Q1. Write a program to search a number from a given list using linear and binary search.

#include<stdio.h>#inlcude<conio.h>void main(){int arr[10];int i,j;int nint snumber;int count=0;

printf("Enter the number of element in array");scanf("%d",&n);printf("\nEnter the element in array");for(i=0;i<n; i++){scanf("%d",&arr[i]);}printf("\nenter the search element");scanf("%d",&snumber);for(i=0;i<n;i++){

if(arr[i]==snumber){

count=1;printf("value is found");break;

}}if(count==0)printf("value is not found");getch();}

1

Output:-Enter the number of element in array 5

Enter the element in array 37895

enter the search element 9value is found

Enter the number of element in array 5

Enter the element in array37894

enter the search element 10value is not found

2

Q2. Write a program to search a number from a given list using binary search recursively.

#include<stdio.h>#include<conio.h>binary(int,int,int);int b[25];

void main(){

int high,low=0,i,j,value,pos,n;clrscr();printf("Enter n number of item");scanf("%d",&n);high=n;printf("Enter the value in sorted form\n");for(i=0;i<n;i++){

scanf("%d",&b[i]);}printf("Enter the value which you have to find");scanf("%d",&value);pos=binary(low,high,value);//printf("\n in location %d th",pos);getch();

}int binary(int low,int high,int value){

if(low==high){

if( b[(low+high)/2]==value){

printf("value is found");return (low+high)/2;

}else{

printf("Value is not found");return;

}}

if( b[(low+high)/2]>value)

3

binary(low,(low+high)/2,value);else if( b[(low+high)/2]<value)

binary(((low+high)/2)+1,high,value);else{

printf("value is found");return(low+high)/2;

}

}

Output

Enter n number of item5Enter the value in sorted form1 2 3 4 5Enter the value which you have to find3value is found

4

Q3. Write the program to find the factorial of number using both recursive and non-recursive method and also compare that performance.

#include<stdio.h>#include<conio.h>void main(){int number;int i,fact=1;clrscr();printf("enter the number");scanf("%d",&number);for(i=number;i>1;i--){ fact=fact*i;}printf("\nfactorial number of %d=%d",number,fact);getch();}

Output

enter the number 5

factorial number of 5=120

(with recursion)

#include<stdio.h>#include<conio.h>int fact(int);int factn=1;void main(){int number;int i,factorial;

clrscr();printf("enter the number");scanf("%d",&number);factorial=fact(number);printf("\nfactorial number of %d=%d",number,factorial);getch();}int fact(int number)

5

{if(number==1)return 1;elsefactn=number*fact(number-1);return (factn);

}

Output

enter the number 5

factorial number of 5=120

6

Q4. Sort the following list in increasing order of number 9,94,45,47,28,98,65,42,8,4,88,6 using selection sort.

#include<stdio.h>#include<conio.h>void main(){int sort[15];int i,j;int n;int temp;clrscr();printf("enter the number of element in array");scanf("%d",&n);printf("\nenter the value in array");for(i=0;i<n;i++)scanf("%d",&sort[i]);

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

for(j=i+1;j<n;j++){ if(sort[i]>sort[j]) {

temp=sort[i];sort[i]=sort[j];sort[j]=temp;

}}

}printf("\nSorted array:");for(i=0;i<n;i++)printf("\n%d",sort[i]);getch();}

7

Output

enter the number of element in array 12

enter the value in array 9 94 45 47 28 98 65 42 8 4 88 6

Sorted array:46892842454765889498

8

Q5. Write a program to sort a list using merge sort by applying divide and conquer method .

