Mysterious World!

Squaring of Number

2015-04-29T14:16:00.001+05:30

Now Squaring of Number will be very easy by the Multiplications method we have learn in previous Posts.

Let me give you few Examples of it.

9 -1

8 / 1

12 +2 97 -3 108 +8 997 -3 1021 +21

14 /4 94 / 09 116 / 64 994/009 1042/441

28 +8 72 +2 72 -8

36 / 64 74 /4 64 /64

x2 x7 x8

72 /64 518 /4 512/64

=784 =518/4

Now Lets see what if Number ends at 5!

The magic comes here!

45*45 = (4*5) /25 = 20 / 25

85*85 = (8*9)/25 = 72 / 25

115*115 = (11*12)/25 = 132 / 25

325*325 = (32*33)/25 = 1056 / 25 ---> i calculate 32*33 directly using Vedic method!

1235*1235 = (123*124)/25 = 15252 / 25 ---> i calculate 123*124 directly using Vedic method!

105*105 = (10*11)/25 = 110 / 25

95*95 = (9*10)/25 = 90 / 25

Now you have learn lots of tricks to multiply and square number very Quick with very easy Methods!
Congratulations!

Don't Forget to share it with your Friends.

Two digit multiplication Easy methods

2015-04-29T00:01:00.001+05:30

Do you tired of teaching your child multiplication tables?

Here i have something for you that you will feel that you do not need to remember Multiplication table more then number 5. Isn't it sounds cool ?

Let me Explain how to multiply 9 x 6 using 10 as base.

1) do 9-10 and write it to the right side of 9

2) do 6-10 and write it to the right side of 6

3) Now left side digit will calculate using Multiplication of right most digits that has been calculated using (1) and (2)

4) And Left Side Digit will calculate using add numbers in cross sides. i.e add 9 to -4 or add 6 to -1 both will give the same result.
Here i have another Example of the same Methods that display another case of getting result in 2 digit right side. and how to handle it.

In such case we simply add right side's first digit to left side.

Now we are Moving towards number having more then 1 digit.

Now i will Show you how to multiply number near 100.

Method will Remain same as above but just the difference is on right side in answer we will have 2 digit as it is multiplication by number near 100.

How Simple it is! Then Why not use it by making daily Practice using it?

Now Lets see number Near 100 but above 100.

113 +13
102 +2
115 / 26

how easy????

Can you Imagine Multiplication can be much faster and easier like this?

Let take another Example of numbers in 1000.

988 -12 1113 +113
998 -2 1002 +2
986 / 024 1115 / 226

What if one number is above our consider Base and another is above it???
Let Understand it by Example.

13 +3
7 -3
10 / -9
=91

because -9 will be subtracted from 10

Let see another Example.

114 +14
97 - 3
111 / -42
=11058
Because 42 will be subtracted from last 100.

Let Take another Example to clear it!

Now What if Number is not near by 10 or not 100 or not 1000????

Don't Worry We have Method for it too!

i.e Example of base 20

24 +4
18 -2
22 / -8
x2 -----> here We Do multiplication by 2 because base is 10x2
44/-8
=432

This Method is Similar for all Number that is come near by some number. Practice of it will make you aware which method we can use for which numbers.

Let Take another Example to Finalize this Method with Practice.

76 +6
68 -2
74 / -12
x7
518/-12
=5168

We have many more methods that we will see in next blog. Till Practice this one!
Have a Good Day.

Multiplication of 3 digit numbers

2015-04-27T10:50:00.000+05:30

If you want to multiply 123 and 456 then consider 1 in 123 as a ,2 in 123 as b and 3 in 123 as c.

so the number 123 will be written as ax2 + bx +c

here x is 10 so

Here i have illustrate how to multiply 3 digit numbers using
Vedic method with Explanation

ax2+bx+c
=1*10*10 + 2*10 + 3
=100 + 20 + 3
=123

in same manner we take 456 as dx2 + ex + f

so now multiplication is between ax2 + bx +c and dx2 + ex + f

so the answer is (ad)x4 + (ae + bd)x3 + (af + be + cd)x2 + (ce+bf)x + cf

If you make this practice then multiplication be very easy for you.

so using vedic trick multiplication is as shown below.

1 2 3

4 5 6

4 3 8 7 8

1 2 2 1

5 6 0 8 8

You can do same method for any digit number for faster Multiplication.

Vedic Maths: Multiplication of two digit numbers

2015-04-26T10:18:00.001+05:30

Learn vedic method of Multiplication of two digit number
and win prize by giving answer till 27/04/2015 9:00 AM

Here i have explained you very easily how we can get multiplication of two digit numbers.

Here i'm also having one skim for you.

All participant can comment your answer with vedic method and have chance to win prize.

You have time till 27/04/2015 9:00 AM. to answer this question.

One among all correct one will be chosen to have prize.

Power of Vedic Mathematics And Why to learn it.

2015-04-26T09:04:00.000+05:30

Vedic Maths is not different kind of maths then we have studied in schools. Then how it make difference and why people learn it?

The difference is vedic math include conclusions based on various sutras(Formula) from our vedas and Upnishdas. That help us to detecting type of problem. And we do calculation based on the problem type. So we do not follow conventional boring method all time and do calculation based on the situation that make our mind very instant to answer.

Sample of how use of Vedic Maths reduces Time and make
maths Easy to learn for students

So the learner of Vedic Maths learn how to apply their knowledge in Maths as well as in other fields too.

My sincere thanks to Jagadguru Sankaracharya Sri Bhakti Krishna Tirthaji Maharaja of Govardhana Matha, Puri for Writing 16 simple Mathematical formula based on Vedas.

Vedic Maths is also popular as High Speed maths or Mental maths Because once you will practice it you can calculate various things by mentally.

Here in this post i'm going to show you how easy Maths become for students using the various methods given in Vedic maths.

At fist extent you might find it difficult but once you will practice it , it will become easy for you.Because it is totally different to use from conventional maths we are using.

I'm lover of Math so i'm starting this online blogging for vedic maths free for all. So all can ask any doubt about maths , vedic maths.

Vedic Maths: Square a Number

2015-04-25T00:14:00.002+05:30

Square a number by the Deficiency Sutra Method

Sometimes we find it difficult to get square of a number by our routine methods. Here's how vedic mathematics can help us!

