+=k1;

l2+=m1;

cout<<"\nThe excess in 2nd coloumn is "<<m1;

cout<<"\nThe excess in 3rd coloumn is "<<k1;

cout<<"\nAdding excess 2nd to the value of coloumn 1 gives "<<x1+x2+m1;

cout<<"\nAdding excess 3rd to the value of coloumn 2 gives "<<y1+y2+k1;

cout<<"\nTherefore the answer is "<<1000*l1+100*l2+10*m2+k2;

cout<<"\nThankyou for using the calculator ";

}

void teacher::subs()

{

sep();

cout<<"\nWe have to find "<<n1<<" - "<<n2;

if(n1>=n2)

{

cout<<"\nSubstracting coloumn wise: ";

cout<<"\n"<<x1<<" - "<<x2<<" = "<<x1-x2;

cout<<"\n"<<y1<<" - "<<y2<<" = "<<y1-y2;

cout<<"\n"<<z1<<" - "<<z2<<" = "<<z1-z2;

cout<<"\n The anser (without conversion) is ";

cout<<x1-x2;

if(y1-y2>0)

cout<<y1-y2;

else

cout<<"("<<y2-y1<<")";

if(z1-z2>0)

cout<<z1-z2;

else

cout<<"("<<z2-z1<<")";

cout<<"\nTo convert, put a 0 in place of (n) and subtract 10*n for tens and n for ones "<<"\n from the resultant number ";

cout<<"\nThe answer in converted form is "<<100*(x1-x2)+10*(y1-y2)+z1-z2;

}

else

{

cout<<"\nThis can be written as "<<" -("<<n2<<" - "<<n1<<")";

cout<<"\nWe will first find "<<n2<<" - "<<n1<<" and then add a minus sign ";

cout<<"\nSubstracting coloumn wise: ";

cout<<"\n"<<x2<<" - "<<x1<<" = "<<x2-x1;

cout<<"\n"<<y2<<" - "<<y1<<" = "<<y2-y1;

cout<<"\n"<<z2<<" - "<<z1<<" = "<<z2-z1;

cout<<"\n The anser (without conversion) is ";

cout<<x2-x1;

if(y2-y1>0)

cout<<y2-y1;

else

cout<<"("<<y1-y2<<")";

if(z2-z1>0)

cout<<z2-z1;

else

cout<<"("<<z1-z2<<")";

cout<<"\nTo convert, put a 0 in place of (n) and subtract 10*n for tens and n for ones "<<"\n from the resultant number ";

cout<<"\nThe converted form is "<<100*(x2-x1)+10*(y2-y1)+z2-z1;

cout<<"\nThefore the answer (with minus sign) is "<<100*(x1-x2)+10*(y1-y2)+z1-z2;

}

}

void teacher::piece()

{

cout<<"\nThis option lets you learn Vedic Mathematics step by step ";

cout<<"\nPress 1 to continue ";

cin>>n1;

if(n1==1)

{

clrscr();

cout<<"\nPress 1 to learn the chapter GIVE AND TAKE ";

cout<<"\nPress 2 to learn the chapter MIGHTY MULTIPLICATION ";

cout<<"\nPress 3 to learn the chapter VERTICALLY AND CROSSWISE(smaller multiplications) ";

cin>>j;

switch(j)

{

case 1:

give_func();

break;

case 2:

mig_func();

break;

case 3:

vert_func();

break;

default:

cout<<"\nInvalid Choice";

}

}

else

{

cout<<"\nPress any key then enter to exit ";

cin>>n2;

exit(n2);

}

}

void teacher::give_func()

{

clrscr();

ofstream Ofile("d:\\giveandtake.txt",ios::out);

clrscr();

cout<<"\nThe first step is GIVE AND TAKE ";

cout<<"\nAddition and substraction always seem to come together. For example if you take some money ";

cout<<"\nout of your pocket it is now in your hand but is no longer in your pocket. So there is no ";

cout<<"\nsurprise there is a Vedic Sutra for this: ";

cout<<"\nBY ADDITION and BY SUBSTRACTION ";

cout<<"\nWe naturally use this when,say,we add 55 and 19.";

cout<<"\NMost people will add 20 and take 1 away, so 55+19=74";

cout<<"\nIn the same way we can subtract easily: For example 44-19, we take 20 and give 1, so 44-19=23";

cout<<"\nThis is what most people do ";

cout<<"\nNow we will apply the vedic sutra for this ";

Ofile<<"\nThe first step is GIVE AND TAKE "

<<"\nAddition and substraction always seem to come together. For example if you take some money "

<<"\nout of your pocket it is now in your hand but is no longer in your pocket. So there is no "

<<"\nsurprise there is a Vedic Sutra for this: "

<<"\nBY ADDITION and BY SUBSTRACTION "

<<"\nWe naturally use this when,say,we add 55 and 19."

