Program covering basics of linked list

#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<iostream.h>
struct node
 {   int num;
      struct node *link;
 };
node *start , *save , *newptr , *ptr;
node * create_node ( int n )
{
      ptr  = ( node * )malloc(sizeof(struct node));
      ptr -> num = n;
      ptr -> link = NULL;
      return ptr;
}
void insert_node ( node *np , int n )
{
     node *tmp;
     if( start == NULL )
          start = np;
     else
     {    tmp = start;
          while (  tmp != NULL)
          { if ( tmp -> num <= n && ( tmp -> link -> num > n || tmp -> link == NULL))
                { np -> link = tmp -> link;
                   tmp -> link = np;
                   return;
                 }
            tmp = tmp -> link;
           }
      }
}
void del_beg()
{
         if ( start == NULL)
             printf("\nUnderflow");
        else
       {
             ptr = start;
             start = start -> link;
             free(ptr);
       }
}
void del (int inf)
{           ptr = start;
             node *tmp;
             while( ptr != NULL )
            {
                     if( ptr-> num == inf)
                    {        if ( ptr == start )
                              {    start = ptr ->link;
                                    free(ptr);
                                     return;
                              }
                             else
                             {      tmp = ptr;
                                     ptr = ptr->link;
                                     free(tmp);
                                     return;
                             }
                     }
                  else
                        ptr = ptr -> link;
              }
}
  void display ( node *np )
{
                 while (np != NULL )
                {     printf("%d" , np -> num );
                       printf("->>");
                       np = np -> link;
                 }
}
void main()
{
       clrscr();
       start = NULL ;
       int inf , ch1;
       char ch ;
       do
       {     printf("\n1.Insertion");
              printf("\n2.Deletion from begining");
              printf("\n3.Display");
              printf("\n4.Deletion of your choice");
              printf("\nEnter your choice:\t");
              scanf("%d" , &ch1 );
              switch(ch1)
             {    case 1:
                   printf("\nEnter the information of the new node:\t");
                   scanf("%d" , &inf );
                   printf("\nCreating new node!!! press Enter to continue");
                   getch();
                   newptr = create_node( inf ) ;
                   insert_node( newptr , inf );
                   display(start);
                   break;
                   case 2:
                   del_beg();
                    break;
                    case 3:
                    display(start);
                     break;
                    case 4:
                     printf("\nEnter the node to be delted:\t");
                     scanf( "%d" , inf );
                     del(inf);
                     break;
             }
              printf("\nDo you want to continue:\t");
              cin>>ch;
       }while ( ch == 'y' || ch == 'Y' );
       getch();
}

Structure solution

0Ans 10.3   Given is the structure declaration

                struct abc

                {              int x;

                                float y;

                                char z;

                }

                Following are the statements which declare and initialize the structure variable :

struct  a , b , c ; // Error The structure variable will not be declared as tag name abc  is         missing             

<!--[if !supportLists]-->II.                  <!--[endif]-->struct abc a , b , c ; //Correct Declaration

<!--[if !supportLists]-->III.                <!--[endif]-->abc  x , y , z ;   //Error the structure variable will not be declared as struct keyword is missing

<!--[if !supportLists]-->IV.                <!--[endif]-->struct abc a[] ;  //Error Array size is  missing

<!--[if !supportLists]-->V.                  <!--[endif]-->struct abc a = { }; // a is assigned a value which is never used

<!--[if !supportLists]-->VI.                <!--[endif]-->struct abc = b , { 1+2 , 3.0 , “xyz”}  //Error syntactically wrong

<!--[if !supportLists]-->VII.              <!--[endif]-->struct abc c = { 4 , 5 , 6 } //Error last member is a character it should be enclosed in single quotes

<!--[if !supportLists]-->VIII.            <!--[endif]-->struct abc a = 4 , 5.0 , “xyz” ; //Error Initialization values should be enclosed in curly braces

Ans 10.4   None of the given statements in the question  is  legal

Ans 10.5  For The given declaration

                struct item_bank

                {              int number;

                                Double cost;