Here are few easy steps to get through!

1 : Find the base of given number. That is maybe, 10, 100, 1000, 10000, ...
2 : Find the difference between base and a given number. That is a deficiency value.
3 : Subtract the given number by deficiency and then multiply by base.
4 : Square the deficiency value.
5 : Add rule 3 result with the rule 4 result. And that's it!

Example: 93 X 93 (Yaavadunam Sutra Method)

1 : 93 base is 100
2 : 100 - 93 = 7
3 : (93 - 7) X 100 = 86 X 100 = 8600
4 : 7 X 7 = 49
5 : = 8600 + 49 = 8649

93 X 93 = 8649

Most common Passwords people use!

2015-04-23T01:06:00.003+05:30

Here is the list of most common passwords people use!
Make sure you don't have anything from this to avoid being hacked by someone!

123456
password
12345678
qwerty
abc123
123456789
111111
1234567
iloveyou
adobe123

Password is the most important thing you should take care of if you have important data in your account. If not then enjoy using simple passwords :D

Jayesh and Sweety Wedding Dance

2015-02-13T15:49:00.001+05:30

Shortest Path Algorithm: Dijkstra

2011-10-14T20:30:00.002+05:30

Aim :- To find shortest path

#include <iostream.h>

#include <conio.h>

class graph

{

public:

char a;

char b;

int e;

};

class dijkstra

{

public:

int n,p,t,f;

char c[10];

void get_data();

void get_connection(graph g[]);

void shortest(graph g[]);

void display(graph g[]);

int minm(int [],int,char []);

int findu(graph g[],int,int,char []);

};

void dijkstra::get_data()

{

cout<<"Enter number of nodes in your network:";

cin>>n;

cout<<endl<<"Write the name of each node"<<endl;

for(int i=0;i<n;i++)

{

cout<<"Name of node "<<i+1<<":";

cin>>c[i];

}

void dijkstra::get_connection(graph g[])

{

clrscr();

cout<<endl<<endl<<"NOW ITS TIME TO CONNECT THE NODES...!";

cout<<endl<<"Enter 1 if there is connection between consicutive nodes otherwise enter 0"<<endl;

p=0;

for(int i=0;i<n;i++)

{

for(int j=i+1;j<n;j++)

{

cout<<c[i]<<" & "<<c[j]<<" :";

int t;

cin>>t;

if(t==1)

{

g[p].a=c[i];

g[p].b=c[j];

cout<<"Enter distance:";

cin>>g[p].e;

p++;

}

void dijkstra::display(graph g[])

{

clrscr();

cout<<"YOUR NETWORK IS LIKE THIS:"<<endl;

for(int i=0;i<p;i++)

{

cout<<g[i].a<<" & "<<g[i].b<<" :"<<g[i].e<<endl;

}

void dijkstra::shortest(graph g[])

{

char s,C[10];

int q,D[10];

cout<<endl<<endl<<"NOW ENTER NODE from which you want to find shortest distance:";

cin>>s;

t=0;

f=0;

//function to get distance & node array (Initialization)

for(int r=0;r<n;r++)

{

if(s!=c[r])

{

C[f]=c[r];

for(int i=0;i<p;i++)

{

if((g[i].a==s && g[i].b==C[f]) || (g[i].a==C[f] && g[i].b==s))

{

D[f]=g[i].e;

q=0;

break;

}

if(q!=0)

{

D[f]=32767;

}

f++;

}

} int w,u,v;

//greedy loop

for(q=0;q<n-2;q++)

{

v=minm(D,f,C);

for(w=0;w<f;w++)

{

if(C[w]!=NULL && C[w]!=C[v])

{

u=findu(g,w,v,C);

D[w]=(D[w]<(u+D[v])?D[w]:(u+D[v]));

}

C[v]=NULL;

}

cout<<"The sortest distance from node "<<s<<":";

for(int z=0;z<f-1;z++)

{

cout<<D[z]<<",";

}

cout<<D[f-1];

}

int dijkstra::findu(graph g[],int w,int v,char C[])

{

for(int z=0;z<p;z++)

{

if((g[z].a==C[w] && g[z].b==C[v]) ||(g[z].a==C[v] && g[z].b==C[w]))

{

return g[z].e;

}

return 32676;

}

int dijkstra::minm(int j[],int t,char c[])

{

int o=32766,q;

for(int i=0;i<t;i++)

{

if(o>j[i] && c[i]!=NULL)

{

o=j[i];

q=i;

}

return q;

}

void main ()

{

clrscr();

graph g[15];

dijkstra d;

d.get_data();

d.get_connection(g);

d.display(g);

d.shortest(g);

getch();

}

Output:

Decimal To Binary Converter

2010-08-23T08:56:00.000+05:30

//This Is The Program To Convert Decimal To Binary But Its Different From Other Ones!
#include<stdio.h>
void printBinary(int n);
void main()
{
int num;
printf("Enter Decimal Number\n");
scanf("%d",&num);
printBinary(num);
}
void printBinary(int n)
{
if (n>0)
{
printBinary(n/2);
printf("%c","01"[n%2]);
}
}

Monoalphabatic cipher encryption and decryption

2010-03-18T22:05:00.005+05:30

//MONOALPHABETIC ENCRYPTION AND DECRYPTION
//copyright jayeshaminnew@gmail.com

//ENCRYPTION OF MONOALPHABAIC CIPHER
#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<iostream.h>
#include<ctype.h>
void main()
{
FILE *f1,*f2,*f3;
int i,j=0,k,asc[50],l,temp,temp1,flag=0;
char plain[50],cipher[50],key[50],m;
clrscr();
f1=fopen("plain.txt","r");
while((l=getc(f1))!=EOF)
{
plain[j++]=l;
}
plain[j]=NULL;
fclose(f1);
j=0;
f2=fopen("key.txt","r");
while((l=getc(f1))!=EOF)
{
key[j++]=l;
}
key[j]=NULL;
fclose(f2);
cout<<"plain text is\n"<<plain;
cout<<"key is \n"<<key;
//logic
k=0;
for(i=0;i<strlen(plain);i++)
{
asc[k++]=plain[i]-97;
}
k=0;
for(i=0;i<strlen(plain);i++)
{
if(asc[i]>=0 && asc[i]<=25)
cipher[k++]=key[asc[i]];
else
cipher[k++]=plain[asc[i]+97];
}
cout<<"\n"<<"cipher text is\n";
cipher[k]=NULL;
cout<<cipher;

f3=fopen("cipher.txt","w");
for(i=0;i<strlen(cipher);i++)
putc(cipher[i],f2);
getch();
}