<<"\NMost people will add 20 and take 1 away, so 55+19=74"

<<"\nIn the same way we can subtract easily: For example 44-19, we take 20 and give 1, so 44-19=23"

<<"\nThis is what most people do "

<<"\nNow we will apply the vedic sutra for this ";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nTo take 357 from 1000 for example, we apply ALL FROM 9 AND THE LAST FROM 10";

cout<<"\nSo we take 3 from 9 to get 6, take 5 from 9 to get 4, and take the 7 from ";

cout<<"\n10 to get 3, so the anser becomes 643.";

cout<<"\nSimilarily 10,000-7621, we take 7,6 and 2 from 9 and 1 from 10 to get the answer 2379.";

cout<<"\nIf the number of figures in the number being subtracted is less than the number of zeroes ";

cout<<"\nin the bas enumber, we just add as many zeroes as needed to make them the same :";

cout<<"\nFor example in 10,000-643 there are four zeroes and only three figures in 643. ";

cout<<"\nSo we write 10,000-0643 so that 0643 has 4 figures ";

cout<<"\nThen 10,000-0643, we take 0,6,4 from 9 and 3 from 10 to get 9357.";

Ofile<<"\nTo take 357 from 1000 for example, we apply ALL FROM 9 AND THE LAST FROM 10"

<<"\nSo we take 3 from 9 to get 6, take 5 from 9 to get 4, and take the 7 from "

<<"\n10 to get 3, so the anser becomes 643."

<<"\nSimilarily 10,000-7621, we take 7,6 and 2 from 9 and 1 from 10 to get the answer 2379."

<<"\nIf the number of figures in the number being subtracted is less than the number of zeroes "

<<"\nin the bas enumber, we just add as many zeroes as needed to make them the same :"

<<"\nFor example in 10,000-643 there are four zeroes and only three figures in 643. "

<<"\nSo we write 10,000-0643 so that 0643 has 4 figures "

<<"\nThen 10,000-0643, we take 0,6,4 from 9 and 3 from 10 to get 9357.";

}

else

cout<<"\nInvalid Choice";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nNow suppose we don't have a base number, but a multiple of it:like 200,3000 etc.";

cout<<"\nSuppose you had 3000-444. We get 2556. The 444 will come off from one of the ";

cout<<"\nthree thousands, so the initial 3 becomes 2. Then applying the sutra to 444 we get 556.";

cout<<"\nSo the anser is 2556. Similarily 700-52, 7 reduces to 6 and applying the sutra to 52 gives 48 ";

cout<<"\nThere is one thing to be careful of though, suppose you had 7000-4430, ";

cout<<"\nWhen you apply the sutra to 430 to have to just apply it to 43, not to 430 and put the ";

cout<<"\n0 as it is. So you get 7000-430=6570 ";

Ofile<<"\nNow suppose we don't have a base number, but a multiple of it:like 200,3000 etc."

<<"\nSuppose you had 3000-444. We get 2556. The 444 will come off from one of the "

<<"\nthree thousands, so the initial 3 becomes 2. Then applying the sutra to 444 we get 556."

<<"\nSo the anser is 2556. Similarily 700-52, 7 reduces to 6 and applying the sutra to 52 gives 48 "

<<"\nThere is one thing to be careful of though, suppose you had 7000-4430, "

<<"\nWhen you apply the sutra to 430 to have to just apply it to 43, not to 430 and put the "

<<"\n0 as it is. So you get 7000-430=6570 ";

}

else

cout<<"\nInvalid Choice";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nNow suppose we had 444-286";

cout<<"\nSubtracting in each coloumn, we get 4-2=2, 4-8=-4, 4-6=-2 ";

cout<<"\nThese negative answers can be written as 2(42) ";

cout<<"\nand 2(42) can be easily converted to 158 as this is just the kind of sum yopu have done: ";

cout<<"\n2(42) means 200-42=158.";

cout<<"\nSimilarily 6769-1908=5(2)6(1), we split it into 5(2) / 6(1) ";

cout<<"\nand 5(2)=50-2=48 also 6(1)=60-1=59 ";

cout<<"\n";

cout<<"\n";

cout<<"Hope this chapter is clear to you. If not please review again ";

Ofile<<"\nNow suppose we had 444-286"

<<"\nSubtracting in each coloumn, we get 4-2=2, 4-8=-4, 4-6=-2 "

<<"\nThese negative answers can be written as 2(42) "

<<"\nand 2(42) can be easily converted to 158 as this is just the kind of sum yopu have done: "

<<"\n2(42) means 200-42=158."

