Evaluation of Binary Expression Tree

Evaluation of Expression Tree

#include<stdio.h>

#include<malloc.h>

#include<math.h>

#include<ctype.h>

struct node

{

            char data;

            struct node *lchild;

            struct node *rchild;

};

typedef struct node * NODE;

NODE getnode()

{

    NODE x;

    x = (NODE)malloc(sizeof(struct node));

    x->lchild =NULL;

    x->rchild = NULL;

    return x;

}

NODE create_tree(char postfix[])

{

            NODE temp, s[70];

            char symb;

            int i, top = -1;

            for(i=0; postfix[i] != '\0' ; i++)

            {

                        symb = postfix[i] ;

                        temp = getnode();

                        temp->data = symb;

                        if(isalnum(symb))

                        {

                                                s[++top] = temp;    //push operand to the stack

                        }

                        else

                        {

                                                temp->rchild = s[top--];    //pop second operand from stack and make it rchild

                                                temp->lchild = s[top--];    // pop first operand from stack and make it lchild

                                                s[++top] = temp;            //push operator node to stack

                        }

            }

            return s[top--];        //return the root of expression tree

}

float evaluate(NODE root)

{

            float num;

            switch(root->data)

            {

                        case '+': return evaluate(root->lchild)+evaluate(root->rchild);

                        case '-': return evaluate(root->lchild)-evaluate(root->rchild);

                        case '*': return evaluate(root->lchild)*evaluate(root->rchild);

                        case '/': return evaluate(root->lchild)/evaluate(root->rchild);

                        case '^':

                        case '$': return  pow(evaluate(root->lchild),evaluate(root->rchild));

                        default: if(isalpha(root->data))

                                    {

                                                printf("\n%c = ", root->data);

                                                scanf("%f", &num);

                                                return num;

                                    }

                                    else

                                                return root->data  - '0';

            }

}

void main()

{

            char postfix[60];

            float res;

            NODE root = NULL;

            printf("\nEnter the postfix expression: ");

            scanf("%s", postfix);

            root = create_tree(postfix);

            res = evaluate(root);

            printf("\nResult of expression is = %f", res);

}

Output:

Case 1:

Enter the postfix expression: 632-5*+1^7+

Result of expression is = 18.000000

Case 2:

Enter the postfix expression: abc-d*+e^f+

e = 1

a = 6

b = 3

c = 2

d = 5

f = 7

Result of expression is = 18.000000

Breadth First Search (BFS)

Breadth First Search (BFS)

#include<stdio.h>

#include<stdlib.h>

void bfs(int v);

int a[50][50], n, visited[50];

int q[20], front = -1,rear = -1;

void creategraph()

{

            int i, j;

            printf("\nEnter the number of vertices in graph:  ");

            scanf("%d",&n);

            printf("\nEnter the adjacency matrix:\n");

            for(i=1; i<=n; i++)

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

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

}

void bfs(int v)

{

            int i, cur;

            visited[v] = 1;

            q[++rear] = v;

            printf("\nNodes reachable from starting vertex %d are: ", v);

            while(front!=rear)

            {

                        cur = q[++front];

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

                        {

                                    if((a[cur][i]==1)&&(visited[i]==0))

                                    {

                                        q[++rear]=i;

                                                visited[i]=1;

                                                printf("%d ", i);

                                    }

                        }

            }

}

int main()

{

            int ch, start, i, j;

            creategraph();

            for(i=1;i<=n;i++)

                        visited[i]=0;

            printf("\nEnter the starting vertex: ");

            scanf("%d", &start);

             bfs(start);

             for(i=1;i<=n;i++)

             {

                                    if(visited[i]==0)

                                                printf("\nThe vertex that is not reachable is %d" ,i);

             }

}

Output:

Enter the number of vertices in graph:  4

Enter the adjacency matrix:

0          1          0          1

0          0          1          0

0          0          0          1

0          0          0          0

Enter the starting vertex: 1

Nodes reachable from starting vertex 1 are: 2 4 3

Enter the number of vertices in graph:  4

Enter the adjacency matrix:

0          1          0          1

0          0          1          0

0          0          0          1

0          0          0          0

Enter the starting vertex: 2

Nodes reachable from starting vertex 2 are: 3 4

The vertex that is not reachable is 1

Depth First Search (DFS)

Depth First Search

#include<stdio.h>

#include<stdlib.h>

void dfs(int v);

int a[50][50], n, visited[50];

int s[20], top = -1, count=0;

void creategraph()

