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

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Chapter 7: Physical-layer Key Encoding for Wireless Physical-layer Secret-key
Generation (WPSG) with Unequal Security Protection (USP)
in the next section.
7.2 Perfect Secrecy in Physical-Layer Key Encoding
for a One-Time-Pad Encryptor
Consider a non-randomized cipher in which elements in the plaintext X = [X 1 , X 2 , ..., X M ],
ciphertext Y = [Y 1 , Y 2 , ..., Y N ], and secret key Z = [Z 1 , Z 2 , ..., Z J ] all take values in an L-
ary alphabet and J = N = M. Suppose that the key is chosen to be completely random,
i.e., P (Z i = z) = L −M , i = 1, 2, ..., M, for all possible values z of the secret key, and that
the enciphering transformation is
Y i = (X i + Z i ) mod L, i = 1, 2, ..., M. (7.1)
Since, for each possible choice x i and y i of X i and Y i , respectively, there is a unique
z i such that Z i = z i satisfies (7.1), it follows that P (Y = y|X = x) = L −M
for every
particular y = [y 1 , y 1 , ..., y M ] and x = [x 1 , x 2 , ..., x M ], no matter what the statistics of X
may be. Thus, X and Y are statistically independent, and hence this system provides
perfect secrecy [31]. The system is called a modulo-L Vernam system or one-time pad and
is used in our model in Fig. 7.1 to combine the message and the encoded key together.
Since, out of the total I K quantized key symbols, I V K symbols are vulnerable symbols,
we can see that, had a one-time pad encryptor been used without physical-key encoding,
I V K ciphertext symbols would have been decrypted by the eavesdropper. Thus, in order
to construct a secure Y 1 , we use the following linear combination for Z 1 .
Z 1 = (K 1 + K 2 + ... + K IV K +1) mod L (7.2)
After substituting (7.2) into (7.1) using i = 1, we can see that even if the set of all
vulnerable key symbols is a subset of {K 1 , K 2 , ..., K IV K +1}, there is still one symbol that is
unknown to the eavesdropper. If every key symbol is statistically independent of others,