//DECRYPTION OF MONOALPHABATIC CIPHER

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<iostream.h>
#include<ctype.h>
void main()
{
FILE *f1,*f2,*f3;
int i,j=0,k,asc[50],l,temp,temp1,flag=0;
char plain[50],cipher[50],key[50],m;
clrscr();
f1=fopen("cipher.txt","r");
while((l=getc(f1))!=EOF)
{
cipher[j++]=l;
}
cipher[j]=NULL;
fclose(f1);
j=0;
f2=fopen("key.txt","r");
while((l=getc(f1))!=EOF)
{
key[j++]=l;
}
key[j]=NULL;
fclose(f2);
cout<<"cipher text is\n"<<cipher;
cout<<"\nkey is \n"<<key;
//logic
k=0;
for(i=0;i<strlen(cipher);i++)
{
for(j=0;j<strlen(key);j++)
{
if(cipher[i]==key[j])
break;
}
plain[k++]=j+97;
}
cout<<"\n"<<"plain text is\n";
plain[k]=NULL;
cout<<plain;

f3=fopen("plain.txt","w");
for(i=0;i<strlen(plain);i++)
putc(plain[i],f2);
getch();
}

Caesar cipher encryption and decryption

2010-03-18T22:04:00.002+05:30

//CAESAR CIPHER ENCRYPTION
//copyright jayeshaminnew@gmail.com
#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<ctype.h>
void main()
{
//declaring variables
FILE *f1,*f2;
int i,j=0,k,l,key,temp,temp1,flag=0;
char plain[50],cipher[50],m;
clrscr();
//reading plain text
f1=fopen("plain.txt","r");
while((l=getc(f1))!=EOF)
plain[j++]=l;
plain[j]=NULL;
fclose(f1);
printf("your plain text is:%s\n",plain);
scanf("%d",&key,printf("enter key:"));
for(i=0;i<j;i++)
{
temp=plain[i];
temp1=temp;
temp1+=key;
if(temp1>122 && islower(temp))
temp1-=26;
else if(temp1>90 && isupper(temp))
temp1-=26;
else if(!((temp>=97 && temp<=122) || (temp>=65 && temp<=90)))
{
flag=1;
}
if(flag==0)
cipher[i]=temp1;
else
cipher[i]=temp;
flag=0;
}
cipher[i]=NULL;
//writing cipher text into file
f2=fopen("cipher.txt","w");
for(i=0;i<strlen(cipher);i++)
putc(cipher[i],f2);
fclose(f2);
printf("your cipher text is:%s",cipher);
getch();
}

//CAESAR CIPHER DECRYPTION

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<ctype.h>
#include<process.h>
void main()
{
FILE *f1,*f2;
int i,j=0,k,l,key,temp,temp1,flag=0;
char plain[50],cipher[50],meaning;
clrscr();
f1=fopen("cipher.txt","r");
while((l=getc(f1))!=EOF)
cipher[j++]=l;
cipher[j]=NULL;
fclose(f1);
j=0;
for(i=0;i<26;i++)
{
for(j=0;j<strlen(cipher);j++)
{
temp=cipher[j];
temp1=temp;
temp1-=i;
if(temp1<97 && islower(temp))
temp1+=26;
else if(temp1<65 && isupper(temp))
temp1+=26;
else if(!((temp>=97 && temp<=122) || (temp>=65 && temp<=90)))
{
flag=1;
}
else
{}
if(flag==0)
plain[j]=temp1;
else
plain[j]=temp;
flag=0;
}
plain[j]=NULL;
printf("Message:%s",plain);
printf("\nis it meaning ful msg?? Y/N?\n");
scanf("%s",&meaning);
if(meaning=='y' || meaning=='Y')
{
printf("your plain text is: %s",plain);
getch();
exit(0);
}
}
getch();
}

Polyalphabatic cipher encryption and decryption

2010-03-18T22:01:00.005+05:30

//POLYALPHABATIC CIPHER ENCRYPTION
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
int start=98,i,temp=97,j,k,m,n,l;
char plain[100],key[100],cipher[100],arr[27][27],ch1,ch2;
FILE *f1,*f2;
clrscr();
//code for matrix generation
for(i=0;i<27;i++)
{
for(j=0;j<27;j++)
{
if(temp>122 )
temp=32;
arr[i][j]=temp++;
if(temp==33)
temp=97;
}
temp=start++;
}
//printing matrix
for(i=0;i<27;i++)
{
for(j=0;j<27;j++)
printf("%c",arr[i][j]);
printf("\n");
}
i=0;
f1=fopen("plain.txt","r");
while((l=getc(f1))!=EOF)
plain[i++]=l;
plain[i]=NULL;
fclose(f1);
i=0;
f1=fopen("key.txt","r");
while((l=getc(f1))!=EOF)
key[i++]=l;
key[i]=NULL;
fclose(f1);
temp=0;
printf("\nplain text: %s\n",plain);
for(i=strlen(key);i<strlen(plain);i++)
{
if(temp>strlen(key))
temp=0;
key[i]=key[temp++];
}
key[i]=NULL;
printf("key: %s\n",key);
for(i=0;i<strlen(plain);i++)
{
if(plain[i]!=32)
m=plain[i]-97;
else
m=26;
n=key[i]-97;
cipher[i]=arr[m][n];
}
cipher[i]=NULL;
f1=fopen("cipher.txt","w");
for(i=0;i<strlen(cipher);i++)
putc(cipher[i],f1);
fclose(f1);
printf("cipher text:%s",cipher);
getch();