<<"\nSimilarily 6769-1908=5(2)6(1), we split it into 5(2) / 6(1) "

<<"\nand 5(2)=50-2=48 also 6(1)=60-1=59 "

<<"\n"

<<"\n"

<<"Hope this chapter is clear to you. If not please review again. ";

cout<<"\nA text file of the chapter has been generated in D drive. You can take a print out ";

}

else

cout<<"\nInvalid Choice";

j=0;

Ofile.close();

}

void teacher::mig_func()

{

ofstream Ofile("d:\\migmul.txt",ios::out);

clrscr();

cout<<"\nThe second step is MIGHTY MULTIPLICATIONS ";

cout<<"\nIn the vedic system there are general methods that always work, and there are special methods that can be applied in special cases. ";

cout<<"\nHere we look at a very powerful special mehod that works when the numbers are near a base number (like 10,100,1000 etc ";

cout<<"\nor a multiple of a base.";

cout<<"\n";

cout<<"\nNUMBERS BELOW A BASE NUMBER ";

cout<<"\nFor example 88 x 97=8536. We notice that both the numbers being multiplied are close to 100 ";

cout<<"\n88 is 12 below 100 and 97 is 3 below it. ";

cout<<"\nSo we write 88=88(3) or 97=97(12) which is equal to 88-3=85 or 97-12=85 (whichever you like)";

cout<<"\nThen we multiply the deficiencies i.e. 12 and 3 to get 12 x 3 =36. So the answer in two parts becomes 85 / 36 ";

cout<<"\n=8536.";

cout<<"\nSimilarily 93 x 96=8928, the deficiencies are 7 and 4 ";

cout<<"\nSo 93(4)=96(7) is equal to 93-4=89 or 96-7=89. Then 7 x 4=28";

cout<<"\nTherefore the answer is 89 / 28 =8928.";

Ofile<<"\nThe second step is MIGHTY MULTIPLICATIONS "

<<"\nIn the vedic system there are general methods that always work, and there are special methods that can be applied in special cases. "

<<"\nHere we look at a very powerful special mehod that works when the numbers are near a base number (like 10,100,1000 etc "

<<"\nor a multiple of a base."

<<"\n"

<<"\nNUMBERS BELOW A BASE NUMBER "

<<"\nFor example 88 x 97=8536. We notice that both the numbers being multiplied are close to 100 "

<<"\n88 is 12 below 100 and 97 is 3 below it. "

<<"\nSo we write 88=88(3) or 97=97(12) which is equal to 88-3=85 or 97-12=85 (whichever you like)"

<<"\nThen we multiply the deficiencies i.e. 12 and 3 to get 12 x 3 =36. So the answer in two parts becomes 85 / 36 "

<<"\n=8536."

<<"\nSimilarily 93 x 96=8928, the deficiencies are 7 and 4 "

<<"\nSo 93(4)=96(7) is equal to 93-4=89 or 96-7=89. Then 7 x 4=28"

<<"\nTherefore the answer is 89 / 28 =8928.";

j=0;

Ofile<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\n876 x 998=874 / 248";

cout<<"\nThe same rule here, but the base number is 1000 so the deficiencies are 1000-876=124 ";

cout<<"\nand by ALL FROM 9 AND LAST FROM 10 sutra 1000-998=2. So the first part ia 876-2=998-124=874 ";

cout<<"\nand the second part is 124 x 2=248. So the answer is 874 / 248=874248.";

cout<<"\n";

Ofile<<"\n876 x 998=874 / 248"

<<"\nThe same rule here, but the base number is 1000 so the deficiencies are 1000-876=124 "

<<"\nand by ALL FROM 9 AND LAST FROM 10 sutra 1000-998=2. So the first part ia 876-2=998-124=874 "

<<"\nand the second part is 124 x 2=248. So the answer is 874 / 248=874248."

