Wednesday, March 9, 2011

Software Daily Updates

Click On The Given Below Link To Download The Software


2 sec Shutdown & reboot



Here is the simple application to shut down r reboot PC in hardly 2 sec. its a promise to u guys just 300 kb

http://www.ziddu.com/download/14124252/SuperfastShutdownReboot.zip.html


Sunday, March 6, 2011

Technical Papers Daily Updates

Click On The Given Link To Download The Abstracts

WIRELESS MOBILE COMMUNICATIONS 4G MOBILE COMMUNICATIONS

LESS POWER TRANSMISSION AND RECEPTION USINGS PSRECTENNA

WIRELESS TECHNOLOGY FUTURE SCOPE

Wireless Transmission of Electricity

WORLD WIDE INTEROPERABILITY FOR MICROWAVE ACCESS WIMAX

Zig-Bee communication protocol

Zone-Routing Protocol

ImageProcessing

HUMAN-COMPUTERINTERFACE

HumanComputerInteractionwithDesign

HumanoidRobot

FIREWALLS

FINGERPRINT_BIOMETRICS_1

FuzzyTopology

HONEYPOTS

HUMANCOMPUTERAPPLICATIONS

GeographicalInformationSystem

HONEYPOTSFORNETWORKSECURITY

HOLOGRAPHICMEMORY

onsinDistributedDataWarehousesusingtheDW STechniq

EVOLUTION_OF_COMPUTER

ExtensibleStylesheetLanguage

Ethicalhacking3

FINGERPRINTAUTHENTICATION

FileCompressionFormat

Ethicalhacking2

Ethicalhacking4

FINGERTRACKINGINREALTimeHUMANCOMPUTERINT ERACTION

Ethicalhacking1

DRIVING_WITHOUT_WHEELS_FLYING_WITHOUT_WI NGS

EllipticCurveCryptosystem

EMBEDDED SPIKING NEURAL NETWORK

DNA_COMPUTING_1

DistributedSystemsArchitectureandImpleme ntation

Digitalsignature

Distributed Virtual Disks

DCOMTechnicalOverview

CRYPTOGRAPHY_IN_SMART_CARDS

DATA COMPRESSION AND EN CODING USING COLORS

CYBORG

CDMA TECHNOLOGY

Cryptography_and_Network_Security_1

BrainComputerInterface

COLLECTIVE INTELLIGENCE

CODEDIVISIONMULTIPLEACCESSSYSTEM

Conditional Access SYSTEM

BIOMETRICS TO AMELIORATE AUTHENTICATION

Blue Ray Disk

BAYOU SYSTEM

Zone Routing Protocol

Zig Bee communication protocol

Friday, March 4, 2011

Computer Programming Week 5 Lab programs


Week 5
a) Write a C program to find both the larges and smallest number in a list of integers.


ALOGORITHM

Step 1: Start
Step 2: Read n
Step 3: Read a[i]
Step 4:repeat step 3 until i=n
Step 5: max=a[0]
           min =a[0]
Step 6: if(max<a[i])
                max=a[i]
                else
                if(min>a[i])
                min=a[i]
step 7: i+1
Step 8: repeat step 6 until i<n
Step 8: Stop

FLOWCHART


Program :
#include <stdio.h>
void main( )
{
        int a[30],max,min,i,n;
        printf(" Enter the Range of Numbers  : ");
        scanf("%d",&n);
        printf("\n Enter %d numbers \n",n);
        for(i=0;i<n;i++)
        scanf(" %d ", &a[ i ]);
        max=a[0];
        min =a[0];
        for(i=0;i<n;i++)
        {
                if(max<a[i])
                max=a[i];
                else
                if(min>a[i])
                min=a[i];
        }
        printf(" Maximum Value is %d  and  Minimum Value  is %d",max,min);
        }
RESULT :
Enter the Range of numbers  : 5
Enter 5 numbers
29  3  10  9  1
Maximum Value is 29 and Minimum Value  is 1


b) Write a C program that uses functions to perform the following:
            i) Addition of Two Matrices
           