{

            int i, j;

            printf("\nEnter the number of vertices in graph:  ");

            scanf("%d",&n);

            printf("\nEnter the adjacency matrix:\n");

            for(i=1; i<=n; i++)

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

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

}

void dfs(int v)

{

            int i;

            visited[v]=1;

            s[++top] = v;

            for(i=1;i<=n;i++)

            {

                        if(a[v][i] == 1&& visited[i] == 0 )

                        {

                                    dfs(i);

                                    count++;

                        }

            }

}

int main()

{

            int ch, start, i, j;

            creategraph();

            for(i=1;i<=n;i++)

                        visited[i]=0;

            printf("\n Enter the starting vertex:\t");

            scanf("%d", &start);

            dfs(start);

            printf("\nNodes reachable from starting vertex %d are:\n", start);

            for(i=1;i<=count;i++)

                        printf("%d\t", s[i]);

}

Output:

Enter the number of vertices in graph:  4

Enter the adjacency matrix:

0          1          0          1

0          0          1          0

0          0          0          1

0          0          0          0

Enter the starting vertex:     1

Nodes reachable from starting vertex 1 are:

2       3       4

Enter the number of vertices in graph:  4

Enter the adjacency matrix:

0          1          0          1

0          0          1          0

0          0          0          1

0          0          0          0

Enter the starting vertex:     2

Nodes reachable from starting vertex 2 are:

3       4

Radix Sort

C program to sort an array elements using Radix Sort

#include<stdio.h>

#include<stdlib.h>

struct node

{

    int info ;

    struct node *link;

};

typedef struct node * NODE;

NODE bucket[10];

int n, a[20];

int largest()

{

    int i, large = a[0];

    for(i=0;i<n;i++)                        /*First find the largest number in the array*/

    {

        if(a[i] > large)

            large = a[i];

    }

    return large;

}

void insert_rear(int item, int rem)

{

      NODE cur, tmp;

      tmp = (NODE)malloc(sizeof(struct node));

      tmp->info = item;

      tmp->link = NULL;

      if(bucket[rem] == NULL)

      {

            bucket[rem] = tmp;

            return;

      }

      else

      {

           cur = bucket[rem];

           while(cur->link != NULL)

          {

                 cur = cur->link;

           }

           cur->link = tmp;

      }

}

void radix_sort()

{

        NODE p;

        int i, m,k, large, pass = 0, rem, divisor=1, j;

        large = largest();      /*largest number in the array*/

        printf("\nLargest Element is %d", large);

        while(large != 0)           /*Finds number of digits in the largest number*/

        {

            pass++;

            large = large/10 ;

        }

        printf("\nNumber of passes are %d", pass);

        for(m=1; m<=pass; m++)

        {

               printf("\n\n::::Pass %d::::", m);

               for(i=0;i<=9;i++)           /*initialize the buckets: Total 9 buckets*/

                      bucket[i] = NULL;

               for(i=0;i<n;i++)

               {

                     rem = (a[i] / divisor) %10;

                     insert_rear(a[i], rem);

               }

               for(i=0; i<=9 ; i++)

              {

                    printf("\nbucket(%d) -> ",i);

                    p = bucket[i];

                    while(p!=NULL)

                   {

                           printf("%d ",p->info);

                           p = p->link;

                    }

              }

            /*Taking the elements of linked lists in array*/

             i = 0;

             for(k=0;k<=9;k++)

             {

                  p = bucket[k];

                  while(p!=NULL)

                 {

                      a[i++] = p->info;

                      p = p->link;

                 }

            }

            divisor = divisor*10;

        }

}

void main()

{

    int i;

    printf("\nEnter the number of elements in the list : ");

    scanf("%d",&n);

    for(i=0;i<n;i++)

    {

        printf("\nEnter element %d : ",i+1);

        scanf("%d",&a[i]);

    }

    printf("\nUnsorted list is :\n");

    for(i=0;i<n;i++)

        printf("%d ",a[i]);

    radix_sort();

    printf("\nSorted list is :\n");

    for(i=0;i<n;i++)

        printf("%d ",a[i]);

}

Output:

Enter the number of elements in the list : 10

Enter element 1 : 179

Enter element 2 : 208

Enter element 3 : 306

Enter element 4 : 93

Enter element 5 : 859

Enter element 6 : 984

Enter element 7 : 55

Enter element 8 : 9

Enter element 9 : 271

Enter element 10 : 33

Unsorted list is :

