TOWER OF HANOI C++ CODE

#include <iostream>
#include <conio.h>
using namespace std;

void tower(int a, char from, char aux, char to)
{
if (a == 1)
{
cout << "Move disc 1 from " << from << " to " << to << "\n";
return;
}
else
{
tower(a - 1, from, to, aux);
cout << "Move disc " << a << " from " << from << " to " << to << "\n";
tower(a - 1, aux, from, to);
}
}

void main()
{
int n;
cout << "Tower of Hanoi\n";
cout << "Enter number of discs : ";
cin >> n;
cout << "\n";
tower(n, 'A', 'B', 'C');
getch();
}

OUTPUT

Tower of Hanoi

Enter number of discs : 3

Move disc 1 from A to C

Move disc 2 from A to B

Move disc 1 from C to B

Move disc 3 from A to C

Move disc 1 from B to A

Move disc 2 from B to C

Move disc 1 from A to C

                 WATER JUG PROBLEM C++ CODE-II

#include<iostream>

#include<iomanip>

#include<math.h>

using namespace std;

int xcapacity;

int ycapacity;

void display(int a, int b,int s);

int min(int d, int f)

{

if (d < f)

return d;

else

return f;

}

int steps(int n)

{

int x = 0, y = 0, step = 0;

int temp;

cout << setw(60) << " Vessel A        Vessel B       Steps" << endl;

while (x != n )

{

if (x == 0)

{

x = xcapacity;

step += 1;

cout << "Fill X     "; display(x, y,step);

}

else if (y == ycapacity)

{

y = 0;

step++; 

cout << "Empty Y    "; display(x, y,step);

}

else

{

temp = min(ycapacity - y, x);

y = y + temp;

x = x - temp;

step++;

cout << "Pour X in Y"; display(x, y, step);

}

}

return step;

}

void display(int a, int b,int s)

{

cout << setw(16) << a << setw(15) << b << setw(15)<<s<<endl;

}

void main()

{

int n, ans;

cout << "Enter the liters(GOAL) of water required to be filled in Vessel 1:";

cin >> n;

cout << "Enter the capacity of the vessel: ";

cin >> xcapacity;

cout << "Enter the capacity of the second vesssel  ";

cin >> ycapacity;

ans = steps(n);

cout << "Steps Required:" << ans;

        system("pause");

}

OUTPUT

Enter the liters(GOAL) of water required to be filled in Vessel 1:2

Enter the capacity of the vessel: 4

Enter the capacity of the second vesssel  3

                        Vessel A        Vessel B       Steps

Fill X                    4              0              1

Pour X in Y               1              3              2

Empty Y                   1              0              3

Pour X in Y               0              1              4

Fill X                    4              1              5

Pour X in Y               2              3              6

Steps Required:6Press any key to continue . . .

                                        WATER JUG PROBLEM C++ CODE

#include<iostream>

using namespace std;

#include<conio.h>

int in1[20], in2[20];

void main()

{

float J1, J2, j1 = 0, j2 = 0, j1_max, j2_max, J1_goal;

int i, j, choice, pos = 0;

in1[0] = 4;

in2[0] = 3;

pos++;

in1[pos] = 0;

in2[pos] = 0;

pos++;

cout << "Enter max volume of JUG1: ";

cin >> j1_max;

cout << "Enter max volume of JUG2: ";

cin >> j2_max;

cout << "Enter Goal state for JUG1: ";

cin >> J1_goal;

cout << "\nInitial State: 0   0";

cout << "\nFinal State: " << J1_goal << "\t n";;

cout << "\nJUG1   JUG2 ";

choice = 1;

cout << "\n  " << j1 << " " << j2;

while (1)

{

J1 = j1;

J2 = j2;

switch (choice)

{

case 1:

{

if (j1 < j1_max)

{ j1 = j1_max; }

choice++;

}

break;

case 2:

{

if (j2 < j2_max)

{ j2 = j2_max; }

choice++;

}

break;

case 3:

{

if (j1>0)

{j1 = 0; }

choice++;

}

break;

case 4:

{

if (j2 > 0)

{ j2 = 0;}

choice++;

}

break;

case 5:

{

if ((j1 + j2 >= j1_max) && (j2 > 0))

{

j2 = j2 - (j1_max - j1);

j1 = j1_max;

}

choice++;

}

break;

case 6:

{

if ((j1 + j2 >= j2_max) && (j1 > 0))