                }

                Following are the correct statements

                (d)          struct item_bank items[10];

Ans 10.6   For the given declaration

                typedef struct abc

                {              char x;

                                int y ;

                                float z[10] ;

                }    ABC ;

Following declarations are invalid

<!--[if !supportLists]-->(a)    <!--[endif]-->struct ABC v3;

Ans 10.8 Purpose of the following is:

<!--[if !supportLists]-->a.       <!--[endif]-->Template

<!--[if !supportLists]-->b.      <!--[endif]-->struct keyword : is used declare , define and initialize a structure variable

<!--[if !supportLists]-->c.       <!--[endif]-->typedef keyword  : is used to declare the alias name for the existing datatype

<!--[if !supportLists]-->d.      <!--[endif]-->sizeof operator :  is used to get the size of the variable which is passed as parameter to the function

<!--[if !supportLists]-->e.      <!--[endif]-->tag name : it’s the name given to a structure defined by us using struct keyword

Ans 10.9  struct                                              // Tag name for structure is missing

                {              int number ;

                                float price ;

                }                                                              //; is missing

Ans 10.10 Use of the following is as given below:

<!--[if !supportLists]-->a.       <!--[endif]-->union -  The concept of union has been partially borrowed from the concept of structure the major difference being the members of union share same memory location whereas the members of structure share different memory location. We use union when we want a group of variables which we are using at different time but want them under a common data type we use union

<!--[if !supportLists]-->b.      <!--[endif]-->Bit fields – There are several instances when data requires much less than 16 bits space. For this we use bit fields which is a set of adjacent bits whose size can be from 1 to 16 bits in length

<!--[if !supportLists]-->c.       <!--[endif]-->The size of operator – this operator is used for determining the size occupied by  the variable in the memory which is provided as operand to the size of operator.

Ans 10.12  For th following declarations and definition

                struct abc

                {              int a;

                                float b ;

                };

                struct xyz

                {              int x;

                                float y; 

                };

                abc         a1  , a2;

                xyz         x1 , x2;

a1 = x1;    //Error two structure variables cannot be copied

<!--[if !supportLists]-->b.      <!--[endif]-->abc.a1 = 10.75; //Error structure variable cannot be initialized in this way

<!--[if !supportLists]-->c.       <!--[endif]-->int m  = a + x ; //Error structure members cannot be accessed in this way

<!--[if !supportLists]-->d.      <!--[endif]-->int n = x1.x + 10 // Correct

<!--[if !supportLists]-->e.      <!--[endif]-->a1 = a2 //Error two structure variables can be  copied only member wise not directly

<!--[if !supportLists]-->f.        <!--[endif]-->if (a.a1 > x.x1)   // Error the member should come after the variable name.

<!--[if !supportLists]-->g.       <!--[endif]-->if( a1.a < x1.x)   //Correct

<!--[if !supportLists]-->h.      <!--[endif]-->if (x1 != x2 )    //Error two structure variables cannot be compared directly

Ans 10.15   Computer stores structures using the concept of “ word boundary”. A  word is the natural/native amount of data fetched frommemory on a single access by a CPU. Now storing the data at the boundary of the word the data can be fetched very fast. The size of the word boundary is machine dependent. Now suppose  that the word boundary of the given machine is 4 byte. Now a structure having three data members char ,  int  ,  float will be stored in the memory as shown below . Now we can easily understand that the empty space will be there  this empty byte in between the two data members is termed as slack byte.

Ans 10.16 The various points to be considered while implementing bit field are as follows:

          i. We cannot take address of a bit field variable. Therefore we cannot use scanf function for reading     them and neither we can use pointer variable for such type of structure.

ii.     Bit fields cannot be arrayed.

iii.    Bit fields should be assigned values that are within the range of the of their size.

