Internet Security - Dang Thanh Binh's Page

62 INTERNET SECURITY
Example 3.1 Assume that a 64-bit key input is K = 581fbc94d3a452ea, including 8
parity bits. Find the first three round keys only: K1, K2, andK3.
The register contents C0 (left) and D0 (right) are computed using Table 3.1:
C0 = bcd1a45
D0 = d22e87f
Using Table 3.2, the blocks C1 and D1 are obtained from the block C0 and D0 by shifting
one bit to the left as follows:
C1 = 79a348b
D1 = a45d0ff
The 48-bit key K1 is derived using Table 3.3 (PC-2) by inputting the concatenated block
(C1||D1) such that K1 = 27a169e58dda.
The concatenated block (C2||D2) is computed from (C1||D1) by shifting one bit to the
left as shown below:
(C2||D2) = f346916 48ba1ff
Using Table 3.3 (PC-2), the 48-bit key K2 at round 2 is computed as K2 = da91ddd7b748.
Similarly, (C3||D3) is generated from shifting (C2||D2) by two bits to the left as follows:
(C3||D3) = cd1a456 22e87fd
Using Table 3.3, we have
K3 = 1dc24bf89768
In a similar fashion, all the other 16-round keys can be computed and the set of entire
DES keys is listed as follows:
3.1.3 DES Encryption
K1 = 27a169e58dda K2 = da91ddd7b748
K3 = 1dc24bf89768 K4 = 2359ae58fe2e
K5 = b829c57c7cb8 K6 = 116e39a9787b
K7 = c535b4a7fa32 K8 = d68ec5b50f76
K9 = e80d33d75314 K10 = e5aa2dd123ec
K11 = 83b69cf0ba8d K12 = 7c1ef27236bf
K13 = f6f0483f39ab K14 = 0ac756267973
K15 = 6c591f67a976 K16 = 4f57a0c6c35b
DES operates on a 64-bit block of plaintext. After initial permutation, the block is split
into two blocks Li (left) and Ri (right), each 32 bits in length. This permuted plaintext