{

j1 = j1 - (j2_max - j2);

j2 = j2_max;

}

choice++;

}

break;

case 7:

{

if ((j1 + j2 <= j1_max) && (j2 > 0))

{

j1 = j1 + j2;

j2 = 0;

}

choice++;

}

break;

case 8:

{

if ((j1 + j2 <= j2_max) && (j1 > 0))

{

j2 = j1 + j2;

j1 = 0;

}

choice++;

}

case 9:

{

if ((j2 == J1_goal) && (j1 == 0))

{

j2 = j1;

j1 = J1_goal;

}

choice++;

}

default:

{

choice = 1;

}

}

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

{

if ((in1[i] == j1) && (in2[i] == j2))

{

j1 = J1;

j2 = J2;

break;;

}

}

if (i == pos)

{

in1[pos] = j1;

in2[pos] = j2;

pos++;

cout << "\n  " << j1 << " " << j2;

}

if (j1 == J1_goal)

{

cout << "\nDONE";

exit(1);

}

}

getch();

}

OUTPUT

Enter max volume of JUG1: 4

Enter max volume of JUG2: 3

Enter Goal state for JUG1: 2

Initial State: 0   0

Final State: 2   n

JUG1   JUG2

  0      0

  4      0

  1      3

  0      3

  3      0

  3      3

  4      2

  0      2

  2      0

DONEPress any key to continue . . .

DEPTH FIRST SEARCH (DFS) & BREADTH FIRST SEARCH (BFS) C++ CODE

#include<iostream> 

#include<conio.h>

using namespace std;

int cost[10][10], k, qu[10], front, rare, visit[10], visited[10], i, j, v, stk[10], top,m,n;

void bfs()

{

cout <<"\nEnter no of vertices: ";

cin >> n;

cout <<"Enter no of edges: ";

cin >> m;

cout <<"\nEnter Edges \n";

for (k = 1; k <= m; k++)

{

cin >> i >> j; 

cost[i][j] = 1; //ASSUME COST BETWEN EDGES i & j IS 1

}

cout <<"Enter Initial Vertex: ";

cin >> v;

cout <<"Visited Vertices\n";

cout << v;

visited[v] = 1;

k = 1;

while (k < n)

{

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

if (cost[v][j] != 0 && visited[j] != 1 && visit[j] != 1)

{

visit[j] = 1; qu[rare++] = j;

}

v = qu[front++];

cout << v << " ";

k++;

visit[v] = 0;

visited[v] = 1;

}

}

void dfs()

{

cout << "Enter no of vertices: ";

cin >> n;

cout << "Enter no of edges: ";

cin >> m;

cout << "\nEnter Edges \n";

for (k = 1; k <= m; k++)

{

cin >> i >> j;

cost[i][j] = 1;

}

cout <<"Enter Initial Vertex: ";

cin >> v;

cout <<"Visited Vertices\n";

cout << v << " ";

visited[v] = 1;

k = 1;

while (k < n)

{

for (j = n; j >= 1; j--)

if (cost[v][j] != 0 && visited[j] != 1 && visit[j] != 1)

{

visit[j] = 1;

stk[top] = j;

top++;

}

v = stk[--top];

cout << v << " ";

k++; visit[v] = 0;

visited[v] = 1;

}

}

void main()

{

int choice;

cout <<"1.DFS\n2.BFS";

cout <<"\nSelect Method: ";

cin >> choice;

if (choice == 1)

dfs();

else

bfs();

getch();