Ans 10.19 

typedef struct product

                {              char name[10];

                                float price ;

                }              PRODUCT products[10] ;                               // Error Immediately after the alias name variable

                                                                                                                cannot be declared

//Correct mode

typedef struct product

                {              char name[10] ;

                                float price;

                }              PRODUCT;

PRODUCT products[10];

Ans 10.20

Output:

8                                                                              Explanation union will occupy the space according to largest member of the same

10                                                                           At a time only one member can use the memory location of the union

1.23                                                                        Same reason as above

Solution to Programing Exercise pointers

                                                    Solution to Programming Exercise Pointers
Guys i think the starting two programs  are quite easy and thus I am not providing the solutions to them
Ans 11.3

#include<stdio.h>
#include<conio.h>
int size;
void main()
{
clrscr();
void insert( int *ptr , int val);
int arr[10] , in , i ;
printf("\nEnter the size of the array:\t");
scanf("%d" , &size);
printf("\nEnter your sorted array:\t");
for( i = 0 ; i <  size ;  i++)
scanf("%d", &arr[i]);
printf("\nEnter the number you want to insert in the array:\t");
scanf("%d", &in);
insert(arr, in);
printf("\nThe array after insertion is:\t");
for ( i = 0 ; i < size ; i++ )
printf("%d " , arr[i] ) ;
getch();
}

void insert( int  *ptr , int val)
{ int pos , i ;
if ( val < *(ptr+0) )
pos = 0 ;
else
{ for( i = 0 ; i < size - 1 ; i++ )
{ if ( *(ptr+i) <= val && val < *(ptr+i+1) )
{ pos = i+1;
break;
}
}
if ( i == size-1 )
pos = size;
}
if ( size == 10)
printf("\nOverflow!");
for ( i = size ; i > pos ; i-- )
*(ptr + i) = *(ptr + (i - 1) ) ;
*(ptr + pos ) = val;
size++;
}

Ans 11.4

#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int * add ( int *a[] , int *b[] );
int i , j , p[3][3] , q[3][3] , r[3][3];
printf("\nEnter a 3*3 Matrix:\n");
for( i = 0 ; i < 3 ; i++ )
for( j = 0 ; j < 3 ; j++ )
scanf("%d", p[i][j] ) ;
printf("\nEnter the second 3*3 Matrix;\n");
for( i =0 ; i < 3 ; i++)
for( j = 0 ; j < 3 ; j++ )
scanf( " %d " , q[i][j]  );
r = add( p , q );
printf("\nThe sum is:\t");
for ( i =0 ; i< 3 ; i++ )
for( j = 0 ; j < 3 ; j++ )
scanf( "%d" , r[i][j] );
getch();
}

int * add ( int *a[3] ,int *b[3] )
{
int  i , j , c[3][3] ;
for( i = 0 ; i < 3 ; i++ )
for( j = 0 ; j < 3 ; j++ )
c[i][j] = *( *(a + i ) + j )  + *( *(b + i ) + j );
       return c;
}

Ans 11.5

#include<stdio.h>
#include<conio.h>
void del ( char *ptr , char itm)
{
int i , pos , size;
for ( i = 0 , size = 0 ; * (ptr + i) != '\0' ; i++ , size++)
{
if( *(ptr + i ) == itm)
pos = i ;
}
for( i = pos ; i < size ; i++ )
*(ptr +i) = *(ptr+i+1);
*(ptr+size) = '\0';

}
void main()
{
clrscr();
char str[10] ,ch;
printf("\nEnter your string:\t");
gets(str);
printf("\nEnter the charcter:\t");
scanf("%c" , &ch);
del(str, ch ) ;
printf("\nYour string after deletion:\t%s", str);
getch();
}

Ans 11.6

#include<stdio.h>

#include<conio.h>

#include<iostream.h>

void main()

{

char * day_name( int no);

clrscr();

int n;

char * day;

printf("\nEnter the number of the day:\t");

scanf("%d" , &n ) ;

day = day_name( n );

printf("\nThe day is:\t%s" , day);

getch();

}

char * day_name ( int no )

{

static char * week_day[7] = { "Monday" , "Tuesday" , "Wednesday", "Thursday" , "Friday" , "Saturday", "Sunday"};

int i , pos = 0;

for ( i = 0 ; i < 7 ; i++ )