<<"\n";

}

else

cout<<"\nInvalid Choice";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nNUMBERS ABOVE THE BASE NUMBER ";

cout<<"\nSuppose 103 x 103=10712";

cout<<"\nHere the excess is 103-100=3 and 104-100=4. So the first part in this case becomes 103+4=107 ";

cout<<"\nor 104+3=107 (whichever you like) and the second part becomes 4 x 3=12. This gives ";

cout<<"\nthe answer 10712. ";

cout<<"\n";

Ofile<<"\nNUMBERS ABOVE THE BASE NUMBER "

<<"\nSuppose 103 x 103=10712"

<<"\nHere the excess is 103-100=3 and 104-100=4. So the first part in this case becomes 103+4=107 "

<<"\nor 104+3=107 (whichever you like) and the second part becomes 4 x 3=12. This gives "

<<"\nthe answer 10712. "

<<"\n";

}

else

cout<<"\nInvalid Choice";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nONE NUMBER ABOVE THE BASE NUMBER AND ONE BELOW IT ";

cout<<"\nSuppose 124 x 98=122(48)=12152 ";

cout<<"\nHere the base number is 100 and the deficiency is 100-98=2 also the excess is 124-100=24 ";

cout<<"\nTherefore the first part becomes 124-2=122 or 98+24=122 (whichever you like)";

cout<<"\nand the second part becomes 24 x minus 2=(48). So the answer is 124(48) ";

cout<<"\n12200-48, now applying ALL FROM 9 AND THE LAST FROM 10 sutra, 122 becomes 121 ";

cout<<"\nand 48 gives 52 therefore the answer is 12152.";

cout<<"\n";

Ofile<<"\nONE NUMBER ABOVE THE BASE NUMBER AND ONE BELOW IT "

<<"\nSuppose 124 x 98=122(48)=12152 "

<<"\nHere the base number is 100 and the deficiency is 100-98=2 also the excess is 124-100=24 "

<<"\nTherefore the first part becomes 124-2=122 or 98+24=122 (whichever you like)"

<<"\nand the second part becomes 24 x minus 2=(48). So the answer is 124(48) "

<<"\n12200-48, now applying ALL FROM 9 AND THE LAST FROM 10 sutra, 122 becomes 121 "

<<"\nand 48 gives 52 therefore the answer is 12152. "

<<"\n";

}

else

cout<<"\nInvalid Choice";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nNUMBERS NEAR A MULTIPLE OF BASE NUMBER ";

cout<<"\nSuppose the numbers to be multiplied are not near a base number but to its multiple ";

cout<<"\nHere we use a sub-Sutra called PROPORTIONATELY ";

cout<<"\nFor example 231 x 203 =43239 ";

cout<<"\nHere excess is 213-200=13 and 203-200=3.";

cout<<"\nWe observe that the numbers are close to 200, a multiple of base number 100";

cout<<"\nThe usual procedure gives us 216 / 39 (because 213+3=216 and 13 x 3=39)";

cout<<"\nNow since our base is 200 which is 100 x 2, we multiply only the left-hand part ";

cout<<"\nof the answer by 2 to get 43239.";

cout<<"\nSimilarily 29 x 28= 3 x 27 / 2= 812";

cout<<"\nDeficiency is 30-29=1 and 30-28=2";

cout<<"\nso 29-2=27 or 28-1=27 and 3 x 27 =81 and 2 x 1=2, therefore the answer becomes 812 ";

cout<<"\n";

cout<<"\n";

cout<<" Hope this chapter is clear to you. If not please review again ";

Ofile<<"\nNUMBERS NEAR A MULTIPLE OF BASE NUMBER "

<<"\nSuppose the numbers to be multiplied are not near a base number but to its multiple "

<<"\nHere we use a sub-Sutra called PROPORTIONATELY "

<<"\nFor example 231 x 203 =43239 "

<<"\nHere excess is 213-200=13 and 203-200=3."

<<"\nWe observe that the numbers are close to 200, a multiple of base number 100"

<<"\nThe usual procedure gives us 216 / 39 (because 213+3=216 and 13 x 3=39)"

<<"\nNow since our base is 200 which is 100 x 2, we multiply only the left-hand part "

<<"\nof the answer by 2 to get 43239."