179   208   306   93   859   984   55   9   271   33

Largest Element is 984

Number of passes are 3

::::Pass 1::::

bucket(0) ->

bucket(1) -> 271

bucket(2) ->

bucket(3) -> 93 33

bucket(4) -> 984

bucket(5) -> 55

bucket(6) -> 306

bucket(7) ->

bucket(8) -> 208

bucket(9) -> 179 859 9

::::Pass 2::::

bucket(0) -> 306 208 9

bucket(1) ->

bucket(2) ->

bucket(3) -> 33

bucket(4) ->

bucket(5) -> 55 859

bucket(6) ->

bucket(7) -> 271 179

bucket(8) -> 984

bucket(9) -> 93

::::Pass 3::::

bucket(0) -> 9 33 55 93

bucket(1) -> 179

bucket(2) -> 208 271

bucket(3) -> 306

bucket(4) ->

bucket(5) ->

bucket(6) ->

bucket(7) ->

bucket(8) -> 859

bucket(9) -> 984

Sorted list is :

9   33   55   93   179   208   271   306   859   984

Address Calculation Sort

C Program to sort array elements using Address Calculation Sort

#include<stdio.h>
#include<stdlib.h>

struct node
{
       int info;
       struct node *link;
};
typedef struct node * NODE;
NODE head[10];
int n,i,a[20];

int hash_fn(int number,  int k)
{
             /*Find kth digit of the number :example: if its 356 then kth digit will be 3 or if its 29 then kth digit will be 2*/
        int digit, i;
        for(i = 1 ; i <=k ; i++)
        {
              digit = number % 10 ;
              number = number /10 ;
        }
        return digit;       //address
}


/* Finds number of digits in the largest element of the list */
int large_dig()
{
       int large = a[0],ndig = 0 ;
       for(i=0;i<n;i++)                        /*First find the largest number in the array*/
      {
              if(a[i] > large)
                      large = a[i];
       }
       printf("\nLargest Element is %d",large);

      while(large != 0)           /*Finds number of digits in the largest number*/
      {
               ndig++;
               large = large/10 ;
      }
      printf("\nNumber of digits in it are %d", ndig);
      return(ndig);
}


/*Inserts the number in linked list in sorted manner*/
void insert(int num, int addr)
{
      NODE cur ,tmp;
      tmp = (NODE)malloc(sizeof(struct node));
      tmp->info = num;

      /*list empty or item to be added in beginning */
       if(head[addr] == NULL || num < head[addr]->info)
      {
            tmp->link = head[addr];
            head[addr] = tmp;
            return;
      }
     else
     {
            cur = head[addr];
            while(cur->link != NULL && cur->link->info < num)
                      cur = cur->link;
            tmp->link = cur->link;
            cur->link = tmp;
     }
}

void addr_sort()
{
       int i, k, dig, addr;
       NODE p;
       k = large_dig();        /*number of digits in the largest number*/
       for(i=0;i<=9;i++)
               head[i]=NULL;
       for(i=0;i<n;i++)
      {
               addr = hash_fn(a[i], k);
               insert(a[i], addr);
       }

      for(i=0; i<=9 ; i++)
      {
              printf("\nhead(%d) -> ", i);
              p = head[i];
              while(p!=NULL)
             {
                    printf("%d ", p->info);
                    p = p->link;
              }
      }
     /*Taking the elements of linked lists in array*/
      i = 0;
      for(k=0;k<=9;k++)
      {
               p = head[k];
               while(p!=NULL)
              {
                       a[i++] = p->info;
                       p = p->link;
               }
      }
}

void main()
{
      int i;
      printf("\nEnter the number of elements in the list : ");
      scanf("%d", &n);
      for(i=0;i<n;i++)
      {
             printf("\nEnter element %d : ",i+1);
             scanf("%d",&a[i]);
      }

      printf("\nUnsorted list is :\n");
      for(i=0;i<n;i++)
            printf("%d ",a[i]);

      addr_sort();
      printf("\nSorted list is :\n");
      for(i=0;i<n;i++)
              printf("%d ",a[i]);
}


Output:

Enter the number of elements in the list : 15
Enter element 1 : 12
Enter element 2 : 345
Enter element 3 : 567
Enter element 4 : 367
Enter element 5 : 789
Enter element 6 : 23
Enter element 7 : 15
Enter element 8 : 56
Enter element 9 : 567
Enter element 10 : 899
Enter element 11 : 123
Enter element 12 : 321
Enter element 13 : 678
Enter element 14 : 67
Enter element 15 : 45
Unsorted list is :
12   345   567  367   789   23   15   56   567   899   123   321   678   67   45
Largest Element is 899
Number of digits in it are 3

head(0) -> 12  15  23  45  56  67
head(1) -> 123
head(2) ->
head(3) -> 321  345  367
head(4) ->
head(5) -> 567  567
head(6) -> 678
head(7) -> 789
head(8) -> 899
head(9) ->
Sorted list is :
12   15   23   45   56   67   123   321   345   367   567   567   678   789  899