{

if( ( i+1 ) == no )

{ pos = i;

break ;

}

}

return week_day[pos];

}

Ans 11.7

#include<stdio.h>

#include<iostream.h>

#include<conio.h>

#include<string.h>

void main()

{

clrscr();

char *name[5] ;

int i ;

void sort ( char  *str[] );

printf("\nEnter any five names:\t");

for ( i = 0 ; i < 5 ; i++ )

gets(name[i]);

sort ( name ) ;

printf("\nThe names in the alphabetical order are:\n");

for ( i = 4 ; i >= 0 ; i-- )

puts(name[i]);

getch();

}

void sort ( char *str[5] )

{

int i , j ;

char *tmp ;

for ( i = 0 ; i < 5 ; i++ )

{ for ( j = 0 ; j < (5-1) - i ; j++ )

{ if ( strcmp( str[i+1] , str[i] )  > 0 )

{ tmp = str[i];

str[i] = str[i+1];

str[i+1] = tmp;

}

}

}

}

Ans 11.9#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

clrscr();

int a[5];

void reverse( int *p ) ;

printf("\nEnter the array:\t");

for (  i = 0 ; i <  5 ; i ++ )

sacnf("%d" , a[i] ) ;

reverse(a);

printf("\nThe reversed array is:\t");

for ( i = 0 i < 5 ; i++)

printf("%d" , a[i] );

getch();

}

void reverse ( int * p )

{

int i , tmp ;

for ( i = 0 , j = 4  ; i < 5/2 ; i++ ,  j-- )

{ tmp = *(p+i);

*(p+i) = *(p+j);

*(p+j) = tmp;

}

}

Ans 11.10

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

clrscr();

int a[5] , b[5] , i , res;

int cmp(int *p , int *q );

printf("\nEnter the array:\t");

for ( i = 0 ; i < 5 ; i++ )

scanf("%d" , &a[i] ) ;

printf("\nEnter the second array:\t");

for ( i = 0 ; i < 5 ; i++ )

scanf( "%d" , &b[i]);

if ( cmp( a , b) == 1)

printf("\nThey are identical");

else

printf("\nThey are not identical");

getch();

}

int cmp ( int *p , int *q)

{

int  i , flag = 1;

for ( i = 0 ; i < 5 ; i++ )

{ if( *(p+i) != *(q+i) )

{ flag = 0;

}

}

return flag;

}

Solution to E Balaguruswamy Chapter - Pointers

                                                               Solution to E Balaguruswamy

                                                                     Chapter  - Pointers

Ans 11.4 Pointer is initialized by keyword NULL or  0  in the following fashion

int  * p = NULL  , * m = 0 ;

 Ans 11.5  Following is the effect of the following statements

i.                     int a, *b = &a;  //Error Pointer being initialized with the address of an uninitialized variable

ii.                   int p ,   *p ;      //Error Multiple Declaration of p

iii.                  char *s;          

iv.                 a = (float * ) &x);

v.                   double  ( *f )( ) ;       //It declares the pointer to a function

Ans 11. 6   int M ,  n ;

    int *P1  , *P2 ;

The following are the errors in the expression given below:

i)                    P1 = &m ; //Error Pointer cannot be initialized with an uninitialized variable

ii)                   P2 = n ; // Error Pointer can only be initialized with the address of the variable

iii)                 P2 = &*&m; // Correct as the first dereferencing operator Is over powered by the second & operator

iv)                  M = P2 – P1 //Correct Pointer Subtraction Is allowed

v)                  P1 = &P2 //Error A simple pointer cannot be initialized with memory address of an another pointer, only a pointer to a pointer can be initialized in this way

  Ans11.7 Difference between the notations  *m[5] and (*m)[3]. Since * has lower precedence than  [] , *m[5] declares m  as an array of 5 pointers while (*m)[3] declares m as a pointer to an array of five elements.

