Longest common subsequence between two strings using Java ..............

here is  the code........

NOTE:- Here the program is printing the length of the longest common           

              subsequence and also the evaluated sequence. 

       
public class LCS {

 public static void main(String[] args) throws Exception {

  Object[] arr = new Object[2];
  arr = evaluateLCS("suman", "paras");
  int t = arr[1].toString().length() - 1;
  System.out.println("length of longest common subsequence is " + Integer.parseInt(arr[0].toString()));
  System.out.println("the required longest subsequence is " + recursion_string(arr[1].toString(), t));
 }

 public static Object[] evaluateLCS(String first, String second) {
  Object[] arr = new Object[2];
  StringBuffer sequence = new StringBuffer();
  int length = 0;
  if (first.length() == 0 || second.length() == 0) {
   arr[0] = new Integer(0);
   arr[1] = "";
   return arr;
  } else {
   if (first.charAt(first.length() - 1) == second.charAt(second
     .length() - 1)) {
    arr = evaluateLCS(first.substring(0, first.length() - 1),
     second.substring(0, second.length() - 1));
    arr[0] = new Integer(1 + Integer.parseInt(arr[0].toString()));
    arr[1] = (Object) sequence.append(
     first.charAt(first.length() - 1)).append(arr[1]);
    return arr;
   } else {
    Object[] arr1 = new Object[2];
    arr = evaluateLCS(first.substring(0, first.length() - 1),
     second);
    arr1 = evaluateLCS(first,
     second.substring(0, second.length() - 1));
    if (Integer.parseInt(arr[0].toString()) >= Integer
     .parseInt(arr1[0].toString()))
     return arr;
    else
     return arr1;
   }

  }

 }

 public static int max(Object[] arr1, Object[] arr2) {
  int max1 = Integer.parseInt(arr1[0].toString());
  int max2 = Integer.parseInt(arr2[0].toString());

  if (max1 >= max2)
   return max1;
  else
   return max2;
 }

 public static String recursion_string(String name, int t) throws Exception {

  if (name.length() == 0) {
   return "";
  }
  if (name.length() == 1) {
   return name;
  }

  String start = name.substring(t);

  String remainder = name.substring(0, t);
  t--;

  return (start + recursion_string(remainder, t));


 }
}

       
 

Breadth First Search of a graph using Java....

NOTE :- while giving input to the adjacency matrix just put "1" against the entry of a[ i ][ j ] if there                  is a path exist between node " i " and " j " or put " 0 " otherwise .

Enter the number of nodes in the graph
4
Enter the adjacency matrix
0 1 0 1
0 0 1 0
0 1 0 1
0 0 0 1
Enter the source for the graph
1
The BFS traversal of the graph is 
1 2 4 3

so here is the code .....

       
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class BFS {
 private Queue < Integer > queue;
 public BFS() {
  queue = new LinkedList < Integer > ();
 }
 public void bfs(int adjacency_matrix[][], int source) {
  int number_of_nodes = adjacency_matrix[source].length - 1;
  int[] visited = new int[number_of_nodes + 1];
  int i, element;
  visited[source] = 1;
  queue.add(source);
  while (!queue.isEmpty()) {
   element = queue.remove();
   i = element;
   System.out.print(i + "\t");
   while (i <= number_of_nodes) {
    if (adjacency_matrix[element][i] == 1 && visited[i] == 0) {
     queue.add(i);
     visited[i] = 1;
    }
    i++;
   }
  }
 }
 public static void main(String[] arg) {
  int number_no_nodes, source;
  Scanner scanner = null;
  try {
   System.out.println("Enter the number of nodes in the graph");
   scanner = new Scanner(System.in);
   number_no_nodes = scanner.nextInt();
   int adjacency_matrix[][] = new int[number_no_nodes + 1][number_no_nodes + 1];
   System.out.println("Enter the adjacency matrix");
   for (int i = 1; i <= number_no_nodes; i++)
    for (int j = 1; j <= number_no_nodes; j++)
     adjacency_matrix[i][j] = scanner.nextInt();
   System.out.println("Enter the source for the graph");
   source = scanner.nextInt();
   System.out.println("The BFS traversal of the graph is ");
   BFS bfs = new BFS();
   bfs.bfs(adjacency_matrix, source);
  } catch (InputMismatchException inputMismatch) {
   System.out.println("Wrong Input Format");
  }
  scanner.close();
 }
}

