Project Euler (P14): Recursion Issues

Question

Hi, I am running the Collatz sequence issue in a Euler project (issue 14). My code works with numbers below 100000, but with numbers more. I get an error.

Is there a way to override the code to use tail recursion or to prevent stack overflows. Code below:

import java.util.*;

public class v4
{

   // use a HashMap to store computed number, and chain size 

   static HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();

   public static void main(String[] args)
   {

      hm.put(1, 1);

      final int CEILING_MAX=Integer.parseInt(args[0]);
      int len=1;
      int max_count=1;
      int max_seed=1;

      for(int i=2; i<CEILING_MAX; i++)
      {
          len = seqCount(i);

          if(len > max_count)
          {
             max_count = len;
             max_seed = i;
          }
      }
      System.out.println(max_seed+"\t"+max_count);
   }

   // find the size of the hailstone sequence for N

   public static int seqCount(int n)
   {

      if(hm.get(n) != null)
      {
         return hm.get(n);
      }

      if(n ==1)
      {
         return 1;
      }
      else
      {
         int length = 1 + seqCount(nextSeq(n));
         hm.put(n, length);
         return length;
      }
   }

   // Find the next element in the sequence

   public static int nextSeq(int n)
   {

      if(n%2 == 0)
      {
         return n/2;
      }
      else
      {
         return n*3+1;
      }
   }

}

Answer 1

Your problem is not with the size of the stack (you are already imagining the values), but with

• the size of some numbers in sequences and
• upper limits of a 32-bit integer.

Prompt:

public static int seqCount(int n)
{
   if(hm.get(n) != null) {
       return hm.get(n);
   }
   if (n < 1) {
      // this should never happen, right? ;)
   } ...
   ...

That should be enough :)

PS you will encounter the need for BigNums in many problems with the project euler ...

Answer 2

, . , :

    for(int i=1;i<=1000000;i+=2)
    {
        steps=1;
        int n=i;
        long current=i;
        while(current!=1)
        {
            if(current%2==0)
            {
                current=current/2;
            }else{
                current=(current*3)+1;
            }
            steps++;
        }
        if(steps>best)
        {
            best=steps;
            answer=n;
        }
    }

, 9

Answer 3

(-, ): Java, , , - JVM, , , - .

Answer 4

Collatz 1000000 Integer. BigInteger .

, , JVM.

Answer 5

, 2 :

• Integer, Integer.Max_VALUE, 2147483647. Long .
• , memoization. , , . "" , do-while for. , - , memoization, "" .

, - . , - . Integer, , Java "" " " , . seqCount (int), n > 0.

Answer 6

, . , int. , , 1, , , stackoverflow

:

public class Collatz {

    public int getChainLength(long i) {
        int count = 1;

        while (i != 1) {
            count++;

            if (i % 2 == 0) {
                i /= 2;
            } else {
                i = 3 * i + 1;
            }
        }

        return count;
    }

    public static int getChainLength(long i, int count) {
        if (i == 1) {
            return count;
        } else if (i % 2 == 0) {
            return getChainLength(i / 2, count + 1);
        } else {
            return getChainLength(3 * i + 1, count + 1);
        }
    }

    public int getLongestChain(int number) {
        int longestChain[] = { 0, 0 };

        for (int i = 1; i < number; i++) {
            int chain = getChainLength(i);

            if (longestChain[1] < chain) {
                longestChain[0] = i;
                longestChain[1] = chain;
            }
        }

        return longestChain[0];
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        System.out.println(new Collatz().getLongestChain(1000000));
    }
}

Answer 7

14:

http://chmu.bplaced.net/?p=265

Answer 8

import java .util.*;
public class file 
  {
 public static void main(String [] args)
  {
   long largest=0;
   long number=0;
    for( long i=106239;i<1000000;i=i+2)
     {
      long k=1;
       long z=i;
      while(z!=1)
       {
        if(z%2==0)
        {
         k++;
         z=z/2;
        } else{
          k++;
          z=3*z+1;
           }
       }    
    if(k>largest)
      {
       number=i;
       largest=k;
       System.out.println(number+" "+largest);
      }
     }//for loop

   }//main
  }