Internet Security - Dang Thanh Binh's Page

164 INTERNET SECURITY
Compute:
2 λ (1 � λ � 10): 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10
2 λ (mod 11) : 2 4 8 5 10 9 7 3 6 1
To initiate communication, the user i chooses Xi = 5 randomly from the integer set
2 λ (mod 11) ={1, 2,...,10} and keep it secret. The user i sends
Yi ≡ α Xi (mod q)
≡ 2 5 (mod 11) ≡ 10
to the user j. Similarly, the user j chooses a random number Xj = 7 and sends
Yj ≡ α Xj (mod q)
≡ 2 7 (mod 11) ≡ 7
to the user i.
Finally, compute their common key Kij as follows:
Kij ≡ Y Xi
j (mod q)
and
≡ 7 5 (mod 11) ≡ 10
Kji ≡ Y Xj
i (mod q)
≡ 10 7 (mod 11) ≡ 10
Thus, each user computes the common key.
Example 5.3 Consider the key exchange problem in the finite field GF(2 m )form = 3.
The primitive polynonial p(x) of degree m = 3 over GF(2) is p(x) = 1 + x + x 3 .Ifα
is a root of p(x) over GF(2), then the field elements of GF(2 3 ) generated by p(α) =
1 + α + α 3 = 0 are shown in Table 5.2.
Table 5.2 Field elements of GF(2 3 )
for q = 7
Power Polynonial Vector
1 1 100
α α 010
α 2 α 2 001
α 3 1 + α 110
α 4 α + α 2 011
α 5 1 + α + α 2 111
α 6 1 + α 2 101