Byte order (or you are both wron

[g-rh at cca]

Subject: Re: Byte order (or you are both wrong)

>In article <17162 at rochester.ARPA> crowl at rochester.UUCP (Lawrence Crowl) writes:
>
>CHALLENGE:  Come up with a scheme for representing numbers, and a sorting
>scheme in which numbers sort naturally.  Your scheme must deal with variable
>length character strings and variable size numbers.  That is, you cannot
>requires strings to be padded with nulls, or numbers to be padded with zeros.

Presented without endorsement:

Addressing nomenclature -- big endian (i.e. msb is bit 0, etc.)

Item is divided into five parts:

Part I - sign bit (0 is negative, 1 is postive)

Part II - Size of field size in unary, i.e. string of 1's terminated by 0

Part III - Field size, in binary (Field size is size of field to left of
           implicit decimal point.

Part IV -  Portion of string to left of implied decimal point, encoded in
	   bytes.

Part V  - Portion of string to right of implied decimal point, encoded in
          ternary, using two bits per ternary digit, as follows:

	00   Sentinel, terminating right field
	01   ternary 0
        02   ternary 1
	03   ternary 2

All numbers are represented as ascii characters.

The point of the ternary representation (or something equivalent) is
that it provides for a sentinel for variable length strings that sorts
lexicographically.

If an integer is to be considered as a number and compared in that fashion
it goes in the left field -- if it is to be considered as a character string
it goes in the right field.

	I believe that this representation meets the challenge.
I sincerely hope that I never have to work with it.

	Richard Harter, SMDS Inc.