Insertion Sort

C Program to sort the elements using Insertion Sort

#include<stdio.h>
int n, a[20];

void insertion_sort()
{
       int i, j, temp;
       for(i=1; i<n ;i++)
      {
              temp = a[i];
              j = i-1;
              while(j>=0 &&  temp < a[j])
             {
                    a[j+1] = a[j];
                    j = j-1;
             }
             a[j+1] = temp;
       }
}

void main()
{
       int i;
       printf("\nEnter the number of elements: ");
       scanf("%d",&n);

      printf("\nEnter the array elements:");
      for(i=0;i<n;i++)
           scanf("%d", &a[i]);

      insertion_sort();

     printf("\nThe array elements after insertion sort: ");
     for(i=0;i<n;i++)
            printf("%d ", a[i]);
}

Output:
Enter the number of elements: 7
Enter the array elements:  89     45      68       90       29       34       17
The array elements after insertion sort: 17       29       34         45        68        89       90

Lab Program 9 Polynomial 15CSL38 Data Structures in C Lab

Lab Program 9

  

Design, Develop and Implement a Program in C for the following operations on Singly Circular Linked List (SCLL) with header nodes.

a. Represent and Evaluate a Polynomial P(x,y,z) = 6x2y2z - 4yz5+3x3yz+2xy5z- 2xyz3

b. Find the sum of two polynomials POLY1(x,y,z) and POLY2(x,y,z) and store the result in POLYSUM(x,y,z)

Support the program with appropriate functions for each of the above operations

#include<stdio.h>

#include<stdlib.h>

#include<math.h>

#define COMPARE(x, y)       ( (x == y) ? 0 : (x > y) ? 1 : -1)

struct node

{

             int coef;

             int xexp, yexp, zexp;

             struct node *link;

};

typedef struct node *NODE;

NODE getnode()

{

            NODE x;

            x = (NODE) malloc(sizeof(struct node));

            if(x == NULL)

            {

                        printf("Running out of memory \n");

                        return NULL;

            }

            return x;

}

NODE attach(int coef, int xexp, int yexp, int zexp, NODE head)

{

            NODE temp, cur;

            temp = getnode();

            temp->coef = coef;

            temp->xexp = xexp;

            temp->yexp = yexp;

            temp->zexp = zexp;

            cur = head->link;

            while(cur->link != head)

            {

                        cur = cur->link;

            }

            cur->link = temp;

            temp->link = head;

            return head;

}

NODE read_poly(NODE head)

{

            int i, j, coef, xexp, yexp, zexp, n;

            printf("\nEnter the no of terms in the polynomial: ");

            scanf("%d", &n);

            for(i=1; i<=n; i++)

            {

                        printf("\n\tEnter the %d term: ",i);

                        printf("\n\t\tCoef =  ");

                        scanf("%d", &coef);

                        printf("\n\t\tEnter Pow(x) Pow(y) and Pow(z): ");

                        scanf("%d", &xexp);

                        scanf("%d", &yexp);

                        scanf("%d", &zexp);

                        head = attach(coef, xexp, yexp, zexp, head);

            }

              return head;

}

void display(NODE head)

{

            NODE temp;

            if(head->link == head)

            {

                         printf("\nPolynomial does not exist.");

                         return;

            }

            temp = head->link;

            while(temp != head)

            {

                         printf("%dx^%dy^%dz^%d", temp->coef, temp->xexp, temp->yexp, temp->zexp);

                         temp = temp->link;

                         if(temp != head)

                                    printf(" + ");

            }

}

int poly_evaluate(NODE head)

{

            int x, y, z, sum = 0;

            NODE poly;

            printf("\nEnter the value of x,y and z: ");

            scanf("%d %d %d", &x, &y, &z);

            poly = head->link;

            while(poly != head)

            {

                        sum += poly->coef * pow(x,poly->xexp)* pow(y,poly->yexp) * pow(z,poly->zexp);

                        poly = poly->link;

            }

            return sum;

}

NODE poly_sum(NODE head1, NODE head2, NODE head3)