<<"\nSimilarily 29 x 28= 3 x 27 / 2= 812"

<<"\nDeficiency is 30-29=1 and 30-28=2"

<<"\nso 29-2=27 or 28-1=27 and 3 x 27 =81 and 2 x 1=2, therefore the answer becomes 812 "

<<"\n"

<<"\n"

<<" Hope this chapter is clear to you. If not please review again";

cout<<"\nA text file of the chapter has been generated in D drive. You can take a print out ";

}

else

cout<<"\nInvalid Choice";

j=0;

Ofile.close();

}

void teacher::vert_func()

{

ofstream Ofile("d:\\vertcros.txt",ios::out);

clrscr();

cout<<"\nThe third step is VERTICALLY AND CROSSWISE ";

cout<<"\nThe Vedic Sutra VERTICALLY AND CROSSWISE enables us to multiply any two numbers together ";

cout<<"\nin one line (from left to right or right to left),no matter how big the numbers are ";

cout<<"\nFor example 21 x 23=483";

cout<<"\nWe put the numbers and multiply:";

cout<<"\nA. vertically on the right, B. then crosswise and add, C. then vertically on the left ";

cout<<"\nmultiply vertically in the right-hand coloumn 1 x 3 =3, so 3 is the last figure of the answer ";

cout<<"\nmultiply crosswise then add, 2 x 3=6, 1 x 2=2 and 6 + 2=8";

cout<<"\nSo 8 is the middle figure of the answer ";

cout<<"\nmultiply vertically in the left-hand coloumn, 2 x 2=4, so 4 is the last figure of the answer ";

cout<<"\nSo the answer becomes 483.";

Ofile<<"\nThe third step is VERTICALLY AND CROSSWISE "

<<"\nThe Vedic Sutra VERTICALLY AND CROSSWISE enables us to multiply any two numbers together "

<<"\nin one line (from left to right or right to left),no matter how big the numbers are "

<<"\nFor example 21 x 23=483"

<<"\nWe put the numbers and multiply:"

<<"\nA. vertically on the right, B. then crosswise and add, C. then vertically on the left "

<<"\nmultiply vertically in the right-hand coloumn 1 x 3 =3, so 3 is the last figure of the answer "

<<"\nmultiply crosswise then add, 2 x 3=6, 1 x 2=2 and 6 + 2=8"

<<"\nSo 8 is the middle figure of the answer "

<<"\nmultiply vertically in the left-hand coloumn, 2 x 2=4, so 4 is the last figure of the answer "

<<"\nSo the answer becomes 483.";

j=0;

cout<<"\nPress 1 to continue ";

cin>>j;

if(j==1)

{

clrscr();

cout<<"\nThe previous example has no carry figure so lets consider 23 x 41=943 ";

cout<<"\nThe three steps give us : 3 x 1=3, 2 x 1 + 3 x 4=14, so we carry 1 to the left coloumn ";

cout<<"\n2 x 4=8 and 8 + the carried 1 gives us 9.";

cout<<"\nSo 23 x 41=943";

cout<<"\nHope this chapter is clear to you. If not please review again ";

Ofile<<"\nThe previous example has no carry figure so lets consider 23 x 41=943 "

<<"\nThe three steps give us : 3 x 1=3, 2 x 1 + 3 x 4=14, so we carry 1 to the left coloumn "

<<"\n2 x 4=8 and 8 + the carried 1 gives us 9."

<<"\nSo 23 x 41=943"

<<"\nHope this chapter is clear to you. If not please review again ";

cout<<"\nA text file of the chapter has been generated in D drive. You can take a print out ";

}

else

cout<<"\nInvalid Choice";

j=0;

Ofile.close();

}

void teacher::might()

{

cout<<"\nEnter two numbers near base number 100 ";

cout<<"\nFor example 98 and 97 or 103 and 92";

enter1();

if(n1>90 && n1<110)

{

if(n2>90 && n2<110)

{

cout<<"\nBase number here is 100";

if(n1>100)

{

cout<<"\nThe excess in the first case is "<<n1-100;

x1=n1-100;

}

else

{

cout<<"\nThe deficiency in the first case is "<<100-n1;

x1=n1-100;

cout<<"\nWhich will be written as ("<<100-n1<<")";

}

if(n2>100)

{

cout<<"\nThe excess in the first case is "<<n2-100;

x2=n2-100;

}

else

{

cout<<"\nThe deficiency in the first case is "<<100-n2;

x2=n2-100;

cout<<"\nWhich will be written as ("<<100-n2<<")";

}

mu();

}

else

cout<<"\nThe entered numbers are not near the same base number ";

