Archive for January 2015

Some Latest Projects of Android Applications being Developed

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1. Android-based Prescription Viewer Application

This project is an electronic-based prescription format for the doctors. The usage of this application is very simple for the doctors as it allows them to enter patient’s name, drugs’ details, and dosage. Once the details like patient’s postal address and area code are fed into the application, it enables the patient to acquire medicines directly from any pharmacy store. This system also provides a facility to send instructions to the patients.

2. Children Tracking System Using Bluetooth MANET Composed of Android Mobile Terminals

Tracking System Application
Tracking System Application
With this application, parents can easily track the whereabouts of their children using Android mobiles. In this system, Android terminals communicate through a Bluetooth MANET. This application also facilitates exchange of information through wireless LAN.

3. Authentication Schemes for Session Passwords using Color and Images by Android

Session Password Application
Session Password Application
Many people often use text-based passwords for the security of the data stored in their mobiles. However, the use of text-based passwords is prone to dictionary attacks and eavesdropping. This project is proposed to develop high security by implementing passwords as colors and images.




4. Android Suburban Railway Ticketing with GPS as Ticket Checker

This project eliminates the need to stand in queue for buying tickets. This particular system enables any user to buy the suburban railway ticket on the smart phone itself, and also provides a reference code for the ticket. This project uses the GPS system of the smart phone to validate and delete the ticket automatically once the user reaches the destination point.

5. Android System Design and Implementation for Telemetric Services

This application enables integration of Android software with the network management functions and media-oriented system transport technologies to use multiple network access. Telemetrics is an integration of both telecommunication and informatics. This system is applicable for traffic safety, road navigation, remote business, etc.

6. Automatic Brightness Control of the Hand-held Device Display with Low Illumination

This project reduces the need to set up brightness of the smart phone based on the ambient light intensity. In this project, the brightness of a Smart phone is controlled particularly in low illumination places. This application is based on using the image details of the user’s face and its background to estimate the contrast ratio and determine the brightness.
Android Based Brightness Control Application
Android Based Brightness Control Application

7. Network-Assisted Mobile Computing with Optimal Uplink Query Processing

Depending upon the queries of users, mobile applications often retrieve data from the remote servers. This affects the life of the batteries of the servers due to slow response time from the remote servers, and also due to a large number of queries. This application provides a solution by deploying mid-network systems with leasing capabilities. It helps reduce the consumption of the battery and also improves the performance of the system or response time.
Network Assisted Mobile Computing
Network Assisted Mobile Computing

8. A Personalized Mobile Search Engine (PMSE)

This project is different from the conventional search engines as it provides the ease of searching based on user preferences. The user preferences are classified as content concepts and location concepts. The project involves arrangement of user preferences in ontology-based multifaceted user profile for rank adaption purpose.
Personalized Mobile Service Engine
Personalized Mobile Service Engine

9. Network Behavior Analysis For Android Malware Detection

Security breaching and malicious attacks are the most common mobile threats that must be reduced to provide security to mobile users. This system detects all of the malware actions by program behaviors. By comparing trace abstractions to reference malicious behaviors, suspicious behaviors can be detected.
The above list of Android projects are implemented for Android OS and are recommended mostly for IT and MCA students as most of the projects require the development of software- based applications.
We hope that our interesting and latest project ideas offer immense help to the students and make them select appropriate projects for their final year project work.

Binary Division

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Problem:
Divide two Binary Numbers
Solution:
#include <iostream>
using namespace std;
long diff( long a,long b);

 int main ()
{
long long int num1,num2,temp=0,quotient=0;
int i=0;
//taking input
do
{
cout<<"Enter Dividend in Binary =";
cin>>num1;
cout<<"Enter Divisor in Binary=";
cin>>num2;
if (num1<num2)
cout<<"Invalid Input! Kindly input correct values\n"<<endl;
} while (num1<num2);

 int num[50];

 while (num1!=0)
{
num[i]=num1%10;
num1/=10;
i++;
}
--i;
while(i>=0)
{
temp=(temp*10)+num[i];
while (temp<num2 && i>=0 )
{
quotient*=10;
i--;
if (i<0)
break;
temp=temp*10+num[i];
}

 if (i>=0)
{
quotient=(quotient*10)+1;
temp=diff(temp,num2);
i--;
}
}

cout<<endl<<endl<<"Quotient = "<<quotient<<"\nRemainder = "<<temp<<endl ;
return 0;
}
long diff( long a,long b)

 {
long int num1[50],i=0,num2[50];
while (a||b)
{
num1[i]=a%10;
num2[i++]=(b%10);
a=a/10;
b=b/10;
}
i--;

 long int diff[50],j,temp;