}

OUTPUT

1.DFS

2.BFS

Select Method: 1

Enter no of vertices: 6

Enter no of edges: 6

Enter Edges

1

2

1

3

1

4

2

5

2

6

3

6

Enter Initial Vertex: 1

Visited Vertices

1 2 5 6 3 4

1.DFS

2.BFS

Select Method: 2

Enter no of vertices: 6

Enter no of edges: 6

Enter Edges

1

2

1

3

1

4

2

5

2

6

3

6

Enter Initial Vertex: 1

Visited Vertices

12 3 4 5 6

INHERITANCE PROGRAM EXAMPLE


#include<iostream>
using namespace std;
#include<conio.h>
#include<string.h>
class student
{
protected:
int roll_no;
char name[20];
public:
void details()
{
cout<<"enter roll number and name of the student:\n";
cin>>roll_no>>name;
}
};
class subject: public student
{
protected:
int thm,prm;
public:
void tp()
{
cout<<"\ntheory marks: ";
cin>>thm;
cout<<"\npractical marks: ";
cin>>prm;
}
};
class sports
{
protected:
int spm;
public:
void sp()
{
cout<<"\nsports marks: ";
cin>>spm;
}
};
class result:public subject,public sports
{
protected:
int total;
public:
int t()
{
total=thm+prm+spm;
return total;
}
void res()
{
cout<<"\n\nName"<<"      "<<"Roll No."<<"   "<<"Theory"<<"   "
<<"Practical"<<"   "<<"Total"<<endl;
cout<<name<<"    "<<roll_no<<"         "<<thm<<"         "
<<prm<<"         "<<total;
}
};
void main()
{
result r;
r.details();
r.tp();
r.sp();
r.t();
r.res();
getch();
}

Output

24 Namrata

theory marks: 89

practical marks: 45

sports marks: 23

Name      Roll No.   Theory   Practical   Total

Namrata    24         89         45         157

Gauss Elimination Method

# include<stdio.h>

# include<conio.h>

void main()

{

 int i,j,k,n;

 float a[10][10],x[10];

 float s,p; 

 printf("Enter the number of equations : ");

 scanf("%d",&n) ;

 printf("\nEnter the co-efficients of the equations :\n\n");

 for(i=0;i<n;i++)

 {

  for(j=0;j<n;j++)

  {

   printf("a[%d][%d]= ",i+1,j+1);

   scanf("%f",&a[i][j]);

  }

  printf("d[%d]= ",i+1);

  scanf("%f",&a[i][n]);

 }

 for(k=0;k<n-1;k++)

 {

  for(i=k+1;i<n;i++)

  {

   p = a[i][k] / a[k][k] ;

   for(j=k;j<n+1;j++)

    a[i][j]=a[i][j]-p*a[k][j];

  }

 }

 x[n-1]=a[n-1][n]/a[n-1][n-1];

 for(i=n-2;i>=0;i--)

 {

  s=0;

  for(j=i+1;j<n;j++)

  {

   s +=(a[i][j]*x[j]);

   x[i]=(a[i][n]-s)/a[i][i];

  }

 }

 printf("\nThe result is:\n");

 for(i=0;i<n;i++)

  printf("\nx[%d]=%f",i+1,x[i]);

 getch();

}

gauss seidel method

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define e 0.01

void main() {
  int i, j, k, n;
  float a[10][10], x[10];
  float sum, temp, error, big;
  printf("Enter the number of equations: ");
  scanf("%d", & n);
  printf("Enter the co-efficients of the equations: \n");
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n + 1; j++) {
      printf("a[%d][%d]= ", i, j);
      scanf("%f", & a[i][j]);
    }
  }
  for (i = 1; i <= n; i++) {
    x[i] = 0;
  }
  do {
    big = 0;
    for (i = 1; i <= n; i++) {
      sum = 0;
      for (j = 1; j <= n; j++) {
        if (j != i) {
          sum = sum + a[i][j] * x[j];
        }
      }
      temp = (a[i][n + 1] - sum) / a[i][i];
      error = fabs(x[i] - temp);
      if (error > big) {
        big = error;
      }
      x[i] = temp;
      printf("\nx[%d] =%f", i, x[i]);
    }
    printf("\n");
  }
  while (big >= e);
  printf("\n\nconverges to solution");
  for (i = 1; i <= n; i++) {
    printf("\nx[%d]=%f", i, x[i]);
  }
  getch();
}

Enter the number of equations: 3

Enter the co-efficients of the equations:

a[1][1]= 2

a[1][2]= 1

a[1][3]= 1

a[1][4]= 5

a[2][1]= 3

a[2][2]= 5

a[2][3]= 2

a[2][4]= 15

a[3][1]= 2

a[3][2]= 1

a[3][3]= 4

a[3][4]= 8

x[1] =2.500000

x[2] =1.500000

x[3] =0.375000

x[1] =1.562500

x[2] =1.912500

x[3] =0.740625

x[1] =1.173437

x[2] =1.999688

x[3] =0.913359

x[1] =1.043477

x[2] =2.008570

x[3] =0.976119

x[1] =1.007655

x[2] =2.004959

x[3] =0.994933

x[1] =1.000054

x[2] =2.001995

x[3] =0.999474

converges to solution

x[1]=1.000054

x[2]=2.001995

x[3]=0.999474