Tower of Hanoi

Tower  :                     x                  z                  y

Solution of Above Tower of Hanoi Problem will be soon..... using c++

#include<conio.h>

#include<iostream.h>

class tower

{

    //Private variables

        int NoDisk;

        char FromTower,ToTower,AuxTower;

    public:

        void hanoi(int,char,char,char);

};

void tower::hanoi(int NoDisk,char FromTower,char ToTower, char AuxTower)

{

     //if only one disk, make the move and return

        if (NoDisk == 1)

        {

            cout<<"\nMove from disk 1 from tower "<<FromTower<<" to tower "<<ToTower;

            return;

        }

            //Move top n1 disks from X to Y, using Z as aux. tower

       

        hanoi(NoDisk - 1,FromTower,AuxTower,ToTower);

       

  //Move remaining disk from X to Z      

        cout<<"\nMove from disk "<<NoDisk<<" from tower "<<FromTower<<" to tower "<<ToTower;

       

  //Move n1 disk from Y to Z using X as aux. tower

        hanoi(NoDisk - 1,AuxTower,ToTower,FromTower);

        return;

       

}

void main()

{

    int No;

    tower Ob;

    clrscr();

    cout<<"\n\t\t\t--- Tower of Hanoi ---\n";

  //Imput the number of disk in the tower

    cout<<"\n\nEnter the number of disks = ";

    cin>>No;

  //We assume that the towers are X, Y and Z

    Ob.hanoi(No,'X','Z','Y');

    getch();

   

}
Output :

Another Solution without recursion Technique :

Move	Binary Number	Interpretation
1	00001	Move disk 1 to empty peg.
2	00010	Move disk 2 to empty peg.
3	00011	Move disk 1 to cover disk 2.
4	00100	Move disk 3 to empty peg.
5	00101	Move disk 1 NOT to cover disk 3.
6	00110	Move disk 2 to cover disk 3.
7	00111	Move disk 1 to cover disk 2.
8	01000	Move disk 4 to empty peg.
9	01001	Move disk 1 to cover disk 4.
10	01010	Move disk 2 NOT to cover disk 4.
11	01011	Move disk 1 to cover disk 2.
12	01100	Move disk 3 to cover disk 4.
13	01101	Move disk 1 NOT to cover disk 3.
14	01110	Move disk 2 to cover disk 3.
15	01111	Move disk 1 to cover disk 2.
16	10000	Move disk 5 to empty peg.
17	10001	Move disk 1 NOT to cover disk 5.
18	10010	Move disk 2 to cover disk 5.
19	10011	Move disk 1 to cover disk 2.
20	10100	Move disk 3 NOT to cover disk 5.
21	10101	Move disk 1 NOT to cover disk 3.
22	10110	Move disk 2 to cover disk 3.
23	10111	Move disk 1 to cover disk 2.
24	11000	Move disk 4 to cover disk 5.
25	11001	Move disk 1 to cover disk 4.
26	11010	Move disk 2 NOT to cover disk 4.
27	11011	Move disk 1 to cover disk 2.
28	11100	Move disk 3 to cover disk 4.
29	11101	Move disk 1 NOT to cover disk 3.
30	11110	Move disk 2 to cover disk 3.
31	11111	Move disk 1 to cover disk 2.

 Think Above Solution for Five Ring Tower. and Consider Cases As Below Detail.

Complexity Consider on (2^5 -1)=31

Move	Binary Number	Interpretation
8	01000	Move disk 4 to empty peg.
13	01011	Move disk 1 to cover disk 2.
25	11001	Move disk 1 to cover disk 4.
18	10010	Move disk 2 to cover disk 5.
10	11010	Move disk 2 NOT to cover disk 4.
17	10001	Move disk 1 NOT to cover disk 5.

check whether given no is odd or even without using any type of arithmetic operator

#include<stdio.h>

void main()
{

    int i,n;

    

    printf("enter the number ");

    scanf("%d",&n);

    i=n&1;

    if(i==0)                                

    {

        printf("Given number is even");

    } 

    else

    {

        printf("Given number is odd ");

    }

}

Cyclic Nature of char data type

What will be output when you will execute following c code?

#include<stdio.h>

void main(){

    char c=256;

    char *ptr="Leon";

    if(c==0)                                

         while(!c)

             if(*ptr++)

                 printf("%+u",c);

             else

                 break;

}

Explanation:

In the above program c is signed (default) char variable. Range of signed char variable in Turbo c is from -128 to 127. But we are assigning 256 which is beyond the range of variable c. Hence variable c will store corresponding cyclic value according to following diagram:

Since 256 is positive number move from zero in clock wise direction. You will get final value of c is zero.

if(c==0)

It is true since value of c is zero.