       
 

Java program to find two numbers in an integer array such that their sum is greater than some integer in O(n) linear time....

problem -find two number 'a' and 'b' such that a+b>sum where sum can be any given integer value...

here is the code ......

logic =you have to execute two passes of bubble sort and then pick the last two elements ...

       
public class findElements {

 public static void main(String[] args) {
  int[] arr = {
   45,
   4,
   23,
   56,
   87,
   27,
   34,
   89
  };
  int temp;
  int sum = 100;
  for (int y = 0; y < 2; y++) {
   for (int i = 0; i < arr.length - y - 1; i++) {

    if (arr[i] > arr[i + 1]) {
     temp = arr[i];
     arr[i] = arr[i + 1];
     arr[i + 1] = temp;
    }
   }
  }
  if (arr[arr.length - 1] + arr[arr.length - 2] > sum)
   System.out.println("required elements are" + arr[arr.length - 1] + " and " + arr[arr.length - 2]);
  else System.out.println("elements not found");
  //for(int s:arr) { System.out.print(s); System.out.print(" ");}
 }

}

time complexity = 2(first loop) + n(second loop) = O(n)

       
 

Dining philosopher's synchronization problem implemented using Java........

here is the code .....

       

public class DiningPhilosopherProblem {
 // Makes the code more readable.
 public static class ChopStick {
  // Make sure only one philosopher can have me at any time.
  Lock up = new ReentrantLock();
  // Who I am.
  private final int id;

  public ChopStick(int id) {
   this.id = id;
  }

  public boolean pickUp(Philosopher who, String where) throws InterruptedException {
   if (up.tryLock(10, TimeUnit.MILLISECONDS)) {
    System.out.println(who + " picked up " + where + " " + this);
    return true;
   }
   return false;
  }

  public void putDown(Philosopher who, String name) {
   up.unlock();
   System.out.println(who + " put down " + name + " " + this);
  }

  @Override
  public String toString() {
   return "Chopstick-" + id;
  }
 }

 // One philosoper.
 public static class Philosopher implements Runnable {
  // Which one I am.
  private final int id;
  // The chopsticks on either side of me.
  private final ChopStick leftChopStick;
  private final ChopStick rightChopStick;
  // Am I full?
  volatile boolean isTummyFull = false;
  // To randomize eat/Think time
  private Random randomGenerator = new Random();
  // Number of times I was able to eat.
  private int noOfTurnsToEat = 0;

  /**
   * **
   *
   * @param id Philosopher number
   *
   * @param leftChopStick
   * @param rightChopStick
   */
  public Philosopher(int id, ChopStick leftChopStick, ChopStick rightChopStick) {
   this.id = id;
   this.leftChopStick = leftChopStick;
   this.rightChopStick = rightChopStick;
  }

  @Override
  public void run() {

   try {
    while (!isTummyFull) {
     // Think for a bit.
     think();
     // Make the mechanism obvious.
     if (leftChopStick.pickUp(this, "left")) {
      if (rightChopStick.pickUp(this, "right")) {
       // Eat some.
       eat();
       // Finished.
       rightChopStick.putDown(this, "right");
      }
      // Finished.
      leftChopStick.putDown(this, "left");
     }
    }
   } catch (Exception e) {
    // Catch the exception outside the loop.
    e.printStackTrace();
   }
  }

  private void think() throws InterruptedException {
   System.out.println(this + " is thinking");
   Thread.sleep(randomGenerator.nextInt(1000));
  }

  private void eat() throws InterruptedException {
   System.out.println(this + " is eating");
   noOfTurnsToEat++;
   Thread.sleep(randomGenerator.nextInt(1000));
  }

