ADA+Lab+All+Programs

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Algorithms Lab Manual

1. Implement Recursive Binary searchand Linear search and determine thetime taken to search an element.

Repeat the experiment for differentvalues of n, the number of elements inthe list to be searched and plot a graphof the time taken versus n.

/* Implementation of recursive binary

search and sequential search */

#include#include#include#include#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.BinarySearch\n 2.Linear Search\n3.Exit\n");

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

switch(op){

case 1:printf("\nEnter the numberof elements \n");

scanf("%d",&n);

printf("\nEnter the elements ofan array in order\n");

for(i=0;i


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break;case 2:printf("\nEnter the number

of elements\n");

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

for(i=0;i


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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,intlow,int high){

int mid;delay(1000);mid=(low+high)/2;if(low>high)return -1;

if(k==a[mid])return(mid);

elseif(k


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return -1;if(k==a[n-1])return(n-1);

elsereturn linsearch(n-1,a,k);

}

OUTPUT

Case 1

.....MENU.....

1.Binary Search

2.Linear Search

3.Exit

Enter your choice

1

Enter the number of elements

3

Enter the elements of an array

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4

8

12

Enter the elements to be searched

12

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 choice

2

Enter the number of elements

4

Enter the elements of an array

3

6

9

12

Enter the elements to be searched

9

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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#include#include

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

int i,j,k,item;for(i=1;ia[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


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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


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6

3

1

the sorted list of elemnts is

1

3

5

6

8

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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 agraph of the time taken versus n.

#include#include

#include#define max 20void mergesort(int a[],int low,inthigh);void merge(int a[],int low,int mid,inthigh);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 ofelements\n");

scanf("%d",&n);

printf("\n\t enter theelements\n");for(i=0;i


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

}

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

int i,j,k,t[max];i=low;j=mid+1;k=low;

while((i


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Enter the elements

6

3

41

9

The sorted array is

1

3

4

69

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 ofelements in the list to be sorted and plot a graph of the

time taken versus n.

#include#include#include

void main(){int i,n,j,min,k,a[20],ch=1;clock_t begin,end;clrscr();while(ch){

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

be sorted\n");for(i=0;i


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if(a[j]


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1

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#include#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 adjacencymatrix\n");for(i=1;i


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while(p


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}}printf("\n topological order

obtained is...\n");for(i=1;i


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5 b. Implement All Pair Shortest paths problem using

Floyd's

algorithm.

#include#include#includeint 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

vertices\n");

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

for(i=1;i


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}getch();

}

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

int i,j,k;for(i=1;i


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OUTPUT

enter the number of vertices

4

enter the adjacency matrix

999 999 3 999

2 999 999 999

999 7 999 1

6 999 999 999

the shortest path obtained is10 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#includeint 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 ofitems\n");

scanf("%d",&n);

for(i=1;i


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enter the weight and profit of the item 4:2 15

enter the capacity of the knapsack5

The table is

0 0 0 0 0 0

0 0 12 12 12 12

0 10 12 22 22 22

0 10 12 22 30 32

0 10 15 25 30 37

The maximum profit is 37

The 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 shortest

paths to other vertices using Dijkstra's algorithm.

#includemain (){

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 notpresent or for theself loop\n");

for (i = 1; i


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}

s[v] = 1;

dist[v] = 0;

for (num = 2; num


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printf (" %d\t %d\t\t %d\n",v, i, dist[i]);

getch();

}

OUTPUT

Enter the vertices pleasen = 5

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

999 1 2 999 9991 999 3 4 999

2 3 999 5 6999 4 5 999 6999 999 6 6 999

Enter the Source vertex please : 1

VERTEX DESTINATION COST1 1 0

1 2 11 3 21 4 51 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 agraph of the time taken versus n.

#include#includevoid 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


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}getch();

}

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

int mid;if(low


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else{

k=a[j];

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

}return j;

}

OUTPUT

enter the number of elements

5

enter the elements to be sorted

8

5

2

4

1

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 graph

using Kruskal's algorithm-

#include#includeint 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


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printf ("%d ---> %d = %d\n",x, y,

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");elseprintf ("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;

elseflag = 1;

}

void find_min (){

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min = 999;for (i = 1; i


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OUTPUT

Enter the number of vertices please

4

Enter the cost of the matrix please

999 1 5 21 999 999 999

5 999 999 3

2 999 3 999

1 ---> 2 = 1

1 ---> 4 = 2

3 ---> 4 = 3

The 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.

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#include

#include

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


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visited[v]=1;

do

{

i=queue[front++];for(j=1;j


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the vertex 1 to 2 is of distance=1



press enter for other source vertex

the starting vertex is 2









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

using DFS method.

#include#includevoid dfs(int n,int cost[10][10],int

u,int s[]){

int v;s[u]=1;for(v=0;v


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printf("\n enter the adjacencymatrix\n");

for(i=0;i


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}

OUTPUT

Case 1:

enter the number of nodes

4


0 0 0 1

0 0 0 0

0 0 1 0

0 0 0 1

graph is not connected

Case 2:

enter the number of nodes

4


1 1 1 1

1 1 1 1

1 1 1 1

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1 1 1 1

graph is connected

11. Find a subset of a given set S == {sl,s2,......sn} of n

positiveintegers 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.

#includeint s[10],d,n,set[10],count=0;void display(int);int flag = 0;

void main(){

int subset(int,int);int i;clrscr();

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printf("Enter the Number of elementsin the set\n");

scanf("%d",&n);

printf("enter the set values\n");for(i=0;id || i>=n) return;else{

set[count]=s[i];

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

}}

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void display(int count){int i;printf("{ ");for(i=0;i


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The Program Output is:

{ 1 2 6 }

{ 1 8 }

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12. a. Implement Horspool algorithm for String

Matching,

#include#includevoid 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


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printf("The position of thepattern is

%d\n",i-m+2);

flag=1;break;}else

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

}

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

the giventext\n");

getch();

}

OUTPUT

Case 1:

Enter the Text

acharya_computer_science_engineering

Enter the Pattern

science

The position of the pattern is 18

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Case 2:

Enter the Text

acharya_computer_sceince_engineering

Enter the Pattern

cj

Pattern is not found in the given text

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

Programming.


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


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

getch();}

OUTPUT

enter the value of n & k

6 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

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13. Find Minimum Cost Spanning Tree of a given

undirected graph

using Prims algorithm.


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

inti,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


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}mincost=minimum;t[1][1]=k;

t[1][2]=l;for(i=1;i


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the cost of minimum spanning tree is=6

14. Compute the transitive closure of a given directed

graph

using Warshall's algorithm.

#include#include

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

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

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

matrix\n");

for(i=1;i


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for(j=1;j


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enter the number of vertices

4

enter the adjacency matrix0 1 0 0

0 0 0 1

1 0 1 0

0 0 0 0

the tranitive closure is

0 1 0 1

0 0 0 11 1 1 1

0 0 0 0

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15. Implement N Queen's problem using Back Tracking

#include

#include#includeint 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


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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


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1 row 3 column

2 row 1 column

3 row 4 column

4 row 2 column

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