Negation operator i.e. ! is always return either zero or one according to following rule:

!0 = 1

!(Non-zero number) = 0

 So,

!c = !0 =1

As we know in c zero represents false and any non-zero number represents true. So

while(!c) i.e. while(1) is always true.

In the above program prt is character pointer. It is pointing to first character of string “Leon” according to following diagram:

In the above figure value in circle represents ASCII value of corresponding character.

Initially *ptr means ‘L’. So *ptr will return ASCII value of character constant ‘L’ i.e. 76

if(*ptr++) is equivalent to : if(‘L’) is equivalent to: if(76) . It is true so in first iteration it will print +0. Due to ++ operation in second iteration ptr will point to character constant ‘e’ and so on. When ptr will point ‘\0’ i.e. null character .It will return its ASCII value i.e. 0. So if(0) is false. Hence else part will execute. 

C Pointer to array of string Actual Representation and allocation

void main(){

char *array[4]={"c","c++","java","sql"};

char *(*ptr)[4]=&array;

printf("%s ",++(*ptr)[2]);

}

Output: ava

Explanation:

In this example

ptr: It is pointer to array of string of size 4.

array[4]: It is an array and its content are string.

Pictorial representation:

Note: In the above figure upper part of box represent content and 

lower part represent memory address. We have assumed arbitrary address.

++(*ptr)[2]

=++(*&array)[2] //ptr=&array

=++array[2]

=++”java”

=”ava” //Since ptr is character pointer so it 

// will increment only one byte

Note: %s is used to print stream of characters 

up to null (\0) character.

Create DOS - DIR Command using c Language

Step 1: Write following code.

#include <stdio.h>

#include <dos.h>

void main(int count,char *argv[])

{

  struct find_t q ;

  int a;

  if(count==1)

  argv[1]="*.*";

  a = _dos_findfirst(argv[1],1,&q);

  if(a==0)

  {

     while (!a)

   {

    printf("  %s\n", q.name);

    a = _dos_findnext(&q);

    }

  }

  else

  {

  printf("File not found");

  }

}

Step 2: Save the as paras.c (You can give any name)

Step 3: Compile and execute the file.

Step 4: Write click on My computer of 
Window XP operating system and select properties.

Step 5: Select Advanced -> Environment Variables

Step 6: You will find following window:

Click on new button (Button inside the red box)

Step 7: Write following:

Variable name: path

Variable value: c:\tc\bin\paras.c  

(the path where you have saved)

Step 8: Open command prompt and write paras and 
press enter button.

Difference Between uninitialized pointer and null pointer .

An uninitialized pointer is a pointer which points unknown memory location while null pointer is pointer which points a null value or base address of segment. For example: 

int *p;   //Uninitialized pointer

int *q= (int *)0;  //Null pointer

#include<stdio.h>

int *r=NULL;   //Null pointer

What will be output of following c program?

#include<string.h>

#include<stdio.h>

void main(){

    char *p;  //Uninitialized pointer

    char *q=NULL;   //Null pointer;

    strcpy(p,"cquestionbank");

    strcpy(q,"cquestionbank");

    clrscr();

    printf("%s  %s",p,q);

    getch();

}

Output: cquestionbank  (null)

Let Know actual Memory address allocation of variable and initialization in c

Suppose

 int a = 7; Than what ?

Memory representation of: 

signed int a=7;            (In Turbo c compiler) 

signed short int a=7   (Both turbo c and Linux gcc compiler) 

Binary equivalent of data 7 in 16 bit:  00000000 00000111 

Data bit: 0000000 00000111 (Take first 15 bit form right side) 

Sign bit: 0                 (Take leftmost one bit) 

First eight bit of data bit from right side i.e. 00000111 will store in the leftmost byte from right to left side and rest seven bit of data bit i.e. 0000000 will store in rightmost byte from right to left side as shown in the following figure:   

Explain wild pointer in c .

A pointer in c which has not been initialized is known as wild pointer. 

Example: 

What will be output of following c program? 

void main(){ 

int *ptr; 

printf("%u\n",ptr); 

printf("%d",*ptr); 

} 

Output: Any address

Garbage value 

Here ptr is wild pointer because it has not been initialized. 

There is difference between the NULL pointer and wild pointer. Null pointer points the base address of segmentwhile wild pointer doesn’t point any specific memory location.