}

//POLYALPHABATIC CIPHER DECRYPTION
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
int start=98,i,temp=97,j,k,m,n,l;
char plain[100],key[100],cipher[100],arr[27][27],ch1,ch2;
FILE *f1,*f2;
clrscr();
//code for matrix generation
for(i=0;i<27;i++)
{
for(j=0;j<27;j++)
{
if(temp>122 )
temp=32;
arr[i][j]=temp++;
if(temp==33)
temp=97;
}
temp=start++;
}
//printing matrix
for(i=0;i<27;i++)
{
for(j=0;j<27;j++)
printf("%c",arr[i][j]);
printf("\n");
}
i=0;
f1=fopen("cipher.txt","r");
while((l=getc(f1))!=EOF)
cipher[i++]=l;
cipher[i]=NULL;
fclose(f1);
i=0;
f1=fopen("key.txt","r");
while((l=getc(f1))!=EOF)
key[i++]=l;
key[i]=NULL;
fclose(f1);
temp=0;
printf("\ncipher text:%s\n",cipher);
for(i=strlen(key);i<strlen(cipher);i++)
{
if(temp>strlen(key))
temp=0;
key[i]=key[temp++];
}
key[i]=NULL;
k=0;
printf("key: %s\n",key);
for(i=0;i<strlen(cipher);i++)
{
for(j=0;j<27;j++)
{
if(cipher[k]==arr[key[i]-97][j])
break;
}
plain[k++]=arr[0][j];
}
plain[k]=NULL;
f1=fopen("plain.txt","w");
for(i=0;i<strlen(plain);i++)
putc(plain[i],f1);
fclose(f1);
printf("plain text: %s\n",plain);
getch();
}

Bresenham's line drawing algorithm

2010-03-18T22:00:00.002+05:30

//This is Bresenham's line drawing algorithm developed in C language
#include <graphics.h>
#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
void jayubre(int,int,int,int);
void main(void)
{
int gdriver = DETECT, gmode,x1,y1,x2,y2;
initgraph(&gdriver, &gmode, "..\\bgi");
printf("Enter starting points(x1,y1):\n"),scanf("%d%d",&x1,&y1);
printf("Enter ending points(x2,y2):\n"),scanf("%d%d",&x2,&y2);
jayubre(x1,y1,x2,y2);
getch();
}
void jayubre(int x1,int y1,int x2,int y2)
{
int p0,dy,dx,sx=1,sy=1,xmax, ymax,x,y;
float m;
putpixel(x1,y1,8);
putpixel(x2,y2,8);
if(x1>x2)
sx=-1;
if(y1>y2)
sy=-1;
dx=abs(x2-x1);
dy=abs(y2-y1);
m=dy/dx;
if(m<1)
{
p0=2*dy-dx;
while(x1!=x2)
{
if(p0<0)
{
p0=p0+2*dy;
x1=x1+sx;
y1=y1;
putpixel(x1,y1,4);
}
else if(p0>=0)
{
p0=p0+2*dy-2*dx;
x1=x1+sx;
y1=y1+sy;
putpixel(x1,y1,4);

}
}
}
else
{
p0=2*dx-dy;
while(y1!=y2)
{
if(p0<0)
{
p0=p0+2*dx;
y1=y1+sy;
x1=x1;
putpixel(x1,y1,4);
}
else if(p0>=0)
{
p0=p0+2*dx-2*dy;
y1=y1+sy;
x1=x1+sx;
putpixel(x1,y1,4);
}
}
}
}

Midpoint circle algorithm

2010-03-18T21:57:00.003+05:30

// This is Midpoint circle algorithm developed in C language

#include <graphics.h>
#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
void jayucircle(int xc,int yc,int r);
void jayuprint(int x,int y,int xc,int yc);
void main()
{
   int xc,yc,p,r,x,y,gdriver = DETECT, gmode, errorcode;
   initgraph(&gdriver, &gmode, "..\\bgi");
   printf("enter value of (x,y,r)"),scanf("%d%d%d",&xc,&yc,&r);
   jayucircle(xc,yc,r);
   getch();
}
void jayucircle(int xc,int yc,int r)
{
int x=0,y=r,p=1-r;
jayuprint(0,r,xc,yc);
while(x<=y)
{
if(p<0)
{
x++;
p+=(2*x)+1;
jayuprint(x,y,xc,yc);
}
else
{
x++;
y--;
p+=(2*x)+1-(2*y);
jayuprint(x,y,xc,yc);
}
}
}
void jayuprint(int x,int y,int xc,int yc)
{
putpixel(xc+x,yc+y,7);
putpixel(xc-x,yc+y,7);
putpixel(xc+x,yc-y,7);
putpixel(xc-x,yc-y,7);
putpixel(yc+y,xc+x,7);
putpixel(yc-y,xc+x,7);
putpixel(yc+y,xc-x,7);
putpixel(yc-y,xc-x,7);
}

2D Modeling: 2D Rotation

2010-03-18T21:55:00.005+05:30

// This is 2D Rotation program using C language

#include<stdio.h>
#include<conio.h>
#include<graphics.h>
#include<math.h>
float ans[3][10];
void multi(float tmat[3][3],float coormat[3][10],int size);
void main()
{
float coormat[3][10],x[10],y[10],temp;
int gd=DETECT,gm,no,sx,sy,sum=0,i,j,k,theta;
float tmat[3][3]={{1,0,0},{0,1,0},{0,0,1}};
clrscr();
initgraph(&gd,&gm,"..\\bgi");
printf("enter theta:"),scanf("%d",&theta);
temp=(float)theta*3.14/180;
tmat[0][0]=cos(temp);
tmat[1][1]=cos(temp);
tmat[0][1]=-1*sin(temp);
tmat[1][0]=-1*sin(temp);
printf("Translation matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%4.1f\t",tmat[i][j]);
printf("\n");
}
printf("enter no. of vertices:\n"),scanf("%d",&no);
for(i=0;i<no;i++)
printf("enter (x%d,y%d):\n",i+1,i+1),scanf("%f%f",&x[i],&y[i]);
for(i=0;i<no;i++)
coormat[0][i]=x[i];
for(i=0;i<no;i++)
coormat[1][i]=y[i];
for(i=0;i<no;i++)
coormat[2][i]=1;
printf("co ordinate matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
printf("%4.2f\t",coormat[i][j]);
printf("\n");
}
//printing starting line
for(j=0;j<no-1;j++)
line(coormat[0][j],coormat[1][j],coormat[0][j+1],coormat[1][j+1]);
line(coormat[0][0],coormat[1][0],coormat[0][j],coormat[1][j]);
multi(tmat,coormat,no);
printf("answer matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
printf("%4.2f\t",ans[i][j]);
printf("\n");
}

for(j=0;j<no-1;j++)
line(ans[0][j],ans[1][j],ans[0][j+1],ans[1][j+1]);
line(ans[0][0],ans[1][0],ans[0][j],ans[1][j]);

getch();
}
void multi(float tmat[3][3],float coormat[3][10],int no)
{
int i,j,k,sum=0;
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
{
sum=0;
for(k=0;k<3;k++)
{
sum+=tmat[i][k]*coormat[k][j];
}
ans[i][j]=sum;
}
}
}