{

    NODE a, b;

    int coef;

    a = head1->link;

    b = head2->link;

    while(a!=head1 && b!=head2)

    {

            while(1)

            {

                        if(a->xexp == b->xexp && a->yexp == b->yexp && a->zexp == b->zexp)

                        {

                                    coef = a->coef + b->coef;

                                    head3 = attach(coef, a->xexp, a->yexp, a->zexp, head3);

                                    a = a->link;

                                    b = b->link;

                                    break;

                        } //if ends here

                        if(a->xexp!=0 || b->xexp!=0)

                        {

                                    switch(COMPARE(a->xexp, b->xexp))

                                    {

                                     case -1 :  head3 = attach(b->coef, b->xexp, b->yexp, b->zexp, head3);

                                                    b = b->link;

                                                    break;

                                    case 0 :    if(a->yexp > b->yexp)

                                                    {

                                                            head3 = attach(a->coef, a->xexp, a->yexp, a->zexp, head3);

                                                            a = a->link;

                                                            break;

                                                     }

                                                     else if(a->yexp < b->yexp)

                                                     {

                                                             head3 = attach(b->coef, b->xexp, b->yexp, b->zexp, head3);

                                                             b = b->link;

                                                             break;

                                                     }

                                                    else if(a->zexp > b->zexp)

                                                    {

                                                            head3 = attach(a->coef, a->xexp, a->yexp, a->zexp, head3);

                                                            a = a->link;

                                                            break;

                                                    }

                                                    else if(a->zexp < b->zexp)

                                                   {

                                                            head3 = attach(b->coef, b->xexp, b->yexp, b->zexp, head3);

                                                            b = b->link;

                                                            break;

                                                   }

                                    case 1 :    head3 = attach(a->coef,a->xexp,a->yexp,a->zexp,head3);

                                                    a = a->link;

                                                    break;

                                    }  //switch ends here

                                    break;

                        } //if ends here

                        if(a->yexp!=0 || b->yexp!=0)

                        {

                                    switch(COMPARE(a->yexp, b->yexp))

                                    {

                                          case -1 :   head3 = attach(b->coef, b->xexp, b->yexp, b->zexp, head3);

                                                          b = b->link;

                                                          break;

                                           case 0 :   if(a->zexp > b->zexp)

                                                          {

                                                                 head3 = attach(a->coef, a->xexp, a->yexp, a->zexp, head3);

                                                                 a = a->link;

                                                                 break;

                                                            }

                                                            else if(a->zexp < b->zexp)

                                                            {

                                                                 head3 = attach(b->coef, b->xexp, b->yexp, b->zexp, head3);

                                                                 b = b->link;

                                                                 break;

                                                            }

                                            case 1 :    head3 = attach(a->coef, a->xexp, a->yexp, a->zexp, head3);

                                                            a = a->link;

                                                            break;

                                    }

                                    break;

                        }

                        if(a->zexp!=0 || b->zexp!=0)

                        {

                                    switch(COMPARE(a->zexp,b->zexp))

                                    {

                                                case -1 :   head3 = attach(b->coef,b->xexp,b->yexp,b->zexp,head3);

                                                                b = b->link;

                                                                break;

                                                case 1 :    head3 = attach(a->coef, a->xexp, a->yexp, a->zexp, head3);

                                                                a = a->link;

                                                                break;

                                    }

                                    break;

                        }

            }

    }

    while(a!= head1)

    {

             head3 = attach(a->coef,a->xexp,a->yexp,a->zexp,head3);

            a = a->link;

    }

    while(b!= head2)

    {

            head3 = attach(b->coef,b->xexp,b->yexp,b->zexp,head3);

            b = b->link;

    }

    return head3;

}

void main()

{

  NODE head, head1, head2, head3;

  int res, ch;

  head =  getnode();     /* For polynomial evalaution */

  head1 = getnode();    /* To hold POLY1 */

  head2 = getnode();    /* To hold POLY2 */

  head3 = getnode();    /* To hold POLYSUM */

  head->link=head;

  head1->link=head1;

  head2->link=head2;

  head3->link= head3;

  while(1)

  {

        printf("\n~~~Menu~~~");

        printf("\n1.Represent and Evaluate a Polynomial P(x,y,z)");

        printf("\n2.Find the sum of two polynomials POLY1(x,y,z)");

        printf("\nEnter your choice:");

        scanf("%d",&ch);

        switch(ch)

        {

            case 1:             printf("\n~~~~Polynomial evaluation P(x,y,z)~~~\n");

                                    head = read_poly(head);