#include<stdio.h>#include<conio.h>int a[100];void merge_sort(int a[],int left,int right);void merge(int a[],int lb,int le,int rb,int re);void main(){int i,num;clrscr();printf("enter the num\n");scanf("%d",&num);printf("Enter the array\n");for(i=0;i<num;i++)scanf("%d",&a[i]);

merge_sort(a,0,num-1);printf("\nSorted list is:\n");for(i=0;i<num;i++)printf("%d\n",a[i]);getch();

}void merge_sort(int a[],int left,int right){

int mid;if(right>left){

mid=(right+left)/2;merge_sort(a,left,mid);merge_sort(a,mid+1,right);merge(a,left,mid,mid+1,right);

}}void merge(int a[],int lb,int le,int rb,int re){

int na,nb,nc,k,c[30];na=lb;nb=rb;nc=lb;while((na<=le)&&(nb<=re)){

if(a[na]<a[nb])c[nc++]=a[na++];

elsec[nc++]=a[nb++];

9

}if(na>le){

while(nb<=re)c[nc++]=a[nb++];

}else{

while(na<=le)c[nc++]=a[na++];

}for(k=lb;k<=re;k++)a[k]=c[k];

}

Output

enter the num5Enter the array54123

Sorted list is:12345

10

Q6. Write a program to sort a list using quick sort by applying divide and conquer method .

#include "stdio.h"

int split ( int*, int, int ) ;void quicksort ( int *, int, int ) ;void main( ){

int arr[10];int i ;int n;printf("enter the number of element in array");scanf("%d",&n);printf("Enter the value of array");for(i=0;i<n;i++)scanf("%d",&arr[i]);

quicksort ( arr, 0, n ) ;

printf ( "\nArray after sorting:\n") ;

for ( i = 0 ; i < n ; i++ )printf ( "%d\t", arr[i] ) ;

getch();}

void quicksort ( int a[ ], int lower, int upper ){

int i ;if ( upper > lower ){

i = split ( a, lower, upper ) ;quicksort ( a, lower, i - 1 ) ;quicksort ( a, i + 1, upper ) ;

}}

int split ( int a[ ], int lower, int upper ){

int i, p, q, t ;

p = lower + 1 ;q = upper ;

11

i = a[lower] ;

while ( q >= p ){

while ( a[p] < i )p++ ;

while ( a[q] > i )q-- ;

if ( q > p ){

t = a[p] ;a[p] = a[q] ;a[q] = t ;

}}

t = a[lower] ;a[lower] = a[q] ;a[q] = t ;

return q ;}

OUTPUTenter the number of element in array 5Enter the value of array 1 4 3 5 2

Array after sorting:1 2 3 4 5

12

Q7. Write a program to perform product of two matrix of order nxn using strassen’s multiplication method.

#include<stdio.h>#include<conio.h>#include<stdlib.h>void main(){int a[4][4],b[4][4],c[4][4],d[7],e[7],f[7],g[7],i,j,choice;do{printf("\n1. 4*4 matrix.");printf("\n2. 2*2 matrix.");printf("\n3. Exit.");printf("\nEnter your choice : ");scanf("%d",&choice);switch(choice){case 1:{printf("\nEnter value for matrix A : ");for(i=0;i<4;i++)for(j=0;j<4;j++)scanf("%d",&a[i][j]);printf("\nEnter value for matrix B : ");for(i=0;i<4;i++)for(j=0;j<4;j++)scanf("%d",&b[i][j]);d[0]=(a[0][0]+a[1][1])*(b[0][0]+b[1][1]);d[1]=(a[1][0]+a[1][1])*b[0][0];d[2]=a[0][0]*(b[0][1]-b[1][1]);d[3]=a[1][1]*(b[1][0]-b[0][0]);d[4]=(a[0][0]+a[0][1])*b[1][1];d[5]=(a[1][0]-a[0][0])*(b[0][0]+b[0][1]);d[6]=(a[0][1]-a[1][1])*(b[1][0]-b[1][1]);e[0]=(a[0][2]+a[1][3])*(b[0][2]+b[1][3]);e[1]=(a[1][2]+a[1][3])*b[0][2];e[2]=a[0][2]*(b[0][3]-b[1][3]);e[3]=a[1][3]*(b[1][2]-b[0][2]);e[4]=(a[0][2]+a[0][3])*b[1][3];e[5]=(a[1][2]-a[0][2])*(b[0][2]+b[0][3]);e[6]=(a[0][3]-a[1][3])*(b[1][2]-b[1][3]);f[0]=(a[2][0]+a[3][1])*(b[2][0]+b[3][1]);f[1]=(a[3][0]+a[3][1])*b[2][0];f[2]=a[2][0]*(b[2][1]-b[3][1]);f[3]=a[3][1]*(b[3][0]-b[2][0]);f[4]=(a[2][0]+a[2][1])*b[3][1];f[5]=(a[3][0]-a[2][0])*(b[2][0]+b[2][1]);