}

}

void teacher::mu()

{

c=0;

cout<<"\nPress 1 to continue ";

cin>>c;

if(c==1)

{

clrscr();

cout<<"\nWe have to find "<<n1<<" x "<<n2;

cout<<"\nAs usual we will add the deficiency/excess of first number to the second one ";

cout<<"\nOr add deficiency/excess of second number to the first number whichever convineint ";

cout<<"\nThis gives us the first part of the answer";

cout<<"\nTherefore the first part is "<<n1<<" + "<<x2<<" = "<<n1+x2;

cout<<"\nOr the first part is "<<n2<<" + "<<x1<<" = "<<n2+x1;

cout<<"\nTo find the second part we will as usual multiply the excess/deficiency of the 2 numbers ";

cout<<"\nThe following points are to be kept in mind ";

cout<<"\n1.) excess X excess is positive ";

cout<<"\n2.) excess X deficiency is negative ";

cout<<"\n3.) deficiency X deficiency is positive ";

cout<<"\nPress 1 to continue ";

c=0;

cin>>c;

if(c==1)

{

clrscr();

cout<<"\nThe negative number will be written as (n) ";

cout<<"\nThe second part must have one less digit than the first part";

cout<<"\nIf it is not so, add as many zeros in front of it to make the number of digits 1";

cout<<"\nlesser than the main part ";

cout<<"\nAlso as you have learnt in the tutorial, (n) would be replaced by a zero everywhere ";

cout<<"\nin the unconverted answer and then 10*n or n would be subtracted from the resulting number";

cout<<"\ndepending upon the place where (n) is substituted, 10*n for tens place and n for ones place";

cout<<"\nPress 1 to continue ";

c=0;

cin>>c;

if(c==1)

{

clrscr();

cout<<"\nSo the first part is "<<n1+x2<<" as calculated before";

cout<<"\nThe second part is ";

if(x1>=0 && x2<=0)

cout<<x1*x2;

else if(x1<0 && x2>0)

cout<<"("<<-(x1*x2)<<")";

else if(x1>0 && x2<0)

cout<<"("<<-(x1*x2)<<")";

else if(x1<0 && x2<0)

cout<<x1*x2;

cout<<"\nSo the answer in unconverted form is "<<n1+x2<<" / ";

if(x1*x2>=0 && x1*x2<=9)

{

cout<<"0"<<x1*x2;

cout<<"\nObserve that a zero has been added for making the no. of digits ";

cout<<"\n1 less than the first part ";

}

else if(x1*x2>9)

cout<<x1*x2;

else if(x1*x2<0 && x1*x2>=-9)

{

cout<<"0("<<-(x1*x2)<<")";

cout<<"\nObserve that a zero has been added for making the no. of digits ";

cout<<"\n1 less than the first part ";

}

else if(x1*x2<-9)

cout<<"("<<-(x1*x2)<<")";

cout<<"\nNow the converted answer becomes ";

cout<<100*(n1+x2)+x1*x2;

}

else

cout<<"\nWrong choice ";

}

else

cout<<"\nWrong choice ";

}

}

//main body

void main()

{

clrscr();

int q;

cout<<"Veda Mathematics teacher (with illustration)--Ashish Chaudhary XII A ";

cout<<"\nEnter your name ";

io i;

i.wel();

cout<<"\n"<<i.name<<", this program helps you learn the basics of the "<<"\n ancient art of Veda Mathematics ";

cout<<"\nPress 1 to continue ";

cin>>q;

if(q==1)

{

clrscr();

cout<<"\nPress 1 for piecewise automated illustartion ";

cout<<"\nPress 2 for addition of two numbers (upto 3 digits) ";

cout<<"\nPress 3 for substraction of two numbers (upto 3 digits) ";

cout<<"\nPress 4 for multiplication of two numbers near the base number 100 ";

}

else

cout<<"\n"<<i.name<<",thank you for using the program, press any key then enter to exit ";

teacher t;

t.choice();

getch();

}

References:1 Vedic Mathematics by Bhararti Krsna Tirthaji, Delhi: Motilal Banarsidass,1965.

ISBN 81-208-0164-4

2 Discover Vedic Mathematics by Kenneth Williams, Inspiration Books, 1984.

ISBN 1869932013

3 Vertically and Crosswise, by Nicholas A. P., K. Williams, J. Pickles.London: Inspiration Books, 1984.

ISBN 1-902517-03-2.