Ans 11.9   int x = 10 ,  y  = 10 ;

                    int  *p1 = &x , *p2 = &y;

                The following Expression generate the following output:

i)                    (*p1)++   Output – 10

ii)                   - - (*p2)   Output  -  9

iii)                 *p1 + (*p2)--   Output  - 20

iv)                   ++(*p2) - *p1  Output  -  1

Ans 11.12  Following Output is  generated by the following code :

                int m = 100 ;

                int *p1 = &m;

                int **p2 = &p1;

                printf(“%d” , **p2);

                Output :

                100

Ans 11. 13   Following is the error in the given code

                                int **p1 , * p2 ;

                                p2 = &p1  // It should be  p1 =&p2

Explanation: p1 is a pointer to a pointer so it should always point a pointer

Ans11.14  Difference between the two expression name + 10; and *(name + 10 ) is that the first expression refers to all the characters after the 10^th position in the array name or rather simply we  can say that it will the array name from the 10^th character whereas the second expression refers specifically to the 10^th character of the array of name .

Ans 11.15  Following is the output of the code given below

                                int m[2] ;

                                *(m+1)

                                *m = *(m + 1);

                                printf(“ %d “ , m[0] );

Output:

                                100

Ans 11.16    int m[2];

                       int *p = m ;

                       m[0] = 100;

                       m[1] = 200;

                       printf(“%d %d “ , ++*p , *p );

                      Output:

                      101  100

Ans 11.17    Output of the following code :

                 int f(char *p );

                main()

                {              char str[] = “ANSI”;

                                printf(“%d” , f(str) );

                }

                int f(char *p )

                {              char *q = p ;

                                while (*++p ) ;

                                return (p – q );                  }

Output :

4

Ans 11.18  Following declarations are not the same:

                1 . void search ( int *m[] , int x) { }

                In this the parameter in the definition is a integer pointer array.

                2. void search ( int **m  ,int x ) {}

                In this the parameter in the definition is a integer pointer to a pointer.

Ans 11.19   Following declaration is are same

                char *s;

                char s[5] ;

                As name of the array is a pointer to the first element of the array it’s as same as declaring a pointers which when incremented can be used to store a list of values same as an array.

Ans 11.20  The following is the correct way to declare a pointer to a function.

                int (*p)(void);

                

FINDING THE MINOR OF 3 X3 MATRIX IN C

//QUERIES ARE WELCOME AS COMMENTS
# include<stdio.h>
# include<conio.h>
#define C 3
int x = 0, y = 0;
void check(int a[][C], int b[][C-1], int c[][C], int m, int n);
void minor(int b[][C-1],int c[][C]);
void main()
{
int a[C][C], b[C-1][C-1], c[C][C], i ,j;
for(i = 0; i < C; i++)
{
for(j = 0; j < C; j++)
{
printf("ENTER YOUR ELEMENT AT %d X %d\n", i , j);
scanf("%d", &a[i][j]);
}
}
for(i = 0; i< C; i++)
{
for(j = 0; j < C; j++)
{
check(a, b, c, i,j);
}
}
for(i = 0; i < C; i ++)
{
for(j = 0; j < C; j++)
{
printf("%d ",c[i][j]);
}
printf("\n");
}
}
void check(int a[][C], int b[][C-1], int c[][C], int m, int n)
{
int i, j, l = 0, k = 0;
for(i = 0; i< C; i++)
{
for(j = 0; j< C; j++)
{
if (i != m && j != n)
{
b[l][k++]  = a[i][j];
if(k == C-1)
{
l++;
k = 0;
}
}
}
}
minor(b, c);
}
void minor(int b[][C-1],int c[][C])
{
int z;
z = b[0][0] * b[1][1] - b[0][1] * b[1][0];
c[x][y++] = z;
if (y == C)
{
x++;
y = 0;
}
}

The Fibonacci Series In C Using Recurssion

//IN CASE OF DOUBT, COMMENT IT
# include<stdio.h>
int fibo(int n);
int r;
void main()
{
int i, j, k, l = 0;
scanf("%d",&j);
for(i = 0; i < j; i++)
{
for( k = 0; k <= i; k++)
{
l = fibo(k);
printf("%d ", l);
}
printf("\n");
}
}
int fibo(int n)
{
if(n == 0)
{
return (0);
}
else if(n == 1)
{
return (1);
}
else
{
r =  fibo(n- 1) + fibo(n-2);
return r;
}
}

