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1 Jordan University of Science and Technology Cryptography and Network Security - CPE 542 Homework #III Handed to: Dr. Lo'ai Tawalbeh By: Ahmed Saleh Shatnawi On: 8/11/2005
2 Review Questions RQ3.3 A block cipher whose block size n is too small in the table (n=4) may be vulnerable to attacks based on statistical analysis. One such attack involves simple frequency analysis of ciphertext blocks. The same plaintext block always produces the same ciphertext block Example (Table 3.1): 4-bit block size has only 16 values use frequency analysis to break it! Note: Table 3.1 n is too small so we have to discuss this situation only RQ3.7 The purpose of the S-Boxes in DES is to complicate the cryptanalysis since they are the only nonlinear part of DES. If S-Boxes were not used, the DES algorithm would be easy to break. Also, welldesigned S-Boxes provide good confusion properties. RQ3.8 Avalanche effect is a desirable property of the encryption algorithms which requires that a small change in either plaintext or key produce a significant change in the ciphertext. Small change in ciphertext can be used to reduce the size of plaintext and key space to be searched and making attempts to home-in by guessing keys impossible.
3 Q3.7 A. Problems First, pass the 64-bit input through PC-1 (Table 3.4a) to produce a 56-bit result. Then perform a left circular shift separately on the two 28-bit halves. Finally, pass the 56-bit result through PC-2 (Table I) to produce the 48-bit K1.: In binary notation: In hexadecimal notation: 0 B B 4 9 A 5 Table I. B. L0, R0 are derived by passing the 64-plaintext through IP (TableII ): L R Table II.
4 C. The E table (Table III) expands R0 to 48 bits: E (R0) = Table III. D. Calculate A = E[R0] XOR K1 A = E. A = S-Box Input S-Box Output F. Concatenate the results of part E. B = G. Apply the permutation to get P(B). Table (IV) P (B) = [B (16) B (7) B (20) B (21) B (29) B (12) B (28)... B(17) B(1) B(15) B(23) B(26) B(5) B(18) B(31)... B(10) B(2) B(8) B(24) B(14) B(32) B(27) B(3)... B(9) B(19) B(13) B(30) B(6) B(22) B(11) B(4) B(25)]; P(B) =
5 Table (IV). H. Calculate R1 = P(B) XOR L0 R1 = I. Write down the ciphertext. L1R1 = Ciphertext = IP-1 (L1R1) IP-1 = [L1R1(40) L1R1(8) L1R1(48) L1R1(16)... L1R1(56) L1R1(24) L1R1(64) L1R1(32) L1R1(39)... L1R1(7) L1R1(47) L1R1(15) L1R1(55) L1R1(23)... L1R1(63) L1R1(31) L1R1(38) L1R1(6) L1R1(46)... L1R1(14) L1R1(54) L1R1(22) L1R1(62) L1R1(30)... L1R1(37) L1R1(5) L1R1(45) L1R1(13) L1R1(53)... L1R1(21) L1R1(61) L1R1(29) L1R1(36) L1R1(4)... L1R1(44) L1R1(12) L1R1(52) L1R1(20) L1R1(60)... L1R1(28) L1R1(35) L1R1(3) L1R1(43) L1R1(11)... L1R1(51) L1R1(19) L1R1(59) L1R1(27) L1R1(34)... L1R1(2) L1R1(42) L1R1(10) L1R1(50) L1R1(18)... L1R1(58) L1R1(26) L1R1(33) L1R1(1) L1R1(41)... L1R1(9) L1R1(49) L1R1(17) L1R1(57) L1R1(25)]; IP -1 (L1R1) = L1 = R0. The ciphertext is the concatenation of L1 and R1. Source: [MEYE82]
6 Q3.9 Encryption (M) Input plaintext to Initial permutation box to get L 0 and R 0 Repeat 15 times with R j =L j-1 f(r j-1,k j ) L j =R j-1 to get L 16 and R 16 Swap them to get R 16 L 16 Put R 16 L 16 to Inverse Permutation box to get ciphertext Decryption (C) Input ciphertext to Initial permutation box to get A16 and B16 Repeat 15 times with Bj-1=Aj f(bj,kj) Aj-1=Bj to get A0 and B0 Swap them to get B0A0 Put B0A0 to Inverse permutation box to get back the plaintext First DES is basically a multi round Feistel cipher that accepts 64 bit plaintext blocks as input and a 56 bit key. I will showed that Feistel decryption is the inverse of Feistel encryption. Therefore, needless to say, DES decryption is the inverse of DES encryption If we prove that an arbitrary round Feistel decryption is the inverse of the same round Feistel en-cryption, then we can show that the whole Feistel decryption is the inverse of the Feistel encryption, since the Feistel process is just stack of each round encryption/decryption. So conceder L0 = AE1BA189HEX and R0 = DC1F10F4HEX Sub-key K2 = 27A1 69E5 8DDAHEX f(r1,k2) =2BA1 536CHEX L0 f(r0,k1)= =85BAF2E5HEX L1 = DC1F 10F4HEX and R1 = 85BA F2E5HEX By applying the Decryption it gives the original text block back L0 = AE1BA189HEX and R0 = DC1F10F4HEX L 0 R 0 L 0 f(r 0, K 1 ) L 1 R 1 Therefore, Feistel decryption, which is the stack of each round Feistel decryption, is the inverse of Feistel encryption, which is the stack of each round Feistel encryption
7 Q3.12 Instead of shift left we will right and we will reverse the order of the keys. Iteration Number #of Right Shifts Q3.16 The corrupted bit is in the transmission of character C N Then the N th block is corrupted along with the next (64/8=8) blocks. In other words, the affected blocks are N, N+1, N+2, N+3, N+4, N+5, N+6, N+7 and N+8. So the total number of affected blocks = 9 plaintext boxes. In the CFB mode if any error occurred in Cj this error will be propagate for the plaintext with the cipher text corrupted and for all packets follow it.
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