13

f[6]=(a[2][1]-a[3][1])*(b[3][0]-b[3][1]);g[0]=(a[2][2]+a[3][3])*(b[2][2]+b[3][3]);g[1]=(a[3][2]+a[3][3])*b[2][2];g[2]=a[2][2]*(b[2][3]-b[3][3]);g[3]=a[3][3]*(b[3][2]-b[2][2]);g[4]=(a[2][2]+a[2][3])*b[3][3];g[5]=(a[3][2]-a[2][2])*(b[2][2]+b[2][3]);g[6]=(a[2][3]-a[3][3])*(b[3][2]-b[3][3]);c[0][0]=d[0]+d[3]-d[4]+d[6];c[0][1]=d[2]+d[4];c[1][0]=d[1]+d[3];c[1][1]=d[0]+d[2]-d[1]+d[5];c[0][2]=e[0]+e[3]-e[4]+e[6];c[0][3]=e[2]+e[4];c[1][2]=e[1]+e[3];c[1][3]=e[0]+e[2]-e[1]+e[5];c[2][0]=f[0]+f[3]-f[4]+f[6];c[2][1]=f[2]+f[4];c[3][0]=f[1]+f[3];c[3][1]=f[0]+f[2]-f[1]+f[5];c[2][2]=g[0]+g[3]-g[4]+g[6];c[2][3]=g[2]+g[4];c[3][2]=g[1]+g[3];c[3][3]=g[0]+g[2]-g[1]+g[5];printf("\nMatrix multiplication is : \n");for(i=0;i<4;i++){for(j=0;j<4;j++){printf("%d",c[i][j]);}printf("\n");}break;}case 2:{printf("\nEnter value for matrix A : ");for(i=0;i<2;i++)for(j=0;j<2;j++)scanf("%d",&a[i][j]);printf("\nEnter value for matrix B : ");for(i=0;i<2;i++)for(j=0;j<2;j++)scanf("%d",&b[i][j]);d[0]=(a[0][0]+a[1][1])*(b[0][0]+b[1][1]);d[1]=(a[1][0]+a[1][1])*b[0][0];d[2]=a[0][0]*(b[0][1]-b[1][1]);d[3]=a[1][1]*(b[1][0]-b[0][0]);d[4]=(a[0][0]+a[0][1])*b[1][1];

14

d[5]=(a[1][0]-a[0][0])*(b[0][0]+b[0][1]);d[6]=(a[0][1]-a[1][1])*(b[1][0]-b[1][1]);c[0][0]=d[0]+d[3]-d[4]+d[6];c[0][1]=d[2]+d[4];c[1][0]=d[1]+d[3];c[1][1]=d[0]+d[2]-d[1]+d[5];printf("\nMatrix multiplication is : \n");for(i=0;i<2;i++){for(j=0;j<2;j++){printf("%d",c[i][j]);}printf("\n");}break;}case 3:{exit(0);}default:{printf("\nInvalid chocie.");printf("\nSelect value from 1-4...");}}}while(choice!=3);getch();}

OUTPUT

15

Q8. Write a program to find minimum spanning tree using Kruskal’s method.

#include<stdio.h>#define INF 1000char vertex[10];int wght[10][10];int span_wght[10][10];int source;struct Sort{int v1,v2;int weight;}que[20];int n,ed,f,r;int cycle(int s,int d){int j,k;if(source==d)return 1;for(j=0;j<n;j++)if(span_wght[d][j]!=INF && s!=j){if(cycle(d,j))return 1;}return 0;}void build_tree(){int i,j,w,k,count=0;for(count=0;count<n;f++){i=que[f].v1;j=que[f].v2;w=que[f].weight;span_wght[i][j]=span_wght[j][i]=w;source=i;k=cycle(i,j);if(k)span_wght[i][j]=span_wght[j][i]=INF;elsecount++;}}void swap(int *i,int *j){int t;t=*i;*i=*j;*j=t;}