PROGRAM:
#include<stdio.h>
#include <stdio.h>
void scan(int a[][5],int m1,int n1)
{
        int i,j;
        for(i=0;i<m1;i++)
                for(j=0;j<n1;j++)
        {
      printf("enter the %d,%d element\t",i,j);
     scanf("%d",&a[i][j]);
        }
}
void print(int a[][5],int m1,int n1)
{

        int i,j;
        for(i=0;i<m1;i++)
        {
                for(j=0;j<n1;j++)
                        printf("%d \t",a[i][j]);
                printf("\n");
        }
}
void add(int a[][5],int b[][5],int c[][5],int m1,int n1)
{        int i,j;
        for(i=0;i<m1;i++)
                for(j=0;j<n1;j++)
                        c[i][j]=a[i][j]+b[i][j];


}
int main()
{        int m1,n1,m2,n2,a[5][5],b[5][5],c[5][5];
        printf("\nEnter order of first matrix ");
        scanf("%d%d",&m1,&n1);

        printf("\nEnter order of second matrix ");
        scanf("%d%d",&m2,&n2);
        if (m1!=n1 || m2!=n2)
        {
          printf("\nMatrix Addition is not possible ");
          exit(0);
        }
        printf("\nEnter elements of first matrix : \n");
        scan(a,m1,n1);

        printf("\nEnter elements of second matrix : \n");
        scan(b,m1,n1);

        add(a,b,c,m1,n1);
        printf("\n The first matrix is\n");
        print(a,m1,n1);

        printf("\n The second matrix is\n");
        print(b,m2,n2);
        printf("\n The sum matrix is\n");
        print(c,m2,n2);
}

RESULT:
Enter order of first matrix 2   2

Enter order of second matrix 2   2

Enter elements of first matrix :
enter the 0,0 element   2
enter the 0,1 element   2
enter the 1,0 element   2
enter the 1,1 element   2

Enter elements of second matrix :
enter the 0,0 element   2
enter the 0,1 element   2
enter the 1,0 element   2
enter the 1,1 element   2

 The first matrix is
2       2
2       2

 The second matrix is
2       2
2       2

 The sum of matrix is
4       4
4       4

ii) Multiplication of Two Matrices
PROGRAM :
#include<stdio.h>
#include <stdio.h>
void scan(int a[ ][5],int m1,int n1)
{
        int i,j;
        for(i=0;i<m1;i++)
                for(j=0;j<n1;j++)
        {
      printf("enter the %d , %d element \t",i,j);
     scanf("%d",&a[ i ] [ j ]);
        }
}
void print(int a[ ][ 5 ],int m1,int n1)
{
        int i,j;
        for(i=0;i<m1;i++)
        {
                for(j=0;j<n1;j++)
                        printf("%d \t",a[ i ] [ j ]);
                printf("\n");
        }
}
void mul(int a[ ][ 5 ],int b[ ][ 5 ],int c[ ][ 5 ],int m1,int n1,int n2)
{        int i,j,k;
        for(i=0;i<m1;i++)
                for(j=0;j<n2;j++)
                 {
                    c[ i ] [ j ]=0;
                    for(k=0;k<n1;k++)
                        c[ i ] [ j ]+=a[ i ] [ k ]*b[ k ] [ j ];

                 }
}
int main( )
{
        int m1,n1,m2,n2,a[5] [5],b[5] [5],c[5] [5];
        printf("\nEnter order of first matrix ");
        scanf("%d%d",&m1,&n1);

        printf("\nEnter order of second matrix ");
        scanf("%d%d",&m2,&n2);
        if (n1!=m2)
        {
          printf("\nMatrix multiplication is not possible ");
          exit(0);
        }
        printf("\nEnter elements of first matrix : \n");
        scan(a,m1,n1);

        printf("\nEnter elements of second matrix : \n");
        scan(b,m2,n2);

        mul(a,b,c,m1,n1,n2);
        printf("\n the first matrix is\n");
        print(a,m1,n1);

        printf("\n the second matrix is\n");
        print(b,m2,n2);
        printf("\n the product matrix is\n");
        print(c,m1,n2);
}