//calculating Difference

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

if (num1[temp]-num2[temp]>=0)
diff[temp]=num1[temp]-num2[temp];
else
{
j=temp;
j++;
while(num1[j]!=1)
num1[j++]=1;
num1[j]=0;
diff[temp]=1;

}

}

 long int ans=0;
while (i>=0)
ans=(ans*10)+diff[i--];

return ans;

}

Binary Addition,Subtraction,Multiplication and Division

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Problem:
Add, Subtract, Multiply and Division of two Binary Numbers 
Solution:
# include <iostream>
# include <cmath>
using namespace std;
int toDec(int binary)
{
int x = 0, a = 0, y = 0;
while (binary != 0)
{
x = binary % 10;
binary = binary / 10;
y = x*pow(2, a) + y;
a++;
}
return y;

 }
int powCheck(int x)
{
for (int v = 0; v<1000; v++)
{
if (pow(2, v) == x)
{
return v;
}
}
}
int main()
{
int bin1, bin2, revBin = 0, conBin = 0;
char choice;
cout << "Please Enter 1st Binary Number : ";
cin >> bin1;
cout << "Please Enter 2nd Binary Number : ";
cin >> bin2;
cout << "Please Enter an operation to be done from below" << endl;
cout << "Enter '+'sign to Add Binary Numbers" << endl;
cout << "Enter '-'sign to Subtract Binary Numbers" << endl;
cout << "Enter '*'sign to Multiply Binary Numbers" << endl;
cout << "Enter '/'sign to Divide Binary Numbers" << endl;
cin >> choice;
int d1 = toDec(bin1); //bin1 to decimal
int d2 = toDec(bin2);
cout << "Decimal of 1st Binary is : " << d1 << endl;
cout << "Decimal of 2nd Binary is : " << d2 << endl;

 int fun;
if (choice == '+')
{
fun = d1 + d2;
}
else if (choice == '-')
{
fun = d1 - d2;
}
else if (choice == '*')
{
fun = d1*d2;
}
else if (choice == '/')
{
fun = d1 / d2;
}
else
{
return 0;
}
int altFun = fun;
while (fun != 0)
{
revBin = revBin * 10 + fun % 2;
fun = fun / 2;

 }
while (revBin != 0)
{
conBin = conBin * 10 + revBin % 10;
revBin = revBin / 10;
}
if (altFun % 2 == 0 && powCheck(altFun))
{
conBin = conBin*pow(10, powCheck(altFun));
cout <<"The result from "<<choice<<" is: "<< conBin << endl;

 }
else if (altFun % 2 == 0)
{
do
{
conBin = conBin * 10;
} while (altFun != toDec(conBin));
cout <<"The result from "<<choice<<" is: "<< conBin << endl;
}
else
{
cout <<"The result from "<<choice<<" is: "<< conBin << endl;
}


return 0;
}

Binary Addition

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Problem:
Add two Binary Numbers...
Solution:
# include <iostream>
using namespace std;
int toDec(int binary)
{
int x = 0, a = 0, y = 0;
while (binary != 0)
{
x = binary % 10;
binary = binary / 10;
y = x*pow(2, a) + y;
a++;
}
return y;
}
int main()
{
int x;
int y;
int a[10];
int b[10];
int result[20];
cout << "Please Enter 1st Binary Number : ";
cin >> x;
cout << "Please Enter 2nd Binary Number : ";
cin >> y;
int x1 = toDec(x);
int x2 = toDec(y);
int x3 = x1 + x2;
for (int j = 0; j < 10; j++)
{
a[j] = x % 10;
b[j] = y % 10;
x = x / 10;
y = y / 10;
}

for (int i = 0; i <10; i++)
{
if (a[i] + b[i]  == 0)
{
result[i] = 0;
}
else if (a[i] + b[i]  == 1)
{
result[i] = 1;
}
else if (a[i] + b[i] == 2)
{
result[i] = 0;
a[i + 1]++;
}
else if (a[i] + b[i] == 3)
{
result[i] = 1;
a[i + 1]++;

 }
else if (a[i] + b[i]  == 4)
{
result[i] = 0;
a[i + 2]++;
}
}

int addResult = 0;
for (int k= 0; k < 10; k++)
{
addResult = addResult * 10 + result[k];
}
while (toDec(addResult) != x3)
{
addResult = addResult / 10;
}
cout <<"The Sum of 2 Binary Numbers is : "<< addResult << endl;

system("pause");
return 0;

