Excess code
The offset binary or surplus code is a binary encoding , with which signed numbers in binary can represent. The coding is based on a value range shift.
Usually positive numbers in the value range 0 to 2 n -1 are coded as n-digit binary numbers as follows (here for the value range 0..7; standard coding):
| shown in decimal | binary mapped |
|---|---|
| 0 | 000 |
| 1 | 001 |
| 2 | 010 |
| 3 | 011 |
| 4th | 100 |
| 5 | 101 |
| 6th | 110 |
| 7th | 111 |
In order to enable the binary representation of negative numbers, the range of values of the numbers is shifted. The extent of the shift is normally in the range k = m - p, where m = 2 n-1 and 0 <= p <= m. One therefore speaks of an excess k code. The excess 0 coding corresponds to the standard coding (see above).
The common excess k codes for three-digit binary numbers are given below.
| Coding | shift | code | |||||||
|---|---|---|---|---|---|---|---|---|---|
| 000 | 001 | 010 | 011 | 100 | 101 | 110 | 111 | ||
| Excess 0 | 0 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th |
| Excess-1 | 1 | -1 | 0 | 1 | 2 | 3 | 4th | 5 | 6th |
| Excess-2 | 2 | -2 | -1 | 0 | 1 | 2 | 3 | 4th | 5 |
| Excess 3 | 3 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4th |
| Excess-4 | 4th | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
The next table lists some possible excess k-codes for binary four-digit numbers.
| Coding | shift | code | |||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0000 | 0001 | 0010 | 0011 | 0100 | 0101 | 0110 | 0111 | 1000 | 1001 | 1010 | 1011 | 1100 | 1101 | 1110 | 1111 | ||
| Excess 0 | 0 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th | 8th | 9 | 10 | 11 | 12 | 13 | 14th | 15th |
| Excess-1 | 1 | -1 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th | 8th | 9 | 10 | 11 | 12 | 13 | 14th |
| Excess-2 | 2 | -2 | -1 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th | 8th | 9 | 10 | 11 | 12 | 13 |
| Excess 3 | 3 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th | 8th | 9 | 10 | 11 | 12 |
| Excess-4 | 4th | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th | 8th | 9 | 10 | 11 |
| Excess 8 | 8th | -8th | -7 | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4th | 5 | 6th | 7th |
The (4-bit) excess 3 code , which is identical to the excess 3 in the table above, has historical significance - it offers advantages when it comes to representing and calculating with decimal numbers.
The excess k code has a very important and special value here, which is shifted by eight places in the above example (i.e. generally k = 2 n-1 , in the example: excess-8). It divides the range of values of the numbers into two equal halves of negative and nonnegative numbers. For binary four-digit codes (decimal 0 to 15) the excess 8 code represents the numbers from -8 to 7, for five-digit codes it would be the excess 16 code and the value range from -16 to 15. One speaks in the case k = 2 n-1 also short of excess coding , so omits the numerical value k. For example, when the excess representation is used for the exponent, this balanced excess code (half negative and half non-negative) is meant in almost all cases.
To encode a number a, one chooses the smallest number b in the range of values and forms the difference: d = | a - b |. The result is then coded as usual.
Conversely, you decode an excess k-coded number by first converting it into a number according to the usual coding and then adding the smallest number in the value range.
Calculation example
The following calculation example only deals with the so to speak balanced excess code (this is the case with k = 2 n-1 ), which divides the numbers evenly into negative and nonnegative ones.
Task: For 8 bits, code the number -79 in excess 128 coding
The code length is n = 8; so for the usual binary representation:
000000002 = 010
and
111111112 = 25510
Since the number is to be coded excess 128, the range of values shifts to:
00000000Exzess-128 = -12810
or.
11111111Exzess-128 = +12710
Coding:
Die zu codierende Zahl ist a = -79. Die kleinste Zahl im Wertebereich ist b = -128 Die Differenz ist d = |-79 - (-128)| = 49 In der Standardcodierung ist d = 4910 = 001100012
So the solution is : a = -79 10 = 00110001 excess-128
The decoding is analogous: 00110001 can be decoded to 49 according to the standard coding. Then the smallest number in the value range is added, here -128, i.e.: 49 - 128 = -79.
application
Excess code is tolerant of binary addition / subtraction and lexical size comparison. In the IEEE-754 standard for representing floating point numbers, the exponent is encoded in a form similar to an excess code. However, integer values are mostly processed in two's complement in the arithmetic unit of modern hardware .