Problem 1- 3n+1 problem

This is Problem1.

You can see the source code(solution) by scrolling little below the problem or try the problem first and see the solution later.

(This is my personal collection)

The 3n + 1 problem

Background

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.

The Problem

Consider the following algorithm:

 
  1.    input n
  2.    print n
  3.    if n = 1 then STOP

  4.       if n is odd then   

  5.       else   

  6.    GOTO 2

Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.) Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16. For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.

The Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0. You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j. You can assume that no operation overflows a 32-bit integer.

The Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

scroll Down for source -code written by me
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||---------------------------------------------------------------------
//Programming Language JAVA
//tool Eclipse IDE
package problem1;
public class Series {
 int lower;
 int upper;
 int maxCycleLength;

 public int getLower() {
  return lower;
 }
 public void setLower(int lower) {
  this.lower = lower;
 }
 public int getUpper() {
  return upper;
 }
 public void setUpper(int upper) {
  this.upper = upper;
 }
 public int getMaxCycleLength() {
  return maxCycleLength;
 }
 public void setMaxCycleLength(int maxCycleLength) {
  this.maxCycleLength = maxCycleLength;
 }

 public Series(){
  this.lower = 0;
  this.upper = 0;
  this.maxCycleLength = 0;
 }
 public Series(int i,int j){
  this.lower = i;
  this.upper = j;
  this.maxCycleLength = 0;
 }
}
//-----------------------------------------2nd Class---------------------------
//MAIN CLASS
package problem1;

import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;
public class CycleLength {
 /**
  * @param args
  */
 public static ArrayList<Series> list = new ArrayList<Series>();

 public static void main(String[] args) {

  int i,j;
  String moreSeries = "y";
  Scanner scanner = new Scanner(new InputStreamReader(System.in));


 do{
      System.out.println("Enter series [Less than 1,000,000 greater than 0]");
     
   i = scanner.nextInt(); j = scanner.nextInt();

   i = (i>0 && i<1000000 )? i :0;
   j = (j>0 && j<1000000 )? j :0;
   if(i ==0 || j ==0){
    System.out.println("Invalid Input.Exiting...");
    System.exit(1);
  
   }
 
   Series s = new Series(i,j);
   list.add(s);
   System.out.println("More inputs?");
   moreSeries = scanner.next();
 }while(moreSeries.equalsIgnoreCase("y"));
   processInput();
  
 }
 static void processInput(){
  System.out.println("Processing input:");
  Series s;
  for(int i=0;i<list.size();i++){
   s = (Series)list.remove(i);
   System.out.println(s.getLower()+"-"+s.getUpper());
   s = processSeries(s);
   list.add(i, s);
  }

  System.out.println("---------OUTPUT---------");
  for(int i=0;i<list.size();i++){
   s = (Series)list.get(i);
   System.out.println(s.getLower()+"-"+s.getUpper()+"-"+s.getMaxCycleLength());
   }

 }
 static Series processSeries(Series s){
  int i,j;
  int n,end,count=0,maxCount = 0;
  i=s.getLower();
  j=s.getUpper();

  if(i<=j){
   n = i;
   end = j;
  }
  else{
   n = j;
   end = i;
  }
 
  do{
  count = getCycleLength(n);
  maxCount = maxCount>count ? maxCount :count;
  n++;
  }while(n<=end);
  s.setMaxCycleLength(maxCount);
  return s;
 }
 static int getCycleLength(int n){
  int count=0;
  System.out.println("**CYCLE FOR "+n+"**");
  do{
   System.out.print(n+" ");
   n = n%2==0 ? n/2 : 3*n+1;
   count++;
  }while(n!=1);
  System.out.println("1");
  ++count;
  System.out.println("-------------->"+count);
  return count;

 }
}
//-------------END-OF-CODE----------------------