}

Swap the two Arrays entered

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Problem:
Define a function swapArrays(…) that takes two arrays and their size as parameters, and swaps the contents of the two arrays. Call this function through main().
Solution:
# include <iostream>
using namespace std;
void swapArrays(int a[],int b[],int size);
int main()
{
const int size=10;
int array1[size];
int array2[size];
swapArrays(array1,array2,size);

 return 0;
}
void swapArrays(int a[],int b[],int size)
{
cout<<"Please enter the "<<size<<" elements of Array 1"<<endl;
for(int i=0;i<size;i++)
{
cin>>a[i];
}
cout<<"Please enter the "<<size<<" elements of Array 2"<<endl;
for(int i=0;i<size;i++)
{
cin>>b[i];
}
for(int i=0;i<size;i++)
{
int temp=a[i];
a[i]=b[i];
b[i]=temp;
}
cout<<"Array 1 after swaping becomes"<<endl;
for(int i=0;i<size;i++)
{
cout<<a[i]<<" ";
}
cout<<endl;
cout<<"Array 2 after swaping becomes"<<endl;
for(int i=0;i<size;i++)
{
cout<<b[i]<<" ";
}
cout<<endl;

}

Calculate Annual Rain Fall and Average Rain Fall a month , and Month with Maximum and Minimum Rain Fall

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Problem:
Write a program that lets the user enter the total rainfall for each of 12 months into an array of doubles. The program should calculate and display the total rainfall for the year, the average monthly rainfall, and the months with the highest and lowest amounts.

Input Validation: Do not accept negative numbers for monthly rainfall figures.
Solution:
# include <iostream>
using namespace std;
void rainInput(double a[],int size)
{
cout<<"Please enter the Rain Fall for "<<size<<" months : "<<endl;
for(int i=0;i<size;i++)
{
cout<<"Please enter Rain Fall for month "<<i+1<<" (Non-Negative): ";
cin>>a[i];
while (a[i]<0)
{
cout<<"The Rain Fall must be Positive"<<endl;
cout<<"Please enter Rain Fall Again for month "<<i+1<<" (Non-Negative): ";
cin>>a[i];
}
}
}
void totalRainFall(double a[],int size)
{
double total=0;
double avg=0;
for(int i=0;i<size;i++)
{
total=total+a[i];
}
avg=total/12;
cout<<"The Total Rain Fall For The Complete Year is : "<<total<<endl;
cout<<"The Average Rain Fall of a Month is : "<<avg<<endl;
}
void findMaxMin(double a[],int size)
{
double max=a[0];
double min=a[0];
int maxPlace=0;
int minPlace=0;
for(int i=1;i<size;i++)
{
if(a[i]>max)
{
max=a[i];
maxPlace=i;
}
}
for(int j=1;j<size;j++)
{
if(a[j]<min)
{
min=a[j];
minPlace=j;
}
}
cout<<"Maximum Rain Fall is in Month "<<maxPlace+1<<endl;
cout<<"Minimum Rain Fall is in Month "<<minPlace+1<<endl;
}

 int main()
{
const int size=12;
double rain[size];
rainInput(rain,size);
totalRainFall(rain,size);
findMaxMin(rain,size);
return 0;

}


Get the Frequency and the Indices of Maximum Number in an Array

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Problem:
Get the Frequency and the Indices of Maximum Number in an Array
Sample Run 1:
Enter 10 Numbers:
10 20 30 87 40 87 86 23 59 78

Total “2” occurrence(s) of the Maximum.
Found at index# 3 5
Solution:
# include <iostream>
using namespace std;
void inputArray(int array[], const int size)
{
cout << "Enter " << size << " Numbers: " << endl;
for (int i=0;i < size;i++)
{
cin >> array[i];
}
}
void getMaxIndex(int array[],int size)
{
int max=array[0];
int count=0;
int maxIndex[10];
for(int i=1;i<size;i++)
{
if(array[i]>max)
{
max=array[i];
}
}
for(int i=0;i<size;i++)
{
if(array[i]==max)
{
maxIndex[count]=i;
count++;
}
}
cout<<"Total "<<count<<" occurrence(s) of the Maximum"<<endl;
for(int j=0;j<count;j++)
{
cout<<maxIndex[j]<<" "<<endl;
}

 }
int main()
{
const int size = 10;
int array[size];
inputArray(array, size);
getMaxIndex(array,size);

 return 0;

}

Get the Index of the First Maximum Number in an Array

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Problem:
Write a program that contains at least two functions.
a) inputArray(…) // It takes input for the array.
b) getMaxIndex(…) // It finds maximum number and returns the index of the first occurrence of the maximum.

