The University of Babylon
Department of Software
LECTURE NOTES ON SUBLIMINAL CHANNEL &
COMMUNICATION SYSTEM
By
Dr. Samaher Hussein Ali
College of Information Technology, University of Babylon, Iraq
Samaher_hussein@yahoo.com
18 December 2012
Definition of Subliminal Channel
 The subliminal channel is a covert communication channel that cannot be read
by those for whom it is not intended.
 Some simple examples : occur in everyday life when we, for instance, give a
certain look to a certain person, or wink, or raise an eyebrow perhaps. Here,
some form of communication is occurring between two (or more) parties, and
those who listen to the conversation but do not observe the communicants will
not see the subliminal channel .
 Subliminal channels can be classified into broadband and narrow-band channel.
The broadband channel uses almost all available bits. Every channel, which uses
less bits, is called a narrow-band channel. The additionally used bits are needed
for further protection, e.g., impersonation.
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Algorithm of Subliminal Channel
INPUT: An n-bit, two prime number P, q (secret), a
Positive integer r such that GCD(r,p)=1 and GCD(r,p)=1.
OUTPUT: Two sets of binary groups C1, C0 such that we
Can choose one element from each set. (X1, X2).
1: Generate 2n bits
2: Divide 2n bits into two groups C0 which represent odd
Bits positions and C1 which represent even bits positions,
C0 = b0, b2, b4 …b2n-2
Where b denotes to a distinct bit position
C1= b1, b3, b5… ,b2n-1
3: Convert set of all possible bits into biquinary code and
These news generated codes are sent via the channel
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Algorithm of Subliminal Channel
3.1: Calculate the meaningful message according to the Following secret equation
3.2: Calculate the meaningful message according to the Following secret equation
3.3: Sift left x1 and X2 certain times according to the following formula:
X11= (x1 shift left (r/n)),
Where r/n represents the number of shifts.
X12=(x2 shift left(r/n))
4: Return x11, x12
5: select new value to r for the next transmission according to the formula
r =r+ α /r
Where α=E (D||T).
6: Go to step 1.
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Example
 The two authentic meaningful messages generated when n=3, p=11,
q=17, and initial value of r=3 is as follows: The numbers of all possible
values are
 000 001 010 011 100 101 110 111
 The even bits C0 = 000 010 100 110
 The odd bits C1 = 001 011 101 111
 By applying equations (1) and (2) we get two meaningful messages
which are
 x1=010 from set C0 and
 x2= 101fromset C1
 which are in different
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Example
The selected meaningful messages are then coded by using biquinary
code(5043210) as illustrated in table1
Decimal
Representation
The original
codes(421)
(biquinary)5043210
0
000
0100001
1
001
0100010
2
010
0100100
3
011
0101000
4
100
0110000
5
101
1000001
6
110
1000010
7
111
1000100
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Applications of Subliminal Channel
 The most obvious application of the subliminal channel is in a spy network. If
everyone sends and receives signed messages, spies will not be noticed sending
subliminal messages in signed documents. Of course, the enemy’s spies can do
the same thing.
 Using a subliminal channel, Alice could safely sign a document under threat.
She would, when signing the document, imbed the subliminal message.

A company can sign documents and embed subliminal messages, allowing
them to be tracked throughout the documents’ lifespans.

The government can “mark” digital cash.
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Types of Subliminal Channel
In General there are two types of Subliminal Channel:
 The Broadband Channel
The easiest way to establish a subliminal channel between sender and receiver using DSA is
to share the authentication key x. One can then use k = m0 as the subliminal message.
Hence, it is impossible to recover the subliminal message m0 without knowing x or to detect
that the subliminal channel is being used, even if the adversary knows the subliminal
message
 The Narrowband Channel
The narrowband channel of DSA eliminates the necessity to share the authentication key.
Simmons describes three narrowband channels in for the
 First narrowband channel sender and receiver have to create a shared secret prime P
with P > q. The public prime p should not be used, because anybody can check the
following explanations with p and therefore would be able to disclose the usage of the
subliminal channel.
 The second narrowband channel is also a 1-bit channel. To use this channel,the
transmitter and the receiver share a random binary sequence B = b1; …..; bt Which can be
used as a one-time key. First both agree on a bit position of r which is used to transport
the subliminal message, e.g., the lowest or the with lowest bit. Then the i-th subliminal
message bit in the i-th signature is XORed withthe i-th random bit bi. This method is much
easier then the _first one, because the complete signature can be used to hide the
subliminal message.
 E.g., the message itself could be chosen as subliminal message channel.
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Example of the Broadband Channel
Step1: Set-Up
 We chose the primes p = 2347, q = 23 and the element z = 1979.
 It follows that the generator g= z^(p-1)/q mod p = 1979^102 mod 2347 =266
 Publishing (p; q; g). We chose the primes p = 2347, q = 23 and the element z =
1979.
Step 2: Key Generation
 We chose the authentication key x = 1468, so that the verification key
 Y=g^x mod p = 266^1468 mod 2347 = 2100.
 Publishing y.
Step 3: Signing
 The message m is 1337 and as hash function we chose for simplicity a modulo
reduction by 107.
 The message hash is then calculated as h=1337 mod 107 = 53.
 We chose randomly the session key k = 17
 r = (g^k mod p) mod q = (266^17 mod 2347) mod 23 = 12
 s = k^-1(h(m) + xr) mod q = 19(53 + 1468 12) mod 23 = 3
 Sending the triple (m = 1337; r = 12; s = 3).
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13
Example of the Broadband Channel
Step 4: Verifying

Receiving the triple (m = 1337; r = 12; s = 3).

h=1337 mod 107 = 53

t=s^-1 mod q = 8

u1=ht mod q = 53 * 8 mod 23 = 10

u2=rt mod q = 12 * 8 mod 23 = 4

v = (g^u1*y^u2 mod p) mod q = (266^10*2100^4 mod 2347) mod 23 = 12

Since v = r it is accepted that the message was signed by the user,
 associated with the public key y.
18 December 2012
Dr. Samaher Hussein Ali
Notes of Lecture 13