11.03.2014 Views

Network Coding and Wireless Physical-layer ... - Jacobs University

Network Coding and Wireless Physical-layer ... - Jacobs University

Network Coding and Wireless Physical-layer ... - Jacobs University

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

108<br />

Chapter 7: <strong>Physical</strong>-<strong>layer</strong> Key Encoding for <strong>Wireless</strong> <strong>Physical</strong>-<strong>layer</strong> Secret-key<br />

Generation (WPSG) with Unequal Security Protection (USP)<br />

priority class can be weakly secure with the eavesdropper’s guessing success probability<br />

of 0.25. If the number of vulnerable key bits is 3, specify the generator matrix of the<br />

physical-<strong>layer</strong> key encoding generating four encoded key bits.<br />

Solution. In this problem, we start from the generator matrix in (7.10) for perfect secrecy<br />

with parameters I V K = 3, <strong>and</strong> n = 4. The overlapping part of the groups of 1-elements<br />

in two adjacent columns consists of two elements such that<br />

⎡<br />

G p =<br />

⎢<br />

⎣<br />

1 1 1 1 0 0 0 0 0 0<br />

0 0 1 1 1 1 0 0 0 0<br />

0 0 0 0 1 1 1 1 0 0<br />

0 0 0 0 0 0 1 1 1 1<br />

⎤<br />

T<br />

. (7.24)<br />

⎥<br />

⎦<br />

For scalable security satisfying the conditions c 2 = 3 <strong>and</strong> p t2 = 0.25, we reduce the<br />

number of 1-elements in each column from 4 to 3. We also shift each row of the transpose<br />

of the generator matrix to the left such that it becomes<br />

⎡<br />

G p =<br />

⎢<br />

⎣<br />

1 1 1 0 0 0<br />

0 1 1 1 0 0<br />

0 0 1 1 1 0<br />

0 0 0 1 1 1<br />

⎤<br />

T<br />

. (7.25)<br />

⎥<br />

⎦<br />

Now, we will verify that this generator matrix results in weak security corresponding<br />

to Definition 7.4. According to eqs. (7.1), (7.2), (7.7), (7.8), <strong>and</strong> (7.24), the ciphertext Y i<br />

can be written as<br />

Y i = K i ⊕ K i+1 ⊕ K i+2 ⊕ X i , i = 1, 2, 3, 4. (7.26)<br />

Next, we will check the first condition in Definition 7.4 where c 2 = 3. The best way<br />

for the eavesdropper to guess three symbols is to add together three adjacent ciphertext<br />

symbols, for example,<br />

3∑<br />

Y i = K 1 ⊕ K 3 ⊕ K 5 ⊕<br />

i=1<br />

3∑<br />

X i . (7.27)<br />

i=1

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!