Databases Reference
In-Depth Information
Each correction requires a single bit, bringing the total number of bits used to 26.
With these corrections, the reconstruction at this stage is
26
0
14
10
0
0
0
0
000
0
000
0
If we go through one more pass, we reduce the threshold value to 4. The coefficients to be
scanned are
6
−
7
7
6
4
4
−
4
4
−
3
2
−
2
−
2
0
The dominant pass results in the following coded sequence:
sp sn sp sp sp sp sn iz iz sp iz iz iz
This pass cost 26 bits, equal to the total number of bits used previous to this pass. The
reconstruction upon decoding of the dominant pass is
26
6
14
10
6
6 66
−
6
−
6 60
0
0 00
The subordinate list is
L
S
={
26
,
13
,
10
,
6
−
7
,
7
,
6
,
4
,
4
,
−
4
,
4
}
By now it should be reasonably clear how the algorithm works. We continue encoding
until we have exhausted our bit budget or until some other criterion is satisfied.
There are several observations we can make from this example. Notice that the encoding
process is geared to provide the most bang for the bit at each step. At each step the bits are used
to provide the maximum reduction in the reconstruction error. If at any time the encoding is
interrupted, the reconstruction using this (interrupted) encoding is the best that the algorithm
can provide using this many bits. The encoding improves as more bits are transmitted. This