MATRIX SOLVER-GAUSS JORDAN METHOD IN C

//KINDLY ASK THE DOUBTS IN COMMENTS
# include<stdio.h>

void subtrac(float b[][6], int k);

float constant(float c[][6], int a, int b);

void divide(float b[][6], int a, float c);

void main()

{

float a[3][6], k = 0.0;

int i, j;

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

{

for( j = 0; j <  6; j++)

{

a[i][j] = 0;

}

}

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

{

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

{

printf("ENTER ELEMENT AT %d x %d\n",j, i);

scanf("%f", &a[j][i]);

}

}

a[0][3] = 1.0;

a[1][4] = 1.0;

a[2][5] = 1.0;

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

{

subtrac(a,i);

}

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

{

k = a[i][i];

divide(a, i, k);

}

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

{

for(j= 3; j < 6; j++)

{

printf("%f ", a[i][j]);

}

printf("\n");

}

}

void subtrac(float b[][6], int k)

{

int i, z;

float j;

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

{

if(i != k)

{

j = constant(b, i , k);

for(z = 0; z < 6; z++)

{

                         b[i][z] = b[i][z] - j * b[k][z];
 
}

printf("R%d -> R%d - (%.2f * R%d)\n",i+1,i+1,j,k+1);

}

}

}

float constant(float c[][6], int a, int b)

{

float con;

con = c[a][b]/ c[b][b];

return (con);

}

void divide(float b[][6], int a, float c)

{

int j;

for(j = 0; j < 6; j++)

{

b[a][j] = b[a][j] / c;

}

printf("R%d -> R%d / %.2f\n", a+1, a+1, c);

}

The N- Queens Problem in C

//FEEL FREE TO DROP ANY QUESTION IN THE COMMENTS
# include<stdio.h>
# include<conio.h>
# define C 8
int check(int k[][C], int m, int n);
void backtrack(int k[][C], int f2[], int m);
void increment(int k[][C], int m, int n);
void main()
{
int a[C][C] = {0},i, j = 0, c, fj[C] = {0}, z = 0, k, count = 0;
clrscr();
for(i = 0; i < C; i++)
{
reloop:
while(j < C)
{
c = check(a, i, j);
if(c == 0)
{
increment(a, i, j);
fj[i] = j;
z = 1;
break;
}
j++;
}
if(z != 1)
{
    backtrack(a, fj, i);
    j = fj[i - 1] + 1;
    i = i - 1;
    if(i < 0)
    {
goto end;
    }
    goto reloop;
}
j = 0;
z = 0;
if(i == (C-1))
{
for(k = 0; k < C; k++)
{
printf("(%d,%d);",k,fj[k]);
}
printf("\n");
getch();
count++;
backtrack(a, fj, i + 1);
j = fj[i] + 1;
    goto reloop;
}

}
end:
printf("%d",count);
getch();
}
int check(int k[][C], int m, int n)
{
int i, j, sum = 0;
for(i = 0; i < C; i++)
{
sum = sum + k[i][n] + k[m][i];
}
if(m == n)
{
for(i = 0; i < C; i++)
{
sum = sum + k[i][i];
}
}
else if(m < n)
{
for(i = 0, j = n - m; j < C; j++, i++)
{
sum = sum + k[i][j];
}
}
else if(m > n)
{
for( i = m - n, j = 0; i < C; i++,j++)
{
sum = sum + k[i][j];
}
}
if((m + n) == C-1)
{
for(i = C - 1, j = 0; j < C; j++, i--)
{
sum = sum + k[i][j];
}
}
else if((m + n) < C - 1)
{
for(i = m + n, j = 0; j <= m + n;  i--, j++)
{
sum = sum + k[i][j];
}
}
else if((m + n) > C-1)
{
for(i = C - 1, j = m + n - C + 1; j < C; i--,j++)
{
sum = sum + k[i][j];
}
}
return sum;
}
void increment(int k[][C], int m, int n)
{
k[m][n] = 1;
}
void backtrack(int k[][C], int f2[], int m)
{
int i, j;
i = m - 1;
j = f2[m-1];
k[i][j] = 0;
}