2D Modeling: 2D scaling

2010-03-18T21:55:00.002+05:30

#include<stdio.h>
#include<conio.h>
#include<graphics.h>
#include<math.h>
float ans[3][10];
void multi(float tmat[3][3],float coormat[3][10],int size);
void main()
{
float coormat[3][10],x[10],y[10];
int gd=DETECT,gm,no,sx,sy,sum=0,i,j,k;
float tmat[3][3]={{1,0,0},{0,1,0},{0,0,1}};
clrscr();
initgraph(&gd,&gm,"..\\bgi");
printf("enter (sx,sy):"),scanf("%d%d",&sx,&sy);
tmat[0][0]=sx;
tmat[1][1]=sy;
printf("Translation matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%4.1f\t",tmat[i][j]);
printf("\n");
}
printf("enter no. of vertices:\n"),scanf("%d",&no);
for(i=0;i<no;i++)
printf("enter (x%d,y%d):\n",i+1,i+1),scanf("%f%f",&x[i],&y[i]);
for(i=0;i<no;i++)
coormat[0][i]=x[i];
for(i=0;i<no;i++)
coormat[1][i]=y[i];
for(i=0;i<no;i++)
coormat[2][i]=1;
printf("co ordinate matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
printf("%4.2f\t",coormat[i][j]);
printf("\n");
}
//printing starting line
for(j=0;j<no-1;j++)
line(coormat[0][j],coormat[1][j],coormat[0][j+1],coormat[1][j+1]);
line(coormat[0][0],coormat[1][0],coormat[0][j],coormat[1][j]);
multi(tmat,coormat,no);
printf("answer matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
printf("%4.2f\t",ans[i][j]);
printf("\n");
}

for(j=0;j<no-1;j++)
line(ans[0][j],ans[1][j],ans[0][j+1],ans[1][j+1]);
line(ans[0][0],ans[1][0],ans[0][j],ans[1][j]);

getch();
}
void multi(float tmat[3][3],float coormat[3][10],int no)
{
int i,j,k,sum=0;
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
{
sum=0;
for(k=0;k<3;k++)
{
sum+=tmat[i][k]*coormat[k][j];
}
ans[i][j]=sum;
}
}
}

2D Modeling: 2D transformation

2010-03-18T21:54:00.002+05:30

#include<stdio.h>
#include<conio.h>
#include<graphics.h>
#include<math.h>
float ans[3][10];
void multi(float tmat[3][3],float coormat[3][10],int size);
void main()
{
float coormat[3][10],x[10],y[10];
int gd=DETECT,gm,no,tx,ty,sum=0,i,j,k;
float tmat[3][3]={{1,0,0},{0,1,0},{0,0,1}};
clrscr();
initgraph(&gd,&gm,"..\\bgi");
printf("enter (tx,ty):"),scanf("%d%d",&tx,&ty);
tmat[0][2]=tx;
tmat[1][2]=ty;
printf("Translation matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%4.1f\t",tmat[i][j]);
printf("\n");
}
printf("enter no. of vertices:\n"),scanf("%d",&no);
for(i=0;i<no;i++)
printf("enter (x%d,y%d):\n",i+1,i+1),scanf("%f%f",&x[i],&y[i]);
for(i=0;i<no;i++)
coormat[0][i]=x[i];
for(i=0;i<no;i++)
coormat[1][i]=y[i];
for(i=0;i<no;i++)
coormat[2][i]=1;
printf("co ordinate matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
printf("%4.2f\t",coormat[i][j]);
printf("\n");
}
//printing starting line
for(j=0;j<no-1;j++)
line(coormat[0][j],coormat[1][j],coormat[0][j+1],coormat[1][j+1]);
line(coormat[0][0],coormat[1][0],coormat[0][j],coormat[1][j]);
multi(tmat,coormat,no);
printf("answer matrix is:\n");
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
printf("%4.2f\t",ans[i][j]);
printf("\n");
}

for(j=0;j<no-1;j++)
line(ans[0][j],ans[1][j],ans[0][j+1],ans[1][j+1]);
line(ans[0][0],ans[1][0],ans[0][j],ans[1][j]);

getch();
}
void multi(float tmat[3][3],float coormat[3][10],int no)
{
int i,j,k,sum=0;
for(i=0;i<3;i++)
{
for(j=0;j<no;j++)
{
sum=0;
for(k=0;k<3;k++)
{
sum+=tmat[i][k]*coormat[k][j];
}
ans[i][j]=sum;
}
}
}