RESULT:
Enter order of first matrix 2
2

Enter order of second matrix 2
2

Enter elements of first matrix :
enter the 0,0 element   2
enter the 0,1 element   2
enter the 1,0 element   2
enter the 1,1 element   2

Enter elements of second matrix :
enter the 0,0 element   2
enter the 0,1 element   2
enter the 1,0 element   2
enter the 1,1 element   2

 The first matrix is
2       2
2       2

 The second matrix is
2       2
2       2

 The product of matrix is
8       8
8       8

Computer Programming Week 4 Lab Programs


Week 4
a) The total distance travelled by vehicle in ‘t’ seconds is given by distance      = ut+1/2at2 where ‘u’ and ‘a’ are the initial velocity (m/sec.) and acceleration (m/sec2). Write C program to find the distance travelled at regular intervals of time given the values of ‘u’ and ‘a’. The program should provide the flexibility to the user to select his own time intervals and repeat the calculations for different values of ‘u’ and ‘a’.


ALGORITHM
STEP1:  START
STEP2:  INITIALIZE x=0
STEP3:  READ u,a,n,t
STEP:4   REPEAT UNTIL x<=t
STEP5:    d=(u*x)+(a*x*x*0.5);
STEP:6  DISPLAY x,d
STEP7: STOP

    FLOW CHART



PROGRAM :
#include<stdio.h>
#include<conio.h>
void main()
{
        int u,t,a,n,x=0,i;
        float d;
        printf(" enter the values of u,a");
        scanf("%d%d",&u,&a);
        printf("\n enter the time interval and the limit ");
        scanf("%d%d",&n,&t);
        while(x<=t)
        {
        d=(u*x)+(a*x*x*0.5);
        printf("\nthe value of displacemnt for the  %d time is   %.2f\n",x,d);
        x+=n;
        }
getch( );
}
 
RESULT:
enter the values of u , a  :  4     5

 enter the time interval and the limit   2     16

the value of displacemnt for the  0 time is   0.00

the value of displacemnt for the  2 time is   18.00

the value of displacemnt for the  4 time is   56.00

the value of displacemnt for the  6 time is   114.00

the value of displacemnt for the  8 time is   192.00

the value of displacemnt for the  10 time is   290.00

the value of displacemnt for the  12 time is   408.00

the value of displacemnt for the  14 time is   546.00

the value of displacemnt for the  16 time is   704.00


a)     Write a C program, which takes two integer operands and one operator form the user, performs the operation and then prints the result. (Consider the operators +,-,*, /, % and use Switch Statement)

ALGORITHM
STEP1: Start
STEP2:  Declare a,b,c
STEP3: Read a,b,c
STEP4: c=’+’ ,display sum of a&b is a+b
               c=’-’ , substraction of a&b is a-b
               c=’*’, multiplication of a&b is a*b
               c=’/’, division of a&b is a/b
               c=’%’,r emainder of a&b is a%b
               else display invalid choice
STEP5: stop
FLOWCHART


 
Program :

#include<conio.h>
#include<stdio.h>
void main( )
{
int a,b,result;
char c;
printf(" enter the two integers \n");
scanf("%d%d",&a,&b);
printf("1.for addition enter + \n");
printf("2.for subtraction enter - \n");
printf("3.for multiplication enter * \n");
printf("4.for division enter / \n");
 