The Time Paradox

since the invention of the clock, the world has wanted us to rotate our universe around it. rise from the clock and set down with it. the concept of the time table is nothing more than a myth , to just bind us to the never stopping flow of the life we have, and which is segregated into the concept time. most of u would have a notion that time is am measurable entity. the rotation of the earth constitutes of the hours, minutes and seconds and the revolution constitutes of the years, months and days. and when the evidences showed that the rotation of the earth fluctuated randomly, a new definition of time was formulated. it was the duration of  9 192 631 770 periods of the radiation corresponding to the transition between the two hyper fine levels of the ground state of the cesium 133 atom. none of us ever sat down to question why only 24 hours, why 60 min, why 60 sec. when the whole worlds obsessed with 100 and its multiples, why bother with a 60. am not going to give you the ans, a magician never reveals his secrets. what confuses me the most that why do human waste so much time perfecting 1 sec when the same concept of time can be used to reveal the future on retell the past. despite a mixed bag of opinions, and some quite sentimental ones, i believe in the theory of time travel, or most precisely at least zonal transfer. i also but the theory that parallel universes exist, despite whatever the heart says. because it makes me happy. now i know that some one some where would have made an opposite choice of what i have taken and would be living that life. if possible i would even like to visit him or her(i really don't know) in their universe. as the mythologist believe that for each yin there is a yang, for each head there is a tail, for each evil there is good, so why not for each choice i make there is an anti choice or the opposite choice making. so i get satisfied, knowing that at least, if i screw up here, i would be doing well in the alternative universe, or some other zone.
well leaving aside the inter dimensional time, what i wanted was, that i have never(read mostly) run out of time. for me a day contains 48 hours and each hour of 30 min. the end output is the same, but i get to sleep more, for like 16 hours. the concepts called, "FOOLING THE MIND". the mind is a divine entity, and controlling it would be predicting whether you are going to get a girl or not. so when you cannot control your monster, you need to train it or either fool it. i decided to fool my mind, into believing that i had twice the time anyone had. so no assignment or project deadline ever bothered me. same way, i conditioned my mind to believe that my phone company was spying on me(or was it not), to just cut down on my phone bills. then there was a dog in our street who was not particularly fond of anyone stepping in his area(who is??), but me being the lost in thought kind of guy would often just go into his den and would end up with an orgy of barks and an occasional chase down the road. i decided that the dog was actually a human transformed by a witch to stay a mad dog. so as common courtesy demands, you don't visit anyone's house(except your own), without an invitation. BOOM. now the neighbors sleep peacefully and the dog also treats me with respect and the occasional cheerful bark, when he is on his way to work. this idea may be childish, or the story supporting it would definitively be, but mind is attracted to oddity. you may see a 1000 beautiful girls in a party and after some time you would be molded to it. addition of some new girls wouldn't create a difference(hopefully). but now if a dog decided to rise to the occasion and accept the invitation to the party, he would be the first thing in your eye. the same thing happens with the mind. this is the essence of life, or the true way to lead it. you believe what you want to, modify the facts according to it, and lead a happy life. of course, some people will call you, back crap crazy, but that is because of the dry life which they lead. they cannot digest the idea of us controlling the output of the life and leaving a bit of it to the random variables of the universe. the food for though may be tough to comprehend, basically because of the non- traditional way of thinking. but try it. this might not be a 100% fail proof idea, but might certainly lead to a happy life.

