ADA LAB MANUAL(MCA IV SEM VTU)

Algorithms Lab Manual

1. Implement Recursive Binary search and Linear search and determine the time taken to search an element. Repeat the experiment for different values of n, the number of elements in the list to be searched and plot a graph of the time taken versus n.

/* Implementation of recursive binary search and sequential search */

#include<stdio.h>#include<conio.h>#include<time.h>#include<stdlib.h>#define max 20

int pos;int binsearch(int,int[],int,int,int);int linsearch(int,int[],int);

void main(){ int ch=1; double t; int n,i,a[max],k,op,low,high,pos; clock_t begin,end; clrscr(); while(ch) {

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printf("\n.....MENU.....\n 1.Binary Search\n 2.Linear Search\n 3.Exit\n");

printf("\nEnter your choice\n"); scanf("%d",&op);

switch(op) { case 1:printf("\nEnter the number of elements \n");

scanf("%d",&n); printf("\nEnter the elements of an array in order\n"); for(i=0;i<n;i++)

scanf("%d",&a[i]); printf("\nEnter the elements to be searched\n"); scanf("%d",&k); low=0;high=n-1; begin=clock(); pos=binsearch(n,a,k,low,high); end=clock(); if(pos==-1) printf("\n\n Unsuccessful

search"); else

printf("\n Element %d is found at position %d",k,pos+1);

printf("\n Time taken is %lf CPU1 cycles\n",(end-begin)/CLK_TCK);

getch();

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break; case 2:printf("\nEnter the number of elements\n");

scanf("%d",&n); printf("\nEnter the elements of

an array\n"); for(i=0;i<n;i++)

scanf("%d",&a[i]); printf("\nEnter the elements to be searched\n");

scanf("%d",&k); begin=clock(); pos=linsearch(n,a,k); end=clock(); if(pos==-1)

printf("\n\n Unsuccessful search");

else printf("\n Element %d is

found at position %d",k,pos+1); printf("\n Time taken is %lf CPU cycles\n",(end-begin)/CLK_TCK);

getch(); break;

default:printf("\nInvalid choice entered\n");

exit(0);

}

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printf("\n Do you wish to run again (1/0) \n"); scanf("%d",&ch); } getch(); }

int binsearch(int n,int a[],int k,int low,int high) { int mid; delay(1000); mid=(low+high)/2; if(low>high) return -1; if(k==a[mid]) return(mid); else if(k<a[mid]) return binsearch(n,a,k,low,mid-1); else return binsearch(n,a,k,mid+1,high);

}

int linsearch(int n,int a[],int k) { delay(1000); if(n<0)

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return -1; if(k==a[n-1]) return(n-1); else return linsearch(n-1,a,k); }

OUTPUT

Case 1

.....MENU..... 1.Binary Search 2.Linear Search 3.Exit

Enter your choice1

Enter the number of elements3

Enter the elements of an array

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4812

Enter the elements to be searched12

Element 12 is found at position 2 Time taken is 1.978022 CPU1 cycles

Case 2

.....MENU..... 1.Binary Search 2.Linear Search 3.Exit

Enter your choice2

Enter the number of elements4

Enter the elements of an array36912

Enter the elements to be searched9

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Element 9 is found at position 3 Time taken is 3.021978 CPU cycles

2. Sort a given set of elements using the Heap sort method and determine the time taken to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n.

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

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void heapcom(int a[],int n){

int i,j,k,item;for(i=1;i<=n;i++){

item=a[i];j=i;k=j/2;while(k!=0 && item>a[k]){

a[j]=a[k];j=k;k=j/2;

}a[j]=item;

}}void adjust(int a[],int n){

int item,i,j;j=1;item=a[j];i=2*j;while(i<n){

if((i+1)<n){

if(a[i]<a[i+1])i++;

}if(item<a[i])

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{a[j]=a[i];j=i;i=2*j;

}elsebreak;

}a[j]=item;

}void heapsort(int a[],int n){

int i,temp;delay(1000);heapcom(a,n);for(i=n;i>=1;i--){

temp=a[1];a[1]=a[i];a[i]=temp;adjust(a,i);

}}void main()

{ int i,n,a[20],ch=1; clock_t start,end; clrscr(); while(ch) {

printf("\n enter the number of elements to sort\n");

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scanf("%d",&n);printf("\n enter the elements to

sort\n");for(i=1;i<=n;i++) scanf("%d",&a[i]);start=clock();heapsort(a,n);end=clock();printf("\n the sorted list of

elemnts is\n");for(i=1;i<=n;i++) printf("%d\n",a[i]);printf("\n Time taken is %lf CPU

cycles\n",(end-start)/CLK_TCK);printf("do u wish to run again

(0/1)\n");scanf("%d",&ch);

}getch();

}

OUTPUTenter the number of elements to sort5

enter the elements to sort85

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631

the sorted list of elemnts is13568

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3. Sort a given set of elements using Merge sort method and determine the time taken to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n.