printf("5.enter modulos to display the remainder \n");
printf(“Enter your CHOICE to perform the calculation \n”);
scanf(" %c",&c);
switch(c)
{
                        case ' + ' :
                        result=a+b;
                        printf(" the sum of  %d and %d is %d",a,b,result);
                        break;
                case ' - ' :
                        result=a-b;
                        printf(" the subraction of %d and %d  is%d",a,b,result);
                        break;
                case ' * ' :
                        result=a*b;
                        printf(" the multiplication of %d and %d is %d",a,b,result);
                        break;
                case ' / ' :
                        result=a/b;
                        printf(" the  divison of %d and %d is %d",a,b,result);
                        break;
                case ' % ' :
                                    result=a%b;
                                    printf(" the remainder of %d and %d id %d",a,b,result);
                                    break;
                        default:
                                    printf(" the enterd choice is not valid");
                                    break;
            }
            getch();
}
            RESULT:
Enter the two integers
29
12
1.For addition enter +
2.For subtraction enter -
3.For multiplication enter *
4.For division enter /
5.Enter modulos to display the remainder

Enter your CHOICE to perform the calculation :  -
The subraction of  29 and 12  is  : 17



Thursday, March 3, 2011

II IT ATTENDANCE UPTO FEB 28TH

HT.No
PPL
OOPS
DBMS
ES
PS
DAA
OOPS LAB
DBMS LAB
TOTAL
%

40
36
35
36
40
39
24
21
271
100
09E51A1201
33
30
31
28
35
33
21
21
232
85.61
09E51A1202
28
28
28
24
33
34
21
18
214
78.97
09E51A1203
23
19
22
20
27
30
21
18
180
66.42
09E51A1205
33
28
30
28
36
33
21
21
230
84.87
09E51A1206
31
26
31
25
33
32
18
21
217
80.07
09E51A1207
26
20
26
23
28
31
18
18
190
70.11
09E51A1208
26
20
28
22
29
30
21
21
197
72.69
09E51A1209
30
26
24
23
32
28
18
18
199
73.43
09E51A1210
24
19
22
23
31
30
21
18
188
69.37
09E51A1211
35
29
30
30
39
31
18
21
233
85.98
09E51A1212
21
20
23
19
24
30
12
18
167
61.62
09E51A1213
36
30
28
30
38
31
21
18
232
85.61
09E51A1214
35
29
25
29
37
31
18
15
219
80.81
09E51A1215
21
18
23
17
23
29
15
21
167
61.62
09E51A1216
14
24
13
13
19
30
9
9
131
48.34
09E51A1217
21
25
22
20
26
27
21
21
183
67.53
09E51A1218
23
16
18
23
30
31
21
15
177
65.31
09E51A1219
33
28
29
29
36
32
21
21
229
84.5
09E51A1220
22
19
18
20
26
30
12
9
156
57.56
09E51A1221
19
21
12
14
24
28
15
9
142
52.4
09E51A1222
33
24
31
28
36
31
21
21
225
83.03
09E51A1223
21
20
13
16
26
31
15
15
157
57.93
09E51A1224
29
20
24
24
31
32
21
9
190
70.11
09E51A1226
29
25
25
23
35
32
18
21
208
76.75
09E51A1227
38
30
30
32
39
33
21
21
244
90.04
09E51A1228
37
31
31
29
34
33
21
21
237
87.45
09E51A1229
38
30
33
29
39
32
21
21
243
89.67
09E51A1230
27
22
28
23
35
29
21
18
203
74.91
09E51A1231
34
23
30
25
35
32
12
21
212
78.23
09E51A1232
36
25
29
28
31
33
18
15
215
79.34
09E51A1233
28
21
26
23
27
28
18
18
189
69.74
09E51A1234
24
20
20
21
33
29
21
15
183
67.53
09E51A1235
32
28
30
28
34
33
18
18
221
81.55
09E51A1236
32
23
27
22
34
27
21
18
204
75.28
09E51A1238
31
26
23
24
34
33
21
18
210
77.49
09E51A1239
27
24
24
24
32
32
21
15
199
73.43
09E51A1240
22
20
21
20
29
31
21
15
179
66.05
08E51A1251
25
23
26
18
23
28
21
15
179
66.05
08E51A1247
22
23
21
23
27
28
15
12
171
63.1
08E51A1209
25
23
23
23
27
28
15
18
182
67.16
08E51A1232 
21
22
23
23
27
28
15
15
174
64.21