Cryptography Reference
In-Depth Information
moments have been widely used in pattern recognition and have proved to be
successful in various applications. These moments can be used to describe the
shape of the information of the image. For the digital image M, considered in
this chapter, we only use the first 4 moments for the luminance component.
That is the four following floating-typed values.
⎧
⎨
⎩
φ
1
= η
20
+ η
02
,
−η
02
)
2
+4η
11
,
φ
2
=(η
20
(11.31)
−3η
21
)
2
,
φ
4
=(η
30
+ η
12
)
2
+(η
03
+ η
21
)
2
,
−3η
12
)
2
+(η
03
φ
3
=(η
30
where
⎧
⎨
η
pq
= µ
pq
/µ
00
;
µ
pq
=
x
0
x
1
(x
0
−x
0
)
p
(x
1
−x
1
)
q
M (x
0
,x
1
);
x
0
= m
10
/m
00
;
x
1
= m
01
/m
00
;
m
pq
=
⎩
x
0
x
1
x
0
x
1
M (x
0
,x
1
) .
11.4.2 O
ine Multipurpose Watermarking
In our system, before we do the online retrieval, we first embed three wa-
termarks into each image in the database. Because these three watermarks
possess different purposes, we call our watermarking a Multipurpose Water-
mark Scheme. The first watermark is a Copyright Watermark, which is used
for copyright protection. The second is the Annotation Watermark, which is
the name of the image or the semantic meaning of the image. The last is the
Feature Watermark, which is composed of the extracted features. These are
all robust watermarks.
In our watermarking system, the famous Dither Modulation Method [30]
is used to embed the three watermarks into the DCT block of each images
luminance component. We transform from RGB space to YUV space, only the
Y component is used, and the U, V components remaining are unchanged. The
watermarks bits are distributed image orientated, so that cropping the image
will crop the watermark in the same way as the host-image. The dither mod-
ulation algorithm encodes each watermark bit in a Middle-Frequency DCT
coe
cient. Assume a DCT coe
cient DCT
i
in an even interval representing
the bit 0 and in an odd interval representing the bit 1. An interval is even,
if, by using a given ∆,⌊DCT
i
/∆⌋= even, or else it is odd. If a DCT coe
cient
is already in an interval which represents its bit, we move it in the middle of
the interval. Alternatively, we move it to the middle of the nearest interval
adjacent to its current interval. Using our system, we embed twelve bits into
each 88DCTblock.
The key problem is how to embed a floating-typed value into the image.
For simplicity, we embed the 4 bytes (i.e., 32 bits) which represents the float-
ing value directly into the image. A problem arising is the error in bits. Even
