Databases Reference
In-Depth Information
Comparing the coefficients against 16, we find 26 is greater than 16 so we send
sp
. The next
coefficient in the scan is 6, which is less than 16. Furthermore, its descendants (13, 10, 6, and
4) are all less than 16. Therefore, 6 is a zerotree root, and we encode this entire set with the
label
zr
. The next coefficient in the scan is
−
7, which is also a zerotree root, as is 7, the final
element in the scan. We do not need to encode the rest of the coefficients separately because
they have already been encoded as part of the various zerotrees. The sequence of labels to be
transmitted at this point is
sp zr zr zr
Since each label requires 2 bits (for fixed-length encoding), we have used up 8 bits from
our bit budget. The only significant coefficient in this pass is the coefficient with a value
of 26. We include this coefficient in our list to be refined in the subordinate pass. Calling the
subordinate list
L
S
,wehave
L
S
={
26
}
The reconstructed value of this coefficient is 1
.
5
T
0
=
24, and the reconstructed bands look
like this:
24
0 00
0
0 00
0
0 00
0
0 00
The next step is the subordinate pass, in which we obtain a correction term for the recon-
struction value of the significant coefficients. In this case, the list
L
S
contains only one element.
The difference between this element and its reconstructed value is 26
−
24 = 2. Quantizing
this with a two-level quantizer with reconstruction levels
4, we obtain a correction term
of 4. Thus, the reconstruction becomes 24+4=28. Transmitting the correction term costs a
single bit, therefore at the end of the first pass we have used up 9 bits. Using only these 9 bits,
we obtain the following reconstruction:
±
T
0
/
28
0 00
0
0 00
0
0 00
0
0 00
We now reduce the value of the threshold by a factor of two and repeat the process. The
value of
T
1
is 8. We rescan the coefficients that have not yet been deemed significant. To
emphasize the fact that we do not consider the coefficients that have been deemed significant