Midpoint ellipse algorithm

2010-03-18T21:37:00.002+05:30

//This is Midpoint ellipse algorithm developed using C language

#include <graphics.h>
#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
#include<math.h>
void jayuellipse(int x,int y,int rx,int ry);
void print(int,int,int,int);
void main()
{
int xc,yc,p,r,rx,ry,gdriver = DETECT, gmode;
initgraph(&gdriver, &gmode, "..\\bgi");
printf("enter values of (x,y,xrad,yrad)"),scanf("%d%d%d%d",&xc,&yc,&rx,&ry);
jayuellipse(xc,yc,rx,ry);
getch();
}
void jayuellipse(int xc,int yc,int rx,int ry)
{
long int p,p2,x,y,dx,dy,ry2,rx2,rx2ry,hfrx2,_2ry2x,_2rx2y;
float x5;
//initialization of values
x=0;
y=ry;
ry2=ry*ry;
rx2=rx*rx;
rx2ry=rx*rx*ry;
hfrx2=.25*rx*rx;
p=ry2-rx2ry+hfrx2;
_2ry2x=2*ry*ry*x;
_2rx2y=2*rx*rx*y;
dx=_2ry2x;
dy=_2rx2y;
//calculation for region 1
while(dx<dy)
{
print(x,y,xc,yc);
if(p<0)
{
x++;
dx+=2*ry2;
p+=dx+ry2;
}
else if(p>=0)
{
x++;
y--;
dx+=2*ry2;
dy-=2*rx2;
p+=dx+ry2-dy;
}
}
//end of region 1
x5=x+.5;
p2=(ry2*x5*x5)+(rx2*(y-1)*(y-1))-(rx2*ry2);
//calculation for region 2
while(y>=0)
{
print(x,y,xc,yc);
if(p2>0)
{
y--;
dx+=-(2*rx2);
p2+=rx2-(dx);
}
else
{
x++;y--;
dx+=-(2*rx2);
dy+=2*ry2;
p2+=rx2-(dx)+(dy);
}
}
}
void print(int x,int y,int xc,int yc)
{
putpixel(x+xc,y+yc,7);
putpixel(x+xc,-y+yc,7);
putpixel(-x+xc,y+yc,7);
putpixel(-x+xc,-y+yc,7);
}

Playfair cipher encyption and decryption

2010-03-10T22:09:00.002+05:30

//PLAYFAIR CIPHER ENCRYPTION

#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
FILE *fp;
int i,m=0,flag=0,j,n,row1,row2,col1,col2;
char ch,d,mat[5][5],key[10],k[10],c[30],a[10],ct[10],pt[10];
clrscr();
fp=fopen("key.txt","r");
fscanf(fp,"%s",key);
fclose(fp);
printf("The key is: %s",key);
//finding new key removing duplicates...
for(i=0;i<strlen(key);)
{
if(key[i]=='j')
key[i]=105;
flag=0;
for(j=0;j<i;j++)
{
if(key[i]==key[j])
flag=1;
}
if(flag==1)
i++;
else
k[m++]=key[i++]; //new array for key
}
k[m]=NULL;
printf("\nThe new key is :\n%s",k);//printing new key
//store key in matrix
m=0;
for(ch=97;ch<=122;)
{
flag=0;
for(i=0;i<strlen(k);i++)
{
if(k[i]==ch)
flag=1;
}
if(flag==1)
ch++;
else if(ch=='j')
ch++;
else
{
c[m++]=ch;
ch++;
}
}
c[m]=NULL;
//construction of 2-D 5*5 matrix
m=0;
n=0;
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
mat[i][j]=k[m++];
if(m>strlen(k))
mat[i][j]=c[n++];
}
}
printf("\n");
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
printf("\t %c",mat[i][j]);
printf("\n");
}
//retrieve plain-text from file
i=0;
fp=fopen("plain.txt","r");
fscanf(fp,"%s",pt);
fclose(fp);
printf("\nThe plain-text is: %s",pt);
//seperation of digram
j=0;
printf("\nAfter Processing Digrams: ");
for(i=0;i<strlen(pt);)
{ if(pt[i]==pt[i+1])
{
a[j++]=pt[i];
a[j++]='x';
i+=1;
}
else
{
a[j++]=pt[i];
a[j++]=pt[i+1];
i+=2;
}
}
a[j]='\0';
printf("%s",a);
printf("\n");
for(i=0;i<strlen(a);i+=2)//Loop for conversion to cipher text
{
for(j=0;j<5;j++)// matrix row
{
for(m=0;m<5;m++)//matrix column
{
//Find position of 1st element of Digram in matrix
if(a[i]==mat[j][m])
{
row1=j;
col1=m;
}
//Find position of 2nd element of Digram in matrix
if(a[i+1]==mat[j][m])
{
row2=j;
col2=m;
}
//Condition For both elements in same row of matrix
if(row1==row2)
{
ct[i]=mat[row1][(col1+1)%5];
ct[i+1]=mat[row1][(col2+1)%5];
}
//Condition For both elements in same column of matrix
else if(col1==col2)
{
ct[i]=mat[(row1+1)%5][col1];
ct[i+1]=mat[(row2+1)%5][col2];
}
else
{
ct[i]=mat[row1][col2];
ct[i+1]=mat[row2][col1];
}
}
}
}
ct[i]='\0';
fp=fopen("cipher.txt","w");
printf("\nCipher Text: ");
//store cipher text in a file
for(i=0;i<strlen(ct);i++)
putc(ct[i],fp);
fclose(fp);
printf("%s",ct); //print cipher text
getch();
}

//PLAYFAIR CIPHER DECRYPTION

#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
FILE *fp;
int i,m=0,flag=0,j,n,row1,row2,col1,col2;
char ch,d,mat[5][5],key[10],k[10],c[30],a[10],ct[10],pt[10];
clrscr();
fp=fopen("key.txt","r");
fscanf(fp,"%s",key);
fclose(fp);
printf("The key is: %s",key);
for(i=0;i<strlen(key);)
{
if(key[i]=='j')
{
key[i]='i';
}
flag=0;
for(j=0;j<i;j++)
{
if(key[i]==key[j])
flag=1;
}

if(flag==1)
i++;
else
k[m++]=key[i++]; //new array for key
}
k[m]=NULL;
// printing new key after elimination of duplicates
printf("\nThe new key is :\n%s",k);
//seperation of key letters from a-z & later stored in an array
m=0;
for(ch=97;ch<=122;)
{
flag=0;
for(i=0;i<strlen(k);i++)
{
if(k[i]==ch)
flag=1;
}
if(flag==1)
ch++;
else if(ch=='j')
ch++;
else
{
c[m++]=ch;
ch++;
}
}
c[m]=NULL;
//construction of 2-D 5*5 matrix
m=0;
n=0;
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
mat[i][j]=k[m++];
if(m>strlen(k))
mat[i][j]=c[n++];
}
}
printf("\n");
for(i=0;i<5;i++)
{
for(j=0;j<5;j++)
{
printf("\t %c",mat[i][j]);
}
printf("\n");
}