Your program takes input from user and displays the position of the maximum number
Sample Run 1:
Enter 10 Numbers:
10 20 30 87 40 87 86 23 23 78
First occurrence of the Maximum is found at index:3.
Solution:
# include <iostream>
using namespace std;
void inputArray(int array[], const int size)
{
cout << "Enter " << size << " Numbers: " << endl;
for (int i=0;i < size;i++)
{
cin >> array[i];
}
}
void getMaxIndex(int array[],int size)
{
int max=array[size-1];
int maxIndex=size-1;
int i=size-2;
while(i>=0)
{
if(array[i]>=max)
{
max=array[i];
maxIndex=i;
}
i--;
}
cout<<"The index of 1st maximum number is : "<<maxIndex<<endl;

 }
int main()
{
const int size = 10;
int array[size];
inputArray(array, size);
getMaxIndex(array,size);

 return 0;

}

Pattern

23:20 0 Comments A+ a-

Problem:
Write a function, which takes starting and ending integer and print the ordered pairs on the screen:
Sample output:
Enter Starting number: 1
Enter Ending number: 5
(1,1) (1,2) (1,3) (1,4) (1,5)
(2,2) (2,3) (2,4) (2,5)
(3,3) (3,4) (3,5)
(4,4) (4,5)

(5,5)
Solution:
# include <iostream>
using namespace std;
void orderedPair(int a,int b)
{
for(int i=a;i<=b;i++)
{
for(int j=i;j<=b;j++)
{
cout<<"("<<i<<","<<j<<")"<<" ";
}
cout<<endl;
}

 }
int main()
{
int startNUm;
int endNum;
cout<<"PLease enter the Starting Number : ";
cin>>startNUm;
cout<<"Please enter the Ending Number : ";
cin>>endNum;
orderedPair(startNUm,endNum);
return 0;

}

Function to convert Lower Case Letters to Upper Case

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Problem:
Write a function which receives a character (‘a’ to ‘z’ or ‘A’ to ‘Z’) and converts the receive character into uppercase. If you receive anything other than English alphabet then simply return it. The function prototype is as follows:

char toUpper( char )
Solution:
# include <iostream>
using namespace std;
char toUpper(char);a
int main()
{
char c;
cout<<"Please enter a Character : ";
cin>>c;
cout<<toUpper(c)<<endl;


 return 0;
}
char toUpper(char c)
{
if(c>='a' && c<='z')
{
return c-32;
}
else 
{
return c;
}


}

Check a function that can take 1,2,3 or 4 arguments at a Time and Display It's Sum

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Problem:
Write function(s) to execute the following main()function.
int main()
{
cout << sum(1,2) << endl; // should display 3
cout << sum(1,2,3) << endl; // should display 6
cout << sum(1,2,3,4) << endl; // should display 10
return 0;

}
Solution:
Note: The green highlighted code is the required Function

 # include <iostream>
using namespace std;
int sum(int a, int b = 0, int c = 0, int d = 0);//while making such fuction define only in prototype if definition is not mentioned alongwith

 int main()
{
cout<<"Sum of 1 and 2 is : "<<sum(1,2)<<endl;
cout<<"Sum of 1 ,2 and 3 is : "<< sum(1,2,3) << endl; 
cout<<"Sum of 1 ,2 ,3 and 4 is : "<< sum(1,2,3,4) << endl; 
return 0;
}

int sum (int a,int b,int c,int d)
{
return a+b+c+d;

}

Generate Random Number's b/w Upper and Lower Limit of your Desire (Commented)

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Problem:
Write a function named ‘generateRandomNumber’ which receives two integral arguments i.e.

lower bound and upper bound and returns a random number within the given range.
Solution:
# include <iostream>
# include <cstdlib>
# include <ctime>
using namespace std;
int generateRandomNumber(int, int); //prototype for generateRandomNumber with 2 int type variables 
int main()
{
int lowerBound; //lower Limit of random number
int upperBound; //Upper Limit of random number
cout << "This Program Will Display Random Number b/w Lower and Upper Limits" << endl;
cout << "Please enter the Upper limit : ";
cin >> upperBound; //input taken and stored in upperBound variable
cout << "Please enter the Lower limit : ";
cin >> lowerBound; //input taken and stored in lowerBound variable
while (lowerBound>upperBound) //condition for check if lower limit is greater than upper limit and takes input of both upper and lower limits until upperBound becomes greater than lowerBound
{
cout << "Lower Limit must be Less Than Upper Limit" << endl;
cout << "Please enter the Upper limit Again : ";
cin >> upperBound;
cout << "Please enter the Lower limit Again : ";
cin >> lowerBound;
}
cout << generateRandomNumber(lowerBound, upperBound) << endl; //display value returned by the function generateRandomNumber by calling it and passing it two int type parameters lowerBound and upperBound and recieve a random number b/w limits

 return 0;
}
int generateRandomNumber(int lower, int upper) //defining of generateRandomNumber function with two int type parameters lower and upper
{
srand(time(0)); //for taking different time for different executions
int diff = (upper - lower) + 1;
return (rand() % diff) + lower; //makes the random number in bound of lower and upper limit provided to function generateRandomNumber and returns it back to statement which called it
}