                                    printf("\nRepresentation of Polynomial for evaluation: \n");

                                    display(head);

                                    res = poly_evaluate(head);

                                    printf("\nResult of polynomial evaluation is : %d \n", res);

                                    break;

            case 2:             printf("\nEnter the POLY1(x,y,z):  \n");

                                    head1 = read_poly(head1);

                                    printf("\nPolynomial 1 is:  \n");

                                    display(head1);

                                    printf("\nEnter the POLY2(x,y,z):  \n");

                                    head2 = read_poly(head2);

                                    printf("\nPolynomial 2 is: \n");

                                    display(head2);

                                    printf("\nPolynomial addition result: \n");

                                    head3 = poly_sum(head1,head2,head3);

                                    display(head3);

                                    break;

                case 3:         exit(0);

        }

    }

}

Output:

~~~Menu~~~

1.Represent and Evaluate a Polynomial P(x,y,z)

2.Find the sum of two polynomials POLY1(x,y,z) and POLY2(x,y,z)

Enter your choice: 1

~~~~Polynomial evaluation P(x,y,z)~~~

Enter the no of terms in the polynomial: 5

        Enter the 1 term:

                Coef =  6

                Enter Pow(x) Pow(y) and Pow(z): 2          2          1

        Enter the 2 term:

                Coef =  -4

                Enter Pow(x) Pow(y) and Pow(z): 0          1          5

        Enter the 3 term:

                Coef =  3

                Enter Pow(x) Pow(y) and Pow(z): 3          1          1

        Enter the 4 term:

                Coef =  2

                Enter Pow(x) Pow(y) and Pow(z): 1          5          1

        Enter the 5 term:

                Coef =  -2

                Enter Pow(x) Pow(y) and Pow(z): 1          1          3

Representation of Polynomial for evaluation:

6x^2y^2z^1 + -4x^0y^1z^5 + 3x^3y^1z^1 + 2x^1y^5z^1 + -2x^1y^1z^3

Enter the value of x,y and z: 1  1  1

Result of polynomial evaluation is : 5

~~~Menu~~~

1.Represent and Evaluate a Polynomial P(x,y,z)

2.Find the sum of two polynomials POLY1(x,y,z) and POLY2(x,y,z)

Enter your choice: 2

Enter the POLY1(x,y,z):

Enter the no of terms in the polynomial: 5

        Enter the 1 term:

                Coef =  6

                Enter Pow(x) Pow(y) and Pow(z): 4          4          4

        Enter the 2 term:

                Coef =  3

                Enter Pow(x) Pow(y) and Pow(z): 4          3          1

        Enter the 3 term:

                Coef =  5

                Enter Pow(x) Pow(y) and Pow(z): 0          1          1

        Enter the 4 term:

                Coef =  10

                Enter Pow(x) Pow(y) and Pow(z): 0          1          0

        Enter the 5 term:

                Coef =  5

                Enter Pow(x) Pow(y) and Pow(z): 0          0          0

Polynomial 1 is:

6x^4y^4z^4 + 3x^4y^3z^1 + 5x^0y^1z^1 + 10x^0y^1z^0 + 5x^0y^0z^0

Enter the POLY2(x,y,z):

Enter the no of terms in the polynomial: 5

        Enter the 1 term:

                Coef =  8

                Enter Pow(x) Pow(y) and Pow(z): 4          4          4

        Enter the 2 term:

                Coef =  4

                Enter Pow(x) Pow(y) and Pow(z): 4          2          1

        Enter the 3 term:

                Coef =  30

                Enter Pow(x) Pow(y) and Pow(z): 0          1          0

        Enter the 4 term:

                Coef =  20

                Enter Pow(x) Pow(y) and Pow(z): 0          0          1

        Enter the 5 term:

                Coef =  3

                Enter Pow(x) Pow(y) and Pow(z): 0          0          0

Polynomial 2 is:

8x^4y^4z^4 + 4x^4y^2z^1 + 30x^0y^1z^0 + 20x^0y^0z^1 + 3x^0y^0z^0

Polynomial addition result:

14x^4y^4z^4 + 3x^4y^3z^1 + 4x^4y^2z^1 + 5x^0y^1z^1 + 40x^0y^1z^0 + 20x^0y^0z^1 + 8x^0y^0z^0

~~~Menu~~~

1.Represent and Evaluate a Polynomial P(x,y,z)

2.Find the sum of two polynomials POLY1(x,y,z) and POLY2(x,y,z)

Enter your choice:3

Credits,

Manoj T