//retrieve cipher-text from file

i=0;
fp=fopen("cipher.txt","r");
fscanf(fp,"%s",ct);
fclose(fp);
printf("\nThe Cipher-text is: %s",ct);

printf("\n");
for(i=0;i<strlen(ct);i+=2)//conversion to plain text
{
for(j=0;j<5;j++) //matrix row
{
for(m=0;m<5;m++)//matrix column
{
//Find position of 1st element of Digram in matrix
if(ct[i]==mat[j][m])
{
row1=j;
col1=m;
}
//Find position of 2nd element of Digram in matrix
if(ct[i+1]==mat[j][m])
{
row2=j;
col2=m;
}
//Condition For both elements in same row of matrix
if(row1==row2)
{
pt[i]=mat[row1][(col1-1+5)%5];
pt[i+1]=mat[row1][(col2-1+5)%5];
}
//Condition For both elements in same column of matrix
else if(col1==col2)
{
pt[i]=mat[(row1-1+5)%5][col1];
pt[i+1]=mat[(row2-1+5)%5][col2];
}
else
{
pt[i]=mat[row1][col2];
pt[i+1]=mat[row2][col1];
}
}
}
}
pt[i]='\0';
fp=fopen("plain.txt","w");
printf("\nPlain Text: ");
//store plain text in a file
for(i=0;i<strlen(pt);i++)
putc(pt[i],fp);
fclose(fp);
printf("%s",pt); //print plain text
fclose(fp);
getch();
}

Synchronous TDM and Asynchronous TDM

2010-03-10T21:30:00.004+05:30

// This is Synchronous TDM developed using C

#include<stdio.h>

#include<string.h>

int main()

{

int i,j,k=0,l,maxlength=0,length[3],flag1=0,flag2=0,flag3=0;

char usr1[10],usr2[10],usr3[10],newusr1[10],newusr2[10],newusr3[10],b[100];

printf("enter data for user 1\n");

gets(usr1);

printf("enter data for user 2\n");

gets(usr2);

printf("enter data for user 3\n");

gets(usr3);

length[0]=strlen(usr1);

length[1]=strlen(usr2);

length[2]=strlen(usr3);

for(j=0;j<3;j++)

{

if(maxlength<length[j])

maxlength=length[j];

}

for(i=0;i<maxlength;i++)

{

if(usr1[i]!=NULL && flag1==0)

b[k++]=usr1[i];

else

{

b[k++]='$';

flag1=1;

}

if(usr2[i]!=NULL && flag2==0)

b[k++]=usr2[i];

else

{

b[k++]='$';

flag2=1;

}

if(usr3[i]!=NULL && flag3==0)

b[k++]=usr3[i];

else

{

b[k++]='$';

flag3=1;

}

b[k]=NULL;

k=0;

for(i=1;i<=maxlength;i++)

{

printf("\nframe %d is ",i);

for(j=0;j<3;j++)

printf("%c",b[k++]);

}

strcpy(newusr1,usr3);

strcpy(newusr2,usr1);

strcpy(newusr3,usr2);

printf("\nTransferring Data To Respective Users...\n\n");

printf("user1 gets %s \n",newusr1);

printf("user2 gets %s \n",newusr2);

printf("user3 gets %s \n",newusr3);

return 0;

}

//This is Asynchronous TDM developed using C

#include<stdio.h>
#include<ctype.h>
#include<string.h>
int main()
{
int i,j,k=0,l,frame,maxlength=0,length_t=0,length[3],flag1=0,flag2=0,flag3=0;
char usr1[10],usr2[10],usr3[10],newusr1[10],newusr2[10],newusr3[10],b[100];
printf("enter data for user 1\n");
gets(usr1);
printf("enter data for user 2\n");
gets(usr2);
printf("enter data for user 3\n");
gets(usr3);
length[0]=strlen(usr1);
length[1]=strlen(usr2);
length[2]=strlen(usr3);
for(j=0;j<3;j++)
{
length_t+=length[j];
if(maxlength<length[j])
maxlength=length[j];
}

if(length_t%2==0)
frame=length_t/2;
else
frame=(length_t/2)+1;
printf("\ntotal frames are %d\n",frame);

for(i=0;i<maxlength;i++)
{
if(isalpha(usr1[i]))
b[k++]=usr1[i];
if(isalpha(usr2[i]))
b[k++]=usr2[i];
if(isalpha(usr3[i]))
b[k++]=usr3[i];
}
b[k]=NULL;
k=0;
for(i=1;i<=frame;i++)
{
printf("\nframe %d is ",i);
for(j=0;j<2;j++)
printf("%c",b[k++]);
}
k=0;
for(i=0;i<frame;i++)
{
if(i<strlen(usr1))
newusr2[i]=b[k++];
if(i<strlen(usr2))
newusr3[i]=b[k++];
if(i<strlen(usr3))
newusr1[i]=b[k++];
}
newusr1[strlen(usr3)]=NULL;
newusr2[strlen(usr1)]=NULL;
newusr3[strlen(usr2)]=NULL;
printf("\nTransferring Data To Respective Users...\n\n");
printf("user1 gets %s \n",newusr1);
printf("user2 gets %s \n",newusr2);
printf("user3 gets %s \n",newusr3);
return 0;
}

Polygon Drawing

2010-02-20T21:44:00.001+05:30

//This program is to draw polygon using C

#include <graphics.h>
#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
#include<iostream.h>
#include<math.h>
void line(float x1,float y1,float x2,float y2);
void main()
{
   int gdriver = DETECT, gmode,i,no;
   float x[100],y[100];
   initgraph(&gdriver, &gmode, "..\\bgi");
   cout<<"how many sides you want";
   cin>>no;
   for(i=0;i<no;i++)
   {
   cout<<"enter x"<<i+1;
   cin>>x[i];
   cout<<"enter y"<<i+1;
   cin>>y[i];
   }
   x[no]=x[0];
   y[no]=y[0];
   for(i=0;i<no;i++)
   {
   line(x[i],y[i],x[i+1],y[i+1]);
   }
   getch();
}
   void line(float x1,float y1,float x2,float y2)
   {
   int i;
   float sign,dx,dy;
   if(abs(x2-x1)>abs(y2-y1))
   sign=abs(x2-x1);
   else
   sign=abs(y2-y1);
   dx=x2-x1;
   dy=y2-y1;
   for(i=0;i<=sign;i++)
   {
   putpixel(x1,y1,15);
   x1+=dx/sign;
   y1+=dy/sign;
   }
   }