Enter a Number (in decimal) and Find Number of 1's in its Binary Pattern (Commented)

22:38 0 Comments A+ a-

Problem:
Write a function named ‘countOnes’, which receives an integral value (base 10) and returns the

number of ones in its binary patterns.
Solution:
# include <iostream>
# include <cmath>
using namespace std;
int countOnes(int dec); //countOnes function prototype written with int dec as a parameter and count number of ones in the binary of the digit entered in decimal
int main()
{
int decimal; //decimal variable taken to store a base 10 number in it
cout << "Please enter an Integer (Base 10) : ";
cin >> decimal; //input stored in decimal variable
cout << "The Number of Ones in the Binary Pattern Of " << decimal << " is : " << countOnes(decimal) << endl; //number of 1's in the binary pattern of digit entered displayed by calling the function countOnes with argument decimal

system("pause");
return 0;

 }
int countOnes(int dec) //defining of function countOnes
{
int revBin = 0,count=0,r=0; //revBin stores the inverted binary pattern of the dec(digit), count variable counts the number of 1's in binary pattern of digit, and r just stores the result of revBin%10 means last digit of revBin

 while (dec != 0) //runs until dec doesn't become 0
{
revBin = revBin * 10 + dec % 2; //Reverse binary of dec(digit) stored in it
dec = dec/ 2; //dec(digit) is divided by 2 to calculate next digit of binary
}
while (revBin != 0) //runs until revBin doesn't become 0
{
r = revBin % 10; //last digit of revBin separated and stored in r
revBin = revBin / 10; //revBin is divided by 10 to remove last digit
if (r == 1) // check if r is equal to 1
{
count++; //if r=1 then count is increased by 1
}
}
return count; //count returned to statement which called it

}

Enter a Number and Find Sum of It's Digits (Commented)

22:35 0 Comments A+ a-

Problem:
Write a program that contains a function int sum_of_digits(int) which takes an integer number
as parameter and returns the sum of the digits of that integer number.

For example, when 234 is passed to the function and it returns 9.
Solution:
# include <iostream>
using namespace std;
int sum_of_digits(int); //prototype for function sum_of_digits with parameter int and calculates the sum of digits of integer passed to it as argument
int main()
{
int integer; //integer is the value to be recieved
cout << "Please Enter the Integer : ";
cin >> integer; //input taken in integer variable
cout << "The Sum of the Digits of Integer entered is : " << sum_of_digits(integer) << endl; //sum of digits in the integer variable shown after calling it with argument integer

 system("pause");
return 0;
}
int sum_of_digits(int num) //defining of sum_of_digits function
{
int sum = 0; //sum of digits of variable num(integer) intialized with 0
while (num != 0) //loop for finding sum of digits of num until it becomes 0
{
sum = sum + num % 10; //last digit in num added to sum
num = num / 10; //last digit of num removed
}
return sum; //sum of digits returned to main function were it is called

}

Find L.C.M of three integers entered

11:07 0 Comments A+ a-

Problem:
Write a program that takes three integers from user and displays the appropriate LCM.

Sample Output

(i)-Enter three integers : 6  36  3
     LCM is : 36

(ii)-Enter three integers : 4  3  6
      LCM is : 12
Solution:
# include <iostream>
using namespace std;
int main()
{
int n1, n2, n3, largest, temp;
cout << "Please enter 1st integer : ";
cin >> n1;
cout << "Please enter 2nd integer : ";
cin >> n2;
cout << "Please enter 3rd integer : ";
cin >> n3;
if (n1>n2 && n1>n3)
{
largest = n1;
}
else if (n2>n1 && n2>n3)
{
largest = n2;
}
else
{
largest = n3;
}
temp = largest;

 if (largest%n1 == 0 && largest%n2 == 0)
{
cout << "L.C.M is : " << largest << endl;
}
else
{
temp++;
for (temp;; temp++)
{
if (temp%n1 == 0 && temp%n2 == 0 && temp%n3 == 0)
{
cout << "L.C.M is : " << temp << endl;
break;
}

 }
}

 system("pause");
return 0;
   }
Sample Output

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