Memory Saver - 1

[Paul McKenna]

Represent a date in the format mmddyyyy in the least amount of memory possible. Solution need not be Java specific, you can use C notations if you want.
[ June 03, 2003: Message edited by: Sriraj Rajaram ]

[Damian Ryan]

The information could be encoded into 23 bits:
4 bits for the month (0-14 covers 1-12 or 0-11)
5 bits for the date (0-31 covers 1-31 or 0-30)
14 bits for the year (0-16383 covers 1-9999)
So the information could be stored in a single int:
000000000mmmmdddddyyyyyyyyyyyyyy
Assuming month, date amd year are int values that have been validated as being in range, you should be able to get the stored value by:

int value = (month << 19) | (date << 14) | year;

to get month, date and year back from the value:

int month = value >> 19;
int date = (value >> 14) & 0x1F;
int year = value & 0x3FFF;

I haven't tried any of that for real, so I might be completely wrong

[Jim Yingst]

It's possible to do it in 22 bits - or less if we restrict the year range to something more realistic. E.g. by limiting years to a span of 178 years, we can get it down to 16 bits, fitting nicely in a short. Here's a 22-bit solution for years 1-9999:

import java.text.*;
import java.util.*;
public class Test
{
  public static int encode(Date date)
  {
    Calendar c = new GregorianCalendar();
    c.setTime(date);
    int val = c.get(Calendar.YEAR) * 366 + c.get(Calendar.DAY_OF_YEAR) - 1;
    return val;
  }
  public static Date decode(int val)
  {
    int year = val / 366;
    int day = val % 366 + 1;
    Calendar c = new GregorianCalendar();
    c.clear();
    c.set(Calendar.YEAR, year);
    c.set(Calendar.DAY_OF_YEAR, day);
    return c.getTime();
  }
  public static void test(String dateStr) throws Exception
  {
    System.out.println("input:\t" + dateStr);
    DateFormat df = new SimpleDateFormat("yyyyMMdd");
    Date d1 = df.parse(dateStr);
    System.out.println("Date:\t" + d1);
    int val = encode(d1);
    System.out.println("val:\t" + val);
    int val2 = val & (-1 >>> 10);
    assert val == val2 : val + " == " + val2;
    Date d2 = decode(val);
    System.out.println("Date:\t" + d2);
    assert d1.equals(d2) : d1 + " equals " + d2;
    String str = df.format(d2);
    assert dateStr.equals(str) :"dateStr.equals(" + dateStr + ")";
    System.out.println();
  }
  public static void main(String[] args) throws Exception
  {
    test("20000228");
    test("20000229");
    test("20000301");
    test("20010228");
    test("20010301");
    test("00010101");
    test("99991231");
    System.out.println("PASSED");
  }
}

[Michael Morris]

E.g. by limiting years to a span of 178 years
Are you trying to start another Y2K incident?

[Jim Yingst]

Hey, Sriraj wants to minimize the memory, I am just offering options. If that means he will have to employ me years from now to sort out the mess made by all the applications he builds on this solution, well, I'm not complaining.

[Paul McKenna]

All your answers were correct. I think 178 years is more than enough for anyone (trying to act like Bill Gates.. )
Given the situation in the IT industry, I definetly would love another Y2K problem. So that demand goes up again. Anyway that thought belongs in a different forum.
[ June 04, 2003: Message edited by: Sriraj Rajaram ]