#include<stdio.h>#include<conio.h>#include<time.h>#define max 20void mergesort(int a[],int low,int high);void merge(int a[],int low,int mid,int high);void main(){

int n,i,a[max],ch=1;clock_t start,end;clrscr();while(ch) {

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printf("\n\t enter the number of elements\n");

scanf("%d",&n); printf("\n\t enter the elements\

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

scanf("%d",&a[i]); start= clock(); mergesort(a,0,n-1); end=clock(); printf("\nthe sorted array is\

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

printf("%d\n",a[i]); printf("\n\ntime taken=%lf",

(end-start)/CLK_TCK); printf("\n\ndo u wish to

continue(0/1) \n"); scanf("%d",&ch);}getch();

}

void mergesort(int a[],int low,int high){

int mid;delay(100);if(low<high){

mid=(low+high)/2;mergesort(a,low,mid);mergesort(a,mid+1,high);

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merge(a,low,mid,high);}

}

void merge(int a[],int low,int mid,int high){

int i,j,k,t[max];i=low;j=mid+1;k=low; while((i<=mid) && (j<=high))if(a[i]<=a[j])t[k++]=a[i++];elset[k++]=a[j++];while(i<=mid)t[k++]=a[i++];while(j<=high)t[k++]=a[j++];for(i=low;i<=high;i++)a[i]=t[i];

}

OUTPUT

Enter the number of elements 5

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Enter the elements 63419

The sorted array is 13469

time taken=0.824176

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4. Sort a given set of elements using Selection sort and hence find the time required to sort elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n.

#include<stdio.h>#include<conio.h>#include<time.h> void main() {

int i,n,j,min,k,a[20],ch=1;clock_t begin,end;clrscr();while(ch){ printf("\n enter the number of

elements\n"); scanf("%d",&n); printf("\n enter the elements to

be sorted\n"); for(i=0;i<n;i++) scanf("%d",&a[i]); begin=clock(); for(i=0;i<=n-2;i++) {

min=i;delay(200);for(j=i+1;j<=n-1;j++){

if(a[j]<a[min])

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min=j;}k=a[i];a[i]=a[min];a[min]=k;

} end=clock(); printf("\n\t the sorted list of

elements are:\n"); for(i=0;i<n;i++) printf("\n%d",a[i]); printf("\n\n\t time taken:%lf",

(end-begin)/CLK_TCK); printf("\n\n do u wish to

continue (0/1)\n"); scanf("%d",&ch);}getch();

}

OUTPUTenter the number of elements5

enter the elements to be sorted8351

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9

the sorted list of elements are: 1 3 5 8 9 time taken:0.824176

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5. a. Obtain the Topological ordering of vertices in a given digraph.

#include<stdio.h>#include<conio.h>#define max 20

int a[max][max],n;void topological_sort();void main(){

int i,j;clrscr();printf("\n enter the number of

vertices\n");scanf("%d",&n);printf("\n enter the adjacency

matrix\n");for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&a[i][j]);topological_sort();getch();

}

void topological_sort(){

int v[max],ver[max],i,j,p=1,flag=0;for(i=1;i<=n;i++)v[i]=0;while(p<=n)

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{j=1;while(j<=n){

flag=0;if(v[j]==0){

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

if((a[i][j]!=0) && (v[i]==0))

{flag=1;break;

}if(flag==0){

v[j]=1;ver[p++]=j;break;

}}j++;if(j>n){

printf("\n topological order is not

possible\n");getch();exit(0);

}}

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}printf("\n topological order

obtained is...\n");for(i=1;i<p;i++)printf("\t%d",ver[i]);getch();

}

OUTPUTenter the number of vertices4

enter the adjacency matrix0 1 1 10 0 0 10 0 0 00 0 1 0

topological order obtained is... 1 2 4 3

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5 b. Implement All Pair Shortest paths problem using Floyd's algorithm.

