ChallengingProgramming. ( 1 )

Programming ( Mathematical )  Practice for Fun:  

 The 3n+1(Collatz conjecture) Problem: 

         Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22:

          22   11   34   17   52   26   13   40   20   10   5    16   8   4   2   1

          
 Question:
          It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers up to at least 1,000,000. For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.

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.

Output
           For each pair of input integers i and j, output i, j in the same order in which they appeared in the input and then the maximum cycle length for integers between and including i and j. These three numbers should be separated by one space, with all three numbers on one line and with one line of output for each line of input.

 Sample Input                      Sample Output

 1    10                            1   10   20           

 100  200                           100   200   125

 201   210                          201   210   89

 900   1000                         900   1000   174

Solution:

Java ( Eclipse ):

import java.util.Scanner;

public class PracRT { 

//PracRT is a random class name.

      public static void main(String[] args) {

            int i,j,cycleCounter,temp;

             Scanner number=new Scanner(System.in);

            System.out.print("Enter first number:");

            i=number.nextInt();

            System.out.print("Enter last number:");

            j=number.nextInt();

            int max=0;

            for (int counter=i;counter<=j;counter=counter+1){

              temp=counter;

              cycleCounter=1;

            while (temp!=1){

              if (temp%2==0){

            temp=temp/2;

            cycleCounter=cycleCounter+1;

              }else{

                    temp=temp*3+1;

                    cycleCounter=cycleCounter+1;

              }

            }

           if(max<cycleCounter){

             max=cycleCounter;

           }

           }

            System.out.print(i+" "+j+" "+max);

      }

}

Visual Basic:

Module Module1

  Sub Main()

        Dim i, j, cycleCounter, counter, temp As Integer

        Console.Write("Enter first number:")

        i = Console.ReadLine()

        Console.Write("Enter last number:")

        j = Console.ReadLine()

        Dim max As Integer = 0

        For counter = i To j

            temp = counter

            cycleCounter = 1

            While temp <> 1

                If temp Mod 2 = 0 Then

                    temp = temp / 2

                    cycleCounter = cycleCounter + 1

                Else

                    temp = temp * 3 + 1

                    cycleCounter = cycleCounter + 1

                End If

            End While

            If max < cycleCounter Then max = cycleCounter

        Next

        Console.Write(i & " " & j & " " & max)

        Console.ReadKey()

    End Sub

 End Module

Collatz's Conjecture: 
          The Collatz's Conjecture states that any number can either be halved (if it is even) or multiplied by three and added one to (if it is odd) and eventually reach 1. The conjecture has not been proven, but most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it.