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The sizes of U and v are 257×256 and 257×1 respectively. An approximate solution
to this over-determined problem in the mean square error sense can be found using
matrix pseudo-inversion:
x
ˆ
=
U
+
v
.
(20)
Let x 0 = 0, x 255 = h 255 and
0
if
x
<
0
t
x
=
h
if
x
>
h
t
=
1
2
,
254
.
(21)
t
t
t
t
int(
x
)
otherwise
t
In this way, a steganographic scheme x is obtained which causes an approximate
least histogram abnormality in the mean square error sense.
The above-described LHA algorithm can be summarized as a four-step process:
1. Pseudo-randomly scramble the bit stream to be embedded, and map each bit to one
pixel in the host image.
2. Count the number of pixels at gray levels 0~255 which differ in LSBs from the
corresponding secret bits to yield a vector h .
3. Calculate x corresponding to an approximate minimum d from (20) and (21).
4. For each gray level j ( j = 0, 1,
, 255), randomly select x j pixels among h j pixels,
decrease them by 1, and increase the other ( h j x j ) pixels by 1.
In this algorithm, although modification to the host pixels may affect the higher bit
planes, the induced distortion to the host image is not increased compared to the simple
LSB replacement technique. Extraction of the embedded information can still be
accomplished by extracting the LSB plane as in the straight LSB method.
4000
20
Original image
Steganographic image
3500
15
3000
10
2500
5
2000
0
1500
-5
1000
-10
500
-15
0
-20
0
50
100
150
200
250
0
50
100
150
200
250
Gray values
Gray values
(a) (b)
Fig. 3. Histograms of the original image Baboon and the stego-image. (a) The two histograms are
hardly distinguishable. (b) Difference between the two histograms in a boosted scale
512 and
the stego-image in which each pixel is used to carry one hidden bit by the proposed
method. The two curves are almost identical as the difference due to steganographic
embedding is extremely small. Fig. 3(b) shows the histogram difference in an expanded
scale.
Fig. 3(a) sketches histograms of an original test image Baboon sized 512
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