#include<stdio.h>#include<conio.h>#include<stdlib.h>int cost[10][10],a[10][10];void all_paths(int [10][10],int [10][10],int);int min1(int,int);

void main(){

int i,j,n;clrscr();printf("\n enter the number of


matrix\n");for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&cost[i][j]);all_paths(cost,a,n);printf("\n\t the shortest path

obtained is\n");for(i=1;i<=n;i++){

for(j=1;j<=n;j++)printf("\t %d",a[i][j]);printf("\n");

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}getch();

}void all_paths(int cost[10][10],int a[10][10],int n){

int i,j,k;for(i=1;i<=n;i++)for(j=1;j<=n;j++)a[i][j]=cost[i][j];for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++)a[i][j]=min1(a[i][j],a[i][k]+a[k]

[j]);}int min1(int a,int b){

return(a<b)?a:b;}

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OUTPUT

enter the number of vertices4

enter the adjacency matrix999 999 3 9992 999 999 999999 7 999 1 6 999 999 999

the shortest path obtained is 10 10 3 4 2 12 5 6 7 7 10 1 6 16 9 10

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6. Implement 0/1 Knapsack problem using dynamic programming.

#include<stdio.h>#include<conio.h>int v[20][20];int max1(int a,int b){

return(a>b)?a:b;}void main(){

int i,j,p[20],w[20],n,max;clrscr();

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printf("\n enter the number of items\n");

scanf("%d",&n);for(i=1;i<=n;i++){

printf("\n enter the weight and profit of the

item %d:",i);scanf("%d %d",&w[i],&p[i]);

}printf("\n enter the capacity of the

knapsack");scanf("%d",&max);for(i=0;i<=n;i++)v[i][0]=0;for(j=0;j<=max;j++)v[0][j]=0;for(i=1;i<=n;i++)for(j=1;j<=max;j++){

if(w[i]>j)v[i][j]=v[i-1][j];elsev[i][j]=max1(v[i-1][j],v[i-1][j-

w[i]]+p[i]);}printf("\n\nThe table is\n");for(i=0;i<=n;i++)

{for(j=0;j<=max;j++)printf("%d\t",v[i][j]);printf("\n");

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}printf("\nThe maximum profit is

%d",v[n][max]);printf("\nThe most valuable subset

is:{");j=max;

for(i=n;i>=1;i--)if(v[i][j]!=v[i-1][j]){

printf("\t item %d:",i);j=j-w[i];

}printf("}");getch();

}

OUTPUT

enter the number of items4

enter the weight and profit of the item 1:2 12



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enter the capacity of the knapsack5

The table is0 0 0 0 0 00 0 12 12 12 120 10 12 22 22 220 10 12 22 30 320 10 15 25 30 37

The maximum profit is 37The most valuable subset is:{ item 4: item 2: item 1:}

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7. From a given vertex in a weighted connected graph, find shortestpaths to other vertices using Dijkstra's algorithm.

#include<stdio.h>main (){ int n, cost[15][15], i, j, s[15], v, u, w, dist[15],

num, min; clrscr(); printf ("Enter the vertices please\n"); scanf ("%d", &n); printf ("Enter the cost of the edges please\n"); printf ("Enter 999 if the edgeis not present or for the

self loop\n"); for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) scanf ("%d", &cost[i][j]); printf ("Enter the Source vertex please\n"); scanf ("%d", &v);

for (i = 1; i <= n; i++) { s[i] = 0; dist[i] = cost[v][i]; }

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s[v] = 1; dist[v] = 0;

for (num = 2; num <= n - 1; num++) { min = 999; for (w = 1; w <= n; w++)

if (s[w] == 0 && dist[w] < min) { min = dist[w]; u = w; }

s[u] = 1;

for (w = 1; w <= n; w++){ if (s[w] == 0) { if (dist[w] > (dist[u] +

cost[u][w]))dist[w] = (dist[u] + cost[u][w]);

}}

}

printf ("VERTEX\tDESTINATION\tCOST\n"); for (i = 1; i <= n; i++) printf (" %d\t %d\t\t %d\n", v, i, dist[i]);

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getch();}

OUTPUT

Enter the vertices pleasen = 5

Enter the cost of the edges pleaseEnter 999 if the edge is not present or for the self loopThe cost of the edges are :

999 1 2 999 9991 999 3 4 9992 3 999 5 6999 4 5 999 6999 999 6 6 999

Enter the Source vertex please : 1

VERTEX DESTINATION COST 1 1 0 1 2 1 1 3 2 1 4 5 1 5 8

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8. Sort a given set of elements using Quick sort method and determine the time taken to sort the elements. Repeat the experiment for different values of n, the number of elements in the list to be sorted and plot a graph of the time taken versus n.

