Calculating Extremely Large Powers of 2

Question

I have made a program in Java that calculates powers of two, but it seems very inefficient. For smaller powers (2^4000, say), it does it in less than a second. However, I am looking at calculating 2^43112609, which is one greater than the largest known prime number. With over 12 million digits, it will take a very long time to run. Here's my code so far:

import java.io.*;

public class Power
{
 private static byte x = 2;
 private static int y = 43112609;
 private static byte[] a = {x};
 private static byte[] b = {1};
 private static byte[] product;
 private static int size = 2;
 private static int prev = 1;
 private static int count = 0;
 private static int delay = 0;
 public static void main(String[] args) throws IOException
 {
  File f = new File("number.txt");
  FileOutputStream output = new FileOutputStream(f);
  for (int z = 0; z < y; z++)
  {
   product = new byte[size];
   for (int i = 0; i < a.length; i++)
   {
    for (int j = 0; j < b.length; j++)
    {
     product[i+j] += (byte) (a[i] * b[j]);
     checkPlaceValue(i + j);
    }
   }
   b = product;
   for (int i = product.length - 1; i > product.length - 2; i--)
   {
    if (product[i] != 0)
    {
     size++;
     if (delay >= 500) 
     {
      delay = 0;
      System.out.print(".");
     }
     delay++;
    }
   }
  }
  String str = "";
  for (int i = (product[product.length-1] == 0) ? 
   product.length - 2 : product.length - 1; i >= 0; i--)
  {
   System.out.print(product[i]);
   str += product[i];
  }
  output.write(str.getBytes());
  output.flush();
  output.close();
  System.out.println();
 }

 public static void checkPlaceValue(int placeValue)
 {
  if (product[placeValue] > 9)
  {
   byte remainder = (byte) (product[placeValue] / 10);
   product[placeValue] -= 10 * remainder;
   product[placeValue + 1] += remainder;
   checkPlaceValue(placeValue + 1);
  }
 }  
}

This isn't for a school project or anything; just for the fun of it. Any help as to how to make this more efficient would be appreciated! Thanks!

Kyle

P.S. I failed to mention that the output should be in base-10, not binary.

Answer 1

The key here is to notice that:

2^2 = 4
2^4 = (2^2)*(2^2)
2^8 = (2^4)*(2^4)
2^16 = (2^8)*(2^8)
2^32 = (2^16)*(2^16)
2^64 = (2^32)*(2^32)
2^128 = (2^64)*(2^64)
... and in total of 25 steps ...
2^33554432 = (2^16777216)*(16777216)

Then since:

2^43112609 = (2^33554432) * (2^9558177)

you can find the remaining (2^9558177) using the same method, and since (2^9558177 = 2^8388608 * 2^1169569), you can find 2^1169569 using the same method, and since (2^1169569 = 2^1048576 * 2^120993), you can find 2^120993 using the same method, and so on...

EDIT: previously there was a mistake in this section, now it's fixed:

Also, further simplification and optimization by noticing that:

2^43112609 = 2^(0b10100100011101100010100001)
2^43112609 = 
      (2^(1*33554432))
    * (2^(0*16777216))
    * (2^(1*8388608))
    * (2^(0*4194304))
    * (2^(0*2097152))
    * (2^(1*1048576))
    * (2^(0*524288))
    * (2^(0*262144))
    * (2^(0*131072))
    * (2^(1*65536))
    * (2^(1*32768))
    * (2^(1*16384))
    * (2^(0*8192))
    * (2^(1*4096))
    * (2^(1*2048))
    * (2^(0*1024))
    * (2^(0*512))
    * (2^(0*256))
    * (2^(1*128))
    * (2^(0*64))
    * (2^(1*32))
    * (2^(0*16))
    * (2^(0*8))
    * (2^(0*4))
    * (2^(0*2))
    * (2^(1*1))

Also note that 2^(0*n) = 2^0 = 1

Using this algorithm, you can calculate the table of 2^1, 2^2, 2^4, 2^8, 2^16 ... 2^33554432 in 25 multiplications. Then you can convert 43112609 into its binary representation, and easily find 2^43112609 using less than 25 multiplications. In total, you need to use less than 50 multiplications to find any 2^n where n is between 0 and 67108864.

Answer 2

Displaying it in binary is easy and fast - as quickly as you can write to disk! 100000...... :D