Problem 92

package javaapplication3; import java.util.ArrayList; /**  * A number chain is created by continuously adding the square of the digits   * in a number to form a new number until it has been seen before.  *    * For example,  * 44 -> 32 ->  13 ->  10 ->  1 ->  1  * 85 -> 89 -> 145 -> 42 -> 20 -> 4 -> 16 -> 37 -> 58 -> 89  *   * Therefore any chain that arrives at 1 or 89 will become stuck in an   * endless loop. What is most amazing is that EVERY starting number will   * eventually arrive at 1 or 89.  *   * How many starting numbers below ten million will arrive at 89?  */ /**  * @author Umang  */ public class Problem92 {     public static ArrayList<Long> getDigits(long no)     {         ArrayList<Long> numbers = new ArrayList<Long>();         long temp = no;         while (temp > 0)         {             numbers.add(temp % 10);             temp /= 10;         }         return numbers;     }     public static long sumSquaresOfNo(long no)     {         ArrayList<Long> numbers = getDigits(no);         long sum = 0;         for (long curunt_no : numbers)         {             sum += (curunt_no * curunt_no);         }         return sum;     }     static boolean doesArrivesAt89(long no)     {         boolean return_val = false;         long squares = sumSquaresOfNo(no);         if (squares == 1 || squares == 89)         {             if (squares == 1)             {                 return_val = false;             }              else if (squares == 89)             {                 return_val = true;             }         } else         {             return_val = doesArrivesAt89(squares);         }         return return_val;     }     public static void main(String[] args)     {         int count = 0;         for (long l = 1; l <= 10000000; l++)         {             if (doesArrivesAt89(l))             {                 count++;             }         }         System.out.println("count is " + count);     } }

Sieve of Eratosthenes - primes

import java.util.LinkedList; import java.util.Scanner; /**  * 2:42 - 3:30  */ /**  * http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes To find all the prime  * numbers less than or equal to a given integer n by Eratosthenes' method:   * 1. Create a list of consecutive integers from 2 to n: (2, 3, 4, ..., n).   * 2. Initially, let p equal 2, the first prime number.   * 3. Starting from p, count  *    up in increments of p and mark each of these numbers greater than p itself   *    in the list. These will be multiples of p: 2p, 3p, 4p, etc.; note that   *    some of them may have already been marked.   * 4. Find the first number greater than p in  *    the list that is not marked. If there was no such number, stop. Otherwise,  *    let p now equal this number (which is the next prime), and   *    repeat from step 3.  */ /**  * @author Umang Bhatt  */ class Node {     private int no;     private boolean marked;     public Node(int no)     {         this.no = no;         marked = false;     }     public int getNo()     {         return no;     }     public void setNo(int no)     {         this.no = no;     }     public boolean isMarked()     {         return marked;     }     public void setMarked(boolean marked)     {         this.marked = marked;     } } public class SieveOfEratosthenes {          public static void main(String[] args)     {         Scanner aScanner = new Scanner(System.in);         System.out.println("Enter an integer: ");         int no = 0;         try         {             no = aScanner.nextInt();         }          catch (java.util.InputMismatchException e)         {             System.out.println("You entered invalid number. Can not eontinue");             System.exit(0);         }         LinkedList<Integer> primeNumberList = getAllPrimesLessThan(no);         System.out.println("");         for (Integer primeNo : primeNumberList)         {             System.out.print(primeNo + " ");         }     }          /**      * takes a number and returns all the primes.      * @param no is the maximum no. the primes will be below this number.      * @return  linked list of integers having only prime numbers      */     public static LinkedList<Integer> getAllPrimesLessThan(int no)     {         LinkedList<Integer> primeNumberList = new LinkedList<Integer>();         LinkedList<Node> allNUmberList = getNodeListWithItemsLessThan(no);         primeNumberList.addAll(getThefirstUnmarkedAndMarkOtherMultiplesIn(allNUmberList));         primeNumberList.addAll(getUnMarkedIntegers(allNUmberList));         return primeNumberList;     }          /**      * This method created a list of nodes with values between 2 and given number no      * @param no is the limit      * @return the list of nodes       */     private static LinkedList<Node> getNodeListWithItemsLessThan(int no)     {         LinkedList<Node> nodeList = new LinkedList<Node>();         for (int i = 2; i <= no; i++)         {             Node aNode = new Node(i);             nodeList.add(aNode);         }         return nodeList;     }          /**      * This method adds the first node. Adds it to the primeNumber list and then      * marks all multiples of that number as marked true      *      * @param allNUmberList is the list of nodes created initially      * @return the list of integers.      */     private static LinkedList<Integer> getThefirstUnmarkedAndMarkOtherMultiplesIn(LinkedList<Node> allNUmberList)     {         LinkedList<Integer> primeNumberList = new LinkedList<>();         for (Node aNode : allNUmberList)         {             if (aNode.isMarked() == false && hasUnmarkedNodes(allNUmberList))             {                     Node curruntNode = getFirstUnmarked(allNUmberList);                     curruntNode.setMarked(true);                     primeNumberList.add(curruntNode.getNo());                     markMultiples(allNUmberList, curruntNode.getNo());             }         }         return primeNumberList;     }          /**      * returns the first un-marked node in the list.      * @param aNodeList is the list of nodes.      * @return the first un-marked node.      */     private static Node getFirstUnmarked(LinkedList<Node> aNodeList)     {         Node returnNode = null;         for (Node aNode : aNodeList)         {             if (!aNode.isMarked())             {                 returnNode = aNode;                 break;             }         }         return returnNode;     }          /**      * checks whether the list has unmarked nodes      * @param aNodeList      * @return boolean stating whether the list has unmarked nodes or not.      */     private static boolean hasUnmarkedNodes(LinkedList<Node> aNodeList)     {         boolean hasUnmarkedEntries = false;         for (Node aNode : aNodeList)         {             if (!aNode.isMarked())             {                 hasUnmarkedEntries = true;                 break;             }         }         return hasUnmarkedEntries;     }     /**      * this method marks all the numbers that are multiples of given number in       * the passed list of nodes.      * @param nodeList is the list of nodes constructed originally      * @param number is the number of which's multiples will be marked.      */     private static void markMultiples(LinkedList<Node> nodeList, int number)     {         for (Node aNode : nodeList)         {             if (aNode.getNo() % number == 0)             {                 aNode.setMarked(true);             }         }     }     /**      * returns all the nodes that are un-marked.      * @param allNUmberList      * @return list of un-marked nodes      */     private static LinkedList<Integer> getUnMarkedIntegers(LinkedList<Node> allNUmberList)     {         LinkedList<Integer> primeNumberList = new LinkedList<>();         for (Node aNode : allNUmberList)         {             if (!aNode.isMarked())             {                 primeNumberList.add(aNode.getNo());             }         }         return primeNumberList;     } }