  // Accessors at the end.
  public int getNoOfTurnsToEat() {
   return noOfTurnsToEat;
  }

  @Override
  public String toString() {
   return "Philosopher-" + id;
  }
 }
 // How many to test with.
 private static final int NO_OF_PHILOSOPHER = 50;
 //private static final int SIMULATION_MILLIS = 1000 * 60 * 8;
 private static final int SIMULATION_MILLIS = 1000 * 10;

 public static void main(String args[]) throws InterruptedException {
  ExecutorService executorService = null;

  Philosopher[] philosophers = null;
  try {

   philosophers = new Philosopher[NO_OF_PHILOSOPHER];

   //As many forks as Philosophers
   ChopStick[] chopSticks = new ChopStick[NO_OF_PHILOSOPHER];
   // Cannot do this as it will fill the whole array with the SAME chopstick.
   //Arrays.fill(chopSticks, new ReentrantLock());
   for (int i = 0; i < NO_OF_PHILOSOPHER; i++) {
    chopSticks[i] = new ChopStick(i);
   }

   executorService = Executors.newFixedThreadPool(NO_OF_PHILOSOPHER);

   for (int i = 0; i < NO_OF_PHILOSOPHER; i++) {
    philosophers[i] = new Philosopher(i, chopSticks[i], chopSticks[(i + 1) % NO_OF_PHILOSOPHER]);
    executorService.execute(philosophers[i]);
   }
   // Main thread sleeps till time of simulation
   Thread.sleep(SIMULATION_MILLIS);
   // Stop all philosophers.
   for (Philosopher philosopher: philosophers) {
    philosopher.isTummyFull = true;
   }

  } finally {
   // Close everything down.
   executorService.shutdown();

   // Wait for all thread to finish
   while (!executorService.isTerminated()) {
    Thread.sleep(1000);
   }

   // Time for check
   for (Philosopher philosopher: philosophers) {
    System.out.println(philosopher + " => No of Turns to Eat =" + philosopher.getNoOfTurnsToEat());
   }
  }
 }
}
       
 

Java program to print reverse number triangle pattern......

Output Pattern-

              123456787654321
                1234567654321
                  12345654321
                    123454321  
                      1234321  
                        12321    
                          121    
                            1      

here is the code .....

       
class printPattern {
 public static void main(String[] args) {
  int ad = 8;
  for (int i = 0; i < 8; i++, ad--) {

   for (int r = 0; r < i; r++) {
    System.out.print(" ");
   }
   for (int k = 1,
    var = ad; k <=
    var; k++) {
    System.out.print(k);

   }
   for (int j = (ad - 1); j >= 1; j--) {
    System.out.print(j);
   }
   for (int r = 0; r < i; r++) {
    System.out.print(" ");
   }

   System.out.println(" ");
  }

 }
}

       
 

Trigonometric Sine function calculation using Java......

//Using the below series here is the code to calculate the Sine value

NOTE-  value of x (the argument of sine function) is given in degree or you can                       modify the code to accept the input in radian 

       
public class sineFunction {
 public static void main(String[] args) {
   int numTerms = 4;
   double degree = 20;
   double argsinRadian;
   argsinRadian = (degree * 3.14) / 180;
   System.out.println("value of sin(" + argsinRadian + ") is -" + evaluateSine(argsinRadian, numTerms));
  } //main

 public static double evaluateSine(double argsinRadian, int numTerms) {
  double functionValue = argsinRadian;
  double value = argsinRadian;
  double squareValue = argsinRadian;
  squareValue = argsinRadian * argsinRadian;
  double temp = 0;
  int p = 3;
  for (int t = 2; t <= numTerms; t++) {
   temp = value * squareValue;
   value = temp;
   temp = temp / fact(p);
   p = p + 2;
   if (t % 2 == 0) {
    temp = temp * -1;
   }
   functionValue = functionValue + temp;
  }
  return functionValue;
 }
 public static int fact(int num) {
  int factorial = 1;
  for (int k = 1; k <= num; k++) {
   factorial = factorial * k;
  }
  return factorial;
 }
}

       
 

Find Transitive relations in a graph represented via adjacency matrix using Java....