16

void main(){int i,j,k=0,temp;int sum=0;clrscr();printf("\n\tEnter the No. of Nodes : ");scanf("%d",&n);for(i=0;i<n;i++){printf("\n\tEnter %d value : ",i+1);fflush(stdin);scanf("%c",&vertex[i]);for(j=0;j<n;j++){wght[i][j]=INF;span_wght[i][j]=INF;}}printf("\n\nGetting Weight\n");for(i=0;i<n;i++)for(j=i+1;j<n;j++){printf("\nEnter 0 if path Doesn't exist between %c to %c : ",vertex[i],vertex[j]);scanf("%d",&ed);if(ed>=1){wght[i][j]=wght[j][i]=ed;que[r].v1=i;que[r].v2=j;que[r].weight=wght[i][j];if(r){for(k=0;k<r;k++)if(que[k].weight>que[r].weight){swap(&que[k].weight,&que[r].weight);swap(&que[k].v1,&que[r].v1);swap(&que[k].v2,&que[r].v2);}}r++;}}build_tree();printf("\n\t\tLIST OF EDGES\n\n");for(i=0;i<n;i++)for(j=i+1;j<n;j++)if(span_wght[i][j]!=INF){printf("\n\t\t%c ------ %c = %d ",vertex[i],vertex[j],span_wght[i][j]);sum+=span_wght[i][j];

17

}printf("\n\n\t\tTotal Weight : %d ",sum);getch();}

OUTPUT

18

Q9. Write a program to find minimum spanning tree using Prim’s method .

Code.#include <stdio.h>

int n; int weight[100][100]; char inTree[100]; int d[100]; int whoTo[100];

void updateDistances(int target) {int i;for (i = 0; i < n; ++i)

if ((weight[target][i] != 0) && (d[i] > weight[target][i])) {d[i] = weight[target][i];whoTo[i] = target;

}}

int main(int argc, char *argv[]) {FILE *f = fopen("dist.txt", "r");fscanf(f, "%d", &n);int i, j;for (i = 0; i < n; ++i)

for (j = 0; j < n; ++j)fscanf(f, "%d", &weight[i][j]);

fclose(f);

/* Initialise d with infinity */for (i = 0; i < n; ++i)

d[i] = 100000;

/* Mark all nodes as NOT beeing in the minimum spanning tree */for (i = 0; i < n; ++i)

inTree[i] = 0;

/* Add the first node to the tree */printf("Adding node %c\n", 0 + 'A');inTree[0] = 1;updateDistances(0);

int total = 0;int treeSize;for (treeSize = 1; treeSize < n; ++treeSize) {

/* Find the node with the smallest distance to the tree */

19

int min = -1;for (i = 0; i < n; ++i)

if (!inTree[i])if ((min == -1) || (d[min] > d[i]))

min = i;

/* And add it */printf("Adding edge %c-%c\n", whoTo[min] + 'A', min + 'A');inTree[min] = 1;total += d[min];

updateDistances(min);}

printf("Total distance: %d\n", total);getch();

return 0;}

20

OUTPUT

21

Q10. Write a program to find single shortest path using dijekstra algorithm.

#include<iostream>#include<conio.h>#include<stdio.h>using namespace std;int shortest(int ,int);int cost[10][10],dist[20],i,j,n,k,m,S[20],v,totcost,path[20],p;main(){int c;cout <<"enter no of vertices";cin >> n;cout <<"enter no of edges"; cin >>m;cout <<"\nenter\nEDGE Cost\n";for(k=1;k<=m;k++){cin >> i >> j >>c;cost[i][j]=c;}for(i=1;i<=n;i++)for(j=1;j<=n;j++)if(cost[i][j]==0)cost[i][j]=31999;cout <<"enter initial vertex";cin >>v;cout << v<<"\n";shortest(v,n); } int shortest(int v,int n){int min;for(i=1;i<=n;i++){S[i]=0;dist[i]=cost[v][i];}path[++p]=v;S[v]=1;dist[v]=0;for(i=2;i<=n-1;i++){k=-1;min=31999;

22

for(j=1;j<=n;j++){if(dist[j]<min && S[j]!=1){min=dist[j];k=j;} }if(cost[v][k]<=dist[k])p=1;path[++p]=k;for(j=1;j<=p;j++)cout<<path[j];cout <<"\n";//cout <<k;S[k]=1;for(j=1;j<=n;j++)if(cost[k][j]!=31999 && dist[j]>=dist[k]+cost[k][j] && S[j]!=1) dist[j]=dist[k]+cost[k][j];}}

OUTPUT

Enter no of vertices6

enter no of edges11

enter

EDGE Cost

1 2 50

1 3 45

1 4 10

2 3 10

2 4 15

3 5 30

4 1 10

4 5 15

5 2 20

5 3 35

6 5 3

enter initial vertex 1

1

14

145

1452

13

23

Q11. Write a program to knapsack problem.