#include<stdio.h>#include<conio.h>void quicksort(int[],int,int);int partition(int[],int,int);void main(){

int i,n,a[20],ch=1;clrscr();while(ch){ printf("\n enter the number of

elements\n"); scanf("%d",&n); printf("\n enter the array

elements\n"); for(i=0;i<n;i++)

scanf("%d",&a[i]); quicksort(a,0,n-1); printf("\n\nthe sorted array

elements are\n\n"); for(i=0;i<n;i++)

printf("\n%d",a[i]); printf("\n\n do u wish to

continue (0/1)\n"); scanf("%d",&ch);

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}getch();

}

void quicksort(int a[],int low,int high){

int mid;if(low<high){

mid=partition(a,low,high);quicksort(a,low,mid-1);quicksort(a,mid+1,high);

}}int partition(int a[],int low,int high){

int key,i,j,temp,k;key=a[low];i=low+1;j=high;while(i<=j){

while(i<=high && key>=a[i])i=i+1;while(key<a[j])j=j-1;if(i<j) {

temp=a[i];a[i]=a[j];a[j]=temp;

}

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else{

k=a[j];a[j]=a[low];a[low]=k;

}}return j;

}

OUTPUT

enter the number of elements5

enter the elements to be sorted85241

the sorted list of elements are: 1 2 4 5 8 time taken:0.824176

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9. Find Minimum Cost Spanning Tree of a given undirected graphusing Kruskal's algorithm-

#include<stdio.h>#include<conio.h>int root[10], flag = 0, count=0, temp, min;int a[20], cost[20][20], n, i, j, k, totalcost = 0, x, y;void find_min (), check_cycle (), update ();main (){ clrscr(); printf ("Enter the number of vertices please\n"); scanf ("%d", &n); printf ("Enter the cost of the matrix please\n"); for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) scanf ("%d", &cost[i][j]); find_min (); while (min != 999 && count != n - 1) { check_cycle (); if (flag)

{ printf ("%d ---> %d = %d\n",

x, y,

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cost[x][y]); totalcost += cost[x][y]; update (); count++;}

cost[x][y] = cost[y][x] = 999; find_min (); }

if (count < n - 2) printf ("The graph is not connected\n"); else printf ("The graph is connected & the min cost is

%d\n", totalcost); getch();}

void check_cycle (){ if ((root[x] == root[y]) && (root[x] != 0)) flag = 0; else flag = 1;}

void find_min (){ min = 999; for (i = 1; i <= n; i++)

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for (j = 1; j <= n; j++) if (min > cost[i][j])

{ min = cost[i][j]; x = i; y = j;}

}

void update (){

if (root[x] == 0 && root[y] == 0) root[x] = root[y] = x;

else if (root[x] == 0) root[x] = root[y];

else if (root[y] == 0) root[y] = root[x];

else { temp = root[y]; for (i = 1; i <= n; i++)

if (root[i] == temp) root[i] = root[x];

}}

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OUTPUT

Enter the number of vertices please4Enter the cost of the matrix please999 1 5 2 1 999 999 999 5 999 999 3 2 999 3 9991 ---> 2 = 11 ---> 4 = 23 ---> 4 = 3The graph is connected & the min cost is 6

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10. a. Print all the nodes reachable from a given starting node in a digraph using Breadth First Search method.

#include<stdio.h>

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#include<conio.h>void distance(int,int);int a[10][10];void main(){

int i,j,n;clrscr();printf("\n Enter the number of vertices in the

diagraph:");scanf("%d",&n);printf("\n Enter the adjacency matrix\n");for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&a[i][j]);for(i=1;i<=n;i++){

printf("\n\t the starting vertex is %d\n",i);distance(i,n);

printf("\n \t press enter for other source vertex\n");

getch();}

}void distance(int v,int n){

int queue[40],visited[20],dis[20],front,rear,i,j;for(i=1;i<=n;i++)visited[i]=dis[i]=0;front=rear=0;queue[rear++]=v;visited[v]=1;

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do{

i=queue[front++];for(j=1;j<=n;j++)if(a[i][j] && !visited[j]){

dis[j]=dis[i]+1;queue[rear++]=j;visited[j]=1;printf("\n\t the vertex %d to %d is of

distance=%d\n",v,j,dis[j]);}

}while(front<rear);

}

OUTPUTEnter the number of vertices in the diagraph:4

Enter the adjacency matrix0 1 1 10 0 0 10 0 0 00 0 1 0

the starting vertex is 1

the vertex 1 to 2 is of distance=1

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press enter for other source vertex

the starting vertex is 2



press enter for other source vertex the starting vertex is 3 press enter for other source vertex the starting vertex is 4 the vertex 4 to 3 is of distance=1 press enter for other source vertex

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10 b. Check whether a given graph is connected or not using DFS method.

