Cryptography Reference
In-Depth Information
MD Encoder, α
0
C
MD Decoder
i
1
index
assignment
Decoder 1
β
1
p
1
inverse
quantizer
X
(1)
i
1
i
1
Channel
1
C
l
1
(•)
i
0
i
X
Quantizer
α
Decoder 0
β
0
inverse
quantizer
X
(0)
i
2
i
2
Channel
2
p
2
C
l
2
(•)
i
2
Decoder 2
β
2
inverse
quantizer
X
(2)
Fig. 12.2.
The structure for MDSQ for two descriptions over two independent
channels with mutually independent breakdown probabilities.
are to number the index assignment matrix from upper-left corner to lower-
right corner and to fill in from the main diagonal outward. In [27], the author
considered a set of index pairs constructed from those that lie on the main
diagonal and on the 2k diagonals closest to the main diagonal. The parameter
k is called
spread
. The index assignment shown in Fig. 12.3 is called the
Nested Index Assignment, where the row and column indices, i
1
,andi
2
,
are transmitted over two independent channels.
The cells of the encoder α, are used in increasing values of i, and are num-
bered from j
0
to j
7
in Fig. 12.3(a), and j
0
to j
21
in Fig. 12.3(b), respectively.
Fig. 12.3(a) is an index assignment scheme with a spread of k = 0. Only eight
samples, or j
0
,j
1
,,j
7
, are the valid scalar samples for transmission. Gen-
erally speaking, the eight samples can be represented by 3-bit strings. Thus,
if j
3
is the scalar sample to be transmitted, then after the index assignment
step l, we obtain i
1
= 011 and i
2
= 011 represented in their binary forms.
The central distortion is the quantization error between the input and the
quantized samples. This configuration with a spread of k = 0 can be regarded
as repetition of samples. It means that a total of 6 bits will be received if
no channel breakdown occurs. Consequently, 6 bits need to be transmitted
over two different channels to describe 3 bits information. It produces a re-
dundancy of
6−3
3
100% = 100%. In decoding the received descriptions, if
both channels are alive, then by calculating the conditional expectation, and
as depicted in [27], the reconstructed image decoded from both descriptions
can be obtained. The conditional probabilities for receiving both descriptions
in Fig. 12.3(a) are:
1,
if t =3;
p (j
t
i
1
= 011,i
2
= 011) =
(12.2)
0,
otherwise.