# include<stdio.h># include<conio.h>

void knapsack(int , float[], float[], float);

void main(){ float weight[20], profit[20], capacity; int n, i ,j; float ratio[20], temp; clrscr();

printf ("\n Enter the no of objects:- "); scanf ("%d", &n);

printf ("\n Enter the profits and weight of each object:- "); for (i=0; i<n; i++) {

scanf("%f %f", &profit[i],&weight[i]); }

printf ("\n Enter the capacity of knapsack:- "); scanf ("%f", &capacity);

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

ratio[i]=profit[i]/weight[i]; }

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

if(ratio[i]<ratio[j]) {

temp= ratio[j];ratio[j]= ratio[i];ratio[i]= temp;

temp= weight[j];weight[j]= weight[i];weight[i]= temp;

24

temp= profit[j];profit[j]= profit[i];profit[i]= temp;

} } }

knapsack(n, weight, profit, capacity); getch(); }

void knapsack(int n, float weight[], float profit[], float capacity) { float x[20], tp=0 ; int i, j, u; u=capacity;

for (i=0;i<n;i++)x[i]=0.0;

for (i=0;i<n;i++) { if(weight[i]>u)

break; else

{x[i]=1.0;tp= tp+profit[i];u=u-weight[i];}

}

if(i<n) x[i]=u/weight[i];

tp= tp + (x[i]*profit[i]);

printf("\n The result vector is:- "); for(i=0;i<n;i++)

printf("%f\t",x[i]);

printf("\n Maximum profit is:- %f", tp);

}

25

Output

Enter the no of objects:- 3

Enter the profits and weight of each object:- 25 1824 1515 10

Enter the capacity of knapsack:- 20

The result vector is:- 1.000000 0.500000 0.000000 Maximum profit is:- 31.500000

26

Q12. Write a program to find the optimal sequence of performing the jobs to get maximum profit.

void JS(int d[],int j[],int n,int p[]){

int profit=0,i,q,m;int k=1,r;j[1]=1;for(i=1;i<=n;i++){ r=k; while((d[j[r]]>d[i]) && (d[j[r]]!=r)) r=r-1; if(d[j[r]]<=d[i] && d[i]>r)

{for(q=(k);q>=(r+1);q--){

if(q==-1)break;

j[q+1]=j[q]; }

j[r+1]=i;k=k+1;

}}

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

profit+=p[m];}printf("\n MAXIMUM PROFIT: %d",profit);

}

void main(){

int q,n,p[10],d[10],i,j[10];clrscr();printf("\n enter no of processes : ");scanf("%d",&n);d[0]=j[0]=0;for(i=1;i<=n;i++){

printf("\n enter profit of process %d:",i);scanf("%d",&p[i]);

}

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

printf("\n enter deadline ");scanf("%d",&d[i]);j[i]=0;

}for(i=n-1;i>0;i--)for(q=1;q<=i;q++)if(p[q]<p[q+1])

27

{int t;t=p[q];p[q]=p[q+1];p[q+1]=t;

}printf("\nProfits : ");for(i=1;i<=n;i++)printf(" %d",p[i]);JS(d,j,n,p);

getch();}

OUTPUT

28

Q13. Write a program to place the queen on a square chessboard using Backtracking approach.