DDA Line drawing algorithm

2010-02-20T21:29:00.001+05:30

// This is Digital differential analyzer(DDA) line drawing algorithm using C

#include <graphics.h>
#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
#include<math.h>
void jayudda(float,float,float,float);
void main()
{
   int gdriver = DETECT, gmode;
   float x1,y1,x2,y2;
   initgraph(&gdriver, &gmode, "..\\bgi");
   printf("enter starting points(x1,y1):"),scanf("%f%f",&x1,&y1);
   printf("enter ending points(x2,y2):"),scanf("%f%f",&x2,&y2);
   jayudda(x1,y1,x2,y2);
   getch();
}
void jayudda(float x1,float y1,float x2,float y2)
{
int i;
float sign;
if(abs(x2-x1)>abs(y2-y1))
   sign=abs(x2-x1);
   else
   sign=abs(y2-y1);
   for(i=0;i<=sign;i++)
   {
   putpixel(x1,y1,7);
   x1+=(x2-x1)/sign;
   y1+=(y2-y1)/sign;
   }
}

Operator precedence parsing algorithm

2009-10-09T22:08:00.000+05:30

//This is operator precedence parsing algorithm using C

char pre[10][10],ss[30],rs[30],stk[10][2];

int i,j,k,n,p,sp=0,br=0;

void push();

void pop();

char chk_pre();

void reduce_ss();

void main()

{

clrscr();

strcpy(pre[0]," +*()]");

strcpy(pre[1],"+><<>>");

strcpy(pre[2],"*>><>>");

strcpy(pre[3],"(<<<=e");

strcpy(pre[4],")>>e>>");

strcpy(pre[5],"[<<<e=");

n=6;

scanf ("%s",ss, printf("Enter the source string : "));

strcpy(rs,"[");

strcat(rs,ss);

strcat(rs,"]");

strcpy(ss,rs);

printf("\n");

for(i=0;i<n;i++)

{

for(j=0;j<strlen(pre[i]);j++)

printf("%c\t",pre[i][j]);

printf("\n");}

while(ss[2]!=']' && br==0)

{

sp=0;

for(i=0;i<strlen(ss);i++)

{

if( (ss[i]>='a' && ss[i]<='z') || (ss[i]>='A' && ss[i]<='Z') )

continue;

push();

if(stk[sp-1][1]=='e' || ss[1]==']' || ss[0]==']')

{ printf("\n\nInvalid string.");

br=1;

break; }

if((stk[sp-1][1]=='>' && stk[sp-2][1]=='<') || stk[sp-1][1]=='=')

{ reduce_ss();

pop(); } } }

if(ss[2]==']')

printf("\nValid String.");

getch();

}

void push()

{ stk[sp][0]=ss[i];

stk[sp][1]=chk_pre();

sp++; }

void pop()

{ char tmp,tmp1;

sp--;

tmp=stk[sp][0];

tmp1=stk[sp][1];

sp--;

stk[sp][0]=tmp;

stk[sp][1]=tmp1;

sp++; }

char chk_pre()

{ char tmp;

if(i==0)

tmp=' ';

else

{ for(j=0;j<n;j++)

for(k=0;k<strlen(pre[j]);k++)

if(stk[sp-1][0]==pre[j][0] && stk[sp][0]==pre[0][k])

tmp=(pre[j][k]); }

return tmp; }

void reduce_ss()

{ char t1[30],t3[30];

int tmp;

if(stk[sp-1][0]==')' && stk[sp-2][0]=='(')

tmp=i-2;

else

tmp=i-3;

for(j=0;j<tmp;j++)

{ t1[j]=ss[j]; }

t1[j]='\0';

tmp=tmp+3;

k=0;

for(j=tmp;j<strlen(ss);j++)

{ t3[k]=ss[j];

k++; }

t3[k]='\0';

strcpy(ss,"");

strcpy(ss,t1);

strcat(ss,"i");

strcat(ss,t3);

}

Database connection in classic ASP

2009-09-17T18:46:00.000+05:30

Dim con
Dim rs
Dim ConnectString
dim accessdb
set con=server.CreateObject("ADODB.Connection")
set rs=server.CreateObject("ADODB.RecordSet")
ConnectString = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=C:\Inetpub\wwwroot\database.mdb"
con.open ConnectString
accessdb="database.mdb"
mysql="select * from table"

rs.open mysql, con, 3, 3

Mysterious World!

Squaring of Number

Now Squaring of Number will be very easy by the Multiplications method we have learn in previous Posts.

Now Lets see what if Number ends at 5!

The magic comes here!

Now you have learn lots of tricks to multiply and square number very Quick with very easy Methods!Congratulations!

Don't Forget to share it with your Friends.

Two digit multiplication Easy methods

how easy????

Can you Imagine Multiplication can be much faster and easier like this?

Let Take another Example to clear it!

Now What if Number is not near by 10 or not 100 or not 1000????

Multiplication of 3 digit numbers

1 2 3

4 5 6

Vedic Maths: Multiplication of two digit numbers

Power of Vedic Mathematics And Why to learn it.

Vedic Maths: Square a Number

Most common Passwords people use!

Jayesh and Sweety Wedding Dance

Shortest Path Algorithm: Dijkstra

Decimal To Binary Converter

Monoalphabatic cipher encryption and decryption

Caesar cipher encryption and decryption

Polyalphabatic cipher encryption and decryption

Bresenham's line drawing algorithm

Midpoint circle algorithm

2D Modeling: 2D Rotation

2D Modeling: 2D scaling

2D Modeling: 2D transformation

Midpoint ellipse algorithm

Playfair cipher encyption and decryption

Synchronous TDM and Asynchronous TDM

Polygon Drawing

DDA Line drawing algorithm

Operator precedence parsing algorithm

Database connection in classic ASP

Now you have learn lots of tricks to multiply and square number very Quick with very easy Methods!
Congratulations!