PHP quiz

PHP quiz » quiz builder

leap year

/**
 * A program to tell whether an year is leap year or not.
 * @author Umang Bhatt
 */
public class LeapYearTester
{

        public static boolean isLeapYear(int year)
        {
                boolean isLeapYear ;
                if (year % 4 == 0)
                {
                        if (year % 100 == 0)
                        {
                                if (year % 400 == 0)
                                {
                                        isLeapYear = true;
                                }
                                else
                                {
                                        isLeapYear = false;
                                }
                        }
                        else
                        {
                                isLeapYear = true;
                        }
                }
                else
                {
                        isLeapYear = false;
                }
                return isLeapYear;
        }

        public static void main(String args[])
        {
                System.out.println(isLeapYear(2003));
        }
}

clear screen in java

Clearing the screen with Java in platform independent way (on both Windows and Linux platform).
Here are the videos that show how to clear the console:

Part 1


Part 2


You can download JANSI from http://jansi.fusesource.org/
You can refer to the API docs from http://jansi.fusesource.org/documentation/api/index.html

one program many ways

Program to check whether a number is positive, negative or is zero: (I did this while one of the teaching sessions in a few minutes to encourage student for coming out with new solutions )many programs same thing:

  1:

#include<stdio.h>

int main()
{
 int a,b;
 
 printf("enter the number");
 scanf("%d",&a);
 if(a==0)
 {
  printf(" %d number is 0",a);
 }
 else 
 {
  if(a>0)
  {
   printf(" %d number is possitive",a); 
  }
  else
  {
   printf("%d number is nagitive",a);
  }
 }
 
 return 0;
}

  2:

#include<stdio.h>

int main()
{
 int a,b;
 
 printf("enter the number");
 scanf("%d",&a);
 if(a>0)
 {
  printf(" %d number is possitive",a); 
 }
 else if(a<0)
 {
  printf("%d number is nagitive",a);
 } 
 else if(a==0)
 {
  printf(" %d number is 0",a);
 }
 
 
 return 0;
}

  3:

#include<stdio.h>

int main()
{
 int a,b;
 
 printf("enter the number");
 scanf("%d",&a);
 if(a>0)
 {
  printf(" %d number is possitive",a); 
 }
 else if(a<0)
 {
  printf("%d number is nagitive",a);
 } 
 else
 {
  printf(" %d number is 0",a);
 }
 
 
 return 0;
}

  4:

#include<stdio.h>

int main()
{
 int a,b;
 
 printf("enter the number");
 scanf("%d",&a);
 if(a>0 || a<0 )
 {
  if(a>0)
  {
   printf(" %d number is possitive",a);   
  }
  else if(a<0)
  {
   printf("%d number is nagitive",a);
  }
 } 
 else
 {
  printf(" %d number is 0",a);
 }
 
 
 return 0;
}

  5:

#include<stdio.h>

int main()
{
 int a,b;
 
 printf("enter the number");
 scanf("%d",&a);
 if(a>0 | a<0 )
 {
  if(a>0)
  {
   printf(" %d number is possitive",a);   
  }
  else if(a<0)
  {
   printf("%d number is nagitive",a);
  }
 } 
 else
 {
  printf(" %d number is 0",a);
 }
 
 
 return 0;
}

  6:

#include<stdio.h>

int main()
{
 int a,b;
 
 printf("enter the number");
 scanf("%d",&a);
 if(a!=0)
 {
  if(a>0)
  {
   printf(" %d number is possitive",a); 
  }
  else
  {
   printf("%d number is nagitive",a);
  }
  
 }
 else 
 {
  printf(" %d number is 0",a); 
 }
 
 return 0;
}

presentation on version control

dear readers
here is a link to one of my presentation that i gave at collage (actually its a recording of the presentation done at home, and is of not great quality)