# include <stdio.h># include <stdlib.h># include <time.h>int N; //For N * N ChessBoardint flag;void printArray(int a[]); /* Just to Print the Final Solution */void getPositions(int a[],int n1,int n2); /* The Recursive Function */int main(){int *a; int ctr=0; printf("\nTHE N QUEENS PROBLEM "); printf("\nNumber Of Rows(N) For NxN Chessboard."); scanf("%d",&N); a=(int *)(malloc(sizeof(int)*N)); printf("\nAll possible Solutions .. \n"); printf("\nIn Each of the solutions the Coordinates of the N-Queens are given (Row,Col) ."); printf("\nNote that the Rows and Colums are numbered between 1 - N :\n"); for(ctr=0;ctr<N;ctr++) getPositions(a,0,ctr); getchar(); getchar();}void printArray(int a[]){ int i,choice; static int counter=0; counter++; printf("\nSOLUTION # %d :",counter); for(i=0;i<N;i++) printf("(%d,%d) ",i+1,a[i]+1); if(counter%10==0) { printf("\nEnter 0 to exit , 1 to continue .");

scanf("%d",&choice); if(choice==0) exit(0); };

}void getPositions(int a1[],int colno,int val){ int ctr1,ctr2; a1[colno]=val; if(colno==N-1)

{ printArray(a1) ; return; };

29

for(ctr1=0;ctr1<N;) { /* This Loop Finds Suitable Column Numbers , in the NEXT ROW */ for(ctr2=0;ctr2<=colno;ctr2++) if(a1[ctr2]==ctr1 || (colno+1-ctr2)*(colno+1-ctr2)==(ctr1-a1[ctr2])*(ctr1-a1[ctr2])) goto miss1; getPositions(a1,colno+1,ctr1); miss1: ctr1++; }}

30

OUTPUT

31

Q14. Write the program to solve Graph Coloring problem using Backtracking approach.

#include<stdio.h>#include<conio.h>void colournext(int k,int colour[],int n,int m,int edge[20][20]){int flag=0,j;do{flag=0;colour[k]=(colour[k]+1)%(m+1);if(colour[k]==0)return;for(j=0;j<n;j++){if((edge[k][j]==1) && (colour[k]==colour[j]))flag=1;}}while(flag==1);}

void main(){int i,t,j,h,n,m,k,maxclique=0;int edge[20][20];int colour[20];printf("\n Enter the no. of vertices in the graph : ");scanf("%d",&n);for(i=0;i<n;i++)colour[i]=0;for(i=0;i<n;i++){for(j=0;j<n;j++)edge[i][j]=0;}for(i=0;i<n;i++){t=0;printf("\n Enter number of adjacent nodes from node %d : ",i);scanf("%d",&t);if(t>maxclique)maxclique=t;printf("\n Enter those adjacent nodes : ");

32

for(j=0;j<t;j++){scanf("%d",&h);edge[i][h]=1;}}m=maxclique;k=0;do{colournext(k,colour,n,m,edge);if(colour[k]==0){k--;colour[k]=colour[k+1]%(m+1);}elsek++;}while( k<n);for(i=0;i<n;i++){printf("\tnode : %d \tcolor no. : %d",i,colour[i]);printf("\n");}}

OUTPUT

33

Q15. Write a program of disjoint set union.

#include<stdio.h>#include<conio.h>int main(){ int nlist,a[10],b[10],i,j,k,l,m; printf("Enter total no of elements including all the list\n"); scanf("%d",&j); for(i=0;i<j;i++) { printf("Enter element\n"); scanf("%d",&a[i]); printf("\nEnter element's parent\n"); scanf("%d",&b[i]); } printf("Elements are:\n"); for(i=0;i<j;i++) { printf("%d\t",a[i]); } printf("\nparents are:\n"); for(i=0;i<j;i++) { printf("%d\t",b[i]); } printf("\nPlease enter the set's element you want to join\n"); scanf("%d %d",&k,&l); for(i=0;i<j;i++) { if((a[i]==l) && (b[i]==-1)) { b[i]=k; } else if((a[i]==l) && (b[i]!=-1)) { m=b[i]; for(i=0;i<j;i++) { if((a[i]==m) && (b[i]==-1)) { b[i]=k; } } } }

34

printf("\n\nElements are:\n"); for(i=0;i<j;i++) { printf("%d\t",a[i]); } printf("\nparents are:\n"); for(i=0;i<j;i++) { printf("%d\t",b[i]); } getch(); return 0; }

OUTPUT

35