Question

I am investigating encoding of date information into a linear barcode that must be as short as possible. One idea we have had is encoding date information rather than using the ASCII characters for the date numerals.

1. The earliest date to be encoded is 01-JAN-2009
2. Date must be encoded using printable ASCII characters

Answer 1

If only the date matters (not the time):

 Take the date to be encoded.
 Subtract 1-1-2009 from it
 Get the amount of elapsed days.

To encode this number as a printable sequence of ascii chars(ascii 48-127) do the following

pseudo code:

 const int ASCIILOWRANGE = 48
 const int ASCIIHIGHRANGE = 126
 const char ASCIIENDMARKER = 127;


 String convertToAscii(int numberOfDays)
 {
       String output = "";

       while(numberOfDays>0)
       {
           output += (char) ASCIILOWRANGE + numberOfDays % (ASCIIHIGHRANGE-ASCIILOWRANGE);
           numberOfDays /= (ASCIIHIGHRANGE-ASCIILOWRANGE);
       }
       output += ASCIIENDMARKER ;
       return output;
 }

 //decoder
 int convertToDays(String ascii)
 {
      char rightmost;
      int numberOfDays = 0;
      while(ascii.length>0)
      {
       rightmost = ascii[0];
           if(rightmost == ASCIIENDMARKER ) break; //ready

           numberOfDays *= (rightmost - ASCIILOWRANGE) * (ASCIIHIGHRANGE-ASCIILOWRANGE);
           ascii = ascii.substr(1); //remove rightmost char from string
      }
      return numberOfDays ;
 }

this way of encoding is the most dense. Since it encodes upto 96 days into the future in just 2 chars. 9216 days, in 3 chars, 884736 days in 4 chars.