#include<stdio.h>#include<conio.h>void dfs(int n,int cost[10][10],int u,int s[]){

int v;s[u]=1;for(v=0;v<n;v++){

if(cost[u][v]==1 && s[v]==0){

dfs(n,cost,v,s);}

}}

void main(){

int n,i,j,cost[10][10],s[10],connected,flag;

clrscr();printf("\n enter the number of

nodes\n");scanf("%d",&n);printf("\n enter the adjacency

matrix\n");for(i=0;i<n;i++)

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

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

}connected=0;for(j=0;j<n;j++){

for(i=0;i<n;i++)s[i]=0;dfs(n,cost,j,s);flag=0;for(i=0;i<n;i++){

if(s[i]==0)flag=1;

}if(flag==0)connected=1;

}

if(connected==1)printf("graph is connected\n");elseprintf("graph is not connected\n");getch();

}

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OUTPUT

Case 1:

enter the number of nodes4

enter the adjacency matrix0 0 0 10 0 0 00 0 1 00 0 0 1graph is not connected

Case 2:


enter the adjacency matrix1 1 1 11 1 1 11 1 1 11 1 1 1graph is connected

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11. Find a subset of a given set S == {sl,s2,......sn} of n positive integers whose sum is equal to a given positive integer d. For example, if S== (1, 2, 5, 6, 8} and d = 9 there are two solutions {1,2,6} and {1,8}. A suitable message is to be displayed if the given problem instance doesn't have a solution.

#include<stdio.h>int s[10],d,n,set[10],count=0;void display(int);int flag = 0;

void main(){

int subset(int,int);int i;clrscr();printf("Enter the Number of elements

in the set\n");scanf("%d",&n);printf("enter the set values\n");

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for(i=0;i<n;++i)scanf("%d",&s[i]);printf("\nEnter the sum\n");scanf("%d",&d);printf(" The Program Output is:\n");subset(0,0);if(flag == 0)printf(" There is no solution \n");getch();

}int subset(int sum,int i){

if(sum == d){

flag = 1;display(count);return;

}if(sum>d || i>=n) return;else{

set[count]=s[i];count++;subset(sum+s[i],i+1);count--;subset(sum,i+1);

}}

void display(int count)

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{int i;printf("{ ");for(i=0;i<count;i++)printf("%d ",set[i]);printf("}\n");

}

OUTPUT

Enter the Number of elements in the set5enter the set values12568

Enter the sum9 The Program Output is:{ 1 2 6 }{ 1 8 }

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12. a. Implement Horspool algorithm for String Matching,

#include<stdio.h>#include<stdlib.h>void main(){

int table[126];char t[100],p[25];int n,i,k,j,m,flag=0;clrscr();printf("Enter the Text\n");gets(t);n=strlen(t);printf("Enter the Pattern\n");gets(p);m=strlen(p);for(i=0;i<126;i++)

table[i]=m;for(j=0;j<=m-2;j++)

table[p[j]]=m-1-j;i=m-1;

while(i<=n-1){

k=0;while(k<=m-1 && p[m-1-k] == t[i-

k])k++;

if(k == m){

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printf("The position of the pattern is

%d\n",i-m+2);flag=1;break;

}else

i=i+table[t[i]];

}

if(!flag)printf("Pattern is not found in

the given text\n");

getch();

}

OUTPUT

Case 1:

Enter the Textacharya_computer_science_engineeringEnter the PatternscienceThe position of the pattern is 18

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Case 2:

Enter the Textacharya_computer_sceince_engineeringEnter the PatterncjPattern is not found in the given text

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12. b. Find the Binomial Co-efficient using Dynamic Programming.

