Cryptography Reference
In-Depth Information
x
1
,y
1
x
2
,y
2
are 2D coordinates of the vertices v
1
and v
2
.Each
vertex pair will be used as an embedding unit. Within a pair, the difference d
and the integer-mean m of two vertices are calculated for x and y, respectively,
where
and
d
i
x
= x
1
−x
2
,
i∈[1,N].
x
1
+x
2
2
m
i
x
=floor
,
d
y
= y
1
−y
2
,
i∈[1,N].
y
1
+y
2
2
m
y
=floor
,
For N vertex pairs, following N -length sequences could be obtained by
the procedure, that is D
x
, M
x
,andD
y
, M
y
, which are the difference sequence
and the integer-mean sequence respectively for coordinates x and y.
D
x
=d
x
,d
x
,,d
x
,
M
x
=m
x
,m
x
,,m
x
.
d
y
,d
y
,,d
y
D
y
=
,
m
y
,m
y
,,m
y
M
y
=
.
The hidden data can then be embedded into x (or y) coordinates by apply-
ing difference expansion to the sequence D
x
(or D
y
). Namely, by left shifting
(expanding) the suitable elements in D
x
(or D
y
), some extra space could be
provided for placing the hidden bits. The selection of the suitable elements
in D
x
(or D
y
) is based on an embedding condition which is related to the
precision tolerance of the original map.
2. Embedding Condition
The maximum distortion that is tolerated in a vector map is called the maps
precision tolerance, and is denoted as τ . To ensure the validity of the map
data, the maximum distortion induced by data hiding should not exceed the
tolerance τ . Taking the situation of embedding data in x coordinates as the
example, a vertex pair could be suitable for hiding data by bitwise shifting the
difference value d
i
x
but only when an embedding condition has been satisfied.
This section considers the case of left shifting 1 bit that is to expand d
i
x
to
double. Suppose a hidden bit w
i
∈0, 1is embedded into the x coordinates
of the vertex pair
′
v
1
,v
2
by difference expansion, the modified difference d
i
x
containing w
i
is then
′
d
i
x
=2d
i
x
+ w
i
.
(6.7)
′
′
′
The watermarked coordinates x
1
and x
2
are calculated by using d
i
x
and m
i
x
in Eqn. (6.6),
⎧
⎨
⎩
d
x
′
+1
2
′
x
1
= m
i
x
+floor
,
(6.8)
d
x
′
2
′
x
2
= m
i
x
−floor
.
