Cryptography Reference
In-Depth Information
30000
20000
10000
2
000 4000 6000 8000 10000 12000 140
00
-10000
-20000
-30000
Figure 14.3: A graph of a .33 seconds piece from a sound file.
7500
5000
2500
100
200
300
400
-2500
-5000
-7500
Figure 14.4: A small section of Figure 14.3 from range (8200,8600).
from each location. The same solution can be used with video data
or any other data source that can withstand small pertubations.
Here are the raw values between 8250 and 8300: 2603, 2556, 2763,
3174, 3669, 4140, 4447, 4481, 4282, 3952, 3540, 3097, 2745, 2599, 2695,
2989, 3412, 3878, 4241, 4323, 4052, 3491, 2698, 1761, 867, 143, -340,
-445, -190, 203, 575, 795, 732, 392, -172, -913, -1696, -2341, -2665,
-2579, -2157, -1505, -729, 6, 553, 792, 654, 179, -548, -1401, -2213 .
The values in the range
(8200
,
8600)
add up to
40
,
813
,oranaver-
age of about
102
.
A basic algorithm encodes one bit by choosing some strength
factor,
, and then arranging for the absolute value of the average
value of the elements to be above
S
S
if the message is a
1
and below
S
if the message is a
0
.
Choosing the right value of
for this basic algorithm is something
of an art that is confounded by the size of the blocks, the strength
of the real signal, the nature of the sound, and several other factors.
Let's imagine
S
S
=10
. If the message to be encoded is
1
, then nothing