#include<stdio.h>#include<conio.h>void main(){

int i,j,k,n,c[50][50];clrscr();printf("\n enter the value of n & k\

n");scanf("%d%d",&n,&k);for(i=0;i<=n;i++)for(j=0;j<=k;j++)c[i][j]=0;for(i=0;i<=n;i++){

c[i][0]=1;c[i][i]=1;

}for(i=2;i<=n;i++)for(j=1;j<=i-1;j++)c[i][j]=c[i-1][j-1]+c[i-1][j];printf("\n the table for valuation

is\n");for(i=0;i<=n;i++){

for(j=0;j<=k;j++)if(c[i][j]!=0)printf("\t%d",c[i][j]);printf("\n");

}

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printf("\n\t the binomial coefficient of C(%d,%d) is %d\n",n,k,c[n][k]);

getch();}

OUTPUT

enter the value of n & k6 3

the table for valuation is 1 1 1 1 2 1 1 3 3 1

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1 4 6 4 1 5 10 10 1 6 15 20

the binomial coefficient of C(6,3) is 20

Dept of MCA 2009 54


13. Find Minimum Cost Spanning Tree of a given undirected graph using Prims algorithm.

#include<stdio.h>#include<conio.h>void main(){

int cost[20][20],t[20][20],near1[20],a[20];

int i,j,n,min,minimum,k,l,mincost,c,b;

clrscr();printf("\n enter the number of

nodes\n");scanf("%d",&n);printf("\n enter the adjacency

matrix\n");for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&cost[i][j]);minimum=cost[1][1];for(i=1;i<=n;i++)for(j=1;j<=n;j++){

if(minimum>=cost[i][j]){

minimum=cost[i][j];k=i;l=j;

}

Dept of MCA 2009 55


}mincost=minimum;t[1][1]=k;t[1][2]=l;for(i=1;i<=n;i++){

if(cost[i][l]<cost[i][k])near1[i]=l;elsenear1[i]=k;

}near1[k]=near1[l]=0;for(i=2;i<=n-1;i++){

min=999;for(j=1;j<=n;j++){

if(near1[j]!=0){

a[j]=cost[j][near1[j]];{

min=a[j];c=near1[j];b=j;printf("\n");

}}

Dept of MCA 2009 56


}mincost=mincost+cost[b][c];near1[b]=0;for(k=1;k<=n;k++)if((near1[k]!=0) &&

(cost[k][near1[k]]>cost[k][b]))

near1[k]=b;}printf("\n\ the cost of minimum

spanning tree is=%d",mincost);getch();

}

OUTPUT


enter the adjacency matrix999 1 5 2 1 999 999 999 5 999 999 3 2 999 3 999

the cost of minimum spanning tree is=6

Dept of MCA 2009 57


14. Compute the transitive closure of a given directed graph using Warshall's algorithm.

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

int a[10][10];void main(){

int i,j,k,n;clrscr();printf("\n enter the number of


matrix\n");for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&a[i][j]);

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

Dept of MCA 2009 58


a[i][j]=a[i][j] || a[i][k] && a[k][j];

printf("\n\t the tranitive closure is\n");

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

for(j=1;j<=n;j++)printf("\t %d",a[i][j]);printf("\n");

}getch();

}

OUTPUTenter the number of vertices

Dept of MCA 2009 59


4

enter the adjacency matrix0 1 0 00 0 0 11 0 1 00 0 0 0

the tranitive closure is 0 1 0 1 0 0 0 1 1 1 1 1 0 0 0 0

Dept of MCA 2009 60


15. Implement N Queen's problem using Back Tracking

#include<stdio.h>#include<conio.h>#include<math.h>int x[20],count=1;void queens(int,int);int place(int,int);

void main(){

int n,k=1;clrscr();printf("\n enter the number of

queens to be placed\n");scanf("%d",&n);queens(k,n);

}void queens(int k,int n){

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

if(place(k,j)){

x[k]=j;if(k==n){

printf("\n %d solution",count);

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

Dept of MCA 2009 61


printf("\n \t %d row <---> %d

column",i,x[i]);getch();

}elsequeens(k+1,n);

}}

}int place(int k,int j){

int i;for(i=1;i<k;i++)if((x[i]==j) || (abs(x[i]-

j))==abs(i-k))return 0;return 1;

}

OUTPUTenter the number of queens to be placed4

1 solution 1 row <---> 2 column 2 row <---> 4 column 3 row <---> 1 column 4 row <---> 3 column 2 solution

Dept of MCA 2009 62


1 row <---> 3 column 2 row <---> 1 column 3 row <---> 4 column 4 row <---> 2 column

Dept of MCA 2009 63

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