Problem:-Using the Java language, have the function TransitivityRelations(strArr) read the strArr parameter being passed which will make up an NxN matrix where the rows are separated by each pair of parentheses (the matrix will range from 2x2 to 5x5). The matrix represents connections between nodes in a graph where each node corresponds to the Nth element in the matrix (with 0 being the first node). If a connection exists from one node to another, it will be represented by a 1, if not it will be represented by a 0. For example: suppose strArr were a 3x3 matrix with input 

["(1,1,1)","(1,0,0)","(0,1,0)"], this means that there is a connection from node 0->0, 0->1, and 0->2. For node 1 the connections are 1->0, and for node 2 the connections are 2->1. This can be interpreted as a connection existing from node X to node Y if there is a 1 in the Xth row and Yth column. Note: a connection from X->Y does not imply a connection from Y->X. 

What your program should determine is whether or not the matrix, which represents connections among the nodes, is transitive. A transitive relationmeans that if the connections 0->1 and 1->2 exist for example, then there must exist the connection 0->2. More generally, if there is a relation xRy and yRz, then xRz should exist within the matrix. If a matrix is completely transitive, return the string transitive. If it isn't, your program should return the connections needed, in the following format, in order for the matrix to be transitive: (N1,N2)-(N3,N4)-(...). So for the example above, your program should return (1,2)-(2,0). You can ignore the reflexive property of nodes in your answers. Return the connections needed in lexicographical order [e.g. (0,1)-(0,4)-(1,4)-(2,3)-(4,1)]. 

Note:-This program is only for 3x3 matrix ,but you can extend to a program generic for any size matrix .

here is the code for it .....

       

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class Transitive {

 public static void main(String[] args) throws Exception {
  // TODO Auto-generated method stub

  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  String matrix = br.readLine();
  //String mat="[\"(1,1,1)\",\"(1,1,1)\",\"(1,1,0)\"]";
  System.out.println("the desired result is " + TransitiveRelations(matrix));

 }

 public static ArrayList < String > TransitiveRelations(String strArr) {

  int[][] mat = new int[3][3];

  ArrayList < String > aka = new ArrayList < String > ();
  String[] arr = strArr.split("\\)");

  String[] a1 = arr[0].split("\\(");
  String[] a2 = arr[1].split("\\(");
  String[] a3 = arr[2].split("\\(");

  String[] r1 = a1[1].split(",");
  String[] r2 = a2[1].split(",");
  String[] r3 = a3[1].split(",");

  for (int k = 0; k < 3; k++)
   mat[0][k] = Integer.parseInt(r1[k]);


  for (int s = 0; s < 3; s++)
   mat[1][s] = Integer.parseInt(r2[s]);


  for (int p = 0; p < 3; p++)
   mat[2][p] = Integer.parseInt(r3[p]);


  for (int h = 0; h < 3; h++) {
   for (int q = 0; q < 3; q++) {
    for (int e = 0; e < 3; e++) {
     if (h != q) {
      if (mat[h][e] == 1) {
       if (mat[e][q] == 1) {
        if (mat[h][q] == 1)
         continue;
        else aka.add(h + "-" + q);
       } else {
        aka.add(e + "-" + q);
        if (mat[h][q] == 1)
         continue;
        else aka.add(h + "-" + q);
       }
      } else {
       aka.add(h + "-" + e);
       if (mat[e][q] == 1) {
        if (mat[h][q] == 1)
         continue;
        else aka.add(h + "-" + q);
       } else {
        aka.add(e + "-" + q);
        if (mat[h][q] == 1)
         continue;
        else aka.add(h + "-" + q);

       }
      }
     }

    }
   }
  }
  ArrayList < String > a = new ArrayList < String > ();
  for (int t = 0; t < aka.size(); t++)
   if (!a.contains(aka.get(t)) == true)
    a.add(aka.get(t));

  return a;
 }

}