The Day I Created Water

Well, i am a great fan of camels. the utility of a hump along with a long neck cannot be expressed in words. you can peek over toilet stands to see what someone is doing. you can the watch the girls who lives upstairs with just a twist of our neck, and a lot more. but the thing that fascinates me the most is the hump. the main function being to store water. come on, who wants to carry a water bottle everywhere, when you can just have a hump installed on your back and get rid of the excess baggage. unfortunately god works in mysterious ways and so does the post office. they have never ever send my mails to god, i can assure you that. otherwise he would have replied.

as we do not have humps, we are in a constant need of water provided that u have frequent outputs. and when you are with me there is always a fight over water. me and my rommie are water people. our body consists of 70 percent water. so to no ones surprise our water tank was cleared up in a couple of days after its arrival. this has been a frequent occurring where we finish a weeks supply of water in a couple of days, and then try to find the alternative source of energy. it ranges from free cold drinks to unannounced guest visits to our friends and relatives. but this week was the different, this week we decided to put our engineering minds to use and try to survive the week without any external influences. we we tied our shoes, packed our bags, filled our wallets and went out on the streets to buy everything necessary, but not the water.

day 1 was as normal as it could be. we managed to bend around the swift punches life threw at us and finished half of the weeks allocated water. day2 was even better, we went for a bit of social service. we went on to a nice restaurant and left the waiter a huge tip. our contribution to the society. when we reached home we had did the regular.finally, we were out of water. but this time was different. we were ready. we had done our math. what we didn't account for was for some unexpected thirsty guests to visit us. they did what we used to do. go to anyone house talk to them and cutoff their supplies. damn then, all ur supplies of extra fluids, food, and watermelons got depleted in serving the guests. i think they were gods messengers, who had come to test us. its a common engineering theorem. "when you are prepared for the worst, there is someway someone makes it even worse". this god god had put us to test. but we are engineers, we cant be defeated.

day 3. no water, no fluids and technically no cooked food. we sat to our work and decided to skip breakfast and directly concentrate on the essentials for the lunch. we cooked noodles without water. don't ask me how we did it. its a secret recipe. it tasted just as tasty. we then got ready for dinner, which was a dozen of lays and wipped cream and a lot of television. trust me, the television fills your stomach more than the food. 

day 4 was the day when the disaster struck. we were out of everything. even the money. the calender was messing with us, so was the bill man. he cut off our electricity, for no particular reason.

stranded in the cold corners of our rooms, in a haunted place which we cold our home, we were hungry , cold, we were angry. they say "neccesity is the mother of invention and the father of tension". we live in the era of smart phone and idiot people, so when the smart phone dies the idiot people feel strandled, alone, depressed and suicidal. we felt the same, but we hung on.

the next day was the living example of hell. nut as the day progressed our minds started kicking in and we started to work on the problems. 1st was that there was no electricity. putting our mastermind skills to work, we hatched a plan. a plan dripping with evil potential, giving a jokerish grin we went to find the source of light, a flashlight. with the flash light in hand, now now had the power of sight. and we cautiously proceeded to the basement of the building. we had the 3d structural analysis of the building in mind, and we located the co-ordinates of the electricity meter box. our plan was to channel the flow of electricity from our rude neighbors wire to our wire and to direct the meter reading into the community meter. so the society pays for our living. our way of taking back what we had given to the society. MM to base camp, the training is complete.
the next step was to get an oven heating. but for that we needed an over, which we didn't have. so we just  wound up some coils in a cylindrical metal sheet and made about 6 to 7 of them, then we took up some dough and added the harsh tap water to it. now we connected the cylindrical metal loops on the dough and connected them to the electric unit. this helped us in heating up the dough and making it crusty and hard. the dough plate had become hot enough for us to advance to the next part of the plan. MM to base camp, camp 1 reached.
now comes the tricky part. we sneaked into the neighbors house disabling their alarm system and tip towed to their kitchen. it was an 007 mission, we had to be very secretiveness. we spied on the neighbors and silently opened their fridge to extract something of great importance. ICE, fresh cold and lots of ice.
we took the ice back to our home, to the electrically heated furnace and put the ice on it. MM to base camp, we see the summit.
as the ice started to melt, we were getting the thing of desire, we were getting WATERRRRR. we had done it, we created water.
and we survived the week........